Marin Marin
Andreas Öchsner
Complements
of Higher
Mathematics
Complements of Higher Mathematics
Marin Marin Andreas Öchsner
•
Complements of Higher
Mathematics
123
Marin Marin
Department of Mathematics and Computer
Science
Transilvania University of Brasov
Brasov
Romania
Andreas Öchsner
Faculty of Mechanical Engineering
Esslingen University of Applied Sciences
Esslingen am Neckar
Germany
ISBN 978-3-319-74683-8
ISBN 978-3-319-74684-5
https://doi.org/10.1007/978-3-319-74684-5
(eBook)
Library of Congress Control Number: 2017964238
© Springer International Publishing AG, part of Springer Nature 2018
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The publisher, the authors and the editors are safe to assume that the advice and information in this
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Preface
It is our belief that this book will help many students and researchers with mastering
applied problems from different fields. The material presented here stems from
courses and seminars on special topics in mathematics given by the authors for
many years at universities, especially at technical, engineering, and economics
faculties. We limit our considerations in this chapter to the basic notions and results
of each domain of mathematics considered in this volume due to the diversity of the
included chapters. Chapter 1 is devoted to complex functions. Here, emphasis is
placed on the theory of holomorphic functions, which facilitate understanding the
role played by the theory of complex variables in mathematical physics, especially
in the modeling of plane problems. Furthermore, we consider the remarkable
importance that the theories of special functions, operational calculus, and variational calculus have. A great part of the book is dedicated to second-order partial
differential equations, since they are widely used to model phenomena in physics
and engineering. In the last chapter, we discuss the basic elements of one of the
most modern areas of mathematics, namely the theory of optimal control. To
achieve a relative independence of the book, each chapter introduces the necessary
mathematical background, i.e., topics from mathematical analysis, topology,
functional analysis, and so on, which are used in the other chapters. For this reason,
the larger part of this book is accessible to students of technical, engineering, and
economics faculties, and researchers working in these fields. Some applications are
included to illustrate the theory discussed. The methods used in the book permit the
analysis of both theoretical and practical cases, thus offering results of interest to the
students of technical, engineering, and economics faculties and, also, for young
researchers, interested of higher mathematics.
Our intention was to help the reader to proceed more easily to the study of
special topics in mathematics, which is usually studied in the second year of all
technical faculties. A number of supplementary topics included in this book have
been chosen particularly in consideration of their use in specialized courses.
For the study of this book, it is sufficient for the reader to be familiar with a
classical course on mathematical analysis and a standard course on differential
geometry and algebra, which usually are included in the first year of most programs.
v
vi
Preface
The authors are aware that there are many more results and even more recent
data regarding the domains which are not presented herein.
Only their simple enumeration, even in a simplified form to become more
accessible, would have considerable enlarged the present book.
Excluding these allowed the authors to present thorough mathematical proofs
of the results presented in the book.
The authors would be grateful for readers’ comments on the content and the
design of the textbook. We would also be pleased to receive any other suggestions
the readers may wish to make.
We express our profound gratitude to Prof. C. Marinescu of the Department of
Mathematics, Transilvania University of Brasov, for his kindness in reading the
manuscript and making pertinent observations, which were taken into
consideration.
We are grateful also for suggesting to write this book with the purpose of
supplying the bibliographical material for students interested in higher mathematics.
Brasov, Romania
Esslingen am Neckar, Germany
October 2017
Marin Marin
Andreas Öchsner
Contents
1
Complex Functions . . . . . . . . . . . . . . . . . . . .
1.1 Complex Functions of Real Variable . . .
1.2 Complex Functions of Complex Variable
1.3 Elementary Complex Functions . . . . . . .
1.4 Complex Integrals . . . . . . . . . . . . . . . . .
1.5 Complex Series . . . . . . . . . . . . . . . . . . .
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1
1
5
11
18
30
2
Special Functions . . . . . . . . . .
2.1 Euler’s Functions . . . . . .
2.2 Bessel’s Functions . . . . .
2.3 Orthogonal Polynomials .
2.4 Legendre’s Polynomials .
2.5 Chebyshev’s Polynomials
2.6 Hermite’s Polynomials . .
2.7 Laguerre’s Polynomials .
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53
53
61
68
71
77
81
85
3
Operational Calculus . . . . . . . . . . . . . . . . . . . . . . . .
3.1 Laplace’s Transform . . . . . . . . . . . . . . . . . . . .
3.2 Operational Methods . . . . . . . . . . . . . . . . . . . .
3.3 Applications . . . . . . . . . . . . . . . . . . . . . . . . . .
3.4 Differential Equations with Constant Coefficients
3.5 Differential Equations with Variable Coefficients
3.6 Integral Equations . . . . . . . . . . . . . . . . . . . . . .
3.7 Partial Differential Equations . . . . . . . . . . . . . .
3.8 Some Improper Integrals . . . . . . . . . . . . . . . . .
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89
89
101
105
106
107
108
108
109
4
Fourier’s Transform . . . . . . . . . . . . .
4.1 Fourier Series . . . . . . . . . . . . . .
4.2 Fourier’s Single Integral Formula
4.3 Fourier’s Transform in L1 . . . . .
4.4 Fourier’s Transform in L2 . . . . .
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111
111
121
130
137
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vii
viii
Contents
5
Calculus of Variations . . . . . . . . . . . . .
5.1 Introduction . . . . . . . . . . . . . . . . .
5.2 Euler’s Equation . . . . . . . . . . . . .
5.3 Generalizations of Euler’s Equation
5.4 Sufficent Conditions for Extremum
5.5 Isoperimetric Problems . . . . . . . . .
5.6 Moving Boundary Problems . . . . .
6
Quasi-linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209
6.1 Canonical Form for n ¼ 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . 209
6.2 Canonical Form for n [ 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . 219
7
Hyperbolical Equations . . . . . . . . . . . . . . . .
7.1 Problem of the Infinite Vibrating Chord
7.2 Initial-Boundary Values Problems . . . . .
7.3 Cauchy’s Problem . . . . . . . . . . . . . . . .
7.4 Problem of the Finite Vibrating Chord .
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223
223
232
237
246
8
Parabolical Equations . . . . . . . . . . .
8.1 The Finite Problem of Heat . . .
8.2 Initial-Boundary Value Problems
8.3 Method of the Green’s Function
8.4 Cauchy’s Problem . . . . . . . . . .
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253
253
263
278
286
9
Elliptic Partial Differential Equations
9.1 Introductory Formulas . . . . . . .
9.2 Potentials . . . . . . . . . . . . . . . .
9.3 Boundary Values Problems . . .
...
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295
295
297
302
10 Optimal Control . . . . . . . . . . . . . . . . . . .
10.1 Preparatory Notions . . . . . . . . . . . . .
10.2 Problems of Optimal Control . . . . . .
10.3 Linear Problems of Control . . . . . . .
10.4 Problems of Quadratic Control . . . . .
10.5 The Synthesis of the Optimal Control
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319
319
323
326
339
348
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147
147
159
169
182
187
201
Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 353
Chapter 1
Complex Functions
This chapter contains the basic results on complex functions of real as well as of
complex variables. For more details and more results, the readers are referred to the
books in the bibliography section.
1.1 Complex Functions of Real Variable
Let us consider a real interval (or a reunion of real intervals) E ⊂ R. Any function
defined on E having complex values, f : E → C, is called a complex function of
real variable. So, for ∀t ∈ R, we have f (t) ∈ C, i.e. f (t) = f 1 (t) + i f 2 (t), where
f 1 (t) and f 1 (t) are real functions of a real variable. Also, i is the well known complex
number such that i 2 = −1.
Due to some clear isomorphisms between R 2 and C, as linear spaces, we can
identify a complex function f (t) by a vectorial function ( f 1 , f 2 ), or by a function in
the classical space of vectors, V2 (for v ∈ V2 , we have v = v1i + v2 j).
As a consequence, the known results from R 2 with regards to the open sets, closed
sets, vicinities, and so on, can be transposed, without difficulties, in the complex space
C. More precisely, by introducing the following distance
∗
, d(z 1 , z 2 ) = |z 1 − z 2 |, ∀z 1 , z 2 ∈ C,
d : C × C → R+
and replacing z 1 = x1 + i y1 , z 2 = x2 + i y2 , it follows that
d(z 1 , z 2 ) =
(x1 − x2 )2 + (y1 − y2 )2 ,
which coincides to the Euclidian distance in R 2 , such that we can identify the metrics
spaces (C, d) and (R 2 , d).
© Springer International Publishing AG, part of Springer Nature 2018
M. Marin and A. Öchsner, Complements of Higher Mathematics,
https://doi.org/10.1007/978-3-319-74684-5_1
1
2
1 Complex Functions
Let a be a fixed complex number, a ∈ C. We define the set
(a, ) = {z ∈ C : d(z, a) < } = {z ∈ C : |z − a| < } ,
which is defined as an open disc of center a and radius .
The boundary of this disc is defined as (a, ) = {z ∈ C : |z − a| = }, such that
the closed disc is
(a, ) = (a, ) ∪ (a, ).
The notions of limit, continuity, derivability, and so on, in the point t0 , are defined
with the help of the open intervals, centred in t0 , and, respectively, relative to the set
of values, with the help of the open discs, centered in a fixed point a ∈ C, with the
radius .
Definition 1.1.1 The function f : E ⊂ R → C is a continuous function in t if and
only if (by definition) the functions f 1 and f 2 are continuous functions in t, where
f (t) = f 1 (t) + i f 2 (t).
It is easy to prove that if the function f is continuous in t0 , then the function | f | is
continuous in t0 , too.
If t0 is an accumulating point of E ⊂ R, then the function f has a limit in t0 if
and only if (by definition) the functions f 1 and f 2 have limits in t0 , and
lim f (t) = lim f 1 (t) + i lim f 2 (t).
t→t0
t→t0
t→t0
Consider the real interval I ⊂ R and define J = I \{0}. Then we can define the ratio
(t) =
f (t) − f (t0 )
, ∀t ∈ J.
t − t0
Definition 1.1.2 The function f is a derivable function in t0 if exists and is a finite
number the limit
f (t) − f (t0 )
.
lim (t) = lim
t→t0
t→t0
t − t0
This limit is denoted by
f (t0 ) = lim
t→t0
f (t) − f (t0 )
.
t − t0
The function f is a derivable function in t0 if and only if (by definition) the functions
f 1 and f 2 are derivable functions in t0 and f (t0 ) = f 1 (t0 ) + i f 2 (t0 ).
Definition 1.1.3 A function F is called an antiderivative of the function f on the
interval I = [a, b] ⊂ R if the derivative of F is equal to f on the interval:
1.1 Complex Functions of Real Variable
3
F (t) = f (t), ∀t ∈ [a, b].
It is quite obvious that if F is an antiderivative of f on [a, b] then the function
F(t) + C, where C is a constant, is also an antiderivative of f because
(F(t) + C) = F (t) + C = F (t) = f (t).
Conversely, if F and F1 are two antiderivatives of f on [a, b] their difference is
necessarily equal to a constant C throughout the interval [a, b]:
F1 (t) = F(t) + C.
Indeed,
(F1 (t) − F(t)) = F1 (t) − F (t) = f (t) − f (t) = 0.
Now, remember that (F1 (t) − F(t)) = 0 implies that there is a constant number C
such that F1 (t) − F(t) = C on [a, b] whence it follows F1 (t) = F(t) + C.
So, we have established the fact that if F is an antiderivative of f on an interval
[a, b], then all the possible antiderivatives of f on that interval are expressed by the
formula F(t) + C where any number can be substituted for C.
Theorem 1.1.1 If f : [a, b] → C is a continuous function, then f admits an
antiderivative on [a, b].
Proof We define the function
F : [a, b] → C, F(t) =
t
f (τ )dτ .
a
So, we have
F(t) =
t
( f 1 (τ ) + i f 2 (τ )) dτ =
a
a
t
f 1 (τ )dτ + i
t
f 2 (τ )dτ .
a
Because f is assumed to be continuous, we deduce that f 1 and f 2 are continuous
functions. But, f 1 and f 2 are real functions such
t that we know for them similar
results. For instance, if we denote by F1 (t) = a f 1 (τ )dτ , then
t
t0
F1 (t) − F1 (t0 )
a f 1 (τ )dτ − a f 1 (τ )dτ
= lim
=
lim
t→t0
t→t0
t − t0
t − t0
t
(t − t0 ) f 1 (c)
t f 1 (τ )dτ
= lim
= f 1 (t0 ),
= lim 0
t→t0
t→t0
t − t0
t − t0
where we used the mean theorem.
4
1 Complex Functions
So, we deduce that F1 is derivable and F1 (t0 ) = f 1 (t0 ). It is quite obvious an
analogous result for f 2 , such that we have F(t) = F1 (t) + i F2 (t), i.e. F is derivable
and
F (t) = F1 (t) + i F2 (t) = f 1 (t) + i f 2 (t) = f (t),
and the theorem is concluded.
Definition 1.1.4 Let us consider the interval I = [a, b] ⊂ R and f : I → C. We
say that the function f is integrable on I if and only if (by definition) the functions
f 1 and f 2 are integrable functions on I and
b
f (t)dt =
a
b
f 1 (t)dt + i
a
b
f 2 (t)dt.
a
It is easy to prove the following properties of the definite integral:
b
b
b
f (t)dt + β
g(t)dt,
(α f (t) + βg(t)) dt = α
a
a
a
b
c
b
f (t)dt =
f (t)dt +
f (t)dt, c ∈ (a, b),
a
a
c
b
b
f (t)dt ≤
| f (t)|dt.
a
a
Example. Let us compute the integral
I1 =
Consider the integral
eαt cos ntdt.
−π
I2 =
too, and then we can write
I1 + i I2 =
π
π
eαt sin ntdt,
−π
π
eαt (cos nt + i sin nt) dt =
eαt eint dt =
−π
−π
π
t (α+in) π
e
1
=
=
eπ(α+in) − e−π(α+in) =
et (α+in) dt =
α + in π
α + in
−π
2
2 eπα − e−πα
(−1)n = (−1)n
sinh απ.
=
α + in
2
α + in
π
1.2 Complex Functions of Complex Variable
5
1.2 Complex Functions of Complex Variable
Consider the set C of all complex numbers as a metrical space endowed with the
already used distance d(z 1 , z 2 ) = |z 1 − z 2 |. Let E be a subset of the space C.
Definition 1.2.1 Any function f : E → C is called a complex function of complex
variable.
If we write the function f (z) in the form f (z) = u(x, y)+iv(x, y), ∀z = x+i y ∈ E,
then any considerations on the complex function f (z) can be reduced on the real
functions u(x, y) and v(x, y), as follows:
(i) if z 0 is an accumulating point of E, then f (z) has a limit in z 0 ⇔ u(x, y) and
v(x, y) have limits in the point (x0 , y0 ), where z 0 = x0 + i y0 , and
lim f (z) =
z→z 0
lim
(x,y)→(x0 ,y0 )
u(x, y) + i
lim
(x,y)→(x0 ,y0 )
v(x, y).
lim | f (z)| = lim f (z) .
z→z 0
z→z 0
Also, we have
(ii) f (z) is continuous in the point z 0 ⇔ u(x, y) and v(x, y) are continuous in the
point (x0 , y0 ).
Definition 1.2.2 The function f : E ⊂ C → C is a monogeneous function in z 0 if
exists and is a finite number the following limit
lim
z→z 0
f (z) − f (z 0 )
.
z − z0
We use the notation
f (z 0 ) = lim
z→z 0
f (z) − f (z 0 )
.
z − z0
Proposition 1.2.1 If f is a monogeneous function in z 0 then f is continuous in z 0 .
Proof We can write the function f (z) in the form
f (z) − f (z 0 )
(z − z 0 ) ⇒
z − z0
f (z)− f (z 0 )
lim f (z) = f (z 0 )+ lim
(z − z 0 ) = f (z 0 )+ f (z 0 ) lim (z − z 0 ) = f (z 0 ),
z→z 0
z→z 0
z→z 0
z −z 0
f (z) = f (z 0 ) +
and the proposition is concluded.
6
1 Complex Functions
Theorem 1.2.1 Consider the open subset E ⊂ C, a fixed point z 0 ∈ E and the
complex function f : E → C, f (z) = u(x, y) + iv(x, y). If the function f is
monogeneous in z 0 , then the functions u and v admit partial derivatives and satisfy
the conditions
∂u
∂v
(z 0 ) =
(z 0 ) ,
∂x
∂y
∂u
∂v
(z 0 ) = −
(z 0 ) .
∂y
∂x
The above relations are called the Cauchy-Riemann’s conditions for monogenity.
Proof Let us consider the ratio
(z) =
f (z) − f (z 0 )
.
z − z0
Since E is an open set and z 0 ∈ E, we deduce that there exists the disc
(z 0 , r ) = {z ∈ C : |z − z 0 | < r } ⊂ E.
In the following we use the notations
0 = \{z 0 }, A0 = {z ∈ 0 : z = x + i y0 } , B0 = {z ∈ 0 : z = x0 + i y}
f (x + i y0 ) − f (x0 + i y0 )
f (x0 + i y) − f (x0 + i y0 )
.
, B0 (z) =
A0 (z) =
x − x0
i(y − y0 )
Since the function f is monogeneous in z 0 , the limits of the last two ratios must
be equal:
u(x, y0 ) − u(x0 , y0 )
v(x, y0 ) − v(x0 , y0 )
+i
, ∀z ∈ A0 ,
x − x0
x − x0
v(x0 , y) − v(x0 , y0 )
u(x0 , y) − u(x0 , y0 )
+i
, ∀z ∈ B0 .
B0 (z) =
i(y − y0 )
i(y − y0 )
A0 (z) =
By equaliting the limits of these ratios we deduce the following relations
∂u(x, y)
∂v(x, y)
(z 0 ) =
(z 0 ) ,
∂x
∂y
∂v(x, y)
∂u(x, y)
(z 0 ) = −
(z 0 ) ,
∂x
∂y
i.e. the Cauchy-Riemann’s conditions and the theorem is proved.
1.2 Complex Functions of Complex Variable
7
Remark. It is easy to see that
∂u(x, y)
∂v(x, y)
1
f (z 0 ) =
(z 0 ) + i
(z 0 ) =
∂x
∂x
i
∂u(x, y)
∂v(x, y)
(z 0 ) + i
(z 0 ) .
∂y
∂y
Theorem 1.2.2 Consider the open subset E ⊂ C, a fixed point z 0 ∈ E and the
complex function f : E → C, f (z) = u(x, y) + iv(x, y). If the functions u and v
admit partial derivatives in a vicinity V of z 0 and these derivatives are continuous
and satisfy the Cauchy-Riemann’s conditions, then the function f is monogeneous
in z 0 .
Proof Because of the hypotheses of theorem, we deduce that the functions u and v
are differentiable. So, we deduce that there are functions α(z) and β(z) such that
lim α(z) = lim β(z) = 0,
z→z 0
z→z 0
and we can write
∂u(x, y)
∂u(x, y)
(x − x0 ) +
(y − y0 ) + |z − z 0 | α(z),
∂x
∂y
∂v(x, y)
∂v(x, y)
v(x, y) − v(x0 , y0 ) =
(x − x0 ) +
(y − y0 ) + |z − z 0 | β(z).
∂x
∂y
u(x, y) − u(x0 , y0 ) =
Clearly, we have
f (z) − f (z 0 ) = u(x, y) + iv(x, y) − u(x0 , y0 ) − iv(x0 , y0 ) =
∂u
∂u
u(x, y) − u(x0 , y0 ) + i [v(x, y) − v(x0 , y0 )] =
(x − x0 ) +
(y − y0 ) +
∂x
∂y
∂v
∂v
+i
(x − x0 ) +
(y − y0 ) + |z − z 0 | [α(z) + iβ(z)] =
∂x
∂y
∂v
∂v
∂u
∂u
+i
+i
=
(x − x0 ) +
(y − y0 ) + |z − z 0 | [α(z) + iβ(z)] =
∂x
∂x
∂y
∂y
∂v
∂u
∂v
∂u
+i
+i
=
(x − x0 ) + −
(y − y0 ) + |z − z 0 | [α(z) + iβ(z)] =
∂x
∂x
∂x
∂x
∂v
∂u
+i
=
[x − x0 + i (y − y0 )] + |z − z 0 | [α(z) + iβ(z)] .
∂x
∂x
So, we can divide by (z − z 0 ) and obtain
f (z) − f (z 0 )
=
z − z0
∂v
∂u
+i
∂x
∂x
(z) +
|z − z 0 |
[α(z) + iβ(z)] .
z − z0
Now, by passing to the limit, by z → z 0 , in the last relation we deduce that the
function f is monogeneous in z 0 .
8
1 Complex Functions
Moreover, it results the formula for the derivative of f :
f (z 0 ) =
∂v
∂u
+i
∂x
∂x
1
(z 0 ) =
i
∂u
∂v
+i
∂y
∂y
(z 0 ).
The theorem is concluded.
Applications 1. Let us prove that the function f : C → C, defined by f (z) = z has
no points of monogenity.
Indeed, for z = x + i y we have f (z) = x − i y ⇒ u(x, y) = x and v(x, y) =
−y. Then, the derivative of these functions does not satisfy the Cauchy-Riemann’s
conditions of monogenity, because
∂u
∂v
∂u
∂v
= 1,
= −1 ⇒
=
.
∂x
∂y
∂x
∂y
2. Let us prove that the function f : C → C, f (z) = e z is monogeneous in the
whole complex plane C.
Indeed, for z = x + i y we have f (z) = e x (cos y + i sin y) ⇒ u(x, y) = e x cos y
and v(x, y) = e x sin y. It is a simple exercise to verify that u and v satisfy the
Cauchy-Riemann’s conditions of monogenity, ∀(x, y) ∈ C.
Moreover, for the formula of derivative of f we have
f (z) =
∂v
∂u
+i
= e x cos y + ie x sin y = e x (cos y + i sin y) = e z .
∂x
∂x
3. Let D be the set D = {z ∈ C : z = x + i y, x > 0}. For z = x + i y consider the
function
y
f (z) = ln x 2 + y 2 + iarctg .
x
By simple calculations we deduce that f satisfies the Cauchy-Riemann’s conditions of monogenity on D and f (z) = 1/z.
Definition 1.2.3 Consider an open complex set E ⊂ C and f : E → C. The
function f is called a holomorphic function on E if f is monogeneous in any point
of E.
Theorem 1.2.3 Let D be a complex domain and the function f : d → C, f (z) =
u(x, y) + iv(x, y) with u, v ∈ C 2 (D). If f is a holomorphic function on D, then the
functions u and v are harmonical functions on D.
Proof Remember that the function u is harmonical if its Laplacian is null, i.e.
u(x, y) =
∂2u
∂2u
+
.
∂x 2
∂ y2
1.2 Complex Functions of Complex Variable
9
Now we use the derivative formula for the function f (z)
∂v
1
∂u
+i
=
f (z) =
∂x
∂x
i
∂u
∂v
+i
∂y
∂y
.
Of course, the last equality can be written in the form
∂u
∂v
∂u
∂v
+i
=− i+
,
∂x
∂x
∂y
∂y
from where, by derivation with regard to x, then to y, it obtained the following two
equalities:
∂2v
∂2v
∂2u
∂2u
i
+
,
+
i
=
−
∂x 2
∂x 2
∂ y∂x
∂x∂ y
∂2u
∂2v
∂2u
∂2v
+i
= −i 2 + i 2 .
∂x∂ y
∂x∂ y
∂y
∂y
By using the definition of two complex number, we deduce
∂2u
∂2u
∂2v
∂2v
+
=
0,
+
= 0,
∂x 2
∂ y2
∂x 2
∂ y2
relations that prove that u and v are harmonic functions and the proof of the theorem
is complete.
Definition 1.2.4 Let E be a complex and open set and the function f : E → C.
(i) A point z 0 ∈ E is a ordinary point of the function f if there exists a disc
(z 0 , ) = {z ∈ E : |z − z 0 | < },
such that f is a holomorphic function on .
(ii) A point z 0 ∈ E is a singular point of the function f if any disc (z 0 , )
contains certain points where f is monogeneous and certain points where f is
not monogeneous.
Definition 1.2.5 In the same conditions on the set E and the function f , a singular
point z 0 ∈ E is called an isolated singular point of the function f if there exists a
disc (z 0 , ) which does not contain any other singular point of f , excepting z 0 .
Definition 1.2.6 Let D be a complex domain and the function f : d → C. A point
a ∈ D is called a zero for the function f if there exists a number α ∈ N ∗ and the
function ϕ : d → C, ϕ holomorphic on D, ϕ(a) = 0, such that
f (z) = (z − a)α ϕ(z), ∀z ∈ D.
10
1 Complex Functions
The number α is called the order of the zero.
Proposition 1.2.2 The order of a zero is unique.
Proof Suppose, by the contrary, that there exists another number β ∈ N ∗ and the
function ψ : d → C, ψ holomorphic on D, ψ(a) = 0 such that
f (z) = (z − a)β ψ(z), ∀z ∈ D.
Let us suppose that β > α. Then we can write
(z − a)α ϕ(z) = (z − a)β ψ(z) ⇒ ϕ(z) = (z − a)β−α ψ(z).
Since the functions ϕ and ψ are holomorphic, it results that they are continuous and
then lim ϕ(z) = ϕ(a) and lim ψ(z) = ψ(a). So, by passing to the limit, by z → a
z→a
z→a
in the above relation we deduce that ϕ(a) = 0, that contradicts the hypothesis on ϕ.
The proposition is concluded.
Theorem 1.2.4 All zeroes of a holomorphic function are isolated.
Proof Let a be a zero of the function f . Then there exists a number α ∈ N ∗ and the
function ϕ : d → C, ϕ holomorphic on D, ϕ(a) = 0, such that
f (z) = (z − a)α ϕ(z), ∀z ∈ D.
Let us prove that there exists a disc (a, ) that contains no other zero of the
function f , i.e. f (z) = 0, ∀z ∈ (a, )\{a}. Suppose, ad absurdum, that ∃z 1 ∈
(a, )\{a} such that f (z 1 ) = 0 ⇔ (z 1 − a)α ϕ(z 1 ) = 0 ⇒ ϕ(z 1 ) = 0. Since ϕ is
holomorphic on D we deduce that ϕ is continuous on D. Since ϕ(a) = 0 we can
suppose that |ϕ(a)| > 0. By using the definition of continuity, we obtain: for an
arbitrary ε ∈ (0, |ϕ(a)|) ∃δ(ε) such that for |z − a| < δ(ε) ⇒ |ϕ(z) − ϕ(a)| < ε.
Consider the disc (a, ) where = δ(ε). If there exists z 1 ∈ (a, δ(ε)) such that
ϕ(z 1 ) = 0 then |z 1 − a| < δ(ε) ⇒ |ϕ(z 1 ) − ϕ(a)| < ε ⇒ |ϕ(a)| < ε and this
contradicts that ε ∈ (0, |ϕ(a)|). The theorem is concluded.
Definition 1.2.7 Let E be an open complex set and the function f : E → C. A
singular point a is called a pole for the function f if there exists a number α ∈ N ∗
and the function ϕ : E ∪ {a} → C, ϕ having a as ordinary point, ϕ(a) = 0, such
that
1
ϕ(z), ∀z ∈ E.
f (z) =
(z − a)α
The number α is called the order of the pole.
1.2 Complex Functions of Complex Variable
11
Proposition 1.2.3 The order of a pole is unique.
Proof We can use the same procedure as in the proof of the Proposition 1.2.
Theorem 1.2.5 All poles of a complex function are isolated.
Proof We can use the same procedure as in the proof of the Theorem 1.5.
1.3 Elementary Complex Functions
First of all, we must state that the rules of derivation, in the case of the complex
function, are the same as in the case of real functions. It is a simple exercise to prove
the following rules:
(c. f (z)) = c. f (z),
( f (z).g(z)) = f (z)g(z) + f (z)g (z),
f (z)
g(z)
=
f (z)g(z) − f (z)g (z)
,
g 2 (z)
( f oϕ) (z) = f (ϕ(z))ϕ (z).
We shall call elementary functions some relative simple functions that are used
to construct the usual functions of the concrete applications.
1. Polynomial Function. The function P : C → C is defined by
P(z) = an z n + an−1 z n−1 + · · · + a1 z + a0 , ∀z ∈ C.
Using the Cauchy-Riemann’s conditions for monogenity, it is easy to prove that
the function f (z) = z = x + i y is holomorphic on whole C, and then by using
the mathematical induction, that the function f : C → C, f (z) = z n , n ∈ N ∗
is holomorphic on whole C, too. So, we deduce that the polynomial function is
holomorphic on the whole complex plane C.
2. Rational Function. The function R(z) is defined by
R(z) =
P(z)
,
Q(z)
where P(z) and Q(z) are polynomial functions.
If we denote by E = {z ∈ C : Q(z) = 0}, we deduce that R(z) is a defined and
holomorphic function on C\E.
12
1 Complex Functions
3. Exponential Function. This function is defined by
f : C → C, f (z) = e z .
In a previous application we already proved that this function is holomorphic on
the whole plane C.
Proposition 1.3.1 The exponential function is a periodic function with the main
period T0 = 2πi and the general period Tk = 2kπi, k ∈ Z \{0, 1}.
Proof Consider a complex number T = T1 + i T2 such that f (z + T ) = f (z), ∀z ∈
C. Since for z = x + i y we have f (z) = e z = e x (cos y + i sin y), it results
e x+T1 [cos(y + T2 ) + i sin(y + T2 )] = e x (cos y + i sin y) , ∀x, y ∈ R.
If we take the modulus in this equality, we deduce e x+T1 = e x and then T1 = 0.
Then the previous equality leads to cos(y + T2 ) = cos y and sin(y + T2 ) = sin y ⇒
y + T2 = y + 2kπ ⇒ T2 = 2kπ. The proposition is concluded.
4. Trigonometric and Hyperbolic Functions. These functions are defined as
cos z =
ei z + e−i z
ei z − e−i z
e z + e−z
e z − e−z
, sin z =
, cosh z =
, sinh z =
.
2
2i
2
2
The main properties of these functions are contained in the following proposition.
Proposition 1.3.2 The trigonometric and hyperbolic functions have the following
properties:
(cos z) = − sin z, (sin z) = cos z, (cosh z) = sinh z, (sinh z) = cosh z;
cos z and sin z are periodic functions with the period Tk = 2kπ
cosh z and sinh z are periodic functions with the period Tk = 2kπi
The relationships between the trigonometric and hyperbolic functions are:
cos i z = cosh z and sin i z = i sinh z.
(5) The trigonometric functions have the same zeros as the correspondent real functions and the hyperbolic functions have the following zeros:
(1)
(2)
(3)
(4)
cosh z = 0 ⇔ z k = i
π
+ kπ , sinh z = 0 ⇔ z k = kπi.
2
Proof (1) By direct calculations:
(cos z) =
ei z + e−i z
2
=
1 iz
i iz
ie − ie−i z =
e − e−i z = − sin z.
2
2
(2) By using the definition of a periodic function, we have
1.3 Elementary Complex Functions
13
cos(z + T ) = cos z ⇔ ei(z+T ) + e−i(z+T ) = ei z + e−i z .
We now multiply both sides of the last equality by ei T :
ei(z+T ) ei T + e−i z = ei(z+T ) + e−i z ei T ⇒ ei T − 1 ei(z+T ) − ei T − 1 e−i z = 0 ⇒
⇒ ei T − 1 ei(z+T ) − e−i z = 0 ⇒ ei(z+T ) = e−i z or ei T = 1.
If the first conclusion is true we deduce that e2i z = e−i T , ∀z ∈ C, i.e. the
exponential function is constant, that is false. Bu using the second conclusion
we deduce that cos T + i sin T = 1, then Tk = 2kπ.
(3) By using the definition of a periodic function, we have
cosh(z + T ) = cos z ⇔ e z+T + e−(z+T ) = e z + e−z .
We now multiply both sides of the last equality by e T :
e z+T e T + e−z = e z+T + e−z e T ⇒ e T − 1 e z+T − ei T − 1 e−z = 0 ⇒
⇒ e T − 1 e z+T − e−z = 0 ⇒ e z+T = e−z or e T = 1.
If the first conclusion is true we deduce that e2z = e−T , ∀z ∈ C, i.e. the exponential function is constant, that is false. Bu using the second conclusion we deduce
that
e T1 (cos T + i sin T ) = 1,
for T = T1 +i T2 . By passing to the modulus we obtain e T1 = 1 such that T1 = 0.
Also, from above, using the fact that T1 = 0, it results cos T + i sin T = 1, such
that cos T2 = 1 and sin T2 = 0, i.e. T2 = 2kπ. Finally, T = 2kπi.
(4) We have
2
2
e−z + e z
ei z + e−i z
=
= cosh z,
cos i z =
2
2
ei z − e−i
sin i z =
2i
2
2
z
=
e−z − e z
e z − e−z
=i
= i sinh z.
2i
2
(5) The zeros of the cos z are the solution of the equation cos z = 0. Then
ei z +e−i z = 0 ⇒ e2i z = −1 ⇒ e2i(x+i y) = −1 ⇒ e−2y (cos 2x +i sin 2x) = −1
By passing to the modulus we obtain
e−2y = 1 ⇒ y = 0,
and then, from the last equality, it results
14
1 Complex Functions
cos 2x + i sin 2x = −1 ⇒ cos 2x = −1, sin 2x = 0 ⇒
⇒ 2x = (2k + 1)π ⇒ x =
π
+ kπ, k ∈ Z .
2
In a similar manner we obtain that for the zeros of the function sin z we have
y = 0 and x = kπ, k ∈ Z , i.e. z k = kπ, k ∈ Z .
Similarly,
cosh z = 0 ⇔ z k = i
π
+ kπ , sinh z = 0 ⇔ z k = kπi, k ∈ Z ,
2
and the proposition is concluded.
It is a simple exercise to prove the following relations:
cos (z 1 ± z 2 ) = cos z 1 cos z 2 ∓ sin z 1 sin z 2 ,
sin (z 1 ± z 2 ) = sin z 1 cos z 2 ± cos z 1 sin z 2 ,
sin 2z = 2 sin z cos z, cos 2z = cosz − sin2 z, sin2 z + cos2 z = 1,
cosh (z 1 ± z 2 ) = cosh z 1 cos z 2 ± sinh z 1 sinh z 2 ,
sinh (z 1 ± z 2 ) = sinh z 1 cosh z 2 ± cosh z 1 sinh z 2 ,
sinh 2z = 2 sinh z cosh z, cosh 2z = cosz + sin2 z, cosh2 z − sinh2 z = 1.
√
5. Radical Function. Let us consider the function f (z) = z. Remember that we
can write z in the form z = (cos θ + i sin θ) = eiθ and then we have
√
z = ±1/2 eiθ/2 .
So, for one z we find two values for the radical function and we have the functions
f 1 (z) = 1/2 eiθ/2 , f 2 (z) = −1/2 eiθ/2 .
Definition 1.3.1 Function f is called a multi-form if almost two values of f (z)
correspond to a single z.
If z is placed on a curve that does not contain the origin, then the functions f 1
and f 2 have the above values. But, if z is placed on a closed curve that contains the
origin, then argz increases by 2π when z moves along this curve, starting from a
point M, which belongs to this curve, and arrives again in M. Hence,
f 1 (z) = 1/2 ei(θ+2π)/2 = 1/2 eiθ/2 eiπ = −1/2 iθ/2 = f 2 .
1.3 Elementary Complex Functions
15
In the same manner we obtain f 2 (z) = f 1 (z). Hence, the two branches of the radical
function pass one to one when z moves along a closed curve that contains the origin.
We say the z = 0 is the critical point or the point of ramification for the radical
function. To make the branches f 1 and f 2 as uniform functions we must make a cut
in the complex plane through a semi-axis starting from the origin. So, the point z
cannot move on a closed curve that contains the origin.
Another procedure to make the branches f 1 and f 2 as uniform functions consists
in a superposition of two identical complex planes, making a cut along a semiaxix starting from the origin. This manner is known as the Riemannian surfaces
method. Analogical considerations are still valid in the case of the more general
√
radical function f (z) = m z, but in this case it is necessary to consider m Riemannian
surfaces.
6. Logarithmic Function. This function is denoted by Lnz and is defined as the
solution of the equation
e f (z) = z.
If we denote f (z) = u + iv and z = (cos θ + i sin θ) then
eu (cos v + i sin v) = (cos θ + i sin θ) ⇒ eu = , v = θ + 2kπ, k ∈ Z ,
u = ln , v = θ + 2kπ, k ∈ Z ⇒ Lnz = ln |z| + i(θ + 2kπ), k ∈ Z ,
where θ is the main argument of z and is denoted by θ =argz. So, we can write the
logarithmic function in the form
Lnz = ln |z| + i(argz + 2kπ).
If we denote by ln z = ln |z|+iargz, that is called the main value of the logarithmic
function, we can write
Lnz = ln z + 2kπi, k ∈ Z .
So, it is easy to see that the logarithmic function is a multi-form function with an
infinite number of branches:
f 0 (z) = ln z,
f 1 (z) = ln z + 2πi,
f 2 (z) = ln z + −πi,
..
.
If z is placed on a closed curve with the origin inside, the f 0 (z) becomes f 1 (z),
f 1 (z) becomes f 2 (z), and so on. So, the branches of a logarithm function are multiform functions, that become uniform functions with the help of a cut in the complex
16
1 Complex Functions
plane along of a semi-axis that starts from the origin. The point z = 0 is called the
point of ramification for the logarithmic function.
The main properties of the logarithmic function are included in the following
proposition.
Proposition 1.3.3 For the main value of the Logarithmic Function we have:
ln (z 1 z 2 ) = ln z 1 + ln z 2 ,
z1
= ln z 1 − ln z 2 ,
ln
z2
ln z n = n ln z,
√
1
ln n z = ln z.
n
Proof All these relations can be proved by using the formula of logarithmic function.
For instance,
ln (z 1 z 2 ) ln |z 1 z 2 | + iarg (z 1 z 2 ) = ln |z 1 | + ln |z 2 | + i (argz 1 + argz 2 ) .
The same procedure can be used for other properties.
Remark. The above properties are still not valid for Lnz. For instance
Ln (z 1 z 2 ) = ln z 1 + Lnz 2 = Lnz 1 + ln z 2 .
7. Power Function. The power function is defined by the formula
f (z) = z α = eαLnz .
Of course, because of the logarithmic function, the power function is a multi-form
function.
Proposition 1.3.4 If α ∈
/ Q then the function f (z) = z α has an infinite number of
branches and if α ∈ Q then the power function has a finite number of branches.
Proof Of course, we can write the power function in the form
f (z) = z α = eαLnz = eα[ln +i(θ+2kπ)] , k ∈ Z .
In order to have a finite number of branches, there must exist two different values
of k which give the same value for z α .
eα[ln +i(θ+2k1 π)] = eα[ln +i(θ+2k2 π)] ⇒ e2k1 πα = e2k2 πα ⇒ e2παi(k1 −k2 ) = 1 ⇒
⇒ 2παi(k1 − k2 ) = 2mπ ⇒ (k1 − k2 )α = m ⇒ α =
and the proposition is concluded.
m
, m, k1 , k2 ∈ Z ,
k1 − k2
1.3 Elementary Complex Functions
17
Application. We want to prove that the expression i i has real values only. Indeed,
i i = eiLni = ei(ln |i|+i(arg(i)+2kπ) = e−(π/2+2kπ) , k ∈ Z .
8. Inverse Trigonometric Functions. By definition, the inverse function of the function cos z is the solution of the equation cos f (z) = z and is denoted by Arccosz. In
the following we want to give an explicit expression for the function Arccosz.
ei f + e−i f = 2z ⇒ e2i f − 2zei f + 1 = 0; = z 2 − 1 = i 2 (1 − z 2 ) ⇒
⇒ ei f = z ± i 1 − z 2 ⇒ i f = Ln z ± i 1 − z 2 ⇒
1
Ln z ± i 1 − z 2 .
i
√
The quantity 1/i ln z + i 1 − z 2 is called the main part of the function Arccosz
and is denoted by arccos z. For the other part, we have
⇒ Arccosz =
z 2 − i 2 (1 − z 2 )
1
= ln
= − ln(z+i 1 − z 2 ).
ln z − i 1 − z 2 = ln
√
√
2
z+i 1−z
z + i 1 − z2
Hence,
Arccosz =
= 2kπ +
1
ln z ± i 1 − z 2 + 2kπ =
i
1
ln z ± i 1 − z 2 = 2kπ ± arccos z.
i
By definition, the inverse function of the function sin z is the solution of the
equation sin f (z) = z and is denoted by Arcsinz. In the following we want to give
an explicit expression for the function Arcsinz.
ei f − e−i f = 2i z ⇒ e2i f − 2i zei f − 1 = 0; = i 2 z 2 + 1 = 1 − z 2 ⇒
⇒ ei f = i z ± 1 − z 2 ⇒ i f = Ln i z ± 1 − z 2 ⇒
1
Ln i z ± 1 − z 2 .
i
√
The quantity 1/i ln i z + 1 − z 2 is called the main part of the function Arccosz
and is denoted by arcsin z. For the other part, we have
⇒ Arcsinz =
ln i z −
i 2 z2 − 1 + z2
−1
1 − z 2 = ln
= ln
√
√
2
iz + 1 − z
i z + 1 − z2
18
1 Complex Functions
= ln(−1) − ln(i z +
1 − z 2 ) = ln(| − 1|) + iarg(−1) − ln(i z +
= πi − ln(i z +
1 − z2) =
1 − z 2 ).
Hence,
Arcsinz =
1
ln i z +
i
1
1 − z 2 + 2kπ = 2kπ + ln i z +
i
1 − z2 ⇒
⇒ Arcsinz = 2kπ + arcsin z.
Arcsinz =
1
ln i z −
i
1
πi − ln i z +
1 − z 2 + 2kπ = 2kπ +
i
= 2kπ + π −
1
ln i z +
i
1 − z2
=
1 − z2 ⇒
⇒ Arcsinz = (2k + 1)π − arcsin z.
1.4 Complex Integrals
Consider γ : [a, b] → C, [a, b] ⊂ R, a parametrized and smooth (i.e. γ ∈ C 1 [a, b])
curve. Let f be a continuous function f : E → C, where E is an open subset of C.
Then ( f oγ)(t)γ (t) is a continuous function, too, and, hence there exists the integral
b
( f oγ)(t)γ (t)dt.
a
Definition 1.4.1 In the above conditions imposed on the curve γ and the function
f we define the complex integral of the function f on the curve γ by
b
f (z)dz =
γ
( f oγ)(t)γ (t)dt.
a
Remarks.
(1) If we write z = x + i y and f (z) = u(x, y) + iv(x, y) then dz = dx + idy and
f (z)dz = (u + iv)(dx + idy) = udx − vdy + i(udy + vdx),
hence,
1.4 Complex Integrals
19
f (z)dz =
γ
[u(x, y)dx − v(x, y)dy] + i
γ
[u(x, y)dy + v(x, y)dx].
γ
(2) If we write γ(t) = α(t) + iβ(t) then γ (t) = α (t) + iβ (t). So,
f (γ(t))γ (t) = [u(α, β) + iv(α, β)] α + iβ =
= u(α, β)α − v(α, β)β + i u(α, β)β + v(α, β)α .
Then the complex integral becomes
f (z)dz =
γ
=
f (γ(t))γ (t)dt =
γ
u(α, β)α − v(α, β)β dt + i
γ
u(α, β)β + v(α, β)α dt.
γ
The main properties of the complex integral are contained in the next proposition.
Proposition 1.4.1 (i) For any two parametrized, smooth and equivalent curves,
γ1 and γ2 , we have
f (z)dz =
f (z)dz.
γ1
γ2
(ii) For any two complex and continuous functions f and g and any two complex
constants α and β we have
(α f (z) + βg(z))dz = α
f (z)dz + β g(z)dz.
γ
γ
γ
(iii) When a point moves on the curve γ in the contrary sense as the positive sense
(established by convention) then
f (z)dz = −
γ−
f (z)dz.
γ
Proof It is easy to prove these assertions by means of similar properties of the real
integral.
20
1 Complex Functions
Application. Let us compute the integral
In =
(z − a)n dz,
γ
where γ is the circle having the origin as center and the radius equal to r . We can
write γ(t) = a + r eit , t ∈ [0, 2π], γ (t) = rieit ,
2π
f (z)dz =
γ
2π
r e rie dt = r
n int
it
n+1
0
i
eit (n+1) dt.
0
Then, for n = −1 ⇒ I−1 = 2πi, and for n = −1 we have
In = r
n+1
2π
eit (n+1) r n+1 2i(n+1)π
e
i
=
− 1 = 0.
i(n + 1) 0
n+1
Definition 1.4.2 If γ : [a, b] → E ⊂ C is a piecewise smooth curve (i.e. γ is a
reunion of a finite number, m, of smooth curves, then the complex integral of the
complex function f on γ is defined as
f (z)dz =
m f (z)dz.
k=1 γ
γ
k
Application. Let us compute the integral
z n dz, n ∈ N .
I =
γ
Here γ = γ1 ∪ γ2 , where γ2 is the semi-circle having the origin as center and the
radius equal to a and γ1 is the interval [−a, a]. We can write γ1 (t) = t, t ∈ [−a, a]
and γ2 (t) = aeit , t ∈ [0, π]. Then
a
f (z)dz =
I =
γ
π
t dt +
n
−a
eitn a n aieit dt.
0
By simple calculations we obtain that I = 0.
In the next theorem we prove a very important result of the theory of complex
functions. The result is due to Cauchy.
1.4 Complex Integrals
21
Theorem 1.4.1 Let be a bounded domain in the complex plane C having the
boundary , where is a reunion of a finite number of closed, simple and smooth
curves. If f is a holomorphic function on = ∪ then
f (z)dz = 0.
Proof Without loss of generality, we suppose that f ∈ C 1 () and that means u, v ∈
C 1 (). We now remember the Riemann-Green’s formula. If P(x, y) and Q(x, y)
are functions of class C 1 () then
P(x, y)dx + Q(x, y)dy =
∂
∂Q
∂P
−
∂x
∂y
dxdy.
Based on the hypothesis, f is a holomorphic function and that means that the
Cauchy-Riemann’s relations are still valid:
∂v
∂u
=
,
∂x
∂x
∂v
∂u
=− .
∂y
∂x
Based on these relations and the Riemann-Green formula we have
∂v
∂v
∂u
∂u
−
−
dxdy +
−
dxdy = 0.
f (z)dz =
∂x
∂y
∂y
∂x
The theorem is concluded.
Corollary 1.4.1 Let D be a simple conex domain and γ a simple, closed and smooth
curve included in D, γ ⊂ D. If f is a holomorphic function on D, then
f (z)dz = 0.
γ
Proof Since γ ⊂ D we deduce that , which is the domain closed by γ, is included
in D. Based on Theorem 1.4.1 we deduce that the integral on the boundary of is
null. But the boundary of is γ such that
f (z)dz = 0.
γ
22
1 Complex Functions
Corollary 1.4.2 Let D be a multi-conex domain having m voids, 1 , 2 , . . . , m ,
bounded by 1 , 2 , . . . , m which are simple, closed and smooth curves included in
D. If 0 is the outside boundary of D, then
f (z)dz =
m 0
Proof. Consider
f (z)dz.
k=1 k
= D\ 1 ∪ 2 ∪ ... ∪ m .
The boundary of is 0 ∪ 1 ∪ ... ∪ m = . Because the voids were avoided, we
deduce that f is holomorphic on and then, based on the Theorem 1.4.1, we deduce
f (z)dz =
f (z)dz = 0.
∂
But the sense on the curves k is negative and then
f (z)dz +
0
Since
f (z)dz +
1−
f (z)dz + · · · +
2−
f (z)dz = −
f (z)dz,
k
k−
we deduce
f (z)dz = 0.
m−
f (z)dz =
m f (z)dz.
k=1 0
k
Application. Let us compute the integral
z2
z
dz,
−1
where is a circle having the origin as center and the radius equal to a = 1.
(1) If a < 1 then the domain bounded by is simple conex and f is holomorphic
such that
z
dz = 0,
2
z −1
based on Theorem 1.4.1
1.4 Complex Integrals
23
(2) If a > 1 the domain bounded by is double conex. Consider the domain
0 = \ 1 ∪ 2 ,
where the points −1 and 1 was isolated by two circle 1 and 2 , respectively.
Because f is holomorphic on 0 we deduce that
f (z)dz = 0.
∂0
But the boundary of 0 is
∂0 = ∪ 1− ∪ 2− .
By using corollary 1.2 we obtain
z
dz =
z2 − 1
If we write
1
z
dz +
z2 − 1
z
1
=
z2 − 1
2
⎡
then
I1 =
1⎣
2
1
1
dz +
z−1
1
2
z
dz = I1 + I2 .
z2 − 1
1
1
+
,
z−1 z+1
⎤
1
1
dz ⎦ = (0 + 2πi) = π,
z+1
2
⎤
⎡
1
1
1
1⎣
I2 =
dz +
dz ⎦ = (2πi + 0) = π.
2
z−1
z+1
2
2
2
Finally, for a > 1 we have I = 2πi.
Theorem 1.4.2 Let D be a simple conex domain in the complex plane C and f :
d → C a holomorphic function. Consider L 1 and L 2 two simple and smooth curves
included in D and having the same extremities. Then
f (z)dz =
L1
f (z)dz.
L2
Proof Firstly, consider the case when the curves L 1 and L 2 have no other common
points, without the extremities. Then L 1 ∪ L 2 is a closed, simple and smooth curve
24
1 Complex Functions
that closes the domain ⊂ D. If the sense on L 1 is positive, on L 2 the sense is
negative. Based in corollary 1.4.1 we have
f (z)dz = 0.
∂
But ∂ = L 1 ∪ L −
2 and then
f (z)dz =
f (z)dz +
∂
L1
⇒
f (z)dz = 0 ⇒
L−
2
f (z)dz =
L1
f (z)dz.
L2
Consider now the case when the curves L 1 and L 2 have other common points.
We can take another curve L 3 that has no other common points both on L 1 and L 2 .
By using the first part of the proof, we deduce
f (z)dz =
L1
f (z)dz,
L3
f (z)dz =
L2
⇒
f (z)dz ⇒
L3
f (z)dz =
L1
f (z)dz.
L2
The theorem is proved.
Remark. The previous theorem afirms that in a simple conex domain the integral of
a holomorphic function is independent of the curve that lies between two complex
points z 0 and z 1 .
Theorem 1.4.3 Let D be a simple conex domain in the complex plane C and f :
d → C a holomorphic function. For a fixed point z 0 ∈ D we define the function
z
F(z) =
f (t)dt, ∀t ∈ D.
z0
Then the function F is holomorphic on D and F (z) = f (z), ∀z ∈ D.
Proof As we already know
z
z
f (t)dt =
z0
z
udx − vdy + i
z0
udy + vdx.
z0
1.4 Complex Integrals
25
Based on the previous theorem the last two integrals are independent of the curves
that lie between the points z 0 and z and then ∃U (x, y), V (x, y) such that dU =
udx − vdy and dV = udy + vdx. So,
z
z
dU + i
F(z) =
z0
But
dV = U + i V.
z0
∂U
∂V
∂V
∂U
= u,
= −v,
= v,
= u, ⇒
∂x
∂y
∂x
∂x
⇒
∂V ∂U
∂V
∂U
=
,
=−
,
∂x
∂x ∂ y
∂x
i.e. U and V satisfy the Cauchy-Riemann’s conditions. So, we deduce that the function F is holomorphic on D and
F (z) =
∂U
∂V
+i
= u + iv = f (z).
∂x
∂x
The theorem is concluded.
Definition 1.4.3 Let D be a simple conex domain in the complex plane C and the
function f : D → C. A function F : D → C is an antiderivative of f if:
(1) F is a holomorphic function on D;
(2) F (z) = f (z), ∀z ∈ D.
Remark. The function F defined in Theorem 1.4.3 is an antiderivative of the
function f .
It is easy to prove the following two properties of the antiderivates, included in
the next proposition.
Proposition 1.4.2 For a given complex function f we have:
(1) If F is an antiderivative of f then F + K is an antiderivative too, for any complex
constant K .
(2) If F1 and F2 are two antiderivatives of the function f then F1 − F2 is constant.
Theorem 1.4.4 (Cauchy’s Formula) Let D be a bounded domain in the complex
plane C with boundary which is a reunion of a finite number of closed, simple
and smooth curves. If f is a holomorphic function and we know the values of f on
the boundary, then we can compute the values of f in any point of D by using the
formula
f (z)
1
dz, ∀a ∈ D.
f (a) =
2πi
z−a
26
1 Complex Functions
Proof Let a be an arbitrary fixed point, a ∈ D. Define the function
g(z) =
f (z)
,
z−a
which is holomorphic on D\{a}.
Consider a disc ω having a as center and a sufficient small radius such that
ω ⊂ D. Denote by γ the boundary of the disc ω and = D\(ω ∪γ). Of course, g is
holomorphic on and, based on the Cauchy’s theorem we have g(z) = 0, where
∂
∂ = ∪ γ − . Hence, we have
g(z)dz + g(z)dz = 0 ⇒ g(z)dz = g(z)dz ⇒
⇒
f (z)
dz =
z−a
⇒
γ
γ−
f (z)
dz =
z−a
γ
f (z)
dz =
z−a
γ
f (z) − f (a) + f (a)
dz ⇒
z−a
f (z) − f (a)
dz + f (a)
z−a
γ
γ
1
dz.
z−a
(1.4.1)
Based on a previous application, we know that
γ
1
dz = 2πi,
z−a
because γ is a circle having a as center.
To arrive at the desired result we must prove that
γ
f (z) − f (a)
dz = 0.
z−a
Since f is holomorphic, we deduce that f is continuous and then ∀ε > 0 ∃δ(ε)
such that ∀z ∈ , |z − a| < δ(ε) we have | f (z) − f (a)| < ε. But we can chose the
radius of the circle γ such that < δ(ε) and then
f (z) − f (a) | f (z) − f (a)|
| f (z) − f (a)|
ε
=
=
< .
z−a
|z − a|
1.4 Complex Integrals
27
Therefore,
f (z) − f (a) dz ≤
z−a
γ
γ
f (z) − f (a) ε
ε
dz <
dz = 2π = 2πε.
z−a
γ
Then, the formula (4.1) becomes
f (z)
dz = 2π f (a),
z−a
and the theorem is proved.
Remark. In the case when the domain is multi-conex, the boundary is given by
= 0 ∪ 1 ∪ 2 ∪ ... ∪ m ,
the Cauchy’s formula becomes
f (a) =
1
2πi
0
m 1 f (z)
f (z)
dz −
dz, ∀a ∈ D.
z−a
2πi k=1
z−a
k
Application. Let us compute the integral
I =
γ
cos πz
dz,
z2 + 1
where γ is a simple, closed and smooth curve that does not pass through −1 and i.
Denote by D the domain closed by γ.
(1) If −i, i ∈
/ D, since f is a holomorphic function we obtain that I = 0.
(2) If −i ∈ D we can write
I =
γ
cos πz
ϕ(z)
dz, ϕ(z) =
,
z+i
z−i
such that, based on the Cauchy’s formula we have
I = 2πiϕ(−i) = −π cos πi = −π cosh π.
(3) If i ∈ D we can write
28
1 Complex Functions
I =
γ
cos πz
ϕ(z)
dz, ϕ(z) =
,
z−i
z+i
such that, based on the Cauchy’s formula we have
I = 2πiϕ(i) = π cos πi = π cosh π.
(4) If −i, i ∈ D we take two circles γ1 , γ2 having the centers −i and i, respectively
and then we can write
I =
f (z)dz = I =
γ
γ1
ϕ(z)
dz +
z+i
γ2
cos πz
cos πz
ψ(z)
dz, ϕ(z) =
, ψ(z) =
.
z−i
z−i
z+i
Theorem 1.4.5 Let E be an open complex set. Consider ϕ a continuous function,
ϕ : E → C and a simple and smooth curve contained in E. Then the function
f n (z) =
ϕ(τ )
dz,
(τ − z)n
is a holomorphic function on C\ and its derivative has the form
f n (z) = n
ϕ(τ )
dz.
(τ − z)n+1
Proof To speed the proof we consider only the case n = 1. Of course, the function
ϕ(τ )/(τ −z) is defined and continuous for any τ ∈ E\{z}. Consider a disc ω bounded
by the circle γ having z as the center and the radius is sufficient small such that
ω ∪ γ ⊂ E\. We must prove that there exists and is a finite number the following
limit
1
f (z + h) − f (z)
1
1
= lim
−
dτ =
ϕ(τ )
lim
h→0
h→0 h
h
τ −z−h
τ −z
1
= lim
h→0 h
h
dτ = lim
ϕ(τ )
h→0
(τ − z − h)(τ − z)
ϕ(τ )
1
dτ .
(τ − z − h)(τ − z)
We can write
1
1
h
=
+
,
(τ − z − h)(τ − z)
(τ − z)2
(τ − z − h)(τ − z)2
and the previous limit becomes
1.4 Complex Integrals
⎡
lim ⎣
29
h→0
ϕ(τ )
dτ + h
(τ − z)2
⎤
ϕ(τ )
dτ ⎦ .
(τ − z)2 (τ − z − h)
Since ϕ is a continuous function we have
∃M = sup |ϕ(τ )|.
τ ∈
Also,
|τ − z| > , |τ − z − h| > |τ − z| − |h| > − |h|.
Then
ϕ(τ )
dτ
(τ − z)2 (τ − z − h) ≤
M
ϕ(τ )
(τ − z)2 (τ − z − h) dτ < 2 ( − |h|) L().
Hence, the last integral is finite. Here L() is the measure of the arc . Finally,
we obtain
f (z + h) − f (z)
ϕ(τ )
=
dτ .
f (z) = lim
h→0
h
(τ − z)2
The theorem is proved.
Theorem 1.4.6 Let D be a complex domain and f a holomorphic function, f :
d → C. Consider a simple, closed and smooth curve that bounds the domain such that = ∪ ⊂ D. Then
f (τ )
n!
dτ , ∀z ∈ .
f (n) (z) =
2πi
(τ − z)n+1
Proof Based on the Cauchy’s formula we have
f (z) =
1
2πi
f (τ )
dτ , ∀z ∈ .
τ −z
According to the previous theorem, we have
f (z) =
1
2πi
f (τ )
dτ , ∀z ∈ .
(τ − z)2
Since f is a monogeneous function we obtain
30
1 Complex Functions
f (z) =
2!
2πi
f (τ )
dτ , ∀z ∈ .
(τ − z)3
The general result can be obtained by using the mathematical induction. The proof
of the theorem is closed.
1.5 Complex Series
Definition 1.5.1 An expression of the form
cn (z − z 0 )n = c0 + c1 (z − z 0 ) + c2 (z − z 0 )2 + · · ·
n≥0
where the numbers cn are complex constants, is called a complex series (more
explicitely, power complex series), the numbers cn being referred to as its coefficients.
In the particular case when z 0 = 0 we have the power series centred in the origin.
As in the real case, we define the set of convergences A, the radius of convergences
and the disc of convergences as
A= z∈C :
cn z n is convergent ,
n≥0
= sup A,
|z|∈R
(0, ) = {z ∈ C : |z| < }.
By using the same procedure as in the case of the real power series, we can prove
the following theorem.
Theorem 1.5.1 Denoting by l the following limit
l = lim
n→∞
n
|cn |,
then
(1) if l ∈ (0, ∞) ⇒ = 1/l;
(2) if l = 0 ⇒ = ∞;
(3) if l = ∞ ⇒ = 0.
Definition 1.5.2 Let D be a complex domain, f a holomorphic function, f : d → C
and z 0 ∈ D. A power series of the form
1.5 Complex Series
31
cn (z − z 0 )n , where cn =
n≥0
1 (n)
f (z 0 ),
n!
is called the Taylor’s series relative to point z 0 .
Theorem 1.5.2 Let D, f , z 0 be as in above definitions. Consider the disc having
z 0 as center and radius a, = {z ∈ D : |z − z 0 | < a}, with boundary . If
= ∪ ⊂ D then the Taylor’s series attached to the function f relative to the
point z 0 is convergent and we have
f (z) =
cn (z − z 0 )n , where cn =
n≥0
1 (n)
f (z 0 ).
n!
Proof As we already know, since f is holomorphic, it admits derivatives of any order
and we have
f (τ )
n!
f (n) (z) =
dτ .
2πi
(τ − z)n+1
Also, by using the Cauchy’s formula, we have
f (z) =
1
2πi
f (τ )
dτ .
τ −z
(1.5.1)
We can write
1
1
1
1
=
=
,
τ −z
τ − z 0 − (z − z 0 )
τ − z0 1 − q
where
q=
Because
z − z0
.
τ − z0
z − z0 = |z − z 0 | = |z − z 0 | < 1,
|q| = τ − z 0 |τ − z 0 |
a
we can write
1 − q n+1 + q n+1
1 − q n+1
q n+1
1
1
=
=
+
=
q n+1 .
qk +
1−q
1−q
1−q
1−q
1
−
q
k=0
n
Therefore
n z − z0 )
1
1
=
τ −z
τ − z 0 k=0 τ − z 0
k
+
1
1−
z−z 0 )
τ −z 0
z − z0 )
τ − z0
n+1
=
32
1 Complex Functions
n
z − z0 )
(z − z 0 )k
1
=
+
k+1
(τ − z 0 )
τ − z τ − z0
k=0
n+1
.
By introducing these estimations in Eq. (1.4.1) we obtain
1
f (z) =
2πi
n
z − z0 )
(z − z 0 )k
1
f (τ )
+
k+1
(τ − z 0 )
τ − z τ − z0
k=0
n+1
dτ =
n
f (τ )
k 1
(z − z 0 )
dτ + Rn (z),
=
2πi
(τ
−
z 0 )k+1
k=0
where
Rn (z) =
1
(z − z 0 )n+1
2πi
f (τ )
dτ .
(τ − z)(τ − z 0 )n+1
Passing to the limit for n → ∞, we obtain Rn (z) → 0 and f (z) becomes
f (z) =
∞
cn (z − z 0 )n ,
n=0
where
cn =
1 (n)
f (z 0 ).
n!
The theorem is concluded.
Applications 1. Consider the function f (z) = e z and an arbitrary fixed point z 0 ∈ C.
Since f (n) (z 0 ) = e z0 , ∀n ∈ N , we have
z − z0
(z − z 0 )2
(z − z 0 )n
1
n
z0
1+
(z−z 0 ) = e
+
+ ··· +
+ ···
e
e =
n!
1!
2!
n!
n=0
z
∞
z0
.
In the particular case when z 0 = 0 we have
ez = 1 +
z2
zn
z
+
+ ··· +
+ ···
1!
2!
n!
2. We want to expand the function f (z) = sin z around the origin. It is easy to prove
that
π
π
(−1)k , for n = 2k + 1
f n (z) = sin z + n
⇒ f n (0) = sin n
=
0, for n = 2k.
2
2
1.5 Complex Series
33
Then the Taylor’s series of the function f (z) = sin z is
sin z =
∞
z 2n+1
n=0
z
z3
z5
(−1)n
z 2n+1
=
−
+
− · · · + +(−1)n
+ ···
(2n + 1)!
1!
3!
5!
(2n + 1)!
By using a similar procedure we obtain
cos z =
∞
n=0
z 2n
z2
z4
(−1)n
z 2n
=1−
+
− · · · + +(−1)n
+ ···
(2n)!
2!
4!
(2n)!
1
= 1 + z + z2 + · · · + zn + · · ·
1−z
1
= 1 − z + z 2 − z 3 + · · · + (−1)n z n + · · ·
1+z
Definition 1.5.3 If the function f is holomorphic on the whole complex plane C
then f is called an integer function.
Remarks.
(1) It is easy to prove that the coefficients of a Taylor’s series satisfy the inequality
|cn | ≤
M(a)
, ∀n ∈ N , where M(a) = sup | f (τ )|, = {z ∈ D : |z−τ | = a}.
an
τ ∈
(2) Based on the above inequality, Liouville proved that an integer function which
is bounded, reduces to a constant.
Definition 1.5.4 A series of the form
∞
cn (z − z 0 )n ,
n=−∞
is called a Laurent’s series, centred in the point z 0 .
We call the main part of a Laurent’s series the following series
−1
cn (z − z 0 )n ,
n=−∞
and the regular part (or Taylor’s part) of a Laurent’s series the following series
∞
n=0
cn (z − z 0 )n .
34
1 Complex Functions
Theorem 1.5.3 Consider the complex domain D, the function f which is holomorphic on D and the corona
= {z ∈ D : R1 < |z − z 0 | < R2 },
having the circles 1 and 2 as boundaries.
If = ∪ 1 ∪ 2 ⊂ D, then the function f admits an expansion in a Laurent’s
series around z 0
∞
f (z) =
cn (z − z 0 )n , cn =
n=−∞
1
2πi
f (τ )
dτ ,
(τ − z 0 )n+1
where is a circle having z 0 as the center and the radius R ∈ [R1 , R2 ].
Proof By using the Cauchy’s formula, we have
1
f (z) =
2πi
1− ∪2
⎞
⎛
f (τ )
f (τ )
f (τ )
1 ⎝
dτ =
dτ −
dτ ⎠ , ∀z ∈ .
τ − z0
2πi
τ − z0
τ − z0
2
1
(1.5.2)
Consider, separately, the ratio 1/(τ − z):
1
1
1
1
=
=
,
τ −z
τ − z 0 − (z − z 0 )
τ − z0 1 − q
where
q=
|z − z 0 |
z − z0
|z − z 0 |
=
⇒ |q| =
< 1.
τ − z0
|τ − z 0 |
R2
Then
n z − z0
1
1
=
τ −z
τ − z 0 k=0 τ − z 0
k
+
1
1−
z−z 0
τ −z 0
n
z − z0
(z − z 0 )k
1
+
=
k+1
(τ − z 0 )
τ − z τ − z0
k=0
z − z0
τ − z0
n+1
=
n+1
.
Therefore
1
2πi
2
1
f (τ )
dτ =
(z − z 0 )k
τ −z
2πi
k=0
n
2
f (τ )
dz + Rn (z),
(τ − z 0 )k+1
1.5 Complex Series
35
where
1
Rn (z) =
2πi
2
f (τ )
τ −z
z − z0
τ − z0
n+1
dτ .
Passing to the limit by n → ∞ it results Rn (z) → 0. Hence
1
2πi
2
∞
f (τ )
(z − z 0 )k ck ,
dτ =
τ −z
k=0
where
ck =
1 (k)
f (z 0 ).
k!
Let us consider the last term in Eq. (1.5.1)
1
−
2πi
1
f (τ )
dτ .
τ −z
Consider, separately, the ratio −1/(τ − z):
−
1
1
1
1
1
=
=
=
,
τ −z
z−τ
z − z 0 − (τ − z 0 )
z − z0 1 − q
where
q=
τ − z0
R1
|τ − z 0 |
=
⇒ |q| =
< 1.
z − z0
|z − z 0 |
z − z0
n τ − z0
1
1
=
−
τ −z
z − z 0 k=1 z − z 0
Then
=
k−1
+
τ − z0
z − z0
n
τ − z0
(τ − z 0 )k−1
1
+
k
(z − z 0 )
z − z0 z − z0
k=1
n
=
n
.
Therefore
1
−
2πi
1
1
1
f (τ )
dτ =
τ −z
2πi (z − z 0 )k
k=1
n
where
Rn (z) =
1
2πi
1
f (τ )
z−τ
f (τ )(τ − z 0 )k−1 dz + Rn (z),
1
τ − z0
z − z0
n
dτ .
36
1 Complex Functions
Passing to the limit by n → ∞ it results Rn (z) → 0. Hence
−
1
2πi
1
∞
f (τ )
1
1
dτ =
k
τ −z
(z − z 0 ) 2πi
k=1
So, we write
−
1
2πi
f (τ )(τ − z 0 )k−1 dz.
1
−∞
f (τ )
dτ =
(z − z 0 )k ck ,
τ −z
k=−1
1
where
ck =
1
2πi
1
f (τ )
dτ .
(τ − z 0 )k+1
The proof of the theorem is closed.
Applications 1. Let us find the Laurent’s series of the function e1/z around the point
z 0 = 0. As we know
ex = 1 +
x
x2
xn
+
+ ··· +
+ ···
1!
2!
n!
such that, by substituting x by 1/z we find
1
ez = 1 +
1 1
11
1 1
+
+ ··· +
+ ···
2
1! z
2! z
n! z n
This series has an infinite number of terms in its main part and only one term in its
regular part.
2. Consider the function
z
,
f (z) =
(z − 2)(z + 1)3
and intend to solve the following two problems:
(i) The Laurent’s series of the function around the point z 0 = −1.
(ii) The Laurent’s series of the function in the corona |z + 1| > 3.
(i). We can write the function in the form
f (z) =
1
g(z),
(z + 1)3
1.5 Complex Series
37
where
g(z) =
z
z−2+2
2
2
2
=
=1+
=1+
= 1 + z+1
=
z−2
z−2
z−2
z+1−3
3( 3 − 1)
2
1
z+1
2
z+1
=1−
+
=
1
−
1
+
z+1
31− 3
3
3
3
2
+
z+1
3
n
+ ··· .
Therefore
f (z) =
=
2
z + 1 (z + 1)2
(z + 1)n
1
1
−
1+
+
+ ··· +
+ ···
3
2
(z + 1)
3
3
3
3n
=
1
2
1
1
2
2 1
2
2
−
− 2
− 3
− 5 (z + 1) − 6 (z + 1)2 − · · ·
3 (z + 1)3
3 (z + 1)2
3 z + 1 34
3
3
It is easy to see that the series has an infinite number of terms in its regular part
and a finite number (three) in its main part.
(ii). We can write the function in the form
f (z) =
1
g(z),
(z + 1)3
where
g(z) =
z−2+2
2
2
z
=
=1+
== 1 +
z−2
z−2
z+1−3
(z + 1)(1 −
2
1
3
2
3
= 1+
+
= 1+
1+
3
z + 1 1 − z+1
z+1
z+1
z+1
2
+
3
z+1
3
)
z+1
n
=
+ ··· .
Therefore
f (z) =
2
2.3
2.33
1
+
+
+
+ ···
(z + 1)3
(z + 1)4
(z + 1)5
(z + 1)6
It is easy to see that the series has an infinite number of terms in its main part and
no one term in its regular part.
Definition 1.5.5 If a complex function has a point that is a singular point and this
is not a pole then it is called an essential singular point.
Theorem 1.5.4 The point z 0 is a pole of order p for the complex function f if and
only if the Laurent’s series of f around z 0 has the form
38
1 Complex Functions
f (z) =
c− p
c− p+1
+
+ · · · + c0 + c1 (z − z 0 ) + c2 (z − z 0 )2 + · · ·
(z − z 0 ) p
(z − z 0 ) p−1
(1.5.3)
where c p = 0.
Proof Sufficiency. If we suppose that the Laurent’s series of f around z 0 has the
form (2.3), then we must prove that z 0 is a pole of order p. Hence, we must find a
holomorphic complex function ϕ, ϕ(z 0 ) = 0 such that
f (z) =
1
ϕ(z).
(z − z 0 ) p
To this, we define the function ϕ by
ϕ(z) = c p + c− p+1 (z − z 0 ) + c− p+2 (z − z 0 )2 + · · · + c0 (z − z 0 ) p + c1 (z − z 0 ) p+1 .
Of course, being a polynomial, ϕ is a holomorphic function. Also, it is easy to
see that ϕ(z 0 ) = c p = 0. By direct calculations we obtain that
1
ϕ(z) = f (z).
(z − z 0 ) p
Necessity. Suppose that z 0 is a pole of order p. Then, there exists a holomorphic
function ϕ such that ϕ(z 0 ) = 0 and
f (z) =
1
ϕ(z).
(z − z 0 ) p
Being a holomorphic function, ϕ can be expanded in a Taylor’s series around z 0 :
ϕ(z) = a0 + a1 (z − z 0 ) + a2 (z − z 0 )2 + · · · + an (z − z 0 )n + · · · ,
where ϕ(z 0 ) = a0 = 0. Multiplying both sides of the above equality by
1/(z − z 0 ) p
we obtain
f (z) =
a1
a0
+
+ · · · + a p + a p+1 (z − z 0 ) + a p+2 + (z − z 0 )2 + · · ·
(z − z 0 ) p (z − z 0 ) p−1
and the theorem is concluded.
Remark. By using this theorem we can say that a singular point is a pole for a
function f if the Laurent’s series of f has a finite number of terms in its main part.
Of course, if the Laurent’s series of f , around z 0 , has an infinite number of terms
in its main part, we deduce that z 0 is an essential singular point of f . For instance, in a
1.5 Complex Series
39
previous application, we proved that the Laurent’a series of the function f (z) = e1/z ,
around z 0 = 0, has an infinite number of terms, such that z 0 = 0 is an essential
singular point of f .
Definition 1.5.6 If all singular points of a function f are poles then the function f
is called a meromorphic function.
By using the definition of a meromorphic function, the readers can prove the
properties included in the following proposition.
Proposition 1.5.1 The following assertions are still valid:
(1) A meromorphic function on a bounded domain has only a finite number of poles.
(2) The sum, product and quotient of two meromorphic functions are also meromorphic functions.
The following two theorems prove two results regarding an integer function and
the point z 0 = ∞.
Theorem 1.5.5 An integer function f has the point z 0 = ∞ as an essential singular
point if and only if f is not a polynomial.
Proof We write the Taylor’s series around the origin z 0 = 0, for |z| < R, with a
high enough R:
f (z) = c0 + c1 z + c2 z 2 + · · · + cn z n + · · ·
Define the function ϕ by
ϕ(τ ) = f (
c2
c1
cn
1
= c0 +
+ 2 + ··· + n + ···
τ
τ
τ
τ
Then the point z 0 = ∞ is an essential singular point of the function f if the point
τ = 0 is an essential singular point of the function ϕ and that is possible if and only
if the Laurent’s series of ϕ has in its main part an infinite number of terms, i.e. f is
not a polynomial and the theorem is concluded.
Theorem 1.5.6 If an integer function f has the point z 0 = ∞ as an ordinary point,
then the function f is a constant.
Proof If the function f has the point z 0 = ∞ as an ordinary point, then the function
ϕ(τ ) = f (1/τ ) has the point τ0 = 0 as an ordinary point. Then the main part of the
Laurent’s series of ϕ is null. But
ϕ(τ ) = c0 +
c2
c1
c1
+ 2 + ··· +
+ ···
τ
τ
τ
Then
cn = 0, ∀n > 1 ⇒ ϕ(τ ) = c0 ⇒ f (z) = c0 .
The theorem is proved.
40
1 Complex Functions
The following result is called the Fundamental Theorem of Algebra and is due to
D Alembert and Gauss.
Theorem 1.5.7 (D Alembert-Gauss) Any polynomial of order n ≥ 1 has at least
one complex root.
Proof Assuming, by contrary, that P(z) = 0, ∀z ∈ C then the function
f (z) =
1
P(z)
is an integer function which has the point z 0 = ∞ as an ordinary point.
Therefore, f reduces to a constant and, as a consequence, the polynomial P
reduces to a constant, that contradicts the assumption that the order of P is n ≥ 1.
The theorem is concluded.
Definition 1.5.7 Let D be a complex domain, f a holomorphic function f : D → C
and a ∈ C a pole or an essential singular point of f . Consider the disc (a, R)
having as boundary the circle (a, R), with the radius R such that \{a} ⊂ D,
where = ∪ .
Then, the value of the integral
1
f (z)dz
2πi
is called the residue of the function f at the point a and is denoted by
1
res( f, a) =
2πi
f (z)dz.
Theorem 1.5.8 Let D be a complex domain, f a holomorphic function f : D → C
and a ∈ C an isolated singular point of f . To compute the residue of the function f
at a we have the following three possibilities:
(1) If we write the Laurent’s series of f around a
f (z) =
∞
cn (z − a)n ,
n=−∞
then res( f, a) = c1 .
(2) If a is a pole of order p of the function f , then
1
ϕ( p−1) (a), ϕ(z) =
res( f, a) =
( p − 1)!
(z − a) p f (z), if z ∈ D\{a}
lim (z − a) p f (z), if z = a.
z→a
1.5 Complex Series
41
(3) If a is a pole or order 1 (called a simple pole) of the function f that is a quotient
of two functions f = g/ h. With other words, a is an ordinary point for g and h
and, in addition, h(a) = 0. Then, the residue can be calculated by the formula
res( f, a) =
g(a)
.
h (a)
Proof 1. Clearly, the formula proposed at point 1. is valid both in the case when a is
a pole and in the case when a is an essential singular point. We write the Laurent’s
series of f around the point a
f (z) =
∞
cn (z − a)n , cn =
n=−∞
1
2πi
f (z)
dz.
(z − a)n+1
Based on the Definition 1.5.7, we obtain
res( f, a) = c1 =
1
2πi
f (z)dz.
2. If a is a pole of order p, then there esists the holomorphic function ϕ with ϕ(a) = 0
such that
1
1
1
1
f (z) =
ϕ(z)
⇒
f
(z)dz
=
ϕ(z)dz.
p
(z − a)
2πi
2πi
(z − a) p
Now, we use the Cauchy’s formula for derivative
1
2πi
( p − 1)!
ϕ(z)
1
dz =
p
(z − a)
( p − 1)! 2πi
ϕ(z)
1
ϕ( p−1) (a).
dz =
p
(z − a)
( p − 1)!
3. Since g and h are holomorphic functions we can write the Taylor’a series around
a such that
c0 + c1 (z − a) + c2 (z − a)2 + · · ·
f (z) =
d0 + d1 (z − a) + d2 (z − a)2 + · · ·
Because a is a pole for f we have h(a) = 0 and then d0 = 0. Thus
f (z) =
c0 + c1 (z − a) + c2 (z − a)2 + · · ·
⇒
d1 (z − a) + d2 (z − a)2 + · · ·
⇒ lim (z − a) f (z) = lim
z→a
z→a
c0
c0 + c1 (z − a) + c2 (z − a)2 + · · ·
g(a)
=
.
=
2
d1 + d2 (z − a) + d3 (z − a) + · · ·
d1
h (a)
The proof of the theorem is closed.
42
1 Complex Functions
Applications 1. Let us compute the residues at points a = 2 and a = −1 for the
function
z
.
f (z) =
(z − 2)(z + 1)3
Since a = 2 is a simple pole, we have
res( f, 2) = lim (z − 2) f (z) = lim
z→2
z→2
z
2
=
.
(z + 1)3
27
Since a = −1 is a pole of order three, we have
1 1
(z − 1)3 f (z) = lim
z→−1 2!
z→−1 2!
res( f, −1) = lim
z
z−2
=−
2
27
2. Let us compute the residue at point a = 0 for the function
1
f (z) = z k e z
Since a = 0 is an essential singular point for the given function, we must write
its Laurent’s series as follows:
1
ez = 1 +
1 1
1
11
1 1
1
+
+ ··· +
+
+ ···
1! z
2! z 2
k! z k
(k + 1)! z k+1
f (z) = z k +
1
1
1 k−1
1
1
z
+ ···
+ z k−2 + · · · + +
1!
2!
k! (k + 1)! z
Then for the residue we obtain
res( f, 0) = c1 =
1
.
(k + 1)!
The following result, called The Theorem of Residues, is a fundamental result in
the theory of complex functions.
Theorem 1.5.9 Let be a bounded domain with boundary which is a simple,
closed and smooth curve. Consider a function f that has a finite number of singular
points (poles or essential singular points) S = {a1 , a2 , . . . , an }. If the function f is
holomorphic on a domain D such that \S ⊂ D, = ∪ , then
f (z)dz = 2πi
n
k=1
res( f, ak ).
1.5 Complex Series
43
Proof Around the points ak , consider the discs δk having as boundaries the circles
γk such that δk ∪ γk = δ k are disjunctive. Since f is a holomorphic function on
\
n
!
δk
k=1
we have
f (z)dz +
n k=1
f (z)dz = 0 ⇒
But, by definition
1
res( f, ak ) =
2πi
Therefore
f (z)dz =
f (z)dz = 2πi
f (z)dz.
k=1 γ
γk−
n k
f (z)dz.
n
res( f, ak )
k=1
and the theorem is concluded.
It is possible that a function f has a large number of singular points and then
we must compute a large number of residues. To avoid this trouble we introduce the
residue of f at z 0 = ∞.
Definition 1.5.8 Let f be a monogeneous function in all points outside to the disc
(0, R0 ) such that the point z 0 = ∞ is an ordinary point of f , or an isolated pole
or essential singular point. The number denoted by r es( f, ∞) and defined by
1
res( f, ∞) = −
2πi
f (z)dz
is called the residue of f to infinity. Here is a circle having the origin as center
and the radius R such that R > R0 .
If we write the Laurent’s series of f in the corona R0 < |z| < R1 , where R1 is
sufficient large
∞
f (z)
1
n
f (z) =
cn z , cn =
dz,
2πi
z n+1
−∞
then the residue of f to infinity is res( f, ∞) = −c1 .
44
1 Complex Functions
Theorem 1.5.10 Let E be a complex set and f : E → C that has a finite number
of singular points. Then the sum of all residues of f is zero:
res( f, ∞) +
n
res( f, ak ) = 0
k=1
where ak are the singular points of f .
Proof Consider the disc having the origin as center and the radius R0 , sufficient
large such that includes all the singular points of f . If the circle is the boundary
of , by using the residues theorem, we have
1
2πi
But
−
f (z)dz =
n
res( f, ak ).
k=1
1
2πi
f (z)dz = res( f, ∞).
Therefore
−res( f, ∞) =
n
res( f, ak ) ⇒ res( f, ∞) +
k=1
n
res( f, ak ) = 0.
k=1
The theorem is concluded.
In the following we will use some procedure to compute some real improper integral
with the aid of the residues.
Firstly, we prove two auxiliary results, included in the following propositions due
to Jordan.
Proposition 1.5.2 (Jordan) Let AB be an arc of the circle |z| = R such that α ≤
ar gz ≤ β. If
lim z f (z) = k, k = constant,
|z|→∞
then
f (z)dz = i(β − α)k.
lim
|z|→∞
AB
Proof We can write z f (z) = k + ϕ(z), where ϕ has the property that ∀ε > 0 we
have |ϕ(z)| < ε, for |z| → ∞, i.e. ϕ(z) → 0, for |z| → ∞. If we write
f (z) =
ϕ(z)
k
+
,
z
z
1.5 Complex Series
45
then
f (z)dz =
AB
k
dz +
z
AB
ϕ(z)
dz.
z
AB
By using the polar coordinates, we obtain
β
k
i Reiθ dθ +
Reiθ
f (z)dz =
α
AB
β
α
ϕ(Reiθ )
i Reiθ dθ =
Reiθ
β
ϕ(Reiθ )dθ.
= ik(β − α) + i
α
Therefore
β
β
iθ
f (z)dz − ik(β − α) = ϕ(Re )dθ ≤ ϕ(Reiθ )dθ < ε(β − α),
α
AB
α
and the proposition is concluded.
Proposition 1.5.3 (Jordan) Let AB be an arc of the circle |z − a| = r such that
α ≤ ar gz ≤ β. If
lim (z − a) f (z) = k, k = constant,
|z|→a
then
f (z)dz = i(β − α)k.
lim
|z|→a
AB
Proof We can write z f (z) = k + ϕ(z), where ϕ has the property that ∀ε > 0 we
have |ϕ(z)| < ε, for |z| → a, i.e. ϕ(z) → 0, for |z| → a. If we write
f (z) =
then
f (z)dz =
AB
AB
ϕ(z)
k
+
,
z−a
z−a
k
dz +
z−a
AB
ϕ(z)
dz.
z−a
By using the polar coordinates, z = a + r eiθ , we obtain
46
1 Complex Functions
β
f (z)dz =
α
AB
k
ir eiθ dθ +
r eiθ
β
α
ϕ(r + aeiθ ) iθ
ir e dθ =
r eiθ
β
ϕ(a + r eiθ )dθ.
= ik(β − α) + i
α
Therefore
β
β
iθ
f (z)dz − ik(β − α) = ϕ(a + r e )dθ ≤ ϕ(a + r eiθ )dθ < ε(β − α).
α
AB
α
The proposition is concluded.
In the last part of this chapter we indicate several real improper integrals which
can be calculate by using the previous two propositions and the Theorem of Residues.
I. Let us consider an integral of the form
α+2π
R(cos θ, sin θ)dθ.
α
By using the substitution
z = eiθ , θ ∈ [α, α + 2π]
we deduce that z lies on the circle having the origin as the center. Also,
cos θ =
1
1
1
dz
1
z+
, sin θ =
z−
, dz = ieiθ dθ ⇒ dθ = .
2
z
2i
z
iz
The given integral becomes
1
1
1
1
1
R
z+
,
z+
iz
2
z
2i
z
where the function R1 is
R1 (z) =
dz = 2πi
1
R(z)
iz
and ak are the singular points of this function.
k
res(R1 , ak ),
1.5 Complex Series
47
Application. Let us compute the integral
π
I1 =
−π
1 + 2 cos θ
dθ.
5 + 4 sin θ
With the above procedure, we have
I1 =
γ
=
γ
1 1 + z + 1/z
=
i z 5 + (z − 1/z)2/i
γ
z2 + z + 1 i z
dz =
i z 2 (5i z + 2z 2 − 2)
z2 + z + 1
dz = 2πi res( f, 0) + res
z(2z 2 + 5i z − 2)
f, −
i
2
.
II. Let us consider an integral of the form
∞
I2 =
R(x)dx, R(x) =
−∞
P(x)
,
Q(x)
where P(x) and Q(x) are polynomials such that Q(x) = 0, ∀x ∈ R and
1+degree(P) <degree(Q). In order to find the value of I2 we write
R
I2 = lim
R→∞
−R
R(x)dx.
We use the domain bounded by the curve determined by the superior semi-circle
having the origin as the center and the radius equal to R together with the segment
[−R, R] and integrate on this curve the function f (z) = R(z):
R
R(z)dz +
−R
R(x)dx = 2πi
res( f, ak ),
(1.5.4)
k
where ak are the singular points of f which lie in the superior half plane y =
I m(z) > 0.
By using the hypotheses and the first Jordan’s results, we obtain
z f (z) =
z P(z)
z P(z)
⇒ lim z f (z) = lim
= 0.
R→∞
R→∞ Q(z)
Q(z)
48
1 Complex Functions
Passing to the limit in Eq. (1.5.3) for R → ∞ we obtain
Eq. (1.5.3) reduces to
∞
I2 =
R(x)dx = 2πi
R(z)dz → 0 such that
res( f, ak ).
k
−∞
Application. Let us compute the integral
∞
I2 =
−∞
x2
dx.
1 + x4
By using the above result, we deduce that
∞
−∞
x2
dx = 2πi [res( f, z 1 ) + res( f, z 2 )] ,
1 + x4
where z 1 and z 2 are the roots of the equation
z4 + 1 = 0
having I m(z) > 0, i.e.
z 1 = eiπ/4 , z 2 = ei(π/4+π/2) .
III. Let us consider an integral of the form
∞
I3 =
x α R(x)dx, α ∈ (−1) ∪ (0, 1), R(x) =
P(x)
,
Q(x)
0
where P(x) and Q(x) are polynomials such that Q(x) = 0, ∀x ∈ [0, ∞) and
1 + α+degree(P) <degree(Q).
In order to find the value of I3 we write
R
I2 =
lim
r →0R→∞
x α R(x)dx.
r
We use the corona bounded by the circle having the origin as the center and the
radius equal to R and the circle γ having the origin as the center and the radius equal
to r . We make a cut in this corona along the x-axis and integrate on this domain the
function
1.5 Complex Series
49
f (z) = z α R(z) :
R
α
r
α
z R(z)dz + e
x R(x)dx+
r
R
+
2πiα
α
z R(z)dz = 2πi
x α R(x)dx +
res( f, ak ),
(1.5.5)
k
γ
where ak are the singular points of f .
By using the hypotheses and the Jordan’s results, we obtain
lim z f (z) = lim
R→∞
R→∞
lim z f (z) = lim
r →0
r →0
z 1+α P(z)
= 0 ⇒ lim
R→∞
Q(z)
z 1+α P(z)
= 0 ⇒ lim
r →0
Q(z)
z α R(z)dz = 0,
z α R(z)dz = 0.
γ
Taking into account these results, by passing to the limit in Eq. (1.5.4) for R → ∞
and r → 0, Eq. (1.5.4) reduces to
1 − e2πiα
∞
x α R(x)dx = 2πi
res( f, ak ).
k
0
Application. Let us compute the integral
∞ √
I3 =
0
x
dx.
1 + x3
Here α = 1/2 and, by using the above result, we deduce that
I3 =
2πi
[res( f, z 1 ) + res( f, z 2 ) + res( f, z 3 )] ,
1 − eπi
where z 1 , z 2 and z 3 are the roots of the equation z 3 + 1 = 0.
IV. Let us consider an integral of the form
∞
I4 =
0
x α R(x) ln(x)dx, α ∈ (−1) ∪ (0, 1), R(x) =
P(x)
,
Q(x)
50
1 Complex Functions
where P(x) and Q(x) are polynomials such that Q(x) = 0, ∀x ∈ [0, ∞) and
1 + α+degree(P) <degree(Q).
In order to find the value of I4 we write
R
I4 =
lim
r →0R→∞
x α R(x) ln(x)dx.
r
We use the corona bounded by the circle having the origin as the center and the
radius equal to R and the circle γ having the origin as the center and the radius equal
to r . We make a cut in this corona along the x-axis and integrate on this domain the
function f (z) = z α R(z) ln(z):
R
α
z R(z) ln(z)dz + e
x R(x) ln(x)dx+
r
r
α
2πiα
x α R(x)[ln(x) + 2πi]dx +
R
α
+
z R(z) ln(z)dz = 2πi
res( f, ak ),
(1.5.6)
k
γ
where ak are the singular points of f .
By using the hypotheses and the Jordan’s results, we obtain
lim z f (z) = lim
R→∞
R→∞
lim z f (z) = lim
r →0
r →0
z 1+α P(z)
ln(z) = 0 ⇒ lim
R→∞
Q(z)
z 1+α P(z)
ln(z) = 0 ⇒ lim
r →0
Q(z)
z α R(z) ln(z)dz = 0,
z α R(z) ln(z)dz = 0.
γ
Taking into account these results, by passing to the limit in Eq. (1.5.5) for R → ∞
and r → 0, Eq. (1.5.5) reduces to
1 − e2πiα
∞
∞
α
x R(x) ln(x)dx − 2πi
0
x α R(x)dx = 2πi
0
where f (z) = z α R(z) ln(z).
Application. Let us compute the integral
∞
I4 =
√
0
ln x
dx, a > 0.
x(x 2 + a 2 )
k
res( f, ak ),
1.5 Complex Series
51
Here α = −1/2 and f is the function
f (z) = √
ln z
.
z(z 2 + a 2 )
By using the above theoretical result, we deduce that
1 − e−πi
∞
√
0
ln x
dx − 2πie−πi
x(x 2 + a 2 )
∞
√
0
1
dx =
x(x 2 + a 2 )
= 2πi [res( f, z 1 ) + res( f, z 2 ) + res( f, z 3 )] .
The integral
∞
J=
√
1
x(x 2
0
+ a2)
dx
can be computed by using the procedure of I3 , by using the function
g(z) = √
1
z(z 2 + a 2 )
such that we obtain
J=
2πi
[res( f, z 1 ) + res( f, z 2 ) + res( f, z 3 )] .
1 − e−πi
Chapter 2
Special Functions
2.1 Euler’s Functions
Euler’s function of first species Definition 2.1.1 Consider the semi-plane 0 = {z ∈ C, z = x + i y : x > 0}.
The complex function : 0 → C defined by
∞
(z) =
t z−1 e−t dt,
0
is called the Euler’s function of first species.
Remark. Since the Euler’s function of first species is defined as an improper integral,
we first must prove that it is well defined. This result and the main properties are
included in the following theorem.
Theorem 2.1.1 The function satisfies the following properties:
(1) is well defined, i.e. ∞
t z−1 e−t dt < ∞;
0
(2) is a holomorphic function on 0 ;
(3) (z +1) = z(z), ∀z ∈ 0 . As a consequence, we have (n +1) = n!, ∀n ∈ N .
Proof (1) We use the well known formula
u v = ev ln u .
Therefore
t z−1 = e(z−1) ln t = e(x−1) ln t+i y ln t =
© Springer International Publishing AG, part of Springer Nature 2018
M. Marin and A. Öchsner, Complements of Higher Mathematics,
https://doi.org/10.1007/978-3-319-74684-5_2
53
54
2 Special Functions
= e(x−1) ln t [cos(y ln t) + i sin(y ln t)] .
Then we obtain
z−1 t = e(x−1) ln t = t x−1 .
∞
∞
t z−1 e−t dt ≤ t z−1 e−t dt =
Hence
0
∞
=
t
0
1
x−1 −t
e dt =
0
t
∞
x−1 −t
e dt +
0
t x−1 e−t dt = I1 + I2 .
1
For the integral I1 we have:
0 < t < 1 ⇒ 0 > −t > −1 ⇒ e−t < 1 ⇒ t x−1 e−t < t x−1 ⇒
1
⇒ I1 ≤
t
x−1
0
1
t x 1
dt = = < ∞.
x 0
x
We make now some estimations on the integral I2 . As we know
et = 1 +
t2
tm
t
+ + ··· +
+ ... ⇒
1! 2!
m!
⇒ et ≥
tm
m!
⇒ e−t ≤ m .
m!
t
If we choose m > x, it results
∞
I2 ≤
m!
t x−1 m dt
t
∞
= m!
1
t x−m−1 dt =
1
∞
t x−m m!
< ∞.
= m!
=
x − m 1
m−x
Finally, we obtain
|(z)| ≤ I1 + I2 ≤
m!
1
+
< ∞.
x
m−x
(2) We write in the usual form of a complex function (z) = u(x, y) + iv(x, y)
and verify the Cauchy–Riemann’s condition:
2.1 Euler’s Functions
55
∂v
∂u
=
,
∂x
∂y
∂v
∂u
=− .
∂y
∂x
By using the Cauchy’s formula for e z we have
∞
(z) =
t x−1 e−t [cos(y ln t) + i sin(y ln t)] dt =
0
∞
=
t
x−1 −t
e
∞
cos(y ln t)dt + i
0
t x−1 e−t sin(y ln t)dt = u(x, y) + iv(x, y).
0
Thus
∂u
=
∂x
∞
t
x−1
ln te
−t
∂v
=
cos(y ln t)dt,
∂y
0
∞
t x−1 ln te−t cos(y ln t)dt,
0
such that
∂u
∂v
=
.
∂x
∂y
In the same manner we can prove the second Cauchy–Riemann’s condition.
(3) Substituting z by z + 1 we obtain
∞
(z + 1) =
t z e−t dt,
0
such that, integrating by parts, it results
(z + 1) =
∞
−t z e−t 0
∞
+
zt
z−1 −t
∞
e dt = z
0
t z−1 e−t dt = z(z),
0
since lim t z /et = 0.
t→∞
In the particular case z = n ∈ N we have
(n + 1) = n(n) = n(n − 1)(n − 1) = · · · = n!(1).
56
2 Special Functions
By using the definition, we have
∞
∞
0 −t
t e dt =
(1) =
0
∞
e−t dt = −e−t 0 = 1
0
and the theorem is concluded.
Euler’s function of second species β
Definition 2.1.2 Let p, q be two complex numbers such that Re( p) > 0 and
Re(q) > 0, i.e. p, q ∈ 0 (see the above definition of 0 ). The function
β : 0 × 0 → C defined by
1
β( p, q) =
t p−1 (1 − t)q−1 dt,
0
is called the Euler’s function of second species.
In the following theorem we prove the main properties of β as well as the connection
between β and .
Theorem 2.1.2 The function satisfies the following properties:
(1) β( p, q) = β(q, p);
(2) pβ( p, q + 1) = qβ( p + 1, q);
(3) β( p, q)( p + q) = ( p)(q).
Proof (1) It is easy to prove this commutative property with the aid of the substitution
1 − t = τ.
(2) By direct calculations we obtain
1
pβ( p, q + 1) = p
1
t
p−1
(1 − t) dt =
q
0
= t (1 −
p
1
t)q 0
p
t (1 − t)q dt =
0
1
+
qt p (1 − t)q−1 dt = qβ( p + 1, q).
0
(3) We start by using the right-hand side term
∞
( p)(q) =
y
0
p−1 −y
e
∞
dy
0
x q−1 e−x dx =
2.1 Euler’s Functions
57
∞∞
=
y p−1 e−y x q−1 e−x dxdy.
0 0
Let us change in the last integral the variables as follows:
x = u 2 ⇒ dx = 2udu. For x = 0 ⇒ u = 0 and for x = ∞ ⇒ u = ∞
y = v 2 ⇒ dx = 2vdv. For y = 0 ⇒ v = 0 and for y = ∞ ⇒ v = ∞.
So, the last integral becomes
∞
( p)(q) =
2 p−1 −u 2
u
e 2udu
0
∞
2 q−1 −v2
v
e 2vdv =
0
∞∞
=4
e−(u
2
+v 2 ) 2 p−1 2q−1
u
v
dudv.
0 0
Now, we use the polar coordinates
D(u, v)
u = cos θ, 0 ≤ θ ≤ π2
= .
⇒
v = sin θ, 0 ≤ < ∞
D(, θ)
Thus, our integral becomes
∞π/2
2
( p)(q) = 4
e− 2 p−1 (cos θ)2 p−1 2q−1 (sin θ)2q−1 dθd = 4 (I1 .I2 ) .
0 0
By using the new variable a defined by a = 2 p ⇒ da = 2d, the integral I1
becomes
∞
1
1
e−a a p+q−1 da = ( p + q).
I1 =
2
2
0
We use now the new variable b defined by b = cos2 θ ⇒ db = −2 sin θ cos θdθ. So,
for θ = 0 ⇒ b = 1 and for θ = π/2 ⇒ b = 0. Thus, the integral I2 becomes
0
I2 =
1
1
=
2
1
0
db
1
b p (1 − b)q
=
cos θ sin θ −2 cos θ sin θ
2
1
b p (1 − b)q
db =
b(1 − b)
2
1
b p (1 − b)q
db =
cos2 θ sin2 θ
0
1
b p−1 (1 − b)q−1 db =
0
1
β( p, q).
2
58
2 Special Functions
Multiplying the integrals I1 and I1 we obtain
1
1
( p)(q) = 4 ( p + q) β( p, q) ⇒
2
2
⇒ β( p, q) =
( p)(q)
.
( p + q)
The theorem is concluded.
Application. Let us compute the integral
b
I =
a
1
dx.
√
(b − x)(x − a)
Introduce a new variable t by
b − x|ab = (b − a)t|01 ⇒ dx = (a − b)dt, x − a = (b − a)(1 − t).
Thus, the integral becomes
0
a−b
dt =
√
t (1 − t)(b − a)
I =
1
1
√
0
1
dt = β
t (1 − t)
1 1
,
2 2
= π.
We now prove a very important result that is known as the complements formula and
that is very useful in many applications.
Theorem 2.1.3 For p ∈
/ Z , the function ( p) satisfies the following very important
relation:
π
( p)(1 − p) =
.
sin pπ
Proof Using the point two of the last theorem, we obtain
( p)(1 − p) = ( p + 1 − p)β( p, 1 − p) = (1)β( p, 1 − p) =
1
=
x
p−1
(1 − x)
−p
1
dx =
0
0
x p−1
dx =
(1 − x) p
1 0
x
1−x
p
We now introduce the variable y by
x 1
y
y
= y|∞
dy.
⇒ dx =
0 ⇒ x =
1−x 0
1+y
(1 + y)2
1
dx.
x
2.1 Euler’s Functions
59
Thus for the product ( p)(1 − p) we find the form
∞
( p)(1 − p) =
y
0
p1
1
+y
dy =
y (1 + y)2
∞
0
y p−1
dy.
1+y
Now, we use the residues theorem to compute this integral. As we already know,
∞
I =
0
2πi
y p−1
z p−1
dy =
⇒
res(
f,
−1),
where
f
(z)
=
1+y
1 − e2 pπi
1+z
⇒ res( f, −1) = (−1) p−1 ⇒ I =
(−1) p−1 2πi
(−1) p 2πi
=
.
1 − e2 pπi
e2 pπi − 1
But, using the formula for u v and the definition of the complex logarithmic function,
we have
(−1) p = e p ln(−1) = e p(ln 1+πi) = e pπi .
Finally, we obtain
( p)(1 − p) =
2πi
e pπi 2πi
2πi
π
=
= pπi
=
.
2
pπi
−
pπi
e
−1
e −e
2i sin pπ
sin pπ
The proof of the theorem is closed.
Application. By using the complements formula, we compute (1/2). Thus
1
z = ⇒ 2
2
√
1
1
=π⇒
= π.
2
2
On the other hand, if we extend the property (1 + n) = n!, we obtain
√
1
1
1
1
1√
π
1
!= 1+
= =
!=
.
π⇒
2
2
2
2
2
2
2
In the final part of this paragraph we indicate a new form for the Euler’s function of
second species.
Theorem 2.1.4 The function β( p, q) can be written in the form
∞
β( p, q) =
0
x p−1
dx =
(1 + x) p+q
∞
0
Proof We use the definition of the function β( p, q)
x q−1
dx.
(1 + x) p+q
60
2 Special Functions
1
β( p, q) =
t p−1 (1 − t)q−1 dt,
0
and introduce a new variable x by
1
x
t 1
⇒ dt =
= x|∞
dx.
0 ⇒t =
1−t 0
1+x
(1 + x)2
Thus the function β( p, q) becomes
∞ β( p, q) =
0
∞
=
0
x
1+x
p−1 1−
x
1+x
q−1
x p−1
1
1
dx =
(1 + x) p−1 (1 + x)q−1 (1 + x)2
∞
0
1
dx =
(1 + x)2
x p−1
dx.
(1 + x) p+q
The theorem is concluded.
Application. As an application of the last theorem, we compute the following integral
∞
I =
0
√
4
x
dx.
(1 + x)2
It is easy to see that we can write the integral in the form
∞
I =
0
x 1/4
dx =
(1 + x)2
∞
0
x 5/4−1
dx = β
(1 + x)5/4+3/4
5 3
,
.
4 4
Now, we use the connection between the functions and β and, then the complements
formula
( 45 )( 43 )
5 3
1
1
β
,
=
=
+1 1−
=
4 4
(2)
4
4
1
= 4
√
1
1
1 π
π 2
1−
=
=
.
4
4
4 sin π/4
4
2.2 Bessel’s Functions
61
2.2 Bessel’s Functions
Consider the differential equation
x 2 y + x y + x 2 − p 2 y = 0,
(2.1)
where the unknown function is y of variable x, y = y(x). Also, p is a complex
parameter. This equation is called the Bessel’s equation.
Definition 2.2.1 By definition, the solutions of Eq. (2.1) are called the Bessel’s functions and are denoted by J p (x) and J− p (x).
We intend to give an explicit form of the solutions of Eq. (2.1), i.e. of the Bessel’s
functions.
Theorem 2.2.1 The functions J p (x) and J− p (x) have the following polynomial form
∞
x
(−1)m
m!(m + p + 1) 2
m=0
J p (x) =
x
2
p
J− p (x) =
x
2
−p
∞
x
(−1)m
m!(m − p + 1) 2
m=0
2m
,
2m
,
where is the Euler’s function of first species.
Proof We are looking for the solution of Eq. (2.1) in the form of an infinite polynomial
∞
y(x) = x r
Ck x k .
(2.2)
k=0
We must find the constant r and the coefficients Ck such that the function y(x) from
Eq. (2.2) verifies Eq. (2.1). By direct calculations, we obtain
∞
y(x) =
∞
Ck x
k+r
⇒ y (x) =
k=0
(k + r )Ck x k+r −1 ⇒
k=0
∞
⇒ y (x) =
(k + r )(k + r − 1)Ck x k+r −2 .
k=0
If we introduce these derivatives in the Bessel’s equation, it follows
∞
∞
(k +r )(k +r −1)Ck x k +
xr
k=0
∞
(k +r )Ck x k +
k=0
∞
Ck x k+2 − p 2
k=0
Ck x k+1 = 0.
k=0
62
2 Special Functions
So, we can write
∞
(k + r )(k + r − 1) + (k + r ) − p 2 Ck x k = −
k=0
∞
Ck x k+2 .
k=0
By identifying the coefficients, it results
k = 0 ⇒ r (r − 1) − p 2 C0 = 0 ⇒ r = ± p
k = 1 ⇒ (r + 1)r + r + 1 − p 2 C1 = 0 ⇒ (r + 1)2 − p 2 C1 = 0 ⇒ C1 = 0
k = 2 ⇒ (r + 2)(r + 1) + r + 2 − p 2 C2 = −C0 ⇒
⇒ (r + 2)2 − p 2 C2 = −C0 ⇒ 1.4(r + 1)C2 = −C0
k = 3 ⇒ (r + 3)(r + 2) + r + 3 − p 2 C3 = −C1 ⇒ C3 = 0 (because C1 = 0).
Thus, we deduce that C2k+1 = 0, ∀k ∈ N and
1.4( p + 1)C2 = −C0
2.4( p + 2)C4 = −C2
3.4( p + 3)C6 = −C4
−−−−−−−−−
m.4( p + m)C2m = −C2m−2 .
By multiplying these relations it follows
C2m =
(−1)m C0
.
m!22m ( p + 1)( p + 2) . . . ( p + m)
The Bessel’s equation is a homogeneous equation and, therefore, its solution is
determined except for a constant. Thus, we are free to choose the coefficient C0
as
1
,
C0 = p
2 ( p + 1)
and, thus, C2m becomes
C2m =
(−1)m
(−1)m
=
.
m!22m+ p ( p + 1)( p + 2) . . . ( p + m)( p + 1)
m!22m+ p (m + p + 1)
In the previous calculations we used the value r = p and the solution becomes
∞
y(x) =
∞
C2m x p+2m =
m=0
m=0
x
(−1)m
+ p + 1) 2
m!22m+ p (m
2m+ p
.
2.2 Bessel’s Functions
63
Therefore, for r = p the Bessel’s function is
J p (x) =
x
2
p
∞
x
(−1)m
m!(m + p + 1) 2
m=0
2m
.
If we take r = − p the Bessel’s function becomes
J− p (x) =
x
2
−p
∞
x
(−1)m
m!(m
−
p
+
1)
2
m=0
2m
.
The theorem is proved.
Application. Let us compute the function J1/2 (x). Using the polynomial form of the
Bessel’s function, we obtain
x
J1/2 (x) =
2
∞
1/2
n=0
x
(−1)n
n!(n + 3/2) 2
2n
.
Using the recurrence relation of the function , it follows
√
3
3
2n + 1
3
π
357
n+
...
.
=
, =
2
222
2
2
2
2
Thus J1/2 (x) becomes
√
π
J1/2 (x) = √
2
∞
n=0
(−1)n
x 2n =
n!2n 3.5.7 . . . (2n + 1)(n + 3/2)
√
2
=√
πx
∞
n=0
(−1)n 2n+1
x
.
(2n + 1)!
So we obtain
J1/2 (x) =
2
sin x.
πx
The Bessel’s equation is of order two. The two above solutions determine the general
solution of the equation if they are linear independent.
In the following theorem we prove that in the case of n ∈
/ N the Bessel’ functions
are linear independent.
Theorem 2.2.2 If the parameter p is not natural n ∈
/ N , then the functions J p (x)
and J− p (x) are linear independent.
64
2 Special Functions
Proof It is a well known the fact that a system of functions is linear independent if
its Wronskian is non zero. In our case, we must prove that
J p (x) J− p (x) = J p (x)J (x) − J (x)J− p (x).
W (x) = W J p (x), J− p (x) = −p
p
J p (x) J− p (x) Let us use the fact the functions J p (x) and J− p (x) satisfy the Bessel’s equation
x 2 J p (x) + x J p (x) + x 2 − p 2 J p (x) = 0,
x 2 J− p (x) + x J− p (x) + x 2 − p 2 J− p (x) = 0.
Multiplying the first relation by J− p (x) and the second by J p (x) and subtracting the
resulting relations, it follows
x 2 J p (x)J− p (x) − J− p (x)J p (x) + J p (x)J− p (x) − J− p (x)J p (x) = 0.
So, we obtain the following differential equation of first order
x W (x) + W (x) = 0 ⇒
dx
C
dW
=−
⇒ W (x) = .
W
x
x
If we prove that C = 0, it results that the Wronskian of the functions J p (x) and
J− p (x) is non zero, such that J p (x) and J− p (x) are linear independent. By using the
polynomial form of the Bessel’s functions, we obtain
x
2
J p (x) =
x
2
J− p (x) =
p
−p
1
p
+ . . . ⇒ J p (x) =
( p + 1)
2
1
−p
+ . . . ⇒ J− p (x) =
(1 − p)
2
x
2
p−1
x
2
1
+ ...
( p + 1)
− p−1
1
+ ...
(1 − p)
Then, the Wronskian becomes
W (x) =
−p
2
x −1
1
−p
+ ··· +
2
( p + 1)(1 − p)
2
=−
x −1
1
+ ··· =
2
( p + 1)(1 − p)
1
2
1
2p
+ ··· = −
+ ...
p( p)(1 − p) x
( p)(1 − p) x
Comparing with the first form of W , we deduce
C =−
2 sin pπ
2
=−
= 0 because p ∈
/ N.
( p)(1 − p)
π
2.2 Bessel’s Functions
65
The theorem is concluded.
Remark. Because the functions J p (x) and J− p (x) are solutions of the Bessel’s
equation and are linear independent, we deduce that the general solution of the
Bessel’s equation is
y(x) = C1 J p (x) + C2 J− p (x), C1 , C2 = constants.
Application 1. Let us solve the equation
1
x y + xy + x −
4
2
2
y = 0.
Since p = 1/2 ∈
/ N , we deduce that this equation has two linear independent
solutions, namely J1/2 (x) and J−1/2 (x). Using the same way as in the case of J1/2 (x),
we obtain that
2
cos x.
J− 21 (x) =
πx
Then, the general solution of our equation is
y(x) = C1 J 21 (x) + C2 J− 21 (x) = C1
2
sin x + C2
πx
2
cos x,
πx
where C1 and C2 are constants.
It is easy to prove, by direct calculations, that in the case n ∈ N the Bessel’s
functions J p (x) and J− p (x) are not linear independent. Namely, we obtain
J p (x) = (−1)n J− p (x).
In this case, we cannot define the general solution of the Bessel’s equation. To solve
this problem, we introduce a new function
N p (x) =
cos pπ J p (x) − J− p (x)
, p∈
/ N,
sin pπ
that is called the Neumann’s function.
It is clear that the functions J p (x) and N p (x) are linear independent since p ∈
/N
and in this case the functions J p (x) and J− p (x) are linear independent.
In the case when p = n ∈ N the Neumann’s functions are defined by the following
limits
cos pπ J p (x) − J− p (x)
.
Nn (x) = lim N p (x) = lim
p→n
p→n
sin pπ
By using the L Hospital’s rule, we obtain
66
2 Special Functions
Nn (x) = lim
−π sin pπ J p (x) + cos pπ J p (x) − J− p (x)
p cos pπ
p→n
=
1 ∂ J p (x)
n ∂ J− p (x)
− (−1)
.
=
π
∂p
∂p
p=n
By direct calculations we obtain that
W (Jn (x), Nn (x)) =
2
= 0,
πx
that is, these functions are linear independent and then the general solution of the
Bessel’s equation is
y(x) = C1 Jn (x) + C2 Nn (x), C1 , C2 = constants.
Remark. Of course, the Neumann’s function is a solution of the Bessel’s equation
because it is a linear combination of two solution of the Bessel’s equation and, the
Bessel’s equation is linear!
Other properties of the Bessel’s functions are contained in the following theorem.
Theorem 2.2.3 The Bessel’s functions satisfy the following properties
(i)
d −p
d p
x J p (x) = x p J p−1 (x),
x J p (x) = −x p J p+1 (x).
dx
dx
(ii)
x J p (x) + p J p (x) = x J p−1 (x), x J p (x) − p J p (x) = −x J p+1 (x).
(iii)
J p−1 (x) − J p+1 (x) = 2J p (x), J p−1 (x) + J p+1 (x) =
2p
J p (x).
x
Proof (i) It is easy to prove these relations by using the polynomial form of the
Bessel’s functions.
(ii) We write (i)1 in detail
px p−1 J p (x) + x p J p (x) = x p J p−1 (x) : x p−1 ⇒
⇒ p J p (x) + x J p (x) = x J p−1 (x)
that is, (ii)1 is proved. If we write (i)2 in detail
2.2 Bessel’s Functions
67
− px − p−1 J p (x) + x − p J p (x) = −x − p J p+1 (x) .x p−1 ⇒
⇒ − p J p (x) + x J p (x) = −x J p+1 (x),
that is, (ii)2 is proved.
(iii) We add (ii)1 to (ii)2 and obtain
2x J p (x) = x J p−1 (x) − J p+1 (x) ⇒
⇒ J p−1 (x) − J p+1 (x) = 2J p (x),
that is (iii)1 is proved. Subtracting (ii)2 from ii)1 it follows
2 p J p (x) = x J p−1 (x) + J p+1 (x) ⇒
⇒ J p−1 (x) + J p+1 (x) =
2p
J p (x),
x
that is (iii)2 is proved and the theorem is concluded.
Application. Let us compute the Bessel’s functions J3/2 (x) and J−3/2 (x).
First, we remember that
J1/2 (x) =
Then
J1/2 (x) =
2
sin x, J−1/2 (x) =
πx
2
sin x
πx
=
2
cos x.
πx
1
2
cos x −
πx
2x
⇒ J1/2 (x) = J−1/2 (x) −
2
sin x ⇒
πx
1
J1/2 (x).
2x
Now, we write (iii)1 for p = 1/2:
J−1/2 (x) − J3/2 (x) = 2J1/2 (x) ⇒
1
⇒ J3/2 (x) =
x
2
sin x −
πx
2
cos x.
πx
In order to obtain J−3/2 (x), we write (iii)2 for p = −1/2:
1
J−3/2 (x) − J1/2 (x) = − J−1/2 (x)
x
from where it results
68
2 Special Functions
J−3/2 (x) = J1/2 (x) −
Thus
J−3/2 (x) =
1
J−1/2 (x).
x
1
2
sin x −
πx
x
2
cos x.
πx
2.3 Orthogonal Polynomials
Consider a sequence of functions { f n }n∈N such that
f n : [a, b] → R, f n ∈ C 0 [a, b], ∀n ∈ N ,
and the function
p = p(x), p : [a, b] → R+ , p ∈ L[a, b],
called the weight function.
Definition 2.3.1 The real number denoted by ( f n , f m ) and defined by
b
( fn , fm ) =
p(x) f n (x) f m (x)dx,
(2.3)
a
is called the scalar product of the functions f n and f m .
It is easy to prove the usual properties of a scalar product.
Proposition 2.3.1 The real scalar product (3.1) has the following properties
(i) ( f n , f m ) = ( f m , f n );
(ii) (λ f n , f m ) = λ ( f m , f n );
(iii) ( f n , f m + f k ) = ( f n , f m ) + ( f n , f k ).
Proof All these properties are obtained based on the respective properties of the
integral. For instance,
b
( fn , fm + fk ) =
p(x) f n (x) [ f m (x) + f k (x)] dx =
a
b
=
b
p(x) f n (x) f m (x)dx +
a
p(x) f n (x) f k (x)dx = ( f n , f m ) + ( f n , f k ) .
a
2.3 Orthogonal Polynomials
69
The readers can easily prove the other properties.
Remark. Based on the above properties of the scalar product, we deduce
(iv) ( f n , λ f m ) = λ ( f m , f n );
(v) ( f n + f m , f k ) = ( f n , f k ) + ( f m , f k ).
Definition 2.3.2 A sequence of functions { f n }n∈N is called orthogonal if
( fn , fm ) =
0, for n = m
cn > 0, for n = m.
If we take f m = f n in the definition of the scalar product, then
b
( fn , fm ) =
p(x) f n2 (x)dx = f n
2
⇒ fn =
( f n , f n ).
a
So, in the definition of a orthogonal sequence, we can take cn = f n 2 .
Definition 2.3.3 A sequence of functions { f n }n∈N is called orthonormal if
( fn , fm ) =
0, for n = m,
1, for n = m.
In other words, an orthonormal sequence is an orthogonal sequence whose every
element has unit norm, i.e. f n = 1, ∀n ∈ N .
The following two propositions establish the connection between an orthogonal
system of functions and a linearly independent system of functions.
Proposition 2.3.2 Any orthogonal system of functions is a linear independent system
of functions.
Proof Consider the following orthogonal system of functions
{ f1 , f2 , . . . , fn } ,
and a linear combination that is null:
α1 f 1 + α2 f 2 + · · · + αn f n = 0.
In order to prove that our system is linear independent, we must prove that αk =
0, k = 1, 2, . . . , n. By multiplying, both members of the above combinations,
scalarly by f k and using the linearity of the scalar product we obtain
( f k , α1 f 1 + α2 f 2 + · · · + αn f n ) = 0 ⇒
⇒ α1 ( f k , f 1 ) + α2 ( f k , f 2 ) + · · · + αk ( f k , f k ) + · · · + αn ( f k , f n ) = 0.
70
2 Special Functions
Since ( f k , f n ) = 0, ∀n = k and ( f k , f k ) > 0, we deduce that αk = 0 and the
proposition is concluded.
Proposition 2.3.3 From every linear independent system of functions one can
extract extract an orthogonal system of functions.
Proof Consider the following linear independent system of fucntions
{ f1 , f2 , . . . , fn } ,
and construct the system {g1 , g2 , . . . , gn } as follows:
(1) define g1 = f 1 ;
(2) define g2 = f 2 + λ1 g1 such that (g2 , g1 ) = 0. So we find
λ1 = −
( f 2 , g1 )
.
(g1 , g1 )
(3) define g3 = f 3 + λ2 g2 + λ1 g1 such that (g3 , g1 ) = 0 and (g3 , g2 ) = 0. So we find
λ1 = −
( f 3 , g1 )
( f3 , f2 )
, λ2 = −
.
(g1 , g1 )
(g2 , g2 )
(4) in the general case we define gn = f n + λ1 g1 + λ2 g2 + · · · + λn−1 gn−1 such that
(gn , g1 ) = 0, (gn , g2 ) = 0,…,(gn , gn−1 ) = 0. So we find
λ1 = −
( f n , g1 )
( f n , g2 )
( f n , gn−1 )
, λ2 = −
, . . . , λn−1 = −
.
(g1 , g1 )
(g2 , g2 )
(gn−1 , gn−1 )
The proposition is proved.
Remark. It is easy to see that this is the Gram–Schmidt orthogonalization procedure.
In the following proposition we indicate, without proof, two properties of a orthogonal system of polynomials.
Proposition 2.3.4 In the case of a orthogonal system of polynomials, we have:
(i) For any orthogonal polynomial all its roots are real, distinct and lying in the
interval of definition.
(ii) For any orthogonal polynomial we have the following recurrence relation
Pn (x) = (An x + Bn ) Pn−1 (x) + Cn Pn−2 (x),
where An , Bn and Cn are constants.
Remark. If in the definition of an orthogonal polynomial we particularize the weight
function and the interval of definition, we obtain different kinds of polynomials, as
follows
2.3 Orthogonal Polynomials
71
(1) [a, b] → [−1, 1], p(x) = 1 ⇒
√ Legendre’s polynomial, Pn (x);
(2) [a, b] → (−1, 1), p(x) = 1/ 1 − x 2 ⇒ Chebyshev’s polynomial, Tn (x);
2
(3) [a, b] → (−∞, ∞), p(x) = e−x ⇒ Hermite’s polynomial, Hn (x);
−x
(4) [a, b] → [0, ∞), p(x) = e ⇒ Laguerre’s polynomial, L n (x).
2.4 Legendre’s Polynomials
First, we remember two well known binomial series
(1+ x)α = 1+
α(α − 1) 2
α
α(α − 1)(α − 2) . . . (α − n + 1) n
x+
x +· · ·+
x +. . .
1!
2!
n!
α(α − 1)(α − 2) . . . (α − n + 1) n
α α(α − 1) 2
x −· · ·+(−1)n
x +. . .
(1−x)α = 1− x+
1!
2!
n!
Consider the function
1
f (x, r ) = √
, |r | < 1,
1 − 2xr + r 2
that is called the generating function of the Legendre’s polynomials.
We expand this function as a power of r series, having the coefficients as functions
of x. By definition, the coefficients of this series are the Legendre’s polynomials. Let
us obtain the form of the Legendre’s polynomials.
Theorem 2.4.1 The expression of the Legendre’s polynomials is
[n/2]
Pn (x) =
k
(−1)k Cn−k
k=0
1.3.5....(2n − 2k − 1) n−2k
x
.
(n − k)!2k
Proof We shall begin with the power of r series of the generating function
√
1
1 − 2xr + r 2
∞
=
Pn (x)r n .
n=0
Denote 2xr − r 2 = u, the above binomial series becomes
−1/2(−1/2 − 1) 2
−1/2
1
u+
u − ··· =
= (1 − u)−1/2 = 1 −
√
1!
2!
1−u
=1+
1.3
1
1.3.5
1.3.5 . . . (2k − 1) k
u + 2 u2 + 3 u3 + · · · +
u + ...
2.1!
2 .2!
2 .3!
2k .k!
(2.4)
72
2 Special Functions
Thus
2
1
1.3 1 2xr − r 2 + 2
2xr − r 2 + · · · +
=1+
√
2
2.1!
2 .2!
1 − 2xr + r
+
k
1.3.5 . . . (2k − 1) 2xr − r 2 + . . .
k
2 .k!
By identifying the coefficients, we obtain, step by step the following polynomials
P0 (x) = 1, P1 (x) = x, P2 (x) =
3 2 1
5
3
x − , P3 (x) = x 3 − x, . . . ,
2
2
2
2
and, in the general case,
[n/2]
Pn (x) =
k
(−1)k Cn−k
k=0
1.3.5....(2n − 2k − 1) n−2k
x
.
(n − k)!2k
The theorem is concluded.
Theorem 2.4.2 The Legendre’s polynomials satisfy the following relation, called
the Olinde-Rodrigues’s relation:
Pn (x) =
n
1 dn 2
x −1 .
n!2n dx n
Proof We shall begin with the equality
n
x2 − 1 =
n
(−1)k Cnk x 2n−2k ,
k=0
from where, by derivative,
n
dn 2
x −1 =
n
dx
[n/2]
(−1)k Cnk (2n − 2k)(2n − 2k − 1) . . . (n − 2k + 1)x n−2k .
k=0
We now make some estimations on the coefficient of x n−2k :
(2n − 2k)!
n!
n!(2n − 2k)! (n − k)!
=
=
k!(n − k)! (n − 2k)!
[(n − k)!]2 k!(n − 2k)!
k
= n!Cn−k
1.3.5 . . . (2n − 2k − 1).2.4 . . . 2(n − k)
1.3.5 . . . (2n − 2k − 1)
k
= n!2n Cn−k
,
2
[(n − k)!]
(n − k)!2k
from where we deduce
2.4 Legendre’s Polynomials
73
n
dn 2
x − 1 = n!2n Pn (x),
n
dx
that is, the Olinde-Rodrigues’s relation and the theorem is concluded.
Theorem 2.4.3 The Legendre’s polynomials satisfy the following recurrence
relation
(n + 1)Pn+1 (x) − (2n + 1)x Pn (x) + n Pn−1 (x) = 0, n = 1, 2, . . .
Proof We shall begin with the equality
√
1
1 − 2xr + r 2
∞
=
Pn (x)r n .
n=0
By derivation with regard to r , it follows
∞
x −r
= 1 − 2xr + r 2
n Pn (x)r n−1 ⇒
√
1 − 2xr + r 2
n=0
∞
⇒ (x − r )
Pn (x)r n = 1 − 2xr + r 2
n=0
∞
n Pn (x)r n−1 .
n=0
In the last equality we identify the coefficient of r n from both sides of the equality:
x Pn (x) − Pn−1 (x) = (n + 1)Pn+1 (x) − 2nx Pn (x) + (n − 1)Pn−1 (x) ⇒
⇒ (n + 1)Pn+1 (x) − (2n + 1)x Pn (x) + n Pn−1 (x) = 0.
The proof of the theorem is closed.
Remark. By using the recurrence relation we can determine step by step the Legendre’s polynomials, starting with P0 , P1 , and so on.
Theorem 2.4.4 The Legendre’s Polynomials satisfy the following differential equation:
2
x − 1 y (x) + 2x y (x) − n(n + 1)y(x) = 0.
Proof We shall begin with the equality
d 2
n
x − 1 = 2nx x 2 − 1 .
x2 − 1
dx
By using the Leibniz’s rule for derivative of superior order, we obtain
74
2 Special Functions
d 2
n
x −1
x −1
dx
2
(n+1)
(n+1)
= 2n x x 2 − 1
.
By direct calculations
dn+2 2 n
dn+1 2 n
d n 2 n
x
x
x −1 =
−1
+2(n+1)x
−1
+n(n+1)
x 2 −1
dx n+2
dx n+1
dx n
n+1
2
n
n
dn 2
d
⇒
= 2n x n+1 x − 1 + (n + 1) n x − 1
dx
dx
dn+2 2
n
n
n
dn+1 2
dn 2
⇒ x2 − 1
x
x
x − 1 = 0.
−
1
+
2x
−
1
−
n(n
+
1)
n+2
n+1
n
dx
dx
dx
We can write the relation in the form
n
n
2 d2
d 2 n
d
d n 2 n
d 2 n
+2x
−n(n+1)
x −1
x
x
x −1 = 0.
−1
−1
dx 2 dx n
dx dx n
dx n
We multiply the both sides of the last equality by 1/2n n!:
d2
x −1
dx 2
2
1 d n 2 n
d
1 d n 2 n
x −1 +2x
x −1 −
2n n! dx n
dx 2n n! dx n
−n(n + 1)
n
1 dn 2
x − 1 = 0,
2n n! dx n
such that, by using the Olinde-Rodrigues’s relation, we obtain
x 2 − 1 Pn (x) + 2x Pn (x) − n(n + 1)Pn (x) = 0.
The theorem is concluded.
Theorem 2.4.5 The Legendre’s polynomials satisfy the following orthogonality
relation:
1
0, if n = m
Pn (x)Pm (x)dx =
2
, if n = m.
2n+1
−1
Proof We use the fact that the Legendre’s polynomials Pn and Pm satisfy its differential equations:
x 2 − 1 Pn (x) + 2x Pn (x) − n(n + 1)Pn (x) = 0,
x 2 − 1 Pm (x) + 2x Pm (x) − m(m + 1)Pm (x) = 0.
2.4 Legendre’s Polynomials
75
Now, we multiply the first equation by Pm and the second by Pn and then subtracting
the resulting relations, it follows
x 2 − 1 Pn (x)Pm (x) − Pm (x)Pn (x) + 2x Pn (x)Pm (x) − Pm (x)Pn (x) −
− n 2 + n − m 2 − m Pn (x)Pm (x) = 0,
and this relation can be, equivalently, written
x 2 − 1 Pn (x)Pm (x) − Pm (x)Pn (x) + x 2 − 1 Pn (x)Pm (x) − Pm (x)Pn (x) =
= (n − m)(n + m + 1)Pn (x)Pm (x),
or,
x 2 − 1 Pn (x)Pm (x) − Pm (x)Pn (x) = (n − m)(n + m + 1)Pn (x)Pm (x).
Now, we integrate this equality
1
x 2 − 1 Pn (x)Pm (x) − Pm (x)Pn (x) dx =
−1
1
= (n − m)(n + m + 1)
Pn (x)Pm (x)dx.
−1
The left-hand side term of the last equality is null such that we obtain
1
(n − m)(n + m + 1)
Pn (x)Pm (x)dx = 0.
−1
If we suppose that n = m it results
1
Pn (x)Pm (x)dx = 0,
−1
that is, the first relation of the theorem.
In the case n = m we use the recurrence relation of the Legendre’s polynomials,
written for Pn and Pn−1 and multiply the first relation by Pn−1 and the second by Pn :
(n + 1)Pn+1 (x) − (2n + 1)x Pn (x) + n Pn−1 (x) = 0| .Pn−1 ,
n Pn (x) − (2n − 1)x Pn−1 (x) + (n − 1)Pn−2 (x) = 0| .Pn ⇒
76
2 Special Functions
2
⇒ n Pn−1
(x) = (2n + 1)x Pn (x)Pn−1 (x) − (n + 1)Pn+1 (x)Pn−1 (x),
n Pn2 (x) = (2n − 1)x Pn (x)Pn−1 (x) − (n − 1)Pn (x)Pn−2 (x).
We shall integrate these relations. By using the first part of the proof, it follows
1
Pn+1 (x)Pn−1 (x)dx = 0,
−1
1
Pn (x)Pn−2 (x)dx = 0,
−1
such that we obtain
1
1
2
Pn−1
(x)dx
n
= (2n + 1)
−1
−1
1
1
Pn2 (x)dx
n
x Pn (x)Pn−1 (x)dx,
= (2n − 1)
−1
x Pn (x)Pn−1 (x)dx, ⇒
−1
1
⇒
Pn2 (x)dx
−1
2n − 1
=
2n + 1
If we use the notation
1
2
Pn−1
(x)dx.
−1
1
In =
Pn2 (x)dx,
−1
then
In =
2n − 1
In−1 .
2n + 1
We write this relation for n = 1, 2, 3, . . ., multiply the resulting relations, such that
after the simplification, it follows
In =
3
I1 .
2n + 1
But, for n = 1 the Legendre’s polynomial is P1 (x) = x such that
2.4 Legendre’s Polynomials
77
1
I1 =
x 2 dx =
−1
and then
In =
Finally,
2
,
3
2
3 2
=
.
2n + 1 3
2n + 1
1
Pn2 (x)dx =
−1
2
,
2n + 1
and the theorem is proved.
2.5 Chebyshev’s Polynomials
Definition 2.5.1 The functions of the form
Tn (x) = cos(n arccos x),
are called the Chebyshev’s polynomials.
Theorem 2.5.1 The Chebyshev’s polynomials have the following expression
[n/2]
Tn (x) =
k
(−1)k Cn2k x n−2k 1 − x 2 .
k=0
Proof We shall begin with the Moivre’s formula
cos nθ + i sin nθ = (cos θ + i sin θ)n = Cn0 cosn θ + iCn1 cosn−1 θ sin θ−
−Cn2 cosn−2 θ sin2 θ − iCn3 cosn−3 θ sin3 θ + Cn4 cosn−4 θ sin4 θ + . . .
Now we equalize the real parts of both sides of this relation:
cos nθ = Cn0 cosn θ − Cn2 cosn−2 θ sin2 θ + Cn4 cosn−4 θ sin4 θ + . . . ⇒
n
⇒ cos nθ =
(−1)k Cn2k cosn−2k θ sin2k θ.
k=0
With the aid of the substitution
78
2 Special Functions
x = cos θ ⇒ θ = arccos x
we obtain
[n/2]
cos(n arccos x) =
k
(−1)k Cn2k x n−2k 1 − x 2 ,
k=0
such that
[n/2]
Tn (x) =
k
(−1)k Cn2k x n−2k 1 − x 2 ,
k=0
such that the proof is closed.
Theorem 2.5.2 The Chebyshev’s polynomials have the function
f (x) =
1 − rx
,
1 − 2xr + r 2
as generating function.
Proof We expand this function as a power series of r and prove that the coefficients
of this series are the Chebyshev’s polynomials, that is
1 − rx
=
1 − 2xr + r 2
∞
Tn (x)r n .
n=0
If we use the substitution x = cos θ it follows
1 − rx
1 − r cos θ
=
=
2
1 − 2xr + r
1 − 2r cos θ + cos2 θ
=
1 − r cos θ
1 − r cos θ
=
=
(r − cos θ − i sin θ)(r − cos θ + i sin θ)
(r − eiθ )(r − e−iθ )
1
1 − r cos θ
=
=
(1 − r eiθ )(1 − r e−iθ )
2
=
1
1
+
1 − r eiθ
1 − r e−iθ
=
1
1+r eiθ +r 2 e2iθ +· · ·+r n eniθ +· · ·+1+r e−iθ +r 2 e−2iθ +· · ·+r n e−niθ +· · · =
2
= 1+r
e2iθ + e−2iθ
eniθ + e−niθ
eiθ + e−iθ
+ r2
+ · · · + rn
+ ··· =
2
2
2
∞
= 1 + r cos θ + r 2 cos 2θ + · · · + r n cos nθ + · · · =
r n cos nθ.
n=0
2.5 Chebyshev’s Polynomials
79
If we substitute θ = arccos x, it follows that the coefficients of the series are
cos nθ = cos(n arccos x) = Tn (x),
and the theorem is proved.
Remark. The Chebyshev’s polynomials do not satisfy an Olinde-Rodrigues’s relation.
Theorem 2.5.3 The Chebyshev’s polynomials satisfy the following recurrence relation
Tn+1 (x) = 2x Tn (x) − Tn−1 (x).
Proof We shall begin with the formula
cos(n + 1)θ + cos(n − 1)θ = 2 cos θ cos nθ.
If we substitute here θ by arccos x it results
cos((n + 1) arccos x) + cos((n − 1) arccos x) = 2x cos(n arccos x),
that is,
Tn+1 (x) = 2x Tn (x) − Tn−1 (x).
The proof of the theorem is closed.
Remark. By using the recurrence relation, we can determine step by step the Chebyshev’s polynomials, starting with P0 , P1 , and so on.
Theorem 2.5.4 The Chebyshev’s polynomials satisfy the following differential equation:
1 − x 2 y (x) − x y (x) + n 2 y(x) = 0.
Proof We derive with regard to x the relation of definition for the Chebyshev’s
polynomials:
Tn (x) = cos(n arccos x)|x ⇒
⇒ Tn (x) = −n sin(n arccos x) √
−1
1 − x2
.
This relation can be written in the form
1 − x 2 Tn (x) = n sin(n arccos x).
We derive here both sides with regard to x
−x
−1
Tn (x) + 1 − x 2 Tn (x) = n 2 cos(n arccos x) √
.
√
2
1−x
1 − x2
80
2 Special Functions
√
Multiplying both sides by − 1 − x 2 , it follows
x Tn (x) − 1 − x 2 Tn (x) = n 2 Tn (x) ⇒
⇒ 1 − x 2 Tn (x) − x Tn (x) + n 2 Tn (x) = 0.
The theorem is concluded.
Theorem 2.5.5 The Chebyshev’s polynomials satisfy the following orthogonality
relation:
⎧
1
⎨ 0, if n = m
1
Tn (x)Tm (x)dx = π2 , if n = m = 0
√
(Tn , Tm ) =
⎩
1 − x2
π, if n = m = 0
−1
because the weight function and the interval of definition for the Chebyshev’s polynomials are
1
, [a, b] → (−1, 1).
p(x) = √
1 − x2
Proof We substitute the expressions of Tn and Tm in the definition of the scalar
product
1
1
Tn (x)Tm (x)dx =
√
(Tn , Tm ) =
1 − x2
−1
1
=
√
−1
1
1 − x2
cos(n arccos x) cos(m arccos x)dx.
Here, we substitute x by cos θ
x = cos θ ⇒ dx = − sin θdθ; x ∈ (−1, 1) ⇒ θ ∈ (π, 0).
Then, we obtain
0
(Tn , Tm ) =
π
π
=
1
cos nθ cos mθ(−dθ) sin θ =
sin θ
1
cos nθ cos mθdθ =
2
0
(i) If n = m then
π
[cos(n + m)θ + cos(n − m)θdθ.
0
2.5 Chebyshev’s Polynomials
1
2
(Tn , Tm ) =
81
sin(n + m)θ π
sin(n − m)θ π
= 0.
+
n + m 0
n − m 0
(ii) If n = m = 0 then
(Tn , Tm ) =
1
2
π
(cos 2nθ + 1)dθ =
1
2
π
π
0
sin 2nθ π
π
π
.
+
θ|
=
0
2n 0
2
(iii) If n = m = 0 then
1
(Tn , Tm ) =
2
2dθ =
0
0
π
dθ = θ = π.
0
So, the theorem is proved.
2.6 Hermite’s Polynomials
Definition 2.6.1 The functions defined by
Hn (x) = e x
2
dn
2
e−x ,
dx n
(2.5)
are called the Hermite’s polynomials.
Remark. As it is easy to see, the Hermite’s polynomials are defined direct by an
Olinde-Rodrigues’s relation.
The first Hermite’s polynomials are
H0 (x) = e x
H1 (x) = e x
H2 (x) = e x
2
ex
2
= ex
2
2
e−x
2
e−x
2
−2xe−x
2
2
(0)
= ex
= e x e−x = 1,
2
2
2
−2xe−x
2
= −2xe x
2
= −2x,
e−x − 2x 2 e−x
2
2
= 4x 2 −2, . . .
The generating function of the Hermite’s polynomials is the function
h(r, x) = e−(r
2
+2xr )
,
that is, if we expand as a power series of r this function, the coefficients of the series
are the Hermite’s polynomials:
82
2 Special Functions
e−(r
2
+2xr )
∞
=
Hn (x)
n=0
1 n
r
n!
Theorem 2.6.1 The Hermite’s polynomials satisfy the following differential equation:
y (x) − 2x y (x) + 2ny(x) = 0.
Proof We derive with regard to x both sides of the relation (6.1)
n
n
n+1
2
x2 d
−x 2
x2 d
−x 2
x2 d
e
e
+
e
e−x ,
=
2xe
Hn (x) = e
n
n
n+1
dx
dx
dx
Hn (x) = 2e x
2
n
n+1
dn
2
2 d
2
2 d
2
e−x + 2x 2 e x
e−x + xe x
e−x +
n
n
dx
dx
dx n+1
+2xe x
2
n+2
d n+1
2
−x 2
x2 d
e
+
e
e−x .
n+1
n+2
dx
dx
Thus
Hn (x) − 2x Hn (x) + 2n Hn (x) = 2e x
+4x 2 e x
2
−2xe x
2
=e
But
x2
2
dn
2
e−x +
dx n
n+1
n
dn
2
2 d
2
2 d
2
e−x + 4xe x
e−x − 4x 2 e x
e−x −
n
n+1
dx
dx
dx n
n
n+1
d n+1
2
−x 2
x2 d
−x 2
x2 d
e
+
2ne
e
+
e
e−x =
n+1
n
n+1
dx
dx
dx
dn
d n+1
d n+2
dn
2
2
2
2
2 n e−x + 2x n+1 e−x + n+2 e−x + 2n n e−x
dx
dx
dx
dx
d n+1
d n+2
2
2
e−x = n+1 −2xe−x =
n+2
dx
dx
dn
d n+1
2
2
= −2 2x n+1 e−x + (n + 1) n e−x
,
dx
dx
such that, we obtain
n
d
d n+1
2
2
Hn (x) − 2x Hn (x) + 2n Hn (x) = e 2 n e−x + 2x n+1 e−x −
dx
dx
x2
.
2.6 Hermite’s Polynomials
83
dn
dn
d n+1
2
2
2
−2x n+1 e−x − 2(n + 1) n e−x + 2n n e−x
dx
dx
dx
= 0.
In conclusion, we obtain the equation
Hn (x) − 2x Hn (x) + 2n Hn (x) = 0,
which is the desired equation and the theorem is proved.
Theorem 2.6.2 The Hermite’s polynomials satisfy the following recurrence relation
Hn+1 (x) + 2x Hn (x) + 2n Hn−1 (x) = 0.
Proof We shall begin with the formula
dn
dn
d n+1
2
−x 2
−x 2
e
=
−2xe
=
−2
xe−x .
dx n+1
dx n
dx n
Now, we apply the Leibniz’s rule for derivative of superior order
dn
dn
d n−1
2
−x 2
−x 2
xe
=
x
e
+
n
e−x ,
n
n
n−1
dx
dx
dx
dn
d n−1
d n+1
2
−x 2
−x 2
e
=
−2x
e
−
2n
e−x ⇒
dx n+1
dx n
dx n−1
⇒ ex
2
n
n−1
d n+1
2
2 d
2
2 d
2
e−x = −2xe x
e−x − 2ne x
e−x .
n+1
n
dx
dx
dx n−1
Using the Olinde-Rodrigues’s relation it follows
Hn+1 (x) = 2x Hn (x) − 2n Hn−1 (x),
such that the theorem is proved.
Theorem 2.6.3 The Hermite’s polynomials satisfy the following orthogonality relation:
∞
0, if√n = m
−x 2
e Hn (x)Hm (x)dx =
(Hn , Hm ) =
2n n! π, if n = m
−∞
because the weight function and the interval of definition for the Hermite’s Polynomials are
2
p(x) = e−x , [a, b] → (−∞, ∞).
Proof Using the Olinde-Rodrigues’s relation, we have
84
2 Special Functions
∞
(Hn , Hm ) =
e
−x 2
−∞
dn
2
Hm (x) n e−x dx =
dx
∞
Hm (x)
−∞
dn
2
e−x dx =
dx n
∞
∞
d n−1
d n−1
2
−x 2 e−x dx =
= Hm (x) n−1 e
−
H
(x)
m
n−1
dx
dx
−∞
−∞
∞
=−
Hm (x)
−∞
d n−1
2
e−x dx,
dx n−1
because we have
lim P(x)e−x = lim
2
x→±∞
x→±∞
P(x)
= 0,
ex 2
for any polynomials P(x).
On the other hand, we have
Hm (x) =
+e x
2
dm
2
e−x
dx m
= 2xe x
2
dm
2
e−x +
dx m
d m+1
2
e−x = 2x Hm (x) + Hm+1 (x),
dx m+1
such that, by using the recurrence relation, we obtain
Hm (x) = −2m Hm−1 (x).
Then
∞
(Hn , Hm ) = 2m
Hm−1 (x)
−∞
d n−1
2
e−x dx.
dx n−1
If we integrate one more time by parts, we are led to
∞
(Hn , Hm ) = 2 (m − 1)m
Hm−2 (x)
2
−∞
d n−2
2
e−x dx,
dx n−2
and, after m steps,
∞
(Hn , Hm ) = 2 m!
m
−∞
d n−m
2
e−x dx.
n−m
dx
2.6 Hermite’s Polynomials
85
If m = n we have
∞
−∞
∞
d n−m
d n−m−1
−x 2
−x 2 e
dx =
e
= 0,
n−m
n−m−1
dx
dx
−∞
that is
(Hn , Hm ) = 0.
If m = n we have
∞
(Hn , Hm ) = 2 m!
m
√
2
e−x dx = 2m m! π,
−∞
and the theorem is proved.
2.7 Laguerre’s Polynomials
Definition 2.7.1 The functions defined by
n
L n (x) =
(−1)k Cnk
k=0
1 k
x ,
k!
are called the Laguerre’s polynomials.
Remark. We remember that the weight function and the interval of definition for the
Laguerre’s polynomials are
p(x) = e−x , [a, b] → [0, ∞).
In the following we indicate (without proof) the main properties of the Laguerre’s
polynomials. We hope that the readers can prove these results in a similar manner
like the other orthogonal polynomials.
Theorem 2.7.1 The Laguerre’s polynomials satisfy the following Olinde-Rodrigues
relation
1 d n n −x x e
.
L n (x) =
n! dx n
Theorem 2.7.2 The generating function for the Laguerre’s polynomials is
f (x, r ) =
1 −r x/(1−r )
e
,
1−r
86
2 Special Functions
that is,
1 −r x/(1−r )
e
=
1−r
∞
L n (x)r n .
n=0
Theorem 2.7.3 The Laguerre’s polynomials are the solutions of the following differential equation
x y (x) + (1 − x)y (x) + ny(x) = 0.
Theorem 2.7.4 The Laguerre’s polynomials satisfy the following recurrence relation
(n + 1)L n+1 (x) − (2n + 1 − x)L n (x) + n L n−1 (x) = 0, ∀n ∈ N .
Theorem 2.7.5 The Laguerre’s polynomials satisfy the following orthogonality
relation:
∞
0, if n = m
(L n , L m ) = e−x L n (x)L m (x)dx =
(n!)2 , if n = m.
0
One of the main reasons for the study of special functions is to solve certain differential equations with variable coefficients. In this sense, we shall consider a few
applications.
Applications 1. Find the solutions of the equation
x 2 y + 2x y + x 3 − 1 y = 0, y = y(x).
Indications. First, we make the change of variable
x 3 = t 2.
Then, we pass to a new unknown function u with the transformation
y(t) = u(t)t α ,
and determine α such that the coefficient of u is t as in the standard form of a Bessel’s
equation. Finally, we make a new change of variable:
2
t = τ,
3
such that the equation received the standard form of a Bessel’s equation.
2. Find a solution of the equation
x 2 − 1 y + 2x y − 6y = 0, y = y(x).
2.7 Laguerre’s Polynomials
87
Solution. It is easy to remark that we have a Legendre equation in the particular case
n = 2 such that a solution is
P2 (x) =
3 2 1
x − .
2
2
3. Find a solution of the equation
1 − x 2 y − x y + 9y = 0, y = y(x).
Solution. It is easy to remark that we have a Chebyshev equation in the particular
case n = 3 such that a solution is
T3 (x) = cos(3arccos x).
4. Find a solution of the equation
y − 2x y = 4y = 0, y = y(x).
Solution. It is easy to remark that we have a Hermite equation in the particular case
n = 2 such that a solution is
H2 (x) = e x
2
d2
2
e−x = −2x.
dx 2
Chapter 3
Operational Calculus
3.1 Laplace’s Transform
An useful instrument in tackling differential equations and partial differential equations is proved to be the Laplace’s transform which we study in this paragraph.
The Laplace’s transform makes the correspondence between two functions set, one
having difficult operations, and second, more accessible. For instance, a differential
equation in the first functions set is transformed in an algebrical equation in the second functions set. This correspondence is made by means of a transformation. We
will deal only with the Laplace’s transform and Fourier’s transform.
Definition 3.1.1 A function f : IR → IR is called original function for the Laplace’s
transform if the following conditions are satisfied:
(i). f (t) and f (t) exist and are continuous throughout the real axis, possible except
a sequence of points {tn }n≥1 in which it can appear discontinuities of first species;
(ii). f (t) = 0, ∀t < 0;
(iii). there exist the constants M > 0, s0 ≥ 0, such that
| f (t)| ≤ Mes0 t , ∀t ∈ IR.
Usually s0 is the growing index of the original. A classic example of original function
is the Heaviside’s function θ, defined by
θ(t) =
0, if t < 0,
1, if t ≥ 0.
If a function f satisfies the conditions (i) and (iii) but does not satisfy the condition
(ii), from the definition of the original function, then we will make the convention
that the function f is multiplied by the Heaviside’s function θ:
© Springer International Publishing AG, part of Springer Nature 2018
M. Marin and A. Öchsner, Complements of Higher Mathematics,
https://doi.org/10.1007/978-3-319-74684-5_3
89
90
3 Operational Calculus
f (t) = f (t)θ(t) =
0, if t < 0,
f (t), if t ≥ 0.
It is easy to see that | f (t)| ≤ 1 = 1.e0.t , such that M = 1 > 0 and the growing index
is s0 = 0.
This convention is made for increasing the set of original functions. We will denote
by O the set of the original functions. In the following theorem we are going to prove
the structure of the set O. More exactly, we will prove that the set O has the structure
of a linear space and even an algebra structure.
Theorem 3.1.1 Consider O the set of the original functions. Then:
1◦ . f + g ∈ O, ∀ f, g ∈ O;
2◦ . λ f ∈ O, ∀ f ∈ O, ∀λ ∈ R;
3◦ . f.g ∈ O, ∀ f, g ∈ O.
Proof 1o . Since f, g ∈ O, we will deduce that f + g satisfies obviously the properties (i) and (ii) of the original functions. Let us check the (iii) condition. If
| f (t)| ≤ M1 es1 t , |g(t)| ≤ M2 es2 t , ∀t ∈ R,
then
| f (t) + g(t)| ≤ | f (t)| + |g(t)| ≤ M1 es1 t + M2 es2 t ≤ M3 es3 t , ∀t ∈ IR,
where s3 = max{s1 , s2 } and m 3 = max{M1 , M2 }.
2o . λ f satisfies obviously the properties (i) and (ii) of the original functions. Let
us check the (iii) condition. Since
| f (t)| ≤ M1 es1 t , ∀t ∈ R
we will deduce that
|λ f (t)| = |λ|| f (t)| ≤ |λ|M1 es1 t , ∀t ∈ R,
that is λ f has the same growing index as f .
3o . With regard to the (iii) condition for the product f.g, we have
| f (t).g(t)| = | f (t)|.|g(t)| ≤ M1 M2 e(s1 +s2 )t , ∀t ∈ R,
and the proof is over, because the properties (i) and (ii) are obvious.
3.1 Laplace’s Transform
91
Remarks.
1o . From the proof of the Theorem 3.1.1, it results that
ind( f + g) = max{ind( f ), ind(g)},
ind( f.g) = ind( f ) + ind(g)
2o . If f i ∈ O, i = 1, 2, ..., n, then
n
λi f i ∈ O, ∀λi ∈ R, or λi ∈ IC, i = 1, 2, ..., n.
i=1
The statement results from the first two points of the Theorem 3.1.1.
3o . If f i ∈ O, i = 1, 2, ..., n, then
n
f i ∈ O.
i=1
The statement can be immediately proved, by applying the point 3o of the theorem.
In the particular case, if f ∈ O then f n ∈ O, ∀n ∈ N ∗ .
4o . The function f (t) = eλt is an original function, ∀λ ∈ IC, λα + iβ, having
the increasing index
0, if α < 0,
s0 =
α, if α ≥ 0.
As a consequence, the following functions are original functions too
sin λt, cos λt, sinh λt, cosh λt.
If we expand the function eλt as power series
eλt = 1 +
λ2 t 2
λn t n
λt
+
+ ... +
+ ..., t ≥ 0
1!
2!
n!
and take into account the fact that
λn t n
< eλt , ∀t ≥ 0,
n!
we will immediately deduce that
tn <
n! λt
e , ∀t ≥ 0,
λn t n
and then we obtain that the function
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3 Operational Calculus
f (t) = t n , t ≥ 0
is an original function.
Based on the above remarks, it follows that the function
f (t) = eλt [P(t) cos αt + Q(t) sin αt]
is an original function, for any two polynomials P and Q.
Definition 3.1.2 If f (t) is an original function, with the increasing index s0 , then
we call it the Laplace’s transform of f , or its image through the Laplace’s transform,
the function F which is defined by
∞
F( p) =
f (t)e− pt dt, ∀ p ∈ C, Re( p) ≥ s0 .
(3.1.1)
0
Let us prove that the image function F from Eq. (3.1.1) is defined on the whole
semiplan [s0 , ∞) and, more, F is an analytic function in this semiplan.
Theorem 3.1.2 If f is an original function with the increasing index s0 , then the
function F : [s0 , ∞) → C has sense for any complex number p for which Re( p) ≥ s0
and F is an analytical function in this semiplan.
Proof Starting from Eq. (3.1.1) we obtain
f (t)e− pt dt ≤
0
∞
M (s0 − p)t ∞
M
s0 t − pt
≤ M
e e dt =
e
=
,
s − s0
s − s0
0
0
|F( p)| ≤
∞
inequality which proves that the function F is well defined.
If Re( p) ≥ s1 ≥ s0 , then we can derive under the integral in Eq. (3.1.1):
∞
F ( p) =
−te− pt dt,
0
and then we find the estimations
∞
|t f (t)|e− pt dt ≤
|F ( p)| ≤
0
∞
∞
≤M
te(s0 − p)t dt ≤ M
te(s0 −s1 )t dt =
0
0
∞
∞
e(s0 −s1 )t M
M
= Mt
+
e(s0 −s1 )t dt =
,
s 0 − s 1 0
s0 − s1 0
(s0 − s1 )2
3.1 Laplace’s Transform
93
after that we had integrated by parts. Since the derivation is bounded, we will deduce
that F is an analytical function in the open semiplan (s0 , ∞).
As a consequence of the Theorem 3.1.2, one can find that
lim |F( p)| = 0.
| p|→∞
It is natural to put ourselves the question if we know a transformation F, which is
the original function whose Laplace’s transform is even F. The answer is given in
the following theorem.
Theorem 3.1.3 Given the Laplace’s transform F, then the original function can be
obtained in each point of continuity t with the aid of the Laplace’s transform through
of following inverse formula:
f (t) =
1
2πi
a+i∞
F( p)e pt d p,
(3.1.2)
a−i∞
where a ∈ R, a ≥ s0 .
Because the proof is arduous, we renounce to give it.
Theorem 3.1.3 asserts that if the Laplace’s transform of an original function is
given, then it is the Laplace’s transform of a single original, that is, the Laplace’s
transform is an one to one correspondence into the set of the originals. The integral
on the right-hand side of Eq. (3.1.2) is an improper integral in the Cauchy sense:
a+i∞
F( p)e pt d p = lim
a+ib
b→∞ a−ib
a−i∞
F( p)e pt d p.
In order to outline the fact that, for a given Laplace’s transform, the original function
is unique determined, we prove the following theorem.
Theorem 3.1.4 Given is the Laplace’s transform F with the properties:
1o . F( p) is an analytical function in the semiplan Re( p) ≥ a > s0 ;
2o .
lim |F( p)| = 0, f or Re( p) ≥ a > s0 ,
| p|→∞
the limit being uniformly with regard to p;
3o . The integral
a+i∞
F( p)e pt d p
a−i∞
is absolutely convergent.
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3 Operational Calculus
Then the function f (t), defined by
1
f (t) =
2πi
a+i∞
F( p)e pt d p,
(3.1.3)
a−i∞
has even the function F( p) as the Laplace’s transform.
Proof First, we make the observation that the Laplace’s transform F( p) will be
denoted also with L( f (t)) or, more simple, L( f ), taking into account that the argument of the original function is denoted by t and the argument of the Laplace’s
transform is denoted by p.
Applying the Laplace’s transform in Eq. (3.1.3):
L( f ) =
∞
0
1
2πi
a+i∞
F( p)e d p e− p0 t dt.
pt
(3.1.4)
a−i∞
Let us prove that L( f ) = F( p0 ), where p0 = a + iσ is arbitrarily fixed in the semiplan [s0 , ∞).
Since
=
a+i∞
a−i∞
a+i∞
F( p)e p0 t d p ≤
a+i∞
|F( p)| e p0 t d p =
a−i∞
|F( p)| eat | eiσt d p =
a−i∞
a+i∞
|F( p)| eat d p,
a−i∞
and the last integral is convergent (see 30 ), we will deduce that in Eq. (3.1.4) we can
commute the integrals:
1
L( f ) =
2πi
a+i∞
F( p)
a−i∞
∞
e
( p− p0 )t
dt d p.
0
∞
Since Re( p − p0 ) = a − s < 0 and e( p− p0 )t 0 = −1, we obtain
L( f ) =
1
2πi
a−i∞
a+i∞
F( p)
d p.
p − p0
Let us consider the circle having the origin as the center and the radius R and consider
the vertical segment between a − ib and a + ib and the arc of circle C R determined
by this segment on the considered circle. Applying the Cauchy’s formula (from the
theory of complex functions), taking into account that p = p0 is a singular pole, it
follows:
3.1 Laplace’s Transform
95
1
F( p0 ) =
2πi
a−ib
a+ib
F( p)
1
dp +
p − p0
2πi
CR
F( p)
d p.
p − p0
(3.1.5)
For the last integral from Eq. (3.1.5) we have the estimation
1
2πi
CR
1
MR
F( p)
2π R
,
d p ≤
p − p0
2π
|R| − | p0 |
where
M R = sup |F( p)|.
p∈C R
Based on the hypothesis 2o we will deduce that M R → 0, for R → ∞. So we deduce
that the last integral from Eq. (3.1.5) converges to zero, for R → ∞.
Therefore, if we take the limit in Eq. (3.1.5) for R → ∞, we obtain
F( p0 ) =
1
2πi
a−i∞
a+i∞
F( p)
1
dp = −
p − p0
2πi
a+i∞
a−i∞
F( p)
d p,
p − p0
that is F( p0 ) = L( f ).
In the following proposition we will prove the main properties of the Laplace’s
transform.
Proposition 3.1.1 If f and g are original functions, having the images, respectively,
F and G, and α, β ∈ IR, then
L(α f (t) + βg(t)) = αF( p) + βG( p).
Proof This result can be immediately obtained, based on the linearity of the Riemann’s integral.
Proposition 3.1.2 If f is an original function, having the image F, and α ∈ IC∗ ,
then
L( f (αt)) =
1 p
F
.
α
α
Proof Using the change of variable αt = τ , we obtain
=
0
∞
∞
f (αt)e− pt dt =
p
p
1 ∞
1
f (τ )e− α τ dτ =
f (τ )e− α τ dτ ,
α
α 0
L( f (αt)) =
0
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3 Operational Calculus
such that the result is proved.
Proposition 3.1.3 If f is an original function, having the image F, then in a point
t in which f is derivable, we have:
L( f (t)) = p F( p) − f (0).
Proof Starting from the definition of the Laplace’s transform, by direct calculations
it follows:
∞
∞
f (t)e− pt dt = e− pt f (t)0 −
L( f (t)) =
0
∞
∞
− pt
−
(− p) f (t)e dt = − f (0) + p
f (t)e− pt dt = p F( p) − f (0)
0
0
such that the result is proved.
Corollary 3.1.1 With regard to the derivative, we can prove a more general result,
as follows
L( f (n) (t)) = p n F( p) − p n−1 f (0) − p n−2 f (0) − ... − f (n−1) (0).
Proof By using the Proposition 3.1.3, we have
L( f (t)) = pL( f (t)) − f (0),
L( f (t)) = pL( f (t)) − f (0),
L( f
(n)
...........................................
(t)) = pL( f (n−1) (t)) − f (n−1) (0).
Now, we multiply the first relation by p n−1 , the second by p n−2 , ..., the last by p 0 .
Then we add the resulting relations and we obtain the desired result.
Proposition 3.1.4 If f is an original function, having the image F, then:
F ( p) = L(−t f (t)).
Proof We already proved that the integral from the definition is convergent. Then,
we can derive under the integral with regard to p:
F( p) =
∞
f (t)e
− pt
dt ⇒ F ( p) =
0
and we arrive at the desired result.
∞
f (t)(−t)e− pt dt,
0
Corollary 3.1.2 With regard to the derivative of the Laplace’s transform we can
prove a more general result, as follows
3.1 Laplace’s Transform
97
F (n) ( p) = L((−t)n f (t)).
Proof This result can be easy obtained by successive derivation under the integral
and then by using the mathematical induction.
As a consequence of this property it immediately follows that
n!
.
p n+1
L(t n ) =
Proposition 3.1.5 Let f be an original function whose Laplace’s transform is F.
Then the integral
t
f (τ )dτ
0
is also an original function, having the same growing index like f . Moreover, the
following formula is still valid
t
L
f (τ )dτ
=
0
1
F( p).
p
Proof It is easy to prove the (i) and (ii) conditions from the definition of an original
function in the case of the integral
t
f (τ )dτ ,
0
taking into account that f satisfies these conditions. Let us denote by g this integral
t
g(t) =
f (τ )dτ .
0
We intend to prove that g satisfies the condition (iii) from the definition of an original
function:
t
t
| f (τ )|dτ ≤ M
es0 τ dτ =
|g(t)| ≤
0
0
M s0 t
M s0 t
e −1 ≤
e ,
=
s0
s0
and it is easy to see that g has the same growing index like f .
On the other hand, since
t
g(t) =
0
f (τ )dτ ,
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3 Operational Calculus
it is readily seen that g(0) = 0 and g (t) = f (t). Therefore
L( f (t)) = L(g (t)) = pG( p) − g(0) = pG( p),
where we used the Laplace’s transform of the derivative and we noted by G the
Laplace’s transform of g, that is
t
G( p) = L(g(t)) = L
f (τ )dτ ,
0
such that the proof is closed.
Proposition 3.1.6 Let f be an original function whose Laplace’s transform is F. If
we suppose that the improper integral
∞
F(q)dq
p
is convergent, then
∞
f (t)
.
t
F(q)dq = L
p
Proof Taking into account the expression of F, we obtain
∞
=
∞
∞
p
−qt
∞
F(q)dq =
p
e
0
p
dq
f (t)dt =
0
∞
=
0
f (t) − pt
e dt = L
t
∞
0
∞
f (t)e
−qt
dt dq =
∞
e−qt f (t)dt =
t p
f (t)
.
t
that is we have obtained the desired result.
Proposition 3.1.7 If the argument of the original function f is “late”, then the
following formula holds
L( f (t − τ )) = e− pτ F( p), ∀τ > 0,
where, as usually, F is the Laplace’s transform of the original function f .
Proof Starting from the definition of the Laplace’s transform, we obtain
∞
L( f (t − τ )) =
0
f (t − τ )e− pt dt,
3.1 Laplace’s Transform
99
such that if we use the change of variable t − τ = u, it follows
L( f (t − τ )) =
=
0
−τ
f (u)e
=e
− pu − pτ
− pτ
e
∞
∞
−τ
du + e
− pτ
f (u)e− pu e− pτ du =
∞
f (u)e− pu du =
0
f (u)e− pu du = e− pτ F( p),
0
since the function f is an original and then f (u) = 0, ∀u < 0.
It has been ascertained that although the product of two original functions is an original function, however one cannot compute the Laplace’s transform for the product.
But, if the usual product is substituted by the product of convolution, one can compute
the Laplace’s transform for this product. We know that the product of convolution for
two functions can be computed in a more general context. In the case of the original
functions, the product of convolution is defined by
t
( f ∗ g)(t) =
f (τ )g(t − τ )dτ .
(3.1.6)
0
Remark. One verifies without difficulty the following properties of the product of
convolution:
– f
– f
– f
– if
∗g = g∗ f;
∗ (g ∗ h) = ( f ∗ g) ∗ h = f ∗ g ∗ h;
∗ (g + h) = f ∗ g + f ∗ h;
f ∗ g = 0 then f ≡ 0 or g ≡ 0.
Proposition 3.1.8 If f and g are original functions, then their product of convolution
(3.1.6) is an original function.
Proof The conditions (i) and (ii) from the definition of the original are immediately
satisfied, taking into account that f and g satisfy these conditions. Since f and g
satisfy the condition (iii), we have
| f (t)| ≤ M1 es1 t , |g(t)| ≤ M2 es2 t ,
such that
t
|( f ∗ g)(t)| ≤
0
If s2 ≤ s1 then
t
| f (τ )||g(t − τ )|dτ ≤ M1 M2
0
es1 τ es2 (t−τ ) dτ .
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3 Operational Calculus
t
|( f ∗ g)(t)| ≤ M1 M2
es1 τ es1 (t−τ ) dτ = M1 M2
0
t
es1 t dτ = M1 M2 tes1 t .
0
It is evident the fact that t + 1 ≤ et ⇒ t ≤ et − 1 ≤ et . Then
|( f ∗ g)(t)| ≤ M1 M2 e(s1 +1)t .
If s1 < s2 , then we change, reciprocally, the function f and g and use the commutativity of the product of convolution.
Proposition 3.1.9 If f and g are original functions, then the Laplace’s transform
of their product of convolution is equal to the usual product of transforms.
L( f ∗ g) = F( p).G( p).
Proof Taking into account Eq. (3.1.6), we obtain
t
f (τ )g(t − τ )dτ =
∞ ∞
∞ t
f (τ )g(t −τ )dτ e− pt dt =
g(t −τ )e− pt dt f (τ )dτ =
=
∞ 0 ∞ 0
∞0 τ
∞
− p(τ+u)
=
g(u)e
du f (τ )dτ =
f (τ )e− pτ
g(u)e− pu dudτ =
0
0
0
0
∞
∞
− pτ
=
f (τ )e G( p)dτ = G( p)
f (τ )e− pτ dτ = G( p).F( p),
L (( f ∗ g)(t)) = L
0
0
0
in which we used the change of variable t − τ = u. The proposition has been
proved.
Corollary 3.1.3 In applications it is useful the following formula called the
Duhamel’s formula
p F( p)G( p) = L
f (t)g(0) +
t
f (τ )g (t − τ )dτ .
0
Proof We denote by h the product of convolution of the functions f and g, that is
t
h(t) = ( f ∗ g)(t) =
f (τ )g(t − τ )dτ .
0
Then, we deduce that h(0) = 0 and
h (t) = f (t)g(0) +
t
0
f (τ )g (t − τ )dτ .
3.1 Laplace’s Transform
101
Applying the Laplace’s transform of the product of convolution then the Laplace’s
transform of the derivative and use the fact that h(0) = 0 and we obtain the Duhamel’s
formula.
3.2 Operational Methods
The Laplace’s transform is a useful instrument to transform the mathematical operations of mathematical analysis, in more accesible operations. For instance, by using
the Laplace’s transform the solution of a differential equation (or of an integral
equation) reduces to the solution of some algebraical equations. So, by applying the
Laplace’s transform a problem becomes more accessible, but the solution will be
obtained in the set of the images, although the initial problem was stated in the set of
the originals. Therefore, we must transpose the solution of the respective problem,
from the set of the images in the set of originals. This is the subject of the so called
“operational formulas”, or, “operational methods”.
We formulate and prove two results in this sense, which are more common.
Theorem 3.2.1 If the series
∞
ck
pk
k=1
(3.2.1)
is convergent for | p| ≥ R, then the function
θ(t)
∞
k=1
ck
t k−1 ,
(k − 1)!
(3.2.2)
is an original function and its Laplace’s transform is the series given in Eq. (3.2.1).
Here, we have noted by θ the Heaviside’s function.
Proof According to the Cauchy’s criterion of convergence, we have
∞
ck
k−1 ck ≤ M R ⇒ θ(t)
t ≤
(k − 1)!
k
k=1
∞
R k |t|k−1
≤M
≤ M Re R|t| ,
(k
−
1)!
k=1
from where we will deduce that the function (2.2) is an original function.
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3 Operational Calculus
For the second statement of the theorem, we use the formula for the transformation
of the function f (t) = t k :
L θ(t)t k−1 =
(k − 1)!
.
pk
Then, based on the linearity of the Laplace’s transform, we have
∞
∞
ck
ck
L θ(t)
θ(t)
t k−1 =
t k−1 e− pt dt =
(k
−
1)!
(k
−
1)!
0
k=1
k=1
∞ ∞
∞
k−1
ck
t
ck dt =
e− pt θ(t)
,
=
(k − 1)!
pk
k=1 0
k=1
∞
that is, even the desired result.
Theorem 3.2.2 Let P and Q be two polynomials such that degr ee P < degr ee Q
and Q has only simple roots p0 , p1 , ..., pn . Then the function
F( p) =
P( p)
Q( p)
is the Laplace’s transform of the function f given by
f (t) =
n
P( pk ) pk t
e .
Q ( pk )
k=0
Proof Taking into account the hypothesis on the polynomial Q, we can write
Q( p) = c( p − p0 )( p − p1 )...( p − pn )
and then we decompose the function F in simple fractions
F( p) =
a1
an
a0
+
+ ... +
.
p − p0
p − p1
p − pn
(3.2.3)
It is easy to see that the function F has the simple poles p0 , p1 , ..., pn . Consider the
circles c j ( p j , r j ) with the centers in the points p j and the radius r j , sufficient small
such that in each closed disc does not lie any other pole, except the center of the
respective circle. The coefficients a j will be determined by integrating the equality
(2.3) on the circles c j :
F( p)d p =
cj
n
k=0
ak
cj
1
d p.
p − pk
(3.2.4)
3.2 Operational Methods
103
According to Cauchy’s theorem, the integrals from the right-hand side of the relation
(3.2.4) are null, excepting the integral corresponding to k = j, for which we have
1
d p = 2πi.
p − pj
cj
then the relation (3.2.4) becomes
F( p)d p = 2πia j .
(3.2.5)
cj
On the other hand, the integral from the right-hand side of the relation (3.2.4) one
can compute with the aid of the residue’s theorem:
F( p)d p = 2πi res(F, p j ) = 2πi
cj
P( p j )
,
Q( p j )
such that substituting in Eq. (3.2.5), we obtain
aj =
P( p j )
.
Q( p j )
Then formula (3.2.3) becomes
F( p) =
n
n
1
P( pk )
P( pk )
L e pk t .
=
( p )
Q
(
p
)
p
−
p
Q
k
k
k
k=0
k=0
Finally, by using the linearity of the Laplace’s transform, we will deduce
n
P( pk ) pk t
e
F( p) = L
,
Q ( pk )
k=0
that leads to the desired result.
Corollary 3.2.1 If one of the roots of the polynomial Q is null, then the original
function becomes
P(0) P( pk ) pk t
+
e ,
Q(0) k=1 R ( pk )
n
f (t) =
(3.2.6)
where R is the polynomial defined such that Q( p) = p R( p).
Proof We suppose that the null root is p0 = 0. Then we write Q( p) = p R( p). Therefore Q ( p) = R( p) + R ( p). For the other roots of Q we have that Q( pk ) = 0 ⇔
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3 Operational Calculus
R( pk ) = 0. Then Q ( pk ) = R( pk ) + pk Q ( pk ) = pk Q ( pk ). Therefore, the desired
result follows with the aid of Theorem 1.6.
Formula (3.2.6) is known as the Heaviside’s formula.
In the final part of this paragraph, we want to find the image through the Laplace’s
transform of two functions, which is useful in many applications. Consider, firstly,
the function f (t) = t α , where α is a complex constant such that Re(α) > −1. If
Re(α) ≥ 0, then f is an original function and then its Laplace’s transform is:
α
∞
L(t ) =
t α e− pt dt.
(3.2.7)
0
If Re(α) ∈ (−1, 0), then
lim f (t) = ∞
t0
and f is not an original, but the integral (3.2.7) is convergent, such that one can
study the integral (3.2.7) for Re(α) > −1. Taking into account the definition of the
function of Euler, from Eq. (3.2.7) we obtain that
L(t α ) =
(α + 1)
.
p α+1
(3.2.8)
We must outline that formula (3.2.8) gives us the possibilty to prove, by using a new
procedure, the connection between the Euler’s functions, and β:
β(x, y) =
(x)(y)
, Re(x) > −1, Re(y) > −1.
(x + y)
Indeed, if we begin from the equalities
L(t x−1 ) =
(x)
(y)
, L(t y−1 ) =
,
x
p
py
and we take into account the Proposition 3.1.6, with regard to for the product of
convolution, we have
(x)(y)
= L t x+y+1
p x+y
∞
θ x−1 (1 − θ) y−1 dθ ,
0
in which we used the change of variable τ = tθ. The last integral is equal to β(x, y)
and
L(t x+y+1 ) =
and then
(x + y)
,
p x+y
3.2 Operational Methods
105
(x)(y) = β(x, y)(x + y).
Now, we consider the Bessel’s function of first species and of the order n ∈ IN, Jn .
It is well known fact that the function Jn admits the integral representation
1
Jn (t) =
2π
2π
ei(t sin θ−nθ) dθ.
0
Jn is a function of the class C 1 on IR and, more,
|Jn (t)| ≤ 1, ∀t ∈ IR, ∀n ∈ IN.
We will deduce such that Jn is an original function with the increasing index s0 = 0.
The image through the Laplace’s transform of the function Jn is:
L(Jn (t)) =
1
2π
2π
e−inθ dθ
0
∞
e(i sin θ− p)t dt.
0
If Re( p) > s0 = 0, then
∞
e
0
(i sin θ− p)t
1
1
, ⇒ L(Jn (t)) =
dt =
p − i sin θ
2π
∞
0
e−inθ
dθ.
p − i sin θ
By using the substitution e−iθ = z, the integral from the right-hand side becomes a
complex integral that can be computed with the aid of the residues theorem, such
that, finally, we obtain
L(Jn (t)) = 1
p2
+ 1( p +
p 2 + 1)n
.
In the particular case when n = 0, we obtain a result very useful for applications:
√
1 1
L(J0 (2 t) = e− p .
p
3.3 Applications
In the final part of this chapter, we study some concrete and useful applications of
the Laplace’s transform.
106
3 Operational Calculus
3.4 Differential Equations with Constant Coefficients
Consider the Cauchy’s problem
an x (n) + an−1 x (n−1) + ... + a1 x (1) + a0 x = f (t),
x(0) = x0 , x (0) = x1 , ..., x (n−1) (0) = xn−1 ,
where the function f (t) and the constants a0 , a1 , ..., an and x0 , x1 , ..., xn−1 are given.
Applying the Laplace’s transform to both sides of the equation, we obtain
an L x (n) + an−1 L x (n−1) + ... + a0 L (x)) = L ( f (t)) .
Using the notations
L (x(t)) = X ( p), L ( f (t)) = F( p),
and the Laplace’s transform for the derivative of an original, we obtain the relations
L x (n) = p n X ( p) − p n−1 x0 + p n−2 x1 + ... + xn−1 ,
L x (n−1) = p n−1 X ( p) − p n−2 x0 + p n−3 x1 + ... + xn−2 ,
..
.
L(x) = X ( p)
Multiplying the first equation by an , the second by an−1 ,..., the last by a0 and adding
the resulting relations, the following relation is obtained
an L x (n) + an−1 L x (n−1) + ... + a0 L (x)) = X ( p) an p n + an−1 p n−1 + ... + a0 −
−x0 an p n−1 + an−1 p n−2 + ... + a1 − x1 an p n−2 + an−1 p n−3 + ... + a2 − ...
that is, an equation of the form
F( p) = X ( p)ϕ( p) − G( p)
with the solution
X ( p) =
X ( p) + G( p)
.
ϕ( p)
The solution of the initial equation becomes
x(t) = L−1 (X ( p)) .
3.4 Differential Equations with Constant Coefficients
107
A similar procedure is used in the case of a system of differential equations. The
Laplace’s transform is applied for each equation of the system and an algebraical system of equations is obtained having as unknown functions the Laplace’s transforms
of the initial unknown functions. After we find the actual unknowns, by applying the
inverse Laplace’s transform we find the solutions of the initial system.
3.5 Differential Equations with Variable Coefficients
There exists some differential equations having variable coefficients which can be
solved by operational methods. For instance, the differential equations for which the
coefficients are polynomials in t can be approached in this manner because such an
equation contains expressions of the form
x, t x, t 2 x, ..., x , t x , t 2 x , ..., x (n) , t x (n) , t 2 x (n) , ...
and we can use the derivative of the Laplace’s transform
L (−t)n f (t) = F (n) ( p), where F( p) = L ( f (t)) .
Example. Let us solve the equation
t x + x + x = 0, where x(0) = 0, x (0) = 1.
Using the Laplace’s transform of the derivative of an original and the derivative of
the Laplace’s transform, we obtain
L x = p X ( p) − x(0) = p X ( p),
L t x = − L x L x = p 2 X ( p) − 1,
= − p 2 X ( p) − 1 = −2 p X ( p) − p 2 X ( p).
Finally, it results the following equation
− p 2 X ( p) − 2 p X ( p) + p X ( p) + X ( p) = 0 ⇒ p 2 X ( p) + ( p − 1)X ( p) = 0.
After simple calculations, we obtain
dx
1− p
1
1
=
d p ⇒ ln X = − − ln p = ln 1/ p
X
p2
p
pe
X ( p) =
1 1
1 1
1
1
1 1
1−
+
=
−
+ ... .
pe1/ p
p
1! p 2! p 2
3! p 3
108
3 Operational Calculus
This equation can be rewritten in the form
L (x(t)) =
1 1
1
1
1 1
1
+
+ ... = L(1) − L (t) + L t 2 − ...
−
p 1! p 2
2! p 3
1!
2!
3.6 Integral Equations
There exists some integral equations which can be solved by operational methods.
For instance, let us consider the differential-integral equation
t
x (t) =
x(τ ) cos(t − τ )dτ , x(0) = 1.
0
Using the Laplace’s transform, it results
p X ( p) − 1 = X ( p)
⇒ X ( p) =
p
p
⇒ X ( p) p − 2
p2 + 1
p +1
=1⇒
1
1
t2
1
p
2
=
+
L
t
=
L
1
+
⇒
L
=
L(1)
+
(x(t))
p2 + 1
p
p3
2
2
such that the solution of the initial equation is
x(t) = 1 +
t2
.
2
3.7 Partial Differential Equations
Consider the following mixed initial boundary value problem
∂2 f
∂f
∂2 f
∂f
+
b
+ c f = g(x, t)
+
a
+ b12
12
11
2
2
∂x
∂x
∂t
∂x
f (x, 0) = h 1 (x), x ∈ [a, b]
∂f
(x, 0) = h 1 (x)
∂x
∂f
∂f
(0, t) + B1
(0, t) + C1 f (0, t) = k1 (t)
A1
∂x
∂t
∂f
∂f
(l, t) + B2
(l, t) + C2 f (l, t) = k2 (t), t ∈ [0, ∞).
A2
∂x
∂t
a11
,
3.7 Partial Differential Equations
109
By using the Laplace’s transform, we obtain
∂f
= p F(x, p) − f (x, 0) = p F(x, p) − h 1 (x)
∂t
∂2 f
= p 2 F(x, p) − p f (x, 0) − f (x, 0) =
L
∂t 2
L
= p 2 F(x, p) − ph 1 (x) − h 2 (x).
In this way, the previous mixt problem becomes
2
(0, p) + b11 p 2 F(0, p) − ph 1 − h 2 +
a11 ddxF2 (0, p) + a12 dF
dx
+b12 [ p F(0, p) − h 1 ] + cF = (0, p) + B1 p 2 F(0, p) − h 1 (0) + C1 F(0, p) = K 1 ( p)
A1 dF
dx
A2 dF
(l, p) + B2 p 2 F(l, p) − h 1 (l) + C2 F(l, p) = K 2 ( p),
dx
that is, a mixt problem for an ordinary differential equation.
3.8 Some Improper Integrals
We consider, directly, an improper integral which is easy calculable by using the
Laplace’s transform.
Let us compute the following integral, well known as the Gauss’s integral
∞
I =
e−x dx.
2
0
As an auxiliary instrument, consider the integral
∞
J (t) =
e−t x dx,
0
such that our initial integral is
I = J (1).
Using the Laplace’s transform, we obtain
2
110
3 Operational Calculus
L (J (t)) = L
⎧∞
⎨
⎩
e−t x dx
2
⎫
⎬
⎭
⎛
⎞
∞ ∞
2
e− pt dt = ⎝ e−t x e− pt dt ⎠ dx =
0
0
∞ ∞
−t x 2
= L e
dx =
0
0
It is known that
0
1
1
1 π
x
dx = √ arctg √ |∞
.
0 = √
2
p+x
p
p
p2
1
L √
πt
1
=√
p
such that we have
L (J (t)) =
Then we deduce that
1
π
L √
2
πt
=L
π
J (t) = √ ,
2 πt
and, consequently,
√
I = J (1) =
π
.
2
π
√
2 πt
.
Chapter 4
Fourier’s Transform
4.1 Fourier Series
Consider the trigonometrical series of the following form
∞
a0 +
(an cos nωx + bn sin nωx) .
2
n=1
(4.1.1)
Since the functions cos nωx and sin nωx are periodical functions having the period
T = 2π/ω we say that the series (4.1.1) is a periodical series.
Let us suppose that the series (4.1.1) is convergent. Denoting by f (x) its sum, we
can write
∞
f (x) =
a0 +
(an cos nωx + bn sin nωx) .
2
n=1
(4.1.2)
Of course, the function f is periodical with the same period, T = 2π/ω. Also, if the
coefficients of the series are changed, without loss the convergence, another sum is
obtained. The reciprocal problem is of interest. If the sum f (x) is fixed, how we can
determine the coefficients with the help of the function f (x). So, in the following
we find the coefficients such that the sum of the series is the function f (x).
Theorem 4.1.1 The coefficients of the series (4.1.2) have the following expressions
a0 =
2
T
2
an =
T
α+T
α
f (x)dx,
α+T
α
f (x) cos nωxdx,
© Springer International Publishing AG, part of Springer Nature 2018
M. Marin and A. Öchsner, Complements of Higher Mathematics,
https://doi.org/10.1007/978-3-319-74684-5_4
(4.1.3)
111
112
4 Fourier’s Transform
2
bn =
T
α+T
f (x) sin nωxdx.
α
Proof We start by integrating the equality (4.1.2) on an interval of length T , say
[α, α + T ]:
α+T
f (x)dx =
α
⇒
α+T
α
α+T
α
f (x)dx =
∞
a0
dx +
2
n=1
α+T
α
(an cos nωx + bn sin nωx) dx ⇒
∞ an
a0
bn
α+T
.
−
T+
sin nωx|α+T
cos
nωx|
α
α
2
nω
nω
n=1
Since
sin nω(α + T ) − sin nωα = 0, cos nω(α + T ) − cos nωα = 0,
we obtain
α+T
α
a0
2
f (x)dx = T ⇒ a0 =
2
T
α+T
α
f (x)dx.
Also, from the above calculations we deduce that the value of α is not important. It is
important the length T of the interval, such that in the following we use the interval
[0, T ].
Multiply the equality (4.1.2) by cos kωx and integrating the resulting equality on
the interval [0, T ], we obtain
T
0
+
∞ n=1
T
a0
cos kωxdx+
2
0
(an cos nωx cos kωx + bn sin nωx cos kωx) dx ⇒
T
0
∞ an
n=1
T
0
+
f (x) cos kωxdx =
T
0
a0
f (x) cos kωxdx =
2
T
cos kωxdx+
0
cos nωx cos kωxdx + bn
0
T
sin nωx cos kωxdx .
(4.1.4)
4.1 Fourier Series
113
The first integral on the right-hand side becomes
T
cos kωxdx =
0
1
sin kωx|0T = 0.
kω
To evaluate the last two integrals in equality (4.1.4), we must consider two cases.
1. n = k. It is easy to see that
T
0
1
cos nωx cos kωxdx =
2
=
T
[cos(n + k)ωx + cos(n − k)ωx]dx =
0
1
1
sin(n + k)ωx|0T +
sin(n − k)ωx|0T = 0.
2(n + k)
2(n − k)
Also,
T
1
2
sin nωx cos kωxdx =
0
=−
=−
T
[sin(n + k)ωx + sin(n − k)ωx]dx =
0
1
1
cos(n + k)ωx|0T −
sin(n − k)ωx|0T = 0 =
2(n + k)
2(n − k)
1
1
(cos(n + k)2π − 1) −
(cos(n − k)2π − 1) = 0.
2(n + k)
2(n − k)
2. n = k In this case the integrals become
T
f (x) cos nωxdx =
0
T
= an
cos nωxdx + bn
0
=
=
an
2
T
2
sin nωx cos nωxdx =
0
T
(1 + cos 2nωx)dx +
0
bn
2
T
sin 2nωx)dx =
0
an
an
bn
an
T+
sin 2nωx|0T −
cos 2nωx|0T = T.
2
4nω
4nω
2
Therefore, we deduce
an =
2
T
0
T
f (x) cos nωxdx.
114
4 Fourier’s Transform
To obtain the coefficients bn we multiply both sides of the equality (4.1.4) by sin kωx.
The resulting equality is integrated on the interval [0, T ], firstly in the case n = k,
then in the case n = k. With the same considerations as in the case of the coefficients
an , we finally obtain
2 T
bn =
f (x) sin nωxdx,
T 0
such that the theorem is concluded.
Remarks
1. The above determined coefficients are called the Fourier coefficients attached
to the periodical function f having the period T .
2. The Fourier coefficients are still valid even in the case the series (4.1.2) is
not convergent. Indeed, in the calculations to obtain the expressions of the Fourier
coefficients, the equality (4.1.2) was considered as a formal relation.
3. The Fourier coefficients received the same expressions if we take another interval, but having the same length T .
Application
Let us compute the Fourier’s coefficients for the periodical function f with the period
T = 2π given by
f : [π, π] → R, f (x) = x.
We have
ω=
2π
2π
=
= 1, α = −π ⇒ [α, α + T ] = [π, π].
T
2π
Using the formulas for the coefficients, we obtain
a0 =
2
an =
2π
1
2π
π
−π
xdx =
1 x2 π
| = 0,
π 2 −π
1 x
1 π
π
x cos nxdx =
sin nxdx =
sin nx|−π −
π n
n −π
−π
π
1
cos nx|π−π = 0,
πn 2
π
x
2
1
1 π
− cos nx|π−π +
x sin nxdx =
cos nxdx =
bn =
2π −π
π
n
n −π
=
π
1
(−1)n
π
1
2
n
n
π
− (−1) − (−1) + 2 sin nx|−π = −2
=
= (−1)n+1 .
π
n
n
n
n
n
Then, the Fourier series of function f (x) = x is
4.1 Fourier Series
115
x=
∞
2
(−1)n+1 sin nx.
n
n=1
The conditions for the convergence of series (4.1.2) are contained in the following
theorem, due to Dirichlet.
Theorem 4.1.2 Consider a periodical function f : R → R, with period T , satisfying the following conditions:
(i) f is bounded;
(ii) f has a finite number of points of discontinuity of first order in any interval of
length T ;
(iii) f has a finite number of interval of monotony on any interval of length T .
Then, the series (4.1.2) is convergent at any point x0 ∈ R, namely, to f (x0 ) if
x0 is a point of continuity for f and, respectively, to
f (x0 + 0) + f (x0 − 0)
,
2
if x0 is a point of discontinuity of first order.
Remark. With simple words, the hope of the Fourier series is to attach to nonperiodic
functions, certain periodical approximations. In the sense of this approximation it is
considered the following trigonometrical polynomial
a0 +
(an cos nωx + bn sin nωx) ,
2
n=1
m
Sm (x) =
that is, the partial sum of order m of the series.
To evaluate the difference between the function f and its polynomial of approximation it is computed the expression
2
E=
T
T
( f (x) − Sm (x))2 dx,
0
from where we deduce
2
a02 2
an + bn2 ≤
+
2
T
n=1
m
T
f (x)2 dx,
0
called the Bessel’s inequality.
Passing to the limit in this inequality, as m → ∞ we are led to
∞
a02 2
2
+
an + bn2 =
2
T
n=1
0
T
f (x)2 dx,
116
4 Fourier’s Transform
called the Parseval’s identity.
In the following we will write the Fourier series for the periodical function having
the particular period 2π.
Theorem 4.1.3 Consider a periodical function f : [−π, π] → R which has the
period T = 2π. Then
(i) if f is an even function, then its Fourier series becomes
∞
f (x) =
where
2
a0 =
π
π
0
a0 +
an cos nx,
2
n=1
2
f (x)dx, an =
π
π
f (x) cos nxdx.
0
(ii) if f is an odd function, then its Fourier series becomes
f (x) =
∞
bn sin nx,
n=1
where
2
bn =
π
π
f (x) sin nxdx.
0
Proof (i) Since f is an even function f (−x) = f (x) we can write
an =
1
=
π
0
−π
2
2π
π
−π
f (x) cos nxdx =
f (x) cos nxdx +
π
f (x) cos nxdx .
0
In the last relation, the first integral reduces to the second using the substitution
x → −x, taking into account that the functions f and cos x are even functions.
Therefore
2 π
f (x) cos nxdx.
an =
π 0
Similarly, it is easy to see that
bn =
1
=
π
0
−π
2
2π
π
−π
f (x) sin nxdx =
π
f (x) sin nxdx +
0
f (x) cos nxdx
= 0,
4.1 Fourier Series
117
since the function sin x is an odd function.
(ii) We can use the same procedure but taking into account that f is an odd function,
that is f (−x) = − f (x). So, we obtain
a0 = 0, an = 0, bn =
1
π
π
f (x) sin nxdx,
0
and the theorem is concluded.
Remark. It is usual to say that an even function has a Fourier’s cosine series and an
odd function has a Fourier’s sine series.
Now, we consider the case of the functions defined on a non-symmetrical interval
of the form [0, π].
Theorem 4.1.4 Consider the function f : [0, π] → R. Then it admits both a
Fourier’s cosine series and a Fourier’s sine series.
Proof To find the Fourier’s cosine series of the function, we construct the following
function
f (x), x ∈ [0, π]
g(x) =
f (−x), x ∈ [−π, 0].
It is a simple matter to verify that g is an even function and then, according to the
Theorem 4.1.3, it admits a cosine series
∞
g(x) =
where
a0 =
2
π
π
a0 +
an cos nx,
2
n=1
g(x)dx, an =
0
2
π
π
g(x) cos nxdx.
0
But on the interval [0, π] the function g(x) is f (x) such that the above series is, in
fact, the series of function f .
To find the Fourier’s sine series of the function, we construct the following function
h(x) =
f (x), x ∈ [0, π]
− f (−x), x ∈ [−π, 0].
It is a simple matter to verify that h is an odd function and then, according to the
Theorem 4.1.3, it admits a sinus series
h(x) =
∞
n=1
where
bn sin nx,
118
4 Fourier’s Transform
bn =
2
π
π
g(x) sin nxdx.
0
But on the interval [0, π] the function h(x) is f (x) such that the above series is, in
fact, the series of function f . So, the theorem is concluded.
In the following we consider the general case of a function defined on an arbitrary
interval [a, b].
Theorem 4.1.5 Consider the function f : [a, b] → R. Then its Fourier series is
∞
f (x) =
a0 +
(an cos nx + bn sin nx) ,
2
n=1
where
b
2
a0 =
f (x)dx,
b−a a
b
2
nπ
(2x − a − b)dx,
an =
f (x) cos
b−a a
b−a
b
2
nπ
bn =
(2x − a − b)dx.
f (x) sin
b−a a
b−a
Proof Let g(x) be the function
g(x) = f
a+b b−a
+
x .
2
2π
In order to find the domain of definition of g we observe that
a≤
a+b b−a
+
x ≤ b,
2
2π
since f is defined on the interval [a, b]. The above inequalities become
2aπ ≤ aπ + bπ + (b − a)x ≤ 2bπ ⇒ x ∈ [−π, π].
Then, based on the Theorem 4.1.3 the Fourier series of g is
∞
g(x) =
where
a0 +
(an cos nx + bn sin nx) ,
2
n=1
4.1 Fourier Series
119
1 π
f (x)dx,
π −π
1 π
f (x) cos nxdx,
an =
π −π
π
1
f (x) sin nxdx.
bn =
π −π
a0 =
But the expression of an is
1
an =
π
π
f
−π
a+b b−a
+
x cos nxdx.
2
2π
Let us make the change of variable
a+b b−a
+
x = y.
2
2π
Then
dx =
2π
dy.
b−a
Also, for x = −π we obtain y = a and for x = π ⇒ y = b. Then
x=
Therefore
1
an =
π
=
a+b
2
y−
b
f (y) cos
a
2
b−a
b
2π
π
(2y − a − b)
.
b−a
b−a
2π
nπ
(2y − a − b)
dy =
b−a
b−a
f (y) cos
a
nπ
(2y − a − b)dy.
b−a
Following the same procedure, the coefficients bn become
bn =
2
b−a
b
f (y) sin
a
such that the theorem is concluded.
nπ
(2y − a − b)dy,
b−a
Remark. In the particular case a = −l and b = l we obtain that f : [−l, l] → R
and its Fourier series has the coefficients
120
4 Fourier’s Transform
1
a0 =
l
l
−l
l
f (x)dx,
1
nπ
f (x) cos
xdx,
l −l
l
π
1
nπ
xdx.
f (x) sin
bn =
π −π
l
an =
In the final part of this paragraph, we give the complex form for the Fourier series.
In the context of complex functions it is a very useful relation
e z = e x+i y = e x (cos y + i sin y) ,
called the Euler’s identity.
Based on the Euler’s identity it is easy to obtain the relations
cos nωx =
einωx − e−inωx
einωx − e−inωx
einωx + e−inωx
, sin nωx =
= −i
.
2
2i
2
Using these relations, the Fourier series becomes
∞
f (x) =
a0 +
(an cos nx + bn sin nx) =
2
n=1
∞ einωx + e−inωx
einωx − e−inωx
a0 an
+
− ibn
=
=
2
2
2
n=1
∞ a0 an − ibn inωx an + ibn −inωx
=
an
.
+
e
e
+
2
2
2
n=1
Let us denote
cn =
an − ibn
.
2
Taking into account the expressions for an and bn we obtain
1 2 T
2 T
f (t) cos nωtdt − i
f (t) sin nωtdt =
cn =
2 T 0
T 0
=
1
T
0
Similarly, we have
T
f (t)[cos nωt − i sin nωt]dt =
1
T
T
0
f (t)e−inωt dt.
4.1 Fourier Series
121
c−n
1
an + ibn
=
=
2
T
T
f (t)einωt dt.
0
Therefore,
∞ a0 2 T
2 T
−inωt
inωx
inωt
−inωx
+
f (x) =
f (t)e
dt xe
+
f (t)e dt xe
=
2
T 0
T 0
n=1
=
∞ 2 T
a0 2 T
+
f (t)einω(x−t) dt +
f (t)e−inω(x−t) dt =
2
T 0
T 0
n=1
=
a0 2
+
2
T
n=1
∞
=
T
f (t)einω(x−t) dt +
0
−∞
2 T
f (t)einω(x−t) dt =
T
0
n=−1
∞
−∞
2 T
2 T
f (t)einω(x−t) dt +
f (t)einω(x−t) dt.
T
T
0
0
n=0
n=−1
Finally, we can write
∞
−∞
2 T
2 T
inω(x−t)
f (x) =
f (t)e
dt +
f (t)einω(x−t) dt,
T
T
0
0
n=0
n=−1
or,
f (x) =
∞ T
2 f (t)einω(x−t) dt.
T n=−∞ 0
4.2 Fourier’s Single Integral Formula
Consider a function f : R → K , where K = R or K = C, having the following
properties:
(1) f is perhaps derivable on R;
(2) In any point of discontinuity of first order t0 f takes the value
f (t0 ) =
f (t0 − 0) + f (t0 + 0)
.
2
(3) f is an absolutely integrable function on R, that is
∞
−∞
| f (t)|dt < ∞.
122
4 Fourier’s Transform
Then the following formula takes place
1
f (t) =
2π
∞
−∞
∞
−∞
f (τ )eiuτ dτ du,
(4.2.1)
which is called the Fourier’s integral formula.
In the following theorem we pass from the complex form of the Fourier’s integral
formula to its real (or trigonometrical) form.
Theorem 4.2.1 In the same conditions imposed to the function f we have the real
form of the Fourier’s integral formula:
f (t) =
1
π
∞
∞
−∞
0
f (τ ) cos u(t − τ )dτ du.
(4.2.2)
Proof Using the well known Euler’s relation
eiu(t−τ ) = cos u(t − τ ) + i sin u(t − τ ),
formula (4.2.1) becomes
1
f (t) =
2π
∞
−∞
∞
−∞
f (τ ) cos u(t − τ )dτ +
∞
−∞
f (τ ) sin u(t − τ )dτ du.
(4.2.3)
Based on the notations:
ϕ(u, t) =
∞
−∞
ψ(u, t) =
∞
−∞
f (τ ) cos u(t − τ )dτ ,
f (τ ) sin u(t − τ )dτ ,
the relation (4.2.3) becomes
f (t) =
1
2π
∞
−∞
ϕ(u, t)du +
i
2π
∞
ψ(u, t)du.
−∞
It is easy to see that
ϕ(−u, t) = ϕ(u, t), ψ(−u, t) = −ψ(u, t),
that is, ϕ(u, t) is an even function and ψ(u, t) is an odd function, with regard to u.
4.2 Fourier’s Single Integral Formula
123
Therefore
ϕ(u, t) =
−∞
f (τ ) cos u(t − τ )dτ =
∞
+
∞
f (τ ) cos u(t − τ )dτ = 2
0
0
−∞
∞
f (τ ) cos u(t − τ )dτ +
f (τ ) cos u(t − τ )dτ ,
0
where we have used the change of variable u → −u on the interval (∞, 0] and taken
into account the parity of the function ϕ. Using the same change of variable and
taking into account the parity of function ψ we obtain
ψ(u, t) =
=
∞
−∞
f (τ ) sin u(t − τ )dτ =
0
∞
f (τ ) sin u(t − τ )dτ +
−∞
f (τ ) sin u(t − τ )dτ = 0.
0
Finally, we deduce
2
π
f (t) =
∞
−∞
ϕ(u, t)du =
1
π
∞
0
∞
f (τ ) cos u(t − τ )dτ du,
−∞
and the theorem is concluded.
Remark. It is interesting to observe the analogy between the Fourier’s series and the
Fourier’s integral formula. Indeed, observing that
cos u(t − τ ) = cos ut cos uτ + sin ut sin uτ ,
the real form of the Fourier’s integral formula becomes
f (t) =
=
1
π
1
π
∞
0
∞
−∞
0
∞
∞
−∞
f (τ ) cos ut cos uτ dτ +
f (τ ) cos uτ dτ cos ut +
∞
−∞
∞
−∞
f (τ ) sin ut sin uτ dτ du =
f (τ ) sin uτ dτ sin ut du.
Therefore, we can write
f (t) =
∞
[A(u) cos ut + B(u) sin ut] dt,
0
where we have used the notations
(4.2.4)
124
4 Fourier’s Transform
A(u) =
1
π
B(u) =
1
π
∞
f (τ ) cos uτ dτ ,
−∞
∞
−∞
f (τ ) sin uτ dτ .
Remember that the general form of the Fourier’s series is
∞
f (t) =
[an cos nt + bn sin nt] ,
0
it is clear that this formula is analogous with Eq. (4.2.4). The sign of integral is
substituted by the sign of the sum.
In the following we obtain a particular form for the Fourier’s integral formula in
the case of a function having parity.
Theorem 4.2.2 Assume satisfied the standard conditions imposed to function f .
Then
(i) if f is an even function then the Fourier’s integral formula becomes:
2
f (t) =
π
∞
∞
cos ut
0
f (τ ) cos uτ dτ du;
(4.2.5)
0
(ii) if f is an odd function then the Fourier’s integral formula becomes:
f (t) =
2
π
∞
sin ut
0
∞
f (τ ) sin uτ dτ du.
(4.2.6)
0
Proof We write the Fourier’s integral formula in the form
f (t) =
∞
[A(u) cos ut + B(u) sin ut] dt,
(4.2.7)
0
where
We write
A(u) =
1
π
A(u) =
1
π
B(u) =
1
π
0
−∞
∞
−∞
∞
−∞
f (τ ) cos uτ dτ ,
f (τ ) sin uτ dτ .
f (τ ) cos uτ dτ +
∞
f (τ ) cos uτ dτ .
0
For the first integral we make the change of variable τ → −τ and this integral is
transformed in the second integral since the function f (τ ) cos uτ is an even function.
4.2 Fourier’s Single Integral Formula
125
Thus
A(u) =
2
π
∞
f (τ ) cos uτ dτ .
0
Taking into account that the function f (τ ) sin uτ is an even function, we obtain
1
π
B(u) =
1
=
π
∞
f (τ ) sin uτ dτ =
−∞
0
−∞
∞
f (τ ) cos uτ dτ +
f (τ ) cos uτ dτ = 0.
0
With these evaluations about A(u) and B(u) the relation (4.2.7) becomes
f (t) =
∞
0
2
A(u) cos utdu =
π
∞
∞
cos ut
0
f (τ ) cos uτ dτ du,
0
that is, the relation (4.2.5) is proved. Using similar calculations it is easy to prove
the relation (4.2.6) and the theorem is concluded.
Application. Let us write the Fourier’s integral formula for the function
⎧
⎨ 1, t ∈ (−a, a)
f : R → R, f (t) = 1/2, t = ±a
⎩
0, t ∈ (−∞, −a) ∪ (a, ∞),
where a is a positive constant. This is the Dirichlet function of discontinuity. It is
easy to see that f (−t) = f (t), that is f is an even function and then
f (t) =
=
2
π
∞
a
cos ut
0
2
π
∞
∞
cos ut
0
f (τ ) cos uτ dτ du =
0
2
=
π
f (τ ) cos uτ dτ du =
0
∞
cos ut
0
sin au
u
2
π
2
du =
π
∞
a
cos ut
0
cos uτ dτ du =
0
0
∞
cos ut sin au
du.
u
At the end of this paragraph we study, in short, the Fourier’s transform, starting from
the Fourier’s integral formula.
Using the complex form of the Fourier’s integral formula, we can write
f (t) =
1
2π
∞
−∞
∞
−∞
f (τ )eiuτ dτ du =
126
4 Fourier’s Transform
=
1
2π
∞
−∞
1
=
2π
∞
f (τ )eiut e−iuτ dτ du =
−∞
∞
e
∞
iut
−∞
f (τ )e−iuτ dτ du.
−∞
In conclusion, we have
1
f (t) = √
2π
∞
eiut
−∞
1
√
2π
∞
−∞
f (τ )e−iuτ dτ du.
(4.2.8)
By definition, the Fourier’s transform is the function
1
F(u) = √
2π
∞
−∞
f (τ )e−iuτ dτ .
From Eq. (4.2.8) we immediately deduce that
1
f (t) = √
2π
∞
−∞
F(u)eiut du,
which is called the inverse Fourier’s transform.
We must outline the analogy between the Fourier’s transform and its inverse,
almost with regard to the kern of the transformation. Let us find the Fourier’s transform in the particular case of functions having parity.
Theorem 4.2.3 In the case f has parity, we have
(i) if f is an even function, then its Fourier’s transform becomes
2
π
Fc (u) =
∞
f (τ ) cos uτ dτ ,
(4.2.9)
0
and it is called the Fourier’s cosine transform. Its inverse transform is
f (t) =
2
π
∞
Fc (u) cos utdu;
0
(ii) if f is an odd function, then its Fourier’s transform becomes
Fs (u) =
2
π
∞
f (τ ) sin uτ dτ ,
0
and it is called the Fourier’s sine transform. Its inverse transform is
f (t) =
2
π
0
∞
Fs (u) sin utdu;
(4.2.10)
4.2 Fourier’s Single Integral Formula
127
Proof (i) Using the Fourier’s integral formula for an even function, we obtain
2
f (t) =
π
=
2
π
∞
∞
cos ut
0
f (τ ) cos uτ dτ du =
0
∞
cos ut
0
Therefore,
Fc (u) =
and then
f (t) =
2
π
2
π
2
π
∞
f (τ ) cos uτ dτ du.
0
∞
f (τ ) cos uτ dτ ,
0
∞
Fc (u) cos utdu.
0
(ii) Similarly, using the Fourier’s integral formula for an odd function, we obtain
f (t) =
=
2
π
2
π
∞
∞
sin ut
0
∞
sin ut
0
Therefore,
Fs (u) =
and then
f (τ ) sin uτ dτ du =
0
f (t) =
2
π
2
π
2
π
∞
f (τ ) sin uτ dτ du.
0
∞
f (τ ) sin uτ dτ ,
0
∞
Fs (u) sin utdu,
0
and the theorem is concluded.
Application. Let us compute the Fourier’s transform for the function
f (t) =
e−at , t ∈ [0, ∞)
eat , t ∈ (−∞, 0),
where a is a positive constant.
It is easy to prove that the given function is an even function. Therefore, it admits
a Fourier’s cosine transform:
∞
∞
2
2
f (τ ) cos uτ dτ =
e−aτ ) cos uτ dτ =
Fc (u) =
π 0
π 0
128
4 Fourier’s Transform
=
=
1 −aτ )
2
u ∞ −aτ )
∞
− e
cos uτ |0 −
e
sin uτ dτ =
π
a
a 0
1 −aτ )
2 1 u
u ∞ −aτ )
∞
− e
sin uτ |0 +
e
cos uτ dτ
.
−
π a
a
a
a 0
Denoting by I the initial integral, we can write
1 u2
a
− 2 ⇒ I = + 2.
a
a
u a
I =
In conclusion,
Fc (u) =
and
f (t) =
2
π
∞
a
2
2
π u + a2
Fc (u) cos utdu =
0
2a
π
∞
0
cos ut
du.
u2 + a2
Let us verify that this function is the initial function. Consider the integrals
1
I1 =
2
I2 =
Then
I1 + I2 =
1
2
1
2
∞
−∞
∞
−∞
∞
−∞
cos ut
du,
u2 + a2
sin ut
du.
u2 + a2
eiut
1
du = .2π res( f, ai),
2
+a
2
u2
based on the theorem of residues. The residue res( f, ai) can be easily computed
res( f, ai) = lim (u − ai)
u→ai
Then
I1 + I2 =
e−at
eiut
=
.
(u − ai)(u + ai)
2ai
e−at
π
1
.2πi
= e−at .
2
2ai
a
Finally, we have
I1 =
π −at
2a π −at
⇒ f (t) =
= e−at , ∀t ∈ [0, ∞).
e
e
2a
π 2a
Using the parity of f we conclude that
4.2 Fourier’s Single Integral Formula
129
e−at , t ∈ [0, ∞)
eat , t ∈ (−∞, 0).
f (t) =
Remark. If we consider the equality
1
g(u) = √
2π
∞
−∞
f (t)e−iut dt,
where the function g and the function f is unknown, then we say that this equality is
an integral equation of Fourier type. Similarly, in the case of functions having parity,
the equalities of the form
g(u) =
g(u) =
2
π
∞
f (t) cos utdt,
0
2
π
∞
f (t) sin utdt,
0
are, also, integral equations of Fourier type. In each case the solutions are determined
with the help of the inverse Fourier’s transform.
Application. Find the solution of the integral equation
∞
f (t) cos utdt = ϕ(u),
0
where f is the unknown function and ϕ is given by
ϕ(u) =
Multiplying by
√
1 − u, u ∈ [0, 1]
0, u ∈ (1, ∞).
2/π both sides of the equation and denoting
g(u) =
2
ϕ(u),
π
we obtain an equation of the above form. With the help of the inverse Fourier’s
transform we find
∞
2
2 ∞
f (t) =
g(u) cos utdu =
ϕ(u) cos utdu =
π 0
π 0
=
2
π
1
0
(1 − u) cos utdu =
2 1 − cos t
, t ∈ [0, ∞).
π
t2
130
4 Fourier’s Transform
4.3 Fourier’s Transform in L 1
We remember, firstly, the fact that a function f : IR → IR belonging to L 1 (IR) and
we write shortly f ∈ L 1 , if
+∞
| f (t)|dt < +∞.
−∞
Definition 4.3.1 If the function f : R → R, f ∈ L 1 then its Fourier’s transform is
defined by
F( f (t))(x) =
+∞
−∞
f (t)ei xt dt,
(4.3.1)
where i is the complex unit, i 2 = −1.
For simplicity, we use the notation F( f (t))(x) = f (x).
Theorem 4.3.1 If f ∈ L 1 , then its Fourier’s transform f is a bounded and continuous function on R. Moreover, we have
f
f (x) ≤ B(R)
≤ f
L 1 (R) ,
(4.3.2)
where we denoted by B(IR) the set of bounded functions R.
Proof Starting from the definition (4.3.1), we obtain
f (x) ≤
=
+∞
−∞
+∞
−∞
| f (t)| ei xt dt =
| f (t)|dt = f
L 1 (IR) .
If in this inequality we pass to the supremum, it results
f
B(IR)
≤ f
L 1 (IR) ,
and this proves that the Fourier’s transform is a bounded function. Moreover, we
already proved the double inequality (4.3.2). Let us prove now that f is a continuous
function. We use the estimations:
4.3 Fourier’s Transform in L 1
131
+∞
f (t) ei(x+h)t − ei xt dt ≤
f (x + h) − f (x) = −∞
+∞
+∞
+∞
i xt i ht i ht e −1 dt =
≤
| f (t)| e
| f (t)| e −1 dt ≤ 2
| f (t)|dt
−∞
−∞
−∞
(4.3.3)
from where we will deduce that the difference from the left-hand side of the inequality
(4.3.3) is bounded by a summable function.
On the other hand,
f (x + h) − f (x) ≤
+∞
−∞
i ht
e − 1 | f (t)|dt
and
lim ei ht − 1 | f (t)| = 0.
h→0
This means that the conditions of the Lebesque’s theorem are valid and we can pass
to the limit under the integral:
lim f (x + h) − f (x) =
h→0
+∞
lim ei ht − 1 | f (t)|dt = 0,
−∞ h→0
that is f (x) is continuous in any point x ∈ IR.
Corollary 4.3.1 If we have a sequence { f n }n≥1 of functions from L 1 (IR) such that
lim f n = f, in L 1 (IR),
n→∞
then
f (x), uniform with regard to x ∈ IR.
lim f n (x) = n→∞
Proof The result immediately follows based on the inequality (4.3.2):
f (x) ≤ f n − f
f n (x) − L 1 (IR) ,
from where it results the conclusion of the corollary.
In the following theorem we will prove the main properties of the Fourier’s transform.
Theorem 4.3.2 If f ∈ L 1 , then its Fourier’s transform f satisfies the following
rules of calculus
132
4 Fourier’s Transform
F( f (t + a)) = e−iax F( f (t)) = e−iax f (x),
ibt
ibt f (t).
f (x + b) = F e f (t) = e
(4.3.4)
Proof We begin from the definition of the Fourier’s transform:
+∞
f (t + a)ei xt dt =
F( f (t + a)) =
−∞
+∞
f (τ )ei xτ dτ = e−iax f (x),
= e−iax
−∞
where we used the change of variable t +a = τ . Thus, we already proved the formula
(4.3.4)1 . In view of formula (4.3.4)2 we begin also from the definition of the Fourier’s
transform
+∞
ibt f (t) =
eibt f (t)ei xt dt =
F(eibt f (t)) = e
−∞
+∞
i(x+b)t
f (t)e
dt = f (x + b),
=
−∞
that is we have obtained (4.3.4)2 .
In the following theorem, due to the great mathematicians Riemann and Lebesque,
we study the behavior of the Fourier’s transform to the infinity.
Theorem 4.3.3 If f ∈ L 1 , then
lim f (x) = lim
x→±∞
+∞
x→±∞ −∞
f (t)ei xt dt = 0.
Proof We can write
+∞
−∞
+∞
−
f (x) = eiπ f (t)eiπ ei xt dt =
f (x) =
−∞
+∞
i x(t+π/x)
f (t)e
dt =
f (τ − π/x)ei xτ dτ ,
−∞
after that we made the change of variable t + π/x = τ .
Then
+∞ π i xt
e dt.
f (t) − f t −
2 f (x) = f (x) − (− f (x)) =
x
−∞
For the last integrant from (4.3.5) we have the estimation
(4.3.5)
4.3 Fourier’s Transform in L 1
133
π π i xt e ≤ | f (t)| + f t −
,
f (t) − f t −
x
x
that is the last integrant from Eq. (4.3.5) is bounded by a summable function (by
hypothesis, f ∈ L 1 ). We can then use the Lebesque’s theorem to pass to the limit
under the integral in Eq. (4.3.5). Taking into account that
π lim f (t) − f t −
= 0,
x→±∞
x
then immediately follows the desired result.
Corollary 4.3.2 If f ∈ L 1 (IR), then
lim
+∞
x→±∞ −∞
f (t) cos xtdt = 0, lim
+∞
x→±∞ −∞
f (t) sin xtdt = 0.
Proof The result immediately follows from the Theorem 4.3.3, by using the Euler’s
formula ei xt = cos xt + i sin xt.
According to the Theorems 4.3.2 and 4.3.3, the Fourier’s transform is a continuous
function on R and has null limits to −∞ and to +∞. Now, we consider the inverse
problem. If we have a function g which is continuous on IR and has null limits to −∞
and +∞, then g is the Fourier’s transform of a functions from L 1 (IR)? The answer
is negative and we prove this by using a counterexample.
Lemma 4.3.1 If a function g has the properties of a Fourier’s transform, then it is
not necessary that g is the image of a function from L 1 (IR).
Proof We define the function g by
⎧
⎨ −g(−x) , if x < 0,
x/e , if 0 ≤ x ≤ e,
g(x) =
⎩
1/ ln x , if x > e.
From the definition, g is symmetric with regard to the origin. Then
1
= 0.
x→∞ ln x
lim g(x) = lim
x→∞
On the other hand, g is continuous, because for x = e, we have
g(e − 0) = g(e + 0) = 1.
Therefore, the function g has the properties of a Fourier’s transform. However, g is
not the image of a function from L 1 (R). We suppose, through absurdum, that there
exists a function f ∈ L 1 (R) such that
134
4 Fourier’s Transform
g(x) =
+∞
−∞
f (t)ei xt dt.
(4.3.6)
Let us compute the following limit:
n
lim
n→∞ e
g(x)
dx = lim
n→∞
x
e
n
1
dx =
x ln x
= lim [ln(ln x)]ne = lim ln(ln n) = ∞.
n→∞
n→∞
(4.3.7)
Therefore, if we begin from the definition of g, we obtain that the limit from (4.3.7)
is infinite. Let us prove that if we use the form (4.3.6) of the function g, then the limit
from (4.3.7) is finite. Indeed, taking into account the form (4.3.6) of the function g,
we obtain
+∞
f (t)e−i xt dt.
g(x) = −g(−x) =
−∞
Summing this relation with (4.3.6) we are led to
2g(x) =
+∞
−∞
f (t) ei xt + e−i xt dt = 2i
+∞
−∞
f (t) sin xtdt.
Then
g(x) = i
0
∞
f (t) sin xtdt + i
f (t) sin xtdt =
−∞
0
∞
=i
[ f (t) − f (−t)] sin xtdt.
0
Therefore, the integral, under the limit from Eq. (4.3.7), becomes
n
e
g(x)
dx = i
x
n
e
0
∞
[ f (t) − f (−t)]
sin xt
dt dx.
x
(4.3.8)
In the last integral we can change the order of integration, because f (t) − f (−t) is
summable (by hypothesis, f ∈ L 1 (R)). Therefore,
n
e
n
∞
g(x)
sin xt
dx = i
dx dt =
[ f (t) − f (−t)]
x
x
0
e
nt
∞
sin ξ
dξ dt < ∞,
=i
[ f (t) − f (−t)]
ξ
0
et
since the integral
4.3 Fourier’s Transform in L 1
135
nt
et
sin ξ
dξ
ξ
is convergent and the function f (t) − f (−t) is summable.
So, we arrive at a contradiction which proves that the function g cannot be the
Fourier’s transform of a function from L 1 (R).
Another natural question with regard to the Fourier’s transform is the following:
f ∈ L 1 (R)? The answer is again, negative and we prove this
If f ∈ L 1 (R) then by using a counter-example.
Lemma 4.3.2 If a function is from L 1 (IR) then its Fourier’s transform is not necessary to be a function from L 1 (IR).
Proof We define the function f by
0 , if t < 0,
e−t , if t ≤ 0.
f (t) =
Since
+∞
−∞
f (t)dt =
+∞
e−t dt = 1,
0
we will deduce that f ∈ L 1 (IR).
But the Fourier’s transform of the function f is
f (x) =
+∞
−t i xt
+∞
e e dt =
0
e(i x−1)t dt =
0
1 + ix
1
=
,
1 − ix
1 + x2
from where, clearly, it follows that f ∈
/ L 1 (IR).
Now, we expose, without proof, two theorems, due to Jordan, which give the connection between the Fourier’s transform and the original function.
Theorem 4.3.4 If f ∈ L 1 (IR) and, more, f is a function with bounded variation
( f ∈ BV (R)), then in the close vicinity of a fixed point u it holds the following
formula of inversion
1
a→∞ 2π
a
lim
−a
1
f (x)e−i xu dx = [ f (u + 0) − f (u + 0)].
2
If u is a point of continuity for the function f , then the formula of inversion becomes
f (u) =
1
2π
∞
−∞
f (x)e−i xu dx,
where f is the Fourier’s transform of the function f .
136
4 Fourier’s Transform
Theorem 4.3.5 If f ∈ L 1 (R) and f ∈ L 1 (R), then in a point u of continuity of the
function f we have the following formula of inversion
f (u) =
1
2π
∞
−∞
f (x)e−i xu dx.
At the end of this paragraph we will provide some considerations on the product of
convolution for the functions from L 1 (R).
By definition, if f, g ∈ L 1 (R), then its product of convolution is
( f ∗ g)(t) =
+∞
−∞
f (t − τ )g(τ )dτ .
(4.3.9)
Theorem 4.3.6 If f, g ∈ L 1 (IR) then its product of convolution is defined piecewise
on IR and is a function from L 1 (IR).
Proof By using the change of variable t − τ = u, we have
+∞
−∞
| f (t − τ )|dτ =
+∞
−∞
| f (u)|du,
such that, taking into account that f ∈ L 1 (IR), we will deduce that the integrant
from the right-hand side of the relation (4.3.9) is a function defined piecewise and
summable. We can, therefore, invert the order of integration:
+∞
−∞
+∞ +∞
|
f
(t
−
τ
)||g(τ
)|dτ
dt =
−∞
−∞
+∞
+∞
= −∞ |g(τ )| −∞ | f (t − τ )|dt dτ =
+∞
= f L 1 −∞ |g(τ )|dτ = f L 1 g L 1 ,
| f ∗g|(t)dt =
and this proves that f ∗ g ∈ L 1 (IR).
Proposition 4.3.1 If f, g ∈ L 1 (IR) then
f ∗g
L1
≤ f
L1
g
L1 .
Proof Since f, g ∈ L 1 (IR), then according to the Theorem 4.3.6, we have that
f ∗ g ∈ L 1 (IR). By using the definition (4.3.9) of the product of convolution and the
definition of the norm in L 1 (IR), we obtain
4.3 Fourier’s Transform in L 1
137
+∞ +∞
dt ≤
f ∗ g| L 1 =
f
(t
−
τ
)g(τ
)dτ
−∞
−∞
+∞ +∞
≤
| f (t − τ )g(τ )|dτ dt =
−∞
−∞
+∞
+∞
=
|g(τ )|
| f (t − τ )|dt dτ =
−∞
−∞
+∞
+∞
|g(τ )|
| f (u)|du dτ =
=
−∞
−∞
+∞
|g(τ )|dτ = f L 1 g L 1 ,
= f L1
−∞
such that the proof is concluded.
Since we already proved that the product of convolution f ∗ g is a function from
L 1 (R), we can compute its Fourier’s transform.
Theorem 4.3.7 If f, g ∈ L 1 (R), then
F(( f ∗ g)(t)) = F( f (t)).F(g(t)).
Proof We take into account the definition of the Fourier’s transform for the functions
from L 1 (R) and the definition of the product of convolution. So, we obtain
F(( f ∗ g)(t)) =
=
+∞
−∞
+∞
−∞
+∞
+∞
−∞
g(τ )
f (t − τ )g(τ )dτ ei xt dt =
+∞
−∞
+∞
f (t − τ )ei xt dt dτ =
g(τ )
f (u)ei xu du ei xτ dτ =
−∞
−∞
+∞
g(τ )ei xτ dτ = f (x).
g (x),
= f (x)
=
−∞
after that we made the change of variable t − τ = u.
4.4 Fourier’s Transform in L 2
The result proved in the following lemma is very useful in this paragraph.
Lemma 4.4.1 For ∀ε > 0 and ∀α ∈ R, we have the following equality
138
4 Fourier’s Transform
+∞
−∞
eiαt e−εt dt =
2
π 1/2
ε
e
−α2
4ε
.
Proof Using the change of variable
x
t=√ ,
ε
we obtain
+∞
−∞
1
2
eiαt e−εt dt = √
ε
+∞
e
iα √xε −x 2
e
dx.
(4.4.1)
−∞
We consider as well known result the value of the Gauss’s integral
+∞
−∞
e−(x+iβ) dx =
2
√
π.
(4.4.2)
This result can be obtained by using the Laplace’s transform, or with the aid of some
procedures in the context of theory of the complex integral.
We can write the integral from Eq. (4.4.2) in the form
+∞
e
−x 2 −2βxi β 2
e dx = e
e
−∞
β2
+∞
e−x e−2βxi dx
2
−∞
and then
+∞
−∞
e−x e−2βxi dx =
2
√ −β 2
πe .
Introducing this result in Eq. (4.3.1) by taking
α
β=− √ :
2 ε
+∞
−∞
and this concludes the proof.
1 √ −α2
2
eiαt e−εt dt = √ πe 4ε ,
ε
In the following theorem we prove a fundamental result, which anticipates the
Fourier’s transform for the functions from L 2 (R).
f , like a Fourier’s
Theorem 4.4.1 Consider the function f ∈ L 1 (R) ∩ L 2 (R). Then transform of a functions from L 1 (R), is a function from L 2 (R). Moreover,
4.4 Fourier’s Transform in L 2
139
f
L 2 (R)
=
√
2π f
L 2 (R) .
f , given by
Proof Since f ∈ L 1 (R), we know that its Fourier’s transform exists, f (x) =
Then
2
f (x) f (x) = f (x) =
+∞
−∞
f (t)ei xt dt.
+∞
f (t)ei xt dt
−∞
+∞
−∞
f (u)e−i xu du.
Multiplying this equality by
e−x
2
/n
and then integrate the resulting equality over R:
I≡
2 x 2
f (x) e− n dx=
+∞
−∞
+∞
x2
e− n
−∞
+∞
−∞
f (t)ei xt dt
+∞
−∞
f (u)e−i xu du dx.
Since f and f are absolute integrable functions, we can invert the order of integration:
I =
+∞
−∞
+∞
x2
f (t)
e− n ei x(t−u) dx dt du =
−∞
−∞
+∞
+∞
√
n(t−u)2
f (u)
f (t)e− 4 dt du,
= πn
f (u)
+∞
−∞
−∞
in which we used the result from the Lemma 4.4.1 with ε = 1/n and α = t − u.
In the last integral we make the change of variable t − u = s and then we denote
again s by t:
I =
√
πn
+∞
f (u)
−∞
+∞ √
= πn
−∞
+∞
−∞
+∞
−∞
f (t + u)e−
nt 2
4
dt du =
f (u) f (t + u)du e−
nt 2
4
dt.
(4.4.3)
We introduce the notation
g(t) =
+∞
−∞
f (u) f (t + u)du.
Let us prove that the function g is continuous in t = 0. Indeed,
(4.4.4)
140
4 Fourier’s Transform
+∞
2
|g(t) − g(0)| = f (u)[ f (t + u) − f (u)]du ≤
−∞
+∞ +∞
| f (t + u) − f (u)|2 du =
≤
f (u) .
−∞
−∞
+∞
| f (t + u) − f (u)|2 du,
= f L2 .
2
−∞
in which we used the inequality of Hőlder.
It is known that any function from L p , p > 1 (in the our case, f ∈ L 2 ) is
continuous in mean and then
+∞
| f (t + u) − f (u)|2 du → 0, for t → 0,
−∞
that proves that
|g(t) − g(0)|2 → 0, for t → 0,
that is, the function g is continuous in the origin.
We come back to the relation (4.4.3) and we write it in the form
+∞
−∞
+∞
2
√
x2 nt 2
e− n e− 4 g(t)dt =
f (x) dx = πn
−∞
+∞
√
2
2
e−τ g √ τ dτ .
=2 π
n
−∞
(4.4.5)
From the definition (4.4.4) of the function g, we will deduce
|g(t)| ≤
+∞
−∞
f (u)2 du
=
f
1/2
.
2 1/2
L2
+∞
−∞
1/2
| f (t + u)|2 du
2 1/2
L2
f
= f
=
2
L2 .
Also, from Eq. (4.4.4) we obtain
g(0) =
+∞
−∞
f (u) f (u)du =
+∞
−∞
| f (u)|2 du = f
Since the function
e
−τ 2
2
g √ τ
n
is superior bounded by a summable function, namely by
2
L2 .
(4.4.6)
4.4 Fourier’s Transform in L 2
141
e−τ
2
2
L2 ,
f
we will deduce that in Eq. (4.4.5) we can use the Lebesque’s theorem which permits
to pass to the limit under the integral. Thus, for n → ∞, from Eq. (4.4.5) it results
√
=2 π f
2
L2
+∞
2
√
f (x) dx = 2 π
−∞
+∞
e−t
−∞
2
√
dt = 2 π f
in which we used the relation (4.4.6).
Therefore,
f
f 2L 2 = 2π f 2L 2 ⇒ L2
+∞
−∞
2
L2
=
e−t g(0)dt =
2
√
π = 2π f
√
2π f
2
L2 ,
L2 ,
that concludes the proof of the theorem.
We make now another step in view of introducing the Fourier’s transform for
the functions from L 2 (R). To this, we introduce the truncated function. Thus, if
f ∈ L 2 (R), then the truncated function f a , associated to f , is defined by
f a (t) =
f (t) , if |t| ≤ a,
0 , if |t| > a.
(4.4.7)
Theorem 4.4.2 If the function f ∈ L 2 (R) then the truncated function f a is a function from f ∈ L 1 (R) ∩ L 2 (R) and, therefore, it admits Fourier’s transform and
f a ∈ L 2 (R). Moreover, for a → 0 we have
f (t), in norm of L 2 .
f a (t) → Proof Firstly, we can observe that from Eq. (4.4.7) it results
| f a (t)| ≤ | f (t)|, ∀t ∈ R ⇒ | f a (t)|2 ≤ | f (t)|2 , ∀t ∈ R.
Thus, by integrating the last inequality, it results
+∞
−∞
| f a (t)|2 dt ≤
⇒ f a (t)
L2
+∞
−∞
| f (t)|2 dt ⇒
≤ f (t)
L2 ,
that proves that f a ∈ L 2 (R).
On the other hand, from the definition of a truncated function we obtain
142
4 Fourier’s Transform
+∞
−∞
| f a (t)| dt =
+a
−a
| f (t)|dt ≤
based on the inequality of Hőlder.
But
+a
−a
and then
| f (t)|2 dt ≤
+∞
−∞
+∞
−∞
√
+a
2a
−a
1/2
| f (t)| dt
2
| f (t)|2 dt = f
| f a (t)| dt ≤
√
,
2
L2
2
L2
2a f
that proves that f a ∈ L 1 (R). Therefore, f a is a function from L 1 (R) and from L 2 (R)
too, that is, it satisfies the hypotheses of the Theorem 4.4.1 and then there exists its
Fourier’s transform in the sense of transform for the functions from L 1 (R):
f a (x) =
+∞
−∞
f a (t)ei xt dt =
+a
f (t)ei xt dt.
−a
Also, from the Theorem 4.4.1 we will deduce that f a ∈ L 1 (R) too. We must prove
2
that f a is convergent in the space L (R). To this we use the Cauchy’s criterion of
fundamental sequence (L 2 (R) is a completed space). For b > 0, we have
2
f a − f a+b 2 ≤ L
+a
−a+b
| f (t)| dt +
a+b
2
| f (t)| dt .
2
a
Therefore, ∀ε > 0, ∃n 0 (ε) such that if a > n 0 (ε) and b > 0, we have
2
f a − f a+b 2 < ε
L
and this proves that the sequence { f a } is convergent in the norm of L 2 (R).
Now, we can define the Fourier’s transform for a function from L 2 (R).
Definition 4.4.1 If the function f ∈ L 2 (R) then one can attach the truncated function f a and to this, as a function from L 1 (R), one can attach the Fourier’s transform
f a (x) =
+∞
−∞
f a (t)e dt =
i xt
+a
−a
f (t)ei xt dt.
By definition
f (x) = lim f a (x), in L 2 .
a→0
The result from the following theorem is due to Parseval.
4.4 Fourier’s Transform in L 2
143
Theorem 4.4.3 If the function f ∈ L 2 (R) then f ∈ L 2 (R) and
f
L2
=
√
2π f
L2
.
Proof The fact that f ∈ L 2 (R) follows from Theorem 4.4.1. Then we have the
inequality
fn L2
fn − − fm L 2 ≤ fm L 2 .
(4.4.8)
In the Theorem 4.4.2 we already proved that the sequence { f n } is convergent and
then
f 2.
fn 2 = lim L
n→∞
L
On the other hand, because f n ∈ L 1 (R) ∩ L 2 (R), for the truncated function we can
write
√
f n 2 = 2π f n L 2 .
L
Thus, that if we pass to the limit we obtain the desired result.
In the following theorem we prove a formula of inversion, due to Plancherel.
Theorem 4.4.4 If the functions f, g ∈ L 2 (R) then we have the following formula
of inversion
+∞
+∞
g (x)dx = 2π
f (x)g(x)dx.
f (x)
−∞
−∞
Proof According to the Parseval’s formula, we can write
2
f +
g L 2 = 2π f + g
2
L2
,
that is,
+∞
−∞
= 2π
f (x) + g (x)
+∞
−∞
g (x) dx =
f (x) + ( f (x) + g(x)) f (x) + g(x) dx.
After simple calculations, again, be using the Parseval’s formula, it follows
= 2π
+∞
−∞
+∞
−∞
g (x)dx +
f (x)
+∞
−∞
f (x)g(x)dx + 2π
g (x) f (x)dx =
+∞
−∞
f (x)g(x)dx.
(4.4.9)
144
4 Fourier’s Transform
We make these calculations substituting g by ig and then
−
= −2πi
+∞
−∞
+∞
−∞
g (x)dx + i
f (x)
+∞
−∞
f (x)g(x)dx + 2πi
f (x)
g (x)dx =
+∞
−∞
f (x)g(x)dx.
Here we simplify by (−i) and add the resulting equality to the equality (4.4.9). So,
we obtain the result of Plancherel.
Another formula of inversion is the result from the following theorem.
Theorem 4.4.5 If the functions f, g ∈ L 2 (R) then we have the following formula
of inversion
+∞
−∞
f (x)g(x)dx =
+∞
−∞
f (x)
g (x)dx.
(4.4.10)
Proof Since the functions f, g ∈ L 2 (R) we will deduce that we can attach the
truncated functions, respectively f n and gk . We already proved that f n , gk ∈ L 1 (R)
and then we can attach its transforms, in the sense of the functions from L 1 (R).
Using the Fourier’s transform for the truncated functions, it results
+∞
+∞
gk (x)
f n (t)ei xt dt dx =
f n (x)gk (x)dx =
−∞
−∞
−∞
+∞
+∞
+∞
f n (x)
gk (t)ei xt dx dt =
f n (t) f k (t)dt.
=
+∞
−∞
−∞
(4.4.11)
−∞
In these calculations we have inverted the order of integration because the integrals
are computed on finite intervals, taking into account the definition of the truncated
functions.
The equality (4.4.11) proves that the formula of inversion (4.4.10) is still valid for
the truncated functions. We now fix f n and use the result from the Theorem 4.4.2,
such that the sequence {
gk } is convergent, piecewise, in the sense of L 2 , to a function
2
f n } is convergent, piecewise,
from L . Similarly it follows the fact that the sequence { in the sense of L 2 , to a function from L 2 . We use then the fact that the two limits are
g and respectively f . Now, based on the Lebesque’s theorem with regard to pass to
the limit under the integral, from (4.4.11) it results (4.4.10), that is, the theorem is
proved.
At the end of this paragraph, we prove the last formula of inversion for the Fourier’s
transform.
Theorem 4.4.6 We consider the function f ∈ L 2 (R) and define the function g by
g(x) = f (x), ∀x ∈ R.
4.4 Fourier’s Transform in L 2
145
Then
f (x) =
1
g (x), ∀x ∈ R.
2π
Proof According to the definition of the norm in L 2 , we have
2 f − 1 g
=
2π 2
+∞
1
1
f (x)− g (x) dx =
g (x) f (x)− 2π
2π
−∞
L
+∞
1
1
g 2L 2 −
= f 2L 2 + 2 f (x)
g (x)dx−
4π
2π −∞
+∞
1
−
f (x)
g (x)dx.
2π −∞
(4.4.12)
By using two times the Parseval’s formula, we obtain
2π
1
g 2L 2 =
g 2L 2 =
4π 2
4π 2
2
1 1 1
2
=
f
f L 2 =
f 2 =
L
2π
2π
2π
2
L2
.
(4.4.13)
On the other hand, using Eq. (4.4.10) and then the Parseval’s formula, it results
1
−
2π
+∞
1
f (x)
g (x)dx = −
f (x)g(x)dx =
2π −∞
−∞
+∞
2
1 1
f (x) f (x)dx = −
f L 2 =
=−
2π −∞
2π
+∞
=−
2π
2π
f
2
L2
=− f
2
L2
(4.4.14)
.
Similarly,
−
1
2π
+∞
−∞
g (x) f (x)dx = −
=− f
2
L2
1
2π
+∞
f (x)
g (x)dx =
−∞
=− f
2
L2
(4.4.15)
.
Here we used the result from Eq. (4.4.14) and the fact that the conjugated of a real
number is equal to itself.
If we take into account formulas (4.4.13)–(4.4.15) in (4.4.12), we obtain the
formula from our theorem.
Chapter 5
Calculus of Variations
5.1 Introduction
The modern engineer often has to deal with problems that require a sound mathematical background and set skills in the use of various mathematical methods. Expanding
the mathematical outlook of engineers contributes appreciably to new advances in
technology. The calculus of variations is one of the most important divisions of
classical mathematical analysis in regards to applications.
At the beginning of this paragraph we remember some basic elementary notions
of the classical mathematical analysis that we will use in this chapter.
1. A linear space R is a normed linear space if to every element x ∈ R there is
associated a nonnegative real number x called the norm of that element, and:
(i) x = 0 only when x = 0;
(ii) αx = |α|x;
(iii) x + y ≤ x + y (triangle axiom for norms).
2. A set M of elements x, y, z, . . . of any nature whatsoever is a metric space
if to each pair of elements x, y ∈ M there is associated a nonnegative real number
(x, y) such that
(i) (x, y) = 0 if and only if x = y (identity axiom);
(ii) (x, y) = (y, x) (symmetry axiom);
(iii) (x, y) + (y, z) ≥ (x, z) (triangle axiom).
The number (x, y) is called the distance between the elements x and y. Of course,
every normed linear space is a metric space. This statement can be immediately
argued if we put (x, y) = x − y.
3. The space C 0 [a, b] is the space of all functions y(x) continuous on [a, b]. The
usual norm for each element y ∈ C 0 [a, b] is
y0 = max |y(x)|,
a≤x≤b
© Springer International Publishing AG, part of Springer Nature 2018
M. Marin and A. Öchsner, Complements of Higher Mathematics,
https://doi.org/10.1007/978-3-319-74684-5_5
147
148
5 Calculus of Variations
where |y(x)| is the modulus of the element y(x) ∈ C 0 [a, b].
4. The space C 1 [a, b] is the space of all functions y(x) continuous on [a, b]
together with their first derivatives. The usual norm for each element y ∈ C 1 [a, b] is
y1 = max |y(x)| + max |y (x)|,
a≤x≤b
a≤x≤b
where y (x) is the first derivative of function y(x).
5. The space C n [a, b] is the space of all functions y(x) continuous on [a, b]
together with their derivatives up to order n inclusive (n is a fixed natural number).
The usual norm for each element y ∈ C n [a, b] is
yn =
n
k=0
max |y (k) (x)|,
a≤x≤b
where y (k) (x) is the derivative of k-th order of function y(x). Sometimes the norm
of the element y(x) ∈ C n [a, b] is defined as follows
y = max |y(x)|, |y (x)|, . . . , |y (n) (x)| .
a≤x≤b
Suppose we have a certain class M of functions y(x). If to each function y(x) ∈ M
there is associated, by some law, a definite number J , then we say that a functional
J is defined in the class M and we write J = J (y(x)).
The class M of functions y(x) on which the functional J = J (y(x)) is defined is
called the domain of definition of the functional.
Example 5.1. Let M = C 1 [a, b] be the class of functions y(x) that have continuous
derivatives on the interval [a, b] and let
J (y(x)) = y (x0 ), x0 ∈ [a, b].
It is clear that J = J (y(x)) is a functional defined in the indicated class of functions
because to each function of this class there is associated a definite number, namely,
the value of the derivative of the function at the fixed point x0 .
If, for instance, a = 1, b = 3 and x0 = 2, then for y(x) = x 2 + 1 we get
J (x 2 ) = 2x|x=2 = 4.
For y(x) = ln(1 + x) we have
1 1
J (ln(1 + x)) =
= .
1 + x x=2
3
5.1 Introduction
149
Example 5.2. Let M = C 0 [−1, 1] be the class of functions y(x) continuous on the
interval [−1, 1] and let ϕ(x, y) be a given function defined and continuous for all
−1 ≤ x ≤ 1 and for all real y. Then
1
J (y(x)) =
ϕ(x, y(x))dx,
−1
will be a functional defined on the indicated class of functions. For instance, if
ϕ(x, y) = x/(1 + y 2 ), then for y(x) = x it follows
1
J (x) =
−1
x
dx = 0,
1 + x2
and for y(x) = 1 + x we have
1
J (1 + x) =
−1
√
x
dx
=
ln
5 − arctan 2.
1 + (1 + x)2
Example 5.3. Let M = C 1 [a, b] be the class of functions y(x) having continuous
derivatives y (x) on the interval [a, b]. Then
J (y(x)) =
b 1 + y 2 (x)dx,
a
is a functional defined on that class of functions. This functional geometrically
describes the arc length of the curve y = y(x) with ends at the points A(a, y(a))
and B(b, y(b)).
Definition 5.1.1 We say that the curves y = y(x) and y = y1 (x) specified on the
interval [a, b] are close in the sense of vicinity of the zero-th order if |y(x) − y1 (x)|
is small on [a, b].
Geometrically, this means that those curves on the interval [a, b] are close in
regards to the coordinates.
[a, b] are
We say that the curves y = y(x) and y = y1 (x) specified on the interval
close in the sense of vicinity of the first order if |y(x) − y1 (x)| and y (x) − y1 (x)
are small on [a, b]. Geometrically, this means that those curves on the interval [a, b]
are close both in regards to the ordinates and to the directions of the tangents at the
appropiate points. More generally, we say that he curves y = y(x) and y = y1 (x)
specified on the interval [a, b] are close in the sens of vicinity of the k-th order if the
moduli
150
5 Calculus of Variations
|y(x) − y1 (x)| , y (x) − y1 (x) , . . . y (k) (x) − y1(k) (x)
are small on [a, b].
Of course, if two curves are close in the sense of vicinity of the k-th order, they
are closer in the sense of vicinity of any smaller order.
Example 5.1. For n sufficiently large, we define on the interval [0, π] the following
functions
sin n 2 x
y(x) =
, y1 (x) = 0, ∀x ∈ [0, π].
n
These functions are close in the sense of vicinity of zero-th order since
sin n 2 x 1
≤ .
|y(x) − y1 (x)| = n n
That is to say, on the entire interval [0, π] this difference is small in modulus if n is
sufficiently large. But, there is no vicinity of the first order since
y (x) − y (x) = n cos n 2 x 1
and, for instance, at the points
2π
,
n2
we have y (x) − y1 (x) = n and, hence, y (x) − y1 (x) can be made arbitrarily
large for sufficiently large n.
x=
Example 5.2. For n sufficiently large, we define on the interval [0, π] the following
functions
sin n 2 x
y(x) =
, y1 (x) = 0, ∀x ∈ [0, π].
n2
These functions are close in the sense of vicinity of the first order since
sin n 2 x ≤ 1
|y(x) − y1 (x)| = n2
2
n
and
y (x) − y (x) = cos nx ≤ 1 ,
1
n
n
are small, for sufficiently large n.
Definition 5.1.2 The distance between the curves y = f (x) and y = f 1 (x) (a ≤
x ≤ b), where f (x) and f 1 (x) are functions continuous on [a, b], is a nonnegative
number equal to the maximum of | f 1 (x) − f (x)| on the interval a ≤ x ≤ b:
5.1 Introduction
151
= [ f 1 (x), f (x)] = max | f 1 (x) − f (x)|
a≤x≤b
The n-th order distance between the curves y = f (x) and y = f 1 (x) (a ≤ x ≤ b),
where the functions f (x) and f 1 (x) have continuous derivatives of order n on the
interval [a, b], is the largest of the maxima of the quantities
| f 1 (x) − f (x)| , f 1 (x) − f (x) , . . . , f 1(n) (x) − f (n) (x)
on the interval [a, b]. We will denote this distance as follows
n = n [ f 1 (x), f (x)] = max max f 1(k) (x) − f (k) (x) .
0≤k≤n a≤x≤b
Example 5.1. Find the first order distance between the curves f (x) = x 2 and f 1 (x) =
x 3 on the interval [0, 1].
We find the derivatives of the given functions, f (x) = 2x and f 1 (x) = 3x 2 , and
we consider the functions
y1 (x) = x 2 − x 3 , y2 (x) = 2x − 3x 2 .
Let us find their maximum values on the interval [0, 1]. We have y1 (x) = 2x − 3x 2 .
Equating this derivative to zero, we find the stationary points of the function y1 (x) :
x1 = 0, x2 = 2/3. Furthermore,
y1 |x=0 = 0, y1 |x= 23 =
4
.
27
The value of y1 (x) at the endpoint is y1 (1) = 0, whence
4
.
0 = max x 3 − x 2 = max x 2 − x 3 =
0≤x≤1
0≤x≤1
27
Let us now find the zero-order distance ˜ between the derivatives f (x) = 2x and
f 1 (x) = 3x 2 :
˜ = max y2 (x) = max 2x − 3x 2 .
0≤x≤1
0≤x≤1
If we construct the graph of the function 2x − 3x 2 then it is evident that ˜ = 1.
Thus, the first order distance 1 between the curves f (x) = x 2 and f 1 (x) = x 3 is
equals to
˜ = 1.
1 = max (0 , )
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5 Calculus of Variations
Definition 5.1.3 The n-th order ε-neighbourhood of a curve y = f (x) (a ≤ x ≤ b)
is defined as the collection of curves y = f 1 (x) whose n-th order distances from the
curve y = f (x) are less than ε:
n = n [ f 1 (x), f (x)] < ε.
A zero-th ε-neighborhood is called a strong ε-neighborhood of the function y =
f (x).
The strong ε-neighborhood of the curve y = f (x) consists of curves located in a
strip of width 2ε about the curve y = f (x).
A first order ε-neighborhood is called a weak ε-neighborhood of the function
y = f (x).
Definition 5.1.4 A functional J (y(x)) defined in a class M of functions y(x) is said
to be continuous for y = y0 (x) in the sense of n-th order vicinity if for any number
ε there exists a number η > 0 such that for admissible functions y = y(x) satisfying
the conditions
|y(x) − y0 (x)| < η, y (x) − y0 (x) < η, . . . , y (n) (x) − y0(n) (x) < η,
the inequality |J (y(x)) − J (y0 (x))| < ε.
In other words,
n [y(x), y0 (x)] < η ⇒ |J (y(x)) − J (y0 (x))| < ε.
A functional that is not continuous in the sense of n − th order vicinity will be
called discontinuous in the sense of the indicated vicinity. Putting
y (k) (x) = y0(k) (x) + αω (k) (x), k = 0, 1, 2, . . . , n
where α is some parameter and ω(x) is an arbitrary function in the class M, we note
that
lim y (k) (x) = y0(k) (x), k = 0, 1, 2, . . . , n
α→0
and the definition of the continuity of the functional when y(x) = y0 (x) may be
written as:
lim J [y0 (x) + αω(x)] = J [y0 (x)] .
α→0
Example 5.1. Let us show that the functional
1
J (y(x)) =
y(x) + 2y (x) dx
0
defined in the space C 1 [0, 1] is continuous on the function y0 (x) = x in the sense of
first order vicinity.
5.1 Introduction
153
Indeed, take an arbitrary number ε > 0 and show that there exists a number η > 0
such that |J (y(x)) − J (x)| < ε as soon as |y(x) − x| < η and |y (x) − 1| < η. We
have
1
y(x) + 2y (x) − x − 2 dx ≤
|J (y(x)) − J (x)| = 0
1
≤
1
|y(x) − x|dx + 2
0
y (x) − 1 dx.
0
We choose η = ε/3. Then for all y(x) ∈ C 1 [0, 1] for which
|y(x) − x| <
ε
ε
and y (x) − x <
3
3
we will have
|J (y(x)) − J (x)| < ε.
Thus, for every ε > 0 there exists an η > 0, for example, η = ε/3, such that as soon
as 1 [y(x), x] < η then |J (y(x)) − J (x)| < ε.
By definition, this means that the given functional is continuous on the function
y0 (x) = x in the sense of first order vicinity. In fact, it is easy to see that this functional
is continuous in the sense of first order vicinity on any curve y(x) ∈ C 1 [0, 1].
Example 5.2. Let us show that the functional
1
J (y(x)) =
y(x) + 2y (x) dx
0
defined in the space C 1 [0, π] is discontinuous on the function y0 (x) = 0 in the sense
of zero-th order vicinity.
Indeed, let y0 (x) = 0 on [0, π] and yn (x) = (sin nx)/n. Then 0 [y0 (x), yn (x)] =
1/n and to0 as n → ∞. On the other hand, the difference
π
J (yn (x)) − J (y0 (x)) =
π
cosn x
dx =
n
2
0
does not depend on n. Thus, as n → ∞, J (yn (x)) does not tend to J (y0 (x)), and,
hence, the given functional is discontinuous in the sense of zero-th order vicinity on
the function y0 (x).
But it is easy to prove that the functional under consideration is continuous on the
function y0 (x) = 0 in the sense of first order vicinity.
154
5 Calculus of Variations
Definition 5.1.5 Let M be a normed linear space of the functions y(x). The functional L(y(x)) defined in the space M is called a linear functional if it satisfies the
following two conditions
(1) L(cy(x)) = cL(y(x)),
where c is an arbitrary constant;
(2) L(y1 (x) + y2 (x)) = L(y1 (x)) + L(y2 (x)),
where y1 (x) ∈ M and y2 (x) ∈ M.
Definition 5.1.6 If the increment in the functional
J = J (y(x) + δ y) − J (y(x))
can be represented as
J = L(y(x), δ y) + β(y(x), δ y)δ y,
where L(y(x), δ y) is a linear functional with respect to δ y and β(y(x), δ y) → 0
as δ y → 0, then the portion of the increment of the functional that is linear with
respect to δ y, that is, L(y(x), δ y) is called the variation of the functional and is
denoted by δ J . In this case the functional J (y(x)) is said to be differentiable at the
point y(x).
Example 5.1. Let us show that the functional
b
J (y(x)) =
y(x)dx
a
specified in the space C 0 [a, b] is differentiable at each point y(x) of that space.
Indeed, by direct calculations
J = J (y + δ y) − J (y) =
b
b
[y(x) + δ y(x)]dx −
=
a
b
y(x)dx =
a
Thus,
δ y(x)dx.
a
b
J =
δ y(x)dx.
a
This is a linear functional with respect to δ y(x). In the given case the entire increment
of the functional reduced to the linear functional with respect to δ y(x). This functional
is differentiable at every point y(x) and its variation is
5.1 Introduction
155
b
δJ =
δ y(x)dx.
a
Now, we give a second definition of the variation of a functional.
Definition 5.1.7 The variation of the functional J (y(x)) at the point y = y(x) is
defined as the value of the derivative of the functional J (y(x)+αδ y(x)) with respect
to the parameter α when α = 0:
∂
J (y(x) + αδ y(x))
δJ =
.
∂α
α=0
If the variation of a functional exists as the principal linear part of its increment,
that is, in the sense of the first definition, then the variation also exists as the value
of the derivative with respect to the parameter α when α = 0, and these variations
coincide.
Example 5.1. Using the second definition, find the variation of the functional
b
J (y(x)) =
y 2 (x)dx.
a
Firstly, the variation of this functional in the sense of the first definition is equal to
b
δu = 2
y(x)δ y(x)dx.
a
Let us find the variation of our functional using the second definition of a variation.
We have
b
J (y(x) + αδ y(x)) = [y(x) + αδ y(x)]2 dx.
a
Then
∂
J (y(x) + αδ y(x)) = 2
∂α
b
[y(x) + αδ y(x)]δ y(x)dx,
a
and, consequently,
b
∂
δJ =
= 2 y(x)δ y(x)dx.
J (y(x) + αδ y(x))
∂α
α=0
a
156
5 Calculus of Variations
So, it is easy to see that the variations of the functional under considerations in the
sense of first and second definitions coincide.
Remark. The second definition of a variation of a functional is somewhat broader
than the first in the sense that there are functionals from the increments of which it is
not possible to isolate a principal linear part, but the variation exists in the sense of
the second definition. We will demonstrate this fact using the example of functions
for which the formulated assertion is equivalent to the fact that the existence of
derivatives in any direction is not sufficient for the existence of a differential of the
function.
So, let us consider the function
f (x, y) = xy
x 2 + y2
=
sin 2ϕ, x 2 + y 2 = 0,
2
where and ϕ are the polar coordinates of the point (x, y).
The partial derivatives ∂ f /∂x and ∂ f /∂ y exist at every point and at the origin
are equal to zero, but the differential d f does not exists at the origin. Indeed, given
the existence of d f , the gradient of the function f at the origin would in this case
be equal to zero, and therefore a derivative in any direction, d f (0, 0)/dl, would be
equal to zero. Yet, as can readily be seen
1
d f (0, 0)
= sin 2ϕ,
dl
2
which is, generally speaking, different from zero. Here, ϕ is the angle formed by the
vector l with the x-axis and the derivative in the direction of l coincides with the
derivative with respect to .
A functional J (x, y) dependent on two elements x and y (lying in a certain linear
space) is called a bilinear functional if, for a fixed x, it is a linear functional of y and,
for a fixed y, it is a linear functional of x. Thus, the functional J (x, y) is bilinear if
J (α1 x1 + α2 x2 , y) = α1 J (x1 , y) + α2 J (x2 , y),
J (x, α1 y1 + α2 y2 ) = α1 J (x, y1 ) + α2 J (x, y2 ).
Putting y = x in a bilinear functional, we get the expresion J (x, x), which is
called a quadratic functional. A bilinear functional in a finite-dimensional space
is called a bilinear form. A quadratic functional J (x, x) is said to be positive definite
if J (x, x) > 0 for any nonzero element x.
Example 5.1. If A(t) is a fixed continuous function, then the expression
b
J (x, y) =
A(t)x(t)y(t)dt
a
5.1 Introduction
157
is a bilinear functional, and the expression
b
A(t)x 2 (t)dt
a
is a quadratic functional in the space C 0 [a, b].
If A(t) > 0 for all t ∈ [a, b], then this quadratic functional will be positive
definite.
Example 5.2. If A(t), B(t) and C(t) are fixed continuous functions, then the expression
b
A(t)x 2 (t) + B(t)x(t)x (t) + C(t)x 2 (t) dt
a
is an example of a quadratic functional defined for all functions in the space C 1 [a, b].
Now, we intend to introduce the second variation of a functional.
Definition 5.1.8 Let J (y) be a functional defined in some normed linear space. We
will say that the functional J (y) has a second variation if its increment
J = J (y + δ y) − J (y),
may be written in the form
J = L 1 (δ y) +
1
L 2 (δ y) + βδ y2 ,
2
where L 1 (δ y) is a linear functional, L 2 (δ y) is a quadratic functional, and β → 0 as
δ → 0.
We will call the quadratic functional L 2 (δ y) the second variation, or second
differential, of the functional J (y) and denote it by δ 2 J .
We can prove that if the second variation of a functional exists, then it is uniquely
defined.
Example 5.1. Let us compute the second variation of the following functional
1
J (y) =
x y 2 + y 2 dx.
0
Using the definition, we obtain
J = J (y + δ y) − J (y) =
158
5 Calculus of Variations
1
=
x(y + δ y)2 + (y + δ y )3 − x y 3 − y 3 dx =
0
1
=
2x yδ y + x(δ y)2 + 3y 2 δ y + 3y (δ y )2 + (δ y )3 dx =
0
1
=
2x yδ y + 3y 2 δ y dx +
0
1
1
2
x(δ y) + 3y (δ y ) dx +
2
0
δ y
3
dx.
0
For a fixed y(x), the first term in the right-hand member of the above relation is a
functional linear with respect to δ y(x); the second term of the right-hand member is
a quadratic functional. Finally, the third term of the right-hand member allows for
the obvious estimate
1
1
1
2 3 2
δ y dx,
δ y dx ≤ δ y
δ y dx ≤ max δ y
0
0
0
(the norm in the sense of the space C 1 [0, 1]), whence it is seen that this term can
be represented in the form β δ y2 , where β → 0 as δ y → 0. According to the
definition, the given functional has the second variation δ 2 J and it is equal to
1
δ J =2
2
x(δ y)2 + 3y (δ y )2 dx.
0
Let us state and demonstrate the necessary condition for the extremum of a functional.
Definition 5.1.9 We say that a functional J (y(x)) attains a maximum on a curve
y = y0 (x) if the values of the functional J (y(x)) on any curve close to y = y0 (x)
do not exceed J (y0 (x)), that is,
J = J (y(x)) − J (y0 (x)) ≤ 0.
If J ≤ 0 and J = 0 only when y(x) = y0 (x), then we say that a strict
maximum is attained on the curve y = y0 (x). The curve y = y0 (x) on which a
minimum is attained can be defined in a similar way. In this case J ≥ 0 on all
curves close to the curve y = y0 (x).
5.1 Introduction
159
Example 5.1. Let us show that the functional
1
J (y(x)) =
x 2 + y 2 dx
0
attains a strict minimum on the curve y(x) = 0.
Indeed, for any functions y(x) continuous on [0, 1] we have
1
J = J (y(x)) − J (0) =
1
x +y
2
0
2
dx −
1
x dx =
y 2 dx ≥ 0,
2
0
0
equality occuring only when y(x) = 0.
5.2 Euler’s Equation
Using a simplified language, the object of study for calculus of variations is to find
the minimum or maximum value of a functional. The results in this division of
mathematics are more general as in the classical theory of functions. But, many
results are equivalent with the corresponding result in the case of functions.
The problems of variational calculus are originated from physics, but in the last
decade the sphere of the problems contains other divisions of sciences.
It is considered that first problem of variational calculus is due to Jean Bernoulli
who has stated in 1696 the following problem.
Consider two points A and B which are not lying on the same vertical and, also,
not on the same horizontal. A power material point falls down on a curve with its
end in A and B. Determine the trajectory of the material point which starts from the
point A such that the time necessary to attain the point B is minimum.
This problem is called the brahistrocronic curve and has been solved after many
years, by Leibniz, Newton and Jacob Bernoulli.
In mechanics, the Hamilton’s principle asserts that a mechanical system chooses,
from all possible trajectories in a certain period of time, that trajectory on which the
mechanical work is minimum.
Let us specify other mathematical problems which need the techniques of variational calculus to find their solutions.
1. Geodesics of surfaces. Given is a surface in the Euclidean space and two
arbitrary points A and B lying in the surface. Find the arc of the curve contained in
the surface that connects A and B in such a way that its length is smallest.
2. Surface of minimum area. Given are two arbitrary points A and B. Find the
arc of the curve that connects A and B in such a way that the area obtained by rotating
this arc about the x-axis is smallest.
160
5 Calculus of Variations
3. Isoperimetric problem. Let A and B be two arbitrary points. In the collection
of the curves having the same length l that connect the points A and B, find one that
together with the segment AB closes a surface of maximum area.
In the following we specify the necessary notions to solve a problem of variational
calculus. We intend to find the extremum only for a functional of the integral type.
The set of definition for the functional will be the space C 1 [a, b], already defined as
C 1 [a, b] = y = y(x), y : [a, b] → R : y is of class C 1 .
This space is endowed by a structure of linear space with regard to the usual operations
of functions and with a structure of normed space with regard to the norm
y1 = max |y(x)| + max |y (x)|.
a≤x≤b
a≤x≤b
The integrant of the functional will be a function L that depends of three variables
and will be called the Lagrangean of the functional:
L : [a, b] × R × R → R, L = L(x, y(x), y (x)).
Now, we can define the functional, denoted by I , for that we will find the extreme
value (minimum or maximum):
b
I : C [a, b] → R, I (y) =
1
L(x, y(x), y (x))dx.
(5.2.1)
a
So, in short, a problem of variational calculus consists in finding of that function of
C 1 class that gives the extreme value (minimum or maximum) of the functional I .
Let us fixe the values of the function y at the ends of the interval, i.e. y(a) = ya
and y(b) = yb , where ya and ya are real known numbers.
In the following lemma we prove an auxiliary, but very important, result, called
the fundamental lemma of variational calculus.
Lemma 5.2.1 Consider y = y(x) a function such that y ∈ C 1 [a, b]. If
b
y(x)η(x)dx = 0
a
for any function η = η(x) such that η ∈ C 1 [a, b] and η(a) = η(b) = 0, then
y(x) = 0, ∀x ∈ [a, b].
Proof We suppose, on the contrary, that ∃x0 ∈ [a, b] such that y(x0 ) = 0. Without
loss of generality, we can suppose that y(x0 ) > 0. The proof is similarly if we should
that y(x0 ) < 0. Since the function y(x) is continuous, according to a classical
5.2 Euler’s Equation
161
theorem, the values of the function y(x) are strict positive in a whole vicinity of x0 ,
that is
y(x) > 0, ∀x ∈ [x0 − ε, x0 + ε] ,
where ε is a positive real number, arbitrarily small.
Because the integral of the hypothesis is null for any function η, we deduce that
statement is true also for a particular η. So, we define the function
⎧
⎨ 0, x ∈ [a, x0 − ε),
2
η0 (x) =
(x − x0 )2 − ε2 , x ∈ [x0 − ε, x0 + ε],
⎩
0, x ∈ (x0 + ε, b].
Clearly, we have (by definition) η0 (a) = η0 (b) = 0. Let us prove that η0 (x) ∈
C 1 [a, b]. It is easy to see that
lim η0 (x) = lim η0 (x) = η0 (x0 − ε),
xx0 −ε
xx0 −ε
such that η0 is a continuous function at the point x0 − ε. Similarly,
lim η0 (x) = lim η0 (x) = η0 (x0 + ε),
xx0 +ε
xx0 +ε
such that η0 is a continuous function at the point x0 + ε. By direct calculations, we
have
⎧
⎨ 0, x ∈ [a, x0 − ε),
η0 (x) = 4(x − x0 ) (x − x0 )2 − ε2 , x ∈ [x0 − ε, x0 + ε],
⎩
0, x ∈ (x0 + ε, b].
Then, we have
lim η0 (x) = lim η0 (x) = 0
xx0 −ε
and, also
xx0 −ε
lim η0 (x) = lim η0 (x) = 0
xx0 +ε
xx0 +ε
such that we conclude that η0 (x) ∈ C 1 [a, b]. Therefore, we must have
b
y(x)η0 (x)dx = 0 ⇒
a
x0 −ε
⇒
x0 +ε
y(x)η0 (x)dx +
a
b
y(x)η0 (x)dx +
x0 −ε
y(x)η0 (x)dx = 0 ⇒
x0 +ε
162
5 Calculus of Variations
x0 +ε
⇒
y(x) (x − x0 )2 − ε2
2
dx = 0.
x0 −ε
But, this is a contradiction, because the integrant of this integral is strict positive,
y(x) > 0, ∀x ∈ [x0 − ε, x0 + ε], (x − x0 )2 − ε2
2
> 0, ∀x ∈ [x0 − ε, x0 + ε].
So, the lemma is concluded.
In the following theorem we prove a necessary condition for the extremum of a
functional. This fundamental result is due to Cauchy and it is very important for the
whole variational calculus.
Theorem 5.2.1 If y(x) is the function where the functional I (y) (defined in Eq.
(5.2.1) attains its extreme value, then y(x) satisfies the following equation
∂L
d
−
∂y
dx
∂L
∂ y
= 0,
(5.2.2)
which is called the Euler’s equation.
Proof In order to evidentiate that the value of the functional I computed at y(x) is
the extremum, we consider a vicinity of first order of the function y(x)
{y(x) + εη(x)}ε , η(x) ∈ C 1 [a, b], η(a) = η(b) = 0.
The last conditions imposed to η(x) mean that any function from this vicinity has
the same ends like y(x).
Let us compute the value of the functional I at an arbitrary representative of this
vicinity:
b
I (y + εη) =
L(y(x) + εη(x), y (x) + εη (x).
(5.2.3)
a
The integral in the right-hand member is a function of ε, that is
I (y + εη) = F(ε).
But, for ε = 0 we obtain I (y) and this value has been supposed to be the extreme
value of the functional I . This means that ε = 0 satisfies the necessary condition for
the extremum of the function F, that is
d F(ε) =0⇒
dε ε=0
5.2 Euler’s Equation
163
b ⇒
a
∂L
∂L
η(x) +
η (x) dx ∂(y + εη)
∂(y + εη )
b ⇒
=0⇒
ε=0
∂L
∂L η(x) + η (x) dx = 0 ⇒
∂y
∂y
a
b
⇒
∂L
η(x)dx +
∂y
a
b
∂L η (x) = 0.
∂ y
(5.2.4)
a
We compute by parts the last integral
b
a
b b
∂L d ∂L
∂L
−
η
(x)
=
η(x)
η(x)dx.
∂ y
∂ y
dx ∂ y
a
a
Due to the conditions η(a) = η(b) = 0 we obtain
b
∂L η (x) = −
∂ y
a
b
d
dx
∂L
∂y
η(x)dx,
a
such that from Eq. (5.2.3) it follows
b
∂L
η(x)dx −
∂y
a
b
d
dx
∂L
∂y
η(x)dx = 0 ⇒
a
b ⇒
∂L
d
η(x) −
∂y
dx
∂L
∂ y
η(x)dx = 0.
a
Now, we can use the fundamental lemma whence it follows
∂L
d ∂L
= 0,
−
∂y
dx ∂ y that is, we obtained Eq. (5.2.2) and the theorem is demonstrated.
164
5 Calculus of Variations
Remark 5.1. Because L = L(x, y(x), y (x)), we have
d
dx
∂L
∂ y
=
∂2 L
∂ 2 L ∂ 2 L +
y + 2 y ,
∂ y ∂x
∂ y∂ y
∂y
and the Euler’s equation becomes
∂2 L
∂L
∂2 L
∂2 L
− − y − 2 y = 0.
∂y
∂ y ∂x
∂y ∂y
∂y
So, we can see that the Euler’s equation is an ordinary differential equation of second
order, such that its general solution depends on two arbitrary real constants. These
constants will be eliminated by using the conditions y(a) = ya and y(b) = yb ,
knowing the fact that the numbers ya and yb are prescribed.
Remark 5.2. We must outline that not any solution of the Euler’s equation is the
extremum value of the functional I . The Euler’s equation assures the contrary result:
if the function is an extreme value of the functional I , then this function satisfies the
Euler’s equation. As in the classical theory of functions, any solution of the Euler’s
equation will be called a stationary point of the functional I .
In order to obtain an effective extremum of the functional I we will give a sufficient
condition, as in the classical theory of functions.
In the following two propositions we obtain two prime integrals for the Euler’s
equation.
Proposition 5.2.1 If the Lagrangean of the functional I does not depend on the
function, then the Euler’s equation admits the following prime integral
∂L
= C,
∂ y
(5.2.5)
where C is an arbitrary constant.
Proof Since the Lagrangean of the functional I does not depend on the function, we
have
∂L
L = L(x, y (x)) ⇒
= 0,
∂y
such that the Euler’s equation reduces to
d
dx
∂L
∂ y
=0⇒
∂L
= C = constant,
∂ y
and the proof of the proposition is concluded.
Proposition 5.2.2 If the Lagrangean of the functional I does not explicitly depend
on the variable x, then the Euler’s equation admits the following prime integral
5.2 Euler’s Equation
165
L − y
∂L
= C,
∂ y
(5.2.6)
where C is an arbitrary constant.
Proof Since the Lagrangean of the functional I does not explicitly depend of variable
x, we have
∂L
L = L(y, y (x)) ⇒
= 0.
∂x
We will prove that the total derivative of the expression
L − y
∂L
∂ y
is null such that this expression will be a constant. By direct calculations, we have
d
dx
L − y
∂L
∂ y
=
∂L
∂ L ∂ L ∂L
d
+
y + y − y − y ∂x
∂y
∂y
∂y
dx
= y
d
dx
∂L
∂ y
−
∂L
∂ y
=
∂L
,
∂y
such that, by using the Euler’s equation, we get
d
dx
whence it follows
L − y
L − y
∂L
∂ y
= 0,
∂L
= C = constant,
∂ y
and the proof of the proposition is concluded.
Remark. Let us consider that the Lagrangean is a linear function of the derivative
of the unknown function, that is
L x, y(x), y (x) = A(x, y)y (x) + B(x, y),
where the functions A(x, y) and B(x, y) satisfy the condition
∂B
∂A
=
.
∂x
∂y
Then we obtain
∂L
∂ A ∂B ∂L
=
y +
,
= A,
∂y
∂y
∂ y ∂ y
166
5 Calculus of Variations
such that the Euler’s equation becomes
d
∂A ∂B
y +
−
(A) = 0 ⇒
∂y
∂y
dx
∂A ∂A ∂A ∂B
y +
−
−
y ≡ 0.
∂y
∂y
∂x
∂y
With other words, the Euler’s equation becomes an identity and it does not have
any solution. This fact can be explained as follows. The functional with the above
Lagrangean has the form
b I (y) =
dy
A(x, y)
+ B(x, y) dx =
dx
a
(b,y(b)
=
[A(x, y)dy + B(x, y)dx] dx.
(a,y(a)
Because of the condition
∂B
∂A
=
∂x
∂y
the above integral depends solely on the initial point (a, y(a)) and terminal point
(b, y(b)) and is independent of the shape of the curve.
Application 1. Let us solve the problem of the brahistocrone curve. Consider the
explicit form y = y(x) for the curve that connects the points O(0, 0) and A(x1 , y1 ).
We know the classical Newton’s law for a moving point (with known notations):
m r̈ = F, F = mg ⇒ m r̈ṙ = Fṙ ⇒ m v̇v = Fṙ ⇒
d
⇒
dt
mv2
2
mv2
= Fṙ ⇒ d
2
= Fṙ = Fdr.
But dr = dxi + dyj such that the last equation becomes
d
mv2
2
= gdy = d(mg y) ⇒
mv2
= mg y ⇒ v 2 = 2g y ⇒ v = 2g y.
2
On the other hand, we have
v=
ds
ds
⇒ dt =
=
dt
v
1 + y 2
,
√
2g y
5.2 Euler’s Equation
167
after we used that the initial
speed is null and the element of arc on the curve y = y(x)
has the formula ds = 1 + y 2 .
We integrate the last equation allong the curve y = y(x) between the points O
and A and obtain
x1 1 + y 2
dx.
t=
√
2g y
0
Of course, considering another curve between O and A we will obtain another value
of the time t, such that we must write
1
t (y) = √
2g
x1 1 + y 2
dx.
y
0
So, we must find the function y(x) for which the functional t (y) has the minimum
value. It is easy to see that
L=
1 + y 2
.
y
Because L = L(y(x), y (x), that is the Lagrangean does not depend explicitely on
x, we can use the prime integral, proved in the Proposition 5.2.2:
L − y
∂L
= C.
∂ y
So, we obtain the equation
y
1 + y 2
1
1 + y 2 − y 2
− y √ =C ⇒
=C ⇒ y
y 1 + y 2
y(1 + y 2 )
⇒
1
y(1 +
y 2 )
1
= C ⇒ y 1 + y 2 = 2 = C1 .
C
The last equation is a parametric equation. If we denote y =cotu, it results
1 + y 2 = 1 + cot 2 u =
⇒ y(u) = C1 sin2 u =
On the other hand, we have
1
⇒
sin2 u
C1
(1 − cos 2u).
2
(5.2.7)
168
5 Calculus of Variations
y =
such that
dy du
dx
dy 1
dy
dy
=
⇒
=
tan u,
=
dx
du dx
du
du y
du
dx
sin u
= C1 sin 2u
= 2C1 sin2 u = C1 (1 − cos 2u).
du
cos u
After a simple integration, we find
x(u) = C1 (u −
C1
sin 2u
=
(2u − sin 2u) + C2 .
2
2
Taking into account this relation together with Eq. (5.2.7), we have the parametric
form of the brahistocrone curve
x(u) = C21 (2u − sin 2u) + C2
(5.2.8)
y(u) = C1 sin2 u = C21 (1 − cos 2u).
The constants C1 and C2 will be determined using the condition that the curve passes
by the points O(0, 0) and A(x1 , y1 ).
In Eq. (5.2.8) we have a family of cycloides.
Application 2. Let us find the point of extreme for the following functional
I (y) =
ln 2
−x 2
e y − e x y 2 dx, y(0) = a, y(ln 2) = b,
0
where a and b are prescribed real numbers.
Using the Euler’s equation, we obtain
−2e x y −
d −x d −x 2e y = 0 ⇒
2e y + e x y = 0,
dx
dx
after we simplified by (−2). We use the rule of the derivation for the product and
multiply by e x :
−e−x y + e−x y + e x y = 0 .e x ⇒ y − y + e2x y = 0.
Using the change of variable e x = t we obtain
e x dx = dt ⇒
⇒ y =
dy dt
dt
dy
= ex = t ⇒ y =
=
= ẏt ⇒
dx
dx
dt dx
dy dt
d
dy =
=
( ẏt) = t ( ÿt + ẏ) ,
dx
dt dx
dt
5.2 Euler’s Equation
169
such that the Euler’s equation becomes
t 2 ÿ + t ẏ − t ẏ + t y = 0 ⇒ ÿ + y = 0.
Since the characteristic equation r 2 + 1 = 0 has the roots ±i we obtain the solution
y(t) = C1 cos t + C2 sin t ⇒ y(x) = C1 cos e x + C2 sin e x ,
where the constants C1 and C2 can be determined by using the initial conditions.
5.3 Generalizations of Euler’s Equation
In this paragraph we will prove a few generalizations of the Euler’s equation, with
regard to a number of functions which appear under the Lagrangean, the order of the
derivative of the function y(x) which appears under the Lagrangean of the functional
and with regard to the number of variables of the unkown function.
In the first following theorem, called the Lagrange-Euler’s system of equations,
we extend the fundamental result of Euler to the case where the functional (also, the
Lagrangean) depends of many unknown functions. Let us consider that the functional
depends on n unknown functions, each depends on a single variable x, that is,
b
I (y1 , y2 , . . . , yn ) =
L x, y1 (x), y2 (x), . . . , yn (x), y1 (x), y2 (x), . . . , yn (x) dx,
a
where yi = yi (x), i = 1, 2, . . . , n.
Theorem 5.3.1 Suppose that the functional I attains its extreme value in the functions (y1 , y2 , . . . , yn ). Then these functions satisfy the following system of equations
∂L
d
−
∂ y1
dx
∂L
d
−
∂ y2
dx
∂L
∂ y1
∂L
∂ y2
= 0,
=0
− − − − − − − − −−
∂L
∂L
d
= 0.
−
∂ yn
dx ∂ yn
(5.3.1)
170
5 Calculus of Variations
Proof Without loss of generality, we suppose that n = 2. Therefore, consider the
functional
b
I (y, z) = L x, y(x), z(x), y (x), z (x) dx,
a
where y = y(x) and z = z(x).
Supposing that the functional attains its extreme value in the functions (y(x), z(x)),
we must prove that these functions satisfy the following system of equations
d
∂L
−
∂y
dx
∂L
d
−
∂z
dx
∂L
∂ y
∂L
∂z = 0,
= 0.
Consider two vicinities of the functions y(x) and z(x), respectively,
{y(x) + ε1 η1 (x)}ε1 , {z(x) + ε2 η2 (x)}ε2 ,
where η1 (x) and η2 (x) are functions of class C 1 on the interval [a, b] and satisfy the
conditions
η1 (a) = η1 (b) = 0, η2 (a) = η2 (b) = 0.
We will compute the value of the functional I for an arbitrary representative of each
vicinity:
I (y + ε1 η1 , z + ε2 η2 ) =
b
=
L x, y(x) + ε1 η1 (x), z(x) + ε2 η2 (x), y (x) + ε1 η1 (x), z (x) + ε2 η2 (x) dx.
a
Of course, we can write
I (y + ε1 η1 , z + ε2 η2 ) = I (ε1 , ε2 ) .
For ε1 = ε2 = 0 the function y(x) + ε1 η1 (x) from first vicinity becomes y(x) and
the function z(x) + ε2 η2 (x) from second vicinity becomes z(x) and the functional I
has its extreme value for the functions y(x) and z(x). So, we can conclude that the
point (0, 0) is the point of the extreme for the function I (ε1 , ε2 ) and must satisfy the
necessary condition of extreme, that is
∂ I (ε1 , ε2 ) ∂ε
1
ε1 =0
∂ I (ε1 , ε2 ) = 0,
∂ε
2
ε2 =0
= 0.
5.3 Generalizations of Euler’s Equation
171
Now, we derivate with respect to ε1 and ε2 the integral that defines the function
I (ε1 , ε2 ):
b a
b a
∂L
∂L
η1 (x) +
η1 (x) dx ∂(y + ε1 η1
∂(y + ε1 η1
= 0,
ε1 =0
∂L
∂L
η2 (x) +
η2 (x) dx ∂(y + ε2 η2
∂(y + ε2 η2
= 0.
ε2 =0
Taking into account that ε1 = 0 and ε2 = 0, we obtain
b b
b
∂L
∂L
∂L ∂L
η1 (x) + η1 (x) dx = 0 ⇒
η1 (x)dx +
η (x)dx = 0,
∂y
∂y
∂y
∂ y 1
a
a
a
and
b b
b
∂L
∂L
∂L ∂L η (x)dx = 0.
η2 (x) + η2 (x) dx = 0 ⇒
η2 (x)dx +
∂z
∂z
∂z
∂z 2
a
a
a
We integrate by parts the last integrals of the above relations. Thus,
b
a
and
b
a
b
∂L d ∂L
∂ L b
η1 (x)dx
η (x)dx =
η1 −
∂ y 1
∂ y a
dx ∂ y a
b
∂L d ∂L
∂ L b
η2 (x)dx.
η
(x)dx
=
η
−
2
∂z 2
∂z a
dx ∂z a
But, by hypothesis, we have
η1 (a) = η1 (b) = 0, η2 (a) = η2 (b) = 0,
and then the integrals become
b
a
∂L η (x)dx = −
∂ y 1
b
a
d
dx
∂L
∂ y
η1 (x)dx
172
5 Calculus of Variations
b
∂L η (x)dx = −
∂z 2
a
b
d
dx
∂L
∂z η2 (x)dx.
a
Finally, the conditions of extreme become
b ∂L
d
−
∂y
dx
∂L
∂ y
η1 (x)dx = 0,
a
and, respectively,
b ∂L
d
−
∂z
dx
∂L
∂z η2 (x)dx = 0.
a
For both integrals in the left-hand member of these relations, we can use the fundamental lemma whence it follows
d ∂L
∂L
= 0,
−
∂y
dx ∂ y ∂L
d
−
∂z
dx
∂L
∂z =0
and the theorem is concluded.
Application 1. Let us find the functions y(x) and z(x) that make the following
integral extremal
I (y, z) =
π
0
2yz − 2y 2 + y 2 − z 2 dx, y(0) = 0, y(π) = 1, z(0) = 0, z(π) = 1.
Taking into account that the Lagrangean is
L x, y(x), z(x), y (x), z (x) = 2yz − 2y 2 + y 2 − z 2 ,
the Euler-Lagrange’s system of equations becomes
d
∂L
−
∂y
dx
2z − 4y −
∂L
∂ y
= 0,
∂L
d
−
∂z
dx
∂L
∂z =0⇒
d d 2y = 0, 2y −
−2z = 0, ⇒
dx
dx
y + 2y − z = 0, z + y = 0, ⇒
5.3 Generalizations of Euler’s Equation
173
y (4) + 2y − z = 0, z = −y ⇒ y (4) + 2y + y = 0.
Since the characteristic equation
r 4 + 2r 2 + 1 = 0,
has the double complex conjugated roots ±i, our differential equation has the general
solution equal to
y(x) = (Ax + B) cos x + (C x + D) sin x.
Using the initial conditions y(0) = 0 and y(π) = 1, we find B = 0 and A = −1/π.
Then the function z(x) can be obtained by using the equation z(x) = y (x) + 2y(x)
and the initial conditions z(0) = 0, z(π) = 1. Finally, we obtain the solutions
y(x) = −
z(x) =
x
cos x + D sin x
π
1
(2 sin x − π cos x) + D sin x,
π
where D is an arbitrary constant.
Another generalization of the Euler’s equation can be obtained considering the
case when the functional I depends on only one function y, but the Lagrangean of
the functional depends on superior derivatives of the function y. The result is due
to Poisson and Euler and is proved in the following theorem. Based on the above
considerations, we will consider the functional I in the form:
b
I (y) =
L x, y(x), y (x), y (x), . . . , y (n) (x) dx.
a
Theorem 5.3.2 If the function y(x) is the function where the functional I , for which
the Lagrangean depends on superior derivatives of y(x), attains its extreme value,
then y(x) satisfies the following equation
d
∂L
−
∂y
dx
∂L
∂ y
+
d2
dx 2
∂L
∂ y + · · · + (−1)n
dn
dx n
∂L
∂ y (n)
= 0,
called the Poisson-Euler’s equation.
Proof For the sake of simplicity of calculations, but without loss of generality, we
consider only the case n = 2. Consequently, the Lagrangean depends, more, on
y (x), that is,
174
5 Calculus of Variations
b
I (y) =
L x, y(x), y (x), y (x) dx.
a
Let us prove that the function y(x) satisfies the equation
∂L
d
−
∂y
dx
∂L
∂ y
d2
+ 2
dx
∂L
∂ y = 0.
Together with function y(x) consider a vicinity of order two that contains functions
of the form
{y(x) + εη(x)}ε ,
where ε is a small parameter and η(x) is a function of class C 2 on the interval [a, b]
satisfying the conditions
η(a) = η(b) = 0, and η (a) = η (b) = 0.
In order to evidentiate that the value of the functional I is extreme for the function
y(x), we compute the value of I for an arbitrary representative of the above vicinity:
b
I (y + εη) =
L x, y(x) + εη(x), y (x) + εη (x), y (x) + εη (x) dx.
a
So, we obtain
I (y + εη) = I (ε).
But for ε = 0 the representative of the vicinity becomes even y(x) and the functional
I attains its extreme value at y(x). So, we conclude that ε = 0 is the point of extreme
for function I (ε) and then, it must satisfy the necessary condition of extreme, that is
d I (ε) = 0.
dε ε=0
We introduce the derivative under the integral because the ends of the integral do not
depend on ε:
b a
∂L
∂L
∂L
η(x) +
η
η
(x)
+
(x)
dx
∂(y + εη)
∂(y + εη )
∂(y + εη )
Using the fact that ε = 0, we obtain
= 0.
ε=0
5.3 Generalizations of Euler’s Equation
b 175
∂L
∂L
∂L
η(x) + η (x) + η (x) dx = 0.
∂y
∂y
∂y
a
Now, we decompose the integral from the left-hand member in three integrals and
compute the last two by parts:
b
∂L
η(x)dx +
∂y
a
b
∂L η (x)dx +
∂ y
a
b
∂ L η (x)dx = I1 + I2 + I3 = 0. (5.3.2)
∂ y a
Integrating by parts the second integral from Eq. (5.3.2), I2 , we obtain
b
I2 =
a
b
∂L d ∂L
∂ L b
η(x)dx.
η
(x)dx
=
η
−
∂ y
∂ y a
dx ∂ y a
Taking into account that η(a) = η(b) = 0, it results
b
I2 =
d
dx
∂L
∂ y
η(x)dx.
a
Now, we integrate by parts the third integral from Eq. (5.3.2):
b
I3 =
a
b
∂L
∂ L d
∂ L b
η (x)dx.
η
(x)dx
=
η
−
∂ y ∂ y a
dx ∂ y a
Taking into account that η (a) = η (b) = 0, it results
b
I3 =
d
dx
∂L
∂ y η (x)dx.
a
Once again, we integrate also by parts
d
I3 = −
dx
∂L
∂ y b b 2 ∂L
d
η +
η(x)dx.
dx 2 ∂ y a
a
But, by hypothesis, we have η(a) = η(b) = 0, and then the integral I3 becomes:
176
5 Calculus of Variations
b
I3 =
d2
dx 2
∂L
∂ y η(x)dx.
a
With the founded form for I2 and I3 , the relation (5.3.2) becomes
b
∂L
η(x)dx −
∂y
b
a
d
dx
∂L
∂ y
b
η(x)dx +
a
b ⇒
d2
dx 2
∂L
∂ y η(x)dx = 0 ⇒
a
∂L
d
−
∂y
dx
∂L
∂ y
d2
η(x) + 2
dx
∂L
∂ y η(x)dx = 0.
a
Now, we can apply the fundamental lemma and the previous relation leads to the
equation
∂L
d2
∂L
d ∂L
+ 2
= 0,
−
∂y
dx ∂ y dx
∂ y which is the Poisson-Euler’s equation and the theorem is proved.
Application 1. As an application of the Poisson-Euler’s equation we will find the
function that makes the following functional extremal
e
I (y) =
x 2 y − 2x 2 y 2 dx, y(1) = 0, y(e) = 1.
1
Taking into account that the Lagrangean of the functional is
L x, y(x), y (x), y (x) = x 2 y − 2x 2 y 2 ,
we obtain
∂L
∂L
∂L
= 0,
= −4x 2 y ,
= x 2,
∂y
∂y
∂ y and then the Poisson-Euler’s equation reduces to
−
d d2 −4x 2 y + 2 x 2 = 0 ⇒ 4 4x 2 y + 2x y + 2 = 0 ⇒
dx
dx
⇒ 2x 2 y + 4x y + 1 = 0.
If we denote by y (x) = z(x), the previous equation reduces to
5.3 Generalizations of Euler’s Equation
177
2x 2 z + 4x z + 1 = 0 ⇒ z +
1
2
z = − 2.
x
2x
The last equation is an ordinary linear differential equation of first order such that
its solution has the form
1 2 ln x
1 2
1
C
1
−2 ln x
C−
,
z(x) = e
e
dx = 2 C −
x dx = 2 −
2x 2
x
2x 2
x
2x
where C is an arbitrary constant. Taking into account the above notation, with regard
to the function y(x) we obtain
y(x) = −
C
1
− ln x + C1 .
x
2
In order to obtain the constants C and C1 we will use the initial conditions y(1) = 0
and y(e) = 1 such that
3e
.
C = C1 =
2(e − 1)
Therefore, the function that makes the given functional extremal is
y(x) =
3e
1
1
1−
− ln x.
2(e − 1)
x
2
The last generalization of the Euler’s equation that we will obtain is regarding to the
number of the independent variables. We will denote by u the unknown function and
suppose that it is dependent on n independent variables, that is,
u = u(x1 , x2 , . . . , xn ).
Also, we suppose that the functional depends only on one unknown function, u, and
the Lagrangean is dependent only on the function u and its first partial derivative:
I (u) =
∂u
L x, u(x),
∂xi
dx,
where x is a vectorial variable x = (x1 , x2 , . . . , xn ) and is a domain in a ndimensional space, as such, we have a multiple integral.
For the sake of simplicity of calculations, but without loss the generality, we
consider only the case n = 2. Consequently, the unknown function u depends only
on two variables, namely, x and y, the Lagrangean depends on the unknown function
u and on the partial derivatives ∂u/∂x and ∂u/∂ y. As such, we have the following
functional
178
5 Calculus of Variations
I (u) =
∂u ∂u
,
dxdy.
L x, y, u(x, y),
∂x ∂ y
The following theorem, due to Ostrogradski and Euler, gives the equation verified
by the function u that makes the above functional extremal.
Let be a bounded domain in the two-dimensional space R 2 having the smooth
boundary = ∂. Consider that the unknown function u and the Lagrangean L are
regular functions, u, L ∈ C 1 () and use the well known Monge’s notations
ux =
∂u
∂u
, uy =
.
∂x
∂y
Consequently, we must make the following functional extremal
I (u) =
L x, y, u(x, y), u x , u y dxdy.
(5.3.3)
Theorem 5.3.3 If the function u extremates the functional I from Eq. (5.3.3) then it
satisfies the following equation
∂L
∂
−
∂y
∂x
∂L
∂u x
∂
−
∂y
∂L
∂u y
= 0,
which is called the Ostrogradski-Euler’s equation.
Proof Together with the function u we consider a vicinity of first order containing
functions of the form
{u(x, y) + εη(x, y)}ε ,
where ε is a small arbitrary parameter and η(x, y) is a function of the class η(x, y) ∈
C 1 () satisfying the condition
η(x, y) = 0, ∀(x, y) ∈ = ∂.
(5.3.4)
The last conditions means that every function from the vicinity has the same ends
like u(x, u).
Let us compute the value of the functional I for an arbitrary representative of this
vicinity.
I (u +εη) =
L x, y, u(x, y)+εη(x, y), u x (x, y)+εηx (x, y), u y (x, y)+εη y (x, y) dxdy.
We obtain a function that depends only on ε, I (ε). For ε = 0 the function from the
vicinity, u(x, y) + εη(x, y) reduces to the function u(x, y) which is the point of the
5.3 Generalizations of Euler’s Equation
179
extreme value for the functional I such that we deduce that ε = 0 must satisfy the
necessary condition for the extremum, that is
d I (ε) = 0.
dε ε=0
Introducing the derivative under the integral, it follows
∂L
∂L
∂L
η(x, y)+
ηx (x, y)+
η y (x, y) dxdy ∂(u +εη)
∂(u x + εηx )
∂(u y +εη y )
= 0.
ε=0
If we take into account the fact that ε = 0, the previous relation becomes
∂L
∂L
∂L
η(x, y) +
ηx (x, y) +
η y (x, y) dxdy = 0.
∂u
∂u x
∂u y
(5.3.5)
Now, we decompose the integral in three integrals and compute by parts the last
two
∂L
∂L
∂L
η(x, y)dxdy +
ηx (x, y)dxdy +
η y (x, y)dxdy =
∂u
∂u x
∂u y
= I1 + I2 + I3 = 0.
With regard to I2 we have the following estimations
I2 =
∂L
ηx (x, y)dxdy =
∂u x
∂
∂x
∂L
∂L
∂
dxdy
η −η
∂u x
∂x ∂u x
and then we can write
∂L
∂L
∂
∂
dxdy.
I2 =
η dxdy −
η
∂x ∂u x
∂x ∂u x
(5.3.6)
Also, with regard to I3 we have the following estimations
I3 =
∂L
η y (x, y)dxdy =
∂u y
and then we can write
∂
∂y
∂L
∂L
∂
dxdy
η −η
∂u y
∂ y ∂u y
180
5 Calculus of Variations
I3 =
∂
∂y
∂L
∂L
∂
dxdy.
η dxdy −
η
∂u y
∂ y ∂u y
(5.3.7)
We remember now the well known Green’s formula, with regard to the connection
between the line integral and double integral. So, if is a bounded domain in the
space R 2 with the boundary which is a closed smooth curve, then the Green’s
formula asserts the following connection
P(x, y)dx + Q(x, y)dy =
∂Q
∂P
−
∂x
∂x
dxdy.
With this formula in mind, we add, member by member, the relations (5.3.6) and
(5.3.7)
I2+I3 =
∂L
∂L
∂L
∂
∂
∂L
η
η
+
dxdy−
ηx +
η y dxdy =
∂u x
∂u y
∂x
∂u x
∂y
∂u y
∂L
∂L
∂
∂
−
η
+η
dxdy.
∂x ∂u x
∂ y ∂u y
On the first integral in the right-hand member of the previous relation we apply the
Green’s formula
∂
∂
∂L
∂L
∂L
∂L
+
dxdy = −η
η
η
dx + η
dy = 0,
∂x
∂u x
∂y
∂u y
∂u y
∂u x
where we used the condition (5.3.4). In this manner, the sum I2 + I3 reduces to
∂
∂
∂L
∂L
I2 + I3 =
η
+η
dxdy.
∂x ∂u x
∂ y ∂u y
Introducing this results in Eq. (5.3.5), we obtain
∂L
∂
−
∂u
∂x
∂L
∂u x
−
∂
∂y
∂L
∂u y
η(x, y)dxdy = 0.
Now, we can use the fundamental lemma such that the previous relation leads to
∂L
∂
−
∂u
∂x
∂L
∂u x
−
∂
∂y
∂L
∂u y
= 0,
5.3 Generalizations of Euler’s Equation
181
which is the Ostrogradski-Euler’s equation, that concludes the proof is
concluded.
Application 1. Let us use the Ostrogradski-Euler’s equation to determine the function
that makes the following functional extremal
I (u) =
∂u
∂x
2
+
∂u
∂y
2 dxdy,
where u = u(x, y), u ∈ C 2 () and is bounded domain in the Euclidean space
R 2 . The Lagrangean of our functional is
∂u ∂u
,
L x, y, u(x, y),
∂x ∂ y
Then
=
∂u
∂x
2
+
∂u
∂y
2
.
∂L
∂u
∂L
∂u
∂L
= 0,
= ux ,
= uy,
=2
=2
∂u
∂u x
∂x
∂u y
∂y
such that the Ostrogradski-Euler’s equation becomes
∂2u
∂2u
∂ ∂
uy = 0 ⇒
+
= 0.
(u x ) +
∂x
∂y
∂x 2
∂ y2
Thus, the function u(x, y) satisfies the Laplace’s equation and, as we know, is called
a harmonic function. It is known that in the Euclidean space R 2 a harmonic function
has the form
1
,
u(x, y) = C ln 2
x + y2
where C is an arbitrary constant. In the Euclidean space R n , n ≥ 3 a harmonic
function has the form
1
u(x) = C ,
r
where x is a vectorial variable, x = (x1 , x2 , . . . , xn ), C is an arbitrary constant and
r is the Euclidean distance, that is,
n
r =
xk2
k=1
182
5 Calculus of Variations
5.4 Sufficent Conditions for Extremum
As we alreay said, a function that satisfies the Euler’s equation is not automatically an
extremal value of the functional. This is only the necessary condition for extremum.
By analogy with the classical theory of functions, to a function that satisfies the
Euler’s equation must impose supplementary conditions to be effective an extremum
of the functional. In the case of the functionals these sufficient conditions will be
obtained with the aid of variation of order two of the functional. Firstly, in the
following theorem, we will obtain the form of variation of order two of a functional
of integral type.
Theorem 5.4.1 Let y(x) be a function that satisfies the Euler’s equation regarding
the functional
b
I (y) = L x, y(x), y (x) dx.
a
Consider a vicinity of the function y(x) which consists of the functions
{y(x) + εη(x)}ε ,
where ε is a small parameter and the functions η(x) satisfy the usual conditions
η ∈ C 1 [a, b], η(a) = 0 and η(a) = 0. If the functional is computed for a function
y(x)εη(x), belonging to the above vicinity, then its variation of order two has the
expression
b 2
∂ L 2
δ 2 I =
η (x)dx.
2
dε ε=0
∂ y 2
a
Proof By direct calculations, we obtain
⎡ b
⎤
δ 2 I (ε) δ2 ⎣
= 2
L x, y(x) + εη(x), y (x) + εη (x) dx ⎦
δε2 ε=0
δε
a
⎤
⎡ b
δ ⎣
∂L
∂L
=
η(x) +
η (x) dx ⎦
δε
∂(y + εη)
∂(y + εη )
a
b =
a
=
ε=0
=
ε=0
2L
2L
∂2 L
∂
∂
2 (x)+2
(x)+
2 (x) dx η(x)η
η
η
∂(y +εη)∂(y +εη )
∂(y +εη)2
∂(y +εη )2
Taking into account that ε = 0, we obtain
ε=0
5.4 Sufficent Conditions for Extremum
183
b 2
∂ L 2
δ 2 I (ε) ∂2 L
∂ 2 L 2
=
η
(x)
+
2
η(x)η
(x)
+
η
(x)
dx,
δε2 ε=0
∂ y2
∂ y∂ y ∂ y 2
a
such that the variation of order two the functional can be restated as follows
b 2
b 2
δ 2 I (ε) ∂ L 2
∂ L
=
η
(x)dx
+
2
η(x)η (x)dx+
δε2 ε=0
∂ y2
∂ y∂ y a
a
b
+
∂ 2 L 2
η (x)dx = I1 + I2 + I3 .
∂ y 2
(5.4.1)
a
We integrate by parts the second integral from Eq. (5.4.1), that is, I2 :
b
a
b b
2 ∂ L
∂2 L
d
∂ 2 L 2 η 2 (x)dx.
η(x)2η
(x)dx
=
η
(x)
−
∂ y∂ y ∂ y∂ y dx
∂
y∂
y
a
a
But η(a) = η(b) = 0 and then I2 becomes
b
I2 = −
d
dx
∂2 L
∂ y∂ y η 2 (x)dx.
a
Introducing this form of I2 in Eq. (5.4.1) it follows
2 !
b 2
∂ L
∂ L
δ 2 I (ε) d
∂ 2 L 2
2
η
=
−
(x)
+
η
(x)
dx.
δε2 ε=0
∂ y2
dx ∂ y∂ y ∂ y 2
a
Let us prove that the first part of the last integral is equal to zero. Starting from the
Euler’s equation, by using the derivative with respect to y, it follows
d
∂L
−
∂y
dx
d
∂2 L
−
⇒
∂ y2
dx
∂L
∂ y
= 0 ⇒
∂2 L
∂ y∂ y y
= 0.
Finally, the second variation of the functional reduces to
184
5 Calculus of Variations
b 2
δ 2 I (ε) ∂ L 2
=
η (x)dx,
2
δε
∂ y 2
ε=0
a
such that the desired result is concluded.
Using the second variation of the functional, we will find a sufficient condition
for the extremum.
In the following theorem we prove the sufficient condition that must satisfy the
function y(x), which verifies the Euler’s equation, to be effectively, the point of
minimum for the functional I . The result is due to Legendre.
Theorem 5.4.2 If the second variation of the functional I is positive, for every
function η(x) such that η(x) ∈ C 1 [a, b], η(a) = η(b) = 0, then
∂2 L
≥ 0.
∂ y 2
Proof According to the hypothesis, we have
δ 2 I (ε) ≥0⇒
δε2 ε=0
b
⇒
∂ 2 L 2
η (x)dx ≥ 0.
∂ y 2
a
Suppose that there exists a point x0 ∈ [a, b] such that
∂2 L
< 0.
∂ y 2
Since L ∈ C 2 , we deduce that
∂2 L
∈ C 0,
∂ y 2
that is, it is a continuous function. Therefore,
∂2 L
∂ y 2
is negative on the whole vicinity of the point x0 . Thus, we can write
α2
∂2 L
, ∀x ∈ [x0 − ε, x0 + ε],
<
−
∂ y 2
2
5.4 Sufficent Conditions for Extremum
185
where ε is a small parameter. The inequality from the hypothesis takes place for any
function η(x) that satisfies the above conditions. That means, the given inequality
takes place for a particular η(x), say η0 (x) that we indicate as follows
⎧
⎨ 0, if x ∈ [a, x0 − ε),
η0 (x) = sin2 (x 2 − ε2 )/ε2 , if x ∈ [x0 − ε, x0 + ε],
⎩
0, if x ∈ (x0 + ε, b].
It is easy to see that, by definition, η0 (a) = η0 (b) = 0.
By direct calculations, we obtain
lim η0 (x) = lim η0 (x) = 0,
xx0 −ε
xx0 −ε
and, similarly,
lim η0 (x) = lim η0 (x) = 0.
xx0 +ε
xx0 +ε
That is, the function η0 (x) is continuous at the points x0 ± ε. Using the Lagrange’s
consequence, one can prove that the function η0 (x) has derivatives at the points x0 ±ε,
as follows
lim η0 (x) = lim η0 (x) = 0,
xx0 −ε
and, similarly,
xx0 −ε
lim η0 (x) = lim η0 (x) = 0.
xx0 +ε
xx0 +ε
These calculations prove that the function η0 (x) ∈ C 1 [a, b]. Taking into account
the above initial conditions satisfied by the function η0 (x) we conclude that η0 (x)
satisfies all conditions from the hypotheses of our theorem. Therefore, we must have
b
∂ 2 L 2
η (x)dx ≥ 0,
∂ y 2
a
the integral being computed for the function y(x) + εη0 (x) belonging to the considered vicinity of the function y(x). We will prove that it is a contradiction.
On the other hand, we have
d
∂2 L
−
2
∂y
dx
∂2 L
∂ y∂ y ∈ C 0.
Also, this function is defined on the closed (thus, bounded) interval [a, b], such that
by using the classical Weierstrass’s theorem we deduce that it is bounded on the
interval [a, b] and there exists M, defined as follows
186
5 Calculus of Variations
2
2 ∂ L
∂ L d
.
M = sup 2 −
dx ∂ y∂ y a≤x≤b ∂ y
Now, we can contradict the hypothesis
δ 2 I (ε) δε2 ε=0
≥ 0.
Indeed, using the above considerations, we obtain
2
δ I (ε) δε2 b 2
2 ! 2
∂
∂
L
L
L
d
∂
=
η 2 (x) + 2 η 2 (x) dx ≤
−
2
∂y
dx ∂ y∂ y
∂y
ε=0
a
b
≤M
x 2 − ε2
α2
sin
dx
−
ε2
2
b
4
a
4 sin2
2
2
2
x 2 − ε2
2 x − ε 4x
cos
dx ≤
ε2
ε2
ε2
a
x0 +ε
≤
x0 −ε
8α2
dx − 2
ε
x0 +ε
x 2 dx = 2Mε − 64α2
x0 −ε
x02
.
ε
Thus, considering, as usual, ε sufficient small, the second variation of the functional,
δ 2 I (ε)
δε2
becomes negative that contradicts the hypothesis.
The theorem is proved.
Application. Using the Legendre’s result, let us find the effective point of minimum
for the following functional
π/2
I (y) =
0
π
y 2 − y 2 dx, y(0) = y( ) = 1.
2
Using the Euler’s equation, we obtain the following differential equation
y + y = 0.
Because the characteristic equation r 2 + 1 = 0 has the complex conjugated roots ±i
we deduce that the solution of the differential equation is
y(x) = A cos x + B sin x.
5.4 Sufficent Conditions for Extremum
187
Using the initial condition y(0) = y( π2 ) = 1, it follows A = 1 and B = 1 such that
y(x) = cos x + sin x.
We will write the Lagrangean of our functional as a function of y (x). By direct
calculations
y (x) = cos x − sin x ⇒ y 2 (x) = 1 − sin 2x.
The Lagrangean becomes
L x, y(x), y (x) = y 2 (x) − y 2 (x) = −2 sin 2x.
But sin 2x = 1 − 2y (x) such that we have
∂L
∂2 L
L x, y, y = 2y 2 − 2 ⇒
=
4y
⇒
= 4 > 0.
∂ y
∂ y 2
According to the Legendre’s result we are led to the conclusion that the function
y(x) = cos x + sin x
is an effective point of minimum for the given functional.
5.5 Isoperimetric Problems
There exists certain problems of variational calculus for which the function, that
must extremate a functional, is subjected to some restrictions. We will call such a
problem as a problem of conditional extremum, or, an isoperimetric problem. From
among all kind of restrictions that can be imposed, we use the following one: Define
a new functional, say J (y(x)), with other Lagrangean, say M(x, y(x), y (x)), and
consider only the functions y(x) along which the new functional assumes a given
value l.
Thus, together with the functional
b
I (y) =
L x, y(x), y (x) dx
a
we consider the new functional
b
J (y) =
a
M x, y(x), y (x) dx
188
5 Calculus of Variations
and an isoperimetric problem can be formulated as follows:
From among all curves y = y(x) ∈ C 1 [a, b] along which the functional J (y)
assumes a given value l, determine the one for which the functional I (y) assumes
an extremal value.
With regard to the Lagrangeans L amd M we assume that they have continuous
first and second partial derivatives for a ≤ x ≤ b and for arbitrary values of y(x)
and y (x).
A well known isoperimetric problem is the Dido’s problem, called also, the
Fisher’s problem. Among closed curves of length l, find the one that bounds the
largest area. In this case, the Lagrangeans L and M are
L x, y(x), y (x) = y(x), M x, y(x), y (x) = 1 + y 2 (x).
Consequently, we must find the curve y = y(x) along which the functional
J (y) =
b 1 + y 2 (x)dx
a
assumes a given value l (the length of the trawl!) and for which the functional
b
I (y) =
y(x)dx
a
assumes an extremal value.
We turn to the general isoperimetric problem and prove the main result in this
context, due to Euler.
Theorem 5.5.1 If a curve y = y(x) extremizes the functional
b
I (y) =
L x, y(x), y (x) dx
a
subject to the conditions
b
J (y) =
M x, y(x), y (x) dx = l, y(a) = ya , y(b) = yb
a
and y = y(x) is not an extremal of the functional J , then there exists a constant λ
such that the curve y = y(x) is an extremal of the functional
5.5 Isoperimetric Problems
I˜(y) =
b
189
L x, y(x), y (x) − λM x, y(x), y (x) dx.
a
Proof Together with the function y(x) consider a vicinity of functions of the form
{y(x) + αη(x) + βγ(x)}α,β .
Every function from this vicinity has the same ends as y(x), that is,
η(a) = η(b) = 0, γ(a) = γ(b) = 0.
If we compute the value of the functional I among one arbitrary representative of
this vicinity, we find a function which depends of α and β:
I (y(x) + αη(x) + βγ(x)) =
b
=
L x, y(x) + αη(x) + βγ(x), y (x) + αη (x) + βγ (x) dx = I (α, β).
a
But α and β are not independent, because
J (y(x) + αη(x) + βγ(x)) =
b
=
M x, y(x) + αη(x) + βγ(x), y (x) + αη (x) + βγ (x) dx = J (α, β).
a
Thus,
J (α, β) = l.
If we assume that J depends on β, we can use the theorem of the implicite functions
such that we obtain the following three statements:
(1) β can be expressed as a function of α, i.e. β = β(α);
(2) if α = 0 then β = 0, that is β(0) = 0;
(3) we can compute the derivative of β as follows
β (α) =
∂ J/∂α
dβ
=−
.
dα
∂ J/∂β
190
5 Calculus of Variations
For α = 0, as such, β = 0, the representative of the vicinity reduces to the curve
y(x) that extremizes the functional I . That means that α = 0 is the extremal of
the function I (α, β) = I (α, β(α)), and, according to the classical condition of the
extremum, we have
d I = 0. ⇒
dα α=0
⇒
b ⇒
a
∂ I dβ
∂I
+
∂α ∂β dα
α=0
=0⇒
∂L
∂L
η(x) +
η (x) dx ∂(y + αη + βγ)
∂(y + αη + βγ )
b ∂L
∂L
dβ +
γ(x) +
γ
(x)
dx
∂(y + αη + βγ)
∂(y + αγ + βγ )
dα a
b ⇒
+
α=0
=0⇒
α=0
b ∂L
∂L
∂L ∂L dβ
η(x) + η (x) dx +
γ(x) + γ (x) dx
=0
∂y
∂y
∂y
∂y
dα
a
a
b
⇒
∂L
η(x)dx +
∂y
a
b
∂L η (x)dx+
∂ y
a
⎛ b
⎞
b
dβ
∂L
∂L γ(x)dx
=0
+⎝
γ (x)dx ⎠
∂y
∂ y
dα
a
a
Integrating by parts, we obtain
b
a
b
∂L d ∂L
∂ L b
η(x)dx,
η
(x)dx
=
η
−
∂ y
∂ y a
dx ∂ y a
and, since η(a) = η(b) = 0, it results
b
a
∂L η (x)dx = −
∂ y
b
a
d
dx
∂L
∂ y
η(x)dx.
(5.5.1)
5.5 Isoperimetric Problems
Similarly,
b
a
191
b
∂L d ∂L
∂ L b
γ(x)dx,
γ
(x)dx
=
γ
−
∂ y
∂ y a
dx ∂ y a
and, since γ(a) = γ(b) = 0, it results
b
∂L γ (x)dx = −
∂ y
b
a
d
dx
∂L
∂ y
γ(x)dx.
a
Taking into account these estimations, the conditions of extremum (5.5.1) become
b
a
∂L
η(x)dx −
∂y
b ⇒
b
a
∂L
d
−
∂y
dx
d
dx
⎛
⎞
b
b ∂L
d ∂L
⎜ ∂L
⎟ dβ
η(x)dx + ⎝
γ(x)dx⎠
γ(x)dx −
=0⇒
∂ y
∂y
dx ∂ y dα
a
∂L
∂ y
⎛
η(x)dx + ⎝
a
a
b ∂L
d
−
∂y
dx
∂L
∂ y
⎞
γ(x) dx ⎠
dβ
= 0.
dα
a
But,
dβ
∂ J/∂α
=−
,
dα
∂ J/∂β
and then the previous relation becomes
b ∂L d
−
∂ y dx
∂L
∂ y
⎛
η(x)dx − ⎝
a
b ∂L d
−
∂ y dx
⎞
∂ J/∂α
∂L
γ(x) dx⎠
= 0.
∂ y
∂ J/∂β
a
(5.5.2)
On the other hand, integrating by parts and taking into account that η(a) = η(b) = 0,
we obtain
b ∂M
d ∂M
∂J
η(x)dx.
=
−
∂α
∂y
dx ∂ y a
In the same manner, taking into account that γ(a) = γ(b) = 0, we obtain
∂J
=
∂β
b a
d
∂M
−
∂y
dx
∂M
∂ y
γ(x)dx.
192
5 Calculus of Variations
Taking into account these estimations in Eq. (5.5.2), we obtain
b ∂L
d
−
∂y
dx
∂L
∂ y
η(x)dx−
a
b
−
a
(b ) ∂ M
a
(b
a
)
∂y
−
d
dx
∂M
∂y
−
d
dx
∂M
∂ y
∂M
∂ y
*
*
η(x)dx γ(x)dx
∂L
d
−
∂y
dx
∂L
∂ y
γ(x)dx = 0.
If we use the notation
(b ) ∂ L
λ=
a
(b
)
a
∂y
−
d
dx
∂M
∂y
−
d
dx
∂L
∂ y
∂M
∂ y
*
γ(x)dx
*
γ(x)dx
the previous relation can be restated as follows
b ∂L
d
−
∂y
dx
∂L
∂ y
−λ
∂M
d
−
∂y
dx
∂M
∂ y
!
η(x)dx = 0.
a
Taking into account that η(x) satisfies the conditions from the fundamental lemma,
we obtain the following equation
∂L
d
−
∂y
dx
∂L
∂ y
∂M
d
−λ
−
∂y
dx
∂M
∂ y
= 0,
which can be restated in the following form
∂
d
(L − λM) −
∂y
dx
∂
= 0.
−
λM)
(L
∂ y
Finally, we observe that this equation is the Euler’s equation for the functional I˜(y),
where
b
˜I (y) = (L − λM) dx,
a
and the theorem is concluded.
5.5 Isoperimetric Problems
193
Remark. The parameter λ is called the Lagrange’s multiplier and it is unknown. We
can determine its value from the equation
b
M x, y(x), y (x) dx = l,
a
after we introduce in the Lagrangean the founded expression of the function y(x).
Application 1. Let us solve the Dido’s problem.
First of all, note that the curve must be convex. Indeed, if that were not so, there
would be a straight line L such that if a portion of the boundary is reflected in the
line, then we obtain a region of greater area than the original region having the same
length of the boundary line.
Further note that any straight line that bisects a closed curve bounding a maximum
area will also divide the area in half. Suppose the opposite line L 1 does not have this
property. Then, by making a mirror reflection about L 1 of that portion of the surface
with the greater area, we would obtain a curve of the same length but it would bound
a greater area.
Choosing for the x-axis any of the straight lines that bisects the curve, we arrive
at the following statement of the problem.
Find a curve y = y(x), y(−a) = y(a) = 0, which together with the segment
−a ≤ x ≤ a, for a given length l > 2a, bounds a maximum area. Thus, the problem
has reduced to seeking the extremum of the functional
a
y(x), y(−a) = y(a) = 0
I (y) =
−a
subjected to the accessory condition that
a J (y) =
1 + y 2 (x)dx = l, l > 2a.
−a
We form the auxiliary Lagrangean
L̃ x, y(x), y = L x, y(x), y + λM x, y(x), y = y(x) + λ 1 + y 2 (x)
and consider the auxiliary functional
I˜(y) =
a
−a
L̃ x, y(x), y dx.
194
5 Calculus of Variations
The Euler’s equation for this functional is
d
dx
whence
+
,
λy = 1,
1 + y 2
λy = x + C, C = constant.
1 + y 2
Solving this equation for y , we get
x + C1
dy
=
.
dx
λ2 − (x + C1 )2
Integrating this equation, we obtain
(x + C1 ) + (y + C2 ) = λ2 ,
that is, a circle of radius λ with center at the point (−C1 , −C2 ). We determine the
constants C1 , C2 and the parameter λ from the boundary conditions y(−a) = y(a) =
0 and the isoperimetric condition, that is, equating the value of the functional J (y)
by l. We obtain
C22 = λ2 − (C1 − a)2
C22 = λ2 − (C1 + a)2
whence
C1 = 0, C2 =
such that
y(x) =
λ2 − a 2 ,
λ2 − x 2 −
y (x) = − √
λ2 − a 2 ,
x
λ2
− x2
.
The isoperimeric condition yields
a
l=
−a
whence
λ
x a
a
= λ arcsin = 2λ arcsin ,
√
2
2
λ
λ
−a
λ −x
a
l
= sin
.
λ
2λ
5.5 Isoperimetric Problems
195
Solving this equation for λ, we find a certain value λ = λ0 and then also the quantity
C2 =
-
λ20 − a 2 .
Let us prove that the last equation always has a solution. Indeed, setting l/2λ = t,
we reduce this equation to the form
sin t =
2a
t,
l
where 2a/l = α < 1 by the statement of the problem. At the point t = 0, the function
y(t) = sin t has the tangent slope π/4, while the function y(t) = αt has a smaller
slope. Hence, the graphs of these functions have at least one point of intersection
other than O(0, 0).
Application 2. Let us solve the geodesics’s problem.
Vector Solution. First of all, we remember that a geodesic of a surface is a curve of
shortest length lying on a given surface and joining two given points of the surface.
In our case, from among all curves on a sphere of radius 1 that joins two given points,
find the shortest curve.
Suppose the sphere is given by the vector equation
r = r(u, v).
The equations of geodesics may be obtained as Euler’s equations corresponding to
the variational problem of finding the shortest distance on a surface between two
given points. Let ϕ be the longitude, θ the latitude of a point on the sphere, and
ϕ = ϕ(θ) be the equation of the desired curve. We then have
r = r(ϕ, θ) = x(ϕ, θ)i + y(ϕ, θ)j + z(ϕ, θ)k.
Let us remember that the element of arc on a surface has the form
ds = Eu 2 + 2Fu v + Gv 2 dθ
and then the length of the curve between the points corresponding to the values θ1
and θ2 of the parameter θ is
J (u, v) =
θ2 Eu 2 + 2Fu v + Gv 2 dθ,
θ1
where E, F, G are coefficients of the quadratic form of the surface, that is,
196
5 Calculus of Variations
E=
∂r ∂r
,
∂u ∂u
, F=
∂r ∂r
,
∂u ∂v
, G=
∂r ∂r
,
∂v ∂v
.
Here, (a, b) is the scalar product of the vectors a and b.
In the case of the sphere,
E = rϕ , rϕ = sin2 θ, G = (rθ , rθ ) = 1, F = rϕ , rθ = 0,
and
θ2 J (ϕ, θ) =
1 + sin2 θ.ϕ2 (θ)dθ.
θ1
The Lagrangean does not contain the function ϕ(θ) such that the Euler’s equation
reduces to its particular prime integral
d
L ϕ = 0,
dθ
where
sin2 θϕ (θ)
,
L ϕ = 1 + sin2 θ.ϕ2 (θ)
such that
sin2 θϕ (θ)
= C1 .
1 + sin2 θ.ϕ2 (θ)
From this we get
ϕ (θ) =
=
-
C1
sin θ sin2 θ − C12
=
C1 d(cot 2 θ)
C1
= −.
sin2 θ (1 − C12 ) − C12 cot 2 θ
(1 − C12 ) − C12 cot 2 θ
Integrating we obtain
C1 cot 2 θ
ϕ(θ) = arccos + C2 ,
1 − C12
that is,
ϕ(θ) = arccos(C cot θ) + C2 , where C = -
C1
1 − C12
,
5.5 Isoperimetric Problems
197
whence
C cot θ = cos[ϕ(θ) − C2 ],
or, equivalently,
cot θ = A cos ϕ(θ) + B sin ϕ(θ)
where
A=
sin C2
cos C2
, B=
.
C
C
Multiplying both members by sin θ, we obtain
cos θ = A cos ϕ sin θ + B sin ϕ sin θ,
from where, passing to Cartesian coordinates, we deduce
z = Ax + By.
This is the equation of the plane passing through the center of the sphere and intersecting the sphere along a great circle. Thus, the geodesic of the sphere is a great
circle.
Cartesian Solution. Let us consider the sphere having the origin as center and the
radius equal to 1. From along all curves that join two arbitrary points A and B on
the sphere, find that of shortest length. Consider the curve in the parametric form
⎧
⎨ x = x(t)
y = y(t)
⎩
z = z(t), t ∈ [a, b].
The ends a and b of the interval will be determined by the fact that the arc of the
curve passes through points A and B. As we know, the element of the arc is
ds =
ẋ 2 + ẏ 2 + ż 2 dt.
The curve between A and B has the length
s(x, y, t) =
b ẋ 2 + ẏ 2 + ż 2 dt.
a
So, we must extremise this functional taking into account that the arc of the curve
must lie on the sphere x 2 + y 2 + z 2 = 1. Based on Theorem 5.5.1, we have the
functional
198
5 Calculus of Variations
I (x, y, z) =
b )
*
ẋ 2 + ẏ 2 + ż 2 − λ x 2 + y 2 + z 2 − 1 dt,
a
that leads to the system of equations
∂L
d
−
∂x
dt
∂L
d
−
∂y
dt
∂L
d
−
∂z
dt
∂L
∂ ẋ
=0
∂L
∂ ẏ
∂L
∂ ż
=0
= 0,
where the Lagrangean L is
L (t, x(t), y(t), z(t), ẋ(t), ẏ(t), ż(t)) =
Since
ẋ 2 + ẏ 2 + ż 2 − λ x 2 + y 2 + z 2 − 1 .
∂L
∂L
∂L
= −2λx,
= −2λy,
= −2λz
∂x
∂y
∂z
and
∂L
ẋ
ẏ
ż
∂L
∂L
=
=
=
,
,
,
2
2
2
2
2
2
2
∂ ẋ
∂ ẏ
∂ ż
ẋ + ẏ + ż
ẋ + ẏ + ż
ẋ + ẏ 2 + ż 2
the above system of equation becomes
d
−2λx −
dt
d
−2λy −
dt
d
−2λz −
dt
+
+
+
,
ẋ
=0
ẋ 2 + ẏ 2 + ż 2
,
ẏ
ẋ 2 + ẏ 2 + ż 2
ż
ẋ 2 + ẏ 2 + ż 2
=0
,
= 0,
which can be restated as follows
,
+
ẋ
d
+ 2λx = 0
dt
ẋ 2 + ẏ 2 + ż 2
5.5 Isoperimetric Problems
199
d
dt
d
dt
+
ẏ
+ 2λy = 0
ẋ 2 + ẏ 2 + ż 2
+
,
,
ż
ẋ 2 + ẏ 2 + ż 2
+ 2λz = 0.
It is strong to solve this system in the variable t, such that we pass to the new variable
s, taking into account that
ds =
ẋ 2 + ẏ 2 + ż 2 dt.
By direct calculations, we obtain
ẋ =
dx ds
ẋ
dx
=
= x ẋ 2 + ẏ 2 + ż 2 ⇒ = x ,
dt
ds dt
ẋ 2 + ẏ 2 + ż 2
ẏ =
dy ds
ẏ
dy
=
= y ẋ 2 + ẏ 2 + ż 2 ⇒ = y,
2
dt
ds dt
ẋ + ẏ 2 + ż 2
ż =
dz ds
ż
dz
=
= z ẋ 2 + ẏ 2 + ż 2 ⇒ = z.
dt
ds dt
ẋ 2 + ẏ 2 + ż 2
Then the above system becomes
d x + 2λx = 0,
dt
d y + 2λy = 0,
dt
d z + 2λz = 0,
dt
that is,
dx ds
+ 2λx = 0,
ds dt
dy ds
+ 2λy = 0,
ds dt
dz ds
+ 2λz = 0,
ds dt
from where we deduce
200
5 Calculus of Variations
x = − y = − 2λ
ẋ 2 + ẏ 2 + ż 2
2λ
ẋ 2
z = − + ẏ 2 + ż 2
2λ
ẋ 2 + ẏ 2 + ż 2
x,
y,
z.
Using the notation
2λ
ẋ 2 + ẏ 2 + ż 2
= β(t),
the last system becomes
⎧ ⎨ x = −β(t)x,
y = −β(t)y,
⎩ z = −β(t)z.
(5.5.3)
On the other hand, by derivation the equation of the sphere, with respect to s, x 2 +
y 2 + z 2 − 1 = 0, we obtain
x x + yy + zz = 0,
and, after one more time derivation
x 2 + y 2 + z 2 + x x + yy + zz = 0.
(5.5.4)
Multiplying the first equation from (5.5.3) by x, the second by y and the third by z
and adding the resulting relations, we obtain
x x + yy + zz = −β(t) x 2 + y 2 + z 2 = −β(t).
Introducing this result in Eq. (5.5.4) it follows
x 2 + y 2 + z 2 = β(t).
On the other hand, using the above proved system of relations
⎧
⎨ ẋ = x ẋ 2 + ẏ 2 + ż 2
ẏ = y ẋ 2 + ẏ 2 + ż 2
⎩
ż = z ẋ 2 + ẏ 2 + ż 2
from where it is easy to see that
(5.5.5)
5.5 Isoperimetric Problems
201
x 2 + y 2 + z 2 = 1
and then from Eq. (5.5.5) we deduce
β(t) = 1,
and the system (5.5.3) becomes
⎧ ⎨ x + x = 0,
y + y = 0,
⎩ z + z = 0.
It is easy to solve the system of differential equations (with the aid of the characteristic
equation) and find the solutions
⎧
⎨ x(s) = A1 cos s + A2 sin s,
y(s) = B1 cos s + B2 sin s,
⎩
z(s) = C1 cos s + C2 sin s,
where A1 , A2 , B1 , B2 , C1 , C2 are constants.
From the first two equations we find the expression for cos s and sin s and introducing in the last equation, we obtain an equation of the form
z = Ax + By, A, B = constants,
that is, the equation of the plane passing through the center of the sphere and intersecting the sphere along a great circle. Thus, the geodesic of the sphere is a great
circle.
5.6 Moving Boundary Problems
In all problems of variational calculus that we already studied, the endpoints of the
function that extremizes a functional there were fixed. In this paragraph we introduce
some notions with regard to moving boundary problems. We will consider only two
cases:
(1) An endpoint of the extremal (for instance, left-hand end of the curve) is fixed
and the other (the right-hand end) is movable along a given curve y = ϕ(x).
(2) The left-hand end of the extremal is movable along a given curve y = ϕ(x) and
the right-hand end is movable along a given curve y = ψ(x).
Of course, there exists more complicated problems when the endpoints of the
extremal are movable along, either a given curve or a certain surface, or, more general,
both endpoints are movable along two given surfaces.
202
5 Calculus of Variations
For the moment, we will consider the problem of finding the shortest distance
between a fixed point A(a, b) and a curve
(C)
y = ϕ(x), x ∈ [α, β].
As such, an endpoint of the extremal will be fixed and the other is movable along
the curve (C). Denote by B(x1 , ϕ(x1 ) the point where it attains the minimum of
distance between point A and the curve (C). In this case the functional that must be
extremized is
x1
I (y) =
L x, y(x), y (x) dx.
(5.6.1)
a
Theorem 5.6.1 If the function y(x) extremizes the functional I (y) from Eq. (5.2.2),
then it satisfies the following system of equations
d ∂L
∂L
=0
−
∂y
dx ∂ y ∂ L dϕ dy
−
L+ = 0.
∂ y dx
dx x=x1
(5.6.2)
(5.6.3)
Proof First of all, we must outline that the relation (5.6.1) is called the transversality
condition.
Consider a vicinity of first order of function y(x)
{y(x) + εη(x)}ε , ε > 0
and compute the value of the functional I (y) for an arbitrary representative of this
vicinity
x1
I (y + εη) =
L x, y(x) + εη(x), y (x) + εη (x) dx = I (ε).
a
For an arbitrary representative of this vicinity (that is, for an arbitrary fixed ε), an
equality of the form
y(x) + εη(x) = ϕ(x),
(5.6.4)
defines the value of the point x1 as an intersection of those curves. Of course, for
arbitrary ε, we obtain x1 = x1 (ε) and the functional can be written in the form
5.6 Moving Boundary Problems
203
x1 (ε)
L x, y(x) + εη(x), y (x) + εη (x) dx = I (ε).
I (y + εη) =
a
It is obvious that for ε = 0 the representative of the vicinity reduces to the function
y(x) that extremizes the functional. Consequently, ε = 0 is the point of extremum
for the function I (ε) and, therefore,
d I (ε) = 0.
dε ε=0
Using the rule of derivation for an integral having parameter, it follows
d I (ε) dx1 (ε) L x1 (ε), y(x1 (ε)), y x1 (ε)
=
dε ε=0
dε
x1 (ε)
+
a
∂L
∂L
η(x) +
η (x) dx ∂(y + εη)
∂(y + εη )
ε=0
+
= 0.
ε=0
Taking into account that ε = 0, we can write
dx1 (0) L x1 (0), y(x1 (0)), y x1 (0) +
dε
x1 (0)
∂L
∂L
η(x) + η (x) dx = 0.
∂y
∂y
a
(5.6.5)
Let us compute, by parts, the last integral in this equation
x1 (0)
a
x1 (0)
x1 (0)
∂L d ∂L
∂L
η(x)dx.
η
(x)dx
=
η(x)
−
∂ y
∂ y
dx ∂ y a
a
Taking into account that η(a) = 0, we have
x1 (0)
∂L ∂L
η (x)dx =
η(x1 (0)) −
∂ y
∂ y
a
x1 (0)
d
dx
∂L
∂ y
η(x)dx.
a
Thus, the equation (5.5.5) becomes
dx1 (0) ∂L
L x1 (0), y(x1 (0)), y x1 (0) + η(x1 (0))+
dε
∂y
204
5 Calculus of Variations
x1 (0)
+
∂L
d
−
∂y
dx
∂L
∂ y
η(x)dx = 0.
(5.6.6)
a
Now, we write the fact that x1 (ε) verifies the equation (5.6.4), that is
y(x1 (ε)) + εη(x1 (ε)) = ϕ(x1 (ε)).
(5.6.7)
Using the derivative with respect to ε in Eq. (5.6.7), it follows
dϕ dx1
dη dx1
dy dx1
+ η(x1 (ε)) + ε
=
⇒
dx1 dε
dx1 dε
dx1 dε
dη dx1
dx1 dϕ
dy
−ε
⇒ η(x1 (ε)) =
−
dε dx1
dx1
dx1 dε
such that, putting ε = 0, we obtain
dx1 (ε)
η(x1 (0)) =
dε
dϕ
dy
−
dx1
dx1
ε=0
.
(5.6.8)
Substituting this result in Eq. (5.6.6), we deduce
dx1 (0)
∂ L dϕ dy
−
+
L x1 , y(x1 ), y x1 + dε
∂ y dx
dx x=x1
x1 +
∂L
d
−
∂y
dx
∂L
∂ y
η(x)dx = 0.
(5.6.9)
a
The equality (5.6.9) holds for any η, because we have computed the value of the
functional I for an arbitrary representative of the vicinity of function y(x). So, we
can choose that η for which η(x1 (0)) = 0 and the relation (5.6.8) becomes
dx1 (ε)
dε
dϕ
dy
−
dx1
dx1
ε=0
= 0.
But, the last term cannot be zero, such that we deduce
dx1 (0)
= 0,
dε
and the equality (5.6.9) reduces to
5.6 Moving Boundary Problems
x1 (0)
205
∂L
d
−
∂y
dx
∂L
∂ y
η(x)dx = 0.
a
Therefore, we can use the fundamental lemma and conclude
∂L
d ∂L
=0
−
∂y
dx ∂ y that is, the Euler’s equation (5.6.2).
Taking into account this conclusion, the relation (5.30) reduces to
L x1 , y(x1 ), y x1
∂L
+ ∂y
dϕ dy
−
dx
dx
= 0,
x=x1
that is, the transversality condition from the enunciation of our theorem. So, both
and the Euler’s equation and the transversality condition are proved such that the
theorem is concluded.
Let us give a geometrical interpretation of the transversality condition. To this,
we prove the following proposition.
Proposition 5.6.1 If the Lagrangean of the functional I (y) has the form
L = L x, y(x), y (x) = h(x, y(x)) 1 + y 2 (x), h(x, y) = 0,
and, as a consequence, the functional is
x1
I (y) =
h(x, y(x)) 1 + y 2 (x)dx,
a
then the transversality condition becomes
y 2 (x) = −
1
,
ϕ2 (x)
where y(x) is the extremal of the functional and ϕ(x) is the curve of one of the
movable endpoints of the extremal.
Proof The general transversality condition
∂L
L x1 , y(x1 ), y x1 + ∂y
becomes, in our case,
dϕ dy
−
dx
dx
= 0,
x=x1
206
5 Calculus of Variations
y (x)
ϕ (x) − y (x)
h(x, y(x)) 1 + y 2 (x) + h(x, y(x)) 1 + y 2 (x)
x=x1
= 0.
By hypothesis, h(x, y) = 0, such that we deduce
1 + y 2 (x) + y (x)ϕ (x) − y 2 (x)
=0⇒
1 + y 2 (x)
⇒ 1 + y (x)ϕ (x) = 0,
such that desired result is proved.
Remark. The transversality condition from the above proposition asserts that the
extremal of the functional and the curve where is movable one of the endpoints of
the extremal, must be two orthogonal curves.
Application. Let us compute the minimum distance between the point A(−1, 5) and
the curve y 2 = x. In this case the functional is
I (y) =
x1
−1
1 + y dx,
x1 being the movable point on the curve
y = ϕ(x) =
√
x.
Instead of the Euler’s equation we can use the prime integral
∂L
= C, C = constant.
∂ y
The transversality condition received the form
y
1 + y + 1 + y 2
From
1
√ − y
2 x
= 0.
x=x1
y
= C ⇒ y = C1 ⇒ y = C1 x + C2 ,
1 + y 2
such that the transversality condition becomes
1
C1
1 + C12 + √ − C1 = 0 ⇒
1 + C12 2 x1
5.6 Moving Boundary Problems
⇒-
207
√
C1
1
+ √ = 0 ⇒ 2 x1 + C1 = 0.
2 x1 1 + C 2
1 + C12
1
1
√
Then, since the points A(−1, 5) and B(x1 , x1 ) belonging to the straight line y =
C1 x + C2 we obtain also two relations to determine C1 = −2, C2 = 3 and x1 = 1.
The straight line becomes y = −2x + 3 and then the minimum distance is
1
−1
1 + y 2 dx =
1
−1
√
√
5dx = 2 5.
Chapter 6
Quasi-linear Equations
6.1 Canonical Form for n = 2
Let be a bounded domain in the n−dimensional Euclidean space IRn . The general
form of a partial differential equation is:
F(x, u, u x1 , . . . , u xn , u x1 x2 , . . . , u xi x j , u x1 x2 ...xi , . . . , u x1 x2 ...xn ) = 0,
(6.1.1)
where by u xi , u xi x j , u xi x j xk , . . . , we have denoted partial derivatives
∂u
∂2u
∂3u
,
,
,...
∂xi ∂xi ∂x j ∂xi ∂x j ∂xk
The unknown function is u(x) = u(x1 , x2 , . . . , xn ), x ∈ . The function F satisfies,
with respect with its arguments, certain hypotheses which permit the mathematical
operations that are necessary to solve the equation.
A real function u(x) defined in the domain , where Eq. (6.1.1) is considered,
which is continuous together with its partial derivatives contained in the equation and
which turns the equation into an identity is called a regular solution of the equation.
We shall study the partial differential equations of the form (6.1.1) which contains
only the partial derivatives until to the order two, inclusively.
Definition 6.1.1 A partial differential equation of order two, in a single unknown
function, depending of n independent variables of the form
n n
i=1 j=1
ai j (x1 , x2 , . . . , xn )
∂2u
= f (x1 , x2 , . . . , xn , u, u x1 , . . . , u xn ) (6.1.2)
∂xi ∂x j
is called a quasi-linear partial differential equation of order two.
The functions ai j = a ji and f are known functions and are generally, assumed to be
continuous with respect to its arguments.
© Springer International Publishing AG, part of Springer Nature 2018
M. Marin and A. Öchsner, Complements of Higher Mathematics,
https://doi.org/10.1007/978-3-319-74684-5_6
209
210
6 Quasi-linear Equations
In all that following we consider only the case of two independent variables,
(n = 2), such that the quasi-linear equation (6.1.2) becomes
a11
∂2u
∂2u
∂2u
∂u ∂u
+
2a
= f (x, y, u,
+
a
,
).
12
22
2
2
∂x
∂x∂ y
∂y
∂x ∂ y
(6.1.3)
The problem of the integration of an equation of the form (6.1.3) is the problem of
the determination of a function u = u(x, y), u : → IR, u ∈ C 2 () which verifies
the given equation. In view of making the integration of Eq. (6.1.3) easier, we make
a change of independent variables: from the variables (x, y) we pass to the variables
(ξ, η), as follows
ξ = ξ(x, y),
η = η(x, y),
such that ξ, η ∈ C 2 () and
∂(ξ, η) ξx ξ y =
∂(x, y) ηx η y = 0, on .
(6.1.4)
(6.1.5)
The change (6.1.4) is a non-degenerate affine transformation of variables and is made
with the hope that in the new partial differential equation, in the variables ξ and η,
also of the form Eq. (6.1.3), one or two coefficients are zero.
Remark. Due to the hypotheses imposed to ξ and η, we can apply at any point from
the theorem of implicit functions. Then, if we arbitrarily fix (x0 , y0 ) ∈ , we can
find the solution of the system (6.1.4) with respect to the unknowns x and y, such
that in a vicinity of the point (x0 , y0 ), we obtain
x = x(ξ, η)
y = y(ξ, η).
(6.1.6)
If we denote by ξ0 = ξ(x0 , y0 ), η0 = η(x0 , y0 ), then we have x0 = x(ξ0 , η0 ) and
y0 = y(ξ0 , η0 ).
In order to obtain the partial differential equation in the new variables ξ and η, we
substitute the derivatives of the unknown function by the derivatives with respect to
the new variables. We have:
∂u
∂u ∂ξ
∂u ∂η
=
+
,
∂x
∂ξ ∂x
∂η ∂x
Then:
∂u
∂u ∂ξ
∂u ∂η
=
+
.
∂y
∂ξ ∂ y
∂η ∂ y
6.1 Canonical Form for n = 2
211
∂ξ 2
∂ 2 u ∂η 2
∂ 2 u ∂ξ ∂η
+ 2
+2
+
∂x
∂ξ∂η ∂x ∂x
∂η
∂x
∂u ∂ 2 η
∂u ∂ 2 ξ
+
,
+
∂ξ ∂x 2
∂η ∂x 2
∂ 2 u ∂ξ ∂ξ
∂ 2 u ∂ξ ∂η
∂ξ ∂η
∂2u
=
+
+
+
∂x∂ y
∂ξ 2 ∂x ∂ y
∂ξ∂η ∂x ∂ y
∂ y ∂x
∂ 2 u ∂η ∂η
∂u ∂ 2 ξ
∂u ∂ 2 η
+ 2
+
+
,
∂η ∂x ∂ y
∂ξ ∂x∂ y
∂η ∂x∂ y
∂ 2 u ∂η 2
∂2u
∂ 2 u ∂ξ 2
∂ 2 u ∂ξ ∂η
+ 2
=
+2
+
∂ y2
∂ξ 2 ∂ y
∂ξ∂η ∂ y ∂ y
∂η
∂y
∂u ∂ 2 ξ
∂u ∂ 2 η
+
+
.
∂ξ ∂ y 2
∂η ∂ y 2
∂2u
∂2u
=
∂x 2
∂ξ 2
(6.1.7)
If we substitute the derivatives from Eq. (6.1.7) in (6.1.3) we obtain
a11
∂2u
∂2u
∂2u
∂u ∂u
+ a22 2 = F(ξ, η, u,
,
),
+ 2a12
2
∂ξ
∂ξ∂η
∂η
∂ξ ∂η
(6.1.8)
where the new coefficients ai j have the expressions:
a11 = a11
a12
∂ξ
∂x
2
∂ξ ∂η
+ a12
= a11
∂x ∂x
a22 = a11
∂η
∂x
2
∂ξ ∂ξ
+ a22
+ 2a12
∂x ∂ y
∂ξ ∂η
∂ξ ∂η
+
∂x ∂ y
∂ y ∂x
∂ξ
∂y
2
,
∂η ∂η
+ a22
+ 2a12
∂x ∂ y
+ a22
∂η
∂y
∂η ∂ξ
,
∂y ∂y
(6.1.9)
2
.
It is clear that the annulment of the coefficients a11 and a22 of the Eq. (6.1.8) is
connected to the solution of the partial differential equation of first order:
a11
∂z
∂x
2
+ 2a12
∂z ∂z
+ a22
∂x ∂ y
∂z
∂y
2
= 0.
(6.1.10)
Indeed, if z = ϕ(x, y) is a solution of Eq. (6.1.10), then using the transformation of
variables
212
6 Quasi-linear Equations
ξ = ϕ(x, y)
η = η(x, y),
where η is an arbitrary variable, but it must satisfy the condition (6.1.5), then, from
Eq. (6.1.9) we deduce that a11 = 0.
If we choose the new variables
ξ = ξ(x, y)
η = ϕ(x, y),
where z = ϕ(x, y) is a solution of Eq. (6.1.10), then from Eq. (6.1.9)3 we obtain that
a22 = 0.
On the other hand, let us observe that the solution of the partial differential equation
(6.1.10) is connected to the solution of the ordinary differential equation
a11 (dy)2 − 2a12 dy dx + a22 (dx)2 = 0,
(6.1.11)
which, formally, can be rewritten in the form:
a11
dy
dx
2
− 2a12
dy
+ a22 = 0.
dx
(6.1.12)
Proposition 6.1.1 (i) Let ϕ(x, y) = C be a prime integral of the Eq. (6.1.11), where
C is an arbitrary constant. Then the function z = ϕ(x, y) is a solution of the
Eq. (6.1.10).
(ii) Let z = ϕ(x, y) be a solution of the Eq. (6.1.10). Then ϕ(x, y) = C, where C is
an arbitrary constant, is a prime integral of the Eq. (6.1.11).
Proof (i). Let ϕ(x, y) = C be a prime integral of the Eq. (6.1.11).
(x, y) = 0 on .
Without restriction of the generality, we can assume that ∂ϕ
∂y
If
∂ϕ
(x, y) = 0, ∀(x, y) ∈ 0 ⊂ ,
∂y
then we continue the study only on \0 .
If
∂ϕ
(x, y) = 0, ∀(x, y) ∈ ,
∂x y
then we change the role of the variables x and y.
If both
∂ϕ
∂ϕ
(x, y) = 0, and
(x, y) = 0, ∀(x, y) ∈ ,
∂x
∂y
then ϕ(x, y) is a constant function and, therefore, the Eq. (6.1.10) has the null solution. Consequently, we can assume that ∂ϕ
= 0. Then in a vicinity of a point (x0 , y0 )
∂y
6.1 Canonical Form for n = 2
for which
Moreover,
∂ϕ
(x0 , y0 )
∂y
213
= 0, we can write y = f (x, c0 ), where c0 = ϕ(x0 , y0 ).
∂ϕ
(x, y)
dy
∂x
= − ∂ϕ
.
dx
(x, y)
∂y
(6.1.13)
We substitute Eq. (6.1.13) in (6.1.12), which is equivalent to Eq. (6.1.11), and we
obtain
dy 2
dy
+ a22
− 2a12
=
0 = a11
dx
dx
(x0 ,y0 )
⎡
⎤
∂ϕ
∂x
= ⎣a11 − ∂ϕ
2
− 2a11 −
∂y
= a11
∂ϕ
∂x
2
∂ϕ
∂x
∂ϕ
∂y
∂ϕ ∂ϕ
+ a22
+ 2a12
∂x ∂ y
+ a22 ⎦
=
(x0 ,y0 )
∂ϕ
∂y
2 1
∂ϕ
∂y
whence it follows
2 ∂ϕ 2
∂ϕ
∂ϕ ∂ϕ
+ 2a12
a11
+ a22
∂x
∂ y ∂x
∂y
2 ,
= 0,
(x0 ,y0 )
for all the possibilities to choosing (x0 , y0 ) in , that is, ϕ(x, y) is a solution for the
Eq. (6.1.10).
(ii). Let us assume that z = ϕ(x, y) is a solution for the Eq. (6.1.10) and show that
ϕ(x, y) = C, where C is an arbitrary constant, is a prime integral for the Eq. (6.1.11).
To this we must show that dϕ(x, y) = 0, that is,
∂ϕ
∂ϕ
dx +
dy = 0,
∂x
∂y
whence it follows
∂ϕ
∂x
∂ϕ
∂y
=−
dy
.
dx
(6.1.14)
If we write the fact that z = ϕ(x, y) is a solution of the Eq. (6.1.10)
a11
∂ϕ
∂x
2
∂ϕ ∂ϕ
+ a22
+ 2a12
∂ y ∂x
∂ϕ
∂y
2
= 0,
214
6 Quasi-linear Equations
and we divide here, formally, by ∂ϕ
, without repeating the considerations regarding
∂y
the theorem of implicit functions, which, obviously, is still valid, from the first part
of the proof, we obtain
∂ϕ
∂x
∂ϕ
∂y
a11
2
∂ϕ
∂x
+ 2a12 ∂ϕ
+ a22 = 0.
∂y
Here, we substitute Eq. (6.1.14) such that the Eq. (6.1.12) is obtained.
Essentially, in the Proposition (6.1.1) it is shown that to find a solution for the
Eq. (6.1.10) means to find the prime integrals of the Eq. (6.1.12).
Consequently, we have shown that to annul the coefficients a11 and a22 of the
Eq. (6.1.8), means to find the prime integrals of the Eq. (6.1.12). The Eq. (6.1.11) is
called the equation of the characteristics and its prime integrals are called characteristics or characteristic curves. Analyzing the equation of the characteristics (6.1.11),
we ascertain that to find its prime integrals, we have three different cases, according
2
− a11 a22 :
to the discriminant of the equation, = a12
o
1 . If > 0, the Eq. (6.1.11) admits two real distinct characteristics curves. Then
the partial differential equation is called the hyperbolical equation.
2o . If = 0, the Eq. (6.1.11) admits only one real characteristic. Then the partial
differential equation is called the parabolical equation.
3o . If < 0, the Eq. (6.1.11) admits two complex conjugated characteristics.
Then the partial differential equation is called the elliptical equation.
The above classification is made regarding the partial differential equation in its
initial form (6.1.3). But, the transformation of coordinates (6.1.4) with the condition
(6.1.5), does not affect the type of the Eq. (6.1.3).
Indeed, if we compute for the form (6.1.8) of the equation, (i.e. the canonical
form), we ascertain that the discriminant has the same sign:
= a12 − a11 a22
2
∂ξ ∂η
2
∂ξ ∂η 2
−
= a12 − a11 a22
,
∂x ∂ y
∂ y ∂x
that is,
∂(ξ, η)
=
∂(x, y)
2
,
(6.1.15)
where we take into account the form (6.1.9) of the coefficients ai j . From Eq. (6.1.15)
we deduce that, if the equation is hyperbolical, parabolical or elliptical, in a system
of coordinates, then it has the same type if we pass to another system of coordinates,
if the transformation of coordinates is non-degenerate, that is, the condition (6.1.5)
holds.
It is easy to ascertain that is a continuous function with respect to the variables
(x, y). It is well known fact that if a continuous function is positive in a point, then it is
6.1 Canonical Form for n = 2
215
positive in an entire vicinity of the respective point. So, we can divide the whole plane
in three different sets. We shall call the domain of hyperbolicity for the Eq. (6.1.3)
the set of the points from the plane IR2 for which the Eq. (6.1.3) is hyperbolical.
Analogically can be define the domains of parabolicity and of ellipticity, respectively.
In the following we intend to find the canonical form of a partial differential equation
in all three cases.
2
− a11 a22 > 0. In this case the Eq. (6.1.11)
1o The hyperbolical case: = a12
has two real and distinct prime integrals: ϕ(x, y) = C1 , ψ(x, y) = C2 , where C1
and C2 are arbitrary constants. Consider the new variables (ξ, η) in the form
ξ = ϕ(x, y), sη = ψ(x, y)
and, based on the Proposition 6.1.1, we will obtain that a11 = 0 and a22 = 0, such
that the canonical form of the hyperbolical equation is
∂2u
= f
a12
∂ξ∂η
∂u ∂u
,
ξ, η, u,
∂ξ ∂η
or, if we divide by a12 (which obviously, can not be null):
∂u ∂u
∂2u
= F ξ, η, u,
,
.
∂ξ∂η
∂ξ ∂η
(6.1.16)
Let us observe that the transformation (6.1.16) is non-degenerate. Indeed,
∂(ξ, η) ∂ϕ ∂ψ
∂ψ ∂ϕ
∂(x, y) = 0 ⇔ ∂ y ∂x − ∂x ∂ y = 0 ⇔
√
√
∂ϕ
∂ψ
a12 + a12 − ∂x
∂x
− ∂ϕ = − ∂ψ ⇔
=
⇔
a11
a11
∂y
∂y
√
√
⇔ = − ⇔ = 0,
which is absurd, taking into account that > 0.
2
− a11 a22 = 0. In this case the characteristic
2o The parabolical case: = a12
equation (6.1.11) has only one real prime integral, ϕ(x, y) = C, where C is an
arbitrary constant. We can use the new variables (ξ, η) as follows
ξ = ϕ(x, y), η = η(x, y),
where η is an arbitrary function of the class C 2 , which together with ϕ assures the
condition that the transformation (6.1.18) is non-degenerate
∂(ξ, η) ∂ϕ ∂η
∂η ∂ϕ
∂(x, y) = ∂x ∂ y − ∂x ∂ y = 0.
(6.1.17)
216
6 Quasi-linear Equations
Since we have chosen ξ = ϕ(x, y), based on the Proposition 6.1.1, we deduce that
a11 = 0. Let us prove that and a12 = 0.
Proposition 6.1.2 If ξ and η are of the form (6.1.18), and satisfy the condition
(6.1.19), then we have a12 = 0.
2
Proof From a12
= a11 a22 we deduce that a11 and a22 have, simultaneous, the same
sign and, without loss of generality we assume that a11 > 0 and a22 > 0. Then
√ √
a12 = ± a11 a22 . According to the Proposition 6.1.1, it results that a11 = 0 and,
therefore
2
∂ξ
∂ξ ∂ξ
+ a11
=
∂x ∂ y
∂y
√ ∂ξ 2
√ ∂ξ 2 √ √ ∂ξ ∂ξ
+
=
a11
± a11 a22
a22
=
∂x
∂x ∂ y
∂x
√ ∂ξ √ ∂ξ 2
a11
.
=
± a22
∂x
∂y
0 = a11
∂ξ
∂x
2
+ 2a12
This implies that
√ ∂ξ √ ∂ξ
± a22
= 0.
a11
∂x
∂y
(6.1.18)
Using Eq. (6.1.9) we deduce
∂ξ ∂η
∂ξ ∂η
∂ξ ∂η
∂ξ ∂η
+ a12
+
+ a22
=
∂x ∂x
∂x ∂ y
∂ y ∂x
∂y ∂y
√ ∂η √ ∂η
√ ∂ξ √ ∂ξ
± a22
± a22
,
=
a11
a11
∂x
∂y
∂x
∂y
a12 = a11
such that, taking into account the relation (6.1.18), we obtain that a12 = 0.
Using the fact that a11 = a12 = 0, we deduce that the parabolical equation has the
canonical form
∂2u
∂u ∂u
,
,
a22 2 = F ξ, η, u,
∂η
∂ξ ∂η
or, equivalently,
∂2u
∂u ∂u
,
.
= G ξ, η, u,
∂η 2
∂ξ ∂η
Remark. If, instead of the transformation (6.1.18), we take the transformation
6.1 Canonical Form for n = 2
217
ξ = ξ(x, y),
η = ϕ(x, y),
where ϕ(x, y) = C is the only one prime integral of the characteristic equation
(6.1.11), and ξ(x, y) is an arbitrary function of the class C 2 and which together with
ϕ(x, y) assures the fact that the transformation is non-degenerate (that is, ξ(x, y) and
ϕ(x, y) satisfies a condition which is analogous to (6.1.19)), then after analogous
calculations as in the Proposition 6.1.2, we obtain the following canonical form
∂u ∂u
∂2u
,
.
=
H
ξ,
η,
u,
∂ξ 2
∂ξ ∂η
2
3o The elliptical case: = a12
− a11 a22 < 0. In this case the characteristic
equation (6.1.11) admits two prime integrals, which are complex conjugated and
which can be written in the form
ϕ(x, y) = C1 ,
ϕ(x, y) = C2 ,
where C1 and C2 are arbitrary constants. Also, we have denoted by ϕ the function
which is a complex conjugated function of the function ϕ. We proceed as in the
hyperbolical case, that is, we take the new variables ξ and η of the form
ξ = ϕ(x, y),
η = ϕ(x, y),
with the condition
∂(ϕ, ϕ) ∂ϕ ∂ϕ ∂ϕ ∂ϕ
∂(x, y) = ∂x ∂ y − ∂x ∂ y = 0.
Then we will obtain a11 = a22 = 0 and therefore the elliptical equation has the
following canonical form
∂u ∂u
∂2u
= F ξ, η, u,
,
.
∂ξ∂η
∂ξ ∂η
Unlike to the hyperbolical case, the last equation is in the case of complex numbers.
We want to find such a transformation of variables to obtain the canonical form
in the set of the real numbers. With this hope in mind, we introduce the functions
α(x, y) and β(x, y) such that
218
6 Quasi-linear Equations
1
(ϕ + ϕ) ,
2
1
β = I m(ϕ) =
(ϕ − ϕ) ,
2i
α = Re(ϕ) =
and the new variables ξ and η are taken of the form
ξ = α + iβ,
η = α − iβ.
(6.1.19)
Proposition 6.1.3 In the case of the elliptical equations we have
ã11 = ã22 , ã12 = 0,
where ãi j are the coefficients of the canonical equation obtained by using the transformation (6.1.19).
Proof It is easy to see that ξ is, in fact, ξ = ϕ(x, y) and then a11 = 0. If we take into
account (6.1.19), then we have
2
∂ξ
∂ξ ∂ξ
+ a22
=
∂x ∂ y
∂y
∂α
∂α
∂β 2
∂β 2
+i
+i
+ a22
+
a11
∂x
∂x
∂y
∂y
∂α
∂β
∂α
∂β
+i
+i
=
+2a12
∂x
∂x
∂y
∂y
2
∂α
∂α
∂α ∂α
+ a22
−
= a11
+ 2a12
∂x
∂x ∂ y
∂y
2 ∂β 2
∂β
∂β ∂β
+ a22
− a11
+ 2a12
+
∂x
∂x ∂ y
∂y
∂α ∂β
∂α ∂β
∂α ∂β
∂α ∂
+ a12
+
+ a22
.
+ 2i a11
∂x ∂x
∂x ∂ y
∂ y ∂x
∂y ∂y
0 = a11 = a11
∂ξ
∂x
2
+ 2a12
This is an equality in the set of the complex numbers and then both the real and
imaginary part are null, whence it follows the result from proposition.
Using the results from the Proposition 6.1.3, we deduce that in the elliptical case the
canonical form of the equation is
∂2u
∂2u
∂u ∂u
,
,
+
= H α, β, u,
∂α2
∂β 2
∂α ∂β
6.1 Canonical Form for n = 2
219
where H is a real function.
We can conclude that, in the domain of ellipticity of the Eq. (6.1.3) does not exist
a characteristic direction. In the domain of hyperbolicity of the Eq. (6.1.3), in each
point there exists two real distinct characteristic directions and in each point of the
domain of parabolicity there exists only one real characteristic direction.
As a consequence, if the coefficients a11 , a12 and a22 of the Eq. (6.1.3) are sufficient regular, the domain of hyperbolicity is a network of two families of characteristic
curves, and the domain of parabolicity is covered by only one such a family.
As an exemplification, let us consider the equation:
ym
∂2u
∂2u
+
= 0,
∂x 2
∂ y2
where m is a natural odd number. In this case the Eq. (6.1.12) receives the form:
y
m
dy
dx
2
+ 1 = 0.
It is easy to see that it does not exist any characteristic direction in the semi-plane
y > 0. But, in each point of the straight line y = 0 and in each point of the semiplane y < 0, there exists a characteristic direction, respectively, two characteristic
directions.
We write the equation of the characteristic curves in the form:
m
dx ± (−y) 2 dy = 0,
from where, by integration, we deduce that the semi-plane y < 0 is covered by two
families of real characteristic curves, described by the equations:
x−
2
m+2
(−y) 2 = C1 ,
m+2
x+
2
m+2
(−y) 2 = C2 ,
m+2
and
where C1 and C2 are real constants.
6.2 Canonical Form for n > 2
In this paragraph we make some considerations on the canonical form of a partial
differential equation of the order two, in the case n > 2.
220
6 Quasi-linear Equations
Let be an open set from IRn and consider the quasilinear equation with partial
derivatives of the order two
n
n i=1
∂2u
ai j (x) =
= f
∂xi ∂x j
j=1
∂u ∂u
∂u
x1 , x2 , . . . , xn , , u,
,
,...,
∂x1 ∂x2
∂xn
,
(6.2.1)
where x = (x1 , x2 , . . . , xn ), ai j = a ji (x) ∈ C().
The function f = f (x1 , x2 , . . . , xn , z, p1 , p2 , . . . , pn ) is defined and is continuous for any (x1 , x2 , . . . , xn ) ∈ and −∞ < z, p1 , p2 , . . . , pn < ∞, and u is the
unknown function, u : → IR, u ∈ C 2 ().
We intend to make a change of variables such that in the new equation (which is
called the canonical equation) some new coefficients, denoted by ai j , as in the case
n = 2, to be null.
Consider the transformation
ξ1 = ξ1 (x1 , x2 , . . . , xn ),
ξ2 = ξ2 (x1 , x2 , . . . , xn ),
................................
ξn = ξn (x1 , x2 , . . . , xn ),
(6.2.2)
with the condition
∂ξ
1
1
∂x
∂ξ ∂ξ2
= ∂x1
∂x ∂ξn
∂x1
∂ξ1
∂x2
∂ξ2
∂x2
∂ξn
∂x2
...
...
∂ξ1
∂xn
∂ξ2
∂xn
...
∂ξn
∂xn
= 0,
(6.2.3)
where ξ = ξ(x) is a vectorial function, ξ : → IRn , ξ ∈ C 2 ().
Due to condition (6.2.3), based on the theory of implicit functions, the system
(6.2.2) can be solved in the vectorial variable x:
x1 = x1 (ξ1 , ξ2 , . . . , ξn ),
x2 = x2 (ξ1 , ξ2 , . . . , ξn ),
................................
xn = xn (ξ1 , ξ2 , . . . , ξn ),
such that at the end, the solution of the Eq. (6.2.1) will be obtained as a function of
x. From Eq. (6.2.2) we have
∂u ∂ξk
∂u
=
, i = 1, 2, . . . , n
∂xi
∂ξk ∂xi
k=1
n
6.2 Canonical Form for n > 2
221
and then
∂ 2 u ∂ξk ∂ξm ∂u ∂ 2 ξk
∂2u
=
+
.
∂xi ∂x j
∂ξk ∂ξm ∂xi ∂x j
∂ξk ∂xi ∂x j
k=1 m=1
k=1
n
n
n
(6.2.4)
We introduce Eq. (6.2.4) in (6.2.1) whence it follows the equation
∂ 2 u ∂ξk ∂ξm
∂u ∂u
∂u
. (6.2.5)
ai j
= G ξ, u,
,
,...,
∂ξk ∂ξm ∂xi ∂x j
∂ξ1 ∂ξ2
∂ξn
k=1 m=1
n n n
n i=1 j=1
We introduce the notation
akm (ξ) =
n
n i=1 j=1
ai j
∂ξk ∂ξm
∂xi ∂x j
(6.2.6)
and then the Eq. (6.2.5) becomes
n
n akm (ξ)
k=1 m=1
∂2u
∂u ∂u
∂u
.
= G ξ, u,
,
,...,
∂ξk ∂ξm
∂ξ1 ∂ξ2
∂ξn
(6.2.7)
fix x 0 = (x10 , x20 , . . . , xn0 ) ∈ and use the notation
λik =
∂ξk 0
(x )
∂xi
such that from Eq. (6.2.6) we deduce
akm (ξ 0 ) =
n n
ai j (x 0 )λik λ jm
(6.2.8)
i=1 j=1
where ξ 0 = ξ(x 0 ).
Using a matrix notation
A = [akm ] , A = ai j , = λi j ,
the Eq. (6.2.8) becomes
A = t A,
where we have denoted by t the transposed of the matrix .
It is well known that if, in Eq. (6.2.9), we make the change of variables
(6.2.9)
222
6 Quasi-linear Equations
= T M,
where by M we have denoted a non-degenerate matrix, and by T an orthogonal
matrix (T t = T −1 ), then the matrix A reduces to the its diagonal form, that is, the
matrix which has nonzero elements only on the principal diagonal.
With regard to the elements on the principal diagonal, we have the following
Sylvester’s rule of inertness:
The number of the positive elements on the principal diagonal is constant. Also,
the number of the negative elements on the principal diagonal is constant. We have
the following variants:
1o If all elements on the principal diagonal are strict positive in a point ξ 0 ∈ ,
then the canonical equation becomes
n
j=1
ajj
∂2u 0
∂u ∂u
∂u
.
(ξ
)
=
G
ξ
,
ξ
,
.
.
.
,
ξ
,
u,
,
,
.
.
.
,
1
2
n
∂ξ1 ∂ξ2
∂ξn
∂ξ 2j
Then, we say that the quasi-linear partial differential equation of the order two is
elliptical in the point ξ 0 ∈ .
2o If does not exist a zero on the principal diagonal, but there exist both positive
elements and negative elements, then we say that the equation is hyperbolical in the
point ξ 0 ∈ . In the particular case in which only one element is strictly positive
and all others are strictly negative, we say that the equation is ultrahyperbolical in
the respective point.
3o If on the principal diagonal there exist certain null elements, then the equation
is parabolical in the respective point.
4o If on the principal diagonal there exist both null elements and non-zero elements
and these have all the same sign then the equation is elliptical-parabolical in the
respective point.
5o If on the principal diagonal there exist null elements and non-zero elements
having different signs then we say that the equation is hyperbolical-parabolical.
It is clear that the benefit for the canonical form of a quasi-linear equation with
partial derivatives of order two is given by the fact that this form of the equation
facilitates its integration.
Chapter 7
Hyperbolical Equations
7.1 Problem of the Infinite Vibrating Chord
The main representative of hyperbolical equations is considered to be the equation
of the vibrating chord, also called the equation of waves.
Firstly, we consider the case of the infinite chord. Properly, the chord is not infinite,
but its length is much more than its cross section. The general aim of this paragraph
is to study the following initial-boundary values problem, attached to the equation
of the infinite chord
∂2u
∂2u
− a 2 2 = f (t, x), ∀(t, x) ∈ (−∞, +∞), ∀t > 0,
2
∂t
∂x
u(0, x) = ϕ(x), ∀x ∈ (−∞, +∞),
∂u
(0, x) = ψ(x), ∀x ∈ (−∞, +∞),
∂t
(7.1.1)
where the functions f (t, x), ϕ(x) and ψ(x) are given and continuous on their domain
of definition. The function u = u(t, x) is the unknown function of the problem and
represents the amplitude of the chord at the moment t, at the point x. The positive
constant a is prescribed for each type of the material of the chord.
We will decompose the Cauchy’s problem (7.1.1) in two other problems, one
homogeneous with regard to the right-hand side of the equation and, second, homogeneous with regard to the initial conditions:
∂2u
∂2u
− a 2 2 = 0, ∀(t, x) ∈ (−∞, +∞), ∀t > 0,
2
∂t
∂x
u(0, x) = ϕ(x), ∀x ∈ (−∞, +∞),
∂u
(0, x) = ψ(x), ∀x ∈ (−∞, +∞),
∂t
© Springer International Publishing AG, part of Springer Nature 2018
M. Marin and A. Öchsner, Complements of Higher Mathematics,
https://doi.org/10.1007/978-3-319-74684-5_7
(7.1.2)
223
224
7 Hyperbolical Equations
and, respectively,
2
∂2u
2∂ u
−
a
= f (t, x), ∀(t, x) ∈ (−∞, +∞), ∀t > 0,
∂t 2
∂x 2
u(0, x) = 0, ∀x ∈ (−∞, +∞),
∂u
(0, x) = 0, ∀x ∈ (−∞, +∞).
∂t
(7.1.3)
Proposition 7.1.1 If the function u 1 (t, x) is a solution of the problem (7.1.2) and
the function u 2 (t, x) is a solution of the problem (7.1.3), then the function
u(t, x) = u 1 (t, x) + u 2 (t, x), ∀(t, x) ∈ (−∞, +∞), ∀t > 0,
(7.1.4)
is a solution of the problem (7.1.1).
Proof Firstly, we verify the initial conditions:
u(0, x) = u 1 (0, x) + u 2 (0, x) = ϕ(x) + 0 = ϕ(x),
∂u 1
∂u 2
∂u
(0, x) =
(0, x) +
(0, x) = ψ(x) + 0 = ψ(x),
∂t
∂t
∂t
where we haven taken into account the initial conditions (7.1.2)2 and (7.1.3)2 , respectively, (7.1.2)3 and (7.1.3)3 .
By using the linearity of the derivative, by derivation in (7.1.4), we obtain
2
∂2u
2∂ u
−
a
=
∂t 2
∂x 2
2
2
2
∂ u2
∂2u1
2 ∂ u1
2 ∂ u2
+
=
−a
−a
∂t 2
∂x 2
∂t 2
∂x 2
= 0 + f (t, x) = f (t, x),
where we haven taken into account Eqs. (7.1.2)1 and (7.1.3)1 .
Now, let us solve the problems (7.1.2) and (7.1.3) and then based on the Proposition 7.1.1, we obtain the solution of the problem (7.1.1).
With regard to the problem (7.1.3) we have the following result.
Theorem 7.1.1 The function U (t, x) defined by
1
U (t, x) =
2a
t x+a(t−τ )
x−a(t−τ )
0
f (τ , ξ)dξ dτ ,
is the solution for the problem (7.1.3).
Proof It is clear that
U (0, x) =
1
2a
0
0
x+a(0−τ )
x−a(0−τ )
f (τ , ξ)dξ dτ = 0.
(7.1.5)
7.1 Problem of the Infinite Vibrating Chord
225
Therefore, by using the rule of derivation of an integral with parameter, it results
x+a(t−τ )
x
t
1
∂U (t, x)
∂
1
=
f (t, ξ)dξ +
f (τ , ξ)dξ dτ =
∂t
2a x
2a 0 ∂t
x−a(t−τ )
t
1
=
a[ f (τ , x + a(t − τ )) + f (τ , x − a(t − τ ))]dτ +
2a 0
t x+a(t−τ )
∂ f (τ , ξ)
1
dξ dτ =
+
2a 0
∂t
x−a(t−τ )
t
1
=
[ f (τ , x + a(t − τ )) + f (τ , x − a(t − τ ))]dτ .
2 0
Now, it is easy to see that
1
∂U
(0, x) =
∂t
2
0
[ f (τ , x + a(0 − τ )) + f (τ , x − a(0 − τ ))]dτ = 0.
0
We use the derivative on the previous relation again with respect to t:
+
1
2
t
0
1
∂ 2U
(t, x) = [ f (t, x + a.0) + f (t, x − a.0)] +
2
∂t
2
∂
[ f (τ , x + a(t − τ )) + f (τ , x − a(t − τ ))] dτ ,
∂t
that is,
a
∂ 2U
(t, x) = f (t, x)+
2
∂t
2
t
∂ f (τ , x +a(t −τ )) ∂ f (τ , x −a(t −τ ))
−
dτ .
∂(x +a(t −τ ))
∂(x −a(t −τ ))
0
(7.1.6)
By derivation with respect to x, by using the rule of derivation of an integral with
parameter, it results
x+a(t−τ )
t
∂U (t, x)
1
∂
=
f (τ , ξ)dξ dτ =
∂x
2a 0 ∂x
x−a(t−τ )
t
1
=
[ f (τ , x + a(t − τ )) − f (τ , x − a(t − τ ))]dτ .
2a 0
Here, we derive one more time, again with respect to x, such that we are led to
1
∂ 2 U (t, x)
=
∂x 2
2a
t
0
∂
[ f (τ , x + a(t − τ )) − f (τ , x − a(t − τ ))] dτ =
∂x
226
7 Hyperbolical Equations
=
t
∂ f (τ , x + a(t − τ )) ∂(x + a(t − τ ))
−
∂(x + a(t − τ ))
∂x
0
∂ f (τ , x − a(t − τ )) ∂(x − a(t − τ ))
dτ .
−
∂(x − a(t − τ ))
∂x
1
2a
Therefore
∂ 2 U (t, x)
1
=
∂x 2
2a
t
∂ f (τ , x +a(t −τ )) ∂ f (τ , x −a(t −τ ))
dτ , (7.1.7)
−
∂(x +a(t −τ ))
∂(x −a(t −τ ))
0
such that, from Eqs. (7.1.6) and (7.1.7), we obtain
∂ 2 U (t, x)
∂ 2 U (t, x)
− a2
= f (t, x),
2
∂t
∂x 2
that is, U (t, x) verifies the Eq. (7.1.3)1 .
Now, let us solve the problem (7.1.2).
Theorem 7.1.2 The solution of the problem (7.1.2) is given by
1
1
u(t, x) = [ϕ(x + at) + ϕ(x − at)] +
2
2a
x+at
ψ(s)ds.
x−at
Proof As a first step, we obtain the canonical form of Eq. (7.1.2)1 .
By using the considerations from the Sect. 1.1 (the Chap. I), the characteristic equation in our case is
dx
dt
2
− a 2 = 0.
We can observe that = a 2 > 0 and, therefore, we are, indeed, in the case of the
hyperbolical equations. One immediately obtains the prime integrals
x + at = C1 ,
x − at = C2 ,
where C1 and C2 are arbitrarily constants. Then we perform the change of variables
ξ = x + at,
η = x − at.
It is easy to see that the transformation (7.1.8) is non-singular, because
∂(ξ, η)
a −a
= 2a > 0.
=
1 1
∂(t, x)
(7.1.8)
7.1 Problem of the Infinite Vibrating Chord
227
With the change of variables (7.1.8), the canonical form is
∂2u
= 0,
∂ξ∂η
that is,
∂
∂η
∂u
∂ξ
=0⇒
∂u
= γ(ξ), γ ∈ C 1 ((−∞, +∞)).
∂ξ
After one more integration, we obtain
u(ξ, η) =
γ(ξ)dξ + β(η) = α(ξ) + β(η),
(7.1.9)
where α is an antiderivative of the arbitrary function γ.
If we suppose that α and β are functions of the class C 1 , then the order of the
above derivation has no importance, according to the Schwarz’s classical criterion.
But, in order to verify the equation with partial derivatives of order two, the functions
α and β must be functions of the class C 2 .
Introducing Eq. (7.1.8) into (7.1.9), it follows
u(t, x) = α(x + at) + β(x − at),
(7.1.10)
where the functions α and β will be determined with the aid of the initial conditions:
ϕ(x) = u(0, x) = α(x) + β(x),
∂u
ψ(x) =
(0, x) = aα (x) − aβ (x).
∂t
This system is equivalent to
α(x) + β(x) = ϕ(x),
1 x
α(x) − β(x) =
ψ(s)ds + C,
a 0
where C is an arbitrary constant of integration. The solution of this system is
x
ϕ(x)
1
C
+
ψ(s)ds + ,
2
2a 0
2
x
ϕ(x)
1
C
β(x) =
−
ψ(s)ds − ,
2
2a 0
2
α(x) =
228
7 Hyperbolical Equations
such that, from Eq. (7.1.10), we obtain
x+at
1
C
ϕ(x + at)
+
ψ(s)ds + +
u(t, x) =
2
2a 0
2
x−at
ϕ(x − at)
1
C
+
−
ψ(s)ds − =
2
2a 0
2
x+at
1
1
= [ϕ(x + at) + ϕ(x − at)] +
ψ(s)ds.
2
2a x−at
This is even the desired result and the theorem is proved.
Remark. Based on the results from the Theorem 7.1.1, the Theorem 7.1.2 and the
Proposition 7.1.1, we will deduce that the solution of the problem (7.1.1) is
u(t, x) =
1
1
[ϕ(x + at) + ϕ(x − at)] +
2
2a
1
+
2a
t 0
x+a(t−τ )
x−a(t−τ )
x+at
ψ(s)ds+
x−at
f (τ , ξ)dξ dτ .
(7.1.11)
In this way, we proved the following result of existence.
Theorem 7.1.3 (of existence) If the given function f (t, x) is assumed to be of the
class C 0 ((0, ∞)×(−∞, +∞)), the given function ϕ(x) is of the class C 2 (−∞, +∞)
and the given function ψ(x) is of the class C 1 (−∞, +∞), then the nonhomogeneous
problem of the infinite chord admits the classical solution (7.1.11).
We call a classical solution a function u = u(t, x) of the class C 2 with respect to
x ∈ (−∞, +∞) and t > 0, which verifies the initial conditions (7.1.1)2 and (7.1.1)3
and if it is replaced in the Eq. (7.1.1)1 , we obtain an identity.
Remark. The form (7.1.11) of the solution of the problem (7.1.10) is also called
D‘Alembert’s formula for the nonhomogeneous problem of the infinite chord.
In the following theorem we prove the uniqueness of the solution of the Cauchy’s
problem (7.1.1).
Theorem 7.1.4 (of uniqueness) The single classical solution of the nonhomogeneous problem of the infinite chord is that given in (7.1.11).
Proof We suppose, through absurd, that the problem (7.1.1) admits two classical
solutions u 1 (t, x) and u 2 (t, x) and then
2
∂2ui
2 ∂ ui
−
a
= f (t, x), ∀(t, x) ∈ (−∞, +∞), ∀t > 0,
∂t 2
∂x 2
u i (0, x) = ϕ(x), ∀x ∈ (−∞, +∞),
∂u i
(0, x) = ψ(x), ∀x ∈ (−∞, +∞),
∂t
(7.1.12)
7.1 Problem of the Infinite Vibrating Chord
229
where i = 1, 2. We define the function v(t, x) by
v(t, x) = u 1 (t, x) − u 2 (t, x).
Then
∂2v
∂2v
∂2u1
∂2u1
− a2 2 =
− a2
=
2
2
∂t
∂x
∂t
∂x 2
= f (t, x) − f (t, x) = 0,
∂2u2
∂2u2
− a2
2
∂t
∂x 2
=
where we have used Eq. (7.1.12)1 .
Therefore,
v(0, x) = u 1 (0, x) − u 2 (0, x) = ϕ(x) − ϕ(x) = 0,
∂u 1
∂u 2
∂v
(0, x) =
(0, x) −
(0, x) = ψ(x) − ψ(x) = 0,
∂t
∂t
∂t
where we have used the initial conditions (7.1.12)2 and (7.1.12)3 . Thus, the function
v satisfies a problem of the form (7.1.1), where f (t, x) = ϕ(x) = ψ(x) = 0. Then,
according to Eq. (7.1.11), we have
v(t, x) = 0 ⇒ u 1 (t, x) = u 2 (t, x),
that concludes the proof of the theorem.
In order to obtain a result of stability with regard to “the right-hand side” and initial
conditions, for the problem (7.1.1), we consider that t ∈ (0, T ], where T suitably
chosen moment.
Theorem 7.1.5 (of stability) We denote by u 1 (t, x), respectively u 2 (t, x), the solutions (unique) of the following two problems
∂2ui
∂2ui
− a 2 2 = f i (t, x), ∀(t, x) ∈ (−∞, +∞), ∀t > 0,
2
∂t
∂x
u i (0, x) = ϕi (x), ∀x ∈ (−∞, +∞),
∂u i
(0, x) = ψi (x), ∀x ∈ (−∞, +∞),
∂t
(7.1.13)
where i = 1, 2 and T is a fixed moment, in a way that we will see below. Then for
any ε > 0, ∃δ = δ(ε) > 0 such that if
| f (t, x)| = | f 1 (t, x) − f 2 (t, x)| < δ,
|ϕ(t, x)| = |ϕ1 (t, x) − ϕ2 (t, x)| < δ,
|ψ(t, x)| = |ψ1 (t, x) − ψ2 (t, x)| < δ,
(7.1.14)
230
7 Hyperbolical Equations
then
|u(t, x)| = |u 1 (t, x) − u 2 (t, x)| < ε.
Proof Based on the Theorems 7.1.3 and 7.1.4, the single classical solutions of the
problems (7.1.13) are the functions u i (t, x), given by
x+at
1
1
ψi (s)ds +
u i (t, x) = [ϕi (x + at) + ϕi (x − at)] +
2
2a x−at
t
x+a(t−τ )
1
+
f i (τ , ξ)dξ dτ ,
2a 0
x−a(t−τ )
where i = 1, 2. Now, we make the difference of these two solutions
1
[ϕ1 (x + at) − ϕ1 (x + at)] +
2
x+at
1
1
+ [ϕ1 (x − at) − ϕ2 (x − at)] +
[ψ1 (s) − ψ2 (s)] ds +
2
2a x−at
t x+a(t−τ )
1
+
[ f 1 (τ , ξ) − f 2 (τ , ξ)] dξ dτ .
2a 0
x−a(t−τ )
u 1 (t, x) − u 2 (t, x) =
If we take the modulus in this equality and use the inequality of the triangle, we
obtain that the modulus of the right-hand side is less than the sum. We use then the
fact that the modulus of an integral is less than the integral of the modulus:
1
|ϕ1 (x + at) − ϕ1 (x + at)| +
2
x+at
1
1
|ψ1 (s) − ψ2 (s)| ds +
+ |ϕ1 (x − at) − ϕ2 (x − at)| +
2
2a x−at
t x+a(t−τ )
1
| f 1 (τ , ξ) − f 2 (τ , ξ)| dξ dτ .
+
2a 0
x−a(t−τ )
|u 1 (t, x) − u 2 (t, x)| ≤
If we take into account Eq. (7.1.14), this inequality leads to
|u 1 (t, x) − u 2 (t, x)| ≤
δ
+
2a
t 0
x+a(t−τ )
x−a(t−τ )
δ
δ
δ
+ +
2 2 2a
dξ dτ = δ + δt +
δ
2a
x+at
ds +
x−at
t
2a(t − τ )dτ =
0
T2
t2
≤δ 1+T +
.
=δ 1+t +
2
2
7.1 Problem of the Infinite Vibrating Chord
231
If we choose T such that
1+T +
ε
T2
< ,
2
δ
we obtain
|u 1 (t, x) − u 2 (t, x)| < ε,
that concludes the proof of the theorem.
At the end of this paragraph we make some comments with regard to the results
of existence, uniqueness and stability from the previous Theorems 7.1.3, 7.1.4 and
7.1.5.
In the case of a problem with initial data, or with boundary data, or, more generally,
in the case of a mixed initial-boundary values problem, there exists the concept of
the well posed, used for the first time by Hadamard. This concept means that for the
corresponding problem we have a theorem of uniqueness of the solution.
A theorem of uniqueness can be proved only for certain classes of functions. In
the case of the problem of the infinite chord, previously exposed we cannot have a
classical solution if we do not suppose that the prescribed functions f, ϕ and ψ are
continuous. Therefore, the class of the continuous functions is the class where one
can put the problem of the uniqueness of the solution.
If we want to prove only the uniqueness of the solution, then it is sufficient to
suppose that the functions f, ϕ and ψ are of the class C 0 on their domains of definition.
If we want to prove the existence of the solution, it is necessary to suppose that the
functions ϕ and ψ are of the class C 1 .
So, it appears the concept of the class of correctness for initial and boundary
conditions. This is the class of the functions where it must be considered the functions
from the initial and boundary conditions such that we have the uniqueness of the
solution for the respective problem.
After we prove the theorem of existence and the theorem of uniqueness, we can
talk about the existence and the uniqueness of the solutions of the problems for which
the function “right-hand side” is given and, also, the functions from the initial and
boundary conditions are prescribed.
A particular solution of a given problem is the solution that uniquely corresponds
(by virtue of the theorem of existence and uniqueness) to the right-hand side, to some
boundary data and to some fixed initial data. Therefore, for each fixed right-hand
side, initial data and boundary data we have a particular solution. In this context, the
general solution will be the family of all particular solutions.
In some cases, there exists certain solutions for which we cannot prove the theorem
of existence and uniqueness. This solution is called a singular solution.
The functions which define the right-hand side, the initial conditions and the
boundary conditions are given by the experiment. In the case of the problem of the
vibrating infinite chord, for the functions f 1 , ϕ1 and ψ1 given by an experimentalist,
we have an unique determined solution u 1 .
232
7 Hyperbolical Equations
If another experimentalist delivers the data f 2 , ϕ2 and ψ2 , for the same phenomenon, the problem will admit the unique determined solution u 2 .
If the data f 1 , ϕ1 and ψ1 is sufficient little different from the data f 2 , ϕ2 and ψ2
we have that the corresponding solutions u 1 and, respectively u 2 , are sufficient near,
we say that the solution is stable.
7.2 Initial-Boundary Values Problems
Let be a bounded domain from the space IRn with the boundary ∂ having a tangent
plane, piecewise continuously varying. As usual, we note by T T the temporal interval
TT = (0, T ] and TT = [0, T ], where T > 0.
Consider the initial boundary values, attached to the equation of the waves
u(t, x) − u tt (t, x) = f (t, x), ∀(t, x) ∈ TT × ,
u(t, y) = α(t, y), ∀x ∈ TT × ∂,
u(0, x) = ϕ(x), ∀x ∈ ,
(7.2.1)
u t (0, x) = ψ(x), ∀x ∈ ,
where the functions f, α, ϕ and ψ are given and continuous on their domains of
definition.
Definition 7.2.1 We call the classical solution of the problem (7.2.1), the function
u = u(t, x) which satisfies the conditions:
- u is a continuous function on TT × ;
- the derivatives u xi xi and u tt are continuous functions on TT × ;
- u satisfies Eq. (7.2.1)1 , the boundary condition (7.2.1)2 and the initial conditions
(7.2.1)3 and (7.2.1)4 .
We use a power method to show that the problem (7.2.1) has only one solution.
Theorem 7.2.1 The initial-boundary values problem (7.2.1) has at the most one
classical solution.
Proof We suppose, ad absurdum, that the problem (7.2.1) has two classical solutions,
u 1 (t, x) and u 2 (t, x). We define the function v by
v(t, x) = u 1 (t, x) − u 2 (t, x).
It is easy to see that the function v satisfies the conditions imposed to a classical
solutions, since u 1 (t, x) and u 2 (t, x) are classical solutions. Also, v satisfies the
problem (7.2.1) in its homogeneous form
7.2 Initial-Boundary Values Problems
233
v(t, x) − vtt (t, x) = 0, ∀(t, x) ∈ TT × ,
v(t, y) = 0, ∀(t, y) ∈ TT × ∂,
v(0, x) = 0, ∀x ∈ ,
vt (0, x) = 0, ∀x ∈ .
(7.2.2)
Now, we attach to the function v, the function E defined by
1
E(t) =
2
n
vt2 (t, ξ)
+
vx2i (t, ξ) dξ,
(7.2.3)
i=1
which is called the power integral.
We will give the proof in two steps. In the first step we will show that E(0) = 0,
and, in the second step, we prove that
d E(t)
= 0,
dt
whence it follows the conclusion that, in fact, E(t) is a constant. But, according to
the first step, E(0) = 0, and then the conclusion will be that E ≡ 0. This conclusion
together with the definition (7.2.3) of the function E lead to the conclusion that
vt = 0, vxi = 0, i = 1, 2, ..., n,
which proves that v is a constant. But, on the boundary, the function v is zero and,
therefore, we will deduce that this constant is null, that is, v ≡ 0 such that u 1 ≡ u 2 .
The first step can be immediately proved. We direct substitute t = 0 and obtain
1
E(0) =
2
n
vt2 (0, ξ) +
vx2i (0, ξ) dξ = 0,
i=1
where we have used the initial conditions (7.2.2)3 and (7.2.2)4 .
Now, we approach the second step. Due to the conditions of regularity is satisfied
by the function v, we can derive in Eq. (7.2.3) under the integral
d E(t)
=
dt
n
vt (t, ξ)vtt (t, ξ) +
vxi (t, ξ)vt xi (t, ξ) dξ.
i=1
But
vxi (t, ξ)vt xi (t, ξ)dξ =
− vt (t, ξ)vxi xi (t, ξ)dξ =
∂
∂
vxi (t, ξ)vt (t, ξ) dξ −
∂xi
vxi (t, ξ)vt (t, ξ) cos αi dσξ −
(7.2.4)
234
7 Hyperbolical Equations
−
vt (t, ξ)vxi xi (t, ξ)dξ = −
vt (t, ξ)vxi xi (t, ξ)dξ,
(7.2.5)
where, firstly, we have used the Gauss–Ostrogradski’s formula (that has been possible, taking into account that the surface ∂ admits tangent plane).
Then we have used the boundary condition (7.2.2)2 .
It results from Eq. (7.2.5) that
n
i=1
vxi (t, ξ)vxi xi (t, ξ)dξ = −
vt (t, ξ)v(t, ξ)dξ,
and then Eq. (7.2.4) becomes
d E(t)
=
dt
vt (t, ξ) [vtt (t, ξ) − v(t, ξ)] dξ = 0,
since v satisfies the homogeneous equation (7.2.2)1 .
We will prove now a result of stability for the solution of the problem (7.2.1), with
regard to the right-hand side of the equation and, also, to the initial conditions.
Theorem 7.2.2 Consider u 1 (t, x) and u 2 (t, x) are the solutions of the problems
∂2ui
(t, x) = f i (t, x), ∀(t, x) ∈ TT × ,
∂t 2
u i (t, y) = α(t, y), ∀x ∈ TT × ∂,
u i (0, x) = ϕi (x), ∀x ∈ ,
∂u i
(0, x) = ψi (x), ∀x ∈ ,
∂t
u i (t, x) −
where i = 1, 2.
We suppose that for ∀ε > 0, ∃δ = δ(ε) > 0 such that
| f 1 (t, x) − f 2 (t, x)|
|ϕ1 (t, x) − ϕ2 (t, x)|
∂ϕ1
∂ϕ1
(t, x) −
(t, x)
∂xi
∂xi
|ψ1 (t, x) − ψ2 (t, x)|
< δ,
< δ,
< δ,
< δ.
Then
|u 1 (t, x) − u 2 (t, x)| < ε.
7.2 Initial-Boundary Values Problems
235
Proof We denote by u(t, x) the difference of those two solutions
u(t, x) = u 1 (t, x) − u 2 (t, x),
and attach the power integral
E(t) =
1
2
n
u 2t (t, ξ) +
u 2xi (t, ξ) dξ.
(7.2.6)
i=1
Due to the conditions of regularity is satisfied by the function u, we can derive under
the integral in Eq. (7.2.6)
d E(t)
=
dt
+
u t (t, ξ) [u tt (t, ξ) − u(t, ξ)] dξ +
n
∂
u t (t, ξ)
u xi (t, ξ) cos αi dσξ ,
(7.2.7)
i=1
after that we have used the Gauss–Ostrogradski’s formula, as in the proof of the
Theorem 7.2.1. But on the boundary we have
∂u 1
∂u 2
∂α
∂α
∂u
=
−
=
−
= 0.
∂xi
∂xi
∂xi
∂xi
∂xi
(7.2.8)
Also,
∂2u1
∂2u2
∂2u
−
u
=
−
u
−
+ u 2 =
1
∂t 2
∂t 2
∂t 2
− f 1 (t, x) + f 2 (t, x).
If we denote by
f (t, x) = f 1 (t, x) − f 2 (t, x)
and take into account Eq. (7.2.8), the derivative from Eq. (7.2.7) becomes
d E(t)
=−
dt
u t (t, ξ) f (t, ξ)dξ.
(7.2.9)
It is easy to prove the inequality
±ab ≤
b2
a2
+ .
2
2
(∗)
236
7 Hyperbolical Equations
Thus, from Eq. (7.2.9) we will deduce
d E(t)
1
≤
dt
2
u 2t (t, ξ)dξ +
1
2
f 2 (t, ξ)dξ.
(7.2.10)
Based on the hypothesis
| f (t, x)| = | f 1 (t, x) − f 2 (t, x)| < δ
we will deduce that the last integral from Eq. (7.2.10) is arbitrarily small. We use the
notation
1
f 2 (t, ξ)dξ.
A(t) =
2 Taking into account Eq. (7.2.6), it is clear that
1
2
u 2t (t, ξ)dξ ≤ E(t),
and then Eq. (7.2.10) becomes
d E(t)
≤ E(t) + A(t),
dt
(7.2.11)
such that, by multiplying it with e−t , it results
d −t
e E(t) ≤ A(t)e−t .
dt
We integrate on the interval [0, t] and obtain
e−t E(t) ≤ E(0) +
t
e−τ A(τ )dτ ,
0
and this relation can be written in the form
t
et−τ A(τ )dτ .
E(t) ≤ et E(0) +
0
Since t ∈ (0, T ], the last inequality leads to
E(t) ≤ e T E(0) +
0
T
e T −τ A(τ )dτ .
(7.2.12)
7.2 Initial-Boundary Values Problems
237
By using the hypothesis of the theorem, we will deduce that E(0) is arbitrarily
small and, since also, A(t) is arbitrarily small too, it results that the integral from
Eq. (7.2.12) is arbitrarily small. Therefore, the function E(t) is superior bounded by
a constant which can be arbitrarily small. To show that u is arbitrarily small, we
define the function E 1 (t) by
E 1 (t) =
1
2
u 2 (t, ξ)dξ.
(7.2.13)
Based on the hypothesis of regularity of the function u, we can derive under the
integral in Eq. (7.2.13) and then obtain
d E 1 (t)
=
u(t, ξ)u t (t, ξ)dξ ≤
dt
1
1
≤
u 2t (t, ξ)dξ +
u 2 (t, ξ)dξ,
2 2 after that we have used again the above inequality (*).
We already proved
d E 1 (t)
≤ E 1 (t) + E(t),
dt
and, using the same procedure as in case of Eq. (7.2.11), it results
E 1 (t) ≤ e T E 1 (0) +
T
e T −τ E(τ )dτ .
0
Since E 1 (0) is arbitrarily small, and also, E(t) is arbitrarily small too, we will deduce
that E 1 (t) is arbitrarily small and then u is arbitrarily small.
7.3 Cauchy’s Problem
The initial-boundary values problems from the previous paragraph contain the conditions imposed on the surface which enclosed the body where the problem is stated.
In this paragraph it is assumed that the surface is to a great distance such that we can
consider that the domain of the problem is the whole space. Therefore, the boundary
condition from the formulation of the problem disappears.
We will consider the problem with initial data, that is, the Cauchy’s problem, in
the Euclidian three-dimensional space IR3 . Therefore, we have the problem
238
7 Hyperbolical Equations
∂2u
(t, x, y, z)−a 2 u(t, x, y, z) = f (t, x, y, z), ∀(t, x, y, z) ∈ (0, +∞)×IR3 ,
∂t 2
(7.3.1)
u(0, x, y, z) = ϕ(x, y, z), ∀(x, y, z) ∈ IR3 ,
∂u
(0, x, y, z) = ψ(x, y, z), ∀(x, y, z) ∈ IR3 ,
∂t
where the functions f, ϕ and ψ are given and continuous on their domains of definition, and a is a positive known constant of material.
We call a classical solution of the problem (7.3.1) a function u = u(t, x, y, z)
which satisfies the conditions:
- u and its derivatives of the first order are continuous functions on
[0, +∞) × IR3 ;
- the homogeneous derivatives of the order two of the function u are continuous
functions on (0, +∞) × IR3 ;
- u verifies Eq. (7.3.1)1 and satisfies the conditions (7.3.1)2 and (7.3.1)3 .
We define the function u(t, x, y, z) by
u(t, x, y, z) = U f (t, x, y, z) + Wψ (t, x, y, z) + Vϕ (t, x, y, z),
(7.3.2)
where the functions U f (t, x, y, z), Wψ (t, x, y, z), Vϕ (t, x, y, z) have, by definition,
the expressions
f (ξ, η, ζ, t − r/a)
dξdηdζ,
r
B(x,y,z,at)
ψ(ξ, η, ζ)
1
Wψ (t, x, y, z) =
dσat ,
4πa 2 ∂ B(x,y,z,at)
t
ψ(ξ, η, ζ)
1 ∂
dσat ,
Vϕ (t, x, y, z) =
4πa 2 ∂t
t
∂ B(x,y,z,at)
U f (t, x, y, z) =
1
4πa 2
(7.3.3)
where
r = |ξx| = 3
(xi − ξi )2 =
(x − ξ)2 + (y − η)2 + (z − ζ)2 .
i=1
Also, in the formulas (7.3.3), B(x, y, z, at) is a ball with the center at the point of
coordinates (x, y, z) and the radius a.t and ∂ B(x, y, z, at) is the boundary of this
ball, that is the sphere with the same center and the same radius.
In the following theorem we show that the function u defined in Eq. (7.3.2), is
effective the classical solution of the Cauchy’s problem (7.3.1). This is the main
result of this paragraph.
7.3 Cauchy’s Problem
239
Theorem 7.3.1 If f ∈ C 2 ((0, +∞)), ϕ ∈ C 3 (IR3 ) and ψ ∈ C 3 (IR3 ), then the
function u defined in Eq. (7.3.2) is the classical solution of the Cauchy’s problem
(7.3.1).
Proof We make the proof in three steps. In the first step we prove that the function
Wψ from Eq. (7.3.3)2 is the solution of the problem
∂ 2 Wψ
(t, x, y, z)−a 2 Wψ (t, x, y, z) = 0, ∀(t, x, y, z) ∈ (0, +∞)×IR3 ,
∂t 2
Wψ (0, x, y, z) = 0, ∀(x, y, z) ∈ IR3 ,
(7.3.4)
∂Wψ
(0, x, y, z) = ψ(x, y, z), ∀(x, y, z) ∈ IR3 .
∂t
In the second step we prove that the function Vϕ from Eq. (7.3.3)3 is the solution of
the problem
∂ 2 Vϕ
(t, x, y, z) − a 2 Vϕ (t, x, y, z) = 0, ∀(t, x, y, z) ∈ (0, +∞) × IR3 ,
∂t 2
Vϕ (0, x, y, z) = ϕ(x, y, z), ∀(x, y, z) ∈ IR3 ,
(7.3.5)
∂Vϕ
(0, x, y, z) = 0, ∀(x, y, z) ∈ IR3 ,
∂t
and, in last step, we prove that the function U f from Eq. (7.3.3)1 is the solution of
the problem
∂ 2U f
(t, x, y, z)−a 2 U f (t, x, y, z) = f (t, x, y, z), ∀(t, x, y, z) ∈ (0, +∞)×IR3 ,
∂t 2
Vϕ (0, x, y, z) = ϕ(x, y, z), ∀(x, y, z) ∈ IR3 ,
(7.3.6)
∂Vϕ
(0, x, y, z) = 0, ∀(x, y, z) ∈ IR3 .
∂t
If we prove the previous three results, then, taking into account Eq. (7.3.2), we will
deduce
∂ 2U f
∂2u
∂ 2 Wψ
− a 2 u =
− a 2 U f +
− a 2 Wψ +
2
2
∂t
∂t
∂t 2
∂ 2 Vϕ
+
− a 2 Vϕ = f (t, x, y, z) + 0 + 0 = f (t, x, y, z).
∂t 2
240
7 Hyperbolical Equations
Then
u(0, x, y, z) = U f (0, x, y, z) + Wψ (0, x, y, z) +
+Vϕ (0, x, y, z) = 0 + 0 + ϕ(x, y, z) = ϕ(x, y, z),
and, finally,
∂U f
∂Wψ
∂u
(0, x, y, z) =
(0, x, y, z) +
(0, x, y, z) +
∂t
∂t
∂t
∂Vϕ
+
(0, x, y, z) = 0 + ψ(x, y, z) + 0 = ψ(x, y, z),
∂t
that is, u from Eq. (7.3.2) effectively verifies the problem (7.3.1) and the proof will
be concluded.
Step I.
We denote by M the point of the coordinates (x, y, z) and then we can write Wψ in
the form
1
ψ(ξ, η, ζ)dσat ,
Wψ (t, x, y, z) =
4πa 2 t ∂ B(M,at)
where dσat is the element of area on the sphere of radius at.
We make the change of variables (ξ, η, ζ) → (α, β, γ):
ξ = x + αat,
η = y + βat,
ζ = z + γat.
Then
α2 + β 2 + γ 2 =
(7.3.7)
(ξ − x)2 + (η − y)2 + (ζ − z)2
= 1,
a2t 2
that is, the point of the coordinates (α, β, γ) is on the unit sphere B(M, 1). As a
consequence, the function Wψ receives the form
t
Wψ (t, x, y, z) =
4π
∂ B(M,1)
ψ(x + αat, y + βat, z + γat)dσ1 .
(7.3.8)
Taking into account that the function ψ has been assumed of the class C 2 and is
defined on a compact set (the unit sphere), we have
|t|c0
Wψ (t, x, y, z) ≤
4π
∂ B(M,1)
dσ1 = tc0 ,
7.3 Cauchy’s Problem
241
and, therefore,
Wψ (t, x, y, z) → 0 as t → 0+ ,
uniformly with regard to x, y, z, that is, Wψ satisfies the condition (7.3.4)2 .
Now, we derive with respect to t, in Eq. (7.3.8):
∂Wψ
1
ψ(x + αat, y + βat, z + γat)dσ1 +
(t, x, y, z) =
∂t
4π ∂ B(M,1)
∂ψ(x + αat, y + βat, z + γat)
at
α
+
+
4π ∂ B(M,1)
∂(x + αat)
∂ψ(x +αat, y +βat, z +γat)
∂ψ(x +αat, y +βat, z +γat)
+β
+γ
dσ1 .
∂(y +βat)
∂(z + ζat)
(7.3.9)
We denote by I2 the last integral from Eq. (7.3.9) and observe that its integrant is a
derivable function in the direction of the normal. Then
∂ψ(x + αat, y + βat, z + γat)
at
|I2 | ≤
dσ1 ≤
4π ∂ B(M,1)
∂ν
atc1
atc1
4π = atc1 ,
dσ1 =
≤
4π ∂ B(M,1)
4π
where c1 is the supremum of the derivative in the direction of the normal, which
exists due to regularity imposed to the function ψ.
Then I2 → 0, as t → 0+ , uniformly with respect to x, y, z.
For the first integral from the right-hand side of the relation (7.3.9), denoted by
I1 , we apply the mean theorem.
Therefore, there exists a point (α∗ , β ∗ , γ ∗ ) ∈ ∂ B(M, 1) such that
I1 =
1
4π
ψ(x + αat, y + βat, z + γat)dσ1 =
1
∗
∗
∗
ψ(x + α at, y + β at, z + γ at)
dσ1 =
=
4π
∂ B(M,1)
∂ B(M,1)
= ψ(x + α∗ at, y + β ∗ at, z + γ ∗ at).
Then, it is clear that I1 → ψ(x, y, z), as t → 0+ , uniformly with respect to x, y, z.
In conclusion, if we pass to the limit in Eq. (7.3.9), with t → 0+ , we obtain
lim
t→0+
∂Wψ
(t, x, y, z) = ψ(x, y, z),
∂t
that is, Wψ verifies the initial condition (7.3.4)3 .
242
7 Hyperbolical Equations
We outline that Eq. (7.3.9) can be rewritten in the form
1
∂Wψ (t, x, y, z) Wψ (t, x, y, z)
=
+
∂t
t
4πat
∂ψ(ξ, η, ζ)
dσat
∂ν
∂ B(M,at)
(7.3.10)
after that we return to the variables (ξ, η, ζ).
In the integral from Eq. (7.3.10) we apply the Gauss–Ostrogradski’s formula such
that Eq. (7.3.10) becomes
Wψ (t, x, y, z)
1
∂Wψ (t, x, y, z)
=
+
∂t
t
4πat
∂ B(M,at)
ψ(ξ, η, ζ)dξdηdζ.
(7.3.11)
We denote by I (t) the integral from Eq. (7.3.11) such that (7.3.11) can be written in
the form
Wψ (t, x, y, z)
1
∂Wψ (t, x, y, z)
=
+
I (t),
∂t
t
4πat
and, after we derive with respect to t, it follows
∂ 2 Wψ (t, x, y, z)
1 I (t).
=
∂t 2
4πat
(7.3.12)
To compute the derivative in I (t), we use the spherical coordinates:
I (t) =
=
0
at
∂ B(M,at)
π 2π
ψ(ξ, η, ζ)dξdηdζ =
ψ(r, θ, ϕ)r sin θdr dθdϕ.
0
0
Then
π
2π
ψ(r, θ, ϕ) sin θdθdϕ =
I (t) = a t
0
0
= a3t 2
ψdσ1 = a
ψ(ξ, η, ζ)dσat .
3 2
∂ B(M,1)
∂ B(M,at)
Therefore, Eq. (7.3.12) becomes
∂ 2 Wψ (t, x, y, z)
1
=
∂t 2
4πt
∂ B(M,at)
ψ(ξ, η, ζ)dσat = a 2 Wψ ,
(7.3.13)
taking into account the Definition (7.3.3)2 for Wψ and the fact that we can derive
under the integral with respect to (ξ, η, ζ), based on the regularity of the function ψ.
7.3 Cauchy’s Problem
243
The relation (7.3.13) shows that Wψ satisfies Eq. (7.3.4)1 and the first step is
completely proved.
Step I I.
Firstly, we can observe that
Vϕ (t, x, y, z) =
∂Wϕ (t, x, y, z)
,
∂t
(7.3.14)
taking into account the Definition (7.3.3)3 for Vϕ and the Definition (7.3.3)2 written
for Wϕ (instead of Wψ ).
Then
∂Wϕ (0, x, y, z)
= ϕ(x, y, z),
∂t
Vϕ (0, x, y, z) =
taking into account the first step, that is, Vϕ verifies the initial condition (7.3.5)2 .
If we derive with respect to t in Eq. (7.3.14), we obtain
∂Vϕ (t, x, y, z)
∂ 2 Wϕ (t, x, y, z)
1 =
I (t),
=
2
∂t
∂t
4πat
after we have used the equality (7.3.12).
Based on the proof from step I, we have
I (t) = a
∂ B(M,at)
ψ(ξ, η, ζ)dσat ,
and then Eq. (7.3.15) becomes
1
∂Vϕ (t, x, y, z)
=
ψ(ξ, η, ζ)dσat =
∂t
4πt ∂ B(M,at)
a2t
=
ψ(x + αat, y + βat, z + γat)dσ1 .
4π ∂ B(M,1)
Thus, we will deduce that
∂Vϕ (t, x, y, z)
→ 0, for t → 0+ .
∂t
Since the integral
∂ B(M,1)
ψ(x + αat, y + βat, z + γat)dσ1
is a bounded function, based on the regularity of the function ϕ.
(7.3.15)
244
7 Hyperbolical Equations
Therefore Vϕ satisfies the initial condition (7.3.5)3 .
Taking into account Eq. (7.3.14), it results
∂ 2 ∂Wϕ
∂Wϕ
∂ 2 Vϕ
2
− a2
− a Vϕ = 2
=
2
∂t
∂t
∂t
∂t
2
∂ ∂ Wϕ
=
− a 2 Wϕ = 0,
∂t
∂t 2
since in the step I we already proved that
∂ 2 Wϕ
− a 2 Wϕ = 0.
∂t 2
In the conclusion, Vϕ satisfies Eq. (7.3.5)1 and the proof steps II is concluded.
Step I I I.
Firstly, from Eq. (7.3.3)1 we immediately can deduce that
1
lim
lim U f (t, x, y, z) =
t→0+
4πa 2 t→0+
B(M,at)
f (ξ, η, ζ, t − r/a)
dξdηdζ = 0,
r
taking into account the regularity of the function f and the fact that, to the limit, the
ball B(M, at) reduces to the point (x, y, z).
Therefore U f satisfies the initial condition (7.3.6)2 . We write now U f in the form
t f (ξ, η, ζ, t − /a)
1
U f (t, x, y, z) =
dσ d =
4πa 2 0
∂ B(M,)
t 1
d.
(7.3.16)
=
f
(x
+α,
y
+β,
z
+γ,
t
−)dσ
4πa 2 0 ∂ B(0,1)
Then
∂U f (t, x, y, z)
1
=
f (x +α, y +β, z +γ, t −)tdσ +
∂t
4πa 2 ∂ B(0,1)
t 1
(7.3.17)
+
f
(x
+α,
y
+β,
z
+γ,
t
−)dσ
d.
4πa 2 0 ∂ B(0,1)
The second integral from Eq. (7.3.17) disappears for t = 0. For the first integral we
use the mean theorem and then this integral becomes the product between t and a
bounded constant and, therefore, tends to zero, for t → 0+ , that is
lim+
t→0
∂U f (t, x, y, z)
= 0,
∂t
7.3 Cauchy’s Problem
245
the limit taking place uniformly with respect to (x, y, z). Therefore U f satisfies the
initial condition (7.3.6)3 .
It remains to prove that U f verifies Eq. (7.3.6)1 . To this we introduce the notation
=
t −τ
4πa 2
U1 (t, τ , x, y, z) =
∂ B(0,1)
f (x +(t −τ )ξ, y +(t −τ )η, z +(t −τ )ζ, t −)dσ1 .
(7.3.18)
Then Eq. (7.3.16) becomes
t
U f (t, x, y, z) =
U1 (t, τ , x, y, z)dτ .
(7.3.19)
0
Starting from Eq. (7.3.18) we obtain, without difficulty, the relations
∂ 2 U1 (t, τ , x, y, z)
− U1 (t, τ , x, y, z) = 0,
∂t 2
U1 (t, t, x, y, z) = 0,
∂U1 (t, t, x, y, z)
= f (t, x, y, z).
∂t
(7.3.20)
Then from Eq. (7.3.19) it results
t 2
∂ 2 U f (t, x, y, z)
∂ U1 (t, τ , x, y, z)
∂U1 (t, t, x, y, z)
+
=
dτ =
2
∂t
∂t
∂t 2
0
t 2
∂ U1 (t, τ , x, y, z)
= f (t, x, y, z) +
dτ .
(7.3.21)
∂t 2
0
On the other hand, we have
t
U f (t, x, y, z) =
U1 (t, τ , x, y, z)dτ =
0
t
=
0
∂ 2 U1 (t, τ , x, y, z)
dτ ,
∂t 2
(7.3.22)
where we have taken into account the relation (7.3.20)1 .
From Eqs. (7.3.21) and (7.3.22), by substrating, it results
∂ 2 U f (t, x, y, z)
− U f (t, x, y, z) = f (t, x, y, z),
∂t 2
that is, U f verifies the nonhomogeneous equation (7.3.6)1 and the proof of the last
is over. In the same time, the proof of the theorem is concluded.
246
7 Hyperbolical Equations
The formula (7.3.2) which gives the form of the solution for the Cauchy’s problem
(7.3.1) is called the Kirchhoff’s formula.
The Kirchhoff’s formula is also useful to prove the uniqueness of the solution of the
Cauchy’s problem. Indeed, if the problem (7.3.1) admits two solutions, u 1 (t, x, y, z)
and u 2 (t, x, y, z), then we denote by u(t, x, y, z) its difference,
u(t, x, y, z) = u 1 (t, x, y, z) + u 2 (t, x, y, z).
It is easy to see that u(t, x, y, z) satisfies a Cauchy’s problem of the form (7.3.1) where
f ≡ 0, ϕ ≡ 0 and ψ ≡ 0. If we write the Kirchhoff’s formula for u, obviously, we
obtain u ≡ 0 from where we will deduce that u 1 ≡ u 2 .
Finally, the Kirchhoff’s formula can be used to prove a result of stability for the
solution of the Cauchy’s problem (7.3.1), with regard to the right-hand side and initial
conditions.
7.4 Problem of the Finite Vibrating Chord
Let us consider the case of the finite chord. The general aim of this paragraph is
to study the following initial-boundary values problem, attached to the equation
(homogeneous, in first instance) of the finite chord
2
∂2u
2∂ u
−
a
= 0, ∀x ∈ [0, l], ∀t > 0,
∂t 2
∂x 2
u(0, x) = ϕ(x), ∀x ∈ [0, l],
∂u
(0, x) = ψ(x), ∀x ∈ [0, l],
∂t
(7.4.1)
where the functions f (t, x), ϕ(x) and ψ(x) are given and continuous on their
domains of definition. The function u = u(t, x) is the unknown function of the
problem and represents the amplitude of the chord at the moment t, at the point x.
The positive constant a is prescribed for each type of the material of the chord and
the constant l represents the length of the chord.
The mixt initial-boundary value problem is complete if we add the boundary
conditions
u(t, 0) = g1 (t), u(t, l) = g2 (t), ∀t > 0,
where the functions g1 (t) and g2 (t) are given and describe the behavior of the ends
of the chord.
For the sake of simplicity we consider only the case g1 (t) = g2 (t) = 0 and we
say that the ends of the chord are fixed.
The procedure to solve the above considered problem is based on the BernoulliFourier’s method, which is called, also, the “separating the variables” method.
7.4 Problem of the Finite Vibrating Chord
247
We try to find a solution of the form
u(t, x) = X (x)T (t)
so that the derivatives become
∂u
∂u
= X T,
= XT
∂x
∂t
∂2u
∂2u
=
X
T,
= X T .
∂x 2
∂t 2
The considered partial differential equation is transformed in an ordinary differential
equation
X T − a 2 X T = 0,
which can be restated in the form
X 1 T =
.
2
a T
X
It is easy to see that both sides of this relation are constants, such that we can write
X − k X = 0, T − ka 2 T = 0,
where the constant k is the common value of the above ratios.
Taking into account the boundary conditions, we obtain
u(t, 0) = 0 ⇒ X (0)T (t) = 0 ⇒ X (0) = 0
u(t, l) = 0 ⇒ X (l)T (t) = 0 ⇒ X (l) = 0.
In this way, with regard to the function X (x) we have the following problem
X − k X = 0,
X (0) = 0, X (l) = 0.
(7.4.2)
The characteristic equation attached to the above differential equation (having constant coefficients) is
r 2 − k = 0,
such that we must consider three cases.
248
7 Hyperbolical Equations
I. k = 0 In this case the equation reduces to X = 0 so that
X (x) = C1 x + C2
and, taking into account the boundary condition from Eq. (7.4.2), the constants C1
and C2 become zero. In conclusion, in this case we obtain the solution X (x) = 0
which does not satisfy our problem (7.4.1).
√
II. k > 0 In this case the characteristic equation has two real roots ± k and the
differential equation has the general solution
X (x) = C1 e
√
kx
√
+ C 2 e−
kx
.
Taking into account the boundary condition from (7.4.1), the constants C1 and C2
become zero. Therefore, also in this case we obtain the solution X (x) = 0 which
does not satisfy our problem (7.4.1).
III. k < 0 Denote k = −λ2 . In this case the characteristic equation has two conjugated complex roots ±iλ and the differential equation has the general solution
X (x) = C1 cos λ + C2 sin λ.
Taking into account the boundary condition from Eq. (7.4.2), in order to determine
the constants C1 and C2 , we obtain the relations C1 = 0 and C2 sin λl = 0. Then
sin λl = 0 ⇒ λ = nπ, n = 0, 1, 2, ...
So, we obtain an infinite number of values for the parameter λ, called proper values
for the problem of the finite vibrating chord
λn =
nπ
, n = 0, 1, 2, ...
l
Corresponding, from the general form of the solution, we find an infinite number of
functions X (x), called proper functions
X n = Cn sin
nπ
x, Cn = constants, n = 0, 1, 2, ...
l
Taking into account the value of the parameter k = −λ2 , for the determination of
the function T (t) we have the equation
T −
nπa l
T = 0,
7.4 Problem of the Finite Vibrating Chord
249
which has the solutions
Tn = Dn cos
nπa
nπa
t + E n sin
t,
l
l
where Dn , E n =constants, n = 0, 1, 2, ...
In conclusion, for our initial mixt problem, there exists an infinite number of
particular solutions
nπ
nπa
nπa t + Bn sin
t sin
x, n = 0, 1, 2, ..
u n (t, x) = X n (x)Tn (t) = An cos
l
l
l
where we used the notation
An = Cn Dn , Bn = Cn E n .
Since our problem is a linear one, its general solution will be a linear combination
of the particular solutions, that is
∞
u(t, x) =
An cos
n=0
nπa
nπa nπ
t + Bn sin
t sin
x.
l
l
l
In order to determine the coefficients An and Bn we can use the initial conditions of
the mixt problem. Firstly, we have
∞
ϕ(x) = u(0, x) =
An sin
n=0
nπ
x.
l
This is the Fourier’s series of the function ϕ(x) and then An are the Fourier’s coefficients of the function ϕ(x). Using the known formula for these coefficients, we
obtain
l
2
nπ
xdx.
ϕ(x) sin
An =
l
l
0
In order to use the other initial condition, we have
∂u
(t, x) =
∂t
nπa
nπa
nπ
nπa
nπa sin
t+
Bn cos
t sin
x.
−An
l
l
l
l
l
n=0
∞
Then, the second initial condition leads to
ψ(x) =
∂u
(0, x) =
∂t
∞
n=0
nπa
nπ
Bn sin
x.
l
l
250
7 Hyperbolical Equations
This is the Fourier’s series of the function ψ(x) and then
naπ
Bn
l
are the Fourier’s coefficients of the function ψ(x). Using the known formula for these
coefficients, we obtain
l
naπ
2
Bn =
l
l
ψ(x) sin
nπ
xdx,
l
0
such that
l
2
Bn =
naπ
ψ(x) sin
nπ
xdx.
l
0
In conclusion, the solution of the homogeneous finite problem of the vibrating chord
is
∞ nπ
nπa
nπa t + Bn sin
t sin
x,
u(t, x) =
An cos
l
l
l
n=0
where the coefficients An and Bn have the expressions
2
An =
l
l
ϕ(x) sin
nπ
xdx,
l
ψ(x) sin
nπ
xdx.
l
0
2
Bn =
naπ
l
0
At the end of this paragraph, we consider the nonhomogeneous finite problem of the
vibrating chord, that is
2
∂2u
2∂ u
−
a
= f (t, x), ∀x ∈ [0, l], ∀t > 0,
∂t 2
∂x 2
u(0, x) = ϕ(x), ∀x ∈ [0, l],
∂u
(0, x) = ψ(x), ∀x ∈ [0, l],
∂t
u(0, t) = u(l, t) = 0.
(7.4.3)
In order to solve this problem, we decompose it in two problems: one having the
differential equation in homogeneous form and nonhomogeneous initial conditions
and, the second having nonhomogeneous differential equation and homogeneous
7.4 Problem of the Finite Vibrating Chord
251
initial conditions:
∂2u
∂2u
− a 2 2 = 0, ∀x ∈ [0, l], ∀t > 0,
2
∂t
∂x
u(0, x) = ϕ(x), ∀x ∈ [0, l],
∂u
(0, x) = ψ(x), ∀x ∈ [0, l],
∂t
u(t, 0) = u(t, l) = 0.
∂2u
∂2u
− a 2 2 = f (t, x), ∀x ∈ [0, l], ∀t > 0,
2
∂t
∂x
u(0, x) = 0, ∀x ∈ [0, l],
∂u
(0, x) = 0, ∀x ∈ [0, l],
∂t
u(t, 0) = u(t, l) = 0.
(7.4.4)
(7.4.5)
Of course, to solve the problem (7.4.4) we use the above exposed procedure. Let us
solve the problem (7.4.5). We will find a particular solution of the form
∞
u p (t, x) =
Cn (t) sin
n=1
nπ
x,
l
that is, the unknown coefficients Cn (t) depend only of t.
It is easy to see that
u p (t, 0) = u p (t, l) = 0.
Also, derivating in Eq. (7.4.6) with respect to t, we have
∂u p
=
∂t
and
∂2u p
=
∂t 2
∞
Cn (t) sin
nπ
x,
l
Cn (t) sin
nπ
x.
l
n=1
∞
n=1
Now, derivating in Eq. (7.4.6) with respect to x, we obtain
∂u p
=
∂x
∞
Cn (t)
n=1
nπ
nπ
cos
x,
l
l
(7.4.6)
252
7 Hyperbolical Equations
and
nπ 2
∂2u p
nπ
x.
=
−
C
(t)
sin
n
∂x 2
l
l
n=1
∞
Then, the equation from the problem (7.4.5) reduces to
naπ 2
nπ
Cn (t) +
x = f (t, x).
Cn (t) sin
l
l
n=1
∞
This relation can be considered as the Fourier’s series of the function f (t, x) and
then the above square brackets are the Fourier’s coefficients of the function f (t, x)
Cn (t)
+
naπ 2
l
2
Cn (t) =
l
l
f (t, x) sin
nπ
x = h n (t),
l
0
where the last equality being a notation.
Since u p (0, x) = 0 we deduce
∞
Cn (0) sin
n=1
Also, since
we deduce
nπ
x = 0 ⇒ Cn (0) = 0.
l
∂u p
(0, x) = 0,
∂t
∞
Cn (0) sin
n=1
nπ
x = 0 ⇒ Cn (0) = 0.
l
Therefore, to find the coefficients Cn (t) we must solve the simple Cauchy’s problem,
attached to an ordinary differential equation
Cn (t) +
naπ 2
Cn (0) = 0,
l
Cn (t) = h n (t),
Cn (0) = 0.
In conclusion, the solution of the problem (7.4.5) is the function (7.4.6) where the
coefficients Cn (t) satisfy the above Cauchy’s problem.
It is a simple matter to show that the solution of the mixt initial-boundary values
problem of the nonhomogeneous finite vibrating chord (7.4.3) is the sum of the
solution of problem (7.4.4) with the solution of the problem (7.4.5).
Chapter 8
Parabolical Equations
8.1 The Finite Problem of Heat
The main exponent of the parabolical equations is the equation of heat conduction
in a body.
The general aim of this paragraph is to study the following initial-boundary values problem, attached to the equation (homogeneous, in first instance) of the heat
conduction in a rod. This is a bar with its cross section small in comparison with the
length.
∂2u
∂u
− a 2 2 = 0, ∀x ∈ [0, l], ∀t > 0,
∂t
∂x
u(0, x) = ϕ(x), ∀x ∈ [0, l],
(8.1.1)
where the functions f (t, x), ϕ(x) and ψ(x) are given and continuous on their
domains of definition. The function u = u(t, x) is the unknown function of the
problem and represents the temperature in the rod at the moment t, at the point x.
The positive constant a is prescribed for each type of the material of the rod and the
constant l represents the length of the rod.
The mixt initial-boundary value problem is complete if we add the boundary
conditions
u(t, 0) = g1 (t), u(t, l) = g2 (t), ∀t > 0,
where the functions g1 (t) and g2 (t) are given and describe the behavior of the ends
of the rod.
For the sake of simplicity we consider only the case g1 (t) = g2 (t) = 0 and we
say that the ends of the rod are free of temperature.
The procedure to solve the above considered problem is based on the Bernoulli–
Fourier’s method, which, also, is called the “separating of the variables” method.
© Springer International Publishing AG, part of Springer Nature 2018
M. Marin and A. Öchsner, Complements of Higher Mathematics,
https://doi.org/10.1007/978-3-319-74684-5_8
253
254
8 Parabolical Equations
We try to find a solution of the form
u(t, x) = X (x)T (t)
so that the derivatives become
∂u
∂u
= X T,
= XT
∂x
∂t
∂2u
= X T.
∂x 2
The considered partial differential equation is transformed in an ordinary differential
equation
X T − a 2 X T = 0,
which can be restated in the form
X 1 T
=
.
2
a T
X
It is easy to see that both sides of this relation are constants, such that we can write
X − k X = 0, T − ka 2 T = 0,
where the constant k is the common value of the above ratios.
Taking into account the boundary conditions, we obtain
u(t, 0) = 0 ⇒ X (0)T (t) = 0 ⇒ X (0) = 0
u(t, l) = 0 ⇒ X (l)T (t) = 0 ⇒ X (l) = 0.
In this way, with regard to the function X (x) we have the following problem
X − k X = 0,
X (0) = 0, X (l) = 0.
(8.1.2)
The characteristic equation attached to the above differential equation (having constant coefficients) is
r 2 − k = 0,
such that we must consider three cases.
8.1 The Finite Problem of Heat
255
I. k = 0 In this case the equation reduces to X = 0 so that
X (x) = C1 x + C2
and, taking into account the boundary condition from Eq. (8.1.2), the constants C1
and C2 become zero. In conclusion, in this case we obtain the solution X (x) = 0
which does not satisfy our problem (8.1.2).
√
II. k > 0 In this case the characteristic equation has two real roots ± k and the
differential equation has the general solution
X (x) = C1 e
√
kx
√
+ C 2 e−
kx
.
Taking into account the boundary condition from Eq. (8.1.2), the constants C1 and
C2 become zero. Therefore, also in this case we obtain the solution X (x) = 0 which
does not satisfy our problem (8.1.2).
III. k < 0 Denote k = −λ2 . In this case the characteristic equation has two conjugated complex roots ±iλ and the differential equation has the general solution
X (x) = C1 cos λ + C2 sin λ.
Taking into account the boundary condition from Eq. (8.1.2), in order to determine
the constants C1 and C2 , we obtain the relations C1 = 0 and C2 sin λl = 0. Then
sin λl = 0 ⇒ λ = nπ, n = 0, 1, 2, ...
So, we obtain an infinite number of values for the parameter λ, called the proper
values for the finite problem of the heat conduction:
λn =
nπ
, n = 0, 1, 2, ...
l
Corresponding, from the general form of the solution, we find an infinite number of
functions X (x), called the proper functions
X n = Cn sin
nπ
x, Cn = constants, n = 0, 1, 2, ...
l
Taking into account the value of the parameter k = −λ2 , for the determination of
the function T (t) we have the equation
T −
nπa l
T = 0,
256
8 Parabolical Equations
from where it results
Tn = Dn e−(
nπa
l
)2 t , Dn = constants, n = 0, 1, 2, ...
In conclusion, for our initial mixt problem, there exists an infinite number of particular
solutions
u n (t, x) = X n (x)Tn (t) = An e−(
nπa
l
)2 t sin nπ x, n = 0, 1, 2, ..
l
where we used the notation
A n = C n Dn .
Since our problem is linear, its general solution will be a linear combination of the
particular solutions, that is
u(t, x) =
∞
An e−(
nπa
l
n=0
)2 t sin nπ x.
l
In order to determine the coefficients An we can use the initial condition of the mixt
problem. So, we have
ϕ(x) = u(0, x) =
∞
n=0
An sin
nπ
x.
l
This is the Fourier’s series of the function ϕ(x) and then An are the Fourier’s coefficients of the function ϕ(x). Using the known formula for these coefficients, we
obtain
l
2
nπ
xdx.
ϕ(x) sin
An =
l
l
0
In conclusion, the solution of the homogeneous finite problem of the heat conduction
is
l
∞
2
nπ
2
nπ
−( nπa
t
)
x, An =
xdx.
u(t, x) =
An e l
sin
ϕ(x) sin
l
l
l
n=0
0
At the end of this paragraph, we consider the nonhomogeneous finite problem of the
heat conduction, that is
∂2u
∂u
− a 2 2 = f (t, x), ∀x ∈ [0, l], ∀t > 0,
∂t
∂x
u(0, x) = ϕ(x), ∀x ∈ [0, l],
u(0, t) = u(l, t) = 0.
(8.1.3)
8.1 The Finite Problem of Heat
257
In order to solve this problem, we decompose it in two problems: one having the differential equation in its homogeneous form and nonhomogeneous initial conditions
and, the second having a nonhomogeneous differential equation and homogeneous
initial conditions:
∂u
∂2u
− a 2 2 = 0, ∀x ∈ [0, l], ∀t > 0,
∂t
∂x
u(0, x) = ϕ(x), ∀x ∈ [0, l],
u(t, 0) = u(t, l) = 0.
∂u
∂2u
− a 2 2 = f (t, x), ∀x ∈ [0, l], ∀t > 0,
∂t
∂x
u(0, x) = 0, ∀x ∈ [0, l],
(8.1.4)
(8.1.5)
u(t, 0) = u(t, l) = 0.
Of course, in order to solve the problem (8.1.4) we use the above exposed procedure.
Let us solve the problem (8.1.5). We will find a particular solution of the form
u p (t, x) =
∞
n=1
Cn (t) sin
nπ
x,
l
that is, the unknown coefficients Cn (t) depend only on t.
It is easy to see that
u p (t, 0) = u p (t, l) = 0.
Also, derivating in Eq. (8.1.6) with respect to t, we have
∞
∂u p
nπ
=
x.
Cn (t) sin
∂t
l
n=1
Now, derivating in Eq. (8.1.6) with respect to x, we obtain
∞
∂u p
nπ
nπ
=
cos
x,
Cn (t)
∂x
l
l
n=1
and
∞
nπ 2
∂2u p
nπ
x.
=
−
C
(t)
sin
n
2
∂x
l
l
n=1
(8.1.6)
258
8 Parabolical Equations
Then, the equation from (8.1.5) reduces to
∞ naπ 2
nπ
Cn (t) +
x = f (t, x).
Cn (t) sin
l
l
n=1
(8.1.7)
We multiply both sides of this equality by the functions
sin
mπ
x
l
and use the fact that these functions are orthogonal:
l
sin
0
l
nπ
nπ
cos(n − m) x − cos(n + m) x dx.
l
l
0
nπ
mπ
1
x sin
xdx =
l
l
2
In order to evaluate the last integral, we must consider two cases. Firstly, if n = m
we obtain
nπ
nπ
1 l
cos(n − m) x − cos(n + m) x dx =
2 0
l
l
1
l
(n − m)π l
(n + m)π l
l
=
sin
x|0 −
sin
x|0 = 0.
2 (n − m)π
l
(n + m)π
l
If n = m we have
l
sin
0
mπ
nπ
x sin
xdx =
l
l
=
l
sin2
0
1
nπ
xdx =
l
2
l
1 − cos
0
2nπ
x dx =
l
1
l
2nπ l
l
2nπ l
l
l
x|l0 −
sin
x|0 = −
sin
x|0 = .
2
2nπ
l
2 4nπ
l
2
In conclusion, the scalar product takes the following value
sin
mπ nπ
x, sin
x =
l
l
l
sin
0
mπ
nπ
x sin
xdx =
l
l
0, n = m
l/2, n = m.
After these calculations, from Eq. (8.1.7) we obtain
l
l
∞ naπ 2
mπ
nπ
mπ
x sin
xdx =
xdx ⇒
Cn (t) +
Cn (t) sin
f (t, x) sin
l
l
l
l
0 n=1
0
l
l
∞ naπ 2
mπ
nπ
mπ
x sin
xdx =
xdx.
Cn (t) +
Cn (t)
sin
f (t, x) sin
⇒
l
l
l
l
0
0
n=1
8.1 The Finite Problem of Heat
259
Using the above result with regard to integral in the left-hand side, we deduce
l
naπ 2
l
nπ
Cn (t) +
Cn (t)
f (t, x) sin
=
xdx,
l
2
l
0
which can be restated in the form
Cn (t) +
naπ 2
l
Cn (t) =
Using the notation
Hn (t) =
2
l
l
2
l
l
f (t, x) sin
0
f (t, x) sin
0
nπ
xdx.
l
nπ
xdx,
l
the above equation can be written in the form
Cn (t) +
naπ 2
l
Cn (t) = Hn (t),
which is a linear and nonhomogeneous differential equation of first order. It is well
known that a linear and homogeneous differential equation
y + P(x)y = Q(x)
has the solution
y(x) = e
x
0
−
P(t)dt
C+
x
Q(t)e
t
0
P(t)dt
dt , C = constant.
0
In our case, we have (taking C = 0):
Cn (y) =
t
e
−(naπ/l)2 t
Hn (s)e
(naπ/l)2 s
t
ds =
0
Hn (s)e−(naπ/l)
0
Then, the particular solution of the nonhomogeneous equation is
u p (t, x) =
∞
Cn (t) sin
0
=
∞ Hn (s)e−(naπ/l)
0
0
=
t
l ∞
0
0
Hn (s)e
2
(t−s)
nπ
x=
l
nπ
xds =
ds sin
l
−(naπ/l)2 (t−s)
nπ
x ds.
sin
l
2
(t−s)
ds.
260
8 Parabolical Equations
We now remember that the general solution of the homogeneous equation is
u 0 (t, x) =
∞
An (t)e−(naπ/l) t sin
2
0
nπ
2
x, An =
l
l
l
ϕ(x) sin
0
nπ
xdx,
l
and this can be restated in the form
u 0 (t, x) =
∞
2
l
0
l
ϕ(s) sin
0
nπ −(naπ/l)2 t
se
ds.
l
Then, taking into account that the general solution of the nonhomogeneous equation
is
u(t, x) = u 0 (t, x) + u p (t, x),
we can write
u(t, x) =
∞ 0
where
2
An =
l
l
0
An +
0
t
nπ
2
2
x,
Hn (s)e(naπ/l) s ds e−(naπ/l) t sin
l
nπ
2
xdx, Hn =
ϕ(x) sin
l
l
l
f (t, x) sin
0
nπ
xdx.
l
In the following we will use an operational method to solve the problem of the
heat conduction in a semi-infinite one-dimensional rod. In fact we use the Laplace’s
transform to solve this problem.
Let us consider a semi-infite one-dimensional rod free of temperature at the initial
moment. The variation of temperature in this rod can be computed by the following
mathematical model
∂2u
1 ∂u
, (t, x) ∈ (0, ∞) × (0, ∞)
= 2
∂x 2
a ∂t
u(0, x) = 0, x > 0
u(t, 0) = u 0 , t > 0.
Here u = u(t, x) is the unknown function of the problem and represents the temperature in the rod at the moment t in the point x of the rod. Also, u 0 is a given constant
and represents the value of the temperature at the end of the rod in any moment.
Denote
U ( p, x) = Lu(t,x) ,
that is, the Laplace’s transform of the function u(t, x). As we know this integral’s
transform is
8.1 The Finite Problem of Heat
261
∞
Lu(t,x) =
u(t, x)e− pt dt.
0
It is easy to prove the following results
L
∂u
∂t
= pLu(t,x) − u(0, x) = pU ( p, x),
=
∂u
∂x
L
∂
∂x
∞
=
∂u − pt
e dt =
∂x
0
∞
u(t, x)e− pt dt =
0
Of course, we can write
∂U
.
∂x
∂U
dU
=
,
∂x
dx
because p is considered as a parameter. Therefore we have
∂u
∂x
L
Similarly,
∂2u
L
∂x 2
=
∂2
∂x 2
in f t y
=
=
in f t y
0
dU
.
dx
∂ 2 u − pt
e dt =
∂x 2
u(t, x)e− pt dt =
0
∂ 2U
d 2U
=
.
∂x 2
dx2
After these calculations the heat conduction equation becomes
1
d 2U
= 2 U ( p, x),
2
dx
a
which is an ordinary differential equation having constant coefficients. Its characteristic equation is
√
p
p
2
r = 2 ⇒r =±
a
a
such that the general solution is
U ( p, x) = C1 e−
√
p
a
x
+ C2 e
√
p
a
x
.
262
8 Parabolical Equations
This solution must by finite when x → ∞ and then we obtain C2 = 0 and the
solution becomes
√
p
U ( p, x) = C1 e− a x .
Now, we can use the condition on the end x = 0
Lu(t,0) = Lu 0 =
that is
C 1 e−
√
p
a 0
=
u0
,
p
u0
u0
⇒ C1 = .
p
p
Finally, the solution of the equation is
u0 − √p x
e a .
p
U ( p, x) =
Then, the solution of the heat conduction equation is
u(t, x) = L−1
u0 − √p x
e a
.
p
In order to compute the above inverse Laplace’s transform, we remember that
(α + 1)
.
p α+1
L (t α ) =
Indeed, starting from the definition, we have
L (t α ) =
∞
t α e− pt dt,
0
where we make the change of variable pt = τ such that
1
dτ
⇒ L (t α ) = α+1
dt =
p
p
∞
τ α e−τ dτ =
0
In the particular case α = −1/2 we obtain
√
(1/2)
π
L t −1/2 = √
=√ ,
p
p
such that we deduce
1
1
√ =L √
p
πt
.
(α + 1)
.
p α+1
8.1 The Finite Problem of Heat
263
Now, we use the well known series
eu = 1 +
where u = −
1
1
1
u + u 2 + ... + u n + ...
1!
2!
n!
√
p
x.
a
e
−
Then
The previous series becomes
√
p
a
x
√
√
1 p 2
1 p
1 p p 3
x+
=1−
x −
x + ...
1! a
2! a 2
3! a a 2
1 − √p x
1
x3 √
x 1
x2
−
p + ...
e a = − √ +
2
p
p
a p 2!a
3!a 3
So, we can write
+ ... =
x
1
u0 − √p x
e a = u 0 L (1) − L √
p
a
πt
x 1
x 1
= u 0 L 1 − √ + ... = L u 0 1 − √ + ... .
a πt
a πt
So, we deduce that the solution (above written as an inverse of the Laplace’s transform) becomes
x 1
u(t, x) = u 0 1 − √ + ... .
a πt
We can write this result in the form
u(t, x) = u 0 1 − erf
x
√
2a t
+ ... ,
where er f is the function of errors, more utilised in the numerical analysis, given by
2
erf(x) = √
π
x
e−τ dτ .
0
8.2 Initial-Boundary Value Problems
Let be a bounded domain from IRn having the boundary ∂ and = ∪ ∂. For
the temporally constant T > 0, arbitrarily fixed, consider the temporally interval TT
given by
TT = {t : 0 < t ≤ T }, TT = {t : 0 ≤ t ≤ T }.
264
8 Parabolical Equations
Then, the equation of heat conduction (shortly, the heat equation) is
u t (t, x) − a 2 u(t, x) = f (t, x), ∀(t, x) ∈ TT × .
(8.2.1)
Here we have used the notation u t = ∂u/∂t, a is a positive given constant, and is
the Laplace’s operator.
Usually, to the Eq. (8.2.1) we add the initial condition in the form:
u(0, x) = ϕ(x), ∀x ∈ .
(8.2.2)
The boundary conditions have the following form:
• Dirichlet’s condition
u(t, y) = α(t, y), ∀(t, y) ∈ TT × ∂;
(8.2.3)
• Neumann’s condition
∂u
(t, y) = β(t, y), ∀(t, y) ∈ TT × ∂;
∂ν
(8.2.4)
• Mixt condition
λ1
∂u
(t, y) + λ2 u(t, y) = γ(t, y), ∀(t, y) ∈ TT × ∂;
∂ν
(8.2.5)
If we consider, for instance, the problem (8.2.1), (8.2.2) and (8.2.3), then we have
the following physical significations:
- u(t, x), which is the unknown function of the problem, represents the temperature
in the body , at any moment t;
- ϕ(x) represents the (known) temperature at the initial moment in all points of the
body (the points of boundary included);
- α(t, y) represents the (known) temperature at any moment on the surface ∂ which
encloses the body.
Therefore, the problem (8.2.1), (8.2.2) and (8.2.3) consists of the determination of
the temperature in all points of the body , at any moment, knowing the temperature
in the body at the initial moment and also, knowing at any moment the temperature
on the surface of the body, ∂.
In all that follows we will study, especially, the problem (8.2.1), (8.2.2) and (8.2.3).
In view of the characterization of this problem we will consider, for the moment, the
following standard hypotheses:
(i) the function f : TT × ∂ → IR is given and f ∈ C(TT × ∂);
(ii) the function ϕ : → IR is given and ϕ ∈ C();
(iii) the function α : TT × ∂ → IR is given and α ∈ C(TT × ∂).
8.2 Initial-Boundary Value Problems
265
We call the classical solution of the problem (8.2.1), (8.2.2) and (8.2.3) a function
u = u(t, x), u : TT × → IR, having the properties:
– u ∈ C(TT × );
– u t , u xi xi ∈ C(TT × );
– u satisfies the Eq. (8.2.1), the initial condition (8.2.2) and the boundary condition
(8.2.3).
In the formulation of the problem (8.2.1), (8.2.2) and (8.2.3), the boundary and initial
values are given on the set TT × ∂ or on the set {0} × .
We define the set by
= (t, x) : (t, x) ∈ TT × ∂ ∪ {0} × ,
(8.2.6)
and we call it the parabolical boundary, which is different of the topological boundary. In fact, to obtain the parabolical boundary we remove from the topological
boundary “the lid” for t = T .
In the following theorem we prove a result with regard to the extreme values, in
the case of the homogeneous parabolical equations.
u t (t, x) − u(t, x) = 0, ∀(t, x) ∈ TT × .
(8.2.7)
Theorem 8.2.1 Consider and TT , defined as above, and consider the function u
such that u ∈ C(T T × ), u t , u xi xi ∈ C(TT × ). If u satisfies the homogeneous
equation (8.1.7), then the extreme values
u(t, x),
sup
(t,x)∈T T ×
inf
(t,x)∈T T ×
u(t, x)
are taken, necessarily, on .
Proof If we make the proof for the supremum, the result for the infimum immediately
follows by substituting u to −u.
Firstly, we must outline that in the conditions of the theorem, u takes effective its
extreme values, according to the classical Weierstrass’s theorem.
We suppose, ad absurdum, that u takes its supremum in the parabolical inside,
not on the boundary . This means that we can suppose that there exists a point
(t0 , x 0 ) ∈ T T × \ such that
M=
u(t, x) = u(t0 , x 0 ).
sup
(t,x)∈T T ×
We denote by m the supremum value of the function u taken on :
m = sup u(t, x).
(t,x)∈
266
8 Parabolical Equations
According to above supposition, we have
M > m.
(8.2.8)
In the following we will prove that Eq. (8.2.8) leads to a contradiction. We define the
function v(t, x) by
v(t, x) = u(t, x) +
n
2
M − m xi − xi0 ,
2
2d
i=1
(8.2.9)
where d is the diameter of the set .
Evaluating the function v on , we obtain
v(t, x)| ≤ m +
M +m
M+M
M −m
=
<
= M.
2
2
2
(8.2.10)
On the other hand,
v(t0 , x 0 ) = u(t0 , x 0 ) +
n
M − m 0
0 2
−
x
= M,
x
i
i
2d 2 i=1
that is v, which verifies the same conditions of regularity like u, takes its largest value
at the point (t0 , x 0 ) like u. Since on the values of v are strictly less than M, we
will deduce that there exists a point (t1 , x 1 ) in the parabolical inside such that
sup
v(t, x) = v(t1 , x 1 ),
(t,x)∈T T ×
while v cannot take its supremum value on . We write the condition of the extremal
value for v(t, x) at the point (t1 , x 1 ):
∂v(t, x) ≥ 0.
∂t (t1 ,x 1 )
(8.2.11)
If t1 ∈ (0, T ) then in Eq. (8.2.11) we have equality whence it follows the Fermat’s
condition. If t1 = T , then the value to the right-hand of T does not exist and then
a extremum point in t1 means that the function v is positive and increasing at the
left-hand of T . On the other hand, the function v(t1 , x), (like only a function of n
spatial variable (x1 , x2 , ..., xn ) takes its supremum on at the point (x11 , x21 , ..., xn1 )
whence it follows the necessary maximum condition
∂ 2 v(t, x) ≤ 0, i = 1, 2, ..., n,
∂xi2 (t1 ,x 1 )
8.2 Initial-Boundary Value Problems
267
from where we obtain
v(t1 , x 1 ) ≤ 0.
(8.2.12)
From Eqs. (8.2.11) and (8.2.12) it follows
(−vt (t, x) + v(t, x))(t1 ,x 1 ) ≤ 0.
(8.2.13)
Starting from the form (8.2.9) of the function v, we obtain
(−vt (t, x) + v(t, x))(t1 ,x 1 ) = (−u t (t, x) + u(t, x))(t1 ,x 1 ) +
(M − m)n
(M − m)n
+
=
> 0,
d2
d2
in which we have taken into account Eq. (8.2.8).
This inequality is the contrary of the inequality (8.2.13), that proves that the
assumption (8.2.8) is falls and the theorem has been proved.
As a direct consequence of the theorem of the extreme values, we will prove the
uniqueness of the solution for an initial boundary values problem.
Theorem 8.2.2 The problem constituted by the Eq. (8.2.1), the initial condition
(8.2.2) and the boundary condition (8.2.3) has at the most one classical solution.
Proof We suppose that the problem (8.2.1), (8.2.2) and (8.2.3) admits two classical
solutions u 1 (t, x) and u 2 (t, x). Then we have
u i (t, x) −
∂u i
∂t
(t, x) = f (t, x), ∀(t, x) ∈ TT × ,
u i (0, x) = ϕ(x), ∀x ∈ ,
(8.2.14)
u i (t, y) = α(t, y), ∀(t, y) ∈ T T × ∂,
where i = 1, 2 and the functions f, ϕ and α are given and continuous where they
are defined.
On the other hand, u 1 and u 2 satisfy the conditions of a classical solution. We
define the function v(t, x) by
v(t, x) = u 1 (t, x) − u 2 (t, x), ∀(t, x) ∈ T T × .
Taking into account the above considerations, we obtain that v satisfies the conditions
of regularity of a classical solution and, more, verifies the problem
∂v
(t, x)
∂t
= 0, ∀(t, x) ∈ TT × ,
v(0, x) = 0, ∀x ∈ ,
v(t, y) = 0, ∀(t, y) ∈ T T × ∂.
v(t, x) −
(8.2.15)
268
8 Parabolical Equations
The function v satisfies all the conditions of the Theorem 8.2.1. Therefore, its extreme
values
v(t, x),
sup
(t,x)∈T T ×
inf
(t,x)∈T T ×
v(t, x)
are, necessarily, reached on . According to Eqs. (8.2.15)2 and (8.2.15)3 , we deduce
that v is null on the parabolical boundary and then
sup
v(t, x) =
(t,x)∈T T ×
inf
(t,x)∈T T ×
v(t, x) = 0,
that is v(t, x) = 0, ∀(t, x) ∈ T T × and de aici u 1 (t, x) ≡ u 2 (t, x).
As a further application of the theorem of the extreme values, we will prove now a
result of stability with regard to initial conditions and boundary conditions, for the
problem (8.2.1), (8.2.2) and (8.2.3).
Theorem 8.2.3 We suppose that the function f (t, x) is given and continuous on
TT × . Let ϕ1 (t, x) and ϕ2 (t, x) be two functions, given and continuous on and
the functions α1 (t, x) and α2 (t, x) given and continuous on TT × ∂.
Consider the following problems
∂u i
(t, x) = f (t, x), ∀(t, x) ∈ TT × ,
∂t
u i (0, x) = ϕi (x), ∀x ∈ ,
u i (t, y) = αi (t, y), ∀(t, y) ∈ T T × ∂,
u i (t, x) −
where i = 1, 2.
If ∀ε > 0, ∃δ = δ(ε) such that
|ϕ(x)| = |ϕ1 (x) − ϕ2 (x)| < δ,
|α(x)| = |α1 (x) − α2 (x)| < δ.
Then
|u(x)| = |u 1 (x) − u 2 (x)| < ε.
Proof The function u(t, x) defined, as in the enunciation of the theorem, by
u(t, x) = u 1 (t, x) − u 2 (t, x),
8.2 Initial-Boundary Value Problems
269
satisfies the conditions of a classical solution. Also, u satisfies the problem
∂u
(t, x) = f (t, x) − f (t, x) = 0,
∂t
u(0, x) = u 1 (0, x) − u 2 (0, x) = ϕ1 (x) − ϕ2 (x) = ϕ(x),
(8.2.16)
u(t, y) = u 1 (t, y) − u 2 (t, y) = α1 (t, y) − α2 (t, y) = α(t, y).
u(t, x) −
Since u satisfies the above conditions of regularity and the homogeneous equation (8.2.16)1 , we will deduce that there are satisfied the conditions of the theorem
of extreme values. Then, the extreme values of the function u are reached on the
parabolical boundary . But on the function u reduces to ϕ or to α and, because
ϕ and α satisfy the conditions |ϕ| < δ, |α| < δ, we obtain the result of the theorem,
taking δ = ε.
A particular solution of the problem consists of Eqs. (8.2.1), (8.2.2) and (8.2.3) is
the solution obtained by fixing the right-hand term f of the equation, of the initial
data ϕ and of the boundary data α.
The family of all particular solutions obtained by the variation of the functions
f , ϕ and α, in the class of the continuous functions, is the general solution of the
problem (8.2.1), (8.2.2) and (8.2.3).
Now, we prove that a particular solution of the homogeneous equation of the heat
conduction is the function V defined by
⎞
⎛ n
(xi − ξi )2
⎟
⎜ i=1
1
⎟.
−
V (t, τ , x, ξ) = √ n √
n exp ⎜
⎝
4(t − τ ) ⎠
2 π
t −τ
(8.2.17)
Proposition 8.2.1 The function V (t, τ , x, ξ), for 0 ≤ τ < t ≤ T , is of the class
C ∞ and satisfies the equations:
∂V (t, τ , x, ξ)
= 0,
∂t
∂V (t, τ , x, ξ)
ξ V (t, τ , x, ξ) +
= 0.
∂τ
x V (t, τ , x, ξ) −
Proof After some elementary calculations, we obtain
∂V (t, τ , x, ξ)
(xi − ξi )
= V (t, τ , x, ξ) −
∂xi
2(t − τ )
=−
∂V (t, τ , x, ξ)
.
∂ξi
Therefore
∂ 2 V (t, τ , x, ξ)
∂ 2 V (t, τ , x, ξ)
(xi −ξi )2
1
=
=
V
(t,
τ
,
x,
ξ)
−
.
4(t − τ )2 2(t − τ )
∂xi2
∂ξi2
270
8 Parabolical Equations
By adding the relations obtained for i = 1, 2, ..., n it follows
n
1
n
2
x V (t, τ , x, ξ) = V (t, τ , x, ξ)
(xi −ξi ) −
=
4(t − τ )2 i=1
2(t − τ )
= ξ V (t, τ , x, ξ).
On the other hand, using the derivative in Eq. (8.2.17) with regard respectively, to t
and τ , it follows:
n
1
n
∂V (t, τ , x, ξ)
= V(t, τ , x, ξ)
(xi − ξi )2 −
=
∂t
4(t − τ )2 i=1
2(t − τ )
=−
∂V (t, τ , x, ξ)
.
∂τ
So, the results from the enunciation of the proposition are immediately obtained.
The fact that the function V (t, τ , x, ξ) is of the class C ∞ can be argued using the
fact that t = τ and, essentially, V (t, τ , x, ξ) is an exponential function.
Remark. It is easy to verify the fact that if x = ξ, then the function V (t, τ , x, ξ) is
superior bounded by an exponential function and
lim V (t, τ , x, ξ) = 0.
t−τ →0+
If x = ξ, then the exponential function disappears and
lim V (t, τ , x, ξ) = +∞.
t−τ →0+
Another important property of the function V (t, τ , x, ξ) will be proved in the
following theorem.
Theorem 8.2.4 The following equalities are true
IRn
V (t, τ , x, ξ)dx = 1,
IRn
V (t, τ , x, ξ)dξ = 1.
Proof We write the volume integral in extension
IRn
1
= √ n
2 π
+∞ ...
−∞
V (t, τ , x, ξ)dξ =
⎞
⎛ n
2
(x
−ξ
)
i
i
+∞
⎟
⎜ i=1
1
⎟ dξ1 dξ2 ...dξn .
−
√
n exp ⎜
⎝
4(t − τ ) ⎠
t −τ
−∞
8.2 Initial-Boundary Value Problems
271
√
We make the change of variable ξi − xi = 2 t − τ ηi and by direct calculations, we
obtain that the Jacobian of the change has the value:
dξ √
n
n
t −τ .
Dη = 2
Then
IRn
n
+∞ +∞ −
ηi2
1
i=1
V (t, τ , x, ξ)dξ = √ n
...
e
dη1 dη2 ...dηn =
π
−∞
−∞
+∞ +∞
1
2
2
2
= √ n
...
e−η1 e−η2 ...e−ηn dη1 dη2 ...dηn =
π
−∞
−∞
+∞
n
1
1 √ n
2
= √ n
e−s ds = √ n
π = 1,
π
π
−∞
where we have used the Gauss’s integral
+∞
−∞
e−s ds =
2
√
π.
Similarly, the other equality from the enunciation can be proved.
In the following, we prove a result which generalizes the results from the
Theorem 8.2.4
Theorem 8.2.5 Consider a bounded domain. If we denote by I the integral
I (t − τ , x) =
V (t, τ , x, ξ)dξ,
then for x ∈ , we have
lim I (t − τ , x) = 1,
t−τ →0+
the limit is still valid uniformly with regard to x, on compact sets from , and, for
x ∈ IRn \ lim I (t − τ , x) = 0,
t−τ →0+
the limit is still valid uniformly with regard to x, on compact sets from IRn \ .
272
8 Parabolical Equations
Proof Firstly, consider the case x ∈ . We use the notations
d0 = dist(x, ), d1 = dist(Q, ∂),
where Q is a compact, arbitrarily fixed in , such that x ∈ Q.
We remember that, by definition, we have
d0 = dist(x, ∂) = sup |x − y|,
y∈∂
d1 = dist(Q, ∂) =
|x − y|.
sup
y∈∂,x∈Q
Consider the balls B(x, d0 ) and B(x, d1 ) and then
B(x, d1 ) ⊂ B(x, d0 ) ⊂ .
(8.2.18)
Using the monotony of the integral and taking into account the inclusion (8.2.8), we
will deduce
I (t − τ , x) =
V (t, τ , x, ξ)dξ ≥
≥
V (t, τ , x, ξ)dξ ≥
V (t, τ , x, ξ)dξ =
(8.2.19)
B(x,d0 )
B(x,d1 )
1
= √ n √
n
2 π
t −τ
B(x,d1 )
⎞
⎛ n
(xi − ξi )2
⎟
⎜ i=1
⎟
exp ⎜
⎝− 4(t − τ ) ⎠ dξ.
We make the change of variable
√
ξi − xi = 2 t − τ ηi , i = 1, 2, ..., n.
As in the proof of the Theorem 8.2.4, the value of the Jacobian of this change is
2n
√
t −τ
n
.
With this change of variable, the last integral from Eq. (8.2.19), becomes:
1
√ n
π
e
B(0, 2√dt−τ )
−
n
i=1
ηi2
dη,
(8.2.20)
8.2 Initial-Boundary Value Problems
273
in which
n
n
√
2
(ξi − xi ) = 2 t − τ
ηi2 .
i=1
i=1
If we pass to the limit in Eq. (8.2.19) with t − τ → 0+ and we take into account
Eq. (8.2.20), we obtain
n
−
ηi2
1
i=1
lim + I (t − τ , x) ≥ lim + √ n
e
dη =
t−τ →0
t−τ →0
π
B(0, 2√dt−τ )
n
−
ηi2
1
i=1
= √ n
e
dη = 1,
π
IRn
in which we have used the Gauss’s integral. Also, we have used the fact that for
t − τ → 0+ , we have
d
→∞
√
2 t −τ
and then the ball
d
B(0, √
)
2 t −τ
becomes the whole space IRn . Also, we already proved that
lim I (t − τ , x) ≥ 1.
t−τ →0+
(8.2.21)
Since ⊂ IRn , we have, obviously, that
lim I (t − τ , x) ≤
t−τ →0+
IRn
V (t, τ , x, ξ)dξ = 1,
and then
lim I (t − τ , x) ≤ 1.
t−τ →0+
(8.2.22)
From Eqs. (8.2.21) and (8.2.22), the first part of the proof is concluded. The limit
takes place uniformly with regard to x, on compact sets from that contains x, since
d used in the above considerations depends only on the compact set that contains x,
not on the choosing of x in the respective compact set.
Now, we approach the case when x ∈ IRn \. Taking into account that has been
assumed to be domain (bounded), using the Jordan’s theorem, we have that IRn \ 274
8 Parabolical Equations
is a domain too. Let us consider a compact set Q ∗ ⊂ IRn \ such that x ∈ Q ∗ and
consider the distances d0∗ = dist(x, ∂), d1∗ = dist(Q ∗ , ∂) and the balls B(x, d0∗ )
and B(x, d0∗ ). Since d0∗ > d1∗ , we will deduce
B(x, d1∗ ) ⊂ B(x, d0∗ ) ⇒
⇒ ⊂IRn \ B(x, d0∗ ) ⊂ IRn \ B(x, d1∗ ).
Corresponding to I (t − τ , x) we have the evaluations
0 ≤ I (t − τ , x) = V (t, τ , x, ξ)dξ ≤
≤ IRn \B(x,d ∗ ) V (t, τ , x, ξ)dξ ≤ IRn \B(x,d ∗ ) V (t, τ , x, ξ)dξ.
0
(8.2.23)
1
We make the change of variable
√
ξi − xi = 2 t − τ ηi , i = 1, 2, ..., n.
Based on the considerations from the first part of the proof, the last integral from
Eq. (8.2.23) becomes
1
√ n
π
e
−
n
i=1
ηi2
dη,
(8.2.24)
D
where the domain of integration D is
d∗
D = R n \ B(0, √
).
2 t −τ
On taking the limit for t − τ → 0+ , the radius
d∗
√
2 t −τ
becomes infinite and then the ball
d∗
)
B(0, √
2 t −τ
becomes the whole space IRn . Therefore, the integral from Eq. (8.2.24) tends to zero
and, comes back to Eq. (8.2.23), it results
0 ≤ lim + I (t − τ , x) ≤
t−τ →0
e
D
−
n
i=1
ηi2
dη = 0,
8.2 Initial-Boundary Value Problems
275
in which the domain of integration D is defined as below. The limit takes place
uniformly with respect to x, on compact sets from that contains x, since d ∗ used
in the above considerations depends only on the compact set that contains x, not on
the choosing of x in the respective compact set.
The usefulness of the results demonstrated in the Theorems 8.2.4 and 8.2.5 follows
from the following theorem.
Theorem 8.2.6 Consider a bounded domain from IRn and we suppose that the
function f is continuous and bounded on .
Then:
• (i). If x ∈ ,
lim +
t−τ →0
V (t, τ , x, ξ) f (ξ)dξ = f (x),
the limit taking place uniformly with respect to x, on compact sets from .
• (ii). If x ∈ IRn \ ,
lim +
t−τ →0
V (t, τ , x, ξ) f (ξ)dξ = 0,
the limit taking place uniformly with regard to x, on compact sets from IRn \ .
Proof (i). Consider Q a compact set arbitrarily fixed, Q ⊂ , such that x ∈ Q. We
have the evaluations
V (t, τ , x, ξ) f (ξ)dξ − f (x) ≤ V (t, τ , x, ξ) f (ξ)dξ−
− f (x) V (t, τ , x, ξ) f (ξ)dξ + f (x) V (t, τ , x, ξ)dξ − f (x) ≤
≤
V (t, τ , x, ξ)| f (x) − f (ξ)|dξ + | f (x)| V (t, τ , x, ξ)dξ − 1 ≤
≤
V (t, τ , x, ξ)| f (x)− f (ξ)|dξ +
V (t, τ , x, ξ)| f (x)− f (ξ)|dξ+
IRn \B(x,δ)
B(x,δ)
+ c0 V (t, τ , x, ξ)dξ − 1 ,
(8.2.25)
where, by c0 we have noted c0 = sup f (x).
x∈
In order to use the continuity of the function f , we take, sufficient small, ε and
then exists η(ε) such that if
|x − ξ| < η(ε) ⇒ | f (x) − f (ξ)| < ε.
276
8 Parabolical Equations
If in the evaluations from Eq. (8.2.25) we take δ < η(ε), it results that
V (t, τ , x, ξ)| f (x) − f (ξ)|dξ < ε
V (t, τ , x, ξ)dξ ≤
B(x,δ)
B(x,δ)
V (t, τ , x, ξ)dξ = ε.
≤ε
IRn
Then
\B(x,δ)
V (t, τ , x, ξ)| f (x) − f (ξ)|dξ < 2c0
\B(x,δ)
V (t, τ , x, ξ)dξ,
and
lim +
t−τ →0
\B(x,δ)
V (t, τ , x, ξ) f (ξ)dξ = 0,
because x ∈
/ \ B(x, δ) and then we can use the second part of the Theorem 8.2.5.
Finally, for the last integral from Eq. (8.2.25), we have
lim + t−τ →0
\B(x,δ)
V (t, τ , x, ξ) f (ξ)dξ − 1 = 0,
because x ∈ and then we can use the first part of the Theorem 8.2.5. If we take
into account these evaluations in Eq. (8.2.25), the point (i) is proved. We outline that
the limit from (i) takes place uniformly with respect to x because the last integrals
from Eq. (8.2.25) tend to zero, uniformly on compact sets from .
(ii). We take, arbitrarily, a compact set Q ∗ such that x ∈ Q ∗ and Q ∗ ⊂ IRn \ .
Since, by hypothesis, f is a bounded function, we have
V (t, τ , x, ξ) f (ξ)dξ ≤
|V (t, τ , x, ξ)| | f (ξ)| dξ ≤
≤ c0
V (t, τ , x, ξ) f (ξ)dξ,
and then
V (t, τ , x, ξ) f (ξ)dξ.
0 ≤ lim + V (t, τ , x, ξ) f (ξ)dξ ≤ c0
t−τ →0
Since x ∈
/ , based on the second part of the Theorem 8.2.5, these inequalities lead
to the conclusion that
lim +
V (t, τ , x, ξ)dξ = 0
(8.2.26)
t−τ →0
8.2 Initial-Boundary Value Problems
277
and then
lim + V (t, τ , x, ξ) f (ξ)dξ = 0,
t−τ →0
the limit taking place uniformly with respect to x, on compact sets from IRn \ ,
because so has been obtained the limit from Eq. (8.2.26).
In the following theorem, we give a generalization of the results from the
Theorem 8.2.6.
Theorem 8.2.7 Consider the function g(τ , ξ) assumed to be continuous and
bounded on TT × . If, more,
lim g(τ , ξ) = g(t, ξ)
τ →t +
the limit taking place uniformly with respect to ξ, on compact sets from , then
– (i). If x ∈ ,
lim
τ →t −
V (t, τ , x, ξ)g(τ , ξ)dξ = g(t, x),
the limit taking place uniformly with respect to x, on compact sets from .
– (ii). If x ∈ IRn \ ,
lim−
τ →t
V (t, τ , x, ξ)g(τ , ξ)dξ = 0,
the limit taking place uniformly with respect to x, on compact sets from IRn \ .
Proof (i). Consider Q a compact set arbitrarily fixed, Q ⊂ , such that x ∈ Q ⊂ .
Then
V (t, τ , x, ξ)g(τ , ξ)dξ − g(t, x) ≤
≤ V (t, τ , x, ξ)[g(τ , ξ)−g(t, ξ)]dξ + V (t, τ , x, ξ)dξ −g(t, x) . (8.2.27)
If in Eq. (8.2.27) we pass to the limit with τ → t − , the first integral from the righthand side tends to zero, based on the hypotheses, and the last integral from Eq. (8.2.27)
tends to zero based on the Theorem 8.2.6. Also, we will deduce that both limits take
place uniformly with respect to x, on compact sets from , based on the hypotheses,
and on the fact that so has been obtained the result from the Theorem 8.2.6.
(ii). In a similarly way, we can prove this result.
278
8 Parabolical Equations
8.3 Method of the Green’s Function
Firstly, we will obtain the Green’s formula for the heat equation. With that end in
view, we define the operators L(τ ,ξ) and M(τ ,ξ) by
∂u
,
∂τ
∂v
.
M(τ ,ξ) v = ξ v +
∂τ
L(τ ,ξ) u = ξ u −
(8.3.1)
Consider a bounded domain whose boundary ∂ admits a tangent plane, piecewise
continuously varying.
In all that follows we will use the function u(t, x) which satisfies the following
standard hypotheses:
– u ∈ C(TT × );
– u xi xi , u t ∈ C(TT × ), for 0 ≤ τ < t ≤ T .
If we amplify (8.3.1)1 by v(τ , ξ) and (8.3.1)2 by u(τ , ξ), we obtain
vLu − uMv = vξ u − uξ v − v
∂v
∂u
−u
∂τ
∂τ
that is
vLu − uMv = vξ u − uξ v −
∂
(uv).
∂τ
(8.3.2)
Proposition 8.3.1 We suppose satisfied the above hypotheses with regard to the
domain and of the function u. If the function v satisfies the hypotheses of the
function u, then there holds the Green’s formula
t
0
t
∂v
∂u
−u
[vLu − uMv]dτ dξ =
v
τ dσξ −
∂ν
∂ν
∂ 0
−
u(t, ξ)v(t, ξ)dξ +
u(0, ξ)v(0, ξ)dξ.
(8.3.3)
Proof We integrate the equality (8.3.2) on the set × [0, t]:
t
[vLu − uMv]dτ dξ =
0
−
0
t
t
[vξ u − uξ v]dτ dξ−
0
∂
(uv) dτ dξ.
∂τ
(8.3.4)
8.3 Method of the Green’s Function
279
By using the well known Gauss–Ostrogradski’s formula, it results
=
t
n
0
t
vξ udτ dξ =
0
∂u
∂ξi
∂
v
∂ξ
i
i=1
dτ dξ =
0
∂
v
n
∂2u
∂ξi2
i=1
n
t v
∂
=
t
0
i=1
v
dτ dξ =
∂u
cos αi dτ dσξ =
∂ξi
∂u
v
dτ dσξ ,
∂νξ
t
0
where ν is the outside unit normal to the surface ∂.
Similarly, we can obtain the following equality
t
uξ vdτ dξ =
0
t
u
∂
0
∂v
dτ dσξ .
∂νξ
Then
=
∂
t
0
∂
(uv) dτ dξ =
∂τ
∂
uv|t0 dξ =
[u(t, ξ)v(t, ξ) − u(0, ξ)v(0, ξ)]dξ.
By using these evaluations in Eq. (8.3.4), we obtain the Green’s formula.
The Green’s formula (8.3.3) can be generalized in the sense that in the form (8.3.1)
of the operators L and M instead of the Laplacean we can take an arbitrarily linear
operator of the order two.
So, we define, the operator L and its adjunct M by
Lu =
n
n i=1 j=1
Mv =
∂2u
∂u
+
bi (x)
+ c(x)u,
∂xi ∂x j
∂x
i
i=1
n
ai j (x)
n n
∂ 2 (ai j (x)v)
i=1 j=1
∂xi ∂x j
−
n
∂(bi (x)v)
i=1
∂xi
+ c(x)u,
(8.3.5)
where ai j = a ji ∈ C 2 (), bi ∈ C 1 () and c ∈ C 0 ().
By using a similar procedure as in Eq. (8.2.1), we construct the operators A and B
by
∂u
,
∂t
∂v
.
Bv = Mv +
∂t
Au = Lu −
(8.3.6)
280
8 Parabolical Equations
Proposition 8.3.2 We suppose to be satisfied the hypotheses from the Proposition 8.3.1 on the domain and of the functions u and v. Moreover, we suppose
that the operator L is elliptical. Then takes place the Green’s formula:
0
t
t
!
∂u
∂v
γ v −u
+buv dτ dσξ−
∂γ
∂γ
[vAu −uBv]dτ dξ =
∂ 0
−
u(t, ξ)v(t, ξ)dξ +
u(0, ξ)v(0, ξ)dξ.
(8.3.7)
Proof We multiply Eq. (8.3.6)1 by v and Eq. (8.3.6)2 by u and subtract the resulting
relations, whence it follows the equality:
vAu − uBv = vLu − u Mv − v
=vLu − u Mv −
∂v
∂u
−u
=
∂t
∂t
∂
(uv) .
∂t
We integrate this equality on the set × [0, t] and, after we use the GaussOstrogradski’s formula, we are led to the Green’s formula (8.3.7).
Consider, again, the operators L and M defined in Eq. (8.3.1). Corresponding,
we will use the Green’s formula in the form (8.3.3). Starting from this form of the
Green’s formula, we intend to find the form of the Riemann–Green’s formula. In this
hope, we use, again, the function V (t, τ , x, ξ) defined by
⎞
⎛ n
(xi − ξi )2
⎟
⎜ i=1
1
⎟.
−
V (t, τ , x, ξ) = √ n √
n exp ⎜
⎝
4(t − τ ) ⎠
2 π
t −τ
(8.3.8)
The unique singular point of the function V (t, τ , x, ξ) is the point (t, x) = (τ , ξ).
To avoid this point, we will consider the domain
{τ ; 0 ≤ τ ≤ t − δ, δ > 0} × .
On this domain we write the Green’s formula (8.3.3) for the pair of functions (v, u),
where v = V (t, τ , x, ξ) and u = u(τ , ξ):
t−δ
[V (t, τ , x, ξ)Lu(τ , ξ) − u(τ , ξ)MV (t, τ , x, ξ)] dτ dξ =
∂u
∂V (t, τ , x, ξ)
V (t, τ , x, ξ) (τ , ξ) − u(τ , ξ)
=
dτ dσξ −
∂ν
∂ν
∂ 0
− V (t, t − τ , x, ξ)u(t − τ , ξ)dξ +
V (t, 0, x, ξ)u(0, ξ)dξ.
0
t−δ
(8.3.9)
8.3 Method of the Green’s Function
281
In this equality we pass to the limit with δ → 0 and we use the Theorem 8.2.7 from
the Sect. 8.2. Thus, if x ∈ , it results
t
u(t, x) = −
V (t, τ , x, ξ)Lu(τ , ξ)dτ dξ+
0
t
∂u
∂V (t, τ , x, ξ)
V (t, τ , x, ξ) (τ , ξ) − u(τ , ξ)
dτ dσξ +
+
∂ν
∂ν
∂ 0
+
V (t, 0, x, ξ)u(0, ξ)dξ.
(8.3.10)
The result proved here will be synthesized in the following theorem.
Theorem 8.3.1 For the heat equation, the Riemann-Green’s formula has the form of
Eq. (8.3.10), where the operators L and M are defined in Eq. (8.3.1) and V (t, τ , x, ξ)
is a function that has the form of Eq. (8.3.8).
Remark. If x ∈ IRn \ then by taking the limit in Eq. (8.3.9) with δ → 0 and by
using the second part of the Theorem 8.2.7 (Sect. 8.2), it results:
0=−
t
+
∂
V (t, τ , x, ξ)
0
t
V (t, τ , x, ξ)Lu(τ , ξ)dτ dξ+
0
∂u
∂V (t, τ , x, ξ)
(τ , ξ) − u(τ , ξ)
dτ dσξ +
∂ν
∂ν
+
V (t, 0, x, ξ)u(0, ξ)dξ.
Now, we consider the initial boundary values problem
Lu(t, x) = f (t, x), ∀(t, x) ∈ TT × ,
u(0, x) = ϕ(x), ∀x ∈ ,
u(t, y) = α(t, y), ∀(t, y) ∈ TT × ∂,
∂u
u(t, y) = β(t, y), ∀(t, y) ∈ TT × ∂.
∂ν
Then, the Riemann-Green’s formula receives the form
u(t, x) = −
+
∂
t
V (t, τ , x, ξ) f (τ , ξ)dτ dξ +
t
t
∂V (t, τ , x, ξ)
α(τ , ξ)dτ dσξ+
V (t, τ , x, ξ)β(τ , ξ)dτ dσξ −
∂ν
∂ 0
0
0
+
V (t, 0, x, ξ)ϕ(ξ)dξ.
(8.3.11)
282
8 Parabolical Equations
The integrals from the right-hand side of the formula (8.3.11) are the associated
potentials of the heat problem, namely,:
t
V (t, τ , x, ξ) f (τ , ξ)dτ dξ
I1 = −
0
is the heat potential of the volume;
I2 =
t
V (t, τ , x, ξ)β(τ , ξ)dτ dσξ
∂ 0
is the surface heat potential of the simple layer;
I3 = −
∂
t
0
∂V (t, τ , x, ξ)
α(τ , ξ)dτ dσξ
∂ν
is the surface heat potential of the double layer;
I4 =
V (t, 0, x, ξ)ϕ(ξ)dξ
is the heat temporal potential.
Therefore, formula (8.3.11) is also called the formula of the heat potentials. As
in the case of the elliptical equations, the heat potentials are used to solve the initial
boundary value problems, in the context of the parabolical equations.
More exactly, the heat potentials permit the transformation of these problems in
integral equations of the Fredholm type.
Let us consider the problem of Dirichlet type
∂u
(τ , ξ) = f (τ , ξ), ∀(τ , ξ) ∈ TT × ,
∂τ
u(0, ξ) = ϕ(ξ), ∀ξ ∈ ,
ξ u(τ , ξ) −
(8.3.12)
u(τ , η) = α(τ , η), ∀(τ , η) ∈ TT × ∂,
where is a bounded domain with the boundary ∂ having the tangent plane piecewise continuously varying. We denote by TT the interval (0, T ] and by TT the closed
interval [0, T ]. The functions f, ϕ and α are given and continuous on the indicated
domains. The condition (8.3.12)3 is called the Dirichlet condition. In a problem of
Neumann type, the condition (8.3.12)3 is replaced by the boundary condition of
Neumann type
∂u
(τ , η) = β(τ , η), ∀(τ , η) ∈ TT × ∂.
∂ν
8.3 Method of the Green’s Function
283
Definition 8.3.1 We call the Green’s function attached to the domain , to the
operator L and to the Dirichlet condition (8.3.12)3 , the function G(t, τ , x, ξ) defined
by
G(t, τ , x, ξ) = V (t, τ , x, ξ) + g(t, τ , x, ξ),
(8.3.13)
where the function V (t, τ , x, ξ) is defined in Eq. (8.3.8) and g(t, τ , x, ξ) has the
properties:
- g(t, τ , x, ξ) is a continuous function with respect to the variables t, τ , x and ξ on
the set TT × TT × × ;
- the derivatives gxi xi and gt are continuous functions on the set TT × TT × × ;
- g(t, τ , x, ξ) satisfies the adjunct homogeneous equation of the heat
Mg(t, τ , x, ξ) = ξ g(t, τ , x, ξ) +
∂
g(t, τ , x, ξ) = 0;
∂τ
- g(t, τ , x, ξ) satisfies the condition g(t, t, x, ξ) = 0.
The Green’s function G(t, τ , x, ξ) satisfies, by definition, the homogeneous
Dirichlet condition
G(t, τ , x, η) = 0, ∀(τ , η) ∈ TT × ∂.
In the following theorem we prove that if the Dirichlet problem (8.3.12) admits a
classical solution, then this solution can be represented with the aid of the Green’s
function.
Theorem 8.3.2 If we suppose that the Dirichlet problem (8.3.12) admits a classical
solution, then it has the form
t
G(t, τ , x, ξ) f (τ , ξ)dτ dξ−
t
∂G(t, τ , x, η)
α(τ , η)dτ dση +
G(t, 0, x, ξ)ϕ(ξ)dξ.
−
∂ν
∂ 0
u(t, x) = −
0
(8.3.14)
Proof We write the Green’s formula (8.3.7) for the pair of functions v = g(t, τ , x, ξ)
and u = u(τ , ξ), where u(τ , ξ) is the solution of the problem (8.3.12):
0=−
t
t
g(t, τ , x, ξ) f (τ , ξ)dτ dξ +
u(τ , ξ)Mg(t, τ , x, ξ)dτ dξ+
0
0
t
∂u(τ , ξ)
∂g(t, τ , x, ξ)
g(t, τ , x, ξ)
− u(τ , ξ)
dτ dσξ −
+
∂ν
∂ν
∂ 0
−
g(t, τ , x, ξ)u(τ , ξ)dξ +
g(t, 0, x, ξ)ϕ(τ , ξ)dξ.
284
8 Parabolical Equations
Based on the hypotheses imposed to the function g, this equality becomes
t
0=−
g(t, τ , x, ξ) f (τ , ξ)dτ dξ + g(t, 0, x, ξ)ϕ(ξ)dξ +
0
t
∂u(τ , ξ)
∂g(t, τ , x, ξ)
g(t, τ , x, ξ)
− α(τ , ξ)
dτ dσξ . (8.3.15)
+
∂ν
∂ν
∂ 0
Now, we write the Riemann-Green’s formula (8.3.10) for the pair of functions v =
V (t, τ , x, ξ) and u = u(τ , ξ), where u(τ , ξ) is the solution of the problem (8.3.12):
t
V (t, τ , x, ξ) f (τ , ξ)dτ dξ + V (t, 0, x, ξ)ϕ(ξ)dξ +
u(t, x) = −
0
t
∂u(τ , ξ)
∂V (t, τ , x, ξ)
V (t, τ , x, ξ)
− α(τ , ξ)
dτ dσξ . (8.3.16)
+
∂ν
∂ν
∂ 0
By adding, term by term the formulas (8.3.15) and (8.3.16), it results
+
∂
t
G(t, τ , x, ξ) f (τ , ξ)dτ dξ+
t
t
∂G(t, τ , x, ξ)
u(τ , ξ)
dτ dσξ −
α(τ , η)dτ dση +
G(t, τ , x, ξ)
∂ν
∂ν
0
∂ 0
+
G(t, 0, x, ξ)ϕ(ξ)dξ.
u(t, x) = −
0
since the Green’s function G(t, τ , x, ξ) becomes null on the boundary (because, by
definition, G(t, τ , x, ξ) satisfies the homogeneous Dirichlet’s condition), we will
deduce that the second integral from the right-hand side of the above equality disappears, and the remained formula is even (8.3.14).
In the following, we make now analogous considerations for the Neumann’s problem which can be deduced from the Dirichlet’s problem (8.3.12) by substituting the
condition (8.3.12)3 with the condition
∂u(τ , η)
= β(τ , η), ∀(τ , η) ∈ TT × ∂.
∂ν
(8.3.17)
The Green’s function for the domain , the operator L and the Neumann’s condition (8.3.17) is given in the formula (8.3.13) from the Definition 8.3.1, but the last
condition from this definition is replaced by
∂G(t, τ , x, η)
= 0, ∀(τ , η) ∈ TT × ∂,
∂ν
(8.3.18)
that is, the function G satisfies the Neumann’s condition in its homogeneous form.
8.3 Method of the Green’s Function
285
Proposition 8.3.3 We suppose that the Neumann’s problem (8.3.12)1 , (8.3.12)2 and
(8.3.17) admits a classical solution. Then, it can be expressed with the aid of the
Green’s function in the form
+
∂
t
G(t, τ , x, ξ) f (τ , ξ)dτ dξ +
t
G(t, τ , x, η)β(τ , η)dτ dση +
G(t, 0, x, ξ)ϕ(ξ)dξ.
u(t, x) = −
0
0
(8.3.19)
Proof We will use the same reasoning as in the proof of the formula (8.3.14). Firstly,
we write the Green’s formula for the pair of functions v = g(t, τ , x, ξ) and u =
u(τ , ξ), where u(τ , ξ) is the solution of our Neumann’s problem. Then, we write the
Riemann-Green’s formula for the pair of functions v = V (t, τ , x, ξ) and u = u(τ , ξ),
where u(τ , ξ) is the solution of the Neumann’s problem. By adding, term by term the
two resulting relations and take into account the conditions imposed to the functions
g(t, τ , x, ξ), G(t, τ , x, ξ), we obtain the formula (8.3.19).
If we examine the formulas (8.3.14) and (8.3.19), we are led to the conclusion that the solutions of the Dirichlet problem and Neumann’s problem, if exists,
can be unique represented with the aid of the Green’s function. Since the function
V (t, τ , x, ξ), from the definition of the Green’s function, is defined in Eq. (8.3.8),
it results that in view of determination of the Green’s function we must determine the function g(t, τ , x, ξ). Apparently, the problem of determining the function
g(t, τ , x, ξ) has the same difficulty like the proper problem of determining the solution of the Dirichlet’s problem or Neumann’s problem, especially, on account to the
conditions of regularity imposed to the function g(t, τ , x, ξ).
But unlike to the classical solution u, the function g(t, τ , x, ξ) satisfies in the case
of the Dirichlet’s problem and in the case of the Neumann’s problem, an homogeneous
equation of heat. Also, in the problem of Dirichlet as well as the problem of Neumann,
the solution u satisfies a boundary condition with α respectively, β arbitrarily. In the
present case, the function g(t, τ , x, ξ) satisfies a boundary condition in which the
right-hand side is perfectly determined, because
g(t, τ , x, η) = −V (t, τ , x, η), ∀(τ , η) ∈ TT × ∂,
and, respectively
∂V (t, τ , x, η)
∂g(t, τ , x, η)
=−
, ∀(τ , η) ∈ TT × ∂,
∂ν
∂ν
where V (t, τ , x, η) is given in Eq. (8.3.8).
These comments prove that the method of the Green’s function can be successfully
used to find the solution of the initial-boundary values problems from the theory of
the parabolical equations.
286
8 Parabolical Equations
In the considerations from this paragraph, the method of the Green’s function
has been used to find the solution of the linear problems. But this method can be
used to find the solution of the nonlinear problems. We want outline that for the
determination of the Green’s function one can use the Laplace’s transform. Applying
the Laplace’s transform on the parabolical equations and on the initial and boundary
conditions we obtain an elliptical boundary value problem, because the Laplace’s
transform proceeds on the temporal variable. Also, an initial-boundary value problem
for parabolical equations, receives some simplifications if we apply the Fourier’s
transform on the spatial variables.
Let us consider the nonlinear problem
u −
∂u
= F(t, x, u, u x1 , u x2 , ..., u xn ), ∀(t, x) ∈ TT × ,
∂t
u(0, x) = ϕ(x), ∀x ∈ ,
(8.3.20)
u(t, y) = α(t, y), ∀(t, y) ∈ TT × ∂.
In tackling of the problem (8.3.20) we can use the same procedure as in the case
of the linear problems. Firstly, we determine the Green’s function attached to the
domain , to the linear operator u − u t and to the boundary conditions (8.3.20)3 .
Assuming that the problem (8.3.20) admits a classical solution, then this solution can
be represented with the aid of the Green’s function in the form
t
G(t, τ , x, ξ)F(τ , ξ, u, u ξ1 , u ξ2 , ..., u ξn )dτ dξ −
u(t, x) = −
0
t
∂G(t, τ , x, η)
α(τ , η)dτ dση + G(t, 0, x, ξ)ϕ(ξ)dξ. (8.3.21)
−
∂ν
∂ 0
We must now determine the conditions that will be imposed to the functions F, α and
ϕ such that the function u from Eq. (8.3.21) is an effective solution for the problem
(8.3.20). One can prove a result of the form: If the function F is continuous in all its
variables and satisfies a Lipschitz condition in the variables u, u x1 , u x2 , ..., u xn , then
u from Eq. (8.3.21) is an effective solution of the problem (8.3.20).
8.4 Cauchy’s Problem
In the considered initial-boundary values problems for the heat equation, in the
previous paragraphs, it is essential to know the temperature on the surface of the
body where the problem has been stated.
In this paragraph we consider that the surface is to a great distance, such that
instead of a bounded domain we will consider the whole space IRn . Therefore, the
boundary conditions disappear and then we have the following Cauchy’s problem
8.4 Cauchy’s Problem
287
u(t, x)−u t (t, x) = f (t, x), ∀(t, x) ∈ TT × IRn ,
u(0, x) = ϕ(x), ∀x ∈ IRn ,
(8.4.1)
where TT is a temporal interval (0, T ] and the functions f and ϕ are given and
continuous on TT × IRn , respectively, on IRn .
The problem (8.4.1) will be complete if it is known the behavior of the function u to
the infinity. It is well known two kind of behavior to the infinity:
– the function u is bounded;
– u asymptotically tends to zero.
In all that follows we will suppose that u is bounded to the infinity.
We call the classical solution for the Cauchy’s problem, a function u which satisfies the conditions:
–
–
–
–
u ∈ C(TT × IRn );
u and u xi are bounded functions on TT × IRn ;
u xi xi , u t ∈ C(TT × IRn );
u satisfies the Eq. (8.4.1)1 and the initial condition (8.4.1)2 .
In tackling of the Cauchy’s problem (8.4.1), we will make two steps. First, assuming
that the problem admits a classical solution, we will find its form with the aid of the
Riemann-Green’s formula.
In the second step, we will show that in certain conditions of regularity imposed
to the functions f and ϕ the founded formula for u is an effective solution for the
problem (8.4.1).
We remember that the fundamental solution V (t, τ , x, ξ) is given by
⎞
⎛ n
(xi − ξi )2
⎟
⎜ i=1
1
⎟.
−
V (t, τ , x, ξ) = √ n √
n exp ⎜
⎝
4(t − τ ) ⎠
2 π
t −τ
(8.4.2)
Theorem 8.4.1 We suppose that the Cauchy’s problem (8.4.1) admits a classical
solution. Then, this solution admits the representation:
t
u(t, x) = −
V (t, τ , x, ξ) f (τ , ξ)dτ dξ +
IRn 0
IRn
V (t, 0, x, ξ)ϕ(ξ)dξ.
(8.4.3)
Proof We arbitrarily fix x ∈ IRn and we take the ball B(0, R) with the center in the
origin and the radius R sufficient big such that the ball contains inside the point x.
We write then the Riemann-Green’s formula on this ball, for the pairs of functions
v = V (t, τ , x, ξ) and u = u(t, x), where u(t, x) is the solution of the problem (8.4.1)
288
8 Parabolical Equations
u(t, x) =−
V (t, τ , x, ξ) f (τ , ξ)dτ dξ +
V (t, 0, x, ξ)ϕ(ξ)dξ+
B(0,R) 0
B(0,R)
t
∂u(τ , ξ)
∂V (t, τ , x, ξ)
V (t, τ , x, ξ)
dτ dσξ .
+
−u(τ , ξ)
∂ν
∂ν
∂ B(0,R) 0
t
(8.4.4)
We have assumed that u and u xi are bounded functions (because u is a classical
solution for the problem (8.4.1)). Then, taking into account the properties of the
function V (t, τ , x, ξ), we can show that if R → ∞, the last integral from Eq. (8.4.4)
tends to zero. With that end in view we write the last integral from Eq. (8.4.4) in the
form
t
∂u(τ , ξ)
dτ dσξ −
V (t, τ , x, ξ)
∂ν
∂ B(0,R) 0
t
∂V (t, τ , x, ξ)
−
u(τ , ξ)dτ dσξ = I1 + I2 .
(8.4.5)
∂ν
∂ B(0,R) 0
Then
|I1 | ≤ c0
∂ B(0,R)
t
0
⎞
⎛ n
(xi − ξi )2
⎟
⎜ i=1
1
⎟ dτ dσξ ,
−
√
n exp ⎜
⎝
4(t − τ ) ⎠
t −τ
where
1
∂u
c0 = √ n sup
∂ν
2 π
and this supremum exists since u is a bounded function.
It is clear that
n
|xk − ξk | ≤ r = |ξx| =
(xi − ξi )2 .
i=1
We can choose the radius of the ball R such that for x arbitrarily fixed, x ∈ I nt B(0, R)
and ξ ∈ ∂ B(0, R), we have |ξx| > R/2.
By using these evaluations, for I1 we obtain
|I1 | ≤ c0
∂ B(0,R)
t
√
0
1
t −τ
R2
n e− 16(t−τ ) dτ dσξ .
8.4 Cauchy’s Problem
289
With regard to the derivative of the function V (t, τ , x, ξ) in the direction of the
normal, we have the estimation
n
n
∂V ∂V ∂V
cos αk ≤ ,
∂ν = ∂x
∂x
k
k
k=1
k=1
Therefore, for I2 we obtain
|I2 | ≤ c1
t
n
|xi − ξi | (t −τ )−
∂ B(0,R) 0 i=1
n+2
2
⎞
⎛ n
(xi − ξi )2
⎟
⎜ i=1
⎟
exp ⎜
⎝− 4(t − τ ) ⎠ dτ dσξ ,
and with the above estimations, it results
|I2 | ≤ c2
∂ B(0,R)
0
≤ c2 R
t
r
∂ B(0,R)
(t − τ )
t
0
r2
(n+2)/2
e− 4(t−τ ) dτ dσξ ≤
1
(t − τ )
R2
e− 16(t−τ ) dτ dσξ ,
(n+2)/2
where c1 proceeds from the supremum of the function u and c2 = nc1 .
If we make the change of variable
t −τ =
R2
R2
⇒
dτ
=
dσ,
16σ 2
8σ 3
then for the increase of I2 we have
|I2 | ≤ c3
1
R n−1
∂ B(0,R)
= c3 ωn
∞
R
√
4 t
∞
R
√
4 t
σ n−1 e−σ dσdσξ =
2
σ n−1 e−σ dσdσξ .
2
An analogous increase follows for I1 too, by using the same change of variable. By
integrating n − 1-times by parts, it will be shown that
∞
lim
R→∞
R
√
4 t
σ n−1 e−σ dσdσξ = 0.
2
Therefore, I1 and I2 tend to zero, for R → ∞. If we pass to the limit with R → ∞
in Eq. (8.4.5), we obtain that the integral from left-hand side tends to zero. So, if we
pass to the limit with R → ∞ in Eq. (8.4.4), we obtain formula (8.4.3).
290
8 Parabolical Equations
The formula (8.4.3) is called the Poisson’s formula for the representation of the
solution of the Cauchy’s problem (8.4.1). With the aid of the Poisson’s formula we
can prove the uniqueness of the classical solution for the problem (8.4.1).
Theorem 8.4.2 The Cauchy’s problem (8.4.1) admits at the most one classical
solution.
Proof We suppose, through absurd, that the problem (8.4.1) admits two classical
bounded solutions, u 1 (t, x) and u 2 (t, x), that is,
∂u i
(t, x) = f (t, x), ∀(t, x) ∈ TT × IRn ,
∂t
u i (0, x) = ϕ(x), ∀x ∈ IRn ,
u i (t, x) −
where i = 1, 2. We define the function v(t, x) by v(t, x) = u 1 (t, x)−u 2 (t, x). Then
∂v
(t, x) = f (t, x) − f (t, x) = 0,
∂t
v(0, x) = u 1 (0, x) − u 2 (0, x) = ϕ(x) − ϕ(x) = 0.
v(t, x) −
(8.4.6)
So, we have obtained a new Cauchy’s problem with f ≡ 0 and ϕ ≡ 0. According
to the Theorem 8.4.1, if a Cauchy’s problem admits a solution, then the solution has
indispensable the form (8.4.3). If we write formula (8.4.3) and take into account
that f ≡ 0 and ϕ ≡ 0, then we obtain v(t, x) = 0, ; ∀(t, x) ∈ TT × IRn such that
u 1 (t, x) = u 2 (t, x).
It remains to prove that the function u from Eq. (8.4.3) is an effective solution
of the Cauchy’s problem (8.4.1). This is the objective of the following theorem of
existence.
Theorem 8.4.3 Assume the following conditions hold
2
∂f
(i) the functions f (t, x), ∂x
(t, x), ∂∂x 2f (t, x) are continuous and bounded on
i
i
TT × IRn , that is,
f (t, x),
∂f
∂2 f
(t, x),
(t, x) ∈ C(TT × IRn ) ∩ B(TT × IRn );
∂xi
∂xi2
(ii) the functions ϕ(t, x),
TT × IRn , that is,
ϕ(t, x),
∂ϕ
(t, x),
∂xi
∂2 ϕ
(t, x)
∂xi2
are continuous and bounded on
∂ϕ
∂2ϕ
(t, x),
(t, x) ∈ C(TT × IRn ) ∩ B(TT × IRn ).
∂xi
∂xi2
Then, the function u from Eq. (8.4.3) is an effective solution of the Cauchy’s
problem (8.4.1), namely, a bounded solution on TT × IRn .
8.4 Cauchy’s Problem
291
Proof We define the integral I1 by
I1 =
IRn
V (t, 0, x, ξ)ϕ(ξ)dξ,
and show that I1 verifies the problem
u(t, x)−
∂u
(t, x) = 0, ∀(t, x) ∈ TT × IRn ,
∂t
u(0, x) = ϕ(x), ∀x ∈ IRn ,
(8.4.7)
and, also, we show that the integral I2
I2 =
IRn
t
V (t, τ , x, ξ) f (τ , ξ)dτ dξ,
0
verifies the problem
u(t, x) −
∂u
(t, x) = f (t, x), ∀(t, x) ∈ TT × IRn ,
∂t
u(0, x) = 0, ∀x ∈ IRn .
Thus, it will be obvious that I1 + I2 , that is, u from Eq. (8.4.3), verifies the Cauchy’s
problem (8.4.1).
Since ϕ is bounded and continuous, we have
|I1 | ≤ ϕ
V (t, 0, x, ξ)dξ = ϕ ,
IRn
that proves that the integral I1 is convergent and, therefore, we can derive under the
integral. Then
I1 −
∂ I1
=
∂t
IRn
V −
∂V
∂t
ϕ(ξ)dξ = 0,
taking into account the properties of the function V (t, τ , x, ξ).
On the other hand, from the properties of the function V (t, τ , x, ξ), we have
lim I1 = lim lim
t→0
t→0 R→∞ IRn
V (t, 0, x, ξ)ϕ(ξ)dξ = ϕ(x).
Since the ball B(0, R) has the radius R sufficient big, such that the point x is contained
inside of the ball.
292
8 Parabolical Equations
As in the case of I1 , it will be shown that the integral from I2 is convergent and
then we can derive under the integral such that
x I2 = −
t
IRn
0
x V (t, τ , x, ξ) f (τ , ξ)dξdτ .
(8.4.8)
In the case of the derivative with respect to t, we have an integral with parameter:
t
∂V (t, τ , x, ξ)
∂ I2
V (t, t, x, ξ) f (t, ξ)dξ −
=−
f (τ , ξ)dξdτ .
∂t
∂τ
IRn
IRn 0
(8.4.9)
For the first integral from the right-hand side of the relation (8.4.9) we have, in fact,
V (t, τ , x, ξ) f (τ , ξ)dξ = f (t, x),
lim lim
τ →t − R→∞
B(0,R)
according to first part of the Theorem 8.2.7 (Sect. 8.2). Thus, Eq. (8.4.9) becomes
∂ I2
= − f (t, x) −
∂t
IRn
t
0
∂V (t, τ , x, ξ)
f (τ , ξ)dξdτ ,
∂τ
relation which, together with Eq. (8.4.8), leads to
∂ I2
= f (t, x)−
∂t
t
∂V (t, τ , x, ξ)
−
x V (t, τ , x, ξ) −
f (τ , ξ)dτ dξ.
∂τ
IRn 0
x I2 −
But
x V (t, τ , x, ξ) −
∂V (t, τ , x, ξ)
= 0,
∂τ
and then the previous relation becomes
x I2 −
∂ I2
= f (t, x).
∂t
Therefore, it is clear that
0
lim I2 =
t→0
0
IRn
V (t, τ , x, ξ) f (τ , ξ)dξdτ = 0,
that concludes the proof of the theorem.
8.4 Cauchy’s Problem
293
At the end of this paragraph, we will solve a Cauchy’s problem, attached to the
equation of the heat, in a particular case.
Let B = [0, T ] × (−∞, ∞) be a band, where T is a positive fixed number, which
can be ∞ too. Consider the equation
∂u
∂2u
−
= 0, ∀(t, x) ∈ B.
2
∂x
∂t
(8.4.10)
If the function u(t, x), defined on the band B, has the continuous derivatives ∂ 2 u/∂x 2
and ∂u/∂t inside of the band and u(t, x) satisfies the Eq. (8.4.10), we call that the
function u(t, x) is a regular solution of the Eq. (8.4.10).
The Cauchy’s problem consists of the determination of a regular solution of the
Eq. (8.4.10) which satisfies the initial condition:
u(0, x) = ϕ(x), ∀x ∈ (−∞, −∞),
(8.4.11)
where the function ϕ(x) is a given real function, which is continuous and bounded.
We will prove that the function u(t, x), defined by
1
u(t, x) = √
2 πt
∞
ϕ(ξ)e−
(ξ−x)2
4t
dξ,
(8.4.12)
−∞
is the solution of the Cauchy’s problem (8.4.10) and (8.4.11).
It is well known that the integral from Eq. (8.4.12) is uniformly convergent in a
vicinity of an arbitrarily point √
(t, x) from the inside of the band B. If we make the
change of variable ξ = x + 2η t, formula (8.3.12) becomes:
1
u(t, x) = √
π
∞
−∞
√
2
ϕ(x + 2η t)e−η dη.
(8.4.13)
Since ϕ is continuous and bounded, we have
sup
−∞<x<∞
|ϕ(x)| < M, M > 0.
The integral from Eq. (8.4.13) is absolute convergent and then
M
|u(t, x)| < √
π
∞
M √
2
e−η dη = √
π = M.
π
−∞
The integrals obtained by derivation under the integral in Eq. (8.4.12), with respect
to x and with t, are uniformly convergent. On the other hand, the function
(ξ−x)2
1
√ e− 4t , t > 0,
t
294
8 Parabolical Equations
satisfies, obvious, the Eq. (8.4.10). These estimations assure that the function u
defined in Eq. (8.4.12) satisfies the Eq. (8.4.10).
Using again the uniformly convergence of the integral in a vicinity of any point
(t, x), with t > 0, from the inside of the band B, we can pass to the limit, as t → 0,
in Eq. (8.4.13), whence it follows
lim u(t, x) = ϕ(x).
t→0
So, we immediately obtain the uniqueness and the stability of the regular solution for
our Cauchy’s problem. One can show that the regular solution of the Eq. (8.4.10), satisfies the inequality m ≤ u(t, x) ≤ M, where m = inf u(0, x) and M = sup u(0, x),
x ∈ (−∞, ∞). One can then use the function v(t, x) = 2t + x 2 , which obvious is a
particular solution of the Eq. (8.4.10).
Chapter 9
Elliptic Partial Differential Equations
9.1 Introductory Formulas
Let us consider the three-dimensinal regular domain D ⊂ R3 bounded by the Liapunov surface S = ∂D.
In the classical mathematical analysis the following formula is proved
S
P(x, y, z)dydz+Q(x, y, z)dzdx+R(x, y, z) dxdy =
∂P ∂Q ∂R
+
+
dxdydz,
∂y ∂z
D ∂x
which is called the Gauss-Ostrogradski-Green’s formula.
In the following we deduce some particular forms of this formula, useful in the
theory of elliptical equations.
Let ϕ be a scalar function, ϕ = ϕ(x, y, z), ϕ ∈ C 1 (D), D ⊂ R3 . To this function
we attach the differential operator, denoted by “grad”
grad ϕ =
∂ϕ ∂ϕ ∂ϕ i+
j+
k
∂x
∂y
∂z
called the gradient of ϕ. Now, consider a vectorial function V = V (x, y, z), V (V1 ,
V2 , V3 ) such that Vi = Vi (x, y, z) ∈ C 1 (D), D ⊂ R3 . To this vectorial function we
can attach two differential operators namely, “the divergence operator”, denoted
“div” and “the rotor operator”, denoted “curl”, as follows
∂V2
∂V3
∂V1
+
+
,
div V =
∂x
∂y
∂z
curl V =
∂V3
∂V2 ∂V3 ∂V1 ∂V1
∂V2
−
−
−
i+
j+
k.
∂y
∂z
∂z
∂x
∂x
∂y
© Springer International Publishing AG, part of Springer Nature 2018
M. Marin and A. Öchsner, Complements of Higher Mathematics,
https://doi.org/10.1007/978-3-319-74684-5_9
295
296
9 Elliptic Partial Differential Equations
Therefore, if we consider the functions P(x, y, z), Q(x, y, z) and R(x, y, z) as the
components of a vectorial function
V (x, y, z) = P(x, y, z)i + Q(x, y, z)j + R(x, y, z)k
then the above Gauss-Ostrogradski-Green’s formula can be restated in the form
V ndσ =
S
div V dv, dv = dxdydz,
(9.1.1)
D
where n is the outward normal of the surface S that bounded the domain D, defined
by
n = (n1 , n2 , n3 ) = (cos α, cos β, cos γ) = (cos(x, n), cos(y, n), cos(z, n)).
For two scalar functions ϕ(x, y, z) and ψ(x, y, z) such that ϕ, ψ ∈ C 2 (D), D ⊂ R3 ,
we can consider the vectorial function V defined by
V = ϕ.grad ψ.
Then, the scalar product V .n becomes
dψ
,
V .n = ϕ.grad ψ n = ϕ.
dn
since
∂ψ
dψ
∂ψ
∂ψ
=
n1 +
n2 +
n3 .
dn
∂x
∂y
∂z
With these calculations, Eq. (9.1.1) becomes
dψ
ϕ
dσ =
dn
S
div (ϕgrad ψ) dv.
(9.1.2)
D
On the other hand, if we use the notation
∇=
∂
∂
∂ i + j + k,
∂x
∂y
∂z
then the above defined operators “grad”, “div” and “curl” can be written in the form
grad ϕ = ∇ϕ, div V = ∇.V , curl V = ∇ × V .
Also, we have the following results
∇.(ϕ.∇ψ) = ∇ϕ.∇ψ + ϕ∇(∇ψ) = grad ϕ.grad ψ + ϕψ,
9.1 Introductory Formulas
297
because
∇(∇ψ) =
∂
∂x
∂ψ
∂x
+
∂
∂y
∂ψ
∂y
+
∂
∂z
∂ψ
∂z
=
∂2ψ
∂2ψ
∂2ψ
+
+ 2 = ψ
2
2
∂x
∂y
∂z
Now, relation (9.1.2) can be restated as follows
dψ
dσ =
ϕ
dn
S
(grad ϕgrad ψ + ϕψ) dv,
(9.1.3)
D
known as the first Green’s formula.
For the vectorial function V = ψgrad ϕ the first Green’s formula becomes
ψ
S
dϕ
dσ =
dn
(grad ψgrad ϕ + ψϕ) dv.
(9.1.4)
D
Subtracting, term by term, Eq. (9.1.4) from (9.1.3), we obtain
dψ
dϕ
ϕ
−ψ
dσ =
(ϕψ − ψϕ) dv,
dn
dn
S
D
(9.1.5)
known as the second Green’s formula.
9.2 Potentials
Let us consider the scalar function ϕ = ϕ(x, y, z) such that ϕ ∈ C 2 (D), D ⊂ R3 and
the domain with the following properties:
(i) is bounded by the regular surface = ∂;
(ii) has the continuous normal outward to the outside of ;
(iii) = ∪ , ⊂ D.
Since ϕ ∈ C 2 (D) and ⊂ D we deduce ϕ ∈ C 2 ().
Definition 9.2.1 By definition, the following integrals
1
4π
1
1 dϕ
dσ, −
r dn
4π
ϕ
d
dn
1
1
1
dσ, −
ϕdσ,
r
4π r
(9.2.1)
are called Single-Layer Potential of surface, Double-Layer Potential of surface and
Volume Potential, respectively.
Here r is the Euclidean distance r = x2 + y2 + z 2 .
Theorem 9.2.1 Let ϕ be a scalar function, ϕ ∈ C 2 (D), the domain having the
above properties and M0 an arbitrary fixed point in . Then, we have
298
9 Elliptic Partial Differential Equations
ϕ(M ) = ϕ(x, y, z) =
1
4π
1 dϕ
1
dσ −
r dn
4π
ϕ
d
dn
1
1
1
dσ −
ϕdv, (9.2.2)
r
4π r
∀M ∈ , where r = (x−x0 ) +(y−y0 )2 +(z−z0 )2 , M0 = M0 (x0 , y0 , z0 ) .
2
Proof Since M ∈ and is a domain we deduce that there exists a ball B(M , )
such that B(M , ) ⊂ , where B(M , ) = B(M , ) ∪ S(M , ). Here we have noted
by S(M , ) the sphere with the center M and radius .
On the domain \ B(M , ) we apply the second Green’s formula (9.1.5) for the
pair of functions ϕ and ψ = 1/r:
dϕ
dψ
−ψ
dσ =
ϕ
(ϕψ − ψϕ) dv.
dn
dn
∂ (\B)
\B
But ∂ \ B = ∪ S. Also, it is well known fact that
ψ = 1
= 0,
r
such that the previous equality becomes
∪S
d 1 1 dϕ
1
ϕ
−
dσ = −
ϕdv.
dn r
r dn
\B r
(9.2.3)
Let us make some evaluations on the left-hand side of Eq. (9.2.3), denoted by IL .
Firstly, we observe that
IL =
=
∪S
d 1 1 dϕ
ϕ
−
dσ =
dn r
r dn
d 1 1 dϕ
d 1 1 dϕ
ϕ
ϕ
−
dσ −
−
dσ.
dn r
r dn
dn r
r dn
S
On the other hand, on the sphere S(M , ) we have
r̄
r = , n̄ = − ,
r
d 1
1
r̄
1 r̄
1
1
= n̄.grad
=−
− 2
= 2 |S = .
dn r
r
r
r r
r
Therefore, we can write
IL =
1
2
ϕdσ −
1
S
dϕ
dσ.
dn
(9.2.4)
9.2 Potentials
299
Since ϕ ∈ C 2 (D) we deduce
dϕ
∈ C 1 (D),
dn
such that we can use the mean theorem in both sides of Eq. (9.2.4) and deduce that
there exists the points Q1 , Q2 ∈ S such that
IL =
1
2
dϕ
1
(Q2 ).
ϕ(Q1 )4π − 4π
dn
Here we can pass to the limit as
→ 0 and obtain
lim 4πϕ(Q1 ) − 4π
→0
dϕ
(Q2 ) = 4πϕ(M ),
dn
because, if → 0 then the sphere S(M , ) reduces to the point M .
Now, we make some evaluation on the right-hand side of Eq. (9.2.3). Firstly,
let us observe that since ϕ ∈ C 2 (D) we deduce that ϕ ∈ C 0 (D), that is, ϕ is
a continuous function. Also, the distance function r is continuous, such that the
function ϕ/r is continuous. According to the Weierstrass’s theorem, a continuous
function defined on a closed set is bounded, such that we can write
1
≤ 1 ϕ dv ≤ M
ϕdv
dv = 4π 3 M .
r
B
B r
B
So, we deduce that
1
− ϕ dv ≤ lim 4π 3 M = 0.
lim
→0 B
→0
r
In conclusion, passing to the limit in Eq. (9.2.3), as lim , we are led to
→0
lim
→0
d
ϕ
dn
= lim
→0
1
1 dϕ
dσ −
dσ − IL =
r
r dn
1
− ϕdv −
r
1
− ϕdv .
r
B
Thus,
d
ϕ
dn
1
1 dϕ
1
dσ −
dσ − 4πϕ(M ) =
− ϕdv,
r
r
dn
r
such that, finally, we have the desired potentials formula
300
9 Elliptic Partial Differential Equations
ϕ(M ) =
1
4π
1
1 dϕ
dσ −
r dn
4π
ϕ
d
dn
1
1
1
dσ −
ϕdv,
r
4π r
and the theorem is proved.
An immediate consequence of this theorem is given in the following corollary.
Corollary 9.2.1 If the scalar function ϕ is a harmonic function, that is, ϕ = 0,
then the potentials formula reduces to
1
ϕ(M ) =
4π
1
1 dϕ
dσ −
r dn
4π
d
ϕ
dn
1
dσ.
r
It is a simple matter to see that this equality is obtained from Eq. (9.2.2) taking into
account that ϕ.
This relation says that to obtain the values of the harmonic function ϕ, in the point
from inside of the domain where ϕ is harmonic, it is sufficient to know the values of
the function on the boundary of this domain.
A very important result which follows from the potentials formulas, also in the
case of harmonic functions, is proved in the following theorem. This result is called
the Gauss’s mean-value formula.
Theorem 9.2.2 Let ϕ be a harmonic function on the domain D ⊂ R3 and an arbitrary point M ∈ D. Consider the ball B(M , R) such that B(M , R) ∪ S(M , R) =
B(M , R) ⊂ D. Then, the value of function ϕ in the point M is the mean of the
values taken by ϕ on the sphere S(M , R):
ϕ(M ) =
1
4πR2
ϕ(P)dσ.
S
Proof In the particular case when the function ϕ is harmonic, the potentials formula
becomes
1
1 dϕ
d 1
1
dσ −
dσ.
(9.2.5)
ϕ
ϕ(M ) =
4π r dn
4π dn r
But, on the sphere S(M , R) we have
r = R, n =
d
dn
n
,
r
1
1
r̄
1 r̄
1
1
= n̄.grad
=
− 2
= − 2 |S = − 2 .
r
r
r
r r
r
R
With the help of this evaluations, Eq. (9.2.5) becomes
1
ϕ(M ) =
4πR
1
dϕ
dσ +
dn
4πR2
ϕdσ.
(9.2.6)
9.2 Potentials
301
Now, we use the second Green’s formula for the ball B(M , R) having as boundary
the sphere S(M , R) and for the pair of functions ϕ, which is harmonic, and ψ ≡ 1:
dψ
dϕ
ϕ
−ψ
dσ = (ϕψ − ψϕ) dv.
dn
dn
B
Since ϕ = 0, ψ = 0 and d ψ/dn = 0 the previous equality leads to
dϕ
dσ = 0.
dn
Substituting this result in Eq. (9.2.6) we obtain
ϕ(M ) =
1
4πR2
ϕ(P)dσ,
S
that is, the Gauss’s formulas and the theorem is concluded.
A very useful result is the min-max principle for harmonic functions which is
proved in the following theorem.
Theorem 9.2.3 If ϕ is a harmonic function on the closed domain = ∪ ∂ and
the function ϕ is not constant, then ϕ takes its minimum value and the maximum
value on the boundary ∂ of the domain .
Proof Suppose that the function ϕ is not constant and let us prove that ϕ takes its
maximum value on the boundary ∂ of the domain . The proof is similar to the case
of the minimum value. Suppose that there exists a point M0 inside of the domain such that the maximum value of ϕ is the value in this point, that is, ϕ(M0 ) > ϕ(M ),
∀M in a vicinity of M0 . Consider a sphere with M0 as center which contains all poins
for what ϕ(M0 ) > ϕ(M ). Then
ϕ(M0 )dσ > ϕ(M )dσ ⇒ ϕ(M0 ) dσ > ϕ(M )dσ.
S
S
S
S
With regards to the last integral we use the Gauss’s formula and deduce
ϕ(M0 )4πR2 > ϕ(M0 )4πR2 ⇒ ϕ(M0 ) > ϕ(M0 ),
which is a contradiction, and the theorem is concluded.
Remark. The restriction imposed to the function ϕ to be not constant is not important,
it is necessary only for the method of demonstration. If the function ϕ is constant
then its value in a point is the same both in the case then the point is inside of the
domain and in the case when the point is on the boundary. So, we can say that its
minimum value and maximum value are taken on the boundary.
302
9 Elliptic Partial Differential Equations
9.3 Boundary Values Problems
The problems for the elliptical partial differential equations do not contain the initial
data because these equations are models for stationary phenomena. In short, in this
context, the problems must determine certain functions which inside of a domain
satisfy an elliptical partial differential equation and on the boundary of the domain
these functions have a known behavior. In many studies dedicated to the elliptical
partial differential equations are well used the following boundary conditions:
(i) the Dirichlet’s condition when the value of the function on the boundary is
prescribed;
(ii) the Neumann’s condition when the value of the normal derivatives of the function on the boundary is prescribed;
(iii) the mixt condition when the value of the function is prescribed on a part of the
boundary and on the rest is prescribed the value of the normal derivatives of
the function.
As a consequence, there exist three boundary value problem:
(i) the Dirichlet’s Problem:
u = f (x, y, z), (x, y, z) ∈ ⊂ R3 ,
u|∂ = g(x, y, z), (x, y, z) ∈ ∂;
(9.3.1)
where the functions f and g are given;
(ii) the Neumann’s Problem:
u = f (x, y, z), (x, y, z) ∈ ⊂ R3 ,
du
|∂ = h(x, y, z), (x, y, z) ∈ ∂;
dn
(9.3.2)
where the functions f and h are given;
(iii) the mixt Problem:
u = f (x, y, z), (x, y, z) ∈ ⊂ R3 ,
u|1 = g(x, y, z), (x, y, z) ∈ 1 ;
du
| = h(x, y, z), (x, y, z) ∈ 2 ;
dn 2
(9.3.3)
where the functions f and g are given and 1 ∪ 2 = ∂.
Let us, first, study the Dirichlet’s problem (9.3.1). Namely, we will prove that
the Dirichlet problem has at most one solution.
9.3 Boundary Values Problems
303
Theorem 9.3.1 The solution of the Dirichlet’s problem is unique.
Proof Suppose that the Dirichlet’s problem has two solutions, say u1 (x, y, z) and
u2 (x, y, z) which correspond to the same right-hand side terms, that is
u1 = f (x, y, z), (x, y, z) ∈ u1 = g(x, y, z), (x, y, z) ∈ ∂,
and
u2 = f (x, y, z), (x, y, z) ∈ u2 = g(x, y, z), (x, y, z) ∈ ∂.
Denote by u(x, y, z) the difference of these solutions, u(x, y, z) = u1 (x, y, z) −
u2 (x, y, z). We intend to prove that
u(x, y, z) = 0, ∀(x, y, z) ∈ .
Firstly, we have
u = u1 − u2 = f − f = 0,
such that the function u is harmonic.
On the other hand, for any (x, y, z) ∈ ∂ we have
u(x, y, z) = u1 (x, y, z) − u2 (x, y, z) = g(x, y, z) − g(x, y, z) = 0.
Let us suppose there exists a point (x0 , y0 , z0 ) such that u(x0 , y0 , z0 ) = 0. Without
restricting the generality, we assume that u(x0 , y0 , z0 ) > 0. As we already proved, u
is a harmonic function such that we can apply the min-max principle for harmonic
functions and deduce that the maximum value of u is taken on the boundary. But
on the boundary the function u takes only the null values. With this contradiction
the theorem is concluded. We must say that in the case that u(x0 , y0 , z0 ) < 0 we use
the same min-max principle but with regard to minimum value. If u is a constant
function then u takes the same value, say C, in all points of including the boundary.
But on the boundary the value of u is zero, therefore C = 0. Thus u = 0 and then
u1 = u2 .
With regard to the Neumann’s problem we will prove that its solution is not unique.
More precisely, we will prove that the difference of any two solutions of the Neumann’s problem is a constant.
Theorem 9.3.2 The solution of the Neumann’s problem is determined until an additive constant.
Proof Let us consider two solutions of the Neumann’s problem, that is
304
9 Elliptic Partial Differential Equations
u1 = f (x, y, z), (x, y, z) ∈ du1
= g(x, y, z), (x, y, z) ∈ ∂,
dn
and
u2 = f (x, y, z), (x, y, z) ∈ du2
= g(x, y, z), (x, y, z) ∈ ∂.
dn
Denote by u(x, y, z) the difference of these solutions, u(x, y, z) = u1 (x, y, z) −
u2 (x, y, z). We intend to prove that
u(x, y, z) = C = constant, ∀(x, y, z) ∈ .
Firstly, we have
u = u1 − u2 = f − f = 0,
such that the function u is harmonic.
On the other hand, for any (x, y, z) ∈ ∂ we have
du1
du2
du
=
(x, y, z) −
(x, y, z) = g(x, y, z) − g(x, y, z) = 0.
dn
dn
dn
Now, remember the first Green’s formulas
dψ
ϕ dσ =
(grad ϕ.grad ψ + ϕψ) dv.
dn
We apply this equality for the pair of functions ϕ = ψ = u and obtain
du
u dσ =
dn
(grad u)2 + uu dv.
But
u = 0, in ,
du
= 0, on ∂,
dn
such that the previous relation reduces to
(grad u)2 dv = 0.
So, we have
grad u = 0 ⇒
∂u ∂u ∂u i+
j+
k=0⇒
∂x
∂y
∂z
9.3 Boundary Values Problems
⇒
305
∂u
∂u
∂u
= 0,
= 0,
= 0.
∂x
∂y
∂z
Thus, the function does not depend on x, y or z, that is u is a constant function
and the theorem is concluded.
In the following we intend to construct the solutions for the Dirichlet’s problem
and the Neumann’s problem, attached to the Laplace’s equation. First, consider the
Dirichlet’s problem
u = 0, in ,
u = 0, on S = ∂.
Taking into account that = 0 and u|S = f the potentials formula reduces to
1
u(x, y, z) =
4π
1
1 du
dσ −
r dn
4π
S
1 d
f
S r dn
1
dσ.
r
So, to know the solution u(x, y, z) we must eliminate, from the above relation,
the expression du/dn which is unknown. To this end, we introduce the function
g = g(x, y, z) which must satisfy the equations
g = 0, in ,
1
g = − , on S.
r
Now, we apply the second Green’s formula for the pair of functions u and g:
g
S
du
du
−u
dσ =
(gu − ug) dv.
dn
dn
But u = 0 and g = 0 such the previous relation reduces to
1 du
dg
du
dσ ⇒
u dσ = g dσ = −
S dn
S dn
S r dn
1 du
dg
dg
⇒
dσ = − u dσ = − f dσ.
r
dn
dn
S
S
S dn
With these evaluations the potentials formula becomes
u(x, y, z) = −
1
4π
f
S
dg
1
dσ −
dn
4π
f
S
d
dn
1
dσ =
r
306
9 Elliptic Partial Differential Equations
=−
1
4π
f
S
1
d
g+
dσ.
dn
r
Function g + 1/r is called the Green’s function attached to the domain and it
is denoted by G, G = g + 1/r. It satisfies the following Dirichlet’s problem
G = 0, in G|S = 0, on S = ∂.
The Green’s function is perfectly determined only by the domain. Therefore, in
order to solve the Dirichlet problem on a domain, firstly, we must determine the
Green’s function for the respective domain and then use the formulas
1
u(x, y, z) = −
4π
f
S
dG
dσ.
dn
Green’s Function for a Sphere
Consider the ball B(0, R) having as boundary the sphere S(0, R). Let M0 be an
arbitrary fixed point in the ball and another point M1 such that
OM0 .OM1 = R2 .
On the sphere we take the point M such that the segment M1 M is tangent to the
sphere. Therefore, we have
OM0 M ˜ OM1 M
and can write
OM
M0 M
OM0
=
.
=
OM
OM1
M1 M
(9.3.4)
Using the notations
OM0 = d , OM1 = d1 , OM = R, M 0 M = r, M 1 M = r 1 ,
the qualities (9.3.4) become
R
r
d
=
= .
R
d1
r1
Define function g by
g=−
R1
,
d r1
(9.3.5)
9.3 Boundary Values Problems
307
or
g(M ) = −
1
OM
.
d |M1 M |
It is easy to see that for M ∈ S(0, R) we have
1
g(M ) = − .
r
Moreover, by direct calculations, it is easy to prove that the function g is harmonic.
Therefore, the Green’s function for the sphere is
G =g+
1 R1
1
= −
.
r
r
d r1
Now, because we know the Green’s function of the sphere, we can find the solution
of the Dirichlet’s problem on the ball, namely
u(x, y, z) = −
1
4π
f
S
dG
dσ.
dn
(9.3.6)
This solution will be completely determinate if we find d G/dn. Thus
dG
1 R1
=
= n.grad G = n
−
dn
r
d r1
R
R
1 r1
1
1 r
= n grad
=n − 2 −
=
− grad
− 2
r
dr1
r r
d
r1 r1
=
nr
R nr 1
+
.
r3
d r13
But, in our case
r = M 0 M = OM − OM0 ⇒ nr = n OM − OM0 = R − d cos θ,
r 1 = M 1 M = OM − OM1 ⇒ nr 1 = n OM − OM1 = R − d1 cos θ,
0 . With these calculations the derivative of the function G
where θ is the angle MOM
becomes
d cos θ − R R R − d1 cos θ
dG
=
+
dn
r3
d
r13 .
308
9 Elliptic Partial Differential Equations
From Eq. (9.3.5) we obtain
r1 =
such that
rR
R2
, d1 =
,
d
d
d cos θ − R R R − R2 /d cos θ d 3
dG
=
+
=
dn
r3
d
1
r 3 R3
d
1
d3
1
1
= 3 d cos θ − R + 2 (d − R cos θ)
= 3 d cos θ − R + (d − R cos θ) =
R
R
r
d
r
=
1
r3R
d cos θ − R2 + d 2 − dR cos θ =
d 2 − R2
.
r3R
Finally, we introduce the derivative of the Green’s function in Eq. (9.3.6) and the
solution of the Dirichlet’s problem is completely determined
r2 − d 2
u(x, y, z) =
4πR
f
S
f (x, y, z)
dσ.
r3
Dirichlet’s Problem for a Circle
In this paragraph we give a complete solution for the Dirichlet’s problem attached
to a circle centered in the origin and having the radius R. With other words, we find
a function u = u(x, y) which is harmonic inside of a circle and has known values on
the circumference of the circle:
u(x, y) = 0, x2 + y2 < R2 ,
u(x, y) = f (x, y), x2 + y2 = R2 ,
where f = f (x, y) is a given function.
First, we write the Laplacean in polar coordinates:
x = r cos θ, y = r sin θ, where r =
y
x2 + y2 , θ = arctg .
x
The first derivatives of the function u are
∂u x ∂u y
∂u
=
−
∂x
∂r r
∂θ r 2
∂u y ∂u x
∂u
=
+
.
∂y
∂r r
∂θ r 2
Then, the second derivatives of the function u become
9.3 Boundary Values Problems
309
y2 ∂ 2 u r 2 − x2 ∂u 2xy ∂u
∂2u
x2 ∂ 2 u 2xy ∂ 2 u
+
+ 4
=
−
+
∂x2
r 2 ∂r 2
r 3 ∂r∂θ
r 4 ∂θ2
r 3 ∂r
r ∂θ
∂2u
y2 ∂ 2 u 2xy ∂ 2 u
x2 ∂ 2 u r 2 − y2 ∂u 2xy ∂u
= 2 2 + 3
+ 4 2 +
− 4
.
2
∂y
r ∂r
r ∂r∂θ
r ∂θ
r 3 ∂r
r ∂θ
The Laplacean in polar coordinates becomes
u =
∂2u
1 ∂ 2 u 1 ∂u
∂2u ∂2u
,
+
=
+
+
∂x2
∂y2
∂r 2
r 2 ∂θ2
r ∂r
such that we find the following form for the Laplace’s equation:
r2
∂2u
∂u ∂ 2 u
+ 2 = 0.
+
r
∂r 2
∂r
∂θ
Consequently, the Dirichlet’s problem can be stated in the form
r2
∂2u
∂u ∂ 2 u
+ 2 = 0, r < R,
+
r
∂r 2
∂r
∂θ
u(r, θ) = f (θ), r = R.
To find the solution of this problem we use the separation of variables method.
Thus, we find the solution in the form
u(r, θ) = V (r).W (θ).
So, for the derivative we find the expressions
∂ 2 u ∂u
= V .W,
V .W,
∂r
∂r 2
∂u
∂2u
= V.W ,
V.W ,
∂θ
∂θ2
such that the Laplace’s equation becomes
r 2 V .W + rV .W + V.W = 0.
From here, dividing by the product V.W , we obtain
r2
V
W V
W V V +r
+
= 0 ⇒ r2
+r
=−
= k, k = constant.
V
V
W
V
V
W
Therefore, from the Laplace’s equation we find two ordinary differential equations
310
9 Elliptic Partial Differential Equations
W − kW = 0,
r2
V V
+r
= −k.
V
V
We study the first equation. Supposing that k > 0 we find the solution
√
W (θ) = C1 e−
kθ
+ C2 e
√
kθ
,
which is not appropriate because it is not periodical. If k = 0, we find
W = 0 ⇒ W (θ) = C1 θ + C2 ,
which also, is not appropriate because it is not periodical.
In the case k < 0 we use the notation k = −λ2 and then
W + λ2 W = 0 ⇒ W (θ) = C1 cos λθ + C2 sin λθ.
This solution is periodical if
W (θ + 2π) = C1 cos λ(θ + 2π) + C2 sin λ(θ + 2π) = C1 cos(λθ + 2nπ) + C2 sin(λθ + 2nπ),
such that λ = n, n = 1, 2, ...
So, we find an infinite number of particular solutions of the form
Wn (θ) = An cos nθ + Bn sin nθ.
(9.3.7)
Let us study the equation for the function V :
r2
V
V +r
+ k = 0,
V
V
which, with the above value of k becomes
r 2 V + rV − n2 V = 0.
This is a differential equation of second order of Euler type for the function V
depending on the variable r. As usual, we make the change of variable r = et and
then
dr
dt
= et ,
= e−t .
dr = et dt ⇒
dt
dr
Consequently,
V =
dv
dv dt
=
= V̇ e−t ,
dr
dt dr
9.3 Boundary Values Problems
V =
311
dv dv −t
=
e = e−t −e−t V̇ + e−t V̈ = e−2t V̈ − V̇ .
dr
dt
Then, the equation of function V becomes
e2t V̈ − V̇ e−2t + et V̇ e−t − n2 V = 0 ⇒ V̈ − n2 V = 0,
having the solution
V (t) = Dn e−nt + Cn ent = Dn et
−n
+ Cn et
n
⇒
⇒ V (r) = Dn r −n + Cn r n .
But the function r −n does not exist in the origin such that we must put Dn = 0,
therefore the particular solutions are
Vn = Cn r n , n = 1, 2, ...
Consequently, taking into account the solution (9.3.7) for the equation of W , the
Laplace’s equation has the particular solutions
un (r, θ) = Wn (θ).Vn (r) = Cn r n (An cos nθ + Bn sin nθ) ,
or, with the convention Cn .An → An , Cn .Bn → Bn ,
un (r, θ) = (An cos nθ + Bn sin nθ) r n .
A linear combination of these particular solutions gives the general solution
u(r, θ) =
∞
(An cos nθ + Bn sin nθ) r n ,
0
or
∞
u(r, θ) =
a0 +
(an cos nθ + bn sin nθ) r n ,
2
1
where a0 = 2A0 , an = An , bn = Bn , n ≥ 1.
The Dirichlet’s condition leads to
∞
u(R, θ) =
a0 n
+
R an cos nθ + Rn bn sin nθ = f (θ).
2
1
The Fourier coefficients of this series are
312
9 Elliptic Partial Differential Equations
2
a0 =
T
1
π
Rn an =
T
T
1
f (t)dt =
π
0
T
f (t)dt
0
f (t) cos ntdt, Rn bn =
0
1
π
2π
f (t) sin ntdt,
0
such that the solution becomes
1
u(r, θ) =
2π
2π
∞
f (t)dt +
0
1
cos nθ
πRn
1
1
+
sin nθ
πRn
2π
2π
f (t) cos ntdt+
0
f (t) sin ntdt r n .
0
We can write the solution in the following forms
1
2π
u(r, θ) =
1
=
2π
1
=
2π
f (t)dt +
0
2π
0
2π
∞
1 r n 2π
f (t) cos n(θ − t)dt =
π R
0
1
∞ r n
1 2π
f (t)dt +
f (t)
cos n(θ − t)dt =
π 1 0
R
2π
0
1
f (t)dt +
π
2π
f (t)
∞ r n
0
R
1
cos n(θ − t)dt.
Finally, we can write
1
u(r, θ) =
π
2π
0
∞
1 r n
+
f (t)
cos n(θ − t) dt.
2
R
1
Now, we make some evaluations on the series
∞ r n
1
R
cos n(θ − t),
by means the series
∞ r n
1
which is convergent because
R
ein(θ−t) =
∞ r
1
R
ei(θ−t)
n
,
(9.3.8)
9.3 Boundary Values Problems
313
r r
ei(θ−t) = < 1.
|q| = R
R
As it is well known the sum of this series is q/(1 − q), that is
∞ r
R
1
ei(θ−t)
n
=
q
.
1−q
Let us evaluate the sum of the series:
rei(θ−t)
q
=
=
1−q
R − rei(θ−t)
r[cos(θ − t) + i sin(θ − t)]
=
R − cos(θ − t) − ir sin(θ − t)
=
=
r[cos(θ − t) + i sin(θ − t)][R − cos(θ − t) + ir sin(θ − t)]
=
[R − r cos(θ − t)]2 + r 2 sin2 (θ − t)
=
r[R cos(θ − t) − r + iR sin(θ − t)]
.
R2 − 2rR cos(θ − t) + r 2
Then our series becomes
∞ r n
n=1
R
cos n(θ − t) = Re
∞ r n
n=1
=
R2
R
ein(θ−t) =
r[R cos(θ − t) − r]
.
− 2rR cos(θ − t) + r 2
For the solution we obtain the following expressions
1
u(r, θ) =
π
=
2π
f (t)
0
1
2π
2π
r[R cos(θ − t) − r]
1
dt =
+ 2
2 R − 2rR cos(θ − t) + r 2
f (t)
0
R2
R2 − r 2
dt.
− 2rR cos(θ − t) + r 2
Finally, we can write the following form of the solution
u(r, θ) =
R2 − r 2
2π
0
2π
R2
f (t)
dt,
− 2rR cos(θ − t) + r 2
which is called the Poisson’s solution for the Dirichlet’s problem.
314
9 Elliptic Partial Differential Equations
Now, we apply the Poisson’s form of the solution to obtain the solution for a
useful application.
Application. Find the distribution of the electrical field in a plane disk knowing that
it is 1 on the superior circumference and 0 on the inferior circumference.
Solution. We apply the Poisson’s formula for the function
f (t) =
1, t ∈ [0.π]
0, t ∈ (π.2π].
Thus, the above relation becomes
R2 − r 2
u(r, θ) =
2π
2π
R2
0
1
dt.
− 2rR cos(θ − t) + r 2
To compute the integral we make the change of variable:
tg
2
t−θ
= τ ⇒ t − θ = 2arctg τ ⇒ dt =
dτ .
2
1 + t2
Also, because t ∈ [0, π], we deduce
θ
θ
τ ∈ −tg , ctg .
2
2
The integrant becomes
R2 − 2rR
1 − τ2
(R − r)2 + τ 2 (R + r)2
2
+
r
=
,
1 + τ2
1 + τ2
such that the solution receives the forms
u(r, θ) =
1
=
π
R2 − r 2
2π
ctg θ/2
−tg θ/2
=
ctg θ/2
−tg θ/2
2
1 + τ2
dτ =
2
1 + τ2
(R − r)2 1 + R+r
τ
R−r
1
R+r
R − r 1 + R+r τ
R−r
2
dτ =
R + r ctg θ/2
1
arctg
τ|
=
π
R − r −tg θ/2
R+r
R+r
1
arctg
ctg θ/2 + arctg
tg θ/2 .
π
R−r
R−r
Finally, we find the solution
u(r, θ) =
r 2 − R2
1
arctg
.
π
2rR sin θ
9.3 Boundary Values Problems
315
Neumann’s Problem
In the following we intend to construct the solution for the Neumann’s problem
defined by the relations
u = 0, in ,
du
= h, on S = ∂.
dn
Taking into account that = 0 and u|S = f the potentials formula reduces to
1
u(x, y, z) =
4π
1
1
h dσ −
r
4π
S
d
u
dn
S
1
dσ.
r
(9.3.9)
So, to know the solution u(x, y, z) we must eliminate, from the above relation
(9.3.9), the expression u which is unknown. Let us write the second Green’s formula
ϕ
S
dϕ
dψ
−ψ
dσ = (ϕψ − ψϕ)dv,
dn
dn
for the pair of functions ϕ = h and ψ = u:
du
du
dh
dh
u −h
dσ = 0 ⇒ h dσ = u dσ.
dn
dn
S
S dn
S dn
Now, consider the function g such that g = 0 in . In the same manner as above
we obtain
du
du
dg
1
dg
g
g
−u
dσ = 0 ⇒
−u
dσ = 0.
dn
dn
4π S
dn
dn
S
If we add this equality, term by term, to the equality (9.3.9) we are led to
1
u(x, y, z) =
4π
1
=
4π
S
S
du
dg
1 du
d
h −u
+
−u
dn
dn
r dn
dn
1
dσ =
r
1
d
1
g+
−u
g+
dσ.
r
dn
r
Function g + 1/r is called the Green’s function attached to the domain and it is
denoted by G, G = g + 1/r. If we impose to the function G to satisfy the condition
4π
dG
=
,
dn
A
where A is the area of the surface S, then the solution receives the form
316
9 Elliptic Partial Differential Equations
1
4π
1
du
udσ =
G dσ −
4π S dn
4π S A
1
1
du
=
G dσ −
udσ.
4π S dn
A S
u(x, y, z) =
The last integral does not depend on the point M = (x, y, z), that is, the integral
is a constant:
1
udσ = C = constant.
−
A S
Taking into account that
du
|S = h,
dn
the solution of the Neumann’s problem is
u(x, y, z) =
1
4π
hGdσ + C,
S
G being the Green’s function of the domain .
Therefore, to determine the function u we must, firstly, determine the Green’s
function of the domain and this, in fact, means the determination of the function g
which is harmonic in the domain and on the surface S satisfies the condition
4π
d 1
dg
=
−
.
dn
A
dn r
Remark. The Neumann’s problem does not have always a solution. To find the
condition when the Neumann’s problem has a solution, we use the first Green’s
formulas
dψ
ϕ dσ =
(grad ϕ.grad ψ + ϕψ) dv
dn
S
for the pair of functions ϕ = 1 and ψ = u:
S
du
dσ =
dn
udv = 0 ⇒
S
du
dσ = 0 ⇒
dn
hdσ = 0.
S
Also, the Neumann’s problem does not have always a solution even in the more
general case when the Laplace’s equation is replaced by the Poisson equation, that
is for the problem
u = f in du
= h on S = ∂.
dn
9.3 Boundary Values Problems
317
Indeed, we use the first Green’s formulas for the pair of functions ϕ = 1 and
ψ = u:
du
dσ =
udv.
S dn
Therefore, the Neumann’s problem attached to the Poisson equation, admits a
solution if
hdσ =
f dv.
S
At the final part of this paragraph we indicate a procedure to solve the Neumann’s
problem in the particular case when the domain is a sphere centered in the origin
and having the radius R. Let us use the spherical coordinates
⎧
⎨ x = r sin θ cos ϕ
y = r sin θ sin ϕ
⎩
z = r cos θ,
where r ∈ [0, R], θ ∈ [0, π], ϕ ∈ [0, 2π].
We consider only the particular case when the data on the boundary depends only
on the angle θ, that is
du
|x2 +y2 +z2 =R2 = h(θ).
dn
We find the solution in the form u = u(r, θ), such that we must write the Laplace’s
equation in polar coordinates
∂
∂r
1 ∂
∂u
∂u
r2
+
sin θ
= 0.
∂r
sin θ ∂θ
∂θ
Now, we can use the method of separation of variables, that is we find the solution
in the form u(r, θ) = R(r)T (θ). The derivatives of the function u become
∂u
∂
= R T ,
= RT ⇒
∂r
∂θ
⇒
∂ 2 1 ∂
r RT +
sin θRT = 0.
∂r
sin θ ∂θ
The Laplace’s equation becomes
2rR T + r 2 R T +
or, equivalently,
1
RT cos θ + RT sin θ = 0,
sin θ
318
9 Elliptic Partial Differential Equations
2r
1
R
R
+ r2
+
R
R
sin θ
T
T cos θ +
sin θ = 0.
T
T
So, we obtain two ordinary differential equations
R
R
+ 2r = k,
R
R
1
T
T cos θ +
sin θ = −k
sin θ T
T
r2
where k is a constant.
These equations can be written in the form
r 2 R + 2rR − kR = 0,
T sin θ + T cos θ + kT sin θ = 0,
and the readers can, easy, find their solutions.
Chapter 10
Optimal Control
10.1 Preparatory Notions
In this section we will introduce some notions and results which are specific to the
functional analysis and are necessary in the whole present chapter. Since we consider
these notions and results being subordinate to the main objectives of this chapter,
we shall renounce to prove the results. Our readers are invited, for more details, to
consult the titles cited in the bibliography dedicated to functional analysis and convex
analysis.
Let us denote by X a real Banach space and by X ∗ its dual, that is the set of all
linear and continuous functionals defined on the space X . By convention, we will
denote by (x ∗ , x) the value of the functional x ∗ in the point x. For an arbitrary fixed
functional x ∗ ∈ X ∗ , we define its seminorm px ∗ by
px ∗ (x) = x ∗ (x) = (x ∗ , x),
such that we get the following family of seminorms:
{ px ∗ }x ∗ ∈X ∗ .
With the aid of this family of seminorms we can introduce a new topology on the
Banach space X , called the weak topology, to make it distinctly of the initial topology
of the space X , called the strong topology. In the weak topology the convergence is
defined by
xn x ⇔ (xn∗ , x) → (x ∗ , x), ∀x ∗ ∈ X ∗ ,
where we have used the notation to designate the weak convergence and → to
designate the strong convergence.
Similarly, for an arbitrarily fixed element x ∈ X , we define the seminorm px by
px (x ∗ ) = x ∗ (x) = (x ∗ , x),
© Springer International Publishing AG, part of Springer Nature 2018
M. Marin and A. Öchsner, Complements of Higher Mathematics,
https://doi.org/10.1007/978-3-319-74684-5_10
319
320
10 Optimal Control
such that we have the family of seminorms { px }x∈X . With the help of this family
of seminorms we can introduce on the Banach space X ∗ a new topology, and, also,
we will called it the weak topology, too. In this weak topology the convergence is
defined by
xn∗ x ∗ ⇔ (xn∗ , x) → (x ∗ , x), ∀x ∈ X,
Let us denote by X ∗∗ the dual space of the space X ∗ . As it is well known, we always
have the inclusion X ⊂ X ∗∗ , since the spaceX can be sunk in X ∗∗ through the
application
x → f x , f x ∈ X ∗∗ , f x (x ∗ ) = x ∗ (x), ∀x ∗ ∈ X ∗ .
If we have the inclusion X ∗∗ ⊂ X , too (and thus, according to the previous statement,
we have X = X ∗∗ ) then we say that X is a reflexive space. A known example of
reflexive space is offered by any Hilbert space. Also, the space L p () for p > 1 is
a reflexive space, but the space L 1 () is not a reflexive space. In all what follows
we will frequently use the notation R̄ to designate the space of all real numbers
completed by ∞, that is
R̄ = R ∪ {∞}.
The function ϕ : X → R̄ is called a convex function if it is subadditive with respect
to any convex combination of elements from the Banach space X , that is
ϕ(λx + (1 − λ)y) ≤ λϕ(x) + (1 − λ)ϕ(y), ∀x, y ∈ X, ∀λ ∈ [0, 1].
We say that the function ϕ : X → R̄ is inferior semi-continuous (and we will
abbreviate it by i. s. c.), if
lim inf ϕ(y) ≥ ϕ(x),
y→x
where, by definition, we have that
lim inf ϕ(y) = sup inf ϕ(x).
y→x
V ∈V(x) y∈V
From this definition we deduce that we always have
ϕ(x) ≥ lim inf ϕ(y)
y→x
and then we can say that the function ϕ is i. s. c. if
lim inf ϕ(y) = ϕ(x).
y→x
10.1 Preparatory Notions
321
An equivalent formulation of the fact that the function ϕ is inferior semicontinuous: the bounded set {x : ϕ ≤ λ} is closed.
We remember that the effective domain of a function ϕ is the set denoted by D(ϕ)
and defined by
D(ϕ) = {x ∈ X : ϕ(x) < ∞} ⊂ X.
Theorem 10.1.1 If X is a Banach space and the function ϕ : X → R̄ is convex and
i. s. c. , then it is inferior bounded by an affine function, that is ∃α ∈ R and x0∗ ∈ X ∗
such that
ϕ(x) ≥ (x0∗ , x) + α, ∀x ∈ X.
Theorem 10.1.2 Let X be a Banach space and the function ϕ : X → R̄ which is a
convex function, i. s. c. and satisfies the condition:
lim ϕ(x) = ∞,
x →∞
where . represents the strong norm of the space X.
Then, the function ϕ actually attains its minimum on the space X .
We remeber that the function ϕ : X → R is a Gateaux differentiable function in the
point x0 if:
• there exists the gradient of ϕ in x0 , denoted by grad ϕ(x0 ) ∈ X ∗ , and:
ϕ(x0 + λx) − ϕ(x0 )
= (grad ϕ(x0 ), x) , ∀x ∈ X.
λ→0
λ
lim
We say that ϕ : X → R is a Frechet differentiable function in the point x0 if
• there exists the gradient of ϕ in x0 , grad ϕ(x0 ) ∈ X ∗ and:
ϕ(x0 + x) − ϕ(x0 ) = (grad ϕ(x0 ), x) + ω(x), ∀x ∈ X,
where the function ω(x) has the property:
lim ω(x) = 0.
x →0
If X is a Banach space and the function ϕ : X → R̄ is a convex and i. s. c. function,
then the subgradient of the function ϕ in the point x0 , denoted by ∂ϕ(x0 ), is defined
by
∂ϕ(x0 ) = x ∗ ∈ X ∗ : ϕ(x0 ) − ϕ(x) ≤ x ∗ , x0 − x , ∀x ∈ X .
322
10 Optimal Control
We say that the function ϕ is subdifferentiable in the point x0 if the subgradient of
the function ϕ computed in x0 is non-empty, that is ∂ϕ(x0 ) = ∅.
From the definition, it is easy to see that the subdifferential of a function is a
multivocal mapping, more exactly, is a multivocal operator.
In the case of a Banach space X , an example of subdifferential is given by application of the duality F of the space X , defined by:
F(x) = x ∗ ∈ X ∗ : x ∗ , x = x
2
X
= x∗
2
X∗
.
It is easy to prove that for the function ϕ : X → R, defined by
ϕ(x) =
1
x
2
2
X,
its subdifferential is just the application of duality, that is ∂ϕ(x0 ) = F(x0 ).
Let X and Y be two Banach spaces and A a multivocal operator, A : X → Y . We
say that A is a monotone operator if:
(y1 − y2 , x1 − 21 ) ≥ 0, ∀x1 , x2 ∈ X, y1 ∈ Ax1 , y2 ∈ Ax2 .
In the particular case when the operator A is univocal, the definition of the monotony
received the following simplified form:
(Ax1 − Ax2 , x1 − 21 ) ≥ 0, ∀x1 , x2 ∈ X.
If the monotone operator A does not admit a proper extension which is also, a
monotone operator, then we say that A is a maximal monotone operator.
An useful example of multivocal operator which is maximal monotone is offered
by the subdifferential of a convex and i. s. c. function. With the help of the subdifferential we can characterize the minimum of a function. More exactly, the function
ϕ : X → R̄ which is convex and i. s. c. . attains its minimum, that is, ∃x0 ∈ X such
that
inf ϕ(x) = ϕ(x0 ),
x∈X
if and only if 0 ∈ ∂ϕ(x0 ), where 0 is the null element of the Banach space X .
To make easy the passing from a problem of minimum to a problem of maximum
for a given function it was introduced the notion of the conjugated function attached
of the respective function. If the function ϕ : X → R̄ is convex and i. s. c. then its
conjugated function, denoted by ϕ∗ , is defined by
ϕ∗ : X ∗ → R, ϕ∗ (x ∗ ) = sup
x ∗ , x − ϕ(x) .
x∈X
The main properties of the conjugated function are:
10.1 Preparatory Notions
323
• 1. ϕ∗ (x ∗ ) + ϕ(x) ≥ (x ∗ , x) , ∀x ∈ X, x ∗ ∈ X ∗ , the equality takes place if and
only if x ∗ ∈ ∂ϕ(x).
• 2. ϕ∗∗ (x) ≤ ϕ(x), ∀x ∈ X, x ∗ ∈ X ∗ , the equality holds if and only if x ∗ ∈ ∂ϕ(x).
• 3. ∂ϕ∗ = (∂ϕ)−1 .
The passing from a problem of minimum to one of maximum and conversely, can
be made based on the following theorem, due to Fenchel.
Theorem 10.1.3 If the functions ϕ, ψ : X → R̄ are convex and i. s. c. and one from
the two following conditions holds:
[I nt D(ϕ)] ∩ D(ψ) = ∅ or [I nt D(ψ)] ∩ I nt D(ϕ) = ∅,
then
inf {ϕ(x) + ψ(x)} = sup −ϕ∗ (x ∗ ) − ψ ∗ (−x ∗ ) .
x∈X
x ∗ ∈X ∗
10.2 Problems of Optimal Control
Suppose that a physical system lies, at a given moment, in the position x0 . From this
position it can continue its evolution on different trajectories. If we impose to the
system some conditions which determine it to displace on such a trajectory which
is more suitable with regard to the proposed purpose, we say that we apply to the
system a optimal command. Therefore, it is possible to interced towards the system
by an external intervention, a command or an optimal control which is imposed to
the system to execute the desired movement. For instance, the system can be ordered
or supervised such that its evolution takes place in an optimal time or by a minimal
consumption of energy.
The mathematical formulation of these problems, in the classical manner, is the
object of the study of the chapter Variational Calculus which has been exposed
already in a previous chapter (Chap. 5).
In the last period of time, in view of the formulation and solution of these problems
has been appeared a new theory, called the Theory of Optimal Control. In all what
follows we intend to expose just the basic notions and results of this theory. Otherwise,
the theory of optimal control is a very abundant and pretentious theory and, therefore,
it is only accessible to a small number of readers.
Essentially, this theory is based on the maximum principle, also called, the Pontreagin’s principle, because this great Russian mathematician is the author of this
principle.
Let us consider a mobil system which can be commanded (or controlled) such
that it evolves according to the desired purpose. Usually, the movement of the system
is mathematical modeled by a system of differential equations of the form:
dxi
= f i (x1 , x2 , ..., xn , u 1 , u 2 , ..., u m ) , i = 1, n,
dt
(10.2.1)
324
10 Optimal Control
in which the quantities xi characterize the position of the mobile and the vector field
x = (x1 , x2 , ..., xn ) is called the state variable. Also, in formula (2.1) u i are the
parameters of the external action which act on the evolution of the system, and the
vector field u = (u 1 , u 2 , ..., u m ) is called variable of command or variable of control.
Let us consider the functional:
T
F (x1 , x2 , ..., xn , u 1 , u 2 , ..., u m ) dt.
(10.2.2)
J=
0
If for t ∈ [0, T ] are given the parameters of control:
u j = u j (t), j = 1, m
(10.2.3)
xi (0) = xi0 , i = 1, n,
(10.2.4)
and the initial conditions:
then the system (10.2.1) possesses only one solution and the integral (10.2.2) has
only one value, which will be the optimal value with respect to the desired purpose.
Suppose that there exists a control u given as in Eq. (10.2.3) which determines
the system, placed in the initial state (10.2.4) to arrive at the final state which we
imposed to it:
xi (T ) = xi1 , i = 1, n.
(10.2.5)
The system will move on a certain trajectory starting from the position (xi0 ) to the
position (xi1 ) and, as a consequence, the functional J from Eq. (10.2.2) takes a certain
value. Corresponding to another imposed command to the system, u, it will evolve
between the two positions on the other trajectory. As a consequence, the functional
J , computed through the new trajectory, will take another value. Thus, it appears the
problem of finding that command (or that control) which determines the evolution
of the system from the initial state to the final state through a trajectory on which the
functional J attains its minimum value.
From a physical point of view, the parameters of control can represent, for instance:
the quantity of fuel delivered to an engine, temperature, the intensity of an electric
current, and so on.
We must specify the fact that in running practical problems the parameters of
control do not take any arbitrary values but these parameters are subjected to certain
restrictions. For instance, in the case of a plane command, if the parameters u 1 and
u 2 are the components of a sub-unit vector field, then the respective command must
carry out the following condition:
u 21 + u 22 ≤ 1.
10.2 Problems of Optimal Control
325
In the general case, it is assumed that (u 1 , u 2 , ..., u m ) ∈ U , where U is a set from a
m−dimensional space and is called the domain of command or the domain of control.
It is possible that, at different moments, the domain of command is different, that is
U = U (t).
Also, in the concrete problems of optimal control there are imposed some restrictions on the state, called restrictions of phase, that is, the state vector field cannot
go out of a certain region of the n−dimensional space R n , that will be written in the
form:
(x1 , x2 , ..., xn ) ∈ G ⊂ R n .
Also, in this case we have G = G(t), t ∈ [0, T ].
More general, it is possible to appear some mixt restrictions which are imposed,
simultaneous, on the command, and, also, on the state:
(x(t), u(t)) ∈ D = D(t) ⊂ R n × R m , t ∈ [0, T ].
If we can write D(t) = G(t) × U (t) we say that the restrictions on the state and on
the command are separately imposed.
A very important particular case is that for what we have
F(x1 , x2 , ..., xn , u 1 , u 2 , ..., u m )= 1,
where F is the Lagrangean of the functional J from the formula (2.2). In this case
we have J = T and then the optimality of the command u(t) reduces to finding
the minimum time in which the mobile system arrives from the position x0 in the
position x1 . We say that we have a control “in optimal time”.
We come back to the system (10.2.1) which can be rewritten in the vectorial form:
dx
= f (x(t), u(t)),
dt
(10.2.6)
in which f (x, ut) is the vector field of the components f 1 (x, u), f 2 (x, u), ...,
f n (x, u). The functions f i depend on the vector field of the state x and on the
vector field of control u. In the form (2.1), or in the form (2.6) of the system, on
the right-hand side, does not appear t, as explicit variable, such that we say that this
system is an autonomous system of differential equations.
If we have a known value for the control u(t), said u 0 (t), then the system (10.2.6)
receives the form:
dx
= f (x(t), u 0 (t)),
dt
which is an usual system of differential equations.
326
10 Optimal Control
Moreover, if we impose the initial condition x(0) = x0 , in the usual conditions of
regularity imposed to the vectorial function f , (see, for instance, Picard’s theorem),
we get an unique solution.
Usually, the functions f i are assumed be continuous with respect to u and continuous differentiable with respect to x. But, in some concrete problems, the control
u = u(t) can have certain points of discontinuity, namely, of first species. If t0 is
such a point, then we will use the notations:
u (t0 − 0) = lim u(t), u (t0 + 0) = lim u(t).
tt0
tt0
For comfortableness, we make the convention that the value of the control in a point
of discontinuity of first species is equal to the value of the limit at left:
u (t0 ) = u (t0 − 0) .
(10.2.7)
For t ∈ [0, T ] we say that u(t) is an accessible control if it satisfies the conditions:
• u is continuous, except a finite number of points;
• u satisfies Eq. (2.7) in the points of discontinuity (of first species);
• u is continuous at the ends of the interval.
10.3 Linear Problems of Control
In the linear case, the evolution of a mobile system can be modeled by the following
Cauchy’s problem:
ẋ(t) = A(t)x(t) + B(t)u(t) + f (t),
x(0) = x0 ,
(10.3.1)
where, for t ∈ [0, T ], A(t) is a matrix whose
components are measurable and
∞
,
a
essential bounded functions,
that
is
A
=
a
ij
i j = ai j (t) ∈ L (0, T ). Also, B(t)
is a matrix, B = bi j , bi j = bi j (t) ∈ C(0, T ). The function f = f (t) is given
and we have f ∈ L 1 (0, T, R n ). The function u = u(t) is the function of control or
command and it is a fixed function and u ∈ L 1 (0, T, R m ). Finally, the function of
the state x = x(t) is the unknown function.
In the above conditions of regularity regarding the coefficients of the system
(10.3.1), the Cauchy’s problem (10.3.1) admits an unique solution which is an absolute continuous function.
In fact, since we have considered the function u(t) as a fixed function, we deduce
that the problem (10.3.1) is just the standard Cauchy’s problem:
ẋ(t) = f (t, x(t))
x(0) = x0
10.3 Linear Problems of Control
327
and it is well known that this problem has a unique solution, if f is a continuous
function with respect to the variables t and x and f is a Lipschitzian function with
regard to the variable x (according to classical theorem due to Picard). On the other
hand, a theorem due to Caratheodory assures that if f is a continuous function with
respect to x and a measurable function with respect to t and if f (t, x) ≤ g(t),
where g ∈ L 1 (0, T ), the above standard Cauchy’s problem has a local solution which
is an absolute continuous function perhaps defined on [0, T ]. In addition, if f is a
Lipschitzian function with respect to x then we obtain the uniqueness of the solution.
We now return back to the problem (10.3.1) and suppose that A(t) is a matrix
of continuous functions. As it is known, in this situation there exists a fundamental
matrix of linear independent solutions, denoted by (t) such that the solution of the
problem (10.3.1) can be represented with the help of (t) in the form:
−1
t
(t)−1 (s) [B(s)u(s) + f (s)] ds, t ∈ [0, T ].
x(t) = (t) (0)x0 +
0
(10.3.2)
It is easy to verify that the function x(t) is an absolute continuous function. We shall
use the notation V (t, s) = (t)−1 (s) and remember that V (t, s) is the matrix of
transition of the differential system from the problem (3.1). It is easy to see that the
solution (10.3.2) can be written in the form:
⎡
x(t) = (t) ⎣−1 (0)x0 +
t
⎤
−1 (s) [B(s)u(s) + f (s)] ds ⎦ , t ∈ [0, T ].
0
In order to deduce this formula, or, its equivalent form (10.3.2), we can use the well
known manner: First, it is attached the homogeneous form of the vectorial equation
d
(t) = A(t)(t),
dt
and then it is attached the initial condition (0) = I , where I is the unit matrix.
A very important problem regarding the differential system (10.3.1) is the problem
of the controllability which, essentially, means the following problem: given the state
x0 , we want to know if we can arrive to another state x1 , which is pre-established,
through one of the trajectories of the system (10.3.1), that is, if exists a control u
which determines the solution of the system to arrive in x1 . If exists such a control,
it is said that the pre-established state x1 is accessible from the state x0 .
Finally, we specify that the stability of the system (10.3.1) means that there exists
a control u for which the system (10.3.1) considered only in the unknown function
x(t), (therefore, the components of the control u(t) are considered as parameters)
is stable in the classical sense of the differential equations and of the systems of
differential equations.
328
10 Optimal Control
Definition 10.3.1 We say that the system (10.3.1) is controllable in the time T if
for any two states x0 and x1 , where x0 , x1 ∈ R n , there exists the control u ∈
L 1 (0, T, R m ) such that the solution of the system (10.3.1), corresponding to this u,
verifies the conditions x(0) = x0 and x(T ) = x1 .
Definition 10.3.2 We say that the system (10.3.1) is null controllable in the time T
if it is controllable starting from the origin, that is we have the definition 3.1 with
x0 = 0.
It is easy to prove that a controllable system is null controllable too. Although it
seems to be surprising, it is a true conversely result.
Proposition 10.3.1 If a system is null controllable, then it is controllable.
Proof Let us consider an arbitrary state x0 and let us prove that there exists a control
u ∈ L 1 (0, T, R m ) which determines the system, which starts from the state x0 , to
arrive, in the time T , in the state x1 where it should arrive if it starts from the origin.
If we take into account the form (10.3.2) of the solution of the system (10.3.1), we
should obtain:
T
−1
x1 = x(T ) = (T ) (0)x0 +
(T )−1 (s) [B(s)u(s) + f (s)] ds,
0
such that we can say that x1 is accessible. But this relation can be written in the form:
T
(T )−1 (s) [B(s)u(s) + f (s)] ds = x1 − (T )−1 (0)x0 .
(10.3.3)
0
But, according to hypothesis that the system is null controllable, that is, starting from
the origin any state is accessible, in particular, also the state
x ∗ = x1 − (T )−1 (0)x0 ,
is accessible, that is the relation (10.3.3) is still valid and, therefore, the state x1 is
attained in the time interval T , starting from the initial state x0 and the proof of the
proposition is closed.
Observation 10.3.1 The controllability of the system (10.3.1) is equivalent with the
following equality of sets:
⎧
⎪
⎨T
⎪
⎩
0
(T )−1 (s)B(s)u(s)ds +
T
0
(T )−1 (s) f (s)ds, u ∈ L 1 (0, T, R m )
⎫
⎪
⎬
⎪
⎭
= Rn .
10.3 Linear Problems of Control
329
Indeed, first let us observe that this equality can be written, equivalently, in the form:
Rn =
⎧ T
⎨
⎩
⎫
⎬
(T )−1 (s)B(s)u(s)ds, u ∈ L 1 (0, T, R m ) ,
⎭
(10.3.4)
0
or
Rn =
⎧ T
⎨
⎩
⎫
⎬
−1 (s)B(s)u(s)ds, u ∈ L 1 (0, T, R m ) .
⎭
0
It is easy to establish that the inclusion {} ⊂ R n is still valid even if the system is not
controllable because in the bracket {} we have elements from R n . Then the inclusion
R n ⊂ {} is assured by the definition of the null controllable system.
Another characterization of the controllability is given in the following theorem.
Theorem 10.3.1 The system (10.3.1) is controllable in the time interval T if and
only if from the equality
−1
B ∗ (t) ∗
(t)x0 = 0, ∀t ∈ [0, T ],
must necessarily result x0 = 0, where by B ∗ and ∗ we have denoted the adjuncts
of the matrix B and .
Proof First, let us remember a result from the theory of the Hilbert spaces: If X is a
Hilbert space of finite dimension then one of its subspace X 0 (which, evidently is of
finite dimension too) coincides with the whole space X if x = 0 is the single vector
which is orthogonal on the subspace X 0 . In view of the proof of our theorem we will
use the characterization (10.3.4) of the controllability. Consider the subspace of R n
given by
⎧ T
⎫
⎨
⎬
X0 =
−1 (s)B(s)u(s)ds, u ∈ L 1 (0, T, R m ) .
⎩
⎭
0
It is obvious that X 0 ⊂ R n . In order to have the equality X 0 = R n , according to
the above result from the theory of the Hilbert spaces, we must demonstrate the
implication
x0 ⊥ X 0 ⇒ x0 = 0.
But the orthogonality relation x0 ⊥ X 0 in our case becomes:
T
0
−1
(s)B(s)u(s), x0 ds = 0, ∀u ∈ L 1 (0, T, R m ),
330
10 Optimal Control
which can be written in the equivalent form:
T −1
u(s)B ∗ (s) ∗
(s), x0 ds = 0, ∀u ∈ L 1 (0, T, R m ).
0
Because this relation is still valid for any u, we deduce that this relation is also still
valid for
−1
(s)x0
u(s) = B ∗ (s) ∗
too, and then we obtain:
T −1
∗ ∗ −1
(s)x0 ds = 0 ⇒ B ∗ (s) ∗
(s)x0 = 0 ⇒ x0 = 0,
B (s) 0
such that the proof of the theorem is closed.
In the particular case when the matrices A(t) and B(t) from the linear system of
control (10.3.1) are constant, i.e. A(t) = A and B(t) = B, a new and easier characterization of the controllability is obtained due to Kalmann.
Theorem 10.3.2 The linear system of control (10.3.1), in which A(t) = A and
B(t) = B, is controllable if and only if
rank B, AB, A2 B, ..., An−1 B = n.
(10.3.5)
Proof From the classical theory of the systems of differential equations it is known
that if the matrix of the coefficients of such a system is constant, then the fundamental
∗
matrix of the solutions (t) is (t) = e At . So, we deduce that ∗ (t) = e A t and
∗
(∗ )−1 (t) = e−A t . Then the condition of controllability from Theorem 10.3.1
becomes
∗
B ∗ e−A t x0 = 0 ⇒ x0 = 0.
∗
If we write the series of powers for the function e−A t , we obtain
∗
e−A t = I −
t n−1 ∗ n−1
t n ∗ n
t ∗ t 2 ∗ 2
A +
A −...+(−1)n−1
A
A +...
+(−1)n
1!
2!
(n − 1)!
(n)!
∗
Then, the equality B ∗ e−A t x0 = 0 ⇒ x0 = 0, ∀t ∈ [0, T ] can be written in the
form:
10.3 Linear Problems of Control
331
⎧ ∗
B x0 = 0
⎪
⎪
⎪ B ∗ A∗ x = 0
⎪
⎪
0
⎪
⎪.
⎪
⎨.
.
⎪ B ∗ (A∗ )n−1 x0 = 0
⎪
⎪
⎪
⎪ B ∗ (A∗ )n x0 = 0
⎪
⎪
⎪
⎩ ..
.
(10.3.6)
In this system only the first n equations are essential because the others can be
obtained from the first by linear combinations. For instance, the matrix (A∗ )n is a
combination of the powers (A∗ )k , in which k ≤ n − 1.
Indeed, based on the known Hamilton–Cayley theorem, any matrix verifies its proper
characteristic equation:
A∗
n
n−1
+ λ1 A ∗
+ ... + λk−1 A∗ + I = 0
and from this equality we deduce:
A∗
n
n−2
n−1
= − λ1 A ∗
+ λ2 A ∗
+ λk−1 A∗ + I .
Therefore, except the first n equations from (10.3.6), the others equations are identically satisfied if the first n are satisfied. So, we take into consideration only the
system of the first n equations from (10.3.6), that is:
⎧ ∗
B x0 = 0
⎪
⎪
⎪
⎨ B ∗ A∗ x = 0
0
..
⎪
.
⎪
⎪
⎩ ∗ ∗ n−1
B (A )
x0 = 0.
(10.3.7)
But in Eq. (10.3.7) we have a linear and homogeneous system and then it admits
only the null solution if and only if the rank of the matrix of the coefficients is n, that
is the number of the unknown functions. With other words, we have
2
n−1 rank B ∗ , B ∗ A∗ , B ∗ A∗ , ..., B ∗ A∗
= n,
which is equivalently to
rank B, AB, A2 B, ..., An−1 B = n
that closes the proof of the theorem.
We return our attention to the system of linear control (10.3.1) and consider it now
in its homogeneous form:
332
10 Optimal Control
ẋ(t) = A(t)x(t) + B(t)u(t),
x(0) = x0 .
(10.3.8)
Suppose that ∀x0 , x1 ∈ R n there exists a control u(t) such that the corresponding
solution of the system (10.3.8) leads x0 in x1 , that is x1 is accessible from x0 . We
denote by xu the solution of the system (10.3.8) corresponding to the control u and
then we have:
xu (0) = x0 , xu (T ) = x1 .
(10.3.9)
We will always suppose that u ∈ where is a convex, closed and bounded set
from L ∞ (0, T, R m ).
The matrix of transition or transference, which has been already defined, is
U (t, s) = (t)−1 (s). We write the solution of the system (10.3.8) corresponding to the control u with the help of the matrix of transition:
T
xu (t, x0 ) = U (t, 0)x0 +
U (t, s)B(s)u(s)ds.
(10.3.10)
0
The set K (T ) = {xu (T, x0 ) : u(t) ∈ , almost everywhere t ∈ [0, T ]} is called the
set of tangibility for the linear system of control.
The essential properties of the set of tangibility are contained in the following
theorem:
Theorem 10.3.3 The set of tangibility K (T ) ⊂ R n is a convex and compact set.
Proof First, we demonstrate the convexity. Let us consider ∀x1 , x2 ∈ K (T ) and
∀λ ∈ [0, 1]. We must show that λx1 +(1−λ)x2 ∈ K (T ). But, because x1 , x2 ∈ K (T ),
we deduce that ∃u 1 , u 2 ∈ such that
T
x1 = U (t, 0)x0 +
U (t, s)B(s)u 1 (s)ds
0
and
T
x2 = U (t, 0)x0 +
U (t, s)B(s)u 2 (s)ds.
0
then
λx1 + (1 − λ)x2 = λU (t, 0)x0 + (1 − λ)U (t, 0)x0 +
T
+λ
T
U (t, s)B(s)u 1 (s)ds + (1 − λ)
0
U (t, s)B(s)u 2 (s)ds =
0
10.3 Linear Problems of Control
333
T
= U (t, 0)x0 +
U (t, s)B(s) [λu 1 (s) + (1 − λ)u 2 (s)] ds.
0
But is a convex set and therefore we have that λu 1 + (1 − λ)u 2 ∈ . So, from
the last equality we deduce that λx1 + (1 − λ)x2 ∈ K (T ). Let us now demonstrate
the fact that the set K (T ) is bounded. For this, we will use the fact that the matrix
of transition is a bounded matrix, U (t, s) ≤ M, ∀t, s ∈ [0, T ], the set is also
bounded too, and, by hypothesis, for the matrix B we have B ∈ L ∞ (0, T ). Then
T
xu (T, x0 ) ≤ U (T, 0)
x0 +
U (T, s)
B(s) u(s) ds.
0
Finally, we will show that the set K (T ) is closed, by using the characterization with
the help of sequences. Let {yn }n∈N be an arbitrary sequence of elements from K (T )
such that yn → y. If we show that y ∈ K (T ), by using the characterization by
sequences of a closed set, we will deduce that the set K (T ) is closed. First, because
yn ∈ K (T ), ∀n ∈ N , we deduce, from the definition of the set K (T ), that there
exists u n ∈ such that
T
yn = U (t, 0)x0 +
U (t, s)B(s)u n (s)ds.
(10.3.11)
0
In functional analysis it is proved the following result:
Any set from the dual of a Banach space is a weak compact set, that is from any
sequence of elements from such a set we can extract a subsequence which is weak
convergent (see Sect. 10.1).
In our case, u n ∈ ⊂ L ∞ and L ∞ is the dual of the Banach space L 1 . On the
other hand, ⊂ L ∞ is bounded and u n ∈ and then, based on the above result
from functional analysis, we deduce that we can extract the subsequence u n k which
is weak convergent, say to u ∈ . Therefore u n k (s) u(s), that is
T
T
u n k (s) f (s)ds 0
u(s) f (s)ds.
0
Corresponding to the subsequence u n k we have the subsequence yn k . Now we write
the relation (10.3.11) corresponding to the subsequences yn k and, respectivelly, u n k
and we obtain that yn k y. But we have assumed that the sequence yn is convergent
to y and, therefore, any of its subsequence is convergent to the same limit, that is
yn → y and the proof of the theorem is closed.
334
10 Optimal Control
In all our previous considerations we have defined and characterized a control. In all
what follows, we shall demonstrate the existence of an optimal control.
We call optimal control that control for what the corresponding solution of the
system (10.3.8) arrives in the state x1 in a minimum time.
Theorem 10.3.4 Consider the state x1 ∈ R n having the property that there exists
T > 0 and an admissible control u such that xu (T, x0 ) = x1 . Then, there exists a
minimum time T ∗ and a corresponding control u ∗ (therefore, an optimal control)
such that xu ∗ (T ∗ , x0 ) = x1 .
Proof For x1 having the property from the hypothesis we have that x1 ∈ K (T ). We
denote by T ∗ the following infimum:
T ∗ = inf {T : x1 ∈ K (T )} .
(10.3.12)
We intend to demonstrate that x1 ∈ K (T ∗ ). From the definition of T ∗ we deduce that
there exists a sequence {Tn }n∈N such that Tn → T ∗ and then evidently x1 ∈ K (Tn ).
We denote by u ∗n the control which corresponds to Tn . We have, therefore
u ∗n (s) =
u n (s), s ∈ [0, Tn ],
0,
s ∈ (Tn , T ].
As a consequence, we can write:
Tn
x1 = U (Tn , 0)x0 +
U (Tn , s)B(s)u ∗n (s)ds.
(10.3.13)
0
By using the above form of u ∗n (s), the relation (10.3.13) receives the form:
Tn
x1 = U (Tn , 0)x0 +
U (Tn , s)B(s)u n (s)ds.
0
The sequence u ∗n (s) is bounded in L ∞ and, therefore, it contains a subsequence
which is weak convergent to an element u from L ∞ . Then, without restriction of the
generality, we suppose that
u ∗n u ∗ in L ∞ (0, T, R n ).
Therefore, we deduce that u ∗ is an admissibil control and u ∗ = 0 for t ∈ [T ∗ , T ].
Passing now to the limit in Eq. (10.3.13), we obtain:
x1 = U (T ∗ , 0)x0 +
T ∗
0
U (T ∗ , s)B(s)u ∗ (s)ds,
10.3 Linear Problems of Control
335
and this shows that x1 ∈ K (T ∗ ) such that the proof of the theorem is closed.
In the following theorem we will give a characterization of the optimal control and of
the corresponding optimal time. This result is known having the name the maximum
principle of Pontreaghin.
Theorem 10.3.5 Consider a problem of the minimum type. Let (x ∗ , u ∗ , T ∗ ) be an
optimal third form, that is the optimal state, the optimal control and the optimal time.
Then, there exists η ∗ and η1 which verify the system:
⎧d ∗
= A(t)x ∗ + B(t)u ∗
⎪
dt (x )
⎪
⎪
∗
⎪
⎨ x (0) = x0
x ∗ (T ∗ ) = x1
⎪
d
⎪
(η ∗ ) = −A∗ (t)η ∗
⎪
⎪
⎩ dt∗ ∗
η (T ) = η1
(10.3.14)
and, in addition, the relation:
B ∗ (t)η ∗ (t) ∈ ∂ I u ∗ (t) , almost everywhere t ∈ [0, T ∗ ].
(10.3.15)
Conversely, if there exists η1 and η which verify the system (10.3.14) and the relation
(10.3.15), then the third form (x ∗ , u ∗ , T ∗ ) is an optimal third form.
Proof First, we must specify that I represents the indicator function of the set which is assumed be a convex and closed set, that is
0, t ∈ ,
I =
+∞, t ∈
/ .
It is easy to show that the indicator function is a convex and s. c. i. function. Also, I
is a subdifferentiable function, therefore it exists ∂ I . According to the definition of
the subdifferential, the relation (10.3.15) can be written, equivalently, in the form:
B ∗ η ∗ , u ∗ − u ≥ 0, ∀u ∈ , almost everywhere t ∈ [0, T ].
The necessity. Suppose that the third form (x ∗ , u ∗ , T ∗ ) is optimal. By accommodating the definition of the set of tangibility, we have
K (T ∗ ) = xu T ∗ , x0 : u ∈ almost everywhere t ∈ [0, T ∗ ] .
First, we demonstrate that
x1 ∈ Fr K (T ∗ ),
(10.3.16)
where Fr K (T ∗ ) is the border of the set K (T ∗ ).
/ FrK (T ∗ ). Then ∀ε > 0 we have that x1 + w ∈
Suppose, by absurdum, that x1 ∈
∗
n
K (T ) for all w ∈ R having the property w < ε. Therefore x1 + w has the form
336
10 Optimal Control
T ∗
∗
x1 + w = U (T , 0)x0 +
U (T ∗ , s)B(s)u(s)ds,
0
where u is an admissible control. We will write x1 in the form
x1 = U (T ∗ − δ, 0)x0 +
T∗ −δ
U (T ∗ − δ, s)B(s)u ∗ (s)ds+
0
T −δ
∗
+
U (T ∗ , s) − U (T ∗ − δ, s) B(s)u ∗ (s)+
0
+ U (T ∗ , 0) − U (T ∗ − δ, 0) x0 +
T ∗
U (T ∗ , s)B(s)u ∗ (s)ds.
T ∗ −δ
Therefore, x1 is of the form x1 = y1 − w, where y1 ∈ K (T ∗ − δ) and w has the
property w < ε from where we deduce that x1 + w = y1 ∈ K (T ∗ − δ) with
w < ε. But this means that x1 ∈ K (T ∗ − δ) which is in the contradiction with
the fact that T ∗ is optimal, that is minimum. This contradiction demonstrates that
x1 ∈ Fr K (T ∗ ). Then, there exists a hyperplane of support for the set K (T ∗ ) in
the point x1 . We denote by η1 the normal to this hyperplane. This normal can be
characterized by:
(η1 , x1 − x) ≥ 0, ∀x ∈ K (T ∗ ) ⇒ η1 , x1 − xu T ∗ , x0 ≥ 0, ∀u ∈ ⇒
⎛
⇒ ⎝η1 ,
T ∗
⎞
U (T ∗ , s)B(s)[u ∗ (s)u(s)]ds ⎠ ≥ 0, ∀u ∈ ⇒
0
T ∗
⇒
∗
u (s) − u(s), B ∗ (s)U ∗ (T ∗ , s)η1 ds ≥ 0, ∀u ∈ ⇒
0
⇒ u ∗ (s) − u(s), B ∗ (s)U ∗ (T ∗ , s)η1 ≥ 0, ∀u ∈ , almost ever ywher e s ∈ [0, T ∗ ].
We use the notation η ∗ = U ∗ (T ∗ , s)η1 and then, by taking into account the definition
of the subdifferential, the last relation can be written in the form:
B ∗ (t)η ∗ (t) ∈ ∂ I u ∗ (t) .
10.3 Linear Problems of Control
337
Since, by definition, we have U (T, s) = (T )−1 (s), we deduce that
−1
(s)∗ (T ∗ ).
U ∗ (T ∗ , s) = ∗
(10.3.17)
Now, we intend to find the fundamental matrix for the equation
d ∗
η = −A∗ η ∗ .
dt
In fact, we will show that ∗ is the fundamental matrix for this equation, knowing
that is the fundamental matrix in the case of the equation having the matrix A.
By using the definition of η ∗ let us show that η ∗ (T ∗ ) = η1 . Taking into account
the form of U , we have:
η ∗ T ∗ = U T ∗ , T ∗ η1 = T ∗ −1 T ∗ η1 = η1 .
With the help of the relation (10.3.17) we will deduce that
d ∗
η = −A∗ η ∗ .
dt
Indeed, we have
d ∗ d ∗ ∗ −1 ∗ ∗ −1
d ∗ ∗ η (s) =
U T , s η1 =
(s) = (s)
(s) (s)
.
ds
ds
ds
But ∗ (s) (∗ (s))−1 = I such that we deduce that
d ∗ ∗ −1
d ∗ −1
(s) (s)
(s)
+ ∗ (s)
= 0.
ds
ds
Now, we use the fact that (∗ )−1 = A ∗ ∗ (s) and therefore we can write
−1 ∗
d ∗ η (s) = − ∗ (s)
A (s)∗ (T ) = −A∗ (s)η ∗ (s),
ds
which is our desired result. The proof of the necessity is closed.
Sufficiency. We can follow the reverse line from the proof of the necessity. However, we relate in detail the fact that η ∗ verifies the equation:
d ∗
η = −A∗ η ∗ .
dt
We have
d ∗ d ∗ −1 ∗ ∗ d ∗ ∗ η (s) =
U T , s η1 =
(s)
(T )η1 =
ds
ds
ds
338
10 Optimal Control
=
d ∗ −1 ∗ ∗
d ∗ −1 ∗ ∗ −1 ∗ ∗ (s)
(s)
(T )η1 =
(s) (s)
(T ) η1 .
ds
ds
Taking into account that (∗ (s))−1 ∗ (T ∗ ) = U ∗ (T ∗ , s), we obtain
d ∗ d ∗ −1 ∗
d ∗ −1 ∗
η (s) =
(s)
(s)
(s)U ∗ (T ∗ , s)η1 =
(s)η ∗ .
ds
ds
ds
But (∗ (s)) = A∗ ∗ and
d ∗ −1 ∗
d ∗ ∗ −1
(s)
(s) (s)
(s) = −
.
ds
ds
In conclusion, we have:
−1 ∗
d ∗ d ∗ ∗ −1 ∗
η (s) = −
(s) (s)
η = −A∗ ∗ ∗ (s)
η = −A∗ η ∗ ,
ds
ds
and the proof of the theorem is closed.
Remarks. 1. As a first conclusion which follows from this theorem is that, now
we can compute, in the conditions of the theorem, both the optimal control and also
optimal time too.
2. The relation (3.15) shows that if u ∗ (t) ∈ I nt then B ∗ (t)η ∗ (t) = 0 and if
∗
B (t)η ∗ (t) = 0, ∀t ∈ [0, T ∗ ) then u ∈ Fr , ∀t ∈ [0, T ∗ ).
Application. We will conclude this paragraph by giving a concrete application.
Consider the system of linear control:
x = x + u
x(0) = x0
and as a domain of control we consider the set = {u : |u| < 1}.
We intend to determine the optimal control which leads 0 in 1. By using Eq.
(10.3.14) we obtain the system
η = −η,
η(T ∗ ) = η1 .
In our case, the relation (10.3.15) becomes: B ∗ η ∗ ∈ ∂ I (u ∗ ) which can be written,
equivalently, in the form:
u ∗ ∈ (∂ I )−1 B ∗ η ∗ .
But (∂ I )−1 = ∂ I∗ . Then we have that I∗ (u) = u and then ∂ I∗ (u) = ∂ u . So,
we obtain:
u ∗ ∈ (∂ I )∗ B ∗ η ∗ = ∂ B ∗ η ∗ = sign B ∗ η ∗ .
In the present case, we have
10.3 Linear Problems of Control
339
u ∗ = sign η ∗ =
η∗
.
|η ∗ |
In conclusion, we must solve the system
⎧ ∗ (x ) = x ∗ + u ∗
⎪
⎪
⎪
⎪
⎨ x ∗ (0) = 0
x ∗ (T ∗ ) = 1
⎪
⎪
(η ∗ ) = −η ∗
⎪
⎪
⎩ ∗ ∗
η (T ) = η1 .
By using the first two relations, we find
x∗ =
t
et−s u ∗ (s)ds.
0
Using the last condition in the last two equations we obtain the solution:
η ∗ (t) = η1 e T
∗
−1
.
First, we suppose that η1 > 0 such that we have η ∗ > 0, therefore u ∗ = η ∗ /|η ∗ | = 1
such that the above solution becomes
t
et−s ds = et − 1.
x ∗ (t) =
0
We impose to this solution to satisfy the condition x ∗ (T ∗ ) = 1 such that we obtain
∗
e T − 1 = 1 and then T ∗ = ln 2. If we suppose that η1 < 0 then and η ∗ < 0 and
then we can follow the above procedure by using u ∗ = −1 such that we will obtain
∗
x ∗ (t) = 1 − et . If we impose the condition x ∗ (T ∗ ) = 1 we are led to 1 − e T = 1,
∗
from where we deduce that e T = 0 which is absurdum. Therefore always we have
η1 > 0.
10.4 Problems of Quadratic Control
Let us consider the system of control
x (t) = A(t)x(t) + B(t)u(t) + f (t)
x(0) = x0 .
We now attach the problem of quadratic minimum:
(10.4.1)
340
10 Optimal Control
⎧ T
⎨
min
⎩
⎫
⎬
[(Q(t)x(t), x(t)) + (N (t)u(t), u(t))] dt, u(t) ∈ , (10.4.2)
⎭
0
where is the set of admissibility which is assumed be a closed, convex and bounded
set. The matrices A and B are quadratic and have the elements from L ∞ , and the
function on the right-hand side f (t) is assumed to be an integrable function. Also,
the quadratic matrices Q and N have the elements from L ∞ and are assumed to be
symmetric and positive definite, therefore
(Qx, x) ≥ w x 2 , ∀x ∈ R n .
A similar relation satisfies also the matrix N too.
We will demonstrate that, based on these hypotheses the problem of minimum
(10.4.2) has a solution and this is even unique.
Theorem 10.4.1 There exists a unique control u and a unique trajectory of the
system (4.1) which verify the problem (10.4.2).
Proof For the proof we will use an already remembered result in the introductory
paragraph:
If the function ϕ : X → R ∪ {∞} is convex and s. c. i. and, in addition, satisfies
the condition
lim ϕ(x) = ∞,
x →∞
then ϕ attains its effective minimum on the Banach space X .
If, in addition, the function ϕ is strict convex, then its minimum is unique. The
function ϕ is strict convex if:
ϕ(λx + (1 − λ)y < λϕ(x) + (1 − λ)ϕ(y), ∀x, y ∈ X, x = y, ∀λ ∈ (0, 1).
For the proof of our theorem we will construct a function which satisfies the above
mentioned conditions and then this will have an unique minimum which will be the
minimum of the problem (10.4.2). We define the function ϕ by:
ϕ(u) =
T
0 [(Qx u , x u ) + (N u, u)] dt, u(t) ∈ , almost everywhere t ∈ [0, T ]
+∞,
elsewhere
(10.4.3)
Here xu represents the solution of the system (10.4.1) corresponding to the control
u, namely
T
xu (t) = (t)−1 (0)x0 +
(t)−1 (s)B(s)u(s)ds.
0
We introduce, as usual, the matrix of transition U (t, s) = (t)−1 (s) and then the
solution xu can be written in the form:
10.4 Problems of Quadratic Control
341
T
xu (t) = U (t, 0)x0 +
U (t, s)B(s)u(s)ds.
0
We take ⊂ L 2 and then we will consider that the function ϕ is defined on the space
L 2 (0, T, R n ), therefore
ϕ : L 2 0, T, R n → R ∪ {∞}.
The fact that ϕ is a convex function can be demonstrated without any difficulty. Let
us show that ϕ is a semi-continuous inferior function. For this we will use the fact
that abounded set of values
of the function ϕ is closed. Consider the bounded set
A = u ∈ L 2 : ϕ(u) ≤ M ⊂ L 2 and let us show that A is a closed set. Let {u n }n∈N
be a sequence of elements from A assumed to be convergent, in the sens of the norm
of the space L 2 , to an element u. Because u n ∈ A we deduce that ϕ(u n ) ≤ M and if
we take into account the definition of the function ϕ, we obtain that u n ∈ almost
everywhere t ∈ [0, T ], therefore:
T
Qxu n , xu n + (N u n , u n ) dt ≤ M.
(10.4.4)
0
Since the set is closed and u n ∈ we deduce that there exists a subsequence
u n k of elements from which is convergent almost everywhere to u 0 ∈ . Then
xu n → xu 0 . Passing now to the limit in Eq. (10.4.4) and we obtain:
T
[(Qxu , xu ) + (N u, u)] dt ≤ M,
0
that is ϕ(u) ≤ M. To close the proof, we must demonstrate that:
lim ϕ(u) = ∞.
u →∞
This is obtained based on the fact that, by hypothesis, the matrices Q and N are
positive defined, and therefore we can write:
ϕ(u) =
0
T
[(Qxu , xu ) + (N u, u)] dt ≥
0
T
w u
2
dt = w u
2
L 2 (0,T,R n )
.
From the last inequality it is clear that ϕ(u) → ∞ if u → ∞. Therefore, the
function ϕ(u) which has been constructed in Eq. (10.4.3) satisfies all the hypotheses
of the theorem mentioned at the beginning of the proof and then the function ϕ attains
its effective minimum and this minimum is the minimum of the problem (10.4.2),
taking into account the mode used for the definition of the function ϕ. This minimum
is effective if we will show that the function ϕ is strict convex. This is an immediate
consequence of the fact that the matrices Q and N , from the definition of the function
342
10 Optimal Control
ϕ, are positive defined and it is known the fact that any quadratic positive defined
form is strict convex.
The final conclusion is that the problem (10.4.2) has a solution and this is unique
and the proof of the theorem is closed.
We now adapt the maximum principle of Pontreaghin to the case of the quadratic
control. In the following theorem, also due to Pontreaghin, we will find the necessary
and sufficient conditions such that the pair (x ∗ , u ∗ ) will be optimal, where x ∗ is the
optimal state and u ∗ is the optimal control.
Theorem 10.4.2 For the problem of minimum (10.4.2) the pair (x ∗ , u ∗ ) is optimal
if and only if there exists a function p(t) which, together with x ∗ and u ∗ , verifies the
system:
⎧ ∗ (x ) = Ax ∗ + Bu ∗ + f
⎪
⎪
⎨ p = −A∗ p + 2Qx ∗
x ∗ (0) = x0
⎪
⎪
⎩
p(T ) = 0,
(10.4.5)
as well as the relation
B ∗ (t) p(t) − 2N (t)u ∗ (t), u ∗ − u ≥ 0, ∀u ∈ .
(10.4.6)
Proof The sufficiency. Suppose that there exists the function p(t) which together
with x ∗ and u ∗ verifies the system (10.4.5) and as well as the relation (10.4.6) and
let us demonstrate that the pair (x ∗ , u ∗ ) is optimal, that is:
T
Qx ∗ , x ∗ + N u ∗ , u ∗ dt ≤
0
T
[(Qx, x) + (N u, u)] dt, ∀u ∈ L 2 (0.T, R n ),
0
(10.4.7)
where it is implied the fact that u ∈ and x is an arbitrary solution of the system
(10.4.1).
In the proof of the sufficiency we will use the following two inequalities:
1 ∗ ∗ 1
Qx , x ≤ (Qx, x) + Qx ∗ , x ∗ − x ,
2
2
(10.4.8)
1 ∗ ∗ 1
N x , x ≤ (N x, x) + N x ∗ , x ∗ − x .
2
2
(10.4.9)
To demonstrate these two inequalities we will use the Schwarz’s inequality:
(Qx, x) ≤
1
[(Qx, x) + (Qy, y)] ,
2
10.4 Problems of Quadratic Control
343
and this, in turn, follows from the inequality:
(Qx, y) ≤
(Qx, x) (Qy, y).
Regarding the inequality (10.4.8) we have:
Qx ∗ , x ∗ = Qx ∗ , x ∗ − x + Qx ∗ , x ≤
1 ∗ ∗ Qx , x + (Qx, x) ,
≤ Qx ∗ , x ∗ − x +
2
from where, by simple calculations, we obtain:
1
1 ∗ ∗ ∗ ∗
Qx , x ≤ Qx , x − x + (Qx, x) .
2
2
Analogous, can be demonstrated the inequality (10.4.9) too. We now write the second
equation from the system (10.4.5) in the form:
2Q(t)x ∗ = p + A∗ p
and multiply both sides of this equality by x ∗ − x. Then, we integrate the resulting
equality on the interval [0, T ]. So, we obtain:
T
2
∗
∗
Qx , x − x dt =
0
T
=
0
p , x ∗ − x + A∗ p, x ∗ − x dt =
p, x ∗ − x dt −
T
p, x ∗ − x dt + p, A x ∗ − x
dt =
0
T
T
=−
0
0
=−
T
T
p, x ∗ − x − p, Ax ∗ − Ax dt = −
0
p, x ∗ − x − Ax ∗ + Ax dt =
p, Ax ∗ + Bu ∗ + f − Ax ∗ + Ax dt = −
0
T
p, B u ∗ − u dt =
0
T
=−
∗
∗
B p, u − u dt ≤ −2
0
T
N u ∗ , u ∗ − u dt.
0
Therefore, we have
2
0
that is:
T
Qx ∗ , x ∗ − x dt + 2
T
0
N u ∗ , u ∗ − u dt ≤ 0,
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10 Optimal Control
T
Qx ∗ , x ∗ − x + N u ∗ , u ∗ − u dt ≤ 0.
(10.4.10)
0
We now add side by side the inequalities (10.4.8) and (10.4.9), then the resulting
inequality is integrated on the interval [0, T ]. So, we obtain:
1
2
≤
1
2
T
T
0
1
Qx ∗ , x ∗ dt +
2
1
2
[(Qx, x)+(N u, u)] dt +
0
T
T
N u ∗ , u ∗ dt ≤
0
Qx ∗ , x ∗ −x + N u ∗ , u ∗ −u dt.
0
(10.4.11)
Finally, from Eqs. (10.4.11) and (10.4.10) we obtain
T
Qx ∗ , x ∗ + N u ∗ , u ∗ dt ≤
0
T
[(Qx, x) + (N u, u)] dt,
0
which is even the inequality (10.4.7) and the sufficiency is demonstrated.
The Necessity. Let us now suppose that the pair (x ∗ , u ∗ ) is optimal and let us
demonstrate that there exists a function p(t) which, together with x ∗ and u ∗ verify
the system (10.4.5) as well as the relation (10.4.6). In order to emphasize that the
pair (x ∗ , u ∗ ) is optimal, we will use the inequality:
T
Qx , x + N u ∗ , u ∗ dt ≤
∗
∗
0
T
[(Qx, x)+(N u, u)] dt, ∀u ∈ L 2 , u ∈ .
0
(10.4.12)
We take the control u and the state x of the form: u = u ∗ + εw and x = x ∗ + εz
such that u ∗ + εw ∈ and x ∗ verifies the system
Then z verifies the system
(x ∗ ) = Ax ∗ + Bu ∗ + f
x ∗ (0) = x0 .
z = Az + Bw
z(0) = 0.
We substitute u and x, which have the expressions above proposed, in (10.4.12) such
that we obtain:
T
0
Qx ∗ , x ∗ + N u ∗ , u ∗ dt ≤
T
0
Qx ∗ +εz, x ∗ +εz + N u ∗ +εw, u ∗ +εw dt.
10.4 Problems of Quadratic Control
345
This inequality can be written in the form:
T
Qx ∗ , x ∗ + N u ∗ , u ∗ dt ≤
0
≤
T
0
!
Qx ∗ , x ∗ + N u ∗ , u ∗ +ε Qx ∗ , z + N u ∗ , w +ε2 [(Qz, z)+(N w, w)] dt,
and this inequality is true ∀ε > 0. From here we deduce:
T
Qx ∗ , z + N u ∗ , w dt ≥ 0,
(10.4.13)
0
and this inequality is true ∀ε > 0 such that u ∗ + εw ∈ .
We will define the function p(t) desired in the enunciation even the solution of
the system
p = −A∗ p + 2Qx ∗
p(T ) = 0.
We write the equation in the form 2Qx ∗ = p + A∗ p, multiply both sides with z,
then we integrate the resulting equality on the interval [0, T ]:
T
2Qx ∗ , z dt =
0
=
T
p + A∗ p, z dt =
0
T
0
( p, z) dt −
T
0
T
p , z + A∗ p, z dt =
0
T
p, z −( p, Az) dt = −
0
p, z − Az dt = −
T
0
( p, Bw) dt
From (10.4.13) we deduce:
T
2Qx ∗ , z + 2N u ∗ , w dt = −
0
T
( p, Bw) − 2N u ∗ , w dt
0
and then we can write:
T
−
B ∗ p, w − 2N u ∗ , w dt ≥ 0.
0
This inequality is equivalently with:
T
0
2N u ∗ − B ∗ p, w dt ≥ 0
346
10 Optimal Control
and this inequality is true for ∀w ∈ L 2 such that u ∗ + εw ∈ . If we will use the
notation v = u ∗ + εw ∈ , then we can write
w=
1
v − u∗ .
ε
Because w has been assumed to be arbitrary, we deduce that v is arbitrary, v ∈ .
Therefore, we have the inequality
T
B ∗ p − 2N u ∗ , u ∗ − v dt ≥ 0, ∀v ∈ .
0
From here we obtain the inequality
B ∗ p − 2N u ∗ , u ∗ − v , ∀v ∈ ,
and this concludes the proof.
At the end of this paragraph, we give a simple application of the Pontreaghin’s
principle in the case of a problem of quadratic control. Consider the system:
x = x + u
x(0) = 1,
(10.4.14)
and, as a set accessibility we will use the set . We take = [0, ∞). Consider the
problem of minimum
min
1
x +u
2
2
"
dt : x ≥ 0, u ∈ .
(10.4.15)
0
According to the maximum principle, in order to determine the optimal pair (x ∗ , u ∗ )
we must solve the system:
⎧ ∗
x = x ∗ + u∗
⎪
⎪
⎨ p = − p + 2x ∗
x ∗ (0) = 1
⎪
⎪
⎩
p(1) = 0,
(10.4.16)
as well as the inequality
p − 2u ∗ , u ∗ − u ≥ 0, ∀u ≥ 0.
We write the inequality (10.4.17) in the form
p ∈ 2u ∗ (t) + ∂ I u ∗ (t) ⇒ p(t) ∈ (2I + ∂ I ) u ∗ (t) ⇒
(10.4.17)
10.4 Problems of Quadratic Control
347
#
∗
⇒ u (t) ∈
1
I + ∂ I
2
$−1 #
$
p(t)
.
2
In all what follows we will show that the expression
#
1
I + ∂ I
2
$−1
p
(10.4.18)
represents the projection of p on the set .
Generally speaking, g(t) is the projection of an operator f on a set if:
| f (t) − g(t)| ≤ | f (t) − ϕ(t)|, ∀ϕ = ϕ(t) ∈ .
In the present case, we must demonstrate that
p − y ≤ p − u , ∀u ∈ ,
where we have denoted by y the projection of p on . Then we have:
p − y ∈ ∂ I (y) ⇔ ( p − y, y − u) ≥ 0, ∀u ∈ .
It is easy to see that this inequality can be written in the form:
( p − y)(y − p) ≥ ( p − y)(y − p) + ( p − y)( p − u), ∀u ∈ .
From here we deduce:
| p − y|2 ≤ ( p − y)( p − u), ∀u ∈ and, after simplification
| p − y| ≤ | p − u|, ∀u ∈ that is, the desired result.
We turn now back to our application. Because Eq. (10.4.18) represents the projection of p on the set , we have:
u ∗ (t) =
0,
p(t) ≤ 0
p(t)/2, p(t) > 0.
The case 1. If p(t) ≤ 0 we have u ∗ (t) = 0 and then x ∗ (t) verifies the system
(x ∗ ) (t) = x ∗ (t)
x ∗ (0) = 1,
348
10 Optimal Control
which evidently has the solution x ∗ (t) = et .
Then the function p(t) verifies the system:
p (t) = − p(t) + 2x ∗ (t)
p(1) = 0,
which has the solution:
p(t) = e
−t
#
$
t
s
C+
2e ds = e−t C + e2t − 1 .
0
By using the condition p(1) = 0 we deduce that C = −e2 + 1 and then the solution
is:
p(t) = e−t e2t − e2 .
The case 2. If p(t) > 0 then u ∗ (t) = p(t)/2. So, the optimal state x ∗ (t) and the
function p(t) satisfy the following system:
⎧ ∗ (x ) (t) = x ∗ (t) + p(t)/2
⎪
⎪
⎨ p (t) = − p(t) + 2x ∗ (t)
x ∗ (0) = 1
⎪
⎪
⎩
p(1) = 0.
We write the characteristic equation of the system:
%
%
1
%
%1 − λ
%
%
2
% 2 −1 − λ % = 0,
√
that is, the equation λ2 − 2 = 0 which has the solutions p = ± 2. Then the solution
of the system is:
%
√
√
% ∗
% x (t) = A1 e√ 2t + A2 e−√ 2t
%
% p(t) = B1 e 2t + B2 e− 2t .
If we impose the initial conditions, x ∗ (0) = 1 and p(1) = 0, we obtain:
A1 = e−2
√
2
√
2
2
√ √
2 + 1 , A2 = 1 − e−2 2
2+1 .
10.5 The Synthesis of the Optimal Control
The problem of the synthesis of the optimal control consists of the determination
of a operator such that the optimal control can be expressed as a function of the
optimal state x(t) by a rule of the form:
10.5 The Synthesis of the Optimal Control
349
u(t) = (t, x(t)).
In this formulation, the optimal control u is expressed as a function of the present
value of the optimal state. There exists also more general formulations in which u
can be expressed as a function of the past values of the optimal state. If we take into
account the relation which is satisfied by the optimal control, namely:
B ∗ p ∈ 2N u ∗ + ∂ I u ∗ ,
then we deduce that in order to express the control as a function of the optimal state,
it is sufficient to express the function p(t) with the help of the optimal state.
We summarily analyse the problem of the synthesis of the control only in the case
of the quadratic control. Suppose that the function p(t) has the form
p(t) = −P(t)x(t) + r (t),
(10.5.1)
in which P(t) is a matrix n × m−dimensional and r (t) is a scalar function. Both
functions, and P(t) and r (t), are unknown and must be determinate.
We can reformulate the problem of the synthesis of the quadratic optimal control
in this way:
Determine the matrix function P(t) and the scalar function r (t) such that the
function p(t) is expressed with the help of P(t) and r (t) by using the rule (10.5.1).
We now remember the system of relations (from the maximum principle) satisfied
by the optimal pair (x, u):
⎧ ⎨ x = Ax + Bu + f
p = −A∗ p + Qx
⎩ ∗
B p ∈ 2N u ∗ + ∂ I (u) .
We substitute in this sistem the expression of p from (10.5.1) such that we are led to
the following system:
− p x − P x + r = −A∗ (−P x + r ) +Qx
x = Ax + B −N −1 B ∗ P x + N −1 B ∗r + f.
(10.5.2)
From here we deduce:
−P − P A + P B N −1 B ∗ P − A∗ P − Q x + r − A∗r − P B N −1 B ∗ r − P f = 0.
We interpret the left-hand side term of this relation as a polynomial with the unknown
variable x. Since the polynomial is null, we deduce that it has only null coefficients.
So, we obtain the following two problems:
p + P A + A∗ P − P D P = −Q
p(T ) = 0
(10.5.3)
350
10 Optimal Control
and, respectivelly:
r = A∗r + P Dr + P f
r (T ) = 0.
(10.5.4)
With the help of first problem we will determine the matrix function P(t), and, with
the help of the second problem we will determine the scalar functionr (t), after we
have determinate the function P(t). We have denoted by D the matrix B N −1 B ∗ .
Conversely, if we make the calculations in the opposite direction, we find that if
the matrix function P(t) satisfies the problem (10.5.3) and the scalar function r (t)
verifies the problem (10.5.4), then the functions P(t) and r (t) verify the relation
(10.5.1) and this is equivalently to the fact that the pair (x(t), u(t)) is optimal.
Therefore, all reduces to show that the problem (10.5.3) has a solution, in the
unknown P(t), since if the function P(t) is determinate, then the problem (10.5.4)
has, without fail, a solution because the equation from the problem (10.5.4) is a linear
equation. It is easy to see that the equation from the problem (10.5.4) is of the Ricatti
type, and the problem is enframed with the general form of a Cauchy problem:
x = f (t, x(t))
x(t0 ) = x0 .
In certain conditions of regularity (for instance, in the conditions of the Picard’s
theorem), the problem (10.5.3) has a unique solution, and, therefore, we obtain the
value of the function P(t) in an interval of the form [T − ε, T ], which can be
prolonged on the whole interval [0, T ].
After we determine the functions P(t) and r (t), the optimal control can be determined with the help of the formula:
u(t) = −N −1 B ∗ P x + N −1 B ∗r,
which is called the synthesis formula of the optimal control.
Remarks 2. The matrix operator P and the scalar function r are depending only
of the system, and do not depend on the chose of the initial conditions.
2. In the case of the quadratic control, the problem of the synthesis reduces to
solve a Cauchy problem attached to a differential equation of the Ricatti type.
Application. Consider the problem of quadratic control:
⎧ ⎨x = x +u +1
x(0) = 1
1
⎩
min 0 x 2 + u 2 dt.
If we take the function p(t) of the form p(t) = −P(t)x(t) + r (t) then we obtain
the following two problems:
10.5 The Synthesis of the Optimal Control
351
p + 2 p − p 2 + 1 = 0, t ∈ [0, 1)
p(1) = 0,
r = r + pr + p, t ∈ [0, 1)
p(1) = 0.
We solve the Ricatti’s equation of the first problem:
dp
= dt ⇒
( p − 1)2 − 2
dp
= t + C,
( p − 1)2 − 2
C1
C2
dp
=
√ +
√ .
2
( p − 1) − 2
p−1− 2
p−1− 2
We obtain:
√
√
1
1
√ ln p − 2 − 1 + √ ln p + 2 − 1 = t + C ⇒
2 2
2 2
⇒
p+
p−
√
√
2−1
2−1
= Ce2
√
2t
.
Now, we add the initial condition p(1) = 0 and then we find p(t).
Bibliography
1. Abd-Alla, A. M., & Mahmoud, S. R. (2008). Elasticity theory and fluid mechanics. Jeddah,
Saudi Arabia: Sci. Algorithm.
2. Abd-Alla, A. M., & Mahmoud, S. R. (2007). Ordinary and partial differential equations.
Jeddah, Saudi Arabia: Sci. Algorithm.
3. Bak, J., & Newman, D. J. (2010). Complex analysis. New York: Springer.
4. Barbu, V. (1993). Boundary value problems for differential equations. Bucharest: Romanian
Academy.
5. Bird, J. (2010). Engineering mathematics. Routledge: Taylor & Francis.
6. Braun, M. (1983). Differential equations and their applications. New York: Springer.
7. Brezis, H. (1983). Analyse fonctioneelle. Paris: Theorie et applications Masson.
8. Berberian, S. K. (1998). Fundamentals of real analysis. New York: Springer.
9. Browder, A. (1996). Mathematical analysis. New York: Springer.
10. Henner, V., Belozerova, T., & Forinash, K. (2009). Mathematical methods in physics. Chapman
& Hall/CRC Press.
11. Henrici, P. (1993). Applied and computational complex analysis. New York: Wiley.
12. Liebeck, M. (2010). A concise introduction to pure mathematics. Chapman & Hall/CRC Press.
13. Marin, M. (1998). Partial differential equations (Edit). Bucharest: Technics.
14. Marin, M. (1997). Cesaro means in thermoelasticity Acta Mechanica (Vol. 122). Wien:
Springer.
15. Marin, M., & Marinescu, C. (1996). Differential and integral equations (Edit). Bucharest:
Technics.
16. Marin, M. (2007). Special mathematics (Edit). University of Brasov.
17. Marin, M., & Arabnia, H. (2004). Problems in algebra and mathematical analysis. Athens,
USA: Elliott & Fitzpatrick Inc.
18. Marin, M., & Arabnia, H. (2010). Equations of evolution. Athens, USA: Elliott & Fitzpatrick
Inc.
19. Marin, M. (2010). Lagrange Identity for microstretch thermoelastic bodies. Journal of Mathematical Analysis and Applications, 363, 1.
20. Meghea, C., & Meghea, I. (1997). Differential and integral calculus (Edit). Bucharest: Technics.
21. Polya, G., & Szego, G. (2009). Problems and theorems in analysis. Berlin, Heidelberg: Springer.
22. Pugh Chapman, C. (2002). Real mathematical analysis. Boston: Birkhauser.
23. Swaminathan, A. (will appeared 2012). Complex analysis and special functions. Chapman &
Hall/CRC Press.
24. Verhulst, F. (1990). Differential equations and dynamical systems. Berlin: Springer.
25. Walter, M. (2011). Mathematic for the environment. Chapman & Hall/CRC Press.
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