An Introduction to Theoretical Fluid Mechanics 9780821848883, 2009028622

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C O U R A N T

19

STEPHEN CHILDRESS

LECTURE NOTES

An Introduction to Theoretical Fluid Mechanics

American Mathematical Society Courant Institute of Mathematical Sciences Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

An Introduction to Theoretical Fluid Mechanics

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Courant Lecture Notes in Mathematics Executive Editor Jalal Shatah Managing Editor Paul D. Monsour Assistant Editor Reeva Goldsmith Copy Editor Will Klump

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http://dx.doi.org/10.1090/cln/019

Stephen Childress Courant Institute of Mathematical Sciences

19

An Introduction to Theoretical Fluid Mechanics

Courant Institute of Mathematical Sciences New York University New York, New York American Mathematical Society Providence, Rhode Island Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

2000 Mathematics Subject Classification. Primary 76–01.

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Library of Congress Cataloging-in-Publication Data Childress, Stephen. An introduction to theoretical fluid mechanics / Stephen Childress. p. cm. — (Courant lecture notes ; v. 19) Includes bibliographical references and index. ISBN 978-0-8218-4888-3 (alk. paper) 1. Fluid mechanics. I. Title. QA901.C526 2009 532.001—dc22 2009028622

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established to ensure permanence and durability. Visit the AMS home page at http://www.ams.org/ 10 9 8 7 6 5 4 3 2 1

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Contents Preface

ix

Chapter 1. The Fluid Continuum 1.1. Eulerian and Lagrangian Descriptions 1.2. The Material Derivative Problem Set 1

1 1 6 10

Chapter 2. Conservation of Mass and Momentum 2.1. Conservation of Mass 2.2. Conservation of Momentum in an Ideal Fluid 2.3. Steady Flow of a Fluid of Constant Density 2.4. Intrinsic Coordinates in Steady Flow 2.5. Potential Flows with Constant Density 2.6. Boundary Conditions on an Ideal Fluid Problem Set 2

13 13 16 18 20 21 22 24

Chapter 3. Vorticity 3.1. Local Analysis of the Velocity Field 3.2. Circulation 3.3. Kelvin’s Theorem for a Barotropic Fluid 3.4. The Vorticity Equation 3.5. Helmholtz’ Laws 3.6. The Velocity Field Created by a Given Vorticity Field 3.7. Some Examples of Vortical Flows Problem Set 3

27 27 29 30 30 32 33 35 41

Chapter 4. Potential Flow 4.1. Harmonic Flows 4.2. Flows in Three Dimensions 4.3. Apparent Mass and the Dynamics of a Body in a Fluid 4.4. Deformable Bodies and Their Locomotion 4.5. Drift Problem Set 4

45 45 51 57 62 65 70

Chapter 5. Lift and Drag in Ideal Fluids 5.1. Lift in Two Dimensions and the Kutta-Joukowski Condition 5.2. Smoothing the Leading Edge: Joukowski Airfoils 5.3. Unsteady and Quasi-Steady Motion of an Airfoil

73 74 77 79

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vi

CONTENTS

5.4. Drag in Two-Dimensional Ideal Flow 5.5. The Three-Dimensional Wing: Prandtl’s Lifting Line Theory Problem Set 5

81 88 94

Chapter 6. Viscosity and the Navier-Stokes Equations 6.1. The Newtonian Stress Tensor 6.2. Some Examples of Incompressible Viscous Flow 6.3. Dynamical Similarity Problem Set 6

97 97 101 106 109

Chapter 7. Stokes Flow 7.1. Solutions of the Stokes Equations 7.2. Uniqueness of Stokes Flows 7.3. The Stokes Solution for Uniform Flow Past a Sphere 7.4. Two Dimensions: Stokes’ Paradox 7.5. Time Reversibility in Stokes Flow 7.6. Stokesian Locomotion and the Scallop Theorem Problem Set 7

111 113 114 114 117 119 121 121

Chapter 8. The Boundary Layer 8.1. The Limit of Large Re 8.2. Blasius Solution for a Semi-Infinite Flat Plate 8.3. Boundary Layer Analysis as a Matching Problem 8.4. Separation 8.5. Prandtl-Batchelor Theory Problem Set 8

123 123 125 132 133 134 137

Chapter 9. Energy 9.1. Mechanical Energy 9.2. Elements of Classical Thermodynamics 9.3. The Energy Equation 9.4. Some Basic Relations for the Nondissipative Case 9.5. Kelvin’s Theorem in a Compressible Medium Problem Set 9

139 139 141 143 145 146 149

Chapter 10. Sound 10.1. One-Dimensional Waves 10.2. The Fundamental Solution in Three Dimensions 10.3. Kirchhoff’s Solution 10.4. Weakly Nonlinear Acoustics in One Dimension Problem Set 10

151 151 152 153 155 158

Chapter 11. Gas Dynamics 11.1. Nonlinear Waves in One Dimension 11.2. Dynamics of a Polytropic Gas 11.3. Simple Waves 11.4. Linearized Supersonic Flow

161 161 162 163 166

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CONTENTS

11.5. Alternative Formulation and Proof of the Drag Formula 11.6. Transonic Flow Problem Set 11

vii

171 173 174

Chapter 12. Shock Waves 12.1. Scalar Case 12.2. Quasi-linear Supersonic Flow 12.3. The Stationary Normal Shock Wave 12.4. Riemann’s Problem: The Shock Tube Problem Set 12

175 175 177 178 183 186

Supplementary Notes

189

Bibliography

195

Index

197

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Preface These notes were prepared for a one-semester graduate course in introductory classical fluid mechanics. The fluid mechanics curriculum at the Courant Institute has traditionally consisted of a two-semester introductory sequence, followed by special topic courses. It was common to treat incompressible fluids, both ideal and viscous, in the first semester, and then to move to compressible flow, gas dynamics, and shock waves in the second. Because of the pressures of time and course scheduling, and the ever-expanding scope of the subject matter, a decision was made to offer instead a one-semester introductory course, which would include at least some of the material on compressible flow, to be followed by a second-semester special topic fluids course that could change from year to year depending upon faculty and student interests. The present course was developed for students with a strong undergraduate mathematics background, but I have assumed no previous exposure to fluid mechanics. The selected material is fairly standard, but it was chosen to emphasize the mathematical methods that have their origin in fluid theory. A central problem of the classical theory is the subtle relation between an ideal fluid and a real fluid of small viscosity (or more precisely, a fluid flow with a large Reynolds number). Many of the crowning achievements of the fluid dynamicists of the nineteenth and twentieth centuries, certainly including Prandtl’s boundary layer theory, airfoil theory, much of the theory of singular perturbations, and the recent developments surrounding triple-deck theory, are all motivated by this problem. I have tried to keep some of these issues front and center when presenting the classical results in potential flow and in models for the lift and drag of bodies in a flow. Some attention is paid to the problem of locomotion in fluids, since it provides an interesting example where both Eulerian and Lagrangian methods play a role. As a course in a mathematics curriculum, fluid mechanics should, in my opinion, be presented as a beautiful, practical subject, involving a moving continuum whose deformations are determined by certain natural physical laws. But the mathematical complexity of the subject is legion; to take one of many examples, the global existence of solutions of the Navier-Stokes equations for an incompressible fluid remains an open question. In an introductory course we must be content with the relatively small number of model problems that convey the flavor of the subject without excessive analysis. The choices made here leave out, or only touch upon, many interesting and important topics. Among these are turbulence, shallow-water theory, rotating fluids and associated geophysical models, water waves, hydrodynamic stability, surface ix Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

x

PREFACE

tension phenomena, and, importantly, computational fluid dynamics. Nevertheless, it is hoped that these notes offer a fair introduction to the classical theory and a preparation for more specialized courses in fluid mechanics. S TEPHEN C HILDRESS

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http://dx.doi.org/10.1090/cln/019/01

CHAPTER 1

The Fluid Continuum Classical fluid mechanics is concerned with a mathematical idealization of common fluids such as air or water. The main idealization is embodied in the notion of a continuum, and our “fluids” will generally be identified with a certain connected set of points in RN , where we will consider dimension N to be 1, 2, or 3. Of course, the fluids will move, so basically our subject is that of a moving continuum. This description is an idealization that neglects the molecular structure of real fluids. Liquids are fluids characterized by random motions of molecules on the scale of 10
7 to 10
8 cm, and by a substantial resistance to compression. Gases consist of molecules moving over much larger distances, with mean free paths of the order of 10
3 cm, and are readily compressed. Both liquids and gases will fall within the scope of the theory of fluid motion that we will develop below. The theory will deal with observable properties such as velocity, density, and pressure. These properties must be understood as averages over volumes that contain many molecules but are small enough to be “infinitesimal” with respect to the length scale of variation of the property. We shall use the term fluid parcel to indicate such a small volume. The notion of a particle of fluid will also be used. It is a point of the continuum, but it should not be confused with a molecule. For example, the time rate of change of position of a fluid particle will be the fluid velocity, which is an average velocity taken over a parcel and is distinct from molecular velocities. The continuum theory has wide applicability to the natural world, but there are certain situations where it is not satisfactory. Usually these will involve small domains where the molecular structure becomes important, such as shock waves or fluid interfaces. 1.1. Eulerian and Lagrangian Descriptions Let the independent variables (observables) describing a fluid be functions of position x D .x1 ; : : : ; xN / in Euclidean space and time t. Suppose that at t D 0 the fluid is identified with an open set S0 of RN . As the fluid moves, the particles of fluid will take up new positions, occupying the set St at time t. We can introduce the map Mt ; S0 ! St to describe this change, and write Mt S0 D St . If a D .a1 ; : : : ; aN / is a point of S0 , we introduce the function x D X .a; t/ as the position of a fluid particle at time t, which was located at a at time t D 0. The function X .a; t/ is called the Lagrangian coordinate of the fluid particle identified by the point a. We remark that the “coordinate” a need not in fact be the initial position 1 Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

2

1. THE FLUID CONTINUUM

of a particle, although that is the most common choice and will be generally used here. But any unique labeling of the particles is acceptable. The Lagrangian description of a fluid emerges from this focus on the fluid properties associated with individual fluid particles. To “think Lagrangian” about a fluid, one must move mentally with the fluid and sample the fluid properties in each moving parcel. The Lagrangian analysis of a fluid has certain conceptual and mathematical advantages, but it is often difficult to apply to realistic fluid flows. Also it is not directly related to experience, since measurements in a fluid tend to be performed at fixed points in space, as the fluid flows past the point. If we therefore adopt the point of view that we will observe fluid properties at a fixed point x as a function of time, we must break the association with a given fluid particle and realize that as time flows different fluid particles will occupy the position x. This will make sense as long as x remains within the set St . Once properties are expressed as functions of x; t we have the Eulerian description of a fluid. For example, we might consider the fluid to fill all space and be at rest “at infinity.” We then can consider the velocity u.x; t/ at each point of space, with limjxj!1 u.x; t/ D 0. Or we might have a fixed rigid body with fluid flowing over it such that at infinity we have a fixed velocity U. For points outside the body the fluid velocity will be defined and satisfy limjxj!1 u.x; t/ D U. It is of interest to compare these two descriptions of a fluid and understand their connections. The most obvious is the meaning of velocity: the definition is ˇ @X ˇˇ D u.X .a; t/; t/: (1.1) xt D @t ˇ a

That is to say, following the particle we calculate the rate of change of position with respect to time. Given the Eulerian velocity field, the calculation of Lagrangian coordinates is therefore mathematically equivalent to solving the initial value problem for the system (1.1) of ordinary differential equations for the function x.t/, with the initial condition x.0/ D a, the order of the system being the dimension of space. The special case of a steady flow, that is, one that is independent of time, leads to a system of autonomous ODEs. E XAMPLE 1.1. In two dimensions (N D 2), with fluid filling the plane, we take u.x; t/ D .u.x; y; t/; v.x; y; t// D .x;
y/. This velocity field is independent of time, hence a steady flow. To compute the Lagrangian coordinates of the fluid particle initially at a D .a; b/ we solve @y @x D x; x.0/ D a; D
y; y.0/ D b; (1.2) @t @t so that X D .ae t ; be
t /. Note that, since xy D ab, the particle paths are hyperbolas; the curves traced out by the particles are independent of time; see Figure 1.1. If we consider the fluid in y > 0 only and take y D 0 as a rigid wall, we have a flow that is impinging vertically on a wall. The point at the origin, where the velocity is 0, is called a stagnation point. This point is a hyperbolic point relative to particle paths. A flow of this kind occurs at the nose of a smooth body placed


We shall often use .x; y; z/ in place of .x1 ; x2 ; x3 /, and .a; b; c/ in place of .a1 ; a2 ; a3 /.

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1.1. EULERIAN AND LAGRANGIAN DESCRIPTIONS

3

F IGURE 1.1. Stagnation point flow.

in a uniform current. Because this flow is steady, the particle paths are also called streamlines. E XAMPLE 1.2. Again in two dimensions, consider .u; v/ D .y;
x/. Then D y and @y D
x. Solving, the Lagrangian coordinates are x D a cos t C @t b sin t; y D
a sin t C b cos t, and the particle paths (and streamlines) are the circles x 2 C y 2 D a2 C b 2 . The motion on the streamlines is clockwise, and fluid particles located at some time on a ray x=y D const remain on the same ray as it rotates clockwise once for every 2
units of time. This is solid-body rotation. @x @t

E XAMPLE 1.3. If instead .u; v/ D .y=r 2 ;
x=r 2 /; r 2 D x 2 C y 2 , we again have particle paths that are circles, but the velocity becomes infinite at r D 0. This is an example of a flow representing a two-dimensional point vortex (see Chapter 3). 1.1.1. Particle Paths, Instantaneous Streamlines, and Streak Lines. The present considerations are kinematic, meaning that we are assuming knowledge of fluid motion, through an Eulerian velocity field u.x; t/ or else Lagrangian coordinates x D X .a; t/, irrespective of the cause of the motion. One useful kinematic characterization of a fluid flow is the pattern of streamlines, as already mentioned in the above examples. In steady flow the streamlines and particle paths coincide. In an unsteady flow this is not the case and the only useful recourse is to consider instantaneous streamlines, at a particular time. In three dimensions the instantaneous streamlines are the orbits of the velocity field u.x; t/ D .u.x; y; z; t/; v.x; y; z; t/; w.x; y; z; t// at time t. These are the integral curves satisfying dy dz dx D D : u v w These streamlines will change in an unsteady flow, and the connection with particle paths is not obvious in flows of any complexity. (1.3)

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1. (a) THE FLUID CONTINUUM

3

3

2

2

1

1

0

0

y

y

4

−1

−1

−2

−2

−3 −5

0 x

5

−3 −5

(b)

0 x

5

F IGURE 1.2. (a) Particle path and (b) streak line in Example 1.4.

Visualization of flows in water is sometimes accomplished by introducing dye at a fixed point in space. The dye can be thought of as labeling by color the fluid particle found at the point at a given time. As each point is labeled it moves along its particle path. The resulting streak line thus consists of all particles that at some time in the past were located at the point of injection of the dye. To describe a streak line mathematically we need to generalize the time of initiation of a particle path. Thus we introduce the generalized Lagrangian coordinate x D X .a; t; ta /, defined to be the position at time t of a particle that was located at a at time ta . A streak line observed at time t > 0, which was started at time t D 0 say, is given by x D X .a; t; ta /; 0 < ta < t. Particle paths, instantaneous streamlines, and streak lines are generally distinct objects in unsteady flows. E XAMPLE 1.4. Let .u; v/ D .y;
x C  cos !t/. For this flow the instantaneous streamlines satisfy dx=y D dy=.
x C  cos !t/, yielding the circles .x
 cos !t/2 C y 2 D const. The generalized Lagrangian coordinates can be obtained from the general solution of a second-order ODE and take the form  cos !t C A cos t C B sin t;
1 ! sin !t C B cos t
A sin t; yD 2 !
1 xD


(1.4)

!2

where ! sin !ta sin ta C a cos ta
1

(1.5)

A D
b sin ta C

(1.6)

 cos !ta cos ta ;
1  cos !ta sin ta B D a sin ta C b cos ta
2 !
1 ! sin !ta cos ta : C 2 !
1 C

!2

!2

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1.1. EULERIAN AND LAGRANGIAN DESCRIPTIONS

5

y

y

y

(a) 2 0 −2 −3

−2

−1

0 (b)

1

2

3

2 0 −2 −3

−2

−1

0 (c)

1

2

3

2 0 −2 −3

−2

−1

0 x

1

2

3

F IGURE 1.3. The oscillating vortex, Example 1.5,  D 1:5, ! D 2. The lines emanate from .2; 1/. (a) Particle path, 0 < t < 20. (b) Streak line, 0 < t < 20. (c) Particle path, 0 < t < 500.

The particle path with ta D 0, ! D 2,  D 1 starting at the point .2; 1/ is given by 7 7 2 1 (1.7) x D
cos 2t C sin t C cos t; y D cos t
sin t C sin 2t; 3 3 3 3 and is shown in Figure 1.2(a). All particle paths are closed curves. The streak line emanating from .2; 1/ over the time interval 0 < t < 2
is shown in Figure 1.2(b). This last example is especially simple since the two-dimensional system is linear and can be integrated explicitly. In general, two-dimensional unsteady flows and three-dimensional steady flows can exhibit chaotic particle paths and streak lines. E XAMPLE 1.5. A nonlinear system exhibiting this complex behavior is the oscillating point vortex: .u; v/ D .y=r 2 ;
.x
 cos !t/=r 2 /. We show an example of particle path and streak line in Figure 1.3. 1.1.2. The Jacobian Matrix. We will, with a few obvious exceptions, be taking all of our functions as infinitely differentiable wherever they are defined. In particular, we assume that Lagrangian coordinates will be continuously differentiable with respect to the particle label a. Accordingly, we may define the Jacobian of the Lagrangian map Mt by the matrix ˇ @xi ˇˇ : (1.8) Jij D @aj ˇt Thus d li D Jij daj is a differential vector that can be visualized as connecting two nearby fluid particles whose labels differ by daj .
If da1    daN is the volume
Here and elsewhere the summation convention is understood: unless otherwise stated, repeated indices are to be summed from 1 to N .

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6

1. THE FLUID CONTINUUM

of a small fluid parcel, then Det.J/da1    aN is the volume of that parcel under the map Mt . Fluids that are incompressible must have the property that all fluid parcels preserve their volume, so that Det.J/ D const D 1 when a denotes initial position, independently of a; t. We may then say that the Lagrangian map is volume preserving. For general compressible fluids Det.J/ will vary in space and time. Another important assumption that we shall make is that the map Mt is always invertible, Det.J/ > 0. Thus when needed we can invert to express a as a function of x; t. 1.2. The Material Derivative Suppose some scalar property P of the fluid can be attached to a certain fluid parcel, e.g., temperature or density. Further, suppose that, as the parcel moves, this property is invariant in time. We can express this fact by the equation ˇ @P ˇˇ D 0; (1.9) @t ˇa since this means that the time derivative is taken with particle label fixed, i.e., taken as we move with the fluid particle in question. We will say that such an invariant scalar is material. A material invariant is one attached to a fluid particle. We now ask how this property should be expressed in Eulerian variables. That is, we select a point x in space and seek to express material invariance in terms of properties of the fluid at this point. Since the fluid is generally moving at the point, we need to bring in the velocity. The way to do this is to differentiate P.x.a; t/; t/, expressing the property as an Eulerian variable, using the chain rule: ˇ ˇ ˇ ˇ @P ˇˇ @xi ˇˇ @P ˇˇ @P.x.a; t/; t/ ˇˇ (1.10) ˇ D 0 D @t ˇ C @t ˇ @x ˇ D Pt C u  rP: @t i t a x a @ C u  r is called the material derivative In fluid dynamics the Eulerian operator @t or substantive derivative or convective derivative. Sometime u  ru is called the “convective part” of the derivative. Clearly it is a time derivative “following the fluid” and expresses the Lagrangian time derivative in terms of Eulerian properties of the fluid.

E XAMPLE 1.6. The acceleration of a fluid parcel is defined as the material 2 ˇ derivative of the velocity u. In Lagrangian variables the acceleration is @@t 2x ˇa , and in Eulerian variables the acceleration is ut C u  ru. Following a common convention we shall often write @ D  C u  r; Dt @t so the acceleration becomes Du=Dt.

(1.11)

E XAMPLE 1.7. We consider the material derivative of the determinant of the Jacobian J. We may divide up the derivative of the determinant into a sum of N

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1.2. THE MATERIAL DERIVATIVE

7

determinants, the first having the first row differentiated, the second having the next row differentiated, and so on. The first term is thus the determinant of the matrix 1 0 @u @u1 @u1 1    @a @a1 @a2 N B @x2 @x2 @x2 C C B B @a1 @a2    @aN C : (1.12) B : :: :: C :: C B :: : : : A @ @xN @a1

@xN @a2



@xN @aN

If we expand the terms of the first row using the chain rule, e.g., @u1 @x1 @u1 @x2 @u1 @xN @u1 D C C  C ; (1.13) @a1 @x1 @a1 @x2 @a1 @xN @a1 we see that we will get a contribution only from the terms involving @u1 =@x1 , since all other terms involve the determinant of a matrix with two identical rows. Thus the term involving the derivative of the top row gives the contribution @u1 Det.J/: @x1 Similarly, the derivatives of the second row gives the additional contribution @u2 Det.J/: @x2 Continuing, we obtain D Det J D div.u/ Det.J/: (1.14) Dt Note that, since an incompressible fluid has Det.J/ D const > 0, such a fluid must satisfy, by (1.14), div.u/ D 0, which is the way an incompressible fluid is defined in Eulerian variables. 1.2.1. Solenoidal Velocity Fields. The adjective solenoidal applied to a vector field is equivalent to “divergence free.” We will use either div.u/ or r  u to denote divergence. The incompressibility of a material with a solenoidal vector field means that the Lagrangian map Mt preserves volume and so whatever fluid moves into a fixed region of space is matched by an equal amount of fluid moving out. In two dimensions the equation expressing the solenoidal condition is @v @u C D 0: (1.15) @x @y If .x; y/ possesses continuous second derivatives we may satisfy (1.15) by setting @ @ ; vD
: (1.16) uD @y @x The function is called the stream function of the velocity field. The reason for the term is immediate: The instantaneous streamline passing through x; y has direction .u.x; y/; v.x; y// at this point. The normal to the streamline at this point is r .x; y/. But we see from (1.16) that .u; v/  r D 0 there, so the lines of constant are the instantaneous streamlines of .u; v/.

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8

1. THE FLUID CONTINUUM

2 n 1

(a)

(b)

F IGURE 1.4. Solenoidal velocity fields. (a) Two streamlines in two dimensions. (b) A stream tube in three dimensions.

Consider two streamlines D i , i D 1; 2 and any oriented simple contour (no self-crossings) connecting one streamline to the other. The claim is then that the flux of fluid across this contour, from left to right seen by an observer facing in the direction of orientation of the contour, is given by the difference of the values of the stream function, 2
1 , if the contour is oriented to go from streamline 1 to streamline R 2; see Figure 1.4(a). R Indeed, oriented as shown the line integral of flux is just .u; v/  .dy;
dx/ D d D 2
1 . In three dimensions, we similarly introduce a stream tube, consisting of a collection of streamlines; see Figure 1.4(b). The flux of fluid across any surface cutting through the tube must be the same. This follows immediately by applying the divergence theorem to the integral of div u over the stream tube. Note that we are referring here to the flux of volume of fluid, not to the flux of mass. In three dimensions there are various “stream functions” used when special symmetries allow them. An example of a class of solenoidal flows generated by two scalar functions takes the form u D r˛  rˇ, where the intersections of the surfaces of constant ˛.x; y; z/ and ˇ.x; y; z/ are the streamlines. Since r˛  rˇ D r  .˛rˇ/, we see that these flows are indeed solenoidal. 1.2.2. The Convection Theorem. Suppose that St is a region of fluid particles and Forming the volume integral over St , F D R let f .x; t/ be a scalar function. dF f dV , we seek to compute . Now x St dt dVx D dx1    dxN D Det.J/da1    daN D Det.J/dVa : Thus

Z d dF D f .x.a; t/; t/ Det.J/dVa dt dt S0 Z Z d d D Det.J/ f .x.a; t/; t/dVa C f .x.a; t/; t/ Det.J/dVa dt dt S0 S0  Z
Df C f div.u/ Det.J/dVa ; D Dt S0

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1.2. THE MATERIAL DERIVATIVE

and so (1.17)

dF D dt

Z


9

 Df C f div.u/ dVx : Dt

St

The result (1.17) is called the convection theorem. We can contrast this calculation with one over a fixed finite region R of space with boundary @R. In that case the rate of change of f contained in R is just Z Z @f d f dVx D dVx : (1.18) dt @t R

R

The difference between the two calculations involves the flux of f through the boundary of the domain. Indeed, we can write the convection theorem in the form  Z
dF @f (1.19) D C div.f u/ dVx : dt @t St

Using the divergence (or Gauss’s) theorem, and considering the instant when St D R, we have Z Z @f dF D dVx C f u  n dSx ; (1.20) dt @t R

@R

where n is the outer normal to the region and dSx is the area element of @R. The second term on the right is flux of f out of the region R. Thus the convection theorem incorporates into the change in f within a region, the flux of f into or out of the region due to the motion of the boundary of the region. Once we identify f with a physical property of the fluid, the convection theorem will be useful for expressing the conservation of this property; see Chapter 2. 1.2.3. Material Vector Fields: The Lie Derivative. Certain vector fields in fluid mechanics, and notably the vorticity field !.x; t/ D r  u (see Chapter 3), can in certain cases behave as a material vector field. To understand the concept of a material vector one must imagine the direction of the vector to be determined by nearby material points. It is wrong to think of a material vector as attached to a fluid particle and constant there. This would amount to a simple translation of the vector along the particle path. Instead, the direction of the vector will be that of a differential segment connecting two nearby fluid particles, d li D Jij daj . Furthermore, the length of the material vector is to be proportional to this differential length as time evolves and the particles move. Consequently, once the particles are selected, the future orientation and length of a material vector will be completely determined by the Jacobian matrix of the flow. Thus a material vector field will have the form (in Lagrangian variables) (1.21)

vi .a; t/ D Jij .a; t/Vj .a/:

Given the inverse a.x; t/ we can express v as a function of x; t to obtain its Eulerian structure.

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10

1. THE FLUID CONTINUUM

D B

C A F IGURE 1.5. Computing the time derivative of a material vector.

Consider now the time rate of change of a material vector field following the fluid parcel. We differentiate v.a; t/ with respect to time for fixed a and develop the result using the chain rule: ˇ ˇ @Jij ˇˇ @ui @vi ˇˇ D Vj .a/ D Vj ˇ ˇ @t @t @a a

a

D

(1.22)

j

@ui @xk @ui Vj D vk : @xk @aj @xk

Introducing the material derivative, a material vector field is seen to satisfy the following equation in Eulerian variables: ˇ @v ˇ Dv D ˇˇ C u  rv
v  ru  vt C Lu v D 0: (1.23) Dt @t x

In differential geometry Lu is called the Lie derivative of the vector field v with respect to the vector field u. The way this works can be understood by moving neighboring points along particle paths. Let v D AB D x be a small material vector at time t; see Figure 1.5. At time t later, the vector has become CD. The curved lines are the particle paths through A; B of the vector field u.x; t/. Selecting A as x, we see that after a small time interval t the point C is A C u.x; t/t and D is the point B C u.x C x; t/t. Consequently, (1.24)

CD
AB D u.x C x; t/
u.x; t/: t

The left-hand side of (1.24) is approximately Dv Dt , and the right-hand side is approximately v  ru, so in the limit x; t ! 0 we get (1.23). A material vector field has the property that its magnitude can change by the stretching properties of the underlying flow, and its direction can change by the rotation of the fluid parcel. Problem Set 1 (1.1) Consider the flow in the .x; y/ plane given by u D
y, v D x C t. (a) What is the instantaneous streamline through the origin at t D 1? (b) What is

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PROBLEM SET 1

11

the path of the fluid particle initially at the origin, 0 < t < 6
? (c) What is the streak line emanating form the origin, 0 < t < 6
? (1.2) The “point vortex ” flow in two dimensions has the velocity field   x
y ; ; x 2 C y 2 ¤ 0; .u; v/ D UL x2 C y2 x2 C y2 where U; L are reference values of speed and length. (a) Show that the Lagrangian coordinates for this flow may be written x.a; b; t/ D R0 cos .!t C 0 /;

y.a; b; t/ D R0 sin .!t C 0 /

where R02 D a2 C b 2 , 0 D arctan . ab /, and ! D UL=R02 . (b) Consider at t D 0 a small rectangle of marked fluid particles determined by the points A.L; 0/, B.L C x; 0/, C.L C x; y/, and D.L; y/. If the points move with the fluid, once point A returns to its initial position what is the shape of the marked region? Since .x; y/ are small, you may assume the region remains a parallelogram. Do in the Jacobian, evaluated at A.L; 0/. Then this, first, by computing the entry @y @a verify your result by considering the “lag” of particle B as it moves on a slightly larger circle at a slightly slower speed relative to particle A for a time taken by A to complete one revolution. (1.3) We have noted that Lagrangian coordinates can use any unique labeling of fluid particles. To illustrate this, consider the Lagrangian coordinates in two dimensions 1 1 x.a; b; t/ D a C e kb sin k.a C ct/; y D b
e kb cos k.a C ct/; k k where k; c are constants. Note here a; b are not equal to .x; y/ for any t0 . By examining the determinant of the Jacobian, verify that this gives a unique labeling of fluid particles provided that b ¤ 0. What is the situation if b D 0? These waves, which were discovered by Gerstner in 1802, represent gravity waves if c 2 D gk where g is the acceleration of gravity. They do not have any simple Eulerian representation. 2x . (1.4) In one dimension, the Eulerian velocity is given to be u.x; t/ D 1Ct (a) Find the Lagrangian coordinate x.a; t/. (b) Find the Lagrangian velocity as a D J as a function of a; t. function of a; t. (c) Find the Jacobean @x @a U , show that a fluid (1.5) For the stagnation point flow u D .u; v/ D L.x;
y/ particle in the first quadrant that crosses the line y D L at time t D 0, crosses L log . UL / on the streamline Uxy . Do this in the line x D L at time t D U L D 2 C v 2 / along a streamline. E two ways. First, consider the line integral of u  ds=.u Second, use Lagrangian variables. (1.6) R Let S be the surface of a deformable body in three dimensions, and let I D S f n dS for some scalar function f , n being the outward normal. Show that Z Z Z @f d f n dS D n dS C .ub  n/rf dS (1.25) dt @t S

S

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12

1. THE FLUID CONTINUUM

where ub is the velocity of the surface of the body. (Hint: First convert to a volume integral between S and an outer surface S 0 that is fixed. Then differentiate and apply the convection theorem. Finally, convert back to a surface integral.)

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http://dx.doi.org/10.1090/cln/019/02

CHAPTER 2

Conservation of Mass and Momentum 2.1. Conservation of Mass Every fluid we consider is endowed with a nonnegative density, usually denoted by , which in the Eulerian setting is a scalar function of x; t. The units of  are mass per unit volume. Water has a density of about 1 gram per cubic centimeter. For air the density is about 10
3 grams per cubic centimeter, but of course pressure and temperature affect air density significantly. The air in a room of a thousand cubic meters (109 cubic centimeters) weighs about a thousand kilograms, or more than a ton! 2.1.1. Eulerian Form. Let us suppose that mass is being added or subtracted from space as a function q.x; t/ having dimensions of mass per unit volume per unit time. The conservation of mass in a fixed region R can be expressed using (1.20) with f D : Z Z Z @ d dVx C u  n dSx :  dVx D (2.1) dt @t R

R

Now d dt

(2.2)

@R

Z

Z  dVx D

R

q dVx R

and if we bring the surface integral in (2.1) back into the volume integral using the divergence theorem we arrive at  Z
@ C div.u/
q dVx D 0: (2.3) @t R

Since our functions are continuous and R is an arbitrary open set in RN , the integrand in (2.3) must vanish, yielding the conservation of mass equation in the Eulerian form: @ C div.u/ D q: @t Note that this last equation can also be written

(2.4)

(2.5)

D C  div u D q: Dt 13

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14

2. CONSERVATION OF MASS AND MOMENTUM

The conservation of mass equation in either of these forms is sometimes called the equation of continuity . The form given in (2.5) shows that the material derivative of the observed density changes in two ways, either by sources and sinks of mass (q > 0 or q < 0, respectively), or else by the nonvanishing of the divergence of the velocity field. A positive value of the divergence, as for u D .x; y; z/, is associated with an expansive flow, thereby reducing local density. This can be examined more closely as follows. Let V be a small volume of fluid where the density is essentially constant. Then V is the mass within this fluid parcel, which is a material invariant D.V /
1 DV D 0. Comparing this with (2.5) we have D 0. Thus D Dt Dt C V Dt 1 DV : V Dt Thus a positive or negative relative dilation of a fluid parcel is equal to the positive or negative divergence of the velocity field at that parcel. (2.6)

div u D

E XAMPLE 2.1. As we have seen in Chapter 1, an incompressible fluid satisfies div u D 0. For such a fluid, free of sources or sinks of mass, we have D D 0; Dt that is, density now becomes a material property. This does not say that the density is constant everywhere in space, only that it is constant at a given fluid parcel. (Note that we use the word parcel here to suggest that we have to average over a small volume to compute the density.) However, a fluid of constant density without mass addition must be incompressible. This difference is important. Sea water is essentially incompressible but density changes due to salinity are an important part of the dynamics of the oceans. (2.7)

2.1.2. Lagrangian Form. If q D 0 the Lagrangian form of the conservation of mass is very simple. If we move with the fluid the observed density changes are due to expansion and dilation of the fluid parcel, which is controlled by Det.J/. Let a parcel have volume V0 initially, with essentially constant initial density 0 . Then the mass of the parcel is 0 V0 and is a material invariant. At later times the density is  and the volume is V0 Det.J/, and V0 Det.J/ D 0 V0 , so conservation of mass is expressed by 0 (2.8) Det J.a; t/ D :  If q ¤ 0 the Lagrangian conservation of mass becomes ˇ @ ˇˇ  Det.J/ D Det.J/q.x.a; t/; t/: (2.9) @t ˇa


The origin of this use of “continuity” lies in the relation between the integral form of a conservation law and its local form. The latter requires the existence of the divergence of a flux and hence a certain smoothness of the fluid flow.

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2.1. CONSERVATION OF MASS

15

It is easy to get from Eulerian to Lagrangian form using (1.14). Assuming q D 0, (2.10)

D D Det.J/=Dt 1 D D C  div u D 0 D C D . Det.J// Dt Dt Det.J/ Det.J/ Dt

and the connection is complete. 2xt E XAMPLE 2.2. Consider the unsteady velocity field u.x; t/ D 1Ct 2 in one dimension. We assume no sources or sinks of mass and set .x; 0/ D x. What is the density field at later times in both Eulerian and Lagrangian forms? First note that this is a reasonable question, since we have a conservation of mass equation with which to evolve the density in time. Deriving the Lagrangian coordinates, we have 2xt dx D ; x.0/ D a: (2.11) dt 1 C t2

The solution is x D a.1 C t 2 /. The Jacobian is then J D 1 C t 2 . The equation of conservation of mass in Lagrangian form, given that 0 .a/ D a, is  D a=.1Ct 2 /: Since a D x=.1 C t 2 /, the Eulerian form of the density is  D x=.1 C t 2 /2 . It is easy to check that this last expression satisfies the Eulerian conservation of mass equation in one dimension t C .u/x D 0. E XAMPLE 2.3. Consider the two-dimensional stagnation point flow .u; v/ D .x;
y/ with initial density 0 .x; y/ D x 2 C y 2 and q D 0. The flow is incompressible, so  is material. In Lagrangian form, .a; b; t/ D a2 C b 2 . To find  as a function of x; y; t, we note that the Lagrangian coordinates of the flow are .x; y/ D .ae t ; be
t /, and so (2.12)

.x; y; t/ D .xe
t /2 C .ye t /2 D x 2 e
2t C y 2 e 2t :

The lines of constant density, which are initially circles centered at the origin, are flattened into ellipses by the flow. 2.1.3. Another Convection Identity. Often fluid properties are most conveniently thought of as densities per unit mass rather than per unit volume. If the conservation of such a quantity, f say, is to be examined, we will need to consider f to get “f per unit volume” and so be able to compute the total amount by integration over a volume. Consider then  Z Z
@f d C div.f u/ dVx : f dVx D (2.13) dt @t St

St

We now assume conservation of mass with q D 0. From the product rule of differentiation we have div.f u/ D f div.u/ C u  rf , and so the integrand splits into a part that vanishes by conservation of mass, and a material derivative of f times the density: Z Z Df d dVx : f dVx D  (2.14) dt Dt St

St

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16

2. CONSERVATION OF MASS AND MOMENTUM

Thus the effect of the multiplier  is to turn the derivative of the integral into an integral of a material derivative. 2.2. Conservation of Momentum in an Ideal Fluid The momentum of a fluid is defined to be u per unit volume. Newton’s second law of motion states that momentum is conserved by a mechanical system of masses if no forces act on the system. We are thus in a position to use (2.14), where the “sources and sinks” of momentum are forces. If F.x; t/ is the force acting on the fluid per unit volume, then we have immediately (assuming conservation of mass with q D 0), 

(2.15)

Du D F: Dt

Since we have seen that Du Dt is the fluid acceleration, (2.15) states Newton’s law that mass times acceleration equals force in both magnitude and direction. Of course, the Lagrangian form of (2.15) is obtained by replacing the acceleration by its Lagrangian counterpart: ˇ @2 x ˇˇ (2.16)  2 ˇ D F: @t a The main issues we face with conservation of momentum are those connected with the forces that act on a parcel of fluid. There are many possible forces to consider: pressure, gravitational, viscous, surface tension, electromotive, etc. Each has a physical origin and a mathematical model with a supporting set of observations and analyses. In the present chapter we consider only an ideal fluid. The only new fluid variable we will need to introduce is the pressure, a scalar function p.x; t/. In general, the force F appearing in (2.15) is assumed to take the form Fi D fi C

(2.17)

@ij : @xj

Here f is a body force (exerted from the “outside”), and
is a second-order tensor called the stress tensor. Integrated over a region R, the force on the region is Z Z Z F dVx D f dVx C
 n dSx ; (2.18) R

R

@R

using the divergence theorem. We can thus see that the effect of the stress tensor is to produce a force on the boundary of any fluid parcel, the contribution from an area element to this force being ij nj dSx for an outward normal n. The remaining body force f will sometimes be taken to be a uniform gravitational field f D g, where g D const. On the surface of the earth, gravity acts toward the Earth’s center with a strength g  980 cm/sec2 . We also introduce a general force potential ˆ such that f D
rˆ.

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2.2. CONSERVATION OF MOMENTUM IN AN IDEAL FLUID

17

2.2.1. The Pressure. An ideal fluid is defined by a stress tensor of the form 0 1
p 0 0 0A (2.19) ij D
pıij D @ 0
p 0 0
p where

( ıij D

1 0

for i D j; otherwise.

Thus when pressure is positive the force on the surface of a parcel is opposite to the outer normal, as intuition suggests. Note that now div
D
rp:

(2.20)

For a compressible fluid the pressure accounts physically for the resistance to compression. But pressure persists as a fundamental source of surface forces for an incompressible fluid, and its physical meaning in the incompressible case is subtle.
An ideal fluid with no mass addition and no body force thus satisfies (2.21)



Du C rp D 0; Dt

together with (2.22)

D C  div u D 0: Dt

The equations in this system for an ideal fluid are also often referred to as Euler’s equations. The term Euler flow for the velocity field is also in wide use. With Euler’s system we have N C 1 equations for the N C 2 unknowns u1 ; : : : ; uN ; ; p. Another equation will be needed to complete the system. One possibility is the incompressibility assumption div u D 0. A useful and common option is to assume constant density. Then  is eliminated as an unknown and the conservation of mass equation is replaced by the incompressibility condition. For gases the missing relation is an equation of state, which brings into our model the thermodynamic properties of the fluid. The pressure force as we have defined it above is isotropic in the sense the pressure is the same independently of the orientation of the area element on which it acts. A simple two-dimensional diagram will illustrate why this is so; see Figure 2.1. Suppose that the pressure is pi on the face of length Li . Equating forces, we have p1 L1 cos  D p2 L2 and p1 L1 sin  D p3 L3 . But L1 cos  D L2 and L1 sin  D L3 , so we see that p1 D p2 D p3 . So indeed the pressure sensed by a face does not depend upon the orientation of the face.


One aspect of the incompressible case should be noted here, namely that the pressure is arbitrary up to an additive function of time. Consequently, it is only pressure differences that matter. This is not the case for a compressible gas.

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18

2. CONSERVATION OF MASS AND MOMENTUM

L1

L3

θ L2 F IGURE 2.1. Isotropicity of pressure.

2.2.2. Lagrangian Form of Conservation of Momentum. The Lagrangian form of the acceleration has been noted above. The momentum equation of an ideal fluid requires that we express rp as a Lagrangian variable. That is, if p is to be a function of a; t, then since r here is actually the x gradient rx , we have rx p D J
1 ra p. This appearance of the Jacobian is an awkward feature of Lagrangian fluid dynamics and is one of the reasons why we shall emphasize Eulerian variables in discussing the dynamics of a fluid. 2.2.3. Hydrostatics: The Archimedean Principle. Hydrostatics is the study of fluids at rest (u D 0), usually in the presence of gravity. We consider here only the case of a fluid stratified in one dimension. To fix the coordinates, let the positive z-axis be vertical up, and g D
giz , where g is a positive constant. We suppose that the density is a function of z alone. This allows, for example, a body of water beneath a stratified atmosphere. Let a solid three-dimensional body (any deformation of a sphere, for example) be submerged in the fluid. Archimedes’ principle says that the force exerted by the pressure on the surface of the body is equal to the total weight of the fluid displaced by the body. We want to establish this principle in the case considered. Now for a fluid at rest Rthe pressure satisfies rp D
g.z/iz . The pressure force is given by Fp D
pn dS taken over the surface of the body. But this surface pressure is just the same as would be acting on a virtual surface within the fluid with no body present. Using the divergence theorem, we may convert this to an integral over the region interior to this surface. Of course, there is no fluid within the body. WeR are just using the math to evaluate the surface integral. The result is Fp D giz  dV . This is a force upward equal to the weight of the displaced fluid, as stated. 2.3. Steady Flow of a Fluid of Constant Density This special case gives us an opportunity to obtain some useful results rather easily in a class of problems of some importance. We shall allow a body force of the form f D
rˆ, so the momentum equation may be written, after division by the constant density, (2.23)

u  ru C 
1 rp C rˆ D 0:

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2.3. STEADY FLOW OF A FLUID OF CONSTANT DENSITY

19

The following vector identity will be useful: (2.24)

A  .r  B/ C B  .r  A/ C A  rB C B  rA D r.A  B/:

Applying this to A D B D u, we obtain the important identity (2.25)

u  ru D

1 rjuj2
u  .r  u/: 2

Using (2.25) in (2.23), we have   1 2
1 (2.26) r  p C ˆ C juj D u  .r  u/: 2 Taking the dot product with u on both sides, we obtain   1 2
1 (2.27) u  r  p C ˆ C juj D 0: 2 The famous Bernoulli theorem for steady flows follows: In the steady flow of an ideal fluid of constant density, the quantity H  
1 p C ˆ C 12 juj2 , called the Bernoulli function, is constant on the streamlines of the flow. The importance of this result is in the relation it gives us between velocity and pressure. Apart from the contribution of ˆ, the constancy of H implies that an increase of velocity is accompanied by a decrease of the pressure. This is not an obvious dynamical consequence of the equations of motion, and it is interesting that we have derived it without referring to the solenoidal property of u. Recall that the latter is implied by the constancy of density when there is no mass added or removed. If we make use of the solenoidal property, then, using the identity r  .A / D r  A C A  r for vector and scalar fields, we see that uH is also solenoidal, and so the flux of this quantity is conserved in stream tubes. This vector field arises from the conservation of mechanical energy, relating changes in kinetic energy to the work done by forces; see Problem 2.2. It is helpful to apply the Bernoulli theorem to flow in a smooth rigid pipe of circular cross section and slowly varying diameter with ˆ D 0. For an ideal (frictionless) fluid we may assume that the velocity is approximately constant over the section, this being reasonable if the slope of the wall of the pipe is small. The velocity may thus be taken as a scalar function u.x/. If the section area is A.x/, then the conservation of mass (and here, volume) implies that uA  Q D const, so that 
1 p C.Q2 =2/A
2 D const. If we consider a contraction, as in Figure 2.2, where the area and velocity go from A1 ; u1 to A2 ; u2 , then the fluid speeds up to satisfy A1 u1 D A2 u2 D Q. To achieve this speedup in steady flow, a force must be acting on the fluid, here a pressure force. Conservation of momentum states the flux of momentum out minus the flux of momentum in must equal the pressure force 2 on the fluid in the pipe between section 1 and section 2. Now H D p C 12 . Q A/ is constant, so (if force is positive to the right) the two ends of the tube give a net pressure force   Q2 1 1
p1 A1
p2 A2 D  2 A2 A1

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20

2. CONSERVATION OF MASS AND MOMENTUM

F IGURE 2.2. Steady flow through a contraction.

acting on the fluid. But there is also a pressure force along the curved part of the tube. This is seen to be   Z A2  2 Z A2 Q2 1 1  Q p dA D
dA D 
: A 2 A2 A1 A1 A1 2 These two contributions are equal in our one-dimensional approximation, and their sum is Q2 .1=A2
1=A1 /. But the momentum out minus momentum in is   1 1 2 2 2
.A2 u2
A1 u1 / D Q A2 A1 and is indeed equal to the net pressure force. Intuitively then, to achieve the speedup of the fluid necessary to force the fluid through a contraction, and to maintain such a flow as steady in time, it is necessary to supply a larger pressure at station 1 than at station 2. Bernoulli’s theorem captures this creation of momentum elegantly, but ultimately the physics comes down to pressure differences accelerating fluid parcels. It should be kept in mind that we are dealing here with an ideal fluid. In actual pipe flow viscous effects at the boundary can substantially change the movement of the fluid there and lead to time-dependent turbulent motions superimposed on a mean flow. 2.4. Intrinsic Coordinates in Steady Flow The one-dimensional analysis just given suggests looking briefly at the relations obtained in an arbitrary steady flow of an ideal fluid using the streamlines as part of the coordinate system. The resulting intrinsic coordinates are revealing of the dynamics of fluid parcels. Let t be the unit tangent vector to an oriented streamline. Then we may write u D qt, q D juj. If s is arc length along the streamline, @q @t @q @u D tCq D t C qn; @s @s @s @s where n is the unit normal,  the streamline curvature, and we have used the first @ , and so we have from Frenet-Serret formula. Now the operator u  r is just q @s (2.28) (2.28)

(2.29)

u  ru D q

@q t C q 2 n: @s

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2.5. POTENTIAL FLOWS WITH CONSTANT DENSITY

21

This shows that the acceleration in steady flow splits into a component along the streamline, determined by the variation of q, and a centripetal acceleration associated with streamline curvature. The equations of motion in intrinsic coordinates (zero body force) are therefore @p @p @q C D 0; q 2 C D 0: (2.30) q @s @s @n What form does the solenoidal condition take in intrinsic coordinates? We consider this question in two dimensions. We have @q C qr  t: (2.31) r  u D r  .qt/ D t  rq C qr  t D @s Let us introduce an angle  so that t.s/ D .cos .s/; sin .s//: Then @ @ @ C cos  D n  r D : (2.32) r  t D
sin  @x @y @n @ is the streamline curvature, @n , which we write as n , is the curvature Since  D @ @s of the coordinate lines normal to the streamlines. Thus the solenoidal condition in two dimensions assumes the form @q C qn D 0: (2.33) @s

2.5. Potential Flows with Constant Density Another important and very large class of fluid flows are the so-called potential flows, defined as flows having a velocity field that is the gradient of a scalar potential, usually denoted by : (2.34)

u D r:

Since then r  u D r  r D 0, a potential flow has the fundamental property of being irrotational. For simplicity we consider here only the case of constant density, but allow a body force
rˆ and permit the flow to be unsteady. Since we now also have that u is solenoidal, it follows that (2.35)

r  r D r 2  D 0:

Thus the velocity field is determined by solving Laplace’s equation (2.35). The momentum equation has not yet been needed, but it must be used to determine the pressure, given u. The momentum equation is   1 2 p (2.36) ut C r juj C C ˆ D u  .r  u/: 2  Since u D r and r  u D 0 we have   1 p 2 (2.37) r t C jrj C C ˆ D 0; 2  or 1 p (2.38) t C jrj2 C C ˆ D h.t/: 2 

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22

2. CONSERVATION OF MASS AND MOMENTUM

The arbitrary function h.t/ may in fact be set equal to 0; otherwise we can replace  R by 
h dt without affecting u. We see that (2.38) determines another “Bernoulli constant,” this time applicable to any region of space where the potential flow is defined. This equation is a “Bernoulli theorem” that allows us to compute the pressure in an unsteady potential flow; see Problem 2.6. 2.6. Boundary Conditions on an Ideal Fluid In Chapter 6 we shall take up the study of viscous fluids and introduce the viscosity of the fluid in the development of a more general stress tensor. The ideal fluid is assumed to have zero viscosity, which endows it with a physical quality that might be described as “slippery.” With zero viscosity there is no force to prevent a fluid from moving along and adjacent to a fixed boundary. Suppose that the pressure varies along a solid wall. The only force that can drive a fluid parcel along the wall is a pressure force associated with the pressure gradient. If the tangential gradient at the wall is nonzero, fluid will be accelerated and there will have to be a tangential component of velocity at the wall. This suggests, since we have no direct control over the pressure, that we cannot place any restriction on the tangential component of velocity at a rigid fixed boundary of the fluid. On the other hand, at a rigid fixed wall the fluid is unable to penetrate the wall, and so we will have to impose the condition n  u D un D 0. There is, however, a subtle point connected with our continuum approximation. It is clear that the fluid cannot penetrate into a rigid wall, but could it not be possible for the fluid to tear off the wall, forming a free interface next to an empty cavity? To see that this cannot be the case for smooth pressure fields, consider the reversed stagnation point flow .u; v/ D .
x; y/. On the upper y-axis we have a Bernoulli function p= C 12 y 2 . The gradient of pressure along this line is indeed accelerating the fluid away from the wall, but the fluid remains at rest at x D y D 0. We cannot really contemplate a pressure force on a particle, only on a parcel. In fact, in this example fluid parcels near the y-axis are being compressed in the x-direction and stretched in the y-direction. Thus, the appropriate boundary condition at a fixed rigid wall adjacent to an ideal fluid is (2.39)

un D 0

on the wall.

For a potential flow, this becomes @ D 0 on the wall. @n We shall find that these conditions at a rigid wall for an ideal fluid are sufficient to (usually uniquely) determine fluid flows in problems of practical importance. Another way to express the appropriate boundary condition on a ideal fluid at a rigid wall is that fluid particles on a wall stay on the wall. This alternative is attractive because it is also true of a moving rigid wall, where the velocity component normal to the wall need not vanish. To obtain the mathematical formulation of the

(2.40)

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2.6. BOUNDARY CONDITIONS ON AN IDEAL FLUID

23

boundary condition on a moving surface, it is convenient to define the surface as a function of time by the equation †.x; t/ D 0. For a particle at position xp .t/ to be on the surface means that †.xp .t/; t/ D 0. Differentiating this expression with respect to time we obtain ˇ @† ˇˇ C u  r† D 0: (2.41) @t ˇ x

For example, let a rigid cylinder of radius a move in the x-direction with velocity U . Then † D .x
Ut/2 C y 2
a2 , and (2.41) becomes
2U.x
Ut/ C 2.x
Ut/u C 2yv D 0. Evaluating this on the surface of the cylinder, we get (2.42)

u cos  C v sin  D U cos  D un :

We remark that the same reasoning can be applied to the moving interface between two fluids. This interface may also be regarded as consisting of fluid particles that remain on the interface. We refer to this generalized boundary condition as a surface condition. Finally, as part of this first look at the boundary condition of fluid dynamics, we should note that for unsteady fluid flows we will need to prescribe initial conditions, insuring that the fluid equations may be used to carry the solution forward in time. E XAMPLE 2.4. Consider potential flow past a body in two dimensions with constant density and no body force. The body is the circular cylinder r D a, and the fluid “at infinity” has fixed velocity .U; 0/. In two-dimensional polar coordinates, Laplace’s equation has solutions of the form ln r, .r n ; r
n /.cos ; sin /, n D 1; 2; : : : . The potential Ur cos  D Ux has the correct behavior at infinity, D0 and so we need a decaying solution that will insure the boundary condition @ @r
1 when r D a. The correct choice is clearly a multiple of r cos  and we obtain   a2 : (2.43)  D U cos  r C r Note that U cos a2 =r is the potential of a flow seen by an observer at rest relative to the fluid at infinity, when the cylinder moves relative ˇ to the fluid with a velocity ˇ D
U cos  as required .
U; 0/. We see that indeed this potential satisfies @ @r rDa by (2.42). Streamlines both inside and outside the cylinder are shown in Figure 2.3. We have found a solution representing the desired flow, but is the solution unique? Perhaps surprisingly, the answer is no. The reason, associated with the fluid region being non-simply-connected, will be discussed in Chapter 4. E XAMPLE 2.5. An interesting case of unsteady potential flow occurs with deep water waves (constant density). The fluid at rest is a liquid in the domain z < 0 of R3 . Gravity acts downward so ˆ D
gz. The space above is taken as having no density and a uniform pressure p0 . If the water is disturbed, waves can form on the surface, which we will assume to be described by a function z D Z.x; y; t/ (no breaking of waves, a condition we assume to be realizable for sufficiently smallwave amplitudes). Under appropriate initial conditions it turns out that we may

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24

2. CONSERVATION OF MASS AND MOMENTUM

F IGURE 2.3. Potential flow past a circular cylinder.

assume the liquid velocity to be a potential flow. Thus our mathematical problem is to solve Laplace’s equation in z < Z.x; y; t/ with a surface condition on  and a pressure condition pzDZ D p0 . For the latter we can use the Bernoulli theorem for unsteady potential flows to obtain
 1 p0 2 D
t
jrj C gz : (2.44)  2 zDZ D .z
Z.x; y; t// D 0 or The surface condition is Dt   (2.45) z
Zt
uZx
vZy zDZ D 0:

The object is to find .x; y; z; t/; Z.x; y; t/, given, e.g., that the water is initially at rest and that the fluid surface is at an initial elevation z D Z0 .x; y/. Problem Set 2 (2.1) For potential flow over a circular cylinder, with pressure equal to the constant p1 at infinity, find the static pressure on the surface of the cylinder as a function of angle from the front stagnation point. (Use Bernoulli’s theorem.) Evaluate the drag force (the force in the direction of the flow at infinity that acts on the cylinder) by integrating the pressure around the boundary. Verify that the drag force vanishes. This is an instance of d’Alembert’s paradox, the vanishing of drag of bodies in steady potential flow. (2.2) For an ideal inviscid fluid of constant density with no gravity, the conservation of mechanical energy is studied by evaluating the time derivative of total kinetic energy in the form Z Z 1 d 2 juj dV D F  n dS: dt 2 D

@D

Here D is an arbitrary fixed domain with smooth boundary @D. What is the vector F? Interpret the terms of F physically.

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PROBLEM SET 2

25

(2.3) An open rectangular vessel of water is allowed to slide freely down a smooth frictionless plane inclined at an angle ˛ to the horizontal in a uniform vertical gravitational field of strength g. Find the inclination to the horizontal of the free surface of the water, given that it is a surface of constant pressure. We assume the fluid is at rest relative to an observer riding on the vessel. (Consider the acceleration of the fluid particles in the water and balance this against the gradient of pressure.) (2.4) Water (constant density) is to be pumped up a hill (gravity = .0; 0;
g/) through a pipe that tapers from an area A1 at the low point to the smaller area A2 at a point a vertical distance L higher. What is the pressure p1 at the bottom, needed to pump at a volume rate Q if the pressure at the top is the atmospheric value p0 ? (Express in terms of the given quantities. Assuming inviscid steady flow, use Bernoulli’s theorem with gravity and conservation of mass. Assume that the flow velocity is uniform across the tube in computing fluid flux and pressure.) (2.5) For a barotropic fluid, pressure is a function of density alone, p D p./. In this case derive the appropriate form of Bernoulli’s theorem for steady flow p without gravity. If p D k where ; k are positive constants, show that q 2 C 2
1  is constant on a streamline, where q D juj is the speed of the fluid. (2.6) Water fills a truncated cone as shown in Figure 2.4. Gravity acts down (the direction
z). The pressure at the top surface, of area A1 , is 0. The height of the container is H . At t D 0 the bottom, of area A2 < A1 , is abruptly removed and the water begins to fall out. Note that at time t D 0C the pressure at the bottom surface is also 0. The water has not moved but the acceleration is nonzero. We may assume the resulting motion is a potential flow. Thus the potential .z; r; t/ D in cylindrical polars has the Taylor series .r; z; t/ D t ˆ.r; z/ C O.t 2 /, so d dt ˆ.r; z/ C O.t/. Using these facts, set up a mathematical problem for determining the pressure on the inside surface of the cone at t D 0C. You should specify all boundary conditions. You do not have to solve the resulting problem, but can you guess what the surfaces ˆ D const would look like qualitatively? What is the force felt at t D 0C by someone holding the cone in the limits A2 ! 0 and A1 ! A2 ?

A

1

A2 F IGURE 2.4. Truncated cone of fluid (Problem 2.6).

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http://dx.doi.org/10.1090/cln/019/03

CHAPTER 3

Vorticity We have already encountered the vorticity field in the identity 1 u  ru D r juj2
u  .r  u/: 2 The vorticity field !.x; t/ is defined from the velocity field by

(3.1)

(3.2)

! D r  u:

A flow field where the vorticity field is not identically zero is said to be rotational and is otherwise irrotational. Any smooth flow is irrotational if and only if it is a potential flow. Potential flows are governed by Laplace’s equation and are determined by boundary conditions. We shall see, however, that vorticity acts as a defining source of fluid flow in the absence of boundary influence, a kind of internal “skeleton” that determines the structure of the flow. According to (3.1) the contribution to the acceleration coming from the gradient of velocity can be split into two components, one having a potential 12 juj2 , the other given as a cross product orthogonal to both the velocity and the vorticity. The latter component in older works in fluid dynamics has been called the vortex force. We remark that, in analogy with stream lines, we shall refer to the instantaneous flow lines of the vorticity field, i.e., the integral curves of the system (3.3)

dy dz dx D D ; !x !y !z

as (instantaneous) vortex lines. Similarly, in analogy with a stream tube in three dimensions, we will refer to a bundle of vortex lines as a vortex tube. This straightforward definition of the vorticity field gives little insight into its importance, either physically or theoretically. This chapter will be devoted to examining the vorticity field from a variety of viewpoints. 3.1. Local Analysis of the Velocity Field The first thing to be noted is that vorticity as defined is fundamentally an Eulerian property since it involves spatial derivatives of the Eulerian velocity field. In a sense the analytical structure of the flow is being enlarged to include the first derivatives of the velocity field. Suppose we expand the velocity field in a Taylor series about the fixed point x: (3.4)

ui .x C y; t/ D ui .x; t/ C yj

@ui .x; t/ C O.jyj2 /: @xj

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28

3. VORTICITY

We can make the division

  1 @ui @uj @uj 1 @ui @ui D C
C : (3.5) @xj 2 @xj @xi 2 @xj @xi @u

@ui C @xj , is often denoted by eij and is the The term first term on the right, 12 Π@x j i rate-of-strain tensor of the fluid. Note that it is symmetric in the indices i; j . It will play a basic role when viscous stresses are considered (Chapter 6). The second @u @ui
@xj , can be seen to be, in three dimensions, the matrix term, 12 Π@x j i 1 0 0
!3 !2 1 0
!1 A : (3.6)  D @ !3 2
! ! 0 2

1

We observe that  is a skew-symmetric matrix formed from the components of the vorticity vector. E XAMPLE 3.1. In two dimensions, since u; v depend only on x; y, only one @v
@u . This is usually written component of the vorticity is nonzero, !3 D @x @y simply as the scalar !. Consider the two-dimensional flow .u; v/ D .y; 0/. In this case     1 1 0 1 0 1 ; D : (3.7) eD 2 1 0 2
1 0 and ! D
1. This is a simple, planar “shear flow,” with horizontal particle paths. Both e and  are nonvanishing. E XAMPLE 3.2. Consider the flow .u; v/ D .
y; x/. This is a simple solidbody rotation in the anticlockwise sense. The vorticity is ! D 2 and e D 0. These examples are a bit atypical because the vorticity is constant, but they emphasize that a close association of the vorticity with fluid rotation, a connection suggested by the skew-symmetric form of , can be misleading. Vorticity is a point property but can only be defined by the limit operations implicit in the definition of the derivatives. It is impossible to attach a physical meaning to “the vorticity of a particle.” We can truncate (3.4) and consider the Lagrangian paths of fluid particles near x. Since e is real symmetric, it may be diagonalized by a rotation to principal axes. Let the eigenvalues along the diagonal be i , i D 1; 2; 3. We may assume our coordinate system is such that e is the diagonal matrix D.x/. Then the Lagrangian coordinates of the perturbed path y satisfy 1 dy D D.x/  y C !.x/  y: dt 2 These equations couple together the rotation associated with the vorticity at x with the straining field described by the first term. Note that the angular velocity associated with the second term is 12 !. The statement “vorticity at x equals twice the angular velocity of the fluid at x” is often heard. But this statement in fact makes (3.8)

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3.2. CIRCULATION

29

no sense, since an angular velocity cannot be attributed to a point. Given the velocity field of a fluid, one can determine the effects of vorticity on the fluid only on a small open set, i.e., a fluid parcel. On the other hand, it is true that when vorticity is continuous and nonzero, and sufficiently large relative to the rate of strain matrix, then there will be a sensible rotation observed in the fluid, and it is true that when one sees “rotation” in the fluid, then vorticity is present. In a sense this is the key to understanding its role, since it forces a definition of “rotation” in a fluid. We now consider a quantity that offers a precise definition of rotation of a fluid parcel.

3.2. Circulation Let C be a simple, smooth, oriented closed contour that is a deformation of a circle, hence the boundary of an oriented surface S . Stokes’ theorem applied to the velocity field states that Z

Z u  dx D

(3.9) C

n  .r  u/dS; S

where the direction of the normal n to S is chosen from the orientation of C by the “right-hand rule.” We can interpret the right-hand side of (3.9) as the flux of vorticity through S . So it must be that the left-hand side is an expression of the effect of vorticity on the velocity field. We thus define the fluid circulation of the velocity field u on the contour C by I (3.10)

C D u  d x: C

The circulation is going to be our measure of the rotation of the fluid. The key “point” is that circulation is defined globally, not at a point. We need to consider an open set containing S in order to make this definition. E XAMPLE 3.3. Potential flows have the property that circulation vanishes on any closed contour as long as u is well behaved in an open set containing S . This is an obvious property of an irrotational flow. 1 .
y=r 2 ; x=r 2 / is a E XAMPLE 3.4. In two dimensions, the flow .u; v/ D 2 point vortex centered at the origin. If C is a simple closed curve encircling the origin, then C is equal to the circulation on a circle centered at the origin, by independence of path since .u; v/ is irrotational everywhere except at the origin. The circulation on a circle, taken counterclockwise, is found to be unity. Indeed, in polar form the velocity is given by ur D 0, u D 21 r . The circulation on the r circle of radius r is thus 2 2 r D 1. This flow may be termed the point vortex of unit strength, as measured by the circulation about its center.

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30

3. VORTICITY

3.3. Kelvin’s Theorem for a Barotropic Fluid In Chapters 9 through 12 we will be taking up the dynamics of general compressible fluids. The present discussion will deal with a restricted class of compressible flows, the barotropic fluids. A barotropic fluid is defined by specifying pressure as a given function of the density, p./. This reduces the dependent variables of an ideal fluid to u; , and so the system of momentum and mass equations is closed. T HEOREM 3.1 (Kelvin’s theorem) Let C.t/ be a simple closed material curve in an ideal fluid with body force
rˆ. Then, if either (i)  D const or (ii) the fluid is barotropic, then the circulation C.t/ of u on C is invariant under the flow: d

C.t/ D 0: dt To prove this, consider a parametrization x.˛; t/ of C.t/, 0  ˛  A. Then  I Z A Z A
@u d @x Du @x d d˛ D  Cu d˛: u  dx D u (3.12) dt dt 0 @˛ Dt @˛ @˛ 0

(3.11)

C

1 Making use of the momentum equation Du Dt C  rp C rˆ D 0 we have   Z A
 @x @u 1 d C D Cu d˛:
rp C rˆ  (3.13) dt  @˛ @˛ 0

This becomes (3.14)

d C D dt

I


  1 2
dp Cd juj
ˆ :  2

C

Now if  is a constant, or if the fluid is barotropic, the integrand may be written R d C as a perfect differential (in the barotropic case a differential of

1 dp d 1 2 2 juj


ˆ). Since all variables are assumed single-valued, the integral vanishes and the theorem is proved. Kelvin’s theorem is a cornerstone of ideal fluid theory since it expresses a global property of vorticity, namely the flux through a surface, as an invariant of the flow. We shall see that it is very useful in understanding the kinematics of vorticity. In fact, it will enable a direct Lagrangian description of the dynamics of vorticity, even though vorticity is defined as an Eulerian vector field. 3.4. The Vorticity Equation In the present section we again assume that either  D const or else that the fluid is barotropic. In either case it is of interest to consider an equation for vorticity, which can be obtained by taking the curl of 1 Du C rp C rˆ D 0: (3.15) Dt 

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3.4. THE VORTICITY EQUATION

31

Under the conditions stated, this will give Du D 0: (3.16) r Dt Recalling u  ru D r 12 juj2
u  !, we use the vector identity (3.17)

r  .A  B/ D B  rA
A  rB C Ar  B
Br  A:

For the case of constant density and no mass addition, both r  u and r  ! vanish, with the result D! D !  ru: (3.18) Dt For a barotropic fluid, we need to bring in conservation of mass to evaluate r  u D

1 D Dt . We then get in place of (3.18) (3.19)

! D D! D !  ru C : Dt  Dt

This can be rewritten as ! D.!=/ D  ru: Dt  Now we want to compare (3.18) and (3.20) with (1.23), and observe that ! in the first case and !  in the second is a material vector field as we defined it in Chapter 1. This is a deep and remarkable property of the vorticity field, which largely accounts for its importance in fluid mechanics. It tells us, for example, that vorticity magnitude can be increased if two nearby fluid particles lying on the same vortex line move apart. One can apply to vorticity all of the intuition associated with the concept of a material vector field, as a field that moves with or is “frozen into” the fluid flow. In three dimensions the term !  ru is sometimes called the vortex-stretching term. It is the possibility of line stretching that will make twoand three-dimensional vorticity behaviors entirely different. (3.20)

E XAMPLE 3.5. In two dimensions !  ru D 0 and so the vorticity ! satisfies D! D 0; (3.21) Dt i.e., in two dimensions, for the cases studied here, vorticity is a scalar material invariant whose value is always the same on a given fluid parcel. We turn now to a Lagrangian form of the vorticity equation, a representation due to Gauss. We can obtain it here by recalling that vi .a; t/ D Jij .a; t/Vj .a/ defines a material vector field. Let us assume that, given the initial velocity and therefore initial vorticity fields, vorticity may be determined uniquely at some time t using Euler’s equations. Then any material vector field assuming the assigned initial values for vorticity must be the unique vorticity field !. However, if the initial vorticity is !0 .x/, then a material vector field that takes on these initial values is J.a; t/  !0 .a/. By uniqueness, we must have (3.22)

!i .a; t/ D Jij .a; t/!0j

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32

3. VORTICITY

C B A

D

F IGURE 3.1. A segment of an oriented vortex tube.

in the constant density case. For the barotropic case, given initial density 0 .x/, the corresponding equation is (3.23)


1 !i .a; t/ D 0
1 .a/Jij .a; t/!0j :

This is Cauchy’s “solution” of the vorticity equation. Of course, nothing has been solved, only represented in terms of the unknown Jacobian. It is, however, a revealing relation that directly ties the changes in vorticity to the deformation experienced by a fluid parcel and thus to the Lagrangian coordinates of the flow. 3.5. Helmholtz’ Laws To understand the behavior of vorticity in a fluid flow, we will want to consider as our basic element a section of a vortex tube as shown in Figure 3.1. Recall that a vortex tube is a bundle of vortex lines, each of the lines being the instantaneous flow lines of the vorticity field. In the mid-nineteenth century Helmholtz laid the foundations for the mechanics of vorticity. His conclusions can be summarized by the following three laws:  Fluid parcels free of vorticity stay free of vorticity.  Vortex lines are material lines.  The strength of a vortex tube (to be defined below) is an invariant of the motion. We have seen that the vorticity field, or the field divided by density in the barotropic case, is a material vector field. The vortex lines are the same in each case if ! is the same. Hence particles on a particular vortex line at one time remain on a line at a later time, and so the line is itself material. Thus the tube segment in Figure 3.1 is bounded laterally by a surface of vortex lines. The small patch D in the surface thus carries no flux of vorticity. The bounding contour of this patch is a material curve, and by Kelvin’s theorem the circulation on the contour is a material invariant. Since this circulation is initially 0 by the absence of flux of vorticity through the patch, it will remain 0. Consequently, the lateral boundary of a vortex tube remains a boundary of the tube. It then follows from the solenoidal property of vorticity and the divergence theorem that the flux of vorticity through the end surface A must equal that through the end surface B. This flux is a property of a vortex tube, called the vortex tube strength. Note that this is independent of the compressibility or incompressibility

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3.6. THE VELOCITY FIELD CREATED BY A GIVEN VORTICITY FIELD

33

of the fluid. The tube strength simply expresses a property of a solenoidal vector field. To establish the third law of Helmholtz, we must show that this strength is a material invariant. But this follows immediately from Kelvin’s theorem, since the circulation on the contour C of Figure 3.1 is a material invariant. This circulation, for the orientation of the contour shown in the figure, is equal to the vortex tube strength by Stokes’ theorem, and we are done. The first of Helmholtz’ laws is also established using Kelvin’s theorem. Suppose that a flow is initially irrotational but at some time a fluid parcel is found where vorticity is nonzero. A small closed contour can then be found with nonvanishing circulation. This contour will have a preimage at the initial time. Applying Kelvin’s theorem to the preimage and using the irrotationality of the initial flow, we obtain a contradiction. Using these laws we may see how changes in the shape of a fluid parcel can change the magnitude of vorticity. In Figure 3.2 we show a segment of a small vortex tube that has changed under the flow from having length L1 and section area A1 to having new values A2 ; L2 . If the density is constant, volume is conserved, A1 L1 D A2 L2 . If the vorticity magnitudes are !1 ; !2 , then invariance of the tube strength implies !1 A1 D !2 A2 . Comparing these expressions, !2 =!1 D L2 =L1 . Consequently, for an ideal fluid of constant density the vorticity is proportional to vortex line length. We understand here that by line length we are referring to the distance between two nearby fluid particles on the same vortex line. Thus the growth or decay of vorticity in an ideal fluid flow is intimately connected to the stretching properties of the Lagrangian map which represents the flow. Fluid turbulence is observed to contain small domains of very large vorticity, presumably created by this stretching. For a compressible fluid the volume of the tube need not be invariant, but mass is conserved. Thus we have, introducing the initial and final densities 1 ; 2 , 1 A1 L1 D 2 A2 L2 ;

(3.24)

!1 A1 D !2 A2 :

It follows that (3.25)

L2 !2 =2 D : !1 =1 L1

Thus we see that it is the magnitude of the material field, whether j!j or 
1 j!j, which is proportional to line length. Notice that in a compressible fluid vorticity may be increased by compressing a tube while holding the length fixed, so as to increase the density. 3.6. The Velocity Field Created by a Given Vorticity Field We have remarked that vorticity has a “skeletal” role as a generator of a flow, and we shall now make that suggestion more precise. Suppose that in R3 the vorticity field is nonzero in some region and vanishes at infinity. What velocity field or fields may be created by this vorticity? It is clear that given a vorticity field ! and a vector field u such that r  u D !, another vector field with the

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34

3. VORTICITY

F IGURE 3.2. Deformation of a vortex tube under a flow.

same property is given by v D u C r for some scalar field , so uniqueness is an issue. This places the potential flow in perspective as a component that must be determined apart from the vorticity of the flow. However, under appropriate conditions a unique construction is possible. T HEOREM 3.2 Let the given vorticity field be smooth and vanish strongly at infinity, e.g., for some R > 0 q (3.26) j!j  C r
N ; r > R; r D x 2 C y 2 C z 2 : Then there exists a unique solenoidal vector field u such that r  u D ! and limr!1 juj D 0. This vector field is given by Z .y
x/  !.y/ 1 dVy : (3.27) uD 4
jx
yj3 R3

To prove this, consider the vector field v defined by Z ! 1 dVy : (3.28) vD 4
jx
yj R3

This field exists and, given (3.26), can be differentiated if ! is a smooth function. Let u D r  v. We have the vector identity (3.29)

r  .r  A/ D r.r  A/
r 2 A:

The right-hand side of (3.28) is the unique solution of the vector equation r 2 v D ! that vanishes at infinity. Also, Z Z 1 1 dVy D
!  ry dVy 4
div v D !  rx jx
yj jx
yj R3 R3 
Z !.y/ dVy D 0 (3.30) D
ry  jx
yj R3

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3.7. SOME EXAMPLES OF VORTICAL FLOWS

35

F IGURE 3.3. Rankine’s combined vortex.

by the divergence theorem and the fact that the integral of j!j over r D R is bounded in R if (3.26) holds. Thus u as defined by (3.27) satisfies r  u D !, as required. Also, this vector field is solenoidal since it is the curl of v and vanishes as jxj ! 1. It is also unique. Indeed, if u0 is another vector field with the same properties, then r  .u
u0 / D 0 and so u
u0 D r for some scalar field whose gradient vanishes at infinity. But by the solenoidal property of u; u0 we see that r 2  D 0, and this implies  D const, giving the uniqueness of u. For compressible flows a general velocity field w with vorticity ! will have the form w D u C r where u is given by (3.27) and  is an arbitrary scalar field. The kernel 1 .y
x/  ./ (3.31) 4
jx
yj3 is interesting for the insight it gives into the creation of velocity as a cross product operation. The velocity induced by a small segment of vortex tube is orthogonal to both the direction of the tube and the vector joining the observation point to the vortex tube segment. A similar law relates the magnetic field created by an electric current, where it is known as the Biot-Savart law. 3.7. Some Examples of Vortical Flows We now consider a few examples of ideal fluid flows with nonzero vorticity. 3.7.1. Rankine’s Combined Vortex. This old example is an interesting use of a vortical flow to model a “bathtub vortex” before the depression of the surface of the fluid develops into an open funnel. It will also give us an example of a flow with a free surface. The fluid is a liquid of constant density  with a free surface given by z D Z.r/ in cylindrical polar coordinates; see Figure 3.3. The pressure above the free surface is the constant p0 . The body force is gravitational, f D
giz . The vorticity is assumed to be constant, and the flow a solid-body rotation in a vertical tube bounded by r D a, z < Z. The only nonzero velocity component is the -component u . In r > a, z < Z, Euler’s equations will be solved by the field of a twodimensional point vortex (actually a line vortex). This will be matched with a rigid rotation for r < a so that velocity is continuous: ( a2 =r when r  a, (3.32) u D r when r < a:

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36

3. VORTICITY

Here  is the angular velocity of the core vortex. Now in the exterior region r > a the flow is irrotational and so we have by the Bernoulli theorem for irrotational flows p0 1 2 4
2 pext D
 a r
gz; (3.33)   2 for z < Z, where we have taken Z D 0 at r D 1. The free surface is thus given for r > a by ZD


(3.34)

2 a 4 : gr 2

Inside the vortex core, the momentum equations in cylindrical polar coordinates (see (3.57)–(3.60) below) reduce to (3.35)

u2 1 @p D  D 2 r;  @r r

1 @p D g:  @z

Thus (3.36)

pcore 1 2 2  r
gz C C  ; 2 

r < a; z < Z:

On the cylinder r D a, with z < Z, we require that the pcore D pext , so (3.37)

1 p0 1 2 2  a
gz C C D
gz
2 a2 : 2  2

Therefore the constant C is given by C D

(3.38) and (3.39)

p0
2 a 2 

  p0 r2 pcore 2 2 D
 a 1
2
gz:   2a

The free surface is then given by 8 4 2 a that is the same as the irrotational flow past a circular cylinder of radius a. That potential flow is easily re-expressed in terms of a harmonic stream function. Thus     a2 a2 ; r > a: (3.43) D Uy 1
2 D U sin  r
r r Setting D h.r/ sin  in r 2 C k 2 D 0 we obtain the ODE for the Bessel functions of order 1. A solution regular in r < a is therefore h D CJ1 .kr/. Thus (3.44)

D C sin J1 .kr/:

Also, (3.45)

! D C k 2 sin J1 .kr/:

We have two constants to determine, and we will do this by requiring that both ! and u be continuous on r D a. The condition on ! requires that J1 .ka/ D 0. We thus choose ka to be the smallest zero of J1 , ka  3:83.

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38

3. VORTICITY

The constant C is determined by the requirement that u be continuous on r D a. Now u D
.sin / y
.cos / x D
r , and ˇ   2 ˇ d
1 d
1 1 dJ0 ˇ J1 .kr/ D
k C J0 J0 .z/ˇ Dk dr dz 2 z dz zDkr (3.46)   zDkr 1 D k
1
J1 C J0 : z zDkr Thus

ˇ ˇ d J1 .kr/ˇˇ D k
1 J0 .ka/: dr rDa

(3.47) The condition that (3.48)

r

be continuous on r D a becomes C D 2k
1

U J0 .ak/

and (3.49)

! D
r 2

D

2kU sin J1 .kr/: J0 .ak/

Since J0 .3:83/ 
0:403, we see that the constant multiplier in this last equation has a sign opposite to that of U . Let us see if this makes sense. If U were negative, then the vorticity in the upper half of the disc would be positive. A positive vorticity implies an eddy rotating counterclockwise. This vorticity induces the vortex in the lower half of the disc to move to the right. Similarly, the negative vorticity in the lower half of the disc causes the upper vortex to move to the right. Thus the vortex dipole propagates to the right, and in the frame moving with the dipole U is negative, as we have found above. 3.7.3. Axisymmetric Flow. We turn now to a large class of vortical flows that are probably the simplest flows allowing vortex stretching, namely the axisymmetric Euler flows. These are solutions of Euler’s equations in cylindrical polar coordinates .z; r; /, under the assumption that all variables are independent of the polar angle . Euler’s equations for the velocity u D .uz ; ur ; u / in cylindrical polar coordinates are (3.50) (3.51) (3.52) where (3.53)

1 @p @uz C u  ruz C D 0; @t  @z u2 1 @p @ur C u  rur
 C D 0; @t r  @r 1 @p ur u @u C u  ru C C D 0; @t r r @  u @ @ @ C ur C ./: u  r./ D uz @z @r r @


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3.7. SOME EXAMPLES OF VORTICAL FLOWS

39

We take the density to be constant, so that the solenoidal condition applies in the form 1 @rur 1 @u @uz C C D 0: (3.54) @z r @r r @ The vorticity vector is given by 
1 @ur 1 @uz @u @ur @uz 1 @ru
;
;
: (3.55) .!z ; !r ; ! / D r @r r @ r @ @z @z @r The vorticity equation is (3.56)

@! h u !r i C u  r!z ; u  r!r ; u  r! C @t r h


!  ruz ; !  rur ; !  ru C

ur ! i D 0: r

In the axisymmetric case we thus have
 1 @p @ @ @uz (3.57) C uz C ur uz C D 0; @t @z @r  @z
 u2 1 @p @ @ @ur (3.58) C uz C ur ur
 C D 0; @t @z @r r  @r
 ur u @ @ @u (3.59) C uz C ur u C D 0; @t @z @r r (3.60) (3.61)

1 @rur @uz C D 0; @z
r r  @u @ur @uz 1 @ru ;
;
: .!z ; !r ; ! / D r @r @z @z @r

If the swirl velocity component u vanishes, the system simplifies further:
 1 @p @ @ @uz (3.62) C uz C ur uz C D 0; @t @z @r  @z
 1 @p @ @ @ur (3.63) C uz C ur ur C D 0; @t @z @r  @r (3.64) (3.65)

1 @rur @uz C D 0; @z r
@r  @ur @uz
: .!z ; !r ; ! / D 0; 0; @z @r

Note that in this case the only nonzero component of vorticity is ! . The vortex lines are therefore all rings with a common axis, the z-axis. The vorticity equation has the form (3.66)

ur ! @! @! @! C uz C ur
D 0: @t @z @r r

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40

3. VORTICITY

The last equation may be rewritten D @ @ @ D ! D 0; D C uz C ur : Dt r Dt @t @z @r Thus ! =r is a material invariant of the flow. We can easily interpret the meaning of this fact. A vortex ring of radius r has length 2
r, and the vorticity associated with a given ring is ! . But the vorticity of a line is proportional to the line length (recall the increase of vorticity by line stretching). Thus the ratio ! =.2
r/ must be constant on a given vortex ring. Since vortex rings move with the fluid, ! =r is a material invariant. To compute axisymmetric flow without swirl, we can introduce the stream function for the solenoidal velocity in cylindrical polar coordinates: (3.67)

(3.68) This

1@ 1@ ; ur D
: r @r r @z is often referred to as the Stokes stream function. Then uz D

@2 1 @2 1 @ : ! D
L. /; L  2 C 2
r @z @r r @r In the steady case, the vorticity equation gives 
1@ @ 1 1@ @
L. / D 0: (3.70) r @r @z r @z @r r 2 (3.69)

Thus a family of steady solutions can be obtained by solving any equation of the form (3.71)

L. / D r 2 f . /;

where f is an arbitrary function, for the stream function . The situation here is closely analogous to the steady two-dimensional case; see the previous subsection. Now turning to axisymmetric flow with swirl, the instantaneous streamline and vortex lines can now be helices and a much larger class of Euler flows results. The same stream function applies. The swirl velocity satisfies, from (3.59), D @ @ @ Dru D 0; D C uz C ur : Dt Dt @t @z @r We can understand the meaning of (3.72) using Kelvin’s theorem. First, note that a ring of fluid particles initially on a given circle C defined by initial values of z; r will stay on the same circular ring as it evolves. The u -component takes the ring into itself, and the .uz ; ur ; 0/ subfield determines the trajectory C.t/ of the ring, and thus the ring evolves as a material curve. Since u is constant on the ring, the circulation on C.t/ is 2
ru . By Kelvin’s theorem, this circulation is a material invariant, and we obtain (3.72). In the case of steady axisymmetric flow with swirl, (3.72) implies that we may take (3.72)

(3.73)

ru D g. /;

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PROBLEM SET 3

41

where the function g is arbitrary. Bernoulli’s theorem for steady flow with constant density gives 1 2 p juj C D H. /; 2 

(3.74)

stating that the Bernoulli function H is constant on streamlines. From the momentum equation in the form rH
u  ! D 0 we get, from the z-component, @H : @z Using the expressions for the components of vorticity and expressing everything in terms of the stream function, (3.73) and (3.75) yield ur !
u !r D

(3.75)

(3.76)

1 dg @ dH @ 1 @ L. / C 2 g D : r 2 @z r d @z d @z

Eliminating the common factor (3.77)

@ @z

and rearranging,

L. / D r 2 f . /
g

dg ; d

f. / D

dH : d

Thus two arbitrary functions f; g are involved and any solution of (3.77) determines a steady solution in axisymmetric flow with swirl. Problem Set 3 (3.1) Consider a fluid of constant density in two dimensions with gravity, and suppose that the vorticity vx
uy is everywhere constant and equal to !. Show that the velocity field has the form .u; v/ D .x C y ; y
x / where  is harmonic and  is any function of x; y (independent of t) satisfying r 2  D
!. Show further that   1 2 p r t C q C ! C C gz D 0 2  where is the stream function for u; i.e., u D . y ;
x / and q 2 D u2 C v 2 . (3.2) Show that, for an incompressible fluid, but one where the density can vary independently of pressure (e.g., salty seawater), the vorticity equation is D! D !  ru C 
2 r  rp: Dt Interpret the last term on the right physically (e.g., what happens if lines of constant p are y D const and lines of constant  are x
y D const?). Explain how the term acts as a source of vorticity, i.e., causes vorticity to be created in the flow. (3.3) For steady two-dimensional flow of a fluid of constant density, we have u  ru C rp D 0; Show that, if u D .

y ;
x /,

r  u D 0:

these equations imply r

 r.r 2 / D 0:

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42

3. VORTICITY

Thus, show that a solution is obtained by giving a function H. / and then solving r 2 D H 0 . /. Show also that the pressure is given by p D H. /
12 .r /2 C const. (3.4) Prove Ertel’s theorem for a fluid of constant density: If f is a scalar material invariant, i.e., Df Dt D 0, then !  rf is also a material invariant, where ! D r  u is the vorticity field. (3.5) A steady Beltrami flow is a velocity field u.x/ for which the vorticity is always parallel to the velocity, i.e., r  u D f .x/u for some scalar function f . Show that if a steady Beltrami field is also the steady velocity field of an inviscid fluid of constant density, then necessarily f is constant on streamlines. What is the corresponding pressure? Show that u D .B sin y C C cos z; C sin z C A cos x; A sin x C B cos y/ is such a Beltrami field with f D
1. (3.6) Another formula exhibiting a vector field u D .u; v; w/ whose curl is ! D .; ; /, where r  ! D 0, is given by Z 1 Z 1 t.tx; ty; tz/dt
y t.tx; ty; tz/dt; uDz 0 0 Z 1 Z 1 t.tx; ty; tz/dt
z t.tx; ty; tz/dt; vDx 0 0 Z 1 Z 1 t.tx; ty; tz/dt
x t.tx; ty; tz/dt: wDy 0

0

Verify this result. (Note that u will not in general be divergence-free, e.g., check  D  D 0,  D x.) (3.7) In this problem the object is to find a two-dimensional propagating vortex dipole structure analogous to that studied in Section 3.7.2. In the present case, the structure will move clockwise on the circle of radius R with angular velocity . Consider a rotating coordinate system and a circular structure of radius a, stationary and with center at .0; R/. Relative to the rotating system the velocity tends to .
y; x/ D .
y 0 ; x/ C R.
1; 0/; y 0 D y
R. It turns out that (assuming constant density), the momentum equation relative to the rotating frame can be reduced to that in the nonrotating frame in that the Coriolis force can be absorbed into the gradient of apmodified pressure. Thus we again take r 2 C k 2 D 0, r 0 < a. Here r 0 D .y 0 /2 C x 2 . A new term proportional to J0 .kr/ must now be included. We require that u and ! be continuous on r 0 D a. Show that, relative to the rotating frame, 8 0. The region above the upper surface of the wedge to the left and the positive xaxis to the right is mapped onto the upper half-plane Y > 0 by the function Z D z =.
˛/ . Thus w.z/ D
Uz =.
˛/ . E XAMPLE 4.6. The map z.Z/ D Z C b 2 =Z maps the circle of radius a > b in the Z-plane onto the ellipse of the semimajor x-axis .a2 C b 2 /=a and the semiminor y-axis .a2
b 2 /=a in the z-plane. The exterior is mapped onto the exterior. Uniform flow with velocity .U; 0/ at infinity past the circular cylinder jZj D a has complex potential W .Z/ D U.Z C a2 =Z/. p Inverting the map and requiring that Z  z for large jzj gives Z D 12 .z C z 2
4b 2 /. Then w.z/ D W .Z.z// is the complex potential for uniform flow past the ellipse. Notice that the dz ! 1 as z ! 1. This insures that infinity maps by the identity map satisfies dZ and so the uniform flow imposed on the circular cylinder is also imposed on the ellipse. 4.1.2. The Circle Theorem. We now state a result that gives the mathematical realization of the physical act of “placing a rigid body in an ideal fluid flow,” at least in the two-dimensional case. T HEOREM 4.1 Let a harmonic flow have complex potential f .z/, analytic in the domain jzj  a. If a circular cylinder of radius a is placed at the origin, then the new complex potential is w.z/ D f .z/ C f .a2 =Nz /. To show this we establish that the analytical properties of the new flow match those of the old, in particular regarding the position of singularities. We also need to verify that the surface of the circle is a streamline. Taking the latter issue first, note that on the circle a2 =Nz D z, so that there w D f .z/ C f .z/, implying D 0,

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4.1. HARMONIC FLOWS

49

and so the circle is a streamline. Next, observe that the added term is an analytic function of z if it is not singular at z. (If f .z/ is analytic at z, so is f .zN /.) As for the location of singularities of w, since f is analytic in jzj  a, it follows that f .a2 =z/ is analytic in jzj  a, and the same is true of f .a2 =Nz /. Thus the only singularities of w.z/ in jzj > a are those of f .z/. E XAMPLE 4.7. If a cylinder of radius a is placed in a uniform flow, then f D Uz and w D Uz C U .a2 =Nz / D U.z C a2 =z/, as we already know. If a cylinder is placed in the flow of a point source at b > a on the x-axis, then we have f .z/ D Q 2 ln.z
b/ and   2 a Q .ln.z
b/ C ln
b w.z/ D 2
zN (4.12)     a2 Q ln.z
b/ C ln z

ln z C C; D 2
b where C is a constant. From this form it may be verified that the imaginary part of w is constant when z D ae i . Note that the image system of the source, with singularities within the circle, consists of a source of strength Q at the image point a2 =b, and a source of strength
Q at the origin. E XAMPLE 4.8. A point vortex at position zk , carrying circulation k , has the k ln.z
zk /: A collection of N such vortices complex potential wk .z/ D
i 2 PN will have the potential w.z/ D kD1 wk .z/. Since vorticity is a material scalar in two-dimensional ideal flow, and the vorticity concentration may be regarded as the limit of a small circular patch of constant vorticity, we expect that each vortex must move with the harmonic flow created at the vortex by the other N
1 vortices. Thus the positions zk .t/ of the vortices under this law of motion is governed by the system of N equations, (4.13)

N
i X k dzj D : dt 2
z
zk kD1 k¤j

Note that the conjugation on the left comes from the identity

dw dz

D u
iv.

4.1.3. The Theorem of Blasius. An important calculation in fluid dynamics is the force exerted by the fluid on a rigid body. In two dimensions and in a steady harmonic flow, this calculation can be carried out elegantly using the complex potential. T HEOREM 4.2 Let a steady uniform flow past a fixed two-dimensional body with bounding contour C be a harmonic flow with velocity potential w.z/. Then, if no external body forces are present, the force .X; Y / exerted by the fluid on the body is given by  I  dw 2 i dz: (4.14) X
iY D 2 dz C

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50

4. POTENTIAL FLOW

Here the integral is taken around the contour in the counterclockwise sense. This formula, due to Blasius, reduces the force calculation to a complex contour integral. Since the flow is harmonic, the path of integration may be distorted to any simple closed contour encircling the body, enabling the method of residues to be applied. The exact technique will depend upon whether there are singularities in the flow exterior to the body. To prove the result, first recall that dX
id Y D p.
dy
idx/ D
ipd zN . Also, Bernoulli’s theorem for steady ideal flow applies, so that ˇ ˇ  ˇˇ dw ˇˇ2 C const; (4.15) pD
ˇ 2 dz ˇ where clearly the constant will play no role. Thus I dw dw i d zN : (4.16) X
iY D 2 dz dz C

However, the contour C is a streamline, so that d have

dw dz

d zN D d wN D dw D

dw dz

D 0 there, and so on C we

dz. Using this in (4.16), we obtain (4.14).

E XAMPLE 4.9. We have found in Problem 2.1 that the force on a circular cylinder in a uniform flow is 0. To verify this using Blasius’ theorem, we set w D U.z C a2 =z/ so that U 2 .1
a2 =z 2 /2 is to be integrated around C . Since there is no term proportional to z
1 in the Laurent expansion about the origin, the residue is 0 and we get no contribution to the force integral. E XAMPLE 4.10. Consider a source of strength Q placed at .b; 0/ and introduce a circular cylinder of radius a < b into the flow. From Example 4.6 we have 1 1 1 dw D C
: (4.17) 2 dz z
b z
a =b z Squaring, we get 1 1 2 1 C C 2C 2 2 2 .z
b/ .z
a =b/ z .z
b/.z
a2 =b/ 2 2
:
z.z
a2 =b/ z.z
b/ The first three terms do not contribute to the integral around the circle jzj D a. For the last three, the partial fraction decomposition is B C A C C ; (4.19) z
b z
a2 =b z (4.18)

2

3

, B D a2 .a2b2
b 2 / , C D where we compute A D .b 22a
a2 /b tions come from the poles within the circle and we have

2.a2 Cb 2 / . a2 b

a2 Q2  i Q2 : 2
i.B C C / D 2 4
2 2
b.b 2
a2 / The cylinder therefore feels a force of attraction toward the source.

(4.20)

X
iY D

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The contribu-

4.2. FLOWS IN THREE DIMENSIONS

51

This introduction to the use of complex variables in the analysis of two-dimensional harmonic flows will provide the groundwork for a discussion of lift and airfoil design, to be taken up in Chapter 5. 4.2. Flows in Three Dimensions We live in three dimensions, not two, and the “flow past body” problem in two dimensions introduces a domain that is not simply connected, with the important consequences discussed above. The relation between two- and three-dimensional ideal flows is particularly significant in the generation of lift, as we shall see in Chapter 5. In the present section we treat topics in three dimensions that are direct extensions of the two-dimensional results just given. They pertain to bodies, such as spheres, which move in an irrotational, harmonic flow. 4.2.1. The Simple Source. The source of strength Q in three dimensions satisfies (4.21)

div u D Qı.x/;

u D r;

.1/ D 0:

Here ı.x/ D ı.x/ı.y/ı.z/ is a very useful distribution, the three-dimensional delta function. The delta function has the following properties: (1) It vanishes if x ¤ 0. (2) Any integral of ı.x/ over an open region containing the origin yields unity. It is helpful, and important in the analysis of distributions, to think of all relations as limits of relations using smooth functions. For example, (4.22)

3

1

2

ı.x/ D lim .
/
2 e
 jxj : !0C

In our case, integrating r 2  D Qı.x/ over a sphere of radius R0 > 0, we get Z @ dS D Q: (4.23) @n RDR0

Since r 2  D 0, x ¤ 0, and since the delta function must be regarded as an isotropic distribution having no exceptional direction, we have (using now r 2  D 2 2 2 2 R
1 d 2 .R/=dR2 ) that  D C R , R D x C y C z , for some constant C . Q . Thus the simple source in three dimensions, of Then (4.23) shows that C D
4 strength Q, has the potential Q 1 : 4
R Note that Q is equal to the volume of fluid per unit time crossing any deformation of a spherical surface, assuming the deformed surface surrounds the origin. (4.24)

D





We indicate how to justify this calculation using a limit operation. We may choose to define 3 1 the three-dimensional delta function by lim !0 ı .R/ where ı D 2 3 1C.R= /3 . Solving   d d R2 r 2  D ı D R
2 dR dR

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52

4. POTENTIAL FLOW

3 2

r/k

1 0

0

1

2

3

z/k

F IGURE 4.3. The Rankine fairing. All lengths are in units of k.

4.2.2. The Rankine Fairing. We consider now a simple source of strength Q placed at the origin in a uniform flow W iz . The combined potential is then Q 1 : (4.25)  D Wz
4
R The flow is clearly symmetric about the z-axis. In cylindrical polar coordinates .z; r; /, r 2 D x 2 C y 2 , we introduce again the Stokes stream function : (4.26)

uz D z D

1@ ; r @r

ur D r D


1@ : r @z

Thus for (4.25) we have (4.27)

Q z 1@ DW C : r @r 4
R3

Integrating, we get

  Q z W r2
C1 : (4.28) D 2 4
R In (4.28) we have chosen the constant of integration to make D 0 on the negative z-axis. We show the stream surface D 0, as well as several stream surfaces > 0, in Figure 4.3. This gives a good example of a uniform flow over a semi-infinite body. An interesting question is whether such a body would experience a force in a potential flow. We will find below that the drag force vanishes for finite bodies in three dimensions in steady potential flow. This is the famous d’Alembert paradox. But it is not obvious that the result applies to bodies that are not finite. under the condition that  vanish at infinity, we obtain Z 1 h 1
i R
2 tan
1 .R3 
3 /
C dR:  D
4
R 2 R For any positive R the integral tends to 0 as  ! 0.

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4.2. FLOWS IN THREE DIMENSIONS

53

S

A z

F IGURE 4.4. Geometry of the momentum integral for computation of the force on the Rankine fairing.

We will use this question to illustrate the use of conservation of momentum to calculate force on a distant contour. In Figure 4.4 the large sphere S of radius R0 is centered at the origin and intersects the fairing at a circle bounding the disc A. Let S 0 be the spherical surface S minus that part within the boundary of A. We are considering steady harmonic flow, and so the momentum equation may be written @ .ui uj C p / D 0: @xj

(4.29)

Let V 0 be the region bounded by S 0 and the piece of fairing enclosed. Integrating (4.29) over V 0 and using the divergence theorem, the contribution from the surface of the fairing is the integral of
np over this surface, where n is the outer normal of the fairing. Thus this contribution is the force F experienced by the enclosed piece of fairing, a force clearly directed along the z-axis and therefore equal to the drag, F D Diz . The remainder of the integral, taking only the z-component, takes the form of an integral over S minus the contribution from A. Thus conservation of momentum gives Z z u  R 1 2 C U
juj2 dS
IA D 0: (4.30) D C  uz R 2 R S

We have made use of the Bernoulli formula for the flow, p C 12 juj2 D 12 U 2 , the pressure at infinity being taken to be 0. Treating first the integral over S , we have Q R UQ z Q2 1 2 2 ; juj D U C C : 4
R3 2
R3 16
2 R4 Thus the integral in question becomes     Z  Q 1 z2 1 Q2 z Q z Uz 1 C UQ 4 C UC
dS: (4.32) 4
R3 R 4
R2 4
R 8
R5 u D U iz C

(4.31)

S

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54

4. POTENTIAL FLOW

F IGURE 4.5. Flow around an airship.

We see that this last integral gives UQ C 12 UQ
12 UQ D UQ. For the 2 , where r contribution IA , we take the limit R0 ! 1 to obtain IA D U 2
r1 1 0 is the asymptotic radius of the fairing as z ! 1. In this limit D ! D, the total drag of the fairing. Thus the momentum integral method gives (4.33)

2 / D 0: D C .UQ
U 2
r1

But from (4.28) we see that the stream surface (4.34)

r 2
12 k 2 ; zDp k2
r 2

D 0 is given by

k2 D

Q :
U

Thus r1 D k, and (4.33) becomes (4.35)

D C .UQ
UQ/ D D D 0;

so the drag of the fairing is 0. E XAMPLE 4.11. The flow considered in the present example provides a model of the flow around an airship. The flow consists of a source of strength Q at position z D 0 on the z-axis and a equalizing sink (source of strength
Q) at the point z D 1 on the z-axis. Since the source strengths cancel, a finite body is so defined when the singularities are placed in the uniform flow U iz . It can be shown (see Problem 4.7 below) that the stream surfaces for the flow are given by constant values of   1
R cos  Q U 2 2 ; cos  C p (4.36) ‰ D R sin 
2 4
R2
2R cos  C 1 where R;  are spherical polars at the origin, with axial symmetry; see Figure 4.5. 4.2.3. The Butler Sphere Theorem. The circle theorem for two-dimensional harmonic flows has a direct analogue in three dimensions.

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4.2. FLOWS IN THREE DIMENSIONS

55

T HEOREM 4.3 Consider an axisymmetric harmonic flow in spherical polar coordinates .R; ; '/, u' D 0, with Stokes stream function ‰.R; / vanishing at the origin:
1 @‰ @‰ ; u D : @ R sin  @R If a rigid sphere of radius a is introduced into the flow at the origin, and if the singularities of ‰ exceed a in distance from the origin, then the stream function of the resulting flow is  2  a R ; : (4.38) ‰s D ‰.R; /
‰ a R (4.37)

uR D

1

R2 sin 

It is clear that ‰s vanishes when R D a, so the surface of the sphere is a stream surface. Also, the added term introduces no new singularities outside the sphere. 2 Thus the theorem is proved if it can be verified that R a ‰.a =R; / represents a harmonic flow. In spherical polars with axial symmetry the only component of vorticity is
 1 @.Ru / @uR
: (4.39) !' D R @R @ Thus the condition on ‰ for an irrotational flow is   2 1 @‰ @ 2@ ‰  LR ‰ D 0: C sin  (4.40) R @R2 @ sin  @ 2 If R a ‰.a =R; / is inserted into (4.40), it is found that the equation is satisfied provided it is satisfied by ‰.R; /; see Problem 4.8. Finally, since ‰.R; / vanishes at the origin at least as R, R‰.a2 =R; / is bounded at infinity and the velocity component must decay as O.R
2 /, so the uniform flow there is undisturbed.

E XAMPLE 4.12. A sphere in a uniform flow U iz has the Stokes stream function (4.41)


 a3 U 2 2 ‰.R; / D R sin  1
3 : 2 R

This translates into the following potential:   1 a3 : (4.42)  D Uz 1 C 2 R3 E XAMPLE 4.13. Consider a source of strength Q placed on the z-axis at z D b. Now place a rigid sphere of radius a < b at the origin. The stream function for this source, vanishing at the origin, is (4.43)

‰.R; / D


Q .cos 1 C 1/; 4


where 1 is defined in Figure 4.6.

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56

4. POTENTIAL FLOW

R R2

R1

θ2

θ

θ1

a2/b

b

a

F IGURE 4.6. A sphere of radius a in the presence of a source at z D b > a.

Now from the law of cosines and Figure 4.6 we have (4.44)

R cos 
b D R1 cos 1 ;

R12 D b 2
2bR cos  C R2 :

Thus R cos 
b : cos 1 D p b 2
2bR cos  C R2

(4.45)

The stream function including the sphere, ‰s , is given by
 R cos 
b Q p C1 ‰s D
4
b 2
2bR cos  C R2 (4.46)
 a2 QR R cos 
b p C1 : C 4
a b 2
2b.a2 =R/ cos  C a4 =R2 Again using the law of cosines, s b 2
2b 2

a4 a2 bR2 cos  C 2 D : R R R 4

Also we may use R2 D R22 C 2 ab R cos 
ab 2 . Then ‰s may be brought into the form 2 


  Q R cos 
b a Q R cos 
ab Q R
R2 : C1
C (4.47) ‰s D
4
R1 b 4
R2 4
a The first term on the right is the source of strength Q at z D b. The second term 2 is another source, of strength ab Q, at the image point z D ab . The last term can be Q , extending from the origin understood as a line distribution of sinks of density 4a to the image point

a2 . b

Indeed, if a point P on this line segment is associated with

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4.3. APPARENT MASS AND THE DYNAMICS OF A BODY IN A FLUID

57

an angle P , the contribution from such a line of sinks would be (4.48)

Q 4
a

Z

a2 b

cos P dz: 0

But dR D
cos P dz, so the integral becomes Z R2 Q Q dR D .R
R2 /: (4.49)
4
a R 4
a 4.3. Apparent Mass and the Dynamics of a Body in a Fluid Although a harmonic flow is an idealization never realized exactly in actual fluids (except in some cases of superfluid dynamics), it is a good approximation in many fluid problems, particularly when rapid changes occur. A good example is the abrupt movement of a solid body through a fluid, for example, the stroke of the hand of a swimmer. We know from experience that an abrupt movement of the hand through water gives rise to a force opposing the movement. It is easy to see why this must be, within the theory of harmonic flows. An abrupt movement through still water causes the fluid to move relative to an observer fixed with the still fluid at infinity. This observer would therefore compute at the instant the hand is moving a finite kinetic energy of the fluid, whereas before the movement began the kinetic energy was 0. To produce this kinetic energy, work must have been done, and so a force with a finite component opposite to the direction of motion must have acted on the hand. We are here dealing only with the fluid, but if the body has mass, then clearly a force is also needed to accelerate that mass. Thus both the body mass and the fluid movement contribute to the apparent force. In a harmonic flow we shall show that, in the absence of external body forces, the force on a rigid finite body is proportional to its acceleration, and further that the force contributed by the fluid can be expressed as an additional, apparent mass of the body. In other words the augmented force due to the presence of the surrounding fluid and the energy it acquires during motion of a body can be explained as an inertial force associated with additional mass and the work done against that force. The terms virtual mass or added mass are also used to denote this apparent mass. For a sphere, which has an isotropic geometry with no preferred direction, the apparent mass is just a scalar to be added to the physical mass. In general, however, the apparent mass associated with the momentum of a body in two or three dimensions will depend on the orientation of the body relative to the direction of motion. It thus must be a second-order tensor, represented by the apparent mass matrix. 4.3.1. The Kinetic Energy of a Moving Body. Consider an ideal fluid at rest and introduce a moving rigid body in two or three dimensions. An observer at rest relative to the fluid at infinity will see a disturbance of the flow that vanishes at infinity. It would R be natural to compute the momentum of this motion by calculating the integral u dV of the region exterior to the body. The problem is that such

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58

4. POTENTIAL FLOW

harmonic flows have an expansion at infinity of the form (4.50)

 a ln r C b
A  rr
2 C O.r
2 /

in two dimensions and (4.51)



in three dimensions. Thus (4.52)

a
A  RR
3 C O.R
3 / R Z



Z n dS;

r dV D S

where S comprises both a surface in a neighborhood of infinity as well as the body surface, is not absolutely convergent as the distant surface recedes. We point out that a D b D 0 in two dimensions if the area of the body is fixed and there is no circulation about the body. In three dimensions a vanishes if the body has fixed volume; see Problem 4.12. But even if a D b D 0 and  D O.R1
N / the value of the integral is only conditionally convergent and will depend on how one defines the distant contour or surface. So the value attributed to the fluid momentum is ambiguous by this calculation. An unambiguous result is, however, possible if we instead focus on the kinetic energy and from it determine the incremental momentum created by a change in velocity. Let us fix the orientation of the body and consider its movement through space, without rotation. This translation is completely determined by a velocity vector U.t/. From the discussion of Section 2.6 we know that a harmonic flow will satisfy the instantaneous boundary condition (4.53)

@ D U.t/  n @n

on the surface of the body. Now r 2  D 0 is a linear equation, and so we see that there must exist a vector potential ˆi such that (4.54)

u D Ui rˆi :

The ˆi determine all possible harmonic flows associated with translation of the body. We may now compute the kinetic energy E of the fluid exterior to the body as Z 1 (4.55) E.t/ D Mij Ui Uj ; Mij D  rˆi  rˆj dV; 2 the integral being over the fluid domain. Clearly the matrix Mij is symmetric, and thus (4.56)

dE D Mij Uj d Ui :

On the other hand, the change of kinetic energy, dE, must equal, in the absence of external body forces, the work done by the force F that the body exerts on the fluid,

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4.3. APPARENT MASS AND THE DYNAMICS OF A BODY IN A FLUID

59

dE D F  U dt. But according to Newton’s second law, the incremental momentum d P is given by d P D F dt. Consequently, dE D U  d P. From (4.55) we thus have dE D Mij d Uj Ui D dPi Ui :

(4.57)

Since this holds for arbitrary U we must have dPi D Mij d Uj . Integrating and using the fact that Mij is independent of time and P D 0 when U D 0, we obtain Pi D Mij Uj :

(4.58)

Thus we have reduced the problem of computing momentum, and then the inertial force, to calculating Mij . Since Mij arises here as an effective mass term associated with movement of the body, it is called the apparent mass matrix. But the calculation of Mij is not ambiguous since the integral for the kinetic energy converges absolutely, and we can deduce Mij once the energy is written in the form (4.55). We write Z Z Z    2 juj dV D .u
U/  .u C U/dV C jUj2 dV: (4.59) ED 2 2 2 V

V

V

The reason for this splitting is to exhibit u
U, whose normal component will vanish on the body by (4.53).  Now u C U D r.  C U  x/ and u
U is solenoidal, so .u
U/  .u C U/ D r  . C U  x/.u
U/ . Thus, remembering that jUj2 is a function only of time, the application of the divergence theorem and use of (4.53) on the inner boundary allows us to reduce (4.59) to Z  . C U  x/.u
U/  n dS C jUj2 .Vo
Vb /; (4.60) ED 2 So

where So is the outer boundary, Vo is the volume contained by So , and Vb is the volume of the body. To compute the integral in (4.60), we need only the leading term of . Referring to (4.50)–(4.51), recall that a D b D 0 for a finite rigid body (and no lift in two dimensions, and even for a flexible body of constant area/volume); see Problem 4.12. Using
A NA  x x Ax (4.61) D
N; uD N C jxj jxj jxjN C2 in (4.60), 
 Z

A NA  x x
A  x  CUx C
U  n dS: (4.62) E 2 jxjN jxjN jxjN C2 So

We are free to choose So to be a sphere of radius Ro . The term quadratic in A in (4.62) is O.Ro1
2N /, and so the contribution is of order Ro
N and will vanish in the limit. The term under the integral quadratic in U yields
jUj2 Vo , thus canceling part of the last term in (4.60). Finally, two of the cross terms in U; A cancel out, the remaining term giving the contribution 2
.N
1/A  U. Thus  (4.63) E D Œ2
.N
1/A  U
Vb jUj2 : 2

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60

4. POTENTIAL FLOW

Since  D ˆi Ui , Ai .t/ D 
1 mij Uj where mij is dependent on body shape but not time. Then 1 (4.64) E D Œ2
.N
1/mij
Vb ıij Ui Uj : 2 From comparing (4.64) and (4.55), the expression for the apparent mass matrix is (4.65)

Mij D 2
.N
1/mij
Vb ıij ;

N D 2; 3:

The apparent mass of a body is thus determined from its volume and the expansion of  in a neighborhood of infinity. Given that we have computed a finite fluid momentum we are in a position to state d’Alembert’s paradox. T HEOREM 4.4 (D’Alembert’s paradox) In a steady flow of a perfect fluid of a solid body in three dimensions, and in steady flow in two dimensions of a solid body with zero circulation, the force experienced by the body is 0. Clearly if the flow is steady ddtP D F D 0, and we are done. This proof hinges on the existence of a finite fluid momentum associated with a single-value potential function. E XAMPLE 4.14. To find the apparent mass matrix of a elliptic cylinder in two dimensions, we may use Example 4.6. In the Z-plane, the complex potential for uniform flow
Q.cos ; sin / past the cylinder of radius a > b is p W .Z/ D
Qe
i Z
Qe i a2 =Z. Since Z D 12 .z C z 2
4b 2 /, we expand at infinity to get # " a2 e i
b 2 e
i
i C  ; (4.66) w.z/
Qe z
Q z so that (4.67)

A D ŒU.a2
b 2 /; V .a2 C b 2 / ;

.U; V / D Q.cos ; sin /:

Now the ellipse intersects the positive x-axis at its semimajor axis ˛ D and the positive y-axis at its semiminor axis ˇ D

a2
b 2 a .

a2 Cb 2 a ,

From (4.67) the apparent

mass matrix becomes     2   2 a4
b 4 1 0 0 0 ˇ a
b2

D
 : (4.68) M D 2
 0 1 0 a2 C b 2 0 ˛2 a2 In particular, for a circular cylinder the apparent mass is just the mass of the fluid displaced by the body. An alternative expression for the apparent mass matrix in terms of an integral over the surface Sb of the body rather than a distant surface is readily obtained in terms of the potential ˆi . The kinetic energy is Z Z   rˆi  rˆj Ui Uj dV D r  ˆj rˆi dV Ui Uj : (4.69) ED 2 2 V

V

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4.3. APPARENT MASS AND THE DYNAMICS OF A BODY IN A FLUID

61

Applying the divergence theorem to the integral between a distant surface So and the body surface Sb , and observing that ˆi rˆj D O.jxj1
2N /, we see that the D U  n on receding surface integral will give zero contribution. Recalling that @ @n @ˆi the body surface, @n D ni , where the normal is directed out of the body surface. In applying the divergence theorem, the normal at the body is into the body, with the result that (4.55) applies with Z (4.70) Mij D
 ˆj ni dS; n directed out of the body: Sb

It follows from (4.58) that the fluid momentum is given by Z (4.71) P D
 n dS: Sb

We can verify the fact that (4.71) gives the fluid momentum by taking its time derivative, using the result of Problem 1.6: Z Z Z @ d n dS C .u  n/r dS: n dS D (4.72) dt @t Sb

Sb

Sb

Using the Bernoulli theorem for harmonic flow, we have  Z Z
p 1 2 d n dS D

juj n
.u  n/u dS: (4.73) dt  2 Sb

Sb

Converting the terms on the right involving u to a volume integral, we observe that the latter converges absolutely at infinity, and so for the integration over the domain exterior to Sb ,  Z Z
1 2 u  ru
rjuj dV D
u  .r  u/dV D 0: (4.74) 2 V

V

Therefore (4.75)

d
 dt

Z

Sb

Z n dS D

pn dS D F; Sb

where F is the force applied by the body to the fluid. Finally, we note again that the inertial force required to accelerate a body in a perfect fluid will contain a contribution from the actual mass of the body, Mb . This mass appears as an additional term Mb ıij in the expression (4.65) for the apparent mass matrix. The total momentum of the body including its apparent mass is thus Pi D Mij Uj C Mb Ui and Newton’s second law becomes d Pi D Fi ; dt where F is the force applied to the body to accelerate it and the surrounding fluid. (4.76)

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62

4. POTENTIAL FLOW

4.3.2. The Case of Time-Dependent Mij . We have used an energy method to define the apparent mass of a body under the assumption that Mij did not depend upon time. Potential flows are, however, determined instantaneously by the boundary conditions, so that our calculation may be regarded as yielding Mij at a particular instant of time. A change of shape or orientation offers an especially simple way to modify the effective mass of a body in a fluid and provides the possibility of inertial modes of locomotion; see Section 4.4. If a time-dependent Mij is assumed, then the differential relation dPi D Fi dt applies to relate Ui and Fi . That is, we may simply adopt Newton’s second law using our computed Mij . The computation of kinetic energy of a deforming body must involve the contribution from the deformation as well as the contribution from translation. Proper accounting of this new component allows the energy method to be generalized, but we omit details. Since we now know how to compute Mij for any shape based on its instantaneous translation, we may now apply (4.76), allowing for a time-dependent apparent mass. 4.3.3. Moment. We have so far restricted the motion of the body to translation, i.e., with no rotation relative to the fluid at infinity. In general, a moment is experienced by a body in translational motion (see Problem 4.14), so that in fact a free body will rotate and thereby give the apparent mass matrix a dependence upon time. We may define, in analogy with (4.71), the apparent angular momentum of the fluid exterior to a body by Z (4.77) PA D
 .x  n/dS; Sb

the normal being out of the body. It may be shown in a manner similar to that used for linear momentum that d PA D T; (4.78) dt where T is the torque applied to the fluid by the body. 4.4. Deformable Bodies and Their Locomotion It might be thought that, in an ideal or, to use our more suggestive adjective, a “slippery” fluid, it would be impossible for a body to locomote, i.e., to “swim” by using some kind of mechanism involving changes of shape. The fact is, however, that inertial forces alone can allow a certain kind of locomotion. The key point is that the flow remains irrotational everywhere. This will have the effect of disallowing the possibility of the body producing an average force on the fluid, by mechanisms to be discussed in the next chapter, which can then accelerate the body. But it is possible to locomote in the sense of getting from point A to point B, without any finite average acceleration, but by a kind of “ratcheting” movement. If the body is assumed to deform periodically over some cycle of configurations,

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4.4. DEFORMABLE BODIES AND THEIR LOCOMOTION

63

then the kind of locomotion we envision is a finite, periodic translation (and possible rotation) of the body, repeated with each cycle of deformation. The Newtonian relationships that we derived above for a rigid body carry over to an arbitrary deformable body, which for simplicity we take to have a fixed area/volume. The idea behind inertial swimming is therefore to deform the body in a periodic cycle in such a way that over one period of the cycle there is a net translation in a fixed direction. To simplify the problem, we consider only a simple translation of a suitably symmetric body along a line, e.g., a body rotationally symmetric about the z-axis, translating with velocity U.t/ along this axis. In general, we cannot expect the velocity to remain of one sign, but over one cycle there will be a positive translation, say in the direction of positive z. Let Um .t/ be the velocity of the center of mass of the body, and let Uv .t/ be the velocity of the center of volume of the body. If the total mass of the body is m, then mUm .t/ is the momentum of the body mass. Consider now the momentum of the fluid. If the apparent mass of the body (now a scalar M.t/) is multiplied by Uv .t/, we get the fluid momentum associated with the instantaneous motion of the shape of the body at time t. Finally, we have the momentum associated with the motion of the boundary of the body relative to the center of volume. Let PD be the momentum of deformation of the body relative to its center of volume. If the potential of this harmonic R flow of deformation is D .t/, then the deformation momentum is PD .t/ D
 Sb D n  i dS . The total momentum of body and fluid is thus mUm .t/ C M.t/Uv .t/ C PD .t/. If initially the fluid and body are at rest, then the total momentum, which is conserved, must vanish. Consider first a body of uniform density so that the center of mass and of volume coincide. The Um D Uv D U and PD .t/ : (4.79) U.t/ D
m C M.t/ The right-hand side of (4.79) need not have zero time average, and when it does not, we call this locomotion by squirming. To see squirming in action it is best to treat a simple case; see Example 4.15 below. Alternatively, we can imagine that the center of mass changes relative to the center of volume without any deformation of the boundary. After this step, a second phase occurs involving deformation, resulting in a new shape. For the third step the center of mass again changes relative to the center of volume, holding the body fixed in the new shape. Finally, the shape returns to its starting configuration. If the two shapes lead to different apparent masses, locomotion occurs by recoil swimming; see Example 4.16. E XAMPLE 4.15. We show in Figure 4.7(a) a squirming body of a simplified kind. The body consists of a thin vertical strip of length L1 .t/ and a horizontal part of length L2 .t/. The length will change as a function of time; think of L2 as being extruded from the material of L1 . We neglect the width w of the strips except when computing mass and volume. The latter are constant, implying that L1 C L2 D L is constant. The density of the material is taken as b , so the total mass is Mb D b wL and the total volume is wL.

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64

4. POTENTIAL FLOW

L1 P

L2

Q

(a)

(b)

F IGURE 4.7. Swimming in an ideal fluid. −0.06 −0.08 −0.1 −0.12 −0.14 −0.16 −0.18 0

F IGURE 4.8.

X L

50

100

150

200

versus for the model squirmer of Figure 4.7(a).

A cycle begins with L1 D L, when L2 begins to grow to the right. If X.t/ is the momentum denotes the position of the point P , then .b wL1 C
L21 =4/ dX dt of the fluid and vertical segment, where we have used the formula for apparent mass of a flat plate in two dimensions. The velocity of the extruded strip varies 2/ at P to d.XCL at Q, so the momentum of the horizontal part is linearly from dX dt dt 1 b wL2 d.X C 2 L2 /=dt, where we neglect the apparent mass of the extruded strip. The first half of the cycle stops when L1 D 0. Assuming the start is from fluid and body at rest, the sum of these momenta remains 0 throughout the half-cycle: !
L21 dX b wL2 dL2 C D 0: (4.80) b wL C 4 dt 2 dt t If we let L2 D Lt T , L1 D L.1
T / where T is the half-period of the cycle, then we may obtain the change X of X over the half-cycle by quadrature: p 1
L 1 : (4.81) .X/ D ln.1 C /
p tan
1 . /; D 4b w We show this relation in Figure 4.8. So we see that at the end of the half-cycle the point P has moved a distance
X to the left. At this point, we imagine another half-cycle in which L1 is created at the expense of L2 , but at the point Q. Observe that at the start of the second half-cycle Q is located a distance L C X from the initial position of P . It can be seen from considerations of symmetry that the point Q will move to the left

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4.5. DRIFT

65

a distance
X in time T over the second half-cycle. Then the cycle is complete, L2 D L, and the midpoint can be relabeled P . Thus the net advance to the right of the point P in time 2T is L C 2X, which from Figure 4.8 always exceeds about 0:68L. E XAMPLE 4.16. Recoil swimming can be illustrated by the two-dimensional model of Figure 4.7(b). Let P denote the center of an elliptical surface of major and minor semiaxes ˛ and ˇ, respectively. Within this body is a mass M on a bar enabling it to be driven to the right or left. The weight of the shell and the mechanism that moves the mass is m. Let the position of the center be X.t/ and the position of the mass be x.t/. At the beginning of the half-cycle the mass lies a distance ˇ2 to the right of P , and the ellipse has its major axis vertical. The mass then moves to the left a distance ˇ. Since momentum is conserved,   d.X C x/ 2 dX CM D 0: (4.82) .m C 
˛ / dt dt Thus over a half-cycle .m C 
˛ 2 /X C M.X C x/ D 0 or, since x D ˇ, (4.83)

X1 D


Mˇ : m C M C 
˛ 2

At this point the surface of the body deforms in a symmetric way, the points (0; ˙ ˛2 ) moving down to (0; ˙ ˇ2 ), and the points (˙ ˇ2 ; 0) moving out to (˙ ˛2 ; 0), so that the major and minor axes get interchanged. There is no movement of P during this process. Now the mass is moved back a distance ˇ to the left. In this second half-cycle the displacement is (4.84)

Mˇ : m C M C 
ˇ 2

X2 D

The displacement over one cycle is then (4.85)

X D X1 C X2 D

Mˇ Mˇ
; 2 m C M C 
ˇ m C M C 
˛ 2

which is positive since ˇ < ˛. 4.5. Drift The Eulerian view of a steady flow emphasizes streamlines but does not keep track of the positions of individual fluid particles. However, it is sometimes of interest to follow particle motions in a flow that is steady. For example, if a passive scalar field is distributed throughout the fluid, the passage of a body through the fluid at a fixed velocity will distort the field and partially mix it. We are interested now in tracking the movement of fluid particles relative to the fluid at infinity when a solid body in potential flow moves along a straight line. Let us assume that the body, in two or three dimensions, moves along the x-axis from C1 to
1 with unit speed. If we place ourselves in a frame moving with

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66

4. POTENTIAL FLOW

the body, we consider the time dt required for a fluid particle to move a distance ds along a streamline. Then (4.86)

dt D

d ds D 2; q q

where q is the speed of the particle and  is the potential of the flow. Comparing the time on the streamline with the time of a particle on a line at infinity, we see that the drift D,  Z C1  1
1 d (4.87) DD q2
1 is the difference between the time spent on the streamline and the time spent on a distant line, taken between two distant values x
; xC of x, in the limit where x
!
1, xC ! C1. Since the speed of the body is unity, D is also the displacement in the x-direction of a fluid particle, caused by the passage of the body from x D
1 to x D C1. This drift will clearly depend upon the initial position of the particle in a plane orthogonal to the direction of motion of the body. If the initial plane of particles is tracked in its entirety, we obtain a drift area or volume by integration of D. In particular, in two dimensions, we may consider the contribution from streamlines between D
and D C on either side of the body. Since D is obtained after the body is passed, and y are interchangeable and the drift area AD is obtained by integrating D as given by (4.87) with respect to from
to C , then letting the latter tend to ˙1: Z C1 D. /d : (4.88) AD D
1

Drift, being a Lagrangian quantity, is fairly difficult to compute for even the simplest flows, and (4.87) is not the most convenient expression for numerical evaluation of D. E XAMPLE 4.17. We calculate drift for a circular cylinder of unit radius by considering the differential equations satisfied relative to the fluid at infinity. The stream function of the fluid perturbation is given by (4.89)

0

D

y ; .x
t/2 C y 2

where the particle in question is at x.t/; y.t/. We solve numerically the differential equations (4.90)

x
t/2
y 2 dx D ; dt Œ.x
t/2 C y 2 2

dy 2.x
t/y D ; dt Œ.x
t/2 C y 2 2

with initial conditions (0; a), a > 0, so that (by symmetry) D.a/ D limt!1 2x.t/, and show the result in Figure 4.9.

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4.5. DRIFT

67

2.5

2

1.5

1

0.5

0 0

1

2

3

4

5

F IGURE 4.9. D.a/ for a circular cylinder of unit radius, compared with approximations for large a (dotted line) and small a (dashed line).

4.5.1. Analysis of Drift for Large and Small Displacements. Some analysis is possible for particles far from the body and those close to it, which we now carry out for the case of a circular cylinder. To seek the behavior of a particle initially at a point .0; a/, where a 1, as the cylinder center moves from .
1; 0/ to .C1; 0/, we expand u; v about .0; a/:

 t2 2t 8a2 t 8a3 6a dx 2C
3 x C
2 C 3 y C  ; (4.91) uD dt A A2 A A A 2ta dy Y
2 C  ; (4.92) vD dt A where A D t 2 C a2 . To leading order (4.93)

t xD
; A

y
a D

a : A

1 2 / D 4a1 2 , so the particle moves on a circle of radius We see that x C .y
a
2a 1 1 2a and center at Y C 2a , to first order. To find the leading expression for the displacements x; y due to passage of the cylinder, note first that v will be odd in t, so that y D 0 identically. Then
   Z C1 
6a 8a2 t t 8a3 a 2t C
dt:
C (4.94) x A2 A3 A A2 A3 A
1

Simplifying, Z (4.95)

C1


x
1

 4a2 2
2 C 3 dt: A A

Letting Z (4.96)

C1

In D
1

dz .1 C z 2 /n

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68

4. POTENTIAL FLOW

and using InC1 D

(4.97)

2n
1 In ; 2n

I1 D
;

we obtain
: 2a3

x

(4.98)

4.5.2. Small a. To get an approximation valid for small a, we shall as before consider the streamlines of a steady flow and determine the difference between the travel time of a Lagrangian particle following a streamline from a point far upstream to a point far downstream, and a point covering the same horizontal distance with velocity .1; 0/. We shall use, however, a slightly different definition of travel time based upon vertical position only. The streamline of constant is given by   1 (4.99) Dy 1
2 : r Let us consider the section of the p streamline D a extending from the point p 1 2 .0; 2 .a C 4 C a // to the point . a= C 1
.a C /2 ; a C /. We shall want a to be small but we will pass to the limit  ! 0 once we have computed the time difference. The time of passage between the points indicated above (which amounts to half of the intended path of the particle) can be written Z

1 2 .aC

p 4Ca2 /

jvj
1 dy D

(4.100) aC

1 2

Z

1 2 .aC

p 4Ca2 /

1

3

1

y 2 .y
a/
2 .1
y.y
a//
2 dy:

aC

We break the integral into two parts in order to extract terms that are not o.1/: Z

1 2 .aC

p 4Ca2 /

Z

p

D

(4.101) aC

Z C

aC

1 p

Co.1/:

Thus
Z (4.102)

p



x D F: P: aC

y 1=2 dy C .y
a/3=2

Z

1 p

2

.y
ay/


1=2

 dy C o.1/:



Here “F.P.” stands for the finite part of the integral as  ! 0. Integration yields (4.103)

x D
log a
2 C 3 log 2 C o.1/ D
log a C 0:0794 C o.1/:

The logarithmic divergence as a ! 0 comes from the long sojourn time of a particle passing near the stagnation points at the front and back of the cylinder. We include in Figure 4.9 these approximations for large and small a.

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4.5. DRIFT

69

4.5.3. Darwin’s Theorem. One could attempt to compute the drift area or volume by first finding the drift function D and then carrying out the integration of this function normal to the direction of motion. However, a beautiful result of Darwin obviates this calculation. T HEOREM 4.5 The drift area or volume obtained in steady motion of a finite rigid body in potential flow is equal to the apparent mass of the body divided by the density of the fluid. To prove this, we may again assume rectilinear motion along the x-axis with unit speed. We also deal only with two dimensions. The three-dimensional extension should be obvious. If the velocity field moving with the body is u; v, then the apparent mass M may be computed directly from the kinetic energy of the perturbations: “ M D Œ.u
1/2 C v 2 dx dy  “ “ 2 (4.104) .u
q 2 /dx dy D I1
2I2 ; D .1
q /dx dy
2 where q 2 D u2 C v 2 . Here all integrations are over some large finite domain between two streamlines, but extending to ˙1 in x. Now dx dy D q
2 d d @x . Consequently, and uq
2 D @  “  Z @x
1 d d D .x
/d D 0 (4.105) I2 D @ since x
 vanishes at x D ˙1. Thus  “  1
1 d d ; (4.106) M D I1 D  q2 which is the desired result. It should be recalled that apparent momentum involves the calculation of a marginally convergent integral, one whose value depends upon how the domain is expanded to infinity. Darwin’s theorem requires a particular limit process, where we first compute the drift function from x D
1 to x D C1, then expand the domain of integration over . It is only in this order that the theorem holds. There is a noteworthy corollary to this result: Under the conditions of Darwin’s theorem, necessarily D.a/  0. To see this, we need only apply the proof of the theorem to two nearby streamlines. The resulting integral is positive, and provided D.a/ is continuous, for a sufficiently small domain all included streamlines will yield D.a/  0. The nonnegativity of D can be understood by considering tracking the xposition of a parcel of fluid as it moves along a streamline in a frame fixed with the body. The perturbation in x-velocity, u
1, must average to 0 when integrated with respect to x. Otherwise the parcels on the streamline would have acquired net R dxx-momentum in the absence of body forces. If now we consider the integral u D t, we observe that the regions where u
1 < 0 will contribute more to t

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70

4. POTENTIAL FLOW

than those regions where u
1 > 0. Consequently, there must be an excess of t relative to an unperturbed parcel, so that D cannot be negative. Problem Set 4 (4.1) (a) Show that the complex potential w D Ue i˛ z determines a uniform flow making an angle ˛ with respect to the x-axis and having speed U . (b) Describe the flow field whose complex potential is given by w D Uze i˛ C

Ua2 e
i˛ : z

(4.2) Recall the system (4.13) governing the motion of point vortices in two dimensions. (a) Using these equations, show that two vortices of equal circulations

, a distance L apart, rotate on a circle with center at the midpoint of the line joining them, and find the speed of their motion. (b) Show that two vortices of circulations and
, a distance L apart, move together on straight parallel lines perpendicular to the line joining them. Again find the speed of their motion. (4.3) Using the method of Blasius, show that the moment of a body in twodimensional potential flow about the axis perpendicular to the plane (positive counterclockwise) is given by
Z    dw 2 1 z dz ; M D
 Re 2 dz C

where Re denotes the real part and C is any simple closed curve about the body. Using this, verify by the residue method that the moment on a circular cylinder with a point vortex of circulation at its center, in uniform flow, experiences zero moment. (4.4) Compute, using the Blasius formula, the force exerted by a point vortex at the point c D be i , b > a, upon a circular cylinder at the origin of radius a. The complex potential of a point vortex at c is
i 2 ln.z
c/. (Use the circle theorem and residues.) Verify that the cylinder is pushed away from the vortex. (4.5) Prove Kelvin’s minimum energy theorem: In a simply connected domain D f on the boundary S of V . (This u is unique V , let u D r, r 2  D 0, with @ @n in a simply connected domain.) If v is any differentiable vector field satisfying r  v D 0 in V and v  n D f on S , then Z Z 2 jvj dV  juj2 dV: V

V

(Hint: Let v D u C w and apply the divergence theorem to the cross term.) (4.6) Establish (4.34) and work through the details of the proof of zero drag of the Rankine fairing using the momentum integral method, as outlined in Section 4.2.2.

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PROBLEM SET 4

71

(4.7) In spherical polar coordinates .r; ; '/ a Stokes stream function ‰ may be defined by @‰ 1
1 @‰ ; u D : uR D 2 R sin  @ R sin  @R Show that in spherical polar coordinates, the stream function ‰ for a source of strength Q, placed at the origin, normalized so that ‰ D 0 on  D 0, is given Q .1
cos /. Verify that the stream function in spherical polars for the by ‰ D 4 airship model consisting of equal source and sink of strength Q, the source at the origin and the sink at R D 1,  D 0, in a uniform stream with stream function 1 2 2 2 UR .sin / , is given by (4.36). (Suggestion: The sink will involve the angle with respect to R D 1,  D 0. Use the law of cosines (c 2 D a2 C b 2
2ab cos  for a triangle with  opposite side c) to express ‰ in terms of R; .) (4.8) In the Butler sphere theorem, we needed the following result: Show that a2 ‰1 .R; /  R a ‰. R ; / is the stream function of an irrotational, axisymmetric flow in spherical polar coordinates provided that ‰.R; / is such a flow. (Hint: a2 Show that LR ‰1 .R; / D R a LR1 ‰.R1 ; /, where R1 D R . Here LR is defined by (4.40).) (4.9) Consider the following model of flow past a circular cylinder of radius a with two eddies downstream of the body. Consider two point vortices, of opposite strengths, the upper vortex having clockwise circulation
(i.e., > 0) located i ln.z
c/ to the complex potential w, at the point c D be i , thus adding a term 2 the other having circulation at the point cN D be
i . Here b > a > 0. Using the circle theorem, write down the complex potential for the entire flow field, and determine by differentiation the complex velocity. Sketch the positions of the vortices and all vortex singularities within the cylinder, indicating their strengths. (4.10) Continuing Problem 4.9, verify that x D ˙a, y D 0 remain stagnation points of the flow. Show that the vortices will remain stationary behind the cylinder (i.e., not move with the flow) provided that U

  N 2 a2 i .c 2
a2 /.b 2
a2 / C .c
c/ a2 : 1
2 D c 2
.c
c/.c N 2
a2 /.b 2
a2 /

Show (by dividing both sides of the last equation by their conjugates and simpli2 fying the result) that this relation implies b
ab D 2b sin , that is, the distance between the exterior vortices is equal to the distance between a vortex and its image vortex. (4.11) Show that the apparent mass matrix for a sphere is 12 M0 ıij where M0 is the mass of fluid displaced by the sphere. (4.12) Show that for a body in three dimensions, which may have a time-dependent shape but is of fixed volume, the quantity a in (4.50)–(4.51) must vanish. (4.13) Using the alternative definition (4.70), show that Mij is a symmetric matrix.

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72

4. POTENTIAL FLOW

(4.14) Let the elliptic cylinder of Examples 4.14 and 5.13 be placed in a steady uniform flow .U; V /. Show, using the result of Problem 4.3, that the moment experienced by the cylinder is

U V .˛ 2
ˇ 2 /, ˛; ˇ being the major and minor semi-axes of the ellipse. (4.15) Prove that the drift function D.a/ for a sphere of unit radius, where a is the initial value of the polar radius r of the fluid particle, behaves like 9 D.a/
a
5 ; a ! 1: 64

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http://dx.doi.org/10.1090/cln/019/05

CHAPTER 5

Lift and Drag in Ideal Fluids We now take up the study of the effects of vorticity on ideal fluid flow. One of the most interesting and subtle properties of ideal fluid flow theory is its relation to the physical properties of real, viscous fluids as the viscosity tends to zero. We will consider viscous fluids in Chapters 6 through 8. But first we comment on some aspects of the role of viscous stresses in determining the relevance of the ideal fluid and the applicability of Euler’s equations when vorticity and circulation do not vanish. Our main point is to draw a distinction between the limit process flow obtained from a real fluid flow in the limit of vanishing viscosity, and the ideal fluid flow theory that results from setting viscosity formally equal to 0. Because of the nature of the mathematical form of viscous stresses involving the spatial rate of change of the velocity, viscous effects can be nonnegligible at arbitrarily small viscosity provided that the spatial gradient of the velocity field is sufficiently large. In ideal fluid theory the fluid velocity is assumed to be tangent to any fixed solid boundary abutting the flow. If this surface undergoes rapid changes in slope, large viscous stresses will develop as the flow tries to “turn the corner.” To relieve these stresses the streamline pattern must change, and we shall give examples of this below. The effect can persist even as viscosity vanishes. In fact, surfaces need not be present. The residual effect of viscosity also remains in the fluid away from boundaries, when the fluid is in turbulent motion. In that case the small spatial scales come from the stretching of vortex tubes by the flow. A main result of such processes is the introduction of vorticity into the fluid near the flow boundaries. When a rigid body moves rapidly through a fluid, it will create free vorticity by the process of vortex shedding. Once this vorticity is created, it is carried about and diffused, embedded in the otherwise irrotational flow. These considerations lead to ideal flow models that incorporate the effects of the viscosity of the real fluid, despite the fact that viscosity has been expelled from the equations of motion. Typically both the boundary conditions on the flow and the position of the vorticity within the flow will be affected. Since the theoretical basis for dealing precisely with the limit process to zero viscosity is not well developed, in particular regarding the shedding of vorticity, the models we will study are fairly crude approximations to reality. Nevertheless, they adequately capture the essential physics and have important applications. The focus of this chapter emphasizes the most important applications of these ideas to the concepts of lift and drag in aerodynamics. A closely related problem 73 Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

74

5. LIFT AND DRAG IN IDEAL FLUIDS

is the generation of thrust in the flapping flight of birds and insects and in the swimming of fish. We begin with the calculation of lift of two-dimensional airfoils, then consider a model of a lifting three-dimensional wing. We also show in the latter case that drag is realized in an ideal (but not irrotational) fluid. Both lift and drag will be associated with vorticity. This vorticity may occur within a body, in which case it is called bound vorticity, or it may be in the flow exterior to the body, where it behaves as a material vector field. A body moving with constant velocity through a quiescent ideal fluid can create a vortical wake stretching out behind it, a familiar example being the trailing vortices behind a high-flying jet. Vortical wakes are also created by birds in forward flight and by swimming fish. Drag is conventionally equal to the component of force experienced by a body, parallel to the direction of motion. Lift is the component normal to the direction of motion, positive if opposite to gravity. It is easy to see how drag might be associated with a vortical wake. If a rigid body is pulled through the fluid with speed U and sheds vorticity at a steady rate, this vorticity is carried off to infinity, and in the absence of viscosity there will be no decay. Consequently, the associated flux of energy Fe is a loss to the system, which must be replenished by the work done against drag D, UD D Fe . A weightless self-propelled body in motion with constant average velocity is, according to Newton, not exerting any average force on the fluid. So in steady flight the drag must be balanced by thrust developed by a propeller, a jet engine, or a flapping fin. (In fact, it is often not easy to separate forces into drag and thrust.) If the net weight of a body, which is the weight minus the buoyancy (Archimedean) force, is nonzero, then in unaccelerated motion a body must create a lift force to compensate this net weight. The equilibrium in the steady translation of a body can thus be expressed as thrust D drag and lift D weight. The energy flux in a wake of ideal fluid represents the work being done on the fluid by the body. A rigid, neutrally buoyant body being towed at a constant velocity will experience a drag, and the average work done against this drag must equal the energy flux in the wake. The vortical wake of a steadily swimming fish or a bird in steady forward flight will also carry energy to infinity, even though there is no average force applied to the fluid. However, this energy will equal the rate of working of the body in the effort of swimming or flying. If a body locomotes steadily in an irrotational fluid (a problem we considered in Chapter 4), then the energy flux in the wake must vanish. 5.1. Lift in Two Dimensions and the Kutta-Joukowski Condition In an ideal fluid any force on a two-dimensional body must be a result of the pressure exerted on the body. According to the Bernoulli theorem for steady flow, the distribution of the pressure force over the surface of the body is directly related to the distribution of velocity there. This viewpoint can then lead to an attempt to understand the creation of lift as being a result of higher velocity over the top of the airfoil than over the bottom. Such an explanation, although mathematically correct, offers no hint of why the observed velocity distribution occurs.

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5.1. LIFT IN TWO DIMENSIONS AND THE KUTTA-JOUKOWSKI CONDITION

75

F IGURE 5.1. Streamlines for uniform flow past a flat plate, ˛ D 12ı . (a) Zero circulation. (b) Circulation determined by the Kutta-Joukowski condition.

To understand the basis of lift in two dimensions, it is helpful to consider the simplest of “airfoils,” namely a simple flat plate. Of course, this choice is special since each endpoint of the plate is an extreme corner where the tangent to the surface changes in direction by 180ı . We will later consider more realistic airfoils. We now utilize conformal mapping to study the aerodynamics of a flat plate. The circle of radius a in the Z-plane is mapped into the doubly covered segment jxj  2a in the (physical) z-plane by z D Z C a2 =Z. We also know that uniform potential flow with velocity .U; V / D Q.cos ˛; sin ˛/ past the circular cylinder is not unique; the general solution is (5.1)

W .Z/ D QŒe
i˛ Z C a2 e i˛ Z
1


i ln Z; 2


where is the circulation about the cylinder. With w.z/ D W .Z.z// we can then consider the streamline pattern about the plate for various . The angle ˛, in the language of aeronautics, is called the angle of attack. We show in Figure 5.1(a) the case of zero circulation D 0. There are points of zero velocity, or stagnation points, on the surface of the plate. The flow is forced around the endpoints so as to maintain the tangency of velocity at the body, and it is easy to see that there are singularities of velocity and pressure at z D .˙2a; 0/. In Figure 5.1(b) we show the same flow with a negative , chosen to move the stagnation point on the upper surface to the point .2a; 0/ and to eliminate the singularity there. The singularity at the point .
2a; 0/ remains. We must now ask, What does one observe in an experiment in a wind tunnel? Under conditions where ideal flow theory should more or less prevail, it is observed that the flow does come away smoothly from the downstream or “trailing” edge of the plate, as in Figure 5.1(b). Experiments also show that, at least at modest angles of attack, the flow at the upstream or “leading” edge of the plate can actually look as shown in Figure 5.2, with separation of the boundary streamline at the leading edge, and perhaps re-attachment further back, a small circulating eddy

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76

5. LIFT AND DRAG IN IDEAL FLUIDS

F IGURE 5.2. Leading-edge separation from a flat plate.

being enclosed by the separating streamline. The flow may also be unsteady. We remark that at larger angles of attack, the flow breaks away from the leading edge completely, irregular shedding of vorticity occurs, and the airfoil is said to stall. What we thus see is a definite upstream-downstream asymmetry of the flow in its response to the singular points of the boundary. The flow seeks to make a smooth flow off the trailing edge, but accommodates itself on the leading edge by the tendency to form a separation bubble that effectively gives a smooth shape to the upstream end of the surface. It is reasonable that a steady flow past our plate exists in the limit of arbitrarily small viscosity. In fact, hydrodynamic instabilities will prevent one from ever observing the limit process as a steady flow, but the assumed limit would presumably apply to the unstable steady branch of solutions of the equations of the real fluid. The condition used here, which selects among all possible values of the unique value that eliminates the singularity at the trailing edge of the plate, is called the Kutta-Joukowski condition. To apply it in the present example of a flat plate, we note that
 i
1 dZ dw
i˛ 2 i˛
2 D QŒe Z
a e Z
dz 2
dz 
p 2 
i
1 z
4a2 C z
i˛ 2 i˛
2 p Z
a e Z
(5.2) : D QŒe 2
2 z 2
4a2 The terms within the first set of large brackets must sum to 0 at Z D a if the singularity at z D 2a is to be removed. Thus we find (5.3)

D
4
Qa sin ˛

from the Kutta-Joukowski condition. Once this condition is applied, the ideal flow theory matches the observations at the trailing edge but fails to account for the flow behavior at the leading edge. To see that the resulting flow gives rise to a lift force, we compute the force on the body from Blasius’ formula (4.14). From the residue of .dw=dz/2 .z/ we find for our flat plate problem (see Problem 5.1(a)) (5.4)

X
iY D  Q.sin ˛ C i cos ˛/:

This is a force of magnitude L D 4
aQ2 sin ˛, which is orthogonal to the free stream velocity for the geometry of Figure 5.1(b), and is upward for positive ˛ by (5.3) ( < 0) , so it is indeed a lift. It is not orthogonal to the plate itself, which raises the paradoxical situation where a pressure force, presumably always
We shall see in Chapter 8 that this asymmetry can be traced to the nature of the partial differential equation for the viscous boundary layer on the surface of the body.

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5.2. SMOOTHING THE LEADING EDGE: JOUKOWSKI AIRFOILS

77

orthogonal to the surface, seems to be in violation of that fact. The resolution of this paradox involves a careful analysis of the singularity near the leading edge. The airfoils considered in the next section have a smooth leading edge, and the flat plate may be regarded as the limit of a family of such smoothed foils. For each member of the family, it is found that the pressure distribution around the smooth nose in fact produces a component of force parallel to the plate, which is precisely the magnitude needed to make the lift vector orthogonal to the free stream velocity. This “leading edge suction force” is preserved in the limit, even though the “edge” disappears, and gives the result (5.4) for the flat plate. The result (5.4) is known as the Kutta-Joukowski theorem. It is central to airfoil theory because it is a general result independent of the actual shape of the airfoil. The reason for this generality can be understood once one sees how the residue computation used to compute the pressure force is carried out for the flat plate. For any foil at angle of attack in a uniform stream, the expansion of w.z/ near infinity will have the form A B i log z C C 2 C    ; (5.5) w D Qze
˛
2
z z which follows from the definition of circulation. The subsequent terms involving A; B; : : : are determined by the particular shape of the airfoil. When we compute
i˛ , and this leads to (5.4). the residue of .dw=dz/2 at infinity we obtain
iQ  e Thus the lift computation for any airfoil really amounts only to a determination of the circulation required by the Kutta-Joukowski (K-J) condition. Although the K-J condition gives a unique circulation and lift for an airfoil, it remains an approximation to reality. Rapid movements of an airfoil can produce momentary flows that differ from those obtained under the K-J condition and may even come close to the flows with zero circulation as in Figure 5.1(a). Also, we have mentioned the “stalling” of an airfoil at higher angles of attack, which completely eliminates the flow field constructed using the K-J condition. In fact, the “true” K-J theory, which would allow the “correct” ideal flow representing the slightly viscous flow under arbitrary movements of a body, remains an important outstanding unsolved problem of fluid dynamics, and lies at the heart of a rigorous theory of vortex shedding from surfaces. 5.2. Smoothing the Leading Edge: Joukowski Airfoils We have noted that the leading edge of a flat plate is not well suited to the smooth flow that we would like to establish around an airfoil by the application of the K-J condition. Airfoil designers therefore prefer a shape that maintains the sharp trailing edge, so as to “force” the K-J condition there, but which also provides a smooth leading edge around which the flow may pass without detachment. Remarkably, such foils can be obtained by a simple modification of the conformal map associated with the flat plate flow. Instead of considering flow around the circular cylinder of radius a and center at the origin, we consider p the flow past a circular cylinder with center at Z0 D  C iı and radius c D .a
/2 C ı 2 , with  < 0, ı > 0. We show in Figure 5.3 the foil shapes that result from various

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78

5. LIFT AND DRAG IN IDEAL FLUIDS

(a)

(b)

(c)

F IGURE 5.3. Joukowski airfoils, a=1. (a)  D
:1, ı D 0. (b)  D 0, ı D 0:1. (c)  D
0:1, ı D 0:1.

Y

Z0 α

β

c a

X

Q

F IGURE 5.4. Geometry of the Z-plane for the Joukowski airfoils. Z0 D  C i ı:

choices of ; ı. Note that  determines the foil thickness, and ı its camber, or the arc the foil makes relative to the x-axis. The geometry of the Z-plane is shown in Figure 5.4. The trailing edge is a cusp. It is not difficult to modify the force calculation to accommodate the Joukowski family of profiles, and there results a lift force orthogonal to the free stream velocity, but with magnitude (see Problem 5.1(b)) ı : a
 Note that the effect of camber is to change the angle of attack at which the lift vanishes. The moment on a Joukowski airfoil can be computed by residue theory using the formula given in Problem 4.3. To work this out we have 

  i Z0 2 c2 i a2 2 dw 2
i˛ i˛
D Qe
2 Qe
1
2 C O.z
3 /; (5.7) dz z 2
z 2
z 2 z (5.6)

D 4
cQ2 sin.˛ C ˇ/;

tan ˇ D

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5.3. UNSTEADY AND QUASI-STEADY MOTION OF AN AIRFOIL

79

and so the residue at infinity of z.dw=dz/2 is i 1 2

:
2Q2 a2 e
21˛
2Q2 c 2
Q Z0 e
i˛

4
2 Thus
  i 1 2 1 2 2
2i˛ 2 2
i˛
2Q c
Q Z0 e
2 : (5.8) M D
< .2
i/
2Q a e 2
4
After substituting Z0 D Z0
a C a D
ce
iˇ C a we obtain (5.9)

M D
2
Q2 Œa2 sin 2˛ C c 2 sin 2.˛ C ˇ/
2ac cos ˛ sin.˛ C ˇ/ :

Recall that moment is positive in the counterclockwise direction and (5.9) refers to a Joukowski airfoil with the trailing edge to the right. If ˇ D 0 and ˛; ı are small, then a  C and M 
4
Q2 a2 ˛ 
aL. This places the center of lift at approximately z D
a. The length of the foil, known as the chord,  4a, so the center of lift is approximately at the quarter-chord point. For many aircraft the position of the center of gravity is located near this point to provide stability to forward flight. If one looks at wind tunnel data for Joukowski airfoils, or for the many other foil designs of a similar kind, it is found that the predictions for lift at small angles of attack are reasonably good, especially in slope. Usually one plots a lift coefficient L CL D 1 2 2 Q 4a / . Realized lift is usually versus ˛. For the Joukowski foils CL D 2
sin.˛Cˇ cos ˇ somewhat smaller than predicted. More dramatic is the failure of the theory to account for airfoil stall, with its time-dependent vortex shedding and abrupt loss of lift. Stall usually begins when the angle of attack is in the range 10–15ı . Aircraft designers introduce modifications of three-dimensional wings, such as twist, so as to reduce the angle of attack at the outboard wing sections, and thus to minimize the control problems and make the stall a more gradual phenomenon. Other modifications of the wing, such as vortex generators placed near the leading edge, help to delay stall by influencing the flow near the wing’s surface.

5.3. Unsteady and Quasi-Steady Motion of an Airfoil Unsteady motion of an airfoil occurs during the takeoff and maneuvering of an airplane and in the flapping of the wings of birds and insects. An interesting thought experiment (in two dimensions) is to imagine an airfoil at positive angle of attack and leading edge to the left, to be suddenly accelerated from rest to the velocity .
Q; 0/. After the flow has settled down, an observer moving with the foil would see a steady flow .Q; 0/ past the foil and would measure a lift, hence a circulation > 0. Now repeat the experiment with a large material contour initially encircling the foil; see Figure 5.5. The initial circulation on this contour is 0 since the fluid is at rest. After the acceleration to a fixed velocity, there exists

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80

5. LIFT AND DRAG IN IDEAL FLUIDS

−Γ Γ

F IGURE 5.5. The starting vortex shed by a lifting foil abruptly accelerated to a constant velocity.

a negative circulation about the foil. However, according to Kelvin’s theorem, the circulation about the image of the large initial contour, now distorted by the motion of the foil, must remain 0. Since we know the foil has negative circulation, there must be other vorticity within the contour contributing positive circulation. Observation of the acceleration of foils shows that this missing vorticity is created during the initial phase of acceleration of the foil. Positive vorticity is rapidly shed at the trailing edge to form a coherent starting vortex whose circulation approximately cancels the circulation bound to the foil in steady flight, as we show in Figure 5.5. This shedding is part of the process of establishing the smooth flow at the trailing edge demanded by the K-J condition. This is an example of the unsteady aerodynamics of an airfoil. Such unsteady motions will generally involve shedding of vorticity from the trailing edge, and the shed vorticity will then influence the flow external to the foil. The shed vorticity moves with the fluid and must be accounted for in calculating the forces on the foil. Under certain conditions, by insuring that the shed vorticity essentially remains in the plane of the foil, it is possible to study analytically the unsteady inviscid flow past an oscillating foil; see [3]. A measure of unsteadiness is a parameter of the form TLU where L is some typical length, T a time over which a cycle of motion is performed, and U a speed of flight. If this dimensionless number is of order unity or larger, then the resulting flow is said to be fully unsteady. If the number is small, the flow is said to be quasi 1 second and have a wing steady. A dragonfly may beat its wings once in T D 40 L chord L D 1 cm. If it moves at U D 40 cm/sec then T U D 1 and the flow should be considered to be unsteady. A pigeon with a wingbeat each 15 sec and a wing chord of 10 cm, flying at 3 m/sec has TLU D 16 , so its flight might be considered quasi-steady. Let us devise a quasi-steady theory of forward fight of a flapping wing. While it is true that birds are flapping their wings to fly, the fact is that the main reason for flapping is to produce thrust, so as to overcome drag. A flapping Joukowski airfoil at angle of attack ˛ D
ˇ produces only thrust. To acquire lift it is only necessary to increase the angle of attack while maintaining the flapping at that angle of attack. But of course to develop lift the wing must be moving! So the initiation of flapping flight is a kind of “bootstrap” operation where special wing movements are needed.

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5.4. DRAG IN TWO-DIMENSIONAL IDEAL FLOW

81

T U

L Q Q U

V

V L (b)

(a) T

F IGURE 5.6. Thrust production by quasi-steady flapping of a flat plate.

To understand thrust production without any net lift consider a simple flat plate in uniform flow, which maintains itself horizontal while moving up and down in a periodic motion; see Figure 5.6. Remember that as the plate moves we assume quasi-steady aerodynamics, so that the instantaneous flow about the plate will be the steady flow corresponding to the instantaneous velocity on the wing, and we assume the Kutta-Joukowski condition also applies at each instant. Thus in Figure 5.6(a) the wing is moving down with speed V , and so sees an effecV . The “lift” vector L is orthogonal to the intive angle of attack ˛, tan ˛ D U stantaneous approach velocity vector .U; V /, which produces a thrust component T D L sin ˛ D 4
aQ2 sin2 ˛. In Figure 5.6(b) the wind moves up, but the same expression for thrust results. The time average yields the positive thrust, T D 4
aQ2 sin2 ˛: We remark that in quasi-steady flapping flight there is a steady stream of vorticity shed from the trailing edge of the foil, but it is swept downstream so fast that its effect on the flow is small. 5.4. Drag in Two-Dimensional Ideal Flow In the present section we give two examples of the modeling of drag in an ideal fluid. Recall that for irrotational flow the drag force will vanish in two or three dimensions. In fact, a body in a real fluid will experience drag. We shall illustrate now how drag in two-dimensional ideal flow can result from vorticity in the fluid. 5.4.1. The Kármán Vortex Street. Experiments with flow past a circular cylinder in a wind tunnel and numerical calculation in two dimensions show that as the velocity of the stream increases, a point is reached where the flow becomes unsteady and vortices are shed into the flow, alternating between the top and bottom of the cylinder; see Figure 5.7(a). These vortices (really patches of vorticity) are then carried by the flow downstream, forming a vortical wake. This wake carries energy downstream, and the cylinder experiences a drag. The time dependence can give rise to an oscillating lateral force, and one manifestation of this is the “singing” of wires in a wind. Von Kármán developed a simple two-dimensional model for such a wake, now called the Kármán vortex street. It consists of a periodic array of point vortices of strengths ˙ , extending from the cylinder to downstream infinity. It can be most conveniently analyzed by extending the street to upstream infinity as well, so as to

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82

5. LIFT AND DRAG IN IDEAL FLUIDS

(a)

(b) b/2 2a

a

0

-b/2 3a/2

a/2

(c)

F IGURE 5.7. (a) Kármán vortex street in the atmosphere due to motion past an island off the Chilean coast (NASA image). These atmospheric motions are very nearly two-dimensional. (b) Schematic of vortex shedding from a circular cylinder. (c) The doubly infinite street. The upper vortices carry circulation of
. The lower vortices carry circulation of > 0.

model the wake well downstream of the cylinder; see Figure 5.7(b,c). To study this flow, consider first a single finite line of vortices of circulation , spaced a distance a apart on the x-axis. The velocity potential is wN (5.10)

CN i X D
log.z
na/ 2
nD
N
 
z N z2 i log … 1
2 2 C const: D
2
a nD1 n a

Using the identity 

(5.11)

z2 sin z D z 1
2


   z2 z2 1
2 2 1
2 2  ; 2
3


we get, in the limit N ! 1 for a suitable additive constant, (5.12)

wn ! w1 D



z i log sin : 2
a

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5.4. DRAG IN TWO-DIMENSIONAL IDEAL FLOW

83

cn Cb C0

F IGURE 5.8. Contours for calculating drag for the vortex street.

For the vortex street shown in Figure 5.7(c), we thus have   sin a .z
ib=2/ i (5.13) w1 D log : 2
sin a .z
a=2 C ib=2/ Since the vortices are being shed by a body in a flow .U; 0/, relative to the body the complete velocity potential is (5.14)

w D Uz C w1 :

To see how the vortices are moving relative to an observer fixed with the body, consider, by symmetry, the velocity at .0; b2 / for the system minus the vortex at that point. The vortices must then move with velocity 
i 1 dw
D U
V: (5.15) lim 2
z
ib=2 z!ib=2 dz Evaluating this limit, we find
b tanh : 2a a In experiments V is considerably less than U , so the vortices move downstream with a speed slightly less than the free stream speed. For a circular cylinder of diameter D the vortices of like sign are shed with a frequency f where fUd  0:2. Thus U f
V D a  5D. The drag force can be computed using the Blasius formula for force, but with an added effect due to the fact that vortices are being steadily added as they are shed from the body. We now outline this computation. Relative to an observer fixed with the vortex street the velocity at infinity is V , and the body is moving to the left with speed U
V . Imagine a rectangular boundary C0 D ABCD surrounding the entire region, as shown in Figure 5.8. The dotted sides of the rectangle will eventually move off to infinity and the solid line will be placed at a position far downstream where the street will be effectively doubly infinite. The small positively oriented contours cn surround the vortices, and the contour Cb is the body contour. The solid right side of the outer rectangular contour does not intersect any vortex. At a particular time the Blasius theorem may be applied to yield   I  I  dw 2 dw 2 dz C dz; (5.17) X
iY D
dz dz P (5.16)

V D

cn

C0

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84

5. LIFT AND DRAG IN IDEAL FLUIDS

where the sum is over the cn within C0 . In this frame the potential seen on AB, wv D V z C w1 ;

(5.18)

will be essentially independent of time if the street is taken as doubly infinite. The contributions from the first integral in (5.17) are seen to contribute only to Y , since i the residues are just ˙2V 2 . (These contributions would allow us to deduce an oscillating vertical force on the body.) The second term in (5.17) gets a contribution in the limit only from the right vertical side of the outer contour, and we obtain from (5.17) the following contribution to the drag:     Z
b
b i Ci1 dwv 2 2 1
tanh : dz D (5.19) D1 D < 2
i1 dz 2
a a a There is also momentum being created as a function of time by the shedding of vortices within C0 . At this point we must make do with an approximate calculation, for the shed vortices break the symmetry of a doubly infinite street. We can approximate the calculation by determining the x-momentum per unit length of the street from w1 , m say, then determine the momentum shed per cycle period T as ma T . Since .U
V /T D a, the contribution will be D2 D
m.U
V / since positive drag contributes negative x-momentum. To compute ma, we need only consider two adjacent vortices of opposite sign. Thus Z dw1 dS ma D  dz Z C1  Z C1 i dy D 2
.x C iy
ib/
1
1  i dx:
2
.x C iy
a=2 C ib/ xDC1
 Z a i C1 (5.20) D dy: log.x C iy
ib/
log x C iy
C ib 2

1 2 xD
1 Now the integrand in (5.20) gets contributions from the change of the argument of the log terms as x goes from
1 to C1. This is seen to give C2
i when jyj < b and 0 when jyj > b. Thus ma D
 b

(5.21) and

D2 D b

(5.22)

U
V : a

The drag of the body is then

 
b
b U
V 2 1
tanh C b 2
a a a a 2   b .U
2V / C : D a 2
a

D D D1 C D2 D (5.23)

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5.4. DRAG IN TWO-DIMENSIONAL IDEAL FLOW

85

C U

B A

p=p

0

B'

C'

F IGURE 5.9. Free streamline flow onto a flat plate, a D 1.

Since V is observed to be well below U2 , both of these terms are positive in any realistic model. The stability of the Kármán vortex street is of considerable interest. It is found that the street is stable to small perturbations of the x-position of the vortices provided that sinh.
ab / D 1, yielding ab D 0:281. Values not too far from this are observed in experiments, but often it is found that ab is not constant in the wake, increasing with distance downstream. 5.4.2. Free Streamline Theory of Flow Normal to a Flat Plate. There is a body of theory for two-dimensional ideal fluid flow that allows for streamlines on which velocity is discontinuous to exist in the flow exterior to any boundaries. These free streamline theories effectively embed free vorticity in an otherwise irrotational flow field. They form a class of models where shed vorticity is concentrated at the streamlines of discontinuity. As such they offer one way of accounting for the phenomenon of separation. Suppose that on one side of such a streamline the fluid is in motion, but on the other side the velocity is identically 0 and pressure is constant. If the flow is steady and Bernoulli’s theorem applies, then on the flow side p C 2 juj2 is constant on the streamline. We now assert that at such a streamline pressure must be continuous. Otherwise a difference of pressure would act across a sheet, with no inertia to support such a force by a finite acceleration. Thus it must be that juj D q is constant on the free streamline. We will now examine a model, due to Kirchhoff, which seeks to represent the detached flow that is observed behind bluff bodies in a uniform stream. The theory will assume a steady flow, even though the observed flows are usually time dependent. The structure is shown in Figure 5.9 in the case of flow broadside onto a finite flat plate. Two free separation streamlines leave the tips of the plate and extend to infinity aft of the body. The region behind the plate, between the free streamlines, is a cavity or “dead water” region, where velocity is 0 and pressure a constant p0 . Well upstream the velocity is .U; 0/ and the pressure is p0 . Thus p C 2 q 2 D p0 C 2 U 2 and in particular q D U on the free streamlines.

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86

5. LIFT AND DRAG IN IDEAL FLUIDS

(a)

w-plane

ψ

C

B

A

φ C'

B' (b) Z-plane -1 B'

C'

(c)

A

log(U/q) B π/2 C C' −π/2 B'

+1 B

C

Q-plane A θ A'

F IGURE 5.10. The conformal mappings for the Kirchhoff solution.

The solution to this flow problem involves a interesting technique in conformal mapping, which exploits a correspondence between identical maps in distinct variables, allowing a direct connection between these variables and an equation determining the complex potential. The procedure is sometimes referred to as a hodograph method, this term indicating that the velocity components appear in the definition of an intermediate complex variable.
We now describe the series of maps involved and the connections between them. We first note that for this flow the complex potential w D .x; y/ C i .x; y/, whatever form it may take, maps the z- or physical plane shown in Figure 5.9 onto the w-plane as shown in Figure 5.10(a). The body is here a streamline D 0. We next map the w-plane onto the Z-plane as shown in Figure 5.10(b). The map is defined by C (5.24) w D Z2 2 where the constant C will need to be chosen to make the points B; B 0 map onto .1; 0/ and .
1; 0/. Finally, consider the variable U (5.25) Q D ln C i; q p :. The Q-plane will be where q D u2 C v 2 and  D tan
1 uv with u
iv D dw dz the hodograph plane. The mapping of Y D 0 to the hodograph plane is especially


The term comes from the Greek hodos, meaning “path.”

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5.4. DRAG IN TWO-DIMENSIONAL IDEAL FLOW

87

simple since either the angle or the speed is constant. Thus we are bound to get a polygon. Since we know how to map a polygon onto the upper half-plane, we can connect Z to Q. We show the Q-plane in Figure 5.10(c). The map from z to Q is a SchwarzChristoffel map, given by 1 dQ  const: D p dZ Z Z2
1

(5.26)

The integral may be calculated using a substitution Z D 1=cosh X. We obtain r
 1 1
1 1 C C2 D C1 log C
1 C C2 D Q.Z/; (5.27) Q D C1 cosh Z Z Z2 where C1;2 are constants. i But Q.e i / D
C1 i
C C2 D
i 2 , Q.1/ D C2 D 2 , giving r   i
1 1 C :
1 C (5.28) Q D log 2 Z Z 2 U we have using (5.28) Since Q D log dw=dz

U

(5.29) Also

dw dZ

p dz D iZ
1 Œ1 C 1
Z 2 : dw

D C Z, so

p dz D iC.1 C 1
Z 2 /: dZ If the width of the plate is L, then   Z C1 Z C1 p dz
2 dZ D iUL D iC : U .1 C 1
Z /dZ D iC 2 C (5.31) dZ 2
1
1 U

(5.30)

This determines C and gives wD

(5.32)

UL 2 Z : 4C


Since we also have U
1

(5.33)

iZ dw p D ; dz 1 C 1
Z2

we have defined implicitly w.z/. We will now show that, because of the cavity, the plate experiences a drag. The drag is given by an integral over the upstream face of the plate, Z Z Z
i .U 2
q 2 /dz: p dy D
i p dz D (5.34) DD 2 plate

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88

5. LIFT AND DRAG IN IDEAL FLUIDS

2 Now on the front face of the plate q 2 D v 2 D @ D
. dw /2 , and so, using @y dz (5.32) and (5.30), we have     Z U 2 L i C1 dw 2 dw
dZ DD 2 2
1 dz dz Z p U 2 L U 2 L C1
.1
1
Z 2 /dZ D 2 4 C

1   2 4

U 2 L
U L (5.35) 1
D : D 2 4C

C4 This drag is not too far from what is observed when a flat plate is placed in a stream and a wake cavity forms. We have already noted that bluff body flows are time dependent and that the cavity breaks up into vortex structures. Nevertheless, the Kirchhoff solution is a classic example of fluid modeling, exhibiting some key features of observed flows and providing a good example of the role of free streamlines in the production of drag. 5.5. The Three-Dimensional Wing: Prandtl’s Lifting Line Theory Airplanes and birds fly in three dimensions. We will now explore how lift and drag arise in the real world. Since d’Alembert’s paradox implies neither lift nor drag is possible in irrotational flow in three dimensions, the occurrence of lift and/or drag must imply the existence of vorticity in the fluid. We start by reviewing the vorticity structure of a two-dimensional airfoil, in particular a flat plate at angle of attack with the Kutta-Joukowski condition applied. We know the complex potential in the flat plate problem from Section 5.1: i ln Z; (5.36) w.z/ D W .Z.z//; W .Z/ D QŒe
i˛ Z C a2 e i˛ Z
1
2
with p 1 (5.37) Z.z/ D Œz C z 2
4a2 ; D
4
a sin ˛: 2 Since the airfoil has zero thickness, vorticity must be concentrated on the line segment jxj < 2a, y D 0. Now v D 0 on the segment, so the vorticity ! D vx
uy is given by
uy , and we shall see that u is discontinuous there. Thus the vorticity of the flat plate is proportional to ı.y/, and the total vorticity at a given value of x must be computed as .x/ D
u.x; 0C/ C u.x; 0
/, with Z 2a .x/dx: (5.38)

D
2a

The plate is therefore said top contain a vortex sheet p of strength .x/. Using (5.36) 2 2 and (5.37), and the fact that z
4a D ˙i 4a2
x 2 when z D .x; 0˙/ we obtain (see Problem 5.5) (5.39)

Q


1

r u.x; 0˙/ D ˙

2a
x sin ˛ C cos ˛; 2a C x

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5.5. THE THREE-DIMENSIONAL WING: PRANDTL’S LIFTING LINE THEORY

89

7 6 5 4 3 2 1 0

0

1

2

F IGURE 5.11. The bound vorticity on a flat plate, in units of
2Q sin ˛, as a function of xa .

z y b x c(y) -b F IGURE 5.12. The three-dimensional wing.

with (5.40)

r .x/ D
2Q

2a
x sin ˛: 2a C x

We show .x/ in Figure 5.11. The vorticity of this foil is bound to the foil. Suppose now that we consider a three-dimensional wing, as shown in Figure 5.12. If the wing is sliced by a plane y D const, we obtain a two-dimensional airfoil section. For example, it might be a Joukowski section, with its chord c, and thickness, camber, and local angle of attack all functions of y. The direction y is called the spanwise direction. The wingspan is here 2b. Since all of the airfoil parameters can vary down the span, we expect the lifting properties of the wing to be a function of y. We also expect that near the center of the wing, the section AB of Figure 5.13(a), the flow should behave as if the section were approximately a two-dimensional airfoil, with vorticity bound to the foil and carrying an associated circulation and lift. However, as we move to the tips of the wing, eventually the section lift must go to 0, if for no other reason than that the section chord goes to 0. Since vorticity is a solenoidal

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90

5. LIFT AND DRAG IN IDEAL FLUIDS

A

B

(a)

y dv

dy

-dv

(b) z γ(y)

y b x

-b (c)

F IGURE 5.13. (a) The vorticity shed from a three-dimensional lifting wing. (b) The origin of the shed vorticity. (c) Prandtl’s lifting line model.

vector, the question has to be, what happens to the vortex lines that were bound to the center section? The answer, suggested in Figure 5.13, is that the decrease in the lift distribution occurs because vortex lines turn and are shed into the wake of the wing, thereby reducing the section circulation. Thus there is a sheet of vorticity emerging from the trailing edge of the wing. For many wings the lift rapidly decreases to 0 near the tips, so that substantial free vorticity is released there. This is the source of the “tip vortices” seen in the wake of high-flying jets. To understand this shedding process, consider Figure 5.13(b). We consider a strip of wing sections of width dy. The wing is assumed to be changing so slowly in the spanwise direction that each such strip acts as if it were a two-dimensional airfoil. On this section the lift will then be l.y/dy where l.y/ is the two-dimensional lift of the local section. Let the pressure on the upper and lower surfaces be p˙ .x/. Then Z Z .pC .x; y/
p
.x; y//dx D
Œp dy D
U .y/; (5.41) l.y/ D
chord

where .y/ is the circulation of the local section and the brackets denote the jump from bottom to top surface. Now consider Z Z DŒv

@Œp

dl D
dx D dx;  (5.42) dy @y Dt chord

chord

where v is the y or spanwise velocity component; we are assuming an ideal fluid of constant density. We further assume that the spanwise acceleration is so small

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5.5. THE THREE-DIMENSIONAL WING: PRANDTL’S LIFTING LINE THEORY

91

that fluid particles near the wing surface, passing over or under the wing, acquire a spanwise velocity that is small compared to U . We may then substitute dx D U dt and evaluate the last integral as a time integral to obtain (5.43)

U Œv :

In this way we compute the jump in y-velocity developed by fluid particles flowing over the top and bottom surfaces, v˙ being the velocity components developed at the trailing edge of the section. Since we expect the lift to decrease as we move toward each tip, the directions of spanwise flow are indicated in Figure 5.13(b) for a piece of the left wing looking upstream from the rear of the wing. Since the pressures are increasing on the upper surface toward the tip, the flow is driven away from the tip. On the bottom surface the flow is in the opposite direction (the dotted arrow in Figure 5.13(b), since on this surface loss of lift is associated with a decrease in pressure. We thus have d dl D U D U Œv

(5.44) dy dy or d D Œv : (5.45) dy Here , the circulation on a section, is positive if the vorticity is in the direction of positive y. Since is the local circulation of a section, (5.45) relates the spanwise change of local lift to the existence of a discontinuity in velocity at the trailing edge of the wing. This discontinuity, Œv , is associated with the production of a vortex sheet at the trailing edge in the x-velocity component !x RD wy
vz . Integrating from z D 0
to z D 0C at the trailing edge, we have !x dz D
Œv . This means that
Œv dy is the incremental vorticity shed into the wake at the trailing edge at section y due to the change of l with y. For the left half-wing, as seen from an observer behind the wing looking upstream, Œv dy is positive if lift increases with y there. Thus
Œv dy is negative, and so the shed vorticity represents a turning downstream of some of the vortex lines bound to the wing, as shown for y < 0 in Figure 5.13(a). Similarly, for the right half-wing the decrease of lift with increasing y causes
Œv dy to be positive. Thus our intuitive picture of bound vortex lines turning downstream to be shed at the trailing edge of the wing is confirmed. Of course this discussion has not provided any theoretical model of the shedding process. Again we must invoke an assumption that ultimately must be established in the inviscid limit, that the jump in v at the trailing edge is continuous at the trailing edge where it becomes free streamwise vorticity. We now examine the model of the three-dimensional wing created by Prandtl, who sought a simple means of deducing the lift and drag of a wing, given the section properties of the wing. This model is sometimes called the lifting line model. The idea is basically to regard the wing as long and thin, and therefore can be defined for a wing plan form quasi-two-dimensional. An aspect ratio

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92

5. LIFT AND DRAG IN IDEAL FLUIDS

(projection onto the xy-plane) by D wingspan2 =wing area D 4b 2 =A where b is the half-span. Mathematically, the Prandtl model is an asymptotic approximation ! 1. The to the fluid dynamics of a three-dimensional wing in the limit situation is as shown in Figure 5.13(c). In the limit the chord is small compared to the wingspan, so the bound vorticity can be thought of as confined to a line, but the circulation about this line becomes a function of y, namely .y/. The variation of .y/ is associated with shed vortex lines downstream of the lifting line and parallel to the x-axis. If, to fix ideas, we take each section of the wing to be a Joukowski foil, then we know from (5.6) that l.y/ D 4
c.y/Q2 sin.˛ C ˇ/. We will, along with Prandtl, make the assumption that the angles ˛; ˇ are small, so that sin.˛ C ˇ/  ˛ C ˇ and Q  U . Then, with the orientation of the coordinate system of 5.13(c) we have, approximately,

.y/ D 4
c.y/U.˛.y/ C ˇ.y//:

(5.46)

We now need to make a crucial reinterpretation of ˛. Owing to the shed vorticity of the wake, the effective angle of attack, that is, the angle made by the oncoming stream at the particular section, will be dependent upon the z-component of velocity induced at that section by the shed vorticity. If this velocity component is w.y/, then the (small) effective angle of attack is given by w (5.47) ˛eff D ˛ C ; U where ˛ is the angle made by the section relative to the velocity at true infinity. In other words, the induced w near the wing, the so-called downwash, changes the apparent “velocity at infinity” from its true value to ˛eff , and each section will “see” a different approach angle. Now w is an as yet unknown function of y, while ˇ; c are given functions of y determined by the section properties. With   w.y/ C ˇ.y/ (5.48)

.y/ D 4
c.y/U ˛ C U we are now in a position to use the Biot-Savart expression for velocity in terms of vorticity to determine w.y/ from .y/. Recall that the shed x-component of  dy. A doubly infinite vortex induces a vorticity at each section is
d .y/ D
d dy velocity given by the two-dimensional point vortex flow. Such a line, carrying unit circulation, parallel to the x-axis at position y D  in the z D 0 plane, will induce a velocity 1 1 2
y


(5.49)

at any point y on the bound vortex. Since the shed vortex is only semi-infinite, this induced vorticity is reduced by a factor 12 . The circulation shed at section  is 

If

d dy

 0 on the left half-wing and

d dy

 0 on the right half-wing, then the shed vorticity is

such as to make w.y/  0 everywhere at the lifting line, hence the term “downwash.”

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5.5. THE THREE-DIMENSIONAL WING: PRANDTL’S LIFTING LINE THEORY

93


ddy dy evaluated at y D , so that we have Z Cb d  ./ 1 dy d: (5.50) wD
4

b y
 Thus, from (5.48) we obtain
Z Cb d  ./  1 dy d ; (5.51)

.y/ D 4
c.y/U ˛ C ˇ.y/
4
U
b y
 which is an integral equation for .y/. The beauty of this model is the direct insight it gives into an important fact about three-dimensional aerodynamics, namely the creation of drag in a perfect fluid model. Observe that whenever the w.y/ is negative, which is generally the case for normal wings, the effective angle of attack is less than ˛. Since our twodimensional airfoil theory tells us that the local lift is perpendicular to the “flow at infinity,” here the apparent or effective flow at infinity, we see that the local lift vector is rotated slightly so as to produce a component in the direction of positive x. This is a drag component, and the summation over all sections will give rise to the wing drag. This drag, since it is caused by the downwash induced at the wing section by the vortical wake, is called the induced drag. We now indicate how to solve the integral equation (5.51) and calculate the lift and induced drag of our three-dimensional wing. We set y D
b cos , 0   
, and suppose that is an even function of y, so it can be represented by a Fourier series 1 X B2nC1 sin.2n C 1/: (5.52)

D Ub nD0

Then 1

X d d dy D d D Ub B2nC1 cos.2n C 1/ d: dy d

(5.53)

nD0

Using this in (5.51) we obtain the definite integral Z  sin m cos m 0 ; d 0 D

(5.54) 0 cos 
cos  sin  0 the verification of which we leave as Problem 5.6. Thus, if c.y/ D C./ and c.y/Œ˛ C ˇ.y/ D D./, (5.51) becomes (5.55) b

1 X

B2nC1 sin  sin.2n C 1/

nD0



C./

1 X

.2n C 1/B2nC1 sin.2n C 1/ D 4
sin D./:

nD0

Given C./ and D./, we are in a position to express all terms as Fourier series in sin.2n C 1/ and solve the resulting linear system for the B2nC1 .

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94

5. LIFT AND DRAG IN IDEAL FLUIDS

Given a solution the lift is Z Z Cb

.y/dy D Ub (5.56) L D U
b

For small

w U,

./ sin  d D 0


U 2 b 2 B1 : 2

then, induced drag is given by Z

(5.57)



1

Cb

Di nd D


w dy D
b

X
2 U 2 b 2 .2n C 1/B2nC1 : 8 nD0

Problem Set 5 (5.1) (a) Verify (5.4) for the flat plate with the Kutta-Joukowski condition applied. (b) Verify that for the Joukowski family of airfoils the lift is given by (5.6) and that the change comes from the new value of the circulation as determined by the K-J condition. (5.2) Consider the Joukowski airfoil with Z0 D bi, a > b > 0. (a) Show that the airfoil is an arc of the circle with center at .0;
.a2
b 2 /i=b/ and radius .a2 C b 2 /=b. (b) With the Kutta condition applied to the trailing edge, at what angle of attack (as a function of b) is the lift 0? (5.3) Let the airfoil parameters other than chord (i.e., k; ˇ) be independent of y. Also, assume the planform is symmetric about the y-axis in the xy-plane. Using Prandtl’s lifting-line theory, show that for a given lift the minimal induced drag occurs for a wing having an elliptical planform. Show in this case that the coefficient of induced drag CDi D 2  drag=.U 2 S / and lift coefficient CL D 2  lift=.U 2 S / are related by CDi D CL2 =.


/:

is the aspect ratio 4b 2 =S . (Some of the World War Here S is the wing area and II fighters, notably the Spitfire, adopted an approximately elliptical wing.) (5.4) This problem will study flow past a slender axisymmetric body whose surface is given (in cylindrical polar coordinates) by r D R.z/, 0  z  L. Here R.z/ is continuous and positive except at 0; L where it vanishes. By “slender” we mean that max0xL R  L. The body is placed in the uniform flow .uz ; ur ; u / D .U; 0; 0/. We are interested in the steady, axisymmetric potential flow past the body. It can be shown that such a body perturbs the free stream by only a small amount, so that in particular, uz  U everywhere. On the other , hand, the flow must be tangent at the body, which implies r .z; R.z//  U dR dz 0 < z < L. We look for a representation of  as a distribution of sources with strength f .z/. Thus Z L f ./ 1 d : .z; r/ D
p 4
0 .z
/2 C r 2

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PROBLEM SET 5

95

(a) Compute @ , and investigate the resulting integral as r ! 0, 0 < z < L. @r Argue that the dominant contribution comes near  D z, and hence show that Z @ 1 f .z/ C1  .1 C s 2 /
3=2 ds for r  L: @r 4
r
1 (b) From the above tangency condition, deduce that f .z/  A.z/ D
R2 is the cross-sectional area of the body.

dA dz

where

(c) By expanding the above expression for  for large z; r, show that in the neighborhood of infinity Z L z 1 A./d ; z 2 C r 2 ! 1: 
4
.z 2 C r 2 /3=2 0 (5.5) Verify (5.40). 0

(5.6) Verify (5.54). (Suggestion: Let z D e i ,  D e i , and convert the integral to one on a contour around the unit circle in the z-plane. You will want to indent the contour at poles on the boundary. Evaluate using residue theory.) (5.7) The Trefftz plane is a virtual plane orthogonal to the x-axis in Figure 5.13 and placed at large x downstream of the wing. The vortical wake in Prandtl’s model may be regarded as intersecting the Trefftz plane on the segment I W jyj  b, z D 0. The induced drag may be calculated in the Trefftz plane as follows. Adopt the energy balance that UDind , the rate of working done by the induced drag, is equal to UE, the flux of wake energy through the Trefftz plane (TP). Here Z  .r/2 dy dz; (5.58) ED 2 TP

where r D .0; v; w/ D .0; y ; z /. That is, in the Trefftz plane the velocity perturbations of the free stream are dominated by the induced velocities of the now doubly infinite line vortices. Use the fact that z D w is continuous on Y but  is discontinuous there to show that Z  Cb wTP ΠTP dy; (5.59) Dind D
2
b where wTP is twice the downwash computed at the lifting line in Prandtl’s model, and Œ TP D .y; 0C/
.y; 0
/ on I . From this result show that (5.57) follows from the definition of circulation.

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http://dx.doi.org/10.1090/cln/019/06

CHAPTER 6

Viscosity and the Navier-Stokes Equations 6.1. The Newtonian Stress Tensor Generally real fluids are not inviscid or ideal. Modifications of Euler’s equations, needed to account for real fluid effects at the continuum level, introduce additional forces into the momentum balance equations. There exists a great variety of real fluids that can be treated at the continuum level. These fluids may differ widely in the forces experienced by a fluid parcel as it is moved about and deformed according to the mathematical description developed above. Because of the molecular structure of various fluid materials, the nature of these forces can vary considerably, and there are many models that attempt to capture the observed properties of fluids under deformation. These models differ in what we shall call their rheology. The simplest of these rheologies, and one applicable to common fluids such as air or water, is the Newtonian viscous fluid. To understand the assumptions let us restrict attention to the determination of a viscous stress tensor at .x; t/, which depends only upon the fluid properties within a fluid parcel at that point and time. One could of course imagine fluids where some local average over space determines stress at a point. Also, it is easy enough to find fluids with a memory, where the stress at a particular time depends upon the stress history at the point in question. It is reasonable to assume that the forces associated with the rheology of the fluid are developed by the deformation of fluid parcels, and hence could be determined by the velocity field. If we allow only point properties, deformation of parcels must involve more than just the velocity itself; first- and higher-order partial derivatives with respect to the spatial coordinates could be important. (The time derivative of velocity has already been taken into consideration in the acceleration terms.) A moment’s thought shows that viscous forces cannot depend on the velocity. The bulk translation of the fluid with constant velocity produces no force. Thus it is the deformation of a small fluid parcel that must be responsible for the viscous force, and the dominant measure of this deformation should come from the first derivatives of the velocity field, i.e., from the components of the velocity derivative matrix @ui =@xj . The Newtonian viscous fluid is one where the stress tensor is linear in the components of the velocity derivative matrix. The specific form of this tensor will depend on other physical conditions. To see why a linear relation of this kind might capture the dominant rheology of many fluids, consider a flow .u; v/ D .u.y/; 0/. Each different plane or lamina 97 Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

98

6. VISCOSITY AND THE NAVIER-STOKES EQUATIONS

u(y) y=y A y=y B

F IGURE 6.1. Momentum exchange by molecules between lamina in a shear flow.

of fluid, y D const, moves with a particular velocity. Now consider the two lamina y D yA ; yB as shown in Figure 6.1, moving at velocities uB < uA . If a molecule moves from B to A, then it is moving from an environment with velocity uB to an environment with a larger velocity uA . Consequently, it must be accelerated to match the new velocity. According to Newton, a force is therefore applied to the lamina y D yA in the direction of negative x. Similarly, a molecule moving from yA to yB must slow down, exerting a force on lamina y D yB in the direction of positive x. Thus these exchanges of molecules would tend to reduce the velocity difference between the two lamina. This tendency to reduce the difference in velocities can be thought of as a force applied to each lamina. Thus if we insert a virtual surface at some position y, a .y/ > 0. force should be exerted on the surface in the positive x-direction if du dy Generally we expect the gradients of the velocity components to vary on a length scale L comparable to some macroscopic scale—the size of a container or the size of a body around which the fluid flows, for example. On the other hand, the scale of the molecular events envisaged above is very small compared to the macroscopic scale. Thus it is reasonable to assume that the force on the lamina is dominated by the first derivative, (6.1)

F .y/ D 

du : dy

The constant of proportionality, , is called the viscosity, and a fluid obeying this law is our Newtonian viscous fluid. We have considered so far only a simple planar flow .u.y/; 0/. In general, all of the components of the velocity derivative matrix need to be brought into the construction of the viscous stress tensor. Let us write (6.2)

ij D
pıij C dij :




Perhaps a more direct analogy would be two boats gliding along on the water on parallel paths, one moving faster than the other. If, at the instant they are side by side, an occupant of the fast boat jumps into the slow boat, the slower boat will speed up, and similarly an occupant of the slow boat can slow up the fast boat by jumping into it.

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6.1. THE NEWTONIAN STRESS TENSOR

99

y

σ n 1j j σ σ

2j n j

x

2j n j σ n 1j j

z

F IGURE 6.2. Showing why 12 D 21 . The forces are per unit area. The area of each face is 2 .

That is, we have simply split off the pressure contribution and exhibited the deviatoric stress tensor dij , which defines the viscous stress and determines the fluid’s rheology. We first show that dij , and hence ij , must be a symmetric tensor. We can do that by considering Figure 6.2. We show a square parcel of fluid of side , and those forces on each face that exert a torque about the z-axis. We see that the torque is 3 .21
12 /, since each face has area 2 , and each of the four forces considered has a moment of

2 about the z-axis. Now this torque must be balanced by the angular acceleration of the parcel about the z-axis. The moment of inertia of the parcel is a multiple of 4 . As  ! 0 we see that the angular acceleration must tend to infinity as 
1 . It follows that the only way to have stability of a parcel is for 21 to be equal to 12 . The same argument applies to moments about the other axes. A final requirement we shall place on dij , and a further condition on the fluids we shall study, is that there should be no preferred spatial direction for stress acting at a parcel, the condition of isotropy. The condition of isotropy of a symmetric matrix of second order requires that dij be a linear combination of derivatives in the two combinations (6.3)

@uj @ui C ; @xj @xi

@uk ıij : @xk

k D r  u.) For a Newtonian fluid the linearity (By the summation convention @u @xk implies that the most general allowable deviatoric stress has the form   @ui @uj 2 @uk @uk C
ıij C 0 ıij ; (6.4) dij D  @xj @xi 3 @xk @xk

for certain scalars ; 0 . Notice that we have divided the two terms so that the first term, proportional to , has zero trace. Thus if 0 D 0, the deviatoric stress contributes nothing to the normal force on an area element; this is given solely by the pressure force. The possibility of a normal force distinct from the pressure force is allowed by the second term of (6.4). We have attached the term viscosity to , so 0 is usually called the second viscosity. Often it is taken as 0, an approximation that is generally valid for liquids. The condition 0 D 0 is equivalent to what is

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100

6. VISCOSITY AND THE NAVIER-STOKES EQUATIONS

sometimes called the Stokes relation. In gases in particular 0 may be positive, in which case the thermodynamic pressure and the normal stresses are distinct. It should be noted that if we had simply taken dij to be proportional to the velocity derivative matrix, then the splitting (3.5) would show that only eij could possibly appear, since otherwise uniform rotation of the fluid would produce a force orthogonal to the rotation axis, which is never observed. The second term in (6.4) then follows as the only isotropic symmetric tensor linear in the velocity derivative that could be included as a contribution to “pressure.” We shall be dealing with two special cases of (6.4). The first is an incompressible fluid, in which case   @ui @uj C (incompressible fluid): (6.5) ij D
pıij C  @xj @xi Note that with the incompressibility @p @ij D
C r 2 ui : @xj @xi

(6.6)

The second case is compressible flow in one space dimension. Then u D .u.x; t/; 0; 0/, and the only nonzero component of the stress tensor is 4 @u ; 00 D  C 0 (one-dimensional gas flow): @x 3 The momentum balance equation in the form

(6.7)

(6.8)

11 D
p C 00



@ij Dui D ; Dt @xj

together with the stress tensor given by (6.4), defines the momentum equation for the Navier-Stokes equations. These are the most commonly used equations for the modeling of the rheology of fluids. They have been found to apply to a wide variety of practical problems, but it is important to realize their limitations. First, for highly rarefied gases, the mean free path of molecules of the gas can become so large that the concept of a fluid parcel, small with respect to the macroscopic scale but large with respect to mean free path, becomes untenable. Also, many common fluids, honey being an everyday example, are non-Newtonian and can exhibit effects not captured by the Navier-Stokes equations. Finally, whenever a flow involves very small domains of transition, the Navier-Stokes model may break down. Examples of this occur in shock waves in gases, where changes take place over distances of only several mean free paths, and in the interface between fluids, which can involve transitions over distances comparable to intermolecular scales. In these problems a multiscale analysis is usually needed, which can bridge the macroscopic-molecular divide. Finally, we point out that the viscosities in this model will generally depend upon temperature, but for simplicity we shall neglect this variation, and in particular for the incompressible case we always take  to be constant. We shall often exhibit the kinematic viscosity  D  in place of . Note that  has dimensions length2 =time, as can be verified from the momentum equations after division by .

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6.2. SOME EXAMPLES OF INCOMPRESSIBLE VISCOUS FLOW

101

6.2. Some Examples of Incompressible Viscous Flow We now take the density as well as the viscosity to be constant and consider several exact solutions of the incompressible Navier-Stokes equations. We shall be dealing with fixed or moving rigid boundaries, and we need the following assumption regarding the boundary condition on the velocity in the Navier-Stokes model: A SSUMPTION (No-slip condition). At a rigid boundary the relative motion of fluid and boundary will vanish. Thus at a nonmoving rigid wall the velocity of the fluid will be 0, while at any point on a moving boundary the fluid velocity must equal the velocity of that point of the boundary. This condition is valid for gases and fluids in situations where the stress tensor is well approximated by (6.4). It can fail in small domains and in rarefied gases, where some slip may occur. 6.2.1. Couette Flow. Imagine two rigid planes y D 0; H where the no-slip condition will be applied. The plane y D H moves in the x-direction with constant velocity U , while the plane y D 0 is stationary. A Newtonian fluid is contained within the two planes. The flow is steady, so the velocity field must be a function of y alone. Assuming constant density, u D .u.y/; 0/, and px D 0, we obtain a momentum balance if (6.9)


uyy D 0:

Thus, given that u.0/ D 0, u.H / D U , we have u D Uy=H: We see that the viscous stress is constant and equal to U H . This is the force per unit area felt by the plane y D 0. No pressure gradient is needed to sustain this stress field, since an equal and opposite stress acts on the plane y D H . Couette flow is the simplest exact solution of the Navier-Stokes equations with nonzero viscous stress. 6.2.2. The Rayleigh Problem. A related unsteady problem results from the time-dependent motion of the plane y D 0 in the x-direction with velocity U.t/. A no-slip condition is applied on this plane. A fluid of constant density occupies the semi-infinite domain y > 0. In this case an exact solution of the Navier-Stokes equations is provided by u D .u.y; t/; 0/, p D 0, with (6.10)

ut
uyy D 0;

u.0/ D U.t/:

For U.t/ D U0 cos !t we see that u.y; t/ D 0 of the Navier-Stokes equations with constant density, satisfying the no-slip condition on y D 0. The vorticity will satisfy  2  @! @ ! @2 ! @! Cv
 C 2 D 0: (6.38) u @x @y @x 2 @y If we set (6.39)

D UL
1 xF .y/, then ! D
UL
1 yF 00 . Insertion in (6.38) gives F 0 F 00
FF 000
Re
1 F 0000 D 0;

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106

6. VISCOSITY AND THE NAVIER-STOKES EQUATIONS 1.4 1.2 1 0.8 0.6 0.4 0.2 0 0

1

2

3

4

F IGURE 6.5. f 0 versus  for the viscous stagnation point flow. 0 where Re D UL

. The boundary conditions are that F .0/ D F .0/ D 0 to make ; u; v vanish on the wall y D 0, and F y as y ! 1, so that we obtain the irrotational stagnation point flow at y D 1. One integration of (6.39) can be carried out to obtain

(6.40)

2

F 0
FF 00
Re
1 F 000 D 1:

With F D Re
1=2 f ./,  D Re1=2 y, (6.40) becomes 2

f 0
ff 00
f 000 D 1;

(6.41)

with conditions f 0 .1/ D 1, f .0/ D f 0 .0/ D 0. We show in Figure 6.5 the This represents a gradual transition through solution f 0 ./ of this ODE problem. p a layer of thickness of order =UL between the null velocity on the boundary and the velocity Ux L at the wall in the irrotational stagnation point flow. We shall be returning to a discussion of such transition layers in Chapter 8. 6.3. Dynamical Similarity In the stagnation point example just considered, the dimensional combination Re D UL

has occurred as a parameter. This parameter, called the Reynolds number in honor of Osborne Reynolds, arose because we chose to exhibit the problem in a dimensionless notation. Consider now the Navier-Stokes equations with constant density in their dimensional form: (6.42)

1 @u C u  ru C rp
r 2 u D 0; @t 

r  u D 0:

We may define dimensionless (starred) variables as follows: (6.43)

u D

u ; U

x D

x ; L

p D

p : U 2

Here U; L are assumed to be a velocity and length characteristic of the problem being studied. In the case of flow past a body, L might be a body diameter and

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6.3. DYNAMICAL SIMILARITY

107

U the flow speed at infinity. In these starred variables it is easily checked that the equations become 1 2  @u C u  r  u C r  p 
r u D 0; r   u D 0: (6.44) @t Re Thus Re survives as the only dimensionless parameter in the equations. For a given value of Re a problem will have a solution or solutions that are fully determined by that value of Re.
Thus we are able to make a correspondence between various problems having different U , L, or , but the same value of Re. We call this correspondence dynamical self-similarity. Two flows that are self-similar in this respect become identical when expressed in the starred, dimensionless variables (6.43). In a sense the statement “the viscosity  is small” conveys no dynamical information. The only meaningful way to state that a fluid is “almost inviscid” is through the Reynolds number, Re 1. If we want to consider fluids whose viscous forces are dominant compared to inertial forces, we should require Re  1. These remarks underline the oft-repeated definition of Re as “the ratio of inertial to viscous forces.” This is because u  r  u u  ru D Re Re (6.45) r 2 u r  2 u since we regard all starred variables as of order unity. Note that we have here included the time derivative as a term of the same order as u  ru. This may not be the case in unsteady problems involving small-amplitude movements at high frequency. E XAMPLE 6.1. The drag D per unit length of a circular cylinder of radius L in a two-dimensional uniform flow of speed U must satisfy D D U 2 LF .Re/ for some function F . Note that we are assuming here that cylinders are fully determined by their radius. In experiments other factors, such as surface material or roughness, slight ellipticity, etc., must be considered. Also, nonuniqueness could lead to functions F with more than one branch. E XAMPLE 6.2. The Navier-Stokes equations are believed to be an adequate model for the study of fluid turbulence. Suppose that energy  per unit volume is created in an infinite expanse of fluid by “homogeneous stirring.” We assume that U; L are scales associated with the mechanism of stirring. We know from observation that sufficiently vigorous stirring produces eddying motions on a spatial scale far smaller than L. It therefore can be expected that on these small scales the fluid will “forget” the stirring mechanism, so that the only parameters that should govern the flow are ; . Now the units of  are those of a time derivative of energy, of length2 =time3 , so that  U 3 =L. Thus  3  14  (6.46)  


It is not always the case that well-formulated boundary value problems for the Navier-Stokes equations have unique solutions. An example of this occurs in the viscous flow out of a diverging two-dimensional channel formed by two planes set at an angle; see Landau and Lifshitz (2003), p. 79.

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108

6. VISCOSITY AND THE NAVIER-STOKES EQUATIONS

TABLE 6.1. Typical values of the Reynolds number

U (cm/sec) L (cm) Re D UL=

Context Swimming bacterium

10
2

10
5

10
5

Swimming spermatozoan

10
2

10
3

10
3

1-mm-diam. sphere translating in air

1

:1

1

Swimming minnow

5

2

103

Automobile at highway speeds

300

300

106

Swimmer Phelps

200

200

4  106

Commercial jet

3000

3000

108

is a length, and u D ./1=3 a velocity. These measure the size of, and velocity change across, a small eddy of dimension . Note that u = D 1, so this is the eddy size associated with a local Reynold number of unity. That is, these eddies will move under the equal influence of inertial and viscous forces. We now suppose that there is a range of eddies much larger than  but much smaller than L, where the local Reynolds number should be so high that the flow is determined by  alone. If l is the eddy size, and u the velocity change across the eddy, then the only way to connect ; l; u dimensionally is l D u3 of u D .l/1=3 . Since L D U 3 , this shows that the energy imparted at the scale L flows through each smaller scale until scale , where viscous stresses act to dissipate the energy into heat. Note that .UL=/ D L4 = 3 D .L= l/4 , or (6.47)

3 l D Re
4 ; L

Re D

UL : 

Thus we have estimated the size of the smallest eddies of Navier-Stokes turbulence, relative to the scale of mixing, in terms of the Reynolds number of mixing. At very large Reynolds number the estimate (6.47) determines the spatial resolution that a computer code would have to achieve in order to compute fluid turbulence reliably. We give in Table 6.1 some typical values of the Reynolds number. We here compute the value in air using  D 0:15 cm2 /sec, and in water using  D 0:01 cm2 /sec, these being the values for a temperature of 20ı C. The value give a rough idea of the wide range of Re encountered in nature, and the large values occurring on human length scales. Note that Reynolds numbers depend very much on how you estimate length and velocity for a given fluid flow. Also, when a movement involves a frequency !, a frequency Reynolds number !L2 = may be the preferred parameter.

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PROBLEM SET 6

109

Q

H

θ

θ 2L

F IGURE 6.6. Bifurcating Poiseuille flow. Assume a parabolic profile in each section.

Problem Set 6 (6.1) Consider the following optimization problem: A Newtonian viscous fluid of constant density flows through a cylindrical tube of radius R1 . The tube then bifurcates into two straight tubes of radius R2 ; see Figure 6.6. A volume flow Q is introduced into the upper tube, which divides into flows of equal flux Q 2 at the bifurcation. Because of the material composition of the tubes, it is desirable that , evaluated at the wall, be the same in both tubes. If L and H the wall stress  du dr are given and fixed, what is the angle  that minimizes the rate of working required to sustain the flow Q? Be sure to verify that you have a true minimum. Neglect all entry effects by assuming Poiseuille flow over the entire length of each section. p (6.2) Look for aR solution of (6.30) of the form ! D t
1 F .r= t/, satisfying 1 !.1; t/ D 0, 2
0 r!.r; t/dr D 1, t > 0. Show, by computing u with u .1; t/ D 0, that this represents the decay of a point vortex of unit strength in a viscous fluid, i.e., 1 ; r > 0: (6.48) lim u .r; t/ D t!0C 2
r (6.3) A Navier-Stokes fluid has constant values of ;  and no body forces. Consider a motion in a fixed bounded domain V with a no-slip condition on its rigid boundary. Show that Z Z juj2 dE D
ˆ; E D dV; ˆ D  .r  u/2 dV: dt 2 V

V

We see that for such a fluid kinetic energy is converted into heat at a rate ˆ.t/. This last function of time gives the net viscous dissipation for the fluid contained in V . (Hint: r  .r  u/ D r.r  u/
r 2 u. Also r  .A  B/ D r  A  B
r  B  A.) (6.4) In two dimensions, with stream function , where .u; v/ D . y ;
x /, show that the incompressible Navier-Stokes equations without body forces for a

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110

6. VISCOSITY AND THE NAVIER-STOKES EQUATIONS

fluid of constant ;  imply .@. ; r 2 / @ 2 r
r 4 D 0:
@t @.x; y/ In terms of , what are the boundary conditions on a rigid boundary if the no-slip condition is satisfied there? (6.5) Find the time-periodic two-dimensional flow in a channel
H < y < H , dp D filled with viscous incompressible fluid, given that the pressure gradient is dx ACB cos.!t/, where A; B; ! are constants. This is an oscillating two-dimensional Poiseuille flow. You may assume that u.y; t/ is even in y and periodic in t with period 2 ! . Discuss the limiting case ! ! 0. (6.6) Verify (6.33). (6.7) The plane z D 0 is rotating about the z-axis with an angular velocity . A Newtonian viscous fluid of constant density and viscosity occupies z > 0 and the fluid satisfies the no-slip condition on the plane; i.e., at z D 0 the fluid rotates with the plane. By centrifugal effect we expect the fluid near the plane to be thrown out radially and a compensating flow of fluid downward toward the plane. Using cylindrical polar coordinates, look for a steady solution of the NavierStokes equations of the form (6.49)

.uz ; ur ; u / D .f .z/; rg.z/; rh.z//:

We assume that the velocity component u vanishes as z ! 1. Show that then df 1 p D
f 2 C F; (6.50)  dz 2 where F is a function of r alone. Now argue that, if h.1/ D 0, i.e., no rotation at infinity, then F must in fact be a constant. From the r- and -components of the momentum equation together with r  u D 0, find equations for f; g; h and justify the following conditions: df D 0; h D ; z D 0; f 0 ; h ! 0; z ! 1: (6.51) f D dz

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http://dx.doi.org/10.1090/cln/019/07

CHAPTER 7

Stokes Flow We have seen in Section 6.3 that the dimensionless form of the Navier-Stokes equations for a Newtonian viscous fluid of constant density and constant viscosity is, now dropping the stars, (7.1)

1 2 @u C u  ru C rp
r u D 0; @t Re

r  u D 0:

The Reynolds number Re is the only dimensionless parameter in the equations of motion. In the present chapter we shall investigate the fluid dynamics resulting from the a priori assumption that the Reynolds number is very small compared to unity, Re  1. Since Re D UL

, the smallness of Re can be achieved by considering extremely small length scales, or by dealing with a very viscous liquid, or by treating flows of very small velocity, the so-called creeping flows. The choice Re  1 is a very interesting and important assumption, for it is relevant to many practical problems, especially in a world where many products of technology, including those manipulating fluids, are shrinking in size. A particularly interesting application is to the swimming of micro-organisms. In all of these areas we shall, with the assumption Re  1, unveil a special dynamical regime that is usually referred to as Stokes flow, in honor of George Stokes, who initiated investigations into this class of fluid problems. We shall also refer to this general area of fluid dynamics as the Stokesian realm, in contrast to the theories of inviscid flow, which might be termed the Eulerian realm. What are the principle characteristics of the Stokesian realm? Since Re is indicative of the ratio of inertial to viscous forces, the assumption of small Re will mean that viscous forces dominate the dynamics. That suggests that we may be able to drop entirely the term Du Dt from the Navier-Stokes equations, rendering the system linear. This will indeed be the case, with some caveats discussed below. The linearity of the problem will be a major simplification. Taking (7.1) in the form   @u C u  ru C rp D r 2 u; r  u D 0; (7.2) Re @t it is tempting to say that the smallness of Re means that we can neglect the left-hand side of the first equation, leading to the reduced (linear) system (7.3)

r 2 u D 0;

r  u D 0:

Indeed, solutions of (7.3) belong to the Stokesian realm and are legitimate. 111 Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

112

7. STOKES FLOW

E XAMPLE 7.1. Consider the velocity field u D .A  R/=R3 in three dimenA , and so sions with A a constant vector and R D .x; y; z/. Note that u D r  R 2
1 r  u D 0 and also r u D 0, R > 0, since R is a harmonic function there. This is, in fact, an interesting example of a Stokes flow. Consider a sphere of radius a rotating in a viscous fluid with angular velocity . On the surface of the sphere the velocity is   R, since the no-slip condition certainly holds when viscous forces are dominant. Comparing this with our example, we see that if A D a3 we satisfy this condition with a Stokes flow. Thus we have solved the Stokes flow problem of a sphere spinning in an infinite expanse of viscous fluid. It is not difficult to see, however, that (7.3) does not encompass all of the Stokes flows of interest. The reason is that, by letting Re ! 0, the pressure has formally been expelled from the equation, whereas there is no physical reason for this. If, in the process of writing the dimensionless equations, we had defined the dimensionless pressure as pL=.U / instead of p=.U 2 /, (7.2) would be changed to   @u C u  ru C rp D r 2 u; r  u D 0; (7.4) Re @t leading in the limit Re ! 0 to (7.5)

rp
r 2 u D 0;

r  u D 0:

Any solution of (7.5) will have the form u D r C v where r 2  D p and r 2 v D 0, r  v D
p. This larger class of flows, valid for Re small, comprises the Stokes flows. The special family of flows with zero pressure form a small subset of all Stokes flows. 7.0.1. Some Caveats. We noted above that the dropping of the inertial terms in Stokes flow might have to be questioned in some cases, and we consider these exceptions now. First, it may be that there is more than one possible Reynolds number that can be formed, involving several distinct lengths and/or frequencies of oscillation. It can then happen that the time derivative of u needs to be kept even though the nonlinear term u  ru may be dropped. An example is a wall adjacent to a viscous fluid, executing a standing wave with amplitude A, frequency !, and wavelength L. If !L2 = is of order unity, and we take U D !L, then the Reynolds number UL= is of order unity and no terms may be dropped. However the actual velocity is of order !A, and if A  L, then the nonlinear terms are negligible. Another unusual situation is associated with the nonuniformity of the Stokes equations in three dimensions near infinity, in steady flow past a finite body. Even through the Reynolds number is small, the fall off of the velocity as R
1 (associated with the fundamental solution of the Stokes equations in three dimensions, see below) means that near infinity the perturbation of the free stream speed U is of order R
1 . Thus the uru term is O.U 2 =R2 // while the viscous term is O.U=R3 /. The ratio is UR=, which means that when R =U the Stokes equations cannot govern the perturbational velocity. The momentum equation needed to replace the @u . We shall remark later on the importance of Stokes equation contains the term U @x

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7.1. SOLUTIONS OF THE STOKES EQUATIONS

113

this new set of equations, the Oseen equations, in connection with two-dimensional Stokes flow. 7.1. Solutions of the Stokes Equations Returning to their dimensional form, the Stokes equations are (7.6)

rp
r 2 u D 0;

r  u D 0:

r 2 u

D 0, using the solenoidal property of From the divergence of rp
2 u, we see that r p D 0, and hence that r 4 u D r 2 r 2 u D 0. The curl of this equation gives also r 2 r  u D 0. The components of u thus solve the biharmonic equation r 4  D 0 as well as the solenoidal condition, and the vorticity is a harmonic vector field. We shall combine these constraints now and set up a procedure for constructing solutions from a scalar biharmonic equation. We first set  2  @  @r 2  2
r ıij aj ; p D  aj ; (7.7) ui D @xi @xj @xj where a is a constant vector. Note that for (7.7) we have @ui =@xi D Œ@r 2 =@xj
@r 2 =@xj aj D 0, so the solenoidal condition on velocity is satisfied. Inserting these expressions into (7.6) we see that the momentum equation is satisfied identically for any a provided that r 4  D 0:

(7.8)

A second class of solutions, having zero pressure, has the form (7.9)

"ij k

@ ak ; @j

r 2  D 0;

for a constant vector a, where "ij k D 1 for subscripts that are an even permutation of 123, and is
1 otherwise. The solutions (7.9) include Example 7.1, with A D a and  D R
1 . By considering linear combinations of solutions of the biharmonic or Laplace equations (finite or infinite), these results allow construction of a large family of Stokes flows. E XAMPLE 7.2. The fundamental solution of the Stokes equations in three dimensions corresponds to a point force Fı.x/ on the right of the momentum equation, F a constant vector: (7.10)

rp
r 2 u D Fı.x/;

r  u D 0:

Setting a D F in (7.7) we must have (7.11)

r 4  D ı.x/:

We know the fundamental solution of r 2  D 0, satisfying r 2  D ı.x/ and van1 in three dimensions. Thus ishing at infinity, is
4R (7.12)

r 2 D


 d 2 R 1 1 D ; 4
R R dR2

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114

7. STOKES FLOW

and so 1 R C A C BR
1 : 8
 The singular component is incompatible with (7.11), and the constant A may be 1 R. Then we find set equal to 0 without changing u, and so  D
8   xi xj 1 ıij 1 xj Fj Fj ; p D C : (7.14) ui D 8
 R3 R 4
R3 The particular Stokes flow (7.14) is often referred to as a Stokeslet. (7.13)

D


7.2. Uniqueness of Stokes Flows Consider Stokes flow within a volume V having boundary S . Let the boundary have velocity uS . By the no-slip condition, the fluid velocity u must equal uS on the boundary. Suppose now that there are two solutions u1;2 to the problem of solving (7.6) with this boundary condition on S . Then v D u1
u2 will vanish on S while solving (7.6). But then Z v  .rp
r 2 v/dV D 0 (7.15)    Z  Z V @ @vi @vi 2 dV; vj p
vi dV C  D @xj @xj @xj V

V

where the solenoidal property of v has been used. The first integral on the right vanishes under the divergence theorem because of the vanishing of v on S . The second is nonnegative (with summation over i; j understood), and vanishes only if v D 0. We remark that the nonnegative term is equal to the rate of dissipation of the kinetic energy of v as a result of viscous stresses. This dissipation, which will cause the fluid to be heated slightly, can vanish only if the velocity is identically 0. The solution of the Stokes equations is not easy in most geometries, and frequently the coordinate system appropriate to the problem will suggest the best formulation. We illustrate this process in the next section. 7.3. The Stokes Solution for Uniform Flow Past a Sphere We consider the classic solution of the Stokes equations representing the uniform motion of a sphere of radius a in an infinite expanse of fluid. We shall first consider this problem using the natural coordinates for the available symmetry, namely spherical polar coordinates. Then we shall re-derive the solution using (7.7). The velocity field in spherical coordinates has the form .uR ; u ; u / D .uR ; u ; 0/ and the solenoidal condition is 1 @ sin u 1 @R2 uR C D 0: R @R sin R @ Introducing the Stokes stream function ‰, @‰ 1 1 @‰ ; u D
: (7.17) uR D 2 R sin  @ R sin  @R (7.16)

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7.3. THE STOKES SOLUTION FOR UNIFORM FLOW PAST A SPHERE

115

The Stokes equations in spherical coordinates are   @ sin u 2uR 2 @p 2 (7.18) D  r uR
2
2 ; @R R R sin  @   u 2 @uR 1 @p 2 (7.19) ; D  r u C 2
R @ R @ R2 sin2  together with (7.16). The vorticity is .0; 0; ! /, where 1 L‰; (7.20) ! D
R sin  and   @2 1 @ sin  @ (7.21) LD : C @R2 R2 @ sin  @ From the form rp C r  r  u D 0 of the momentum equation, we have the alternative system  @ @p D
! sin ; (7.22) @R R sin  @  @ 1 @p D R! : (7.23) R @ R @R Eliminating the pressure and using (7.20) we obtain @ 1 @ 1 @ 1 @ L‰ C L‰ D 0: (7.24) 2 R @ sin  @ @R sin  @R We seek to solve (7.24) with the conditions 1 (7.25) uR D u D 0; R D a; ‰ .R2 sin2  /U; R ! 1: 2 We may separate variables in the form ‰ D sin2 f .R/;

(7.26) to obtain from (7.24) (7.27)



2 @2
@R2 R2

2 f D 0:

Trying f D R we get the indicial equation . 2
1/.
2/.
4/ D 0 and therefore the general solution of (7.27) is A (7.28) f D C BR C CR2 C DR4 : R From the behavior needed for large R, D D 0, C D U2 . The two conditions at R D a, namely f .a/ D f 0 .a/ D 0, then require that 3 1 (7.29) A D Ua3 ; B D
Ua: 4 4 Thus  3  a 1 2
3aR C 2R sin2 : (7.30) ‰D U 4 R

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116

7. STOKES FLOW

7.3.1. Drag. To find the drag on the sphere, we need the following stress components evaluated on R D a:    @uR @uR @ u ; R D R C : (7.31) RR D
p C 2 @R @R R R @ Given these functions the drag D is calculated from Z 2 Z  2 ŒRR cos 
R sin  sin  d d: (7.32) DDa 0

0

From (7.23) the pressure is determined by

   @ 2 1 @p 2 D
sin  fRR
2 f ; R @ R sin  @R R

(7.33) or, using (7.30),

cos  3 p D
Ua 2 C p1 : 2 R

(7.34) Also,

  @‰ U cos  a3 1 2 D
3aR C 2R ; uR D 2 R sin  @ 2R2 R U sin  1 @‰ D
.
a3 =R
3a C 4R/: u D
R sin  @R 4R

(7.35) (7.36) Thus (7.37) D D 2
a

2

Z

 0

   cos 
3a3 3a 3 Ua 2
p1 C 2 cos 
cos  2 R R4 R RDa    3a U sin2  3a2 C 2 sin  d: C 4 R4 R RDa





This reduces to D D 3
aU „ (7.38)

Z



0

2

Z



cos  sin  d C 3
aU sin3  d ; 0 ƒ‚ … „ ƒ‚ …

pressure

viscous

D 2
aU C 4
aU D 6
aU:

That is, one-third of the drag is due to pressure forces, two-thirds to viscous forces. 7.3.2. An Alternative Derivation. Now knowing the drag, we can re-derive Stokes’ solution for a sphere by realizing that at large distances from the sphere the flow field must consist of a uniform flow plus the fundamental solution for a force
6
Uai. To this must be added a term or terms that will account for the finite sphere size. Given the symmetry we try a dipole term proportional to r.x=R3 /. We thus postulate     i i xR 6
aU xR CC C
3 5 ; (7.39) u D Ui
8
 R3 R R3 R

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7.4. TWO DIMENSIONS: STOKES’ PARADOX

117

where C remains to be determined. By inspection we see that C D
14 a2 U makes u D 0 on R D a, so we are done. The pressure is as given previously, p D
32 Uax=R3 C p1 , and is entirely associated with the fundamental part of the solution. We remark that the R
1 decay of velocity exhibited by the Stokeslet is typical of the disturbance caused by a translating finite body in three-dimensional Stokes flow. Indeed, the Stokeslet associated with the drag of any such body gives the dominant far-field term for the velocity. 7.4. Two Dimensions: Stokes’ Paradox The fundamental solution of the Stokes equations in two dimensions may be studied by the procedure given in Example 7.2. If radial symmetry is again assumed, we may try to solve for the Stokes flow past a circular cylinder of radius a. If the pressure is eliminated from the Stokes equations in two dimensions, we get (7.40)

r 2 ! D r 4

D0

in terms of the two-dimensional stream function . We set separate variables in polar coordinates, leading to 
2 1 2 1 d d
C f D 0: (7.41) dr 2 r dr r 2

D sin f .r/ to

The no-slip condition at the surface of the cylinder requires that .a/ D @@r .a/ D 0, while the attaining of a free stream u D .U; 0/ at infinity requires that f Ur, r ! 1. By quadrature we can find the most general solution of (7.41) as (7.42)

f .r/ D Ar 3 C Br ln r C C r C Dr
1 :

The condition at infinity requires that A D B D 0. The no-slip conditions then yield (7.43)

C a C Da
1 D 0;

C
Da
2 D 0;

which implies C D D D 0. There is no satisfactory steady solution of the twodimensional Stokes equations representing flow of an unbounded fluid past a circular cylinder. This result, known as Stokes’ paradox, underlines the profound effect that dimension can play in the properties of a fluid flow. What is the reason for this nonexistence? We can get some idea of what is going on by introducing a finite circle r D R on which we make u D .U; 0/. Then there does exist a function f .r/ satisfying f .a/ D f 0 .a/ D 0, f .R/ D R, f 0 .R/ D 1. We shall obtain an asymptotic approximation to this solution for large R a by setting A D 0 in (7.42) and by satisfying the conditions at r D a with the remaining terms. Then we obtain  
 1 a2 1 rC : (7.44) f B r ln r
ln a C 2 2 r
If y D U; x D 0 on a circle r D R, then f =r C .f =r/0 .y 2 =r/ D U; .f =r/0 .xy=r/ D 0 when r D R, by differentiation of sin f . Thus f .R/ D UR and f 0 .R/ D U .

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118

7. STOKES FLOW

0 We make f .R/ UR, R a ! 1, by setting B D U= ln.R=a/. Then also f .R/ U C o.1/, R a ! 1, so all conditions are satisfied exactly or asymptotically for R large a . Thus
     r 1 1 a2 U r ln
C : (7.45) f ln.R=a/ a 2 2 r

At a fixed value of ar > 1 we see that f ! 0 as R a ! 1. It is only when ln.r=a/= ln.R=a/ becomes O.1/ that order UR values of f , and hence order U values of velocity, are realized. Thus a cylindrical body creeping through a viscous fluid will tend to carry with it a relatively large stagnant body of fluid. There is no solution of the boundary value problem for an infinite domain in Stokes flow, for then the stagnant region extends to infinity. This paradox results from a failure to properly account for the balance of forces in a viscous fluid at large distances from the translating cylinder, however small the Reynolds number of translation may be. The remedy for this paradox involves a problem of singular perturbation. The regions distant from the cylinder effectively “see” a disturbance from a point force. Let the velocity at some point a distance R a from the body be q. Then the inertial forces at this point will be, given the variation as log R, approximately Uq=R, (since we should linearize u  ru about the free stream velocity). Also, the viscous forces are of order q=R2 . These

1 two estimates are comparable when R a Ua D Re . Thus when Re < 1 and we try to apply the Stokes equations, there are always points, at a distance of order Re
1 times the cylinder size, where the neglect of the inertial terms fails to be valid. In the case of three dimensions, we have Stokes’ solution for a sphere, and 1 / the perturbation velocity caused by the sphere we know that at distances O. Re is small, of order Re. An analogous estimate, mentioned in Section 7.0.1, shows that again the Stokes approximation fails at a distance Re
1 . However, there the O.R
1 / decay of the Stokeslet in three dimensions insures that the free stream velocity is essentially unperturbed, and there is no Stokes paradox. In two dimen1 . sions, the perturbation caused by the cylinder persists out to distances of order Re Thus the Stokes equations fail to be uniformly valid in a domain large enough to allow necessary conditions at infinity to be satisfied. The remedy for this paradox in two dimensions involves a proper accounting for the singular nature of the limit Re ! 0 in the neighborhood of infinity. At distances r Re
1 the appropriate equations are found to be (7.46)

U

@u C rp
r 2 u D 0; @x

r  u D 0;

which are Oseen equations. Oseen proposed them as a way of approximately accounting for fluid inertia in problems where there is an ambient free stream U i. Their advantage is of course that they comprise a linear system of equations. The fact remains that they arise rigorously to appropriately treat viscous flow in the limit of small Reynolds numbers in a way that expels any paradox associated with large distances.

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7.5. TIME REVERSIBILITY IN STOKES FLOW

119

To summarize, in creeping flow the Stokes model works well in three dimensions; near the body the equations are exact, and far from the body the nonuniformity, leading to the replacement of the Stokes equations by the Oseen equations, is of no consequence to first approximation, and Stokes’ solution for a sphere is valid. In two dimensions the distant effect of a cylinder must be determined from Oseen’s model. It is only by looking at that solution, expanded near the position of the cylinder, that we can determine near the cylinder the appropriate solution of Stokes’ equations in two dimensions. 7.5. Time Reversibility in Stokes Flow Consider a viscous fluid contained in some finite region V bounded by a surface @V . If Stokes flow prevails, and if the boundary moves, each point of @V being assigned a boundary velocity ub , then we have a boundary value problem for the Stokes equations whose solution will provide the instantaneous velocity of every fluid particle in V . We assume the existence of this solution, which we have seen to be unique. In Stokes flow the instantaneous velocity of a fluid particle at P is determined uniquely by the instantaneous velocities of all points on the boundary of the fluid domain. Let us then assume a motion of the boundary through a sequence of configurations C.t/. Each C represents a point in configuration space, and the motion can be thought of as a path in configuration space with time as a parameter. Indeed, “time” has no dynamical significance. A path from A to B in configuration space can be taken quickly or slowly. In general, let the configuration at time t be given by C. .t//, where  .0/ D 0,  .1/ D 1, but is otherwise an arbitrary differentiable function of time. If the point P has velocity uP .t/, 0  t  1, when  .t / D t, then in general uP .t/ D P .t/uP . .t//. The vector displacement of the point P under this sequence of configurations is Z tD1 Z 1 P .t/uP . .t//dt D u. /d  (7.47) P D tD0

0

and so is independent of the choice of  . Another way to say this is that the displacement depends upon the ordering of the sequence of configurations but not on the timing of the sequence. The displacement does, however, depend in general on the path taken in configuration space in going from configuration C0 to C1 . We now give an example of this dependence. E XAMPLE 7.3. We must find two paths in configuration space having the same starting and finishing configurations (i.e., the boundary points coincide in each case), but for which the displacement of some fluid particle is not the same. Consider then a two-dimensional geometry with fluid contained in the circular annulus a < r < b. Let the inner cylinder of radius a rotate with time so that the angle made by some fixed point on the cylinder is .t/ relative to a reference axis. The D 0 and that the velocity is u .r/, the outer circle r D b is fixed. Given that @p @ function u .r/ satisfies (from the Stokes form of (6.15)) Lu D 0. Integrating and

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120

7. STOKES FLOW

applying boundary conditions, the fluid velocity in the annulus is (7.48)

u D

aP ab 2 P
1 r
r : a2
b 2 a2
b 2

Consider now two paths that leave the position of the point of the inner circle unchanged. In the first,  rotates from 0 to 4 in one direction, then from 4 back to 0 in the other direction. Clearly every fluid particle will return to its original position after these two moves. For the second path, rotate the cylinder through 2
. Again every point of the boundary returns to its starting point, but now every point of fluid in a < r < b moves through an angle  that is positive and less than 2
. Thus only the points on the two circles r D a; b are in their starting positions at the end of the rotation. Note that in this example the first path, returning all fluid particles to their starting positions, is special in that the sequence of configurations in the second movement is simply a reversal of the sequence of configurations in the first movement (a rotation through angle 4 ). The zero particle displacement is a necessary consequence of this kind of path—a sequence followed by the reverse sequence. Recall that the timing of each of these sequences may be different. The second path, a full rotation of the inner circle, involves no such reversal. In fact, if the direction of rotation is reversed, the fluid points move in the opposite direction. If we now let these two paths be repeated periodically, say every one unit of time t, then in the first case fluid particles move back and forth periodically with no net displacement, while in the second case particles move on circles with a fixed displacement for each unit of time. Notice now an important difference in the time symmetry of these two cases. If time is run backwards in the first case, we again see fluid particles moving back and forth with no net displacement. In the second case, reversal of time leads to steady rotation of particles in the opposite direction. We may say that the flow in the first case exhibits time-reversal symmetry, while in the second case it does not exhibit this symmetry. In general, a periodic boundary motion exhibiting time-reversal symmetry cannot lead to net motion of any fluid particle over one period, as determined by the resulting time-periodic Stokes flow. On the other hand, if net motion over one cycle is observed, the boundary motion cannot be symmetric under time reversal. However, the converse of the last statement does not hold. A motion that is not symmetric under time reversal may in fact not produce any displacement of fluid particles over one cycle. E XAMPLE 7.4. In the previous example, let both circles rotate through 2
with P b D ba Pa . The boundary motion does not then exhibit time-reversal symmetry, and in fact the fluid can be seen to be in a solid body rotation. Thus every fluid particle returns to its starting position. T HEOREM 7.1 In Stokes flow, time-reversal symmetry of periodic boundary motion is sufficient to insure that all fluid particles return periodically to their starting positions. If particles do not return periodically to their starting position, the boundary motion cannot be time symmetric.

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PROBLEM SET 7

121

7.6. Stokesian Locomotion and the Scallop Theorem One of the most important and interesting applications of Stokes flow hydrodynamics is to the swimming of micro-organisms. Most micro-organisms move by a periodic or nearly periodic motion of organelles such as cilia and flagella. The aim of this waving of organelles is usually to move the organisms from point A to point B, a process complementary to a variable boundary that moves the fluid about but does not itself locomote. Indeed, time-reversal symmetry plays a key role in the selection of swimming strategies. We now state without proof the following: T HEOREM 7.2 (Scallop theorem) Suppose that a small swimming body in an infinite expanse of fluid is observed to execute a periodic cycle of configurations, in a coordinate system moving with constant velocity U relative to the fluid at infinity. Suppose that the fluid dynamics is that of Stokes flow. If the sequence of configurations is indistinguishable from the time-reversed sequence, then U D 0 and the body does not locomote. The idea behind the proof is that actual time reversal of the swimming motions would lead to locomotion with velocity
U. But if the two motions are indistinguishable, then U D
U and so U D 0. The name of the theorem derives from the nonlocomotion of a scallop in Stokes flow, given that the scallop simply opens and closes its shell periodically. In Stokes flow this would lead to a back and forth motion along a line (assuming suitable symmetry of shape of the shell), with no net locomotion. In nature the breaking of time-reversal symmetry takes many forms. Flagella tend to propagate waves from head to tail. The wave direction gives the arrow of time, and it reverses, along with the swimming velocity, under time reversal. Cilia also execute complicated forward and return strokes that are not time symmetric. Problem Set 7 (7.1) Consider the uniform slow motion with speed U of a viscous fluid past a spherical bubble of radius a, filled with air. Do this by modifying the Stokes flow analysis for a rigid sphere as follows. The no-slip condition is to be replaced on r D a by the condition that both ur and the tangential stress r vanish. (This latter condition applies since the air within the bubble cannot support a sensible stress.) Show that U ‰ D .r 2
ar/ sin2  2 and that the drag on the bubble is D D 4
Ua. (7.2) Prove that Stokes flow past a given rigid body is unique as follows: Show if p1 ; u1 and p2 ; u2 are two solutions of rp
r 2 u D 0;

r  u D 0;

satisfying ui D
Ui on the body and     @ui 1 1 ; ; p O 2 u O r @xj r

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122

7. STOKES FLOW

as r ! 1, then the two solutions must agree. (Hint: Consider the integral of @=@xi .wj @wj =@xi / over the region exterior to the body, where w D u1
u2 .) (7.3) Two small spheres of radius a and density s are falling in a viscous fluid with centers at P and Q. The line PQ has length L a and is perpendicular to gravity. Using the Stokeslet approximation to the Stokes solution past a sphere, and assuming that each sphere sees the unperturbed Stokes flow of the other sphere, show that the spheres fall with the same speed  2   a ka CO ; U  Us 1 C L L2 and determine the number k. Here Us D 2a2 g=9.s =
1/ is the settling speed of a single sphere in Stokes flow. (7.4) Let z D x C iy, z  D x
iy, and define @ @ @ @ @ @ D
i ; Ci : D  @z @x @y @z @x @y Then, if w D .x; y/ C i .x; y/, show that

implies that  and

@2 w D0 @z  2 are biharmonic functions of x; y. Verify that consequently w D z  f .z/ C g.z/;

where f and g are analytic functions of z. Assuming f and g are constant multiples of e iz , derive the corresponding family of real biharmonic functions. (These functions may be used to construct Stokes flows periodic in x and bounded in the half-space y  0.) (7.5) Prove the following reciprocal theorem for Stokes flow: Let u1 ; 1 and u2 ; 2 be the velocity and stress tensors of two Stokes flows external to a rigid three-dimensional body with boundary S . Assume that u1;2 ! 0; Then we have

r ! 1;

u1;2 D v1;2 on S: Z

Z v1  2  n dS D S

v2  1  n dS: S

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http://dx.doi.org/10.1090/cln/019/08

CHAPTER 8

The Boundary Layer The physical concept of the boundary layer is a classic example of a result of applied science greatly influencing the development of mathematical methods of wide applicability. The theory most often associated with the boundary layer concept was introduced in a 10-minute address in 1904 by Ludwig Prandtl, then a 29-year-old professor in Hanover, Germany. Prandtl had done experiments in the flow of water over bodies and sought to understand the effect of the viscosity on the flow. Realizing that the no-slip condition had to apply at the surface of the body, his observations led him to the conclusion that the flow was brought to rest in a thin layer adjacent to the rigid surface. His reasoning suggested that the NavierStokes equations should have a somewhat simpler form owing to the thinness of this layer. This led to the equations of the viscous boundary layer. Boundary layer methods now occupy a fundamental place in many asymptotic problems for partial differential equations. 8.1. The Limit of Large Re Let us consider the steady viscous two-dimensional flow over a flat plate that is aligned with a uniform stream .U; 0/. In dimensionless variables the steady Navier-Stokes equations in two dimensions may be written 1 2 r u D 0; Re 1 2 r v D 0; u  rv C py
Re

u  ru C px


(8.1) (8.2)

ux C vy D 0:

(8.3)

We are dealing with the geometry of Figure 8.1, where we show only the boundary layer in y > 0. The layer is seen to grow in thickness as x moves from 0 to L. This suggests that the term u  ru in (8.1) has been properly estimated as of order U 2 =L

y

u=U

0

L

x

F IGURE 8.1. Boundary layer on a finite flat plate. 123 Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

124

8. THE BOUNDARY LAYER

y

U(x) x

F IGURE 8.2. Boundary layer over a general body with varying U.x/.

in the dimensionless formulation, and so should be taken as O.1/ at large Re in (8.1). If this term is to balance the viscous stress term, then the natural choice, since the boundary layer on the plate is observed to be so thin, is to assume that 1 uyy . Thus it makes the y-derivatives of u are so large that the balance is with Re p sense to define a “stretched” variable yN D Re y. If we now apply the stretched variable to (8.3), still taking ux as of order unity, then in order to keep this essential equation intact we must compensate the stretched variable yN by a stretched form of the y-velocity component: p (8.4) vN D Re v: Prandtl would have been comfortable with this last definition. The boundary layer on the plate was so thin that there could have been only a small velocity component normal to its surface. Thus the continuity equation will survive our limit Re ! 1: (8.5)

ux C vN yN D 0:

Returning now to consideration of (8.1), we also retain the pressure term px as O.1/. Then the simplified equation, obtained in the limit Re ! 1 in the stretched 1 uxx : variables, amounts to dropping the term Re (8.6)

N yN C px
uyN yN D 0: uux C vu

Finally, using these stretched variables in (8.2), we have 1 1 (8.7) pyN D
.uvN x C vN vN yN
vN yN yN / C 2 vN xx : Re Re Thus in the limit Re ! 1 the vertical momentum equation reduces to (8.8)

pyN D 0:

We thus see from (8.8) that the pressure does not change as we move vertically through the thin boundary layer. That is, the pressure throughout the boundary layer at a station x must be the pressure outside the layer. At this point a crucial contact is made with inviscid fluid theory. The “pressure outside the boundary layer” should be determined by the inviscid theory for the Euler flow past the given body, since the boundary layer is thin and will presumably not disturb the inviscid flow very much. In particular, for a flat plate the Euler flow is the uniform stream— the plate has no effect—and so the pressure has its constant free-stream value. Prandtl’s striking insight is clearer when we consider flow going past a general smooth body, as in Figure 8.2. Since the boundary layer is again taken as thin in the neighborhood of the body, curvilinear coordinates may be introduced, with x the arc length along curves paralleling the body surface and y the coordinate

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8.2. BLASIUS SOLUTION FOR A SEMI-INFINITE FLAT PLATE

125

normal to these curves. In the stretched variables, and in the limit for large Re, it turns out that we again get (8.6)–(8.8), only now (8.8) must be interpreted to mean that the pressure is what would be computed from the inviscid flow past the body (cf. Problem 8.3). If p0 is the dimensionless free-stream value of p, then the dimensionless Bernoulli’s theorem for steady flow along the body surface yields 1 1 2
U .x/ D p.x/: 2 2 By (8.8) it is this p.x/ that now applies in the boundary layer. Thus the inviscid flow past the body determines the pressure variation, which is then imposed on the boundary layer through the function px . The system of equations (8.6)–(8.8) comprises what are usually called the Prandtl boundary layer equations. We are giving here the essence of Prandtl’s idea without any indication of possible problems in implementing it for an arbitrary body. The main problem that will arise is that of boundary layer separation. It turns out that the function p.x/, which is determined by the inviscid flow past the body, may lead to a boundary layer that cannot be continued indefinitely along the surface of the body using Prandtl’s equations. What can and does occur is a breaking away of the boundary layer from the surface, the ejection of vorticity into the free stream, and the creation of free separation streamline similar to the free streamline of the Kirchhoff flow we considered in Chapter 6. Separation is part of the stalling of an airfoil at high angles of attack, for example. The possibility of separation raises the issue of just what the “inviscid flow past the body” really is. Consider, for example, steady two-dimensional flow past a circular cylinder at large Re. If the inviscid flow is taken to be the classical potential flow without circulation, the Prandtl boundary layer equations can be applied using the corresponding pressure distribution. It is found that the boundary layer solution breaks down near an angle of about 55ı from the upstream stagnation point. This is indicative of separation near that point. A more appropriate “inviscid flow” would thus appear to be an analogue of the Kirchhoff flow of Chapter 6, involving detachment of free streamlines from the separation point. Solutions of this kind exist and presumably represent an improved pressure distribution upstream of separation. This can lead to an iteration between inner and outer approximations, and to a condition on the separating streamline that determines the “correct” Euler flow. Numerical calculations are usually necessary, and special approximations are needed near the separation point; see Section 8.4. (8.9)

peuler D p0 C

8.2. Blasius Solution for a Semi-Infinite Flat Plate We now study the famous Blasius solution of the boundary layer past a semiinfinite flat plate; geometrically the problem is that of Figure 8.1 with L D 1. The fact that the plate is infinite will mean that the boundary layer extends to infinity. We will comment on this later. For the moment simply note that we have now expelled the length L from the problem, even though we used it previously to define a Reynolds number, which number we then let tend to infinity. Without a

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126

8. THE BOUNDARY LAYER

length, however, the problem becomes much simpler to solve, because the no-slip condition applies on the entire line x > 0, yN D 0. We recall that for the aligned flat plate the pressure in Prandtl’s boundary layer equations is 0, so we seek to solve (8.10)

N yN
uyN yN D 0; uux C vu

ux C vN yN D 0;

subject to the conditions u D vN D 0, yN D 0, x > 0, and u ! 1 as yN ! 1, x > 0. We can satisfy the solenoidal condition in the usual way with a boundary p layer stream function N D Re such that u D N yN , vN D
N x . We then observe that our problem has a self-similar structure in the following sense. The equations and conditions are invariant under the group of “stretching” transformations (8.11)

x ! Ax;

yN ! B y; N

N ! C N;

provided that A D B 2 and B D C . Indeed, the condition u D 1 transforms to C 2 2 B D 1, so we must have B D C . Also the term uux scales like C =AB while uyN yN scales like C =B 3 , and the equality of these two factors requires A D BC . The remaining terms follow suit and so (8.10) and the boundary conditions p are invariant under the stated conditions A D B 2 D C 2 . The combinationp D y= x is then invariant under (8.11), and therefore so is the equation N D xF ./ for any function F . If we assume a N of this form and substitute it into (8.12)

N yN N x yN
N x N yN yN
N yxyxyx D 0;

it is straightforward to show that we get
 1 1 FF 00 C F 000 D 0: (8.13)
x 2 The conditions to be satisfied are then (8.14)

F .0/ D F 0 .0/ D 0;

F 0 ! 1;  ! 1:

The simplest way to solve this problem is to replace it by the following initial value problem: 1 GG 00 C G 000 D 0; G.0/ D G 0 .0/ D 0; G 00 .0/ D 1: 2 When this problem is solved (a simple matter using ODE45 in MATLAB on an interval 0 <  < 5 say) we obtain values of G 0 ./ similar to Figure 8.3 (the solution of the actual problem) but asymptoting to c D 2:0854 instead of 1. However, if G./ is a solution of our equation, so is AG.A/ for any constant A. Since G c C o./,  ! 1, we set (8.15)

(8.16)

1

1

F ./ D c
2 G.c
2 /:

This gives the curve for F 0 ./ shown in Figure 8.3. One finds (8.17)

F ./ 
1:7208 C o.1/;

 ! 1;

and also F 00 .0/ D c
3=2 D 0:332. Note that our solution is consistent with the “initial condition” u ! U as x ! 0C, y > 0.

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8.2. BLASIUS SOLUTION FOR A SEMI-INFINITE FLAT PLATE

127

1 0.8 0.6 0.4 0.2 0 0

1

2

3

4

5

6

7

F IGURE 8.3. The Blasius boundary layer velocity profile,

dF d

versus .

8.2.1. Discussion of the Blasius Solution. Recalling that the dimensional form of the stream function is UL , the dimensional form of is UL D ULR
1=2 N . In terms of dimensional x; y,  p 1 N D .x=L/ 2 F pRy=L : .x=L/ Thus with dimensions fully restored, the stream function may be written  r  p U ; UxF y (8.18) x confirming the fact that the problem we have solved is free of a length L. From the asymptotic behavior of F for large , we then have the dimensional stream function for large y in the form p (8.19) Uy
1:7208 Ux C o.1/; y ! 1: p This combination of terms vanishes when y D 1:7208 x=U . This shows that well away from the plate the streamlines look like those over a thin parabolic cylinder. This process of “lifting” the distant streamlines makes the plate look like it has p some thickness, which grows downstream as x. This thickness, which has been given the term displacement thickness, can be understood from the nature of the volume flux in the boundary layer. As the boundary layer grows with increasing x, more and more fluid parcels originally moving with the free-stream velocity U are found to be moving more slowly. This depleted volume flux near the wall, which increases with x, must be compensated by an outward flux of volume away from the wall. It is this outward flux that lifts the streamlines to their parabolic form. The displacement thickness can be given a precise definition as follows: Let  Z 1 u dy: 1
(8.20) ı.x/ D U 0

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128

8. THE BOUNDARY LAYER

Then U d ı.x/ D V .x/dx where V .x/ is the compensating upward velocity, equal to the integral of U
u through the layer, or to the reduced p volume flux through the boundary layer. But according to (8.18), u D UF 0 .y U=.x// and so, since F ./ 
1:7208 C o.1/, we have r r x x lim .
F .// D 1:7208 : (8.21) ı.x/ D U !1 U Thus r (8.22)

0

V .x/ D 1:7208Uı .x/ D 0:8604

dy U D ; x dx

p where y.x/ D 1:7208 x=U determines the zero streamline of the “effective body” whose thickness we may now identify with the displacement thickness as defined. It is interesting to ask what error is being made if we substitute the Blasius solution into the full Navier-Stokes equations and look at the remainder. We consider here only the dimensionless form of the x-momentum equation. There the terms 1 uxx and px . Substituting we dropped to get the boundary layer equation were R 0 u D F ./ we obtain the exact equation

  1 1  00 000 00 000 FF C F 3F C F
2 D 0: (8.23) px
x 2 4x Re We see that the second bracketed term fails to be smaller than the first when xRe D O.1/. Thus near the front edge of the plate the boundary layer equations are not uniformly valid. In a small circular domain of order =U in radius about the origin, the full Navier-Stokes equations can be shown to govern the fluid flow. This small nonuniformity does not affect the validity elsewhere, however. We can assert this because of the existence of the Blasius solution, and the fact that experimental measurements confirm its validity at large Re. As a final remark concerning the Blasius solution, we note that the finite flat plate, of length L, can be approached with exactly the same apparatus. Although the Prandtl boundary layer equations fail to hold near x D L as well as near x D 0, the development of the layer on the plate is otherwise unaltered to first order. In particular, the drag on the plate, accounting for both sides, is given by Z Lr Z L U 00 F .0/dx: uy .x; 0/dx D 2U (8.24) D D 2 x 0 0 This yields r (8.25)

D D 2U  2

U2 UL  .0:332/ D 1:328 p ;  R

and the friction drag on a plate is O.R
1=2 / at large R, at least in a laminar, nonturbulent flow.

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8.2. BLASIUS SOLUTION FOR A SEMI-INFINITE FLAT PLATE

129

1.4 1.2 1 m=.3

0.8

0 −.0904

0.6 0.4 0.2 0 0

2

4

6

8

F IGURE 8.4. The Falkner-Skan profiles for various m, F 0 versus .

8.2.2. The Falkner-Skan Family of Boundary Layers. An immediate generalization of the Blasius solution is to boundary layers whose pressure gradient is some power of x. From the Bernoulli equation for steady flow, a gradient px D
mA2 x 2m
1 results from an external stream with velocity U.x/ D Ax m . We remark that for positive m there is an associated physical Euler flow problem. Such a velocity variation occurs on the surface of an infinite wedge aligned with m
. Then a constant free stream, provided that the half-angle of the wedge is mC1 x is measured along the surface of the wedge, and y is measured perpendicular to the surface. So again there is no length in the problem. The equations to be solved are then, in dimensional form, (8.26)

uux C vuy
mA2 x 2m
1
uyy D 0;

ux C vy D 0:

Since there is no length, we are led to look for a similarity solution. If we try D x ˛ F .y=x ˇ /, then the factors of x coming from insertion into (8.26) will cancel, leaving an ordinary differential equation, provided that 1
m 1Cm ; ˇD : (8.27) ˛D 2 2 Setting r 1Cm y  2 ; F ./;  D ; KD (8.28) D AKx 1m .m C 1/A Kx 2 the equation that results (see Problem 8.1) is 1 m 2 .1
F 0 / D 0: (8.29) F 000 C FF 00 C 2 1Cm The boundary conditions are again F .0/ D F 0 .0/ D 0, F 0 .1/ D 1. We show in Figure 8.4 several profiles for various m. For m positive, existence and uniqueness of the solution have been established, and the profiles become somewhat steeper. The cases m > 0 are said to correspond to a favorable pressure gradient, U 0 .x/ > 0 and p 0 .x/ < 0. The boundary layer can be said to respond favorably to a pressure that decreases in the streamwise direction. When m becomes negative, the story is significantly different. Uniqueness of the profile can be lost,

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130

8. THE BOUNDARY LAYER

y

M x

F IGURE 8.5. A two-dimensional laminar jet emerges from a slit in a wall.

although profiles such that u  0 for all  can be shown to be unique. In Figure 8.4 we display the limiting case of such nonnegative profiles, occurring when vanishes m D
0:0904. Note that F 00 .0/ D 0 for this profile. This implies du dy at the wall, and so the viscous friction force is 0 there. Positive pressure gradients are said to be unfavorable since they can lead to separation of the boundary layer. Here the suggestion is that m < 0:0904 would lead to a boundary layer that has a negative value of uy at the wall, and so would involve a region of reversed flow; the streamline D 0 must actually bifurcate from the wall, so the term “separation” is appropriate. The placement of the bifurcating streamline relative to the point of zero skin friction must be determined by an asymptotic theory for large Re distinct from the Prandtl boundary layer limit; see Section 8.4. We may summarize the general picture of high Reynolds number flow, as provided by the boundary layer concept, as follows. For a general finite body in a flow, there should be a portion of the surface of the body, upstream of any point or points of separation of the boundary layer, where the flow is that of an inviscid fluid except within a small layer adjacent to the body, called the boundary layer. Within the boundary layer, the pressure gradient is imposed by the inviscid exterior flow. At the same time the boundary layer modifies the inviscid flow slightly due to its displacement thickness. The picture is clouded by separation and the tendency of high Reynolds number flows to be unstable and hence time dependent. E XAMPLE 8.1. We give here an application of boundary layer ideas to a different physical problem. The idea is to model a laminar two-dimensional steady jet issuing from a small slit in a wall; see Figure 8.5. We are going to treat the jet as “thin” when Re 1, and so apply Prandtl’s reasoning to obtain again his boundary layer equations. Since pyN D 0 the pressure is invariant through the jet, and assuming that at yN D 1 we have uniform conditions, we may assert that p is independent of x, as in Blasius’ semi-infinite plate problem. There is no length in the problem (ignoring the small width of the slit), so again we are led to try a solution of the form D x ˛ F .y=x ˇ /. The condition that u  ru and uyy have common factors of x requires that ˛ C ˇ D 1. We do not have a nonzero value assumed by F 0 at infinity, as in the Blasius problem. However, there is a new physical constraint, given the constant pressure. There are no forces available to cause

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8.2. BLASIUS SOLUTION FOR A SEMI-INFINITE FLAT PLATE

131

the net flux of x-momentum to vary as a function of x. Consequently, the integral giving the momentum flux (omitting a constant factor of ), Z C1 u2 dy; (8.30) M D
1

must be independent of x. This requires that ˇ D 2˛, so that ˛ D 1=3; ˇ D 2=3. Substituting D x 1=3 F ./;  D y=x 2=3 into the dimensional equation for , (8.31)

y

xy




x

yy




yyy

D 0;

we get the equation 1 F 000 C .FF 0 /0 D 0: 3

(8.32)

We require that F 00 ; F 0 ! 0 as  ! 1 and Z C1 2 F 0 d D M: (8.33)
1

Integrating twice leads to (8.34)

1 2 1 2 F C F 0 D F1 ; 6 6

F1 D F .1/:

The integral yields  F1  : F D F1 tanh 6 

(8.35)

Applying the condition (8.33) we obtain 3 2F1 D M; 9

(8.36)

which determines F1 in terms of M . The velocity component u, which dominates in the jet, is given by (8.37)

uD

2 1 F1 1 : 1=3 2 6x cosh F6

Note that the jet spreads as x 2=3 and velocity decays as x
1=3 . In practice, it is difficult to establish a laminar jet of this kind because of instabilities, and the jets obtained in the laboratory are usually turbulent.
To get this by a stretching group, x ! Ax; y ! By; ! C , the momentum equation requires A D BC as in the Blasius solution, but the momentum flux constraint is invariant when C 2 D B, so then C 3 D A. Thus y=x 2=3 is invariant, and must be proportional to x 1=3 .

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132

8. THE BOUNDARY LAYER

8.3. Boundary Layer Analysis as a Matching Problem We now digress somewhat to indicate some of the mathematical ideas that have grown out of Prandtl’s approach to high Reynolds number flow. We have suggested that there is a kind of interaction at work between an “outer,” inviscid flow, and an “inner” boundary layer flow. That is, the pressure gradient is fundamentally an outer property imposed on the boundary layer. On the other hand, the boundary layer modifies somewhat the streamlines well away from the body in the inviscid flow. We now explore a model problem in one space dimension, involving a singular perturbation of an ordinary differential equation. The small parameter  will 1 , and the problem is not one of fluid dynamics; nevertheless, there will replace Re be an inner solution and an outer solution that will be analogous to our viscous boundary layer and outer inviscid flow. We suggest that the model indicates how a more formal approach to boundary layer theory might proceed, although we shall not pursue this further here. The model problem is the following: let f .x/ D f .x; / satisfy (8.38)

f 00 C f 0 D a;

0 < a < 1; 0  x  1; f .0/ D 0; f .1/ D 1:

The “singular” adjective is usually applied to problems where the limiting operation, in this case  ! 0, reduces the order of the differential equation, in our case from order 2 to order 1. We first define our “outer problem,” analogous to the inviscid Euler flow. We bound x away from 0, 0 < A  x  1, and apply the limit  ! 0 to the model equation. This gives the reduced system (8.39)

f 0 D a:

We apply the condition at x D 1 to the solution of this reduced equation, yielding (8.40)

fouter D ax C 1
a:

We see that fouter does not satisfy the condition on f at x D 0. This adjustment will happen in a boundary layer near x D 0. So we consider with Prandtl how to deal with the combination fxx . If derivatives become large this combination need not be small. On the other hand fx can also be large, so that it is tempting to suppose that at least minimally fxx and fx must be the same size. This suggests a function of x , so we define the stretched variable xN D x .
Using the stretched variable, our equation takes the form (8.41)

fxN xN C fxN D a:

We now consider the limit  ! 0 with 0  xN < B < 1, obtaining the limiting equation (8.42)

fxN xN C fxN D 0:




The fact that we do not have a square root in defining a stretched variable, as we did for the Reynolds number in the Prandtl boundary layer, reflects the vast difference in the fluid equations and the model equation.

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8.4. SEPARATION

133

LRe-3/8 upper deck

LRe-3/8

main deck

LRe-1/2

lower deck

LRe-5/8

Wall

Entering boundary layer

F IGURE 8.6. The triple deck.

This is our model of the Prandtl boundary layer equations. We require that its solution vanish at xN D 0, so that (8.43)

finner D C.1
e
xN /:

Here C is an undetermined constant. Note that finner ! C as xN ! 1, so that we have the model equivalent of obtaining the “velocity at infinity” for the viscous boundary layer. Since f is supposed to be represented by fouter away from x D 0, it is natural to identify C with the limit of fouter for small x. This yields C D 1
a:

(8.44)

This is usually stated as a matching condition: (8.45)

lim finner D lim fouter :

x!1 N

x!0

An approximation to f .x; / that applies to the entire interval can be obtained by adding the inner and outer solutions, provided we account for any terms that are common to both. The common part in our problem is just 1
a. We define the approximate composite solution by (8.46)

x

fcomp D finner C fouter
1 C a D ax C .1
a/.1
e
 /:

It is interesting to compare our approximation with the exact solution of the model problem, namely, x .1
a/ .1
e
 /: (8.47) f .x; / D ax C 1 1
e
 The difference is of order e
1= uniformly over the domain. 8.4. Separation One of the great accomplishments of twentieth-century fluid dynamics was an understanding of the fundamental mechanisms of separation of a boundary layer in the limit of large Re. This work led to a full description of the mechanism of separation in a class of problems of wide applicability. The main result of this effort has been the so-called triple deck theory. The name applies to the layering of domains of different orders of magnitude, in the neighborhood of the point of separation. We show the structure of the triple deck in Figure 8.6. The main point to be made in discussing the triple deck is that the

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134

8. THE BOUNDARY LAYER

F IGURE 8.7. A viscous flow with closed contours.

layered structure results from a nonlinear interaction between the boundary layer and the pressure gradient. In other words, separation represents a breaking of the inner-outer separation of the pressure gradient from the boundary layer responding to the pressure gradient. Within the triple-deck region, the boundary layer is modifying the pressure gradient, which in turn is affecting the boundary layer. Entering from the left of the main deck is the profile of the boundary layer as it has evolved through a length we call L in the figure. Thus the main deck has thickness LR
1=2 . The thinner lower deck is a region where the full boundary layer equations apply, with viscous stress important and reversal of the flow occurring following separation. Over a x of order LR
3=8 the boundary layer is essentially raised by the same order, forming the upper layer. During this lifting the main deck profile is unchanged by viscosity, since it is traversing such a small domain. This lifting of the boundary layer modifies the pressure gradient locally, and this penetrates down to the lower layer, providing the feedback that completes the cycle. Unfortunately, this brief description of separation does not do justice to the analysis involved, nor to the insight that was needed to determine the construction of the triple deck, nor to the many related questions that have been tackled with this machinery. 8.5. Prandtl-Batchelor Theory We have often commented on the absence of nonturbulent fluid flow at large Reynolds numbers. Nevertheless, the mathematical problem of studying the analytic structure of solutions of the Navier-Stokes equations in the limit of infinite Reynolds numbers is of some interest. There are very few examples of such studies outside of boundary layer theory, and we deal with one of these in this section. We consider incompressible flow in two dimensions, and so exhibit again the vorticity equation 1 2 r ! D 0: (8.48) !t C u  r!
Re 8.5.1. Steady Flow. If we set Re D 1 and consider only steady flow, then we obtain u  r!0 D 0, where the subscript 0 indicates the formal limit. In terms of the stream function, u D y , v D
x , (8.49)

!0 D
r 2

D F. /

for some function F . We now ask, For what choice of F , and corresponding solution of (8.49), do we obtain a limit of a steady Navier-Stokes flow as Re ! 1? We try to answer this question for a region of closed streamlines; see Figure 8.7. We

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8.5. PRANDTL-BATCHELOR THEORY

135

may think of the domain as bounded by a rigid wall where the tangential velocity is prescribed. We may write the equation as   1 r! D 0 (8.50) r  u!
Re and integrate this over the domain D enclosed by a given streamline C . Since u  n D 0 for the normal vector n to the bounding streamline, we obtain by the divergence theorem I @ 1 ds D 0: (8.51) Re @n C

We now assume that the streamline C is well outside of any boundary layer at the outer wall. In this case ! D !0 D F . / and I I @ 0 ds D F . / q ds D 0; (8.52) @n C

C

where q D juj, and the contour integral is taken in the counterclockwise direction. Because of the nature of the assumed streamline pattern, the integral over C > 0. Thus F . / must in fact be a constant throughout the region exterior to the boundary layer. The implication is that for such a flow the viscosity of the fluid acts, irrespective of how large Re may be, to slowly bring ! toward a constant, much as the temperature in a room will slowly equilibrate under the absence of sources and sinks of heat. We have therefore identified those Navier-Stokes limits in two dimensions that have a single region of closed streamlines. There is an implicit assumption here, connected with the invariance of the basic topology of smooth nested streamlines, as the limit for large Re is taken. It could be, for example, that the streamlines develop a singularity in the limit of large Re. Also, a natural question is, How can the constant value of !0 within the region be determined? We propose that the boundary values of the velocity must determine !0 , through a solution of a “circulating” boundary layer problem at the outer wall. In general, this is a difficult problem since, even if the necessary solution exists, the pressure variation along the boundary layer makes it difficult to calculate. An exceptional case is when the outer wall is circular and angular symmetry of the solution is assumed; see Problem 8.7. 8.5.2. Time Dependence. We consider now the analogous time-dependent two-dimensional flow. Let us suppose that we have, for large Re, a development of the form (8.53)

!.x; t/ D !0 . ; t/ C !1 C    ;

u D u0 C u1 C    ;

where !0 D
r 2 , and we regard the time dependence of !0 as “slow,” i.e., j@!0 =@tj  1. If we refer to subscript 1 terms as “first-order,” then the first-order equation for ! has the form ˇ ˇ @!0 ˇˇ @!0 ˇˇ @ 1 2
u1  r!0 C r !0 :
(8.54) u0  r!1 D
ˇ ˇ @t @ t @t Re

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136

8. THE BOUNDARY LAYER

We define the contour-averaging operator by I  ds: (8.55) hiC D ju0 j The application of this operator to the left-hand side of (8.54) gives 0, producing a compatibility condition on our solution from the operator applied to the left-hand side. To see what this yields, we first note the identity “ Z  dA: (8.56) hiC d D D

We leave the proof of this to Problem 8.6. In particular, then h1iC D A , where A. ; t/ is the area within C at time t. We also note that h

(8.57)

t iC

D
At :

To prove (8.57), consider local tangential and normal coordinates s; n on C . By t we mean the partial hold s; n fixed. Now ˇ ˇ ˇ @ ˇˇ @n ˇˇ @ ˇˇ D0D C n ˇ : (8.58) @t ˇ @t ˇ @t ;s

s;n

;s

Applying hiC to this last expression, we obtain (8.57). We are now prepared to obtain our compatibility condition. Using the preceding relations, we have from (8.54)
 I I @!0 @A @!0 1 @ @!0 @!0 @A C
q0 ds D 0: u1  n ds C (8.59)
@t @ @ @t @ Re @ @ C

However,

C

I u1  n ds D

(8.60)

@A ; @t

C

since the zeroth-order streamlines behave as material lines under perturbation by u1 . Thus there is cancellation and we obtain
I  1 @ @!0 @!0 @A
D 0: q0 ds (8.61) @t @ Re @ @ C

If we regard !0 as a function of A and t, (8.61) can be written
 1 @ @!0 @!0
D.A; t/ D 0; (8.62) @t Re @A @A where I I ju0 jds ju0 j
1 ds: (8.63) D.A; t/ D C

C

This is a nonlinear diffusion equation in one space variable. If steady flow prevails, 0 D C . Since u0 should vanish as A1=2 near we may integrate to obtain D @! @A

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PROBLEM SET 8

137

the center of the eddy, necessarily C D 0, so for a regular limit we must have !0 D const over the eddy. This recovers our earlier result for steady flow. Problem Set 8 (8.1) Verify (8.27) and (8.29). (8.2) The Oseen equations, mentioned in Chapter 7, are sometimes also proposed as a model of the Navier-Stokes equations in the study of steady viscous flow past a body. Oseen’s equations, for a flow with velocity .U; 0; 0/ at infinity, are U

1 @u C rp
r 2 u D 0; @x 

r  u D 0:

(a) Show that in this model, if viscous stresses are neglected, the vorticity is a function of y; z alone. (b) For the Oseen model, and for a flat plate aligned with the flow, carry out Prandtl’s simplifications for deriving the boundary layer equations in two dimensions, given that the boundary condition of no slip is retained at the body. That is, find the form of the boundary layer on a flat plate of length L aligned with the flow at infinity, according to Oseen’s model, and show that in the boundary layer the the x-component of velocity, u, satisfies U

@2 u @u
 2 D 0: @x @y

What are the boundary conditions on u for qthe flat-plate problem? Find the solu-

tion, by assuming that u is a function of y

U

x ,

for 0 < x < L.

(c) Using the Oseen model, compute the drag coefficient (drag divided by U 2 L ) of one side of the plate, and compare it with the drag computed from the Blasius solution. (8.3) In polar coordinates, what are the boundary layer equations for the boundary layer on the front portion of a circular cylinder of radius a, when the free stream velocity is .
U; 0; 0/? (Use cylindrical polar coordinates.) What is the role of the pressure in the problem? Be sure to include the effect of the pressure as an explicit function in your momentum equation, the latter being determined by the potential flow past a circular cylinder studied previously. Show that, by defining x D a, p yN D .r
a/ R in the derivation of the boundary layer equations, the equations are equivalent to a boundary layer on a flat plate aligned with the free stream in rectangular coordinates, but with pressure a given function of x. (8.4) For a cylindrical jet emerging from a hole in a plane wall, we have a problem analogous to the laminar two-dimensional jet. Consider only the boundary layer limit. (a) Show that 1 @  @ @ 2 .uz / C .rur uz /
@z r @r r @r



@uz r @r

 D 0;

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138

8. THE BOUNDARY LAYER

and hence that the momentum M is a constant, where Z 1 ru2z dr: M D 2
 0

(b) Letting .uz ; ur / D . 1r /. r ;
z / where .0; z/ D 0, show that we must have D zf ./,  D r 2 =z 2 : Determine the equation for f and thus show that the boundary layer limit has the form  ; f D 4  C 0 where 0 is a constant. Express 0 in terms of M , the momentum flux of the jet defined above. (8.5) Consider the Prandtl boundary layer equations with U.x/ D x1 , so that D p.x/= D p1
1=.2x 2 /: Verify that the similarity solution has the form f ./,  D yx . Find the equation for f . Show that there is no continuously differentiable solution of the equation that satisfies f .0/ D f 0 .0/ D 0 and f 0 ! 1, f 00 ! 0 as  ! 1. (Hint: Obtain an equation for g D f 0 .) H (8.6) Prove (8.56) by considering the incremental area A D n ds and introducing

n  ju0 j. (8.7) It is sometimes useful to transform p Prandtl’s steady boundary layer equations into Von Mises coordinates x; D Re in place of x; y. Thus, for examN . Show that in these variables the boundary ple, uyN D uu , ux jyN D ux j
vu layer x-momentum equation with px D 0 becomes @u 1 @u2
D 0: @x 2 @ 2 Using this result show that for steady Prandtl-Batchelor flow within a circular cylinder, with !0 a constant up to the boundary layer, we must have 1=2
Z 2 2 1 2 u d ; !0 D r0 2
0 wall where r0 is the radius of the cylinder.

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http://dx.doi.org/10.1090/cln/019/09

CHAPTER 9

Energy 9.1. Mechanical Energy We recall the two conservation laws.  conservation of mass: @ C r  .u/ D 0: (9.1) @t  conservation of momentum: @.ui uj / Dui @ij @.ui / C D Fi C D : (9.2) @t @xj Dt @xj Here ij are the components of the viscous stress tensor:
 @ui @uj 2 (9.3) ij D
pıij C  C
ıij r  u : @xj @xi 3 We now want to use these results to compute the rate of change of total kinetic energy 12 u2 within a fixed fluid volume V : Z 1 2 d u dV dt 2 V  Z
1 @ 2 @ui C u dV ui D @t 2 @t V   Z 
@ij 1 @uj 2 @ui C C Fi ui
u dV
uj D xj @xj 2 @xj V  Z
Z 1 2 (9.4)
uj u C ui ij nj dS; D Œui Fi C pr  u dV
ˆ C 2 V

S

Here S is the boundary of V and ˆ is the total viscous dissipation in V :
   Z 1 @ui @uj 2 2 C
.r  u/2 : (9.5) ˆ D dV;  D  2 @xj @xi 3 

We remark that  is a nonnegative function; see Problem 9.1.


For simplicity we here assume the Stokes relation; see Chapter 6. 139

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140

9. ENERGY

The contributions to the energy balance from the right-hand side of (9.4) are, taken in order, the work done by body forces, the work done by pressure in compression, the rate of viscous heating, the flux of kinetic energy through S , and the work done by stresses on S . We refer to (9.4) as the mechanical energy balance equation, since we have used only conservation of mass and momentum. To put this expression into a different form, we complete the fluid equations by assuming a barotropic fluid, p D p./. Then Z Z Z pr  u dV D pu  n dS
u  rp dV: (9.6) V

But

S

Z

Z u  rp dV D V

Z Z 

D

(9.7)

1 0 p ./d dV 

u  r Z

V

1 0 p ./du  n dS C 

S

Define g./ by

Z

@ @t

Z

1 0 p ./d dV: 

V

Z

1 0 p ./d D g 0 ./; 

(9.8)

g.0/ D 0;

and define e by g D e:

(9.9) Noting that (9.10)

  Z Z 1 0 1 0 d p
 p ./d D
p ./d D
g 0 D
.e/0 ; d  

it follows that

Z

1 0 p ./d D
e: 

p


(9.11) Thus

Z

Z pr  u dV D


(9.12) V

Z eu  n dS


S

@ .e/0 dV: @t

V

Using this in (9.4) we obtain Z Z Z Z d E dV C Eu  n dS D
ˆ C ui Fi dV C ui ij nj dS; (9.13) dt V

where (9.14)

S

V

 1 2 E D eC u : 2 

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S

9.2. ELEMENTS OF CLASSICAL THERMODYNAMICS

141

Note that if  D 0 and Fi D 0, then (9.13) reduces to a conservation law of the form Z Z d E dV C .E C p/u  n dS D 0: (9.15) dt V

S

We see that E is a total energy, combining e and the kinetic energy, so we shall refer to e as the internal energy of the fluid (per unit mass). Then (9.13) can be viewed as an expression of the first law of thermodynamics E D Q
W , where E is the change of energy of an isolated system (no flux of energy through the boundary), Q is the heat added to the system, and W is the work done by the system. The form of (9.13) can be used as a model for formulating a more general energy equation, and we shall do this after first reviewing some of the basic concepts of reversible thermodynamics. 9.2. Elements of Classical Thermodynamics Thermodynamics deals with transformations of energy within an isolated system. These transformations are determined by thermodynamic variables. These come in two types: Extensive variables are proportional to the amount of material involved. Examples are internal energy, entropy, and heat. Intensive variables are not proportional to quantity. Examples are pressure, density, and temperature. We have just introduced two new scalar fields, the absolute temperature T and the specific entropy s. We shall also make use of specific volume v, defined by v D 1 . Turning to the thermodynamics of gases, we shall assume the existence of an equation of state of the gas, connecting p, , and T . An important example is the equation of state of an ideal or perfect gas, defined by (9.16)

pv D RT:

Here R is a constant associated with the particular gas. In general, all thermodynamic variables are determined by , p, and T . Given an equation of state, in principle we can regard any variable as a function of two independent variables. We can now view our thermodynamic system as a small volume of static gas that can do work by changing volume, can absorb and give off heat, and can change its internal energy. The first law then takes the differential form (9.17)

dQ D de C p dv:

It is important to understand that we are considering here small changes that take place in such a way that irreversible dissipative processes are not present. For example, when the volume of a parcel of gas changes, the fluid acquires some velocity, and viscous dissipation could occur. We are assuming here that the operations are performed so that such effects are negligible. We then say that the system is reversible. If the changes are such that dQ D 0, so that no heat is exchanged with the external environment, we say that the system is adiabatic.

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142

9. ENERGY

We introduce now the specific heats of the gas: The specific heat at constant pressure is defined by ˇ     @Q ˇˇ @e @v D Cp : (9.18) cp D ˇ @T dpD0 @T p @T p @v /p D R. Note that for an ideal gas p. @T The specific heat at constant volume is defined by ˇ   @Q ˇˇ @e D : (9.19) cv D ˇ @T @T v

dvD0

We will make use of these presently. The second law of thermodynamics for reversible systems establishes the existence of the thermodynamic variable s, the specific entropy, such that dQ D T ds:

(9.20)

Thus we have the basic thermodynamic relation T ds D de C p dv:

(9.21)

We now make use of (9.21) to establish an important property of an ideal gas, namely that its internal energy is a function of T alone. To see this, note from (9.21) that     @e @e D T; D
p: (9.22) @s v @v s Thus (9.23)

 R

@e @s



 Cv v

@e @v

 D 0; s

and so e is a function of s
R ln v alone. Then, by the first equation in (9.22), T D e 0 .s
R ln v/, implying s
R ln v is a function of T alone, and therefore e is also a function of T alone. Thus the derivative of e with respect to T at constant volume is the same as the derivative at constant pressure. By the definition of the specific heats, we then have (9.24)

cp
cv D R:

That is, for an ideal gas cp and cv differ by a constant. If both specific heats are constants, so that e D cv T , it is customary to define that ratio cp : (9.25) D cv For air is about 1:4. The case of constant specific heats gives rise to a useful model gas. Indeed, we then have dv dT CR : (9.26) ds D cv T v

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9.3. THE ENERGY EQUATION

143

Note that here the right-hand side explicitly verifies the existence of the differential ds. Using the equation of state of an ideal gas, the last equation may be integrated to obtain p D k.s/ ;

(9.27)

k.s/ D Ke s=cv ;

where K is a constant. The relation p D k defines a polytropic gas. 9.3. The Energy Equation The fundamental variables of compressible fluid mechanics of ideal gases are u; ; p; T . We have three momentum equations, one conservation of mass equation, and an equation of state. We need one more scalar equation to complete the system, and this will be an equation expressing conservation of energy. Guided by the mechanical energy equation, we are led to introduce the total energy per unit mass as e C 12 u2 D E , and express energy conservation by the following relation: Z Z d E dV C Eu  n dS D (9.28) dt Z Z Z V S ui ij nj dS C Fi ui dV C rT  n dS: S

V

S

We have on the right the working of body and surface forces and the heat flux to the system. The latter is based upon the assumption of Fourier’s law of heat conduction in an isotropic medium, stating that heat flux is proportional to minus the gradient of temperature. We have introduced as the factor of proportionality. Given that heat flows from higher to lower temperature, as defined is a positive function, most often of ; T . We now use (9.4) to eliminate some of the terms involving kinetic energy. Note the main idea here. Once we recognize that the energy of the fluid involves both kinetic and internal parts, we are prepared to write the first law as above. Then we make use of (9.4) to move to a more “thermodynamic” formulation. So (9.28) becomes (9.29)

d dt

Z

Z e dV C

V

Z eu  n dS D

S

Z rT  n dS C ˆ


pr  u dV:

S

This implies the local equation De
r  rT
ˆ C pr  u D 0: (9.30)  Dt Using T ds D de C p dv and the equation of conservation of mass, the last equation may be written Ds D r  rT C ˆ: (9.31) T Dt This is immediately recognizable as having on the right precisely the heat inputs associated with changes of entropy.

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144

9. ENERGY

There are other forms taken by the energy equation in addition to (9.30) and (9.31). These are easiest to derive using Maxwell’s relations. To get each such relation we exhibit a principle function and from it obtain a differentiation identity, by using T ds D de C p dv in a form exhibiting the principle function. For example, if e is the principle function, then     @e @e D T; D
p: (9.32) @s v @v s Then the Maxwell relation is obtained by cross differentiation:     @p @T D
: (9.33) @v s @s v We define the next principle function by h D e C pv, the specific enthalpy. Then T ds D dh
v dp. Thus     @h @h D T; D v; (9.34) @s p @p s giving the relation (9.35)



@T @p



 D s

@v @s

 : p

The principle function and the corresponding Maxwell relation in the two remaining cases are  the free energy F D e
T s, yielding     @s @p D ; (9.36) @T v @v T  the free enthalpy G D h
T s, yielding the relation     @v @s D
: (9.37) @p T @T p We illustrate the use of these relations by noting that     @s @s dT C dp ds D @T p @p T (9.38)

dT
D cp T



@v @T

 dp; p

@v /p D R where we have used (9.37). Now for a perfect gas . @T p , so that (9.31) may be written Dp DT
D r  rT C : (9.39) cp Dt Dt In particular, if cp , , and  are known functions of temperature say, then we have with the addition of (9.39) a closed system of six equations for u; p; ; T .

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9.4. SOME BASIC RELATIONS FOR THE NONDISSIPATIVE CASE

145

9.4. Some Basic Relations for the Nondissipative Case If  D D 0 in (9.28), local conservation of energy may be written @E C r  .uE/ D u  F
r  .up/: @t

(9.40)

Using conservation of mass we have   D.e C 12 u2 / 1 p D D u  F
u  rp C Dt   Dt   1 @p D p D uFC
:  @t Dt  Thus (9.41)

    1 2 p 1 @p D eC u C D uFC : Dt 2   @t

If now the flow is steady, and F D
r‰, then we obtain a Bernoulli function H in the form p 1 H  e C u2 C C ‰ D const 2 

(9.42)

on streamlines of the flow. An appropriate term for H is total enthalpy. To see how H changes from streamline to streamline in steady flow, observe that     1 2 1 2 u ChC‰ Dd u C ‰ C T ds C v dp; (9.43) dH D d 2 2 so that we may write  (9.44)

rH D r

 1 1 2 u C ‰ C T rs C rp: 2 

But for inviscid steady flow we have u  ru C rp D
r‰, or   1 1 2 u C ‰ C rp D u  !; (9.45) r 2  where ! D r u is the vorticity vector. Using the last equation in (9.44) we obtain Crocco’s relation: for an ideal compressible fluid: (9.46)

rH
T rs D u  !:

A flow in which Ds Dt D 0 is called isentropic. From (9.31) we see that  D D 0 implies isentropic flow. If in addition s is constant throughout space, the flow is said to be homentropic. We see from (9.46) that in homentropic flow we have (9.47)

rH D u  !:

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146

9. ENERGY

Note also that in homentropic flow the Bernoulli relation (9.42) becomes (since now dh D v dp) Z 2 1 c d C u2 C ‰ D const (9.48) H D  2 on streamlines, where c2 D

(9.49)



@p @

 s

is the speed of sound in the gas. 9.5. Kelvin’s Theorem in a Compressible Medium Following the calculation of the rate of change of circulation that we carried out in the incompressible case, consider the circulation integral over a material contour C : I I d @x d d˛; u  dx D u (9.50) dt dt @˛ C.t/

C.t/

where ˛ is a Lagrangian parameter for the curve. then I I I Du d  dx C u  d u: u  dx D (9.51) dt Dt C.t/

Using (9.52)

Du Dt

C

C

D
rp 
r‰, we get after disposing of perfect differentials, I Z 1 d u  dx D .r  rp/  n dS: dt 2 C.t/

S

Here S is any oriented surface spanning C . In a perfect gas, T ds D cv d T Cp dv, so that 1 T p  pr D
2 rp  r: (9.53) T rT  rs D r R   Thus (9.54)

d dt

I

Z u  dx D

C.t/

.rT  rs/  n dS: S

E XAMPLE 9.1. Consider the equations of acoustics, i.e., the propagation of sound. The disturbances of the air are so small that viscous and heat conduction effects may be neglected to first approximation, and the flow taken as homentropic. Since disturbances are small, we write  D 0 C0 , p D p0 Cp 0 , u D u0 , where the subscript 0 denotes constant ambient conditions. If the ambient speed of sound is   ˇ @p ˇˇ D c02 ; (9.55) ˇ @ s 0

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9.5. KELVIN’S THEOREM IN A COMPRESSIBLE MEDIUM

147

we assume 0 =0 ; p 0 =p0 , and ku0 j=c0 are all small. Also, we see that p 0  c02 0 : With no body force, the mass and momentum equations give us c2 @0 @u0 C 0 r0 D 0; C 0 r  u0 D 0: @t 0 @t Here we have neglected terms quadratic in primed quantities. Thus we obtain acoustics as a linearization of the compressible flow equations about a homogeneous ambient gas at rest. Combining (9.56) we obtain the wave equation   2 @ 2 2 .0 ; u0 / D 0:
c r (9.57) 0 @t 2

(9.56)

If sound waves arise from still air, Kelvin’s theorem guarantees that u0 D r, where  will also satisfy the wave equation, with 1 @ : (9.58) p 0 D c02 0 D
0 @t We shall study sound propagation in more detail in Chapter 10. E XAMPLE 9.2. Another example of compressible flow is two-dimensional steady isentropic flow of a polytropic gas with  D D ‰ D 0. Then 1 (9.59) u  ru C rp D 0; r  .u/ D 0:  Let u D .u; v/; q 2 D u2 C v 2 . The Bernoulli relation now takes the form Z dp 1 2 q C D const (9.60) 2  on streamlines. For a polytropic gas we have Z 1 2 dp Dk 
1 D c : (9.61) 
1
1 In component form, the momentum equations are c2 c2 x D 0; uvx C vvy C y D 0;   and conservation of mass may be written     y x Cv D 0: (9.63) ux C vy C u   Substituting for the -terms using (9.62), we obtain (9.62)

(9.64)

uux C vuy C

.c 2
u2 /ux C .c 2
v 2 /vy
uv.vx C uy / D 0:

If we assume irrotational flow, vx D uy , so that .u; v/ D .x ; y /, then we have the system (9.65) (9.66)

.c 2
x2 /xx C .c 2
y2 /yy
2x y xy D 0; x2 C y2 C

2 2 c D const:
1

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148

9. ENERGY

E XAMPLE 9.3. Meteorologists as well as pilots utilize the so-called standard atmosphere as a model of the average structure of a static atmosphere. We now consider this model and use it to attach some numbers to our thermodynamic variables. The principal approximation made in the standard atmosphere is the assumption of a standard lapse rate for temperature, which is roughly valid in the troposphere (altitudes up to 10 000 meters). This lapse rate is intended to represent typical conditions of atmospheric moisture, and it is taken as L D 0:0065ı C per meter or L D 1:981ı C per 1000 feet. Temperature will be in Kelvin, where K = ı C C 273:15. Thus the assumed temperature in the troposphere is  z  K (9.67) T D T0
1:981 1000 when the altitude z above the reference line (in this case mean sea level) is expressed in feet, T0 being the reference temperature. The next basic assumptions of the model are that the atmosphere is an ideal gas and that the pressure is hydrostatic: p D RT;

(9.68)

dp D
g; dz with the gas constant R D 287:04 m2 /s2 /K, and the surface acceleration of gravity, g D 9:80665 m/s2 . Now if (9.68) and (9.67) are used in (9.69), we have (9.69)

gp dp D
: dz R.T0
Lz/

(9.70) Integrating, we get (9.71)

p D p0



T T0

g  RL

  g Lz RL D 1
; T0

where p0 is the reference pressure. Inverting this last equation,    RL  p g T0 1
: (9.72) zD L p0 Another expression yields z in terms of density. From (9.68) we have . 0 /. TT0 /,

p p0

D

and if (9.71) is used we have p D p0

(9.73)



 0



1 1 RL g

:

Using this in (9.72) we obtain (9.74)

   g1   RL 1 T0 1
: zD L 0

The 1976 Standard Atmosphere, shown in Figure 9.1, utilizes the following reference values: p0 D 1013:25 mb D 29:92 inches of mercury, T0 D 15ı C D

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PROBLEM SET 9

149

30 Pressure in inches of mercury 20

Density*10000 gm/cm3 10 Temperature in oC 0

-10

-20

-30

0

2000

4000

6000

8000 10000 Altitude in feet

12000

14000

16000

18000

F IGURE 9.1. The 1976 Standard Atmosphere.

288:15K. To get 0 in g/cm3 , we have 0 D 1013:25=287:04=288:15=10 D 0:001225. The extra factor of 10 is needed to resolve cm and m units. Problem Set 9 2 (9.1) We have asserted that  D 2 .@ui =@xj
@uj =@xi /2
2 3 .r  u/ is nonnegative. Prove this. (9.2) Show that for a perfect gas, another form of the energy equation for a viscous, heat-conducting fluid is p D DT
D  C r  krT: cv Dt  Dt

@S

d T C @S dv.) (Hint: Start with dS D @T @v T v (9.3) Show that for an inviscid gas with zero heat conductivity, the energy equation may be written

@e C r  .ue/ D
pr  u: @t (9.4) (a) Show that



@p cp
cv D T @T

  v

@v @T

 : p

@s @s /v d T C . @v /T dv, from (Hint: Regarding s as a function of T and v, ds D . @T @s @s @s /v D cv , which we can get an expression for . @T /p : Now use T . @T /p D cp , T . @T and a Maxwell relation.) Verify that, for a perfect gas, this relation gives R D cp
cv .

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150

9. ENERGY

(b) Show that for a perfect gas   @p D RT: @ s Note that this quantity equals c 2 where c is the speed of sound under isentropic conditions. (9.5) For the study of thermal convection in water, it is usually assumed that the fluid density is a function of temperature alone. The energy equation is then usually approximated as a temperature equation of the form DT
r  krT D 0: cp Dt Justify this as an approximation to the full energy equation,   @v DT Dp
T D  C r  krT: cp Dt @T p Dt You should make use of the data for water at 20ı C; see, e.g., [1, pp. 596–597]. Assume a characteristic fluid speed is U  1 cm/sec in a fluid layer of thickness L  1 cm. Use U; L to estimate terms involving spatial derivatives and velocity. Also take L=U as a characteristic timescale, one dyne/cm2 as a characteristic pressure, and 10ı C as typical of T and its variations (1 joule = 107 dyne-cm.) The heat conduction term is retained, even though relatively small, because it can be important in boundary layers. R (9.6) For a perfect gas, cv ; cp are functions of T alone and so e D cv d T . Also then cp
cv D R D const. Using these facts show R that for steady flow of a 1 2 perfect gas, Bernoulli’s theorem may be written 2 u C cp .T /d T C ‰ D const on streamlines.

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http://dx.doi.org/10.1090/cln/019/10

CHAPTER 10

Sound The study of acoustics is of interest as the fundamental problem of linearized gas dynamics, and we have seen in Chapter 9 that the linear wave equation governs the variation of the fluid variables. In the present chapter we study some of the properties of solutions of this wave equation, @2 
c 2 r 2  D 0; @t 2

(10.1)

where c is a constant phase speed of sound waves.

10.1. One-Dimensional Waves We first consider the one-dimensional case and the initial value problem on
1 < x < C1. The natural initial conditions are for the gas velocity and the pressure or density, implying that both  and t should be supplied initially. Thus the problem is formulated as follows: (10.2)

2 @2  2@ 
c D 0; @t 2 @x 2

.x; 0/ D f .x/; t .x; 0/ D g.x/:

@ @ @ @ C c @x /. @t
c @x / the general solution is seen to have From the factorization . @t the form

 D F .x
ct/ C G.x C ct/:

(10.3)

Using the initial conditions to solve for F; G we obtain d’Alembert’s solution: (10.4)

.x; t/ D

1 1 Œf .x
ct/ C f .x C ct/ C 2 2c

Z

xCct

g.s/ds: x
ct

In the xt-plane, a given point .x0 ; t0 / in t > 0 is influenced only by the initial data on that interval of the x-axis lying between the points of intersection with the axis of the two lines x
x0 D ˙c.t
t0 /. This interval is called the domain of dependence of .x0 ; t0 /. Conversely, a given point .x0 ; t0 / in t  0 can influence only the points within the wedge bounded by the two lines x
x0 D ˙c.t
t0 / with t
t0  0. This wedge is called the range of influence of .x0 ; t0 /. These two lines are also known as the characteristics through the point .x0 ; t0 /. 151 Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

152

10. SOUND

10.2. The Fundamental Solution in Three Dimensions In three dimensions, under the condition of spherical symmetry, the wave equation has the form   2 2 @ @2  2 @  D 0:
c C (10.5) @t 2 @r 2 r @r Here r 2 D x 2 C y 2 C z 2 . Note that we can rewrite this as (10.6)

.r/t t
.r/rr D 0;

reducing the three-dimensional problem to the one-dimensional problem. We seek to solve the three-dimensional wave equation with a distribution as a forcing function, and also with null initial conditions. In particular, we seek the solution of (10.7)

t t
c 2 r 2  D ı.x/ı.t/;

with .x; 0
/ D t .x; 0
/ D 0. Here ı.x/; ı.t/ are, respectively, three-dimensional and one-dimensional delta functions. Since the three-dimensional delta function imposes no deviation from spherical symmetry, we assume this symmetry and solve the problem as a one-dimensional problem. When t > 0 we see from the one-dimensional problem that
    r r 1 F t
CG t C : (10.8) D r c c (The change r
ct to t
rc is immaterial but will be convenient.) The term G represents “incoming” signals propagating toward the origin from 1. Such waves are unphysical for the problem considered. Think of the delta function as a disturbance localized in space and time, like a firecracker set off at the origin. The explosion should produce only outgoing signals, and we set G D 0. Also, near the origin F .t
rc /  F .t/, so the ı.x/ı.t/ distribution would result, using r 2 . 1r / D 4
ı.x/, provided     1 r r D : ı t
(10.9) F t
c 4
c 2 c As we saw for Laplace’s equation, another way to obtain this is to integrate the left-hand side of (10.7) over the ball r  , use the divergence theorem, and let  ! 0. The fundamental solution of the three-dimensional wave equation is therefore defined by   r 1 ı t
: (10.10) ˆ.x; t/ D 4
c 2 r c E XAMPLE 10.1. To illustrate a solution in three dimensions we consider the problem of a bursting balloon. The balloon will be spherical so we can adopt

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10.3. KIRCHHOFF’S SOLUTION

153

radially symmetric initial conditions. We assume that the pressure perturbation p satisfies ( p if 0 < r < rb  N.r/: (10.11) p.x; 0/ D b 0 if r > rb Here pb is a positive constant representing the initial pressure in the balloon. Now pt D c 2 t D
c 2 0 r 2 . If the velocity of the gas is to be zero initially, as we must assume in the case of a fixed balloon, then (10.12)

pt .x; 0/ D 0:

Since rp satisfies the one-dimensional wave equation and p is presumably bounded at r D 0, we extend the solution to negative r by making rp an odd function of r, still denoted by N . The initial value problem for rp is thereby defined as a solution of a one-dimensional problem: (10.13)

1 rp D ŒN.r
ct/ C N.r C ct/ : 2

Note that we have both incoming and outgoing waves since the initial condition is over a finite domain. For large time, however, the incoming wave does not contribute and the pressure is a decaying “N ” wave of width 2rb centered at r D ct. 10.3. Kirchhoff’s Solution We now take up the solution of the general initial value problem for the wave equation in three dimensions: (10.14)

t t
c 2 r 2  D 0;

.x; 0/ D f .x/;

t .x; 0/ D g.x/:

This can be accomplished in two steps. First, note that if  solves the wave equation with the initial conditions f D 0, g D h, then t solves the wave equation with f D h, g D 0. Indeed, t t D c 2 r 2  tends to 0 as t ! 0 since this is a property of . The second step is to note that the solution  with f D 0, g D h, is given by Z 1 h.x0 /dS 0 : (10.15) .x; t/ D 4
tc 2 S.x;t/

The meaning of S is indicated in Figure 10.1. To verify that this is the solution, note first that we are integrating over a spherical surface of area 4
c 2 t 2 . Given that h is bounded, division by t still leaves a factor t, so we obtain 0 in the limit as t ! 0. Also Z t h.x C ytc/dSy (10.16) D 4
jyjD1

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154

10. SOUND

S(x,t)

ct x

O F IGURE 10.1. Definition of S.x; t / in the Kirchhoff solution.

by a simple change of variable.Thus Z Z 1 t h.x C ytc/dSy C (10.17) t D 4
4
jyjD1

cy  rh.x C ytc/dSy :

jyjD1

The first term on the right clearly tends to h.x/ as t ! 0, while the second term tends to 0 provided that h is a sufficiently well-behaved function. We now show that (10.16) solves the wave equation. Using the divergence theorem we can write (10.17) in the form Z Z 1 1 h.x C ytc/dSy C r 2 h.x0 /dV 0 ; (10.18) t D 4
4
ct jyjD1

V .x;t/

where V .x; t/ denotes the sphere of radius ct centered at x. Then Z Z ct Z 1 1 h.x C ytc/dSy C r 2 h.y/dSy d; (10.19) t D 4
4
ct 0 SR.x/

jyjD1

where SR.x/ is the spherical surface of radius R , centered at x. We then compute Z Z c 1 y  rh.x C ytc/dSy
r 2 h.x0 /dV 0 t t D 4
4
ct 2 jyjD1

C

1 4
t

Z

V .x;t/

r 2 h.y/dSy

S.x;t/

1 D 4
ct 2

Z

r 2 h.x0 /dV 0


V .x;t/

1 C 4
t

Z

1 4
ct 2

Z

r 2 h.x0 /dV 0

V .x;t/

r 2 h.y/dSy D

S.x;t/

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10.4. WEAKLY NONLINEAR ACOUSTICS IN ONE DIMENSION

α O

155

ct Ut

F IGURE 10.2. Supersonic motion of a sound source.

D

Z

1 4
t

r 2 h.y/dSy :

S.x;t/

(10.20)

D

c2t 4


r

2

Z

h.x C ytc/dSy D c 2 r 2 

jyjD1

and we are done. Given these facts we may write down Kirchhoff’s solution to the initial value problem with initial data f; g as follows: Z Z @ 1 1 0 0 g.x /dS C f .x0 /dS 0 : (10.21) .x; t/ D 4
tc 2 @t 4
tc 2 S.x;t/

S.x;t/

Although we have seen that the domain of dependence of a point in space at a future time is in fact a finite segment of the line in one dimension, the corresponding statement in three dimensions, that the domain of dependence is a finite region of 3-space, is false. The actual domain of dependence is the surface of a sphere of radius ct, centered at x. This fact is known as Huygens’ principle. Moving sound sources give rise to different sound fields depending on whether the source is moving slower or faster than the speed of sound. In the latter case, a source moving to the left along the x-axis with a supersonic velocity U > c will produce sound waves having a conical envelope; see Figure 10.2. 1 where M D Uc is the Mach number. The nose of a Here sin ˛ D Uc D M slender body moving through a compressible fluid at a supersonic velocity can be thought of as a sound source. The effect of the body is then confined to within the conical surface of Figure 10.2. This surface is called the Mach cone. 10.4. Weakly Nonlinear Acoustics in One Dimension We have seen that sound propagation in one dimension involves the characteristics x ˙ ct D const, representing two directions of propagation. If a sound pulse traveling in one of these directions is followed, over time weak nonlinear effects can become important, and a nonlinear equation is needed to describe the compressive waves. In this section we shall derive the equation that replaces the

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156

10. SOUND

simple linear wave equation t ˙ cx D 0 associated with the two families of characteristics; we also all the gas to have a nonzero viscosity . We shall suppose that the disturbance is moving to the right, i.e., is in linear theory a function of x
ct alone. The characteristic coordinates  D x
ct;

(10.22)

 D x C ct;

can be used in place of x; t provided c > 0. Then @ @ @ D
c C c ; @t @ @

(10.23)

@ @ @ D C : @x @ @

Thus with the linear theory our right-moving disturbance is annihilated by the operator   1 @ @ @ D C : (10.24) @ 2 c@t @x We shall therefore be looking for a compressive wave which, owing to nonlinearity, has a nonzero but small variation with respect to . The variation with respect to  will involve small effects, both from nonlinearity and from the viscous stresses. If the variables are again 0 C 0 , p0 C p 0 , and u0 , the exact conservation of mass equation is @.0 u0 / @u0 @0 C 0 D
: @t @x @x To get the proper form of the momentum equation, we expand the pressure as a function of , assuming that we have a polytropic gas. With p D h we have the Taylor series (10.25)

p D p0 C c 2  0 C

(10.26)

.
1/ 2 .0 /2 C  : c 0 2


1

Here we have used c 2 D k0 . Thus the momentum equation takes the form, through terms quadratic in primed quantities, (10.27)

0

4 @2 u0 @u0 @0 @u0 @u0
1 2 0 @0 C c2 D
0
0 u0
C c  : @t @x @t @x 0 @x 3 @x 2

Note that the viscous stress term comes from the difference 2
23  in the co0 in the one-dimensional stress tensor. We assume here that  is a efficient of @u @x constant. To derive a nonlinear equation for the propagating disturbance we proceed in two steps. First, eliminate the -differentiations from the linear parts of the two equations. This will yield an equation with a first derivative term in , along with the viscosity term and a collection of quadratic nonlinearities in u0 ; 0 . Then we use the approximate linear relation between u0 and 0 to eliminate 0 in favor of u0 in these quadratic terms. The result will be a nonlinear equation for u0 in which all terms are small but comparable.

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10.4. WEAKLY NONLINEAR ACOUSTICS IN ONE DIMENSION

157

The linear relation used in the nonlinear terms comes from @u0 @0 D
c 2 : (10.28) 0 @t @x Since dependence upon  is weak, the last relation expressed in ;  variables becomes @u0 @0 D
c 2 ; (10.29)
c0 @ @ so that 0 (10.30) 0  u0 c in the nonlinear terms as well as in any derivative with respect to . In characteristic coordinates the linear parts of the equations take the forms @ @.0 u0 / @ .
c0 C 0 u0 / C .c0 C 0 u0 / D
; @ @ @x @ @ .
c0 u0 C c 2 0 / C .c0 u0 C c 2 0 / D    ; (10.32) @ @ where the right-hand side consists of nonlinear and viscous terms. Dividing the second of these by c and adding the two equations, we get

(10.31)

@.0 u0 / 1 0 @u0 0 0 @u0 @ .c0 C 0 u0 / D


u @ @x c @t c @x (10.33) 0 2 0 4 @ u
1 0 @ C c :
0 @x 3c @x 2 Since the left-hand side of (10.33) involves now only the -derivative, we may use (10.30) to eliminate 0 , and similarly with all terms on the right-hand side. Also, we may express x- and t-derivatives on the right in terms of . Thus we have 2

0 0 @u0 0 0 @u0 0 @u0 @u0 D
2 u C u
C u0 @ c @ c @ c @ (10.34) 0 2 0 4 @ u .
1/0 0 @u u C :
c @ 3c @ 2 Combining terms, we get 40

C 1 0 @u0 2 @2 u0 @u0 C u
D 0: @ 2 @ 30 @ 2 Recalling the definition of the -derivative (10.24) we have ˇ ˇ 2 @2 u0 @u0 ˇˇ C 1 0 @u0 @u0 ˇˇ u
C c C D 0: (10.36) @t ˇ @x ˇ 2 @ 3 @ 2

(10.35)

2c

x

0

t

This involves a linear operator describing the time derivative relative to an observer moving with the speed c, which is just the time derivative holding  fixed. Thus ˇ 2 @2 u0 C 1 0 @u0 @u0 ˇˇ u
C D 0: (10.37) @t ˇ 2 @ 3 @ 2 

0

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158

10. SOUND

The velocity perturbation u0 is that relative to the fluid at rest at infinity. Traveling with the wave, the gas is seen to move with velocity u D u0
c, so u0 really denotes u C c where u is the velocity seen by the moving observer. What we have in (10.37) is the viscous form of Burgers’ equation. It is a nonlinear dissipative wave equation in u0 ; ; t. By rescaling u0 and , it may be brought into the form (10.38)

ut C uux
uxx D 0:

If the viscous term is dropped we have the inviscid Burgers wave equation ut C uux D 0:

(10.39)

This equation is a prototypical nonlinear wave equation and will be studied in the next chapter. Problem Set 10 (10.1) Sound waves may be treated using a potential  satisfying the wave equation t t
c 2 r 2  D 0 with velocity u D r and pressure p D
0 t . Here c 2 D dp=d.0 /. Find the potential of the three-dimensional, spherically symmetric sound field emitted by an oscillating sphere. Assume oscillations are of amplitude a about a radius r0 , where a  r0 , so the boundary condition to be satisfied is @ .r0 ; t/ D a sin !t D a=.e i!t / @r approximately. (Why is the last expression approximately correct?) Assume also that only outgoing waves appear in the solution. From your expression for  comD r . The result should involve the dimenpute the radial velocity field ur D @ @r !r0 sionless parameter  D c . (The frequency of the standard note A above middle C is 440 Hertz (cycles per second), in which case ! D 880
. Taking r0 D 1 cm and c D 350 m/sec we get !r0 =c  0:08. We may thus take the parameter  to be of order unity.) The length scale R D !c  12 cm separates the “near” and “far” fields of the sound pattern. Discuss the r-variation of ur r in the two cases r0 C a < r  R and r R. (10.2) A semi-infinite tube is closed at the end x D 0. A perfect gas is confined to the region L < x < 2L at pressure, density, and temperature p1 ; 1 ; T0 . Elsewhere the ambient state is p0 ; 0 ; T0 . At t D 0 the barriers at x D L; 2L are broken. If p1 D 1:1p0 , and the approximation of acoustics is used, what is the pressure distribution in the tube at time t D 7L=4c0 ? (Hint: The domain may be extended to the real line by making pressure even and velocity odd in x.) (10.3) Establish uniqueness of the solution of the IVP for the wave equation in N dimensions using the energy method: (a) Show that if u is twice continuously differentiable and ut t
c 2 r 2 u D 0, 2 2 for t > 0, all x, with r 2 D @ 2 C    C @ 2 ; and also either u or ut is identically @x1

@xN

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PROBLEM SET 10

159

0 outside a sphere of some large radius R centered at the origin, then the energy Z C1 Z 1 C1 ::: .u2t C c 2 .ru/2 /dx1    dxN E.t/ D 2
1
1 is constant in time. Use the fact that then the range of influence of the initial data lies inside the sphere of radius R C ct at time t. (b) Obtain uniqueness by considering the null initial conditions ut D 0, u D 0. (10.4) Solve the bursting balloon problem of Example 10.1 using the Kirchhoff equation (10.21). (10.5) Let B./ be a given differentiable function, B 0 ¤ 0. Show that if u.x; t/ is defined implicitly by u D B.x
ut/, so u.x; 0/ D B.x/, then u.x; t/ satisfies Burgers’ equation ut C uux D 0 for t > 0. (10.6) Consider the viscous Burgers equation (10.38). The Cole-Hopf transformation is given by u D
2 x . Show that under this transformation, Burgers’ equation reduces to
 @ t xx
 D 0: @x Hence show that a solution of Burgers’ equation is generated by taking 1

x2

D 1 C t
2 e
4t : p p Describe this solution by sketching u t= as a function of  D x=.2 t / for various t. (10.7) This is a problem for exploring the interesting phenomenon of sound generated by fluid flow (e.g., the bubbling of a brook or the sound of an air jet). To study the sound generated aerodynamically in a fluid, assume that the density is  D 0 C 0 and similarly for p, where 0 ; p0 are constants and 0  p 0 =c02 . For an inviscid gas, show that the equations of mass and momentum may then be combined to obtain a equation of the form 1 @2 p
r 2 p D F: c02 @t 2 Find the form of F in terms of  and u and their derivatives. What form is taken by F if we have   0 and u is taken to have zero divergence?

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http://dx.doi.org/10.1090/cln/019/11

CHAPTER 11

Gas Dynamics 11.1. Nonlinear Waves in One Dimension The simplest scalar wave equation can be written in the conservation form (11.1)

@ @u C F .u/ D 0 @t @x

and is equivalent to @u @u C v.u/ D 0; v.u/ D F 0 .u/: @t @x These relations can be regarded as stating that an observer moving with the velocity v.u/ observes that u does not change. The particle path of the observer is called a characteristic curve. Since u is constant on the characteristic and the velocity v is a function of u alone, we see that the characteristic is a straight line in the xt-plane. If u.x; 0/ D u0 .x/, the characteristics are given by the family (11.2)

(11.3)

x D v.u0 .x0 //t C x0 :

Here x0 acts like a fluid Lagrangian coordinate, marking the intersection of the characteristic with the initial line t D 0. E XAMPLE 11.1. As an example of the solution of the initial value problem using characteristics, consider the equation 8 ˆ 1. First observe that the characteristics are vertical lines in x < 0, so that u D 0 in x < 0, t > 0. Similarly, the characteristics are the line x D t C x0 when x0 > 1, so that u D 1 when x > 1 C t. Solving x D x02 t C x0 for x0 .x; t/, we arrive at the following solution in the middle region 0 < x < 1 C t: p
1 C 1 C 4xt : (11.5) u.x; t/ D t0 .x; t/ D 2t E XAMPLE 11.2. Now we modify the initial condition to 8 ˆ if x < 0, . 161 Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

162

11. GAS DYNAMICS

Since the characteristics in the middle region are x D . x 0 /2 t C x0 , the solution is r 
4xt 
1 C 1 C 2 : (11.7) u.x; t/ D 2t  Letting  ! 0 in (11.7) we obtain r u!

(11.8)

x : t

We shall refer to this solution, existing in the wedge 0 < xt < 1 of the xt-plane, as an expansion fan. (The name derives from the analogous structure in gas dynamics; see below.) Given the discontinuous initial condition ( 0 if x < 0, (11.9) uD 1 if x > 0, we can solve for the expansion fan directly by noting that u must be a function of  D xt . Substituting u D f ./ in our equation, we obtain (11.10)


f 0 C f 2 f 0 D 0;

p implying f D ˙ . The positive sign is needed to make the solution continuous at the edges of the fan. 11.2. Dynamics of a Polytropic Gas We have the following equations for a polytropic gas in one dimension in the absence of dissipative processes and with a constant entropy: (11.11)

ut C uux C

c2 x ; 

t C ux C ux D 0:

Here c 2 D k 
1 . If we define the column vector Œu  T D w,the system may be written wt C A  wx where   u c 2 = : (11.12) AD  u To find analogues of the characteristic lines x ˙ ct D const that arose in acoustics in one space dimension, we look for curves on which some physical quantity is invariant. Suppose that v is a right eigenvector of AT (transpose of A), AT v D v. We will show that the eigenvalue plays a role analogous to the acoustic sound velocity. Indeed, we see that (11.13) v T  Œwt C A  wx D v T  wt C AT  v  wx D v T  wt C v T  wx D 0: Now suppose that we can find an integrating factor  such that v T  dw D dF . We would then have (11.14)

Ft C Fx D 0:

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11.3. SIMPLE WAVES

163

Thus dx D would define a characteristic curve in the xt-plane on which F D dt const. Thus can be appropriately termed a characteristic velocity. The quantity F is called a Riemann invariant. We thus solve the eigenvalue equation ˇ ˇ ˇ ˇ u
 T ˇ D 0: ˇ (11.15) det.A
I / D ˇ 2 c = u
ˇ Then .u
/2 D c 2 or D u ˙ c  ˙ :

(11.16)

We see that the characteristic velocities are indeed related to sound velocity, but now altered by the Doppler shift introduced by the fluid velocity. (Unlike light through empty space, the speed of sound does depend upon the motion of the observer. Sound moves relative to the compressible fluid through which it passes.) Thus the following eigenvectors are obtained:  
c  T v D 0; vC (11.17) D Πc ; C W c 2 =
c C   c  T v D 0; v
(11.18) D Œ
c ;
W c 2 = c
Now  must be chosen so that Œ

(11.19)


 du ˙ c

D dF˙ : d

Since c is a function of , we may take  D 1 to obtain Z c d; (11.20) F˙ D u ˙  which may be brought into the form 2 c: (11.21) F˙ D u ˙
1 Thus the Riemann invariants u ˙ dx dt

D u ˙ c:

(11.22)




2 
1 c

@ @ C .u ˙ c/ @t @x

are constant on the characteristic curves


u˙

 2 c D 0:
1

11.3. Simple Waves Any region of the xt-plane that is adjacent to a region where all fluid variables are constant (i.e., a region at a constant state), but that is not itself a region of constant state, will be called a simple wave region, or SWR. The characteristic families of curves associated with ˙ will be denoted by C˙ . Curves of both families will generally propagate through a region. In a simple wave region one family of characteristics penetrates into the region of constant state, so that one of the two invariants F˙ will be known to be constant over a SWR. Suppose that

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164

11. GAS DYNAMICS

c+

t

c-

t

x

x (b)

(a)

F IGURE 11.1. Simple expansion waves, the curves indicating the direction of particle paths. (a) Forward facing. (b) Backward facing.

F
is constant over the SWR. Then any CC characteristic in the SWR not only carries a constant value of FC but also a constant value of F
, and this implies a constant value of u C c (see the definitions (11.21) of F˙ ). Thus in a SWR where F
is constant, the CC characteristics are straight lines, and similarly for the C
characteristics over a SWR where FC is constant. Let us suppose that a simple wave region involving constant F
involves u > 0, so fluid particles move upward in the xt-plane. All of the CC characteristics have positive slope. They may either converge on diverge. In the latter case we have the situation shown in Figure 11.1(a). Since u C c > u, fluid particles must cross the CC characteristics from right to left. Moving along this path, a fluid particle experiences steadily decreasing values of u C c. We assume now that > 1. Since u D 2c
1 C const by the constancy of F
, we see that c, and hence , is decreasing on this path. Thus the fluid is becoming less dense, or expanding. We have in Figure 11.1(a) what we shall call a forward-facing expansion wave. When the CC characteristics emerge from a point, the structure reduces to an expansion fan. Similarly in Figure 11.1 FC is constant in the SWR, and u
c is constant on each C
characteristic. These are again expansion waves, and we term them backward facing. Forward- and backward-facing compression waves are similarly obtained when CC characteristics converge and C
characteristics diverge. E XAMPLE 11.3. We consider the “pullback” movement of a piston in a tube with gas to the right; see Figure 11.2. The motion of the piston is described by 2 < 0. Setting X.t/ D
at2 , x D X.t/, the movement being to the left, dX dt u D
at on the piston. We assume that initially u D 0,  D 0 in the tube. On the C
characteristics, we have u
(11.23)

c D c0 C

2c 
1

D F
D


2c0 
1 ,

or


1 u: 2

Also on CC characteristics we have u C 2c
1 constant. By this fact and (11.23) we 0 is constant there. But since u D up D
at at the piston surface, this have 2uC 2c
1 determines the constant value of u. Let the CC characteristic in question intersect

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11.3. SIMPLE WAVES

165

Ct

C+

x=X(t)

x gas x=0

F IGURE 11.2. Pullback of a piston, illustrating a simple wave region.

the piston path at t D t0 . Then the equation of this characteristic is (11.24)

C1 C1 dx DuCc D u C c0 D
at0 C c0 : dt 2 2

Thus a 2 C1 t0 t C t C c0 .t
t0 /: 2 2 0 Solving the last equation for t0 .X; T / we obtain s " #   1 at. C 1/ at. C 1/ 2 c0 C
C 2a .x
c0 t/ : c0 C (11.26) t0 D a 2 2

(11.25)

x D
a

Then u D
at0 .x; t/ in the simple wave region, c being given by (11.23). 2 c0 . Note that, according to (11.23), c D 0 when t D t  , where at  D 
1 This piston speed is the limiting speed the gas can obtain. For t > t  the piston pulls away from a vacuum region bounded by an interface moving with velocity
at  . If we consider the case of instantaneous motion of the piston with velocity up , the CC characteristics emerge from the origin as an expansion fan. Their equation is C1 x D u C c D c0 C u; (11.27) t 2 so that
 x 2
2c0 : (11.28) uD C1 t To compute the paths .t/ of fluid particles in this example, we solve
 2  d D
2c0 : (11.29) dt C1 t

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166

11. GAS DYNAMICS

A particle begins to move when the rightmost wave of the expansion fan, namely the line x D c0 t, meets the initial particle position. Thus (11.29) must be solved with the initial condition .t0 / D c0 t0 . The solution is   2 C 1 t C1
2c0 t C c0 t0 : (11.30) .t/ D
1
1 t0 For the location of the C
characteristics we must solve  3
3
x dx Du
c D u
c0 D
c0
c0 ; (11.31) dt 2 C1 t with the initial condition x.t0 / D c0 t0 . There results   3 C 1 t 1
2c0 t C c0 t0 : (11.32) x.t/ D
1
1 t0 11.4. Linearized Supersonic Flow We have seen in Chapter 9 that two-dimensional irrotational inviscid homentropic flow of a polytropic gas satisfies the system (11.33)

.c 2
x2 /xx C .c 2
y2 /yy
2x y xy D 0;

2 c 2 D const:
1 This system determines a single nonlinear partial differential equation for . An important problem of inviscid compressible aerodynamics is that of uniform flow past a given rigid body. We are interested here in an approximate analysis of this problem, appropriate to bodies which are thin in the sense that they perturb only slightly the imposed uniform flow. The assumption of small perturbations then leads to a theory of linearized compressible flow. This theory draws upon many of the techniques developed in the linear theory of acoustics. We assume that the air moves with a speed U0 past the body, from left to right in the direction of the x-axis. The potential is taken to have the form

(11.34)

(11.35)

x2 C y2 C

 D U0 x C  0 ;

where jx0 j  U0 . It is easy to derive the linearized form of (11.33), since the second-derivative terms must be primed quantities. The other factors are then evaluated at the ambient conditions, c  c0 , x  U0 . Thus (11.36)

0 0
yy : .M 2
1/xx

Here U0 c0 is the Mach number of the ambient flow. Note that in the linear theory the pressure is obtained from 1 @p 0 @u0 C ; (11.38) U0 @x 0 @x

(11.37)

M D

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11.4. LINEARIZED SUPERSONIC FLOW

167

y

y=y (x) + x y

y-y- (x) dy/dx=m (x) +

U0

dy/dx=m-(x)

α x

F IGURE 11.3. Thin airfoil geometry.

or (11.39)

p 0 
U0 0 x0 :

We may now drop the prime from  0 . The corresponding density perturbation is 0 D
c0
2 U0 x . 11.4.1. Thin Airfoil Theory. We consider first the two-dimensional supersonic flow over a thin airfoil. Linearized supersonic flow results when M > 1, linearized subsonic flow when M < 1. The transonic regime M  1 is distinct and needs to be examined as a special case. The geometry of a thin airfoil is shown in Figure 11.3. We assume that the slopes dy˙ =dx and the angle of attack ˛ are small. In this case dy˙ : (11.40) m˙ .x/ 
˛ C dx Let the chord of the airfoil be l, so we consider 0 < x < l. We note that Z 1 l .mC C m
/dx D
2˛; l 0 Z Z 1 l 2 1 l 2 2 .mC C m2
/dx D
2˛ 2 C .y C y
/dx: (11.41) l 0 l 0 C The analysis now makes use of the following fact, analogous to the our analysis of the linear wave equation in one dimension: the linear operator factors as 

p @ p 2 @ @ @ 2
C : M
1 M
1 (11.42) @x @y @x @y p p Thus  D f .x
y M 2
1/ C g.x C y M 2
1/, where f; g are arbitrary functions. Physical reasoning must be used to choose the right form of solution. In linearized supersonic flow past an airfoil, the disturbances made by the foil propagate out relative to the fluid at the speed of sound but are simultaneously carried downstream with speed U0 . In supersonic flow the foil cannot therefore

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168

11. GAS DYNAMICS

cause disturbances p of the fluid upstream of the body. Consequently, the characteristic lines x ˙ y M 2
1 D const, which carry disturbances away from the foil, must always point downstream. Thus in the half-space above the foil the correct p 2 choice is  D fp .x
y M
1/, while in the space below it the correct choice is  D g.x C y M 2
1/. To determine these functions, we impose the condition that the flow is tangent to the foil surface. Since we are dealing with thin airfoils and small angles, the condition of tangency can be applied, approximately, at y D 0. The approximate tangency conditions are then ˇ p y ˇˇ (11.43) D m .x/ D
M 2
1 U0
1 f 0 .x/; C U0 ˇyD0C ˇ p y ˇˇ (11.44) D m .x/ D M 2
1U0
1 g 0 .x/:
U0 ˇyD0
Consider now the lift and the drag of the foil. These may be computed knowing the pressures 0 .x/ D
U0 0 u0 .x; 0C/ D
U0 0 f 0 .x/; (11.45) pC

Thus 0 D ˙p p˙

(11.46)

U02 0 M2
1


˛ C

0 p
.x/ D
U0 0 g 0 .x/:

 dy˙ ; dx

and Z (11.47)

l

Lift D 0

Z (11.48)

Drag D 0

l

2˛0 U02 l 0 0 ; .p

pC /dx D p M2
1 0 0 .pC mC
p
m
/dx


     Z
 0 U02 l 1 l dyC 2 dy
2 2 2˛ C C dx : Dp l 0 dx dx M2
1 Inviscid theory thus yields a small but positive drag, second order in the slope of the foil surface. We recall that for incompressible potential flow we obtained zero drag (d’Alembert’s paradox). In supersonic flow, the characteristics carry finite signals to infinity. Disturbances are being created so that the rate of increase of kinetic energy per unit time is just equal to the drag times U0 . This drag is often called wave drag because it is associated with characteristics, usually called Mach waves in this context, and these waves propagate to infinity. What happens if we solve for compressible flow past a body in the subsonic case M < 1? In the case of thin airfoil theory, it is easy to see that we must get zero drag. The reason ispthat the equation we are now solving may be written xx C yN yN D 0 where yN D 1
M 2 y. The boundary conditions are at y D yN D 0, so in the new variables we have a problem equivalent to that of an incompressible potential flow.

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11.4. LINEARIZED SUPERSONIC FLOW

169

z-βr < 0 0< z-βr < 1 α

z-βr>1

F IGURE 11.4. Steady supersonic flow past a slender body of revolution. p Here tan ˛ D 1= M 2
1.

In fact, compressible potential flow past any finite body will give zero drag so long as the flow field velocity never exceeds the local speed of sound; i.e., the fluid stays locally subsonic everywhere, and the fluid remains irrotational. In that case no shock waves can form, there is no dissipation, there are no free streamlines, and d’Alembert’s paradox remains. 11.4.2. Slender Body Theory. Another case of interest is the steady supersonic flow past a slender body of revolution. If the ambient flow is along the z-axis in cylindrical polar coordinates z; r, the body we consider is a slender body of revolution about the z-axis; see Figure 11.4. It is easy to show that the appropriate wave equation, coming from the linearized equations (11.49)

0 U0

@u0 @0 C c02 D 0; @z @z

U0

@0 C 0 r  u0 D 0; @z

is p 1 ˇ 2 zz
rr
r D 0; ˇ D M 2
1: r To find a fundamental solution of this equation, note that

(11.50)

(11.51)

a2 xx C b 2 yy C c 2 xx D 0

has, upon rescaling variables, a “sinklike” solution Œ. xa /2 C . yb /2 C . zc /2
1=2 . This holds for arbitrary complex numbers a; b; c. From the fact that r r C 1r r is the two-dimensional symmetric Laplacian in polar coordinates, it follows that a solution of (11.50) is given by (11.52)

S.z; r/ D p

1 z2


ˇ2r 2

:

This is a real solution only if ˇr < z, where S is singular. Since a disturbance originating from the origin should not affect the region ˇ > z, we complete the definition of S by setting (11.53)

S.z; r/ D 0;

ˇr > z:

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170

11. GAS DYNAMICS

Suppose now that we superimpose these solutions by distributing them on the interval .0; l/ of the z-axis, Z

l

f ./ d : p .z
/2
ˇ 2 r 2

D

(11.54)

0

Notice that if we are interested in the solution on the surface z
ˇr D C , then there can be no contributions from values of  exceeding C . We therefore propose a potential

(11.55)

8R z
ˇr p f ./ < 0 d .z
/2
ˇ 2 r 2 .z; r/ D R l : p f ./ d 0 .z
/2
ˇ 2 r 2

if 0 < z
ˇr < l, if z
ˇr > l.

where we now require f .0/ D 0. It can be shown that (11.55) gives us a solution of (11.50) for any admissible f ; see Problem 11.2. Consider now the behavior of  near the body. When r is small the main contribution comes from the vicinity of  D z, so we may extract f .z/ and use the change of variables  D z
ˇr cosh to obtain Z

cosh1 .z=ˇr/

d D f .z/ cosh

(11.56)   f .z/


1

0



z ˇr



  2z f .z/ log : ˇr

Let the body be described by r D R.z/, 0 < z < l. The tangency condition is then ˇ f .z/ dR r ˇˇ 
DR : (11.57) r ˇ U0 rDR dz U0 If A.a/ denotes cross-sectional area, then we have (11.58)

f .z/ D


dA 1 U0 : 2
dz

The calculation of drag is more complicated than for the two-dimensional foil, and we give the result only for the case where R.0/ D R.l/ D 0 and also A0 .0/ D A0 .l/ D 0. Then (11.59)

0 U02 Drag D
4


Z 0

l

Z

l

A00 .z/A00 ./ log jz
jdz d :

0

The form of this equation emphasizes the importance of having a smooth distribution of area in order to minimize wave drag.

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11.5. ALTERNATIVE FORMULATION AND PROOF OF THE DRAG FORMULA

171

11.5. Alternative Formulation and Proof of the Drag Formula To prove (11.59) it is convenient to reformulate the problem in terms of a stream function. We go back to the basic equations for steady homentropic potential flow in cylindrical polar coordinates: (11.60)

@ur @uz
D 0; @z @r

@ruz @rur C D 0; @z @r 2 c 2 D const: u2z C u2r C
1

(11.61)

From the second of (11.60) we introduce the stream function

,

@ @ ; rur D
: @r @z We then expand the equations as follows: ruz D

(11.62)

(11.63)

D

0 U0 2 r C 2

0

 D 0 C  0 ;

;

uz D U0 C u0z ;

and linearize. Making use of  0 =0 D
U0 u0z =c02 , the result is the equation for @2 0 @ 1@ 0
ˇ2 D 0: @r r @r @z 2 Now the boundary condition in terms of the stream function is that on the slender body r D R.z/. Approximately, the condition is

(11.64)

0:

r

equals 0

U0 2 r C 0 .z; 0/  0: 2 We also want no disturbance upstream, so 0 and z0 should vanish on z D 0, r > 0. A solution of this problem may be related to the solution of (11.50) and is given by Z q 0 U0 z
ˇr 0 D
.z
/2
ˇ 2 r 2 A00 d ; 0 < z
ˇr < l: (11.66) 2
0 (11.65)

0

It is easy to see that the equation and upstream conditions are satisfied under the conditions that R.0/ D 0. For the boundary condition we have Z Z 0 U0 z 0 U0 z 0 U0 0 00 .z; 0/ D
.z
/A d  D
A d  D
0 r 2 : (11.67) 2
0 2
0 2 Now the drag is given by Z (11.68)

l

DD

p 0 A0 ./d ;

0

where the linear theory gives p 0 
0 U0 uz . However, it turns out that the ur velocity components become sufficiently large near the body to make a leadingorder contribution to drag. So we need to use the expanded representation 0 2 u C  : (11.69) p 0 
0 U0 u0z
2 r

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172

11. GAS DYNAMICS

Note that uz D

(11.70)

U0 0 1 @ 0
p; 0 r @r 0 c02

which, since p 0 D
0 U0 u0z , implies
ˇ 2 uz D

(11.71) Thus uz D


(11.72)

U0 2


Z

1 @ 0 : 0 r @r

z
ˇr

p

0

A00 ./ .z
/2
ˇ 2 r 2

d :

Similarly, ur D

(11.73)

U0 2
r

Z

z
ˇr 0

A00 ./ .z
/ p d : .z
/2
ˇ 2 r 2

Letting r ! R  0 in (11.73) we have, since R.0/ D 0, ur 

(11.74)

U0 A0 .z/ : 2
R.z/

Also,


 Z z
ˇr 00 U0 00 z A ./
A00 .z
ˇr/ A .z
ˇr/ cosh
1 C p d 2
ˇr .z
/2
ˇ 2 r 2 0
 Z z 00 2z A .z/
A00 ./ U0 00 A log
d : 
2
ˇR z
 0 We are now in a position to compute D: Z Z z 00 Z z
ˇR
0 U02 l 0 2z A .z/
A00 ./ 00
d A .z/ A .z/ log DD 2
0 ˇR z
 0 0 (11.75)  1
1 0 2
A .z/.A / .z/ dz: 4 uz 


After an integration by parts (using A0 .0/ D A0 .l/ D 0, and a cancellation, we have 
Z Z z 00 0 U02 l 0 A .z/
A00 ./ 00 d  dz: A .z/ A .z/ log z
(11.76) DD 2
0 z
 0 Our last step is to show that (11.76) agrees with (11.59). Now if B.z/ D A0 .z/, Z lZ l B 0 .z/B 0 ./ log jz
jd  dz
0

0

Z

l

(11.77)

0 l

D2 0

l

B .z/B.z/ log z dz C 2

D
2 Z

Z

0

B 0 .z/B.z/ log z dz
2

Z

Z

0

0 l

0

Z

z

B.z/ 0

z

B .z/

0

B.z/
B./ d  dz z


B 0 .z/
B 0 ./ d  dz; z


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11.6. TRANSONIC FLOW

173

which proves the agreement of the two expressions. Here we have used d dz

Z

z 0

B.z/
B./ d z


ˇ Z z B.z/
B./ ˇˇ .z
/B 0 .z/
B.z/ C B./ D C d D ˇ z
 .z
/2 0 !z
 Z z 0 .z
/B 0 .z/
B.z/ C B./ z B .z/
B 0 ./ 0 C d D B .z/ C .z
/2 z
 0 0 Z z 0 B.z/ B .z/
B 0 ./ (11.78) D d C : z
 z 0 11.6. Transonic Flow Linearized two-dimensional theory as represented by (11.36) clearly breaks down as M ! 1 in the so-called transonic regime of flight. When M is near 1 we must re-examine the neglected nonlinear terms to determine an equation valid in the transonic regime. We thus refer to (9.65) and (9.66) for a polytropic gas and perturb around an ambient sound speed c0 . Let (11.79)

c D c0 C ıc;

 D c0 x C ı:

Retaining terms of first and second order in ı in (9.65), we have (11.80)

c02 ıyy C 2c0 .ıc
ıx /ıxx
2c0 ıy ıxy  0:

Now ıc can be expressed in terms of ı using the first-order terms of (9.66): 2 ıc C ıx  0: (11.81)
1 The equation for ı then becomes (11.82)


. C 1/ıx ıxx
2ıy ıxy C c0 ıyy  0:

To balance the first- and second-order terms in ı, we must regard the y
partials as of order ı 1=2 , which renders the middle term of (11.82) of order ı 3 and therefore negligible. Thus our approximate transonic equation is (11.83)

. C 1/ıx ıxx D c0 ıyy :

This equation is of elliptic type if ıx < 0 and of hyperbolic type if ıx > 0. It thus is said to be of mixed type. As a nonlinear equation of mixed type, it is difficult to solve in useful geometries, but see Problem 11.3. Even without solutions we can obtain some information from (11.83) concerning transonic similitude. Consider flows past a slender symmetric foil at zero angle of attack given by the contours y D ˙LY.x=L/,   1. We want to compare flows for various Mach numbers, foil thicknesses, and gases. We can eliminate all variable data except a single dimensionless parameter K by the substitutions (11.84) (11.85)

2

1

.x; y/ D L 3 . C 1/
3 c0   .x  ; y  /; 1 y x x D ; y  D Œ. C 1/ 3 : L L

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174

11. GAS DYNAMICS

Indeed, we now have (11.86)

x
x
x
D y
y


 y C .x  ; 0/ D Y 0 .x  /;

the second of which expresses the tangency of the flow to the foil, and also (11.87)

x
D K;

y
D 0

at infinity. Here the dimensionless parameter K, .M
1/. C 1/1=3 ;  2=3 where M is the Mach number of the unperturbed flow, is the determining parameter. Increasing the thickness of the foil by a factor of 8 and increasing M
1 by a factor of 4 results in the same flow in starred variables within the linearized theory. (11.88)

KD

Problem Set 11 (11.1) Find the solution of Burgers wave equation ut C uux D 0 with the initial conditions u.x; 0/ D 0, x < 0; u.x; 0/ D x 2 , 0  x  1; u.x; 0/ D 1, x > 1. (11.2) Suppose that a steady, two-dimensional flow from left to right of a polytropic gas at a Mach number of 2, adjacent to a plane wall, encounters an abrupt 5ı clockwise bend in the wall. Using linearized theory, what is the approximate change in the pressure as the fan is crossed by a fluid parcel? (11.3) Verify by direct differentiation that (11.55) gives us a solution of (11.50) for any smooth f satisfying f .0/ D 0. (11.4) The hodograph plane for two-dimensional flow of a compressible gas utilizes the velocity components u; v in place of x; y as the independent variables. The transformation u.x; y/; v.x; y/ to x.u; v/; y.u; v/ relates first derivatives of each set by inversion formulas, one of which is xu D j
1 vy , j D ux vy
uy vx . Consider the rescaled transonic flow equation x xx D yy in the hodograph plane, using the system uux D vy , uy D vx . Show that, under the hodograph transformation, this system leads to a transonic potential equation of the form uˆvv D ˆuu for a suitable ˆ.u; v/. This linear partial differential equation of mixed type is the Euler-Tricomi equation.

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http://dx.doi.org/10.1090/cln/019/12

CHAPTER 12

Shock Waves 12.1. Scalar Case We have seen that the equation ut C uux D 0 with initial condition u.x; 0/ D 1
x on the segment 0 < x < 1 produces a family of characteristics (12.1)

x D .1
x0 /t C x0 :

This family of lines intersects at .x; t/ D .1; 1/. If the initial condition is extended as ( 1 if x < 0, (12.2) u.x; 0/ D 0 if x > 1, we observe that u.x; t/ D 1
x0 D x
t 1
t for t < x < 1 and that a discontinuity develops in u as a function of x at t D 1. We shall see that, following their formation, conservation laws imply that such discontinuities propagate for later times as shock waves. We consider first the general scalar wave equation in conservation form, ut C .F .u//x D 0. This equation is assumed to come from a conservation law of the form Z b d u dx D F .u.a; t//
F .u.b; t//: (12.3) dt a Suppose now that there is a discontinuity present at position .t/ 2 .a; b/. Then we study the conservation law by breaking up the interval so that differentiation under the integral sign is permitted:
Z   Z b d u dx C u dx D F .u.a; t//
F .u.b; t//: (12.4) dt a  We assume that within each interval of integration u satisfies the general scalar wave equation. Differentiating under the integral and eliminating the time derivatives of u we obtain d D Œu.C; t/
u.
; t/ D F .u.C; t//
F .u.
; t//: (12.5) dt Thus we have an expression for the propagation velocity of the shock wave: (12.6)

ŒF xD d D ; dt Œu xD

where here Œ means “jump in.” This rule for shock velocity must hold for all times exceeding the moment of shock formation. It will be seen that in general the 175 Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

176

12. SHOCK WAVES

t

x F IGURE 12.1. Example of shock formation and propagation.

magnitude of the jump will vary as a function of time following formation. Note that the direction you take the jump is immaterial provided that you do the same in numerator and denominator. E XAMPLE 12.1. Let ut C uux D 0, 8 ˆ ˆ0 ˆ 12 .

The characteristic family of x D .1 C x0 /t C x0 is associated with the interval
1 < x < 0, while x D .1
2x0 /t C x0 is associated with 0 < x < 12 . The shock first occurs at .x; t/ D . 12 ; 12 /. To the right of the shock u D 0, while to the left the former family gives 1Cx x
t D : (12.8) u D 1 C x0 .x; t/ D 1 C 1Ct 1Ct Then   1 1 C1 1 1 d D Œu.
; t/ C u.C; t/ D ;  D : (12.9) dt 2 21Ct 2 2 Thus r 3p 1 C t
1: (12.10) .t/ D 2 We show the xt-diagram for this in Figure 12.1. The jump across q pthe shock folC1 lowing formation is given by u.C; t/
u.
; t/ D 1Ct D 32 = 1 C t. So the shock decays as its velocity slows to 0. 12.1.1. A Cautionary Note. One peculiarity of shock propagation theory is that it is strongly tied to the physics of the problem. Suppose that u  0 solves ut C uux D 0. Then v D u2 will also solve vt C ŒG.v/ x D 0 where G D 23 v 3=2 and v D u2 . In the former case the shock wave propagation speed is 12 .uC C u
/,

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12.2. QUASI-LINEAR SUPERSONIC FLOW

177

while in the latter it is 23 .u2C C uC u
C u2
/=.uC C u
/, which is different. Why the difference? The point is that ut C uux = 0 is based fundamentally on a conservation law involving F .u/ D u2 =2. In actual physical problems the conservation laws will be known and have to be respected. Another way to say this is that equivalent partial differential equations can arise from different conservation laws. It is the proper conservation law that determines the relevant shock velocity. E XAMPLE 12.2. We illustrate this with a simple example from the continuum theory of traffic flow. Consider a single-lane highway with n.x; t/ cars per mile as the traffic density. The cars are assumed to move at a speed determined by the local density, equal to u D U.1
nn0 / where U is the maximum velocity and n0 is the density of full packing and zero speed. The flux of cars is F .n/ D nu D U n.1
nn0 /, and the corresponding conservation of car number yields the PDE nt C ŒF .n/ x D 0. This is equivalent to vt C ŒG.v/ x D 0 where v D n2 and G D U Œv
3n20 v 3=2 . However, the conservation law associated with v; G makes no physical sense. We know how the speed of the cars depends upon n, and conservation of number (if indeed that is what happens) dictates the former law. 12.2. Quasi-linear Supersonic Flow We seek now to modify linearized supersonic flow in two dimensions by nonlinear effects sufficient to capture the formation of weak shock waves. Returning to equations (9.65) and (9.66) set  D Ux C ı, c D c0 C ıc, and retain the quadratic nonlinear terms. This yields    @ @ @ @ C ˇ C ı
2Mc0
1 ıy ıxy (12.11)
ˇ @x @y @x @y C 2.c0
1 ıc
Mc0
1 ıx /ıxx C 2c0
1 ıcıyy  0;

p where again ˇ D M 2
1 where M D cU0 is the free-stream Mach number. We now study the pattern of Mach waves set up in y > 0 by a thin foil lying along the x-axis. This pattern is to be a function predominantly of  D x
ˇy, but will also have a separate weak dependence upon y. With ı D ıc0 ˆ.; / and also ıy D  in (12.11), the y-derivatives become of order ı. Neglecting terms of order ı 2 , the equation reduces to   C1 ˆ ˆ  0: (12.12) ˆ C Mˇ 2

With ˆ D u and  D Mˇ  C1 2 , this takes the form of the Burgers wave equation, (12.13)

u C uu D 0:

We can thus study the supersonic wave pattern as an initial value problem of a familiar kind in the  -plane. Note that as M nears unity from above, the y-variable becomes “stretched,” with the nonlinear wave formation occurring very far above the foil and slightly downstream. Here we make contact with the inviscid weakly

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178

12. SHOCK WAVES

1 0.5 0 5

0

0.5

1

1.5

2

2.5

3

3.5

4

3

2

1

0

0

1

2

3

F IGURE 12.2. The initial value problem for (12.13). Top: the initial condition u.; 0/. Bottom: the solution in the  -plane. The dotted lines indicate the shocks.

nonlinear acoustics described by (10.37). The “sonic boom” heard on the ground from a high-flying supersonic jet is caused by the pressure perturbation of this wave pattern. Typically a fluid parcel experiences first a pressure rise (as it is deflected upward by the forward part of a supersonic foil, for example), followed by a pressure decrease. We will represent this by the triangular wave for u as shown in Figure 12.2. (Recall that in linearized flow a positive pressure perturbation is associated with a negative perturbation in u.). In Figure 12.3 we show the Mach waves in the physical plane for M D 3, D 1:4, and ı D 0:1. Note the formation of two shocks as the wave pattern moves outward. Once these shocks are established, the pressure perturbation as a function of x is an “N” wave, consisting of an abrupt rise, a linear decrease to an opposite and equal negative pressure, and a final abrupt rise to zero. It is this generic disturbance that one hears as a sonic boom.

12.3. The Stationary Normal Shock Wave Shock waves occur in many compressible media, and in particular in air. We shall now consider a stationary planar shock wave in gas dynamics, without viscosity or heat conduction. Our task is to understand how conservation of mass, momentum, and energy constrains the changes that can occur across a normal shock. We assume that constant conditions prevail on either side of the shock, denoted by the subscripts 1; 2; see Figure 12.4.

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12.3. THE STATIONARY NORMAL SHOCK WAVE

F IGURE 12.3. The physical plane corresponding to equation (12.12), with M D 3, D 1:4, ı D 0:1.

shock

1

2

F IGURE 12.4. The stationary normal shock wave.

The conservation laws are: (12.14) (12.15)

Mass: 1 u1 D 2 u2 : Momentum: p1 C 1 u21 D p2 C 2 u22 :

Recall the following statement of conservation of energy: (12.16)

@e C r  .eu/ D
pr  u: @t

From conservation of mechanical energy we also have (12.17)

1 D u2 D
u  rp
u2 r  u: Dt 2 2

Combining these two equations we obtain (cf. 9.28) (12.18)

@E C r  .E C p/u D 0; @t

  1 2 E D eC u : 2

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179

180

12. SHOCK WAVES

At the shock we must therefore require continuity of u.e C p C 12 u2 /, and since u is continuous we have that e C p C 12 u2 D h C 12 u2 is continuous: 1 2 1 u1 D h2 C u22 : 2 2 Let us write m D u as the constant mass flux, and let v D 1 . Then the conservation of energy may be rewritten 1 (12.20) h2
h1 D m2 .v12
v22 /: 2 Also, conservation of momentum becomes p1
p2 : (12.21) m2 D v2
v1 Thus 1 (12.22) h1
h2 D .v1 C v2 /m2 .v2
v1 /; 2 which is equivalent to 1 (12.23) h1
h2 D .v1 C v2 /.p1
p2 /: 2 Written out, this means 1 (12.24) e1
e2 C p1 v1
p2 v2 D .v1 C v2 //p1
p2 / 2 or 1 (12.25) e1
e2 D .v2
v1 /.p1 C p2 /: 2 The relations (12.23) and(12.25), involving the values of the primitive thermodynamic quantities on either side of the shock, are called the Rankine-Hugoniot relations. For a polytropic gas we have 1 pv: (12.26) eD
1 This allows us to write (12.25) as 2 .p1 v1
p2 v2 / C .v1
v2 /.p1 C p2 / D 0 (12.27)
1 or C1 .p1 v1
p2 v2 /
v2 p1 C p2 v1 D 0: (12.28)
1 (12.19)

Energy: h1 C


1 . (Of course,  has no relation whatsoever to We now introduce  by 2 D  C1 viscosity.) According to (12.28), if the state p1 ; v1 exists upstream of a shock, the possible downstream states p; v satisfy

(12.29)


p1 v1 C pv C 2 vp1
2 pv1 D 0

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12.3. THE STATIONARY NORMAL SHOCK WAVE

181

p

(v1 ,p ) 1 v -µ2p 1 µ 2v 1 F IGURE 12.5. The Hugoniot of the stationary normal shock wave.

or (12.30)

.p C 2 p1 /.v
2 v1 / C .4
1/p1 v1 D 0:

The hyperbola in the vp-plane determined by (12.30) is called the Hugoniot curve. We show the Hugoniot for a polytropic gas in Figure 12.5. The allowed transitions from v1 ; p1 to v; p must obey the general thermodynamical law for irreversible physical systems that entropy not decrease. It may then be shown that under this condition the only allowed transition states are upward along the Hugoniot from the point .v1 ; p1 /, as indicated by the arrow in the figure. We prove this now for a polytropic gas. @s D Recall that in the polytropic case s D cv log.pv  / C const, so that cv
1 @v @p
1
1 p @v C v . Computing this on the Hugoniot, we obtain, using pD

(12.31)

.1
4 /p1 v1
2 p1 ; v
2 v1

the relation (12.32)

cv
1


2 .v
v1 /2 ds : D dv vŒ.1 C 4 /vv1
2 .v 2 C v12 /

Since pressure is nonnegative in the polytropic case we have 2 v1  v  
2 v1 ds and consequently see from (12.32) that dv  0 on the Hugoniot. We note that (12.32) shows that for weak shocks, where any quantity changes by only a small fraction ı across the shock, the increase in entropy is of third order in this change, or O.ı 3 /.

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182

12. SHOCK WAVES

12.3.1. Prandtl’s Relation. For a polytropic gas the energy conservation may be rewritten as 1
2 2 p D c : (12.33) hD
1 22 Then conservation of energy then becomes (12.34)

.1
2 /c12 C 2 u21 D .1
2 /c22 C 2 u22  c2 :

Here c is called the critical velocity, since it is the velocity of the gas at the sonic speed u D c. It is an important quantity for the following reason. Constancy of .1
2 /c 2 C 2 u2 implies that .1
2 /.u2
c 2 / D u2
c2 . Since  < 1, this last relation shows that u > c iff u > c and u < c iff u < c. Thus the constant c is the division point between subsonic and supersonic flow along a streamtube where u and c are simultaneously varying. Prandtl’s relation asserts that u1 u2 D c2 :

(12.35)

This implies that on one side of the shock u > c and hence u > c; i.e., the flow is supersonic relative to the shock position, and on the other side it is subsonic. Since density increases as u decreases, the direction of transition on the Hugoniot indicates that the transition must be from supersonic to subsonic as the shock is crossed by a fluid parcel. To prove Prandtl’s relation, observe that .1 C 2 /p D .1
2 /c 2 since
1 : Then, if P D u2 C p, 2 D C1 (12.36) 2 P C p1 D 2 u21 1 C .1 C 2 /p1 D 1 Œ2 u21 C .1
2 /c12 D 1 c2 : Similarly 2 P C p2 D c2 2 . Thus p1
p2 D c2 .1
2 /; or (12.37)

c2 D

p1
p2 p1
p2 1 m2 D 1 D D u1 u2 ; 1   1
2 1 2 2
1 2 1

where we have used (12.21). We now give an example of fitting a shock wave into the solution of a problem in gas dynamics. E XAMPLE 12.3. Suppose that a piston is driven at a velocity up through a tube containing polytropic gas. We seek to see under what conditions a shock will be formed. We assume the shock to have formed and that it moves with fixed velocity U . In treating a moving shock our relations for the stationary shock remain valid provided that u
U replaces u. Thus Prandtl’s relation becomes (12.38)

.u1
U /.u2
U / D c2 D 2 .u1
U /2 C .1
2 /c12 ;

where the gas velocities are relative to the laboratory, not the shock. Rearranging, we have (12.39)

.1
2 /.u1
U /2 C .u1
U /.u2
u1 / D .1
2 /c12 :

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12.4. RIEMANN’S PROBLEM: THE SHOCK TUBE

183

t x=up t 2 shock 1 x F IGURE 12.6. Shock fitting in the piston problem.

Consider now the flow as shown in Figure 12.6. The gas ahead of the shock is at rest, u1 D 0, with ambient sound speed c1 D c0 . Behind the shock the gas moves with the piston speed, u2 D up . Thus we have a quadratic for U : (12.40)

.1
2 /U 2
up U D .1
2 /c02 :

Solving, up C (12.41)

q

U D

u2p C 4.1
2 /c02 2.1
2 /

:

We see that a shock forms for any piston speed. If up is small compared to c0 , the shock speed is approximately c0 , but slightly faster, as we expect. To get the density p behind the shock in terms of the density 0 of the ambient air, we make use of mass conservation, .up
U /p D
U0 , or (12.42)

p D

U 0 : U
up

12.4. Riemann’s Problem: The Shock Tube Riemann considered a basic initial value problem of gas dynamics in one space dimension, where constant conditions of one kind prevail initially in x < 0, u1 ; p1 ; 1 say, and a different state u2 ; p2 ; 2 exists in x > 0. Since, if u1 ¤ u2 , the values must be assumed to be connected by a shock as determined by the Hugoniot, five of these quantities are independent. Riemann’s problem is of practical interest, as we discuss in this section, but it is also fundamental as a tool for dealing with piecewise constant initial conditions. Since smooth functions can be approximated by piecewise constant functions, the Riemann solutions have important applications to numerical schemes as well as to the analysis of global properties of solutions.

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12. SHOCK WAVES

ns

ion

fan

4

nuity

pa

conti

CK

O

SH

1

ct dis conta

3

ex

184

2

F IGURE 12.7. The shock tube Riemann problem.

The most interesting case of Riemann’s problem is when u1 D u2 D 0, i.e., when the gas is initially in two different rest states of density and pressure. This situation is realized in gas dynamics laboratories by the shock tube. This is a long hollow tube, sealed at both ends, and divided into two chambers by a thin metal membrane placed across the tube. The gas on one side of the membrane is compressed relative to that on the other. When the membrane is broken, a flow ensues, leading to the formation of a shock wave moving away from the compressed side. We now study this shock tube problem in detail, formulating the Riemann problem with u1 D u2 D 0 and p2 > p1 . Thus the gas to the left of the membrane is driven to the left, and we anticipate that a shock wave will form and move to the left. In essence, the interface between the two initial states plays the role of a piston. At the same time the movement of the interface appears to the gas on the right as a piston being moved to the left, so an expansion fan must be formed. Thus we propose the sketch of the xt-plane as shown in Figure 12.7. The region is divided into five parts, the two initial states, the regions 3 and 4, and the expansion fan. We now explore the conditions in each. We remark first that the connection through the shock from state 1 to state 3 is totally independent of the connection from state 2 to state 4 through the expansion fan, except for the important conditions that the pressures and the fluid velocities be the same, p3 D p4 ; u3 D u4 . Thus there must be a line of discontinuity separating states, generally of different density and entropy. This discontinuity, separating regions 3 and 4, is a new kind of discontinuity, called a contact discontinuity. We have encountered such a line previously in Example 11.3, marking the vanishing of the density. It is a discontinuity of a property of the fluid attached to a fluid parcel, i.e., a Lagrangian invariant. If u3 D u4 D uc , then the contact discontinuity is the D uc . straight line dx dt Let us first consider the shock, with U < 0 the shock velocity. The velocity upstream of the shock, relative to the shock, is
U and the speed of sound there is c1 . If M1 is the Mach number of the flow upstream of a stationary normal shock, 2 2 3 the ratio of pressure after to pressure before, in our case p p1 , is just .1 C  /M1


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12.4. RIEMANN’S PROBLEM: THE SHOCK TUBE

185

2 ; see Problem 12.4. Thus U2 p3 D .1 C 2 / 2
2 : p1 c1

(12.43)

The velocity relative to the shock in region 3 is
U Cuc , so from the constancy 2 of h C u2 across a normal shock we have, by the Prandtl relation, (12.44)


U.
U C uc / D c2 D .1
2 /c12 C 2 U 2 :

Thus C1 U uc
c12 D 0; (12.45) U 2
2

C1 uc
U D 4

s 

C1 4

2

u2c C c12 :

Consider now the transition from state 2 to state 4 across the expansion fan. 2 c is constant Here the flow is isentropic, and the Riemann invariant F
D u

1 on the C
characteristics. Thus uc


(12.46)

2 2 c4 D
c2
1
1

or
1 uc c4 D1C : c2 2 c2

(12.47)

From the definition of c 2 for a polytropic gas c4 D c2

(12.48)



p4 p2

 1 2 ;

and so   2
1 uc 1 p4 D 1C : p2 2 c2

(12.49)

Since p3 D p4 , (12.43) and (12.49) imply (12.50)

  2 U2 p2
1 uc 1 2 : 1C .1 C  / 2
 D p1 2 c2 c1 2

Looking now at (12.45) and (12.50), we have two equations for U and uc . It can be seen that for p2 > p1 and > 1 there is a unique solution satisfying U < uc < 0, so Figure 12.7 correctly represents the situation. Knowing U and uc we can construct the complete flow as follows. The exD c1 and dx D uc C c4 where pansion fan is bounded by the characteristics dx dt dt 
1 p4 1= c4 D c2 C 2 uc . Also, 4 D 2 . p2 / . Finally, 3 is obtained from mass conservation, 1 U D 3 .U
uc /.

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186

12. SHOCK WAVES

Problem Set 12 (12.1) For the simple scalar wave equation ut C uux D 0 the characteristics .t; a/ ¤ 0. Use this x.t; a/; x.0; a/ D a do not intersect at t; a provided that @x @a fact to show that for the solution of the IVP with initial values u0 .x/, a shock will 0 < 0 somewhere; show also that the time T of the first shock occur only if du dx formation is given by  
1 : T D min x du0 =dx When and where do shocks appear if U0 .x/ D sin x? (12.2) The equations of shallow-water theory are ut C uux C vuy C ghx D 0;

ht C .uh/x D 0:

Here u.x; t/ is the water velocity and h.x; t/ is the height of the free surface. The density  and the acceleration of gravity g are constants. Show that shallow-water theory is mathematically equivalent to gas dynamics of an isentropic, polytropic gas with D 2. Using this analogy, work out the characteristics and Riemann invariants, and solve the following initial value problem, which corresponds to the breaking of a dam onto dry land: ( H > 0; x < 0, u.x; 0/ D 0; h.x; 0/ D 0; x > 0. (12.3) Consider, in shallow-water theory, the analogue of a normal shock. This is called a hydraulic jump. Let u1 ; h1 be the state of the left of the jump, u2 ; h2 on the right, all quantities positive. Show that the jump conditions are 1 1 h1 u21 C gh21 D h2 u22 C gh22 : 2 2 (Recall that p D g.h
y/ in shallow water.) Show that the energy flux on each side of the jump is given by Q D 12 m.gh C u2 /. When the rate of working by the pressure is included, the mechanical energy balance would imply that

(i)

u1 h1 D u2 h2 D m;

1 1 Q1 C gmh1 D Q2 C gmh2 : 2 2 In fact, this last equation does not hold. Given (i), it follows that energy is dissipated. Show that (ii)

(iii)

1 gm .h1
h2 /3 : Q2
Q1 C gm.h2
h1 / D 2 4h1 h2

Since energy can only be lost, conclude that across the jump the height of the water must increase. (Note the analogy with the third-order increase in entropy in gas dynamics.) The energy loss in a hydraulic jump occurs through wave propagation and viscous dissipation in turbulence. This can be easily demonstrated in a simple experiment, by running a small steady stream of water from a faucet onto a flat surface.

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PROBLEM SET 12

187

(12.4) For a normal shock in a polytropic gas, prove the relation
1 p2 ; D .1 C 2 /M12
2 ; 2 D p1 C1 where M1 D

u1 c1

is the Mach number ahead of the shock.

(12.5) Work out the details and indicate on an x; t diagram the explicit solution of the Riemann problem for the following initial states: 1 ; p1 given and positive and u1 D 0 for x < 0. In x > 0, p2 D 4p1 , 2 D 41 , and u2 D 0. Express speeds in terms of c1 (i.e., as Mach numbers). Also, give the densities in the steady state regions on either side of the contact discontinuity. Solve the resulting simultaneous equations numerically. (12.6) Work out the Mach wave diagram shown in Figure 12.3 for M D 1:5, D 1:4, ı D 0:1.

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Supplementary Notes Chapters 1 and 2. Much of the background material in these notes is covered in the excellent classical texts [1, 5, 8]. Our discussion of compressible flow is based for the most part on [2]. The other selected reference texts above contain useful supplementary material. In particular [7] begins with a thorough treatment of fluid kinematics. The classic work of L. Prandtl [9] is a rich source of ideas basic to modern fluid dynamics. Also to be recommended is the review by James Serrin, “Mathematical principles of classical fluid mechanics,” Handbuch der Physik, vol. VIII/I, pp. 125–263, Springer, 1959. While the basic laws of mechanics that underlie our subject are due to Newton (1642–1727), the mathematical formulation of classical fluid dynamics was largely the creation of J. L. d’Alembert (1717–1783), J. and D. Bernoulli (1667–1748 and 1700–1782), and especially L. Euler (1707–1783). An early description of the “Eulerian” viewpoint was put forward in 1749 by d’Alembert, but as Lamb [4] notes, Euler generalized this approach and also introduced the “Lagrangian” form of fluid kinematics. Two of three fundamental papers in hydrodynamics, written by Euler in 1755 and published in 1757, are translated in the recent volume celebrating the 250th anniversary of Euler’s work (see [10]). The reader interested in the history of fluid dynamics will want to consult this valuable reference. In particular, the paper “From Newton’s mechanics to Euler’s equations” by O. Darrigol and U. Frisch in [10] provides an excellent discussion of the emergence of crucial ideas such as that of fluid pressure and the convective component of acceleration from the dynamical models of fluid motion prevalent in that day. Chapter 3. We refer to the monograph of Truesdell [21] for a discussion of the history and evolution of the theory of vorticity and circulation. For an excellent survey of many problems of vortex dynamics, see the monograph of P. G. Saffman [19]. The work of Hermann von Helmholtz (1821–1894) and W. Thomson (Lord Kelvin, 1824–1907) laid out the basic laws of vortex theory, although the vector field r  u appears in d’Alembert’s work. However, d’Alembert wrongly assumed that steady flows of an ideal incompressible fluid were necessarily irrotational. An important outstanding problem of classical fluid mechanics concerns the evolution of the vorticity field in the initial value problem for Euler flow of an ideal incompressible fluid when the initial flow field is smooth. What is the largest ultimate rate of growth in time of the magnitude of the vorticity? How fast does R the ! 2 dV ultimately grow with time? Is it possible that vorticity could become infinite somewhere in a finite time? The question is discussed in papers in [10], where many references to recent work may be found. 189 Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

190

SUPPLEMENTARY NOTES

Chapter 4. Paul R. H. Blasius (1883–1970) was one of the first students of Prandtl. The famous Blasius formulas appeared in 1910 (Z. Math. Phys. 58, pp. 90– 110). Locomotion in an ideal fluid was first studied by P. G. Saffman in 1968 (J. Fluid Mech. 28, pp. 385–389). Calculations of the virtual mass of bodies is discussed in most texts; see especially [1, 5, 8]. Darwin’s theorem was the work of Charles Galton Darwin (1887–1962), a physicist and grandson of Charles Darwin. He derived the theorem in 1953 (Proc. Camb. Phil. Soc. 49, pp. 342–354). Chapter 5. Calculations of drag lift of bodies moving in fluids have been central to our subject since the time of Newton. D’Alembert’s finding of zero drag in potential flow accounted for the early view of theoretical fluid dynamics as unphysical and predictive of absurd conclusions. The reconciliation of d’Alembert’s paradox with the modern theory of viscous fluids is a continuing theme in fluid mechanics; see the excellent review by K. Stewartson, “D’Alembert’s paradox,” SIAM Review, vol. 23, no. 3, pp. 308–343, 1981. It is fair to say that the Kutta-Joukowski theory is incomplete in a larger sense of providing a theory of the inviscid limit. The choice of the proper selection rules for inviscid flows, accounting under various conditions for the effects of small viscosity, remains an unsolved problem of fluid dynamics. Vol. III of [3] contains a section by Witoszy´nski and Thompson on “The theory of single burbling,” a largely forgotten attempt to improve on the lift computed in K-J theory by a modification of conditions at the trailing edge. Recent efforts have tried to generalize the K-J condition to unsteady flow; see D. G. Crighton, Ann. Rev. Fluid Mech. 17, pp. 411– 445, 1985, and also Allan D. Pierce, J. Acoust. Soc. Am. 109, no. 5, pp. 2469–2470, 2001. For a useful collection of articles covering the aerodynamic theory of lift and drag used in the design of early aircraft and many details concerning airfoil design, see [3]. In particular, we remark that the cusped trailing edge of the Joukowski foils is not well suited to wing fabrication, and foils having a finite tangent angle at the trailing edge are desirable. These are provided by the Kármán-Trefftz family. Writing our mapping z D Z C a2 =Z in the form   ZCa 2 z C 2a D ; z
2a Z
a the Kármán-Trefftz theory generalizes this to   ZCa n z C na D : z
na Z
a If n D 2
 , it can be shown that the tangents at the trailing edge form an angle . For an extended discussion of wakes and drag, including experimental results, see [14]. Many additional examples of free-streamline flow calculations can be found in [8]. A discussion of Prandtl’s lifting-line theory as a singular perturbation problem for large aspect ratio is described in [22]. Erich Trefftz (1888–1937) was a German mathematician and aerodynamicist. He studied at Aachen and then at Göttingen, where he was a student of Hilbert and

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SUPPLEMENTARY NOTES

191

Prandtl. His research included key problems of hydrodynamics, but ranged widely in applied and numerical mathematics. Chapters 6 and 7. The concept of stress is due to A. L. Cauchy (1789– 1857) from a paper in 1823, although the physics of fluid friction goes back to Newton. Claude-Louis Navier (1785–1836) and George G. Stokes (1819–1903) developed the equations we study here for Newtonian viscous flow. Jean Louis Marie Poiseuille (1797–1869) was a French physician and physiologist whose experiments in 1840 studied viscous flow through long, thin tubes. J. M. Burgers (1895–1981) was a Dutch physicist whose research included many fundamental aspects of fluid dynamics. A variety of examples of viscous flow may be found in the review by R. Berker, “Intégration des équations du mouvement d’un fluide visqueux incompressible,” Handbuch der Physik, vol. VIII/II, Springer (1963). For a very complete text on creeping flows with emphasis on flows past bodies, see J. Happel and H. Brenner, Low Reynolds number hydrodynamics, 2nd ed., Springer (1983). Stokes noted in 1851 the nonexistence of a solution of his equations appropriate to a circular cylinder in a uniform infinite flow. Carol Wilhelm Oseen (1879–1944) introduced his equations and succeeded in 1910 in showing that they allowed a solution of this problem for small Reynolds numbers. A discussion of the solution may be found in [4]. Oseen went on to publish his famous monograph Neue Methoden und Ergebnisse in der Hydrodynamik, Akademische Verlagsgesellschaft, Leipzig, in 1927, where many problems involving fluid inertia are worked out in the linear setting provided by his equations. The modern analysis of Stokes’ paradox is based on the seminal paper of S. Kaplun, “Low Reynolds number flow past a circular cylinder,” J. Math. Mech. 6, pp. 595–603, 1957; see also [22]. In this work the Stokes equations emerge from an inner limit and the Oseen equations from an outer limit. A formal procedure of matching of the two sets of solutions completely resolves the singular nature of the low Reynolds number limit in the neighborhood of infinity. For a discussion of locomotion in Stokes flow, see the delightful paper of E. M. Purcell, “Life at low Reynolds number,” Am. J. Phys. 45, pp. 3–11, 1977, and also M. J. Lighthill, Mathematical Biofluiddynamics, SIAM (1987), and S. Childress, Mechanics of Swimming and Flying, Cambridge University Press (1981). The scallop theorem applies to a single locomoting body, but not to multiple interacting Stokesian swimmers; see E. Lauga and D. Bartolo, “No many-scallop theorem: Collective locomotion of reciprocal swimmers,” Phys. Rev. E 78, 030901(R) (2008). Also, the constraint posed by the scallop theorem is broken at finite Reynolds number, however small, but can lead for certain symmetric movements to a bifurcation to locomotion at a positive critical Reynolds number; see S. Childress et al., “Symmetry breaking leads to forward flapping flight,” J. Fluid Mech. 506, pp. 147–155 (2004). Chapter 8. For a discussion of early examples of the boundary layer concept, see M. Van Dyke, "Nineteenth-century roots of the boundary-layer idea," Siam Review 36, pp. 415–424, 1994. For discussion of a range of topics in classical boundary layer theory, see [14] and especially H. Schlichting and K. Gersten,

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192

SUPPLEMENTARY NOTES

Boundary-layer theory, 8th ed., Springer (2000). The numerical solution of the Prandtl equations was given by Blasius in his 1907 dissertation; see Zeitschr. Math. Phys. 56, pp. 1–37, 1908. The solutions of Falkner and Skan appeared in Philos. Mag. 12, pp. 865–896, 1931. For an interesting alternative treatment of the Blasius and Falkner-Skan problems, see H. Weyl, “On the differential equations of the simplest boundary-layer problems,” Ann. of Math. (2) 43, pp. 381–407, 1942. The one-dimensional boundary layer model of Section 8.3 appeared in Fluid dynamics, K. O. Friedrichs and R. von Mises, Springer (1971). The modern treatment of fluid boundary layer theory as a singular perturbation problem is discussed in [22]; see also [16]. A landmark paper by S. Kaplun, “The role of coordinate systems in boundary-layer theory,” Z. Angew. Math. Physik 5, pp. 111–135, 1954, showed that there exist optimal coordinate systems for capturing the flow due to displacement thickness within the boundary layer limit. The triple-deck theory emerged from the work of many researchers, including Goldstein, Kaplun, Messiter, Neiland, Smith, Stewartson, and Sychev. For reviews of the triple-deck theory of laminar separation, see F. T. Smith, “On the high Reynolds number theory of laminar flows,” IMA J. Appl. Math 28, pp. 207–281, 1982, and R. E. Meyer, “A view of the triple deck,” SIAM J. Appl. Math. 43, pp. 639–663, 1983. For a survey of rigorous mathematical results for the Prandtl equations, see K. Nickel, “Prandtl’s boundary-layer theory from the viewpoint of a mathematician,” Ann. Rev. Fluid Mech. 5, pp. 405–428, 1973. For the origins of Prandtl-Batchelor theory, see L. Prandtl, “Über Flüssigkeitsbewegung bei sehr kleiner Reibung,” Gesammelte Abhandlungen II, pp. 575–584, Springer (1961), and G. K. Batchelor, “A proposal concerning laminar wakes behind bluff bodies at large Reynolds number,” J. Fluid Mech. 1, pp. 388–398, 1956.

Chapters 9–12. The general solution of the initial value problem for the wave equation in three dimensions was given by S. D. Poisson (1781–1840) in 1820 and was obtained in the form given here by G. Kirchhoff (1824–1887) in 1876; see Rayleigh’s The theory of sound, vol. II, Dover, p. 97 (1945). Sir James Lighthill (1924–1998) and Theodore von Kármán (1881–1963) are two giants of twentiethcentury fluid mechanics who contributed substantially to the development of the theory of compressible flow. The books of Lighthill [17] and Whitham [25] are excellent sources for the theory of linear and nonlinear waves in compressible fluids. In connection with viscous and weak nonlinear effects on sound waves, we mention the important review by Lighthill, “Viscosity effects in sound waves of finite amplitude,” in Surveys in mechanics, pp. 250–351, Cambridge University Press (1956). The theory of aerodynamic generation of sound, introduced in Problem 10.7, was developed by Lighthill in the 1950s. The theory was subsequently developed and expanded by J. E. Ffowcs Williams and others, and an excellent survey is given by Ffowcs Williams in “Hydrodynamic noise,” Annu. Rev. Fluid Mech. 1, pp. 197–222, 1969, and “Aeroacoustics,” Annu. Rev. Fluid Mech. 9, pp. 447– 468, 1977. For a detailed account of linearized compressible flow, see [24]. For a discussion of transonic similitude, see W. D. Hayes, “Pseudotransonic similitude

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SUPPLEMENTARY NOTES

193

and first-order wave structure,” J. Aero. Sci. 21, pp. 721–730, 1954. Hayes (1919– 2001) was a student of von Kármán at CalTech. His 1947 dissertation, “Linearized supersonic flow” (see also his “Linearized supersonic flows with axial symmetry,” Quart. Appl. Math. 4, pp. 255–261, 1946), contained the theoretical basis for the area rule, later used in the design of transonic aircraft. Hayes’s monograph [15] contains a full analysis of the Hugoniot curve under various physical assumptions. For discussion of the water waves, and in particular a nice treatment of shallowwater theory and its applications, see the book by J. J. Stoker [20].

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Bibliography Selected Reference Texts [1] Batchelor, G. K. An introduction to fluid dynamics. Cambridge University Press, Cambridge, 1967. [2] Courant, R.; Friedrichs, K. O. Supersonic flow and shock waves. Interscience, New York, 1948. [3] Durand, W. F., ed. Aerodynamic theory. 6 vols. Dover, New York, 1963. [4] Lamb, H. Hydrodynamics. 6th ed. Cambridge University Press, Cambridge, 1932. [5] Landau, L. D.; Lifshitz, E. M. Course of theoretical physics. Vol. 6. Fluid mechanics. 2nd ed. Butterworth-Heinemann, Oxford, 2003. [6] Lighthill, M. J. An informal introduction to theoretical fluid mechanics. Clarendon, Oxford, 1986. [7] Meyer, R. E. Introduction to mathematical fluid dynamics. Pure and Applied Mathematics, 24. Wiley-Interscience, New York, 1971. [8] Milne-Thomson, L. M. Theoretical hydrodynamics. 5th ed. Dover, New York, 1996. [9] Prandtl, L. Essentials of fluid dynamics. Blackie, London, 1952.

Other References [10] Eyink, G.; Frisch, U.; Moreau, R.; Sobolevski˘ı, A., eds. Euler equations: 250 Years On— Proceedings of an International Conference, Aussois, France, 18–23 June 2007. Phys. D 237, nos. 14–17, 2008. [11] Flanders, H. Differential forms with applications to the physical sciences. Academic, New York–London, 1963. [12] Garabedian, P. R. Partial differential equations. Wiley, New York–London–Sydney, 1964. [13] Germain, P. Mécanique des milieux continus. Masson, Paris, 1962. [14] Goldstein, S., ed. Modern developments in fluid dynamics. 2 vols. Clarendon, Oxford, 1938. [15] Hayes, W. D. Gas dynamic discontinuities. Princeton University Press, Princeton, N.J., 1960. [16] Lagerstrom, P. A. Matched asymptotic expansions. Ideas and techniques. Applied Mathematical Sciences, 76. Springer, New York, 1988. [17] Lighthill, M. J. Waves in fluids. Cambridge University Press, Cambridge, 1978. [18] Majda, A. J.; Bertozzi, A. L. Vorticity and incompressible flow. Cambridge Texts in Applied Mathematics, 27. Cambridge University Press, Cambridge, 2002. [19] Saffman, P. G. Vortex dynamics. Cambridge Monographs on Mechanics and Applied Mathematics. Cambridge University Press, New York, 1992. [20] Stoker, J. J. Water waves: The mathematical theory with applications. Pure and Applied Mathematics, 4. Interscience, New York–London, 1957. [21] Truesdell, C. The kinematics of vorticity. Indiana University Publications in Science Series, 19. Indiana University Press, Bloomington, 1954. [22] Van Dyke, M. Perturbation methods in fluid mechanics. Applied Mathematics and Mechanics, 8. Academic, New York–London, 1964. [23] von Mises, R. Mathematical theory of compressible fluid flow. Applied Mathematics and Mechanics, 3. Academic, New York, 1958. 195 Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

196

BIBLIOGRAPHY

[24] Ward, G. N. Linearized theory of steady high-speed flow. Cambridge University Press, Cambridge, 1955. [25] Whitham, G. B. Linear and nonlinear waves. Pure and Applied Mathematics. WileyInterscience, New York–London–Sydney, 1974.

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Index

Absolute temperature, 141 Acoustics, 146, 151 weakly nonlinear, 155 Airfoil unsteady motion of an, 79 Airship flow around an, 54 Angle of attack, 75 Angle of attack effective, 92 Apparent angular momentum, 62 Apparent mass, 57 Apparent mass matrix, 57, 59 alternative representation, 60 of an elliptic cylinder, 60 time-dependent, 62 Archimedean principle, 18 Aspect ratio, 91 Axisymmetric flow, 38 with swirl, 40 without swirl, 40

two-dimensional jet solution, 130 Burgers vortex, 104 Burgers’ equation, 158 Bursting balloon problem, 152 Butler sphere theorem, 54 Camber, 78 Cauchy-Riemann equations, 47 Characteristic velocity, 163 Characteristics, 151, 161, 164 Chord, 79 Circle theorem, 48 Circulation, 29, 46, 58, 75 Coefficient of lift, 79 Complex potential, 47 Complex variables, 47 Configuration space, 119 Conformal map, 48, 75 Conservation of momentum, 16 of mass, 13 Continuity equation, 14 Convection identity, 15 Convection theorem, 8 Coordinate Lagrangian, 1 Couette flow, 101 Critical velocity, 182 Crocco’s relation, 145

Barotropic fluid, 30, 140 Bernoulli function, 145–147, 150 Bernoulli theorem for steady flow, 19 for unsteady flow, 22 Biot-Savart law, 92 Blasius solution, 125, 127 Blasius’ theorem, 49, 83 Body force, 16 Bound circulation, 80 Boundary conditions, 22 Boundary layer, 123 Blasius solution, 125, 127 Falkner-Skan solutions, 129 matching, 132 on an aligned flat plate, 123 Prandtl equations, 125 separation, 125

D’Alembert’s paradox, 52, 60, 168 Darwin’s theorem, 69 Delta function, 51, 152 Derivative material, 6 Deviatoric stress tensor, 99 general form of, 99 Displacement thickness, 127 Domain of dependence, 151 197

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198

Downwash, 92, 93 Drag induced, 93 in the Kármán vortex street model, 81 in the Oseen model for a flat plate, 137 of an aligned flat plate, 128 of a circular cylinder, 107 of a flat plate in Kirchhoff’s model, 87 of a slender body in supersonic flow, 170 of a sphere in Stokes flow, 116 on a Rankine fairing, 52 Drift, 65 area or volume, 66 calculations for a circular cylinder, 66 Dynamical similarity, 106 Energy, 139 free, 144 internal, 141 kinetic, 139 mechanical, 139 Energy equation, 143, 149 Enthalpy free, 144 total, 145 Entropy, 141 increase across a normal shock, 181 Equation of continuity, 14 Equation of state, 141 Euler flow, 17 Euler’s equations, 17 Euler-Tricomi equation, 174 Eulerian description, 2 Eulerian realm, 111 Expansion wave, 164 Expansion fan, 162, 165 Falkner-Skan family of boundary layers, 129 Favorable pressure gradient, 129 First law of thermodynamics, 141 Flapping flight quasi-steady theory of, 80 Flat plate flow normal to a, 85 Flat plate foil, 75 Flow compressible, nondissipative, 145 down an incline, 103 homentropic, 145 irrotational, 147 isentropic, 145 isentropic, polytropic gas, 147 stagnation point, 105

INDEX

with circular streamlines, 104 Fluid acceleration, 6 barotropic, 30 boundary conditions for real case, 22 compressible, 17 continuum, 1 ideal, 16 incompressible, 6, 17 of constant density, 18 parcel, 1 particle, 1 velocity, 1 Force experienced by a cylinder in two-dimensional potential flow, 50 on a cylinder in the presence of a source, 50 Fourier’s law of heat conduction, 143 Free energy, 144 Free enthalpy, 144 Free streamline theory, 85 Gas constant, 141 Gases, 1 Harmonic flows, 45 Helmholtz’ laws, 32 Hodograph method, 86 Hugoniot curve, 181 Huygens’ principle, 155 Hydraulic jump, 186 Hydrostatics, 18 Ideal fluid, 16 Ideal gas, 141 Incline flow down an, 103 Incompressible fluid, 6 Induced drag, 93 Internal energy, 141 Intrinsic coordinates, 20 Invariant material, 6 Irrotational flow, 21, 27 Isotropicity of pressure, 17 Isotropy of the stress tensor, 99 Jacobian, 18 Jacobian matrix, 5 Joukowski airfoils, 77

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INDEX

Kármán-Trefftz family of airfoils, 190 Kármán vortex street, 81 Kelvin’s theorem, 30, 80 in a compressible fluid, 146 Kinematic viscosity, 100 Kinetic energy calculated for potential flow past a body, 58 of fluid about a moving body, 57 Kirchhoff flow, 125 Kirchhoff model, 85 Kutta-Joukowski condition, 76, 77, 81 theorem, 77 Lagrangian coordinate, 1 Laminar jet cylindrical, 137 two-dimensional, 130 Leading edge suction, 77 Lie derivative, 9 Lift in three dimensions, 88 Lift coefficient, 79 Lifting line theory, 88 Limit process, 73 Linearized supersonic flow, 166 Liquids, 1 Locomotion by a deformable body in potential flow, 62 by recoil, 63 by squirming, 63 Mach cone, 155 Mach number, 155, 166 Mach waves, 168 Mass conservation of, 13 Eulerian form, 13 Lagrangian form, 14 Matching, 132 Material derivative, 6 Maxwell relations, 144 Mechanical energy, 139 Moment acting on a body in potential flow, 62 on a Joukowski airfoil, 78 Momentum conservation of Eulerian form, 16 Lagrangian form, 16, 18 Moving sound source, 155 Navier-Stokes equations, 100, 128

199

in cylindrical coordinates, 102 Newtonian viscous fluid, 97, 98 No-slip condition, 101 Nonuniqueness of flow past a circular cylinder, 47 Oseen equations, 118, 137 Particle path, 1, 10, 165 Perfect gas, 141 Peristaltic pump, 43 Piston shock formation by a, 182 Point vortices equations for a system of, 49 Point vortex, 3, 47 Poiseuille flow, 102 entry effect, 103 Polytropic gas, 143, 180 isentropic flow of, 147, 162 Potential flow, 45 in three dimensions, 51 of constant density, 21 past a circular cylinder, 23 past a sphere, 55 uniqueness of, 46 Prandtl lifting line theory of, 88 Prandtl boundary layer equations, 125 Prandtl’s relation, 181 Prandtl-Batchelor theory steady flow, 134 time dependence, 135 Pressure, 16, 17 favorable gradient, 129 Quasi-steady flow, 80 Range of influence, 151 Rankine fairing, 52 Rankine’s combined vortex, 35 Rankine-Hugoniot relations, 180 Rayleigh problem, 101 Reciprocal theorem for Stokes flow, 122 Reversible system, 141 Reynolds number, 111 typical values, 108 Rheology, 97 Riemann invariant, 163 Riemann problem, 183 Rotational flow, 27 Scallop theorem, 121 Second viscosity, 99

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200

Self-similarity of the Blasius boundary layer, 126 Separation, 125, 130, 133 from the leading edge of an airfoil, 76 Shallow-water theory, 186 Shear flow, 28 Shock fitting, 183 Shock tube, 183 Shock velocity dependence on conservation law, 176 scalar case, 175 Shock wave stationary normal, in gas dynamics, 178 Shock waves scalar case, 175 Similarity, 106 Simple wave, 163 produced by a piston, 164 region, 163 Simply connected domain, 46 Single-valued function, 46 Singular perturbation, 118 Slender body theory, 169 Solid-body rotation, 28 Sonic boom, 178 Sound, 45, 146, 151 Sound waves one-dimensional, 151 Source in three dimensions, 51 Specific enthalpy, 144 Specific entropy, 141 Specific heats, 142 Sphere potential flow in presence of a source, 55 potential flow past a, 55 Stagnation point, 2, 75 Stagnation point flow, 105 Stall, 76 Standard atmosphere, 148 Starting vortex, 80 Stokes equations, 113 solutions of, 113 Stokes flow, 111 fundamental solution, 113 past a sphere, 114 time reversibility, 119 uniqueness, 114 Stokes relation, 100 Stokes stream function, 52 Stokes’ paradox, 117 Stokes’ theorem, 29 Stokesian realm, 111

INDEX

Streak line, 4 Streamline, 3 instantaneous, 3 Stream function, 8, 37 Stokes, 40, 52 Stress tensor, 16, 17, 97 deviatoric , 99 Stretched variable, 124 Supersonic flow quasilinear, 177 Symmetry of deviatoric stress tensor, 99 Thermal convection, 150 Thermodynamic variables extensive and intensive, 141 Thermodynamics classical, 141 first law of, 141 Thin airfoil theory subsonic flow, 168 supersonic flow, 167 Three-dimensional wing, 89 Thrust in flapping flight, 80 Time reversibility, 119 Time-reversal symmetry, 120, 121 Traffic flow, 177 Transonic flow, 173 Transonic similitude, 173 Triple deck, 133 Turbulence, 107 Uniqueness of potential flow, 46 Unsteady motion of an airfoil, 79 Variables Eulerian, 2 Vector field material, 9 Velocity critical, 182 derivative matrix, 97 supersonic, 155 Velocity field associated with a given vorticity field, 34 local analysis of, 27 solenoidal, 7 Viscosity, 98 kinematic, 100 limit of small, 73 second, 99

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INDEX

Viscous stress tensor, 139 Von Mises coordinates, 138 Vortex Burgers, 104 Vortex dipole, 37 Vortex force, 27 Vortex lines, 27 Vortex shedding, 73 Vortex sheet, 88 strength, 88 Vortex street, 81 Vortex tube, 27 Vorticity shed, 80, 90 shedding of, 73 Vorticity equation, 30 Lagrangian form, 31 Vorticity field, 27 Wake energy flux in, 74 Water waves, 23, 45 Wave drag, 168 Wave equation, 151, 161 d’Alembert’s solution in one dimension, 151 fundamental solution in three dimensions, 152 Kirchhoff’s solution of initial value problem in three dimensions, 153 Wingspan, 89

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201

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Titles in This Series 19 Stephen Childress, An introduction to theoretical fluid mechanics, 2009 18 Percy Deift and Dimitri Gioev, Random matrix theory: Invariant ensembles and universality, 2009 17 Ping Zhang, Wigner measure and semiclassical limits of nonlinear Schr¨ odinger equations, 2008 16 S. R. S. Varadhan, Stochastic processes, 2007 15 14 13 12

Emil Artin, Algebra with Galois theory, 2007 Peter D. Lax, Hyperbolic partial differential equations, 2006 Oliver B¨ uhler, A brief introduction to classical, statistical, and quantum mechanics, 2006 J¨ urgen Moser and Eduard J. Zehnder, Notes on dynamical systems, 2005

11 V. S. Varadarajan, Supersymmetry for mathematicians: An introduction, 2004 10 Thierry Cazenave, Semilinear Schr¨ odinger equations, 2003 9 Andrew Majda, Introduction to PDEs and waves for the atmosphere and ocean, 2003 8 Fedor Bogomolov and Tihomir Petrov, Algebraic curves and one-dimensional fields, 2003 7 S. R. S. Varadhan, Probability theory, 2001 6 Louis Nirenberg, Topics in nonlinear functional analysis, 2001 5 Emmanuel Hebey, Nonlinear analysis on manifolds: Sobolev spaces and inequalities, 2000 3 Percy Deift, Orthogonal polynomials and random matrices: A Riemann-Hilbert approach, 2000 2 Jalal Shatah and Michael Struwe, Geometric wave equations, 2000 1 Qing Han and Fanghua Lin, Elliptic partial differential equations, 2000

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An Introduction to Theoretical Fluid Mechanics STEPHEN CHILDRESS

This book gives an overview of classical topics in fluid dynamics, focusing on the kinematics and dynamics of incompressible inviscid and Newtonian viscous fluids, but also including some material on compressible flow. The topics are chosen to illustrate the mathematical methods of classical fluid dynamics. The book is intended to prepare the reader for more advanced topics of current research interest.

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