Solving Problems in Mathematical Analysis, Part III (Problem Books in Mathematics) 3030385957, 9783030385958

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Problem Books in Mathematics

Tomasz Radożycki

Solving Problems in Mathematical Analysis, Part III Curves and Surfaces, Conditional Extremes, Curvilinear Integrals, Complex Functions, Singularities and Fourier Series

Problem Books in Mathematics Series Editor: Peter Winkler Department of Mathematics Dartmouth College Hanover, NH 03755 USA

More information about this series at http://www.springer.com/series/714

Tomasz Rado˙zycki

Solving Problems in Mathematical Analysis, Part III Curves and Surfaces, Conditional Extremes, Curvilinear Integrals, Complex Functions, Singularities and Fourier Series

Tomasz Rado˙zycki Faculty of Mathematics and Natural Sciences, College of Sciences Cardinal Stefan Wyszy´nski University Warsaw, Poland

Scientific review for the Polish edition: Jerzy Jacek Wojtkiewicz Based on a translation from the Polish language edition: “Rozwiazujemy ˛ zadania z analizy ´ matematycznej” cz˛es´c´ 3 by Tomasz Rado˙zycki Copyright ©WYDAWNICTWO OSWIATOWE “FOSZE” 2015 All Rights Reserved. ISSN 0941-3502 ISSN 2197-8506 (electronic) Problem Books in Mathematics ISBN 978-3-030-38595-8 ISBN 978-3-030-38596-5 (eBook) https://doi.org/10.1007/978-3-030-38596-5 Mathematics Subject Classification: 00-01, 00A07, 40-XX, 30-XX, 41-XX © Springer Nature Switzerland AG 2020 This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors, and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Switzerland AG. The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

Preface

This book is the third and last part of the collection of problems for students in their first 2 years of undergraduate mathematical analysis. It is intended to include all the material required from the students in the three–four semesters, although more ambitious readers will probably discover some gaps (e.g., Lebesgue’s integral, Fourier’s transform, distributions). Due to the limited volume of the book, however, I had to make a choice. This last part is intended for second-year students who have already gained significant experience in operating various formulas. Very detailed transformations, which appeared in the previous two parts, would lead to excessive growth of the volume. Therefore, when possible, I omit some computational details, referring to the results obtained in the previous volumes. I also assume that the reader is already familiar with some known integrals (e.g., Gaussian) and that they do not require any separate derivation. The chapters devoted to differential forms and oriented integrals are presented in a “parallel” way, so that the reader can easily see the relation between the vector description and that using forms. Warsaw, Poland

Tomasz Rado˙zycki

v

Contents

1

Examining Curves and Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Finding Curvature and Torsion of Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Examining k-Surfaces in N Dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Examining Ruled Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4 Exercises for Independent Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 2 18 31 37

2

Investigating Conditional Extremes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Using the Method of the Lagrange Multipliers . . . . . . . . . . . . . . . . . . . . . 2.2 Looking for Global Extremes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Exercises for Independent Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

41 42 56 65

3

Investigating Integrals with Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67 3.1 Examining Limits and Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69 3.2 Differentiating with Respect to Parameters . . . . . . . . . . . . . . . . . . . . . . . . . 76 3.3 Integrating over Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97 3.4 Exercises for Independent Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101

4

Examining Unoriented Curvilinear Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1 Finding Area of Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Calculating Various Curvilinear Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 Exercises for Independent Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

103 104 114 121

5

Examining Differential Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 Studying the Exterior Forms Operating on Vectors. . . . . . . . . . . . . . . . . 5.2 Performing Various Operations on Differential Forms . . . . . . . . . . . . . 5.3 Calculating Exterior Derivatives. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4 Looking for Primitive Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5 Finding Potentials in R3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.6 Exercises for Independent Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

123 125 133 145 149 164 176

vii

viii

Contents

6

Examining Oriented Curvilinear Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1 Calculating Integrals over Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 Calculating Integrals over Surfaces. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3 Using Stokes’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4 Exercises for Independent Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

179 181 189 198 217

7

Studying Functions of Complex Variable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1 Examining the Holomorphicity of Functions . . . . . . . . . . . . . . . . . . . . . . . 7.2 Finding Domains of Convergence of Complex Series . . . . . . . . . . . . . . 7.3 Calculating Contour Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4 Using Cauchy’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.5 Looking for Images of Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.6 Exercises for Independent Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

219 220 233 237 245 258 264

8

Investigating Singularities of Complex Functions . . . . . . . . . . . . . . . . . . . . . . 8.1 Identifying the Types of Singularities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2 Expanding Functions into Laurent Series . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3 Using the Residue Theorem to Calculate Definite Integrals . . . . . . . 8.4 Using Residue Theorem to Find Sums of Series. . . . . . . . . . . . . . . . . . . . 8.5 Exercises for Independent Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

267 268 278 289 304 312

9

Dealing with Multivalued Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.1 Analytically Continuing Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2 Calculating Integrals Involving Functions with Branch Points . . . . 9.3 Exercises for Independent Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

315 316 327 357

10

Studying Fourier Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.1 Examining Expandability of Functions into Fourier Series . . . . . . . . 10.2 Finding Fourier Coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.3 Exercises for Independent Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

359 360 363 373

Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 375

Definitions and Notation

In this book series the notation and conventions adopted in the former two parts are used. Several additional indications are given below. • Partial derivatives are often denoted in the following brief way, e.g., ∂/∂x := ∂x , ∂/∂ϕ := ∂ϕ , etc. • The versors of the coordinate axes are denoted as ex , ey , ez . Similarly for the spherical variables the following notation is used: er , eθ , eϕ defined with the formulas (5.1.22), (5.1.23), and (5.1.24). ¯ • The complex plane is denoted with C and the compactified plane with C. • In the chapters dealing with the functions of the complex variable (i.e., 7, 8, and 9), in contrast to the other chapters of the book and to the previous volumes, the symbol “ln” denotes the natural logarithm, in order to avoid collision with the symbol “log” reserved for the complex logarithm.

ix

Chapter 1

Examining Curves and Surfaces

The present chapter is concerned with basic properties of curves and surfaces. By a curve in RN , we understand the continuous function γ : [a, b] → RN (in R3 we will write γ (s) or r(s)). Thus, one can say that the curve is the image of the interval [a, b]. If the function γ is differentiable, the curve is also called differentiable. If a given curve has cusps, it can be piecewise differentiable. If the function γ is of the class C n , the curve is also called of the class Cn . If a curve is of sufficiently high class for a given problem, it is called a smooth curve. Naturally it may also be piecewise smooth. A curve in R3 is often defined as an intersection of two surfaces. Unless stated otherwise, it will be assumed below  that N = 3.   The curvature of a curve is defined as κ = d T /ds  = T  (s), where s ∈ [a, b] denotes the parameter and T (s) is the vector tangent to the curve at the point labeled by s, normalized to unity. With the appropriate choice of s, one has T (s) = γ  (s). The vector N (s) = κ −1 T  (s) is the normal vector again normalized to unity. The so-called binormal vector is defined as B = T × N and is the unit vector too.  constitutes The orthonormal system composed of these three vectors (T , N , and B) the Frenet frame. In turn, the planes designated by the Frenet frame, respectively called the normal plane (that perpendicular to T ), the straightening plane (that  create perpendicular to N ), and the strictly tangent plane (that perpendicular to B), the Frenet trihedron. In some textbooks, however, the Frenet trihedron refers to  and B and not planes. vectors T , N, The following system of Frenet’s equations can be derived (see Problem 2 for the details):  T  (s) = κ N(s),

(1.0.1a)

 N  (s) = −κ T (s) + τ B(s),

(1.0.1b)

B  (s) = −τ N (s).

(1.0.1c)

Naturally κ and τ are in general s-dependent. © Springer Nature Switzerland AG 2020 T. Rado˙zycki, Solving Problems in Mathematical Analysis, Part III, Problem Books in Mathematics, https://doi.org/10.1007/978-3-030-38596-5_1

1

2

1 Examining Curves and Surfaces

The torsion of the curve (denoted with τ ) is defined by the formula: τ = −N · B  or B  = −τ N . The notions of curvature and torsion are addressed in more detail in the following section, when solving concrete problems. A k-surface in RN is such subset V ⊂ RN that locally (i.e., at least on some neighborhood of each point cut to V ) constitutes the graph of a certain function. If these functions are of the class C 1 , this surface will be called smooth. A k-surface can also be described in the parametric way, via the relations: ⎧ x1 = g1 (τ1 , τ2 , . . . , τk ), ⎪ ⎪ ⎨ x2 = g2 (τ1 , τ2 , . . . , τk ), ⎪ ... ⎪ ⎩ xN = gn (τ1 , τ2 , . . . , τk ),

(1.0.2)

which can be collectively written in the form: x = G(τ ),

(1.0.3)

for τ ∈ D ⊂ Rk . This representation should be treated in the local sense too. The condition for (1.0.3) to define a certain smooth k-surface is rank(G ) = k, where G is the Jacobian matrix. A ruled surface in R3 is a surface that has the parametrization in the form:  r(u, v) = α (u) + v β(u),

(1.0.4)

 where α (u) is called the base curve or directrix and β(u) the director curve. If a given surface has two distinct (inequivalent) parametrizations like (1.0.4), then it is called doubly ruled.

1.1 Finding Curvature and Torsion of Curves Problem 1  Given the curve on the plane in the parametric form: x(ς ) = 3ς , y(ς ) = 2 ς 3 , where ς ∈ [0, ∞[. Its curvature at the point of coordinates (3, 2) will be found.

Solution In order to imagine what quantity we mean by talking about “curvature” let us refer to the elementary knowledge of physics. We know that a body performing the circular motion has the so-called centripetal acceleration equal to

1.1 Finding Curvature and Torsion of Curves

ad =

3

v2 , R

(1.1.1)

where v is the constant speed of the body and R is the circle’s radius. It is R that interests us with the difference that, in general, it will change from point to point. It can be calculated from (1.1.1):   1 ad 1  d v  (1.1.2) = 2 = 2  . R v v dt Since the speed is constant, this expression can be saved in the form      1 d v   1 d( 1 v /v)     = . = · · R  v 2 dt   v dt 

(1.1.3)

The velocity vector v(t) is, of course, tangent to the trajectory at the temporary position of the body. In turn, the vector v/v is a unit tangent vector (surely, it is assumed here, that v = 0). It will be denoted with the symbol T in order to avoid collision with time for which the lowercase letter t is reserved. Therefore, one has    1 d T  1   (1.1.4) = · . R  v dt  The distance s overpassed by the body on the circle or along any curve is related to the speed v in the obvious way: v=

ds . dt

(1.1.5)

Hence, instead of calculating the derivatives with respect to time t, one can equally well calculate the derivatives over s, using the formula for the differentiation of a composite function. First, denoting the temporary position of the body with r(t), one can write that v d r/dt d r T = = = , v ds/dt ds

(1.1.6)

and then     1 d T ds   1 d T 1    = · · = · R  v ds dt   v ds

     d T     v =  .   ds 

(1.1.7)

The vector d T /ds which appears here is nothing other than a vector perpendicular to the curve at the same point. In fact, it is proportional to the change of the vector T , which by definition has the fixed length equal to 1. Thereby, d T may not

4

1 Examining Curves and Surfaces

Fig. 1.1 The translation of the tangential vector along the curve

have any component tangent to the trajectory but only perpendicular ones. It is easy to show this in a formal way by writing T · T = 1 ⇒

  T · T = 2T · T  = 0.

(1.1.8)

The considered vector will appear repeatedly in our deliberations, so let us introduce the symbol: d T N = R , ds

(1.1.9)

where the presence of R ensures its normalization: N · N = 1. The result (1.1.7) becomes natural if one looks at the situation sketched in Fig. 1.1, where the trajectory of a moving body is shown. At the two very close moments of time, the body is located successively at points denoted as A and B. The unit vectors tangent to this trajectory at these two points are TA and TB . The normal vectors (the so-called “principal” ones that are lying in the plane spanned by TA and TB ) are depicted too. The angle ϕ between two tangential vectors and two normal vectors is the same because they are orthogonal to each other in pairs. Imagine now that from the points A and B the two straight lines along perpendicular vectors NA and NB are drawn. If only the curve is actually “curved” then these two lines intersect at a certain point (marked in the figure with O). The distance |OA| (and in fact also |OB|, since we consider the limit B → A) is the radius R of curvature measured at the point A. Let us denote with s the length of the arc contained between the points A and B. Then one has φ =

s . R

(1.1.10)

On the other hand, the same angle can be obtained from the vectors TA and TB too, the second one having been shifted to A in parallel (the shifted vector is indicated with the dashed line). Since φ → 0, so

1.1 Finding Curvature and Torsion of Curves

φ =

5

|T | |TB − TA | . = 1 |TA |

(1.1.11)

Comparing these two formulas, it can be seen that    T  s 1   = |T |, or =  = |T  |,  R R s  s→0

(1.1.12)

i.e., again the formula (1.1.7) is obtained. The quantity 1/R is henceforth denoted with the symbol κ and called the “curvature.” The milder the curve is bent, the smaller the value of the curvature κ has, and the larger the curvature radius R is, and vice versa: for a tightly bent curve, the curvature is large and the radius of curvature is small. In particular for a straight line one obtains κ = 0 and R = ∞ because, regardless of the distance s at which one will slide along it, the angle φ, by which tangent vector is rotated, will be equal to zero. Incidentally, this is just the sense of the (local) curvature radius: the shift along the curve at the infinitesimal distance δs generates the rotation of the tangential vector by the angle δφ = δs/R. This vector would be rotated by the same angle if it was shifted (at the same infinitesimal distance) along the arc of the circle of radius R instead of the curve in question. This situation is schematically presented in Fig. 1.2. After these general considerations, we can now proceed to solve the specific problem. The above results indicate that one needs to find the tangential vector T and then calculate its derivative with respect to the parameter s, introduced below. As we know, s has the meaning of the distance traveled along the curve. Such quantities were dealt with in Sect. 3.1 of the second part of this book series. Using the formula (3.1.7) given at that time and denoting the integration variable with σ , one can write

y

Fig. 1.2 The circle tangent to the curve at some point (x(ς0 ), y(ς0 )) of the same curvature radius R

R

circle

curve y 0 x 0

x

6

1 Examining Curves and Surfaces

s(ς ) =

ς 

x  (σ )2 + y  (σ )2 dσ =

0

ς

√

9 + 9σ dσ

0

ς =3

√

ς  1 + σ dσ = 2(1 + σ )3/2  = 2(1 + ς )3/2 − 2, 0

(1.1.13)

0

primes in the first line denoting derivatives with respect to σ . This relation can be very easily reversed, yielding s 2/3 ς (s) = 1 + − 1. (1.1.14) 2 As a consequence, one finds s 2/3 x(s) = 3ς (s) = 3 1 + − 3, 2  3/2 s 2/3 y(s) = = 2(ς (s))3/2 = 2 1 + −1 2

(1.1.15)

and after having differentiated:    d r = (1 + s/2)−1/3 1, (1 + s/2)2/3 − 1 . T (s) = ds

(1.1.16)

When calculating the length of this vector, the reader can easily ascertain that it is in fact a unit vector (T · T = 1). Now one has to further differentiate over s in accordance with the formula (1.1.7):    d T 1 = (1 + s/2)−4/3 −1, 1/ (1 + s/2)2/3 − 1 , (1.1.17) ds 6 and finally it is obtained that    d T  1   = κ= = R  ds 



1

.

(1.1.18)

6(1 + s/2) (1 + s/2)2/3 − 1

Something more than what the text of the problem had required has been found: the curvature can be known at any point of the curve. In particular the point of coordinates (3, 2) corresponds to the value of the parameter ς0 = 1, or alternatively √ using (1.1.13), the point corresponds to s0 = 2(2 2 − 1). By inserting this value into the formula (1.1.18), one gets κ=

1 √ , 12 2

or

√ R = 12 2.

(1.1.19)

1.1 Finding Curvature and Torsion of Curves

7

Problem 2 The curvature and torsion of the curve given in the parametric form: x(ς ) = sin ς,

y(ς ) = cos ς,

z(ς ) = cosh ς,

(1.1.20)

where ς ∈ [0, ∞[, will be found.

Solution If a curve does not lie in a plane as it did in the previous exercise but is a truly spatial curve, in addition to the curvature, it can still undergo some twisting. The fact that it is described with the use of three coordinates, as in (1.1.20) and not two, does not determine the spatial character of the curve yet. It can still happen that the curve is flat while the plane in which it is contained gets inclined. For example, if one relocates a circle—which undoubtedly is a flat curve—so that it will be tilted with respect to all three coordinate axes, it will not become the spatial curve! A truly spatial curve is such that it cannot be contained in any plane, which means that it is subject to a torsion. The convenient parameter τ , which measures this quantity, was defined in the theoretical summary. Below we will be concerned with this notion in detail, starting with some general considerations. In the previous problem, two important vectors were introduced: the tangent one T and the normal one N . To complete the set, let us now append the third vector (called binormal): B = T × N . By definition it is orthogonal to the curve (as it is orthogonal to T ) and to N and, therefore, also to the plane in which (locally!) the curve is lying. When the curve is being twisted, this plane is rotating together with the binormal vector. This situation is depicted in Fig. 1.3. Note that thanks to the prior normalization of the vectors T and N as well as their orthogonality one has  · (T × N)  = |T |2 |N |2 − (T · N )2 = 1. B · B = (T × N)      =1

=1

(1.1.21)

=0

The torsion of the curve should be connected with the rotation of the binormal vector, because this means the identical rotation of the plane, in which the curve is lying locally, i.e., the plane defined by the vectors T and N . In view of the normalization condition (1.1.21), any change of the vector B indicates its rotation (the same was pointed out in the preceding problem relative to the vector T ). Let us denote    d B    (1.1.22) |τ | =  ,  ds 

8

1 Examining Curves and Surfaces

Fig. 1.3 Modification of the tangential, normal, and binormal vectors when moving along the curve

T

B

N

B

N T

the parameter s (the distance traveled along the curve) having been introduced in the previous exercise. What is the interpretation of the parameter τ ? Well, one knows that if a given vector u is rotating by an infinitesimal angle d α (the vector symbol refers here to the direction of the rotation axis in accordance with the right-hand screw rule), then the change in the vector u is given by d u = d α × u.

(1.1.23)

In our case the vector B is rotating around the perpendicular axis, designated by T . Hence it can be inferred from (1.1.23) that    d u    = | u|. (1.1.24)  dα  We are interested in the twist of the vector u = B along the curve. So we have      d B   d B ds       = 1 ⇒  ·   = |B|  = 1,  dα   ds dα 

(1.1.25)

i.e.,    dα  |τ | =   . ds

(1.1.26)

1.1 Finding Curvature and Torsion of Curves

9

Thus, the needed interpretation has been found: the parameter τ tells us by what angle the vector B is rotated if one moves along the curve by the distance ds. This parameter is called the torsion. One can additionally fix the sign of τ . The vector  B(s) is constantly orthogonal to T (s), so B  (s) is as well. In turn, the constant  imposes the orthogonality condition between B  (s) and B(s).  normalization of B(s)       it Consequently, B (s) must be proportional to N(s), since within the trio T , N , B, is the only remaining linearly independent vector. Usually, one assumes B  (s) = −τ N (s),

(1.1.27)

which stays in harmony with (1.1.22), but also fixes the sign of τ .   As we already know, the vectors T (s), N(s), and B(s) for each value of s constitute the complete orthonormal system. Any other vector must be their linear combination, which has been exploited above to write down B  (s). Naturally the same applies also to the derivatives T  (s) and N  (s). From the Eq. (1.1.9), one already has  T  (s) = κ N(s).

(1.1.28)

Additionally N · N  = 0 due to normalization, so  N  (s) = a T (s) + bB(s).

(1.1.29)

The constants a and b are easily determined from the conditions of orthogonality. For one has N · T = 0 ⇒ [N · T ] = 0 ⇒ N  · T = −N · T  = −κ|N |2 = −κ.

(1.1.30)

Similarly,  =0 N · B = 0 ⇒ [N · B] ⇒ N  · B = −N · B  = τ |N |2 = τ

(1.1.31)

 N  (s) = −κ T (s) + τ B(s).

(1.1.32)

and finally

Equations (1.1.27), (1.1.28), and (1.1.32) are called Frenet’s equations and were given in the theoretical summary at the beginning of this chapter. Now let us have a look at the applications of the above in the current problem. One has to start by tying the parameter s with ς using the following formula:

10

1 Examining Curves and Surfaces

s(ς ) =

ς 

x  (σ )2 + y  (σ )2 + z (σ )2 dσ =

0

ς =

ς 

cos2 σ + sin2 σ + sinh2 σ dσ

0



ς 1 + sinh σ dσ = 2

0

ς  = sinh σ  = sinh ς,

 cosh2 σ dσ = cosh σ dσ ς

0

0

(1.1.33)

0

which is the generalization of (1.1.13) for a curve in three dimensions. Reversing this relation, one obtains ς (s) = log(s +



s 2 + 1),

(1.1.34)

where the function on the right-hand side is nothing other than arsinh s, i.e., the inverse of the hyperbolic sine (see formula (14.1.54) in Part I). The curve parametrized with this new parameter will now be defined in the following way:   r(s) = sin ς (s), cos ς (s), s 2 + 1 ,

(1.1.35)

where for the first two coordinates the explicit expression ς (s) is not substituted so as not to overcomplicate the formula, and the last coordinate derives from the hyperbolic version of the Pythagorean trigonometric identity. By calculating the derivative of (1.1.35) with respect to s, the normalized tangent vector can be found:  d r   = ς (s) cos ς (s), −ς  (s) sin ς (s), s/ s 2 + 1 T (s) = ds 1 = √ [cos ς (s), − sin ς (s), s], (1.1.36) 2 s +1 √ where we have used the derivative: ς  (s) = 1/ s 2 + 1. It is easy to find out that this vector is in fact normalized to unity. For one has T (s)·T (s) =

s2

1 1 (cos2 ς (s)+sin2 ς (s)+s 2 ) = 2 (1+s 2 ) = 1. +1 s +1

(1.1.37)

As it has been argued, the curvature is given by    d T  1   κ= = ,  R ds 

(1.1.38)

1.1 Finding Curvature and Torsion of Curves

11

so the derivative of the vector (1.1.36) has to be found. We obtain   d T 1 − s cos ς (s) − s 2 + 1 sin ς (s), = ds (s 2 + 1)3/2   s sin ς (s) − s 2 + 1 cos ς (s), 1 ,

(1.1.39)

and then d T d T 2 . · = 2 ds ds (s + 1)2

(1.1.40)

Thereby, the curvature κ is found to be: √ 2 1 . κ(s) = = 2 R(s) s +1

(1.1.41)

It naturally depends on the parameter s and, therefore, on the point of the curve. It becomes clearer if one looks at the curve presented in Fig. 1.4. It is also obvious that this is in fact a truly spatial curve and, therefore, one should get the nonzero value of the torsion (but still dependent on the parameter s). The torsion τ will be found by calculating the derivative of the binormal vector  so first the vector itself should be found from the formula: B,   B(s) = T (s) × N(s), Fig. 1.4 The curve defined with the parametric equations (1.1.20). Obviously it is not a flat curve

(1.1.42) z

y x

12

1 Examining Curves and Surfaces

where N is the vector (1.1.39) normalized to unity, i.e.,   1 − s cos ς (s) − s 2 + 1 sin ς (s), N (s) = √ √ 2 s2 + 1   s sin ς (s) − s 2 + 1 cos ς (s), 1 .

(1.1.43)

Calculating the cross product does not present major difficulties and the vector B is obtained in the form:   1  s cos ς (s) − s 2 + 1 sin ς (s), B(s) = √ √ 2 s2 + 1   −s sin ς (s) − s 2 + 1 cos ς (s), −1 .

(1.1.44)

As it was previously indicated, this vector is found in the normalized form, and the reader is encouraged to check it by direct calculation. The last step is to calculate the derivative of the above expression with respect to s:   s d B = √ − s cos ς (s) − s 2 + 1 sin ς (s), ds 2(s 2 + 1)3/2   s s sin ς (s) − s 2 + 1 cos ς (s), 1 = 2 N (s), s +1

(1.1.45)

from which it occurs that the torsion is given by the formula: τ (s) = −

s . s2 + 1

(1.1.46)

In particular at the very start of the curve, i.e., for s = 0, one gets κ(0) =

√ 2,

or

1 R(0) = √ , 2

τ (0) = 0.

(1.1.47)

At this point, the curvature reaches the largest value (κ(s) is a decreasing function, as can be seen from (1.1.41)), but the graph is not twisted there. In turn the torsion of the curve is maximal for s = 1 (this is the maximum of the function τ (s)) and is equal to τ (1) = −1/2. Then also 1 κ(1) = √ , 2

or

R(1) =

√ 2.

(1.1.48)

It is still worth seeing at the end how the Frenet frame and trihedron look at some point. By substituting, for example, s = 0 into the formulas (1.1.36), (1.1.43), and (1.1.44), we obtain

1.1 Finding Curvature and Torsion of Curves

T (0) = [1, 0, 0], √ √  N(0) = [0, −1/ 2, 1/ 2], √ √  B(0) = [0, −1/ 2, −1/ 2].

13

(1.1.49)

All these vectors are clearly orthogonal. The point on the curve, for which s = 0, has the coordinates (0, 1, 1). In Fig. 1.4, this point is seemingly located on the y-axis, but this is only an illusion resulting from the logarithmic scale adopted for the z-axis, which must reflect the fact that the variable z grows exponentially (strictly speaking, as the hyperbolic cosine). In fact, the curve starts at a point located above the y-axis, i.e., for z = 1. As the reader surely knows, the equation of the plane passing through a point (x0 , y0 , z0 ) and perpendicular to a given vector v can be written in the form: v · [x − x0 , y − y0 , z − z0 ] = 0.

(1.1.50)

Thus, for the normal plane one gets the equation: T · [x − 0, y − 1, z − 1] = 0 ⇒ x = 0.

(1.1.51)

Similarly for the straightening plane one finds N · [x − 0, y − 1, z − 1] = 0 ⇒ y − z = 0,

(1.1.52)

and for a strictly tangent plane: B · [x − 0, y − 1, z − 1] = 0 ⇒ y + z = 2.

(1.1.53)

Problem 3 The curvature and torsion of the curve representing the intersection of the sphere x 2 + y 2 + z2 = 9 with the surface of the cylinder (x − 3/2)2 + y 2 = 9/4 (the so-called Viviani curve) will be found.

Solution In the first place, the convenient parametrization for our curve should be chosen. We are going to refer below to the well-known physical notions involving the motion of a point-like mass along a curve, so unlike previous exercises, the parameter will be

14

1 Examining Curves and Surfaces

denoted with the symbol t and called “time.” It seems to be most suitable if, for x and y, the parametrization related to the cylindrical surface is adopted, i.e., x(t) =

3 (1 + cos t), 2

y(t) =

3 sin t, 2

(1.1.54)

and the variable z is obtained from the equation of the sphere: 9 9 (1 + cos t)2 + sin2 t + z2 = 9. 4 4

(1.1.55)

This entails that z2 =

9 (1 − cos t) = 9 sin2 (t/2), 2

(1.1.56)

where the known formula 1 − cos α = 2 sin2 (α/2) has been used. As a result, one gets z(t) = 3 sin(t/2),

(1.1.57)

where the sign has been chosen arbitrarily since it does not affect the solution of the problem—when flipping the sign, the two “loops” of the Viviani curve shown in Fig. 1.5 simply swap their roles. Due to the presence of 1/2 under the sine, in order to obtain the entire curve, one must allow for t ∈ [0, 4π ].

z

x

Fig. 1.5 The Viviani curve defined in the text of the exercise

y

1.1 Finding Curvature and Torsion of Curves

15

At this point in the previous problems, we were changing the parametrization. Instead of the original parameter, the “traveled” distance measured along the curve was introduced. Such a parametrization is called the arc length parametrization. In this exercise, however, we will proceed in a different manner: the “time” parameter t still will be used. Let us start with the tangent vector, i.e., the velocity vector: v(t) = r˙ (t) =

3 [− sin t, cos t, cos(t/2)]. 2

(1.1.58)

The vector of acceleration 3 a (t) = v˙ (t) = − [2 cos t, 2 sin t, sin(t/2)] 4

(1.1.59)

does not need to be orthogonal to the curve (as opposed to T  (s)) because v is not normalized to any particular value. During the motion, the velocity can be changed, as refers to both its direction and its value. So, in general, the vector of acceleration can have both components: perpendicular and tangent to the trajectory. Let us now find the derivative of the acceleration: 3 a˙ (t) = v¨ (t) = [4 sin t, −4 cos t, − cos(t/2)]. 8

(1.1.60)

Note now that the determinant of the matrix created of the vectors v, a , and a˙ does not identically vanish. For one has ⎤ ⎤ ⎡ v(t) − sin t cos t cos(t/2) 27 81 det ⎣ a (t) ⎦ = − det ⎣ 2 cos t 2 sin t sin(t/2) ⎦ = cos(t/2), 64 32 a˙ (t) 4 sin t −4 cos t − cos(t/2) (1.1.61) ⎡

which means that these three vectors are in general linearly independent, i.e., the curve has actually the spatial character. For a flat curve, the vector a˙ would have to lie in the plane spanned by v and a and, therefore, would constitute their linear combination. The other way of expressing the spatial character of the curve is a˙ · ( v × a ) = 0.

(1.1.62)

Now the question arises, how to construct the curvature κ out of the vectors introduced above. The answer is provided by the Eq. (1.1.2), where in place of the centripetal acceleration ad , the component of the acceleration that is perpendicular to the velocity (i.e., a⊥ ) should be substituted. Denoting the parallel component with the symbol a , one gets from the properties of the cross product v v v v × a = × ( a + a⊥ ) = × a + × a⊥ , v v v    v =0

(1.1.63)

16

1 Examining Curves and Surfaces

which implies that a⊥ := | a⊥ | =

| v × a⊥ | | v × a | = . v v

(1.1.64)

Now making use of the formula (1.1.2), one can write κ=

| v × a | 1 a⊥ . = 2 = R v v3

(1.1.65)

The vectors v and a are given by the formulas (1.1.58) and (1.1.59), so in order to find the curvature, first one has to calculate the product: v × a =

9 [2 sin t cos(t/2) − cos t sin(t/2), −2 cos t cos(t/2) − sin t sin(t/2), 2], 8 (1.1.66)

and then 3 √ | v | = √ cos t + 3, 2 2  9 4 cos2 (t/2) + sin2 (t/2) + 4 | v × a | = 8 9 √ = √ 3 cos t + 13. 8 2

(1.1.67)

Now the formula (1.1.65) may be used, which entails  2 κ= · 3

3 cos t + 13 . (cos t + 3)3

(1.1.68)

Note that this expression is always positive and, therefore, the curvature does not vanish anywhere (the curve nowhere “straightens out”). In particular, let us consider the point of coordinates x = 3 and y = z = 0. This is the point on the Viviani curve obtained for t = 0, t = 2π , or t = 4π . As it seems from the figure, at these points, the curvature should be equal to the curvature of the sphere, i.e., R = 3 (κ = 1/3). Substituting into the formula (1.1.68) the values 0, 2π , or 4π for t, one can easily verify that actually κ = 1/3. Now let us pass to the torsion of the curve. The formula for τ analogous to (1.1.65) has the form τ=

( v × a ) · a˙ . | v × a |2

(1.1.69)

Note that the numerator of this expression is in fact the determinant (1.1.61) dealt with earlier, so it is not a surprise that (1.1.69) constitutes a measure of the spatial

1.1 Finding Curvature and Torsion of Curves

17

nature of the curve. In order to strictly justify it, one should first note that the acceleration of a body (as a vector) has components tangent to the trajectory and perpendicular to it, which can be written in the form v2  a = v˙ T + N. R

(1.1.70)

The velocity instead is fully tangent, i.e., v = v T . It is easy to calculate the cross product; one needs   v3  v2    N = B, v × a = v T × v˙ T + R R

(1.1.71)

 It entails also that | since T × T = 0, and T × N = B. v × a | = v 3 /R. When differentiating the expression (1.1.70) with respect to time in order to obtain a˙ , one can see that several terms emerge. However, only one of them is needed: exclusively that whose scalar product with B does not vanish. The others do not contribute to the numerator of (1.1.69). The term in question is that in which the derivative is acting on the vector N at the very end of the formula (1.1.70).  As a result, it is obtained Other vectors, i.e., T , T˙ , and N , are orthogonal to B. from (1.1.69): τ=

˙ v 3 /R B · (v 2 /R N) 1 d B  1 · N, = B · N˙ = − B˙ · N = − 3 2 v v ds (v /R)

(1.1.72)

i.e., the expression defined in the previous problem by the formula (1.1.27). When  the fact of transferring the above derivative over time from the vector N onto B, their orthogonality has been used: 0=

d   ˙ [B · N ] = B˙ · N + B · N, dt

(1.1.73)

and when replacing the derivative over t with that over s, the relation ds d d d = · =v . dt dt ds ds

(1.1.74)

After making use of (1.1.58), (1.1.59), and (1.1.60), one finally finds τ=

4 cos(t/2) . 3 cos t + 13

(1.1.75)

It should be stressed at the end that the formulas (1.1.65) and (1.1.69) obtained when solving this problem allow us to find the curvature and torsion without introducing the “arc length” parameter.

18

1 Examining Curves and Surfaces

1.2 Examining k-Surfaces in N Dimensions Problem 1 It will be examined whether the elliptic hyperboloid defined by the equation x 2 /4 + y 2 /9 − z2 = 1

(1.2.1)

is a surface in R3 .

Solution The set defined by the Eq. (1.2.1) cut with any “vertical” plane, i.e., that parallel to the z-axis, becomes a hyperbola, and with any “horizontal” plane (i.e., z = const)— an ellipse. Hence the appropriate name is used: elliptic hyperboloid. In order to examine whether this shell is a surface, one must first recall what is understood by k-dimensional surface in RN . As it was told in the theoretical introduction, this notion means a certain subset (denoted below with V ) of the space RN such that in some neighborhood of each of its points (strictly speaking the intersection of the neighborhood in RN with V ) is a graph of some mapping. For example, in R3 as a 2-surface—and this is just the case of this exercise—one classifies such set, that everywhere constitutes a graph of one of the functions: z(x, y), y(x, z), or x(y, z). What requires the emphasis is that such function does not need to be defined on the entire set V , but rather one expects the existence of at least one of them for each open and small “patch” defined for any point of the set V . In addition, the smoothness of these functions will be demanded, i.e., the existence and continuity of all partial derivatives. If this condition is fulfilled, the surface will be called smooth. In our further considerations, we will only deal with such smooth surfaces (later there will also occur piecewise smooth surfaces), hence it is assumed by default, even if it is not explicitly articulated. So, when can one be sure that the equation F (x, y, z) = 0 defines locally at least one of the functions: z(x, y), y(x, z), or x(y, z)? In order to answer this question, it is best to refer to the implicit function theorem, known to the reader from Chap. 8 of the second part of this book series. There, the independent variables were denoted with the symbol X (for example, this can be the variables x and y), and the dependent variables with the symbol Y (e.g., z). The theorem stated that if F is of the class C 1 , then for the existence of the function Y (X) (i.e., z(x, y) in our example), it is necessary and sufficient that  ∂F  = 0. ∂Y (X0 ,Y0 )

(1.2.2)

1.2 Examining k-Surfaces in N Dimensions

19

The symbol (X0 , Y0 ) means here simply (x0 , y0 , z0 ), i.e., a point belonging to the set V , the neighborhood of which is of interest to us. Since it has not been decided which variable will be treated as the dependent one (and this is in fact irrelevant for our findings), it is sufficient simply that one of the following conditions be met:  ∂F  = 0, ∂x (x0 ,y0 ,z0 )

 ∂F  = 0, ∂y (x0 ,y0 ,z0 )

 ∂F  = 0, ∂z (x0 ,y0 ,z0 )

(1.2.3)

which is equivalent to the requirement that at each point (x0 , y0 , z0 ) belonging to V , the Jacobian matrix   ∂F ∂F ∂F , , (1.2.4) F = ∂x ∂y ∂z have the rank equal to 1. In our case, after plugging in the expression (1.2.1) one finds   1 2 F = x, y, −2z , (1.2.5) 2 9 and it is obvious that the rank equals zero only when x = y = z = 0. This point of the space R3 does not, however, belong to V , since the Eq. (1.2.1) is not satisfied there. Thus, it has been shown that the rank of the matrix F  on the set V is equal to 1, which entails that V is the two-dimensional surface in R3 (Fig. 1.6).

z

Fig. 1.6 The surface described by the Eq. (1.2.1)

x

y

20

1 Examining Curves and Surfaces

One could terminate the solution of this problem here, but we will try below to look at this issue from the other side. Not always does one have to deal with the situation in which the subset is defined by an equation F (x1 , x2 , . . . , xn ) = 0 (or a system of equations of this type). Relatively common is to face the parametric description of the set V by the following system of relations as mentioned at the beginning of this chapter: ⎧ x1 = g1 (τ1 , τ2 , . . . , τk ), ⎪ ⎪ ⎨ x2 = g2 (τ1 , τ2 , . . . , τk ), ⎪ ... ⎪ ⎩ xN = gN (τ1 , τ2 , . . . , τk ),

(1.2.6)

collectively written in the form: x = G(τ ),

(1.2.7)

where τ belongs to a certain set D ⊂ Rk , and x ∈ RN . It is again assumed that the function G is smooth. The number of parameters (i.e., k), if a subset V turns out to be a surface, will determine its dimension. Naturally in this case, it is subject to the condition k ≤ N . Equations (1.2.6) should be treated locally, which means that for each “patch” of the set V , the function G can be distinct. Not always can any universal parametrization for an entire set be found. In such a case, the procedure, that we deal with below, has to be carried out for each of the functions G and for each “patch” separately. In order to determine whether the set is a surface, one can now—rather than examining the function F —analyze the function G. The appropriate theorem— probably known to the reader from the lecture—says that if the rank of the Jacobian matrix G is maximal at a given point, then on its appropriate neighborhood the set V is the graph of a certain mapping and constitutes, therefore, a surface. The Jacobian matrix G has N rows and k columns, so (since k ≤ N ) the maximal rank equals k. What is the geometrical interpretation of this condition? Well, if we change only one of the parameters (e.g., τi ), and fix the remaining k − 1 ones, then the equations (1.2.6) specify the curve in RN . The quantity ti = ∂x/∂τi is nothing other than a vector tangent to the curve. The matrix G has the form: G = [t1 , t2 , . . . , tk ] ,

(1.2.8)

and, therefore, its columns constitute the tangential vectors. The requirement of the maximal rank is in fact equivalent to the requirement that all tangent vectors are linearly independent. If this is not the case, one “loses” some dimension and the shell gets degenerated (“cusps” or “self-intersections” may appear as in Fig. 1.8).

1.2 Examining k-Surfaces in N Dimensions

21

In order to examine the set V in this manner, one must first propose some parametrization. The structure of the expression x 2 /4 + y 2 /9 suggests as the first choice: x = 2r cos ϕ,

y = 3r sin ϕ,

(1.2.9)

so that the equation F = 0 takes the form: r 2 − z2 = 1.

(1.2.10)

Then, if we recall the hyperbolic version of the Pythagorean trigonometric identity, we can write r = cosh w,

z = sinh w

(1.2.11)

and get the following result: ⎤ ⎡ ⎤ ⎤ ⎡ g1 (w, ϕ) x(w, ϕ) 2 cosh w cos ϕ ⎣ y(w, ϕ) ⎦ = ⎣ g2 (w, ϕ) ⎦ = ⎣ 3 cosh w sin ϕ ⎦ . z(w, ϕ) sinh w g3 (w, ϕ) ⎡

(1.2.12)

As it is easy to find out, this is the universal parametrization for the entire set V : for each point in V , one finds the values of the parameters w and ϕ in a unique way. What remains is to examine the rank of the matrix G . Calculating the derivatives with respect to parameters, one obtains ⎤ 2 sinh w cos ϕ −2 cosh w sin ϕ G = ⎣ 3 sinh w sin ϕ 3 cosh w cos ϕ ⎦ . cosh w 0 ⎡

(1.2.13)

In order to be convinced that the rank of this matrix is maximal, i.e., equal to 2, one needs to subsequently calculate the two-dimensional minors. Out of G , three square 2 × 2 submatrices can be created if the first, the second, or the third row is deleted. The following determinants are then obtained (indices denote columns that remained after the deletion): d12 = 6 sinh w cosh w cos2 ϕ + 6 sinh w cosh w sin2 ϕ = 6 cosh w sinh w, d13 = 2 cosh2 w sin ϕ, d23 = −3 cosh2 w cos ϕ.

(1.2.14)

It is sufficient that locally at least one of them does not vanish, and the rank of matrix G will be maximal. Note that (d23 /3)2 + (d13 /2)2 = cosh4 w > 0.

(1.2.15)

22

1 Examining Curves and Surfaces

These two minors cannot, therefore, simultaneously vanish (the function cosh w never equals zero) and consequently the rank is equal to 2. Again, we conclude that the subset V is a two-dimensional surface. It should be noted at the end that it can sometimes happen that at a certain point the matrix G will not have the maximal rank, but it will merely be a consequence of choosing the wrong parametrization (at a given point). Then, one should try another parametrization on the neighborhood of this point. As an example, let us look at the standard parametrization of a sphere: x(θ, ϕ) = R sin θ cos ϕ, y(θ, ϕ) = R sin θ sin ϕ, z(θ, ϕ) = R cos θ and the point which can conventionally be called the “north pole” (for θ = 0). Obviously at this point nothing unusual happens on the sphere—it does not differ from other points. It is the parametrization which gets “degenerated” there because the value of the parameter ϕ is not unequivocally defined. Another example of this kind appears in Problem 4c at the end of this chapter.

Problem 2 It will be examined if the intersection of two spheres: 

(x − 1)2 + y 2 + z2 = 4, (x + 1)2 + y 2 + z2 = 4

(1.2.16)

3 . If so, vectors that span the tangent planes at the points A(0, is a surface √in R √ and B(0, 3/2, 3/2) will be found.

√

3, 0)

Solution If one defines the functions f1 and f2 with the formulas: f1 (x, y, z) = (x − 1)2 + y 2 + z2 − 4, f2 (x, y, z) = (x + 1)2 + y 2 + z2 − 4,

(1.2.17)

it becomes clear that the function F from the previous exercise has now the form:  f1 (x, y, z) , F (x, y, z) = f2 (x, y, z) 

(1.2.18)

and the equation F (x, y, z) = 0 is a brief recording of the system of equations (1.2.16). One can expect, at least from the formal point of view, that these equations will allow us to determine two of the variables x, y, z as the functions of

1.2 Examining k-Surfaces in N Dimensions

23

the third one. Thereby, there are two type-Y variables and one type-X variable. If each point of the set defined by the equation F (x, y, z) = 0 (marked below as V ) has a neighborhood on which it is possible, it is a graph of some mapping Y (X) and one can call it a surface. From the implicit function theorem, we know that this takes place if at each of the points of the set V , the condition det

∂F = 0 ∂Y

(1.2.19)

is satisfied. As we remember, the function F has two components and the symbol Y also refers to two variables, so the matrix symbolically denoted as ∂F /∂Y is a square 2 × 2 matrix. There is no obstacle in the calculation of its determinant. The condition (1.2.19) simply means that the matrix is nonsingular. It should be stressed, however, that this condition is local in nature and in the neighborhoods of different points other variables can be considered as dependent ones (i.e., as being of the type Y ). To take account for this possibility, let us first calculate the entire Jacobian matrix: ⎡ ∂f ∂f 1 1 ⎢ ∂x ∂y  F = ⎣ ∂f ∂f 2 2 ∂x ∂y

∂f1 ⎤   ∂z ⎥ = 2(x − 1) 2y 2z . ∂f2 ⎦ 2(x + 1) 2y 2z ∂z

(1.2.20)

When choosing two variables, one actually picks out two columns of this matrix and such a reduced array—i.e., after the deletion of the unnecessary column—must be nonsingular. This corresponds to calculating two-dimensional minors, of which at least one should be (locally) nonzero. Using the language of algebra, one could simply say that (1.2.20) must have the rank equal to 2. Calculating the subsequent minors (indices refer to the labels of columns remaining after deleting one), one finds d12 = 4(x − 1)y − 4(x + 1)y = −8y, d13 = 4(x − 1)z − 4(x + 1)z = −8z, d23 = 4yz − 4yz = 0.

(1.2.21)

For all three being simultaneously equal to zero, the condition y = z = 0 would have to be met. After plugging these values into the equations (1.2.16), one, however, comes to the contradiction: (x − 1)2 = 4 ∧ (x + 1)2 = 4.

(1.2.22)

Upon adding these equations to each other, one obtains x 2 = 3, and upon subtracting: x = 0. Thereby, in the set V , no point where all three minors (1.2.21)

24

1 Examining Curves and Surfaces

z

Fig. 1.7 The one-dimensional surface (i.e., the curve) which is the intersection of the two spheres (1.2.16)

x

y

would vanish can be found. This means that the set V is a surface. It is depicted in Fig. 1.7 with the thick line. One still has to find the tangent space by which we understand the set of tangent vectors (i.e., the vector space) at a given point. Let us start with the point A. First, one needs to make sure that√the given point is located on the surface. Substituting its coordinates x = 0, y = 3, z = 0 into the equations (1.2.16), it is easily seen that both are satisfied. The tangent vectors that are looked for span the kernel of the matrix F  :  (1.2.23) F  A · v = 0. The rows of this matrix are in fact gradients of functions f1 and f2 (see formula (2.0.1)), and hence the vectors are normal to the surfaces defined by the  may not have any component tangent equations f1 = const and f2 = const (∇f to the surface f = const, since when moving along it, the value of the function f does not change by definition). The condition (1.2.23) simply means that the wanted vector v must be orthogonal to both of these normal vectors. When plugging the coordinates of the point A into (1.2.20) and writing the vector v in the form v = [a, b, c], one obtains the equation: ⎡ ⎤ √  a    −2 2√3 0 ⎣ ⎦ 0 , (1.2.24) b = 2 2 30 0 c from which it immediately follows that a = b = 0. The third coordinate, c, remains undetermined, but one can set it as equal to 1 since the normalization of this vector is inessential (all normalizations are equally good). As a result one can see that the vector tangent to the surface of V at the point a is

1.2 Examining k-Surfaces in N Dimensions

25

⎡ ⎤ 0 ⎣ vA = 0 ⎦ . 1

(1.2.25)

It spans the one-dimensional vector space which can be denoted as TA (V ). The point B belongs to V too, which can be easily checked. The following equation is required to be fulfilled at this point:  F  B · v = 0,

(1.2.26)

i.e., 

⎡ ⎤ √ √  a   −2 √6 √6 ⎣ ⎦ 0 . b = 6 6 2 0 c

(1.2.27)

This time, one gets a = 0 and b = −c. Setting c = 1, we have ⎤ 0 vB = ⎣ −1 ⎦ . 1 ⎡

(1.2.28)

This vector spans the space TB (V ). The reader could note at this point that instead of looking for the kernel of the Jacobian matrix, the knowledge acquired when investigating curves might be used. Then, the tangent vectors were simply derivatives of r(ς ) with respect to the parameter ς . Below we will see that the tangent space can actually be found with this method too. First, however, one needs to dispose of a parametric description of V . It can be obtained relatively simply. If the two equations (1.2.16) are subtracted from each other, one immediately gets x = 0. After inserting this result into any of the equations one obtains in turn y 2 + z2 = 3. Now the parametric relations can be easily proposed: x(ς ) = 0,

y(ς ) =

√

3 cos ς,

z(ς ) =

√ 3 sin ς,

(1.2.29)

and then the tangent vector is found: ⎡

⎤ √0 v(ς ) = ⎣ − 3 sin ς ⎦ . √ 3 cos ς

(1.2.30)

Since the point A corresponds to the value ς = 0 and the point B to the value ς = π/4, after inserting them into (1.2.30), the tangent vectors are immediately obtained in the form:

26

1 Examining Curves and Surfaces

⎡ ⎤ 0 √ ⎣ vA = 3 0 ⎦ , 1

vB =

⎤ ⎡ 0 3⎣ −1 ⎦ . 2 1

(1.2.31)

They differ from the previously found ones only by normalization factors.

Problem 3 It will be examined for which positive values of parameters r0 and R the torus defined by (ρ − R)2 + z2 = r02 ,

(1.2.32)

 where ρ = x 2 + y 2 , is the surface in R3 . Then the vectors spanning the tangent space at the point of coordinates (R − r0 , 0, 0) will be found.

Solution This time the function F dealt with in the previous exercises has the form: F (x, y, z) = (ρ − R)2 + z2 − r02 . Let us find the Jacobian matrix:       R R  2y 1 − 2z . F = 2x 1 − ρ ρ

(1.2.33)

(1.2.34)

It is obvious that all partial derivatives exist, apart from ρ = 0, which corresponds to the straight line x = y = 0. In order to check whether points of these coordinates are located on the torus, let us substitute these values into the Eq. (1.2.32) and see whether it is possible to be satisfied. In this way we get the condition: z2 = r02 − R 2 ,

(1.2.35)

and if the right-hand side is nonnegative, i.e., for r0 ≥ R, the solutions for z do exist. The condition r0 ≥ R means, however, that the torus has self-intersections. In order not to appeal solely to the reader’s imagination, in Fig. 1.8 the cross sections of the torus for r0 ≥ R and r0 < R are depicted. The points where the surface is intersected with itself do not have neighborhoods on which the shell would be the graph of a function. Thus, it is seen that in this case the set of points described with

1.2 Examining k-Surfaces in N Dimensions

27

Fig. 1.8 The cross sections of a torus in the case of self-intersections (on the left) and without self-intersections (on the right)

the Eq. (1.2.32) is not a surface in R3 . On the other hand for r0 < R, there are no intersections, and the rank of the Jacobian matrix (1.2.34) always equals 1, which means that one is dealing with a surface. The same conclusion emerges by means of the parametric description. The parametrization of a torus was introduced in Part II (see formula (13.2.21)): x = (R + r0 cos θ ) cos ϕ,

y = (R + r0 cos θ ) sin ϕ,

z = r0 sin θ.

(1.2.36)

These relations can be collectively written, when defining the function G: ⎡

⎤ (R + r0 cos θ ) cos ϕ G(θ, ϕ) = ⎣ (R + r0 cos θ ) sin ϕ ⎦ . r0 sin θ

(1.2.37)

If the set is a surface, then, as we know, the partial derivatives with respect to parameters are tangent vectors. They are columns of the Jacobian matrix: ⎡

∂x ⎢ ∂θ ⎢ ∂y ⎢  G =⎢ ⎢ ∂θ ⎣ ∂z ∂θ

∂x ∂ϕ ∂y ∂ϕ ∂z ∂ϕ

⎤ ⎥ ⎡ −r sin θ cos ϕ −(R + r cos θ ) sin ϕ ⎤ ⎥ 0 0 ⎥ ⎣ ⎥ = −r0 sin θ sin ϕ (R + r0 cos θ ) cos ϕ ⎦ . ⎥ ⎦ 0 r0 cos θ

(1.2.38)

When calculating the subsequent minors, one sees that the rank of this matrix is maximal (i.e., equal to 2): d12 = −(R + r0 cos θ )r0 sin θ, d13 = (R + r0 cos θ )r0 cos θ sin ϕ, d23 = −(R + r0 cos θ )r0 cos θ cos ϕ, (1.2.39)

28

1 Examining Curves and Surfaces

since it is easy to check that 2 2 2 + d13 + d23 = (R + r0 cos θ )2 r02 , d12

(1.2.40)

and if r0 < R, all three minors cannot simultaneously vanish. On the other hand for r0 ≥ R, the matrix G gets degenerated in a visible way for cos θ = −R/r0 , i.e., at the points of self-intersections. One can be certain of it by substituting this value into the formulas (1.2.36) and obtaining x = y = 0,

 z = ± r02 − R 2 .

(1.2.41)

These are the same points that have already been found in the first part of our solution. Assuming below that r0 < R, we are going to find the tangent space at the point of coordinates (R − r0 , 0, 0). As written above, the tangent vectors can be read off from the columns of the matrix (1.2.38) and the simultaneous substitution of the appropriate values of the parameters θ and ϕ. For the point in question, one has θ = π,

ϕ = 0,

(1.2.42)

which leads to the following tangent vectors: ⎡ ⎤ 0 v1 = ⎣ 0 ⎦ , 1

⎡ ⎤ 0 v2 = ⎣ 1 ⎦ , 0

(1.2.43)

the inessential normalization constants having been omitted. The reader is encouraged to check that these vectors actually span the kernel of the matrix (1.2.34), if in place of coordinates (x, y, z) one substitutes (R − r0 , 0, 0).

Problem 4 It will be examined whether the intersection of the hyperboloid x 2 +y 2 +z2 −w 2 = 1 and the hyperplane x − y − z = 1 is√ a surface in R4 . If so, the vectors spanning the tangent space at the point (1, 1, −1, 2) will be found.

Solution This time the issue takes place in four-dimensional space. However, there are two conditions provided in the text of the problem which restrict the number of

1.2 Examining k-Surfaces in N Dimensions

29

independent variables to two (4 − 2 = 2). Therefore, we are going to examine whether the set V defined through the above equations is a 2-surface in R4 . However, before starting, one should ensure that this set is not empty. To this end, let us examine whether the system of equations 

x 2 + y 2 + z2 − w 2 = 1, x−y−z=1

(1.2.44)

has any solutions. Solving the second equation for z and plugging it into the first one, one gets x 2 + y 2 + (x − y − 1)2 − w 2 = 1,

(1.2.45)

w 2 = x 2 + y 2 + (x − y − 1)2 − 1.

(1.2.46)

i.e.,

It is clear that for arbitrarily large values of the variables x and y the right-hand side is positive, and the equation has the solutions for w. It is also obvious that the solutions of the equation x − y − z = 1 for z always exist. This leads to the conclusion that the equations that define V are consistent. The function F introduced in the previous problems this time has the form: 

 x 2 + y 2 + z2 − w 2 − 1 F (x, y, z, w) = , x−y−z−1

(1.2.47)

and its derivative: F =



 2x 2y 2z −2w . 1 −1 −1 0

(1.2.48)

In order to determine the rank of the Jacobian matrix, one should now calculate successively the determinants of submatrices 2 × 2 obtained by the deletion of two columns. Even though there are as many as six matrices, writing out the respective minors does not pose any problem: d12 = −2(x + y), d13 = −2(x + z), d23 = −2(y − z), d14 = −d24 = −d34 = 2w.

(1.2.49)

For the rank of the matrix (1.2.48) to be smaller than 2, all of them should simultaneously vanish. Let us consider if in the set V there exists a point with these

30

1 Examining Curves and Surfaces

properties. First of all it should be required that w = 0. Then the equations resulting from the condition of the vanishing of d12 , d13 , and d23 would lead to y = z = −x. Upon inserting these values into the equations defining the set V , one gets the system 

3x 2 = 1, 3x = 1,

(1.2.50)

which is manifestly inconsistent. The conclusion which follows may be stated in the form: at each of the points of the set V , the rank of the matrix F  equals 2, which entails that on some neighborhood of each point, the set V constitutes the graph of a certain mapping, although in neighborhoods of distinct points generally different mappings can be defined. If so, V is the 2-surface in R4 . Leaving to the reader to propose the appropriate parametrization and to solve the problem using the second approach, we are going to find below the tangent vectors as those belonging to the kernel of the Jacobian matrix. Let us √ denote such general vector with the symbol v = [a, b, c, d]. At the point (1, 1, −1, 2) one has F





√  2 2 −2 −2 2 = , 1 −1 −1 0 

√ (1,1,−1, 2)

(1.2.51)

and the requirement that  F  (1,1,−1,√2) · v = 0

(1.2.52)

leads to the two equations for the coordinates a, b, c, d: 

√ 2a + 2b − 2c − 2 2 d = 0, a − b − c = 0.

(1.2.53)

Now the values of c and d can be found in an easy way: c = a − b,

d=

√

2 b,

(1.2.54)

and finally we are able to write down a vector v as the linear combination of two tangent base vectors: ⎡

⎤ ⎡ ⎤ ⎤ ⎡ a 1 0 ⎢ b ⎥ ⎢0⎥ ⎢ 1 ⎥ ⎥ ⎢ ⎥ ⎥ ⎢ v=⎢ ⎣ a − b ⎦ = a ⎣ 1 ⎦ + b ⎣ −1 ⎦ . √ √ 2b 2 0

(1.2.55)

1.3 Examining Ruled Surfaces

31

1.3 Examining Ruled Surfaces Problem 1 It will be examined whether the cone described by the equation z2 = x 2 + y 2 is a ruled surface.

Solution We will now be limited to two-dimensional surfaces in R3 . As we know, they can be described parametrically through the relation r(u, v), where symbols u and v denote the parameters. Of course there exist various parametrizations, but this section is devoted to those that can be given the following form:  r(u, v) = α (u) + v β(u),

(1.3.1)

spoken of in the theoretical summary. For any fixed value of the parameter u ( α and β will then become some constant vectors) this formula describes the ordinary straight line in R3 . Now, if one allows the parameter u to evolve too, then the formula (1.3.1) will mean that the whole surface has to be constructed of straight lines only. They come out of the curve α (u) in the direction defined by  the vector β(u). Such a surface is called a ruled surface. This kind of surface can play an important role in architecture (e.g., describe different types of vaults) or in the construction of boats, when it becomes necessary because of the material used for the hull. Working to find the solution of this exercise, one sees that the shell (i.e., the cone) is axially symmetric. Because of it, let us write first x(ρ, ϕ) = ρ cos ϕ,

y(ρ, ϕ) = ρ sin ϕ,

(1.3.2)

and after plugging that into the cone equation, we find z2 = ρ 2 cos2 ϕ + ρ 2 sin2 ϕ = ρ 2 ,

(1.3.3)

which leads to z = ±ρ. If “+” were chosen, only the upper half of the cone sketched in Fig. 1.9 would be described, and for “−” only the lower half. In order to describe the entire cone, let us set z = ρ, but unlike the regular polar parametrization, both positive and negative values of ρ will be allowed. The range of variability of the parameters is, therefore, as follows: ρ ∈ R,

ϕ ∈ [0, 2π [.

(1.3.4)

32

1 Examining Curves and Surfaces

Fig. 1.9 The cone as a ruled surface

This parametrization can be now easily given the desirable form (1.3.1): ⎤ ⎡ ⎤ ⎤ ⎡ 0 x(ρ, ϕ) cos ϕ r(ϕ, ρ) = ⎣ y(ρ, ϕ) ⎦ = ⎣ 0 ⎦ +ρ ⎣ sin ϕ ⎦ . 0 z(ρ, ϕ) 1       ⎡

α

(1.3.5)

β

One can say that by changing the parameter ϕ, various straight lines are chosen, and by modifying ρ, one moves along a given line. The cone referred to in the text of this exercise is shown in Fig. 1.9 together with several exemplary straight lines (1.3.1). As one can see, the base curve (see theoretical introduction) is degenerated here to a single point. If the parameter ρ was shifted by a certain number, for example by 1 (ρ → ρ+1), then instead of (1.3.5) the following parametrization would be obtained: ⎤ ⎤ ⎡ ⎤ ⎡ cos ϕ x(ρ, ϕ) cos ϕ r(ϕ, ρ) = ⎣ y(ρ, ϕ) ⎦ = ⎣ sin ϕ ⎦ +ρ ⎣ sin ϕ ⎦ . 1 z(ρ, ϕ) 1       ⎡

α

(1.3.6)

β

This parametrization is equally good leading to the same straight lines (that is, it has nothing to do with the doubly ruled surfaces dealt with in the next problem), and the base curve becomes now the circle x(ϕ) = cos ϕ, y(ϕ) = sin ϕ lying in the plane z = 1.

1.3 Examining Ruled Surfaces

33

Problem 2 It will be examined whether the one-sheeted hyperboloid x 2 + y 2 − z2 = 1 is a ruled and doubly ruled surface.

Solution For the hyperboloid described with the formula given in the text of this problem it is natural to use the following parametrization: x(u, ϕ) = cosh u cos ϕ,

y(u, ϕ) = cosh u sin ϕ,

z(u, ϕ) = sinh u,

(1.3.7)

where u ∈ R,

ϕ ∈ [0, 2π [.

(1.3.8)

It stems from the observation that the expression in question depends on x and y only through the combination x 2 + y 2 =: r 2 , which suggests the use of the Euclidean trigonometric identity, and then it depends on r and z through r 2 − z2 , which in turn hints at its hyperbolic counterpart. The reader can easily be convinced by inserting (1.3.7) into the equation of the hyperboloid, and it will be automatically satisfied. Unfortunately, the chosen parametrization, although getting imposed, does not have the required form (1.0.4). This is not to say, of course, that the considered hyperboloid is not a ruled surface since there can exist different parametrizations. At this point, it might help to look at a figure. If a parametrization in terms of straight lines (1.0.4) is to exist, these lines should look as it is illustrated in Fig. 1.10. Surely, having found this type of straight lines, we know that a second family of them, inclined in the opposite direction, must exist. Obviously it stems from the symmetry of the surface and would simultaneously mean that it is doubly ruled. The figure suggests that the lines one is looking for have to go out of the circle x 2 + y 2 = 1 located in the plane z = 0. It is the circle of the strongest contraction of the hyperboloid. If so, that is to say, the base curve of the surface has been found. Using the notation of the Eq. (1.0.4), one can write ⎤ ⎡ cos θ (1.3.9) α (θ ) = ⎣ sin θ ⎦ . 0  Let us then denote β = [a, b, c]. Then one must One still has to find the vector β. have ⎤ ⎤ ⎡ ⎡ ⎡ ⎤ cos θ x(t, θ ) a ⎦ ⎣ ⎦ ⎣ ⎣ r(θ, t) = y(t, θ ) = sin θ + t b ⎦ . (1.3.10) 0 z(t, θ ) c       α

β

34

1 Examining Curves and Surfaces

Fig. 1.10 The hyperboloid as a ruled surface

Let us now check if it is possible to adjust the numbers a, b, c so that (1.3.10) represents the parametrization of the hyperboloid The straight line, we are looking for (and, therefore, the vector β as well) at the point at which it intersects the base curve, i.e., at (cos θ, sin θ, 0), must be perpendicular to the vector of coordinates [cos θ, sin θ, 0]. This is due to the reflection symmetry z → −z of the solid. Thus the following condition must hold: β · [cos θ, sin θ, 0] = 0 ⇒ a cos θ + b sin θ = 0.

(1.3.11)

One can choose a = − sin θ , b = cos θ , and c remains undetermined for a while. Any other choice of the parameters a and b leads only to the rescaling of t in the Eq. (1.3.10). One should now use the fact that the coordinates x, y, z satisfy the hyperbola equation. This requirement allows us to determine the value of c. Therefore, the following equation has to be satisfied: (cos θ − t sin θ )2 + (sin θ + t cos θ )2 − t 2 c2 = 1,

(1.3.12)

which, after some simple transformations, implies that c2 = 1, i.e., c = ±1. Thus, we were able to pursue our plan. The two different parametrizations that were looked for are: ⎤ ⎤ ⎡ ⎤ ⎡ cos θ x(t, θ ) − sin θ r(θ, t) = ⎣ y(t, θ ) ⎦ = ⎣ sin θ ⎦ + t ⎣ cos θ ⎦, 0 z(t, θ ) ±1       ⎡

α

(1.3.13)

β

the former being demonstrated in the figure. Thereby the hyperboloid has proved to be the doubly ruled surface.

1.3 Examining Ruled Surfaces

35

Problem 3 It will be examined whether the hyperbolic paraboloid defined through the relation z = 2xy is a ruled surface.

Solution One can start the search of the parametrization in the form of straight lines with the observation that the equation z = 2xy is linear, separately, in y and in z. Therefore, if the value of x was fixed, for example by putting x = u, an equation which defines a plane perpendicular to the x-axis, the obtained condition z = 2uy describes a plane as well. The intersection of the two planes is a straight line. Now, changing the value of u, one will get various lines contained in the paraboloid, i.e., the needed parametrization. In order to find the base curve, let us consider the intersection of the family of straight lines:  x = u, (1.3.14) z = 2uy with the plane z = 0. Obviously the result is x = u, y = z = 0, and thus the base curve is α (u) = [u, 0, 0].  For each fixed u, it must be parallel to the Now, one has to find the vector β. straight line (1.3.14). In order to determine the vector, it is sufficient to select the two points on this line, for example, (u, 0, 0) and (u, 1, 2u). By subtracting their coordinates, one gets ⎡ ⎤ 0  (1.3.15) β(u) = ⎣ 1 ⎦. 2u This allows us to write the needed parametrization in the form: ⎤ ⎡ ⎤ ⎡ ⎡ ⎤ u x(u, v) 0 r(u, v) = ⎣ y(u, v) ⎦ = ⎣ 0 ⎦ + v ⎣ 1 ⎦, 0 z(u, v) 2u       α

(1.3.16)

β

and thus conclude that the surface in question is a ruled one. By inserting the above expressions for x(u, v), y(u, v), and z(u, v) into the equation of the paraboloid, one can easily verify that it is automatically satisfied. In Fig. 1.11, the examined surface is depicted along with a few straight lines of the type (1.3.16).

36

1 Examining Curves and Surfaces

Fig. 1.11 Hyperbolic paraboloid as a ruled surface

Problem 4 Given the point P of coordinates (2, 1, 1) on the one-sheeted hyperboloid defined by the equation x 2 + y 2 − 4z2 = 1. All generating lines passing through this point will be found.

Solution This problem can be solved with the use of the parametrization similar to (1.3.13). The considerations leading to it will not be repeated, because the modification of the formulas is obvious: due to the presence of “4” in the hyperboloid equation, the coordinate z is now simply multiplied by 1/2: ⎤ ⎡ ⎤ ⎤ ⎡ − sin θ cos θ x(t, θ ) r(θ, t) = ⎣ y(t, θ ) ⎦ = ⎣ sin θ ⎦ + t ⎣ cos θ ⎦ . ±1/2 0 z(t, θ ) ⎡

(1.3.17)

Let us first consider the parametrization obtained by choosing “+.” In the first place, one needs to determine what the values of the parameters t and θ at P are. To this end, the following set of equations is to be solved: ⎧ ⎨ cos θ − t sin θ = 2, sin θ + t cos θ = 1, ⎩ t/2 = 1.

(1.3.18)

We immediately get t = 2, and plugging this value into the first two equations we find: sin θ = −3/5, cos θ = 4/5. The value of the angle θ does not need to be known and, moreover, it is not any “nice” number. It is sufficient to use the obtained values of the trigonometric functions. By inserting these numbers into (1.3.17), one gets the equation for the first line:

1.4 Exercises for Independent Work

37

Fig. 1.12 The straight lines contained in the hyperboloid x 2 + y 2 − 4z2 = 1 passing through the point P (2, 1, 1)

P

⎤ ⎡ ⎤ ⎤ ⎡ 3/5 4/5 x(t) r(t) = ⎣ y(t) ⎦ = ⎣ −3/5 ⎦ + t ⎣ 4/5 ⎦ . 1/2 0 z(t) ⎡

(1.3.19)

In order to find the second one, we again use the parametrization (1.3.17), but now choose −1/2. The requirement for the straight line to pass through the point P leads to the system: ⎧ ⎨ cos θ − t sin θ = 2, sin θ + t cos θ = 1, ⎩ −t/2 = 1.

(1.3.20)

This time one has t = −2, sin θ = 1, cos θ = 0 and the needed equation becomes ⎤ ⎡ ⎤ ⎡ ⎤ −1 0 x(t) r(t) = ⎣ y(t) ⎦ = ⎣ 1 ⎦ + t ⎣ 0 ⎦ . −1/2 0 z(t) ⎡

(1.3.21)

Both lines, along with the discussed surface, are shown in Fig. 1.12.

1.4 Exercises for Independent Work Exercise 1 Assuming that a curve is given parametrically by r(ς ), demonstrate the following equations for the curvature and torsion: κ=

|r  × r  | , |r  |3

τ=

(r  × r  ) · r  . |r  × r  |2

38

1 Examining Curves and Surfaces

Exercise 2 Given a curve in the parametric form: x(ς ) = cos ς , y(ς ) = sin ς , z(ς ) = ς 2 , where ς ∈ R. Find the curvature and the torsion of the curve at the points, for which ς = π/2 and ς = 1.

Answers κ(π/2) = (5√+ π 2 )1/2 /(1 + π 2 )3/2 , τ (π/2) = π/(5 + π 2 ); κ(1) = 3/(5 5), τ (1) = 2/9. √ Exercise 3 Given a curve in the parametric form: x(ς ) = ς , y(ς ) = log ς , z(ς ) = ς , where ς ∈]0, ∞[. Find the curvature and the torsion at the points on the curve, for which ς = 1 and ς = 1/4.

Answers

√ √ κ(1) = 2 2/9, τ (1) = −1/9; κ(1/4) = 1/(3 2), τ (1/4) = −16/81.

Exercise 4 Verify whether (a) The set defined with the equation x 3 = z2 − y 2 − 2y, (b) The cone x 2 + y 2 = z2 , (c) The set defined parametrically: x(u, v) = 1/2 sinh u cos v, y(u, v) = 1/3 sinh u sin v, z(u, v) = cosh u, where u ∈ [0, ∞[ and v ∈ [0, 2π [, is a surface in R3 .

Answers (a) Yes. (b) No (because of the point (0, 0, 0)). (c) Yes.

Exercise 5 Find the vectors spanning the space tangent to the surface: √ (a) x 2 + 2y 2 + 3z2 = 6 in R3 at the point (0, 3/2, 1), 2 2 2 2 4 (b) x + y − z − w = 1 in R at the point (1, 1, 0, 1).

1.4 Exercises for Independent Work

39

Answers

√ (a) v1 = [1, 0, 0], v2 = [0, 3/2, −1]. (b) v1 = [1, 0, 0, 1], v2 = [0, 1, 0, 1], v3 = [0, 0, 1, 0].

Exercise 6 Check whether the given surfaces are ruled ones and if so, find the appropriate parametrizations: (a) The cylinder x 2 + y 2 = 4, (b) The set defined with the equation 4z = x 2 − y 2 .

Answers (a) Ruled surface, parametrization: r(θ, t) = [2 cos θ, 2 sin θ, 0] + t[0, 0, 1]. (b) Ruled surface, parametrization: r(u, t) = [u, u, 0] + t[1, −1, u].

Chapter 2

Investigating Conditional Extremes

The second chapter is concerned with the so-called conditional extremes, i.e., those appearing due to the presence of some additional conditions to be fulfilled. This concept is explained in full detail when solving the first problem. Below some principal notions are collected. Given a differentiable scalar function f : RN ⊂ D → R, where D is an open set and (x1 , x2 , . . . , xN ) ∈ RN denote the Cartesian coordinates. The gradient of this function is the vector field composed of all partial derivatives, i.e., in the form:  := gradf := ∇f



 ∂f ∂f ∂f . , ,..., ∂x1 ∂x2 ∂xn

(2.0.1)

This vector at any point of D indicates the direction of the fastest increase of the value of the function f (x1 , x2 , . . . , xN ). The Lagrange multipliers method of finding the conditional extremes of the function f (x1 , x2 , . . . , xN ) subject to the set of additional conditions gi (x1 , x2 , . . . , xN ) = 0, i = 1, . . . , r,

(2.0.2)

which describe r smooth surfaces in RN , consists of the following steps. 1. A certain new function F (x1 , x2 , . . . , xN ) called the Lagrange function is defined according to the relation: F (x1 , x2 , . . . , xN ) = f (x1 , x2 , . . . , xN ) −

r !

λi gi (x1 , x2 , . . . , xN ),

(2.0.3)

i=1

where λi for i = 1, . . . , r are certain unknown real numbers.

© Springer Nature Switzerland AG 2020 T. Rado˙zycki, Solving Problems in Mathematical Analysis, Part III, Problem Books in Mathematics, https://doi.org/10.1007/978-3-030-38596-5_2

41

42

2 Investigating Conditional Extremes

2. All stationary points (see Chap. 7 of Part II) are found by solving the set of N equations written in short as ∇F (x1 , x2 , . . . , xN ) = 0

(2.0.4)

supplemented by r conditions (2.0.2). One has in common N + r equations and N + r unknowns (x1 , . . . , xN , λ1 , . . . , λr ). 3. The behavior of the function F at a given stationary point P is examined by  analyzing the sign of the quadratic form hT · F  P · h. This method is known from the second part of the book series with the slight modification that now the allowed vectors h have to be tangent to all surfaces (2.0.2) at P , i.e., the following conditions are to be satisfied:  gi P · h = 0, i = 1, . . . , r.

(2.0.5)

The justification and application of this method are discussed comprehensively below.

2.1 Using the Method of the Lagrange Multipliers Problem 1 The local extremes of the function f (x, y) = x + y on the circle x 2 + y 2 = 4 will be found.

Solution In Part II in Chap. 7, we were looking for extremes of multivariable functions. It was then found that at an extremal point all partial derivatives must be equal to zero, i.e.,  = 0. the gradient (2.0.1) of the function vanishes: ∇f However, it can happen—and this is just the case we are concerned with now— that one is interested not in an extreme of a given function as such, but only as subject to a certain condition. Let us imagine for example that we participate in a mountain excursion along a tourist trail. The path on which we are moving can raise and lower, which we perceive as achieving certain local maxima or minima, although in fact, we reach neither the top of any hill nor the lowest point of any valley. The path is merely leading us this way, and one cannot get off the trail and move perpendicularly to it. Let us look at the exemplary in Fig. 2.1. It shows a fragment of some map with a few terrain contours sequentially labeled (with the increasing height): h1 , h2 , up to h5 . In the considered area, there is no mountaintop.

2.1 Using the Method of the Lagrange Multipliers

43

Fig. 2.1 The exemplary trail on a mountain side marked with the dashed line. The symbols h1 , h2 , etc. refer to the increasingly “higher” contours. The point P constitutes the conditional maximum

However, when walking along the trail marked with a dashed line, we achieve the maximum at the point P where our path reaches the highest contour. The “maxima” or “minima” one is overcoming are just the so-called conditional extremes mentioned in the theoretical introduction at the beginning of this chapter (they are met under the condition that one is going along the trail). When solving this problem, we will learn how to find them. It should be stressed, however, that we will be primarily interested only in cases in which the condition (conditions) cannot be explicitly solved and, thereby, the problem cannot be boiled down to that of the unconditional extremes of a function of fewer variables. The examples dealt with in this section will be treated as such that the elimination of any variable is impossible or unviable. It is easy to see that the function given in the text of this problem, i.e., f (x, y) = x + y, has the graph which constitutes an inclined plane and, therefore, it does not have extremes anywhere. Both partial derivatives are constantly equal to unity and it is clear that they cannot vanish. If we were looking for the unconditional extremes, then our conclusion would be: the function f has no local extremes. On the other hand, however, if one is allowed to move only along the circle x 2 + y 2 = 4 and at the end we return to the starting point, as in this exercise, then certainly one has to meet somewhere the maximal and minimal values of the function (they even can be met several times). The question arises, how to find these points and what should be modified within the procedure used in the previous part of the book. The answer seems to be quite obvious. Now it is not the gradient of the function that must vanish, but only its component tangent to the path! For, we are not interested if and how the value of the function varies in a direction perpendicular to the trail, because one cannot get off it. It is still important to stress that by the “path” or “trail” the curve lying in the plane xy is understood (i.e., contained in the domain of the function), just as it would be plotted on the flat map. Any rising and lowering when moving is included in the values read off the contours of the function f during the excursion.

44

2 Investigating Conditional Extremes

Fig. 2.2 The decomposition of the gradient of the function f onto components tangent and perpendicular to the path. In the general case, this path obviously does not have to be closed

Let us suppose that the two-dimensional gradient of the function f looks on the plane xy, as it is drawn in Fig. 2.2. It can be decomposed onto the component tangent to the trajectory and perpendicular to it. We are interested in the former one. It can be expressed in a natural way as

 ∇f



 − ∇f  = ∇f . ⊥

(2.1.1)

 on the right-hand side can readily be calculated, since the The vector ∇f formula for the function f is known. However, how can one find its perpendicular component? Well, one should use the equation that defines the path which has the  one knows that the vector perpendicular to form g(x, y) = 0. When calculating ∇g,  cannot the curve g is obtained. This is because of the obvious observation that ∇g have any tangential component, since the function g does not change its value (i.e., is constantly equal to zero), when moving along

the curve. It was already mentioned  in Sect. 1.2. Consequently the vector ∇f must be proportional to the gradient ⊥ of g (these vectors may differ only by a fixed ratio). One can, therefore, write   ∇f = λ∇g, (2.1.2) ⊥

with λ denoting a constant. This means that at extremes, we are looking for, the following condition must be satisfied:   − λ∇g  = ∇(f  − λg) = 0. ∇f = 0 ⇒ ∇f (2.1.3) 

This latter form suggests that maybe instead of dealing with the function f (x, y), from the very beginning one should introduce the auxiliary function with the formula: F (x, y; λ) := f (x, y) − λg(x, y),

(2.1.4)

2.1 Using the Method of the Lagrange Multipliers

45

 = 0), and examining its extremes as “unconditional” ones (with the requirement ∇F determining the additional unknown λ using the condition g(x, y) = 0. This method is called the Lagrange multipliers method as it was mentioned in the theoretical introduction. In the present case, the multiplier is only one—it is just the number λ—because the surface (curve), to which the domain of the function was reduced, is defined by one condition: g(x, y) = 0. However, one can easily generalize this formula—and the reader has certainly encountered the appropriate expression during lectures of analysis—for the case when things take place in RN , and the surface is an intersection of r sets described with the equations: g1 (x1 , x2 , . . . , xN ) = 0, g2 (x1 , x2 , . . . , xN ) = 0, . . . , gr (x1 , x2 , . . . , xN ) = 0. (2.1.5) Then one will have r Lagrange multipliers, and the function F is defined as (see (2.0.3)) F (x1 , x2 , . . . , xN ; λ1 , λ2 , . . . , λr ) := f (x1 , x2 , . . . , xN ) − λ1 g1 (x1 , x2 , . . . , xn ) −λ2 g2 (x1 , x2 , . . . , xN ) − · · · − λr gr (x1 , x2 , . . . , xN ).

(2.1.6)

Thereby, in our problem r additional unknowns appear, but fortunately we have at our disposal the very same number of additional equations (2.1.5). Applying the method described above to the current exercise, we first define the function F : F (x, y; λ) = x + y − λ(x 2 + y 2 − 4),

(2.1.7)

and then require that all its derivatives vanish. Including the additional condition the following set of equations is obtained: ⎧ ⎨ 1 − 2λx = 0, 1 − 2λy = 0, ⎩ 2 x + y 2 = 4.

(2.1.8)

Subtracting the two former equations from each other, one gets 2λ(x − y) = 0.

(2.1.9)

The quantity λ cannot be equal to zero, as this would be in conflict with the first and second equations,√so the only possibility remains x = y. The last equation (2.1.8) then gives x = ± 2. Thereby, two points A and B suspected of being the extremes are found: √ √ A( 2, 2),

√ √ B(− 2, − 2).

(2.1.10)

46

2 Investigating Conditional Extremes

√ At these points the √values of the Lagrange multipliers can be found: λA = 1/(2 2) and λB = −1/(2 2). One can now proceed to examine whether the points A and B are actually local extremes. The method, well known to us (see the second part of the book), based on the investigation of the matrix F  can be used. This matrix has the form: ⎡

∂ 2F ⎢ ∂x 2 F  = ⎢ ⎣ ∂ 2F ∂x∂y

⎤ ∂ 2F   ∂x∂y ⎥ ⎥ = −2λ 0 . ∂ 2F ⎦ 0 −2λ ∂y 2

(2.1.11)

It now becomes clear why we needed the values of the Lagrange multiplier. Even after having found the points (2.1.10) but without knowing λ, one would be unable to assess whether or not they are extremes and of which √ kind. Considering the first point, after setting λ = 1/(2 2), one easily sees that the matrix is negative-definite: ⎤ ⎡ 1   ⎢ − √2 0 ⎥ ⎥ F   = ⎢ ⎣ 1 ⎦. A 0 −√ 2

(2.1.12)

It is obvious since both its eigenvalues are negative. Such a situation corresponds to a maximum of the function. Remember, however, that it would be sufficient that the matrix be negative-definite only in the subspace of vectors that are tangent to the path. In this example, one was not obliged to make this further reduction because the sign of the matrix F  is determined for any vector. However, one should not always count on such a comfortable result, so we are going to show below how to perform the appropriate reduction. As we know from Sect. 7.2 of Part II, examining the sign on the matrix F  corresponds to examining that of the quadratic form:   h · F  · h, T

 

A

 hx . h= hy 

where

(2.1.13)

This time, however, only tangent vectors h come into play, i.e., those fulfilling the condition:  √ √  ∇g  · h = 0, i.e., 2 2hx + 2 2hy = 0, (2.1.14) A

which implies that hy = −hx. By inserting this result into the quadratic form (2.1.13), one gets

2.1 Using the Method of the Lagrange Multipliers

⎡ ⎤ 1  0 ⎥  ⎢−√   T   2 ⎢ ⎥ hx h · F  · h = hx , hy ⎣ 1 ⎦ hy A 0 −√ 2 √ 1 2 = − √ (hx + h2y ) = − 2 h2x . 2

47

(2.1.15)

Since hx may not equal zero (otherwise hy = 0 and in consequence h would become a null vector), the above expression is always negative, √ which means that the function f has at A the conditional maximum (fmax = 2 2). As to the point B, we proceed in an analogous way. The matrix of the second derivatives this time has the form: ⎤ ⎡ 1  0 √  ⎥ ⎢ 2 ⎥ F   = ⎢ (2.1.16) ⎣ 1 ⎦ B 0 √ 2 and is positive-definite, which can be seen already at this stage. Nonetheless, reducing it to the subspace of the tangent vectors, we first require   ∇g  · h = 0,

i.e.

√ √ − 2 2hx − 2 2hy = 0,

(2.1.17)

B

which entails that hy = −hx . Then for the (reduced) quadratic form one finds ⎡ ⎤ 1    0 √   ⎢ 2 ⎥ hx T   ⎢ ⎥ h · F  · h = hx , hy ⎣ 1 ⎦ hy B 0 √ 2 √ 1 = √ (h2x + h2y ) = 2 h2x > 0. 2

(2.1.18)

It is then obvious that the function has at this point the conditional minimum √ (fmin = −2 2). At the end, it is worth adding that if the equation g = 0 describes a compact set (as in the present case), examining of the second derivatives may be abandoned. In order to determine whether a given function has extremes, it is sufficient to find and compare its values achieved at all “suspected” points (see Sect. 7.1 in Part II).

48

2 Investigating Conditional Extremes

z

x

extremes

y

Fig. 2.3 The conditional extremes of function f (x, y). The graph is shown “from underneath”

Problem 2 The critical points of the function f (x, y, z) = x 2 + 2y 2 + z on the surface g(x, y, z) = x 2 + y 2 − z2 + 2z = 0 will be found and explored.

Solution This time we are dealing with a function of three variables, so it would be difficult to draw its graph. However, the method of the Lagrange multipliers can still be used. To do this, one should define the auxiliary function F (x, y, z; λ), according to formulas (2.0.3) or (2.1.6): F (x, y, z; λ) = x 2 + 2y 2 + z − λ(x 2 + y 2 − z2 + 2z).

(2.1.19)

 = 0 together with the condition g(x, y, z) = 0 leads to The requirement that ∇F the system of equations: ⎧ 2x − 2λx = 2(1 − λ)x = 0, ⎪ ⎪ ⎨ 4y − 2λy = 2(2 − λ)y = 0, ⎪ 1 + 2λz − 2λ = 0, ⎪ ⎩ 2 x + y 2 − z2 + 2z = 0.

(2.1.20)

The first equation yields two possibilities: λ = 1 or x = 0. Let us consider them in turn. 1. λ = 1. Upon inserting this value into the second and third equations, one immediately obtains y = 0 and z = 1/2. However, with these values, the last equation

2.1 Using the Method of the Lagrange Multipliers

49

cannot be satisfied, since one obtains a contradiction: x 2 + 3/4 = 0. Thereby, the possibility λ = 1 must be rejected. 2. x = 0. The second equation offers now two options: λ = 2 or y = 0. Assuming the first possibility, from the third equation one obtains z = 3/4, but the fourth one again leads to a contradiction: y 2 + 15/16 = 0. We are then left with y = 0. After having inserted the values x = y = 0, the fourth equation takes the form z(2 − z) = 0, which implies that z = 0 or z = 2. Summarizing, the two “suspected” points have been found: A(0, 0, 0), λA =

1 2

and

1 B(0, 0, 2), λB = − , 2

(2.1.21)

the values of the Lagrange multiplier having been established with the use of the third equation (2.1.20). Now one needs to analyze the matrix of second derivatives. Simple calculations give ⎤ ⎡ 2(1 − λ) 0 0 F  = ⎣ (2.1.22) 0 2(2 − λ) 0 ⎦ . 0 0 2λ When examining the first point, the value 1/2 should be plugged in for λ. Then one gets the matrix ⎤ ⎡  100  F   = ⎣ 0 3 0 ⎦ , (2.1.23) A 001 which obviously is positive-definite. Even without any reduction to the subspace of the vectors tangent to g, it can be seen that at the point A the function f has a minimum. Nevertheless, let us find the tangent vectors. If such vector is denoted as h = [hx , hy , hz ], the following condition must be satisfied: ⎡ ⎤   hx   ∇g  · h = 0, i.e. [2x, 2y, −2z + 2] · ⎣ hy ⎦ A A hz ⎡ ⎤ hx = [0, 0, 2] · ⎣ hy ⎦ = 2hz = 0, hz

(2.1.24)

which implies that hz = 0, and hx , hy are arbitrary. This means that there are two linearly independent tangent vectors, e.g., [1, 0, 0] and [0, 1, 0]. This should not be surprising because in the present problem the condition g = 0 describes a twodimensional surface, so at each point there is a two-dimensional tangent plane.

50

2 Investigating Conditional Extremes

The reduced quadratic form: ⎤⎡ ⎤ ⎡  hx  100  hT · F   · h = hx , hy , 0 ⎣ 0 3 0 ⎦ ⎣ hy ⎦ A 001 0 = h2x + 3h2y

(2.1.25)

is positive-definite and, therefore, it corresponds to the minimum of the function f . In the case of the second “suspected” point, the value −1/2 should be substituted for λ into (2.1.22), giving ⎤ ⎡  30 0  F   = ⎣ 0 5 0 ⎦ . B 0 0 −1

(2.1.26)

This matrix is neither positive nor negative, since it has eigenvalues of different signs. Therefore, the reduction to the two-dimensional subspace is necessary. This time any tangent vector must satisfy the requirement:   ∇g  · h = 0,

(2.1.27)

B

or   [2x, 2y, −2z + 2]

B

⎡ ⎤ ⎤ hx hx · ⎣ hy ⎦ = [0, 0, −2] · ⎣ hy ⎦ = −2hz = 0, hz hz ⎡

(2.1.28)

leading again to the conclusion that hz = 0, and hx , hy are arbitrary. The reduced quadratic form is ⎤⎡ ⎤ ⎡  hx  30 0  hT · F   · h = hx , hy , 0 ⎣ 0 5 0 ⎦ ⎣ hy ⎦ B 0 0 −1 0 = 3h2x + 5h2y .

(2.1.29)

As can be seen, after the reduction it has become positive-definite, and thus one is again dealing with a minimum of the function f . It may seem puzzling to the reader how it is possible that the function has only two minima without any maxima or even saddle points in between. It seems to be counterintuitive. However, a careful look at the set defined with the equation g(x, y, z) = 0 explains this apparent paradox. This area consists of two separate sheets (it is simply a two-sheeted hyperboloid) and on each of them the function f has one minimum. A similar situation will be also encountered in the next problem.

2.1 Using the Method of the Lagrange Multipliers

51

Problem 3 The critical points of the function f (x, y, z) = 2/x 2 +y +4z2 on the surface defined with the system of equations: 

x 2 − y 2 + z2 = 0, x 2 + y 2 + z2 = 2

(2.1.30)

will be found and examined.

Solution Let us introduce the notation: g1 (x, y, z) = x 2 − y 2 + z2

and

g2 (x, y, z) = x 2 + y 2 + z2 − 2.

(2.1.31)

Both these functions are defined on R3 so each of the conditions gi (x, y, z) = 0 defines a two-dimensional surface, and both of them together—a curve (or a sum of curves) being their intersection. Thereby, we will be looking for the extremes of the function f (x, y, z) on a certain curve. It is easy to see that due to the condition x = 0 resulting from the domain of the function, (2.1.30) does not define closed curves. Due to the presence of two conditions, the two Lagrange multipliers will be introduced: F (x, y, z; λ1 , λ2 ) = f (x, y, z) − λ1 g1 (x, y, z) − λ2 g2 (x, y, z) =

2 + y + 4z2 − λ1 (x 2 − y 2 + z2 ) − λ2 (x 2 + y 2 + z2 − 2). x2

(2.1.32)

The requirement of the vanishing of the gradient of the function F leads, together with the conditions (2.1.30), to the set of equations: ⎧ ⎪ −4/x 3 − 2λ1 x − 2λ2 x = 0, ⎪ ⎪ ⎪ ⎪ ⎨ 1 + 2y(λ1 − λ2 ) = 0, 8z − 2λ1 z − 2λ2 z = 2z(4 − λ1 − λ2 ) = 0, ⎪ ⎪ ⎪ x 2 − y 2 + z2 = 0, ⎪ ⎪ ⎩ x 2 + y 2 + z2 = 2.

(2.1.33)

Solving this type of equations can be tedious and cumbersome, so a good idea is to start with the simplest one, i.e., the third one, thanks to its product form. It entails that z = 0 or λ1 + λ2 = 4. Let us examine these possibilities.

52

2 Investigating Conditional Extremes

1. z = 0. Plugging this value into the last two equations leads to a subsystem which is easy to solve:  2  2 x − y 2 = 0, x = 1, ⇒ (2.1.34) x 2 + y 2 = 2, y 2 = 1, the equations on the right having been obtained by adding and subtracting those on the left. Their solutions can be written out: x = ±1 and y = ±1. One still has to check if for these values of x, y, and z the first two of the equations (2.1.33) can be satisfied as well, and to find the corresponding values of the parameters λ1 and λ2 . This task is very easy, since the equations obtained by using the Lagrange multipliers method are always linear in these quantities. In this way the following set of “suspected” points is obtained: 5 3 A(1, 1, 0), λ1A = − , λ2A = − , 4 4 5 3 B(−1, 1, 0), λ1B = − , λ2B = − , 4 4 3 5 C(1, −1, 0), λ1C = − , λ2C = − , 4 4 3 5 D(−1, −1, 0), λ1D = − , λ2D = − . 4 4

(2.1.35)

2. λ1 + λ2 = 4. Inserting this value into the first of the equations (2.1.33) leads to the condition −4/x 3 − 8x = 0, which cannot be fulfilled (both terms on the left-hand side are positive—for negative x—or negative—for positive x). The points A, B, C, D are then the only “suspected” ones. Let us now examine them in detail. First one has to calculate the matrix of the second derivatives: ⎤ ⎡ 12 − 2λ − 2λ 0 0 1 2 ⎥ ⎢ x4 ⎥, F  = ⎢ (2.1.36) ⎦ ⎣ 0 2λ − 2λ 0 1

0

2

0

8 − 2λ1 − 2λ2

and then all the points are dealt with in turn. 1. The point A(1, 1, 0), λ1A = −5/4, λ2A = −3/4. In this case, one has ⎤ ⎡  16 0 0  F   = ⎣ 0 −1 0 ⎦ . A 0 0 12

(2.1.37)

2.1 Using the Method of the Lagrange Multipliers

53

This matrix is neither positive- nor negative-definite (it has eigenvalues of different signs), but remember that we are interested in its sign only after the reduction to the subspace of tangent vectors. This subspace is naturally one dimensional, as the conditions (2.1.30) describe a curve (i.e., a one-dimensional “surface”) in R3 . So one has to find one tangential vector h = [hx , hy , hz ] satisfying the set of equations: ⎧   ⎪ ⎪ ⎨ g1  · h = 0, A  ⎪ ⎪ ⎩ g2  · h = 0,

(2.1.38)

A

i.e., 

2(hx − hy ) = 0, 2(hx + hy ) = 0.

(2.1.39)

This system immediately implies that hx = hy = 0 and hz is arbitrary (but nonzero). The reduced quadratic form is, therefore, ⎤⎡ ⎤ ⎡  0 16 0 0   hT · F   · h = 0, 0, hz ⎣ 0 −1 0 ⎦ ⎣ 0 ⎦ B hz 0 0 12 = 12 · h2z > 0.

(2.1.40)

The conclusion can be drawn that at the point A the function f (x, y, z) has a conditional minimum. 2. The point B(−1, 1, 0), λ1B = −5/4, λ2B = −3/4. The matrix of the second derivatives is identical as that found for A. One only needs to perform the appropriate reduction. As usual it is required that ⎧   ⎪ ⎪ ⎨ g1  · h = 0, B  ⎪  ⎪ ⎩ g2  · h = 0,

(2.1.41)

B

i.e., 

−2(hx + hy ) = 0, −2(hx − hy ) = 0.

(2.1.42)

The same conclusion is then obtained: hx = hy = 0 and hz is arbitrary. The reduced quadratic form is again given by (2.1.40), which implies that at the point B the function reaches a conditional minimum.

54

2 Investigating Conditional Extremes

3. The point C(1, −1, 0), λ1C = −3/4, λ2C = −5/4. The conditions ⎧   ⎪ ⎪ ⎨ g1  · h = 0, C  ⎪  ⎪ ⎩ g2  · h = 0,

(2.1.43)

C

in the same way as before lead to hx = hy = 0 and the reduced quadratic form is ⎤⎡ ⎤ ⎡  0 16 0 0   hT · F   · h = 0, 0, hz ⎣ 0 1 0 ⎦ ⎣ 0 ⎦ B hz 0 0 12 = 12 · h2z > 0.

(2.1.44)

Thereby, a new conditional minimum has been found. 4. The point D(−1, −1, 0), λ1D = −3/4, λ2D = −5/4. The results are the same as for C. In conclusion, one can see that four minima and no maxima have been found. It might seem that this stays in contradiction with the observation made in the first exercise. Indeed, if we go for an excursion along a certain path and at the end we come back to the starting point, it is inevitable to reach both the smallest and the largest height. So why this time only minima have been encountered? The answer is similar to the one at the end of the previous problem: in the case of the above path we do not come back to the starting point! Having a closer look at the conditions (2.1.30), it can be concluded that they do not represent a closed curve but four open fragments (semicircles without ends). It was mentioned at the very beginning. This is a consequence of the fact that the plane yz is excluded from the domain of the function. As a result of the curve, which is the intersection of the sphere with the cone surface and is defined with the equations (2.1.30)—this is simply two circles—four points are removed: (0, 1, 1), (0, 1, −1), (0, −1, 1), and (0, −1, −1).

Problem 4 The extremes of the function given by f (x1 , x2 , . . . , xN ) = −

N !

xi log xi ,

(2.1.45)

i=1

defined on the set ]0, 1[N , will be found, subject to the condition x1 + x2 + . . . + xN = 1.

2.1 Using the Method of the Lagrange Multipliers

55

Solution The function (2.1.45) is called the information entropy and the variable xi is the probability of the occurrence of an ith event from among the set of N possible ones. The probability of whatever event to happen is, naturally, equal to 1 and hence the side condition. The problem encountered in this exercise can be then stated as follows: at what distribution of probabilities will the entropy take its extreme values? If necessary, the domain of the function f can be easily extended to the set [0, 1]N , by the appropriate limits. Since lim x log x = 0,

(2.1.46)

x→0+

if either of the following numbers x1 , x2 , . . . , xN equals zero, the corresponding term will simply disappear from the sum (2.1.45). Now let us define the function F , introducing the Lagrange multiplier λ: F (x1 , x2 , . . . , xN ; λ) = −

N !

N ! xi log xi − λ( xi − 1).

i=1

(2.1.47)

i=1

After differentiating over consecutive variables, one gets the set of equations: ⎧ ⎪ ⎪ log x1 + 1 + λ = 0, ⎪ ⎪ ⎪ ⎨ log x2 + 1 + λ = 0, ··· ⎪ ⎪ ⎪ log xN + 1 + λ = 0, ⎪ ⎪ ⎩ x + x + . . . + x = 1. 1 2 N

(2.1.48)

These equations ensure that at the critical point the values of all variables are identical. Taking into account the last equation, one obtains x1 = x2 = . . . = xN =

1 . N

(2.1.49)

One still has to determine the nature of this point. It is easy to calculate the matrix of the second derivatives: ⎤ ⎡ 1 − 0 ··· 0 ⎥ ⎡ −N 0 · · · 0 ⎤ ⎢ x1 ⎥ ⎢ 1 ⎥ ⎢ 0 −N · · · 0 ⎥ ⎢ 0 − ··· 0 ⎥ ⎢  ⎥. (2.1.50) F =⎢ ⎥=⎢ x2 ⎣ ⎦ ⎥ ⎢ · · · · · · · · · · · · ⎢ ··· ··· ··· ··· ⎥ ⎣ 0 0 · · · −N 1 ⎦ 0 0 ··· − xN

56

2 Investigating Conditional Extremes

It is negative-definite in an obvious way, even without any reduction to the subspace of tangent vectors. This means that at the critical point the function has a maximum and f (1/N, 1/N, . . . , 1/N) = log N . This result indicates that the entropy achieves the largest value if the probabilities are uniformly distributed. This simultaneously means that our knowledge about the system is the smallest. Therefore, one can say that the entropy is a measure of our ignorance. Any inequality between the values of probabilities x1 , x2 , . . . , xN means that we have gained some additional knowledge (as compared to the complete ignorance) and the value of the entropy gets smaller. It can easily be seen that in the extreme case, when our knowledge about the system becomes full—that is, if one of the xi ’s equals 1 (then the others must vanish) and one can be sure that a particular event occurs—the entropy vanishes. This zero is the minimal value of the entropy since, because of the condition xi ≤ 1, the expression (2.1.45) is nonnegative.

2.2 Looking for Global Extremes Problem 1 The global extremes of the function given with the formula f (x, y) = x 2 + (x − y)2

(2.2.1)

A = {(x, y) ∈ R2 | x 2 ≥ y 2 ∧ |x| ≤ 4}.

(2.2.2)

will be found on the set

Solution In Sect. 7.1 of Part II, we were concerned with maxima and minima of functions on compact sets. Now we come back to this topic, using the Lagrange multipliers method introduced in this chapter. As we remember, searching for global extremes of functions on compact sets consists of two steps. First one had to identify all critical points located in the interior of a given set, and second one needed to examine the behavior of the function on the boundary. And this is just the place where conditional extremes enter into play. We need to start by drawing the set A. The set of equations: 

x2 ≥ y2, |x| ≤ 4

(2.2.3)

2.2 Looking for Global Extremes

57

Fig. 2.4 Graphical representation of the set A

can be rewritten in the form of an alternative: (y ≤ x ∧ y ≥ −x ∧ 0 ≤ x ≤ 4) ∨ (y ≥ x ∧ y ≤ −x ∧ −4 ≤ x ≤ 0) . (2.2.4) It is now easy to demonstrate graphically the obtained predicate, as well as the set A. It is made up of two triangles, as shown in Fig. 2.4. Let us then proceed to investigate the function f . 1. The interior of A. In this case we deal with unconditional extremes. By following exactly the same way as in Sect. 7.1 of the previous part of this book series, where this procedure is described in detail, one finds the critical points, and, without checking if they correspond to extremes or saddle points, the appropriate values of the function are calculated. Later this list will be compared to the values achieved at “suspected” points located on the boundary. Thus one has the set of equations: ⎧ ∂f ⎪ ⎪ = 4x − 2y = 0, ⎪ ⎨ ∂x ⎪ ⎪ ∂f ⎪ ⎩ = 2y − 2x = 0, ∂y

(2.2.5)

58

2 Investigating Conditional Extremes

whose only solution is x = y = 0. This point, however, is not situated in the interior of A. It means that “suspected” points do not appear in the interior and the global extremes are located on the border. 2. The border of A. As shown in the figure, the border is composed of two segments contained in the vertical straight lines x = ±4 and the two contained in the inclined lines, described jointly with the equation x 2 − y 2 = 0. In the first case, one simply plugs x = ±4 into the formula for the function f and obtains a one variable function to examine. And this is what is done below. In the second case, one could do the same. The side condition is simple enough that it could be accomplished without difficulties. Remember, however, that not always will it be possible and one needs to learn how to solve more complex cases as well. For this reason we denote g(x, y) = x 2 − y 2 and treat this part of the problem as an exercise for finding extremes subject to the condition g(x, y) = 0 and, therefore, for using the Lagrange multipliers. We start, however, with the easier part. With x = 4, one gets the quadratic trinomial in the variable y in the form f (4, y) = 16 + (4 − y)2 ,

(2.2.6)

from which it is visible that the minimum (vertex) corresponds to y = 4. The point of coordinates (4, 4) belongs to the set A. The value of f (4, 4) = 16 is the first one that should be noted as the potential extremal value. In turn for x = −4, one gets f (−4, y) = 16 + (4 + y)2 ,

(2.2.7)

and the minimum (again equal to 16) is achieved at the point (−4, −4). One has yet to record the values obtained at two other “corners” of the set A: f (4, −4) = f (−4, 4) = 80. As we know from Part II, these values could not be found with the use of the differential approach. Now we are going to use the method of the Lagrange multipliers for two other edges of the set A. Let us define the function F (x, y; λ) = f (x, y) − λg(x, y) = x 2 + (x − y)2 − λ(x 2 − y 2 )

(2.2.8)

and require its gradient to vanish. Together with the side condition, the following system of equations is obtained: ⎧ ⎨ (2 − λ)x − y = 0, x − (1 + λ)y = 0, ⎩ 2 x − y 2 = 0.

(2.2.9)

2.2 Looking for Global Extremes

59

One obvious solution of this system is x = y = 0, so the new value should be appended to our list: f (0, 0) = 0. Since the function defined in the content of the problem is nonnegative, this value certainly corresponds to the global minimum. Moreover, this minimum is unique. To find it out, it is sufficient to insert y = x and y = −x into the first two equations. In the first case, one gets  (1 − λ)x = 0, ⇒ x = 0, (2.2.10) −λx = 0, and in the second



(3 − λ)x = 0, (2 + λ)x = 0,

⇒ x = 0.

(2.2.11)

If one does not wish to resolve the condition x 2 − y 2 = 0, which may not be possible in some more complicated case, the same result can be obtained upon multiplying the first two equations of (2.2.9) by each other (after having separated the terms with x and y on opposite sides) and then using the fact that x 2 = y 2 . This leads to the condition (1−2λ)x 2 = 0, and hence either λ = 1/2 or x = 0 (and consequently y = 0). This former case, however, again leads to the result x = y = 0, since after substituting into the first two equations one obtains 5 4 x = 0. In conclusion, the largest value assumed by the function f on the set A equals 80 (at the points (4, −4) and (−4, 4)), and the global minimum corresponds to the value 0 (at the point (0, 0)).

Problem 2 The global extremes of the function f (x, y, z) = x + y − z on the set A = {(x, y, z) ∈ R3 | x 2 + y 2 + z2 ≤ 9}

(2.2.12)

will be found.

Solution This time the set A is a closed ball. To establish that no critical points of the function f appear in its interior is not a problem. When calculating the partial derivatives, one finds that all of them are nonzero ∂f ∂f ∂f = =− = 1. ∂x ∂y ∂z

(2.2.13)

60

2 Investigating Conditional Extremes

Thereby, only the border of the set A has to be examined, which leads to the standard issue of looking for conditional extremes with the side condition in the form: g(x, y, z) := x 2 + y 2 + z2 − 9 = 0.

(2.2.14)

As usual the function F is defined as F (x, y, z; λ) = f (x, y, z)−λg(x, y, z) = x+y−z−λ(x 2 +y 2 +z2 −9)

(2.2.15)

 = 0, we get the set of equations: and after having imposed the requirement ∇F ⎧ 2λx = 1, ⎪ ⎪ ⎨ 2λy = 1, ⎪ 2λz = −1, ⎪ ⎩ 2 x + y 2 + z2 = 9,

(2.2.16)

the side condition having been appended. The values of x, y, and z easily obtained from the first three equations can be plugged into the last one: 

1 2λ

2

 +

1 2λ

2

 +

−1 2λ

2 = 9.

(2.2.17)

Solving it for λ, one finds √ √ 3 3 3 ∨ λ=− . = 9 ⇒ λ = 6 6 4λ2

(2.2.18)

Having found these values of the Lagrange multiplier, one can now get from the first three equations “suspected” points. There are two of √ √(2.2.16) √ the coordinates √ of√the √ them: P ( 3, 3, − 3) and Q(− 3, − 3, 3). All that remains is to calculate and compare the values of the function f at these points: √ √ √ √ √ √ √ √ (2.2.19) f ( 3, 3, − 3)) = 3 3, f (− 3, − 3, + 3)) = −3 3. We conclude that at the point P the function f has the global maximum and at Q the global minimum.

Problem 3 The global extremes of the function 2  z2 2 2 − 2xy f (x, y, z) = 6(x + y ) + 3z − 4xy − 3x + 3y + 2 2

2

2

(2.2.20)

2.2 Looking for Global Extremes

61

will be found on the set A = {(x, y, z) ∈ R3 | x 2 + y 2 + z2 /4 ≤ 1}.

(2.2.21)

Solution The former examples were relatively easy to solve due to the simplicity of the functions. At the end of this section we will, therefore, be concerned with a somewhat more intricate case. The procedure to be followed, however, remains the same: to find critical points in the interior of the set A (the closed ellipsoid) and then to examine the behavior of the function at the border using the method of the Lagrange multipliers. 1. The interior of the set A. At the critical points, all three partial derivatives must be equal to zero. One, therefore, obtains the set of equations: ⎧ ∂F ⎪ ⎪ = 4(3x − y)(1 − 3x 2 − 3y 2 − z2 /2 + 2xy) = 0, ⎪ ⎪ ∂x ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ∂F = 4(3y − x)(1 − 3x 2 − 3y 2 − z2 /2 + 2xy) = 0, (2.2.22) ⎪ ∂y ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ∂F ⎪ ⎩ = 2z(3 − 3x 2 − 3y 2 − z2 /2 + 2xy) = 0. ∂z The fulfillment of these equations requires the vanishing of the relevant factors on their left-hand sides. One has in total 2 × 2 × 2 = 8 possibilities, which will be examined one by one, starting with the simplest one. (a) y = 3x, x = 3y, z = 0. These conditions mean that x = y = z = 0. The critical point P1 (0, 0, 0) belongs to the set A, so it must be included in the list of “suspected” points. We have f (0, 0, 0) = 0 and this is the first one from among the values that should be noted. (b) 3x 2 + 3y 2 + z2 /2 − 2xy = 1, x = 3y z = 0. Substituting x = 3y and z = 0 into√ the first condition, we obtain 24y 2 = 1, which implies that y = ± 6/12. In √ this way √ two critical points√are found, both lying in the interior of A: P ( 6/4, 6/12, 0) and 2 √ P3 (− 6/4, − 6/12, 0). When calculating the value of the function f for both of them, one finds √ √ √ √ f ( 6/4, 6/12, 0) = f (− 6/4, − 6/12, 0) = 1. These values should be noted too.

(2.2.23)

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2 Investigating Conditional Extremes

(c) y = 3x, 3x 2 + 3y 2 + z2 /2 − 2xy = 1, z = 0. It is visible that, as compared to the previous case, x and √ y simply √ swap their roles.√Then there are two further critical points: P ( 6/12, 6/4, 0) and 4 √ P5 (− 6/12, − 6/4, 0). Because of the symmetry of the formula (2.2.20) when replacing x ↔ y, we know that at both of them the value of the function equals 1. (d) 3x 2 + 3y 2 + z2 /2 − 2xy = 1, 3x 2 + 3y 2 + z2 /2 − 2xy = 1, z = 0. Two of the above conditions overlap, so the solutions will constitute the entire curve and not discrete points only. After eliminating z, one gets the condition, which can be given the form 3x 2 + 3y 2 − 2xy = 1.

(2.2.24)

If one now introduced some new variables defined with the relations ξ = x + y and η = x − y in place of x and y, the Eq. (2.2.24) would become an ellipse equation: ξ 2 + 2η2 = 1.

(2.2.25)

Because of the linear nature of relations between variables x, y and ξ, η, the axes of the new system are only inclined respective to the old ones. In particular, the ξ axis, which corresponds to the condition η = 0, is just the straight line y = x, and η, for which ξ = 0, is the line y = −x. The axes of both systems, together with the curve (2.2.25) lying in the plane z = 0, are depicted in Fig. 2.5. The intersection of the set A with the plane z = 0 is also drawn. The figure shows that the ellipse (2.2.25) is entirely contained in the set A. That this is really the case, one can also easily be convinced in an analytical way, rewriting the Eq. (2.2.24) in the form x2 + y2 +

1 1 (x − y)2 = , 2 2

(2.2.26)

which obviously entails that x 2 +y 2 < 1, i.e., the ellipse lies inside the set A. In order to obtain the value of the function f at any point of this curve, it is sufficient to arrange the terms in (2.2.20) in the following way: f (x, y, z) = 2[3(x 2 + y 2 ) − 2xy ] + 3  z2    =1

=0

  1 2 2 2 2 z − 3x + 3y − 2xy +  = 1.   2  =1

(2.2.27)

=0

Thereby, the value of 1 is obtained as that of the possible global extreme.

2.2 Looking for Global Extremes

63

Fig. 2.5 The curve (2.2.24) together with the intersection of the set A with the plane z=0

(e) y = 3x, x = 3y, 3x 2 + 3y 2 + z2 /2 − 2xy = 3. As in the √ item (a), one immediately gets x = y = 0, which implies that z = ± 6. The points of these coordinates are located, however, away from the set A. (f) All other cases lead to contradictions, since they require that the following conditions are simultaneously satisfied: 3x 2 + 3y 2 + z2 /2 − 2xy = 1 and 3x 2 + 3y 2 + z2 /2 − 2xy = 3. 2. The border of the set A. Let us now denote g(x, y, z) = x 2 + y 2 + z2 /4 − 1 and define the function: F (x, y, z; λ) = f (x, y, z) − λg(x, y, z).

(2.2.28)

We are not going to explicitly write out its form, but rather to provide the equations obtained after the differentiation: ⎧ ∂F ⎪ = 4(3x − y)(1 − 3x 2 − 3y 2 − z2 /2 + 2xy) − 2λx = 0, ⎪ ⎪ ⎪ ∂x ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ∂F ⎪ ⎪ ⎪ = 4(3y − x)(1 − 3x 2 − 3y 2 − z2 /2 + 2xy) − 2λy = 0, ⎨ ∂y ⎪ ⎪ ⎪ ∂F ⎪ ⎪ ⎪ = 2z(3 − 3x 2 − 3y 2 − z2 /2 + 2xy) − λz/2 = 0, ⎪ ⎪ ∂z ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ 2 x + y 2 + z2 /4 = 1.

(2.2.29)

64

2 Investigating Conditional Extremes

This system is quite intricate, but one has to be able to deal with such a system as well. To start with, let us try to simplify it a bit. First of all, the quantity z2 can be removed from the brackets in the first three equations thanks to the side condition. This yields after some minor transformations ⎧ 2(3x − y)[(x − y)2 + 1] + λx = 0, ⎪ ⎪ ⎨ 2(3y − x)[(x − y)2 + 1] + λy = 0, ⎪ z[(x − y)2 − 1 + λ/4] = 0, ⎪ ⎩ 2 x + y 2 + z2 /4 = 1.

(2.2.30)

Now the first two equations can be replaced by taking their sum and difference. It should be pointed out that the operations which are performed on the equations do not constitute any universal procedure; they are merely the result of observing that a simpler system is obtained. So if the reader ever meets the necessity of solving this kind of a complex system, the first step should be to carefully look at the structure of the equations and use one’s imagination to think of a way to simplify it. Then, we obtain ⎧ (x + y)[4(x − y)2 + 4 + λ] = 0, ⎪ ⎪ ⎨ (x − y)[8(x − y)2 + 8 + λ] = 0, ⎪ z[4(x − y)2 − 4 + λ] = 0, ⎪ ⎩ 2 x + y 2 + z2 /4 = 1.

(2.2.31)

Now let us observe that the expressions in the brackets in the first two equations cannot simultaneously vanish. Therefore, the following possibilities come into play (a) x + y = 0 and x − y = 0. These conditions mean that x = y = 0 and the last two equations give z = ±2, λ = 4. Thus, one gets two successive “suspected” points: P6 (0, 0, 2)

and

P7 (0, 0, −2).

(2.2.32)

Let us still note the values of the function f at these points: f (0, 0, 2) = f (0, 0, −2) = 8.

(2.2.33)

(b) x − y = 0 and 4(x − y)2 + 4 + λ = 0. We immediately get λ = −4, and z = 0 from the third of the equations (2.2.31). The side condition gets simplified to 2x 2 = 1 and, in consequence, the new “suspected” points are found: √ √ P8 (1/ 2, 1/ 2, 0)

and

√ √ P9 (−1/ 2, −1/ 2, 0).

(2.2.34)

2.3 Exercises for Independent Work

65

The values of the function at these points are as follows: √ √ √ √ f (1/ 2, 1/ 2, 0) = f (−1/ 2, −1/ 2, 0) = 0.

(2.2.35)

(c) x + y = 0 and 8(x − y)2 + 8 + λ = 0. This implies that λ = −32x 2 − 8 and the third equation of (2.2.31) can be given the form: − z(16x 2 + 12) = 0.

(2.2.36)

The only solution is now obviously z = 0, and consequently the side condition requires 2x 2 = 1. This means that two more interesting points have been found: √ √ P10 (1/ 2, −1/ 2, 0)

and

√ √ P11 (−1/ 2, 1/ 2, 0)

(2.2.37)

with the function values: √ √ √ √ f (1/ 2, −1/ 2, 0) = f (−1/ 2, 1/ 2, 0) = −8.

(2.2.38)

At the end, the outcome needs to be reviewed. It shows that the largest value (equal to 8) is adopted by the function √ f at the √ points (0, 0, ±2), √ and√the smallest value (equal to −8) at the points (1/ 2, −1/ 2, 0) and (−1/ 2, 1/ 2, 0).

2.3 Exercises for Independent Work Exercise 1 Find conditional extremes of the functions: (a) f (x, y) = x 2 + y 2 on the set A = {(x, y) ∈ R2 | x 4 − y 4 = 1}, (b) f (x, y, z) = x + y + z on the set A = {(x, y, z) ∈ R3 | x 2 + y 2 + z2 = 9}, (c) f (x, y) = x(y + z) on the set A = {(x, y, z) ∈ R3 | yz = 1 ∧ x + y + z = 0}.

Answers (a) Local minima at the points (±1, √ 0).√ √ (b) Local maximum at the point ( 3, 3, 3), local minimum at the point √ √ √ (− 3, − 3, − 3). (c) Local maxima at the points (−2, 1, 1), (2, −1, −1).

66

2 Investigating Conditional Extremes

Exercise 2 Find global extremes of the functions: (a) f (x, y, z) = x 2 − y 2 , on the set A = {(x, y, z) ∈ R3 | x 2 /4 + y 2 + z2 ≤ 1}, (b) f (x, y, z) = x + y + z2 , on the set A = {(x, y, z) ∈ R3 | x 2 + y 2 ≤ 1 ∧ 0 ≤ z ≤ 1}.

Answers (a) Global minima at the points (0, ±1, 0), global maxima at the points (±2, 0, 0). √ √ (b) Global minimum at the point (− 2/2, − 2/2, 0), global maximum at √ √ the point ( 2/2, 2/2, 1).

Chapter 3

Investigating Integrals with Parameters

In this chapter we are interested in functions defined as integrals

b F (t) =

f (x, t) dx,

(3.0.1)

a

where a and b are some numbers but may be infinite too. Our main concern constitutes the limits of such functions, their continuity, differentiability, or integrability. These properties are also applied to find the values of certain integrals. The following continuity theorem holds: Given a function f (x, t) continuous as a function of two variables on the rectangle [a, b] × [A, B]. Then the function F (t) defined with the formula (3.0.1) is continuous on [A, B], which also means that

b lim F (t) =

b lim f (x, t) dx =

t→t0

f (x, t0 ) dx = F (t0 ),

t→t0 a

(3.0.2)

a

for any t0 ∈ [A, B]. In the case when b = ∞, the above theorem takes the following form: Given a function f (x, t) continuous as a function of two variables on the set [a, ∞] × [A, B] and the integral

∞ F (t) =

f (x, t) dx,

(3.0.3)

a

uniformly convergent with respect to t ∈ [A, B]. Then the function F (t) is continuous on [A, B]. © Springer Nature Switzerland AG 2020 T. Rado˙zycki, Solving Problems in Mathematical Analysis, Part III, Problem Books in Mathematics, https://doi.org/10.1007/978-3-030-38596-5_3

67

68

3 Investigating Integrals with Parameters

The notion of the uniform convergence for sequences and series is known to the reader from Chap. 15 in Part I. Here, in turn, it refers to the continuous variable b. The integral (3.0.2) is called uniformly convergent if the function "b G(b, t) := a f (x, t)dx is uniformly convergent to the function F (t) given with the formula (3.0.3) for b → ∞, i.e., if ∀>0 ∃K>0 ∀b>K |G(b, t) − F (t)| < .

(3.0.4)

As regards the differentiability of the function (3.0.1), the so-called Leibniz rule holds: Given a function f (x, t) continuous in x on [a, b] for any fixed t ∈ [A, B]. Then, if the partial derivative ∂f/∂t exists and is continuous as a function of two variables on the rectangle [a, b] × [A, B], the function F (t) defined with the formula (3.0.1) is differentiable for any t ∈]A, B[, and d F (t) = dt

b

∂ f (x, t) dx. ∂t

(3.0.5)

a

This theorem is sometimes reformulated so that the continuity of f (x, t) as a function of two variables is assumed. It can be shown that this formulation is equivalent. For the case of the integral (3.0.3), the appropriate theorem called again the Leibniz rule has the form: If the integrand function f (x, t) and partial derivative ∂f (x, t)/∂t are continuous on the set [a, ∞[×[A, B], the improper integral (3.0.3) is convergent for any t ∈ [A, B] and the improper integral

∞

∂ f (x, t) dx ∂t

(3.0.6)

a

is uniformly convergent on [A, B], then the function F (t) is differentiable for t ∈ ]A, B[ and d F (t) = dt

∞

∂ f (x, t) dx. ∂t

(3.0.7)

a

As to the integrability, this issue was considered in Chap. 12 of the second part. For our purposes, the following Fubini theorem is sufficient: for a continuous (i.e., integrable) function f (x, t) both iterated integrals exist and are equal, so one can write

3.1 Examining Limits and Continuity

b

69

B dt f (x, t) =

dx a

B

A

b dt

dx f (x, t).

(3.0.8)

a

A

In the case of the improper integral over x, the following formula still can be used:

∞

B dt f (x, t) =

dx a

B

A

∞ dt

A

dx f (x, t),

(3.0.9)

a

provided, apart from the above assumptions, the integral

∞ f (x, t) dx

(3.0.10)

a

is uniformly convergent.

3.1 Examining Limits and Continuity Problem 1 The limit lim F (t), where the function F (t) is given by the formula: t→1

1 F (t) = 0

xt dx, 1+x

(3.1.1)

will be found.

Solution As already mentioned above, in this chapter we are concerned with integrals dependent on a certain additional parameter, and in particular we are interested in such properties of emerging functions like continuity or differentiability with respect to this variable. There are many cases in which, when dealing with a two-variable function, one is able to explicitly carry out the integration and then examining the function F (t) does not pose any problems. This kind of example remains beyond our interest below. We will be focused only on those for which are needed to determine the properties of the function F (t) without executing the integral.

70

3 Investigating Integrals with Parameters

The first and simplest task standing in front of us is how to perform the limit t → 1 in the expression (3.1.1). To this end, one would be willing to simply put t = 1 "1 on the right-hand side and to perform the emerging easy integral 0 x/(1 + x) dx. In this way, however, one would not calculate limt→1 F (t) but rather F (1). As the reader might remember, both of these values must be actually equal, provided the function F (t) is continuous. Hence, if one wants to perform the limit in this way, one must first make sure of the continuity of F (t) at least on a certain neighborhood of the point t = 1. The following statement, already formulated in the theoretical introduction above, comes to our aid: if a given function f (x, t) is continuous as a function of two variables on the set [a, b] × [A, B], then the function

b F (t) =

f (x, t) dx

(3.1.2)

a

is continuous (naturally as a function of one variable t) on the interval [A, B]. When we write “continuous as a function of two variables,” we have in mind not the continuity in each variable separately, but the genuine continuity on the plane dealt with in Chap. 4 of the second part of this book series. In our example, one has [a, b] = [0, 1], and as [A, B] any closed interval may be chosen, provided it contains in its interior the essential point t = 1. Of course there is no need to consider an interval that would be too broad, for example, containing the negative values of t, since the easy and nice integral (3.1.1) would be converted into a much more troublesome improper integral. Therefore, our simple choice is [A, B] = [1/2, 3/2]. According to the above-mentioned theorem, one has to verify the twodimensional continuity of the function f (x, t) on the set D := [0, 1] × [1/2, 3/2]. Using the method from Part II, one can estimate that for any (x1 , t1 ) ∈ D and (x2 , t2 ) ∈ D:      x t2 x1t1   (1 + x1 )x2t2 − (1 + x2 )x1t1   2 |f (x2 , t2 ) − f (x1 , t1 )| =  −  =   1 + x2 1 + x1   (1 + x2 )(1 + x1 )   ≤ (1 + x1 )x2t2 − (1 + x2 )x1t1    = (1 + x1 )(x2t2 − x2t1 + x2t1 ) − (1 + x2 )(x1t1 − x2t1 + x2t1 )   = (1 + x1 )(x2t2 − x2t1 ) − (1 + x2 )(x1t1 − x2t1 ) + x2t1 (x1 − x2 )       ≤ |1 + x1 | x2t2 − x2t1  + |1 + x2 | x1t1 − x2t1  + x2t1  |x1 − x2 | . (3.1.3) Now, if (x1 , t1 ) → (x2 , t2 ), all three above terms converge to zero. The first one due to the (one-dimensional) continuity of the exponential function of the base x2 , the second one because of the (again one-dimensional) continuity of the power function with positive (!) exponent t1 , and the last one thanks to the presence of |x1 − x2 |.

3.1 Examining Limits and Continuity

71

As one can see, the assumptions of the theorem are met and, thereby, the function F (t) is continuous in the neighborhood of the unity. Then, the limit (3.1.1) can be calculated as follows:

1 lim F (t) = F (1) =

t→1

0

x dx = 1+x

1  0

1 1− 1+x

1  = (x − log(1 + x))  = 1 − log 2.

 dx

(3.1.4)

0

Problem 2 The continuity of the function F (y) defined with the formula:

π F (y) =

sin2 (x 2 y) dx

(3.1.5)

0

will be examined, for any y ∈ R.

Solution This exercise is very similar to the previous one, with the difference that now we are interested in the continuity of the function F (y) not at a given point but for any real value of y. If one wished to apply directly the theorem used in the previous example, one would need to reduce the problem to testing the continuity on a closed interval. It can be done very easily. Let us first choose any y, then fix the values of two constants A and B, so that A < y < B, and finally examine the continuity of the integrand function in both variables on the rectangle [0, π ] × [A, B]. If this continuity was demonstrated, the continuity of the function F (y) on the interval [A, B] would be proved as well. As long as no specific values for y, A, and B have been chosen, the same reasoning remains legitimate for any y ∈ R. Thereby, it would mean that the continuity of the function F for each real argument is proved. It is assumed, therefore, that y ∈ [A, B], A and B are not subject to any restriction except that A < B. Let us now examine the (two-dimensional) continuity of the function f (x, y) = sin2 (x 2 y)

(3.1.6)

72

3 Investigating Integrals with Parameters

on the rectangle [0, π ] × [A, B]. To this end, let us transform the expression as follows:     |f (x2 , y2 ) − f (x1 , y1 )| = sin2 (x22 y2 ) − sin2 (x12 y1 ) =

 1   1     1 − cos(2x22 y2 ) − 1 + cos(2x12 y1 ) = cos(2x12 y1 ) − cos(2x22 y2 ) , 2 2 (3.1.7)

where the well-known formula: sin2 α =

1 − cos(2α) 2

(3.1.8)

has been used. Now, if one applies the formula cos α − cos β = −2 sin

α+β α−β sin , 2 2

(3.1.9)

the following estimate can be done:       |f (x2 , y2 ) − f (x1 , y1 )| = sin(x12 y1 + x22 y2 ) sin(x12 y1 − x22 y2 )    ≤1

    ≤ sin(x12 y1 − x22 y1 + x22 y1 − x22 y2 )     = sin(x12 y1 − x22 y1 ) cos(x22 y1 − x22 y2 ) + cos(x12 y1 − x22 y1 ) sin(x22 y1 − x22 y2 )             ≤ sin[(x12 − x22 )y1 ] cos[x22 (y1 − y2 )] + cos[(x12 − x22 )y1 ] sin[x22 (y1 − y2 )]       ≤1

≤1

            ≤ sin[(x12 − x22 )y1 ] + sin[x22 (y1 − y2 )] ≤ x12 − x22  |y1 | + x22 |y1 − y2 |     −→ 0. (3.1.10) ≤ max{|A|, |B|} x12 − x22  + π 2 |y1 − y2 | (x2 ,y2 )→(x1 ,y1 )

The continuity of the function f (x, y) on the rectangle [0, π ] × [A, B] is then demonstrated, and hence the continuity of the function F (y) on the interval [A, B] as well. As to A and B nothing special was assumed, so in line with the comments made at the beginning of this exercise, this implies the continuity of the function defined with the integral (3.1.5) on the entire real axis.

3.1 Examining Limits and Continuity

73

Problem 3 The limit lim (t) will be found, where (t) is given by the formula: t→n

∞ (t) =

x t−1 e−x dx,

(3.1.11)

0

with t > 0, n being a natural number.

Solution The function (t) defined above bears the name of the Euler’s Gamma function and is one of the most important special functions applicable in many branches of mathematics and physics. As it will be seen in a moment, it constitutes the extension of the operation called the factorial beyond natural numbers. When looking at the formula (3.1.11), one notices the difficulty, which was not encountered in the previous problems: the integral on the right-hand side is now an improper integral for two reasons. First, the integration interval is infinite, and second the function can diverge in the lower limit of the integration (and it is really the case for t < 1). Therefore, one cannot directly use the theorem applied in the preceding problems and instead one has to rely on another one formulated in the theoretical introduction. Before recalling the needed theorem, it is a good idea to slightly prepare the expression (3.1.11), in order not to have to simultaneously deal with both the above difficulties. Let us then rewrite the function (t) in the form:

1 (t) =

x

∞

t−1 −x

dx +





0



e

1



x t−1 e−x dx , 

(3.1.12)



2 (t)

1 (t)

thanks to which in the expression for 1 (t) only the lower limit of integration, and in that for 2 (t) only the upper limit will be of interest for us. 1. 1 [t]. In order to get rid of the trouble, let us integrate by parts the expression for 1 (t):

1 1 (t) =

x

t−1 −x

e

1 dx = t

0

1

 t  −x x e dx

0



1

1 1 1 t −x 1 1 1 t −x x e dx = x t e−x dx. = xe  + + t t te t 0 0

0

(3.1.13)

74

3 Investigating Integrals with Parameters

The reader might ask what has been gained through this operation. The answer is the following: in the integrand expression, the unity in the exponent disappeared (one has now x t instead of x t−1 ) and the function is no longer divergent at x → 0+ even for small values of t. For, we remember that t is positive. The separated term 1/(te) is continuous for t > 0 in a visible way. The continuity of the function 1 (t) is, therefore, conditioned by the continuity of "1 the integral 0 x t e−x dx, to which the theorem used in the previous problems can be applied. For this purpose, as we remember, it is sufficient to check the continuity of the expression x t e−x as the function of two variables on a certain rectangle [a, b] × [A, B]. According to the text of the exercise, we are interested in the limit t → n, where n ∈ N, so for convenience the rectangle may be chosen in the form [0, 1] × [n − 1/2, n + 1/2]. Now let (x1 , t1 ) and (x2 , t2 ) be two points of this rectangle. The following estimate can be done:  t2 −x  x e 2 − x t1 e−x1  2 1   = x2t2 e−x2 − x2t2 e−x1 + x2t2 e−x1 − x2t1 e−x1 + x2t1 e−x1 − x1t1 e−x1    = x2t2 (e−x2 − e−x1 ) + (x2t2 − x2t1 )e−x1 + (x2t1 − x1t1 )e−x1        ≤ x2t2 (e−x2 − e−x1 ) + (x2t2 − x2t1 )e−x1  + (x2t1 − x1t1 )e−x1  −→

(x2 ,t2 )→(x1 ,t1 )

0.

(3.1.14)

This last result has been obtained, as: • e−x2 − e−x1 −→ 0, pursuant to the continuity of the exponential function, x2 →x1

• x2t2 − x2t1 −→ 0, pursuant to the continuity of the exponential function, and t2 →t1

• x2t1 − x1t1 −→ 0, pursuant to the continuity of the power function. x2 →x1

The integrand function is, therefore, continuous as the function of two variables on the rectangle concerned, which entails the conclusion that 1 (t) is continuous on the interval [n − 1/2, n + 1/2] and the limit t → n may be found by simply plugging t = n into the integral. 2. 2 [t]. In order to verify the continuity of the second integral in (3.1.12), let us invoke the following theorem formulated at the beginning of this chapter: if the function f (x, t) is continuous as the function of two variables on the set [a, ∞[×[A, B] and if the integral

∞ F (t) =

f (x, t) dx a

(3.1.15)

3.1 Examining Limits and Continuity

75

is uniformly convergent with respect to the parameter t ∈ [A, B], then the function F (t) is continuous on this interval. The first condition can be easily checked in the same way as above, so we will be focused on the second one. The notion of the uniform convergence for discrete variable n appeared in the last chapter of Part I. For a continuous parameter t, it was formulated in the theoretical introduction. Let us, however, recall it here. The integral (3.1.15) "b is said to be uniformly convergent if the function G(b, t) = a f (x, t)dx is uniformly convergent to the function F (t) given by (3.1.15) for b → ∞, i.e., if ∀>0 ∃K>0 ∀b>K |G(b, t) − F (t)| < .

(3.1.16)

The reader might object, pointing to the fact that to demonstrate (3.1.16) may not be easy either. Fortunately, the following Weierstrass test (or, in other words, the theorem) comes to the aid: if there exists an integrable majorant, i.e., a function ϕ(x) satisfying |f (x, t)| ≤ ϕ(x)

(3.1.17)

for all x ≥ a and arbitrary t ∈ [A, B], then the integral (3.1.15) is uniformly convergent. So let us look for a majorant for the function f (x, t) = x t−1 e−x for x ≥ 1 and t ∈ [n − 1/2, n + 1/2]. Note that if x ≥ 1 the following inequality holds: x t−1 ≤ x n+1/2−1 = x n−1/2 .

(3.1.18)

This last exponent is positive, as n ≥ 1, so one has f (x, t) ≤ x n−1/2 e−x .

(3.1.19)

A candidate for a majorant is, therefore, ϕ(x) = x n−1/2 e−x ,

(3.1.20)

but one needs to make sure that it is integrable. The “know-how” gained on this subject when solving the problems of Chap. 2 in the second part of the book series allows us to immediately draw a conclusion in this case. In the lower limit of the integration interval, no problems emerge, since ϕ(x) is bounded. In turn for x → ∞, the function exponentially decreases, i.e., sufficiently quickly. The continuity of (3.1.20) for x ≥ 1 does not raise any doubts either, since we are dealing with the product " ∞ of the power and exponential functions. Thus we come to the conclusion: 1 ϕ(x)dx exists. Consequently the Weierstrass test ensures the uniform convergence of the integral (3.1.15) and the continuity of the function 2 (t) as well.

76

3 Investigating Integrals with Parameters

Since the continuity of the functions 1 (t) and 2 (t) on the interval [n−1/2, n+ 1/2] has been proved, so the function (t) is continuous too. One, therefore, finds

∞ lim (t) = (n) =

t→n

x n−1 e−x dx,

(3.1.21)

0

and after n − 1 integrations according to the pattern below, one comes to

∞ (n) =

x

n−1 −x

e

∞ dx = −

0

= −x

x n−1 [e−x ] dx

0

∞

∞

∞  n−2 −x n−2 −x   + (n − 1) x e dx = −(n − 1) x [e ] dx

n−1 −x 

e

0

0

0

∞ = . . . = (n − 1)(n − 2) · . . . · 1 · 0



e−x dx = (n − 1)! 

=1

(3.1.22)



This result justifies the treatment of the Euler’s Gamma function as a generalization of the concept of the factorial beyond natural numbers as mentioned previously.

3.2 Differentiating with Respect to Parameters Problem 1 The integral:

π/2 F (ξ ) =

arctan(ξ tan x) dx, tan x

(3.2.1)

0

will be calculated for ξ > 0.

Solution In this problem we will make use of the so-called Leibniz rule, formulated at the beginning of this chapter. Let us recall its form below. Assume that a given function f (x, t) is continuous in the variable x on the interval [a, b] for any t ∈ [A, B].

3.2 Differentiating with Respect to Parameters

77

Besides, it is assumed that the partial derivative ∂f/∂t is continuous as the function of two variables on the set [a, b] × [A, B]. Then in the interval ]A, B[, there exists the derivative dF /dt, and it can be calculated according to the formula: d F (t) = dt

b

∂ f (x, t) dx. ∂t

(3.2.2)

a

Now one needs to check whether the above assumptions are met for f (x, ξ ) =

arctan(ξ tan x) . tan x

(3.2.3)

Inside the interval [0, π/2] the above function is, naturally, continuous in the variable x pursuant to the continuity of the functions tangent and arcus tangent. The problem emerges only at the ends, where it is not defined at all. In order to take account of these points, let us modify the definition of the function f : ⎧ ⎨ arctan(ξ tan x)/tan x for x ∈]0, π/2[, f (x, ξ ) = ξ for x = 0, ⎩ 0 for x = π/2.

(3.2.4)

The inclusion of these end points has no consequences for the value of the integral (3.2.1), but at the same time ensures that the integrand function is continuous in the variable x (for any ξ ) on the entire interval [0, π/2]. One can be easily convinced about that, when using the methods of Part I (see Sect. 8.2). Now, as required by the Leibniz rule given above, we must address the partial derivative ∂f/∂ξ : ∂f (x, ξ ) = g(x, ξ ) := ∂ξ



1/(ξ 2 tan2 x + 1) for x ∈ [0, π/2[, 0 for x = π/2,

(3.2.5)

the case x = 0 having been covered by the upper formula. One needs to make sure that such a function is continuous as the function of the variables x, ξ on the set [0, π/2] × [A, B], with no restrictions on A and B except that 0 < A < B < ∞. This task is not difficult. For x = π/2, it is enough to choose any sequence of points (xn , ξn ) convergent to (x, ξ ) and write     1 1   |g(xn , ξn ) − g(x, ξ )| =  2 2 − 2 2 ξn tan xn + 1 ξ tan x + 1         2 2 ξ 2 tan2 x − ξn2 tan2 xn   ≤ ξ tan x − ξ 2 tan2 xn  −→ 0, = 2 2 n n→∞ (ξn tan xn + 1)(ξ 2 tan2 x + 1) 

(3.2.6)

78

3 Investigating Integrals with Parameters

thanks to the continuity of the power and tangent functions. At the point (π/2, ξ ) the function g is continuous too, since one has   1 1 ≤ − 0 −→ 0,  2 2 2 ξn tan xn + 1 (ξ/2) tan2 xn + 1 n→∞ (3.2.7) where the fact that (for respectively large n) one has ξn > ξ/2 (once ξn −→ ξ )) n→∞ has been made use of. The partial derivative has turned out to be continuous as the function of two variables on the set [0, π/2] × [A, B] and, therefore, the formula (3.2.2) may be used, bearing in mind that here t = ξ :   |g(xn , ξn ) − g(π/2, ξ )| = 

π/2



F (ξ ) = 0

=

1 dx 2 2 ξ tan x + 1

1 1 − ξ2

∞  0

∞ =

(u2

u=tan x 0

1 ξ2 − 2 2 2 u +1 ξ u +1

1 du + 1)(ξ 2 u2 + 1)

 =

∞  1  u − ξ arctan(u ξ )) (arctan  2 1−ξ 0

π = . 2(1 + ξ )

(3.2.8)

The value of the derivative having been found, one now obtains by simple integration: F (ξ ) =

π log(1 + ξ ) + C. 2

(3.2.9)

It remains only to determine the value of the constant C. The above formula easily shows that C = F (0). So, if one could independently find the value F (0), our problem would be solved. At this place, one might like to make use of (3.2.1), writing

π/2 C = F (0) = lim F (ξ ) = lim ξ →0

ξ →0 0

arctan(ξ tan x) dx = tan x

π/2 0 dx = 0,

(3.2.10)

0

however, it is not obvious whether the limit may be inserted under the symbol of the integration, that is, whether the function F is continuous at zero. Its differentiability (which also entails the continuity) has been verified only for ξ > 0. We know, however, from the previous section that it would be sufficient to examine the twodimensional continuity of the function f (x, ξ ) on the set [0, π/2] × [A, B], where this time we can accept 0 ≤ A < B < ∞. Again the sequence of points (xn , ξn ) convergent to (x, ξ ) is chosen (all these points lying in the set [0, π/2] × [A, B]) and for x = 0, π/2 the following estimate can be done:

3.2 Differentiating with Respect to Parameters

79

   arctan(ξn tan xn ) arctan(ξ tan x)    |f (xn , ξn ) − f (x, ξ )| =  −  tan xn tan x    tan x arctan(ξn tan xn ) − tan xn arctan(ξ tan x)   =   tan xn tan x    tan x[arctan(ξn tan xn )−arctan(ξ tan x)]+[tan x −tan xn ]arctan(ξ tan x)   =   tan xn tan x ≤

|tan x| · |arctan(ξn tan xn ) − arctan(ξ tan x)| |tan xn | · |tan x| +

|tan x − tan xn | · |arctan(ξ tan x)| −→ 0. n→∞ |tan xn | · |tan x|

(3.2.11)

As it can be easily checked, both of the terms tend to zero due to the continuity of the tangent and arcus tangent functions. For x = 0 and x = π/2 to show the two-dimensional continuity is not a problem either. In the first case, one has      arctan(ξn tan xn )   arctan(ξn tan xn )−ξ tan xn   |f (xn , ξn )−f (0, ξ )| =  −ξ  =   tan xn tan xn    arctan(ξn tan xn ) − ξn tan xn + (ξn − ξ )tan xn   =   tan xn      arctan(ξn tan xn ) − ξn tan xn   tan xn     −→ 0.  ≤ (3.2.12)  + |ξn − ξ | ·  tan x  n→∞ tan xn n The reader should not have any difficulties with the justification of this result. The last term goes to zero in an obvious way. The same is true for the first one, which can be easily established by using the Taylor expansion of the arcus tangent function (this method of finding limits was discussed in Part I in Sect. 11.3) in accordance with the formula: arctan y = y −

y3 y5 + + ... . 3 5

(3.2.13)

One can see that the first term in the numerator in (3.2.12) cancels, and only the second, i.e., (ξn tan xn )3 “survives.” In turn, for x = π/2, our job is even easier (this time we, obviously, have (xn , ξn ) −→ (π/2, ξ )): n→∞

     arctan(ξn tan xn )   π/2    −→ 0,   |f (xn , ξn ) − f (π/2, ξ )| =  − 0 ≤  tan xn tan xn  n→∞ (3.2.14) since −π/2 < arctan y < π/2, and the tangent in the denominator diverges to infinity.

80

3 Investigating Integrals with Parameters

Thereby, the two-dimensional continuity of the function f (x, ξ ) on the set [0, π/2] × [A, B] has been demonstrated, which implies the continuity of the function F (ξ ) at zero. Pursuant to (3.2.10), the constant C = 0 and we find

π/2

arctan(ξ tan x) π dx = log(1 + ξ ). tan x 2

(3.2.15)

0

Problem 2 The integral:

∞ I=

e−x

2 −1/x 2

dx

(3.2.16)

0

will be calculated.

Solution In this problem some new elements are encountered. First, one has to deal with the improper integral, so the Leibniz rule as it stands in the previous examples is not adequate. Second, (3.2.16) is not an integral with a parameter, but simply a number. However, since this exercise appears in this chapter, the reader probably can guess that this parameter shall appear in a moment. Indeed, rather than directly calculating the integral (3.2.16), one can try to explicitly find the function:

∞ F (ξ ) =

e−x

2 −ξ/x 2

dx.

(3.2.17)

0

It might seem to be illogical. In general it should be easier to find a single number than the entire function. Why to complicate our task, then? Well, the problem lies in the fact that one does not know how to directly calculate the integral I in the given form. Introducing the parameter ξ will enable us to perform some additional operations like differentiation over ξ (if only the assumptions of the appropriate theorem are met), which can help to transform the integrand expression. Now let us restate the theorem formulated in the theoretical introduction, which will be helpful. It specifies when one is allowed to differentiate the function

∞ F (t) =

f (x, t) dx a

(3.2.18)

3.2 Differentiating with Respect to Parameters

81

by applying the formula d F (t) = dt

∞

∂ f (x, t) dx, ∂t

(3.2.19)

a

i.e., by differentiating the integrand expression. Suppose that one is interested in the dependence F (t) for t lying within a certain interval [A, B]. The conditions to be met are the following: 1. The integrand function f (x, t) and its partial derivative ∂f (x, t)/∂t are continuous on the set [a, ∞[×[A, B]. 2. The improper integral (3.2.18) is convergent for each t ∈ [A, B]. 3. The improper integral (3.2.19) is uniformly convergent on [A, B]. Let us examine whether these conditions are satisfied by our function # f (x, ξ ) =

e−x 0

2 −ξ/x 2

for x > 0, for x = 0,

(3.2.20)

whose definition, as in the previous problem, has been extended to the case of x = 0 as well. Let us note at the beginning that finally we are interested in the value ξ = 1. Therefore, our discussion can be limited to the interval [A, B] = [1/2, 3/2]. The choice of these specific numbers is of no particular significance. It allows, however, to leave outside the interval the point ξ = 0, which could cause some unnecessary problems. Now we are going to examine the continuity of the function f (x, ξ ). We start with considering the case x > 0. Let us choose a point (x, ξ ) and a certain arbitrary sequence (xn , ξn ) convergent to it. Without loss of generality one can assume here and in further considerations that ξn ∈ [1/4, 7/4] ⊃ [1/2, 3/2]. Let us perform the following transformations:  2  2 2 2  |f (xn , ξn ) − f (x, ξ )| = e−xn −ξn /xn − e−x −ξ/x    2 2  2 2 2 2 2  = e−x −ξ/x ex −xn e(ξ −ξn )/xn eξ(1/x −1/xn ) − 1  2 2  2 2 2   (3.2.21) ≤ ex −xn e(ξ −ξn )/xn eξ(1/x −1/xn ) − 1 . Note that if n → ∞, each of the exponential factors goes to 1. Since x = 0, and consequently xn = 0 for sufficiently large n, no problem with vanishing denominators arise. In the consequence the entire expression under the absolute value converges to 0. This ensures the continuity of the function f (x, ξ ) on the set ]0, ∞[×[1/2, 3/2].

82

3 Investigating Integrals with Parameters

In the case when x = 0, one again has  2  2   |f (xn , ξn ) − f (0, ξ )| = e−xn −ξn /xn − 0 = e−xn e−ξn /xn −→ 0, 2

2

n→∞

(3.2.22)

because the former exponential factor tends to 1 and the latter to 0 (remember that ξn > 1/4). Now let us turn to the derivative with respect to the parameter: ∂f (x, ξ ) = g(x, ξ ) := ∂ξ

#

− x12 e−x 0

2 −ξ/x 2

for x > 0, for x = 0.

(3.2.23)

The presence of the additional factor 1/x 2 does not make our estimate more complicated. One has (again for x > 0)   1 1 2 2 2 2 |g(xn , ξn ) − g(x, ξ )| =  2 e−xn −ξn /xn − 2 e−x −ξ/x  xn x  2   1 2 2 x 2 2 2 2 2 = 2 e−x −ξ/x  2 ex −xn e(ξ −ξn )/xn eξ(1/x −1/xn ) − 1 −→ 0, n→∞ x xn

(3.2.24)

similarly as before, since x 2 /xn2 −→ 1. n→∞ Now the case x = 0 should be addressed. One has    1 −x 2 −ξ /x 2  n  n n |g(xn , ξn ) − g(0, ξ )| =  2 e − 0 xn =

1 −xn2 −ξn /xn2 1 2 e e ≤ 2 e−(1/4)/xn , xn2 xn

(3.2.25)

since e−xn ≤ 1, and ξn ≥ 1/4. The obtained expression leads to the limit of the type ∞ · 0. It is, however, a product of a power factor and an exponential factor, and as we know from Part I, in such a case the latter determines the result. Consequently the limit equals zero, as is needed. In conclusion, one can say that both the function f (x, ξ ) and its derivative with respect to ξ are continuous on the set [0, ∞[×[1/2, 3/2]. The first step of our proof is then completed. The second step requires examining the convergence of the improper integral (3.2.17) for an arbitrary value of ξ (but belonging to the given interval). Since the integrand function is continuous in the variable x, so the eventual doubts may concern only the upper limit of the integration. They can be immediately dispelled thanks to the experience gained from the previous volume of the book (see 2 Chap. 2). For, the function f (x, ξ ) is the product of a bounded factor (e−ξ/x ) and a 2

3.2 Differentiating with Respect to Parameters

83

factor decreasing exponentially—and thus sufficiently quickly—to zero (e−x ). The convergence of the integral in this case is evident. In the last point, one has to investigate the uniform convergence of the integral: 2

∞

1 −x 2 −ξ/x 2 e dx. x2

(3.2.26)

0

As we know from the previous section, for this purpose one has to find a majorant (see formula (3.1.17)) of the function: g(x, ξ ) =

1 −x 2 −ξ/x 2 e . x2

(3.2.27)

Since it is decreasing in the variable ξ and 1/2 ≤ ξ ≤ 3/2, one has: g(x, ξ ) ≤

1 −x 2 −(1/2)/x 2 e =: ϕ(x). x2

(3.2.28)

The function on the right-hand side does not depend on ξ any longer and is integrable on the interval [0, ∞[. This can be clearly seen if this interval is broken down for example onto [0, 1] and ]1, ∞[. On the former of these intervals, the integral exists, since the continuous function is being integrated on the compact 2 set (the majorant ϕ(x) at zero is given the value 0), and on the latter |ϕ(x)| ≤ e−x occurs, so the integrability is obvious. Thereby, the integral (3.2.26) is uniformly convergent. In this way all the necessary conditions have been verified, so one can proceed with the implementation of our plan of calculating dF (ξ )/dξ rather than I . At the same time, it becomes clear why the parameter ξ was introduced into the integral. We have 

∞

F (ξ ) = −

1 −x 2 −ξ/x 2 e dx x2

= √ u= ξ /x

0

1 −√ ξ

∞

e−u

2 −ξ/u2

1 du = − √ F (ξ ). ξ

0

(3.2.29) Clearly we have succeeded in deriving a simple differential equation for the function F (ξ ): 1 F  (ξ ) = − √ F (ξ ), ξ

(3.2.30)

which can be easily solved by separating the variables: dξ dF = −√ F ξ

⇒ F (ξ ) = Ce−2

√

ξ

.

(3.2.31)

84

3 Investigating Integrals with Parameters

The final step is to determine the integration constant C. One might be tempted to simply put ξ = 0 on both sides of (3.2.17), but it is not known, however, if the function F (ξ ) is continuous at zero (in our previous considerations it was assumed for convenience that ξ ∈ [1/2, 3/2]). If one now wished to demonstrate this continuity by using the theorem formulated previously (see formulas (3.0.3), (3.1.15), etc.) and by extending the interval for example to [0, 2], one would encounter a serious difficulty: it is unclear how to demonstrate the continuity of the function f (x, ξ ) at the point (0, 0). In the case when ξn −→ 0, our claim expressed in n→∞ the formula (3.2.22) breaks down. What is more, this problem comes not from the inadequacy of our proof, but simply from the fact that the function f (x, ξ ) is actually not continuous at the origin. In order to somehow deal with this, let us split up the integral in the following way:

 F (ξ ) =

e

∞

−x 2 −ξ/x 2

dx +





0



e−x





2 −ξ/x 2



(3.2.32)

dx , 

F2 (ξ )

F1 (ξ )

where in a moment  is going to converge to zero. The continuity of the function F2 for ξ → 0 does not raise any suspicion. The integrand expression is continuous on the set [, ∞[×[0, 2], and the integrable majorant can be easily indicated in the 2 form e−x . One can, therefore, write C = lim F (ξ ) = lim (F1 (ξ ) + F2 (ξ )) = lim F1 (ξ ) + F2 (0), ξ →0

ξ →0

ξ →0

(3.2.33)

which implies that

 |C − F2 (0)| = lim

e

ξ →0

−x 2 −ξ/x 2

 dx ≤ lim

1 dx = .

ξ →0

0

(3.2.34)

0

Letting  go to zero, one gets the well-known Gaussian integral:

∞ C = lim F2 (0) = lim →0

e

→0 

−x 2

∞ dx =

e−x dx = 2

π , 2

(3.2.35)

0

and the final result is F (ξ ) =

π −2√ξ π e ⇒ I = F (1) = 2 . 2 2e

(3.2.36)

3.2 Differentiating with Respect to Parameters

85

Problem 3 The integral:

∞ I (a, b) =

e−ax − cos bx dx, x2 2

(3.2.37)

0

will be calculated for a ∈ [a1 , a2 ], where 0 < a1 < a2 < ∞ and b ∈ [b1 , b2 ], where 0 < b1 < b2 < ∞.

Solution This time the integral depends on two parameters a and b, but one need not worry about this. The behavior of the function can be examined, fixing one parameter, if necessary. In the present problem, however, one can easily get rid of the second parameter, by the appropriate rescaling of the integration variable. To this goal, let us define the new variable y = bx, getting the result:

∞ I (a, b) = b

e−(a/b

2 )y 2

− cos y

y2

dy.

(3.2.38)

0

Introducing temporarily the symbol c = a/b2 , one obtains

∞ I (a, b) = bF (c),

where

F (c) =

e−cy − cos y dy, y2 2

(3.2.39)

0

which means that the issue has been transformed to a problem with one parameter c. Now the procedure of the previous exercise can be applied. It can be seen that the possibility of the differentiation of the integrand function with respect to c would be very promising, since then the denominator y 2 would disappear, and the integral would become easy to calculate. Let us denote ⎧ −cy 2 − cos y ⎨e for y > 0, f (y, c) = y2 ⎩ 1/2 − c for y = 0,

(3.2.40)

this lower value having been established with the use of the limit e−cy − cos y 1 = − c. 2 y2 2

lim

y→0+

(3.2.41)

86

3 Investigating Integrals with Parameters

Let us now examine the continuity of (3.2.40) on the set [c1 , c2 ] × [0, ∞[, where 0 < c1 < c2 < ∞. First, suppose that y = 0 and let (yn , cn ) be a sequence of points convergent to (y, c). At least for large values of n one has yn = 0 and it is clear that   2  e−cn yn2 − cos y e−cy − cos y  n  |f (yn , cn ) − f (y, c)| =  − (3.2.42)  −→ 0.  n→∞  yn2 y2 In turn for y = 0 one can make the following estimate:     e−cn yn2 − cos y 1 n   |f (yn , cn ) − f (0, c)| =  + c −    2 yn2    e−cn yn2 − 1 + cy 2 cos yn − 1 − yn2 /2   n = −    yn2 yn2    e−cn yn2 − 1 + cy 2   cos y − 1 − y 2 /2  n  n n  −→ 0, ≤  +   n→∞   yn2 yn2

(3.2.43)

since each of the numerators for yn → 0 behaves as yn4 (this results from the simple Taylor expansion). The next step requires the verification whether for arbitrary c ∈ [c1 , c2 ] the improper integral (3.2.39) exists. As the integrand function is continuous (the point y = 0 included) which has been shown above, only the upper limit of the integration needs to be considered. Because of the presence of the denominator y 2 , the convergence of the integral is ensured (the function in the numerator is bounded). As we remember from Part II, the power y 1+ would be sufficient. In the last step, the uniform convergence of the following integral should be checked:

∞ g(y, c) dy,

where

g(y, c) :=

∂f (y, c) 2 = −e−cy . ∂c

(3.2.44)

0

Due to the fact that c ∈ [c1 , c2 ] with c1 > 0, the natural majorant is ϕ(y) = e−c1 y , 2

(3.2.45)

clearly integrable over the variable y on the interval [0, ∞[. All assumptions having been met, and one can now differentiate the integrand function with respect to c, obtaining the Gaussian integral: 

∞

F (c) = 0

∂f (y, c) dy = − ∂c

∞ 0

e−cy dy = − 2

1 2

π . c

(3.2.46)

3.2 Differentiating with Respect to Parameters

87

Now it is straightforward to find F (c): F  (c) = −

1 2

π c

√ ⇒ F (c) = − π c + D.

(3.2.47)

The constant D still remains to be determined. To this end one would like to put c = 0, but so far the continuity of the function F (c) at zero has not been demonstrated (formula (3.2.47) is by now valid only for c > 0). Therefore, one is not allowed yet to write that

∞ F (0) =

$ √ % 1 − cos y − dy = lim F (c) = lim π c + D = D. ↑ c→0 c→0 y2

(3.2.48)

0

The equality marked with an arrow still requires justification. We will try to do it in the same way as in the previous exercise. First let us split the integral (3.2.39) into the sum of two terms:



∞ −cy 2 2 e−cy − cos y e − cos y F (c) = dy + dy , 2 y y2  0      

(3.2.49)

F2 (c)

F1 (c)

and then both of them will be examined separately. The function F2 (c) is continuous at zero, since first the function f (y, c) is continuous on the set [, ∞[×[0, c2 ] (the result (3.2.42) still applies), and second the integral

∞

e−cy − cos y dy y2 2

(3.2.50)



is uniformly convergent with respect to c because there exists an integrable majorant, for example:    e−cy 2 − cos y  1 + 1 2   = 2. ≤  2 2   y y y

(3.2.51)

One can then write D = lim F (c) = lim (F1 (c) + F2 (c)) = lim F1 (c) + F2 (0), c→0

c→0

c→0

(3.2.52)

88

3 Investigating Integrals with Parameters

which entails          −cy 2  −cy 2    e  e − cos y − 1 + 1 − cos y  = lim   |D − F2 (0)| = lim  dy dy    2 2 c→0  y y  c→0   0 0 ⎤ ⎡   

 −cy 2

    1 − cos y  e − 1     dy ⎦ . ≤ lim ⎣  dy + (3.2.53)      c→0 y2 y2 0

0

If one uses now the two known inequalities (we are interested in the case x > 0): 1 − e−x ≤ x

1 − cos x ≤ x 2 ,

and

(3.2.54)

easily demonstrated by methods of Sect. 10.1 of the first part of this book series, one can write: ⎤ ⎡ 

 2

  2   cy  y  |D − F2 (0)| ≤ lim ⎣  2  dy +  2  dy ⎦ = lim (c + 1) = . c→0 c→0 y y 0

0

(3.2.55) Letting  go to zero, one obtains the result:

∞ D = lim F2 (0) = lim →0

→0

⎡ = lim ⎣ →0

1 − cos y dy = lim →0 y2



 cos y − 1 ∞  + y 

∞

⎤

 

∞ −1  (1 − cos y) y 

sin y ⎦ π dy = , y 2

(3.2.56)



the value of the well-known integral

∞

π sin x dx = , x 2

(3.2.57)

0

having been used here. The reader has certainly encountered it during lectures of analysis, and its value will be calculated in Sect. 7.4. Going back to the formulas (3.2.39) and (3.2.47), it is finally found   a √ π πb a I (a, b) = bF 2 = b − π 2 + . = − πa + 2 2 b b

(3.2.58)

3.2 Differentiating with Respect to Parameters

89

Problem 4 The integral:

∞ I=

arctan(αx) − arctan(βx) dx, x

(3.2.59)

0

will be calculated for positive values of the parameters α and β.

Solution Integrals of the type (3.2.59) are called the Froullani’s integrals. It is a common term for such expressions as

∞ K(α, β) =

f (αx) − f (βx) dx. x

(3.2.60)

0

Upon appropriate assumptions regarding the function f and the constants α, β, one is able to show that   α K(α, β) = f (0+ ) − f (+∞) log , β

(3.2.61)

the expression in square brackets being understood as the appropriate limits. Let us consider the specific example given in the content of this problem. All of the necessary theorems have already been formulated in the previous exercises, so one can immediately start to study the integral. We assume that α ∈ [αmin , αmax ],

β ∈ [βmin , βmax ],

(3.2.62)

where αmin , βmin > 0. We would like to differentiate the integral (3.2.59) with respect to one of the parameters (e.g., α), so we need to go through the procedure described on page 81. This differentiation will make it easier to find the integral I , since it will transform the “unpleasant” integrand function into a rational function, for which, as the reader already knows, the indefinite integral can always be calculated, in so far as one knows how to decompose the polynomial in the denominator into elementary factors. Here the situation is even simpler:

90

3 Investigating Integrals with Parameters

dI = dα

∞ 0

 

∞ ∂ arctan(αx) − arctan(βx) 1 dx = dx 2 2 ∂α x α x +1

∞  1 π = arctan(αx) = . α 2α 0

0

(3.2.63)

At this point one cannot yet be certain, however, if the above formula is true. For that purpose all the required conditions must be verified. 1. We check the continuity of the integrand function and of its derivative over α on the set [0, ∞[×[αmin , αmax ]. Denoting ⎧ ⎨ arctan(αx) − arctan(βx) for x > 0, f (x, α) := x ⎩α − β for x = 0

(3.2.64)

and selecting a sequence (xn , αn ) −→ (x, α), we will estimate the expression: n→∞

|f (xn , αn ) − f (x, α)| .

(3.2.65)

We are not going to differentiate or take any limits over the parameter β, so its value can be treated as fixed and, therefore, it is not explicitly written as an argument of the function f . One can, of course, work vice versa: to fix α and perform all operations on the parameter β. First, let us assume that x > 0. Then, at least for large n, one must have xn > 0 as well. Now the expression (3.2.65) can be given the form:    f (xn , αn ) − f (x, α) = 1 x(arctan(αn xn ) − arctan(αx)) xxn + (x − xn )arctan(αx) − x(arctan(βxn ) − arctan(βx))  (3.2.66) + (xn − x)arctan(βx). The transformations that lead to this form are very similar to those of the previous problems, so their details have been omitted. Roughly speaking, they consist of adding and subtracting identical terms, so as to obtain (3.2.66). When n → ∞, all the terms under the sign of the modulus vanish, regardless of the specific form of (αn , xn ), which entails the two-dimensional continuity of the function f . The particular attention requires only the point x = 0, dealt with below. Let us estimate:       f (xn , αn ) − f (0, α) =  arctan(αn xn ) − arctan(βxn ) − α + β    xn

3.2 Differentiating with Respect to Parameters

≤

  arctan(αn xn ) − αxn  xn

91

+

  arctan(βxn ) − βxn  xn

−→ 0.

n→∞

(3.2.67) One can easily be convinced that the above result is correct, for example by using the Taylor expansion of the function arctan (as a result, both numerators will behave as xn3 ). Thus the continuity of the function f (x, α) on the set [0, ∞[×[αmin , αmax ] has been established. The continuity of the partial derivative 1 ∂f = 2 2 ∂α α x +1

(3.2.68)

is obvious, so the reader will not encounter any problems when checking it. 2. We check the convergence of the indefinite integral (3.2.59) for each α ∈ [αmin , αmax ]. In order to investigate the convergence of the integral, a certain known trigonometric formula will be applied, made use of in the first two parts of the book series (correct for u · v > −1): arctan u − arctan v = arctan

u−v . 1 + uv

(3.2.69)

The modulus of the numerator in (3.2.59) can now be given the form:    (α − β)x  |α − β|x  |arctan(αx) − arctan(βx)| = arctan = arctan .  2 1 + αβx 1 + αβx 2 (3.2.70) Since α ≥ αmin > 0 and β ≥ βmin > 0, it is clear that, for large values of x, the behavior of the argument of the arcus tangent (and with it the function itself) is not worse than const/x, i.e., |α − β|x |α − β| 1 · , ≤ 2 αβ x 1 + αβx

(3.2.71)

which ensures the convergence of the integral (3.2.59) in the upper limit. As to the lower limit, i.e., zero, no problem emerges, since it was previously shown that the function f (x, α) is continuous at this point. 3. We check the uniform convergence of the integral (3.2.63) on the interval [αmin , αmax ]. Demonstrating the uniform convergence of the integral (3.2.63) is painless, as it has the obvious integrable majorant (thanks to the fact that αmin = 0):

92

3 Investigating Integrals with Parameters

    ∂ arctan(αx) − arctan(βx)  1 1 =   α 2 x 2 + 1 ≤ α 2 x 2 + 1 =: ϕ(x).  ∂α x min (3.2.72) All three conditions having been verified, and the formula (3.2.63) becomes true. Integrating over α the equation dI π = , dα 2α

(3.2.73)

one comes to I (α) =

π log α + C, 2

(3.2.74)

where the constant C, independent of α, may of course depend on the second parameter, i.e., β. In order to find its value, let us set α = β. Because the differentiability of the function I (α) has already been shown, it must also be continuous and one is allowed to make the transition α → β. In this way one gets π log β + C = lim F (α) = F (β) = 0, α→β 2

(3.2.75)

which implies that C=−

π log β, 2

(3.2.76)

and the final result takes the form: I = π/2 log(α/β).

(3.2.77)

At the end, it should be noted that this expression is antisymmetric when replacing α ↔ β, as it was expected from (3.2.59). This type of checking whether the given result retains the symmetries of the initial expression, it is always desirable, as it allows us to avoid some calculational mistakes.

Problem 5 The integral

∞ I=

e−2x − e−x cos x dx x

0

will be calculated, by introducing an appropriate parameter.

(3.2.78)

3.2 Differentiating with Respect to Parameters

93

Solution When looking at the integral to be calculated, it is seen that the main factor that bothers us is x in the denominator. The integrals of the functions such as eax cos(bx) can be easily calculated by parts, whereas no primitive function for the integrands like eax /x or cos(bx)/x can explicitly be found since they are expressed through special functions. How x can be removed from the denominator, then? The answer should be imposed upon the reader: to temporarily introduce into one of the exponents an additional parameter (this idea appeared already in Problem 2) and next to differentiate with respect to it, in which case x will disappear from the denominator. Therefore, instead of (3.2.78) let us find:

∞ F (α) =

e−2x − e−αx cos x dx, x

(3.2.79)

0

where the parameter α will be assumed to belong to the interval [1/2, 5/2]. The only requirement for this interval is that it must contain the number 1, because this value of the parameter will be of interest to us (since one has I = F (1)). At the end, it will become clear that it is good to have the number 2 inside the interval as well. Let us define the function ⎧ ⎨ e−2x − e−αx cos x for x > 0, (3.2.80) f (x, α) = x ⎩ α−2 for x = 0 and verify all the conditions set out on page 81. The first two create no special difficulties. The examining of the continuity of the function f and its partial derivative ∂f/∂α on the set [0, ∞[×[1/2, 5/2] is similar as in the previous examples, so the appropriate calculations are left to the reader. In turn, the integral

∞ f (x, α) dx

(3.2.81)

0

is clearly convergent in the upper limit, due to the presence of the exponential factors, and in the lower limit because of the continuity of the integrand function (3.2.80). Therefore, for a moment we will stop only at the third point, which requires to investigate the uniform convergence of the integral

94

3 Investigating Integrals with Parameters

∞

∂f (x, α) dx. ∂α

(3.2.82)

0

As we remember, it is sufficient to indicate an integrable majorant for ∂f/∂α. It does not pose any particular difficulty either. Thanks to the condition α ≥ 1/2 and having in mind that the function e−αx is decreasing in α (with x > 0 fixed), one can write    ∂f (x, α)  −αx   | cos x| ≤ e−αx ≤ e−x/2 , (3.2.83)  ∂α  = e and this last function constitutes the majorant that is looked for. These results justify the formula: 

∞

F (α) =

∂f (x, α) dx = ∂α

0

∞

e−αx cos x dx =

0

α , 1 + α2

(3.2.84)

where the latter result can be easily obtained by (twice) integrating by parts (see Sect. 14.1 in Part I). Now one can integrate both sides of the Eq. (3.2.84), obtaining the formula for the function F : F (α) =

1 log(1 + α 2 ) + C. 2

(3.2.85)

In order to determine the value of the constant C, we put α = 2. The integrand function in (3.2.79) then vanishes, so F (2) = 0. Because the function F is continuous at this point (it is, after all, differentiable there), one can write  0 = F (2) = lim F (α) = lim α→2

α→2

 1 log 5 log(1 + α 2 ) + C = + C, 2 2

(3.2.86)

and C=−

log 5 . 2

Thereby, the constant C is found. The last step requires letting α go to 1 in order to obtain the value of I . In the same way as above, one gets I = F (1) =

log 2 log 5 log(5/2) − =− . 2 2 2

(3.2.87)

3.2 Differentiating with Respect to Parameters

95

Problem 6 It will be proved that the functions: √ F1 (t) = π

t e

−x 2

∞ dx

and

F2 (t) =

0

sin(tx) −x 2 /4 dx, e x

(3.2.88)

0

for t ∈ R, are identically equal.

Solution This last problem in this section slightly differs from the former ones. First, our goal is not to calculate both integrals, but only to check whether they are equal (in fact, they cannot be calculated in an explicit way, i.e., by expressing them through elementary functions). Second, in one of the functions, t does not appear as an argument but at the upper limit of the integration. The idea of the solution of this problem is the following: if the functions f1 (t) and f2 (t) were to be identically equal (i.e., for any t ∈ R), by subtracting them from each other one would get a constant (strictly speaking zero), which would entail the condition d [F1 (t) − F2 (t)] = F1 (t) − F2 (t) = 0. dt

(3.2.89)

And vice versa: if the above equation is valid, the functions may differ from each other at most by a constant. Our aim is thus, primarily, the calculation of both derivatives. To calculate F1 (t) is not a problem: it is just the integrand function, where t is put in place of the integration variable: F1 (t) =

√

πe−t . 2

(3.2.90)

This is because we know that the indefinite integral is simply the primitive function for the integrand expression. When dealing with F2 (t), one needs to be more careful. In order to achieve our goal it is necessary to differentiate under the integral, which requires to ensure that it is allowed. Again the conditions formulated on page 81 should be verified. As a range of variability of t, we accept the interval [A, B], where A < B, but no other restrictions are imposed on A and B.

96

3 Investigating Integrals with Parameters

Let us define the function:

⎧ ⎨ sin(tx) −x 2 /4 e for x > 0, f (x, t) = x ⎩t for x = 0.

(3.2.91)

Then one has

∞ F2 (t) =

(3.2.92)

f (x, t)dx. 0

Both the function f (x, t) and its derivative ∂f/∂t are continuous on the set [0, ∞[×[A, B]. The verification of this fact proceeds in the same way as in the previous problems so it is omitted here and left to the reader. The improper integral (3.2.92) is convergent for each t ∈ [A, B], due to the presence of the exponential factor in the function f (x, t). It remains, therefore, to find an integrable majorant for ∂f/∂t. This task is not a problem either, for one has    ∂f (x, t)  2 −x 2 /4   ≤ e−x /4 =: ϕ(x). (3.2.93)  ∂t  = |cos(tx)| e All the necessary assumptions are met, so one can write F2 (t) =

∞

∂f (x, t) dx = ∂t

0

∞

cos(tx)e−x

2 /4

(3.2.94)

dx.

0

Now, in order to calculate the integral (3.2.94), we are going to solve an auxiliary problem. In the same way as above, one can be convinced that (3.2.94) can be differentiated under the integral as well: d  F (t) = dt 2

∞

∂ 2 cos(tx)e−x /4 dx = − ∂t

0

∞

sin(tx)xe−x

2 /4

dx.

(3.2.95)

0

The additional factor x is easily removed upon writing xe−x

2 /4

= −2d(e−x

2 /4

)/dx,

thanks to which the integration by parts can be applied: d  F (t) = 2 dt 2

∞ 0

d −x 2 /4 e sin(tx) dx = −2t dx

∞

cos(tx)xe−x

2 /4

dx = −2tF2 (t).

0

(3.2.96)

3.3 Integrating over Parameters

97

A very simple differential equation for the function F2 (t) (it does not matter whether the function is called y(x), f (z), F2 (t), or whatever) has been obtained, with the obvious solution: ˜ −t 2 . F2 (t) = Ce

(3.2.97)

˜ one can plug t = 0 into the Eq. (3.2.94). In order to determine the constant C, The integral in (3.2.94) then becomes purely Gaussian and one gets F2 (0)

= C˜ =

∞

e−x

2 /4

dx =

√ 1√ 4π = π , 2

(3.2.98)

0

1/2 originating from the integration over the half of the interval ] − ∞, ∞[. Thus, we have obtained the derivative F2 (t) =

√ −t 2 πe ,

(3.2.99)

i.e., the expression identical to (3.2.90). As a consequence, one has F1 (t) − F2 (t) =

d [F1 (t) − F2 (t)] = 0 ⇒ F1 (t) = F2 (t) + C. dt

(3.2.100)

The final step is to show that C = 0, i.e., that F1 (t) = F2 (t). In order to achieve it, let us simply find the value of both functions at one selected point t. The easiest is to do it for t = 0. It is clear that both F1 (0) = 0 and F2 (0) = 0. Hence the constant C vanishes and the equality between the two functions F1 and F2 is established.

3.3 Integrating over Parameters Problem 1 The integral:

1 I (α, β) = 0

will be calculated, where β > α > 0.

xα − xβ dx, log x

(3.3.1)

98

3 Investigating Integrals with Parameters

Solution In the previous sections we were considering the issue, under what conditions one is allowed to switch the order of performing a limit or differentiation with respect to the parameter t with the integration over the variable x. Now we move on to the integration over t and the question arises if and when one can use the formula:

b

B dt f (x, t) =

dx a

B

A

b dt

dx f (x, t).

(3.3.2)

a

A

The answer should already be known after carefully studying Part II of this book series, where we were dealing with concrete examples of the two-dimensional integrability (see Sect. 12.1). There, we learned that if a given function is integrable on the set [a, b] × [A, B], then both iterated integrals exist and are equal to each other, which means that the above formula is true. In particular, this happens when the integrand function is continuous on the set [a, b] × [A, B]. Our main concerns below are the applications of this formula to the calculation of certain one-dimensional integrals such as (3.3.1). In order to convert it into a double integral, let us use the known formula:

x t dt =

xt log x

(3.3.3)

plus an irrelevant constant. Now one can easily observe that

β x t dt = α

 x t β xβ − xα , =  log x α log x

(3.3.4)

thanks to which the expression (3.3.1) can be given the form:

1 I (α, β) = −

β dt x t .

dx 0

(3.3.5)

α

Then one can try to use the theorem recalled above and at the beginning of the chapter. Unfortunately the integrand expression f (x, t) = x t

(3.3.6)

is not continuous on the set concerned, i.e., [0, 1] × [α, β], even if one supplement its definition at the origin in some way. The problem is that going to the point (0, 0) along both axes of the coordinate system, one gets two different limits: 1, along the

3.3 Integrating over Parameters

99

x-axis, and 0, along the t-axis. Thus the effort spared on proving the (nonexistent) continuity of the function f would be wasted. The origin is the only point of the set [0, 1] × [α, β] that causes difficulties. In all other points, the continuity of the function f is clear and we are not going to deal with them. In Problem 3 in Sect. 12.1 of the preceding part of this book series, we were able to cope with this kind of troublesome point by separating it from the rest of the set with a curve γ , which later was contracted to zero (see Fig. 12.1). It was essential for the integral on this small area to be convergent to zero together with the contraction of γ to a point. The same is true in the current problem as well because the expression x t is bounded on our set. This consequently means that the function is integrable on the rectangle [0, 1]×[α, β] and both iterated integrals exist and are equal. In (3.3.5), both integrations may now be “legally” swapped, which yields

β I (α, β) = −

1 dx x = −

dt α

β =− α

β t

α

0

1 x t+1  dt t + 1 0

β    1 α+1  dt = − log(t + 1) = log . t +1 β +1 α

(3.3.7)

The fraction under the logarithm is of course smaller than 1 (as α < β), and therefore the integral I (α, β) < 0. This was expected because the integrand expression in (3.3.1) is negative due to the fact that log x < 0 in the interval ]0, 1[. As one can see, the integrals cumbersome to calculate in one order of integrations have proved to be very simple in the other. This constitutes the clue of the method presented in this section. It is also closely related to the method of the differentiation over the parameter, as described in the previous section. The reader is encouraged to find the result (3.3.7) by calculating ∂I /∂α.

Problem 2 Assuming that a, b, c > 0, the integral:

∞ I (a, b, c) = 0

will be calculated.

e−ax − e−bx sin(cx) dx x

(3.3.8)

100

3 Investigating Integrals with Parameters

Solution This problem is a variation of the fifth exercise of the previous section, in which we were concerned with the differentiation with respect to the parameter under the integral. This time we would like to change the order of integrations. To achieve this goal let us write e−ax − e−bx = x

b

e−xt dt,

(3.3.9)

a

thanks to which the cumbersome denominator is removed. After inserting the righthand side into (3.3.8) one obtains

∞ I (a, b, c) =

b dx sin(cx)

dt e

−xt

a

0

b =

∞ dt

a

dx sin(cx)e−xt ,

(3.3.10)

0

but the “legitimacy” of the last equality still has to be established. The reader has certainly noticed that the new element in relation to the previous problem constitutes the improper character of the x integration. In this case, the appropriate theorem, recalled in the theoretical introduction, requires that apart from the continuity of the integrand function f (x, t) = sin(cx)e−xt

(3.3.11)

on the set [0, ∞[×[a, b], the uniform convergence of the integral

∞ (3.3.12)

f (x, t) dx 0

must be proved. The continuity of the function f is obvious and it can be demonstrated in the same way as in the previous problems. In turn the uniform convergence of (3.3.12) stems from the existence of the integrable majorant (remember that t ≥ a):   |f (x, t)| = sin(cx)e−xt  ≤ e−ax =: ϕ(x).

(3.3.13)

Thereby, the formula (3.3.10) has become “legal” and one is left with executing the integrals in the given order. That with respect to x is already well known to the reader: it can be calculated by integrating by parts twice, which gives

∞ 0

sin(cx)e−xt dx =

c . t 2 + c2

(3.3.14)

3.4 Exercises for Independent Work

101

Finally, calculating the integral over t, one obtains the result:

b I (a, b, c) = a

 t b a c 1 b dt = c arctan  = arctan − arctan . 2 2 c c a c c t +c

(3.3.15)

3.4 Exercises for Independent Work Exercise 1 Demonstrate the continuity of the function

f (ξ ) =

2 

x 2 + ξ 2 dx

−2

on the interval ] − 1, 1[ and of the function

2 g(ξ ) = −1

x2 dx + ξ 2x4

1 + x2

on R. Exercise 2 Differentiating with respect to the parameter, calculate the integrals:

π/2 (a) log(sin2 x + ξ 2 cos2 x) dx, where ξ > 0, 0

π log(1 − 2ξ cos x + ξ 2 ) dx, where ξ > 1.

(b) 0

Answers (a) π log[(1 + ξ )/2]. (b) 2π log ξ .

Exercise 3 Integrating over the parameter, calculate the integrals:

1 (x α − x β )

(a) 0

sin(log x) dz for α, β > 0, log x

102

3 Investigating Integrals with Parameters

∞

2 2 e−a/x − e−b/x dx for a, b > 0. (b) 0

Answers (a) arctan[(β √ √ −√α)/(αβ + α + β + 2)]. (b) π( b − a).

Chapter 4

Examining Unoriented Curvilinear Integrals

In this chapter, we are concerned with the calculations of unoriented integrals in R3 such as areas, center-of-mass locations, or moments of inertia of the sheets of surfaces (curved, in general). The following formulas will be of use. Let the surface be defined with the equation z = f (x, y), where (x, y) ∈ D ⊂ R2 are Cartesian coordinates and f is a certain differentiable function. Then the surface area of the curved sheet corresponding to the domain D may be calculated from the formula: 

 2  2 ∂f ∂f S= + + 1 dx dy. (4.0.1) ∂x ∂y D

This expression is more carefully analyzed when solving the first problem. If the surface is defined in the parametric way via the relation r(u, v), where r = [x, y, z], the appropriate formulas for our choice have the form:

S=

  tu × tv  dudv =

 tu2 tv2 − (tu tv )2 dudv

D



 =

D

∂ r ∂u

2 

∂ r ∂v

2

 −

∂ r ∂ r · ∂u ∂v

2 dudv,

(4.0.2)

D

where tu = ∂ r/∂u, tv = ∂ r/∂v are vectors tangent to the surface and associated with the parameters u, v ∈ D.

© Springer Nature Switzerland AG 2020 T. Rado˙zycki, Solving Problems in Mathematical Analysis, Part III, Problem Books in Mathematics, https://doi.org/10.1007/978-3-030-38596-5_4

103

104

4 Examining Unoriented Curvilinear Integrals

Let us introduce the metric tensor g: ˆ ⎤ ⎡  2 ∂ r ∂ r ∂ r · ⎢ ∂u ∂u ∂v ⎥ ⎥ gˆ = ⎢ ⎣ ∂ r ∂ r  ∂ r 2 ⎦ , · ∂u ∂v ∂v

(4.0.3)

which defines the distance between adjacent points (calculated along the surface), for which the parameters differ by du and dv, according to   du ds 2 = [du, dv] · gˆ · . (4.0.4) dv Then the formula (4.0.2) may be given the form: S=

 det gˆ dudv.

(4.0.5)

D

A certain function ρ(x, y) (for instance, the mass or charge density) can be integrated over the surface following the formula 



M=



ρ(x, y)

∂f ∂x

2

 +

∂f ∂y

2 + 1 dx dy

(4.0.6)

D

or in the parametric formulation

  M= ρ(u, v) tu × tv  dudv,

(4.0.7)

D

where the symbol M refers here to the mass of the sheet, but may denote any other quantity expressible in the form of a surface integral of a scalar function. Either variant of (4.0.2) may be used as well.

4.1 Finding Area of Surfaces Problem 1 The surface area of the sheet: S = {(x, y, x) ∈ R3 | x 2 = zy}, cut out with the planes: z = 0, y = 0, and 2y + z = 2, will be found.

(4.1.1)

4.1 Finding Area of Surfaces

105

Fig. 4.1 The projection of a curved surface onto the plane

Solution If one wishes to calculate the area of a curved sheet constituting the graph of a (smooth) function f (x, y), where (x, y) ∈ D, the formula (4.0.1) may be used. It can be easily justified if one looks at Fig. 4.1. Let us start with mentally dividing the curved surface into “patches” so small that each of them can be considered as flat. In such a case, the area will be simply the sum of individual patches, and the formula will be exact when they become infinitesimally small. This naturally means that in place of the sum an integral appears. One of such patches is marked in the figure in gray. The vector normal to it is inclined to the z-axis by the angle θ (in general, different for various patches). The area of the i-th patch projected onto the plane xy (be it σi proj ) is expressed by the area of the patch itself (i.e., σi ) through the formula: σi proj = σi cos θi .

(4.1.2)

Then, the area to be found can be given the form: S=

! i

σi =

! σi proj i

cos θi

.

(4.1.3)

Note that with our assumptions cos θi never equals zero. Otherwise this would mean that the normal to the patch is horizontal, and thus one of the tangents would parallel to the z-axis. The function f with these properties would be non-

106

4 Examining Unoriented Curvilinear Integrals

differentiable, which was excluded at the beginning. Such cases will be covered in the following exercises, and now we notice that presently the problem of the vanishing denominator of (4.1.3) does not emerge. By choosing the patches corresponding to the infinitesimal rectangles on the xyplane (i.e., with dimensions dx × dy), in place of (4.1.3) one gets the integral:

S=

1 dx dy. cos θ

(4.1.4)

D

In order to determine cos θ , we are going to find the normal vector shown in  is orthogonal Fig. 4.1. As one knows from the previous chapters, the vector ∇g to the surface of g(x, y, z) = const. It is then sufficient to rewrite the equation z = f (x, y) in the form z − f (x, y) = 0,   

(4.1.5)

  ∂f ∂f  n = ∇g(x, y, z) = − , − , 1 . ∂x ∂y

(4.1.6)

=:g(x,y,z)

to easily obtain

The cosine of the angle at which it is inclined with respect to the z-axis is given by the scalar product: cos θ =

n · ez 1 . = 2 | n| (∂f/∂x)2 + (∂f/∂y)2 + 1

(4.1.7)

After plugging this result into (4.1.4), one gets the needed formula (4.0.1). Now one can move on to applying the obtained result to the present exercise. In Fig. 4.2, the surface defined in the text of this problem is drawn. It is true that it does not constitute the graph of the function x(y, z), but thanks to the symmetry, when replacing x → −x, it will suffice to simply limit oneself to the case x > 0 (then the graph shall represent a function), and multiply the obtained surface area by 2. Surely, equally well one could (and the reader is encouraged to do it) in place of the function x(y, z) consider z(x, y) or y(x, z). The choice having been made, one has √ x(y, z) = + yz

(4.1.8)

4.1 Finding Area of Surfaces

107

x

z

y Fig. 4.2 The surface given by the formula (4.1.1)

which constitutes just the function f of the formula (4.0.1). Thus, the expression for the area will take the form 

 2  2 ∂x ∂x + + 1 dydz. (4.1.9) S=2 ∂y ∂z D

Stemming from the content of the exercise, the following conditions must be simultaneously satisfied: y > 0,

2y + z < 2.

(4.1.10)

0 < z < 2(1 − y),

(4.1.11)

z > 0,

They entail that 0