Advanced Engineering Mathematics with Modeling Applications (Solutions, Instructor Solution Manual) [1 ed.] 084939533X, 9780849395338

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SOLUTIONS MANUAL FOR Advanced Engineering Mathematics with Modeling Applications

by S. Graham Kelly

SOLUTIONS MANUAL for

ADVANCED ENGINEERING MATHEMATICS WITH MODELING APPLICATIONS by

S. GRAHAM KELLY

PREFACE This manual provides solutions to 148 problems from the text Advanced Engineering Mathematics with Modeling Applications. Most problems are solved in detail, reaching a final answer. Only a few solutions provide partial answers. Consistent with the book MATHCAD is used as a tool to solve problems which require a large amount of computation. The book is oriented toward those with a bachelor’s degree in engineering with an interest in advanced mathematics and those with a degree in mathematics with an interest in engineering applications. The book is well suited for use as a textbook in an advanced engineering mathematics course taught within either an engineering program or an applied mathematics program. The end-of-chapter problems were chosen to help readers achieve the two goals which I stated in the book’s preface. These goals are to prepare readers for graduate coursework in engineering and to prepare readers for research in engineering. The former goal is achieved by presenting mathematics that is used in graduate coursework in a consistent framework. Most engineering classes will present the mathematics needed to solve the problems specific to the topic without presenting the underlying theory for the mathematics. A background in this theory is often necessary to understand engineering literature. There are several changes to equations and conditions in certain problems from those printed in the text. The fault is solely mine for missing these changes when reviewing the proofs. The enclosed solutions reflect these changes. I am also sure that I have made errors during solution of these problems. There will be typographical errors and computational errors. I tell my students that they are more capable using calculators than I. However I apologize for these errors in advance and hope that you will contact me when you find one. In addition please contact me if you wish to learn of errata in the book and solutions manual. ii

The changes which I have made to the problem statements in preparing this solutions manual are • Problem 2.21: The inner product should be defined as , 3 2 . • Problem 3.14: The equation for the rate at which heat is transferred over the surface by convection should be . 2 ℓ • Problem 3.25: The first boundary condition should be 0.5 2. • Problem 3.29: The first boundary condition should be 2 1. • Problems 3.30 and 3.31: The first boundary condition in each should be 0.5 3. • Problems 3.37 and 3.38: The problems should ask only to find one solution. • Problem 5.28: The end at x=1 is attached to a linear spring. • Problem 5.29: The end at x=0 is fixed. • Problem 5.30: Equation (a) should be 0. • Problem 5.31: The second boundary condition should be 0. • Problems 5.35-5.44: The first boundary condition, Equation (b) should be 2 0. • Problem 6.2: The boundary condition specified in the ,0 0. problem statement should be • Problem 6.5: The last boundary condition should be ,1 1.2 Θ , 1 0. • Problem 6.7: The first boundary condition should be Θ 0, , 0. iii

• Problem 6.8: The first boundary condition should be Θ 0, , 1. • Problem 6.9: The first boundary condition should be Θ 0, , 0. • Problems 6.17-6.24: The governing partial differential . equation is • Problem 6.23: The governing partial differential equation should be . Thank you for using this book. I would appreciate any comments.

S. Graham Kelly [email protected]

December, 2008

iv

TABLE OF CONTENTS I.

INTRODUCTION Problem 1.1 Problem 1.2 Problem 1.3 Problem 1.4 Problem 1.5 Problem 1.6 Problem 1.7 Problem 1.8 Problem 1.9 Problem 1.10 Problem 1.11

1 2 6 11 14 16 18 20 22 24 25 26

II.

LINEAR ALGEBRA Problem 2.1 Problem 2.2 Problem 2.3 Problem 2.4 Problem 2.5 Problem 2.6 Problem 2.7 Problem 2.8 Problem 2.9 Problem 2.10 Problem 2.11 Problem 2.17 Problem 2.18 Problem 2.19 Problem 2.20 Problem 2.21 Problem 2.22 Problem 2.23

31 32 33 34 35 36 37 38 39 41 42 43 44 46 47 48 49 50 51 v

Problem 2.24 Problem 2.25

52 53

III. ORDINARY DIFFERENTIAL EQUATIONS 55 Problem 3.1 56 Problem 3.3 57 Problem 3.4 58 Problem 3.5 59 Problem 3.6 60 Problem 3.7 61 Problem 3.8 62 Problem 3.9 63 Problem 3.10 65 Problem 3.11 66 Problem 3.12 67 Problem 3.13 69 Problem 3.14 71 Problem 3.15 74 Problem 3.16 79 Problem 3.18 81 Problem 3.19 84 Problem 3.20 86 Problem 3.21 88 Problem 3.22 91 Problem 3.23 93 Problem 3.24 94 Problem 3.25 95 Problem 3.27 96 Problem 3.28 97 Problem 3.29 98 Problem 3.33 99 Problem 3.36 101 Problem 3.37 103 Problem 3.38 105 Problem 3.39 107 vi

Problem 3.40 Problem 3.41 Problem 3.42 Problem 3.43

108 109 110 111

IV. VARIATIONAL METHODS Problem 4.1 Problem 4.2 Problem 4.3 Problem 4.4 Problem 4.5 Problem 4.6 Problem 4.7 Problem 4.8 Problem 4.9 Problem 4.10 Problem 4.11 Problem 4.12 Problem 4.13 Problem 4.14 Problem 4.15 Problem 4.16 Problem 4.17 Problem 4.18 Problem 4.19 Problem 4.20 Problem 4.22

112 113 116 118 120 123 125 127 131 133 136 139 142 145 148 151 154 157 160 164 168 172

V.

176 177 180 184 186 188 191

EIGENVALUE PROBLEMS Problem 5.1 Problem 5.2 Problem 5.3 Problem 5.4 Problem 5.5 Problem 5.6 vii

Problem 5.7 Problem 5.8 Problem 5.9 Problem 5.10 Problem 5.11 Problem 5.12 Problem 5.13 Problem 5.14 Problem 5.15 Problem 5.16 Problem 5.17 Problem 5.18 Problem 5.19 Problem 5.22 Problem 5.23 Problem 5.24 Problem 5.25 Problem 5.26 Problem 5.27 Problem 5.28 Problem 5.29 Problem 5.30 Problem 5.31 Problem 5.33 Problem 5.35 Problem 5.36 Problem 5.37 Problem 5.38 Problem 5.39 Problem 5.40 Problem 5.41 Problem 5.43 Problem 5.44 Problem 5.45

194 196 199 201 203 205 207 210 213 217 219 220 222 224 228 233 235 237 239 241 243 245 250 253 255 258 260 265 270 276 277 280 284 290 viii

VI. PARTIAL DIFFERENTIAL EQUATIONS Problem 6.1 Problem 6.2 Problem 6.3 Problem 6.4 Problem 6.5 Problem 6.6 Problem 6.7 Problem 6.8 Problem 6.9 Problem 6.10 Problem 6.12 Problem 6.13 Problem 6.17 Problem 6.18 Problem 6.19 Problem 6.20 Problem 6.22 Problem 6.23 Problem 6.24 Problem 6.26 Problem 6.27

ix

291 292 295 298 302 306 309 312 318 324 328 330 334 337 341 345 349 353 358 361 364 367

CHAPTER 1 INTRODUCTION

1

1.1 An elastic bar has a static displacement due to an applied axial load F. The force is suddenly removed resulting in free oscillations of the bar. Derive the problem governing the displacement u(x,t) of a particle along the axis of the bar. (a)Specify the governing differential equation as well as all boundary and initial conditions necessary to solve for u(x,t). (b) Non-dimensionalize the problem by introducing x* =

x t and t * = where T is L T

to be chosen for convenience Solution: (a) Assume the displacement is a function only of x, a coordinate along the axis of the bar and time t. This assumption implies that every particle the same distance from the left end has the same displacement. Consider a differential element of a slice of thickness dx. The area of the slice is A(x). Let σ represent the normal stress acting on the face of the element. The resultant force acting on the face is σA , as illustrated on the free-body diagram of Figure 1.1a. The mass of the differential element is

dm = ρdV = ρAdx . Let u ( x, t ) represent the displacement of the left face of the element .Application of Newton’s second law to the free-body diagram of the slice gives − (σA)x + (σA)x + dx = ρAdx

∂ 2u ∂t 2

(a)

The force on the right face can be expressed as

(σA)x+ dx = (σA)x +

∂ (σA)dx ∂x

(b)

Substitution of Equation b into Equation a leads to − (σA)x + (σA)x +

2 ∂ (σA)dx = ρAdx ∂ u2 ∂x ∂t

∂ ∂ 2u (σA ) = ρA 2 ∂x ∂t

(c)

2

The normal strain is defined as ε =

∂u . If the material follows a linear ∂x

constitutive equation, σ = Eε where E is the elastic modulus, then ∂u ∂x

(d)

∂ ⎛ ∂u ⎞ ∂ 2u ⎜ EA ⎟ = ρA 2 ∂x ⎝ ∂x ⎠ ∂t

(e)

σ =E Use of Equation (d) in Equation (c) leads to

Equation e governs the displacement of the bar. If the bar is uniform and the elastic modulus is constant, Equation e reduces to ∂ 2u ρ ∂ 2u = ∂x 2 E ∂t 2

(f)

The bar is constrained from motion at x=0, thus

u (0, t ) = 0

(g)

Once the force is removed the end of the bar at x=L is free with no stress E

∂u ( L, t ) = 0 ∂x

(h)

The initial condition is developed from the static displacement of a bar with a force F at its free end,

u ( x,0) =

F x AE

(i)

The initial velocity of every particle on the bar is zero, thus

∂u ( x,0) = 0 ∂t (b) Non-dimensional variables are introduced according to

3

(j)

x* x= L Noting that

u* u= L

t* t= T

(k)

∂ ∂ dt * 1 ∂ ∂ ∂ dx * 1 ∂ = = = * = and use of Equation k in ∂t ∂t * dt T ∂t * ∂x ∂x dx L ∂x *

Equation e leads to

( )

( )

ρ 1 ∂ 2 Lu * 1 ∂ 2 Lu * = E T 2 ∂t *2 L2 ∂x *2 ∂ 2u * ρL2 ∂ 2 u * = ∂x *2 ET 2 ∂t *2

(l)

For convenience choose

ρL2 ET 2

=1

T=L

ρ E

(m)

Use of Equation m in Equation l and dropping the *s from non-dimensional variables leads to

∂ 2 u ∂u = ∂x 2 ∂t 2

(n)

The non-dimensional forms of the boundary conditions are

u (0, t ) = 0 ∂u (1, t ) = 0 ∂x

(o) (p)

Nondimensionalization of the initial conditions leads to

Lu * ( x,0) =

F (Lx * ) AE

(q)

Dropping the *s from the non-dimensional variables in Equationq leads to

u ( x,0) =

F x AE

4

(r)

The second non-dimensional initial condition is

∂u ( x,0) = 0 ∂t

5

(s)

1.2 The elastic bar of Figure P1.2 is fixed at x=0 and has a particle of mass m attached at x=L. The particle is attached to a spring in parallel with a viscous damper. The differential equation governing the displacement of a particle along the axis of the bar is

EA

∂ 2u ∂ 2u ρ = A ∂x 2 ∂t 2

(a)

(a) Specify the boundary conditions at x=0 and x=L. (b) The particle of mass m is displaced a distance δ to the right, held in this position and released. Specify the initial conditions for the resulting vibrations. (c) Introduce non-dimensional variables, x* =

x * u t , u = and t * = . Substitute the L L T

non-dimensional variables into Eq. (a), the boundary conditions and initial conditions. Choose T such that the resulting partial differential equation is

∂ 2u ∂ 2u = ∂x 2 ∂t 2

(b)

where all variables in Eq. (b) are non-dimensional. (d) Identify and physically define all non-dimensional parameters which appear in the non-dimensional formulation of the boundary conditions. (e) If the inertia of the elastic bar is small compared to the inertia of the particle the system of Fig. P1.2a can be modeled by the discrete system of Fig. P1.2b. The inertia effects of the bar are approximated by adding a particle of mass

1 ρAL is 3

added to the existing particle. Derive the differential equation governing the displacement y(t) of the particle. Express the appropriate initial conditions when the particle is displaced as in part (b). (f) Non-dimensionalize the differential equation obtained in part (e).

6

(g) Discuss a strategy to determine the accuracy of the approximation using the mass ratio

m . ρAL

Solution: (a) The bar is fixed at x=0, thus

u (0, t ) = 0

(c)

Application of Newton’s law to the free-body diagram of the block, drawn at an arbitrary instant and illustrated in Figure P1.2c leads to − EA

∂u ∂u ∂ 2u ( L, t ) − ku ( L, t ) − c ( L, t ) = m 2 ( L, t ) ∂x ∂t ∂t

(d)

(b) Assume the static displacement of the bar is a linear function of x. Thus if the block is statically displaced a distance δ to the right then

δ

u ( x,0) =

L

x

(e)

Since the bar is released from rest ∂u ( x,0) = 0 ∂t

(f)

(c) Substitution of the non-dimensional variables into Equation (a) leads to

( )

( )

1 ∂ 2 Lu * EA1 ∂ 2 Lu * ρ A = L2 ∂x *2 T 2 ∂t *2 ρL2 ∂ 2 u * ∂ 2u * = ET 2 ∂t *2 ∂x *2

(g)

For convenience choose

ρL2 ET 2

=1

T=L

7

ρ E

(h)

Use of Equation h in Equation l and dropping the *s from non-dimensional variables leads to ∂ 2 u ∂u = ∂x 2 ∂t 2

(i)

Non-dimensionalization of the boundary conditions leads to u (0, t ) = 0 −

( )

( )

( )

EA ∂ Lu c ∂ Lu m ∂ Lu (1, t ) − kLu * (1, t ) − (1, t ) = 2 (1, t ) * * L ∂x T ∂t T ∂t *2 *

*

2

(j)

*

(k)

Rearranging and dropping the *s from Equation k leads to

−

Dividing Equation l by

−

EA ∂u c (1, t ) − ku (1, t ) − L ∂x L

E ∂u mE ∂ 2 u (1, t ) = 2 2 (1, t ) ρ ∂t ρL ∂t

(l)

EA leads to L m ∂ 2u c ∂u kL ∂u (1, t ) − (1, t ) = (1, t ) u (1, t ) − ρAL ∂t 2 EA ∂x A ρE ∂t

(m)

Non-dimensionalization of the initial conditions leads to u ( x,0) =

δ

x

(n)

∂u ( x,0) = 0 ∂t

(o)

L

(d) The non-dimensional parameters in the boundary conditions are •

m which is the ratio of the mass of the block to the mass of the bar ρAL

•

kL which is the ratio of the force in the spring to the resultant force from the EA normal stress

8

•

c A ρE

which is the ratio of the energy dissipated by the viscous damper to the

strain energy in the bar (e) The differential equation governing the displacement of the particle when a discrete model is used is 2 1 dy ⎛ ⎞d y ⎜ m + ρAL ⎟ 2 + c + ky = 0 3 dt ⎝ ⎠ dt

(p)

Equation p is subject to the initial condition y (0) = δ

(q)

(f) Define non-dimensional variables by x* = t* =

x

(r)

δ t Tˆ

(s)

Substitution of Equations r and s into Equation p leads to 2 * 1 cδ dy * ⎛ ⎞ δ d y m + ρ AL + + kδy * = 0 ⎜ ⎟ 2 *2 * ˆ 3 T dt ⎝ ⎠ T dt

(t)

Dropping the *s from non-dimensional variables in Equation t and rearranging leads to d2y + dt 2

dy kTˆ 2 + y=0 ⎛ ρAL ⎞ dt ⎛ ρAL ⎞ m⎜1 + m⎜1 + ⎟ ⎟ 3m ⎠ 3m ⎠ ⎝ ⎝ cT

(u)

The characteristic time is chosen as Tˆ =

Define the mass ratio

9

m k

(v)

μ=

ρAL 3m

(w)

Use of Equation v and Equation w in Equation u leads to d2y c dy 1 + + y=0 2 dt (1 + μ ) mk dt 1 + μ

(x)

Equation x is supplemented by the non-dimensional initial condition

y (0) = 1

(y)

(g) The accuracy of the approximation can be determined by finding the exact solution of the partial differential equation subject to the derived boundary and initial conditions. The lowest natural frequency is a function of the mass ratio μ. Plot the exact natural frequency as a function of μ against the approximate natural frequency as a function of μ.

10

1.3 The shaft of Figure P1.3a is non-uniform such that its polar moment of inertia varies with x, J=J(x). Let θ ( x, t ) represent the angular displacement of the plane a distance x from the left support. A differential element of thickness dx is illustrated in Fig. P1.3b. (a) Use the free-body diagram to derive the partial differential equation governing the torsional oscillations of the shaft. Specify appropriate boundary conditions. (b) Introduce non-dimensional variables x* = function α ( x) =

t x and t * = and a non-dimensional T L

J ( x) where J 0 is the polar moment of inertia of the shaft at x=0. J0

Substitute these variables into the differential equation and choose T such that the governing differential equation contains no parameters.

(c) The shaft is circular with a radius which varies according to r ( x) = r0 (1 + μx ) where r0 is the shaft’s radius at x=0 and μ is the rate of increase (or decrease if μ is negative) of the radius and has dimensions of (length ) . Write the non-dimensional differential −1

equation obtained in part (b) for this case. Solution: (a) The polar moment of inertia of the differential element is dI = ρJdx .

Defining θ ( x) as the angular displacement of a plane along the axis of the shaft, the ∂ 2θ angular acceleration of the differential disk is α = 2 . Application of the moment ∂t

equation

∑ M = Iα to the free-body diagram of the differential element leads to − M ( x) + M ( x + dx) = ρJdx

11

∂ 2θ ∂t 2

(a)

The moment acting across the cross section is the resultant of a shear stress distribution in the cross section. The shear strain in the cross section is γ = r

∂θ . Where r is the ∂x

distance from the axis to a point in the cross section. If the bar is elastic then the shear stress is τ = Gγ . The resultant moment from the shear stress distribution is M = ∫ τrdA = G A

∂M ∂θ 2 ∂θ . Then noting that M ( x + dx) = M ( x) + r dA = JG dx , ∫ ∂x ∂x A ∂x

Equation a is rewritten as ∂ 2θ ∂ ⎛ ∂θ ⎞ ⎟ = ρJ 2 ⎜ JG ∂x ⎝ ∂x ⎠ ∂t

(b)

Equation b is subject to the boundary condition for a fixed end at x=0,

θ (0, t ) = 0

(c)

and the boundary condition for a free end (stress free) at x=L, ∂θ ( L, t ) = 0 ∂x

(d)

(b) Introduction of the non-dimensional variables into Equation a gives 1 ∂ 2θ G ∂ ⎡ * 1 ∂θ ⎤ * J 0α ( x ) = ρJ 0α ( x ) 2 *2 L ∂x * ⎢⎣ L ∂x * ⎥⎦ T ∂t

(e)

Dropping the *s from non-dimensional variables and rearranging Equation e leads to ∂ ⎛ ∂θ ⎞ ρL2 ∂ 2θ α ⎟= ⎜α ∂x ⎝ ∂x ⎠ GT 2 ∂t 2

Choosing T = L

ρ G

(f)

leads to the non-dimensional governing equation as ∂ 2θ ∂ ⎛ ∂θ ⎞ ⎟ =α 2 ⎜α ∂x ⎝ ∂x ⎠ ∂t

12

(g)

(c) The polar moment of inertia of a circular shaft is J= =

π 2

π

2

r4 r04 (1 + μx )

4

(h)

The non-dimensional function α(x) is

π α (x* ) = 2

(

r04 1 + μLx *

)

4

π

r04 2 4 = (1 + μLx )

(i)

where the * has been dropped from the non-dimensional variable. Use of Equation I in Equation g leads to 2 ∂ ⎡ 4 ∂θ ⎤ 4 ∂ θ ( ) ( ) Lx Lx 1 1 μ μ + = + ∂x ⎥⎦ ∂x ⎢⎣ ∂t 2

13

(j)

1.4. Consider the shaft of Figure P1.3a when it is subject to a harmonic torque M (t ) = M 0 sin (ωt ) applied at its free end. The radius of the shaft varies linearly as described in Problem 1.3(c). Let Θ represent the steady-state amplitude of the end of the shaft. The amplitude is a function of the parameters ρ , G, L, r0 , μ , M 0 and ω . That is Θ = Θ(ρ , G, L, r0 , μ , M 0 , ω ) . Determine a set of non-dimensional parameters

π 1 , π 2 , π 3and π 4 for a non-dimensional formulation of the relationship between parameters Θ = Θ(π 1 , π 2 , π 3 , π 4 ) . Solution: There are 8 parameters in the problem, seven independent parameters

ρ , G, L, r0 , μ , M 0 , ω and the dependent parameter Θ . The parameters involve three basic dimensions, mass, length and time. Application of the Buckingham PI Theorem reveals that the non-dimensional relation should be in terms of five dimensionless parameters. The steady-state amplitude of angular displacement, Θ , is already dimensionless. Thus the task is to determine four dimensionless parameters involving the seven independent dimensional parameters. Noting the basic dimensions of each parameter

ρ G L r0

M L3 F ML M = 2 2 = 2 LT L T L L L

M0 F⋅L =

μ ω

14

1 L 1 T

ML2 T2

(a)

Two non-dimensional parameters are apparent from Equation a, the ratio of the base radius to the length of the shaft

π1 =

r0 L

(b)

π 2 = μL

(c)

and the rate of taper times the shaft length

Inspection leads to two additional non-dimensional parameters as M0 ω 2 L2 ρL2 π4 = Gω

π3 =

Equation b, Equation c, Equation d and Equation e provide one choice of a set on nondimensional parameters. Note that the choice of parameters is not unique.

15

(d) (e)

1.5 The steady-state amplitude of the mass attached to the bar of Fig. P1.5 is X=

F0 L ⎛ ⎛ ρ ⎞⎟ ρ ⎞⎟ 2 ⎜ ωL sin − ω AωL Eρ cos⎜⎜ ωL m L ⎜ E ⎟⎠ E ⎟⎠ ⎝ ⎝

(a)

Eq. (a) shows that X = X (F0 , A, L, m, ω , E , ρ ) . (a) Determine a set of non-dimensional parameters which can be used to formulate Eq. (a) in a non-dimensional form. (b) Rewrite Eq. (a) in terms of these non-dimensional parameters. Solution: The relationship suggested by Equation a involves seven independent

parameters, F0 , A, L, m, ω , E , ρ and one dependent parameter X. The basic dimensions of the parameters are X

L

F0

F=

A L m

L2 L M 1 T F M = 2 L LT 2 M L3

ω E

ρ

ML T2

(b)

There are three independent dimensions in the eight parameters. Thus the Buckingham Pi Theorem implies that a non-dimensional formulation of Equation a involves five dimensionless parameters, one of which involves the dependent parameter X while the other four are independent of X. An obvious choice of the dependent parameter is

16

π1 =

X L

(c)

Choices for the dependent parameters may be

π2 = π3 =

m ρAL

(d)

ρL2ω 2

(e)

E

A L2 F π5 = 0 EA

π4 =

(f) (g)

Dividing Equation a by L and using Equations c-e in the result leads to

π1 =

π 5 ( EA) 1 2

(h)

( )

( )

⎛π E ⎞ ρALπ 2 LE A⎜⎜ 3 2 ⎟⎟ L Eρ cos π 3 − sin π 3 ρL2 ⎝ ρL ⎠

Dividing the numerator and denominator of Equation h by EA leads to

π1 =

π5

( )

( )

π 3 cos π 3 − π 2 sin π 3

17

(i)

1.6 The temperature distribution in the extended surface of Fig. P1.6 is ⎡ ⎛ 2h ⎞ ⎛ 2h ⎞ ⎛ 2h ⎞ ⎤ T ( x) = T∞ + (T1 − T∞ ) ⎢cosh⎜ x ⎟ − tanh⎜ L ⎟ sinh ⎜ x ⎟⎥ ⎝ kw ⎠ ⎝ kw ⎠ ⎝ kw ⎠⎦ ⎣

(a)

The rate of heat transfer from the base of the extended surface is Q = − kwl

dT ( 0) dx

(b)

Develop a non-dimensional relationship between the heat transfer at the base and appropriate non-dimensional parameters. Speculate on the physical meanings of the nondimensional parameters. Solution: The heat transfer at the base is determined by using Equation a in Equation b

leading to ⎛ 2h ⎞ ⎛ 2hL ⎞ Q = −kwl(T1 − T∞ )⎜ − ⎟ tanh⎜ ⎟ ⎝ kw ⎠ ⎝ kw ⎠

(c)

The argument of the hyperbolic tangent function is dimensionless; call it A. Manipulation of Equation c yields Q=

kl w (T1 − T∞ )A tanh( A) 2 L

(d)

The non-dimensional form of Equation d is 2Q w = A tanh( A) kl(T1 − T∞ ) L

(e)

The three dimensionless parameters involved in Equation e are A, the ratio of the rate of convective heat transfer to conductive heat transfer,

w a length scale which is also the L

ratio of the area over which conduction occurs to the area over which convection occurs,

18

and

2Q which represents the ratio of the rate of heat transfer from the base to a kl(T1 − T∞ )

combination of conduction and convection terms.

19

1.7 The differential equation governing the transverse deflection of a pipe with an inviscid fluid flowing with a velocity U is 2 ∂2w ∂w ∂2w ∂4w 2 ∂ w + ( M + m) 2 = 0 + ( M + m) g + 2MU EI 4 + MU ∂t ∂x ∂x∂t ∂x 2 ∂x

(a)

where E is the elastic modulus of the pipe, I is the area moment of inertia of the pipe’s cross section, M is the mass of the fluid in the pipe per unit length and m is the mass of the pipe per unit length. Define the following non-dimensional variables 1

x w * ⎡ EI ⎤ 2 t x = w* = t =⎢ 2 L L ⎣ M + m ⎥⎦ L *

(b)

Non-dimensionalize Eq. (a) through introduction of the non-dimensional variables of Eq. Write the non-dimensional equation in terms of the non-dimensional parameters 1

M ⎡ M ⎤2 β= and u = ⎢ ⎥ UL . M +m ⎣ EI ⎦

Solution: Introduction of the defined non-dimensional variables into Equation a leads to

(

)

EI ∂ 4 ( Lw * ) MU 2 ∂ 2 Lw * 2 MU ∂ 2 ( Lw * ) ( M + m) g ∂ ( Lw* ) + 2 + + + LT ∂x * ∂t * L L4 ∂x *4 L ∂x *2 ∂x * M + m ∂ 2 ( Lw * ) =0 T2 ∂t *2

(c)

Dropping the *s from the non-dimensional variables in Equation c, substituting for T and multiplying by

L3 leads to EI 1

∂ 4 w MU 2 L2 ∂ 2 w 2 MUL3 ⎛ EI ⎞ 2 ∂ 2 w ( M + m) gL3 ∂w ∂ 2 w + + + + =0 ⎜ ⎟ ∂x ∂t 2 EI EI ∂x 4 ∂x 2 EIL :2 ⎝ M + m ⎠ ∂x∂t

Rewriting Equation d in terms of u and β gives

20

(d)

1

1

∂ 4 w ML2 u 2 EI ∂ 2 w 2 ML3 ⎛ EI ⎞ 2 u ⎛ EI ⎞ 2 ∂ 2 w gL3 β ∂w ∂ 2 w + + + + =0 ⎜ ⎟ ⎜ ⎟ EI ML2 ∂x 2 EIL2 ⎝ M + m ⎠ L ⎝ M ⎠ ∂x∂t EI M ∂x ∂t 2 ∂x 4 2 ∂4w ∂ 2 w gL3 β ∂w ∂ 2 w 2 ∂ w + u + 2 u + + =0 (e) β ∂x∂t EI M ∂x ∂t 2 ∂x 4 ∂x 2

21

1.8 The general form of the momentum equation for the flow of a viscous fluid near a free surface is

ρ

Dv = μ∇ 2 v − ∇p + ρgj Dt

(a)

where v = ui + vj + wk is the velocity vector, ρ is the mass density of the fluid, p is the fluid pressure, μ is the dynamic viscosity, and g is the acceleration due to gravity. Let L be a characteristic length in the flow, V be a characteristic velocity, and p∞ a characteristic pressure and define non-dimensional variables as v V V t* = t L p − p∞ p* = ρV 2 x y z x* = , y * = , z * = L L L v* =

(b) (c) (d) (e)

(a) Use Eqs. (b)-(e) in Eq. (a) to obtain Dv 1 2 1 = −∇p + ∇ v+ j Dt Re Fr

(f)

where Re is the Reynolds number and Fr is the Froude number and all variables are taken as non-dimensional. (b) Define the Froude number, mathematically and physically. (c) Write the component equations represented by Eq. (f).

Solution: (a) Use of the non-dimensional variables, Equation b, Equation c, Equation d, and Equation e, in Equation a leads to

22

ρ D (Vv * ) L V

Dt

*

=

μ 2

L

( )

∇ 2 Vv * −

[

(

)]

1 ∇ ρV 2 p * + p ∞ + ρgj L

(g)

Rearranging Equation g while dropping *s from non-dimensional quantities leads to gL μ D (v ) = ∇ 2 (Vv ) − ∇p + 2 j ρVL V Dt

(h)

(b) Comparison of Equation h with Equation f leads to Fr =

V2 gL

(i)

The Froude number is the ratio of the inertia force to the gravity force (c) The non-dimensional velocity vector is written as . v = ui + vj + wk . The component equations derived from Equation f are ∂u ∂u ∂u ∂u 1 ⎛ ∂ 2 u ∂ 2 u ∂ 2 u ⎞ ∂p ⎟− ⎜ +u +v +w = + + ∂t ∂x ∂y ∂z Re ⎜⎝ ∂x 2 ∂y 2 ∂z 2 ⎟⎠ ∂x

(j)

∂v ∂v ∂v ∂v 1 ⎛ ∂ 2 v ∂ 2 v ∂ 2 v ⎞ ∂p 1 ⎜ ⎟− +u +v +w = + + + ∂y ∂z Re ⎜⎝ ∂x 2 ∂y 2 ∂z 2 ⎟⎠ ∂y Fr ∂t ∂x

(k)

∂u ∂w ∂w ∂w 1 ⎛ ∂ 2 w ∂ 2 w ∂ 2 w ⎞ ∂p ⎜ ⎟− +u +v +w = + + ∂x ∂y ∂z Re ⎜⎝ ∂x 2 ∂y 2 ∂z 2 ⎟⎠ ∂x ∂t

(l)

23

1.9 The momentum equation for the flow of a with free convection is

ρ

Dv = μ∇ 2 v − ρβ (T − T∞ ) gj Dt

(a)

where, in addition to the variables defined in Prob. 1.8, β is the coefficient of thermal expansion and T is temperature. The non-dimensional variables defined in Problem 1.8 are used as well as Θ =

T − T∞ where T1 and T∞ are reference temperatures. T1 − T∞

(a) Non-dimensionalize Eq. (a) (b) The non-dimensional formulation of Eq. (a) contains two non-dimensional parameters, Re and Gr, the Grashopf number. Define the Grashopf number, both mathematically and physically. Solution: (a) Using the non-dimensional variables defined in Equation b, Equation c, Equation d and Equation e of Problem 1.8 as well as the non-dimensional temperature in Equation a leads to

ρ D (Vv * ) L V

Dt

*

=

μ L2

( )

∇ 2 Vv * − ρβg (T1 − T∞ )Θj

(b)

Rearranging Equation b while dropping the *s from the non-dimensional variables leads to

β gL(T1 − T∞ ) D (v ) 1 2 = ∇ (v ) − Θj Dt Re V2

(c)

(b) The Grashopf number is defined as Gr =

βg (T1 − T∞ ) V2

The Grashopf number is the ratio of the rate of convective heat transfer to the rate of momentum transfer. 24

(d)

1.10 Use Lagrange’s equations to derive the differential equations governing the motion of the system of Figure P1.10. Solution: The kinetic energy of the system at an arbitrary instant is

1 1 1 T = m x&12 + m2 x&22 + m3 x&32 2 2 2

(a)

The potential energy of the system at an arbitrary instant is V =

1 2 1 1 1 1 2 2 kx1 + (2k )(x 2 − x1 ) + (2k )x 22 + k ( x3 − x 2 ) + kx32 2 2 2 2 2

(b)

Application of Lagrange’s equations to the lagrangian, L=T-V leads to d ⎛ ∂L ⎞ ∂L ⎜ ⎟− =0 dt ⎜⎝ ∂x&1 ⎟⎠ ∂x1 d ⎡1 1 ⎤ ⎡1 ⎤ m(2 x&1 )⎥ − ⎢ k (2 x1 ) + (2k )(2)( x 2 − x1 )(−1)⎥ = 0 ⎢ dt ⎣ 2 2 ⎦ ⎣2 ⎦ m&x&1 + 3kx1 − 2kx 2 = 0

(c)

d ⎛ ∂L ⎜ dt ⎜⎝ ∂x& 2

⎞ ∂L ⎟⎟ − =0 ⎠ ∂x 2 d ⎡1 1 1 ⎤ ⎡1 ⎤ 2m(2 x& 2 )⎥ − ⎢ (2k )(2)( x 2 − x1 ) + (2k )(2 x 2 ) + k (2)( x3 − x 2 )(−1)⎥ = 0 ⎢ dt ⎣ 2 2 2 ⎦ ⎣2 ⎦ 2m&x&2 − 2kx1 + 5kx 2 − kx3 = 0 (d) d ⎛ ∂L ⎞ ∂L ⎟− ⎜ =0 dt ⎜⎝ ∂x& 3 ⎟⎠ ∂x3 1 d ⎡1 ⎤ ⎡1 ⎤ m (2 x& 3 )⎥ − ⎢ (2k )(2)( x 2 − x1 ) + k (2 x3 )⎥ = 0 ⎢ 2 dt ⎣ 2 ⎦ ⎣2 ⎦ m&x&3 − kx2 + 2kx3 = 0

25

(e)

1.11 A non-uniform bar of length L is in a medium of ambient temperature T∞ . One end of the bar is insulted while its other end is maintained at a temperature T0 . The bar is made of a smart material which produces an internal heat generation proportional to the gradient of the temperature. Assuming one-dimensional heat generation along the length of the bar and defining x as a coordinate along the length of the bar, the differential equation governing the temperature distribution in the bar is k

d ⎛ dT ⎞ dT + hP( x)(T − T∞ ) = 0 ⎜ A( x) ⎟ +η dx ⎝ dx ⎠ dx

(a)

where k is the thermal conductivity of the bar, h is the heat transfer coefficient between the bar and the ambient and η is a parameter related to the internal heat generation. The appropriate boundary conditions are T (0) = T0

(b)

dT ( L) = 0 dx

(c)

(a) Non-dimensionalize Equation a, Equation b and Equation c through introduction of appropriate non-dimensional variables. (b) Identify mathematically and physically all parameters appearing in the nondimensional formulation. (c) Write a non-dimensional formulation of the problem when

x⎞ ⎛ A( x) = A0 ⎜1 − μ ⎟ L⎠ ⎝

2

(d)

and x⎞ ⎛ P ( x) = 2 w + 2b0 ⎜1 − μ ⎟ L⎠ ⎝ 26

(e)

(d) Suppose the ratio of the rate of conduction to the rate of convection is assumed to be small,

1 = O(ε ) where Bi is the Biot number and ε is a small non-dimensional Bi

parameter. Speculate whether a boundary layer exists. If so determine the thickness of the boundary layer and which terms physically interact within the boundary layer.

Solution: (a) Define the following non-dimensional variables and functions x L T − T∞ Θ= T0 − T∞ x* =

( )

α (x * ) =

A x* A0

β (x * ) =

P x* P0

( )

(f) (g) (h) (i)

Use of Equation f, Equation g, Equation h and Equation i in Equation a leads to

d [(T0 − T∞ )Θ + T∞ ] ⎞ η d [(T0 − T∞ )Θ + T∞ ] k d ⎛ + hP0 β ( x * )(T0 − T∞ )Θ = 0 (j) ⎜ A0α ( x * ) ⎟+ 2 * * * L dx ⎝ dx dx ⎠ L

Rearranging Equation j while dropping the *s from non-dimensional variables leads to

kA0 (T0 − T∞ ) d ⎛ dΘ ⎞ η (T0 − T∞ ) dΘ + hP0 β ( x)(T0 − T∞ )Θ = 0 ⎜ α ( x) ⎟+ 2 dx ⎠ L dx L dx ⎝

(k)

Dividing Equation k by the coefficient of its highest derivative term leads to d ⎛ dΘ ⎞ ηL dΘ hP0 L2 + β ( x )Θ = 0 ⎟+ ⎜ α ( x) dx ⎝ dx ⎠ kA0 dx kA0

27

(l)

The appropriate non-dimensional boundary conditions are Θ(0) = 1 dΘ (1) = 0 dx

(m) (n)

(b) The three non-dimensional parameters in Equation l are the Biot number Bi =

hL k

(o)

which is the ratio of the rate of convection heat transfer to the rate of conduction heat transfer and a parameter defined as W =

ηL

(p)

kA0

and a scaling ratio

λ=

P0 L A0

(q)

The parameter W represents that ratio of the rate of internal heat generation to the rate of heat transfer by conduction. (c) The non-dimensional functions become

( ) ⎞⎟

⎛ μ Lx * A0 ⎜⎜1 − L * α (x ) = ⎝ A0

(

= 1 − μx *

28

)

2

2

⎟ ⎠ (r)

⎛ L* x ⎞ ⎟ 2 w + 2b0 ⎜⎜1 − μ L ⎟⎠ ⎝ * β ( x) = 2 w + 2b0

λA0 = L

− μx *

λA0 L

= 1 − νx

*

(s)

where

ν=

μL λA0

(t)

Thus the governing equation becomes d ⎡ (1 − μx )2 dΘ ⎤⎥ + W dΘ + λBi(1 − νx )Θ = 0 ⎢ dx ⎣ dx ⎦ dx (d) Suppose Bi =

1

ε

(u)

. Then Equation u is rewritten as

ε

d ⎡ (1 − μx )2 dΘ ⎤⎥ + εW dΘ + λ (1 − νx )Θ = 0 ⎢ dx dx ⎣ dx ⎦

(v)

Since a small parameter multiplies the highest order derivative a boundary layer may exist. Its existence depends upon the magnitude of W. Assume a boundary layer variable of the form

ϕ=

x

(w)

εk

Use of Equation w in Equation v leads to

ε 1− 2 k

d dϕ

⎡ k ⎢ 1 − με φ ⎣

(

)

2

dΘ ⎤ dΘ + ε 1− k W + λ 1 − νε k ϕ Θ = 0 ⎥ dφ ⎦ dφ

(

29

)

(x)

A boundary layer may occur when 1-2k=1 or k=1/2 if W=O(1). Thus the boundary

⎛ 1⎞ layer thickness is O⎜⎜ ε 2 ⎟⎟ ⎝ ⎠

30

CHAPTER 2 LINEAR ALGEBRA

31

2.1 The triangle inequality for all vectors in R 3 states that if A=B+C then A ≤ B + C . Prove the triangle inequality holds for all vectors A,B, and C in R 3 .

Solution: Consider u + v where the norm is the standard norm for R 3 . Using the 2

definition of the norm in terms of the dot product

u+v

2

= (u + v ) ⋅ (u + v ) = u ⋅ u + 2u ⋅ v + v ⋅ v

(a)

Using the Cauchy-Schwartz inequality, (u ⋅ v ) ≤ (u ⋅ u )(v ⋅ v ) in Equation a leads to 2

u+v

2

≤ u ⋅ u + 2 (u ⋅ u)( v ⋅ v ) + v ⋅ v =

[ u ⋅u +

v⋅v

]

(b)

Taking the positive square root of the inequality of Equation b leads and using the definition of the standard norm leads to u+v ≤ u + v

Thus the triangle inequality is proved.

32

(c)

2.2 Prove that the triangle inequality, stated in Problem 2.1, holds for all vectors in R n for any finite value of n where the norm is the standard inner product generated norm. Solution: Consider u + v where the norm is an inner product generated norm. Using 2

the definition of an inner product generated norm u+v

2

= (u + v, u + v )

= (u, u ) + (u, v ) + ( v, u) + ( v, v)

(a)

Using the Cauchy-Schwartz inequality, (u, v) 2 ≤ (u, u)( v, v) in Equation a leads to u+v

2

≤ (u, u ) + 2 (u, u)( v, v) + ( v, v ) =

[ (u, u +

(v, v ) ]

2

(b)

Taking the positive square root of the inequality of Equation b leads and using the definition of inner product generated norms leads to u+v ≤ u + v

Thus the triangle inequality is proved.

33

(c)

2.3 Let A=2i+3j-k and B=3j+2k. Let V be the set of vectors in the span of A and B (a) Show that V is a subspace of R 3 . (b) Determine an orthonormal basis for V with

respect to the standard inner product for R 3 . Solution: (a) Since V is a subset of vectors in R 3 and consists of all vectors which are

linear combinations of A and B, by definition V is closed under both vector addition and scalar multiplication. The zero vector is clearly in V and if C is in V then –C is also in V. This is sufficient to show that V is a subspace of R 3 . (b) Clearly A and B are linearly independent (A cannot be written as a multiple of B as A has a non-zero k component). Thus the dimension of V is 2. Normalizing A v1 =

A = A

2i + 3 j - k

(2)2 + (3)2 + (−1) 2

1

=

14

(2i + 3j - k )

(a)

The Gram-Schmidt process is used to determine a vector in V orthogonal to A w 2 = B − (v 1 , B )v 1 ⎡⎛ 2 ⎞ ⎛ 3 ⎞ ⎛ −1 ⎞ ⎤ 1 = 3 j + 2k − ⎢⎜ (2i + 3j − k ) ⎟(0) + ⎜ ⎟(3) + ⎜ ⎟(2)⎥ ⎝ 14 ⎠ ⎝ 14 ⎠ ⎦ 14 ⎣⎝ 14 ⎠ 9 3 = i + j+ k 2 2

(b)

which is then normalized leading to

v2 =

w2 = w2

i+

9 3 j+ k 2 2 2

(1)2 + ⎛⎜ 9 ⎞⎟ + ⎛⎜ 3 ⎞⎟ ⎝2⎠ ⎝2⎠

2

=

1 94

34

(2i + 9 j + 3k )

(c)

2.4 Let V be the set of vectors in R 3 such that if u = [u1

u2

u 3 ] is in V then

u1 + 2u 2 − u 3 = 0 . Prove or disprove that V is a subspace of R 3 . If V is a subspace of R 3 , what is its dimension? Solution: To prove the V is a subspace of R 3 it is necessary to show closure under vector addition, closure under scalar multiplication, the presence of a zero vector in V, and for every u in V the presence of –v in V. Consider u and v, both in V. That is

u1 + 2u 2 − u 3 = 0 and v1 + 2v 2 − v3 = 0 . Now consider w=u+v= [u1 + v1 , u 2 + v 2 , u 3 + v3 ] . Then w1 + 2w2 − w3 = (u1 + v1 ) + 2(u 2 + v 2 ) − (u 3 + v3 ) = u1 + 2u 2 − u 3 + v1 + 2v 2 − v3 = 0 . Hence V is closed under vector addition. If α is an arbitrary scalar then z= αu= [αu1 , αu 2 , αu 3 ] and z 1 + 2 z 2 − z 3 = αu1 + 2(αu 2 ) − αu 3 = α (u1 + 2u 2 − u 3 ) = 0 . Thus V is closed under scalar multiplication. Clearly the zero vector 0=[0,0,0] is in V. For every u in V define –u= [−u1 ,−u 2 ,−u 3 ] . Then − u 2 + 2(−u 2 ) − (−u 3 ) = −(u1 + 2u 2 − u 3 ) = 0 and –u is in V. Hence V is a subspace of R 3 . The dimension of V is 2 as [0,1,2] and [1,0,1] form a basis for V.

35

2.5 The boundary conditions satisfied by a non-dimensional temperature distribution,

θ ( x) in an extended surface subject to a constant temperature at x=0 and insulated at x=1 are θ (0) = 0 and

dθ (1) = 0 . Define S as the set of all functions in C 2 [0,1] which satisfy dx

these boundary conditions. Show that S is a subspace of C 2 [0,1] . Solution: Since C 2 [0,1] is a vector space in its own right, in order to show that S is a subspace of C 2 [0,1] it is only necessary to show that S is closed under vector addition and scalar multiplication and that the zero vector is in S. Since the zero function satisfies both boundary conditions it is an element of S. Consider two functions f(x) and g(x), both of which satisfy the boundary conditions. If S is closed under vector addition then f(x)+g(x) must be in S, that is f(x)+g(x) satisfies both boundary conditions. To this end

f (0) + g (0) = 0 + 0 = 0 . Hence f(x)+g(x) is in S. If S is closed under scalar multiplication then for any scalar α, αf(x) is in S. To this end consider αf (0) = 0 and thus S is closed under scalar multiplication. S is a subspace of C 2 [0,1] .

36

2.6 Let V be the space of functions which are in both P 4 [0,1] , the space of all polynomials of degree four or less, and the space S, defined in Problem 2.5. Determine a basis for V. Solution: The general form of a polynomial of degree four or less is

p( x) = a 0 + a1 x + a 2 x 2 + a3 x 3 + a 4 x 4

(a)

If p(x) is in V it must also satisfy the boundary conditions of Problem 2.5. Thus p(0) = 0 ⇒ a 0 = 0

(b)

and dp (1) = 0 ⇒ a1 + 2a 2 + 3a3 + 4a 4 = 0 ⇒ a1 = −(2a 2 + 3a3 + 4a 4 ) dx

(c)

Equations (b) and (c) imply that three coefficients may be chosen arbitrarily and independently leading to three basis functions. One possible set is

a 2 = 1, a3 = 0, a 4 = 0 ⇒ p1 ( x) = −2 x + x 2

(d)

a 2 = 0, a3 = 1, a 4 = 0 ⇒ p 2 ( x) = −3x + x 3

(e)

a 2 = 0, a3 = 0, a 4 = 1 ⇒ p3 ( x) = −4 x + x 4

(f)

37

2.7 Determine an orthonormal basis with respect to the standard inner product on C 2 [0,1] for the vector space V defined in Problem 2.6. Solution: The Gram-Schmidt procedure is used. Normalizing the first function leads to 1

⎡1 ⎤2 2 p1 ( x) = ⎢ ∫ − 2 x + x 2 dx ⎥ = 0.730 ⎣0 ⎦

(

)

(a)

Then

v1 (x) =

p1 (x) = −2.739x + 1.369x 2 p1(x)

(b)

Continuing with Gram-Schmidt, w2 ( x) = p 2 ( x) − (v1 , p 2 )v1 ( x) ⎡1 ⎤ = −3 x + x 3 + ⎢ ∫ − 3 x + x 3 − 2.739 x + 1.369 x 2 dx ⎥ − 2.739 x + 1.369 x 2 ⎣0 ⎦ 3 2 = x − 1.906 x + 0.8125 x

(

)(

) (

)

w2 ( x) = 14.379 x 3 − 27.411x 2 + 11.683x w2 ( x)

(c)

Normalization leads to v 2 ( x) =

The third orthogonal basis element is calculated as w3 ( x) = p3 ( x) − ( p3 , v1 )v1 − ( p3 , v 2 )v 2 = x 4 − 2.277 x 3 + 1.573 x 2 − 0.3143x

(d)

Nornmalization leads to v3 ( x) =

w3 ( x) = 80.714 x 4 − 183.780 x 3 + 126.925 x 2 − 25.367 w3 ( x)

38

(e)

2.8 Expand f ( x) = x 2 (1 − x) 2 in terms of (a) the basis vectors obtained in the solution of Problem 2.6 and (b) the orthonormal basis obtained in Problem 2.7. Solution: (a) Using the basis developed in Problem 2.6

f ( x) = α 1 p1 ( x) + α 2 p 2 ( x) + α 3 p3 ( x)

(a)

Taking the inner product of Equation a with each of the basis functions leads to a set of simultaneous equations of the form ⎡ ( p1 , p1 ) ⎢( p , p ) ⎢ 2 1 ⎢⎣( p3 , p1 )

( p1 , p 2 ) ( p1 , p3 )⎤ ⎡α 1 ⎤ ⎡ ( p1 , f )⎤ ( p 2 , p 2 ) ( p 2 , p3 )⎥⎥ ⎢⎢α 2 ⎥⎥ = ⎢⎢( p 2 , f )⎥⎥ ( p3 , p 2 ) ( p3 , p3 )⎥⎦ ⎢⎣α 3 ⎥⎦ ⎢⎣( p3 , f )⎥⎦

(b)

where, for example,

( p1 , p 2 ) = ∫ (− 2 x + x 2 )(− 3x + x 3 )dx = 1.017 1

(c)

0

( p1 , f ) = ∫ (− 2 x + x 2 )[x 2 (1 − x) 2 ]dx = −0.024 1

(d)

0

After performing the remaining integrations the system of Equation b becomes ⎡0.533 1.017 1.476 ⎤ ⎡α 1 ⎤ ⎡− 0.024⎤ ⎢1.017 1.943 2.825⎥ ⎢α ⎥ = ⎢− 0.044⎥ ⎥ ⎥⎢ 2 ⎥ ⎢ ⎢ ⎢⎣1.476 2.825 4.111⎥⎦ ⎢⎣α 3 ⎥⎦ ⎢⎣ − 0.063⎥⎦

(e)

Simultaneous solution of Equation e leads to α 1 = 1, α 2 = −2, α 3 = 1 such that x 2 (1 − x) 2 = −2 x + x 2 − 2(−3x + x 3 ) − 4 x + x 4

(f)

(b) The orthonormal basis of Problem 2.7 is

v1 (x) = −2.739x +1.369x2

(g)

v2 (x) = 14.379x3 − 27.411x2 +11.683x

(h)

v3 (x) = 80.714x4 −183.780x3 +126.925x2 − 25.36

(i)

An expansion is sought of the form

39

f ( x) = α 1v1 ( x) + α 2 v 2 ( x) + α 3 v3 ( x)

(j)

Since the basis is othonormal

α 1 = ( f , v1 ) = ∫ x 2 (1 − x) 2 [− 2.739 x + 1.369 x 2 ]dx = −0.0326

(k)

α 2 = ( f , v2 ) = ∫ x 2 (1 − x) 2 [14.379 x 3 − 27.411x 2 + 11.683 x ]dx = 0.0193

(l)

1

0

1

0

α 3 = ( f , v2 ) = ∫ x 2 (1 − x) 2 [80.714 x 4 − 183.780 x 3 + 126.925 x 2 − 25.36]dx = 0.0124 1

(m)

0

Thus

(

)

(

x 2 (1 − x) 2 = −0.0326 1.369 x 2 − 2.739 x + 0.0193 14.379 x 3 − 27.411x 2 + 11.683x

(

+ 0.0124 80.714 x 4 − 183.780 x 3 + 126.925 x 2 − 25.36

40

)

) (n)

2.9 Let p( x) = a n x n + a n −1 x n −1 + ... + a1 x + a 0 and q ( x) = bn x n + bn −1 x n −1 + ... + b1 x + b0 be n

elements of P n [0,1] . Show that ( p, q ) = ∑ ai bi is a valid inner product defined for i =0

P n [0,1] . Solution: It is necessary to check to see if the proposed inner product satisfies the four

properties of an inner product over the field of real numbers. n

n

i =1

i =1

(a) Commutivity ( p, q ) = ∑ ai bi = ∑ bi ai = (q, p ) n

n

i =1

i =1

(b) Scalar multiplication (αp, q ) = ∑ (αai )bi = α ∑ ai bi = α ( p, q ) n

n

n

i =1

i =1

i =1

(c) Distributive property ( p + q, w) = ∑ (ai + bi )ci = ∑ a i ci + ∑ bi ci = ( p, w) + (q, w) n

n

i =1

i =1

(d) Non-negative property ( p, p) = ∑ ai ai = ∑ ai2 Clearly ( p, p) ≥ 0 and that (0,0)=0. Also since the inner product of p with itself is the sum of non-negative numbers, the sum can equal zero only if all numbers in the sum equal zero. If this is the case then p(x)=0.

41

2.10 Determine an orthonormal basis with respect to the inner product defined in Problem 2.9 for the vector space V defined in Problem 2.6. Solution: Consider the basis of Problem 2.6

p1 ( x) = −2 x + x 2

(a)

p 2 ( x ) = −3 x + x 3

(b)

p3 ( x) = −4 x + x 4

(c)

Applying Gram-Schmidt using the inner product of Problem 2.9 leads to

[

p1 ( x) = ( p1 , p1 ) = (0) + (−2) + (1) 1 2

v1 ( x) =

2

2

]

1 2 2

= 2.236

p1 ( x) = 0.4472x 2 − 0.8944x 2.236

(d)

w2 ( x) = p 2 ( x) − ( p 2 , v1 )v1

= −3 x + x 3 + [( −3)(−0.8944) + (0)(0.4472) + (1)(0)](0.4472 x 2 − 0.8944 x ) = x 3 − 1.2 x 2 − 0.6 x

(e)

[

w2 = (1) 2 + (−1.2) 2 + (−0.6) 2 v 2 ( x) =

]

1 2

= 1.673

w2 ( x) = 0.5976x 3 − 0.7171x 2 − 0.3596x 1.6723

w3 ( x) = p3 ( x) − ( p3 , v1 )v1 − ( p3 , v 2 )v 2

(

= −4 x + x 4 − [( −4)(−.8944)] 0.4472 x 2 − 0.8944 x

(

)

− [(−4)(−0.3596)] 0.5976 x 3 − 0.7171x 2 − 0.3596 x

)

= x 4 + 0.8571x 3 + 0.5714 x 2 − 7.7142 x

w3

= [(1)

v3 ( x ) =

2

+ (0.8571) + (0.5714) + (−7.7142) 2

2

(g)

]

1 2 2

= 7.847

w3 ( x) = 0.1274x 4 + 0.1092x 3 + 0.0728x 2 − 0.9831x 7.847

42

(f)

(h)

2.11 Expand f ( x) = x 2 (1 − x) 2 in terms of the basis vectors obtained in Problem 2.10. Solution: The orthogonal basis determined in Problem 2.10 is

v1 ( x) = 0.4472 x 2 − 0.8944 x

(a)

v 2 ( x)0.5976 x − 0.7171x − 0.3596 x 3

2

(b)

v3 ( x) = 0.1274 x + 0.1092 x + 0.0728 x − 0.9831x 4

3

2

(c)

An expansion is sought of the form f ( x ) = α 1v1 ( x ) + α 2 v 2 ( x ) + α 3 v3 ( x )

(j)

where

(

)

f ( x) = x 2 1 − 2 x + x 2 = x 2 − 2 x 3 + x 4

(k)

Since the basis is othonormal

α 1 = ( f , v1 ) = [(0.4472)(1)] = 0.4472 α 2 = ( f , v 2 ) = [(1)(−0.7171) + (−2)(0.5976)] = - 1.912 α 3 = ( f , v 2 ) = [(1)(0.0728) + (−2)(0.1092) + (1)(0.1274)] = −0.018

(l) (m) (n)

Thus

(

)

(

x 2 (1 − x) 2 = 0.4472 0.4472 x 2 − 0.8944 x − 1.912 0.5976 x 3 − 0.7171x 2 − 0.3596 x

(

− 0.018 0.1274 x + 0.1092 x + 0.0728 x − 0.9831x 4

3

2

43

)

) (o)

2.17 The boundary conditions for a the deflection, w(x) , of a fixed-free Euler-Bernoulli beam with an axial load are w(0) = 0,

dw d 2w d 3w dw (0) = 0, 2 (1) = 0, 3 (1) + ε (1) = 0 . Let dx dx dx dx

Q be the subspace of P 6 [0,1] such that if f(x) is in Q , f (0) = 0,

df d2 f d3 f df (0) = 0, 2 (1) = 0, 3 (1) + 2 (1) = 0 . Determine a basis for Q. dx dx dx dx

Solution: The general form of a vector in P 5 [0,1] is

p( x) = a 0 + a1 x + a 2 x 2 + a 3 x 3 + a 4 x 4 + a5 x 5

(a)

If the vector is in Q it must also satisfy the specified boundary conditions. To this end p ( 0) = 0 ⇒ a 0 = 0

(b)

dp (0) = 0 ⇒ a1 = 0 dx

(c)

d2p (1) = 0 = 2a 2 + 6a3 + 12a 4 + 20a5 = 0 dx 2 d 3w dw (1) + ε (1) = 0 = 6a3 + 24a 4 + 60a5 + ε (2a 2 + 3a3 + 4a 4 + 5a5 ) = 0 3 dx dx

(d)

(e)

There are a number of ways to pursue the remaining algebra. Equation d and Equation e can be used to solve for two of the coefficients in terms of the remaining two. The latter coefficients are arbitrary and independent. Two linearly independent vectors which form a basis for Q are generated through two independent choices of these coefficients. As an example Equation d and Equation e are used to determine a 2 and a3 in terms of a 4 and a5 . From Equation (d) a2 = −3a3 − 6a4 − 10a5 Use of Equation f in Equation e leads to

44

(f)

2ε (− 3a3 − 6a 4 − 10a5 ) + (6 + 3ε )a3 + (24 + 4ε )a 4 + (60 + 5ε )a5 = 0

(6 − 2ε )a3 + (24 − 8ε )a4 + (60 − 15ε )a5 = 0 (24 − 8ε )a4 + (60 − 15ε )a5 a3 =

2ε − 6

(g)

Use of Equation g in Equation f gives ⎡ (24 − 8ε )a 4 + (60 − 15ε )a5 ⎤ a 2 = −3⎢ ⎥ − 6a 4 − 10a5 2ε − 6 ⎣ ⎦ −1 [(72 − 24ε )a 4 + (180 − 45ε )a5 + 6(2ε − 6)a 4 + 10(2ε − 6)a5 ] = 2ε − 6 −1 = [(36 − 12ε )a 4 + (120 − 25ε )a5 ] (h) 2ε − 6 Choosing a 4 = 1 and a5 = 0 gives a3 = −4 and a 2 = 6 . Hence one polynomial in Q is p1 ( x) = 6 x 2 − 4 x 3 + x 4 Choosing a 4 = 0 and a5 = 1 gives a3 =

(i)

15(4 − ε ) 5(5ε − 24) and a 2 = . A second 2ε − 6 2ε − 6

polynomial in Q is p 2 ( x) =

5(5ε − 4) 2 15(4 − ε ) 3 x + x + x5 2ε − 6 2ε − 6

The polynomials given in Equation i and Equation j are an example of a basis for Q.

45

(j)

2.18 Determine an orthonormal basis for the vector space Q, defined in Problem 2.17, with respect to the standard inner product on C 4 [0,1] .

Solution: The basis vectors generated in Problem 2.17 are p1 ( x) = 6 x 2 − 4 x 3 + x 4

p2 (x) =

(a)

5(5ε − 4) 2 15(4 − ε ) 3 5 x + x +x 2ε − 6 2ε − 6

(b)

Using the standard inner product on C 4 [0,1] 1

2 ⎡1 2⎤ p1 ( x) = ⎢ ∫ 6 x 2 − 4 x 3 + x 4 ⎥ = 1.52 ⎣0 ⎦ p ( x) v1 ( x) = 1 = 0.6578 x 4 − 2.6312 x 3 + 3.9468 x 2 1.52

(

)

(c)

An orthogonal polynomial is determined using the Gram-Schmidt process w2 ( x) = p 2 ( x) − ( p 2 , v1 )v1 =

5(5ε − 4 ) 2 15(4 − ε ) 3 x + x + x5 2ε − 6 2ε − 6

⎡ 1 ⎛ 5(5ε − 4) 2 15(4 − ε ) 3 ⎤ ⎞ − ⎢∫ ⎜ x + x + x 5 ⎟ 0.6578 x 4 − 2.6312 x 3 + 3.9468 x 2 dx ⎥ 2ε − 6 ⎠ ⎣ 0 ⎝ 2ε − 6 ⎦ 4 3 2 0.6578 x − 2.6312 x + 3.9468 x

(

(

)

)

(d)

Using ε=2 the above becomes w2 ( x) = x 5 + 22.17 x 4 − 163.71x 3 + 118.07 x 2 w2 ( x) = 8.156 v 2 ( x) =

w2 ( x ) = 0.122 x 5 + 2.719 x 4 − 20.072 x 3 + 14.476 x 2 8.156

46

(e)

2.19 For what value(s) of c3 are the vectors a,b and c linearly independent? ⎡2⎤ a = ⎢⎢− 2⎥⎥ ⎢⎣ − 1 ⎥⎦

⎡4⎤ b = ⎢⎢− 1⎥⎥ ⎢⎣ 3 ⎥⎦

⎡ − 2⎤ c = ⎢⎢ − 1 ⎥⎥ ⎢⎣ c3 ⎥⎦

Solution: The vectors are linearly independent if αa + βb + γc = 0 implies that

α = β = γ = 0 . Thus ⎡2⎤ ⎡4⎤ ⎡ − 2 ⎤ ⎡0 ⎤ ⎢ ⎥ ⎢ ⎥ α ⎢− 2⎥ + β ⎢− 1⎥ + γ ⎢⎢ − 1⎥⎥ = ⎢⎢0⎥⎥ ⎢⎣ − 1 ⎥⎦ ⎢⎣ 3 ⎥⎦ ⎢⎣ c3 ⎥⎦ ⎢⎣0⎥⎦ 4 − 2⎤ ⎡α ⎤ ⎡0⎤ ⎡ 2 ⎢ − 2 − 1 − 1 ⎥ ⎢ β ⎥ = ⎢0 ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢⎣ − 1 3 c3 ⎥⎦ ⎢⎣ γ ⎥⎦ ⎢⎣0⎥⎦

(a)

A non-trivial solution of Equation a exists only if the determinant of the coefficient matrix is zero. Thus the vectors are linearly independent unless 2 4 −2 − 2 − 1 − 1 = 0 = 6c3 + 24 ⇒ c3 = −4 − 1 3 c3

The vectors are linearly independent unless c3 = −4 .

47

(b)

2.20 For what value(s) of c3 are the vectors b and c of Problem 2.19 orthogonal with respect to the standard inner product for R 3 ? Solution: The vectors b and c are orthogonal with respect to the standard inner product

on R 3 if

(b, c ) = 0 = (4)(−2) + (−1)(−1) + (3)(c3 ) = −7 + 3c3 ⇒ c3 = 7

3

48

(a)

2.21 For what value(s) of c3 are the vectors b and c of Problem 2.19 orthogonal with respect to the inner product defined by (a, b ) = a1b1 + 3a 2 b2 + 2a3 b2 ? Solution: The vectors are orthogonal with respect to the given inner product if

(a, b ) = 0 = (4)(−2) + 3(−1)(−1) + 2(3)(c3 ) = −5 + 6c3 ⇒ c3 = 5 6

49

(a)

⎡1 3 2⎤ 2.22 What is the adjoint of the matrix ⎢⎢2 2 1⎥⎥ with respect to (a) the standard inner ⎢⎣4 2 1⎥⎦

product for R 3 ? and (b) with respect to the inner product (x, y ) = 2 x1 y1 + 2 x 2 y 2 + 3x3 y 3 ? Solution: First note that ⎡1 3 2⎤ ⎡ x1 ⎤ ⎡ x1 + 3 x 2 + 2 x3 ⎤ Ax = ⎢⎢2 2 1 ⎥⎥ ⎢⎢ x 2 ⎥⎥ = ⎢⎢2 x1 + 2 x 2 + x3 ⎥⎥ ⎢⎣4 2 1 ⎥⎦ ⎢⎣ x3 ⎥⎦ ⎢⎣4 x1 + 2 x 2 + x3 ⎥⎦

(a)

Then using the definition of the proposed inner product

(Ax, y ) = 2(x1 + 3x2 + 2 x3 ) y1 + 2(2 x1 + 2 x2 + x3 ) y 2 + 3(4 x1 + 2 x2 + x3 ) y3

(b)

Equation b is rearranged to

(Ax, y ) = (2 y1 + 4 y 2 + 12 y3 )x1 + (6 y1 + 4 y 2 + 6 y3 )x2 + (4 y1 + 2 y 2 + 3 y3 )x3 2 ⎛4 ⎞ = 2( y1 + 2 y 2 + 6 y 3 )x1 + 2(3 y1 + 2 y 2 + 3 y 3 )x 2 + 3⎜ y1 + y 2 + y 3 ⎟ x3 3 ⎝3 ⎠

(c)

If the adjoint matrix is defined as ⎡ ⎢1 * A = ⎢3 ⎢4 ⎢ ⎣3 then ( Ax, y ) = (x, A * y ) as given in Equation c.

50

⎤ 6⎥ 2 3⎥ ⎥ 2 1⎥ 3 ⎦ 2

(d)

2.23 Determine sufficient conditions for a 2x2 matrix to be self adjoint with respect to the standard inner product on R 2 . Solution: A matrix is self-adjoint with respect to the standard inner product if it is symmetric. Thus, for a 2x2 matrix A this requires a1, 2 = a 2,1 .

51

2.24 Determine sufficient conditions for a 2x2 matrix to be self adjoint with respect to the inner product (x, y ) = 2 x1 y1 + x 2 y 2 . ⎡ a1,1 Solution: Consider a matrix of the form A = ⎢ ⎣a 2,1

a1, 2 ⎤ . Then for the given inner a 2, 2 ⎥⎦

product

(Ax, y ) = 2(a1,1 x1 + a1, 2 x 2 )y1 + (a 2,1 x1 + a2,2 x2 )y 2 = (2a1,1 y1 + a 2,1 y 2 )x1 + (2a1, 2 y1 + a 2, 2 )x 2

1 ⎛ ⎞ = 2⎜ a1,1 y1 + a 2,1 ⎟ x1 + (2a1, 2 y1 + a 2, 2 )x 2 2 ⎝ ⎠ Equation a shows that (Ax, y ) = (x, Ay ) using the defined inner product if a1, 2 =

52

(a) 1 a 2,1 2

d2 f 2.25 For what values of ε is the operator Lf = − 2 positive definite with respect to dx the standard inner product for C 2 [0,1] if the domain of L is specified such that f(0)=0 and df (1) + εf (1) = 0 ? dx Solution: Consider the inner product 1

⎛

f⎞ ⎟ g ( x)dx dx ⎟⎠

(Lf , g ) = ∫ ⎜⎜ − d 0

⎝

2

(a)

2

where f(x) and g(x) satisfy both boundary conditions. Applying integration by parts with u = g (x) and dv = −

d2 f dx leads to dx 2 x =1

(Lf , g ) = ⎡⎢− g ( x) df ⎤⎥ + ∫ df dg dx dx ⎦ x =0 0 dx dx ⎣ A second integration by parts with u = x =1

1

(b)

df dg dx leads to and dv = dx dx

x =1

1 ⎛ d 2g ⎞ df ⎤ ⎤ ⎡ dg ⎡ (Lf , g ) = ⎢− g ( x) ⎥ + ⎢ f ( x)⎥ + ∫ f ( x)⎜⎜ − 2 ⎟⎟dx dx ⎦ x =0 ⎣ dx ⎦ x =0 0 ⎣ ⎝ dx ⎠ x =1

x =1

df ⎤ ⎤ ⎡ dg ⎡ f ( x ) ⎥ + ( f , Lg ) = ⎢− g ( x) ⎥ + ⎢ dx ⎦ x =0 ⎣ dx ⎦ x =0 ⎣

(c) x =1

x =1

df ⎤ ⎡ ⎡ dg ⎤ Equation c shows that L is self adjoint if ⎢− g ( x) ⎥ + ⎢ f ( x)⎥ = 0 . Tothis end dx ⎦ x =0 ⎣ dx ⎣ ⎦ x =0

consider x =1

x =1

df ⎤ df df dg dg ⎡ ⎡ dg ⎤ ⎢− g ( x) dx ⎥ + ⎢ dx f ( x)⎥ = − g (1) dx (1) + g (0) dx (0) + f (1) dx (1) − f (0) dx (0) (d) ⎣ ⎦ x =0 ⎣ ⎦ x =0

Noting that f(0)=0, g(0)=0,

df dg (1) = −εf (1) and (1) = −εg (1) leads to dx dx

53

x =1

x =1

df ⎤ ⎤ ⎡ dg ⎡ ⎢− g ( x) dx ⎥ + ⎢ dx f ( x)⎥ = − g (1)(− εf (1) ) + f (1)(− εg (1) ) = 0 ⎦ x =0 ⎦ x =0 ⎣ ⎣

(e)

Equation e shows that L is self adjoint for all values of ε. The positive definitesness of L is examined by setting g=f in Equation b, leading to x =1

(Lf , f ) = ⎡⎢− f ( x) df ⎤⎥ + ∫ df df dx dx ⎦ x =0 0 dx dx ⎣ First note that the integral is non-negative unless

1

(f)

df = 0 for all x between 0 and 1. In this dx

case f(x)=C, a constant. However since f(0)=0, this constant must be zero. Now consider the boundary terms x =1

df df df ⎤ ⎡ ⎢− f ( x) dx ⎥ = − f (1) dx (1) + f (0) dx (0) ⎣ ⎦ x =0

Noting that f(0)=0 and

(g)

df (1) = −εf (1) , Equation g becomes dx x =1

df ⎤ ⎡ 2 − f ( x ) = − f (1)(− εf (1) ) = ε [ f (1)] ⎢ ⎥ dx ⎦ x =0 ⎣

(h)

Equation h shows that the boundary terms are positive for ε>0. Hence L is self adjoint for all ε and positive definite for ε ≥ 0 .

54

CHAPTER 3 ORDINARY DIFFERENTIAL EQUATIONS

55

3.1 Prove that the homogeneous solution space of a linear differential operator L is a vector space. Solution: It is necessary to show that the ten axioms which define a vector space hold.

However solutions of the differential equation are themselves members of the vector space C n [a, b] . Thus it is sufficient to show that the proposed vector space is closed under addition and scalar multiplication and that the zero vector is in the space. Suppose u1 , and u 2 are homogeneous solutions of a differential equation defined by linear operator

L. The usual definitions of addition of functions and multiplication by a scalar apply. The

axiom requiring closure under addition implies that the sum of two homogeneous solutions is also a homogeneous solution. To this end using the linearity of L, L(u1 + u 2 ) = Lu1 + Lu 2 . However since u1 , and u 2 are homogeneous solutions Lu1 = 0

and Lu 2 = 0 . Hence L(u1 + u 2 ) = 0 and u1 + u 2 is by definition a homogeneous solution. Closure under scalar multiplication is shown using linearity by L(αu1 ) = αLu1 = α (0) = 0 . The zero vector is homogeneous solution due to linearity.

Thus the homogeneous solution space is a subspace of C n [a, b] and thus a vector space in its own right.

56

3.3

d2y − 4 y = 0 with y (0) = 2 and y (1) = 0 . dx 2

Solution: A homogeneous solution is assumed as y (t ) = Aeαx which when substituted

into the differential equation leads to

α2 −4 = 0

(a)

whose solutions are α = −2,2 . Thus the form of the general solution is y (t ) = C1 cosh(2 x) + C 2 sinh( 2 x)

(b)

Application of the boundary conditions to Equation b leads to y (0) = 2 ⇒ C1 = 2

(c)

and y (1) = 0 ⇒ 2 cosh(2) + C 2 sinh( 2) = 0 ⇒ C 2 = −2 coth(2)

(d)

Thus the general solution is y ( x) = 2[cosh(2 x) − coth(2) sinh( 2 x)]

57

(e)

3.4

dy d2y + 16 y = 0 with y (0) = 2 and (1) + 3 y (1) = 0 2 dx dx

Solution: A homogeneous solution is assumed as y (t ) = Aeαx which when substituted

into the differential equation leads to

α 2 + 16 = 0

(a)

The solutions of Equation a are α = ±4i . Thus the general solution of the differential equation is y ( x) = C1 cos(4 x) + C 2 sin( 4 x)

(b)

Application of boundary conditions to Equation b leads to y (0) = 2 ⇒ C1 = 2

(c)

and dy (1) + 3 y (1) = 0 ⇒ −8 sin( 4) + 4C 2 cos(4) + 6 cos(4) + 3C 2 sin(4) = 0 dx 8 sin(4) − 6 cos(4) 8 tan(4) − 6 ⇒ C2 = = 4 cos(4) + 3 sin( 4) 3 tan(4) + 4

(d)

The solution of the differential equation which satisfies the given boundary conditions is y ( x) = 2 cos(4 x) +

8 tan(4) − 6 sin(4 x) 3 tan(4) + 4

58

(e)

3.5

d2y dy + 4 + 4y = 0 2 dx dx

Solution: A homogeneous solution is assumed as y (t ) = Aeαx which when substituted

into the differential equation leads to

α 2 + 4α + 4 = 0

(a)

Equation a has only one solution, α = −2 . Thus the general solution of the differential equation is y ( x) = C1e -2x + C 2 xe −2 x

59

(b)

3.6

d2y dy + 5 − 14 y = 0 2 dx dx

Solution: A homogeneous solution is assumed as y (t ) = Aeαx which when substituted

into the differential equation leads to

α 2 + 5α − 14 = 0

(a)

The solutions of Equation a are α = −7,2 . Thus the general solution of the differential equation is y ( x) = C1e -7x + C 2 e 2 x

(b)

Application of the boundary conditions leads to dy (0) = 0 ⇒ −7C1 + 2C 2 = 0 dx y (1) = 1 ⇒ C1e −7 + C 2 e 2 = 1 Simultaneous solution of Equations c and d leads to C1 = 0.6547 and C 2 = 0.1875

60

(c) (d)

3.7

dy d2y dy + 5 + 14 y = 0 with (0) = 0 and y (1) = 1 2 dx dx dx

Solution: A homogeneous solution is assumed as y (t ) = Aeαx which when substituted

into the differential equation leads to

α 2 + 5α + 14 = 0 The solutions of Equation a are α =

(

(a)

)

1 − 5 ± i 31 . Thus the general solution of the 2

differential equation is y ( x) = e

5 − x 2

[C1 cos(2.784 x) + C 2 sin(2.784 x)]

(b)

Application of the given boundary conditions to Equation b leads to dy 5 (0) = 0 ⇒ − C1 + 2.784C 2 = 0 dx 2 and y (1) = 1 ⇒ C1 cos(2.784) + C 2 sin( 2.784) = 1

(d)

Simultaneous solution of Equation c and Equation d leads to C1 = −1.606 and C 2 = − − 1.447 . Thus the solution of the differential equations which satisfies the given

boundary conditions is y ( x) = −e −2.5 x [1.606cos(2,784x) + 1.447sin(2.784x)]

61

(e)

3.8

dy dy d2y dy + 2 + 10 y = 0 with (0) = 0 and (1) = 1 2 dx dx dx dx

Solution: A homogeneous solution is assumed as y (t ) = Aeαx which when substituted

into the differential equation leads to

α 2 + 2α + 10 = 0

(a)

The solutions of Equation a are α = −1 ± 3i . Thus the general solution of the differential equation is y ( x) = e -x [C1 cos(3 x) + C 2 sin(3x)]

(b)

Application of the given boundary conditions to Equation b leads to dy (0) = 0 ⇒ −C1 + 3C 2 = 0 dx

(c)

and dy (1) = 1 ⇒ −e −1 [cos(3) + 3sin(3)]C1 + e −1 [− sin(3) + 3 cos(3)]C 2 = 1 dx Simultaneous solution of Equation c and Equation d leads to C1 = −

C2 = −

(d)

e and 4 sin(3)

e . Thus the solution of the differential equations which satisfies the given 12 sin(3)

boundary conditions is e1− x [3cos(3x) + sin(3x)] y ( x) = − 12 sin(3)

62

(e)

3.9

d2y − y = 3e − 2 x with y (0) = 0 and y(1) = 0 2 dx

Solution: A homogeneous solution is assumed as y (t ) = Aeαx which when substituted

into the differential equation leads to

α 2 −1 = 0

(a)

The solutions of Equation a are α = −1,1 . Thus the homogeneous solution of the differential equation is y h ( x) = C1e − x + C 2 e 2

(b)

The method of undetermined is used and a particular solution is assumed of the form y p ( x) = Ae −2 x

(c)

Substitution of Equation c into the differential equation leads to 4 Ae −2 x − Ae −2 x = 3e −2 x ⇒ A = 1

(d)

The general solution of the differential equation is y ( x) = C1e − x + C 2 e x + e −2 x

(e)

Application of the boundary conditions to Equation e leads to y (0) = 0 = C1 + C 2 + 1 ⇒ C1 + C 2 = −1

(f)

and y (1) = 0 ⇒ C1e −1 + C 2 e1 + e −2 = 0 ⇒ C1 + C 2 e 2 = −e −1 Simultaneous solution of Equation f and Equation g leads to C1 = −

C2 =

(g) 1 − e −3 and 1 − e −1

e −1 − e −3 . Thus the solution of the differential equation which satisfies the given 1 − e −1

boundary conditions is

63

y ( x) =

[(

)

(

) ]

1 e −1 − e −3 e x − 1 − e −3 e − x + e − 2 x −1 1− e

64

(h)

3.10

dy d2y + 3 y = 4e − 3 x + 5 with y (0) = 0 and (1) = 0 2 dx dx

Solution: A homogeneous solution is assumed as y (t ) = Aeαx which when substituted

into the differential equation leads to

α2 +3= 0

(a)

The solutions of Equation a are α = ±i 3 . Thus the homogeneous solution of the differential equation is

( )

( )

y h ( x) = C1 cos 3 x + C 2 sin 3 x

(b)

The method of undetermined is used and a particular solution is assumed of the form y p ( x) = Ae −3 x + B

(c)

Substitution of Equation c into the differential equation leads to 9Ae -3x + 3Ae -3x + 3B = 4e −3 + 5 ⇒ A = 3, B = 5

(d)

The general solution of the differential equation is

( )

( )

y ( x) = C1 cos 3 x + C 2 sin 3 x + 3e −3 x + 5

(e)

Application of the boundary conditions to Equation e leads to y (0) = 0 ⇒ C1 + 3 + 5 = 0 ⇒ C1 = -8

(f)

and

( ) ( )

( ) ( )

dy (1) = 0 ⇒ − 3 (− 8)sin 3 + 3C 2 cos 3 − 9e −3 = 0 dx ⇒ C 2 = 3 3e −3 csc 3 − 8 cot 3

(g)

Thus the solution of the differential equation which satisfies the given boundary conditions is

( ) [

( )

( ) ( )]

y ( x) = −8 cos 3 x + 3 3e −3 csc 3 − 8 cot 3 sin 3 x + 3e −3 x + 5

65

(h)

3.11

d2y + 9 y = 0.1sin(4 x) with y (0) = 1 and y (1) = 0 dx 2

Solution: A homogeneous solution is assumed as y (t ) = Aeαx which when substituted

into the differential equation leads to

α2 +9 = 0

(a)

The solutions of Equation a are α = ±3i . Thus the homogeneous solution of the differential equation is

y h ( x) = C1 cos(3 x ) + C 2 sin (3 x )

(b)

A particular solution is assumed as

y p ( x) = A sin( 4 x) + B cos(4 x)

(c)

Use of Equation c in the differential equation leads to

− 14 A sin(4 x) − 16 B cos(4 x) + 9 A sin(4 x) + 9 B cos(4 x) = 0.1sin(4 x) − 5 A sin(4 x) − 5 B cos(4 x) = 0.1sin( 4 x) Equation d implies A =

(d)

0.1 = −0.02 and B = 0 . Thus the general solution is −5

y ( x) = C1 cos(3 x) + C 2 sin(3 x) − 0.02 sin( 4 x)

(e)

Application of the boundary conditions to Equation e leads to y (0) = 0 ⇒ C1 = 0

(f)

And

y (1) = 0 → C 2 sin(3) − 0.02 sin(4) = 0 ⇒ C 2 = −

0.02 sin(4) = 0.1073 sin(3)

(g)

Thus the solution of the differential equation which satisfies the boundary conditions is

y ( x) = 0.1073 sin(3 x) − 0.02 sin(4 x)

66

(h)

3.12

d2y − 3 y = 2e − 3 x sin(2 x) with y (0) = 1 and y (1) = 0 2 dx

Solution: A homogeneous solution is assumed as y (t ) = Aeαx which when substituted

into the differential equation leads to

α2 −3= 0

(a)

The solutions of Equation a are α = ± 3 . Thus the homogeneous solution of the differential equation is

y h ( x) = C1e −

3x

+ C2 e

3x

(b)

An alternate form of the homogeneous solution is

( )

( )

y ( x) = D1 cosh 3 x + D2 sinh 3 x

(c)

A particular solution is assumed as y p ( x) = Ae −3 x sin( 2 x) + Be −3 x cos(2 x)

(d)

Use of Equation d in the differential equation leads to

[

] [

A 5e −3 x sin(2 x) − 12e −3 x cos(2 x) + B 5e −3 x cos(2 x) + 12e −3 x sin(2 x)

]

− 3 Ae −3 x sin(2 x) − 3Be − 2 x cos(2 x) = 2e −3 x sin(2 x)

(2 A + 12 B )e −3 x sin(2 x) + (− 12 A + 2 B )e −3 x cos(2 x) = 2e −3 x sin(2 x)

(e)

Equation e implies

⎡ 2 12⎤ ⎡ A⎤ ⎡2⎤ ⎢− 12 2 ⎥ ⎢ B ⎥ = ⎢0⎥ ⎣ ⎦⎣ ⎦ ⎣ ⎦ Simultaneous solution of the system of Equation f leads to A =

(f) 1 6 and B = . Thus the 7 7

general solution of the differential equation is

( )

( )

1 6 y ( x) = D1 cosh 3 x + D2 sinh 3 x + e −3 x sin( 2 x) + e −3 x cos(2 x) 7 7

67

(g)

Application of the boundary conditions to Equation g leads to

y (0) = 0 ⇒ D1 +

6 6 = 0 ⇒ D1 = − 7 7

(h)

and

( )

( )

6 1 6 y (1) = 0 ⇒ − cosh 3 + D2 sinh 3 + e −3 sin(2) + e −3 cos(2) = 0 ⇒ D2 = 0.1480 (i) 7 7 7

The solution of the differential equation which satisfies the boundary conditions is

( )

( )

1 6 y ( x) = −0.8571 cosh 3 x + 0.1480 sinh 3 x + e −3 x sin(2 x) + e −3 x cos(2 x) 7 7

68

(j)

3.13

dy d2y dy + 3 = 3x + 2e − 3 x with y (0) = 0 and (1) = 0 2 dx dx dx

Solution: A homogeneous solution is assumed as y (t ) = Aeαx which when substituted

into the differential equation leads to

α 2 + 3α = 0

(a)

The solutions of Equation a are α = 0,−3 . Thus the homogeneous solution of the differential equation is

y h ( x) = C1 + C 2 e −3 x

(b)

The appropriate form of the particular solution is y p = Ax 2 + Bx + Cxe −3 x

(c)

Substituting Equation c into the differential equation leads to

(

) [

)]

(

2 A + C 9 xe −3 x − 6e −3 x + 3 2 Ax + B + C e −3 x − 3xe −3 x = 3x + 2e −3 x

(d)

4 2 Equation d requires 6 A = 3, 2 A + 3B = 0 and − 3C = 2 or A = 2, B = − and C = − . 3 3 The general solution is the sum of the homogeneous solution and the particular solution

y ( x) = C1 + C 2 e −3 x + 2 x 2 −

4 2 x − xe −3 x 3 3

(e)

Application of the boundary conditions to Equation e leads to

Equations f and g lead to C1 = −

(

y (0) = 0 ⇒ C1 + C 2 = 0

(f)

dy 4 2 (1) = 0 ⇒ −3C 2 e −3 + 4 − − e −3 + 2e −3 dx 3 3

(g)

)

(

)

1 3 1 8e + 2 and C 2 = 8e 3 + 2 . The solution of the 9 9

differential equation which satisfies the boundary conditions is

69

y ( x) =

(

)(

)

4 2 1 3 8e + 2 e −3 x − 1 + 2 x 2 − x − xe −3 x 3 3 9

70

(h)

3.14 The non-dimensional equation for the temperature distribution in the extended surface of Fig. P3.14 is

d 2θ − Biθ = 0 dx 2 where Bi =

(a)

2hL2 (l + b ) . The face at x=0 is maintained at a constant non-dimensional klb

temperature of 1, θ (0) = 1 and is insulated at its right end

dθ (1) = 0 . dx

(a) Determine θ ( x) (b) Determine the rate at which heat is transferred from the base of the surface, q0 = −

klb(T0 − T∞ ) dθ ( 0) L dx

(c) Determine the rate at which heat is transferred over the surface by convection 1

q c = L ∫ 2h(l + b)(T0 − T∞ )θ ( x)dx 0

(d) Compare the answers to (b) and (c) and explain. (e) The efficiency of the extended surface is defined as the ratio of total heat transfer from the surface of the fin as calculated in part (c) to the heat transfer from the surface if were held at the base temperature. Determine the efficiency for the extended surface. Solution: (a) The general solution of Equation a is

(

)

(

θ ( x) = C1 cosh Bi x + C 2 sinh Bi x Application of the boundary conditions leads to

71

)

(b)

θ (0) = 1 ⇒ C1 = 1 dθ dx

(c)

( Bi ) + = − tanh ( Bi )

(1) = 0 ⇒ BiC1 sinh

⇒ C2

BiC 2 s cosh

( Bi ) (d)

The temperature distribution in the extended surface is

(

)

( ) (

θ ( x) = cosh Bi x − tanh Bi sinh Bi x

)

(e)

(b) The rate of heat transfer across the base is klb(T0 − T∞ ) dθ (0) L dx klb(T0 − T∞ ) =− − Bi tanh L

q0 = −

(

( Bi ))

(f)

(c) The rate at which heat is transferred across the surface through convection is 1

q c = L ∫ 2h(l + b)(T0 − T∞ )θ ( x)dx 0

1

[ (

)

= 2hL(l + b )(T0 − T∞ )∫ cosh

Bi x − tanh

( Bi )sinh (

)]

Bi x dx

0

= =

2hL(l + b )(T0 − T∞ ) Bi 2hL(l + b )(T0 − T∞ ) Bi

(d) Noting that Bi =

q0 =

[sinh ( tanh

)

Bi x − tanh

( Bi )cosh (

( Bi )

Bi x

)]

x −1 x =0

(g)

2hL2 (l + b ) , Equation f can be rewritten as klb

klb(T0 − T∞ ) Bi tanh L

( Bi )

⎛ 2hL2 (l + b ) ⎞ klb(T0 − T∞ ) 2hL2 (l + b ) ⎟ = tanh⎜ ⎜ ⎟ l L klb k b ⎝ ⎠ ⎛ 2hL2 (l + b ) ⎞ ⎟ = (T0 − T∞ ) 2hklb(l + b ) tanh⎜ ⎜ ⎟ l k b ⎝ ⎠

Substituting for the Biot number in Equation g leads to

72

(h)

qc =

=

2hL(l + b )(T0 − T∞ ) Bi

2hL(l + b )(T0 − T∞ ) 2hL2 (l + b ) klb

tanh

( Bi )

⎛ 2hL2 (l + b ) ⎞ ⎟ tanh⎜ ⎟ ⎜ klb ⎠ ⎝

⎛ 2hL2 (l + b ) ⎞ ⎟ = (T0 − T∞ ) 2hklb(l + b ) tanh⎜ ⎟ ⎜ k l b ⎠ ⎝

(i)

Equations h and i show that the rate of heat transfer at the base is equal to the rate of heat transfer across the surface by convection. This is a result of an energy balance over the entire fin. (e) The total heat transfer across the surface if the surface is held at a uniform temperature equal to the base temperature θ ( x) = 1 is 1

q c = L ∫ 2h(l + b)(T0 − T∞ )dx 0

= 2hL(l + b)(T0 − T∞ )

(j)

The efficiency of the fin is

⎛ 2hL2 (l + b ) ⎞ ⎟ (T0 − T∞ ) 2hklb(l + b ) tanh⎜ ⎟ ⎜ l k b ⎠ ⎝ η= 2hL(l + b)(T0 − T∞ ) =

1 Bi

tanh

( Bi )

(k)

73

3.15 Rework Problem 3.14 if the surface is subject to an internal heat generation per volume, u(x) such that the non-dimensional equation for the temperature distribution is d 2θ − Biθ = −Λu ( x) dx 2

where Λ =

umax L2 . Let Bi=0.2 and Λ = 0.5 . Solve parts (a)-(c) of Problem 3.12 if (i) klb

u ( x) = x(1 − x) (ii) u ( x) = sin (πx ) and (iii) u ( x) = e − x . Solution: The homogeneous solution of Equation a is

(

)

(

θ ( x) = C1 cosh Bi x + C 2 sinh Bi x

)

(b)

(i) A particular solution is assumed as

θ p ( x) = Ax 2 + Bx + C

(c)

Which when substituted into the differential equation leads to

(

)

(

2 A − Bi Ax 2 + Bx + C = Λ x − x 2 Equation d is satisfied if A = C=

)

(d)

Λ 0.5 Λ = = 2.5 , B = − = −2.5 and Bi 0.2 Bi

2Λ 2(0.5) = = 25 . Thus the particular solution is 0.04 Bi 2 y p ( x) = 2.5 x 2 − 2.5 x + 25

(e)

The general solution of the differential equation is

θ ( x) = C1 cosh (0.4472 x ) + C 2 sinh(0.4472 x) + 2.5 x 2 − 2.5 x + 25

(f)

Application of the boundary conditions to Equation f leads to

θ (0) = 1 ⇒ C1 + 25 = 0 ⇒ C1 = −25 dθ dx

(g)

(1) = 0 ⇒ −25(0.4472) sinh(0.4472) + C 2 (0.4472) cosh(0.4472) + 5 − 2.5 = 0 ⇒ C 2 = 5.416

(h)

74

The temperature distribution is

θ ( x) = −25 cosh (0.4472 x ) + 5.416 sinh(0.4472 x) + 2.5 x 2 − 2.5 x + 25 Noting that

(i)

dθ (0) = 5.416(0.4472) − 2.5 = −0.0782 the rate of heat transfer from dx

the base of the fin is q 0 = −

klb(T0 − T∞ ) dθ klb(T0 − T∞ ) (0) = 0.0782 . L dx L

Noting that 2 ∫ θ ( x)dx = ∫ [− 25 cosh(0.4472 x ) + 5.416 sinh(0.4472 x) + 2.5x − 2.5x + 25]dx 1

1

1

0

x =1

⎧ ⎫ 25 8.221 2.5 x 3 2.5 x 2 = ⎨− − + 25 x ⎬ sinh(0.4472 x) + cosh(0.4472 x) + 0.4472 3 2 ⎩ 0.4472 ⎭ x =0 5.416 [cosh(0.4472) − 1] − 25 sinh(0.4472) + 2.5 − 2.5 + 25 = −0.0270 (k) = 0.4472 0.4472 3 2

the heat transfer by convection across the surface is 1

q c = L ∫ 2h(l + b)(T0 − T∞ )θ ( x)dx = −0.0540 Lh(l + b)(T0 − T∞ )

(l)

0

(ii) A particular solution is assumed as

θ p ( x) = A sin(πx) + B cos(πx)

(c)

Which when substituted into the differential equation leads to − π 2 [ A sin(πx) + B cos(πx )] − 0.2[A sin(πx) + B cos(πx )] = 0.5 sin (πx )

75

(d)

Equation d is satisfied if A = −

0.5 = −0.0497 and B = 0 . Thus the particular π 2 0.2

solution is

θ p ( x) = −0.0497 sin(πx)

(e)

The general solution of the differential equation is

θ ( x) = C1 cosh (0.4472 x ) + C 2 sinh(0.4472 x) − 0.0497 sin(πx)

(f)

Application of the boundary conditions to Equation f leads to

θ (0) = 1 ⇒ C1 = 0 dθ dx

(g)

(1) = 0 ⇒ C 2 (0.4472) cosh(0.4472) − .0497 cos(π ) = 0 ⇒ C 2 = −0.1009

(h)

The temperature distribution is

θ ( x) = −0.1009 sinh(0.4472 x) − 0.0497 sin(πx) Noting that

(i)

dθ (0) = −0.1009(0.4472) − 0.0497(π ) = −0.2013 the rate of heat dx

transfer from the base of the fin is q 0 = −

klb(T0 − T∞ ) dθ klb(T0 − T∞ ) (0) = −0.2013 . L dx L

Noting that 1

1

1

0

∫ θ ( x)dx = ∫ [− 0.1009 sinh(0.4472 x) − 0.0497 sin(πx)]dx x =1

0.0497 ⎧ 0.1009 ⎫ = ⎨− cosh(0.4472 x) + cos(πx)⎬ π ⎩ 0.4472 ⎭ x =0 = −0.0546

the heat transfer by convection across the surface is 1

q c = L ∫ 2h(l + b)(T0 − T∞ )θ ( x)dx = −0.1092 Lh(l + b)(T0 − T∞ ) 0

76

(l)

(iii) A particular solution is assumed as

θ p ( x) = Ae − x

(c)

which when substituted into the differential equation leads to Ae − x − 0.2 Ae − x = 0.5e − x Equation d is satisfied if A =

(d)

0.5 = 0.625 . Thus the particular solution is 0.8

θ p ( x) = 0.625e − x

(e)

The general solution of the differential equation is

θ ( x) = C1 cosh (0.4472 x ) + C 2 sinh(0.4472 x) + 0.625e − x

(f)

Application of the boundary conditions to Equation f leads to

θ (0) = 1 ⇒ C1 + 0.625 = 0 ⇒ C1 = -0.625 dθ dx

(g)

(1) = 0 ⇒ −0.625(0.4472) sinh(0.4472) + C 2 (0.4472) cosh(0.4472) − 0.625e −1 = 0 ⇒ C 2 = 0.7289

(h)

The temperature distribution is

θ ( x) = −0.625 cosh(0.4472 x) + 0.7289 sinh(0.4472 x) + 0.625e − x Noting that

dθ (0) = 0.7289(0.4472) − 0.625 = −.2990 the rate of heat transfer dx

from the base of the fin is q 0 = −

klb(T0 − T∞ ) dθ klb(T0 − T∞ ) (0) = 0.2990 . L dx L

Noting that −x ∫ θ ( x)dx = ∫ [− 0.625 cosh(0.4472 x) + 0.7289 sinh(0.4472 x) + 0.625e ]dx 1

1

1

0

x =1

0.7289 ⎧ 0.625 ⎫ = ⎨− sinh(0.4472 x) + cosh(0.4472 x) − 0.625e − x ⎬ 0.4472 ⎩ 0.4472 ⎭ x =0 = −0.2865

77

the heat transfer by convection across the surface is 1

q c = L ∫ 2h(l + b)(T0 − T∞ )θ ( x)dx = −0.5630 Lh(l + b)(T0 − T∞ ) 0

78

(l)

3.16 Rework Problem 3.14 if the tip of the fin is not perfectly insulated such that the boundary condition at the right end is dθ hL (1) + θ (1) = 0 dx k For computational purposes take

hL = 0.5 and Bi=2. k

Solution:(a) The general solution of Equation a is

(

)

(

θ ( x) = C1 cosh Bi x + C 2 sinh Bi x

)

(b)

Application of the boundary conditions leads to (c) θ (0) = 1 ⇒ C1 = 1 dθ (1) + 0.5θ (1) = 0 ⇒ Bi sinh Bi +C 2 Bi cosh Bi + 0.2 cosh Bi + 0.2C 2 sinh Bi = 0

( )

dx

⇒ C2 = −

( ) ( ) Bi sinh ( Bi ) + 0.2 cosh ( Bi ) = −0.7299 Bi cosh ( Bi ) + 0.2 sinh ( Bi )

( )

(d)

The temperature distribution in the extended surface is

θ ( x) = cosh (0.4472 x ) − 0.7299 sinh (0.4472 x )

(e)

(b) The rate of heat transfer across the base is klb(T0 − T∞ ) dθ (0) L dx klb(T0 − T∞ ) (− 0.7299)(0.4472) =− L klb(T0 − T∞ ) = 0.3264 L

q0 = −

(c) The rate at which heat is transferred across the surface through convection is

79

(f)

1

q c = L ∫ 2h(l + b)(T0 − T∞ )θ ( x)dx 0

1

= 2hL(l + b )(T0 − T∞ )∫ [cosh (0.4472 x ) − 0.7299 sinh (0.4472 x )]dx 0

2hL(l + b )(T0 − T∞ ) = [sinh (0.4472 x ) − 0.7299 cosh(0.4472 x )]xx−=10 0.4472 = 1.73542hL(l + b )(T0 − T∞ )

(g)

(e) The total heat transfer across the surface if the surface is held at a uniform temperature equal to the base temperature θ ( x) = 1 is 1

q c = L ∫ 2h(l + b)(T0 − T∞ )dx 0

= 2hL(l + b)(T0 − T∞ )

(h)

The efficiency of the fin is klb(T0 − T∞ ) L η= 2hL(l + b)(T0 − T∞ ) 0.3264

=

0.1632klb hL2 (l + b )

(i)

80

3.18 The transverse deflection of the uniform beam of Fig. P3.18 is governed by the differential equation d 4w EI 4 = f ( x) dx

(a)

where f(x) is the distributed load per length. Determine the transverse deflection of a pinned-pinned beam which is subject to the boundary conditions w(0) = 0,

w( L) = 0 and

d 2w (0) = 0 , dx 2

d 2w ⎛ 3πx ⎞ ( L) = 0 for (a) f ( x) = F0 x( L − x) and (b) f ( x) = Fm sin ⎜ ⎟. 2 dx ⎝ L ⎠

Solution: The homogeneous solution of Equation a is

w( x) = C1 x 3 + C 2 x 2 + C 3 x + C 4

(b)

(a) A particular solution is assumed of the form w p ( x) = Ax 6 + Bx 5

(c)

which when substituted into Equation a leads to

(

)

EI 360 Ax 2 + 120 Bx = F0 Lx − F0 x 2 Equation d is satisfied if A = −

(d)

F0 FL and B = 0 . The general solution is 360 EI 120 EI

w( x) = C1 x 3 + C 2 x 2 + C 3 x + C 4 −

F0 FL x6 + 0 x5 360 EI 120 EI

Application of the boundary conditions to Equation e leads to

81

(e)

w(0) = 0 ⇒ C 4 = 0

(f)

d 2w (0) = 0 ⇒ C 2 = 0 dx 2

(g)

F0 L6 F0 L6 + w( L) = 0 ⇒ C1 L + C 3 L − 360 EI 120 EI F L6 ⇒ C1 L3 + C 3 L = − 0 (h) 180 EI 30 F0 L4 20 F0 L4 d 2w ( ) = ⇒ − + =0 0 6 L C L 1 360 EI 120 EI dx 2 F0 L3 ⇒ C1 = (i) 72 EI 3

Equations h and i lead to C 3 = −

w( x) =

7 F0 L5 and the deflection is 320 EI F0 ⎡ x 3 L3 7 xL5 x6 x5 L ⎤ − − + ⎢ ⎥ 320 360 120 ⎦ EI ⎣ 72

(j)

(b) A particular solution is assumed as ⎛ 3πx ⎞ ⎛ 3πx ⎞ wm ( x) = A sin ⎜ ⎟ + B cos⎜ ⎟ ⎝ L ⎠ ⎝ L ⎠

(k)

which when substituted into the differential equation leads to ⎛ 3π ⎞ ⎡ ⎛ 3πx ⎞ ⎛ 3πx ⎞⎤ ⎛ 3πx ⎞ EI ⎜ ⎟ ⎢ A sin ⎜ ⎟ + B cos⎜ ⎟⎥ = Fm sin ⎜ ⎟ ⎝ L ⎠ ⎣ ⎝ L ⎠ ⎝ L ⎠⎦ ⎝ L ⎠ 4

(l)

Fm L4 Equation l is satisfied if A = and B=0. The general solution becomes 81π 4 EI w( x) = C1 x 3 + C 2 x 2 + C 3 x + C 4 +

Fm L4 ⎛ 3πx ⎞ sin ⎜ ⎟ 4 81π EI ⎝ L ⎠

Application of the boundary conditions to Equation e leads to

82

(m)

w(0) = 0 ⇒ C 4 = 0

(n)

2

d w (0) = 0 ⇒ C 2 = 0 dx 2 w( L) = 0 ⇒ C1 L3 + C 3 L = 0 d 2w (L ) = 0 ⇒ 6C1 L = 0 dx 2

(o) (p) (q)

Equations p and q imply C 3 = 0 and the beam deflection is Fm L4 ⎛ 3πx ⎞ w( x) = sin ⎜ ⎟ 4 81π EI ⎝ L ⎠

83

(r)

3.19 Repeat Problem 3.18 for a fixed-pinned beam which has boundary conditions dw d 2w w(0) = 0, (0) = 0, w( L) = 0 and 2 ( L) = 0 . dx dx Solution: The homogeneous solution of Equation a is

w( x) = C1 x 3 + C 2 x 2 + C 3 x + C 4

(b)

(a) A particular solution is assumed of the form w p ( x) = Ax 6 + Bx 5

(c)

which when substituted into Equation a leads to

(

)

EI 360 Ax 2 + 120 Bx = F0 Lx − F0 x 2 Equation d is satisfied if A = −

(d)

F0 FL and B = 0 . The general solution is 360 EI 120 EI

w( x) = C1 x 3 + C 2 x 2 + C 3 x + C 4 −

F0 FL x6 + 0 x5 360 EI 120 EI

(e)

Application of the boundary conditions to Equation e leads to w(0) = 0 ⇒ C 4 = 0

(f)

dw (0) = 0 ⇒ C3 = 0 dx F0 L6 F L6 + 0 360 EI 120 EI F0 L6 3 ⇒ C1 L + C 3 L = − 180 EI 2 30 F0 L4 20 F0 L4 d w ( ) L C L C = ⇒ + − + =0 0 6 2 1 2 360 EI 120 EI dx 2 F L4 ⇒ 6C1 L + 2C 2 = 0 12 EI

(g)

w( L) = 0 ⇒ C1 L3 + C 2 L2 −

84

(h)

(i)

17 F0 L3 Equations h and i are solved simultaneously leading to lead to C1 = and the 720 EI C2 = −

7 F0 L4 deflection is 240 EI w( x) =

F0 EI

⎡17 x 3 L3 7 x 2 L4 x6 x5 L ⎤ − − + ⎢ ⎥ 240 360 120 ⎦ ⎣ 720

(j)

(b) A particular solution is assumed as ⎛ 3πx ⎞ ⎛ 3πx ⎞ wm ( x) = A sin ⎜ ⎟ + B cos⎜ ⎟ ⎝ L ⎠ ⎝ L ⎠

(k)

which when substituted into the differential equation leads to ⎛ 3π ⎞ ⎡ ⎛ 3πx ⎞ ⎛ 3πx ⎞⎤ ⎛ 3πx ⎞ EI ⎜ ⎟ ⎢ A sin ⎜ ⎟ + B cos⎜ ⎟⎥ = Fm sin ⎜ ⎟ ⎝ L ⎠ ⎣ ⎝ L ⎠ ⎝ L ⎠⎦ ⎝ L ⎠ 4

(l)

Fm L4 Equation l is satisfied if A = and B=0. The general solution becomes 81π 4 EI w( x) = C1 x 3 + C 2 x 2 + C 3 x + C 4 +

Fm L4 ⎛ 3πx ⎞ sin ⎜ ⎟ 4 81π EI ⎝ L ⎠

(m)

Application of the boundary conditions to Equation e leads to w(0) = 0 ⇒ C 4 = 0

(n)

Fm L3 d 2w ( ) 0 = 0 ⇒ = C 3 27π 3 EI dx 2

(o)

Fm L4 =0 w( L) = 0 ⇒ C1 L + C 2 L + 27π 3 EI d 2w (L ) = 0 ⇒ 6C1 L = 0 dx 2 3

Equations o and p imply C 2 = −

2

(p) (q)

Fm L2 and the beam deflection is 27π 3 EI

Fm L4 w( x) = 81π 4 EI

⎡ 3π 2 3π ⎛ 3πx ⎞ ⎤ ⎢− L2 x + L x + sin ⎜ L ⎟ ⎥ ⎝ ⎠⎦ ⎣

85

(r)

3.20 Repeat Problem 3.19 for a fixed-pinned beam with a concentrated load of magnitude F0 applied at the midspan. One approach to solve this problem is to define w1 ( x) as the transverse deflection between x=0 and x=L/2 and w2 ( x) as the deflection between x=L/2 and L. The boundary conditions of Problem 3.18 apply appropriately to w1 (0) and w2 ( L) . ⎛L⎞ ⎛L⎞ Matching conditions at x=L/2 are that the deflection is continuous, w1 ⎜ ⎟ = w2 ⎜ ⎟ , the ⎝2⎠ ⎝2⎠ slope of the elastic curve is continuous,

continuous,

dw1 ⎛ L ⎞ dw2 ⎛ L ⎞ ⎜ ⎟= ⎜ ⎟ , the bending moment is dx ⎝ 2 ⎠ dx ⎝ 2 ⎠

d 2 w1 ⎛ L ⎞ d 2 w2 ⎛ L ⎞ ⎜ ⎟= ⎜ ⎟ and the shear force is discontinuous at x=L/2 due to dx 2 ⎝ 2 ⎠ dx 2 ⎝ 2 ⎠

the presence of the concentrated load, EI

d 3 w1 ⎛ L ⎞ d 3 w2 ⎛ L ⎞ − EI ⎜ ⎟ ⎜ ⎟ = F0 . dx 3 ⎝ 2 ⎠ dx 3 ⎝ 2 ⎠

Solution: The homogeneous solutions for the deflections are

w1 ( x) = C1 x 3 + C 2 x 2 + C 3 x + C 4

0≤x≤

L 2

(b)

w2 ( x) = D1 x 3 + D2 x 2 + D3 x + D4

L ≤x≤L 2

(c)

Application of the boundary conditions for a fixed-pinned beam leads to w1 ( x) = 0 ⇒ C 4 = 0 dw1 (0) ⇒ C 3 = 0 dx w2 ( L) = 0 ⇒ D1 L3 + D2 L2 + D3 L + D4 = 0 d 2 w2 ( L) = 0 ⇒ 6 D1 L + 2 D2 = 0 dx 2

Application of the matching conditions leads to

86

(d) (e) (f) (g)

3

2

3

2

⎛ L⎞ ⎛ L⎞ ⎛ L⎞ ⎛ L⎞ ⎛ L⎞ ⎛ L⎞ ⎛ L⎞ w1 ⎜ ⎟ = w2 ⎜ ⎟ ⇒ C1 ⎜ ⎟ + C 2 ⎜ ⎟ = D1 ⎜ ⎟ + D2 ⎜ ⎟ + D3 ⎜ ⎟ + D4 ⎝ 2⎠ ⎝ 2⎠ ⎝2⎠ ⎝ 2⎠ ⎝ 2⎠ ⎝ 2⎠ ⎝ 2⎠ 2

2

dw1 ⎛ L ⎞ dw2 ⎛ L ⎞ ⎛ L⎞ ⎛ L⎞ ⎛ L⎞ ⎛ L⎞ ⎜ ⎟= ⎜ ⎟ ⇒ 3C1 ⎜ ⎟ + 2C 2 ⎜ ⎟ = 3D1 ⎜ ⎟ + 2 D2 ⎜ ⎟ + D3 dx ⎝ 2 ⎠ dx ⎝ 2 ⎠ ⎝ 2⎠ ⎝ 2⎠ ⎝2⎠ ⎝ 2⎠ 2 2 d w1 ⎛ L ⎞ d w2 ⎛ L ⎞ ⎛ L⎞ ⎛ L⎞ ⎟= ⎟ ⇒ 6C1 ⎜ ⎟ + 2C 2 = 6 D1 ⎜ ⎟ + 2 D2 2 ⎜ 2 ⎜ dx ⎝ 2 ⎠ dx ⎝ 2 ⎠ ⎝ 2⎠ ⎝2⎠ d 3 w1 ⎛ L ⎞ d 3 w2 ⎜ ⎟− dx 3 ⎝ 2 ⎠ dx 3

(h)

F ⎛ L ⎞ F0 ⇒ 6C1 − 6 D1 = 0 ⎜ ⎟= EI ⎝ 2 ⎠ EI

(i) (j) (k)

Equations f-k are solved simultaneously resulting in

w1 ( x) =

F0 ⎛ 11 3 3 2⎞ ⎜ x − Lx ⎟ EI ⎝ 96 32 ⎠

w2 ( x ) =

F0 ⎛ 5 3 5 1 1 3⎞ x + Lx 2 − L2 x + L ⎟ ⎜− EI ⎝ 96 32 8 48 ⎠

0≤x≤

87

L 2

L ≤x≤L 2

(l) (m)

3.21 Repeat Problem 3.18 for a beam pinned at x=0, but with a linear spring of stiffness k d 2w attached at x=L. The appropriate boundary conditions are w(0) = 0, 2 (0) = 0, dx d 2w d 3w ( L ) = 0 and EI ( L) + kw( L) = 0 . dx 2 dx 3 Solution: The homogeneous solution of Equation a is

w( x) = C1 x 3 + C 2 x 2 + C 3 x + C 4

(b)

(a) A particular solution is assumed of the form w p ( x) = Ax 6 + Bx 5

(c)

which when substituted into Equation a leads to

(

)

EI 360 Ax 2 + 120 Bx = F0 Lx − F0 x 2 Equation d is satisfied if A = −

(d)

F0 FL and B = 0 . The general solution is 360 EI 120 EI

w( x) = C1 x 3 + C 2 x 2 + C 3 x + C 4 −

F0 FL x6 + 0 x5 360 EI 120 EI

(e)

Application of the boundary conditions to Equation e leads to

w(0) = 0 ⇒ C 4 = 0

(f)

d 2w (0 ) = 0 ⇒ C 2 = 0 (g) dx 2 30 F0 L4 20 F0 L4 d 2w (L ) = 0 ⇒ 6 C 1 L − + =0 360 EI 120 EI dx 2 F0 L3 (h) ⇒ C1 = 72 EI

88

d 3w EI 3 ( L) + kw( L) = 0 ⇒ dx ⎛ F0 L6 ⎛ F L3 ⎞ 120 F0 L3 60 F0 L3 F0 L6 F0 L5 ⎞ ⎟⎟ = 0 ⎜ k C L 6⎜⎜ 0 ⎟⎟ − + + ⎜ + 3 − + EI EI EI 72 360 120 72 360 120 ⎠ ⎝ ⎠ ⎝ 2 5 FL 7F L ⇒ C2 = − 0 − 0 12k 360 EI

(i)

The deflection is w( x) =

F0 L3 3 ⎛ F0 L2 7 F0 L5 ⎞ F0 FL ⎟⎟ x − x − ⎜⎜ + x6 + 0 x5 72 EI 360 EI 120 EI ⎝ 12k 360 EI ⎠

(j)

(b) A particular solution is assumed as ⎛ 3πx ⎞ ⎛ 3πx ⎞ wm ( x) = A sin ⎜ ⎟ + B cos⎜ ⎟ ⎝ L ⎠ ⎝ L ⎠

(k)

which when substituted into the differential equation leads to ⎛ 3π ⎞ ⎡ ⎛ 3πx ⎞ ⎛ 3πx ⎞⎤ ⎛ 3πx ⎞ EI ⎜ ⎟ ⎢ A sin ⎜ ⎟ + B cos⎜ ⎟⎥ = Fm sin ⎜ ⎟ ⎝ L ⎠ ⎝ L ⎠ ⎣ ⎝ L ⎠ ⎝ L ⎠⎦ 4

Equation l is satisfied if A =

(l)

Fm L4 and B=0. The general solution becomes 81π 4 EI

w( x) = C1 x 3 + C 2 x 2 + C 3 x + C 4 +

Fm L4 ⎛ 3πx ⎞ sin ⎜ ⎟ 4 81π EI ⎝ L ⎠

(m)

Application of the boundary conditions to Equation e leads to w(0) = 0 ⇒ C 4 = 0

(n)

2

d w (0) = 0 ⇒ C 2 = 0 dx 2 d 2w (L ) = 0 ⇒ 6C1 L = 0 dx 2 F L F d 3w EI 3 ( L) + kw( L) = 0 ⇒ − m + k (C 3 L ) = 0 ⇒ C 3 = m 3π 3kπ dx The deflection is

89

(o) (p)

(q)

Fm Fm L4 ⎛ 3πx ⎞ sin ⎜ x+ w( x) = ⎟ 4 3kπ 81π EI ⎝ L ⎠

90

(r)

3.22 The transverse deflection of a beam on an elastic foundation (a Winkler foundation), Fig. P3.22, is governed by the non-dimensional differential equation d 4w + η w = Λf ( x ) dx 4

(a)

Determine w(x) for a fixed-free beam when (a) f ( x) = 1 − x and (b) f ( x) = sin (πx ) . The appropriate boundary conditions are w(0) = x,

dw d 3w d2w (0) = 0, 2 (1) = 0 and 3 (1) = 0 . Use dx dx dx

η = 4 and Λ = 2 . Solution: A homogeneous solution is assumed of the form w( x) = e αx . When substituted

into Equation a, the algebraic equation α 4 + 2 = 0 results. The solutions of the equation are

α = ±0.8409(1 ± i )

(b)

The resulting homogeneous solution is w( x) = C1 cosh (0.8409 x ) cos(0.8409 x) + C 2 cosh(0.8409 x) sin(0.8409 x) + C 3 sinh(0.8409 x) cos(0.8409 x) + C 4 sinh(0.8409 x) sin(0.8409 x) (c) (i) Using the method of undetermined coefficients a particular solution is determined as w p ( x) = 1 − x . Thus the general solution is w( x) = C1 cosh (0.8409 x ) cos(0.8409 x) + C 2 cosh(0.8409 x) sin(0.8409 x) + C 3 sinh(0.8409 x) cos(0.8409 x) + C 4 sinh(0.8409 x) sin(0.8409 x) + 1 − x

(d)

Application of the boundary conditions leads to w(0) = 0 ⇒ C1 + 1 = 0 ⇒ C1 = −1

(e)

dw (0) = 0 ⇒ 0.8409C 2 + 0.8409C 3 − 1 = 0 dx d 2w (1) = 0 ⇒ −C1 sinh(0.8409) sin(0.8409) + C 2 sinh(0.8409) cos(0.8409) dx 2 − C 3 cosh(0.8409) sin(0.8409) + C 4 cosh(0.8409) cos(0.8409) = 0

91

(f)

(g)

d 3w (1) = 0 ⇒ −C1 [sinh(0.8409) cos(0.8409) + cosh(0.8409) sin(0.8409)] dx 3 + C 2 [cosh(0.8409) cos(0.8409) − sinh(0.8409) sin(0.8409)] − C 3 [sinh(0.8409) sin(0.8409)

+ cosh(0.8409) cos(0.8409)] + C 4 [sinh(0.8409) cos(0.8409) − cosh(0.8409) sin(0.8409)] = 0

(h)

Equations e-f are solved simultaneously to obtain values of the constants of integration.

(ii) The method of undetermined coefficients is used to determine a particular solution as w p ( x) =

2 sin (πx ) = 0.0201sin(πx) leading to a general solution of π +2 4

w( x) = C1 cosh (0.8409 x ) cos(0.8409 x) + C 2 cosh(0.8409 x) sin(0.8409 x) + C 3 sinh(0.8409 x) cos(0.8409 x) + C 4 sinh(0.8409 x) sin(0.8409 x) + 0.0201sin(πx) Application of the boundary conditions leads to

w(0) = 0 ⇒ C1 = 0

(j)

dw (0) = 0 ⇒ 0.8409C 2 + 0.8409C 3 − (0.0201)π = 0 dx d 2w (1) = 0 ⇒ −C1 sinh(0.8409) sin(0.8409) + C 2 sinh(0.8409) cos(0.8409) dx 2 − C 3 cosh(0.8409) sin(0.8409) + C 4 cosh(0.8409) cos(0.8409) = 0

(k)

(l)

d 3w (1) = 0 ⇒ −C1 [sinh(0.8409) cos(0.8409) + cosh(0.8409) sin(0.8409)] dx 3 + C 2 [cosh(0.8409) cos(0.8409) − sinh(0.8409) sin(0.8409)] − C 3 [sinh(0.8409) sin(0.8409) + cosh(0.8409) cos(0.8409)] + C 4 [sinh(0.8409) cos(0.8409) − cosh(0.8409) sin(0.8409)] + 0.0201π 3 = 0

(m)

Equations k-m are solved to determine the remaining constants of integration.

92

(i)

3.23 Determine the static deflection of a fixed-free beam on a Winkler foundation when a force is applied at the free end of the beam such that its deflection is δ. The problem is dw d 2w d 4w + ηw = 0 subject to w(0) = 0, (0) = 0, 2 (1) = 0 and w(1) = δ . Use η = 4 . What dx 4 dx dx is the value of the required force which is equal to

d 3w (1) . dx 3

Solution: A homogeneous solution is assumed of the form w( x) = e αx . When substituted

into Equation a, the algebraic equation α 4 + 4 = 0 results. The solutions of the equation are

α = ±(1 ± i)

(b)

The resulting homogeneous solution is w( x) = C1 cosh ( x ) cos( x) + C 2 cosh( x) sin( x) + C 3 sinh( x) cos( x) + C 4 sinh( x) sin( x) (c) Application of the boundary conditions leads to w(0) = 0 ⇒ C1 = 0

(d)

dw (0) = 0 ⇒ C 2 + C 3 = 0 (e) dx w(1) = δ ⇒ C1 cosh(1) cos(1) + C 2 cosh(1) sin(1) + C 3 sinh(1) cos(1) + C 4 sinh(1) sin(1) = δ (f) d 2w (1) = 0 ⇒ −C1 sinh(1) sin(1) + C 2 sinh(1) cos(1) dx 2 − C 3 cosh(1) sin(1) + C 4 cosh(1) cos(1) = 0 Equations e-f are solved simultaneously leading to C 2 = −0.690δ , C 3 = 0.690δ and

(g)

C 4 = 1.6δ . The deflection is

w( x) = −0.690δ cosh( x) sin( x) + 0.690δ sinh( x) cos( x) + 1.6δ sinh( x) sin( x) (h) The applied load is calculated as F=

d 3w (1) = 2δ [0.690(cosh(1) sin(1) + sinh(1) cos(1) + cosh(1) cos(1) − sinh(1) sin(1) )] dx 3 + 3.2δ [sinh(1)cos(1) - cosh(1)sin(1)] = 0.3308δ (i)

93

3.24 Solve

d ⎛ 2 dw ⎞ 1 ⎜r ⎟ + w = 0 subject to w(1)=4. dr ⎝ dr ⎠ 2

Solution: The differential equation has the form of a Cauchy-Euler equation for which

substitution of an assumed solution of w = r α leads to

α (α − 1) + 2α + α 2 +α +

1 =0 2

1 =0 2

Use of the quadratic formula to determine the roots of Equation a leads to α =

(a) 1 (− 1 ± i ) . 2

Since the roots are complex the appropriate general solution of the differential equation is

94

3.25 Solve

d ⎛ 2 dw ⎞ 1 ⎜r ⎟ − w = 0 subject to w(0.5)=2 and w(1)=4 dr ⎝ dr ⎠ 2

Solution: The differential equation has the form of a Cauchy-Euler equation for which

substitution of an assumed solution of w = r α leads to

α (α − 1) + 2α − α 2 +α −

1 =0 2

1 =0 2

(a)

Use of the quadratic formula to determine the roots of Equation a leads to

α=

(

)

1 − 1 ± 3 . The general solution of the differential equation is 2 w = C1 r (−1− 3 )/ 2 + C 2 r (−1+ 3 )/ 2

(b)

Requiring w(1)=4 leads to 4 = C1 + C 2 . Requiring w(0.5)=2 then leads to 2 = 2.577C1 + 0.7759C 2 . Simultaneous solution of these equations gives C1 = −0.6127, C 2 − 4.6127 and the solution becomes w = −0.6127 r (−1− 3 )/ 2 + 4.6127 r (−1+ 3 )/ 2

95

(c)

3.27 Solve

d ⎛ 2 dw ⎞ ⎜r ⎟ + 3w = 0 subject to w(1)=2 and w(2)=-2 dr ⎝ dr ⎠

Solution: The differential equation has the form of a Cauchy-Euler equation for which

substitution of an assumed solution of w = r α leads to

α (α − 1) + 2α + 3 = 0 α 2 +α + 3 = 0

(a)

Use of the quadratic formula to determine the roots of Equation a leads to

α=

(

)

1 − 1 ± i 11 . The general solution of the differential equation is 2 w(r ) = C1 r

−

1 2

1 − ⎞ ⎛ 11 ⎞ ⎛ 11 cos⎜⎜ ln(r ) ⎟⎟ + C 2 r 2 sin ⎜⎜ ln(r ) ⎟⎟ ⎠ ⎝ 2 ⎠ ⎝ 2

(b)

Application of the boundary conditions leads to w(1) = 2 → C1 = 2

(c)

and 1 1 ⎞ ⎛ 11 ⎞ ⎛ 11 − − ln(2) ⎟⎟ + C 2 (2) 2 sin ⎜⎜ w(2) = 2 ⇒ (2)(2) 2 cos⎜⎜ ln(2) ⎟⎟ ⇒ C 2 = 2.2031 ⎠ ⎝ 2 ⎠ ⎝ 2

(d)

Thus the solution of the differential equation which satisfies the boundary conditions is w(r ) = 2r

−

1 2

1 − ⎞ ⎛ 11 ⎞ ⎛ 11 cos⎜⎜ ln(r ) ⎟⎟ + 2.2031r 2 sin ⎜⎜ ln(r ) ⎟⎟ ⎠ ⎝ 2 ⎠ ⎝ 2

96

(e)

3.28 Solve r 2

d 2w dw + 3r + 2w = 0 subject to w(1)=2 and w(2)=1 2 dr dr

Solution: The differential equation has the form of a Cauchy-Euler equation for which

substitution of an assumed solution of w = r α leads to

α (α − 1) + 3α + 2 = 0 α 2 + 2α + 2 = 0

(a)

Use of the quadratic formula to determine the roots of Equation a leads to α = −1 ± i . The general solution of the differential equation is w(r ) = C1 r −1 cos(ln(r ) ) + C 2 r −1 sin (ln(r ) )

(b)

Application of the boundary conditions leads to w(1) = 2 → C1 = 2

(c)

and w(2) = 1 ⇒ 1 = (2)(2 ) cos(ln(2) ) + C 2 (2 ) sin (ln(2) ) ⇒ C 2 = 1.2953 −1

−1

(d)

Thus the solution of the differential equation which satisfies the boundary conditions is w(r ) = 2r −1 cos(ln(r ) ) + 1.2953r −1 sin (ln(r ) )

97

(e)

3.29 Solve r 2

d 2w dw − 6r + 9w = 0 subject to w(2)=-1 and w(1)=0 2 dr dr

Solution: The differential equation has the form of a Cauchy-Euler equation for which

substitution of an assumed solution of w = r α leads to

α (α − 1) − 6α + 9 = 0 α 2 − 7α + 9 = 0

(a)

Use of the quadratic formula to determine the roots of Equation a leads to

α=

(

)

1 7 ± 13 = 1.6972,5.3028 . The general solution of the differential equation is 2 w(r ) = C1 r 1.6972 + C 2 r 5.3028

(b)

Application of the boundary conditions leads to w(1) ==⇒ C1 + C 2 = 0

(c)

and w(2) = −1 ⇒ −1 = (2)1.6972 C1 + (2)

5.3028

C 2 ⇒ 3.2428C1 + 39.4725C 2 = −1

(d)

Simultaneous solution of Equations c and d leads to C1 = 0.0276 and C 2 = −0.0276 Thus the solution of the differential equation which satisfies the boundary conditions is

(

w(r ) = 0.0276 r 1.6972 − r 5.3208

98

)

(e)

3.33 Use the power series method to determine two linearly independent solutions to d2y + 4y = 0 dx 2

(a)

Solution: The point x=0 is an ordinary point of the differential equation. A series

solution is assumed to be of the form ∞

y ( x) = ∑ c k x k

(a)

k =0

Substituting Equation a into the given differential equation leads to ∞

∑c k =2

∞

k

(k )(k − 1) x k − 2 + 4∑ c k x k = 0

(b)

k =1

Renaming the index of summation of the firs summation in Equation b to m and then setting k=m-2 leads to ∞

∞

k =0

k =0

∑ ck +2 (k + 2)(k + 1) x k + 4∑ ck x k = 0

(c)

Equation c is satisfied for all x if and only if (k + 1)(k + 2)c k + 2 + 4c k = 0

k = 0,1,2,...

(d)

Equation d is rearranged to ck +2 =

−4 ck (k + 1)(k + 2)

Evaluation of Equation e for even values of k leads to

99

(e)

c2 =

−4 c0 (1)(2)

c4 =

−4 42 c2 = c0 (3)(4) (3)(4)(2)(1)

−4 − 43 − 43 c6 = c4 = c0 = c0 (5)(6) (6)(5)(4)(3)(2)(1) 6! ck

(− 1) =

k 2

(− 1) 2 2 k c 4 c0 = 0 (k )! k! k 2

k

(f) k −1

It is similarly shown that for odd values of k that c k

(− 1) 2 =

2k

k!

c1 . Thus the general

solution of the differential equation is y ( x) = c0

∞

∑ (− 1) 2

k = 0 , 2 , 4 ,...

k

∞ k −1 2 k 2k k x + c1 ∑ (− 1) 2 xk k! k ! k =1, 3, 5 ,......

It is noted that the expansions in Equation g are those for cos(2x) and sin(2x) respectively.

100

(g)

3.36 Use a series solution to obtain two linearly independent solutions to d2y − 4y = 0 . dx 2 Solution: The point x=0 is an ordinary point of the differential equation. A series

solution is assumed to be of the form ∞

y ( x) = ∑ c k x k

(a)

k =0

Substituting Equation a into the given differential equation leads to ∞

∑c k =2

∞

k

(k )(k − 1) x k − 2 − 4∑ c k x k = 0

(b)

k =1

Renaming the index of summation of the firs summation in Equation b to m and then setting k=m-2 leads to ∞

∞

k =0

k =0

∑ ck +2 (k + 2)(k + 1) x k − 4∑ ck x k = 0

(c)

Equation c is satisfied for all x if and only if (k + 1)(k + 2)c k + 2 − 4c k = 0

k = 0,1,2,...

(d)

Equation d is rearranged to ck +2 =

4 ck (k + 1)(k + 2)

Evaluation of Equation e for even values of k leads to

101

(e)

c2 =

4 c0 (1)(2)

c4 =

4 42 c2 = c0 (3)(4) (3)(4)(2)(1)

4 43 43 c6 = c4 = c0 = c0 (5)(6) (6)(5)(4)(3)(2)(1) 6! ck =

k 2

4 2k c0 = c0 (k )! k!

(f)

It is similarly shown that for odd values of k that c k =

2k c1 . Thus the general solution of k!

the differential equation is y ( x) = c0

∞

∞ 2k k 2k k x + c1 ∑ x k =1, 3, 5,...... k! k = 0 , 2 , 4 ,... k!

∑

It is noted that the expansions in Equation g are those for cosh(2x) and sinh(2x) respectively.

102

(g)

3.37 Use the method of Frobenius to determine solutions of d2y dy x 2 + 3 + xy = 0 dx dx

(a)

Solution: x=0 is regular singular point of Equation a. An expansion for y(x) is assumed

as ∞

y ( x ) = xν ∑ c k x k

(b)

k =0

Substituting Equation b into Equation a leads to ∞

∑ c (k + ν )(k + ν − 1)x k =0

k +ν −1

k

∞

∞

k =0

k =0

+ 3∑ c k (k + ν ) x k +ν −1 + ∑ c k x k +ν +1 = 0

(c)

Factoring xν from every term and rearranging Equation c leads to

(

)

(

)

∞

c0 ν 2 + 2ν x −1 + c1 ν 2 + 4ν + 3 + ∑ [c k + 2 (k + ν + 2)(k + ν + 4) + c k ]x k +1

(d)

k =0

The indicial equation is ν 2 + 2ν = 0 whose solutions are ν = 0 and ν = −2 . Since the two solutions differ by an integer Case 3 of the method of Frobenius applies. The first solution is obtained by setting ν = 0 and c1 = 0 in Equation d leading to ∞

∑ [c (k + 2)(k + 4) + c ]x k =0

k +2

Setting the coefficients of powers of x to zero independently leads to

103

k

k +1

(e)

1 c0 (2)(4) 1 1 c4 = − c2 = c0 (4)(6) (2)(4)(4)(6) 1 −1 c6 = − c4 = c0 (6)(8) (2)(4)(4)(6)(6)(8) 1 c0 =− 6 2 (1)(2)(2)(3)(3)(4) 1 c0 =− 6 2 4!3! c2 = −

(f)

Equation f suggests that for even k a general form for the coefficients are obtained as

(− 1) 2 k

ck =

⎛k⎞⎛k ⎞ 2 ⎜ ⎟!⎜ + 1⎟! ⎝2⎠ ⎝2 ⎠

c0

(g)

k

Equation g is confirmed through mathematical induction. One solution of Equation a is

y ( x) =

k 2

(−1) xk ⎞ k = 0 , 2 , 4 ,... k ⎛ k ⎞ ⎛ k 2 ⎜ ⎟!⎜ + 1⎟! ⎝2⎠ ⎝2 ⎠ ∇

∑

∞

=∑

(− 1)k

2k k!(k + 1)! k =0 2

x 2k

A second solution is obtained using Case 3 of the method of Frobenius.

104

(h)

3.38 Use the method of Frobenius to determine solutions of d2y dy x − 2x + x 2 y = 0 2 dx dx 2

(a)

Solution: x=0 is regular singular point of Equation a. An expansion for y(x) is assumed

as ∞

y ( x ) = xν ∑ c k x k

(b)

k =0

Substituting Equation b into Equation a leads to ∞

∑ c (k + ν )(k + ν − 1)x k =0

k

k +ν

∞

∞

k =0

k =0

− 2∑ c k (k + ν ) x k +ν + ∑ c k x k +ν + 2 = 0

(c)

Factoring xν from every term and rearranging Equation c leads to

(

)

(

)

∞

c0 ν 2 − 3ν x + c1 ν 2 − 2 − 2 x + ∑ [c k + 2 (k + ν + 2 )(k + ν − 1) + c k ]x k + 2

(d)

k =0

The indicial equation is ν 2 − 3ν = 0 whose solutions are ν = 0 and ν = 3 . Since the two solutions differ by an integer Case 3 of the method of Frobenius applies. The first solution is obtained by setting ν = 3 and c1 = 0 in Equation d leading to ∞

∑ [c (k + 5)(k + 2) + c ]x k =0

k +2

Setting the coefficients of powers of x to zero independently leads to

105

k

k +1

(e)

1 c0 (5)(2) 1 1 c4 = − c2 = c0 (7)(4) (2)(5)(7)(4) 1 −1 c6 = − c4 = c0 (9)(6) (2)(4)(6)(5)(7)(9) 1 =− 3 c0 2 (1)(2)(3)(5)(7)(9) 1 =− 3 c0 2 3! π (m ) c2 = −

(f)

m=5, 7 , 9

Equation f suggests that for even k a general form for the coefficients are obtained as

(− 1) 2 k

ck =

⎛k⎞ (m ) 2 ⎜ ⎟! π ⎝ 2 ⎠ m =5, 7 ,9...k +3 k 2

c0

(g)

Equation g is confirmed through mathematical induction. One solution of Equation a is

(− 1) 2 k

∇

∑

y ( x) =

k = 0 , 2 , 4 ,...

⎛k⎞ (m ) 2 ⎜ ⎟! π ⎝ 2 ⎠ m =5,7 ,9...k +3

(− 1)k

∞

=∑ k =0

k 2

k

2 k!

(m ) m =5, 7 , 9...2 k + 3 π

xk

x 2k

A second solution is obtained using Case 3 of the method of Frobenius.

106

(h)

3.39-3.41 Determine the solution of each equation in terms of Bessel functions. 3.39

d ⎛ 3 dy ⎞ 2 ⎜x ⎟ + 4x y = 0 dx ⎝ dx ⎠

Solution: The solution is

⎛ 1⎞ ⎛ 1⎞ y ( x) = C1 x −1 J 2 ⎜⎜ 4 x 2 ⎟⎟ + C 2 x −1Y2 ⎜⎜ 4 x 2 ⎟⎟ ⎝ ⎠ ⎝ ⎠

107

3.40

d ⎛ dy ⎞ ⎜ x ⎟ − 16 xy = 0 dx ⎝ dx ⎠

Solution: The solution is

y ( x) = C1 I 0 (4 x ) + C 2 K 0 (4 x )

108

3.41

d2y + 2x2 y = 0 2 dx

Solution: The solution is 1 ⎛ 2 2⎞ ⎛ 2 2⎞ ⎜ ⎟ y ( x) = C1 x J 1 ⎜ x ⎟ + C1 x 2 Y1 ⎜⎜ x ⎟⎟ 2 2 4⎝ 4⎝ ⎠ ⎠ 1 2

109

3.42 Evaluate the integral

∫ [J ( 1

6n

)]

2

λ r rdr

0

Solution: The integral is evaluated as

∫ [J ( 1

6n

0

λr

)]

2

[ ( λ r )]

1 ⎡ 2 2 rdr = ⎢(λr − 36n ) J 6 n 2λ ⎣⎢ =

1 2λ

2

⎡d + r ⎢ J 6n ⎣ dr 2

( )

[ ( λ )] + ⎡⎢⎣ drd J ( λ r )⎤⎥⎦

⎡ 2 ⎢(λ − 36n ) J 6 n ⎣

110

2

6n

r =1

2 ⎤ ⎤ λr ⎥ ⎥ ⎦ ⎦⎥ r =o

2⎤ + 36n 2 [J 6 n (0)] ⎥ r =1 ⎦

3.43 Evaluate the integral 1

∫ rJ (ar )dr 0

0

Solution: A basic identity of Bessel functions is

[

]

d ν x Jν ( x ) = x v Jν −1 ( x) dx

(a)

Application of Equation a for ν = 1 leads to d [xJ1 (x )] = xJ 0 ( x) dx

(b)

Let x = ar . Equation b is rewritten as 1 d [arJ 1 (ar )] = arJ 0 (ar ) a dr

(c)

Equation c is rearranged to rJ 0 (ar ) =

1 d [rJ 1 (ar )] a dr

(d)

Thus the integral is written as 1

1

0

0

∫ rJ 0 (ar )dr = ∫

1 d [rJ 1 (ar )]dr a dr

1 [rJ 1 (ar )]rr ==10 a J (a ) = 1 a =

111

(e)

CHAPTER 4 VARIATIONAL METHODS

112

4.1 (a) Find the least squares approximation to f ( x) = xe −2 x over the interval − 1 ≤ x ≤ 1 from the span of 1, x, x 2 using the standard inner product on C[-1,1].

(b)Calculate the error in the approximation using the inner product generated norm. (c) Determine the Taylor series expansion for f(x) through the quadratic term. Calculate the error in this approximation and compare to the error in the least squares approximation. Solution: The calculations are illustrated in the accompanying MATHCAD output Solution to Problem 4.1

f ( x) := x⋅ exp( −2x)

Function to be approximated

p1( x) := 1

p2( x) := x

2

p3( x) := x

Basis vectors

⎛ p1( x) ⎞ p ( x) := ⎜ p2( x) ⎟ ⎜ ⎟ ⎝ p3( x) ⎠

Inner products of f with basis vectors i := 1 .. 3

⌠ b := ⎮ i ⌡

1

⎛ −1.949 ⎞ b = ⎜ 1.678 ⎟ ⎜ ⎟ ⎝ −1.245 ⎠

f ( x) ⋅ p ( x) dx i

−1

j := 1 .. 3

⌠ a := ⎮ i, j ⌡

1

p ( x) ⋅ p ( x) dx i

Least squares matrix

j

−1

0 0.667 ⎞ ⎛ 2 ⎜ 0 0.667 0 ⎟ −a 1= α := a ⋅ b⎜ ⎟ 0.4 ⎠ ⎝ 0.667 0

113

Coefficient vector

⎛ 0.142 ⎞ α = ⎜ 2.517 ⎟ ⎜ ⎟ ⎝ −3.349 ⎠

g ( x) := α ⋅ p1( x) + α ⋅ p2( x) + α ⋅ p3( x) 1

2

3

x := 0 , .02.. 1

1

0.5 f ( x) g( x)

0

− 0.5

−1

0

0.2

0.4

0.6

0.8

1

x

Error in approximation

⎡⌠ 1 ⎤ ⎢ 2 ⎥ ⎮ e := ⎢⌡ ( f ( x) − g( x) ) dx⎥ ⎣ −1 ⎦

0.5

e = 0.637

(c) The Taylor series approximation, taken through the quadratic term is df x2 d 2 f (0) T ( x) = f (0) + x (0) + 2 dx 2 dx It is noted that

114

(a)

f ( x) = xe −2 x ⇒ f (0) = 0 df df = −2 xe − 2 x + e − 2 x ⇒ (0) = 1 dx dx d2 f d2 f −2 x −2 z = 4 xe − e ⇒ (0) = −1 dx 2 dx 2 Thus T ( x) = x −

(b) (c) (d)

1 2 x 2

Comparison with Taylor series expansion 2

T( x) := x − 0.5⋅ x

1

0.5 T ( x) f ( x)

0

g( x) − 0.5 −1

0

0.2

0.4

0.6 x

115

0.8

1

4.2 Repeat Problem 4.1 using the Legendre polynomials P0 ( x), P1 ( x), P2 ( x ) as basis functions.

Solution: The solution is developed in the accompanying MATHCAD program Solution to Problem 4.2

f ( x) := x⋅ exp( −2x)

Function to be approximated

p1( x) := 1

p2( x) := x

3 2 1 ⋅x − 2 2

Legendre polynomials as basis vectors

⎛ p1( x) ⎞ p ( x) := ⎜ p2( x) ⎟ ⎜ ⎟ ⎝ p3( x) ⎠

i := 1 .. 3 ⌠ b := ⎮ i ⌡

p3( x) :=

1

f ( x) ⋅ p ( x) dx

Inner products of f with basis vectors

i

−1

j := 1 .. 3

a

⌠ := ⎮ i, j ⌡

1

p ( x) ⋅ p ( x) dx i

Least squares matrix

j

−1

⎛ −1.949 ⎞ b = ⎜ 1.678 ⎟ ⎜ ⎟ ⎝ −0.893 ⎠

⎛2 0 0 ⎞ a = ⎜ 0 0.667 0 ⎟ ⎜ ⎟ ⎝ 0 0 0.4 ⎠

α := a

−1

⋅b

Coefficient vector

⎛ −0.974 ⎞ α = ⎜ 2.517 ⎟ ⎜ ⎟ ⎝ −2.233 ⎠

g ( x) := α ⋅ p1( x) + α ⋅ p2( x) + α ⋅ p3( x) 1

2

3

x := 0 , .02.. 1

116

1

0.5 f ( x) g( x)

0

− 0.5

−1

0

0.2

0.4

0.6

0.8

1

x

Error in approximation

⎤ ⎡⌠ 1 ⎢ 2 ⎥ e := ⎮ ( f ( x) − g ( x) ) dx ⎢⌡ ⎥ ⎣ −1 ⎦

0.5

e = 0.637

Comparison with Taylor series expansion 2

T( x) := x − 0.5⋅ x

1

0.5 T ( x) f ( x)

0

g( x)

− 0.5 −1

0

0.2

0.4

0.6 x

117

0.8

1

4.3 Find the least squares approximation for f ( x) = cos 2 (2 x) over the interval 0 ≤ x ≤ 2π using 1, x 2 , x 4 as basis functions and using the standard inner product on

C[0,2π].

Solution: The solution is developed in the attached MATHCAD program Solution to Problem 4.3

f ( x) := cos ( 2⋅ x)

2

Function to be approximated 2

p2( x) := x

p1( x) := 1

Basis vectors

⎛ p1( x) ⎞ p ( x) := ⎜ p2( x) ⎟ ⎜ ⎟ ⎝ p3( x) ⎠

i := 1 .. 3 ⌠ b := ⎮ i ⌡

4

p3( x) := x

2⋅ π

f ( x) ⋅ p ( x) dx

Inner products of f with basis vectors

i

0

j := 1 .. 3

a

⌠ := i, j ⎮ ⌡

2⋅ π

p ( x) ⋅ p ( x) dx i

Least squares matrix

j

0

3 ⎛ 6.283 82.683 1.959 × 10 ⎞⎟ ⎜ 3 4⎟ a = ⎜ 82.683 1.959 × 10 5.523 × 10 ⎟ ⎜ ⎜ 3 4 6⎟ ⎝ 1.959 × 10 5.523 × 10 1.696 × 10 ⎠

α := a

−1

⎛ 3.142 ⎞ ⎜ ⎟ b = ⎜ 41.734 ⎟ ⎜ 3⎟ ⎝ 1.01 × 10 ⎠

⋅b

Coefficient vector

0.52 ⎛ ⎞ ⎜ ⎟ −3 α = ⎜ −6.203 × 10 ⎟ ⎜ −4 ⎟ ⎝ 1.967 × 10 ⎠

g ( x) := α ⋅ p1( x) + α ⋅ p2( x) + α ⋅ p3( x) 1

x := 0 , .02.. 2⋅ π

2

3

118

1 0.8 f ( x) 0.6 g( x) 0.4 0.2 0

0

2

4

6

x

119

8

4.4 Repeat Problem 4.3 using a set of orthonormal functions which have the same span as 1, x 2 , x 4 as basis functions.

Solution: Gram-Schmidt is first used to determine an orthonormal basis. Then since the basis vectors are orthonormal the least squares coefficients are simply inner product evaluations. The details are presented in the accompanying MATHCAD program. Problem 4.4

f ( x) := cos ( 2⋅ x)

2

Function to be approximated 2

p2( x) := x

p1( x) := 1

4

p3( x) := x

Given basis

Gram Schmidt process

⎛ ⌠ 2⋅ π ⎞ ⎜ 2 ⎟ norm1 := ⎮ p1( x) dx ⎜⌡ ⎟ ⎝ 0 ⎠

norm1 = 2.507

p1( x)

v1( x) :=

⌠ ip1 := ⎮ ⌡

0.5

norm1 2⋅ π

p2( x) ⋅ v1( x) dx

ip1 = 32.986

0

w2( x) := p2( x) − ip1⋅ v1( x)

⎛ ⌠ 2⋅ π ⎞ ⎜ 2 ⎟ norm2 := ⎮ w2( x) dx ⎜⌡ ⎟ ⎝ 0 ⎠

v2( x) :=

0.5

norm2 = 29.503

w2( x) norm2

120

inner product

⌠ ip2 := ⎮ ⌡

2⋅ π

⌠ ip3 := ⎮ ⌡

2⋅ π

p3( x) ⋅ v1( x) dx

ip2 = 781.339

0

p3( x) ⋅ v2( x) dx

ip3 = 998.358

0

w3( x) := p3( x) − ip2⋅ v1( x) − ip3⋅ v2( x)

⎛ ⌠ 2⋅ π ⎞ ⎜ 2 ⎟ norm3 := ⎮ w3( x) dx ⎜⌡ ⎟ ⎝ 0 ⎠

v3( x) :=

0.5

norm3 = 297.653

w3( x) norm3

Least squares coefficients ⌠ α1 := ⎮ ⌡

2⋅ π

f ( x) ⋅ v1( x) dx α1 = 1.253

0

⌠ α2 := ⎮ ⌡

2⋅ π

⌠ α3 := ⎮ ⌡

2⋅ π

f ( x) ⋅ v2( x) dx

α2 = 0.013

0

f ( x) ⋅ v3( x) dx

α3 = 0.059

0

121

g ( x) := α1⋅ ( v1( x) + α2⋅ v2( x) ) + α3⋅ v3( x) x := 0 , .02.. 2⋅ π 1

0.8

f ( x) 0.6 g( x) 0.4

0.2

0

0

2

4

6

x

122

8

4.5 Find the least squares approximation to J 0 ( x) using 1, x 2 , x 4 as basis functions and 1

using the inner product defined by ( f , g ) = ∫ f ( x) g ( x) xdx . 0

Solution: The solution is in the attached MATHCAD file. Solution to Problem 4.3

f ( x) := J0( x)

p1( x) := 1

Function to be approximated 2

p2( x) := x

4

p3( x) := x

Basis vectors

⎛ p1( x) ⎞ p ( x) := ⎜ p2( x) ⎟ ⎜ ⎟ ⎝ p3( x) ⎠

i := 1 .. 3 1

⌠ b := ⎮ f ( x) ⋅ p ( x) ⋅ x dx i i ⌡

Inner products of f with basis vectors

0

j := 1 .. 3

1

a

⌠ := ⎮ p ( x) ⋅ p ( x) ⋅ x dx i, j i j ⌡

Least squares matrix

0

⎛ 0.5 0.25 0.167 ⎞ a = ⎜ 0.25 0.167 0.125 ⎟ ⎜ ⎟ ⎝ 0.167 0.125 0.1 ⎠

α := a

⎛ 0.44 ⎞ b = ⎜ 0.21 ⎟ ⎜ ⎟ ⎝ 0.137 ⎠

−1

⋅b

⎛ 1 ⎞ α = ⎜ −0.25 ⎟ ⎜ ⎟ ⎝ 0.015 ⎠

Coefficient vector

123

g ( x) := α ⋅ p1( x) + α ⋅ p2( x) + α ⋅ p3( x) 1

2

3

x := 0 , .02.. 1

1

0.9 f ( x) g( x) 0.8

0.7

0

0.2

0.4

0.6

0.8

1

x

Error in approximation

⎤ ⎡⌠ 1 ⎢ 2 ⎥ e := ⎮ ( f ( x) − g ( x) ) ⋅ x dx ⎢⌡ ⎥ ⎣0 ⎦

0.5 −6

e = 5.621 × 10

124

4.6 Find the least squares approximation to J 1 ( x) using x, x 3 , x 5 as basis functions and 1

using the inner product defined by ( f , g ) = ∫ f ( x) g ( x) xdx . 0

Solution: The solution is in the attached MATHCAD file. Solution to Problem 4.3

f ( x) := J1( x)

Function to be approximated 3

p2( x) := x

p1( x) := x

5

p3( x) := x

Basis vectors

⎛ p1( x) ⎞ p ( x) := ⎜ p2( x) ⎟ ⎜ ⎟ ⎝ p3( x) ⎠ i := 1 .. 3 1

⌠ b := ⎮ f ( x) ⋅ p ( x) ⋅ x dx i i ⌡

Inner products of f with basis vectors

0

j := 1 .. 3

1

⌠ a := ⎮ p ( x) ⋅ p ( x) ⋅ x dx i, j i j ⌡

Least squares matrix

0

⎛ 0.25 0.167 0.125 ⎞ a = ⎜ 0.167 0.125 0.1 ⎟ ⎜ ⎟ ⎝ 0.125 0.1 0.083 ⎠

α := a

⎛ 0.115 ⎞ b = ⎜ 0.076 ⎟ ⎜ ⎟ ⎝ 0.056 ⎠

−1

⋅b

Coefficient vector

0.5 ⎛ ⎞ ⎜ ⎟ −0.062 ⎟ α=⎜ ⎜ 2.513 × 10− 3 ⎟ ⎝ ⎠

125

g ( x) := α ⋅ p1( x) + α ⋅ p2( x) + α ⋅ p3( x) 1

2

3

x := 0 , .02.. 1

0.5 0.4 f ( x) 0.3 g( x) 0.2 0.1 0

0

0.2

0.4

0.6

0.8

1

x

Error in approximation

⎤ ⎡⌠ 1 ⎢ 2 ⎥ ⎮ e := ⎢⌡ ( f ( x) − g ( x) ) ⋅ x dx⎥ ⎣0 ⎦

0.5 −7

e = 3.769 × 10

126

4.7 Determine the least squares quadratic fit for y as a function of x for the data X

0

0.25

0.5

0.75

1.0

1.25

1.5

1.75

2.0

Y

3.0

2.2

2.0

1.9

2.5

3.4

5.1

7.5

10.0

Solution: Define an inner product by ( f , g ) = f (0) g (0) + f (0.25) g (0.25) + f (0.5) g (0.5) + f (0.75) g (0.75) + f (1) g (1) + + f (1.25) g (1.25) + f (1.5) g (1.5) + f (1.75) g (1.75) + f (2) g (2)

(a)

The basis functions for a quadratic least squares fit are p1 ( x) = 1. p 2 ( x) = x, p3 ( x) = x . 2

Using the inner product of Equation a the matrix is a1,1 = (1,1) = 9

a1, 2 = (1, x ) = 0 + 0.25 + 0.5 + 0.75 + 1 + 1.25 + 1.5 + 1.75 + 2 = 9 a1,3 = (1, x 2 ) = (0) 2 + (0,25) 2 + (0.5) 2 + (0.75) 2 + (1) 2 + (1.25) 2 + (1.5) 2 + (1.75) 2 + (2) 2 = 12.75 a 2, 2 = ( x, x) = (0) 2 + (0,25) 2 + (0.5) 3 + (0.75) 2 + (1) 2 + (1.25) 2 + (1.5) 2 + (1.75) 2 + (2) 2 = 12.75

( ) = (x , x ) = (0)

a 2,3 = x, x 2 = (0) 3 + (0,25) 3 + (0.5) 3 + (0.75) 3 + (1) 3 + (1.25) 3 + (1.5) 3 + (1.75) 3 + (2) 3 = 20.25 a 3, 3

2

2

4

+ (0,25) 4 + (0.5) 4 + (0.75) 4 + (1) 4 + (1.25) 4 + (1.5) 4 + (1.75) 4 + (2) 4 = 34.26

The symmetric matrix is 9 12.75 ⎤ ⎡ 9 ⎢ 12.75 20.25⎥⎥ A=⎢ 9 ⎢⎣12.75 20.25 34.26⎥⎦

The components of the right-hand side vector are b1 = ( y,1) = 3 + 2.2 + 2 + 1.9 + 2.5 + 3.4 + 5.1 + 7.5 + 10 = 37.6

b2 = ( y , x ) = (3)(0) + (2.2)(0.25) + (2)(0.5) + (1.9)(0.75) + (2.5)(1) + (3.4)(1.25) + (5.1)(1.5) + (7.5)(1.75) + (10)(2) = 50.5

(

)

b3 = y , x 2 = (3)(0) 2 + (2.2)(0.25) 2 + (2)(0.5) 2 + (1.9)(0.75) 2 + (2.5)(1) 2 + (3.4)(1.25) 2 + (5.1)(1.5) 2 + (7.5)(1.75) 2 + (10)(2) 2 = 83.96

127

The solution of the system Ac=b is c = [3.115 − 4.6985 4.0693] . The quadratic least squares approximation to the data is p( x) = 3.115 − 4.6985 x + 4.0693 x 2

128

4.8 The following table presents the data for lattice sums of an atom above a carbon sheet X (nm)

1.0

1.5

2.0

2.5

3.0

3.5

4.0

S

2.40

0.50

0.17

0.076

0.040

0.025

0.015

By observation it is determined that the data may fit a curve of the form S = ax b . Taking the natural logarithm of the proposed regression equation leads to ln( s ) = ln(a) + b ln( x) . Determine the linear least squares approximation for ln(s) as a function of ln(x). Use the regression to approximate S for x=2.8 nm and extrapolate S for x=5.0 nm.

Solution: A linear regression is to be performed on the following set of data ln(X)

0

0.4055 0.6931 0.9163 1.0986 1.2538

ln(S)

0.8755 -0.693

-1.77

-2.557

-3.219

-3.669

1.3863 -4.200

Let z and x be the basis functions for a liner regression and define ( f , g ) = f (0) g (0) + f (0.4055) g (0.4055) + f (0.6931) g (0.6931) + F (0.9163) g (0.9163) + f (1.0986) g (1.0986) + f (1.2358) g (1.2358) + f (1.3863) g (1.3863) The elements of the symmetric 2x2 least squares matrix are a1, 2 = (1,1) = 7 a1, 2 = ( x, x) = 0 + 0.4055 + 0.6931 + 0.9163 + 1.0986 + 1.2538 + 1.3863 = 5.7526 a 2, 2 = ( x, x) = (0) 2 + (0.4055) 2 + (0.6931) 2 + (0.9163) 2 + (1.0986) 2 + (1.2538) 2

+ (1.3863) 2 = 6.1826 The components of the right-hand side vector are b1 = (1, ln(s )) = 0.8755 − 0.6931 − 1.77 − 2.557 − 3.219 − 3.669 − 4.200 = −15.326 b2 = ( x, ln(s )) = (0)(0.8755) + (0.4055)(−0.6931) + (0.6931)(−1.77) + (0.9163)(−2.557) + (1.0986)(−3.219) + (1.2538)(−3.669) + (1.3863)(−4.200) = −17.0861

129

The solution of the system of Equations Ac=b is c = [0.3388 − 3.0716] . The T

appropriate regression is ln(s)=0.3348-3.0716ln(x) or

s = e 0.3348 x −3.0716 = 1.3977 x −3.0716 . Then s (2.8) = 1.3977(2.8) −3.0716 = 0.0591 and s (5) = 1.3977(5) −3.0716 = 0.0100

130

4.8 The first ten zeroes of the Bessel function of the first kind of order zero J 0 ( x) are 2.405, 5.530, 8.654, 11.792, 14.931, 18.071, 21.212, 24.352, 27.493 and 30.653. Use a quadratic least squares regression to predict the zeroes as a function of the number of the zero.

Solution: Define an inner product by ( f , g ) = f (1) g (1) + f (2) g (2) + f (3) g (3) + f (4) g (4) + f (5) g (5) + + f (6) g (6) + f (7) g (7) + f (8) g (8) + f (9) g (9) + f (10) g (10

(a)

The basis functions for a quadratic least squares fit are p1 ( x) = 1. p 2 ( x) = x, p3 ( x) = x 2 . Using the inner product of Equation a the least squares matrix is a1,1 = (1,1) = 10

a1, 2 = (1, x ) = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 55 a1,3 = (1, x 2 ) = (1) 2 + (2) 2 + (3) 2 + (4) 2 + (5) 2 + (6) 2 + (7) 2 + (8) 2 + (9) 2 + (10) 2 = 385 a 2, 2 = ( x, x ) = (1) 2 + (2) 2 + (3) 2 + (4) 2 + (5) 2 + (6) 2 + (7) 2 + (8) 2 + (9) 2 + (10) 2 = 385

( ) = (x , x ) = (1)

a 2,3 = x, x 2 = (1) 3 + (2) 3 + (3) 3 + (4) 3 + (5) 3 + (6) 3 + (7) 3 + (8) 3 + (9) 3 + (10) 3 == 3025 a 3, 3

2

2

4

+ (2) 4 + (3) 4 + (4) 4 + (5) 4 + (6) 4 + (7) 4 + (8) 4 + (9) 4 + (10) 4 = 25333

The symmetric matrix is 55 385 ⎤ ⎡ 10 ⎢ A = ⎢ 55 385 3025 ⎥⎥ ⎢⎣385 3025 25333⎥⎦

The components of the right-hand side vector are b1 = ( y,1) = 2.405 + 5.530 + 8.654 + 11.792 + 14.931 + 18.071 + 21.212 + 24.352 + 27.493 + 30.653 = 168.1

b2 = ( y, x ) = (2.405)(1) + (5.530)(2) + (8.654)(3) + (11.792)(4) + (14.931)(5) + (18.071)(6) + (21.212)(7) + (24.352)(8) + (27.493)(9) + (30.653)(10) = 1196.8

(

)

b3 = y , x 2 = (2.405)(1) 2 + (5.530)(2) 2 + (8.654)(3) 2 + (11.792)(4) 2 + (14.931)(5) 2 + (18.071)(6) 2 + (21.212)(7) 2 + (24.352)(8) 2 + (27.493)(9) 2 + (30.653)(10) 2 = 9504.4

131

The solution of the system Ac=b is c = [0.1799 2.5326 0.0700]. The quadratic least squares approximation to the data is p ( x) = 0.1799 + 2.5326 x + 0.0700 x 2

132

Problems 4.9-4.17 refer to the system of Fig. P4.9. The differential equation for the nondimensional steady-state temperature in the bar which is subject to an external heat generation over its length is d ⎡ dθ ⎤ α ( x) ⎥ − β ( x)θ = q( x) ⎢ dx ⎣ dx ⎦

(a)

4.9 Consider a uniform bar such that α ( x) = 1, β ( x) = 0.5 with q ( x) = x(1 − x) . The bar is subject to a constant temperature at x=0 and is insulated at x=1 such that

θ (0) = 0,

dθ (1) = 0 . Let S be the vector space which is the intersection of dx

P 3 [0,1] with the subspace of C[0,1] of all functions satisfying the homogeneous boundary conditions. Find the best least squares approximation for the temperature distribution in the bar from S.

Solution: The general form of a function in P 3 [0,1] is p ( x) = a 0 + a1 x + a 2 x 2 + a3 x 3

(b)

Requiring p(x) to satisfy the given boundary conditions leads to p (0) = 0 ⇒ a 0 0

(c)

dp (1) = 0 ⇒ a1 + 2a 2 + 3a3 = 0 dx

(d)

Rearranging Equation d leads to a1 = −2a 2 − 3a3 which when substituted into Equation b gives

(

)

(

p ( x) = a 2 x 2 − 2 x + a3 x 3 − 3x Equation e suggests the choice of two basis functions as

133

)

(e)

p1 ( x) = x 2 − 2 x

(f)

p 2 ( x) = x − 3 x

(g)

3

The following MATCHAD file was developed to provide a least squares solution of Equation a using the basis functions of Equations f and g. Solution of Problem 4.9 2

p1( x) := x − 2⋅ x

Basis functions 3

p2( x) := x − 3⋅ x Ls1( x) :=

d

2 2

p1( x) − 0.5p1( x)

dx

Ls2( x) :=

d

2 2

L( p ( x) p2( x) − 0.5p2( x)

dx

q ( x) := x⋅ ( 1 − x)

⎛ Ls1( x) ⎞ ⎟ ⎝ Ls2( x) ⎠

Ls( x) := ⎜

Non homogeneous function

⎛ p1( x) ⎞ ⎟ ⎝ p2( x) ⎠

p ( x) := ⎜

i := 1 .. 2

1

⌠ b := ⎮ p ( x) ⋅ q ( x) dx i i ⌡

right hand side vector

0

j := 1 .. 2

1

⌠ := ⎮ Ls( x) ⋅ Ls( x) dx i, j i j ⌡

⎛ −0.117 ⎞ ⎟ ⎝ −0.217 ⎠

b=⎜

Least squares matrix

L

0

⎛ 5.467 8.754 ⎞ ⎟ ⎝ 8.754 17.286⎠

L=⎜

−1

A := L

⋅b

⎛ −6.715 × 10− 3 ⎞ ⎟ A=⎜ ⎜ −3⎟ ⎝ −9.134 × 10 ⎠ 134

Least squares coefficients

Approximation θ( x) := A ⋅ p1( x) + A ⋅ p2( x) 1

2

x := 0 , .02.. 1

Least squares approximation for Problem 4.9 0.03

0.02 θ ( x)

0.01

0

0

0.2

0.4

0.6

0.8

x

135

1

4.10 Consider a uniform bar such that α ( x) = 1, β ( x) = 0.5 with q ( x) = x(1 − x) . The bar is subject to a constant temperature at x=0 and is insulated at x=1 such that θ (0) = 0,

dθ (1) = 0 . Let S be the vector space which is the intersection of dx

P 3 [0,1] with the subspace of C[0,1] of all functions satisfying the homogeneous boundary conditions. Find the best Rayleigh-Ritz approximation for the temperature distribution from S.

Solution: The general form of a function in P 3 [0,1] is p ( x) = a 0 + a1 x + a 2 x 2 + a3 x 3

(b)

Requiring p(x) to satisfy the given boundary conditions leads to p (0) = 0 ⇒ a 0 0

(c)

dp (1) = 0 ⇒ a1 + 2a 2 + 3a3 = 0 dx

(d)

Rearranging Equation d leads to a1 = −2a 2 − 3a3 which when substituted into Equation b gives

(

)

(

p ( x) = a 2 x 2 − 2 x + a3 x 3 − 3x

)

(e)

Equation e suggests the choice of two basis functions as p1 ( x) = x 2 − 2 x

(f)

p 2 ( x) = x 3 − 3 x

(g)

The following MATCHAD file was developed to provide a Rayleigh-Ritz approximation for the solution of Equation a using the basis functions of Equations f and g.

136

Solution of Problem 4.10 2

p1( x) := x − 2⋅ x

Basis functions 3

p2( x) := x − 3⋅ x Ls1( x) := −

d

2 2

p1( x) + 0.5p1( x)

dx

Ls2( x) := −

d

2 2

L( p ( x) p2( x) + 0.5p2( x)

dx

q ( x) := −x⋅ ( 1 − x)

⎛ Ls1( x) ⎞ ⎟ ⎝ Ls2( x) ⎠

Ls( x) := ⎜

Non homogeneous function

⎛ p1( x) ⎞ ⎟ ⎝ p2( x) ⎠

p ( x) := ⎜

i := 1 .. 2

1

⌠ b := ⎮ p ( x) ⋅ q ( x) dx i i ⌡

right hand side vector

0

j := 1 .. 2 1

⌠ L := ⎮ Ls( x) ⋅ p ( x) dx i, j i j ⌡

Rayleigh Ritz matrix

0

L=

⎛ 1.6 3.008 ⎞ ⎜ ⎟ ⎝ 3.008 5.771 ⎠

⎛ 0.117 ⎞ ⎟ ⎝ 0.217 ⎠

b=⎜

−1

A := L

⋅b

A=

⎛ 0.117 ⎞ ⎜ ⎟ ⎝ −0.023 ⎠

137

Rayleigh Ritz coefficients

Approximation

θ( x) := A ⋅ p1( x) + A ⋅ p2( x) 1

2

x := 0 , .02.. 1

Rayleigh-Ritz approximation for PRoblem 4.10 0

− 0.02

θ ( x) − 0.04

− 0.06

− 0.08

0

0.2

0.4

0.6 x

138

0.8

1

4.11 Repeat Problem 4.10 but using an orthonormal basis for S with respect to the energy inner product.

Solution: A basis for S is determined in the solution of Problem 4.10 as p1 ( x) = x 2 − 2 x

(b)

p 2 ( x) = x 3 − 3 x

(c)

The following MATHCAD program determines an orthonormal basis which spans S and then uses inner product evaluations to determine the Rayleigh-Ritz approximation.

Solution of Problem 4.11 2

p1( x) := x − 2⋅ x

Basis functions 3

p2( x) := x − 3⋅ x

q ( x) := −x⋅ ( 1 − x)

Gram Schmidt process using energy inner product

L1( x) := −

d

2 2

p1( x) + 0.5p1( x)

dx

L2( x) := −

d

2 2

L( p ( x) p2( x) + 0.5p2( x)

dx 1

⌠ pn1 := ⎮ L1( x) ⋅ p1( x) dx ⌡

pn1 = 1.6

Energy inner product (p1,p1)

0

v1( x) :=

p1( x)

First basis function

pn1 1

⌠ I1 := ⎮ L2( x) ⋅ v1( x) dx ⌡

I1 = 2.378

Energy inner product (v1,p2)

0

w2( x) := p2( x) − I1⋅ v1( x)

139

Lw2( x) := −

d

2 2

w2( x) + 0.5⋅ w2( x)

Energy inner product (w2,w2)

dx 1

⌠ n2 := ⎮ Lw2( x) ⋅ w2( x) dx ⌡ 0

v2( x) :=

w2( x)

Second basis function

n2

Check for orthogonality 1

⌠ − 15 ⎮ Lw2( x) ⋅ v1( x) dx = 5.676 × 10 ⌡ 0

Rayleigh Ritz coefficients 1

⌠ A := ⎮ v1( x) ⋅ q ( x) dx 1 ⌡ 0

1

⌠ A := ⎮ v2( x) ⋅ q ( x) dx 2 ⌡ 0

A=

0.092 ⎛ ⎞ ⎜ −3⎟ ⎝ −7.931 × 10 ⎠

Approximation

θ( x) := A ⋅ v1( x) + A ⋅ v2( x) 1

2

x := 0 , .02.. 1

140

Rayleigh-Ritz approximation for Problem 4.11 0

− 0.02

θ ( x) − 0.04

− 0.06

− 0.08

0

0.2

0.4

0.6 x

141

0.8

1

4.12 Consider a uniform bar such that α ( x) = 1, β ( x) = 0.5 with q ( x) = x(1 − x) . The bar is subject to a constant temperature at x=0 and is open for heat transfer at x=1 such that θ (0) = 0,

dθ (1) + 0.5θ (1) = 0 . Let S be the vector space which is the intersection dx

of P 3 [0,1] with the subspace of C[0,1] of all functions satisfying the homogeneous boundary conditions. Find the best Rayleigh-Ritz approximation for the temperature distribution in the bar from S.

Solution: The general form of a function in P 3 [0,1] is p ( x) = a 0 + a1 x + a 2 x 2 + a3 x 3

(b)

Requiring p(x) to satisfy the given boundary conditions leads to p(0) = 0 ⇒ a 0 0

(c)

dp (1) + 0.5 p (1) = 0 ⇒ a1 + 2a 2 + 3a3 + 0.5(a1 + a 2 + a 3 ) = 0 dx

(d)

5 7 Rearranging Equation d leads to a1 = − a 2 − a3 which when substituted into Equation 3 3 b gives 5 ⎞ 7 ⎞ ⎛ ⎛ p ( x) = a 2 ⎜ x 2 − x ⎟ + a3 ⎜ x 3 − x ⎟ 3 ⎠ 3 ⎠ ⎝ ⎝

(e)

Equation e suggests the choice of two basis functions as 5 p1 ( x) = x 2 − x 3 7 p 2 ( x) = x 3 − x 3

(f) (g)

The following MATCHAD file was developed to provide a Rayleigh-Ritz approximation for the solution of Equation a using the basis functions of Equations f and g.

142

Solution of Problem 4.12 5

2

p1( x) := x −

3

3

p2( x) := x −

Ls1( x) := −

d

2 2

⋅x

Basis functions 7 3

⋅x

p1( x) + 0.5p1( x)

dx Ls2( x) := −

d

2 2

L( p ( x) p2( x) + 0.5p2( x)

dx

q ( x) := −x⋅ ( 1 − x)

⎛ Ls1( x) ⎞ ⎟ ⎝ Ls2( x) ⎠

Ls( x) := ⎜

Non homogeneous function

⎛ p1( x) ⎞ ⎟ ⎝ p2( x) ⎠

p ( x) := ⎜

i := 1 .. 2

1

⌠ b := ⎮ p ( x) ⋅ q ( x) dx i i ⌡

right hand side vector

0

j := 1 .. 2 1

⌠ L := ⎮ Ls( x) ⋅ p ( x) dx i, j i j ⌡

Rayleigh Ritz matrix

0

L=

⎛ 1.146 2.106 ⎞ ⎜ ⎟ ⎝ 2.106 3.979 ⎠

⎛ 0.089 ⎞ ⎟ ⎝ 0.161 ⎠

b=⎜

−1

A := L

⋅b

A=

⎛ 0.116 ⎞ ⎜ ⎟ ⎝ −0.021 ⎠

143

Rayleigh Ritz coefficients

Approximation

θ( x) := A ⋅ p1( x) + A ⋅ p2( x) 1

2

x := 0 , .02.. 1

Rayleigh-Ritz approximation for Problem 4.12 0

− 0.02 θ ( x) − 0.04

− 0.06

0

0.2

0.4

0.6 x

144

0.8

1

4.13 Use the Rayleigh-Ritz method to approximate the temperature distribution in a 3 2

surface with α ( x) = (1 − 0.1x) , and q ( x) = x(1 − x) . The end x=0 is maintained at a constant temperature, θ (0) = 0 while its other end is insulated

dθ (1) = 0 . Use a basis dx

for the intersection of P 4 [0,1] with the subspace of C[0,1] of all functions which satisfy the boundary conditions.

Solution: The general form of a function in P 4 [0,1] is p ( x) = a 0 + a1 x + a 2 x 2 + a 3 x 3 + a 4 x 4

(b)

Requiring p(x) to satisfy the given boundary conditions leads to p (0) = 0 ⇒ a 0 0

(c)

dp (1) = 0 ⇒ a1 + 2a 2 + 3a3 4a 4 = 0 dx

(d)

Rearranging Equation d leads to a1 = −2a 2 − 3a3 − 4a 4 which when substituted into Equation b gives

(

5 ⎞ 7 ⎞ ⎛ ⎛ p ( x) = a 2 ⎜ x 2 − x ⎟ + a3 ⎜ x 3 − x ⎟ + a 4 x 4 − 4 x 3 ⎠ 3 ⎠ ⎝ ⎝

)

(e)

Equation e suggests the choice of three basis functions as p1 ( x) = x 2 − 2 x

(f)

p 2 ( x) = x − 3x

(g)

p3 ( x) = x 4 − 4 x

(h)

3

The following MATCHAD file was developed to provide a Rayleigh-Ritz approximation for the solution of Equation a using the basis functions of Equations f-h.

145

Solution of Problem 4.13 2

p1( x) := x − 2⋅ x

Basis functions 3

p2( x) := x − 3⋅ x 4

p3( x) := x − 4⋅ x

f ( x) := ( 1 − 0.1⋅ x)

1.5

d ⎡ ⎛d ⎞⎤ Ls1( x) := − ⎢f ( x) ⋅ ⎜ p1( x) ⎟⎥ + 0.5⋅ p1( x) dx⎣ ⎝ dx ⎠⎦

d ⎡ ⎛d ⎞⎤ Ls2( x) := − ⎢f ( x) ⋅ ⎜ p2( x) ⎟⎥ + 0.5⋅ p2( x) dx⎣ d x ⎝ ⎠⎦

L( p ( x) )

d ⎡ ⎛d ⎞⎤ Ls3( x) := − ⎢f ( x) ⋅ ⎜ p3( x) ⎟⎥ + 0.5⋅ p3( x) dx⎣ ⎝ dx ⎠⎦

q ( x) := −x⋅ ( 1 − x)

⎛ Ls1( x) ⎞ Ls( x) := ⎜ Ls2( x) ⎟ ⎜ ⎟ ⎝ Ls3( x) ⎠

Non homogeneous function

⎛ p1( x) ⎞ p ( x) := ⎜ p2( x) ⎟ ⎜ ⎟ ⎝ p3( x) ⎠

i := 1 .. 3

1

⌠ b := ⎮ p ( x) ⋅ q ( x) dx i i ⌡ 0

⎛ 0.117 ⎞ b = ⎜ 0.217 ⎟ ⎜ ⎟ ⎝ 0.31 ⎠

j := 1 .. 3

146

right hand side vector

1

⌠ L := ⎮ Ls( x) ⋅ p ( x) dx i, j i j ⌡

⎛ 1.551 2.904 4.18 ⎞ L = ⎜ 2.904 5.549 8.07 ⎟ ⎜ ⎟ ⎝ 4.18 8.07 11.808⎠

0

−1

A := L

Rayleigh Ritz matrix

⎛ −0.019 ⎞ A = ⎜ 0.176 ⎟ ⎜ ⎟ ⎝ −0.087 ⎠

⋅b

Rayleigh Ritz coefficients

Approximation

θ( x) := A ⋅ p1( x) + A ⋅ p2( x) + A ⋅ p3( x) 1

2

3

x := 0 , .02.. 1

Rayleigh-Ritz approximation for Problem 4.13 0

− 0.02

θ ( x) − 0.04

− 0.06

− 0.08

0

0.2

0.4

0.6 x

147

0.8

1

4.14 Use the Rayleigh-Ritz method to approximate the temperature distribution in a 3 2

surface with α ( x) = (1 − 0.1x) , β ( x) = (1 − 0.1x) and q ( x) = x(1 − x) . The end x=0 3

is maintained at a constant temperature, θ (0) = 0 while its other end is insulated dθ (1) = 0 . Use a basis for the intersection of P 4 [0,1] with the subspace of C[0,1] of dx all functions which satisfy the boundary conditions.

Solution: The general form of a function in P 4 [0,1] is p ( x) = a 0 + a1 x + a 2 x 2 + a 3 x 3 + a 4 x 4

(b)

Requiring p(x) to satisfy the given boundary conditions leads to p (0) = 0 ⇒ a 0 0

(c)

dp (1) = 0 ⇒ a1 + 2a 2 + 3a3 4a 4 = 0 dx

(d)

Rearranging Equation d leads to a1 = −2a 2 − 3a3 − 4a 4 which when substituted into Equation b gives

(

5 ⎞ 7 ⎞ ⎛ ⎛ p ( x) = a 2 ⎜ x 2 − x ⎟ + a3 ⎜ x 3 − x ⎟ + a 4 x 4 − 4 x 3 ⎠ 3 ⎠ ⎝ ⎝

)

(e)

Equation e suggests the choice of three basis functions as p1 ( x) = x 2 − 2 x

(f)

p 2 ( x) = x 3 − 3x

(g)

p3 ( x) = x 4 − 4 x

(h)

The following MATCHAD file was developed to provide a Rayleigh-Ritz approximation for the solution of Equation a using the basis functions of Equations f-h. Solution of Problem 4.14 2

p1( x) := x − 2⋅ x

Basis functions 3

p2( x) := x − 3⋅ x

148

4

p3( x) := x − 4⋅ x

f ( x) := ( 1 − 0.1⋅ x)

3

g ( x) := ( 1 − 0.1⋅ x)

1.5

d ⎡ ⎛d ⎞⎤ Ls1( x) := − ⎢f ( x) ⋅ ⎜ p1( x) ⎟⎥ + 0.5⋅ p1( x) dx⎣ ⎝ dx ⎠⎦

d ⎡ ⎛d ⎞⎤ Ls2( x) := − ⎢f ( x) ⋅ ⎜ p2( x) ⎟⎥ + 0.5⋅ p2( x) dx⎣ d x ⎝ ⎠⎦

L( p ( x) )

d ⎡ ⎛d ⎞⎤ Ls3( x) := − ⎢f ( x) ⋅ ⎜ p3( x) ⎟⎥ + 0.5⋅ p3( x) dx⎣ ⎝ dx ⎠⎦

q ( x) := −x⋅ ( 1 − x)

⎛ Ls1( x) ⎞ Ls( x) := ⎜ Ls2( x) ⎟ ⎜ ⎟ ⎝ Ls3( x) ⎠

Non homogeneous function

⎛ p1( x) ⎞ p ( x) := ⎜ p2( x) ⎟ ⎜ ⎟ ⎝ p3( x) ⎠

i := 1 .. 3

1

⌠ b := ⎮ p ( x) ⋅ q ( x) dx i i ⌡

right hand side vector

0

j := 1 .. 3 1

⌠ L := ⎮ Ls( x) ⋅ p ( x) dx i, j i j ⌡

⎛ 0.117 ⎞ b = ⎜ 0.217 ⎟ ⎜ ⎟ ⎝ 0.31 ⎠ Rayleigh Ritz matrix

0

149

⎛ 1.504 2.807 4.032 ⎞ L = ⎜ 2.807 5.342 7.751 ⎟ ⎜ ⎟ ⎝ 4.032 7.751 11.314⎠ −1

A := L

⎛ −0.038 ⎞ A = ⎜ 0.199 ⎟ ⎜ ⎟ ⎝ −0.095 ⎠

⋅b

Rayleigh Ritz coefficients

Approximation

θ( x) := A ⋅ p1( x) + A ⋅ p2( x) + A ⋅ p3( x) 1

2

3

x := 0 , .02.. 1

Rayleigh-Ritz approximation for Problem 4.14 0

− 0.02

θ ( x) − 0.04

− 0.06

− 0.08

0

0.2

0.4

0.6 x

150

0.8

1

4.15 Use the Rayleigh-Ritz method to approximate the temperature distribution in a surface with α ( x) = (1 − 0.1x) , β ( x) = 1 and q ( x) = x(1 − x) . The end x=0 is maintained at a constant temperature, θ (0) = 0 while its other end is insulated dθ (1) = 0 . Use a basis for the intersection of dx

P 4 [0,1] with the subspace of C[0,] of

all functions which satisfy the geometric boundary condition only. Use the energy formulation of the inner products.

Solution: Using the energy formulation of the inner products the natural boundary condition at x=1 is not satisfied. A set of basis functions from P 4 [0,1] which satisfy p1 ( x) = x, p 2 ( x) = x 2 , p 3 ( x) = x 3 , p 4 ( x) = x 4 . The energy formulation is

p(0)=0 is

obtained through integration by parts of

( f , g )L = ∫ ⎡⎢− d ⎛⎜ α ( x) df ⎞⎟ f ( x)⎤⎥ g ( x)dx dx ⎠ ⎣ dx ⎝ ⎦ 1

0

1

= ∫ α ( x) 0

1

df dg df dx + ∫ f ( x) g ( x)dx − α (1) g (1) (1) dx dx dx 0

The Rayleigh-Ritz approximation is obtained in the following MATHCAD file. Solution of Problem 4.15 p1( x) := x

Basis functions 2

p2( x) := x

3

p3( x) := x

4

p4( x) := x

f ( x) := ( 1 − 0.1⋅ x)

q ( x) := −x⋅ ( 1 − x)

151

Non homogeneous function

⎛⎜ p1( x) ⎞⎟ p2( x) ⎟ p ( x) := ⎜ ⎜ p3( x) ⎟ ⎜ p4( x) ⎟ ⎝ ⎠

i := 1 .. 4

1

⌠ b := ⎮ p ( x) ⋅ q ( x) dx i i ⌡

right hand side vector

0

j := 1 .. 4 1

⌠ ⎤ ⎛d ⎞ ⎡d L := ⎮ f ( x) ⋅ ⎢ p ( x) ⎥ ⋅ ⎜ p ( x) ⎟ dx + i, j ⎮ i j ⎣dx ⎦ ⎝ dx ⎠ ⌡

(

)

0

⎛⎜ 1.283 1.183 L=⎜ ⎜ 1.125 ⎜ 1.087 ⎝

−1

A := L

1

⌠ ⎮ p ( x) i⋅ p ( x) j dx ⌡ 0

1.183 1.125 1.087 ⎞

⎟ ⎟ 1.547 1.793 1.954 ⎟ ⎟ 1.61 1.954 2.197 ⎠

⎛⎜ −0.083 ⎞⎟ −0.05 ⎟ b=⎜ ⎜ −0.033 ⎟ ⎜ −0.024 ⎟ ⎝ ⎠

1.433 1.547 1.61

⎛⎜ −0.124 ⎞⎟ −0.01 ⎟ A=⎜ ⎜ 0.155 ⎟ ⎜ −0.08 ⎟ ⎝ ⎠

⋅b

Approximation

θ( x) := A ⋅ p1( x) + A ⋅ p2( x) + A ⋅ p3( x) + A ⋅ p ( x) 1

2

3

4

4

x := 0 , .02.. 1

152

Rayleigh Ritz coefficients

Rayleigh-Ritz approximation for Problem 4.15 0

− 0.02 θ ( x) − 0.04

− 0.06

0

0.2

0.4

0.6 x

153

0.8

1

4.16 Use the finite-element method to solve Problem 4.10. Use five equally spaced elements between x=0 and x=1.

Solution: The finite-element solution is in the following MATHCAD file Problem 4.16

The interval from x=0 to x=1 is divided into five elements of equal length leading to definition of the basis functions as φ0 ( x) := ( 1 − 5⋅ x) ( Φ( x) − Φ( x − 0.2) ) φ1 ( x) := 5⋅ x⋅ ( Φ( x) − Φ( x − 0.2) ) + ( 2 − 5⋅ x) ⋅ ( Φ( x − 0.2) − Φ( x − 0.4) ) φ2 ( x) := φ1 ( x − 0.2) φ3 ( x) := φ1 ( x − 0.4) φ4 ( x) := φ1 ( x − 0.6) φ5 ( x) := ( 5⋅ x − 4) ⋅ Φ( x − 0.8)

where (x) is the Heaviside function defined as 0 for x0. It is also called the unit step function u(x) x := 0 , .01.. 1

Basis functions 1

φ0 ( x)0.8 φ1 ( x) φ2 ( x)0.6 φ3 ( x) φ4 ( x)0.4 φ5 ( x) 0.2

0

0

0.2

0.4

0.6

0.8

1

x

The geometric boundary condition at x=0 satisfied by setting the coefficient of φ0 ( x) tozero

154

⎛ φ1 ( x) ⎞ ⎜ ⎟ φ2 ( x) ⎜ ⎟ ϕ ( x) := ⎜ φ3 ( x) ⎟ ⎜ φ4 ( x) ⎟ ⎜ ⎟ ⎝ φ5 ( x) ⎠

Basis functions satisfying geometric boundary condition

q ( x) := −x⋅ ( 1 − x)

Inner product evaluations

i := 1 .. 5 1

⌠ b := ⎮ q ( x) ⋅ ϕ ( x) dx i i ⌡ 0

j := 1 .. 5 1

⌠ ⎞ d ⎛d A := ⎮ ϕ ( x) ⋅ ⎜ ϕ ( x) ⎟ dx i , j ⎮ dx i dx j ⎝ ⎠ ⌡ 0

1

⌠ C := ⎮ ϕ ( x) ⋅ ϕ ( x) dx i, j i j ⌡ 0

L

i, j

:= A

i, j

+ ( 0.5C)

i, j

0 0 ⎞ ⎛ 10.067 −4.983 0 ⎜ ⎟ −4.983 10.067 −4.983 0 0 ⎜ ⎟ L=⎜ 0 −4.983 10.067 −4.983 0 ⎟ ⎜ 0 0 −4.983 10.067 −4.97 ⎟ ⎜ ⎟ 0 0 −4.97 5.04 ⎠ ⎝ 0

155

⎛ −0.027 ⎞ ⎜ −0.048 ⎟ ⎜ ⎟ D = ⎜ −0.062 ⎟ ⎜ −0.066 ⎟ ⎜ ⎟ ⎝ −0.067 ⎠

−1

D := L

⋅b

(

)

w ( x) := D ⋅ φ1 ( x) + D ⋅ φ2 ( x) + D ⋅ φ3 ( x) + D ⋅ φ4 ( x) + D ⋅ φ5 ( x) 1

2

3

4

5

z := 0 , .01.. 1 z := 0 , .01.. 1

0

− 0.02

w ( z )− 0.04

− 0.06

− 0.08

0

0.2

0.4

0.6 z

156

0.8

1

4.17 Use the finite-element method to solve Problem 4.15. Use five equally spaced elements between x=0 and x=1.

Solution: The finite-element solution is presented in the following MATHCAD file. Problem 4.17

The interval from x=0 to x=1 is divided into five elements of equal length leading to definition of the basis functions as φ0 ( x) := ( 1 − 5⋅ x) ( Φ( x) − Φ( x − 0.2) ) φ1 ( x) := 5⋅ x⋅ ( Φ( x) − Φ( x − 0.2) ) + ( 2 − 5⋅ x) ⋅ ( Φ( x − 0.2) − Φ( x − 0.4) ) φ2 ( x) := φ1 ( x − 0.2) φ3 ( x) := φ1 ( x − 0.4) φ4 ( x) := φ1 ( x − 0.6) x := 0 , .01.. 1 φ5 ( x) := ( 5⋅ x − 4) ⋅ Φ( x − 0.8)

where (x) is the Heaviside function defined 0 for x0. It is also called the unit step Basis as functions function u(x) 1

φ0 ( x)0.8 φ1 ( x) φ2 ( x)0.6 φ3 ( x) φ4 ( x)0.4 φ5 ( x) 0.2

0

0

0.2

0.4

0.6

0.8

1

x

The geometric boundary condition at x=0 satisfied by setting the coefficient of φ0 ( x) tozero f ( x) := 1 − 0.1⋅ x q ( x) := −x⋅ ( 1 − x)

157

⎛ φ1 ( x) ⎞ ⎜ ⎟ ⎜ φ2 ( x) ⎟ ϕ ( x) := ⎜ φ3 ( x) ⎟ ⎜ φ4 ( x) ⎟ ⎜ ⎟ ⎝ φ5 ( x) ⎠

Basis functions satisfying geometric boundary condition

Inner product evaluations i := 1 .. 5

1

⌠ b := ⎮ q ( x) ⋅ ϕ ( x) dx i i ⌡ 0

j := 1 .. 5 1

⌠ ⎞ d ⎛d ϕ ( x) ⋅ ⎜ ϕ ( x) ⎟ ⋅ f ( x) dx A := ⎮ i , j ⎮ dx i dx j ⎝ ⎠ ⌡ 0

1

⌠ := ⎮ ϕ ( x) ⋅ ϕ ( x) dx i, j i j ⌡

C

0

L

i, j

:= A

i, j

+ ( 0.5C)

i, j

0 0 ⎞ ⎛ 9.867 −4.833 0 ⎜ ⎟ 0 ⎜ −4.833 9.667 −4.733 0 ⎟ L=⎜ 0 −4.733 9.467 −4.633 0 ⎟ ⎜ 0 0 −4.633 9.267 −4.521 ⎟ ⎜ ⎟ 0 0 −4.521 4.59 ⎠ ⎝ 0 −1

D := L

⋅b

⎛ −0.027 ⎞ ⎜ ⎟ −0.049 ⎜ ⎟ D = ⎜ −0.063 ⎟ ⎜ −0.068 ⎟ ⎜ ⎟ ⎝ −0.069 ⎠

158

(

)

w ( x) := D ⋅ φ1 ( x) + D ⋅ φ2 ( x) + D ⋅ φ3 ( x) + D ⋅ φ4 ( x) + D ⋅ φ5 ( x) 1

2

3

4

5

z := 0 , .01.. 1

0

− 0.02

w ( z )− 0.04

− 0.06

− 0.08

0

0.2

0.4

0.6

0.8

z

159

1

Problems 4.18-4.24 refer to the approximate solution of the non-dimensional differential equation for the static deflection of a stretched beam on an elastic foundation and subject to transverse loading, as illustrated in Fig. P4.18. In all problems the subspace S refers to the intersection of P 6 [0,1] with a subspace of C 4 [0,1] which consists of all functions which satisfy the homogeneous boundary conditions, d2 ⎡ d 2w⎤ d 2w + ηw = f ( x ) ⎢α ( x) 2 ⎥ − ε dx 2 ⎣ dx ⎦ dx 2

(a)

Use f ( x) = 2 x 2 (1 − x) 2 for all problems. 4.18 Use the least squares method to approximate the deflection of a uniform, α ( x) = 1 dw d 2w fixed-pinned beam w(0) = 0, (0) = 0, w(1) = 0, 2 (1) = 0 with ε = 0 and η = 2 . dx dx Use basis functions for S as trial functions.

Solution: A general element of P 6 [0,1] is of the form p ( x) = a 0 + a1 x + a 2 x 2 + a 3 x 3 + a 4 x 4 + a5 x 5 + a 6 x 6

(b)

Requiring p(x) to satisfy the boundary conditions leads to p ( 0) = 0 ⇒ a 0 = 0 dp (0) = 0 ⇒ a1 = 0 dx d 2w (1) = 0 ⇒ 2a 2 + 6a3 + 12a 4 + 20a5 + 30a 6 = 0 dx 2 d 3w (1) = 0 ⇒ 6a3 + 24a 4 + 60a5 + 120a 6 = 0 dx 3

(c) (d) (e) (f)

Equations e and f are rearranged to a 2 = 6a 4 + 20a5 + 45a 6

(g)

a3 = −4a 4 − 10a 5 − 20a6

(h)

Substitution f Equations g and h into Equation b leads to

160

(

)

(

)

(

p ( x) = a 4 x 4 − 4 x 3 + 6 x 2 + a5 x 5 − 10 x 3 + 20 x 2 + a 6 x 6 − 20 x 3 + 45 x 2

)

(i)

A choice of basis functions for S is obtained from Equation I as p1 ( x) = x 4 − 4 x 3 + 6 x 2

(j)

p 2 ( x) = x 5 − 10 x 3 + 20 x 2

(k)

p3 ( x) = x − 20 x + 45 x

(l)

6

3

2

The following MATHCAD file determines the least squares approximation to the solution of Equation a using the basis functions of Equations j-l.

Problem 4.18

Basis functions 4

3

2

p1( x) := x − 4⋅ x + 6⋅ x 5

3

2

6

3

2

p2( x) := x − 10⋅ x + 20⋅ x

p3( x) := x − 20⋅ x + 45⋅ x

⎛ p1( x) ⎞ p ( x) := ⎜ p2( x) ⎟ ⎜ ⎟ ⎝ p3( x) ⎠

Operator acting on basis functions

L1( x) :=

d

4 4

p ( x) + 2⋅ p ( x) 1

dx

L2( x) :=

d

4 4

p ( x) + 2⋅ p ( x) 2

dx

L3( x) :=

d

1

4 4

dx

2

p ( x) + 2⋅ p ( x) 3

3

161

⎛ L1( x) ⎞ Lp( x) := ⎜ L2( x) ⎟ ⎜ ⎟ ⎝ L3( x) ⎠

Non homogeneous term 2

f ( x) := 2⋅ x ⋅ ( 1 − x)

2

Inner product evaluations

i := 1 .. 3

1

⌠ b := ⎮ f ( x) ⋅ p ( x) dx i i ⌡ 0

j := 1 .. 3 1

A

⌠ := ⎮ Lp( x) ⋅ Lp( x) dx i, j i j ⌡ 0

⎛ 700.444 1.89 × 103 3.933 × 103 ⎞ ⎜ ⎟ ⎜ 3 3 4 A = 1.89 × 10 6.431 × 10 1.463 × 10 ⎟ ⎜ ⎟ ⎜ 3 4 4⎟ ⎝ 3.933 × 10 1.463 × 10 3.491 × 10 ⎠

C := A

⎛ 0.075 ⎞ b = ⎜ 0.267 ⎟ ⎜ ⎟ ⎝ 0.623 ⎠ ⎛ −1.115 × 10− 5 ⎞ ⎜ ⎟ ⎜ ⎟ − 5 C = 3.012 × 10 ⎜ ⎟ ⎜ −6 ⎟ ⎝ 6.486 × 10 ⎠

−1

⋅b

Least squares approximation

162

3

w ( x) :=

∑ (Ci⋅p(x)i)

i=1

x := 0 , .02.. 1

Least squares approximation for Problem 4.18

−4

5×10

−4

4×10

−4

3×10 w ( x)

−4

2×10

−4

1×10

0

0

0.2

0.4

0.6 x

163

0.8

1

4.19 Use the Rayleigh-Ritz method to approximate the deflection of a uniform, α ( x) = 1 fixed-pinned beam w(0) = 0,

dw d 2w (0) = 0, w(1) = 0, 2 (1) = 0 with ε = 0 and η = 2 . dx dx

Use basis functions for S as trial functions.

Solution: A general element of P 6 [0,1] is of the form p ( x) = a 0 + a1 x + a 2 x 2 + a 3 x 3 + a 4 x 4 + a5 x 5 + a6 x 6

(b)

Requiring p(x) to satisfy the boundary conditions leads to p ( 0) = 0 ⇒ a 0 = 0

(c)

dp (0) = 0 ⇒ a1 = 0 dx w(1) = 0 ⇒ a 2 + a 3 + a 4 + a5 + a 6 = 0

(d) (e)

d 2w (1) = 0 ⇒ 2a 2 + 6a3 + 12a 4 + 20a5 + 30a 6 = 0 dx 2

(f)

Equations e and f are rearranged to 3 7 a 4 + a5 + 68a 6 2 2 5 9 a3 = − a 4 − a5 − 7a 6 2 2

a2 =

(g) (h)

Substitution f Equations g and h into Equation b leads to

(

5 3 ⎞ 9 7 ⎞ ⎛ ⎛ p ( x) = a 4 ⎜ x 4 − x 3 + x 2 ⎟ + a5 ⎜ x 5 − x 3 + x 2 ⎟ + a 6 x 6 − 7 x 3 + 6 x 2 2 2 ⎠ 2 2 ⎠ ⎝ ⎝

)

(i)

A choice of basis functions for S is obtained from Equation I as 5 3 3 2 x + x 2 2 9 7 p 2 ( x) = x 5 − x 3 + x 2 2 2 6 3 p3 ( x) = x − 7 x + 6 x 2 p1 ( x) = x 4 −

164

(j) (k) (l)

The following MATHCAD file determines the Rayleigh-Ritz approximation to the solution of Equation a using the basis functions of Equations j-l. Problem 4.19

Basis functions 4

3

2

5

3

2

p1( x) := x − 2.5⋅ x + 1.5⋅ x

p2( x) := x − 4.5⋅ x + 3.5⋅ x 6

3

2

p3( x) := x − 7⋅ x + 6⋅ x

⎛ p1( x) ⎞ p ( x) := ⎜ p2( x) ⎟ ⎜ ⎟ ⎝ p3( x) ⎠

Operator acting on basis functions

L1( x) :=

d

4 4

p ( x) + 2⋅ p ( x) 1

dx

L2( x) :=

d

4 4

p ( x) + 2⋅ p ( x) 2

dx

L3( x) :=

d

1

4 4

2

p ( x) + 2⋅ p ( x) 3

dx

3

⎛ L1( x) ⎞ Lp( x) := ⎜ L2( x) ⎟ ⎜ ⎟ ⎝ L3( x) ⎠ 2

f ( x) := 2⋅ x ⋅ ( 1 − x)

2

Non homogeneous term

165

Inner product evaluations

i := 1 .. 3

1

⌠ b := ⎮ f ( x) ⋅ p ( x) dx i i ⌡ 0

j := 1 .. 3 1

A

⌠ := ⎮ Lp( x) ⋅ p ( x) dx i, j i j ⌡ 0

⎛ 1.815 5.042 9.508 ⎞ A = ⎜ 5.042 14.26 27.222⎟ ⎜ ⎟ ⎝ 9.508 27.222 52.421⎠

C := A

−1

⎛ 6.746 × 10− 3 ⎞ ⎜ ⎟ b=⎜ ⎟ 0.019 ⎜ ⎟ ⎝ 0.035 ⎠

⎛ −2.35 × 10− 3 ⎞ ⎜ ⎟ ⎜ C = 5.854 × 10− 3 ⎟ ⎜ ⎟ ⎜ −3⎟ ⎝ −1.946 × 10 ⎠

⋅b

⎛ −2.35 × 10− 3 ⎞ ⎜ ⎟ ⎜ C = 5.854 × 10− 3 ⎟ ⎜ ⎟ ⎜ −3⎟ ⎝ −1.946 × 10 ⎠

166

Rayleigh Ritz approximation 3

w ( x) :=

∑ (Ci⋅p(x)i)

i=1

x := 0 , .02.. 1

Rayleigh-Ritz approximation for Problem 4.19

−4

5×10

−4

4×10

−4

3×10 w ( x)

−4

2×10

−4

1×10

0

0

0.2

0.4

0.6 x

167

0.8

1

4.20 Use the Rayleigh-Ritz method to approximate the deflection of a uniform, α ( x) = 1 fixed-free beam w(0) = 0,

dw d 2w d 3w (0) = 0, 2 (1) = 0, 3 (1) = 0 with ε = 1 and η = 2 . dx dx dx

Use basis functions for S as trial functions.

Solution: A general element of P 6 [0,1] is of the form p ( x) = a 0 + a1 x + a 2 x 2 + a 3 x 3 + a 4 x 4 + a5 x 5 + a6 x 6

(b)

Requiring p(x) to satisfy the boundary conditions leads to

p ( 0) = 0 ⇒ a 0 = 0

(c)

dp (0) = 0 ⇒ a1 = 0 dx d 2w (1) = 0 ⇒ 2a 2 + 6a3 + 12a 4 + 20a5 + 30a 6 = 0 dx 2 d 3w (1) = 0 ⇒ 6a3 + 24a 4 + 60a5 + 120a 6 = 0 dx 3

(d) (e) (f)

Equations e and f are rearranged to a 2 = 6a 4 + 20a5 + 45a 6

(g)

a3 = −4a 4 − 10a5 − 20a6

(h)

Substitution f Equations g and h into Equation b leads to

(

)

(

)

(

p ( x) = a 4 x 4 − 4 x 3 + 6 x 2 + a5 x 5 − 10 x 3 + 20 x 2 + a6 x 6 − 20 x 3 + 45 x 2

)

(i)

A choice of basis functions for S is obtained from Equation I as p1 ( x) = x 4 − 4 x 3 + 6 x 2

(j)

p 2 ( x) = x 5 − 10 x 3 + 20 x 2

(k)

p3 ( x) = x 6 − 20 x 3 + 45 x 2

(l)

The following MATHCAD file determines the least squares approximation to the solution of Equation a using the basis functions of Equations j-l. Note that as formulated

168

the problem is not self-adjoint. Thus the Rayleigh-Ritz matrix is not symmetric. This is really a Galerkin approximation. Problem 4.20

Basis functions 4

3

2

p1( x) := x − 4⋅ x + 6⋅ x 5

3

2

6

3

2

p2( x) := x − 10⋅ x + 20⋅ x

p3( x) := x − 20⋅ x + 45⋅ x

⎛ p1( x) ⎞ p ( x) := ⎜ p2( x) ⎟ ⎜ ⎟ ⎝ p3( x) ⎠

Operator acting on basis functions

L1( x) :=

4

2

d p ( x) ) ) − p1( x) + 2⋅ p ( x) 2 4 (( 1 1 dx dx d

L2( x) :=

d

4 4

p ( x) − 2

dx

L3( x) :=

d

4 4

dx

d

2 2

p2( x) + 2⋅ p ( x)

dx

p ( x) − 3

d

2 2

dx

2

p3( x) + 2⋅ p ( x)

3

⎛ L1( x) ⎞ Lp( x) := ⎜ L2( x) ⎟ ⎜ ⎟ ⎝ L3( x) ⎠

169

Non homogeneous term 2

f ( x) := 2⋅ x ⋅ ( 1 − x)

2

Inner product evaluations

i := 1 .. 3

1

⌠ b := ⎮ f ( x) ⋅ p ( x) dx i i ⌡ 0

j := 1 .. 3 1

A

⌠ := Lp( x) ⋅ p ( x) dx i, j ⎮ i j ⌡ 0

⎛ 31.708 114.716 268.81 ⎞ ⎜ ⎟ A = ⎜ 113.716 413.039 970.155 ⎟ ⎜ 3⎟ ⎝ 264.81 964.155 2.268 × 10 ⎠

C := A

⎛ 0.075 ⎞ b = ⎜ 0.267 ⎟ ⎜ ⎟ ⎝ 0.623 ⎠

⎛ 0.018 ⎞ ⎜ ⎟ −0.011 ⎟ C=⎜ ⎜ −3⎟ ⎝ 2.982 × 10 ⎠

−1

⋅b

Rayleigh-Ritz approximation

3

w ( x) :=

∑ (Ci⋅p(x)i)

i=1

x := 0 , .02.. 1

170

Rayleigh-Ritz approximation for Problem 4.20

−3

8×10

−3

6×10

w ( x)

−3

4×10

−3

2×10

0

0

0.2

0.4

0.6 x

171

0.8

1

4.22

Repeat Problem 4.20, but with α ( x) = 1 + 0.2 x .

Solution: A general element of P 6 [0,1] is of the form

p( x) = a0 + a1 x + a 2 x 2 + a3 x 3 + a 4 x 4 + a5 x 5 + a6 x 6

(b)

Requiring p(x) to satisfy the boundary conditions leads to p ( 0) = 0 ⇒ a 0 = 0

(c)

dp (0) = 0 ⇒ a1 = 0 dx d 2w (1) = 0 ⇒ 2a 2 + 6a3 + 12a 4 + 20a5 + 30a 6 = 0 dx 2 d 3w (1) = 0 ⇒ 6a3 + 24a 4 + 60a5 + 120a 6 = 0 dx 3

(d) (e) (f)

Equations e and f are rearranged to a 2 = 6a 4 + 20a5 + 45a 6

(g)

a3 = −4a 4 − 10a5 − 20a6

(h)

Substitution f Equations g and h into Equation b leads to

(

)

(

)

(

p ( x) = a 4 x 4 − 4 x 3 + 6 x 2 + a5 x 5 − 10 x 3 + 20 x 2 + a6 x 6 − 20 x 3 + 45 x 2

)

(i)

A choice of basis functions for S is obtained from Equation I as p1 ( x) = x 4 − 4 x 3 + 6 x 2

(j)

p 2 ( x) = x − 10 x + 20 x

2

(k)

p3 ( x) = x − 20 x + 45 x

2

(l)

5

6

3

3

The following MATHCAD file determines the least squares approximation to the solution of Equation a using the basis functions of Equations j-l. Note that as formulated the problem is not self-adjoint. Thus the Rayleigh-Ritz matrix is not symmetric. This is really a Galerkin approximation. Problem 4.22

Basis functions

172

4

3

2

p1( x) := x − 4⋅ x + 6⋅ x 5

3

2

p2( x) := x − 10⋅ x + 20⋅ x

6

3

2

p3( x) := x − 20⋅ x + 45⋅ x

⎛ p1( x) ⎞ p ( x) := ⎜ p2( x) ⎟ ⎜ ⎟ ⎝ p3( x) ⎠ α( x) := 1 + 0.2⋅ x

Operator acting on basis functions

L1( x) :=

⎛⎜ ⎞⎟ d2 2 d − α( x) ⋅ p ( x) p1( x) + 2⋅ p ( x) 2 2⎜ 2 1⎟ 1 dx ⎝ dx ⎠ dx d

2

⎛⎜ ⎞⎟ d2 2 d − α( x) ⋅ p ( x) p1( x) + 2⋅ p ( x) 2 2⎜ 2 2⎟ 2 dx ⎝ dx ⎠ dx d

L2( x) :=

L3( x) :=

2

⎛⎜ ⎞⎟ d2 2 d − α( x) ⋅ p ( x) p1( x) + 2⋅ p ( x) 2 2⎜ 2 3⎟ 3 dx ⎝ dx ⎠ dx d

2

⎛ L1( x) ⎞ Lp( x) := ⎜ L2( x) ⎟ ⎜ ⎟ ⎝ L3( x) ⎠

Non homogeneous term 2

f ( x) := 2⋅ x ⋅ ( 1 − x)

2

173

Inner product evaluations

i := 1 .. 3

1

⌠ b := ⎮ f ( x) ⋅ p ( x) dx i i ⌡ 0

j := 1 .. 3 1

A

⌠ Lp( x) ⋅ p ( x) dx := i, j ⎮ i j ⌡ 0

277.639 ⎞ ⎛ 32.668 118.373 ⎜ ⎟ 3 A = ⎜ 122.73 446.047 1.048 × 10 ⎟ ⎜ 3 3⎟ ⎝ 289.972 1.056 × 10 2.485 × 10 ⎠

C := A

−1

⎛ 0.566 ⎞ C = ⎜ −0.396 ⎟ ⎜ ⎟ ⎝ 0.102 ⎠

⋅b

Rayleigh Ritz approximation

3

w ( x) :=

∑ (Ci⋅p(x)i)

i=1

x := 0 , .02.. 1

174

Rayleigh-Ritz approximation for Problem 4.22

−3

8×10

−3

6×10

w ( x)

−3

4×10

−3

2×10

0

0

0.2

0.4

0.6 x

175

0.8

1

CHAPTER 5 EIGENVALUE PROBLEMS

176

5.1 The differential equations governing the free vibrations of the system of Figure. P5.1 are ⎡m 0 0 ⎤ ⎡ &x&1 ⎤ ⎡ 4k ⎢ 0 2m 0 ⎥ ⎢ &x& ⎥ + ⎢− 3k ⎥⎢ 2 ⎥ ⎢ ⎢ ⎢⎣ 0 0 m ⎥⎦ ⎢⎣ &x&3 ⎥⎦ ⎢⎣ 0

− 3k 5k − 2k

0 ⎤ ⎡ x1 ⎤ ⎡0⎤ − 2k ⎥⎥ ⎢⎢ x2 ⎥⎥ = ⎢⎢0⎥⎥ 2k ⎥⎦ ⎢⎣ x3 ⎥⎦ ⎢⎣0⎥⎦

(a)

Recall that the natural frequencies of a discrete system are the square roots of the eigenvalues of M −1K and that the mode shapes are the corresponding eigenvectors (Example 5.2) (a) Determine the natural frequencies for the system (b) Determine the normalized mode shapes for the system. Solution: The natural frequencies are the eigenvalues of the matrix

⎡1 ⎢ k M −1 K = ⎢ 0 m⎢ ⎣0

⎡ 4 0 0⎤ ⎡ 4 − 3 0 ⎤ ⎢ 3 ⎥⎢ 1 k 0⎥ ⎢− 3 5 − 2⎥⎥ = ⎢− 2 m⎢ 2 0 1⎥⎦ ⎢⎣ 0 − 2 2 ⎥⎦ ⎣ 0

−3 0 ⎤ ⎥ 5 − 1⎥ 2 − 2 2 ⎥⎦

(b)

and are calculated by solving 4φ − λ 3 M −1 K − λ I = 0 = − φ 2 0

− 3φ 5 φ −λ 2 − 2φ

0 −φ

(c)

2φ − λ

Expansion of the determinant of Equation (c) leads to

(4φ − λ )⎛⎜ 5 φ − λ ⎞⎟(2φ − λ ) − (− 2φ )(− φ )(4φ − λ ) − (2φ − λ )(− 3φ )⎛⎜ − 3 φ ⎞⎟ = 0 ⎝2

⎠

⎝ 2 ⎠

33 2 φ λ + 3φ 3 = 0 2 17 33 −β3 + β2 − β +3= 0 2 2

− λ3 + 6φλ2 −

(d)

where β = λ / φ . The solutions of Equation d are β = 0.2024,2.6022,5.6954 . The natural frequencies are the square roots of the eigenvalues, thus

177

ω1 = 0.450

k k k , ω 2 = 1.613 , ω 3 = 2.387 m m m

(e)

(a) The mode shape vector corresponding to a natural frequency ω is obtained by finding a non-trivial solution of M −1 KX = ω 2 X or ⎡ 4φ ⎢ 3 ⎢− φ ⎢ 2 ⎣ 0

− 3φ 5 φ 2 − 2φ

0 ⎤⎡ X ⎤ ⎡ X1 ⎤ 1 ⎥⎢ ⎥ 2⎢ − φ ⎥ ⎢ X 2 ⎥ = ω ⎢ X 2 ⎥⎥ ⎥ ⎢⎣ X 3 ⎥⎦ 2φ ⎦ ⎢⎣ X 3 ⎥⎦

(f)

The first and third equations in the system described by Equation f give 3φ X2 4φ − ω 2 2φ X3 = X2 2φ − ω 2 X1 =

(g) (h)

Keeping X 2 arbitrary, substitution of natural frequencies from Equation e into Equations g and h lead to mode shape vectors of ⎡0.7900⎤ X 1 = c1 ⎢⎢ 1 ⎥⎥ ⎢⎣1.1126 ⎥⎦

⎡ 2.1462 ⎤ ⎥ X 2 = c 2 ⎢⎢ 1 ⎥ ⎢⎣− 3.3212⎥⎦

⎡ − 1.7695 ⎤ ⎥ X 3 = c3 ⎢⎢ 1 ⎥ ⎢⎣− 0.5412⎥⎦

(i)

The mode shape vectors are normalized by requiring (X i , X i )M = 1 . Normalizing the mode shape vector corresponding to the lowest natural frequency leads to

(X1 , X1 )M

=1

X 1T MX 1 = 1

⎡m 0 0 ⎤ ⎡0.7900⎤ c [0.7900 1 1.1126]⎢⎢ 0 2m 0 ⎥⎥ ⎢⎢ 1 ⎥⎥ = 1 ⎢⎣ 0 0 m⎥⎦ ⎢⎣1.1126 ⎥⎦ ⎡0.7900m⎤ 2 c1 [0.7900 1 1.1126]⎢⎢ 2m ⎥⎥ = 1 ⎢⎣1.1126m ⎥⎦ 2 1

178

(j)

Evaluation of Equation j leads to c12 [(0.7900 )(0.7900m ) + (1)(2m) + (1.1136)(1.1126m)] = 1

c12 (3.8619m ) = 1 c1 =

0.5089

(k)

m

The normalized mode shape for the first mode is ⎡0.4020⎤ 1 ⎢ X1 = 0.5089⎥⎥ ⎢ m ⎢⎣0.5062⎥⎦

(l)

Normalization of the mode shapes for the second and third modes leads to ⎡ 0.5111 ⎤ 1 ⎢ X2 = 0.2381 ⎥⎥ ⎢ m ⎢⎣− 0.7908⎥⎦

⎡ − 0.7598⎤ 1 ⎢ X3 = 0.4294 ⎥⎥ ⎢ m ⎢⎣− 0.2324⎥⎦

179

(m)

5.2 The initial conditions for the system of Figure P5.1 and Problem 5.1 are ⎡ x&1 (0) ⎤ ⎡0⎤ ⎡ x1 (0) ⎤ ⎡ 0 ⎤ ⎢ x ( x)⎥ = ⎢ 0 ⎥ and ⎢ x& ( x)⎥ = ⎢0⎥ ⎢ 2 ⎥ ⎢ ⎥ ⎢ 2 ⎥ ⎢ ⎥ ⎢⎣ x& 3 ( x) ⎥⎦ ⎢⎣0⎥⎦ ⎣⎢ x3 ( x) ⎦⎥ ⎣⎢δ ⎦⎥

(b)

Let ω1 , ω 2 and ω 3 be the natural frequencies of the system of Problem 5.1 and let X 1 , X 2 and X 3 be their corresponding normalized mode shapes. Let x represent the

solution of Eq. (a) of Problem 5.1 subject to the initial conditions of Eq.(b). At any time x can be expanded in terms of the normalized mode shapes, x(t ) = α 1 (t ) X 1 + α 2 (t ) X 2 + α 3 (t ) X 3

(c)

(a) Substitute Equation c into Equation a of Problem 5.1, take the standard inner product of both sides with respect to X j for an arbitrary j. Use the properties of mode shape orthogonality to derive uncoupled differential equations for α i (t ) (b) Substitute Equation c into the initial conditions of Equation b to derive initial conditions for α i (t ) . (c) Solve the differential equations to solve for α i (t ) (d) Determine x(t) using Eq. (c)

Solution: (a) Substitution of Equation c into Equation a leads to − 3k 5k − 2k

⎡ 4k ⎡m 0 0 ⎤ ⎢ 0 2m 0 ⎥{α&& X + α&& X + α&& X } + ⎢− 3k 1 2 3 3 ⎢ ⎥ 1 1 ⎢ ⎢⎣ 0 ⎢⎣ 0 0 m⎥⎦

0 ⎤ − 2k ⎥⎥{α 1 (t ) X 1 + α 2 (t ) X 2 + α 3 (t ) X 3 } 2k ⎥⎦

⎡0 ⎤ = ⎢⎢0⎥⎥ ⎢⎣0⎥⎦

Taking the standard inner product of Equation with X i leads to

180

(d)

α&&1 X Ti MX 1 + α&&2 X Ti MX 2 + α&&3 X Ti MX 3 + α 1 X Ti KX 1 + α 2 X Ti KX 2 + α 31 X Ti KX 3 = 0

(e)

Equation e is rewritten using energy inner products as

α&&1 (X i , X 1 )M + α&&2 (X i , X 2 )M + α&&3 (X i , X 3 )M + α 1 (X i , X 1 )K + α 2 (X i , X 2 )K + α 3 (X i , X 3 )K = 0 (f) Use of orthonormality in Equation f leads to

α&&i + ω i2α i = 0

(g)

(b) Substitution of Equation c in the first set of initial conditions leads to ⎡0⎤ α 1 (0) X 1 + α 2 (0) X 2 + α 3 (0) X 3 = ⎢⎢ 0 ⎥⎥ ⎢⎣δ ⎥⎦

(h)

Taking the kinetic energy inner product of Equation h with X i leads to

α1 (0)(Xi, X1 )M + α 2 (0)(Xi, X 2 )M + α 3 (0)(Xi, X3 )M

⎡0⎤ = XTi M⎢⎢0⎥⎥ ⎢⎣ ⎥⎦

(i)

Use of mode shape orthonormality in Equation i leads to ⎡0⎤ α i (0) = X Ti M ⎢⎢ 0 ⎥⎥ ⎢⎣δ ⎥⎦

(k)

The normalized mode shape vectors are obtained during the solution of Problem 5.1 as ⎡ 0.5111 ⎤ ⎡0.4020⎤ 1 ⎢ 1 ⎢ ⎥ X1 = 0.5089⎥ X 2 = 0.2381 ⎥⎥ ⎢ ⎢ m m ⎢⎣− 0.7908⎥⎦ ⎢⎣0.5062⎥⎦

Then

181

⎡ − 0.7598⎤ 1 ⎢ X3 = 0.4294 ⎥⎥ ⎢ m ⎢⎣− 0.2324⎥⎦

(l)

α 1 (0) =

⎡m 0 0 ⎤ ⎡ 0 ⎤ [0.4020 0.5089 0.5062]⎢⎢ 0 2m 0 ⎥⎥ ⎢⎢ 0 ⎥⎥ = 0.5062 mδ m ⎣⎢ 0 0 m⎦⎥ ⎣⎢δ ⎦⎥

1

⎡m 0 0 ⎤ ⎡ 0 ⎤ α 2 (0) = [0.5111 0.2381 − 0.7908]⎢⎢ 0 2m 0 ⎥⎥ ⎢⎢ 0 ⎥⎥ = −0.7908 mδ m ⎢⎣ 0 0 m⎥⎦ ⎢⎣δ ⎥⎦ ⎡m 0 0 ⎤ ⎡ 0 ⎤ 1 [− 0.7598 0.4294 − 0.2324]⎢⎢ 0 2m 0 ⎥⎥ ⎢⎢ 0 ⎥⎥ = −0.2324 mδ α 3 (0) = m ⎢⎣ 0 0 m⎥⎦ ⎢⎣δ ⎥⎦ 1

(m)

(n)

(o)

It is similarly shown that

α&1 (0) = 0

α& 2 (0) = 0

α&13 (0) = 0

(p)

(c) The natural frequencies obtained during the solution of Problem 5.1 are

ω1 = 0.450

k k k , ω 2 = 1.613 , ω 3 = 2.387 m m m

(q)

Thus the problem for α 1 (t ) is

α&&1 + 0.2024

k α1 = 0 m

α 1 (0) = 0.5062 mδ

α&1 (0) = 0

(r)

The general solution of the differential equation in Equation r is ⎛

α 1 (t ) = C1 cos⎜⎜ 0.450 ⎝

⎛ k ⎞ k ⎞ t ⎟⎟ + C 2 sin ⎜⎜ 0.450 t ⎟⎟ m ⎠ m ⎝ ⎠

(s)

Application of initial conditions to Equation s leads to

α 1 (0) = 0.5062 mδ ⇒ C1 = 0.5062 mδ α& 2 (0) = 0 ⇒ C 2 = 0

(t) (u)

leading to ⎛

α 1 (t ) = 0.5062 mδ cos⎜⎜ 0.450 ⎝

182

k ⎞ t⎟ m ⎟⎠

(v)

The same solution procedure is used to obtain

⎛

α 2 (t ) = −0.7908 mδ cos ⎜⎜1.613

k ⎞⎟ t m ⎟⎠

⎝ ⎛ k ⎞ α 3 (t ) = −0.2324 mδ cos ⎜⎜ 2.387 t ⎟⎟ m ⎝ ⎠

(w) (x)

(d) Use of Equation c, Equation l, Equation v, Equation w and Equation x leads to ⎡ 0.5111 ⎤ ⎡0.4020⎤ ⎛ ⎛ k ⎞ 1 ⎢ k ⎞ 1 ⎢ ⎥ t ⎟⎟ t ⎟⎟ x(t ) = 0.5062 mδ cos⎜⎜ 0.450 0.5089⎥ − 0.7908 mδ cos⎜⎜1.613 0.2381 ⎥⎥ ⎢ ⎢ m ⎠ m m ⎠ m ⎝ ⎝ ⎢⎣− 0.7908⎥⎦ ⎢⎣0.5062⎥⎦ ⎡ − 0.7598⎤ ⎛ k ⎞ 1 ⎢ t ⎟⎟ − 0.2324 mδ cos⎜⎜ 2.387 0.4294 ⎥⎥ ⎢ m ⎠ m ⎝ ⎢⎣− 0.2324⎥⎦ ⎧⎡0.2035⎤ ⎫ ⎡− 0.4042⎤ ⎡ 0.1760 ⎤ ⎛ ⎛ ⎛ k ⎞ ⎢ k ⎞ ⎢ k ⎞⎪ ⎪⎢ ⎥ ⎥ ⎥ t ⎟⎟ + ⎢ − 0.1883⎥ cos⎜⎜1.613 = δ ⎨⎢0.2576⎥ cos⎜⎜ 0.450 t ⎟⎟ + ⎢− 0.0997 ⎥ cos⎜⎜ 2.387 t ⎟⎟⎬ (y) m m m ⎝ ⎝ ⎝ ⎠ ⎠ ⎠⎪ ⎪⎢0.2562⎥ ⎢⎣ 0.0254 ⎥⎦ ⎢⎣ 0.0540 ⎥⎦ ⎦ ⎩⎣ ⎭

183

5.3 The differential equations governing the motion of the system of Fig. P5.3 are ⎡m 0 ⎤ ⎡ &x&1 ⎤ ⎡ k ⎢ 0 2m ⎥ ⎢ &x& ⎥ + ⎢− k ⎣ ⎦⎣ 2 ⎦ ⎣

− k ⎤ ⎡ x1 ⎤ ⎡0⎤ = k ⎥⎦ ⎢⎣ x 2 ⎥⎦ ⎢⎣0⎥⎦

Determine the natural frequencies and normalized mode shapes for the system. Solution: The natural frequencies of this discrete system are the square roots of the

eigenvalues of M −1 K =

where φ =

k ⎡1 ⎢ m ⎢0 ⎣

− 1⎤ ⎡ 2 − 2⎤ 1 ⎥ = φ⎢ ⎥ ⎣− 1 1 ⎦ 2 ⎥⎦

0 ⎤ ⎡ 1 − 1⎤ k ⎡ 1 1 ⎥⎢ = ⎢ 1 − 1 1 ⎥⎦ m ⎢− ⎥ ⎣ ⎣ 2 2⎦

(a)

k . The eigenvalues are the values of λ which solve 2m M −1 K − λ I = 0 =

2φ − λ − 2φ −φ φ −λ

(b)

The determinant in Equation b is expanded leading to

(2φ − λ )(φ − λ ) − (− 2φ )(− φ ) = 0 λ2 − 3φλ = 0

(c)

The solutions of Equation c are λ = 0,3φ . The natural frequencies are the square roots of the eigenvalues, thus

ω1 = 0

ω2 =

3k 2m

(d)

A mode shape vector corresponding to a natural frequency ω solves M −1 KX = ω 2 X or ⎡ 2φ ⎢− φ ⎣

− 2φ ⎤ ⎡ X 1 ⎤ ⎡X ⎤ = ω2⎢ 1⎥ ⎢ ⎥ ⎥ φ ⎦⎣ X 2 ⎦ ⎣X 2 ⎦

(e)

The first equation of the equations represented in Equation e leads to 2φ − ω 2 X2 = X1 2φ

184

(f)

Setting X 1 to an arbitrary constant and substituting the natural frequencies of Equation d into Equation f leads to mode shape vectors of ⎡1⎤ X 1 = c1 ⎢ ⎥ ⎣1⎦

⎡1⎤ X 2 = c2 ⎢ ⎥ ⎣− 1⎦

(g)

The mode shape vectors are normalized by requiring (X i , X i )M = 1 . To this end X 1T MX 1 = 1

⎡m 0 ⎤ ⎡1⎤ c12 [1 1]⎢ ⎥⎢ ⎥ = 1 ⎣ 0 2m⎦ ⎣1⎦ c12 (3m) = 1 c1 =

A similar calculation leads to c 2 =

X1 =

1 3m

1

(h)

3m

. Thus the normalized mode shape vectors are

1 ⎡1⎤ ⎢⎥ 3m ⎣1⎦

185

X2 =

1 ⎡1⎤ ⎢ ⎥ 3m ⎣− 1⎦

(i)

5.4 The stress tensor at a point in a solid is ⎛ 200 150 100 ⎞ ⎟ ⎜ S = ⎜ 150 − 200 300 ⎟ kPa ⎜ 100 300 100 ⎟ ⎠ ⎝

Determine the principal stresses and unit vectors normal to the places on which they act. Solution: The principal stresses are the eigenvalues of the stress tensor. Let σ represent a

principal stress, then 200 − σ 150

150 − 200 − σ

100 300

100

300

100 − σ

=0

− σ 3 + 100σ 2 + 1.62 x10 5 σ − 1.32 x10 7 = 0

(a)

The principal stresses are the roots of the polynomial of Equation a

σ 1 = −395.53 kPa

σ 2 = 80.77 kPa

σ 3 = 414.77 kPa

(b)

The components of the vectors normal to the principal planes are the eigenvectors corresponding to the principal stresses. For example, for σ 1 = −395.53 kPa the components of the normal vector are obtained from 100 ⎤ ⎡ n1 ⎤ ⎡0⎤ ⎡595.53 150 ⎢ 150 195.53 300 ⎥⎥ ⎢⎢n 2 ⎥⎥ = ⎢⎢0⎥⎥ ⎢ ⎢⎣ 100 300 495.53⎥⎦ ⎢⎣ n3 ⎥⎦ ⎢⎣0⎥⎦

(c)

The first two equations represented by the matrix system of Equation c can be written as 595.53n1 + 150n2 = −100n3

(d)

150n 1 + 595.53n2 = −300n3

(e)

Equation d and Equation e are solved simultaneously to yield n1 = 0.2709n3 and n2 = −1.7421n3 . A unit vector normal to the principal plane is obtained by requiring

186

n12 + n22 + n32 = 1

(0.2709n3 )2 + (− 1.7421n3 )2 + n32 = 1 4.1083n32 = 1 n3 = 0.4934

(f)

Hence a unit vector normal to the lane on which the minimum normal stress acts is n 1 = 0.1336i − 0.8595 j + 0.8891k

(g)

Unit vectors normal to the other principal planes are calculated in a similar fashion leading to n 2 = 0.7744i − 0.2201j − 0.5932k

(h)

n 3 = 0.6184i + 0.4613 j + 0.6362k

(i)

187

5.5 Recall that the stress vector acting on a plane whose unit normal is n can be calculated as Sn. The normal stress acting on a plane σ is the component of the stress vector normal to the plane,

σ = (Sn, n )

(a)

where the inner product is the standard inner product on R 3 (the dot product). Prove that the maximum normal stress at a point is the largest principal stress and the minimum normal stress is the smallest principal stress by following these steps. The expansion theorem implies that the unit normal can be written as a linear combination of the unit normals to the principal planes. (a) Assume a linear combination for the unit normal to a plane, n = α 1n 1 + α 2 n 2 + α 3n 3 , noting that since the vector is a unit vector the sum of the squares of the coefficients in the linear combination is one. Substitute the expansion into Eq. (a), (b)Use properties of inner products and orthonormality of the unit vectors normal to the principal planes to reduce the resulting equation, (c) The resulting expression is a function of the expansion coefficients. The normal stress is stationary if dσ = 0 =

∂σ ∂σ ∂σ dα 1 + dα 2 + dα 3 ∂α 1 ∂α 2 ∂α 3

(b)

Since the coefficients are independent each of the partial derivatives in Eq. (b) must be zero. Calculate the partial derivatives and set them to zero, (d) Determine the solution(s) of the equations obtained in part (c) and use them to determine the extreme values of the normal stress. Solution: (a) Substitution of the linear combination into Equation a and using the

properties of inner products gives

188

σ = α 12 (Sn 1 , n 1 ) + α 1α 2 (Sn 1 , n 2 ) + α 1α 3 (Sn 1 , n 3 ) + α 2α 1 (Sn 2 , n 1 ) + α 22 (Sn 2 , n 2 ) + α 2α 3 (Sn 2 , n 3 ) + α 3α 1 (Sn 3 , n 1 ) + α 3α 2 (Sn 3 , n 2 ) + α 32 (Sn 3 , n 3 )

(b)

The definition of principal planes implies that Sn i = σ i n i . Substitution into Equation b and rearranging leads to

σ = σ 1 [α 12 (n 1 , n 1 ) + α 1α 2 (n 1 , n 2 ) + α 1α 3 (n 1 , n 3 )] + σ 2 [α 2α 1 (n 2 , n 1 ) + α 22 (n 2 , n 2 )

[

]

+ α 2α 3 (n 2 , n 3 )] + σ 3 α 3α 1 (n 3 , n 1 ) + α 3α 2 (n 3 , n 2 ) + α 32 (n 3 , n 3 )

(c)

(b) Orthonormality of the principal planes implies that (n i , n j ) = δ i , j . Equation c thus reduces to

σ = σ 1α 12 + σ 2α 22 + σ 3α 32

(d)

(c) The normal stress is stationary if ∂σ = 0 = 2α 1σ 1 ∂α 1

(e)

∂σ = 0 = 2α 2σ 2 ∂α 2

(f)

∂σ = 0 = 2α 3σ 3 ∂α 3

(g)

(d) Unless one of the principal stresses is zero the three solutions of Equation e, Equation g and Equation h are (i) α 1 = 1, α 2 = 0, α 3 = 0 , (ii) α 1 = 0, α 2 = 1, α 3 = 0 and (iii)

α 1 =, α 2 = 0, α 3 = 1 . If a principal stress, say σ 1 , is zero, the n the solutions of Equation e, Equation g and Equation h are of the form (i) α 2 = 0 and α 12 + α 32 = 1 and (ii)

α 3 = 0 and α 12 + α 22 = 1 . The maximum normal stress is the maximum value of σ and occurs when σ is stationary. Evaluation of the normal stress for each of the cases when σ is stationary leads

189

to a principal stress. Thus if σ 3 ≥ σ 2 ≥ σ 1 the maximum normal stress is σ 3 and the minimum normal stress is σ 1 . This is also true when a principal stress is zero.

190

5.6 Using the notation of Problem 5.5 the shear stress vector acting on a plane whose unit normal is n is calculated by

τ = Sn − σn

(c)

The magnitude of the shear stress is

τ = (τ ,τ ) 2 = (Sn − σn, Sn − σn ) 2 1

1

(d)

Expand the unit normal in terms of the unit vectors to the principal planes. (a) Use the expansion in Equation d to prove that the maximum normal stress acts on a plane which makes an angle of

π 4

to the plane of the largest normal stress.

(b) Calculate the maximum shear stress and a unit vector normal to the plane on which it acts for the sate of stress described by the stress tensor of Equation a of Problem 5.4. Solution: (a) Using the properties of inner products, Equation d can be rearranged to

τ

2

= (Sn − σn, Sn − σn ) = (Sn, Sn ) − 2σ (Sn, n ) + σ 2 (n, n)

(e)

Using the expansion n = α 1n 1 + α 2 n 2 + α 3n 3 where n i represent the unit vectors normal to the principal planes and noting that, by definition, Sn i = σ i n i where σ i is the corresponding principal stress and that since n is a unit vector, (n,n)=1 Equation e becomes

τ

2

= ∑∑ σ iσ j α iα j (n i , n j ) − 2σ ∑∑ σ iα iα j (n i , n j ) + σ 2 3

3

i =1 j =1

3

3

i =1 j =1

Orthonormality of the principal planes implies that (n i , n j ) = δ i , j . Thus Equation f reduces to

191

(f)

τ

2

3

3

i =1

i =1

= ∑ α i2σ i2 − 2σ ∑ σ iα i2 + σ 2

The magnitude of the shear stress is rendered stationary by requiring

(g) ∂τ

2

∂α k

= 0 for

k=1,2,3. To this end 3 ∂α i2 ∂α i2 ∂σ ⎛ 3 ⎞ ∂σ − 2σ ∑ σ i − 2⎜ ∑ σ iα i2 ⎟ + 2σ =0 σ ∑ ∂α k ∂α k ∂α k i =1 i =1 ⎝ i =1 ⎠ ∂α k 3

2 i

Noting that

(h)

∂α i = δ i ,k , Equation h reduces to ∂α k

∂σ ⎛ 3 ⎞ ∂σ 2σ k2α k − 2σσ k α k − 2⎜ ∑ σ iα i2 ⎟ + 2σ =0 ∂α k ⎝ i =1 ⎠ ∂α k

(i)

Note that σ is the normal stress which also depends upon the values of α’s, σ = (Sn, n ) . Using the same steps in the solution of Problem 5.5 it is shown that

σ = σ 1α 12 + σ 2α 22 + σ 3α 32 . Hence

∂σ = 2σ k α k and Equation i becomes ∂α k ⎛

3

3

⎞

⎛

⎠

⎝ i =1

3

⎞

σ k2α k − σ k α k ∑ σ iα i2 − 2⎜ ∑ σ iα i2 ⎟σ k α k + 2⎜ ∑ σ iα i2 ⎟σ k α k = 0 ⎝ i =1

i =1

⎠

(j)

or 3

σ k2α k − σ k α k ∑ σ iα i2 = 0 i =1

α k σ k [σ k − (σ 1α 12 + σ 2α 22 + σ 3α 32 )] = 0 Evaluation of Equation j for k=1,2,3 leads to

α 1 [σ 12 − σ 1 (σ 1α 12 + σ 2α 22 + σ 3α 32 ) − 2] Equation i reduces to, for k=1

192

(k)

3 3 ⎧⎪⎛ ⎤ ⎞ ⎡⎛ ⎞ 2α 1 ⎨⎜⎜ σ 1 − ∑ σ j α 2j ⎟⎟ ⎢⎜⎜ σ 1 − ∑ σ j α 2j ⎟⎟ − 2σ 1α 12 ⎥ − j =1 j =1 ⎪⎩⎝ ⎥⎦ ⎠ ⎢⎣⎝ ⎠ 3 3 ⎛ ⎛ ⎞ ⎞⎪⎫ 2σ 1α 22 ⎜⎜ σ 2 − ∑ σ j α 2j ⎟⎟ − 2σ 1α 32 ⎜⎜ σ 3 − ∑ σ j α 2j ⎟⎟⎬ = 0 j =1 j =1 ⎝ ⎠ ⎝ ⎠⎪⎭

(l)

One solution of Equation l is α 1 = 1, α 2 = 0, α 3 = 0 which corresponds to the case of minimum shear stress. Similar solutions exist which correspond to the other principal planes. Further algebra leads to the equation

σ 1 − 2σ 1α 12 = 0 which leads to α 1 = ±

angle of

π 4

2 implying that the plane of maximum shear stress makes an 2

to the plane of maximum shear stress.

193

(m)

5.7 The stress vector defining the state of stress at a point is − 50 − 100 ⎞ ⎛ 100 ⎜ ⎟ S = ⎜ − 50 − 100 200 ⎟ kPa ⎜ − 100 200 200 ⎟⎠ ⎝

(a)

The principal stresses and unit vectors normal to the principal planes are ⎡− 0.4082⎤ σ 1 = 350 kPa n1 = ⎢⎢ 0.4082 ⎥⎥ ⎢⎣ 0.8165 ⎥⎦ ⎡0.9129⎤ σ 2 = 50 kPa n 2 = ⎢⎢0.1826⎥⎥ ⎢⎣ 03651 ⎥⎦

(b)

⎡ 0 ⎤ σ 3 = −200 kPa n 3 = ⎢⎢ 0.8944 ⎥⎥ ⎢⎣− 04472⎥⎦ A small change in loading leads to a stress vector of ⎛100 + 10ε ⎜ S = ⎜ − 50 + 2ε ⎜ − 100 ⎝

− 50 + 2ε − 100 + ε 200

− 100 ⎞ ⎟ 200 ⎟ kPa 200 ⎟⎠

(c)

where ε is a small dimensionless parameter. The principal stresses for the new state of stress and the normal vectors are expanded as

σˆ i = σ i + εμ i nˆ i = n i + εm i

(d) (e)

Determine the perturbations in principal stresses, μ i , i=1,2,3. Solution: The problem for the perturbation in principal stresses is

(S 0 + εS 1 )(n i + εm i ) − (σ i + εμ i )(n i + εm i ) = 0 The two lowest hierarchal equations are

194

(f)

S 0n i − σ i n i = 0

(g)

S 0 m i − σ i m i = μ i n i − S 1n i

(h)

The given principal stresses and principal planes solve Equation g. A solution of Equation h exists only if a solvability condition is satisfied. The non-homogeneous terms on the right-hand side of Equation h must be orthogonal to the unit vector for the principal plane,

(μ i n i − S 1n i , n i ) = 0 μ i (n i , n i ) − (S 1n i , n i ) = 0 μ i = (S 1n i , n i )

(i)

Noting that

⎛10 2 0 ⎞ ⎜ ⎟ S1 = ⎜ 2 1 0 ⎟ ⎜ 0 0 0⎟ ⎝ ⎠

(j)

application of Equation i leads to

μ1 = (S 1n 1 , n 1 ) ⎡10 2 0⎤ ⎡− 0.4082⎤ = [− 0.4082 0.4082 0.8165]⎢⎢ 2 1 0⎥⎥ ⎢⎢ 0.4082 ⎥⎥ ⎢⎣ 0 0 0⎥⎦ ⎢⎣ 0.8165 ⎥⎦ = 1.1664

(k)

⎡10 2 0⎤ ⎡0.9129⎤ μ 2 = [0.9129 0.1826 0.3651]⎢⎢ 2 1 0⎥⎥ ⎢⎢0.1826⎥⎥ = 9.034 ⎢⎣ 0 0 0⎥⎦ ⎢⎣ 03651 ⎥⎦

(l)

⎡10 2 0⎤ ⎡ 0 ⎤ μ 3 = [0 0.8944 − 0.4472]⎢⎢ 2 1 0⎥⎥ ⎢⎢ 0.8944 ⎥⎥ = 0.800 ⎢⎣ 0 0 0⎥⎦ ⎢⎣− 04472⎥⎦

(m)

195

5.8 The differential equations governing the motion of the system of Fig P5.8 are ⎡m 0 0 0 ⎤ ⎡ &x&1 ⎤ ⎡ 3k ⎢ 0 2m 0 0 ⎥ ⎢ &x& ⎥ ⎢− k ⎢ ⎥⎢ 2 ⎥ + ⎢ ⎢ 0 0 m 0 ⎥ ⎢ &x&3 ⎥ ⎢ 0 ⎢ ⎥⎢ ⎥ ⎢ ⎣ 0 0 0 m ⎦ ⎣ &x&4 ⎦ ⎣ 0

−k

0 −k

2k −k

3k − 2k

0

The lowest natural frequency for the system is 0.4344

normalized eigenvector of X =

0 ⎤ ⎡ x1 ⎤ ⎡0⎤ 0 ⎥⎥ ⎢⎢ x 2 ⎥⎥ ⎢⎢0⎥⎥ = − 2 k ⎥ ⎢ x 3 ⎥ ⎢0 ⎥ ⎥⎢ ⎥ ⎢ ⎥ 2 k ⎦ ⎣ x 4 ⎦ ⎣0 ⎦

(a)

k which has a corresponding m

1 [0.1397 0.4005 0.5541 0.5938] . m

(a) Determine the perturbation in the lowest natural frequency if the particle of mass 2m is adjusted to 2m + εm where ε is a small independent dimensionless parameter. (b) Determine the perturbation in the lowest natural frequency if the stiffness of the spring connecting the left most particle of mass m with the particle of mass 2m changes to a stiffness of k − εk where ε is a small dimensionless parameter. ~ ˆ = M + εM where Solution: (a) Define M

⎡0 0 0 0 ⎤ ⎥ ⎢ ~ ⎢0 m 0 0 ⎥ M= ⎢0 0 0 0 ⎥ ⎥ ⎢ ⎣0 0 0 0 ⎦

(b)

Let ωˆ 1 = ω1 + εω~ be the lowest natural frequency of the perturbed problem with a ~ ˆ = X + εX . The perturbed problem is corresponding mode shape vector of X

ˆ = ωˆ 2 M ˆX ˆ KX

(c)

Substitution of the perturbed quantities into Equation c leads to

(

)

(

)(

~ ~ ~ 2 K X + εX = (ω1 + εω~ ) M + εM X + εX

196

)

(d)

The hierarchal equations resulting from Equation d are KX − ω 12 MX = 0 ~ ~ ~ KX − ω12 MX = 2ω 1ω~MX + ω 12 MX

(e) (f)

Since Equation e has a non-trivial solution, X, Equation f has a solution only if a solvability condition is satisfied. Pre-multiplying both sides of Equation f by M −1 leads to ~ ~ ~ M −1 KX − ω12 X = 2ω1ω~X + ω12 M −1 MX

(g)

The appropriate solvability condition is that the right-hand side of Equation g is orthogonal to X with respect to the kinetic energy inner product,

(2ω ω~X + ω M M~ X, X) ω ~ ω~ = − (M MX, X ) 2 2 1

1

1

−1

M

=0

−1

M

=−

ω1 2

~ X T MX

(h)

Calculations lead to ⎡0 0 ⎢0 m k 1 ~ [0.1397 0.4005 0.5541 0.5938]⎢ ω = −0.4344 ⎢0 0 mm ⎢ ⎣0 0 ⎡ 0 ⎤ ⎢0.4005⎥ k ⎥ = −0.4344 [0.1397 0.4005 0.5541 0.5938]⎢ ⎢ ⎥ 0 m ⎢ ⎥ ⎣ 0 ⎦ = -.0697

k m

Thus ωˆ = (0.4344 − 0.0697ε )

0 0⎤ ⎡0.1397 ⎤ 0 0⎥⎥ ⎢⎢0.4005⎥⎥ 0 0⎥ ⎢ 0.5541⎥ ⎥⎢ ⎥ 0 0⎦ ⎣0.5938⎦

(i)

k . m

~ ˆ = K + εK where (b) Define K

197

⎡− k ⎢ k ~ K=⎢ ⎢ 0 ⎢ ⎣ 0

k −k 0 0

0 0 0 0

0⎤ 0⎥⎥ 0⎥ ⎥ 0⎦

(j)

Following the same steps as in part (a), the hierarchal equation at the first order becomes ~ ~ ~ KX − ω12 MX = 2ω1ω~MX − KX

(k)

~ ~ ~ M −1 KX − ω12 X = 2ω1ω~X − M −1KX

(l)

Equation k is rewritten as

The solvability condition for Equation l is

(2ω ω~X − M 1 (M ω~ = 2ω

) ~ KX, X )

−1

1

−1

~ KX, X

M

=0

M

1

=

1 ~ X T KX 2ω1

⎡− k ⎢k 1 m 1 [0.1397 0.4005 0.5541 0.5938]⎢ = ⎢ 0 2(0.4344) k m ⎢ ⎣ 0

k −k 0 0

0 0 0 0

0⎤ ⎡0.1397⎤ 0⎥⎥ ⎢⎢0.4005⎥⎥ 0⎥ ⎢ 0.5541⎥ ⎥ ⎥⎢ 0⎦ ⎣0.5938⎦

⎡ 0.2608 ⎤ ⎢− 0.2608⎥ k ⎥ [0.1397 0.4005 0.5541 0.5938]⎢ = 1.1510 ⎥ ⎢ 0 m ⎥ ⎢ 0 ⎦ ⎣ = 0.0783

k m

(m)

198

5.9 (a) Determine all eigenvalues and eigenvectors for the matrix ⎡ 3 −1 0 ⎤ A = ⎢⎢− 1 2 − 1⎥⎥ ⎢⎣ 0 − 1 2 ⎥⎦

(b) Normalize the eigenvectors and then demonstrate orthonormality of eigenvectors. Solution: The eigenvalues are determined by setting A − λI = 0 3−λ −1 0

−1 2−λ −1

0 −1 = 0 2−λ

(a)

Expansion of the determinant in Equation a leads to the polynomial equation − λ3 + 7λ2 − 14λ + 7 = 0

(b)

The eigenvalues of A are the roots of Equation b which are

λ1 = 0.7530

λ2 = 2.445

λ3 = 3.8019

(c)

The eigenvector corresponding to an eigenvalue is obtained from ⎡3 − λ ⎢ −1 ⎢ ⎢⎣ 0

−1 2−λ −1

0 ⎤ ⎡ a ⎤ ⎡0 ⎤ − 1 ⎥⎥ ⎢⎢b ⎥⎥ = ⎢⎢0⎥⎥ 2 − λ ⎥⎦ ⎢⎣ c ⎥⎦ ⎢⎣0⎥⎦

(d)

The first of the equations in Equation d leads to a=

1 b 3−λ

(e)

while the third of equations in Equation d leads to c=

1 b 2−λ

(f)

Setting λ1 = 0.7530 in Equations e and f leads to a = 0.4450b and c = 0.8019b , leading to an eigenvector of the form 199

⎡0.4450⎤ u 1 = ⎢⎢ 1 ⎥⎥b ⎢⎣0.8019⎥⎦

(g)

for any non-zero value of b. The matrix A is symmetric, hence it is self adjoint with respect to the standard inner product. It is normalized by requiring that

(u 1 , u 1 )2 = 1 = (0.4450b )2 + (b) 2 + (0.8019b) 2 = 1.8411b 2 ⇒ b =

1 1.8411

= 0.7370

The normalized eigenvector is thus ⎡0.4450⎤ ⎡0.3280⎤ u 1 = 0.7370 ⎢⎢ 1 ⎥⎥ = ⎢⎢0.7370⎥⎥ ⎢⎣0.8019⎥⎦ ⎢⎣0.5910⎥⎦

(h)

The eigenvectors for the larger eigenvalues are determined and normalized in a similar manner leading to ⎡ 0.5910 ⎤ u 2 = ⎢⎢ 0.3280 ⎥⎥ ⎢⎣− 0.7370⎥⎦

⎡ 0.7370 ⎤ u 3 = ⎢⎢− 0.5910⎥⎥ ⎢⎣ 0.3280 ⎥⎦

(i)

(b) Orthogonality of the eigenvectors is demonstrated by showing that (u i , u j ) = 0 for i ≠ j and I,j=1,2,3. For example ⎡ 0.5910 ⎤ (u 2 , u 3 ) = u u 2 = [0.7370 − 0.5910 0.3280]⎢⎢ 0.3280 ⎥⎥ ⎢⎣− 0.7370⎥⎦ = (0.7370)(0.5910) + (−0.5910)(0.3280) + (0.3280)(−0.7370) T 3

=0

200

⎡1 6⎤ 5.10 (a) Determine all eigenvalues and eigenvectors of the matrix A = ⎢ ⎥. ⎣3 4⎦ (b) Determine all eigenvalues and eigenvectors of A T (c) Demonstrate the principle of biorthogonality of the eigenvectors of A and A T . Solution: (a) The eigenvalues of A are determined from A − λI = 0 or 1− λ 3

6 =0 4−λ

(1 − λ )(4 − λ ) − (3)(6) = 0 λ2 − 5λ − 14 = 0 (λ − 7 )(λ + 2) = 0

(a)

The eigenvalues of A are -2 and 7. The eigenvectors are determined by finding nontrivial solutions of ⎡1 − λ ⎢ 3 ⎣

6 ⎤ ⎡ a ⎤ ⎡0 ⎤ = 4 − λ ⎥⎦ ⎢⎣b ⎥⎦ ⎢⎣0⎥⎦

(b)

The first equation represented by the matrix representation of Equation b is

(1 − λ )a + 6b = 0 ⇒ b = λ − 1 a . For 6

λ = −2, b = −

1 a and for λ = 7, b = a . 2

Arbitrarily taking a=1, the eigenvectors of A are ⎡ 1 ⎤ ⎢ 1⎥ ⎢⎣− 2 ⎥⎦

⎡1⎤ ⎢1⎥ ⎣⎦

(b)

(b) The eigenvalues of A T are determined from A T − λI = 0 or 1− λ 6

3 =0 4−λ

(1 − λ )(4 − λ ) − (3)(6) = 0 λ2 − 5λ − 14 = 0 (λ − 7 )(λ + 2) = 0

201

(c)

The eigenvalues of A T are -2 and 7. The eigenvectors are determined by finding nontrivial solutions of ⎡1 − λ ⎢ 6 ⎣

3 ⎤ ⎡ a ⎤ ⎡0 ⎤ = 4 − λ ⎥⎦ ⎢⎣b ⎥⎦ ⎢⎣0⎥⎦

(d)

The first equation represented by the matrix representation of Equation d is

(1 − λ )a + 3b = 0 ⇒ b = λ − 1 a . For λ = −2, b = −a and for λ = 7, b = 2a . Arbitrarily 3

taking a=1, the eigenvectors of A are ⎡1⎤ ⎢− 1⎥ ⎣ ⎦

⎡1 ⎤ ⎢ 2⎥ ⎣ ⎦

(b)

(c) The principle of biorthogonality implies that an eigenvector of A T corresponding to an eigenvalue μ is orthogonal to all eigenvectors of A which do not correspond to an eigenvalue λ = μ . Checking biorthogonality for this problem ⎛ ⎡ 1 ⎤ ⎡1 ⎤ ⎞ ⎜ ⎢ 1 ⎥, ⎟ = (1)(1) + ⎛⎜ − 1 ⎞⎟(2) = 0 ⎥ ⎢ − ⎜ ⎢ ⎥ ⎣ 2⎦ ⎟ ⎝ 2⎠ ⎝ ⎣ 2⎦ ⎠ ⎛ ⎡1⎤ ⎛ 1 ⎞ ⎞ ⎜ ⎢ ⎥, ⎜⎜ ⎟⎟ ⎟ = (1)(1) + (1)(−1) = 0 ⎜ 1 −1 ⎟ ⎝ ⎣ ⎦ ⎝ ⎠⎠

Thus the principle of biorthogonality is satisfied.

202

(c)

5.11 An valid inner product defined for R 2 is (x, y ) = x1 y1 + 2 x 2 y 2 . (a) Show that the matrix A in Problem 5.10 is self adjoint with respect to this inner product. (b) Demonstrate orthogonality of eigenvectors of A. (c) Since A is self adjoint with respect to a valid inner product, its eigenvalues are all real. Use this idea to deduce sufficient conditions for which a 2x2 matrix has all real eigenvalues. Hint: Is there any inner ⎡ 1 6⎤ product for which ⎢ ⎥ is self adjoint? ⎣ − 3 4⎦ Solution: The matrix is self-adjoint with respect to the inner product if for all u and v

(Au, v ) = (u, Av) . To this end consider ⎡1 6⎤ ⎡ u1 ⎤ ⎡ u1 + 6u 2 ⎤ Au = ⎢ ⎥ ⎥⎢ ⎥ = ⎢ ⎣3 4⎦ ⎣u 2 ⎦ ⎣3u1 + 4u 2 ⎦

(a)

Then, using the definition of the given inner product,

(Au, v ) = v1 (u1 + 6u 2 ) + 2v 2 (3u1 + 4u 2 ) = u1v1 + 6v1u 2 + 6v 2 u1 + 8v 2 u 2

= u1 (v1 + 6v 2 ) + 2u 2 (3v1 + 4v 2 ) = (u, Av )

(b)

(b) The eigenvectors of A are determined in the solution of Problem 5.10 as ⎡ 1 ⎤ u1 = ⎢ 1 ⎥ ⎢⎣− 2 ⎥⎦

⎡1⎤ u2 = ⎢ ⎥ ⎣1⎦

( c)

Using the given definition of the inner product

(u 1 , u 2 ) = (1)(1) + 2⎛⎜ − 1 ⎞⎟(1) = 0 ⎝ 2⎠

203

(d)

(c) An inner product of a 2x2 matrix is of the form (x, y ) = αx1 y1 + βx 2 y 2 where α and β are positive real numbers. Then for a 2x2 matrix A,

(Ax, y ) = αy1 (a1,1 x1 + a1,2 x 2 ) + βy 2 (a 2,1 x1 + a 2, 2 x 2 ) ⎞ ⎛α β ⎛ ⎞ = αx1 ⎜ a1,1 y1 + a 2,1 y 2 ⎟ + β x 2 ⎜⎜ a1, 2 y1 + a 2, 2 y 2 ⎟⎟ α ⎝ ⎠ ⎠ ⎝β

(e)

The matrix is self adjoint with respect to such an inner product if

β a 2,1 = a1, 2 α

(f)

Since α and β are positive real numbers a1, 2 and a 2,1 must be of the same sign in order for Equation f to be possible for appropriate choices of α and β .

204

5.12 A valid inner product defined for R 3 is (x, y ) = x1 y1 + x 2 y 2 + 3x3 y 3 . (a) Give an example of a 3x3 matrix A which is self-adjoint with respect to this inner product. (b) Determine the eigenvalues and eigenvectors of this matrix and demonstrate orthogonality of the eigenvectors. (c) Determine the eigenvectors of A T . Note A T = A * with respect to the standard inner product for R 3 . Demonstrate the biorthogonality principle. (d) Using this problem as a guide, determine sufficient conditions for a 3x3 matrix to have all real eigenvalues. Solution: (a) Consider the matrix ⎡ 4 1 3⎤ A = ⎢⎢1 2 6⎥⎥ ⎢⎣1 2 3⎥⎦

(a)

Note that a1,3 = 3a3,1 , a 2,3 = 3a 3, 2 and a1, 2 = a 2,1 . This matrix is self adjoint with respect to the defined inner product. (b) The eigenvalues and eigenvectors of A are obtained as

λ1 = −1.0319

λ2 = 2.8232

λ3 = 7.2007

⎡ 0.1003 ⎤ u 1 = ⎢⎢ 0.8812 ⎥⎥ ⎢⎣− 0.4620⎥⎦ ⎡ 0.8825 ⎤ u 2 = ⎢⎢− 0.4231⎥⎥ ⎢⎣− 0.2051⎥⎦ ⎡0.6218⎤ u 3 = ⎢⎢0.6398⎥⎥ ⎣⎢0.4518⎦⎥

Orthogonality with respect to the given inner product is illustrated by

(u 1 , u 2 ) = (0.1003)(0.8825) + (0.8812)(−0.4231) + 3(−0.4620)(−0.2051) = 0 (c) The eigenvalues and eigenvectors of A T are determined as

205

(b)

λ1 = −1.0319

λ2 = 2.8232

λ3 = 7.2007

⎡ 0.0610 ⎤ v 1 = ⎢⎢ 0.5355 ⎥⎥ ⎢⎣− 0.8423⎥⎦ ⎡ 0.7633 ⎤ v 2 = ⎢⎢− 0.3660⎥⎥ ⎢⎣ − 0.5323⎥⎦ ⎡0.3832⎤ u 3 = ⎢⎢0.3943⎥⎥ ⎢⎣0.8353⎥⎦

(c)

Biorthogonality with respect to the standard inner product is illustrated by the example of

(u 1 , v 2 ) = (0.1003)(0.7633) + (0.8812)(−0.3660) + (−0.4620)(−0.5323) = 0

206

(d)

5.13-5.16 For the differential system (a) Specify an inner product for which the system is self adjoint. (b) Determine all eigenvalues and eigenvectors for the system, (c) Normalize the eigenvectors and (d) Demonstrate orthogonality of the eigenvectors. 5.13

d2y + λy = 0 dx 2 dy (0) = 0 dx dy y (1) + 2 (1) = 0 dx

Solution: (a) The system is self-adjoint with respect to the standard inner product on

C 2 [0,1]] 1

( f ( x), g ( x)) = ∫ f ( x) g ( x)dx

(a)

0

(b) The general solution of the differential equation is

( )

( )

y ( x) = C1 cos λ x + C 2 sin λ x

(b)

Application of the boundary conditions to Equation b leads to dy (0) = 0 ⇒ − λ C1 sin(0) + λ C 2 cos(0 ) = 0 ⇒ C 2 = 0 dx y (1) + 2

( )

( )

( )

dy 1 (1) = 0 ⇒ C1 cos λ − λ C1 sin λ = 0 ⇒ = tan λ dx λ

(c)

(d)

There are an infinite, but countable, number of solutions of Equation d, 0 < λ1 < λ 2 < ... < λ k −1 < λ k < λ k +1 < ... Each eigenvector has an eigenvalue of the form

(

y k ( x) = C k cos λ k x

(c) The eigenvectors are normalized by requiring

207

)

(e)

1

( y k , y k ) = 1 = ∫ [ y k ( x)]

2

0

=

1

dx = ∫ C sin 2 k

2

(

⎛ ⎜ x − 1 cos 2 λ k x ⎜ 2 λk ⎝

)

(

C λ k x dx = k 2

0

x =1

⎞ ⎟ ⎟ ⎠ x =0

)

⎤ Ck ⎡ 1 1 cos 2 λ k ⎥ + ⎢1 − 2 ⎣⎢ 2 λ k 2 λ k ⎥⎦

(

)

2

Ck =

(f)

⎡ ⎤ 1 1 cos 2 λ k ⎥ + ⎢1 − ⎢⎣ 2 λ k 2 λ k ⎥⎦

(

)

(d) Consider the inner product

(y , y ) = ∫ C 1

i

j

i

(

)

(

)

cos λi x C j cos λ j x dx

(g)

0

A trigonometric identity is used for the integrand of Equation g leading to

[(

)]

(

)

[(

)]

x =1

⎧ sin λi − λ j x sin λi + λ j x ⎫⎪ (yi , y j ) = Ci C j ⎪⎨ + ⎬ 2 ⎪ − + λ λ λ λ ⎪⎭ x =1 i j i j ⎩ =

(

sin λi + λ j C i C j ⎧⎪ sin λi − λ j + ⎨ 2 ⎪ λi − λ j λi + λ j ⎩

)⎫⎪ ⎬ ⎪⎭

(h)

Further use of trig identities in Equation h leads to i

i

j

( )

⎧⎪ sin λi cos λ j − cos λi sin λ j sin λi cos λ j + cos λi sin λ j + ⎨ 2 ⎪ λ λ λi + λ j − i j ⎩ Ci C j = 2 λi sin λi cos λ j − 2 λ j cos λi sin λ j (i) 2(λi − λ j )

(y , y ) = C C

j

[

]

Equation c, the transcendental equation defining the eigenvalues, is rearranged as cos λi = λi sin λi Use of Equation j in Equation i leads to

208

(j)

⎫⎪ ⎬ ⎪⎭

(y , y ) = i

j

Ci C j

2(λi − λ j )

[2

λi λ j sin λi sin λ j − 2 λ j λi sin λi sin λ j

=0

] (k)

Thus eigenvectors corresponding to distinct eigenvalues are mutually orthogonal with respect to the standard inner product.

209

d2y dy 5.14 + 6 + λy = 0 2 dx dx y (0) = 0 dy (1) = 0 dx Solution: (a) The differential equation is put into the self adjoint Sturm-Liouville form of

−

1 d ⎛ 6 x dy ⎞ ⎜e ⎟ = λy dx ⎠ e 6 x dx ⎝

(a)

The eigenvectors are orthogonal with respect to the inner product defined as 1

( f , g )r = ∫ f ( x) g ( x)e 6 x dx

(b)

0

(b)The general solution to the differential equation is obtained by assuming y = e αx which leads to α 2 + 6α + λ = 0 . The quadratic formula is used to obtain α = −3 ± 9 − λ . Since the Sturm-Liouville problem is positive definite the eigenvalues are all positive. Also it can be shown that λ > 9 such that the general solution of the differential equation is

y ( x ) = C1 e −3 x cos

(

)

(

λ − 9 x + C 2 e −3 x sin λ − 9 x

)

(c)

Application of the boundary conditions to Equation c leads to y (0) = 0 ⇒ C1 = 0

(d)

and

[

(

)

(

dy (1) = 0 ⇒ C 2 − 3e −3 sin λ − 9 + e −3 λ − 9 cos λ − 9 dx λ −9 tan λ − 9 = 3

(

)

210

)] (e)

The eigenvalues λ k , k = 1,2,... are the solutions of Equation e. The corresponding eigenvectors are

(

y k ( x) = C k e −3 x sin λ − 9 x

)

(f)

(c) The eigenvectors are normalized by requiring ( y k . y k )r = 1 which leads to

∫ [C e 1

−3 x

k

)]

(

2

sin λ − 9 x e 6 x dx = 1

0

[ (

1

)]

2 ∫ C k sin λ − 9 x dx = 1 2

0

C k2 2

∫ [1 − cos(2

C k2 2

⎡ ⎤ 1 sin 2 λ − 9 x ⎥ = 1 ⎢x − 2 λ −9 ⎣ ⎦ x =1

C k2 2

⎡ ⎤ 1 sin 2 λ − 9 ⎥ = 1 ⎢1 − ⎣ 2 λ −9 ⎦

1

)]

λ − 9 x dx = 1

0

(

)

(

x =1

)

(g)

Use of the double angle formula and Equation e in Equation g leads to

(

) (

)

C k2 2

⎡ ⎤ 1 sin λ − 9 cos λ − 9 ⎥ = 1 ⎢1 − λ −9 ⎣ ⎦

C k2 2

⎡ 1 2 ⎢⎣1 − 3 cos

Ck =

(

)⎤

λ −9 ⎥ =1 ⎦

6 3 − cos

2

(

λ −9

)

(d) Consider the inner product

211

(h)

(y , y ) = ∫ [C e 1

i

j r

−3 x

i

)][

(

) (

)

(

)]

sin λi − 9 x C j e −3 x sin λ j − 9 x e 6 x dx

0

1

( ⎧⎪ sin [(

= ∫ C i C j sin λi − 9 x sin λ j − 9 x dx 0

)]

[(

)]

sin λi − 9 + λ j − 9 ⎫⎪ λi − 9 − λ j − 9 − ⎬ ⎨ 2 ⎪ 9 9 9 9 λ λ λ λ − − − − + − ⎪⎭ i j i j ⎩ C i C j ⎧⎪ sin λi − 9 cos λ j − 9 − cos λi − 9 sin λ j − 9 = ⎨ 2 ⎪ λi − 9 − λ j − 9 ⎩ =

Ci C j

(

−

[(

)

(

(

) (

)

Ci C j

(λ

i

− λj )

(

)

) (

)

sin λi − 9 cos λ j − 9 + cos λi − 9 sin λ j − 9 ⎫⎪ ⎬ λi − 9 + λ j − 9 ⎪⎭

) ( ( Noting from Equation e that sin ( λ − 9 ) = =

) (

) (

) ( λ − 9 )cos( λ − 9 )sin ( λ λ − 9) cos( λ − 9 ) , Equation (i) is

λ j − 9 sin λi − 9 cos λ j − 9 −

i

i

j

−9

)] (i)

i

i

i

3

rewritten as

(y , y ) i

j r

=

Ci C j

3(λi − λ j )

[(

λi − 9

)(

)

(

) (

λ j − 9 cos 2 λi − 9 −

=0

λi − 9

)(

)

λ j − 9 cos 2

(

λi − 9

)]

(j)

Eigenvector orthogonality is confirmed by Equation j.

212

dy d2y + 4 x + λy = 0 5.15 x 2 dx dx 2

y (1) = 0 dy (2) = 0 dx

Solution: (a) The Sturm-Liouville form of the differential equation is obtained by noting x2

p ( x) = e

∫ 4 x dx

= x 4 and r ( x) =

p ( x) = x 2 . Thus the differential equation is rewritten as 2 x 1 d ⎛ 4 dy ⎞ ⎟ + λx = 0 ⎜x x 2 dx ⎝ dx ⎠

(a)

The eigenvectors are orthogonal with respect to the inner product defined by 2

( f , g )r = ∫ f ( x) g ( x) x 2 dx

(b)

1

(b) The differential equation is of the Cauchy-Euler form. A solution is assumed as y = x α which when substituted into the differential equation leads to

α (α − 1) + 4α + λ = 0 or α 2 + 3α + λ = 0 . The roots are obtained using the quadratic formula as

λ=

− 3 ± 9 − 4λ 2

(c)

Assuming 9 − 4λ < 0 (the Sturm-Liouville operator is positive definite) the roots are imaginary and the general solution of the differential equation is y ( x) = C1 x

−

3 2

3 − ⎡ 4λ − 9 ⎤ ⎡ 4λ − 9 ⎤ ln( x)⎥ ln( x)⎥ + C 2 x 2 sin ⎢ cos ⎢ 2 2 ⎣ ⎦ ⎣ ⎦

Application of boundary conditions to Equation d leads to y (1) = 0 ⇒ C1 = 0 and

213

(d)

dy (2) = 0 dx 3 ⎧⎪ 3 − 5 ⎡ 4λ − 9 ⎤ ⎡ 4λ − 9 ⎤ ⎫⎪ 4λ − 9 ⎞ 1 − ⎛ ⎟ cos ⎢ ⇒ C 2 ⎨− (2) 2 sin ⎢ ln(2)⎥ + (2) 2 ⎜⎜ ln( 2 ) ⎥⎬ = 0 ⎟2 2 2 2 ⎪⎩ 2 ⎣ ⎦ ⎠ ⎣ ⎝ ⎦ ⎪⎭

(e)

Equation e is rearranged to ⎡ 4λ − 9 ⎤ ln(2)⎥ = tan ⎢ 2 ⎣ ⎦

4λ − 9 3

(f)

There are an infinite, but countable, number of solutions of Equation f. The eigenvector corresponding to an eigenvalue λ k is y k ( x) = C k x

−

3 2

⎡ 4λ k − 9 ⎤ ln( x)⎥ sin ⎢ 2 ⎥⎦ ⎣⎢

(g)

2

The eigenvectors are normalized by requiring

∫y

k

( x) y k ( x) x 2 dx = 1 or

1

2

3 ⎧⎪ ⎡ 4λ k − 9 ⎤ ⎫⎪ − ln( x)⎥ ⎬ x 2 dx 1 = ∫ ⎨C k x 2 sin ⎢ 2 ⎢⎣ ⎥⎦ ⎪⎭ 1⎪ ⎩ 2

2

⎤ ⎫⎪ 1 ⎧⎪ ⎡ 4λ k − 9 ln( x)⎥ ⎬ dx = C k2 ∫ ⎨sin ⎢ x ⎪ ⎢⎣ 2 ⎥⎦ ⎪⎭ 1 ⎩ 2

The integral is evaluated by defining u = ln(x) such that du =

214

dx . This leads to x

(h)

ln( 2 )

1= C

2 k

∫ 0

C k2 = 2

2

⎡ ⎛ 4λ k − 9 ⎞ ⎤ ⎢sin ⎜ u ⎟ ⎥ du ⎟⎥ ⎜ 2 ⎢⎣ ⎝ ⎠⎦

∫ [1 − cos (

)]

ln( 2 )

4 λ k − 9 u du

0

C2 = k 2

⎡ ⎢u − ⎢⎣

C k2 = 2

⎡ sin ⎢ ln( 2 ) − ⎢⎣

1 4λ k − 9

(

sin

(

u = ln( 2 )

⎤ 4 λ k − 9u ⎥ ⎥⎦ u = 0

)

)

4 λ k − 9 ln( 2 ) ⎤ ⎥ 4λ k − 9 ⎥⎦

(i)

Thus the normalized eigenvectors are y k ( x) =

ln(2) −

3 ⎡ 4λk − 9 ⎤ − 2 x 2 sin ⎢ ln( x)⎥ 2 sin 4λk − 9 ln(2) ⎢⎣ ⎥⎦

(

)

(j)

4λk − 9

(d) Orthogonality of the eigenvectors is shown by

(y

3 3 2⎧ ⎡ 4λ j − 9 ⎤ ⎫⎪⎧⎪ ⎡ 4λ k − 9 ⎤ ⎫⎪ 2 − − ⎪ 2 2 ⎢ ⎥ ) , sin ln( ) sin ln( ) y C x x C x x = ⎢ ⎥ ⎬ x dx ⎬⎨ k j k r ∫1 ⎨ j 2 2 ⎢ ⎥ ⎪ ⎢ ⎥⎦ ⎪⎭ ⎪⎩ ⎣ ⎣ ⎦ ⎪⎭⎩ 2 ⎡ 4λ j − 9 ⎤ ⎡ 4λ − 9 ⎤1 k ln( x)⎥ sin ⎢ ln( x)⎥ dx = C j C k ∫ sin ⎢ 2 2 ⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦ x 1

The integral is evaluated by defining u = ln(x) such that du =

(y

j , y k )r = C j C k

ln( 2 )

∫ 0

Defining v j =

4λ j − 9 2

dx . This leads to x

⎡ 4λ j − 9 ⎤ ⎡ 4λ − 9 ⎤ k sin ⎢ u ⎥ sin ⎢ u ⎥ du 2 2 ⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦

evaluation of the integral of Equation l leads to

215

(k)

(l)

(y

j

[

]

[

]

C j C k ⎧⎪ sin (v j − v k )ln(2) sin (v j + v k )ln(2) ⎫⎪ − ⎬ ⎨ 2 ⎪⎩ v j − vk v j + vk ⎪⎭ C j Ck {(v j + vk ) sin v j ln(2) cos[vk ln(2)] − cos v j ln(2) sin[vk ln(2)] − = 2 v 2j − v k2

, y k )r =

(

)

[ [

[

]

[

]

[

]

] [

]

]

⎡sin v j ln(2) cos[v k ln(2)] + cos v j ln(2) sin v j ln(2) ⎤ ⎫⎪ − v k )⎢ ⎥⎬ ⎣ ⎦ ⎪⎭ C j Ck {vk sin v j ln(2) cos[vk ln(2)] − vk cos v j ln(2) sin[vk ln(2)]} = 2 v j − v k2

(v

(

j

)

[

]

[

[

]

Noting from Equation f that sin v j ln(2) =

[

]

]

2 v j cos v j ln(2) , it is easily shown that the 3

right-hand side of Equation m is zero for j ≠ k .

216

(m)

d2y 5.16 + λy = 0 dx 2 dy (0) = 0 dx dy (1) = 0 dx Solution: (a) The eigenvectors are orthogonal with respect to the standard inner product

on C 2 [0,1] . (b) The general solution of the differential equation for λ > 0 is

( )

( )

y ( x) = C1 cos λ x + C 2 sin λ x

(a)

Application of the boundary conditions leads to dy (0) = 0 ⇒ C 2 = 0 dx dy (1) = 0 ⇒ λ C1 sin λ = 0 dx

(b)

( )

(c)

λ k = k 2π 2 k = 0,1,2,...

(d)

y k ( x) = C k cos(kπx )

(e)

( )

Equation c implies that sin λ = 0 which leads to

The eigenvectors are of the form

Note that the problem is positive semi-definite and λ = 0 is an eigenvalue with a corresponding eigenvector y 0 ( x) = C 0

(f)

(b) Normalization of eigenvector requires ( y k , y k ) = 1 . For k=0, 1

∫C

2 0

dx = 1 ⇒ C 0 = 1 ⇒ y 0 ( x) = 1

0

217

(g)

For k>0, 1

∫ [C

cos(kπx )] dx = 1 2

k

0

1

C k2 2

∫ [1 + cos(2kπx )]dx = 1 0

x =1

C ⎡ 1 ⎤ x+ sin (2kπx )⎥ = 1 ⎢ 2 ⎣ 2kπ ⎦ x =0 2 k

C k = 2 ⇒ y k ( x) = 2 cos(kπx)

(h)

(d) To demonstrate orthogonality first consider

( y 0 , y k ) = ∫ (1)[ 1

]

2 cos(kπx ) dx

0

=

2

[sin (kπx )]xx==10

kπ 2 [sin(kπ ) − sin(0)] = kπ 2 = ( 0 − 0) = 0 kπ

(i)

Now consider

(y , y ) = ∫ [ 1

i

j

][

2 cos(iπx )

]

2 cos( jπx ) dx

0

x =1

⎧ sin[(i − j )πx] sin[(i + j )πx] ⎫ + =⎨ ⎬ (i + j )π ⎭ x =0 ⎩ (i − j )π

(j)

Since i and j are integers and the sine of an integer multiple of π is zero, the inner product evaluation of Equation j is zero. Equations i and j demonstrate mode shape orthogonality.

218

dy d2y + 4x 2 + λy = 0 5.17 x 2 dx dx 3

Solution: (a) The differential equation is put in its self adjoint form by defining ∫

4 x2

dx

= e 4 ln( x ) = x 4 p ( x) = e x p( x) r ( x) = 3 = x x 3

(a) (b)

Thus the self-adjoint form of the eigenvalue problem is 1 d ⎛ 4 dy ⎞ ⎟ + λy = 0 ⎜x x dx ⎝ dx ⎠

(c)

The resulting eigenvectors are orthogonal with respect to the inner product 1

( f , g ) r = ∫ f ( x) g ( x) xdx

(d)

0

(b) The Bessel function solution of the differential equation is 3 1 3 1 − − ⎞ − − ⎞ ⎛ 2 ⎛ 2 y ( x) = C1 x 2 J 3 ⎜⎜ − λ x 2 ⎟⎟ + C 2 x 2 Y3 ⎜⎜ − λ x 2 ⎟⎟ ⎝ 3 ⎠ ⎝ 3 ⎠

219

(e)

Problems 5.18-5.23 refer to the following problem: The differential equation for determination of the natural frequencies and mode shapes for the longitudinal motion of a non-uniform bar is dy ⎤ d ⎡ α ( x ) ⎥ + ω 2α ( x ) y = 0 ⎢ dx ⎦ dx ⎣

(a)

5.18 Find the first five natural frequencies and normalized mode shapes for a uniform bar, α ( x) = 1 which is fixed at both ends, y (0) = 0 and y (1) = 0 .

Solution: The differential equation for the uniform bar is d2y +ω2y = 0 2 dx

(b)

y ( x) = C1 cos(ωx ) + C 2 sin (ωx )

(c)

The general solution of Equation b is

Application of the boundary conditions to Equation c leads to y (0) = 0 ⇒ C1 = 0

(d)

y (1) = 0 ⇒ C 2 sin (ω ) = 0

(e)

and

The general form of the natural frequencies is ω k = kπ with corresponding mode shapes y k ( x) = C k sin (kπx ) . The mode shapes are normalized by requiring

220

( yk , yk ) = 1 1

∫ [C

sin (kπx )] dx = 1 2

k

0

C k2 2

1

∫ [1 − cos(2kπx )]dx = 1 0

x =1

C ⎡ 1 ⎤ x− sin (2kπx )⎥ = 1 ⎢ 2 ⎣ 2kπ ⎦ x =1 2 k

Ck = 2

(f)

Thus the first five natural frequencies and normalized mode shapes are

ω1 = π

y1 ( x) = 2 sin (πx )

ω 2 = 2π

y 2 ( x) = 2 sin (2πx )

ω 3 = 3π

y 3 ( x) = 2 sin (3πx )

ω 4 = 4π

y 4 ( x) = 2 sin (4πx )

ω 5 = 5π

y 5 ( x) = 2 sin (5πx )

221

5.19 Find the first five natural frequencies and normalized mode shapes for a uniform bar, α ( x) = 1 which is fixed at x=0 and has a linear spring attached at x=1 such that the boundary conditions are y (0) = 0 and

dy (1) + 0.25 y (1) = 0 . dx

Solution: The differential equation for the uniform bar is d2y +ω2y = 0 2 dx

(b)

y ( x) = C1 cos(ωx ) + C 2 sin (ωx )

(c)

The general solution of Equation b is

Application of the boundary conditions to Equation c leads to y (0) = 0 ⇒ C1 = 0

(d)

and dy (1) + 0.25 y (1) = 0 ⇒ ωC 2 cos(ω ) + 0.25C 2 sin(ω ) = 0 ⇒ tan (ω ) = −4.0ω dx

(e)

The natural frequencies ω k k = 1,2,3,... are the solutions of Equation e. The mode shapes are of the form y k ( x) = C k sin (ω k x ) . The mode shapes are normalized by requiring

( yk , yk ) = 1 1

∫ [C

sin (ω k x )] dx = 1 2

k

0

1

C k2 2

∫ [1 − cos(2ω x )]dx = 1

C k2 2

⎡ ⎤ 1 sin (2ω k )⎥ = 1 ⎢1 − ⎣ 2ω k ⎦

k

0

Use of the double angle property and Equation e in Equation f leads to

222

(f)

C k2 2

⎡ ⎤ 1 sin (ω k ) cos(ω k )⎥ = 1 ⎢1 − ⎣ ωk ⎦

[

]

C k2 1 + 4 cos 2 (ω k ) = 1 2 2 Ck = 1 + 4 cos 2 (ω k ) The first five solutions of Equation e lead to

ω1 = 1.716 ω1 = 4.765 ω 3 = 7.886 ω 4 = 11.018 ω 5 = 14.155

y1 ( x) = 1.359 sin (1.716 x )

y 2 ( x) = 1.407 sin (4.765 x ) y 3 ( x) = 1.411sin (7.886 x )

y 4 ( x) = 1.413 sin (11.018 x )

y1 ( x) = 1.413 sin (14.155)

223

(g)

5.22 Find the first five natural frequencies and normalized mode shapes for a nonuniform bar with α ( x) = (1 − 0.5 x ) which is fixed at x=0 and free at x=1 such that 2

y (0) = 0 and

dy (1) = 0 . dx

Solution: The differential equation is dy ⎤ d ⎡ 2 (1 − 0.5 x) 2 ⎥ + ω 2 (1 − 0.5 x ) y = 0 ⎢ dx ⎦ dx ⎣

(a)

The mode shapes will be orthogonal with respect to the inner product 1

( f , g ) r = ∫ f ( x) g ( x)(1 − 0.5 x ) dx 2

(b)

0

Let z = (1 − 0.5 x ) . Rewriting Equation a using z as the independent variable leads to d ⎛ 2 dy ⎞ 2 2 ⎟ + 4ω z y = 0 ⎜z dz ⎝ dz ⎠

(c)

The solution of Equation c is −

1 2

y ( z ) = C1 z J

−

1 2

(2ωz ) + C 2 z

−

1 2

Y

−

1 2

(2ωz )

(d)

An alternate form of the solution using spherical Bessel functions is y ( z ) = C1 j 0 (2ωz ) + C 2 y 0 (2ωz )

(e)

Noting that when x=0, z=1 and when x=1, z=0.5, application of the boundary conditions to Equation e leads to y (1) = 0 ⇒ C1 j 0 (2ω ) + C 2 y 0 (2ω ) = 0 ⇒ C 2 = −

j 0 (2ω ) y 0 (2ω )

(f)

and dy (z = 0.5) = 0 ⇒ 2ωC1 j0′ (ω ) + 2ωC 2 y 0′ (ω ) = 0 dz

224

(g)

Noting that j 0′ ( z ) = − j1 ( z ) and y 0 (z ) = − y1 ( z ) and using Equation f in Equation g leads to j 0 (2ω ) y1 (ω ) − y 0 (2ω ) j1 (ω ) = 0

(h)

There are an infinite, but countable, number of solutions of Equation h. The mode shape for a natural frequency ω k is

⎧ ⎫ j (2ω ) yk (x) = Ck ⎨ jo (2ωk (1 − 0.5x)) − 0 k yo (2ωk (1 − 0.5x))⎬ y0 (2ωk ) ⎩ ⎭

(i)

The mode shapes are normalized by requiring

( y k , y k )r

⎧ ⎫ j (2ω k ) 2 = 1 = ∫ C ⎨ j o (2ω k (1 − 0.5 x) ) − 0 y o (2ω k (1 − 0.5 x) )⎬ (1 − 0.5 x ) dx y 0 (2ω k ) ⎩ ⎭ 0 2

1

2 k

⎧ ⎫ j (2ω k ) = 2 ∫ C ⎨ j o (2ω k z ) − 0 y o (2ω k z )⎬ z 2 dz y 0 (2ω k ) ⎩ ⎭ 0.5 2

`1

2 k

MATHCAD is used to generate the five lowest natural frequencies and their corresponding normalized mode shapes.

Solution to Problem 5.22

Eigenvalue relation

f ( ω) := js ( 0 , 2⋅ ω) ⋅ ys ( 1 , ω) − ys ( 0 , 2⋅ ω) ⋅ js ( 1 , ω) ω := 1.7, 1.71.. 15

225 ωn := root ( f ( ω) , ω) ωnωnωn= =2.029 =7.979 4.913 11.086 ω := 4321 12 4.5 6.5 1.55 1 32graph 4 Guess value of ω based upon

(j)

0.1

0.05

f(ω)

0

− 0.05

− 0.1

0

5

10

15

20

ω

ω := 14 ωn := root ( f ( ω) , ω) 5

ωn = 14.207 5

Normalization constants

i := 1 , 2 .. 5

2

C := i

⌠ ⎮ ⎮ ⎮ ⎮ ⌡

1

2 js ( 0 , 2⋅ ωn ) ⎞ 2 ⎛⎜ i ⎟ js 0 , 2⋅ ωn ⋅ z) − ⋅ ys ( 0 , 2⋅ ωn ⋅ z) ⋅ z dz i i ⎟ ⎜ ( ys ( 0 , 2⋅ ωn ) i ⎝ ⎠

0.5

⎛ 6.393 ⎞ ⎜ ⎟ 25.088 ⎜ ⎟ C = ⎜ 43.404⎟ ⎜ 61.449⎟ ⎜ ⎟ ⎝ 79.382⎠

(( ((

226

)) 1)2)

jsjs0 ,02,⋅2ωn ⋅ ωn Mode shape ⎡ ⎡plots ⎤⎤ 12 y1y2 ( x()x:= ) :=CC⋅ ⎢⋅js⎢js 0 , 0 2 , ⋅ 2 ωn ⋅ ωn ⋅ ( ⋅ 1 ( 1 − − 0.5 0.5 ⋅ x ⋅ ) x ) − − ⋅ ys ⋅ ys 0⎡,02,⋅2ωn ⋅ ωn⋅ (⋅1( 1− −0.5 0.5 ⋅ x⋅)⎤x⎥)⎤⎥ ⎡ ⎡ ⎤ ⎤ ⎡ 1 2⎢ ⎢⎣ ⎣ 12 12 ⎦ ⎦ ysys0 ,02,⋅2ωn ⎣⎣ ⎦⎥⎦⎥ ⋅ ωn

⎣⎣

⎦⎦

⎡ 3⎢ ⎣ ⎣

y3( x) := C ⋅ ⎢js ⎡0 , 2⋅ ωn ⋅ ( 1 − 0.5⋅ x)⎤ −

⎦

3

⎡ 1⎢ ⎣ ⎣

y4( x) := C ⋅ ⎢js ⎡0 , 2⋅ ωn ⋅ ( 1 − 0.5⋅ x)⎤ −

⎡ 5⎢ ⎣ ⎣

⎦

4

y5( x) := C ⋅ ⎢js ⎡0 , 2⋅ ωn ⋅ ( 1 − 0.5⋅ x)⎤ −

⎦

5

⎤ ( 3) ⋅ ys ⎡0 , 2⋅ ωn ⋅ ( 1 − 0.5⋅ x)⎤⎥ 3 ⎣ ⎦⎥ ys ( 0 , 2⋅ ωn ) 3 ⎦ js 0 , 2⋅ ωn

⎤ ( 4) ⋅ ys ⎡0 , 2⋅ ωn ⋅ ( 1 − 0.5⋅ x)⎤⎥ 4 ⎣ ⎦⎥ ys ( 0 , 2⋅ ωn ) 4 ⎦ js 0 , 2⋅ ωn

⎤ ( 5) ⋅ ys ⎡0 , 2⋅ ωn ⋅ ( 1 − 0.5⋅ x)⎤⎥ 5 ⎣ ⎦⎥ ys ( 0 , 2⋅ ωn ) 5 ⎦ js 0 , 2⋅ ωn

x := 0 , .01.. 1

10

y1 ( x)

5

y2 ( x) y3 ( x) 0 y4 ( x) y5 ( x) −5

− 10

0

0.2

0.4

0.6 x

227

0.8

1

5.23 Find the first five natural frequencies and normalized mode shapes for a nonuniform bar with α ( x) = (1 − 0.5 x ) which is fixed at x=0 and has a discrete particle and a 2

linear

spring

y (0) = 0 and α (1)

attached

at

x=1

such

that

the

boundary

conditions

are

dy (1) + 0.25 y (1) = 0.5ω 2 y (1) . dx

Solution: The differential equation is dy ⎤ d ⎡ 2 (1 − 0.5 x) 2 ⎥ + ω 2 (1 − 0.5 x ) y = 0 ⎢ dx ⎦ dx ⎣

(a)

The mode shapes will be orthogonal with respect to the inner product 1

( f , g ) r = ∫ f ( x) g ( x)(1 − 0.5 x ) dx + 0.5 f (1) g (1) 2

(b)

0

Let z = (1 − 0.5 x ) . Rewriting Equation a using z as the independent variable leads to d ⎛ 2 dy ⎞ 2 2 ⎜z ⎟ + 4ω z y = 0 dz ⎝ dz ⎠

(c)

The solution of Equation c is −

1

y ( z ) = C1 z 2 J

1 (2ωz ) + C 2 z −

−

1 2

Y

2

−

1 2

(2ωz )

(d)

An alternate form of the solution using spherical Bessel functions is y ( z ) = C1 j 0 (2ωz ) + C 2 y 0 (2ωz )

(e)

Noting that when x=0, z=1 and when x=1, z=0.5, application of the boundary conditions to Equation e leads to y (1) = 0 ⇒ C1 j 0 (2ω ) + C 2 y 0 (2ω ) = 0 ⇒ C 2 = −

and

228

j 0 (2ω ) y 0 (2ω )

(f)

dy dy (1) + 0.25 y (1) = 0.5ω 2 y (1) ⇒ 0.5( −0.5) ( z = 0.5) + 0.25 y ( z = 0.5) = 0.5ω 2 y ( z = 0.5) dz dx 2 − 0.25[2ωj 0′ (ω )C1 + 2ωyj0 (ω )C 2 ] + 0.25 − 0.5ω [C1 j 0 (ω ) + C 2 y 0 (ω )] = 0 (g)

α (1)

(

)

Noting that j 0′ ( z ) = − j1 ( z ) and y 0 (z ) = − y1 ( z ) and using Equation f in Equation g leads to 2ω [ j1 (ω ) y 0 (2ω ) − y1 (ω ) j 0 (2ω )] +

(1 − 2ω )[ j (ω ) y (2ω ) − y (ω ) j (2ω )] = 0 2

0

0

0

(h)

0

There are an infinite, but countable, number of solutions of Equation h. The mode shape for a natural frequency ω k is

⎧ ⎫ j (2ω ) yk (x) = Ck ⎨ jo (2ωk (1 − 0.5x)) − 0 k yo (2ωk (1 − 0.5x))⎬ y0 (2ωk ) ⎩ ⎭

(i)

The mode shapes are normalized by requiring

( y k , y k )r

⎫ ⎧ j (2ω k ) 2 y o (2ω k (1 − 0.5 x) )⎬ (1 − 0.5 x ) dx = 1 = ∫ C ⎨ j o (2ω k (1 − 0.5 x) ) − 0 y 0 (2ω k ) ⎭ ⎩ 0 2

1

2 k

⎫ ⎧ j (2ω k ) y o (ω k ) )⎬ + 0.5C ⎨ j o (ω k ) − 0 y 0 (2ω k ) ⎭ ⎩

2

2 k

⎫ ⎧ j (2ω k ) y o (2ω k z )⎬ z 2 dz = 2 ∫ C ⎨ j o (2ω k z ) − 0 y 0 (2ω k ) ⎭ ⎩ 0.5 2

`1

2 k

⎫ ⎧ j (2ω k ) y o (ω k ) )⎬ + 0.5C ⎨ j o (ω k ) − 0 y 0 (2ω k ) ⎭ ⎩ 2 k

2

(j)

MATHCAD is used to generate the five lowest natural frequencies and their corresponding normalized mode shapes. Solution to Problem 5.23

Eigenvalue relation

g ( ω) := 2( ω) ⋅ ( js ( 1 , ω) ⋅ ys ( 0 , 2⋅ ω) − ys ( 1 , ω) ⋅ js ( 0 , 2⋅ ω) )

(

)

2

f ( ω) := g ( ω) + 1 − 2⋅ ω ⋅ ( js ( 0 , ω) ⋅ ys ( 0 , 2⋅ ω) − 229 ys ( 0 , ω) ⋅ js ( 0 , 2⋅ ω) ) ω := .7, .71.. 15

3

2

f(ω) 1

0

−1

0

5

10

15

ω

Guess value of ω based upon graph

ω := 1.55 ωn := root ( f ( ω) , ω) 1

ωn = 1.345 1

ω := 4.5 ωn := root ( f ( ω) , ω) 2

ωn = 3.461 2

ω := 6.5 ωn := root ( f ( ω) , ω) 3

ωn = 6.443 3

ω := 10 ωn := root ( f ( ω) , ω) 4

ωn = 9.531 4

ω := 12 ωn := root ( f ( ω) , ω) 5

ωn = 12.646 5

Normalization constants

i := 1 , 2 .. 5 2 js ( 0 , 2⋅ ωn ) ⎛⎜ ⎞⎟ i a := 0.5⋅ js ( 0 , ωn ) − ⋅ ys ( 0 , ωn ) i i i ⎟ ⎜ ys ( 0 , 2⋅ ωn ) i ⎝ ⎠

230

20

2

C := i

⌠ ⎮ ⎮ ⎮ ⎮ ⌡

1

2 js ( 0 , 2⋅ ωn ) ⎛⎜ ⎞⎟ 2 i js 0 , 2⋅ ωn ⋅ z) − ⋅ ys ( 0 , 2⋅ ωn ⋅ z) ⋅ z dz + a i i ⎟ i ⎜ ( ys ( 0 , 2⋅ ωn ) i ⎝ ⎠

0.5

⎛ 2.357 ⎞ ⎜ 12.056⎟ ⎜ ⎟ C = ⎜ 31.885⎟ ⎜ 50.737⎟ ⎜ ⎟ ⎝ 69.114⎠

Mode shape plots

⎡ 1⎢ ⎣ ⎣

y1( x) := C ⋅ ⎢js ⎡0 , 2⋅ ωn ⋅ ( 1 − 0.5⋅ x)⎤ −

⎦

1

⎡ 2⎢ ⎣ ⎣

⎤ ( 1) ⋅ ys ⎡0 , 2⋅ ωn ⋅ ( 1 − 0.5⋅ x)⎤⎥ 1 ⎣ ⎦⎥ ys ( 0 , 2⋅ ωn ) 1 ⎦ js 0 , 2⋅ ωn

y2( x) := C ⋅ ⎢js ⎡0 , 2⋅ ωn ⋅ ( 1 − 0.5⋅ x)⎤ −

⎦

2

⎡ 3⎢ ⎣ ⎣

y3( x) := C ⋅ ⎢js ⎡0 , 2⋅ ωn ⋅ ( 1 − 0.5⋅ x)⎤ −

⎦

3

⎡ 1⎢ ⎣ ⎣

y4( x) := C ⋅ ⎢js ⎡0 , 2⋅ ωn ⋅ ( 1 − 0.5⋅ x)⎤ −

⎡ 5⎢ ⎣ ⎣

⎦

4

y5( x) := C ⋅ ⎢js ⎡0 , 2⋅ ωn ⋅ ( 1 − 0.5⋅ x)⎤ − 5

⎦

⎤ ( 2) ⋅ ys ⎡0 , 2⋅ ωn ⋅ ( 1 − 0.5⋅ x)⎤⎥ 2 ⎣ ⎦⎥ ys ( 0 , 2⋅ ωn ) 2 ⎦ js 0 , 2⋅ ωn

⎤ ( 3) ⋅ ys ⎡0 , 2⋅ ωn ⋅ ( 1 − 0.5⋅ x)⎤⎥ 3 ⎣ ⎦⎥ ys ( 0 , 2⋅ ωn ) 3 ⎦ js 0 , 2⋅ ωn

⎤ ( 4) ⋅ ys ⎡0 , 2⋅ ωn ⋅ ( 1 − 0.5⋅ x)⎤⎥ 4 ⎣ ⎦⎥ ys ( 0 , 2⋅ ωn ) 4 ⎦ js 0 , 2⋅ ωn

⎤ ( 5) ⋅ ys ⎡0 , 2⋅ ωn ⋅ ( 1 − 0.5⋅ x)⎤⎥ 5 ⎣ ⎦⎥ ys ( 0 , 2⋅ ωn ) 5 ⎦ js 0 , 2⋅ ωn

231

x := 0 , .01.. 1

6

4 y1 ( x) y2 ( x)

2

y3 ( x) 0 y4 ( x) y5 ( x)− 2 −4 −6

0

0.2

0.4

0.6 x

232

0.8

1

Problems 5.24-5.29 refer to the following general problem. The natural frequencies ω and mode shapes w(x) of a uniform, stretched beam on an elastic foundation are governed by the differential equation d 4w d 2w −ε + ηw − ω 2 w = 0 4 2 dx dx

(a)

5.24 Determine the first five natural frequencies and mode shapes for a beam with

ε = 0 and η = 1 which is pinned at x=0 and x=1 such that d2y d2y y (0) = 0, 2 (0) = 0, y (1) = 0, 2 (1) = 0 . dx dx

Solution: The differential equation is d 4w + w − ω 2w = 0 4 dx

(b)

A solution of Equation b is assumed to be of the form w( x) = e αx which when substituted into Equation b leads to α 4 + 1 − ω 2 = 0 whose solutions are

α = ± (ω 2 − 1)4 , ± i (ω 2 − 1)4 1

1

(c)

The general solution of Equation b is w( x) = C1 cosh (μx ) + C 2 sinh (μx ) + C 3 cos(μx ) + C 4 sin (μx )

(d)

where μ = (ω 2 − 1)4 . Application of the boundary conditions to Equation d leads to 1

y (0) = 0 → C1 + C 3 = 0

(e)

d2y (0) = 0 ⇒ μ 2 C1 − μ 2 C 3 = 0 2 dx y (1) = 0 ⇒ C1 cosh μC + C 2 sinh μ + C 3 cos μ + C 4 sin μ = 0 d2y (1) = 0 ⇒ μ 2 C1 cosh μ + μ 2 C 2 sinh μ − C 3 μ 2 cos μ − C 4 μ 2 sin μ = 0 2 dx

233

(f) (g) (h)

The only solutions which satisfy Equation e and Equation f for μ ≠ 0 require that C1 = C 3 = 0 . Equations g and h arte then summarized in matrix form by ⎡sinh μ ⎢sinh μ ⎣

sin μ ⎤ ⎡C 2 ⎤ ⎡0⎤ = − sin μ ⎥⎦ ⎢⎣C 4 ⎥⎦ ⎢⎣0⎥⎦

(i)

A non-trivial solution exists only if the determinant of the matrix in Equation f is zero, sinh μ sinh μ

sin μ = 0 ⇒ −2 sin μ sinh μ = 0 ⇒ sin μ = 0 ⇒ μ k = kπ − sin μ

(j)

Substitution into Equation i leads to C 4 = 0 . Hence the natural frequencies are

ω k = k 4π 4 + 1

(k)

wk ( x) = C k sin (kπx )

(l)

Their corresponding mode shapes are

The first five natural frequencies and mode shapes are

ω1 = 9.9201 ω 2 = 39.4911 ω 3 = 88.8321 ω 4 = 157.9168 ω5 = 246.7421

w1 ( x) = C1 sin(πx) w2 ( x) = C 2 sin(2πx) w3 ( x) = C3 sin(3πx) w4 ( x) = C 4 sin(4πx) w5 ( x) = C 5 sin(5πx)

234

5.25 Determine the first five natural frequencies and mode shapes for a beam with

ε = 2 and η = 0 which is pinned at x=0 and x=1 such that y (0) = 0,

d2y d2y ( 0 ) = 0 , ( 1 ) = 0 , (1) = 0 . y dx 2 dx 2

Solution: The differential equation is

d 4w d 2w − 2 2 − ω 2w = 0 4 dx dx

(b)

A solution of Equation b is assumed to be of the form w( x) = eαx which when substituted into Equation b leads to α 4 − 2α 2 − ω 2 = 0 whose solutions are

α = ± 1± 1+ ω2

(c)

Define μ = 1 + 1 + ω 2 and ν = − 1 + 1 + ω 2 . The general solution of Equation b is

w( x) = C1 cosh (μx ) + C 2 sinh (μx ) + C 3 cos(νx ) + C 4 sin (νx )

(d)

Application of the boundary conditions to Equation d leads to y (0) = 0 → C1 + C 3 = 0

(e)

2

d y (0) = 0 ⇒ μ 2 C1 − ν 2 C 3 = 0 2 dx y (1) = 0 ⇒ C1 cosh μC + C 2 sinh μ + C 3 cosν + C 4 sinν = 0

(f) (g)

2

d y (1) = 0 ⇒ μ 2 C1 cosh μ + μ 2 C 2 sinh μ − C 3ν 2 cosν − C 4ν 2 sinν = 0 2 dx

(h)

The only solutions which satisfy Equation e and Equation f for ω ≠ 0 require that C1 = C 3 = 0 . Equations g and h arte then summarized in matrix form by ⎡ sinh μ ⎢ μ 2 sinh μ ⎣

sinν

⎤ ⎡C 2 ⎤ ⎡0⎤ = − ν sinν ⎥⎦ ⎢⎣C 4 ⎥⎦ ⎢⎣0⎥⎦ 2

A non-trivial solution exists only if the determinant of the matrix in Equation f is zero,

235

(i)

sinh μ μ sinh μ

sinν = 0 ⇒ − μ 2 + ν 2 sin μ sinhν = 0 ⇒ sinν = 0 ⇒ ν k = kπ − ν sinν

(

)

(j)

Substitution into Equation i leads to C 4 = 0 . Hence the natural frequencies are

ωk =

(k π 2

2

)

2

+1 −1

(k)

Their corresponding mode shapes are

wk ( x) = C k sin (kπx ) The first five natural frequencies and mode shapes are

ω1 = 10.8235 ω 2 = 40.4661 ω3 = 88.8209 ω 4 = 158.9105 ω5 = 247.7381

w1 ( x) = C1 sin(πx) w2 ( x) = C 2 sin(2πx) w3 ( x) = C3 sin(3πx) w4 ( x) = C 4 sin(4πx) w5 ( x) = C 5 sin(5πx)

236

(l)

5.26 Determine the first five natural frequencies and mode shapes for a beam with

ε = 2 and η = 1 which is pinned at x=0 and fixed at x=1 such that y (0) = 0,

d2y dy (0) = 0, y (1) = 0, (1) = 0 . 2 dx dx

Solution: The differential equation is

d 4w d 2w − 2 2 + w − ω 2w = 0 4 dx dx

(b)

A solution of Equation b is assumed to be of the form w( x) = eαx which when substituted into Equation b leads to α 4 − 2α 2 + 1 − ω 2 = 0 whose solutions are

α = ± 1± ω2

(c)

Define μ = 1 + ω 2 andν = ω 2 − 1 . The general solution of Equation b is

w( x) = C1 cosh (μx ) + C 2 sinh (μx ) + C 3 cos(νx ) + C 4 sin (νx )

(d)

Application of the boundary conditions leads to

y (0) = 0 → C1 + C 3 = 0

(e)

2

d y (0) = 0 ⇒ μ 2 C1 − ν 2 C 3 = 0 2 dx y (1) = 0 ⇒ C1 cosh μC + C 2 sinh μ + C 3 cosν + C 4 sinν = 0

(f) (g)

dy (1) = 0 ⇒ μC1 sinh μ + μC 2 cosh μ − C 3ν sinν + C 4ν cosν = 0 dx

(h)

The only solutions which satisfy Equation e and Equation f for ω ≠ 0 require that C1 = C 3 = 0 . Equations g and h arte then summarized in matrix form by sinν ⎤ ⎡C 2 ⎤ ⎡0⎤ ⎡ sinh μ ⎢ μ cosh μ ν cosν ⎥ ⎢C ⎥ = ⎢0⎥ ⎣ ⎦⎣ 4 ⎦ ⎣ ⎦

A non-trivial solution exists only if the determinant of the matrix in Equation f is zero,

237

(i)

sinh μ sinν = 0 ⇒ ν cosν sinh μ − μ sinν cosh μ μ cosh μ ν cosν

(j)

Equation j leads to the transcendental equation

μ tanν = ν tanh μ

(k)

There are an infinite but countable number of solutions of Equation k leading to the natural frequencies ω k . Equation i leads to C 4 = −

sinh μ C 2 such that the mode shapes sinν

are ⎫ ⎧ sinh (1 + ω k2 ) 2 wk ( x) = C k ⎨sinh (1 + ω k )x − sin (ω k2 − 1)x ⎬ 2 sin (ω k − 1) ⎭ ⎩

[

]

[

]

The first five natural frequencies determined using MATHCAD are

ω1 = 4.002, ω 2 = 7.129, ω 3 = 10.254, ω 4 = 13.386 and ω 5 = 16.552

238

(l)

5.27 Determine the first five natural frequencies and mode shapes for a beam with

ε = 0 and η = 0 which is fixed at x=0 and has a linear spring attached at x=1 such that y (0) = 0,

dy d2y d3y (0) = 0, 2 (1) = 0, 3 (1) − 0.5 y (1) = 0 . dx dx dx

The governing differential equation is

d 4w − ω 2w = 0 4 dx

(b)

The general solution of Equation b is

w( x) = C1 cosh( μx) + C 2 sinh (μx ) + C 3 cos( μx) + C 4 sin (μx )

(c)

where μ = ω . Application of the boundary conditions to Equation c leads to y (0) = 0 ⇒ C1 + C 3 = 0

(d)

dy (0) = 0 ⇒ μC 2 + μC 4 = 0 dx d2y (1) = 0 ⇒ μ 2 C1 cosh μ + μ 2 C 2 sinh μ − μ 2 C 3 cos μ − μ 2 C 4 sin μ dx 2 d3y (1) − 0.5 y (1) = 0 ⇒ C1 μ 3 sinh μ − 0.5 cosh μ + C 2 μ 3 cosh μ − 0.5 sinh μ 3 dx + C 3 μ 3 sin μ − 0.5 cos μ + C 4 − μ 3 cos μ − 0.5 sin μ = 0

(

)

(

)

(

(

(e) (f)

)

)

(g)

Equations d and e imply that C 3 = −C1 and C 4 = −C 2 . Substituting into Equations f and g and rewriting in matrix form leads to cosh μ + cos μ sinh μ + sin μ ⎡ ⎤ ⎡ C1 ⎤ ⎡0⎤ ⎢ μ 2 (sinh μ − sin μ ) − 0.5(cosh μ − cos μ ) μ 2 (cosh μ + cos μ ) − 0.5(sinh μ − sin μ )⎥ ⎢C ⎥ = ⎢0⎥ (h) ⎣ ⎦⎣ 2 ⎦ ⎣ ⎦

A non-trivial solution of Equation h requires

239

cosh μ + cos μ sinh μ + sin μ =0 2 μ (sinh μ − sin μ ) − 0.5(cosh μ − cos μ ) μ (cosh μ + cos μ ) − 0.5(sinh μ − cos μ ) 2

μ 2 (cosh 2 μ + 2 cosh μ cos μ + cos 2 μ − sinh 2 μ + sin 2 μ ) − 0.5(cosh μ sinh μ − cosh μ sin μ + sinh μ cos μ − cos μ sin μ + cosh μ sinh μ − sinh μ cos μ + cosh μ sin μ − cos μ sin μ ) = 0 2 μ 2 (1 + cosh μ cos μ ) − (cosh μ sinh μ − cos μ sin μ ) = 0

(i)

Equation i leads to an infinite, but countable, number of natural frequencies. For any

μ k Equation h leads to ⎫ ⎧ cos μ k cosh μ k (sinh (μ k x ) − sin (μ k x ))⎬ wk ( x) = C k ⎨cosh (μ k x ) − cos(μ k x ) − sinh μ k + sin μ k ⎭ ⎩

The five lowest natural frequencies and their corresponding mode shapes are determined using MATHCAD.

240

(j)

5.28 Determine the first five natural frequencies and mode shapes for a beam with

ε = 0.5 and η = 1 which is fixed at x=0 and free at x=1 such that y (0) = 0,

dy d2y d3y dy (0) = 0, 2 (1) = 0, 3 (1) + 0.5 (1) = 0 . dx dx dx dx

Solution: The differential equation is

d 4w d 2w − 0.5 2 + w − ω 2 w = 0 4 dx dx

(b)

A solution of Equation b is assumed to be of the form w( x) = eαx which when substituted into Equation b leads to α 4 − 0.5α 2 + 1 − ω 2 = 0 whose solutions are

α =±

Define μ =

1 15 ± ω2 − 4 64

(c)

1 15 1 15 + ω2 − and ν = − + ω 2 − . The general solution of Equation 4 64 4 64

b is

w( x) = C1 cosh (μx ) + C 2 sinh (μx ) + C 3 cos(νx ) + C 4 sin (νx )

(d)

Application of the boundary conditions to Equation d leads to y (0) = 0 → C1 + C 3 = 0

(e)

d2y (0) = 0 ⇒ μ 2 C1 − ν 2 C 3 = 0 (f) dx 2 d2y (1) = 0 ⇒ μ 2 C1 cosh μ + μ 2 C 2 sinh μ − C 3ν 2 cosν − C 4ν 2 sinν = 0 (g) 2 dx d3y dy (1) + 0.5 (1) = 0 ⇒ C1 μ 3 sinh μ + 0.5μ sinh μ + C 2 μ 3 cosh μ + 0.5μ cosh μ 3 dx dx + C 3 ν 3 sinν − 0.5ν sinν + C 4 − ν 3 cosν + 0.5ν cosν = 0 (h)

(

(

)

)

(

(

)

The only solutions which satisfy Equation e and Equation f require that C1 = C 3 = 0 . Equations g and h arte then summarized in matrix form by

241

)

⎡ μ 2 sinh μ ⎢ 3 ⎣ μ + 0.5μ cosh μ

(

)

⎤ ⎡C 2 ⎤ ⎡0⎤ ⎥⎢ ⎥ = ⎢ ⎥ − ν − 0.5ν cosν ⎦ ⎣C 4 ⎦ ⎣0⎦

(

− ν 2 sinν

)

3

(i)

A non-trivial solution exists only if the determinant of the matrix in Equation I is zero,

− ν 2 sinν μ 2 sinh μ =0 (μ 3 + 0.5μ )cosh μ − (ν 3 − 0.5ν )cosν

(

)

(

)

− μ 2 ν 3 − 05ν sinh μ cosν + ν 2 μ 3 + 0.5μ cosh μ sinν = 0

(j)

Equation j is rearranged to

ν ν − 0.5 2

tanν =

μ μ + 0.5 2

tanh μ

(k)

Equation k is a transcendental equation with an infinite, but countable, number of solutions for the natural frequencies ω k . For a specific natural frequency, Equation j is used to determine C4 =

μ 2 sinh μ ν 2 sinν

(l)

The mode shapes are given by

⎡ 15 1 ω k2 − + ⎢ ⎞ ⎛ 1 15 64 4 ⎛⎜ 2 ⎟ ⎜ wk ( x) = C k ⎢sinh x + sin + ωk − ⎜ ⎜ 4 64 ⎟ ⎢ 15 1 2 ⎝ ⎠ ⎝ ωk − − ⎢ 64 3 ⎣

⎤ ⎞ 15 1 ⎟⎥⎥ ω k2 − − x 64 4 ⎟⎥ ⎠⎥ ⎦

MATHCAD is used to determine the natural frequencies and mode shapes

242

(m)

5.29 Determine the first five natural frequencies and mode shapes for a beam with

ε = 0 and η = 1 which is fixed at x=0 and has an attached mass at x=1 such that. y (0) = 0,

dy d2y d3y (0) = 0, 2 (1) = 0, 3 (1) = 0.5ω 2 y (1) = 0 . dx dx dx

Solution: The differential equation is d 4w + w − ω 2w = 0 4 dx

(b)

A solution of Equation b is assumed to be of the form w( x) = e αx which when substituted into Equation b leads to α 4 + 1 − ω 2 = 0 whose solutions are

α = ±(ω − 1) 2

1 4

, ± i (ω − 1) 2

1 4

(c)

The general solution of Equation b is w( x) = C1 cosh (μx ) + C 2 sinh (μx ) + C 3 cos(μx ) + C 4 sin (μx )

(d)

where μ = (ω 2 − 1)4 . Application of the boundary conditions to Equation d leads to 1

y (0) = 0 → C1 + C 3 = 0

(e)

2

d y (0) = 0 ⇒ μC 2 + μC 4 = 0 (f) dx d2y (1) = 0 ⇒ μ 2 C1 cosh μ + μ 2 C 2 sinh μ − C 3 μ 2 cos μ − C 4 μ 2 sin μ = 0 (g) dx 2 d3y (1) − 0.5ω 2 y (1) = 0 ⇒ C1 μ 3 sinh μ − 0.5ω 2 cosh μ + C 2 μ 3 cosh μ − 0.5ω 2 sinh μ 3 dx + C 3 μ 3 sin μ − 0.5ω 2 cos μ + C 4 − μ 3 cos μ − 0.5ω 2 sin μ = 0 (h)

(

(

)

)

(

(

)

)

Equations d and e imply that C 3 = −C1 and C 4 = −C 2 . Substituting into Equations f and g and rewriting in matrix form leads to

243

cosh μ + cos μ sinh μ + sin μ ⎡ ⎤ ⎡ C1 ⎤ ⎢ μ 2 (sinh μ − sin μ ) − 0.5ω 2 (cosh μ − cos μ ) μ 2 (cosh μ + cos μ ) − 0.5ω 2 (sinh μ − sin μ )⎥ ⎢C ⎥ ⎣ ⎦⎣ 2 ⎦ ⎡0 ⎤ =⎢ ⎥ (i) ⎣0 ⎦ A non-trivial solution of Equation h requires cosh μ + cos μ sinh μ + sin μ =0 2 2 μ (sinh μ − sin μ ) − 0.5ω (cosh μ − cos μ ) μ (cosh μ + cos μ ) − 0.5ω 2 (sinh μ − cos μ ) 2

μ 2 (cosh 2 μ + 2 cosh μ cos μ + cos 2 μ − sinh 2 μ + sin 2 μ ) − 0.5ω 2 (cosh μ sinh μ − cosh μ sin μ + sinh μ cos μ − cos μ sin μ + cosh μ sinh μ − sinh μ cos μ + cosh μ sin μ − cos μ sin μ ) = 0 2μ 2 (1 + cosh μ cos μ ) − ω 2 (cosh μ sinh μ − cos μ sin μ )

(j)

Equation j leads to an infinite, but countable, number of natural frequencies. For any

μ k Equation i leads to ⎫ ⎧ cos μ k cosh μ k (sinh (μ k x ) − sin (μ k x ))⎬ wk ( x) = C k ⎨cosh (μ k x ) − cos(μ k x ) − sinh μ k + sin μ k ⎭ ⎩

The five lowest natural frequencies and their corresponding mode shapes are determined using MATHCAD

244

(k)

5.30 One end of the non-uniform bar of Fig. P5.30 is fixed while its other end is attached to a spring whose other end is attached to a particle. The differential equations governing the displacement of the bar u(x,t) and the displacement of the particle y(t) are

∂ ⎛ ∂u ⎞ ∂ 2u − =0 ρ A EA ⎜ ⎟ ∂x ⎝ ∂x ⎠ ∂t 2 u (0, t ) = 0 ∂u EA( L) ( L, t ) − k [ y (t ) − u ( L, t )] = 0 ∂x 2 d y m 2 + k [ y (t ) − u ( L, t )] = 0 dt

(a) (b) (c) (d)

(a) Non-dimensionalize the problem through introduction of appropriate nondimensional variables. (b) Introduce the normal mode solution

u ( x , t ) = U ( x ) e i ωt y (t ) = Ye

i ωt

(e) (f)

into the non-dimensional equation and determine an eigenvalue-eigenvector problem whose solution leads to the natural frequencies and mode shapes.

⎡ f ( x)⎤ (c) Define the vector space V as the set of vectors of the form g = ⎢ ⎥ where f(x) ⎣ a ⎦ is in the subspace of C 2 [0,1] of all vectors f(x) such that f(0)=0 and du (1) + ηu (1) = 0 where η is an appropriate non-dimensional parameter. Define an dx operator L whose domain is V by ⎡ 1 d ⎛ df ⎞⎤ ⎡ f ( x)⎤ ⎢− ⎜α ⎟⎥ = L⎢ α x dx dx ( ) ⎥ ⎝ ⎠⎥ ⎣ a ⎦ ⎢ ( ) μα f ν 1 − ⎦ ⎣

245

(g)

where μ and ν are appropriate non-dimensional constants. Show that the eigenvalue problem is of the form

Lg = ω 2 g

(h)

(d) Determine an appropriate inner product for which L is self adjoint. (e) Show that L is positive definite with respect to this inner product (f) Determine the first five natural frequencies and mode shapes for a uniform bar assuming all non dimensional constants are one. (g) Repeat part (f) if α ( x) = (1 − 0.1x) 3

Solution: The following non-dimensional variables are introduced x* =

x L

u* =

u L

y* =

y L

t* =

t T

(i)

and the following non-dimensional functions defined

α ( x) =

EA( x) E 0 A0

β (x) =

ρA(x) ρA 0

(j)

Use of Equations i and j in Equations a-d leads to

( )

* L ∂ 2 ( Lu * ) 1 ∂ ⎡ * ∂ Lu ⎤ * =0 ⎢ E 0 A)α ( x ) ⎥ + ρ 0 A0 β ( x ) 2 ∂x * ⎦ ∂t *2 L2 ∂x * ⎣ T Lu 0 ,t * = 0

( )

E 0 A0α (1)

( ) [

]

1 ∂ Lu * − k Ly * − Lu * (1, t * ) = 0 L ∂x *

( )

[

(k) (l) (m)

]

m d 2 Ly * (1, t ) + k Ly * − Lu * (1, t * ) = 0 2 2 T dt

Choosing T = L

ρ0 E0

, rearranging and dropping *’s from non-dimensional variables,

Equations k-n are written in non-dimensional form as

246

(n)

∂ ⎡ ∂u ⎤ ∂ 2u α ( x) ⎥ − β ( x) 2 = 0 ∂x ⎢⎣ ∂x ⎦ ∂t u (0, t ) = 0

α (1)

(p)

∂u (1) − η [ y − u (1, t )] = 0 ∂x

d2y + φ [ y − u (1, t )] = 0 dt 2 where η =

(o)

(q) (r)

kL kT 2 and φ = . E 0 A0 m

(b) Use of the normal-mode solution in Equations o-r leads to the following problems for U(x) and Y d ⎡ dU ⎤ + ω 2 β ( x)U = 0 α ( x) ⎢ ⎥ dx ⎣ dx ⎦ U (0) = 0 dU (1) − η [Y − U (1)] dx - ω 2Y + φ [Y − U (1)] = 0

α (1)

(s) (t) (u) (v)

(c) The problem is formulated using an eigenvalue problem as 1 d d ⎤U ⎡ 0 ⎤ ⎡U ⎤ ⎢− β ( x) dx [α ( x)] dx 0 ⎥ ⎡ ⎤ − ⎡ = ω2⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎥ ⎢ ⎣Y ⎦ 0 φ ⎦ ⎣ Y ⎦ ⎣φU (1)⎦ ⎣

(w)

⎡a ( x)⎤ ⎡b( x)⎤ = (d) Define two vectors in V by a = ⎢ and b ⎢ s ⎥ Consider the kinetic energy ⎥ ⎣ r ⎦ ⎣ ⎦ inner product defined by 1

(a, b ) = ∫ a( x)b( x)α ( x)dx + φrs

(x)

0

It can be shown using integration by parts that for all a and b in the vector space V in which a(x) and r must satisfy the boundary conditions determined from Equations t and u

247

that the operator defined in Equation w is self adjoint with respect to the inner product of Equation x. (e) Integration by parts can be used to show that the operator is positive definite. (f) Assuming a uniform bar with all non-dimensional constants equal to one leads to the problem

d 2U + ω 2U = 0 2 dx U (0) = 0 dU (1) − [Y − U (1) ] dx - ω 2 Y + [Y − U (1) ] = 0

(y) (z) (aa) (bb)

The general solution of Equation y is U ( x) = C1 cos(ωt ) + C 2 sin (ωt )

(cc)

Application of Equation z to Equation cc leads to C1 = 0 . Then application of Equations aa and bb leads to C 2ω cos ω − Y + C 2 sin ω = 0

(dd)

− ω Y + Y − C 2 sin ω = 0

(ee)

2

Equations dd and ee are combined leading to Y=

sin ω C2 1− ω2

sin ω ⎤ ⎡ C 2 ⎢ω cos ω + sin ω + =0 1 − ω 2 ⎥⎦ ⎣

(ff) (gg)

The natural frequencies are the solutions of tan ω = The mode shapes are of the form

248

ω 2 −ω2

(hh)

⎡sin(ω k x)⎤ ⎡U k ( x)⎤ ⎢ sin ω k ⎥ C = k ⎢ Y ⎥ ⎢ 2 ⎥ k ⎣ ⎦ ⎣ 1 − ωk ⎦

(ii)

MATHCAD is used to determine the first five natural frequencies and mode shapes as

(g) The differential equations for this case are

(

)

d ⎡ dU ⎤ 1 − 0.1x 3 + ω 2 (1 − 0.1x) 3 U = 0 ⎢ ⎥ dx ⎣ dx ⎦ U ( 0) = 0 dU (1) − [Y − U (1)] dx - ω 2Y + [Y − U (1)] = 0

(0.9) 3

(jj) (kk) (ll) (mm)

Define z=1-.1x. Equation jj is rewritten using z as the dependent variable as d ⎛ 3 dU ⎞ 2 3 ⎜z ⎟ + 100ω z U = 0 dz ⎝ dz ⎠

249

(nn)

5.31 Consider the boundary value problem d2y + n 2 y = 2x 2 dx dy (0) = 0 dx dy (π ) = 0 dx

(a) (b) (c)

where n is an integer.

(a) Use a solvability condition to determine for what values of n a solution of Eq. (a) subject to Eqs. (b) and (c) a solution exists. (b) Solve Eq. (a) subject to Eqs. (b) and (c) to confirm the results of part (a). (c) Determine y(x) when a solution exists.

Solution: (a) The homogeneous solution of Equation e is y ( x) = C1 cos(nx) + C 2 sin( nx)

(d)

Application of the boundary conditions to Equation (d) leads to dy (0) = C 2 n ⇒ C 2 = 0 if n ≠ 0 dx

(e)

and dy (π ) = 0 =− C1n sin (nπ ) dx

(f)

However Equation f is identically satisfied. Thus a non-trivial solution exists for the homogeneous problem which implies that a solvability condition must be satisfied for the non-homogeneous solution to exist. The case n=0 is a special case. It also has a no-trivial homogeneous solution. The solvability condition is that the non-homogeneous term must be orthogonal to all non-trivial homogeneous solutions. To this end

250

π

(cos(nx),2 x ) = ∫ 2 x cos(nx )dx 0

x =π

x ⎡1 ⎤ = 2 ⎢ 2 cos(nx ) + sin (nx )⎥ n ⎣n ⎦ x =0 1 ⎡1 ⎤ = 2 ⎢ 2 (cos nπ − 1) + sin nπ ⎥ n ⎣n ⎦ 2 n = 2 (− 1) − 1 n 0 even n ⎧⎪ =⎨ 4 n odd ⎪⎩ n 2

[

]

(g)

Equation g implies that a solution of Equation a exists for even n, but not for odd n. Consider the special case of n=0. The non-trivial homogeneous solution is simply a constant. The existence of a solution to Equations a-c is determined from π

(1,2 x ) = ∫ 2 xdx 0

=x =π

2 x =π x =1

2

(h)

Equation h implies that there is no solution of Equations a-c for n=0. (b) The particular solution of Equation a, for n ≠ 0 is of the form y p ( x) = Ax + B Use of Equation i in Equation a leads to n 2 ( Ax + B ) = 2 x which implies A =

(i) 2 and B=0. n2

Thus the general solution of Equation a is y ( x) = C1 cos(nx) + C 2 sin( nx) +

2 x n2

(j)

Application of the boundary conditions to Equation j leads to 2 2 dy (0) = 0 ⇒ nC 2 + 2 = 0 ⇒ C 2 = − 3 dx n n 251

(k)

and dy (π ) = 0 ⇒ −nC1 sin(nπ ) + n⎛⎜ − 23 ⎞⎟ cos(nπ ) + 22 = 0 ⇒ cos(nπ ) = 1 dx n ⎝ n ⎠

(l)

Equation l is true for even n and false for odd n. This confirms the solvability condition. The resulting solution is of the form y ( x) = C1 cos(nx) −

2 2 sin(nx) + 2 x 3 n n

(m)

Consider the special case of n=0. Direct integration leads to y ( x) =

x3 + C1 x + C 2 3

(n)

Application of the boundary conditions to Equation n leads to dy (0) = 0 ⇒ C1 = 0 dx

(o)

and dy (π ) = 0 ⇒ π 2 = 0 dx Obviously Equation p is false and thus a solution does not exist for n=0.

252

(p)

5.33 Use the modal analysis method derived in Example 5.9 to determine the response of a two-degree-of-freedom mechanical system whose motion is described by the differential equation 0 − k ⎤ ⎡ x1 ⎤ ⎡ ⎤ = ⎢ ⎢ ⎥ ⎥ k ⎦ ⎣ x 2 ⎦ ⎣ F0 sin(ωt )⎥⎦

⎡m 0 ⎤ ⎡ &x&1 ⎤ ⎡ 2k ⎢ 0 2m⎥ ⎢ &x& ⎥ + ⎢− k ⎣ ⎦⎣ 2 ⎦ ⎣

And with all initial conditions equal to zero.

Solution: The natural frequencies are the square roots of the eigenvalues of M −1K . To this end

0 ⎤ ⎡ 2 − 1⎤ k ⎡ 2 − 1⎤ 1 ⎥⎢ = ⎢ 1 1⎥ − 1 1 ⎥⎦ m ⎢ − ⎥ ⎣ 2⎦ ⎣ 2 2 ⎥⎦ 2φ − λ −φ ⎛φ ⎞ ⎛ φ⎞ φ φ M −1 K − λ I = = (2φ − λ )⎜ − λ ⎟ − (− φ )⎜ − ⎟ − −λ ⎝2 ⎠ ⎝ 2⎠ 2 2 φ2 5 2 = λ − φλ + 2 2

M −1 K =

k ⎡1 ⎢ m ⎢0 ⎣

(a)

The eigenvalues are the roots of the polynomial of Equation a, λ1 = 0.2192φ and

λ 2 = 2.2808φ . Thus the natural frequencies are ω1 = 0.4682

k k and ω 2 = 1.5102 . m m

The mode shape vectors are determined from ⎡2φ − λ ⎢ φ ⎢ − ⎣ 2

− φ ⎤ ⎡ X ⎤ ⎡0 ⎤ ⎥ ⎢ 1 ⎥ = ⎢ ⎥ ⇒ X 2 = (2φ − λ ) X 1 φ − λ ⎥ ⎣ X 2 ⎦ ⎣0 ⎦ 2 ⎦

(b)

Equation b leads to ⎡ 1 ⎤ X 1 = c1 ⎢ ⎥ ⎣1.7818⎦

⎡ 1 ⎤ X 2 = c2 ⎢ ⎥ ⎣− 0.808⎦

The mode shape vectors are normalized by requiring (X i , X i )M = 1 . The results are

253

(c)

X1 =

1 ⎡0.7357 ⎤ ⎥ ⎢ m ⎣1.3430 ⎦

X2 =

1 ⎡ 0.9654 ⎤ ⎥ ⎢ m ⎣− 0.2711⎦

(d)

Using modal analysis the forced response is determined using the expansion theorem x(t ) = c1 (t ) X 1 + c 2 (t ) X 2

(e)

The differential equations for the coefficients are c&&1 + ω12 c1 = (X 1 , F )

(f)

c&&2 + ω c = (X 2 , F ) 2 2 2

(g)

where F = [0 F0 sin (ωt )] . Substitution leads to T

1.7818 F0 k c1 = sin (ωt ) m m 0.2711F0 k c&&2 + 2.2808 c 2 = − sin(ωt ) m m

c&&1 + 0.2192

(h) (i)

The solutions of Equations h and i subject to all initial conditions of zero are c1 (t ) =

1.7818F0 ⎡ m ⎛ k ⎞⎤ ω t ⎟⎥ sin ⎜⎜ 0.4682 ⎢sin (ωt ) − m ⎟⎠⎥⎦ 0.4682 k m ⎢⎣ ⎝

(j)

c 2 (t ) =

− 0.2711F0 ⎡ m ⎛ k ⎞⎤ ω t ⎟⎟⎥ sin ⎜⎜1.5102 ⎢sin (ωt ) − k m 1 . 5102 m ⎝ ⎠⎥⎦ ⎣⎢

(k)

The response of the system is obtained by substituting Equations j and k in Equation e. Problems 5.35-5.44 refer to the problem to find the eigenvalues and eigevectors of the system d ⎛ 3 dy ⎞ ⎟ = λxy ⎜x dx ⎝ dx ⎠ y (2) = 0 y (1) + 2 y ′(1) = 0

−

254

(a) (b) (c)

5.35 Using the exact solution determine the five lowest eigenvalues and their corresponding eigenvectors.

Solution: The differential equation is rewritten as d2y dy + 3x 2 + λxy = 0 2 dx dx d2y dy x 2 2 + 3 x + λy = 0 dx dx x3

(d)

Equation d is of the Cauchy-Euler type for which a solution of the form y = x α leads to

α 2 + 2α + λ = 0 . The solutions of the quadratic equation are α = −1 ± 1 − λ . Since the Sturm-Liouville Problem of Equations a-c is positive definite, assume λ > 1 . Then the solution of Equation b is written as y ( x) = C1

[

]

[

cos λ − 1 ln( x) sin λ − 1 ln( x) + C2 x x

]

(e)

Application of the boundary conditions leads to

[

]

y (1) + 2 y ′(1) = 0 ⇒ C1 + 2 − C1 + λ − 1C 2 = 0 y (2) = 0 ⇒ C1

[

]

[

cos λ − 1 ln(2) sin λ − 1 ln(2) + C2 2 2

]

(f) (g)

Non-trivial solutions to Equations f and g exist only if −1 cos λ − 1 ln(2)

2 λ −1 =0 sin λ − 1 ln(2)

[ ] [ tan [ λ − 1 ln(2)] = −2 λ − 1

Equation f implies that for a λ satisfying Equation h, C2 =

]

(h) 1 2 λ −1

. There are an infinite,

but countable, number of eigenvalues 0 < λ1 < λ 2 < ... which satisfy Equation h. For an eigenvalue λ k the corresponding eigenvector is

255

(

)

(

)

⎡ cos λ k − 1 ln( x) sin λ k − 1 ln( x) ⎤ 1 + y k ( x) = C k ⎢ ⎥ x x 2 λk − 1 ⎣⎢ ⎦⎥

(i)

The inner product for which the differential operator is self adjoint is obtained from Equation a as 2

( f , g ) r = ∫ f ( x) g ( x) xdx

(j)

1

The eigenvectors are normalized by requiring

( y k , y k )r

=1

(

)

(

⎡ cos λ k − 1 ln( x) sin λ k − 1 ln( x) 1 + ∫1 C ⎢⎢ x x 2 λk − 1 ⎣ 2

2 k

)⎤⎥

2

xdx = 1

⎥⎦

(k)

The integral of Equation k is evaluated using a change of variable, u = ln( x) such that du =

1 dx . Using this change of variable, Equation k becomes x ln(2)

C

2 k

∫ 0

2

⎡ ⎤ 1 sin λ k − 1u ⎥ du = 1 ⎢cos λ k − 1u + 2 λk − 1 ⎣⎢ ⎦⎥

(

)

(

)

(l)

The first five solutions of Equation h are

λ1 = 7.482, λ 2 = 48.649, λ3 = 130.827, λ4 = 254.083 and λ5 = 418.423 . The corresponding normalized mode shapes are

256

sin (2.546 ln( x ) )⎤ ⎡ cos(2.546 ln( x ) ) y1 ( x ) = 1.584⎢ + 0.196 ⎥ x x ⎣ ⎦ sin (6.903 ln( x ) )⎤ ⎡ cos(6.903 ln( x ) ) y 2 ( x ) = 1.682⎢ + 0.072 ⎥ x x ⎣ ⎦

(m) (n)

sin (11.344 ln( x ) ) ⎤ ⎡ cos(11.344 ln( x ) ) y 3 ( x ) = 1.692⎢ + 0.044 (o) ⎥ x x ⎣ ⎦ sin (15.909 ln( x ) ) ⎤ ⎡ cos(15.909 ln( x ) ) y 4 ( x) = 1.695⎢ + 0.31 ⎥ (p) x x ⎣ ⎦ sin (20.431 ln( x) ) ⎤ ⎡ cos(20.431 ln( x ) ) y 5 ( x) = 1.697 ⎢ + 0.024 ⎥ x x ⎣ ⎦

257

(q)

5.36 Choose two polynomials of order 3 or less which satisfy the boundary conditions. Use Rayleigh’s quotient to approximate the lowest eigenvalue.

Solution: Two polynomials which satisfy the boundary conditions are p1 ( x) = x 2 − x − 2

(a)

p 2 ( x) = x 3 + x − 10

(b)

The inner product for which the eigenvectors are orthogonal is 2

( f , g ) r = ∫ f ( x) g ( x) xdx

(c)

1

Evaluation of Rayleigh’s quotient for these polynomials leads to

(Lp1 , p1 )r ( p1 , p1 )r

R ( p1 ) =

2

=

⎡ 1 d ⎛

∫ ⎢⎣− x dx ⎜⎝ x 1

3

dp1 ⎞⎤ ⎟ p1 xdx dx ⎠⎥⎦

2

∫ p p xdx 1

1

1

∫ [- (8 x

][

2

=

3

]

− 3 x 2 ) x 2 − x − 2 dx

1

∫ (x 2

2

)

2

− x − 2 xdx

1

= 8.679

(d)

Similarly

258

R( p 2 ) = 2

=

(Lp 2 , p 2 )r ( p1 , p1 )r ⎡ 1 d ⎛

∫ ⎢⎣− x dx ⎜⎝ x 1

3

dp 2 dx

⎞⎤ ⎟⎥ p 2 xdx ⎠⎦

2

∫p

2

p 2 xdx

1

∫ [-( 15 x

][

2

=

4

]

+ 3 x 2 ) x 3 + x − 10 dx

1

∫ (x 2

3

)

2

+ x − 10 xdx

1

= 9.83

(e)

259

5.37 Choose the elements of a basis for the subspace of C 2 [0,1] defined by the intersection of P 4 [0,1] with the set of all functions satisfying boundary conditions (b) and (c).Use these functions as a basis for a Rayleigh-Ritz approximation of the lowest eigenvalues of the system.

Solution: An arbitrary element of P 4 [0,1] is of the form p ( x) = a 0 + a1 x + a 2 x 2 + a 3 x 3 + a 4 x 4

(d)

Requiring p(x) to satisfy the boundary conditions leads to p(2) = 0 ⇒ a 0 + 2a1 + 4a 2 + 8a 3 + 16a 4 = 0 p(1) + 2 p ′(1) = 0 ⇒ a 0 + a1 + a 2 + a3 + a 4 + 2(a1 + 2a 2 + 3a 3 + 4a 4 = 0 ) ⇒ a 0 + 3a1 + 5a 2 + 7 a3 + 9a 4 = 0

(e) (f)

Equations e and f are rearranged as a 0 = −2a 2 − 10a 3 − 30a 4

(g)

a1 = a3 − a 2 + 7 a 4

(h)

A general form of p(x) which satisfies the boundary conditions is

(

)

(

)

(

p ( x) = a 2 x 2 − x − 2 + a3 x 3 + x − 10 + a 4 x 4 + 7 x − 30

)

(i)

Equation i suggests a choice of basis functions as p1 ( x) = x 2 − x − 2

(j)

p 2 ( x) = x + x − 10

(k)

p3 ( x) = x + 7 x − 30

(l)

3

4

The differential operator is self adjoint with respect to the inner product defined as 2

( f , g ) r = ∫ f ( x) g ( x) xdx

(m)

1

The following MATHCAD code is used to obtain the Rayleigh-Ritz approximation for the eigenvalues and eigenvectors.

260

Problem 5.37

Basis functions 2

p1( x) := x − x − 2

3

p2( x) := x + x − 10

4

p3( x) := x + 7⋅ x − 30

Operator acting on p

Lp1( x) :=

−1 ⎡d ⎡ 3 ⎛ d ⎞⎤⎤ ⋅ ⎢ ⎢x ⋅ ⎜ p1( x) ⎟⎥⎥ x ⎣dx⎣ ⎝ dx ⎠⎦⎦

−1 ⎡d ⎡ 3 ⎛ d ⎞⎤⎤ ⋅ ⎢ ⎢x ⋅ ⎜ p2( x) ⎟⎥⎥ x ⎣dx⎣ ⎝ dx ⎠⎦⎦

Lp2( x) :=

Lp3( x) :=

−1 ⎡d ⎡ 3 ⎛ d ⎞⎤⎤ ⋅ ⎢ ⎢x ⋅ ⎜ p3( x) ⎟⎥⎥ x ⎣dx⎣ ⎝ dx ⎠⎦⎦

j :=1, 2..3 ⎛ p1( x) ⎞

p ( x) := ⎜ p2( x) ⎟ ⎜ ⎟ ⎝ p3( x) ⎠

⎛ Lp1( x) ⎞ Lp( x) := ⎜ Lp2( x) ⎟ ⎜ ⎟ ⎝ Lp3( x) ⎠

2

⌠ Construction of Rayleigh-Ritz matrices B := ⎮ p ( x) ⋅ p ( x) ⋅ x dx i, j i j ⌡ 1 i := 1 , 2 .. 3 2

j := 1 , 2 ⌠ .. 3 A := ⎮ Lp( x) p ( x) ⋅ x dx i, j i j ⌡

(

)

⎛ 2.183 8.86 24.715 ⎞ B = ⎜ 8.86 35.958 100.337⎟ ⎜ ⎟ ⎝ 24.715 100.337 280.05 ⎠

1

⎛ 18.95 78.007 220.979 ⎞ ⎜ ⎟ A = ⎜ 78.007 321.625 912.655 ⎟ 2 ⌠ ⎜ 220.979 3⎟ A ⎝ := ⎮ Lp912.655 ( x) ( p ( x)2.595 ⋅ x dx× 10 ⎠ i, j i j) ⌡ ji := 1 , 2 .. 13

261

⎛ 3.885 × 103 1.945 × 104 6.744 × 104 ⎞ ⎜ ⎟ ⎜ 3 3 4 C = −1.99 × 10 −9.906 × 10 −3.417 × 10 ⎟ ⎜ ⎟ ⎜ 3 3 ⎟ ⎝ 370.739 1.836 × 10 6.301 × 10 ⎠

−1

C := B

⋅A

Eigenvalues of C

⎛ 222.637⎞ q = ⎜ 50.174 ⎟ ⎜ ⎟ ⎝ 7.487 ⎠

q := eigenvals ( C)

Eigenvectors of C

(

q1 := eigenvec C, q

(

)

3

q2 := eigenvec C, q

(

q3 := eigenvec C, q

⎛ −0.924 ⎞ q1 = ⎜ 0.378 ⎟ ⎜ ⎟ ⎝ −0.056 ⎠

)

2

⎛ −0.91 ⎞ q2 = ⎜ 0.409 ⎟ ⎜ ⎟ ⎝ −0.066 ⎠

T

c1 := q1 ⋅ B⋅ q1

T

c2 := q2 ⋅ B⋅ q2

)

1

⎛ −0.897 ⎞ q3 = ⎜ 0.435 ⎟ ⎜ ⎟ ⎝ −0.077 ⎠

262

T

c3 := q3 ⋅ B⋅ q3

Eigenvector approximations

1

y1( x) :=

1

y2(x) :=

1 c3

1

(

1

2

3

⋅ q2 ⋅p1(x) + q2 ⋅p2(x) + q2 ⋅p3(x)

c2

y3( x) :=

(

⋅ q1 ⋅ p1( x) + q1 ⋅ p2( x) + q1 ⋅ p3( x)

c1

(

2

3

⋅ q3 ⋅ p1( x) + q3 ⋅ p2( x) + q3 ⋅ p3( x) 1

2

3

)

)

)

263

x := 1 , 1.02.. 2

2

1 y1 ( x) y2 ( x) 0 y3 ( x) −1

−2

1

1.5

2 x

264

2.5

5.38 Choose the elements of a basis for the subspace of C 2 [0,1] defined by the intersection of P 4 [0,1] with the set of all functions satisfying boundary condition (b) (the geometric boundary condition). Use these functions as a basis for a Rayleigh-Ritz approximation of the lowest eigenvalues of the system.

Solution: An arbitrary element of P 4 [0,1] is of the form p ( x) = a 0 + a1 x + a 2 x 2 + a 3 x 3 + a 4 x 4

(d)

Requiring p(x) to satisfy the geometric boundary condition leads to p (2) = 0 ⇒ a 0 + 2a1 + 4a 2 + 8a3 + 16a 4 = 0 ⇒ a 0 = −2a1 − 4a 2 − 8a 3 − 16a 4

(e)

A general form of p(x) which satisfies the geometric boundary condition is

(

)

(

)

(

p ( x) = a1 ( x − 2 ) + a 2 x 2 − 4 + a3 x 3 − 8 + a 4 x 4 − 16

)

(f)

Equation i suggests a choice of basis functions as p1 ( x) = x − 2

(g)

p 2 ( x) = x − 4

(h)

2

p3 ( x) = x 3 − 8

(i)

p 4 ( x) = x 4 − 16 The differential operator is self adjoint with respect to the inner product defined as 2

( f , g ) r = ∫ f ( x) g ( x) xdx 1

The following MATHCAD code is used to obtain the Rayleigh-Ritz approximation for the eigenvalues and eigenvectors.

265

(j)

Problem 5.38

Basis functions p1( x) := x − 2 2

p2( x) := x − 4

3

p3( x) := x − 8

4

p4( x) := x − 16

Operator acting on p

Lp1( x) :=

−1 ⎡d ⎡ 3 ⎛ d ⎞⎤⎤ ⋅ ⎢ ⎢x ⋅ ⎜ p1( x) ⎟⎥⎥ x ⎣dx⎣ ⎝ dx ⎠⎦⎦

Lp2( x) :=

−1 ⎡d ⎡ 3 ⎛ d ⎞⎤⎤ ⋅ ⎢ ⎢x ⋅ ⎜ p2( x) ⎟⎥⎥ x ⎣dx⎣ ⎝ dx ⎠⎦⎦

Lp3( x) :=

−1 ⎡d ⎡ 3 ⎛ d ⎞⎤⎤ ⋅ ⎢ ⎢x ⋅ ⎜ p3( x) ⎟⎥⎥ x ⎣dx⎣ ⎝ dx ⎠⎦⎦

Lp4( x) :=

−1 ⎡d ⎡ 3 ⎛ d ⎞⎤⎤ ⋅ ⎢ ⎢x ⋅ ⎜ p4( x) ⎟⎥⎥ x ⎣dx⎣ ⎝ dx ⎠⎦⎦

⎛⎜ p1( x) ⎞⎟ p2( x) ⎟ p ( x) := ⎜ ⎜ p3( x) ⎟ ⎜ p4( x) ⎟ ⎝ ⎠

⎛⎜ Lp1( 1) ⎞⎟ Lp2( x) ⎟ Lp( x) := ⎜ ⎜ Lp3( x) ⎟ ⎜ Lp4( x) ⎟ ⎝ ⎠

266

Construction of Rayleigh-Ritz matrices i := 1 , 2 .. 4 j := 1 , 2 .. 4

⎤ ⎡⎢⌠ 2 3 ⎡d ⎤ ⎛d ⎞ ⎥ A := ⎢⎮ x ⋅ ⎢ ( p ( x) )⎥ ⋅ ⎜ p ( x) ⎟ dx⎥ i, j i ⎮ ⎦ ⎝ dx j ⎠ ⎥ ⎢⌡1 ⎣dx ⎣ ⎦ 2

⌠ B := ⎮ p ( x) ⋅ p ( x) ⋅ x dx i, j i j ⌡ 1

⎛ 3.75 12.4 ⎜ 12.4 42 A=⎜ ⎜ 31.5 108.857 ⎜ ⎝ 72.571 255

⎞ ⎟ 108.857 255 ⎟ 286.875 681.333 ⎟ 3⎟ 681.333 1.637 × 10 ⎠ 31.5

⎛⎜ 0.417 1.367 B=⎜ ⎜ 3.433 ⎜ 7.81 ⎝

⎛⎜ 4.362 × 103 2.056 × 103 ⎜ 3 −4.038 × 10 −993.69 ⎜ C= ⎜ 3 −44.77 ⎜ 1.626 × 10 ⎜ −240.468 85.47 ⎝

−1

C := B

72.571

⋅A

⎞⎟ 4.5 11.343 25.875⎟ 11.343 28.675 65.578⎟ ⎟ 25.875 65.578 150.3 ⎠ 1.367 3.433

4

5⎞

⎟ ⎟ 2.405 × 10 1.115 × 10 ⎟ 4 4⎟ −1.195 × 10 −5.284 × 10 ⎟ 3 3 ⎟ 2.167 × 10 9.209 × 10 ⎠ −2.095 × 10

−1.027 × 10

4

5

Eigenvalues of C

⎛⎜ 393.938⎞⎟ 174.269⎟ q=⎜ ⎜ 8.657 ⎟ ⎜ 51.053 ⎟ ⎝ ⎠

q := eigenvals ( C)

Eigenvectors of C

(

q1 := eigenvec C, q

(

)

3

q2 := eigenvec C, q

)

4

⎛⎜ 0.692 ⎞⎟ −0.67 ⎟ q1 = ⎜ ⎜ 0.265 ⎟ ⎜ −0.038 ⎟ ⎝ ⎠ ⎛⎜ 0.642 ⎞⎟ −0.696 ⎟ q2 = ⎜ ⎜ 0.316 ⎟ ⎜267 ⎟ ⎝ −0.051 ⎠

7.81

T

c1 := q1 ⋅ B⋅ q1

T

c2 := q2 ⋅ B⋅ q2

(

q3 := eigenvec C, q

⎛⎜ 0.79 ⎞⎟ −0.588 ⎟ q3 = ⎜ ⎜ 0.171 ⎟ ⎜ −0.014 ⎟ ⎝ ⎠

)

2

(

q4 := eigenvec C, q

T

c3 := q3 ⋅ B⋅ q3

⎛⎜ 0.671 ⎞⎟ −0.677 ⎟ q4 = ⎜ ⎜ 0.298 ⎟ ⎜ −0.048 ⎟ ⎝ ⎠

)

1

T

c4 := q4 ⋅ B⋅ q4

Eigenvector approximations y1( x) :=

1

(

⋅ q1 ⋅ p1( x) + q1 ⋅ p2( x) + q1 ⋅ p3( x) + q1 ⋅ p4( x)

c1

1

1

y2( x) :=

1

1 c4

1

(

1

4

)

2

3

4

⋅ q3 ⋅ p1( x) + q3 ⋅ p2( x) + q3 ⋅ p3( x) + q3 ⋅ p4( x)

c3

y4( x) :=

(

3

⋅ q2 ⋅ p1( x) + q2 ⋅ p2( x) + q2 ⋅ p3( x) + q2 ⋅ p4( x)

c2

y3( x) :=

2

(

2

3

4

⋅ q4 ⋅ p1( x) + q4 ⋅ p2( x) + q4 ⋅ p3( x) + q4 ⋅ p4( x) 1

2

3

4

268

)

)

)

x := 1 , 1.02.. 2

2 1 y1 ( x)

0

y2 ( x) y3 ( x)− 1 y4 ( x) −2 −3 −4

1

1.5

2 x

269

2.5

5.39 Use a finite-element method to approximate the lowest eigenvalues and eigenvectors for the system. Use five equally spaced elements.

Solution: The following MATHCAD file contains the solution

The interval from x=1 to x=2 is divided into five elements of equal length leading to definition of the basis functions as p0( x) := [ 1 − 5⋅ ( x − 1) ] ( Φ( x − 1) − Φ( x − 1.2) ) p1( x) := 5⋅ ( x − 1) ⋅ ( Φ( x − 1) − Φ( x − 1.2) ) + [ 2 − 5⋅ ( x − 1) ] ⋅ ( Φ( x − 1.2) − Φ( x − 1.4) ) φ1 ( x) := 5⋅ x⋅ ( Φ( x) − Φ( x − 0.2) ) + ( 2 − 5⋅ x) ⋅ ( Φ( x − 0.2) − Φ( x − 0.4) ) p2( x) := p1( x − 0.2) p3( x) := p1( x − 0.4) p4( x) := p1( x − 0.6) p5( x) := [ 5⋅ ( x − 1) − 4] ⋅ Φ( x − 1.8)

where (x) is the Heaviside function defined as 0 for x0. It is also called the unit step function u(x)

270

⎛ p0( x) ⎞ ⎜ p1( x) ⎟ ⎜ ⎟ p ( x) := ⎜ p2( x) ⎟ ⎜ p3( x) ⎟ ⎜ ⎟ ⎝ p4( x) ⎠ x := 1 , 1.01.. 2

Basis functions 1.5

p0 ( x) p1 ( x)

1

p2 ( x) p3 ( x) p4 ( x) p5 ( x)0.5

0

1

1.5

2 x

The geometric boundary condition at x=2 is imposed by ignoring p5(x)

271 ⎡⎢⌠2 ⎥⎤ 3 ⎡d ⎤ ⎞ ⎛ d ⌠ ⎮ A := ⎢ p (xx)⋅ ⎢⋅ p ( xp) (⋅xx)dx⎥ ⋅ ⎜ p ( x) ⎟ dx⎥ B i , j := j i , j ⎮⎮ i⎣dx j i ⎦ ⎝ dx ⎠ ⎥ ⌡ ⎢ ⌡ ji :=The 1 , 2 ..⎣geometric 151 Construction of Rayleigh-Ritz boundary condition matrices ⎦ at x=2 is imposed by ignoring p5(x) 2

(

)

2.5

Construction of Rayleigh-Ritz matrices i := 1 , 2 .. 5 j := 1 , 2 .. 5

⎡⎢⌠ 2 ⎥⎤ 3 ⎡d ⎤ ⎞ ⎛ d ⎮ A := ⎢ x ⋅ ⎢ ( p ( x) )⎥ ⋅ ⎜ p ( x) ⎟ dx⎥ i, j i ⎮ d ⎦ ⎝ dx j ⎠ ⎥ ⎢⌡1 ⎣ x ⎣ ⎦ 2

⌠ p ( x) ⋅ p ( x) ⋅ x dx := i, j ⎮ i j ⌡

B

1

0 0 0 ⎞ ⎛ 6.71 −6.71 ⎜ −6.71 ⎟ 17.76 −11.05 0 0 ⎜ ⎟ A = ⎜ 0 −11.05 28 −16.95 0 ⎟ ⎜ 0 0 −16.95 41.6 −24.65 ⎟ ⎜ ⎟ 0 0 −24.65 59.04 ⎠ ⎝ 0 −1

C := B

⋅A

⎛ 137.059 ⎜ −78.649 ⎜ C = ⎜ 19.569 ⎜ −4.894 ⎜ ⎝ 1.155

0 0 ⎞ ⎛ 0.07 0.037 0 ⎜ ⎟ 0.037 0.16 0.043 0 0 ⎜ ⎟ B = ⎜ 0 0.043 0.187 0.05 0 ⎟ ⎜ 0 0 0.05 0.213 0.057 ⎟ ⎜ ⎟ 0 0 0.057 0.24 ⎠ ⎝ 0 −192.66

−28.882

⎞ ⎟ ⎟ −109.426 221.61 −179.129 60.005 ⎟ 27.364 −140.187 281.953 −208.005⎟ ⎟ −6.461 33.101 −169.283 295.115 ⎠ 184.79

272

76.845

−146.695 55.135

9.675

−18.469

Eigenvalues of C

⎛ 571.989⎞ ⎜ 326.724⎟ ⎜ ⎟ q = ⎜ 159.137⎟ ⎜ 8.731 ⎟ ⎜ ⎟ ⎝ 53.945 ⎠

q := eigenvals ( C)

Eigenvectors of C

(

q1 := eigenvec C, q

)

4

(

q2 := eigenvec C, q

(

⎛ 0.657 ⎞ ⎜ ⎟ 0.57 ⎜ ⎟ q1 = ⎜ 0.412 ⎟ ⎜ 0.25 ⎟ ⎜ ⎟ ⎝ 0.11 ⎠

q3 := eigenvec C, q

)

5

)

3

⎛ −0.745 ⎞ ⎜ ⎟ −0.252 ⎜ ⎟ q2 = ⎜ 0.312 ⎟ ⎜ 0.456 ⎟ ⎜ ⎟ ⎝ 0.274 ⎠ ⎛ 0.76 ⎞ ⎜ ⎟ −0.268 ⎜ ⎟ q3 = ⎜ −0.416 ⎟ ⎜ 0.222 ⎟ ⎜ ⎟ ⎝ 0.358 ⎠

273

T

c1 := q1 ⋅ B⋅ q1

T

c2 := q2 ⋅ B⋅ q2

T

c3 := q3 ⋅ B⋅ q3

(

q4 := eigenvec C, q

⎛ −0.671 ⎞ ⎜ ⎟ 0.58 ⎜ ⎟ q4 = ⎜ −0.297 ⎟ ⎜ −0.145 ⎟ ⎜ ⎟ ⎝ 0.323 ⎠

)

2

(

q5 := eigenvec C, q

T

c4 := q4 ⋅ B⋅ q4

⎛ 0.295 ⎞ ⎜ −0.355 ⎟ ⎜ ⎟ q5 = ⎜ 0.503 ⎟ ⎜ −0.59 ⎟ ⎜ ⎟ ⎝ 0.431 ⎠

)

1

T

c5 := q5 ⋅ B⋅ q5

Eigenvector approximations y1( x) :=

1

(

⋅ q1 ⋅ p1( x) + q1 ⋅ p2( x) + q1 ⋅ p3( x) + q1 ⋅ p4( x) + q1 ⋅ p5( x)

c1

1

1

y2( x) :=

1

1 c4

y5( x) :=

1

(

1

4

5

)

2

3

4

5

⋅ q3 ⋅ p1( x) + q3 ⋅ p2( x) + q3 ⋅ p3( x) + q3 ⋅ p4( x) + q3 ⋅ p5( x)

c3

y4( x) :=

(

3

⋅ q2 ⋅ p1( x) + q2 ⋅ p2( x) + q2 ⋅ p3( x) + q2 ⋅ p4( x) + q2 ⋅ p5( x)

c2

y3( x) :=

2

(

2

3

4

5

⋅ q4 ⋅ p1( x) + q4 ⋅ p2( x) + q4 ⋅ p3( x) + q4 ⋅ p4( x) + q4 ⋅ p5( x)

1 c5

1

(

2

3

4

5

)

)

)

⋅ q5 ⋅ p1( x) + q5 ⋅ p2( x) + q5 ⋅ p3( x) + q5 ⋅ p4( x) + q5 ⋅ p5( x) 1

2

3

4

274

5

)

x := 1 , 1.02.. 2

3 2 y1 ( x) y2 ( x) 1 y3 ( x) 0 y4 ( x) y5 ( x)− 1 −2 −3

1

1.5

2 x

275

2.5

Problems 5.40-5.47 still refer to the systems of Problems 5.35-5.39 but with boundary condition (c) replaced by y (1) + 2 y ′(1) = 0.5λy (1)

(d)

5.40 Develop the appropriate form of the inner product for orthogonality of the eigenvectors.

Solution: The appropriate inner product is defined by 2

( f , g ) = ∫ f ( x) g ( x) xdx + 0.5 f (1) g (1) 1

276

(e)

5.41 Use the exact solution to determine the five lowest eigenvectors.

Solution: The differential equation is rewritten as d2y dy x + 3x 2 + λxy = 0 2 dx dx d2y dy x 2 2 + 3 x + λy = 0 dx dx 3

(d)

Equation d is of the Cauchy-Euler type for which a solution of the form y = x α leads to

α 2 + 2α + λ = 0 . The solutions of the quadratic equation are α = −1 ± 1 − λ . Since the Sturm-Liouville Problem of Equations a-c is positive definite, assume λ > 1 . Then the solution of Equation b is written as y ( x) = C1

[

]

[

cos λ − 1 ln( x) sin λ − 1 ln( x) + C2 x x

]

(e)

Application of the boundary conditions leads to

[

]

y (1) + 2 y ′(1) = 0.5λy (1) ⇒ C1 + 2 − C1 + λ − 1C 2 = 0.5λC1 y (2) = 0 ⇒ C1

[

]

[

cos λ − 1 ln(2) sin λ − 1 ln(2) + C2 2 2

[

]

y (1) + 2 y ′(1) = 0 ⇒ C1 + 2 − C1 + λ − 1C 2 = 0 y (2) = 0 ⇒ C1

[

]

[

cos λ − 1 ln(2) sin λ − 1 ln(2) + C2 2 2

]

]

(f) (g) (f) (g)

Non-trivial solutions to Equations f and g exist only if − 1 − 0.5λ

2 λ −1 =0 sin λ − 1 ln(2)

[ ] [ 2 λ −1 tan [ λ − 1 ln(2)] = − 1 + 0.5λ cos λ − 1 ln(2)

277

]

(h)

Equation f implies that for a λ satisfying Equation h, C 2 =

1 + 0.5λ 2 λ −1

. There are an

infinite, but countable, number of eigenvalues 0 < λ1 < λ 2 < ... which satisfy Equation h. For an eigenvalue λk the corresponding eigenvector is

(

)

(

⎡ cos λ k − 1 ln( x) 1 + 0.5λ sin λ k − 1 ln( x) y k ( x) = C k ⎢ + x x 2 λk − 1 ⎢⎣

)⎤⎥ ⎥⎦

(i)

The inner product for which the differential operator is self adjoint is obtained from Equation a as 2

( f , g ) r = ∫ f ( x) g ( x) xdx + 0.5 f( 1 )g( 1 )

(j)

1

The eigenvectors are normalized by requiring

( y k , y k )r

=1

(

)

(

)

2

⎡ cos λ k − 1 ln( x) + 0.5γ 1 sin λ k − 1 ln( x) ⎤ 2 C + ⎢ ⎥ xdx + 0.5C k = 1 ∫1 ⎢ x x 2 λk − 1 ⎥⎦ ⎣ 2

2 k

(k)

The integral of Equation k is evaluated using a change of variable, u = ln(x) such that du =

1 dx . Using this change of variable, Equation k becomes x ln(2)

C

2 k

∫ 0

2

⎡ ⎤ 1 sin λ k − 1u ⎥ du + 0.5C k2 = 1 ⎢cos λ k − 1u + 2 λk − 1 ⎥⎦ ⎣⎢

(

)

(

)

(l)

The first five solutions of Equation h are

λ1 = 12.939, λ2 = 72.394, λ3 = 174.68, λ4 = 318.327 and λ5 = 503.138 . The corresponding normalized mode shapes are

278

sin (3.455 ln( x ) ) ⎤ ⎡ cos(3.455 ln( x) ) y1 ( x ) = 0.843⎢ + 1.081 ⎥ x x ⎣ ⎦ sin (8.45 ln( x ) ) ⎤ ⎡ cos(8.45 ln( x) ) y 2 ( x) = 0.614⎢ + 2.201 ⎥ x x ⎣ ⎦

(m) (n)

sin (13.179 ln( x ) ) ⎤ ⎡ cos(13.179 ln( x ) ) y 3 ( x ) = 0.453⎢ + 3.352 ⎥ x x ⎣ ⎦ sin (17.814 ln( x) ) ⎤ ⎡ cos(17.814 ln( x) ) y 4 ( x) = 0.354 ⎢ + 4.496 ⎥ x x ⎣ ⎦

sin (22.408 ln( x) ) ⎤ ⎡ cos(22.408 ln( x) ) y 5 ( x) = 0.289 ⎢ + 5.636 ⎥ x x ⎣ ⎦

279

(o) (p) (q)

5.43 Choose the elements of a basis for the subspace of C 2 [0,1] defined by the intersection of P 4 [0,1] with the set of all functions satisfying boundary condition (b) (the geometric boundary condition).Use these functions as a basis for a Rayleigh-Ritz approximation of the lowest eigenvalues of the system.

Solution: An arbitrary element of P 4 [0,1] is of the form p ( x) = a 0 + a1 x + a 2 x 2 + a 3 x 3 + a 4 x 4

(d)

Requiring p(x) to satisfy the geometric boundary condition leads to p (2) = 0 ⇒ a 0 + 2a1 + 4a 2 + 8a3 + 16a 4 = 0 ⇒ a 0 = −2a1 − 4a 2 − 8a 3 − 16a 4

(e)

A general form of p(x) which satisfies the geometric boundary condition is

(

)

(

)

(

p ( x) = a1 ( x − 2 ) + a 2 x 2 − 4 + a3 x 3 − 8 + a 4 x 4 − 16

)

(f)

Equation i suggests a choice of basis functions as p1 ( x) = x − 2

(g)

p 2 ( x) = x − 4

(h)

2

p3 ( x) = x 3 − 8

(i)

p 4 ( x) = x 4 − 16 The differential operator is self adjoint with respect to the inner product defined as 2

( f , g ) r = ∫ f ( x) g ( x) xdx + 0.5 f( 1 )g( 1 ) 1

The following MATHCAD code is used to obtain the Rayleigh-Ritz approximation for the eigenvalues and eigenvectors.

Problem 5.43

280

(j)

Basis functions p1( x) := x − 2 2

p2( x) := x − 4

3

p3( x) := x − 8

4

p4( x) := x − 16

⎛⎜ p1( x) ⎞⎟ p2( x) ⎟ p ( x) := ⎜ ⎜ p3( x) ⎟ ⎜ p4( x) ⎟ ⎝ ⎠ Construction of Rayleigh-Ritz matrices i := 1 , 2 .. 4 j := 1 , 2 .. 4

⎤ ⎡⎢⌠ 2 3 ⎡d ⎤ ⎛d ⎞ ⎥ A := ⎢⎮ x ⋅ ⎢ ( p ( x) )⎥ ⋅ ⎜ p ( x) ⎟ dx⎥ i, j i ⎮ d ⎦ ⎝ dx j ⎠ ⎥ ⎢⌡1 ⎣ x ⎣ ⎦ 2

⌠ := ⎮ p ( x) ⋅ p ( x) ⋅ x dx + 0.5⋅ p ( 1) ⋅ p ( 1) i, j i j i j ⌡

B

1

⎛ 3.75 12.4 ⎜ 12.4 42 A=⎜ ⎜ 31.5 108.857 ⎜ ⎝ 72.571 255

⎞ ⎟ ⎟ 286.875 681.333 ⎟ 3⎟ 681.333 1.637 × 10 ⎠ 31.5

72.571

108.857

255

⎛⎜ 0.917 2.867 6.933 15.31 ⎞⎟ 2.867 9 21.843 48.375 ⎟ B=⎜ ⎜ 6.933 21.843 53.175 118.078⎟ ⎜ 15.31 48.375 118.078 262.8 ⎟ ⎝ ⎠

281

−1

C := B

⎛⎜ −1.849 × 103 −9.492 × 103 ⎜ 3 4 1.976 × 10 1.019 × 10 ⎜ C= ⎜ 3 ⎜ −924.752 −4.787 × 10 ⎜ 159.838 829.713 ⎝

⋅A

4

5⎞

⎟ ⎟ 3.669 × 10 1.141 × 10 ⎟ 4 4⎟ −1.731 × 10 −5.396 × 10 ⎟ 3 3 ⎟ 3.009 × 10 9.386 × 10 ⎠

−3.401 × 10

−1.054 × 10

4

5

Eigenvalues of C

⎛⎜ 289.133⎞⎟ 90.29 ⎟ q=⎜ ⎜ 26.996 ⎟ ⎜ 3.581 ⎟ ⎝ ⎠

q := eigenvals ( C)

Eigenvectors of C

(

q1 := eigenvec C, q

(

q2 := eigenvec C, q

(

0.841 ⎛ ⎞ ⎜ −0.528 ⎟ ⎟ q1 = ⎜ 0.121 ⎜ ⎟ ⎜ −3⎟ ⎝ −6.587 × 10 ⎠

)

3

0.92 ⎛ ⎞ ⎜ −0.386 ⎟ ⎟ q2 = ⎜ T 0.067 ⎜ ⎟ c2 := q2 ⋅ B⋅ q2 ⎜ −3⎟ ⎝ −3.171 × 10 ⎠

)

4

q3 := eigenvec C, q

T

c1 := q1 ⋅ B⋅ q1

)

2

⎛⎜ 0.693 ⎞⎟ −0.665 ⎟ q3 = ⎜ ⎜ 0.276 ⎟ ⎜ −0.042 ⎟ ⎝ ⎠

282

T

c3 := q3 ⋅ B⋅ q3

(

q4 := eigenvec C, q

⎛⎜ 0.654 ⎞⎟ −0.687 ⎟ q4 = ⎜ ⎜ 0.314 ⎟ ⎜ −0.053 ⎟ ⎝ ⎠

)

1

T

c4 := q4 ⋅ B⋅ q4

Eigenvector approximations 1

y1( x) :=

(

⋅ q1 ⋅ p1( x) + q1 ⋅ p2( x) + q1 ⋅ p3( x) + q1 ⋅ p4( x)

c1

1

1

y2( x) :=

1

1 c4

1

(

1

4

)

2

3

4

⋅ q3 ⋅ p1( x) + q3 ⋅ p2( x) + q3 ⋅ p3( x) + q3 ⋅ p4( x)

c3

y4( x) :=

(

3

⋅ q2 ⋅ p1( x) + q2 ⋅ p2( x) + q2 ⋅ p3( x) + q2 ⋅ p4( x)

c2

y3( x) :=

2

(

2

3

4

⋅ q4 ⋅ p1( x) + q4 ⋅ p2( x) + q4 ⋅ p3( x) + q4 ⋅ p4( x) 1

2

3

4

)

)

)

x := 1 , 1.02.. 2

2

y1( x)

1

y2( x) y3( x) 0 y4( x) −1

−2

1

1.5

2 x

283

2.5

5.44 Use the finite-element method to approximate the natural frequencies and mode shapes. Use five equally spaced elements.

Solution: The approximation is obtained using the following MATHCAD file Problem 5.44

The interval from x=1 to x=2 is divided into five elements of equal length leading to definition of the basis functions as p0( x) := [ 1 − 5⋅ ( x − 1) ] ( Φ( x − 1) − Φ( x − 1.2) ) p1( x) := 5⋅ ( x − 1) ⋅ ( Φ( x − 1) − Φ( x − 1.2) ) + [ 2 − 5⋅ ( x − 1) ] ⋅ ( Φ( x − 1.2) − Φ( x − 1.4) ) φ1 ( x) := 5⋅ x⋅ ( Φ( x) − Φ( x − 0.2) ) + ( 2 − 5⋅ x) ⋅ ( Φ( x − 0.2) − Φ( x − 0.4) ) p2( x) := p1( x − 0.2) p3( x) := p1( x − 0.4) p4( x) := p1( x − 0.6) p5( x) := [ 5⋅ ( x − 1) − 4] ⋅ Φ( x − 1.8)

where (x) is the Heaviside function defined as 0 for x0. It is also called the unit step function u(x)

x := 1 , 1.01.. 2

284

Basis functions 1.5

p0 ( x) p1 ( x)

1

p2 ( x) p3 ( x) p4 ( x) p5 ( x)0.5

0

1

1.5

2 x

285

2.5

The geometric boundary condition at x=2 is imposed by ignoring p5(x)

⎛ p0( x) ⎞ ⎜ ⎟ p1( x) ⎜ ⎟ p ( x) := ⎜ p2( x) ⎟ ⎜ p3( x) ⎟ ⎜ ⎟ ⎝ p4( x) ⎠ Construction of Rayleigh-Ritz matrices i := 1 , 2 .. 5 j := 1 , 2 .. 5

⎤ ⎡⎢⌠ 2 3 ⎡d ⎤ ⎛d ⎞ ⎥ A := ⎢⎮ x ⋅ ⎢ ( p ( x) )⎥ ⋅ ⎜ p ( x) ⎟ dx⎥ i, j i ⎮ d ⎦ ⎝ dx j ⎠ ⎥ ⎢⌡1 ⎣ x ⎣ ⎦ 2

⌠ B := ⎮ p ( x) ⋅ p ( x) ⋅ x dx + 0.5⋅ p ( 1) ⋅ p ( 1) i, j i j i j ⌡ 1

0 0 0 ⎞ ⎛ 6.71 −6.71 ⎜ −6.71 ⎟ 17.76 −11.05 0 0 ⎜ ⎟ A = ⎜ 0 −11.05 28 −16.95 0 ⎟ ⎜ 0 0 −16.95 41.6 −24.65 ⎟ ⎜ ⎟ 0 0 −24.65 59.04 ⎠ ⎝ 0 −1

C := B

⋅A

⎛ 44.943 ⎜ −56.012 ⎜ C = ⎜ 13.937 ⎜ −3.485 ⎜ ⎝ 0.823

0 0 ⎞ ⎛ 0.195 0.037 0 ⎜ ⎟ 0.037 0.16 0.043 0 0 ⎜ ⎟ B = ⎜ 0 0.043 0.187 0.05 0 ⎟ ⎜ 0 0 0.05 0.213 0.057 ⎟ ⎜ ⎟ 0 0 0.057 0.24 ⎠ ⎝ 0 −63.176

25.199

−9.471

152.971 −134.004 50.365 −101.509 218.452 −177.942 25.384 −5.994

Eigenvalues of C

286

−139.398 281.656 32.915

−169.213

⎞ ⎟ −16.871 ⎟ 59.607 ⎟ −207.905⎟ ⎟ 295.092 ⎠ 3.173

⎛ 558.767⎞ ⎜ ⎟ 271.472 ⎜ ⎟ q = ⎜ 117.447⎟ ⎜ 6.648 ⎟ ⎜ ⎟ ⎝ 38.779 ⎠

q := eigenvals ( C)

Eigenvectors of C

(

q1 := eigenvec C, q

)

4

(

q2 := eigenvec C, q

(

q3 := eigenvec C, q

(

⎛ 0.706 ⎞ ⎜ 0.549 ⎟ ⎜ ⎟ q1 = ⎜ 0.376 ⎟ ⎜ 0.222 ⎟ ⎜ ⎟ ⎝ 0.097 ⎠ ⎛ −0.667 ⎞ ⎜ 0.07 ⎟ ⎜ ⎟ q2 = ⎜ 0.489 ⎟ ⎜ 0.491 ⎟ ⎜ ⎟ ⎝ 0.265 ⎠

)

5

⎛ 0.443 ⎞ ⎜ ⎟ −0.652 ⎜ ⎟ q3 = ⎜ −0.261 ⎟ ⎜ 0.391 ⎟ ⎜ ⎟ ⎝ 0.397 ⎠

)

3

q4 := eigenvec C, q

)

2

⎛ −0.234 ⎞ ⎜ 0.649 ⎟ ⎜ ⎟ q4 = ⎜ −0.559 ⎟ ⎜ −0.069 ⎟ ⎜ ⎟ ⎝ 0.455 ⎠

287

T

c1 := q1 ⋅ B⋅ q1

T

c2 := q2 ⋅ B⋅ q2

T

c3 := q3 ⋅ B⋅ q3

T

c4 := q4 ⋅ B⋅ q4

(

q5 := eigenvec C, q

⎛ 0.074 ⎞ ⎜ ⎟ −0.279 ⎜ ⎟ q5 = ⎜ 0.51 ⎟ ⎜ −0.648 ⎟ ⎜ ⎟ ⎝ 0.486 ⎠

)

1

T

c5 := q5 ⋅ B⋅ q5

Eigenvector approximations y1( x) :=

1

(

⋅ q1 ⋅ p1( x) + q1 ⋅ p2( x) + q1 ⋅ p3( x) + q1 ⋅ p4( x) + q1 ⋅ p5( x)

c1

1

1

y2( x) :=

1

1 c4

y5( x) :=

1

(

1

4

5

)

2

3

4

5

⋅ q3 ⋅ p1( x) + q3 ⋅ p2( x) + q3 ⋅ p3( x) + q3 ⋅ p4( x) + q3 ⋅ p5( x)

c3

y4( x) :=

(

3

⋅ q2 ⋅ p1( x) + q2 ⋅ p2( x) + q2 ⋅ p3( x) + q2 ⋅ p4( x) + q2 ⋅ p5( x)

c2

y3( x) :=

2

(

2

3

4

5

⋅ q4 ⋅ p1( x) + q4 ⋅ p2( x) + q4 ⋅ p3( x) + q4 ⋅ p4( x) + q4 ⋅ p5( x)

1 c5

1

(

2

3

4

5

)

)

)

⋅ q5 ⋅ p1( x) + q5 ⋅ p2( x) + q5 ⋅ p3( x) + q5 ⋅ p4( x) + q5 ⋅ p5( x) 1

2

3

4

x := 1 , 1.02.. 2

288

5

)

2

y1 ( x) 1 y2 ( x) y3 ( x) 0 y4 ( x) y5 ( x) −1

−2

1

1.5

2 x

289

2.5

5.45 Consider the eigenvalue problem d2y dy λ 1− x − 2 x + 30 y = y 2 dx 1− x2 dx

(

2

)

which is to be solved on the interval − 1 ≤ x ≤ 1 . (a) Determine the self-adjoint form of the differential operator. (b) What boundary conditions must be specified at x=-1 and x=1? (c) Determine the inner product for eigenvector orthogonality. (d) Determine the eigenvalues and eigenvectors for this problem (Hint: 30=(5)(6)).

Solution: (a) The differential equation is put into the standard Sturm-Liouville form with −2 x

p ( x) = e

∫ 1− x 2

2 = e ln( (1− x ) = 1 − x 2

(a)

leading to −

(

)

(

)

1 d ⎡ dy ⎤ 1− x2 + 30 1 − x 2 y = λy 2 ⎢ dx ⎥⎦ 1 − x dx ⎣

(b)

(b) Since p(x)=0 at x=1 and x=-1, the only conditions requires are that p is finite at x=1 and p is finite at x=-1. (c) Equation b is similar to Equation 3.128, associated Legendre’s equation, with n-=5. The solution is bounded at x=1 and x=-1 if and only if

λ = n 2 , n = 0,1,...

(c)

The bounded solution is of the form y n ( x) = C n P5n ( x )

290

(d)

CHAPTER 6 PARTIAL DIFFERENTIAL EQUATIONS

6.1 The non-dimensional natural frequencies ω and mode shapes φ ( x, y ) of a rectangular membrane are determined from ∂ 2φ ∂ 2φ + 2 + ω 2φ = 0 2 ∂x ∂y

(a)

Determine the natural frequencies and normalized mode shaped for a membrane which is clamped along all along its edge. The appropriate boundary conditions are

φ (0, y ) = 0, φ (1, y ) = 0, φ ( x,0) = 0 and φ ( x, α ) = 0 where α is the ratio of the lengths of the sides of the membrane.

Solution: A product solution is assumed of the form φ ( x, y ) = X ( x)Y ( y ) which when substituted into Equation a leads to X ′′( x)Y ( y ) + X ( x)Y ′′( y ) + ω 2 X ( x)Y ( y ) = 0

(b)

Equation b is rearranged to X ′′ Y ′′ +ω2 = − X Y

(c)

The usual separation argument is used on Equation c, specifying λ as the separation constant. The results are

(

)

X ′′( x) + ω 2 − λ X = 0 Y ′′( y ) + λY = 0

(d) (e)

Use of the separation argument in the boundary conditions leads to X (0) = 0 X (1) = 0 Y(0) = 0 Y(α ) = 0

(f) (g) (h) (i)

The general solution of Equation e is

(

)

(

Y ( y ) = C1 cos λ y + C 2 cos λ y 292

)

(j)

Application of Equations h and I to Equation j leads to Y (0) = 0 ⇒ C1 = 0

(

(k)

)

Y (α ) = 0 ⇒ C 2 sin λα = 0

Non-trivial solutions of Equation l occur if and only if

(l)

λα = kπ where k is a positive

integer. Thus the solutions of Equation e subject to Equations h and i are of the form ⎛ kπ ⎞ Yk ( y ) = C k sin ⎜ y⎟ ⎝α ⎠

(m)

The separation constants are ⎛ kπ ⎞ λk = ⎜ ⎟ ⎝α ⎠

2

(n)

For each λk , a solution of Equation d subject to equations f and g is sought. For convenience define β k = ω 2 − λ k . The general solution of Equation d is

(

)

(

X k ( x) = D1 cos β k x + D2 sin β k x

)

(o)

Application of Equations f and g to Equation o leads to X (0) = 0 ⇒ D1 = 0

(

)

X (1) = 0 ⇒ D2 sin β k = 0

(p) (q)

Thus non-trivial solutions exist if and only if β k = nπ where n is a positive integer. For each k there are an infinite number of solutions of the form X n ( x) = Dn sin (nπx )

(r)

where

β k ,n = (nπ )2 The natural frequencies are

293

(s)

ω k ,n = β k ,n + λ k

(nπ )

=

2

⎛ kπ ⎞ +⎜ ⎟ ⎝α ⎠

2

(t)

The mode shapes are of the form

φ k ,n ( x, y ) = X n ( x)Yk ( y ) ⎛ kπ ⎞ = C k Dn sin ⎜ y ⎟ sin (kπx ) ⎝α ⎠

(u)

Let A k,n = C k Dn The mode shapes are normalized by requiring

(φ

k ,n

, φ k ,n ) = 1

α 1

2

⎡ ⎤ ⎛ kπ ⎞ ∫0 ∫0 ⎢⎣ Ak ,n sin⎜⎝ α y ⎟⎠ sin (nπx )⎥⎦ dxdy = 1 ⎡α 2 ⎛ kπ ⎞ ⎤ ⎡ 1 2 ⎤ A ⎢ ∫ sin ⎜ y ⎟dy ⎥ ⎢ ∫ sin (nπx )dx ⎥ = 1 ⎝ α ⎠ ⎦⎣ 0 ⎣0 ⎦ ⎛ α ⎞⎛ 1 ⎞ Ak2,n ⎜ ⎟⎜ ⎟ = 1 ⎝ 2 ⎠⎝ 2 ⎠ 2 k ,n

(v)

Thus the normalized mode shapes are

φ k , n ( x, y ) =

⎛ kπ ⎞ sin ⎜ y ⎟ sin (kπx ) α ⎝α ⎠

2

294

(w)

6.2 Solve problem 6.1 if the membrane is free along the edge y=0 such that its boundary condition is

∂φ ( x,0) = 0 . ∂y

Solution: A product solution is assumed of the form φ ( x, y ) = X ( x)Y ( y ) which when substituted into Equation a leads to X ′′( x)Y ( y ) + X ( x)Y ′′( y ) + ω 2 X ( x)Y ( y ) = 0

(b)

Equation b is rearranged to X ′′ Y ′′ +ω2 = − X Y

(c)

The usual separation argument is used on Equation c, specifying λ as the separation constant. The results are

(

)

X ′′( x) + ω 2 − λ X = 0 Y ′′( y ) + λY = 0

(d) (e)

Use of the separation argument in the boundary conditions leads to X (0) = 0 X (1) = 0 Y ′(0) = 0 Y(α ) = 0

(f) (g) (h) (i)

The general solution of Equation e is

(

)

(

Y ( y ) = C1 cos λ y + C 2 cos λ y

)

(j)

Application of Equations h and i to Equation j leads to Y (0) = 0 ⇒ C 2 = 0

(

)

Y (α ) = 0 ⇒ C1 cos λα = 0

295

(k) (l)

Non-trivial solutions of Equation l occur if and only if

λα =

(2k − 1)π 2

where k is a

positive integer. Thus the solutions of Equation e subject to Equations h and i are of the form ⎡ (2k − 1)π Yk ( y ) = C k cos ⎢ ⎣ 2α

⎤ y⎥ ⎦

(m)

The separation constants are ⎡ (2k − 1)π ⎤ λk = ⎢ ⎣ 2α ⎥⎦

2

(n)

For each λk , a solution of Equation d subject to equations f and g is sought. For convenience define β k = ω 2 − λ k . The general solution of Equation d is

(

)

(

X k ( x) = D1 cos β k x + D2 sin β k x

)

(o)

Application of Equations f and g to Equation o leads to X (0) = 0 ⇒ D1 = 0

(

)

X (1) = 0 ⇒ D2 sin β k = 0

(p) (q)

Thus non-trivial solutions exist if and only if β k = nπ where n is a positive integer. For each k there are an infinite number of solutions of the form X n ( x) = Dn sin (nπx )

(r)

where

β k ,n = (nπ )2 The natural frequencies are

296

(s)

ω k ,n = β k ,n + λ k

(nπ )

=

2

⎡ (2k − 1)π ⎤ +⎢ ⎣ 2α ⎥⎦

2

(t)

The mode shapes are of the form

φ k ,n ( x, y ) = X n ( x)Yk ( y ) ⎡ (2k − 1)π = C k Dn cos ⎢ ⎣ 2α

⎤ y ⎥ sin (kπx ) ⎦

(u)

Let A k,n = C k Dn The mode shapes are normalized by requiring

(φ

k ,n

, φ k ,n ) = 1

α 1

2

⎡ ⎤ ⎛ (2k − 1)π ⎞ ∫0 ∫0 ⎢⎣ Ak ,n cos⎜⎝ 2α y ⎟⎠ sin (nπx )⎥⎦ dxdy = 1 1 ⎡α ⎤ ⎞ ⎤⎡ 2 ⎛ ( 2k − 1)π A ⎢ ∫ cos ⎜ y ⎟dy ⎥ ⎢ ∫ sin 2 (nπx )dx ⎥ = 1 ⎝ 2α ⎠ ⎦⎣ 0 ⎣0 ⎦ ⎛ α ⎞⎛ 1 ⎞ Ak2,n ⎜ ⎟⎜ ⎟ = 1 ⎝ 2 ⎠⎝ 2 ⎠ 2 k ,n

(v)

Thus the normalized mode shapes are

φ k , n ( x, y ) =

⎛ (2k − 1)π ⎞ cos⎜ y ⎟ sin (kπx ) α ⎝ 2α ⎠

2

297

(w)

6.3 Determine the steady-state temperature distribution, Θ( x, y ) in the thin square slab of Figure P6.3 . The problem is governed by Laplace’s equation subject to the boundary conditions Θ(0, y ) = 1,

∂Θ (1, y ) = 0, Θ( x,0) = 0 and ∂Θ ( x,1) = 0 . ∂x ∂y

Solution: A product solution is assumed of the form Θ( x, y ) = X ( x)Y ( y ) which when substituted into Equation a leads to X ′′( x)Y ( y ) + X ( x)Y ′′( y ) = 0

(b)

Equation b is rearranged to X ′′ Y ′′ =− X Y

(c)

The usual separation argument is used on Equation c, specifying λ as the separation constant. The results are X ′′( x) − λX = 0 Y ′′( y ) + λY = 0

(d) (e)

Use of the separation argument in the boundary conditions leads to X ′(1) = 0

(f)

Y(0) = 0 Y ′(1) = 0

(g) (h)

The general solution of Equation e is

(

)

(

Y ( y ) = C1 cos λ y + C 2 sin λ y

)

(i)

Application of Equations g and h to Equation j leads to Y (0) = 0 ⇒ C1 = 0

( )

Y ′(1) = 0 ⇒ C 2 λ cos λ = 0

298

(j) (k)

λ=

Non-trivial solutions of Equation l occur if and only if

(2k − 1)π where k is a 2

positive integer. Thus the solutions of Equation e subject to Equations h and i are of the form ⎛ (2k − 1)π ⎞ Yk ( y ) = C k sin ⎜ y⎟ 2 ⎝ ⎠

(l)

The separation constants are ⎛ (2k − 1)π ⎞ λk = ⎜ ⎟ 2 ⎝ ⎠

2

(m)

1

The eigenvectors are normalized by requiring

∫ [Y ( y)] dx = 1leading to 2

k

0

⎛ (2k − 1)π ⎞ Yk ( y ) = 2 sin ⎜ y⎟ 2 ⎝ ⎠

(n)

For each λk , a solution of Equation d is sought. The general solution of Equation d is

(

)

(

X k ( x) = D1 cosh λ k x + D2 sinh λ k x

)

(o)

Application of Equation f to Equation o leads to

( )

( )

( )

X ′(1) = 0 ⇒ D1 λ k sinh λ k + D2 λ k cosh λ k = 0 ⇒ D2 = − tanh λ k D1

(p)

and Equation o becomes

[ (

)

(

) (

) (

)]

X k ( x) = Dk cosh λ k x − tanh λ k x sinh λ k x

)]

(q)

The product solution has led to

[ (

)

(

Θ k ( x, y ) = Dk cosh λ k x − tanh λ k x sinh λ k x

299

⎛ (2k − 1)π ⎞ 2 sin ⎜ y⎟ 2 ⎝ ⎠

(r)

Equation r satisfies the partial differential equation and three boundary conditions. The final boundary condition is applied by forming the general solution of ∞

[ (

)

(

) (

θ ( x, y ) = ∑ Dk cosh λk x − tanh λk x sinh λk x k =1

)]

⎛ (2k − 1)π 2 sin ⎜ 2 ⎝

⎞ y⎟ ⎠

(s)

Application of the remaining boundary condition to Equation s leads to

[

(

∞ ∂Θ (1, y ) = 1 = ∑ Dk λk − tanh λk x ∂x k =1

)]

⎛ (2k − 1)π ⎞ 2 sin ⎜ y⎟ 2 ⎝ ⎠

(t)

An eigenvector expansion leads to ∞

1 = ∑ (1, Yk ( y ) )Yk ( y )

(u)

k =1

where 1

(1, Yk ( y) ) = ∫ (1) 0

⎡ (2k − 1)π 2 sin ⎢ 2 ⎣

2 2 =− (2k − 1)π =− =

2 2 (2k − 1)π

⎤ y ⎥ dy ⎦ y =1

⎡ ⎛ (2k − 1)π ⎞⎤ y ⎟⎥ ⎢cos⎜ 2 ⎠⎦ y =0 ⎣ ⎝

⎡ ⎛ (2k − 1)π ⎢cos⎜ 2 ⎣ ⎝

2 2 (2k − 1)π

⎤ ⎞ ⎟ − cos(0)⎥ ⎠ ⎦ (v)

Equations t, u and v lead to Dk = −

4 2

(2k − 1) π tanh⎛⎜ (2k − 1)π ⎞⎟ 2 ⎝ ⎠ 2

2

Use of Equation w in Equation s leads to

300

(w)

θ ( x, y ) =

8

π

2

∞

1

∑ (2k − 1) k =1

2

(

)

⎧⎡ ⎤ ⎛ (2k − 1)π ⎞ ⎛ (2k − 1)π ⎞ ⎛ (2k − 1)π ⎞ x ⎟ − coth ⎜ x ⎟ sinh λ k x ⎥ ⎟ cosh⎜ ⎨⎢sinh ⎜ 2 2 2 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎦ ⎩⎣

⎛ (2k − 1)π ⎞⎫ sin ⎜ y ⎟⎬ 2 ⎝ ⎠⎭

(x)

301

6.4 Solve problem 6.3 if the boundary conditions are Θ(0, y ) = y (1 − y ),

∂Θ (1, y ) = 0, Θ( x,0) = 0 and ∂Θ ( x,1) = 0 . ∂x ∂y

Solution: A product solution is assumed of the form Θ( x, y ) = X ( x)Y ( y ) which when substituted into Equation a leads to X ′′( x)Y ( y ) + X ( x)Y ′′( y ) = 0

(b)

Equation b is rearranged to X ′′ Y ′′ =− X Y

(c)

The usual separation argument is used on Equation c, specifying λ as the separation constant. The results are X ′′( x) − λX = 0 Y ′′( y ) + λY = 0

(d) (e)

Use of the separation argument in the boundary conditions leads to X ′(1) = 0

(f)

Y(0) = 0 Y ′(1) = 0

(g) (h)

The general solution of Equation e is

(

)

(

Y ( y ) = C1 cos λ y + C 2 sin λ y

)

(i)

Application of Equations g and h to Equation j leads to Y (0) = 0 ⇒ C1 = 0

( )

Y ′(1) = 0 ⇒ C 2 λ cos λ = 0

302

(j) (k)

λ=

Non-trivial solutions of Equation l occur if and only if

(2k − 1)π where k is a 2

positive integer. Thus the solutions of Equation e subject to Equations h and i are of the form ⎛ (2k − 1)π ⎞ Yk ( y ) = C k sin ⎜ y⎟ 2 ⎝ ⎠

(l)

The separation constants are ⎛ (2k − 1)π ⎞ λk = ⎜ ⎟ 2 ⎝ ⎠

2

(m)

1

The eigenvectors are normalized by requiring

∫ [Y ( y)] dx = 1leading to 2

k

0

⎛ (2k − 1)π ⎞ Yk ( y ) = 2 sin ⎜ y⎟ 2 ⎝ ⎠

(n)

For each λk , a solution of Equation d is sought. The general solution of Equation d is

(

)

(

X k ( x) = D1 cosh λ k x + D2 sinh λ k x

)

(o)

Application of Equation f to Equation o leads to

( )

( )

( )

X ′(1) = 0 ⇒ D1 λ k sinh λ k + D2 λ k cosh λ k = 0 ⇒ D2 = − tanh λ k D1

(p)

and Equation o becomes

[ (

)

(

) (

) (

)]

X k ( x) = Dk cosh λ k x − tanh λ k x sinh λ k x

)]

(q)

The product solution has led to

[ (

)

(

Θ k ( x, y ) = Dk cosh λ k x − tanh λ k x sinh λ k x

303

⎛ (2k − 1)π ⎞ 2 sin ⎜ y⎟ 2 ⎝ ⎠

(r)

Equation r satisfies the partial differential equation and three boundary conditions. The final boundary condition is applied by forming the general solution of ∞

[ (

)

(

) (

θ ( x, y ) = ∑ Dk cosh λk x − tanh λk x sinh λk x k =1

)]

⎛ (2k − 1)π 2 sin ⎜ 2 ⎝

⎞ y⎟ ⎠

(s)

Application of the remaining boundary condition to Equation s leads to

[

(

∞ ∂Θ (1, y ) = y (1 − y ) = ∑ Dk λk − tanh λk x ∂x k =1

)]

⎛ (2k − 1)π ⎞ 2 sin ⎜ y⎟ 2 ⎝ ⎠

(t)

An eigenvector expansion leads to ∞

y (1 − y ) = ∑ ( y (1 − y ), Yk ( y ) )Yk ( y )

(u)

k =1

where 1

( y (1 − y ), Yk ( y )) = ∫ y (1 − y ) 0

⎡ (2k − 1)π 2 sin ⎢ 2 ⎣

⎡1 ⎡ (2k − 1)π = 2 ⎢ ∫ y sin ⎢ 2 ⎣ ⎣0

⎤ ⎡ (2k − 1)π y ⎥ dy − ∫ y 2 sin ⎢ 2 ⎦ ⎣ 0 1

⎧⎪⎡ 4 ⎡ (2k − 1)π sin ⎢ = 2 ⎨⎢ 2 2 2 ⎪⎩⎣ (2k − 1) π ⎣ ⎡ ⎢ 8y ⎡ (2k − 1)π sin ⎢ −⎢ 2 2 ⎢ (2k − 1) π 2 ⎣ ⎢ ⎢⎣ =

⎤ y ⎥ dy ⎦ ⎤ ⎤ y ⎥ dy ⎥ ⎦ ⎦

2y ⎤ ⎡ (2k − 1)π y⎥ − cos ⎢ 2 ⎣ ⎦ (2k − 1)π

⎡ (2k − 12 )π 2 ⎤ 2 ⎢ ⎥y − 2 4 ⎤ ⎣ ⎡ (2k − 1)π ⎦ y⎥ − cos ⎢ 3 2 ⎦ ⎣ ⎡ (2k − 1)π ⎤ ⎥⎦ ⎢⎣ 2

⎡ ⎤ 4 k ( 1 ) − + ⎢ (2k − 1)π ⎥⎦ (2k − 1) π 2 ⎣ 4 2

2

y =1

⎤⎤ y⎥⎥ ⎦ ⎦ y =0 y =1 ⎤ ⎫ ⎥ ⎪ ⎤⎥ ⎪ y⎥ ⎬ ⎦⎥ ⎪ ⎥ ⎥⎦ y =0 ⎪⎭

(v)

Equations t, u and v lead to Dk =

⎡ ⎤ 4 (−1) k + ⎢ (2k − 1)π ⎥⎦ (2k − 1)3 π 3 tanh⎛⎜ (2k − 1)π ⎞⎟ ⎣ 2 ⎝ ⎠ 8 2

304

(w)

Use of Equation w in Equation s leads to

θ ( x, y ) =

16

π

3

∞

1

∑ (2k − 1) k =1

3

⎡ ⎛ (2k − 1)π ⎢sinh ⎜ 2 ⎝ ⎣

⎡ ⎤ ⎧ ⎛ (2k − 1)π ⎞ 4 k y⎟ ⎢( −1) + (2k − 1)π ⎥ ⎨sin ⎜ 2 ⎠ ⎣ ⎦⎩ ⎝ ⎞ ⎛ (2k − 1)π x ⎟ − coth ⎜ 2 ⎠ ⎝

305

⎞ ⎛ (2k − 1)π ⎟ cosh⎜ 2 ⎠ ⎝

(

)

⎤⎫ ⎞ x ⎟ sinh λ k x ⎥ ⎬ (x) ⎠ ⎦⎭

6.5 Solve problem 6.3 if the boundary conditions are Θ(0, y ) = y (1 − y ),

∂Θ (1, y ) = 0, Θ( x,0) = 0 and ∂Θ ( x,1) + 1.2Θ( x,1) = 0 ∂x ∂y

Solution: A product solution is assumed of the form Θ( x, y ) = X ( x)Y ( y ) which when substituted into Equation a leads to X ′′( x)Y ( y ) + X ( x)Y ′′( y ) = 0

(b)

Equation b is rearranged to X ′′ Y ′′ =− X Y

(c)

The usual separation argument is used on Equation c, specifying λ as the separation constant. The results are X ′′( x) − λX = 0 Y ′′( y ) + λY = 0

(d) (e)

Use of the separation argument in the boundary conditions leads to X ′(1) = 0

(f)

Y(0) = 0 Y ′(1) + 1.2Y(1) = 0

(g) (h)

The general solution of Equation e is

(

)

(

Y ( y ) = C1 cos λ y + C 2 sin λ y

)

(i)

Application of Equations g and h to Equation j leads to Y (0) = 0 ⇒ C1 = 0

( )

( )

Y ′(1) + 1.2Y (1) = 0 ⇒ C 2 λ cos λ + 1.2C 2 sin λ = 0

(j) (k)

Non-trivial solutions of Equation l occur if and only if

( )

tan λ = −0.867 λ

306

(l)

There are an infinite, but countable , number of solutions of Equation l, indexed by λ k for k=1,2,…. Thus the solutions of Equation e subject to Equations h and i are of the form

(

Yk ( y ) = C k sin λ k y

)

(l)

1

The eigenvectors are normalized by requiring

∫ [Y ( y)] dx = 1leading to 2

k

0

∫ [C 1

k

)]

(

2

sin λ k y dy = 1

0

C k2 2

∫ [1 − cos(2

C k2 2

⎡ ⎤ 1 sin 2 λ k ⎥ = 1 ⎢1 − ⎢⎣ 2 λ k ⎥⎦

1

)]

λ k y dy = 1

0

Ck =

(

)

4 λk

(

2 λ k − sin 2 λ k

)

(m)

For each λk , a solution of Equation d is sought. The general solution of Equation d is

(

)

(

X k ( x) = D1 cosh λ k x + D2 sinh λ k x

)

(o)

Application of Equation f to Equation o leads to

( )

( )

( )

X ′(1) = 0 ⇒ D1 λ k sinh λ k + D2 λ k cosh λ k = 0 ⇒ D2 = − tanh λ k D1

(p)

and Equation o becomes

[ (

)

(

) (

X k ( x) = Dk cosh λ k x − tanh λ k x sinh λ k x

)]

(q)

The product solution has led to

[ (

)

(

) (

)] (

Θ k ( x, y ) = C k Dk cosh λ k x − tanh λ k x sinh λ k x sin λ k y

307

)

(r)

Equation r satisfies the partial differential equation and three boundary conditions. The final boundary condition is applied by forming the general solution of

[ (

∞

)

(

) (

)] (

θ ( x, y ) = ∑ C k Dk cosh λk x − tanh λk x sinh λk x sin λk y k =1

)

(s)

Application of the remaining boundary condition to Equation s leads to

[

)] (

(

∞ ∂Θ (1, y ) = y (1 − y ) = ∑ C k Dk λk − tanh λk x sin λk y ∂x k =1

)

(t)

An eigenvector expansion leads to ∞

y (1 − y ) = ∑ ( y (1 − y ), Yk ( y ) )Yk ( y )

(u)

k =1

where 1

[

]

[

]

Ak = ( y (1 − y ), Yk ( y ) ) = ∫ y (1 − y )C k sin λ k y dy 0

[

]

1 ⎤ ⎡ = C k ⎢ ∫ y sin λ k y dy − ∫ y 2 sin λ k y dy ⎥ 0 ⎣0 ⎦ 1

y =1

⎧⎪⎡ 1 ⎤ y cos λ k y ⎥ = 2 ⎨⎢ sin λ k y − λk ⎪⎩⎢⎣ λ k ⎥⎦ y =0

[

]

(

)

y =1 ⎡ ⎤ ⎫ 2 ⎪ λ y −2 2y cos λ k y ⎥⎥ ⎬ − ⎢⎢ sin λ k y − k 3 λ ⎢⎣ k ⎥⎦ y =0 ⎪ λk2 ⎭

[

]

[

]

⎤ ⎛ 32 ⎞ Ck ⎡ = 2 ⎢2 λ k − ⎜⎜ λ k + λ2k − λ k − 2 ⎟⎟ sin λ k ⎥ λk ⎣⎢ ⎥⎦ ⎝ ⎠

( )

(v)

Equations t, u and v lead to Dk = −

308

Ak

( )

λ k tanh λk

(w)

6.6 The unsteady temperature distribution in an extended surface with a large heat transfer coefficient is governed by ∂Θ ∂ 2Θ − (Bi )Θ = 2 ∂t ∂x

(a)

The surface has an initial temperature distribution when the ambient temperature suddenly changes to Θ1 . The left end of the surface is maintained at the new temperature while the right end of the surface is insulated. The boundary conditions are Θ(0, t ) = 0 and

∂Θ (1, t ) = 0 Assume the initial condition is Θ( x,0) = Θ 0 cos(πx ) . ∂x

Solution: A product solution is assumed of the form Θ( x, t ) = X ( x)T ( y ) which when substituted into Equation a leads to X ′′( x)T (t ) − BiX ( x)T (t ) = X ( x)T ′(t )

(b)

Equation b is rearranged to −

X ′′ ⎛T′ ⎞ = −⎜ + Bi ⎟ X ⎝T ⎠

(c)

The usual separation argument is used on Equation c, specifying λ as the separation constant. The results are X ′′( x) + λX = 0 T ′(t ) + (λ + Bi )T (t ) = 0

(d) (e)

Use of the separation argument in the boundary conditions leads to X (0) = 0 X ′(1) = 0

(f) (g)

The general solution of Equation d is

( )

( )

X ( x) = C1 cos λ x + C 2 sin λ x

309

(h)

Application of Equations f and g to Equation h leads to X (0) = 0 ⇒ C1 = 0

(i)

( )

X ′(1) = 0 ⇒ C 2 λ cos λ = 0

Non-trivial solutions of Equation j occur if and only if

(j)

λ=

(2k − 1)π where k is a 2

positive integer. Thus the solutions of Equation d subject to Equations f and g are of the form ⎡ (2k − 1)π X k ( x) = C k sin ⎢ 2 ⎣

⎤ x⎥ ⎦

(k)

The eigenvectors are normalized by 2

⎡ ⎛ (2k − 1)π ⎞⎤ ∫0 ⎢⎣C k sin⎜⎝ 2 x ⎟⎠⎥⎦ dx = 1 ⇒ C k = 2 1

(l)

Thus the normalized eigenvectors are ⎡ (2k − 1)π X k ( x) = 2 sin ⎢ 2 ⎣

⎤ x⎥ ⎦

(m)

For each λk , a solution of Equation d is sought. The general solution of Equation e is Tk (t ) = Dk e − (λk + Bi )t

(n)

The product solution has led to ⎡ (2k − 1)π Θ k ( x, t ) = 2 sin ⎢ 2 ⎣

⎤ x ⎥ Dk e −(λk + Bi )t ⎦

Equation r satisfies the partial differential equation and the boundary conditions. The initial condition is applied by forming the general solution of

310

(o)

∞ ⎡ (2k − 1)π Θ( x, t ) = ∑ 2 sin ⎢ 2 ⎣ k =1

⎤ x ⎥ Dk e −(λk + Bi )t ⎦

(p)

⎤ x ⎥ Dk ⎦

(q)

Application of the initial condition to Equation p leads to ∞ ⎡ (2k − 1)π Θ(0, t ) = Θ 0 cos(πx ) = ∑ 2 sin ⎢ 2 ⎣ k =1

An eigenvector expansion leads to ∞ ⎡ (2k − 1)π Θ 0 cos(πx ) = ∑ Ak 2 sin ⎢ 2 ⎣ k =1

⎤ x⎥ ⎦

(r)

where ⎡ (2k − 1)π Ak = (Θ 0 cos(πx ), X k ( x) ) = ∫ Θ 0 cos(πx) 2 sin ⎢ 2 ⎣ 0 1

⎧ ⎡ (2k − 3)π ⎪⎪ cos ⎢ 2 ⎣ = 2Θ 0 ⎨− (2k − 3)π ⎪ ⎪⎩ =

⎤ ⎡ (2k + 1)π x ⎥ cos ⎢ 2 ⎦− ⎣ (2k + 1)π

⎤ x⎥ ⎦

2 2Θ 0 (k − 1) (2k − 3)(2k + 1)π

⎤ x ⎥ dy ⎦ x =1

⎫ ⎪⎪ −⎬ ⎪ ⎪⎭ x =0 (s)

Equations q, r and s lead to Dk =

2 2Θ 0 (k − 1) (2k − 3)(2k + 1)π

(t)

and Θ( x, t ) =

4Θ 0

π

∞

⎡ (2k − 1)π 2

k −1

∑ (2k − 3)(2k + 1) sin ⎢⎣ k =1

311

⎤ x ⎥ Dk e −(λk + Bi )t ⎦

(u)

6.7 The thin slab of Fig P6.7 initially is at a uniform temperature when a heat flux is applied to one side. Determine the resulting unsteady temperature distribution is ∂ 2 Θ ∂ 2 Θ ∂Θ + = subject to the boundary governed by the partial differential equation ∂t ∂x 2 ∂y 2 conditions Θ(0, y, t ) = 0,

∂Θ ⎛ πy ⎞ (1, y, t ) = sin ⎜ ⎟, Θ(x,0, t) = 0 and Θ( x, α , t ) = 0 and the ∂x ⎝α ⎠

initial condition Θ( x, y,0) = 0 .

Solution: The specified problem is non-homogeneous due to the non-homogeneous boundary condition at x=1. The solution is a superposition of a transient solution and a steady-state solution Θ( x , y , t ) = Θ s ( x , y ) + Θ t ( x , y , t )

(a)

The problem governing the transient solution is ∂ 2 Θ t ∂ 2 Θ t ∂Θ t + = ∂t ∂x 2 ∂y 2

(b)

subject to Θ t (0, y, t ) = 0

(c)

∂Θ t (1, y, t ) = 0 ∂x Θ t (x,0, t) = 0

(e)

Θ t ( x, α , t ) = 0

(f)

Θ t ( x, y,0) = −Θ s ( x, y )

(g)

(d)

The problem governing the steady-state solution is ∂ 2Θ s ∂ 2Θ s + =0 ∂x 2 ∂y 2 subject to

312

(h)

Θ s (0, y ) = 0

(i)

∂Θ s ⎛ πy ⎞ (1, y ) = sin ⎜ ⎟ ∂x ⎝α ⎠ Θ s (x,0) = 0

(k)

Θ s ( x, α ) = 0

(l)

(j)

The problem for the transient response is solved first using separation of variables. Assuming a product solution Θ s ( x, y ) = X ( x)Y ( y ) , substituting into Equation h and using the usual separation argument with a separation parameter of λ leads to X ′′( x) − λX = 0 Y ′′ + λY = 0

(m) (n)

Application of the product solution to the boundary conditions of Equations i, k and l leads to X (0) = 0

(o)

Y (0) = 0 Y (α ) = 0

(p) (q)

The eigenvalues and normalized eigenvectors of the eigenvalue problem defined by Equations n,p and q are ⎛ kπ ⎞ λk = ⎜ ⎟ ⎝α ⎠ Yk ( y ) =

2

(r)

⎛ kπ ⎞ sin ⎜ y⎟ α ⎝α ⎠ 2

(s)

The resulting solution of Equation m which satisfies Equation o is ⎛ kπ ⎞ X k ( x) = Dk sinh ⎜ x⎟ ⎝α ⎠

(t)

The general solution is of the form ∞

Θ s ( x, y ) = ∑ k =1

⎛ kπ ⎞ ⎛ kπ Dk sin ⎜ y ⎟ sinh ⎜ α ⎝α ⎠ ⎝α 2

313

⎞ x⎟ ⎠

(u)

However application of Equation j leads to

α

1 2 ⎛π ⎞ sinh ⎜ ⎟ ⎝α ⎠ Dk = 0 k ≠ 1 D1 =

(v)

(w)

and thus ⎛π ⎞ ⎛π sin ⎜ y ⎟ sinh ⎜ ⎝α ⎠ ⎝α Θ s ( x, y ) = ⎛π ⎞ sinh ⎜ ⎟ ⎝α ⎠

⎞ x⎟ ⎠

(x)

The transient solution is obtained using separation of variables with a product solution of Θ t ( x, y, t ) = Φ ( x, y )T (t ) which when substituted into Equation b, rearranging and using the usual separation argument with a separation parameter μ leads to ∂ 2Φ ∂ 2Φ + 2 + μΦ = 0 ∂x 2 ∂y dT + μT = 0 dt

(y) (z)

Use of the product solution in the boundary conditions, Equations c-f leads to Φ (0, y ) = 0 ∂Φ (1, y ) = 0 ∂x Φ (x,0) = 0 Φ ( x, α ) = 0

(aa) (bb) (cc) (dd)

Separation of variables is used to determine the solution of the eigenvalue problem of Equations y and aa-dd. To this end Φ ( x, y ) = X ( x)Y ( x) which when substituted into Equation y, rearranging and using the usual separation argument with a separation parameter κ leads to

314

X ′′ + (μ − κ )X = 0 Y ′′ + κY = 0

(ee) (ff)

Substitution of the product solution into the boundary conditions leads to X (0) = 0 X ′(1) = 0

(gg) (hh)

Y (0) = 0 Y (α ) = 0

(ii) (jj)

The eigenvalues and normalized eigenvectors obtained through solving Equation ff subject to Equations ii and jj are

⎛ kπ ⎞ κk = ⎜ ⎟ ⎝ α ⎠ Yk ( y ) =

2

(kk)

⎛ kπ ⎞ y⎟ sin ⎜ α ⎝ α ⎠ 2

(ll)

Defining β = μ − κ , the solutions of the eigenvalue problem defined by Equations ff-hh are

⎡ (2n − 1)π ⎤ βn = ⎢ ⎥⎦ 2 ⎣ ⎡ (2n − 1)π X n ( x) = 2 sin ⎢ 2 ⎣ 2

(mm) ⎤ x⎥ ⎦

(nn)

Thus the eigenvalues and eigenvectors for Φ( x, y ) are of the form

⎛ kπ ⎞ ⎡ (2n − 1)π ⎤ μ k ,n = ⎜ ⎟ + ⎢ ⎥⎦ 2 ⎝α ⎠ ⎣ 2 ⎡ (2n − 1)π sin ⎢ Φ k,n ( x, y ) = 2 α ⎣ 2

2

(oo) ⎤ ⎛ kπ ⎞ x ⎥ sin⎜ y⎟ ⎦ ⎝α ⎠

(pp)

The solution of Equation z is Tk ,n = Dk ,n e

315

− μ k , nt

(qq)

Thus the general form of the transient solution is ∞

∞

Θ t = ∑∑ k =1 n =1

⎡ (2n − 1)π Dk ,n sin ⎢ 2 α ⎣

2

⎤ ⎛ kπ ⎞ − μ k , nt x ⎥ sin ⎜ y ⎟e ⎦ ⎝α ⎠

(rr)

Application of the initial condition leads to ⎛π ⎞ ⎛π ⎞ sin ⎜ y ⎟ sinh ⎜ x ⎟ ∞ ∞ 2 ⎡ (2n − 1)π ⎤ ⎛ kπ ⎞ ⎝α ⎠ ⎝α ⎠ − = ∑∑ Dk ,n sin ⎢ x ⎥ sin ⎜ y⎟ 2 ⎛π ⎞ α ⎣ ⎦ ⎝α ⎠ k =1 n =1 sinh ⎜ ⎟ ⎝α ⎠ From Equation ss it is clear that Dk ,n = 0 for all k>1. For k=1 ⎛π ⎞ sinh ⎜ x ⎟ ∞ 2 ⎡ (2n − 1)π ⎝α ⎠ = D1,n sin ⎢ − ∑ 2 ⎛π ⎞ ⎣ n =1 α sinh ⎜ ⎟ ⎝α ⎠

⎤ x⎥ ⎦

(ss)

(tt)

An eigenvector expansion leads to ⎛π ⎞ ∞ sinh ⎜ x ⎟ = ∑ Bk X n ( x) ⎝ α ⎠ n =1

(uu)

where

⎡ (2n − 1)π ⎛π ⎞ Bk = ∫ sinh⎜ x ⎟ 2 sin ⎢ 2 ⎣ ⎝α ⎠ 0 1

=

⎤ x ⎥ dx ⎦

2 ⎡ (− 1)n+1 π sinh⎛⎜ π ⎞⎟ + (2n − 1)π cosh⎛⎜ π ⎞⎟⎤⎥ ⎢ 2 μ1,n ⎣ α ⎝α ⎠ ⎝ α ⎠⎦

(vv)

Then from Equations tt and uu Di ,n = −

=−

α

Bn 2 ⎛π ⎞ sinh ⎜ ⎟ ⎝α ⎠

α ⎡ μ1,n

⎢(− 1) ⎣

n +1

π (2n − 1)π ⎛ π ⎞⎤ coth⎜ ⎟⎥ + 2 α ⎝ α ⎠⎦

and

316

(ww)

1 ⎡ (− 1)n+1 π + (2n − 1)π coth⎛⎜ π ⎞⎟⎤⎥ sin ⎡⎢ (2n − 1)π ⎢ α 2 2 ⎝ α ⎠⎦ ⎣ n =1 μ1, n ⎣ ∞

Θ t ( x, y , t ) = − 2 ∑

⎤ ⎛ π ⎞ −μ t x ⎥ sin ⎜ y ⎟e 1, n (xx) ⎦ ⎝α ⎠

Finally 1 ⎡ (− 1)n +1 π + (2n − 1)π coth⎛⎜ π ⎞⎟⎤⎥ sin ⎡⎢ (2n − 1)π ⎢ 2 2 α ⎝ α ⎠⎦ ⎣ n =1 μ 1, n ⎣ ∞

Θ( x , y , t ) = − 2 ∑ ⎛π sin ⎜ α + ⎝

⎞ ⎛π y ⎟ sinh ⎜ ⎠ ⎝α ⎛π ⎞ sinh ⎜ ⎟ ⎝α ⎠

⎞ x⎟ ⎠

⎤ ⎛ π ⎞ −μ t x ⎥ sin ⎜ y ⎟e 1, n ⎦ ⎝α ⎠

(yy)

317

6.8 A thin slab is initially at a uniform temperature when the temperature of one side is suddenly changed. Determine the resulting unsteady temperature distribution which is ∂ 2 Θ ∂ 2 Θ ∂Θ + = subject to the boundary governed by the partial differential equation ∂t ∂x 2 ∂y 2 conditions Θ(0, y, t ) = 1,

∂Θ (1, y, t ) = 0, Θ(x,0, t) = 0 and Θ( x, α , t ) = 0 and the initial ∂x

condition Θ( x, y,0) = 0 .

Solution: The specified problem is non-homogeneous due to the non-homogeneous boundary condition at x=1. The solution is a superposition of a transient solution and a steady-state solution Θ( x , y , t ) = Θ s ( x , y ) + Θ t ( x , y , t )

(a)

The problem governing the transient solution is ∂ 2 Θ t ∂ 2 Θ t ∂Θ t + = ∂t ∂x 2 ∂y 2

(b)

subject to Θ t (0, y, t ) = 0

(c)

∂Θ t (1, y, t ) = 0 ∂x Θ t (x,0, t) = 0

(e)

Θ t ( x, α , t ) = 0

(f)

Θ t ( x, y,0) = −Θ s ( x, y )

(g)

(d)

The problem governing the steady-state solution is ∂ 2Θ s ∂ 2Θ s + =0 ∂x 2 ∂y 2 subject to

318

(h)

Θ s (0, y ) = 1

(i)

∂Θ s (1, y ) = 0 ∂x Θ s (x,0) = 0

(k)

Θ s ( x, α ) = 0

(l)

(j)

The problem for the transient response is solved first using separation of variables. Assuming a product solution Θ s ( x, y ) = X ( x)Y ( y ) , substituting into Equation h and using the usual separation argument with a separation parameter of λ leads to X ′′( x) − λX = 0 Y ′′ + λY = 0

(m) (n)

Application of the product solution to the boundary conditions of Equations i, k and l leads to X ′(1) = 0

(o)

Y (0) = 0 Y (α ) = 0

(p) (q)

The eigenvalues and normalized eigenvectors of the eigenvalue problem defined by Equations n,p and q are ⎛ kπ ⎞ λk = ⎜ ⎟ ⎝α ⎠ Yk ( y ) =

2

(r)

⎛ kπ ⎞ sin ⎜ y⎟ α ⎝α ⎠ 2

(s)

The resulting solution of Equation m which satisfies Equation o is ⎡ ⎡ kπ X k ( x) = Dk ⎢cosh ⎢ ⎣α ⎣

⎛ kπ ⎤ x ⎥ − tanh⎜ ⎦ ⎝α

The general solution is of the form

319

⎞ ⎛ kπ ⎟ sinh ⎜ ⎠ ⎝α

⎞⎤ x ⎟⎥ ⎠⎦

(t)

⎛ kπ ⎞ ⎡ ⎡ kπ Dk sin ⎜ y ⎟ ⎢cosh ⎢ α ⎝ α ⎠⎣ ⎣α

∞

⎤ ⎛ kπ ⎞ ⎛ kπ ⎞⎤ x ⎥ − tanh⎜ x ⎟⎥ ⎟ sinh ⎜ ⎦ ⎝α ⎠ ⎝ α ⎠⎦

2

Θ s ( x, y ) = ∑ k =1

(u)

However application of Equation j leads to ⎛ kπ ⎞ Dk sin ⎜ y⎟ α ⎝α ⎠

∞

2

1= ∑ k =1

(v)

An eigenvector expansion leads to ⎛ kπ ⎞ sin ⎜ y⎟ α ⎝α ⎠

∞

2

1 = ∑ Ak k =1

(w)

where α

Ak = ∫ (1) 0

⎛ kπ ⎞ y ⎟dy sin ⎜ α ⎝α ⎠ 2

2 α =− α kπ =−

y =1

⎡ ⎛ kπ ⎞⎤ ⎢cos⎜ α y ⎟⎥ ⎠⎦ y =1 ⎣ ⎝

2α (−1) k − 1 kπ

[

]

(x)

Noting from Equations v and w that Dk = Ak the steady-state solution is Θ s ( x, y ) =

2

π

∞

∑ k =1

[1 − (−1) ] sin⎛⎜ kπ y ⎞⎟⎡cosh ⎡ kπ x⎤ − tanh⎛⎜ kπ ⎞⎟ sinh⎛⎜ kπ x ⎞⎟⎤ k

k

⎝α

⎢ ⎠⎣

⎢α ⎣

⎥ ⎦

⎝α ⎠

⎝α

⎥ ⎠⎦

(v)

The transient solution is obtained using separation of variables with a product solution of Θ t ( x, y, t ) = Φ ( x, y )T (t ) which when substituted into Equation b, rearranging and using the usual separation argument with a separation parameter μ leads to ∂ 2Φ ∂ 2Φ + 2 + μΦ = 0 ∂y ∂x 2 dT + μT = 0 dt

320

(y) (z)

Use of the product solution in the boundary conditions, Equations c-f leads to Φ (0, y ) = 0 ∂Φ (1, y ) = 0 ∂x Φ (x,0) = 0 Φ ( x, α ) = 0

(aa) (bb) (cc) (dd)

Separation of variables is used to determine the solution of the eigenvalue problem of Equations y and aa-dd. To this end Φ ( x, y ) = X ( x)Y ( x) which when substituted into Equation y, rearranging and using the usual separation argument with a separation parameter κ leads to X ′′ + (μ − κ )X = 0 Y ′′ + κY = 0

(ee) (ff)

Substitution of the product solution into the boundary conditions leads to X (0) = 0 X ′(1) = 0 Y (0) = 0 Y (α ) = 0

(gg) (hh) (ii) (jj)

The eigenvalues and normalized eigenvectors obtained through solving Equation ff subject to Equations ii and jj are

⎛ kπ ⎞ κk = ⎜ ⎟ ⎝ α ⎠ Yk ( y ) =

2

⎛ kπ ⎞ y⎟ sin ⎜ α ⎝ α ⎠ 2

(kk) (ll)

Defining β = μ − κ , the solutions of the eigenvalue problem defined by Equations ff-hh are

321

⎡ (2n − 1)π ⎤ βn = ⎢ ⎥⎦ 2 ⎣ ⎡ (2n − 1)π X n ( x) = 2 sin ⎢ 2 ⎣ 2

(mm) ⎤ x⎥ ⎦

(nn)

Thus the eigenvalues and eigenvectors for Φ ( x, y ) are of the form

⎛ kπ ⎞ ⎡ (2n − 1)π ⎤ μ k ,n = ⎜ ⎟ + ⎢ ⎥⎦ 2 ⎝α ⎠ ⎣ 2 ⎡ (2n − 1)π sin ⎢ Φ k,n ( x, y ) = 2 α ⎣ 2

2

(oo) ⎤ ⎛ kπ ⎞ x ⎥ sin⎜ y⎟ α ⎦ ⎝ ⎠

(pp)

The solution of Equation z is Tk ,n = Dk ,n e

− μ k , nt

(qq)

Thus the general form of the transient solution is ∞

∞

Θ t = ∑∑

⎡ (2n − 1)π Dk ,n sin ⎢ 2 α ⎣

2

k =1 n =1

⎤ ⎛ kπ ⎞ − μ k , nt x ⎥ sin ⎜ y ⎟e ⎦ ⎝α ⎠

(rr)

Application of the initial condition leads to −

2

π

∑

[1 − (−1) ]sin⎛⎜ kπ y ⎞⎟⎡cosh ⎡ kπ x⎤ − tanh⎛⎜ kπ ⎞⎟ sinh⎛⎜ kπ x ⎞⎟⎤

∞

∞

∞

k

k

k =1

= ∑∑ k =1 n =1

⎝α

⎢ ⎠⎣

⎡ (2n − 1)π Dk ,n sin ⎢ 2 α ⎣

2

⎢⎣ α

⎥⎦

⎝α ⎠

⎤ ⎛ kπ ⎞ x ⎥ sin ⎜ y⎟ ⎦ ⎝α ⎠

⎝α

⎥ ⎠⎦ (ss)

An eigenvector expansion can be developed as ⎡ ⎡ kπ ⎢cosh ⎢ α ⎣ ⎣

⎛ kπ ⎞⎤ ∞ ⎡ (2n − 1)π ⎤ ⎛ kπ ⎞ x ⎥ − tanh⎜ x ⎟⎥ = ∑ Bk ,n 2 sin ⎢ ⎟ sinh ⎜ 2 ⎦ ⎝α ⎠ ⎝ α ⎠⎦ n =1 ⎣

where

322

⎤ x⎥ ⎦

(tt)

⎡ ⎡ kπ Bk ,n = ∫ ⎢cosh ⎢ ⎣α 0 ⎣ 1

=

⎡ (2n − 1)π ⎛ kπ ⎞⎤ ⎤ ⎛ kπ ⎞ x ⎥ − tanh ⎜ x ⎟⎥ 2 sin ⎢ ⎟ sinh ⎜ 2 ⎣ ⎝ α ⎠⎦ ⎦ ⎝α ⎠

⎤ x ⎥ dx ⎦

2 ⎧ kπ ⎛ kπ ⎞ n +1 ⎡ 2 ⎛ kπ ⎞ ⎤ ⎟⎥ ⎟ − sinh ⎜ ⎨ (− 1) ⎢cosh ⎜ μ k ,n ⎩ α ⎝ α ⎠⎦ ⎝α ⎠ ⎣ +

(2n − 1)π 2

⎡ ⎛ kπ ⎢sinh ⎜ α ⎝ ⎣

⎞ ⎛ kπ ⎟ − cosh⎜ ⎠ ⎝α

⎞⎤ ⎫ ⎟⎥ ⎬ ⎠⎦ ⎭

(uu)

Then from Equations ss and tt Dk , n = −

2

π

[

2 Bk , n 1 − (−1) k α k

]

(vv)

Finally ⎡ (2n − 1)π ⎤ ⎛ kπ ⎞ − μ k , nt + Dk ,n sin ⎢ x ⎥ sin ⎜ y ⎟e 2 ⎣ ⎦ ⎝α ⎠ k =1 n =1 α 2 ∞ 1 − (−1) k ⎛ kπ ⎞ ⎡ ⎡ kπ ⎤ ⎛ kπ ⎞ ⎛ kπ ⎞⎤ sin ⎜ y ⎟ ⎢cosh ⎢ x ⎥ − tanh⎜ x ⎟⎥ ⎟ sinh ⎜ ∑ π k =1 k ⎝ α ⎠⎣ ⎣α ⎦ ⎝α ⎠ ⎝ α ⎠⎦ ∞

∞

Θ t = ∑∑

[

2

]

323

(ww)

6.9 A thin slab is initially at a temperature f(x,y) when the temperature of two parallel sides is changed to the same temperature while the other two sides remain insulated. Determine the resulting unsteady temperature distribution which is governed by the partial differential equation

Θ(0, y, t ) = 0, Θ(1, y, t ) = 0,

∂ 2 Θ ∂ 2 Θ ∂Θ + = subject to the boundary conditions ∂t ∂x 2 ∂y 2 ∂Θ ∂Θ (x,0, t) = 0 and ( x, α , t ) = 0 and the initial condition ∂y ∂y

Θ( x, y,0) = f ( x, y ) .

Solution: A product solution is assumed as Θ( x, y, t ) = Φ ( x, y )T (t ) which when substituted into the partial differential equation, after rearranging and using the usual separation argument with a separation constant of μ leads to ∂ 2Φ ∂ 2Φ + 2 + μΦ = 0 ∂y ∂x 2 dT + μT = 0 dt

(a) (b)

Use of the product solution in the boundary conditions leads to Φ (0, y ) = 0

(c)

Φ (1, y ) = 0

(d)

∂Φ ( x,0) = 0 ∂y ∂Φ ( x, α ) = 0 ∂y

(e) (f)

A product solution is applied to determine the eigenvalues and eigenvectors of the problem defined by Equation a and Equations c-f. Assume Φ ( x, y ) = X ( x)Y ( y ) . Substituting the product solution into Equation a, rearranging and using the usual separation argument with κ as the separation constant leads to

324

X ′′ + κX = 0 Y ′′ + ( μ − κ )Y = 0

(g) (h)

Use of the product solution in the boundary conditions leads to X (0) = 0 X (1) = 0 Y ′(0) = 0 Y ′(α ) = 0

(i) (j) (k) (l)

The eigenvalues and normalized eigenvectors of the problem defined by Equations g, i and j are

κ k = (kπ )2

(m)

X k ( x) = 2 sin (kx )

(n)

Defining β = μ − κ the solutions of Equation h subject to Equations k and l are 1

Y0 ( x) =

(o)

α ⎛ nπ ⎞ cos⎜ y ⎟ n = 1,2,... α ⎝α ⎠ 2

Yn ( x) =

(p)

with ⎛ nπ ⎞ ⎟ ⎝α ⎠

βn = ⎜

2

n = 0,1,2,...

(q)

⎛ nπ ⎞ +⎜ ⎟ ⎝α ⎠

(r)

Thus the eigenvalues are

μ k ,n = (kπ )

2

with corresponding normalized eigenvectors

325

2

2

Φ k ,0 =

α

sin( kπx )

⎛ nπ ⎞ y⎟ sin( kπx ) sin ⎜ α ⎝ α ⎠

2

Φ k ,n =

(s)

The solution of Equation b, for a specific eigenvalue is Tk ,n (t ) = Dk ,n e

− μ k , nt

(t)

The most general solution of the partial differential equation obtained form the product solution is ∞

Θ( y , y , t ) = ∑ D k , 0 k =1

∞

sin (kπx )e −(kπ ) t

2

2

α

⎛ nπ ⎞ sin( kπx) sin ⎜ y ⎟e α ⎝α ⎠

∞

2

+ ∑∑ Dk ,n k =1 n =1

2 ⎡ ⎛ nπ ⎞ ⎤ − ⎢ ( kπ ) 2 + ⎜ ⎟ ⎥t ⎢⎣ ⎝ α ⎠ ⎥⎦

(t)

Application of the initial condition to Equation t leads to ∞

2

f ( x, y ) = ∑ Dk , 0

α

k =1

∞

sin (kπx )

∞

+ ∑∑ Dk ,n k =1 n =1

⎛ nπ ⎞ sin( kπx ) sin ⎜ y⎟ α ⎝α ⎠

2

(u)

An eigenvector expansion of f(x,y) leads to ∞

2

f ( x, y ) = ∑ B k , 0

α

k =1

∞

∞

sin (kπx )

+ ∑∑ Bk ,n k =1 n =1

⎛ nπ ⎞ sin(kπx) sin ⎜ y⎟ α ⎝α ⎠

2

where

326

(v)

α

Bk , 0 = ∫

0

α

Bk , n = ∫

0

1

∫

0 1

∫

0

f ( x, y ) f ( x, y )

2

α

sin( kπx)dxdy

⎛ nπ ⎞ sin(kπx) sin ⎜ y ⎟dxdy α ⎝α ⎠

2

Finally it is noted from Equations u and v that Dk ,n = Bk ,n .

327

(w) n>0

(x)

6.10 The thin shaft of Fig. P6.10 is initially subject to a torque at its end which is suddenly removed. Determine the resulting response of the shaft which is governed by ∂ 2Θ ∂ 2Θ = 2 subject to boundary the non-dimensional partial differential equation ∂x 2 ∂t conditions Θ(0, t ) = 0 and

constant and

∂Θ (1, t ) = 0 and the initial conditions Θ( x,0 ) = Cx where C is a ∂x

∂Θ ( x,0) = 0 . ∂t

Solution: A product solution is assumed as Θ( x, t ) = X ( x)T (t ) . Substitution of the product solution into the partial differential equation with subsequent rearrangement and use of the usual separation argument with λ as the separation constant leads to X ′′( x) + λX = 0 T ′′ + λT = 0

(a) (b)

Use of the product solution in the boundary and initial conditions leads to X (0) = 0 X ′(1) = 0 T ′(0) = 0

(c) (d) (e)

The solutions of Equation a satisfying Equations c and d are ⎡ (2k − 1)π X k ( x) = 2 sin ⎢ 2 ⎣ ⎡ (2k − 1)π ⎤ ⎥⎦ 2

⎤ x⎥ ⎦

(f)

2

λk = ⎢ ⎣

(g)

For each k the solution of Equation b which satisfies Equation e is ⎡ (2k − 1)π Tk (t ) = Dk cos ⎢ 2 ⎣ The general solution, formed from the product solution, is

328

⎤ t⎥ ⎦

(h)

∞ ⎡ (2k − 1)π Θ( x, t ) = ∑ 2 Dk sin ⎢ 2 ⎣ k =1

⎤ ⎡ (2k − 1)π x ⎥ cos ⎢ 2 ⎦ ⎣

⎤ t⎥ ⎦

(i)

Application of the remaining initial condition leads to ∞ ⎡ (2k − 1)π Cx = ∑ 2 Dk sin ⎢ 2 ⎣ k =1

⎤ x⎥ ⎦

(j)

⎤ x⎥ ⎦

(k)

An eigenvector expansion leads to ∞ ⎡ (2k − 1)π Cx = ∑ 2 Bk sin ⎢ 2 ⎣ k =1

where ⎡ (2k − 1)π Bk = ∫ (Cx ) 2 sin ⎢ 2 ⎣ 0 1

⎤ x ⎥ dx ⎦

4 2C (− 1) = 2 π (2k − 1) 2

k +1

(l)

Comparison of Equation j with Equation k shows that Dk = Bk . The resulting solution is Θ( x, t ) =

(−1) k +1 ⎡ (2k − 1)π sin ⎢ 2 ∑ 2 2 π k =1 (2k − 1) ⎣ 8C

∞

329

⎤ ⎡ (2k − 1)π x ⎥ cos ⎢ 2 ⎦ ⎣

⎤ t⎥ ⎦

(m)

Problems 6.12-6.16 refer to the system of Fig. P6.12 The beam is subject to a concentrated load at its end which leads to a static deflection of the form ⎛ x 2 x3 ⎞ w0 ( x) = α ⎜⎜ − ⎟⎟ ⎝ 2 6⎠

(a)

When the load is removed the resulting vibrations are governed by

∂4w ∂2w =0 + ∂x 4 ∂t 2

(b)

The boundary conditions are w(0, t ) = 0 ∂w (0, t ) = 0 ∂x ∂3w ∂2w t + w t + ( 1 , ) η ( 1 , ) μ (1, t ) 1 ∂x 3 αt 2 ∂2w ∂3w (1, t ) + μ 2 ∂x 2 ∂x∂t 2

(c) (d) (e) (f)

The initial conditions are ⎛ x2 x3 ⎞ w( x,0) = α ⎜⎜ − ⎟⎟ 6 ⎠ ⎝ 2 ∂w (x,0) = 0 ∂t

(g) (h)

6.12 Determine w(x,t) if η = μ1 = μ 2 = 0 .

Solution: Separation of variables is used with a product solution of w( x, t ) = X ( x)T (t ) . Substitution into Equation b, rearranging and using the usual separation argument with a separation parameter of λ leads to

330

X iv − λX = 0 T ′′ + λT = 0

(i) (j)

Use of the product solution in the boundary and initial conditions leads to X (0) = 0 X ′(0) = 0 X ′′(1) = 0 X ′′′(1) = 0 T ′(0) = 0

(k) (l) (m) (n) (o)

The general solution of Equation i is X ( x) = C1 cosh (βx ) + C 2 sinh( βx) + C 3 cos( β x) + C 4 sin( β x)

(p)

1 4

where β = λ . Use of the boundary conditions in Equation p leads to X (0) = 0 ⇒ C1 + C 3 = 0 X ′(0) = 0 ⇒ β C 2 + β C 4 = 0

(q) (r)

X ′′(1) = 0 ⇒ β 2 C1 cosh( β ) + β 2 C 2 sinh (β ) − β 2 C 3 cos(β ) − β 2 C 4 sin (β ) = 0

(s)

X ′′′(1) = 0 ⇒ β 3 C1 sinh( β ) + β 3 C 2 cosh( β ) + β 3C 3 sin( β ) − β 3C 4 cos(β ) = 0

(t)

Equation q implies C 3 = −C1 while Equation r implies C 4 = −C 2 . Equations s and t are then summarized in a matrix form as ⎡cosh (β ) + cos(β ) sinh (β ) + sin (β ) ⎤ ⎡ C1 ⎤ ⎡0⎤ ⎢ sinh (β ) − sin (β ) cosh (β ) + cos(β )⎥ ⎢C ⎥ = ⎢0⎥ ⎣ ⎦⎣ 2 ⎦ ⎣ ⎦

(u)

A non-trivial solution of Equation u exists if and only if the determinant of the coefficient matrix is zero. To this end cosh (β ) + cos(β ) sinh (β ) + sin (β ) = 0 = [cosh (β ) + cos(β )][cosh (β ) + cos(β )] sinh (β ) − sin (β ) cosh (β ) + cos(β ) − [sinh (β ) + sin (β )][sinh (β ) − sin (β )]

cosh β cos β = −1

(v)

331

There are an infinite, but countable, number of solutions of Equation v, β k k=1,2,…. For a specific β k , it is noted from Equation u that

C2 = −

coshβ k + cosβ k C1 sinhβ k + sin β k

(w)

The eigenvector is thus ⎤ ⎡ cosh β k + cos β k (sinh β k x − sin β k x )⎥ X k( x) = C k ⎢cosh β k x − cos β k x − sinh β k + sin β k ⎦ ⎣

(x)

The eigenvectors are normalized by requiring Ck =

1 2

⎡ ⎤ cosh β k + cos β k ∫0 ⎢⎣cosh β k x − cos β k x − sinh β k + sin β k (sinh β k x − sin β k x )⎥⎦ dx 1

(y)

The solution of Equation j is

(

)

(

T k ( t ) = D 1, k cos β k2 t + D 2 , k sin β k2 t

)

(z)

Application of the homogeneous initial condition to Equation z leads to Tk′ (0) = 0 ⇒ D2,k = 0

(aa)

The general form of the solution obtained from the product solution is ∞

∑D k =1

1, k

⎤ ⎡ cosh β k + cos β k (sinh β k x − sin β k x )⎥ cos(β k2 t ) C k ⎢cosh β k x − cos β k x − sinh β k + sin β k ⎦ ⎣

(bb)

Application of the remaining initial condition leads to ⎤ ⎡ ⎛ x2 x3 ⎞ ∞ cosh β k + cos β k (sinh β k x − sin β k x )⎥ − ⎟⎟ = ∑ D1,k C k ⎢cosh β k x − cos β k x − 6 ⎠ k =1 sinh β k + sin β k ⎝ 2 ⎦ ⎣

α ⎜⎜

Use of an eigenvector expansion leads to

332

(cc)

1 ⎤ ⎛ x2 x3 ⎞ ⎡ cosh β k + cos β k (sinh β k x − sin β k x )⎥ dx Dk = ∫ α ⎜⎜ − ⎟⎟C k ⎢cosh β k x − cos β k x − 6 ⎠ ⎣ sinh β k + sin β k ⎝ 2 ⎦ 0

333

(dd)

6.13 Determine w(x,t) if η = 2 and μ1 = μ 2 = 0 .

Solution: Separation of variables is used with a product solution of w( x, t ) = X ( x)T (t ) . Substitution into Equation b, rearranging and using the usual separation argument with a separation parameter of λ leads to X iv − λX = 0 T ′′ + λT = 0

(i) (j)

Use of the product solution in the boundary and initial conditions leads to X (0) = 0 X ′(0) = 0 X ′′(1) = 0 X ′′′(1) + 2 X (1) = 0 T ′(0) = 0

(k) (l) (m) (n) (o)

The general solution of Equation i is X ( x) = C1 cosh (βx ) + C 2 sinh( βx) + C 3 cos( β x) + C 4 sin( β x)

(p)

1 4

where β = λ . Use of the boundary conditions in Equation p leads to X (0) = 0 ⇒ C1 + C3 = 0 X ′(0) = 0 ⇒ βC 2 + β C 4 = 0

(q) (r)

X ′′(1) = 0 ⇒ β 2 C1 cosh( β ) + β 2 C 2 sinh (β ) − β 2 C 3 cos(β ) − β 2 C 4 sin (β ) = 0

(s)

X ′′′(1) + 2 X (1) = 0 ⇒ β 3 C1 sinh( β ) + β 3 C 2 cosh( β ) + β 3 C 3 sin( β ) − β 3C 4 cos(β ) + 2C1 cosh (β ) + 2C 2 sinh (β ) + 2C 3 cos( β ) + 2C 4 sin( β ) = 0

Equation q implies C 3 = −C1 while Equation r implies C 4 = −C 2 . Equations s and t are then summarized in a matrix form as

334

(t)

cosh (β ) + cos(β ) sinh (β ) + sin (β ) ⎡ ⎤ ⎡ C1 ⎤ ⎢ β 3 [sinh (β ) − sin (β )] + 2[cosh (β ) − cos(β )] β 3 [cosh (β ) + cos(β )] + 2[sinh (β − sin (β ))]⎥ ⎢C ⎥ ⎣ ⎦⎣ 2 ⎦ ⎡0 ⎤ =⎢ ⎥ (u) ⎣0 ⎦

A non-trivial solution of Equation u exists if and only if the determinant of the coefficient matrix is zero. To this end cosh (β ) + cos(β ) sinh (β ) + sin (β ) =0 β 3 [sinh (β ) − sin (β )] + 2[cosh (β ) − cos(β )] β 3 [cosh (β ) + cos(β )] + 2[sinh (β − sin (β ))]

{

}

= [cosh (β ) + cos(β )] β 3 [cosh (β ) + cos(β )] + 2[sinh (β − sin (β ))]

{

}

− [sinh (β ) + sin (β )] β 3 [sinh (β ) − sin (β )] + 2[cosh (β ) − cos(β )]

β 3 [1 + cosh β cos β ] = 2[cosh β sin β − sinh β cos β ]

(v)

There are an infinite, but countable, number of solutions of Equation v, β k k=1,2,…. For a specific β k , it is noted from Equation u that C2 = −

cosh β k + cos β k C1 sinh β k + sin β k

(w)

The eigenvector is thus ⎤ ⎡ cosh β k + cos β k (sinh β k x − sin β k x )⎥ X k( x) = C k ⎢cosh β k x − cos β k x − sinh β k + sin β k ⎦ ⎣

(x)

The eigenvectors are normalized by requiring Ck =

1 2

⎡ ⎤ cosh β k + cos β k ∫0 ⎢⎣cosh β k x − cos β k x − sinh β k + sin β k (sinh β k x − sin β k x )⎥⎦ dx 1

(y)

The solution of Equation j is

(

)

(

T k ( t ) = D 1, k cos β k2 t + D 2 , k sin β k2 t

335

)

(z)

Application of the homogeneous initial condition to Equation z leads to Tk′ (0) = 0 ⇒ D2,k = 0

(aa)

The general form of the solution obtained from the product solution is ∞

∑D k =1

1, k

⎤ ⎡ cosh β k + cos β k (sinh β k x − sin β k x )⎥ cos β k2 t C k ⎢cosh β k x − cos β k x − sinh β k + sin β k ⎦ ⎣

( )

(bb)

Application of the remaining initial condition leads to ⎤ ⎡ ⎛ x2 x3 ⎞ ∞ cosh β k + cos β k (sinh β k x − sin β k x )⎥ − ⎟⎟ = ∑ D1,k C k ⎢cosh β k x − cos β k x − 6 ⎠ k =1 sinh β k + sin β k ⎝ 2 ⎦ ⎣

α ⎜⎜

(cc)

Use of an eigenvector expansion leads to ⎤ ⎛ x2 x3 ⎞ ⎡ cosh β k + cos β k (sinh β k x − sin β k x )⎥ dx Dk = ∫ α ⎜⎜ − ⎟⎟C k ⎢cosh β k x − cos β k x − 6 ⎠ ⎣ sinh β k + sin β k ⎝ 2 ⎦ 0 1

336

(dd)

Problems 6.17-6.24 refer to the non-dimensional temperature distribution in a cylinder which, in general, is governed by ∂ 2 Φ 1 ∂Φ 1 ∂ 2 Φ ∂ 2 Φ ∂Φ = + + + ∂t ∂r 2 r ∂r r 2 ∂θ 2 ∂z 2 Non-dimensional variables are introduced such that 0 ≤ z ≤ 1 , 0 ≤ r ≤

(a) R = α and L

0 ≤ θ ≤ 2π .

6.17 Determine the steady temperature distribution when the temperature at each end is prescribed as different constants and the circumference is insulated. The temperature is independent of θ. The boundary conditions for Φ (r , z ) are

∂Φ (α , z ) = 0 , Φ (r ,0) = 0 and ∂r

Φ (r ,1) = 1

Solution: The differential equation reduces to ∂ 2 Φ 1 ∂Φ ∂ 2 Φ + + 2 =0 ∂r 2 r ∂r ∂z

(b)

Using separation of variables a product solution of the form Φ (r , z ) = R(r ) Z ( z ) is assumed. Substitution into Equation b leads to 1 R ′′Z + R ′Z + RZ ′′ = 0 r ⎛ R ′′ 1 R ′ ⎞ Z ′′ −⎜ + ⎟= ⎝ R r R⎠ Z

(c)

The usual separation argument is applied with a separation parameter λ leading to rR ′′ + R ′ + λrR = 0 Z ′′ − λZ = 0 Use of the product solution in the boundary conditions leads to

337

(d) (e)

R ′(α ) = 0 Z (0) = 0

(f) (g)

The general solution of Equation d is R (r ) = C1 J 0

( λ r )+ C Y ( λ r )

(h)

2 0

The temperature if bounded at r=0, however the Bessel function of the second kind is undefined at r=0. Thus C 2 = 0 and R (r ) = C1 J 0

( λr)

(i)

Application of the boundary condition, Equation f to Equation i leads to J 0′

(

)

λα = 0

(j)

There are an infinite number of solutions of Equation j, but they can be indexed as λk , k=1,2,…. For each k, the solution of Equations d and f is Rk ( r ) = C k J 0

(

λk r

)

(k)

The eigenvectors are normalized by requiring α

∫C

2 k

[J ( λ r )] rdr = 1 2

(l)

0

0

Recalling a result from Example 3.10, modified such that the upper limit of the integral is β β

{

}

x=β

⎡ x2 ⎤ 2 = − ( ) ( ) ( ) ( ) x [ J α x ] dx [ J α x ] [ J α x J α x ] 1 1 + − n n n n ⎢ ⎥ ∫0 ⎣2 ⎦ x =0 2

Applying Equation m to the Equation l leads to

338

(m)

{[ ( λ r )] − J ( λ r )J ( )} ⎡α ⎤ C ⎢ {[J ( λ α )] − J ( λ α )J ( λ α )}⎥ ⎣ 2 ⎦

r =α

⎡r2 C ⎢ J0 ⎣2

k

(

)

−1

k

1

2

2

2 k

However J 0′

⎤ λk r ⎥ = 1 ⎦ r =0

2

2 k

k

0

−1

k

1

(n)

k

( )

λα = J 1 λ r which by Equation j is identically zero. Thus Ck =

2

(

αJ 0 λ k α

)

(o)

And the normalized eigenvectors are

Rk (r) =

( αJ ( λ α ) 2

0

J 0 λk r

)

(p)

k

For a specific k, the solution of Equation e subject to Equation g is

(

Z k ( z ) = Dk sinh λ k z

)

(q)

The general solution is ∞

Φ ( r.z ) = ∑ k =1

2 Dk

(

αJ 0 λ k α

)J ( 0

) (

)

(r)

) ( )

(s)

λk r sinh λk z

Application of the remaining boundary condition leads to ∞

1= ∑ k =1

2 Dk

αJ 0

(

λk α

)J ( 0

λk r sinh λ k

An eigenvector expansion for 1 leads to ∞

1= ∑ k =1

2 Bk

(

αJ 0 λ k α

where

339

)J ( 0

λk r

)

(t)

α

Bk = ∫ 0

(

2

αJ 0 λ k α

)J ( 0

)

λk r rdr

(u)

Then Dk =

Bk

( )

sinh λ k

340

(v)

6.18 Determine the steady temperature distribution in the cylinder when the temperature at each end is maintained at the same value and a heat flux q(θ , z ) is applied to the circumference of the cylinder. The boundary conditions for Φ (r , θ , z ) are ∂Φ (α , θ , z ) = q(θ , z ) , Φ(r , θ ,0) = 0 and Φ (r , θ ,1) = 0 . ∂r

Solution: The governing partial differential equation is ∂ 2 Φ 1 ∂Φ 1 ∂ 2 Φ ∂ 2 Φ =0 + + + ∂r 2 r ∂r r 2 ∂θ 2 ∂z 2

(a)

A product solution of Equation a is assumed of the form Φ (r , θ , z ) = R(r )Θ(θ ) Z ( z ) . Substitution into Equation a leads to 1 1 R ′′ΘZ + R ′ΘZ + 2 RΘ′′Z + RΘZ ′′ = 0 r r Multiplying Equation b by

(b)

r2 and rearranging leads to RΘZ

r2

R′′ R′ Z ′′ Θ′′ + r + r2 =− R R Z Θ

(c)

Application of the usual separation argument with a separation parameter λ leads to Θ′′ + λΘ = 0 1 R ′′ + R ′ r − λ = − Z ′′ R Z r2

(d) (e)

Application of the usual separation argument to Equation e with a separation parameter μ gives Z ′′ + μZ = 0

(f)

1 ⎛λ ⎞ R ′′ + R ′ − ⎜ 2 + μ ⎟ R = 0 r ⎝r ⎠

(g)

341

Application of the product solution to the boundary conditions leads to Z (0) = 0

(h) (i)

Z(1) = 0 The general solution of Equation d is

(

)

(

Θ(θ ) = C1 cos λθ + C 2 cos λθ

)

(j)

The temperature distribution is to be obtained in a full circular cylinder. The only conditions which Equation j must satisfy are single-valuedness conditions. That is the temperature distribution and the rate of heat transfer must be single valued. This leads to periodicity conditions of the form Θ(θ + 2π ) = Θ(θ ) Θ′(θ + 2π ) = Θ′(θ )

(k) (l)

This problem is thoroughly discussed in Chapter 5. The separation constants and the normalized solutions are

λn = n 2

n = 0,1,2,...

(m)

1

Θ 0 (θ ) = Θ1n (θ ) = Θ 2n (θ ) =

2π 1

π 1

π

(n) cos(nθ )

(o)

sin( nθ )

(p)

For each n>0 there are two linearly independent eigenvectors. The solution of Equation f subject to Equations h and i is

μ k = (kπ )2

(q)

Z k ( z ) = 2 sin (kπz )

(r)

Equation g becomes

[

]

r 2 R ′′ + rR ′ − (kπ ) r 2 + n 2 R = 0

342

2

(s)

The general solution of Equation s is in terms of modified Bessel functions Rn ,k = Dn ,k I n (kπr ) + E n ,k K n (kπr )

(t)

The temperature is finite at r=0 only if En ,k =0. The general solution developed from the product solution is ∞

Φ(r ,θ , z ) = ∑

Do , k

π

k =1

sin (kπz )I 0 (kπr )

∞

∞

+ ∑∑ n =1 k =1

2

π

sin (kπz )I n (kπr )[Dn ,k ,1 cos(nθ ) + Dn ,k , 2 sin( nθ )]

(u)

Application of the remaining boundary condition leads to ∞

q (θ , z ) = ∑ k =1

Do , k

π

sin (kπz )I 0 (kπα ) ∞

∞

+ ∑∑ n =1 k =1

2

π

sin (kπz )I n (kπα )[Dn ,k ,1 cos(nθ ) + Dn ,k , 2 sin( nθ )]

(v)

An eigenvector expansion leads to ∞

q (θ , z ) = ∑ k =1

Bo ,k

π

∞

∞

sin (kπz ) + ∑∑ n =1 k =1

2

π

sin (kπz )[Bn ,k ,1 cos(nθ ) + Bn ,k , 2 sin( nθ )]

(w)

where 1 2π

[

]

⎡ 1 ⎤ B0,k = ∫ ∫ q(θ , z )⎢ ⎥ 2 sin (kπz ) dθdz ⎣ 2π ⎦ 0 0 1 2π

(x)

[

]

[

]

⎡ 1 ⎤ Bn ,k ,1 = ∫ ∫ q(θ , z )⎢ cos(nθ )⎥ 2 sin (kπz ) dθdz ⎣ π ⎦ 0 0 1 2π

⎡ 1 ⎤ Bn ,k , 2 = ∫ ∫ q(θ , z )⎢ sin( nθ )⎥ 2 sin (kπz ) dθdz ⎣ π ⎦ 0 0 Comparison of Equations v and w leads to

343

(y) (z)

D0 , k =

B0 , k

I 0 (kπα ) Bn ,k ,1

Dn ,k ,1 =

I n (kπα )

Dn , k , 2 =

Bn , k , 2

344

I n (kπα )

(aa) (bb) (cc)

6.19 The cylinder is at a uniform temperature when the temperature of the surrounding medium is suddenly changed. Both ends of the cylinder are insulated, but heat is transferred by convection to the surrounding medium from the circumferential surface of the cylinder. The unsteady temperature is independent of θ. The boundary conditions for

Φ(r , z , t ) are

∂Φ ∂Φ ∂Φ (α , z , t ) + BiΦ (α , z , t ) = 0 , (r ,0, t ) = 0 and (r ,1, t ) = 0 . The initial ∂r ∂z ∂z

condition for the system is Φ (r , z ,0) = 1 .

Solution: The governing partial differential equation is ∂ 2 Φ 1 ∂Φ ∂ 2 Φ ∂Φ + + 2 = ∂t ∂r 2 r ∂r ∂z

(a)

A product solution is assumed in the form of Φ (r , z , t ) = Q(r , z )T (t ) which when substituted into the partial differential equation leads to ⎡ ∂ 2 Q 1 ∂Q ∂ 2 Q ⎤ + ⎢ 2 + ⎥T (t ) = Q(r , z )T ′(t ) r ∂r ∂z 2 ⎦ ⎣ ∂r

(b)

Rearranging and application of the usual separation argument with a separation parameter λ leads to T ′ + λT = 0 ∂ 2 Q 1 ∂Q ∂ 2 Q + + = −λ Q ∂r 2 r ∂r ∂z 2

(c) (d)

A product solution of the form of Q(r , z ) = R (r ) Z ( z ) is used in Equation d leading to 1 ⎞ ⎛ ⎜ R ′′ + R ′ ⎟ Z + RZ ′′ + λRZ = 0 r ⎠ ⎝

(e)

Rearranging and application of the usual separation argument with a separation parameter μ leads to

345

Z ′′ + (λ − μ )Z = 0 1 R ′′ + R ′ + μR = 0 r

(f) (g)

Use of the product solutions in the boundary conditions leads to R ′(α ) + BiR (α ) = 0 Z ′( 0 ) = 0 Z ′(1) = 0

(h) (i) (j)

Equation g is Bessel’s equation of order zero with solutions of R (r ) = C1 J 0

( μ r )+ C Y ( μ r )

(k)

2 0

Since the cylinder is solid, the temperature must be finite at r=0 and since Y0 (0) is undefined, C 2 = 0 leading to R (r ) = C1 J 0

( μr)

(l)

Application of Equation h to Equation l leads to

μ J 0′

(

)

μα + BiJ 0

(

)

μα = 0

(m)

There are an infinite, but countable, number of solutions of Equation m which can be indexed as μ k k = 1,2,... . The corresponding solutions are Rk ( r ) = C k J 0

(

μk r

)

(n)

The eigenvectors of Equation n are normalized by requiring

[ (

⎧α C k = ⎨∫ J 0 ⎩0

μk r

)]

2

⎫ rdr ⎬ ⎭

−

1 2

(o)

The integral of Equation o can be evaluated using the formula derived in Example 3.10. Defining β = λ − μ the solution of Equation f is

(

)

(

Z ( z ) = D1 cos β z + D2 sin β z

346

)

(p)

Application of Equations i and j lead to Z ′(0) = 0 ⇒ D2 = 0

(q)

( )

Z ′(1) = 0 ⇒ D1 β sin β = 0

(r)

Equation r leads to

β n = (nπ )2

n = 0,1,2,...

(s)

The normalized eigenvectors are Z 0 ( z) = 1

(t)

Z n ( z ) = 2 cos(nπz )

(u)

The separation constants are of the form

λ k , n = μ k + (n π )2

k = 1,2,...

n = 0,1,2, , , , ,

(v)

The solution of Equation c is Tk ,n (t ) = E k ,n e

− λk , n t

(w)

The general solution, formed from the product solution is ∞

Φ(r , z , t ) = ∑ C k J 0 k =1

(

)⎡

μ k r ⎢ E k ,0 e

∞

+ ∑ 2 E k ,n cos(nπz )e

− λk , o t

⎣

n =1

− λk , n t

⎤ ⎥ ⎦

(x)

Application of the initial condition leads to ∞

1 = ∑ Ck J 0 k =1

(

)⎡

∞

⎤

n =1

⎦

μ k r ⎢ E k ,0 + ∑ 2 E k ,n cos(nπz )⎥ ⎣

(y)

The constants are obtained from the eigenvector expansion ∞

1 = ∑ Ck J 0 k =1

(

)⎡

∞

⎤

n =1

⎦

μ k r ⎢ Bk , 0 + ∑ 2 Bk ,n cos(nπz )⎥ ⎣

where

347

(z)

1α

Bk , 0 = ∫ ∫ (1)C k J 0

(

μ k r rdrdz

)

(

μk r

(aa)

0

1α

Bk ,n = ∫ ∫ (1)C k J 0 0

348

)[

]

2 cos(nπz ) rdrdz

(bb)

6.20 The cylinder has a steady temperature distribution of f (r ,θ , z ) when the temperature of the surrounding medium is suddenly changed. Both ends of the cylinder are insulated, but heat is transferred by convection to the surrounding medium from the circumferential surface of the cylinder. The boundary conditions for

Φ(r ,θ , z , t ) are

∂Φ ∂Φ ∂Φ (α , θ , z , t ) + BiΦ (α , θ , z , t ) = 0 , (r , θ ,0, t ) = 0 and (r , θ ,1, t ) = 0 . ∂r ∂z ∂z

The initial condition for the system is Φ (r , θ , z ,0) = f (r , θ , z ) .

Solution: A product solution is assumed as U (r , θ , z )T (t ) which when substituted into the governing differential equation, rearranging and using the usual separation argument with a separation parameter λ leads to

T ′(t ) + λT = 0

(b)

∂ 2U 1 ∂U 1 ∂ 2U ∂ 2U + λU = 0 + + + ∂r 2 r ∂r r 2 ∂θ 2 ∂z 2

(c)

A product solution is assumed for Equation c in the form

U (r ,θ , z ) = R(r )Θ(θ ) Z ( z ) which when substituted into Equation c leads to 1 1 R ′′ΘZ + R ′ΘZ + 2 Θ′′RZ + Z ′′RΘ + λRΘZ = 0 r r

(d)

The usual separation argument is applied to Eq. (d) with a separation constant of μ . The resulting equations are d 2Θ + μΘ = 0 dθ 2

(e)

and

(

)

r d ⎛ dR ⎞ r 2 d 2 Z + λr 2 − μ = 0 ⎜r ⎟+ 2 R dr ⎝ dr ⎠ Z dz

349

(f)

Equation f is divided by r 2 and rearranged to 1 d ⎛ dR ⎞ μ 1 d 2Z −λ ⎜r ⎟+ 2 = − rR dr ⎝ dr ⎠ r Z dz 2

(g)

The usual separation argument is used on Equation g with a separation constant of − κ resulting in d 2Z + (λ − κ )Z = 0 dz 2

(h)

and r

(

)

d ⎛ dR ⎞ 2 ⎜r ⎟ + κr − μ R = 0 dr ⎝ dr ⎠

(i)

Use of the product solution in the boundary conditions leads to R ′(α ) + BiR(α ) = 0 Z ′( 0 ) = 0 Z ′(1) = 0

(j) (k) (l)

The temperature distribution is to be obtained in a full circular cylinder. The only conditions which Equation j must satisfy are single-valuedness conditions. That is the temperature distribution and the rate of heat transfer must be single valued. This leads to periodicity conditions of the form Θ(θ + 2π ) = Θ(θ ) Θ′(θ + 2π ) = Θ′(θ ) This problem is thoroughly discussed in Chapter 5. The separation constants and the normalized solutions are

350

(m) (n)

μn = n2

n = 0,1,2,...

Θ 0 (θ ) =

1

Θ1n (θ ) = Θ 2n (θ ) =

2π 1

π

(o) (p)

cos(nθ )

(q)

sin( nθ )

(r)

1

π

Equation i is Bessel’s equation of order n with solutions of R (r ) = C1 J n

( κ r )+ C Y ( κ r )

(s)

2 n

Since the cylinder is solid, the temperature must be finite at r=0 and since Yn (0) is undefined, C 2 = 0 leading to R(r ) = C1 J n

( κ r)

(t)

Application of Equation h to Equation t leads to

(

)

(

)

κ J n′ κ α + BiJ 0 κ α = 0

(u)

For each n, there are an infinite, but countable, number of solutions of Equation m which can be indexed as κ k ,n k = 1,2,... . The corresponding solutions are Rk ( r ) = C k J 0

( κ r)

(v)

k ,n

The eigenvectors of Equation n are normalized by requiring

C k ,n

[ ( κ r )]

⎧α = ⎨∫ J 0 ⎩0

2

k ,n

⎫ rdr ⎬ ⎭

−

1 2

(w)

The integral of Equation o can be evaluated using the formula derived in Example 3.10. Defining β = λ − κ the solution of Equation f is

(

)

(

Z ( z ) = D1 cos β z + D2 sin β z

Application of Equations k and l lead to 351

)

(x)

Z ′(0) = 0 ⇒ D2 = 0

(y)

( )

Z ′(1) = 0 ⇒ D1 β sin β = 0

(z)

Equation z leads to

β p = ( pπ )2

p = 0,1,2,...

(aa)

The normalized eigenvectors are Z 0 ( z) = 1

(bb)

Z p ( z ) = 2 cos( pπz )

(cc)

The separation constants are of the form

λk ,n , p = κ k ,n + ( pπ )2

k = 1,2,... n = 0,1,2, , , , , p = 0,1,2,...

(dd)

The solution of Equation b is Tk ,n , p (t ) = E k ,n , p e

− λk , n , p t

(ee)

The general solution, formed from the product solution is

[

∞ ⎧ ⎛ 1 ⎞ ⎟⎟(1) C k ,0 J 0 Φ (r , z , t ) = ∑ ⎨ E k ,0,0 ⎜⎜ k =1 ⎩ ⎝ 2π ⎠ ∞

∑E p =1

k , 0,0

[

][

⎛ 1 ⎞ ⎜⎜ ⎟⎟ 2 cos( pπz ) C k ,0 J 0 ⎝ 2π ⎠

[

⎧⎛ 1 ⎞ ⎟⎟(1) C k ,n J n n =1 k =1 ⎩⎝ π ⎠ ∞

∞

∑∑ ⎨⎜⎜

[

∞ ⎛ 1 ⎞ + ∑ ⎜⎜ ⎟⎟(1) C k ,n J n p =1 ⎝ π ⎠

(

(

)]

μ k ,n r e

(

− λk , n , 0 t

)]

(

)]

μ k ,0 r e

)]

μ k ,0 r e

[E

μ k ,n r cos( pπz )e

k , n , 0 ,1

− λk , n , p t

− λk , 0 , 0 t

− λk , 0 , o t

+

⎫ ⎬ + ⎭

cos(nθ ) + E k ,n , 0, 2 sin(nθ )]

[E

k , n , p ,1

⎫ cos(nθ ) + E k ,n , p , 2 sin( nθ ) ⎬ ⎭

]

The unknown constants in Equation ff are determined from application of the initial condition and use of an eigenvector expansion. An example is

∫ ∫ f (r ,θ , z )[

1 2π α

E k ,n , p ,1 = ∫

0 0 0

]

[

⎡ 1 ⎤ 2 cos( pπz ) ⎢ cos(nθ ⎥ C k ,n J n ⎣ π ⎦

352

(

)]

μ k ,n r rdrdθdz

(gg)

(ff)

6.22 The cylinder has a uniform temperature when the circumferential surface is suddenly subject to a heat flux q (z ) . Both ends of the cylinder are insulated The boundary conditions for Φ(r , z , t ) are ,

∂Φ (α , z, t ) = q( z ) , ∂Φ (r ,0, t ) = 0 and ∂r ∂z

∂Φ (r ,1, t ) = 0 . The initial condition for the system is Φ (r , z,0) = 1 . ∂z

Solution: A superposition is assumed in which the temperature is the sum of a transient solution and a steady-state solution. To this end Φ(r , , z , t ) = Φ t (r , , z , t ) + Φ s (r , , z )

(b)

The transient solution is chosen to solve ∂ 2 Φ t 1 ∂Φ t ∂ 2 Φ t ∂Φ t = + + r ∂r ∂t ∂z 2 ∂r 2

(c)

subject to ∂Φ t (α , z, t ) = 0 ∂r

(d)

∂Φ t (r ,0, t ) = 0 ∂z

(e)

∂Φ t (r ,1, t ) = 0 ∂z

(f)

and Φ t (r , z ,0) = 1 − Φ s (r , z )

(g)

The steady-state solution solves ∂ 2 Φ s 1 ∂Φ s ∂ 2 Φ s 0 + + r ∂r ∂z 2 ∂r 2

subject to

353

(h)

∂Φ s (α , z ) = q(z) ∂r

(i)

∂Φ s (r ,0) = 0 ∂z

(j)

and ∂Φ s (r ,1) = 0 ∂z

(k)

First consider the steady-state solution. A steady state governed by Equations h-k cannot exist. Since the left and right ends of the cylinder are insulated and heat is added at its circumference, there is no boundary through which heat is transferred. Thus a reformulation must occur in that Φ s (r , z ) = A(r , z ) + B(t ) such that the resulting equation is

∂ 2 A 1 ∂A ∂ 2 A dB + + = dt ∂r 2 r ∂r ∂z 2

(l)

An argument similar to the separation argument is used to separate Equation l into ∂ 2 A 1 ∂A ∂ 2 A + + =C ∂r 2 r ∂r ∂z 2 dB =C dt

(m) (n)

A further superposition is employed in the solution of Equation m

A(r , z ) = f (r ) + g (r , z ) where

354

(o)

d 2 f 1 df + =C dr 2 r dr df (α ) = 0 dr ∂ 2 g 1 ∂g ∂ 2 g + + =0 ∂r 2 r ∂r ∂z 2 ∂g (α , z ) = q( z ) ∂r ∂g (r ,0) = 0 ∂z ∂g (r ,1) = 0 ∂z

(p) (q) (r) (s) (t) (u)

Thus Φ(r , θ , z, t ) = Φ t (r , θ , z , t ) + B(t ) + f (r ) + g (r , z )

(v)

Each function is determined as follows. dB = C ⇒ B(t ) = Ct + F dt

(w)

It can be shown that F is arbitrary and can be taken as zero.

d 2 f 1 df d ⎛ df ⎞ df C 3 df C 2 F + = C ⇒ ⎜ r ⎟ = Cr 2 ⇒ r = r +G ⇒ = r + 2 r dr dr ⎝ dr ⎠ dr 3 dr 3 r dr

(x)

Then f (r ) =

C 3 r + F ln(r ) + G 9

(y)

The temperature is finite at r=0 only if F=0. Then applying the boundary condition at r=1

Using separation of variables a product solution of the form Φ s (r , z ) = R(r ) Z ( z ) is assumed. Substitution into Equation b leads to

355

1 R ′′Z + R ′Z + RZ ′′ = 0 r ′ ′ ⎛ R 1 R ′ ⎞ Z ′′ −⎜ + ⎟= ⎝ R r R⎠ Z

(l)

The usual separation argument is applied with a separation parameter λ leading to rR ′′ + R ′ − λrR = 0 Z ′′ + λZ = 0

(m) (n)

Use of the product solution in the boundary conditions leads to Z ′(0) = 0 Z ′(1) = 0

(o) (p)

The solution of Equation n is

( )

( )

Z ( z ) = D1 cos λ z + D2 sin λ z

(q)

Application of Equations o and p lead to Z ′(0) = 0 ⇒ D2 = 0

(r)

( )

Z ′(1) = 0 ⇒ D1 λ sin λ = 0

(s)

Equation s leads to

λ p = ( pπ )2

p = 0,1,2,...

(t)

The normalized eigenvectors are Z 0 ( z) = 1

(u)

Z p ( z ) = 2 cos( pπz )

(v)

For each k, Equation m has a solution of Rk (r ) = C k I o

(

)

λ k r + Dk K 0

(

λk r

)

(w)

The temperature at the center of the cylinder is finite only if Dk = 0 which leads to Rk (r ) = C k I o

356

(

λk r

)

(x)

Note that k=0 is a special case in which Equation m becomes ′ rR0′′ + R0′ = 0 ⇒ (rR ′) = 0 ⇒ rR ′ = A ⇒ R = A ln(r ) + C 0

(y)

However, the temperature is finite at r=0 only if A=0. The general solution is ∞

Φ s (r , z ) = C 0 + ∑ 2C k sin( kπz ) I 0 (kπr )

(z)

k =1

The final boundary condition is imposed by ∞

q( z ) = ∑ 2C k kπ sin( kπz ) I 0′ (kπα ) k =1

357

(aa)

6.23 The cylinder has a steady temperature distribution of f (r ,θ , z ) when an internal heat generation u(z) begins in the cylinder. The right end of the cylinder is insulated, but heat is transferred by convection to the surrounding medium from the circumferential surface of the cylinder. The governing partial differential equation is modified to ∂ 2 Φ 1 ∂Φ 1 ∂ 2 Φ ∂ 2 Φ ∂Φ + + + = + u(z) . The boundary conditions for Φ (r , θ , z , t ) are ∂t ∂r 2 r ∂r r 2 ∂θ 2 ∂z 2 ,

∂Φ ∂Φ ∂Φ (α , θ , z , t ) + BiΦ (α , θ , z , t ) = 0 (r , θ ,0, t ) = 0 and (r , θ ,1, t ) = 0 . The initial ∂r ∂z ∂z

condition for the system is Φ (r , θ , z ,0) = f (r , θ , z ) .

Solution: A superposition is assumed in which the temperature is the sum of a transient solution and a steady-state solution. To this end Φ(r , θ , z , t ) = Φ t (r , θ , z , t ) + Φ s (r , θ , z )

(b)

The transient solution is chosen to solve ∂ 2 Φ t 1 ∂Φ t 1 ∂ 2 Φ t ∂ 2 Φ t ∂Φ t = + + + r ∂r ∂t r 2 ∂θ 2 ∂z 2 ∂r 2

(c)

whereas the steady-state solution solves ∂ 2 Φ s 1 ∂Φ s 1 ∂ 2 Φ s ∂ 2 Φ s = u(z) + + 2 + r ∂r r ∂θ 2 ∂z 2 ∂r 2

(d)

The boundary and initial conditions for the transient solution are ∂Φ t (α ,θ , z , t ) + BiΦ t (α ,θ , z , t ) = 0 ∂r

∂Φ t (r , θ ,0, t ) = 0 ∂z ∂Φ t (r , θ ,1, t ) = 0 ∂z Φ t (r , θ , z ,0) = f (r , θ , z ) − Φ s (r , θ , z )

358

(e)

(f) (g) (h)

Once the steady-state solution is obtained, the transient solution can be obtained by direct application of separation of variables. The boundary conditions for the steady-state response are ∂Φ s (α , θ , z ) + BiΦ s (α , θ , z ) = 0 ∂r ∂Φ s (r , θ ,0) = 0 ∂z ∂Φ s (r ,θ ,1) = 0 ∂z

(i)

(j)

(k) .

A superposition is applied to obtain the steady-state distribution as Φ s (r , θ , z ) = A(r , θ , z ) + B( z )

(l)

The partial differential equation for A is ∂ 2 A 1 ∂A 1 ∂ 2 A ∂ 2 A =0 + + + ∂r 2 r ∂r r 2 ∂θ 2 ∂z 2

(m)

while the ordinary differential equation for B is d 2B = u( z) dz 2

(n)

The boundary conditions which A must satisfy are ∂A (α , θ , z ) + BiA(α , θ , z ) = − BiB( z ) ∂r ∂A (r , θ ,0) = 0 ∂z ∂A (r , θ ,1) = 0 ∂z The boundary conditions chosen for B(z) are

359

(o)

(p)

(q)

dB (0) = 0 dz dB (1) = 0 dz

(r) (s)

Equations n, r and s are used to determine B(z). Then Equations m,o,p and q are used to determine A using separation of variables. The eigenvalue problem is in the z direction. Since B is independent of θ, A is also independent of θ.

360

6.24 The cylinder has the steady-state temperature distribution obtained in the solution of problem 6.18 when an internal heat generation begins in the cylinder. The boundary conditions remain the same as problem 6.18 but the governing partial differential equation is modified to

∂ 2 Φ 1 ∂Φ 1 ∂ 2 Φ ∂ 2 Φ ∂Φ = + u(z) . Solve the problem to + + + ∂t ∂r 2 r ∂r r 2 ∂θ 2 ∂z 2

determine Φ (r , θ , z , t ) .

Solution: The boundary conditions are

∂Φ (α , θ , z ) = q (θ , z ) , Φ(r , θ ,0) = 0 and ∂r

Φ (r , θ ,1) = 0 . The solution is a superposition of a transient solution and a steady-state

solution, Φ (r , θ , z , t ) = Φ t (r , θ , z , t ) + Φ s (r , θ , z )

(b)

The transient solution is chosen to solve ∂ 2 Φ t 1 ∂Φ t 1 ∂ 2 Φ t ∂ 2 Φ t ∂Φ t = + + + r ∂r ∂t r 2 ∂θ 2 ∂z 2 ∂r 2

(c)

whereas the steady-state solution solves ∂ 2 Φ s 1 ∂Φ s 1 ∂ 2 Φ s ∂ 2 Φ s = u(z) + + 2 + r ∂r r ∂θ 2 ∂z 2 ∂r 2

(d)

The boundary and initial conditions for the transient solution are

∂Φ t (α ,θ , z, t ) = 0 ∂r

(e)

Φ t (r , θ ,0, t ) = 0

(f)

Φ t (r , θ ,1, t ) = 0

(g)

Φ t (r , θ , z ,0) = f (r , θ , z ) − Φ s (r , θ , z )

361

(h)

where f (r , θ , z ) is the temperature distribution obtained in Problem 6.18. Once the steady-state solution is obtained, the transient solution can be obtained by direct application of separation of variables. The boundary conditions for the steady-state response are ∂Φ s (α , θ , z ) = q(θ , z ) ∂r

(i)

Φ s (r , θ ,0) = 0

(j)

Φ s (r , θ ,1) = 0

(k) .

A superposition is applied to obtain the steady-state distribution as Φ s (r , θ , z ) = A(r , θ , z ) + B( z )

(l)

The partial differential equation for A is ∂ 2 A 1 ∂A 1 ∂ 2 A ∂ 2 A =0 + + + ∂r 2 r ∂r r 2 ∂θ 2 ∂z 2

(m)

while the ordinary differential equation for B is d 2B = u( z) dz 2

(n)

The boundary conditions which A must satisfy are ∂A (α , θ , z ) = q(θ , z ) ∂r A(r , θ ,0) = 0 A(r , θ ,1) = 0

(o) (p) (q)

The boundary conditions chosen for B(z) are

362

B(0) = 0

(r)

B(1) = 0

(s)

Equations n, r and s are used to determine B(z). Then Equations m,o,p and q are sued to determine A using separation of variables.

363

6.26 The disk at the end of the thin shaft of Fig. P6.11 is at rest in its equilibrium position when a time-dependent uniformly distributed torque is applied across the length of the shaft. Determine the resulting response of the shaft which is governed by the nondimensional partial differential equation

∂ 2Θ ∂ 2Θ = 2 + φx(1 − x) sin ωt subject to ∂x 2 ∂t

boundary conditions Θ(0, t ) = 0 where μ and C and initial conditions Θ( x,0 ) = 0 and

∂Θ ∂ 2Θ (1, t ) = − μ 2 (1, t ) and the ∂x ∂t

∂Θ ( x,0) = 0 where μ and φ are constants. ∂t

Solution: Since the non-homogeneous term is a function t, a superposition method is not applicable. Instead the non-homogeneous term will be expanded in an eigenvector expansion. Consider the eigenvalue problem which would occur in the x direction if separation of variables were applicable, X ′′ + λX = 0 subject to X (0) = 0 and X ′(1) − λμX (1) = 0 . The general solution of the differential equation is

( )

( )

X ( x) = C1 cos λx + C 2 sin λx

(a)

Application of the boundary conditions to Equation a leads to X (0) = 0 ⇒ C1 = 0

(b)

X ′(1) − μλX (1) = 0 ⇒ C 2 λ cos λ − μλC 2 sin λ = 0

(c)

Rearrangement of Equation c leads to the characteristic equation tan λ =

1

(d)

μ λ

There are an infinite, but countable, number of solutions of Equation d, λ1 < λ 2 < ... > The eigenvector corresponding to an eigenvalue λk is

(

X k ( x) = C k sin λ k x

364

)

(e)

Eigenvectors corresponding to distinct eigenvalues are orthogonal with respect to the inner product 1

( f , g ) = ∫ f ( x) g ( x)dx + μf (1) g (1)

(f)

0

Mode shapes are normalized by requiring

(X k , X k ) = 1

∫ [C 1

k

(

)]

[

[

(

)]

2

sin λ k x dx + μ C k sin λ k

]

2

=1

0

⎫ ⎧1 1 C ⎨ ∫ 1 − cos 2 λ k x dx + μ sin 2 λ k ⎬ = 1 ⎭ ⎩2 0 ⎤ C k2 ⎡ 1 sin 2 λ k + 2μ sin 2 λ k ⎥ = 1 ⎢1 − 2 ⎢⎣ 2 λ k ⎥⎦ 2 k

(

C k2 2

)

⎤ ⎡ 1 sin λ k cos λ k + 2μ sin 2 λ k ⎥ = 1 ⎢1 − λk ⎥⎦ ⎢⎣

( )

(g)

Use of Equation d in Equation g leads to

[

]

C k2 1 + 2 μ sin 2 λ k = 1 2 2 Ck = 1 + 2μ sin 2 λ k

(h)

Since Θ( x, t ) satisfies the same boundary conditions as X(x), it has an eigenvector expansion of the form ∞

Θ( x, t ) = ∑ ci (t ) X i ( x)

(i)

i =1

Use of Equation I in the governing partial differential equation leads to ∞

∞

i =1

i =1

∑ ci (t ) X i′′( x) = ∑ c&&i (t ) X i ( x) + φx(1 − x) sin (ωt ) Noting that X i′′ = −λi X i equation j is rearranged to

365

(j)

∞

∑X i =1

i

( x)[c&&i + λi ci ] = φx(1 − x) sin (ωt )

(k)

Taking the inner product defined in Equation f of both sides of Equation k with X k (x) for an arbitrary k leads to ∞

∑ [c&& + λ c ]( X i =1

i

i i

i

(

)

, X k ) = φ x − x 2 , X k sin (ωt )

(l)

Orthonormality of eigenvectors implies ( X i , X k ) = δ i ,k . Thus Equation l reduces to c&&k + λ k c k = bk sin (ωt )

(m)

where

(

bk = φ x − x 2 , X k ( x)

)

) (

)

⎡1 ⎤ = φC k ⎢ ∫ x − x 2 sin λ k x dx + μ (1 − 1) sin λ k ⎥ ⎣0 ⎦

(

(n)

The given initial conditions imply c k (0) = 0 and c&k (0) = 0 , The solution of Equation m satisfying these initial conditions is c k (t ) =

bk λk − ω 2

⎡ ω sin λ k t ⎢sin (ωt ) − λk ⎢⎣

(

)⎤⎥

(o)

)⎤⎥ sin (

λk x

)

(p)

⎥⎦

The system response is ∞

bk C k 2 k =1 λ k − ω

Θ( x, t ) = ∑

⎡ ω sin λ k t ⎢sin (ωt ) − λk ⎢⎣

(

366

⎥⎦

6.27 Determine the steady state temperature distribution in a sphere whose surface has a temperature distribution f (θ ) which is independent of φ .

Solution: The differential equation governing the unsteady temperature distribution in the sphere is ∂ ⎛ 2 ∂Λ ⎞ 1 ∂ 2Λ 1 ∂ ⎛ ∂Λ ⎞ + ⎜⎜ sin φ ⎟=0 ⎜r ⎟+ 2 2 ∂r ⎝ ∂r ⎠ sin φ ∂θ sin φ ∂φ ⎝ ∂φ ⎟⎠

(a)

The non-homogeneous boundary condition is at r=1. Thus a product solution is assumed as Λ (r ,θ ,φ ) = R(r )U (θ ,φ )

(b)

Substitution of Eq. (a) into Eq. (b) leads to U

∂U ⎞ d ⎛ 2 dR ⎞ R ∂ 2U R ∂ ⎛ + ⎜⎜ sin φ ⎟=0 ⎜r ⎟+ 2 2 sin φ ∂φ ⎝ ∂φ ⎟⎠ dr ⎝ dr ⎠ sin φ ∂θ

(c)

Dividing Eq. (c) by RU and rearranging leads to

∂ ⎛ ∂U ⎞ 1 d ⎛ 2 dR ⎞ 1 ∂ 2U 1 − ⎜⎜ sinφ ⎟ ⎜r ⎟=− 2 2 ∂φ ⎟⎠ R dr ⎝ dr ⎠ U sin φ ∂θ U sinφ ∂φ ⎝

(d)

Applying the usual separation argument to Eq. (d) and defining the separation constant as λ leads to d 2R dR r +r − λR = 0 2 dr dr 1 ∂ 2U 1 ∂ ⎛ ∂U − − ⎜⎜ sin φ 2 2 sin φ ∂θ sin φ ∂φ ⎝ ∂φ 2

(e) ⎞ ⎟⎟ = λU ⎠

(f)

It is noted that Eq. (e) is a Cauchy-Euler equation. A product solution for U is assumed as U (θ ,φ ) = Θ(θ )Φ (φ )

Substitution of Eq. (g) into Eq. (f) and rearranging gives 367

(g)

−

1 d 2Θ sin φ d ⎛ dΦ ⎞ = ⎜⎜ sin φ ⎟⎟ − λ sin 2 φ 2 Θ dθ Φ dφ ⎝ dφ ⎠

(h)

The usual separation argument is used on Eq. (h) with μ defined as the separation constant resulting in dΦ ⎞ ⎛ 1 d ⎛ μ ⎞ ⎜⎜ sin φ ⎟⎟ + ⎜⎜ λ - 2 ⎟⎟Φ = 0 dφ ⎠ ⎝ sin φ ⎠ sin φ dφ ⎝ d 2Θ + μΘ = 0 dθ 2

(i) (j)

Solutions of Eqs. (e), (i) and (j) are required to be finite and single valued at all points in the sphere. The coordinates (r ,θ + 2nπ , φ ) for any integer value of n refer to the same point as the coordinates (r ,θ ,φ ) . Thus, in order for the temperature and rate of heat transfer to be single valued, Θ(θ ) and

dΘ must be periodic of period 2π. Equation (j) with the dθ

periodicity conditions is the eigenvalue problem which defines the trigonometric Fourier series. Its eigenvalues, obtained in Section 5.5 and used in Section 6.2, are μ n = n 2 for n = 0,1,2,... . The normalized eigenvector corresponding to n=0 is Θ0 =

1 2π

(k)

Each eigenvalue for n>0 has two linearly independent normalized eigenvectors given by Θ n ,1 (θ ) =

1

Θ n , 2 (θ ) =

1

π π

cos(nθ )

(l)

sin (nθ )

(m)

Equation (i) is written as an eigenvalue problem for each value of n,

368

1 d ⎛ dΦ n ⎞ n2 = λΦ n ⎜⎜ sin φ ⎟⎟ + sin φ dφ ⎝ dφ ⎠ sin 2φ

−

Eq. (n) is of the form of a Sturm-Liouville problem with p (φ ) = − sin φ , q(φ ) =

(n) n2 and sin 2 φ

r (φ ) = sin φ . For each n, there are an infinite, but countable, number of eigenvlaues

λn , m for m=1,2,…. The eigenvectors Φ n , m and Φ n , p respectively corresponding to λn , m and λn , p satisfy the orthogonality condition (Φ n , m , Φ n , p )φ = 0 where π

(Φ

n, m

, Φ n, p

) = ∫ Φ (φ )Φ (φ )sin φdφ 2

φ

n, m

n, p

(o)

0

Since 0 ≤ φ ≤ π and p(0) = p(π ) = 0 the eigenvectors must only satisfy the conditions that they are finite at φ = 0 and φ = π . Using the change of independent variable, s = cos φ , Eq. (n) is rewritten as ⎛ d ⎡ n2 ⎞ 2 dΦ n ⎤ ⎜ ⎟Φ n = 0 − + − s 1 μ ds ⎢⎣ 1 − s 2 ⎟⎠ ds ⎥⎦ ⎜⎝

(

)

(o)

Eq. (o) is the associated Legendre’s equation which is solved in Sec. 3.7. Unless

μ = m(m + 1) , its solution is not finite at s = ±1 which correspond to φ = 0 and φ = π . For μ = m(m + 1) the general solution of Eq. (o) is Φ n, m = C1Pmn (s ) + C1Qmn (s ) where Pmn (s ) and Qmn (s ) are the associated Legendre functions of

the first and second kinds of order m and index n. Noting that Qmn (s ) is not finite at s = ±1 , C2 = 0 . Thus Φ n , m (φ ) = Cn , m Pmn (cos φ )sin n (φ )

The eigenvectors are normalized by choosing

369

(p)

Cn , m

⎧π n ⎫ 2 = ⎨∫ Pm (cos φ ) sin 2 n +1 φdφ ⎬ ⎩0 ⎭

[

]

−

1 2

(q)

The differential equation for R(r) now becomes d 2 Rm dR r + r m − m(m + 1) Rm = 0 2 dr dr 2

(r)

A solution of Eq. (r) is assumed as Rm = r α which when substituted into Eq. (r) leads to a quadratic equation whose roots are the values of α. The general solution is then attained as Rm (r ) = Am r m + Bm r − ( m +1)

(s)

The solution must remain finite at the center of the sphere, r=0 requiring Bm = 0 . Using Eqs. (b),(g),(k),(l),(m),(p) and (s) the general solution is ∞ C ⎧ Λ(r ,θ ,φ ) = ∑ ⎨ A0, m 0, m Pm (cos φ )r m 2π m=0 ⎩ ∞ C + ∑ n , m Pmn (cos φ )sin n (φ )r m ( An , m cosθ + Bn , m sin θ ) n =1

(t)

π

The boundary condition Λ (1,θ , φ ) = f (θ ) is applied using an eigenvector expansion for f (θ ,φ ) . The constants in Eq. (t) are obtained as ⎛ C 0,m ⎞ C 0,m A0,m = ⎜⎜ f , Pm (cos φ )⎟⎟ = 2π 2π ⎝ ⎠

2π π

∫ ∫ f (θ ) P

(cos φ ) sin n φdφdθ

m

(cos φ ) sin n (φ ) cos θ sin n +1 φdφdθ

(u)

0 0

⎛ C n,m n ⎞ C n,m An , m = ⎜⎜ f , P m (cos φ ) sin n (φ ) cos θ ⎟⎟ = π π ⎝ ⎠

2π π

⎛ C n,m n ⎞ C n,m Bn ,m = ⎜⎜ f , P m (cos φ ) sin n (φ ) sin θ ⎟⎟ = π π ⎝ ⎠

2π π

370

m

∫ ∫ f (θ ) P

n

0 0

∫ ∫ f (θ ) P

n m

0 0

(cos φ ) sin θ sin n +1 φdφdθ

(w)

(v)

371