V S Vladimirov (auth ), V S Vladimirov (eds ) - A Collection of Problems on the Equations of Mathematical Physics (1986, Springer) [10 1007_978-3-662-05558-8] - libgen li

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A Collection of Problems 
on the Equations 
of Mathematical Physics 
Edited by V.S.Vladimirov 
Translated from the Russian by Eugene Yankovsky 
Springer-Verlag Berlin Heidelberg GmbH 
Edited by Academician V. S. Vladimirov 
Contributors: V. S. Vladimirov, V. P. Mikhailov, 
A. A. Vasharin, Kh. Kh. Karimova, Yu. V. Sidorov, 
M. 1. Shabunin 
First published 1986 
Revised from the 1982 Russian edition 
With 4 Figures 
ISBN 978-3-662-05560-1 ISBN 978-3-662-05558-8 (eBook) 
DOI 10.1007/978-3-662-05558-8 
© I13J\aTeJIbCTBO «HaYKa». fJIaBHaH peJ\aKI\HH tPH3HIw-MaTeMaTH'leCKoü: 
JIHTepaTypbI, 1982 
© Springer-Verlag Berlin Heidelberg 1986 
Originally published by English translation, Mir Publishers, in 1986. 
Softcover reprint ofthe hardcover 1st edition 1986 
Preface 
The extensive application of modern mathematical teehniques to 
theoretical and mathematical physics requires a fresh approach to 
the course of equations of mathematical physics. This is especially 
true with regards to such a fundamental concept as the 80lution of 
a boundary value problem. The concept of a generalized solution 
considerably broadens the field of problems and enables solving 
from a unified position the most interesting problems that cannot be 
solved by applying elassical methods. To this end two new courses 
have been written at the Department of Higher Mathematics at the 
Moscow Physics anrl Technology Institute, namely, "Equations of 
Mathematical Physics" by V.S. Vladimirov and "Partial Differential 
Equations" by V.P. Mikhailov (both books have been translated 
into English by Mir Publishers, the first in 1984 and the second 
in 1978). 
The present collection of problems is based on these courses and 
amplifies them considerably. Besides the classical boundary value 
problems, we have ineluded a large number of boundary value 
problems that have only generalized solutions. Solution of these 
requires using the methods and results of various branches of modern 
analysis. For this reason we have ineluded problems in Lebesgue in-
tegration, problems involving function spaces (especially spaces of 
generalized differentiable functions) and generalized functions (with 
Fourier and Laplace transforms), and integral equations. 
The book is aimed at undergraduate and graduate students in the 
physical sciences, engineering, and applied mathematics who have 
taken the typical "methods" course that includes vector analysis, 
elementary complex variables, and an introduction to Fourier series 
and boundary value problems. Asterisks denote the more difficult 
problems. 
6 Preface 
We would like to express our gratitude to all who helped with 
constructive comment to improve this book, our colleagues at the 
Department of Higher Mathematics at the Moscow Physics and 
Technology Institute, and especially T.F. Volkov, Yu.N. Drozhzhi-
nov, A.F. Nikiforov, and V.1. Chekhlov. 
V.S. Vladimirov 
V.P. Mikhailov 
A.A. Vasharin 
Kh.Kh. Karimova 
Yu. V. Sidorov 
M.I. Shabunin 
Contents 
Prefaee 5 
Symbols and Definitions 9 
Chapter I Statement of Boundary Value Problems in Mathe-
matieal Physies 12 
1 Deriving Equations of Mathematieal Physies 12 
2 Classifieation of Seeond-order Equations 35 
Chapter I I Funetion Spaees and Integral Equations 41 
3 Measurable Funetions. The Lebesgue Integral 41 
4 Funetion Spaees 48 
5 Integral Equations 67 
Chapter I II Generalized Funetions 88 
6 Test and Generalized Funetions 88 
7 Differentiation of Generalized Funetions 95 
8 The Direet Produet and Convolution of Generalized 
Funetions 104 
9 The Fourier Transform of Generalized Funetions of Slow 
Growth 114 
10 The Laplaee Transform of Generalized Funetions 121 
11 Fundamental Solutions of Linear DifferentialOperators 125 
Chapter IV The Cauehy Problem 134 
12 The Cauehy Problem for Seeond-order Equations of Hyper-
bolie Type 134 
13 The Cauehy Problem for the Heat Conduction Equation 157 
14 The Cauchy Problem for Other Equations and Goursat 's 
Problem 167 
Chapter V Boundary Value Problems for Equations of Elliptie 
Ty~ W 
15 The Sturm-Liouville Problem 181 
16 Fourier's Method for Laplaee's and Poisson's Equations 190 
17 Green 's Funetions of the Diriehlet Problem 205 
18 The Method of Potentials 211 
19 Variation al Methods 230 
Chapter VI Mixed Problems 239 
20 Fourier's Method 239 
21 Other Methods 269 
8 Contents 
Appendix Examples of Solution Techniques for Some Typical 
Problems 277 
At Method of Characteristics 277 
A2 Fourier's Method 279 
A3 Integral Equations with aDegenerate Kernel 281 
A4 Variational Problems 283 
References 284 
Subject Index 287 
Symbols and Definitions 
1. We denote areal n-dimensional Euclidean space by Rn and its 
points by x = (Xl' x 2, . .. , xn),y = (YI' Y2" .. , Yn), ~ and the like. 
2. dx = dXI dX2 ... dxn, 
~ f (x) dx = ) f (Xi' ... , x n) dX1 ••• dxn· 
Rn 
3. a = (al> a 2 , ••• , an) is a multi-index, with the aj nonnegative 
integers. We will also use the abbreviations 
4. (x, y) = X1Yl + X2Y2 + ... + XnYn 
7= lxi = V(X, x)=Vx~+x;+ •.. +x~. 
5. U (xo; R) = {x: ! x - Xo I < R} is an open ball of radius R 
centered at point xo, and S (xo; R) = {x: I x - Xo I = R} is a sphere 
of radius R centered at Xo; UR = U (0, R) and SR = S (0; R). 
6. A set A will be said to be lying strictly in a region GeRn 
(this is denoted by A ~ G) if it is bounded and A c G. 
7. A function f (x) is said to be locally integrable in a region G-
if it is absolutely integrable in every subregion G' ~ G. Functions 
that are locally integrable in Rn will be said to be simply locally 
integrable. 
8. Daf (x) = 81al / (Xl, X2' ••• , Xn) 
8x~1 ax~2 ..• ax~n 
9. CP (G) is a class of functions t that are continuous together 
with the derivatives Dat, I a I ::s; p (0 ::s; p < 00) in the region 
GeRn. The functions f of class CP (G) for which all the derivatives 
Dat, I a I ::s; p, allow continuous continuation into the closure G 
form the class CP (G); C (G) = Co (G), C (G) = Co (G). We denote 
10 Symbols and Definitions 
the class of functions belonging to CP (G) for all p's by Coo (G); the 
class Coo (G) is defined in a similar manner. 
10. The uniform convergence of a sequence of functions {/k} to 
a function f on a set A is denoted by 
xEA 
fk (x) => f (x), k -+ 00 
11. A U B is the union of sets A and B, A n B is the interseetion 
of A and B, A ""B is the complement of B with respect to A, and 
A X B is the direct product (or simply product) of A and B (the set 
of pairs (a, b) with a E A and bEB). 
12. The support of a continuous function f is denoted by supp f 
and is the closure of the set of all points x for which f (x) is nonzero. 
If a function f (x) that is measurable on a region G vanishes almost 
everywhere in G" GI, where GI ~ G, then it is finite in G; a function 
that is finite in Rn is said to be simply finite. 
a2 a2 82 • , 
13. VZ = "2 + "2 + ... + --;;-"'2 IS Laplace s operator; 
vXl uX2 uXn 
a2 o = at2 - a2y2 is the wave operator; [Ji = 0; 
:t - a2'\72 is the heat conduction operator. 
14. r+ = {x, t: at> lxi} is a future cone. 
~ 
15. <l> m = :2,,; ~ e- z2/ 2 dz 
-00 
. { Cee-e2/(e2 -lxI2) lxi ~e, 
1(i. (t)e(x)= 0, Ixl>e, 
1 
where Cp,=e-nx ! = ~ e- 1/(1-x2) dx, 
o 
«(t) e is the averaging kernei, or "cap"). 
17. C is the complex plane. 
18. e (x) is the Heaviside unit function: 
{ 1 if x~O, e (x) = 0 if x< O. 
19. (J ds - is the surface area of the unit sphere ~ 2,,;n/2 n = - r (n/2) 
SI 
S1 in Rn. 
20. In CP (G) the norm is 
11 f IlcP(G) = ~ max IDa.f (x) I 
1a.I~p xeG 
Symbols and Definitions 11 
21. The totality of (measurable) functions 1 (x) for which IfIP 
is integrable on G is denoted by L p (G). In L p (G) the norm is 
11I IILp(G) = [I I/IPdx T'p" 1~p< 00. 
G 
11 1 liLoo(G) == vrai sup 1I (x) I, p = 00 
xEG 
The scalar product in L 2 (G) is introduced thus: 
(f,g)=\/gdx,l,gELt,(G). 
J 
G 
22. Let p (x) be a continuous positive-valued function in a region 
G. The totality of (measurable) functions 1 (x) for which p (x) I 1(x) 12 
is integrable on G is denoted by L 2 ,p (G) and constitutes a Hilbert 
space with the scalar product 
(I, g)L (G) = r pli dx. 
2t p J 
23. Cylinder functions. 
(a) Bessel functions: 
00 
G 
" (-1)k ( x )2k+ V 
J..,(X)=L,J f(k+v+1)f(k+1) 2" ' 
h.=O 
-oo<x<oo; 
(b) Neumann 's Bessel functions: 
1 
N..,(x) = " [J..,(x)cosnv-J_..,(x)], v=!=n, 
Sin :n;v 
N ( ) _.!. [ ßJ.., (x) _ (_1)n ßL.., (x) _. 
11 X - n ßv ßv ,v - n, 
(c) Hankel functions: 
H~ll (x) = J.., (x) + iN.., (x), H~2> (x) = J.., (X) - iN y (x); 
(d) modified BesseI and HankeI functions: 
1t " 1t 
I ( -T..,tJ ( K () ni T ViH<1l (" ) .., x) = e .., ix), y x =""2 e v zx. 
Chapter I 
Statement of Boundary Value 
Problems in Mathematical Physics 
1 Deriving Equations of Mathematical Physics 
We start by introducing the following notation: 
p (x) = P is the density of a material (linear, surface, or volume); 
T 0 is the tensile force acting on astring or membrane; 
E is Young's modulus; 
k is the coefficient of elasticity of astring or rod with elastically 
fixed ends or of a membrane; 
S is the cross-sectional area of a rod, shaft, and the like; 
l' = cp/cv is the specific heat ratio; 
p and Po are press ures in a fluid; 
m and mo denote mass; 
g is the gravitational acceleration near the earth 's surface; 
u) is angular velocity; 
k, k (x) and k (x, u) are internal thermal conductivities; 
IX is the extern al thermal conductivity (the heat-exchange coef-
ficient); 
D is the diffusion coefficient. 
Here are some examples dealing with deriving equations. 
Example 1: Transverse vibrations 0/ strings. Astring of length l 
with a tensile force T 0 is in the horizontal state of equilibrium. At 
time t = 0 the string is displaced with an initial velocity. We 
wish to state the problem of determining the small transverse vib-
rations of the string at t > 0 if the ends of the string are (a) rigidly 
fixed, (h) free (Le. can freely move along straight lines parallel to 
the direction of displacement u), (c) fixed elastically (Le. each end 
is under a resistance force from the fixture, and t11is force is pro-
portional to the displacement and is directed opposite Lo the dis-
placement), or (d) move in the transverse direction according to 
specified laws. We ignore the force of gravity and the resistance of 
the medium in which the string moves. 
Solution. We send the x axis along the string in t11e equilibrium 
position. We assume that the string is a thin cord that does not 
res ist bending not related to changes in length. This means that if 
we mentally cut the string at a point x, the action of one portion of 
the string on the other (or simply the tensile force T) is directed 
along the tangent to the string at that point. To derive the equation 
Deriving Equations of Mathematical Physics 13 
of motion (vibrations) of the string we examine the portion of the 
string between x and x + ~x and project all the forces acting on this 
portion (including the forces of inertia) on the axes of coordinates. 
According to D'Alembert's principle, the sum of the projections of 
all forces on each axis must be zero. Since we are studying only 
transverse vibrations, we can assurne that the external forces and 
the force of inertia act along the U axis. We will consider only small 
u 
T(x+.1x) 
x+.1x x 
Fig.1 
vibrations, which means that in deriving our equation we will 
neglect squares of U x (x, t). The length S of the arc AB is expressed 
x+ Ale 
by the integral S =) (1 + U~)1/2 dx ~ ~x. This means that 
le 
no portion of the string changes its length during vibrations and, 
hence, by Hooke 's law the magnitude of the tensile force T 0 = I T I 
depends neither on time nor on x. Let us find the projections of all 
the forces at time t onto the U axis. Up to first-order infinitesimals. 
the projection of the tensile force is (see Fig. 1) 
To [sin (X (x + ~x) - sin (X (x)] 
tan Gt (x) ] 
y1+tans Gt (x) 
Ux (x. t) ] 
Y1+ui, (x, t) 
~ Toulele (x, t) ~x. 
Let p (x, t) be the continuous linear density of the external forces 
acting on the string. Then the force that acts on the portion AB along 
the U axis is p (x, t) ~x. To find the force of inertia acting on AB, 
we recall the expression -mutt, with m the mass of that portion of 
the string. If p (x) is the continuous linear density of the material 
of the string, then m = p ~x. Thus the projection of the force of 
inertia on the U axis is -pUtt ~x, and the projection of all the forces 
on the same axis obeys the equation 
rToun + P (x. t) - p (x) Utt1 ~x = O. (1.1) 
Hence. 
Toulex - r (x) Utt + P (x, t) = O. 
14 Statement of Boundary Value Problems 
which is the equation 01 lorced vibrations 01 astring. If p is constant 
all along the string, the equation takes the form 
Utt = a2uxx + g (x, t), 
wherea 2 = To/p and g (x, t) = p (x, t)/p. Thefunction u (x, t) satis-
fies the initial conditions u It=o = er (x) and Ut It=o = 1p (x), where 
er (x) and 1p (x) are given functions. We now turn to deriving the 
boundary conditions: 
(a) If ends of the string are rigidly fixed, U Ix=o = ° and U Ix=1 = 
= 0. 
(b) If the ends are free, then to obtain the boundary condition at 
x = ° we project the fore es that act on the portion KM onto the u 
u 
T(OJ K 
o x 
Fig. 2 
axis (Fig. 2). Since the tensile force at point x = ° can act only 
parallel to the x axis, the projer-tion of this force on KM is T OU x (~x, t). 
The projection of the external force on this axis is p (0, t) ~x, while 
the projection of the force of inertia is -pUtt (0, t) ~x. Nullifying 
the sum of these projections yields the following equation: 
T OUx (~x, t) + p (0, t) ~x - pUtt (0, t) ~x = O. (1.2) 
Let us send ~x to zero. In view of the continuity and boundedness of 
the functions in this equation we arrive at the bOllndary condition 
U x 1'tC:o = O. Similarly, Ux Ix=1 = 0. 
(c) We take the case where the left fixture acts on the left end 
of the string with an elastic force -ku (0, t). We nullify the sum of 
the projections of all the fore es acting on the portion KM onto the 
U axis. The left-hand side of Eq. (1.2) has an additional term equal 
to -ku (0, t). The resulting equation is 
T oUx (~x, t) - ku (0, t) + p (0, t) ~x - pUtt (0, t) ~x = 0. 
which transforms into the following equation as ~x -+ 0: 
(ux - hu)lx=o = 0, h = k/To 
At the right end of the string (see Fig. 3) the sum of the projections 
of all the forces must obey the equation 
-Toux (l- ~x, t) - ku (l, t) + p (l, t) ~x - pUu (l, t) ~x = 0, 
since sin a (l - ~x) ~ U x Ix=I-ßx' If we take the limit as ~x -+ 0, 
we obtain (ux + hu) Ix=1 = 0. 
Deriving Equations of Malhematical Physics 15 
(d) u Ix=o = fLl (t) and u Ix=! = fL2 (t), where the functions 
fLl (t) and fL2 (t) determine the law by which the two ends of the 
string move (fLl (0) = <p (0) and fL2 (0) = <p (l». 
Example 2: Vibrations 0/ a rod. An elastic straight rod of length 1 
is taken out of the state of rest by imparting sm all longitudinal 
u 
T( [-/lx) 
--Q--.... TU) 
o I-Ax x 
Fig.3 
displacements and velocities to its cross sections at time t = O. 
Assuming that during the motion all cross sections remain parallel 
to a plane that is normal to the rod 's axis, we wish to state the 
problem of determining the small longitudinal vibrations of the 
rod at t> O. The ends of the rod may (a) be fixed rigidly, (b) move 
p p, 
--oF Yx YX+/lX I, ,.. x 
q q, 
Fig.4 
in the longitudinal directions according to specified laws, (c) move 
freely, or (d) be fixed elastically (Le. each end is under a longitudinal 
force from the fixture, and this force is proportional to the displace-
ment and is directed opposite to the displacement). 
Solution. We send the x axis along the axis of the rod (see Fig. 4), 
with x the coordinate of the cross section pq in the state of rest. We 
are interested only in small longitudinal vibrations, which means 
that all external forces and the forces of inertia are directed along 
the x axis. If u (x, t) denotes thedisplacement of this cross section 
at time t, then (und er our assumptions) the displacement of the 
section at point x + dx will be 
u (x+ dx. t)~ u (x, t) + U x (x. t) dx. 
For this reason the elongation (per unit length of the rod) at section x 
will be U x (x, t). Hooke 's law then states that the tensile force in 
this cross section is T = ESux (x, t), with S the cross-sectional area 
and E the elastic modulus of the material of the rod. We can obtain 
the vibration equation if we nullify the sum of all forces acting on the 
portion from section pq to section Plql, including the forces of inertia. 
16 Statement of Boundary Value Problems 
The sum of the tensile forces is 
T (x + Ax) - T (x) = ES [ux (x + Ax, t) - Ux (x, t)] 
~ ESuxx (x, t) Ax. 
Let p (x, t) be the volume density of the external forces acting on the 
rod. There are two other forces acting on the chosen section: the 
extern al force Sp (x, t) Ax and the force of inertia -p(x)SUtt X 
X (x, t) Ax. The sum of all three forces must be zero, according to 
D'Alembert's principle, that is, 
[ESuxx (x, t) + p (x, t) S - P (x) SUtt (x, t)] Ax = O. (1.3) 
Whence 
p (x) Utt (x, t) = Euxx (x, t) + P (x, t); (1.4) 
in addition, u (x, t) satisfies the initial conditions u It=o = cp (x) 
and Ut 1 t=o = 'IjJ (x), with cp (x) and'IjJ (x) given functions. If p (x) = 
= pis constant (i.e. the rod is homogeneous), the equation takes the 
form 
where 
a2 = Elp, g (x, t) = P (x, t)/p (1.5) 
Now we must derive the boundary conditions: 
(a) In the case of rigidly fixed ends we can write ulx=o = 
= u Ix=l = O. 
(b) u Ix=o = !l1 (t) and u Ix=1 = !l2 (t), where !l1 (t) and !l2 (t) 
are the functions that define the law of the motion of the rod 's ends 
(!lI (0) = cp (0) and !l2 (0) = cp (1». 
(c) When both ends are free, we must construct the balance equation 
for the forces acting at the ends. At the left end the resultant of the 
elastic tensile forces is T (Ax) = ESux (Ax, t), the external force is 
Sp (0, t) Ax, and the force of inertia is -pSUtt (0, t) Ax. The sum of 
all forces acting on the chosen portion must be zero, that is, 
ESux (Ax, t) + P (0, t) S Ax - pSUtt (0, t) Ax = 0, (1.6) 
and the limit as Ax -+ 0 is U x Ix=o = O. Similarly, at the right end 
U x Ix=l = O. 
(d) To the 1eft-hand side of Eq. (1.6) we must add the force 
-ku (0, t). If we now take the limit as Ax -+ 0, we obtain 
ESux (0, t) - ku (0, t) = 0, or (ux - hu) Ix=o = 0, 
where h = kiES. At the right end we have 
-T (l - Ax) = -ESux (l - Ax, t), 
the extern al force Sp (l, t) Ax, and the force of inertia 
-p (x) SUtt (l, t) Ax. The resultant must be zero, or 
-ESux (l - Ax, t) - ku (1, t) + Sp (l, t) Ax - Utt (l, t) Sp (x) Ax 
= 0, 
Deriving Equations of Mathematical Physics 17 
and the limit as L1x -+ ° is (ux + hu) Ix=l = 0, which constitutes 
the second boundary condition. 
Example 3: Vibrations oj a membrane. A membrane is a stretched 
film that resists stretching but does not resist twisting. The work 
produced by an external force that is applied to the membrane and 
changes the surface area of aselected portion is proportional to the 
change in area. The positive proportionality factor T depends neither 
on the shape of the portion nor on its position and is called a tensile 
force. 
Let us derive the equation that determines the equilibrium of the 
membrane, assuming that at time t = ° the membrane coincided 
with a region G of the Xl' X2 plane bounded by a sufficiently smooth 
curve L. The work performed by internal elastic fore es is equal in 
magnitude to the work of the external forees on the membrane but 
has an opposite sign. Let f (x) be the density of external fore es at 
point X acting perpendicularly to the Xl' x 2 plane. As a result the 
membrane occupies a new position given by the equation u = u (x). 
We will assurne that the membrane is not bent too strongly, so that 
we can ignore u~, and u~. when deriving the equation. An additional 
assumption is that under external forces the points of the membrane 
move along norm als to the Xl' x 2 plane, which means that the coor-
dinates (Xl' x 2 ) of an arbitrary point of the membrane do not change. 
The work performed by an external force that takes the membrane 
out of its initial position (u = 0, X E G) into a position given by 
the equation u = u (x), X E G, is equal to the integral J' f (x) u (x) dx. 
G 
Under this displacement the area of the membrane changes by 
) ev 1+ ui, + ui. - 1) dx, while the work performed by the 
G 
internal elastic fore es is 
-T ~ [V1+ui1+ui2-1jdx ~ -; ~ (ui1+ui2)dx. 
G G 
Hence, the sum of all works is 
A (u) = i [ - ; (ui1 + ui2) + fuJ dx. 
G 
The variation of functional (1. 7) is given by the formula 
öA (u) = ~ [- T (U:r:lÖU:r:, + U:r:2ÖUX2) + jöu] dx. 
G 
( 1.7) 
According to the virtual work principle, in the state of equilibrium 
we have (u = U (x» öA (u) = ° for all admissible Öu (x). Sinee 
) (UX1ÖUX1 + U:r:2ÖUX2) dx = ~ ~: Öu dl- ) V 2uöu dx, 
G L G 
2-0176 
18 Statement of Boundary Value Problems 
where n is the outward unit normal vector to contour L, we have 
öA(u)= -T ~ ~: öudl+ ~ (TV2U+/)öudx=0. (1.8) 
L G 
Since every function that is continuously differentiable in G and 
vanishes at the boundary is admissible and the functions u (x) and 
1 (x) can be assumed to be sufficiently smooth, Eq. (1.8) yields 
TV 2u = -I (x), x E G. (1.9) 
N ow we go over to the boundary conditions: 
(a) A fixed membrane. If the edge of the membrane is rigidly fixed, 
the boundary L does not experience any deflections and u IL = O. 
(b) If the membrane's edge is free, it can move along the ver-
tical lateral surface of a cylinder with base L. In this case öu can be 
arbitrary both on G and on L, and condition (1.8) yields (au/on) IL = 
= O. 
(c) If a force with a linear density 11 is applied to the edge of the 
membrane, the line integral in (1.8) must be replaced by 
J [-T (au/on) + 11] öu dl. Since öu is arbitrary on L, we obtain 
L 
[-T (au/on) + 11] IL = O. 
(d) If the edge of the membrane is elastically fixed, the force that 
acts on the edge has a density -ku, where k characterizes the rigidity 
with which the edge is fixed. To obtain the required boundary con-
dition, we must substitute -ku for 11 in the boundary condition 
[-T (au/on) + 11] IL = O. We then have 
(:: +hu) IL =0, where h=k/T 
Now we wish to derive the equation that describes the vibrations 
of the membrane. Let u = u (x, t) be the equation that describes 
the position of the membrane at time t. According to D'Alembert's 
principle, the function u (x, t) satisfies the equation T V 2u = 
= -(I - pUtt), with 1 = 1 (x, t) the density of the external force 
and -p (x) Utt the density of the force of inertia. Thus, the equation 
of vibrations of the membrane is 
a2V 2u-utt =F(x, t), where a2=T/p and F=-/(x, t)/p. (1.10) 
From the physics of the problem it is clear that to give an unam-
biguous description of the vibration process we mnst, besides using 
Eq. (1.10) and one of the boundary conditions (a)-(d), specify the 
initial position of the membrane (i.e. its shape at t = 0) and the 
initial velocities of its points. 
We have thus arrived at the following problem involving 
Eq. (1.10): to find a twice differentiable solution u (x, t), x E G, t> 0, 
that is continuously differentiable in G at t > 0 and satisfies the 
following boundary value problem: 
a2V2u-utt=F(x, t), ult=o=cp(x), utlt=o='IjJ(x), 
Deriving Equations of Mathematical Physics 19 
where er (x) and 'Ijl (x) are given functions. In addition, depending 
on the mode in which the edge of the membrane is fixed, we must 
select one of the boundary conditions (a)-(d). 
Example 4: The continuity equation, the flow problem, and the 
equations 0/ acoustics. We start by examining the motion of an ideal 
fluid, that is, a fluid in which viscosity does not manifest itself. 
(The motion of the fluid is studied in Eulerian coordinates). Suppose 
v = (VI' V 2 , V3 ) is the velocity of the fluid, p(x, t) itsdensity, and 
/ (x, t) the source strength. We isolate a volume Q of the fluid bounded 
by a surface S. Then the rate at which the mass of fluid inside Q 
changes is 
:t J pdx= J ~~ dx. 
Q Q 
On the other hand, the rate must be equal to the amount QI of fluid 
produced by the sources inside Q minus the amount Q2 flowing out 
of Q through S. Obviously, 
Q1= J t(x, t)dx, Qz= J p(v, n)ds= J div(pv)dx, 
g S Q 
where n is the outward unit normal vector. Thus, 
J [pt+div(pv)-f] dx=O. 
Q 
Since Q is taken arbitrary and the integrand is continuous, we can 
write 
Pt + div (pv) = / (x, t). (1.11) 
This equation is known as the equation 0/ continuity of an ideal 
fluid. 
N ow let us turn to the problem of studying the flow of a homo-
geneous incompressible liquid around asolid Q with a boundary S. 
The liquid has a given velocity Vo at infinity and no sources inside 
it. Since p = const and / = 0, the problem results in solving the 
equation 
div v = ° (1.12) 
with the condition that 
vnls = 0, (1.13) 
where Vn = (v, n) and n is the outward unit normal vector. Let u 
be the velo city potential, that is, v = grad u. Then Eq. (1.12) takes 
the form div grad u = V 2u = ° and the boundary condition becomes 
(au/an) Is = 0, since 
8u 
vn = (v, n) = (grad u, n) = an" 
The physics of the problem implies that v (x) must tend to Vo as 
I x 1-+ 00, where Vo is the flow velocity at infinity. 
2* 
20 Statement of Boundary Value Problems 
Thus the problem has reduced itself to the following boundary 
value problem: 
V2u=O, x Et"ri, 
~u I = 0, !im grad u = vo. 
vß 8 Ixl-"" 
Finally, we derive the equations 0/ acoustics. Suppose that a portion 
of ideal gas within a certain volume performes small vibrations 
about the position of equilibrium due to certain external forces of 
a density F (x, t) and that these vibrations are adiabatic, that is, 
the pressure P (x, t) of the gas and its density P (x, t) are related by 
the following formula (the equation of state): 
L=(L)'I', 
Po Po 
(1.14) 
where Po and Po are the initial pressure and density, and the specific 
heat ratio y is positive. 
Let us denote the vector by which the gas is displaced with respect 
to the equilibrium position by u (x, t) = (UI (x, t), U 2 (x, t), U3 (x, t)) 
and the velo city vector by v (x, t) = (VI (x, t), V2 (x, t), Va (x, t)), 
with 
i)u 
7jt=v. ( 1.15) 
Under the assumptions we have made, that is, that P - Po, u, v, 
and their derivatives are smalI, we can rewrite Eq. (1.14) thus: 
(1.16) 
while the continuity equation (1.11) can be rewritten in the form 
Pt + po div v = 0 (1.17) 
(we assurne that the source strength is zero). 
According to the first law of Newton the balance of the forces 
acting on a small volume ~ V of the gas is zero, i.e. p (8v/8t) ~ V + 
+ grad p ~ V = F ~ V, whence after we substitute Po for p (within 
the framework of our assumption) we have 
i)v 
P0at= F-grad p. (1.18) 
Differentiating (1.18) with respect to t and employing Eqs. (1.16) 
and (1.17), we find an equation for the velocity vector v, namely, 
i)2v _ 2 d d· + 1 i)F 
ßi2-a gra IV v Po at' (1.19) 
with a2 = Poy/po. 
If we assurne that at t = 0 div u = -1, then from Eqs. (1.17) 
and (1.15) we find that in all subsequent moments of time p + 
+ po div u = O. This together with (1.15), (1.16), and (1.18) yields 
Deriving Equations of Mathematical Physics 21 
an equation for the displaeement veetor u, namely, 
~u 1 
ßtZ = a2 grad div U + Po F. (1.20) 
Finally, if we differentiate Eq. (1.17) with respeet to time and employ 
(1.16) and (1.18), we obtain the equations for p and p: 
(1.21) 
Equations (1.19)-(1.21) are known as the equations of acoustics. 
Example 5: Propagation of heat. The derivation of the he at eon-
duetion equation is based on Fourier's law, aeeording to whieh the 
amount of heat passing during a time interval M through a sm all 
area IlS lying inside the objeet being studied is given by the formula 
IlQ = - k (x, u) ou ~:' t) /).S M, (1.22) 
where n is the unit normal veetor pointing in the direetion of the heat 
flow, k (x, u) the internal thermal eonduetivity, and u (x, t) the 
temperature of the objeet at point x = (Xl' X2' x3) and at time t. 
Let us assume that the objeet is isotropie in relation to heat eon-
duetivity, whieh me ans that k (x, u) does not depend on the orien-
tation of the area. To derive an equation that u (x, t) would obey we 
isolate inside the objeet a volume Q bounded by a surfaee S. Aeeord-
ing to Fourier's law, the amount of heat flowing into Q through S 
during an interval of time from t l to t 2 is 
t2 t2 i dt 1 k :~ ds= i dt ) div (kgrad u) dx. 
t1 S t1 Q 
If F (x, t) is the density of the he at sourees, then the amount of 
heat produeed by them in Q over the speeified time interval is 
12 
~ dt i F (x, t) dx. 
t Q 
The total amount of heat flowing into Q during the time from t1 
to t 2 ean be ealeulated from the temperature differenee: 
t2 
J cp [u (x, t2)-u (x, t 1)] dx = ~ dt ~ cp ~~ dx, 
g t1 Q 
with c (x) and p (x) the speeifie heat and density of the substanee. 
Henee, 
t2 i dt ~ (cp ~~ -div(kgradu)-F(x, t») dx=O (1.23) 
t1 g 
(here we assume that the integrand is a continuous funetion). Since 
both Q and the time interval from t 1 to t 2 are arbitrary, Eq. (1.23) 
22 Statement of Boundary Value Problems 
yields 
CpUt - div (k grad u) = F (x, t), (1.24) 
which is known as the heat conduction equation. 
If the thermal conductivity k is independent of temperature u, 
or k (x, u) = k (x), then Eq. (1.24) is linear. If the object is homo-
geneous, then c (x) == const, p = const, k = const, and Eq. (1.24) 
takes the form 
(1.25) 
where a2 = k/cp and f (x, t) = F (x, t)/cp. The physics of the prob-
lem implies that to deseribe the heat propagation process unam-
biguously we must, besides writing Eq. (1.24) or (1.25), state the 
initial temperature u Ibo = (j) (x) and the temperature (or heat) 
condition at the boundary. If the boundary r of the object is kept 
at a given temperature, the boundary condition is 
ulr = 'ljJ. 
If a heat flux q is specified at the boundary, the boundary condition is 
where h = q/k and n is an outward unit normal vector. In particular, 
if the object is thermally isolated at its boundary, then 
!.!!..I = O. an r 
When the temperature of the surrounding medium is known, we 
may ass urne that heat exchange occurs according to Newton 's law, 
that is, q Ir = a (Ul - u)r, where q is the heat flux, a the external 
thermal conductivity (the heat-exchange coefflcient), and Ul the 
temperature of the surrounding medium. On the other hand, Fourier's 
law states that a unit area of the boundary r pro duces every second 
a heat flux ql = k (au/an) pointed within the object. The two fluxes 
must be equal, that is, 
k ;~ Ir=a(ul-u)lr, or (:~ +hu)lr=(j)dS). 
Example 6: Diffusion. We wish to derive the equation that de-
scribes the diffusion of a substance in a medium at rest that occupies 
a limited region Q with a boundary f. We assurne that we know the 
density F (x, t) of the sources and that diffusion takes place along 
with absorption (for instance, particles of the diffusing substance 
react chemically with the medium). Also, let us ass urne that the 
absorption rate at each point of space x that belongs to Q is pro-
portional to the density u (x, t) of the diffusing substance. Besides 
finding this equation, we must state the boundary conditions, which 
may be as follows: 
(a) A given density u is maintained at the boundary. 
(b) The boundary 1S impenetrable. 
Deriving Equations of Mathematical Physics 23 
(c) The boundary is semipermeable, with diffusion of particles 
through the boundary obeyingalawthatis similar to Newton's law 
for heat exchange by convection. 
We derive the diffusion equation starting from Nernst's law, 
according to which the amount of substance passing through a small 
area !1S during a time interval M is 
ßu 
!1Q= -D(x)7ill!1S M, 
where D (x) is the diffusion coefficient, and n the unit vectornormal 
to !1S and pointing in the direction of the flow of the substance. 
Let p (x) be the density of the medium. J ust as in the case with the 
heat conduction equation, we isolate a volume Q with a boundary S 
and calculate the balance of the amount of substance that is trans-
ferred into Q during a time interval from t 1 to t 2• 
The amount of substance transferred into Q during this time 
interval is, according to Nernst's law, 
t2 t2 
~ dt ) D (x) :: ds = ) dt ) div (D grad u) dx. 
ti S t1 Q 
The amount of substance that is produced within Q by sources is 
t2 
~ dt .\ F (x, t) dx. 
t, Q 
The amount of substance dissipated in Q due to absorption by the 
medium is 
t2 
~ dt ~ q (x) u (x, t) dx, 
t1 Q 
where q (x) is the absorption coefficient. Since, on the other hand, 
the amount of substance in Q changes during the same time interval 
by 
) p(x) [u(x, t2 )-u(x, tj)ldx= ~2 dt ~ P ~~ dx, 
Q t1 Q 
we can write 
t2 
~ dt j (put-div(Dgradu)-F+qu)dx=O (1.26) 
t1 Q 
(the integrand is assumed to be continuous). Since both Q and the 
time interval are arbitrary, Eq. (1.26) yields 
PUt + qu = div (D grad u) + F. (1.27) 
This is known as the diffusion equation. The physics of the problem 
implies that to describe the diffusion process unambiguously we 
24 Statement of Boundary Value Problems 
must know the initial distribution of the density u It=o = cP (x), 
xE Q, and the diffusion mode at the boundary. Just as in Example 5, 
the boundary conditions are 
(a) Ulli'=uo• 
(b) :: Ir= 0, 
(c) D :: Ir=a(ut-u)lr, 
where Uo and Ul are given functions, and a the penetrabilHy of the 
boundary f. 
1.1. Find the static sag of astring with its ends fastened and under 
a continuously distributed load (per unit length). 
1.2. Derive the equation of sm all transverse vibrations of astring 
that has a sm all ball of mass m attached to it at an inner point Xo. 
1.3. Derive the equation of vibrations of astring in an elastic 
medium. 
1.4. The periodic motion of a rod or shaft in which the cross seetions 
are rotated with respect to each other and about the axis is known 
as torsional vibrations. Derive the equation of small torsional vibra-
tions for a homogeneous cylindrical rod. Consider the cases when 
(a) the ends of the rod are free, (b) the ends are rigidly fixed, and 
(c) the ends are elastically fixed. 
1.5. An elastic homogeneous rectanguJar bar has one of Hs ends 
rigidly fixed and the other free. At time t = 0 sm all transverse 
deflections and velocities parallel to the vertical lateral symmetry 
plane of the bar are imparted to the points of the bar. State the 
boundary value problem of determining the small transverse devia-
tions of the points of the bar at t > 0 assuming that the bar performs 
small transverse vibrations. 
1.6. A pipe filled with an ideal gas and open at one end is in trans-
lational motion in the direction of its axis with a constant yelocity v. 
At time t = 0 the pipe suddenly stops. State the boundary value 
problem of determining the displacement of the gas inside the pipe 
at a distance x from the closed end. 
1. 7. An ideal gas occupying a cylindrical pipe performs sm all 
lateral vibrations. The planar cross sections, consisting of gas par-
ticles, do not deform and all gas particles move parallel to the axis 
of the cylinder. State the boundary value problem of determining 
the displacement u (x, t) of the gas particles when the ends of the 
pipe are (a) closed with impenetrable partitions, (b) open, and (c) closed 
by small pistons of infinitesimal masses, the pistons being mounted 
on springs with a stiffness v and moving without friction inside the 
pipe. 
Deriving Equations of Mathematical Physics 25 
1.8. At time t = 0 one end of a straight elastic homogeneous rod 
begins to perform longitudinal vibrations according to a certain 
law, while the other end is under a force <D (t) directed along the 
axis of the rod. Before t = 0 the cross sections of the rod were at 
rest and in nondeflected positions. State the boundary value problem 
of determining the small longitudinal deflections of the points of the 
rod at t> O. 
1.9. State the boundary value problem of the small transverse 
vibrations of astring in a medium with a resistance proportional to 
the first power of velocity. Both ends of the string are fixed. 
1.10. Construct the equation of longitudinal vibrations of a rod 
whose cross-sectional area is a given function of x, assuming that 
the material of the rod is homogeneous. 
1.11. State the boundary value problem of the lateral vibrations of 
an elastic rod in the form of a frustum of a cone. Both ends of the 
rod are fixed, and the rod is taken out of the state of rest by imparting 
initial velocities and longitudinal deflections at time t = O. The 
]ength of the rod is I, the radii of the bases are Rand r (R > r), the 
material of the rod is homogeneous, and all deformations of the 
cross sections are ignored. 
1.12. A light string is rotating in the horizontallplane with a con-
stant angular velocity (ü around a vertical axis, with one end of the 
string heing fixed to a point on the axis and the other being free. 
At time t = 0 small deflections and velocities normal to the plane 
are imparted to the points of the string. State the boundary value 
problem of determining the deflections of the points of the string 
from the plane of equilibrium motion. 
1.13. Suppose a ball of mass mo is fixed at point x = 0 of an infinite 
homogeneous string. The initial velocities and deflections of the 
string's points are zero. State the boundary value problem of deter-
mining the deflections of the points of the string from the position 
of equilibrium when (a) a force F = F 0 sin Qt begins to act on the 
ball at time t = 0, (b) momentum Po is imparted to the ball trans-
versely at time t = 0, and (c) the ball is elastically fixed with an 
effective rigidity k 2 with case (b) otherwise being effective. 
1.14. State the boundary value problem of small lateral vibrations 
of an elastic homogeneous rod one end of which is rigidly fixed and 
the other moves with resistance proportional to the first power of 
velo city. The resistance of the medium can be ignored. 
1.15. Masses m;, i = 1, 2, ... , n, are concentrated at the internal 
points x = Xi, i = 1, 2, .. . ,n, of astring with fixed ends. State 
the boundary value problem of small transverse vibrations of the 
string with arbitrary initial conditions. 
1.16. Two sem i-infinite elastic homogeneous rods with the same 
cross sections are connected rigidly at their ends and constitute an 
26 Statement of Boundary Value Problems 
infinite rod. Let Pt and EI be the density and elasticity modulus of 
one rod and P2 and E 2 the respective quantities of the other. State 
the boundary value problem of the deflections of the cross sections 
of the infinite rod from their positions of equilibrium if at time 
t = 0 certain longitudinal deflections and velocities are imparted 
to the cross sections. 
1.17. A heavy homogeneous cord of length 1 is fixed by its upper 
end (x = l) to a vertical axis and rotates about the axis with a con-
stant angular velocity 00. Prove that the equation of small vibrations 
of the cord ab out its vertical position of equilibrium is 
02U 0 ( OU), 2 ---ai2 = g OX x 7iX T 00 u. 
1.18. State the boundary value problem of transverse vibrations of 
a heavy homogeneous string about its vertical position of equilibrium 
if the upper end of the string is rigidly fixed and the lower end is free. 
1.19. State the problem of finding the magnetic field inside and 
outside of a cylindrical conductor with a current J travelling along 
the surface of the conductor. 
1.20. An electric cable with a potential Vo is grounded at one of its 
ends through a lumped capacitance (or inductance), while the other 
end is insulated. State the problem of determining the electric current 
flowing through the cable. 
1.21. The end x = 0 of a round homogeneous shaft is fixed, while 
theend at x = 1 is rigidly attached to a disk with a moment of iner-
tia J o. The disk is twisted through an angle a and at time t = 0 
is let go without any initial velo city imparted to it. State the bound-
ary value problem of determining the angles of rotation of the cross 
sections of the shaft at t > O. 
1.22. A heavy rod is hung up and is jammed in such a manner that 
the deflection of each of its points is zero. At time t = 0 the rod is 
released. State the boundary value problem of the forced vibrations 
of the rod. 
1.23. Let all conditions of the previous problem be fulfilled except 
that at the lower end, namely, a load Q is attached to the lower end 
of the rod, with the unstressed state of the rod being the equilibrium 
position (e.g. at time t = 0 a support is taken out from und er the 
load and the rod begins to stretch). 
1.24. State the problem of the motion of a semi-infinite string 
(0 :::;;; x < 00) at t > 0 if at t < 0 a wave u (x, t) = f (x + at) travels 
along the string; the end x = 0 is rigidly fixed. 
1.25. State the boundary value problem of small radial vibrations 
of a homogeneous ideal gas occupying a cylindrical pipe of radius R; 
the pipe is so long that it can be considered to extend infinitely in 
both directions. The initial deflections and initial velocities are 
given functions of r. 
Deriving Equations of Mathematical Physics 27 
1.26. State the problem of the steady state flow of an ideal fluid 
around a sphere (potential flow). Give an eleetrostatie analog of this 
proeess. 
1.27. State the boundary value problem of small radial vibrations of 
a homogeneous ideal gas oeeupying a spherieal vessel of radius R 
if the initial velocities and deflections are given functions of r. 
1.28. State the boundary value problem of transverse vibrations of 
a membrane to whieh apressure P is applied (per unit surfaee area 
and normal to the surface of the membrane) if in the unperturbed 
state the membrane is flat and the surrounding medium does not 
res ist the membrane vibrations. Consider the eases where (a) the 
membrane is rigidly fixed at the edge L, (b) the membrane is free at 
L, and (e) a part LI of the edge is rigidly fixed while the remaining 
part L 2 is free. 
1.29. State the boundary value problem of the vibrations of a round 
homogeneous membrane fixed at its edge and plaeed in a medium 
whose resistanee is proportional to the first power of velo city. At 
time t = 0 an external force of density f (r, <p, t) is applied to the 
surfaee of the membrane in the direetion normal to the plane of the 
unperturbed membrane. The initial veloeities and defleetions of the 
points of the membrane are zero. 
1.30. A homogeneous rectangular membrane is fixed at its edge. 
At time t = 0 it receives a shoek in the vieinity of its eentral point, 
so that 
lim ~ Vo (x) dx = A, x = (Xi' X2), 
10 ..... 0 U 
e 
where A is a eonstant, and Vo (X) is the initial velocity. State the 
boundary value problem of free vibrations of this membrane. 
1.31. Suppose an eleetric eircuit eonsists of a resistance R, a self-
induetance L, and a eapaeitanee C. At time t = 0 an emf E o is 
introduced into the eireuit. Show that the eurrent i (t) in the circuit 
obeys the following equation: 
t 
Li'(t)+Ri(t)+ b ~ i(-c)d-c=Eo, t>O. 
o 
1.32. Let us eonsider an electromagnetie field in a medium. Starting 
with Maxwell 's equations, derive equations whose solutions are 
the various eomponents of the electrie and magnetie fields for the 
cases where (a) the charge density p = 0, I': = const, 'A = eonst, 
/-t = eonst, J = 'AE (Ohm's law), and (b) the medium is a vaeuum 
and no currents exist. 
1.33. State the problem of the penetration hy a magnetie field into 
the right half-space filled by a medium with eonduetivity (J if starting 
28 Statement of Boundary Value Problems 
from time t = 0 a magnetic field H = Ho sin Qt is maintained at 
the surfaee x = 0 (the magnetic field is direeted parallel to the 
surface). 
1.34. State the boundary value problem of determining the tem-
perature of a rod 0 ~ x ~ l with heat-insulated lateral surfaee. Con-
sider the cases where (a) the ends of the rod are kept at a given tem-
perature, (b) at the ends of the rod a given heat flux is maintained, 
and (e) at the ends of the rod there is convective heat exchange by 
Newton's law with a medium whose temperature is given. 
1.35. Derive the equation of diffusion in a medium at rest assuming 
that the planes perpendieular to the x axis are at eaeh moment of 
time t the surfaces of constant density. Write the boundary conditions 
assuming that diffusion oecurs in the flat layer 0 ~ x ~ land con-
sider the cases where (a) the concentration of the diffusing substance 
is kept equal to zero at the boundary planes, (h) the boundary planes 
are impenetrable, and (c) the boundary surfaces are semipermeable, 
with diffusion through these planes governed by a law similar to 
Newton 's law for eonvective he at exchange. 
1.36. Derive the equation of diffusion for a disintegrating gas (the 
number of moleeules decaying every second at a given point is pro-
portional to the density with a positive proportionality factor a). 
1.37. Suppose we have a thin homogeneous rod of length l whose 
initial temperature is f (x). State the boundary value problem of 
determining the temperature of the rod if a constant temperature Uo 
is maintained at the end x = 0, while the lateral surface and the 
end x = l are involved in convective heat exchange (governed by 
Newton's law) with the surrounding medium whose temperature is 
zero. 
1.38. State the problem of determining the temperature distribution 
in a thin heat-insulated rod of infinite length along whieh a point 
heat source begins to move at time t = O. The source moves in the 
positive direction, its velo city is vo, and it generates q units of heat 
every second. 
1.39. State the boundary value;problem of the cooling off of a thin 
homogeneous ring of radius R whose surface is involved in convective 
he at exchange with the surrounding medium whose temperature is 
known. The nonuniformity of the temperature distribution in the 
ring ean be ignored. 
1.40. Derive the equation that describes the. diffusion of suspended 
particles and allows for precipitation. Assurne that the velocity of 
the particles caused by gravity is constant and the density of the 
particles depends only on the height z and the time t. Write the 
boundary condition that corresponds to an impenetrable partition. 
1.41. State the boundary value problem of the cooling off of a uni-
formly heated rod in the form of a frustum of a eone (the warpage 
Deriving Equations of Mathematical Physics 29 
of isothermal surfaces can be ignored) if the ends of the rod are ther-
mally insulated and the lateral surface is involved in convective 
heat exchange with a medium whose temperature is zero. 
1.42. A solute with an initial concentration Co = const diffuses from 
the solution that occupies the space between two planes, x = 0 and 
x = h, into the solvent that occupies the space between the planes 
x = hand x = l. State the boundary value problem of the process 
of equalization of the density assuming that the boundaries at 
x = 0 and x = l are impenetrable for the solute. 
1.43. Heat sources that are distributed with a constant density Q 
within a homogeneous ball begin to operate at time t = O. State 
the boundary value problem of determining the temperature dis-
tribution within the ball at t > 0 if the initial temperature of every 
point of the ball depends only on the distance between this point 
and the ball 's center. Consider the cases where (a) the temperature of 
ball' s surface is maintained at zero, and (b) the ball' s surface is in-
volved in convective heat exchange (obeying Newton's law) with the 
surrounding medium whose temperature is zero. 
1.44. Let us take a uniform ball of radius R whose initial tempera-
ture is zero. State the boundary value problem of determining the 
temperaturedistribution inside the ball at t > O. Suppose that 
(a) the ball is heated uniformly over its entire surface by a constant 
heat flux q, and (b) the surface of the ball is involved in convective 
heat exchange with the surrounding medium whose temperature 
depends only on time 
1.45. Suppose that the initial temperature of an unlimited plate of 
thickness 2h is zero. State the boundary value problem of fmding the 
temperature distribution at t > 0 over the thickness if (a) the plate 
is heated from both sides by equal he at fluxes q, and (b) a he at source 
with a constant density Q begins to operate inside the plate at t = 0 
while the temperature of both sides of the plate is kept constant and 
equal to zero. 
1.46. An infinitely long cylinder of radius R has an initial tem-
perature f (r). State the boundary value problem of the radial pro-
pagation of heat in the cylinder if (a) the temperature of the lateral 
surface is maintained constant, and (b) the lateral surface emits 
heat into a medium whose temperature is zero. 
1.47. Given a thin rectangular plate with edges land m and with 
a known initial temperature distribution, state the boundary value 
problem of heat propagation in the plate if the lateral sides of the 
plate are heated to the following temperatures: 
U /y=o = 1P1 (x), U /y=m = 1P2 (x), 
u/x=o = 'l/Ji (x), u/x=t = 'l/J2 (y). 
1.48. The initial temperature distribution in a homogeneous ball 
is given by the function f (r, e, IP). State the boundary value problem 
30 Statement of Boundary Value Problems 
of heat distribution in the ball if the temperature of the ball 's surfaee 
is maintained constant and equal to Uo. 
1.49. Two semi-infinite straight rods made out of different materials 
are brought into eontaet at time t = 0 through their ends. State 
the boundary value problem of he at distribution in the new infinitely 
long rod if the initial temperatures of both rods are known. 
1.50. State the boundary value problem of the time-independent 
temperature distribution in a thin reetangular plate OAGB with 
edges OA = a and OB = b if (a) the temperatures of the lateral 
surfaees of the plate are known, and (b) known heat fluxes eross the 
edges OA and OB while the edges BG and AG are thermally insu-
lated. 
1.51. A time-independent transverse load with a density f (x, y) 
is applied to a flat membrane bounded by a closed eurve L. State the 
boundary value problem of determining the deviations of the points 
of the membrane from the plane if (a) the edge of the membrane is 
fixed, (b) the edge is free, and (e) the edge is elastieally fixed. 
1.52. Given a eylinder with a base of radius Rand altitude h, state 
the boundary value problem of the time-independent temperature 
distribution inside the eylinder if the temperature of the upper and 
lower bases is a given funetion of r, while (a) the lateral surfaee is 
thermally insulated, (b) a temperature that depends only on z is 
maintained at the surfaee, and (e) the surfaee freely gives off heat 
to a medium whose temperature is zero. 
1.53. State the boundary value problem of finding the time-indepen-
dent temperature distribution inside a hemisphere if the spherieal 
surface is kept at a given temperature f (cp, 8), while the temperature 
of the hase of the hemisphere is maintained eonstant and equal to 
zero. 
1.54. A ball of radius R is heated by a plane-parallel heat flux of 
density q that falls on its surfaee, and gives its heat off to the sur-
rounding medium in aeeordanee with Newton's law. State the 
boundary value problem of the temperature distribution inside 
the ball. 
1.55. Let n (x, s, t) be the time-dependent density at point x of par-
ticles that fly with a eonstant veJocity v in the direction speeified 
by the vector s = (81,82,83), a (x) the absorption eoeffieient at point 
x, and h (x) the multiplieation faetor at point x. Assuming that the 
scattering at eaeh point x is isotropie, show that n (x, s, t) satisfies 
the integro-differential transport equation 
+ !; +(sgradn)+a(x)n= ß4~) j n(x, s', t)ds'+F, 
18'1=1 
where F (x, s, t) is the souree density, and ß (x) = a (x) h (x). 
Deriving Equations of Mathematical Physics 31 
1.56. State the boundary value problem for the equation of Problem 
1.55 with a given initial density distribution and a given incident 
flux of particles onto the boundary S of a region G. 
1.57. Show that for the solution n (x, s) of the time-independent 
boundary value problem 
(s, grad n) + a (x) n = ß4~) I n (x, s') ds' + F (x), 
\8'1=1 
with n 18 = ° if (s, n) < 0, where n is the outward unit vector normal 
to S, the average density no (x) = 1n ~ n (x, s) ds satisfies Peierls's 
/ si = 1 
integral equation 
1 
e-/x-x'l S a [tx+(1-t)x'J dt 
no (x) = 4~ ~ °IX_X'12 [ß(x')no(x')+F(x')ldx'. 
1.58. Expanding the solution n (x, s) of the time-independent bound-
ary value problem 1.57 in aseries in spherical functions of sand 
retaining only terms with the zeroth and first harmonics, show that 
the function 
no (x) = 4~ i n (x, s) ds 
\81=1 
is the solution of the boundary value problem (the diffusion ap-
proximation) 
- ; div ( ! grad ho) +(1-h)ho= ~ , 
(no+ 3! ~:o) 18= 0. 
Answers to Problems of Sec. 1 
1.1. Tuxx + f (x) = 0, 0< x< l, u Ix=o = U Ix=l = 0, where 
f (x) is the load density. 
1.2. pUtt = T OU xx , ° < x< I, x =1= xo, t > 0, U fx=o = U fx=l = 0, 
U (xo + 0, t) = u (xo - 0, t), U x (xo + 0, t) - U x (xo - 0, t) = 
= ; Utt (xo, t). 
o 
1.3. pUtt = Tuxx - au, ° < x< l, t> 0, where a is the elasticity 
of the medium. 
1.4. 8 tt = a28xx , ° < x< I, 0< t< 00, 8 (x, 0) = f (x), 
8 t (x, 0) = F (x), ° ~ x ~ I, where 8 (x, t) is the angle through 
which the cross section with coordinate x has turned at time t, 
a2 = GJ/(J) , G is the shear modulus, J the polar moment of inertia 
of the cross section in relation to the point at which the axis inter-
sects the cross section, and <D is the axial moment of inertia per 
32 Statement of Boundary Value Problems 
unit length of rod. The boundary eonditions are: (a) 8x (0, t) = 
= 8x (l, t) = 0, (b) 8 (0, t) = 8 (l, t) = 0, (e) (8 x - h8) Ix=o = ° 
and (8 x + h8) Ix=l = 0, where h = k/GJ, and k is the rigidity of 
the elastie fixt ure. 
1.5. Utt + a2uxxxx = 0, ° < x< l, t > 0, U (x, 0) = t (x), 
Ut (x, 0) = F (x), ° ~ x ~ l, U (0, t) = Ux (0, t) = Uxx (l, t) = 
= Uxxx (l, t) = 0, where a2 = EJ/pS, and J is the geometrie moment 
of inertia of the eross seetion in relation to its center line perpen-
dieular to the plane of vibrations. 
1.6. Utt = a2uxx , ° < x< l, t > 0, where a2 =YPo/Po is the speed of 
sound, U (x, 0) = 0, Ut (x, 0) = v, ° ~ x ~ l, U (0, t) = 0, 
U x (l, t) = 0, t > 0. 
1.7. Utt = a2uxx , a2 = YPo/Po, ° < x< l, t> 0, U (x, 0) = t (x), 
u,' (x, 0) = F (x), ° ~ x ~ l. The boundary eonditions are 
(a) U (0, t) = U (l, t) = 0, (b) Ux (0, t) = Ux (l, t) = 0, (e) (ux -
- hu) Ix=o = ° and (ux + hU)x=1 = 0, where h = v/sYPo, and 
S is the pipe's eross-seetional area. 
1.8. Utt = a2uxx , ° < x< l, t> 0, U (0, t) = cp (t), Ux (l, t) = 
= <I> (t)IES, t > 0, U (x, 0) = 0, Ut (x, 0) = 0, ° ~ x ~ 1, a2 = Elp. 
1.9. Utt = a2uxx - 2V2Ut, ° < x< l, t> 0, U (x, 0) = cp (x), 
Ut (x, 0) = 'ljJ (x), ° ~ x ~ I, U (0, t) = U (l, t) = 0, t > 0, where 
2v2 = klp, and k is the frietion eoeffieient. 
1.10. ~ [S(x) ~~J=a2 :~, a2 =pS/E. 
1 12 iJ"u - 2 iJ ( 2 iJu ) ° 1 ° 2 2/ ') . . ift2- a 7iX x ax' < x <, t>, a = (t) '-', 
1 U (0, t) I< 00, U (l, t) = 0, t> 0, U (x, 0) = f (x), Ut (x, 0) = 
= F (x), ° ~ x ~ I. 
1.13. Utt = a2u;x, x =i= 0, t > 0, a2 = To/p, U (x, 0) = 0, 
Ut (x, 0) = 0, x =i= 0; the eondition at point x = ° is (a) 
-mOUtt (0, t) + T o (ux (+ 0, t) - Ux (-0, t)J + F 0 sin Qt = 0, t> 
> 0, (b) U (-0, t)= U (+0, t), -moUtt (0, t) + T o (u x (+ 0, t) -
- U x (-0, t)J = 0, t> 0, U (-0, 0) = U (+ 0, 0) = 0, 
moUt (-0, 0) = mOUt (+0, 0) = Po, (e) U (-0, t) = U (+0, t), 
t> 0, moutt (0, t) + T o [ux (+0, t) - Ux (-0, t)J - k 2u (0, t) = 0, 
moUt (-0, 0) = mOUt (+0, 0) = Po, U (-0, 0) = U (+0, 0)= 0. 
1.14. Utt = a2uxx , ° < x< l, t> 0, a2 = Elp, U (x, 0) = f (x), 
Ut (x, 0) = g (x), ° ~ x ~ l, U (0, t) = 0, (ESux - kUt) Ix=l = 0, 
t> 0, where k is the frietion eoeffieient for the end of the rod at 
x = 1. 
1.15. Utt = a2uxx , X=i=Xi' i = 1, ... , n, o<x< 1, t>O, 
U (0, t) = U (l, t) = 0, U (Xi - 0, t) = U (Xi + 0, t), Ux (Xi + 0, t) 
- Ux (Xi - 0, t) = m; Utt (Xi' t), t> 0, i = 1, ••. , n; U It=o 
= f (x), Ut It=o = F (x), ° ~ x ~ 1. 
Deriving Equations of Mathematical Physics 33 
1.16. u}t = aiu1x, -00 < x< o} t> ° u1 (0 t) = 2 (0 t) 
2 22 ° + "U" Utt = a2U XX , < X < 00 
E1ui, (0, t) = E 2ui (0, t), t > 0, U1 (X, 0) = ! (X), U} (X, 0) = F (X), 
-00 < X< 0, U2 (X, 0) = ! (X), U~ (X, 0) = F (X), X> 0, where 
U 1 and u2 are the deflections of the points of the left and right rods, 
and a~ = E;lPi! i = 1, 2. 
ß2U ß ( ßU) 1.18. ----at2=g7iX x7iX ' O<x<l, t>O, lu(O, t)1 <00, 
u (I, t) = 0, t > 0, U It=o = ! (x), Ut It=o = F (x), ° ~ X ~ I. 
1.19. V2<D(I) = 0, r > R, V2<D(i) = 0, O~r < R, grad<D = H, 
«.(1) ffi(i) ffi(l) (ffi(i) 4Jt. ) I . ~r Ir=R=~r Ir=R' ~<plr=R= ~<p +-Jsur ,1<D(1)(O,t)l< 
C r=R 
< =, jsur = JI2nR is the surface current density, and <D(;) and 
<D(l) are the magnetic field potentials inside and outside the con-
ductor, respectively. 
1.20. J x = -CVt, Vx = -LJh O<x<l, t>O, Vlt=o=vo, 
1 
1 ~ 
v (0, t) = - \ J dt at the grounded end and Vx (l, t) = ° at the 
C • 
o 
insulated end. 
1.21. ett = a2exx , ° < x< I, t> 0, e It=o = axll, 8t It=o = 0, 
cD o ~ X ~ I, 8 Ix=o = 0, 8x Ix=1 = - JG 8tt , where the constants, 
a2, <D, J, and G have the same meaning as in Problem 1.4. 
1.22. Utt = a2uxx + g, ° < x< I, t> 0, U (x, 0) = Ut (x, 0) = 0, 
o ~ x ~ I, U (0, t) = 0, Ux (l, t) = 0, t > 0, a2 = Elp. 
1.23. Utt = a2uxx + g, ° < x< I, t> 0, U It=o = Ut It=o = 0, 
Q ° ~ x ~ 1, U 1 .. =0 = 0, - Utt Ix=1 = -ESux Ix=1 + Q. g 
1.25. utt=a2 (urr ++ur ) , O<r<R, t>O, u(r, 0) = !(r), 
Ut (r, 0) = F (r), ° ~ r ~ R, 1 U (0, t) I< =, Ur Ir=R = 0. 
1.26. V2rp=0, r>R, t>O, ßß<Jl/ =0, t>O, lim V =limgradrp=vo, 
r r=R r-+oo T-OO 
where Vo is the flow velo city at infinity. 
1.27. Utt=az(urr + ~ ur), O~r<R, t>O, u(r, 0) = !(r), 
Ut It=u = F (r), ° ~ r ~ R, 1 U (0, t) I< =, Ur Ir=R = 0, where 
a2 = "(Polpo. 
1.29. Utt + kUt = a2 V 2u+ ! (r, rp, t)/p, ° ~ r< R, O~rp < 2n, 
t> 0, U It=o = Ut It=o = 0, 1 U (0, rp, t) I< =, U (R, rp, t) = 0, 
where a2 = Tlp, k = alp, and a is the coefficient of the elastic 
resistance of the medium. 
3-0176 
34 Statement of Boundary Value Problems 
1.32. (a) Utt - a2v2u + 4nA. Ut = 0, a2 = ~ (b) (~- a2V2 ) m = 
8 8ft ' at2 '1'0 
= _ 4ne2 p (~_a2'r72) m=O ~ a<po -divm=O where E= 
8 2ft ' at2 - v 'I' , e at '1" 
= (Eh E 2 , E a) is the electric field strength, H = (H1 , H 2 , Ha) the 
magnetic field strength, (l (x) the charge density, e the dielectric 
constant of the medium, /A. the magnetic permeability of the medium, 
and I (x, t) = (11' 1 2 , 1 3) the conduction current. In the case (a) 
the components of E and of H obey the same telegrapher's equation. 
In tae case (b) one must introduce the four-component electromagnet-
ic potential (<Po, cp), with cp = (<PI' <P2, 'Pa)' Then the solution to 
Maxwell's equations can be sought in the form E = grad 'Po -
- ..!. a<p and H = ..!. curl <po 
e at ft 
4ncr 1 ° 1.33. Hxx=-:a-Ht+-,;Htt, x> , t>O, Hlt=o=O, Htlt=o, e e 
x> 0, H 1=0 = Ho sin Qt, t> 0, where c is the speed of light. 
1.34. Ut = a2uxx , ° < x< 1, t> 0, u (x, 0) = f (x), 0:::;:; x :::;:; 1; 
the boundary conditions are (a) u Ix=o = 'PI (t), U Ix=1 = <P 2 (t), 
t> 0, (b) -kSux Ix=o = ql (t), kSux Ix=1 = 'P2 (t), t> 0, 
(c) Ux Ix=o = h Cu (0, t) - 'PI (t)l, Ux Ix=l = -h Cu (l, t) - 'P2 (t»), 
a2 = k/cp is the specific heat, 'PI (t) and 'P2 (t) are the temperatures 
of the ends of the rod in the case (a) 01' the temperatures of the sur-
rounding medium at the two ends in the case (b), and the qi are the 
heat fluxes at the ends of the rod. 
1.35. Ut = Duxx , ° < x< l, t > 0, u (x, 0) = f (x), 0:::;:; x :::;:; l; 
the boundary conditions are (a) u (0, t) = u (l, t) = 0, t> 0, 
(b) U x (0, t) = Ux (l, t) = 0, t>O, (c) ux Ix=o = h [u (0, t) -'PI (t)], 
t> 0, Ux Ix=/ = -h fu (l, t) - 'P2 (t)], where alD = h, and a is 
the penetrahiJity at the eIlds. 
1.36. Ut = D V 2u - au, t> 0, x = (Xl' x 2 , X3) E R 3• 
1.37. ut=a2uxx - asp u, O<x<l, t>O, ult=o=f(x), O~x:::;;;l, ep 
U Ix=o = Uo, (ux + hu) Ix=1 = 0, t> 0, P is the perimeter of the 
cross section of the rod, h = alk, and a2 = k/cp. 
1.38. ut=a2uxx +l..ö(x-vot), -oo<x<+oo, t>O, u(x, 0)= c 
='P(x), a2 =k/cp. 
1.39. Ut = a2uxx - b (u - uo), ° < x< l, t> 0, u (x, 0) = f (x), 
0:::;:; x :::;:; l, u Ix=o = U Ix=l, Ux Ix=o = U x Ix=h a2 = k/cp, b = 
=aP/cpS, where P is the perimeter of the cross section of the ring, 
x = Re, and e is the angular coordinate. 
1.40. Ut = Duzz - vUz, Z >zo, t> 0, (Duz - lJu) Iz=zo = 0, t> 
> 0, where v is the precipitation rate. 
Deriving Equations of Mathematical Physics 35 
1.41. (1_-=-)2 .!!!:.... = a2 ~ [( 1_~)2 !..!::..-J _ 2a (1-x/H) U 0< 
H iJt iJx H iJx CprO cos y , 
< X< 1, t > 0, U It=o = Uo, ° ~ X ~ l, Ux Ix=o = Ux Ix=/ = 0, 
t> 0, where a2 = klep, H is the overall height of the cone, y the half 
of the opening span of the angle of the cone, ro the radius of the 
larger base, and 1 the height of the frustum of the cone. 
{ Co, O<x< h, 1.42. Ct = Dexx, ° < x< 1, t > 0, C (x, 0) = ° h 1 , <x<. 
I 2) Q 1.43. ut=a2Iurr+-ur +-, O~r<R, t>O, Ult=o=t(r), 
\ r er ° ~ r ~ R, 1 U (0, t) I< 00; the boundary conditions are 
(a) U (R, t) = 0, (b) (ur + Hu) !r=R = 0, H = alk, and a2 = klcp. 
1.44. Ut= a2 (Urr++ Ur), O~J< R, t >0, ult=o =0, O~r~R; 
the boundary conditions are (a) 1 U (0, t) I< 00, Ur (R, t) = qlk, 
t > 0, (b) 1 U (0, t) I< 00, (Ur + Hu) Ir=R = <p (t), t> 0, H = 
= alk, a2 = klcp. 
1.45. (a) Ut = a2uxx , -h< x< h, t> 0, U It=o = 0, 
(kux + q) Ix=-h = 0, (-kux + q) Ix=h = 0, (b) Ut = a2uxx + 
+ Qlcp, -h< x< h, t> 0, U It=o = 0, U Ix= ±h = 0, a2 = 
= klep. 
1.49. ut=a(x)uxx , x*O, t>O, u(x,O)=t(x), u(-O, t)= 
{ ai, x< 0, =u(+O, t), k 1uX (-0, t)=k2ux (+0, t), a(x)= 2 ° a2 , x> , 
2 Classification of Sccond-order Equations 
The equation 
n 
LJ aij (x) U x .x . + <D (x, u, grad u) = ° 
i, j=l l ] 
can be reduced at each point Xo to canonical form by applying the 
nonsingular linear transformation S = BTX , where B is a matrix 
such that the transformation y = BT) reduces the quadratic form 
n 
~ aij (xo) YiY j to canonical form. (Every quadratic form can be 
i,i=1 
reduced to canonical form by, say, isolating perfect squares.) 
2.1. Reduce the following equations to canonical form: 
1. Uxx + 2uxy - 2uxz + 2uyy + 6uzz = 0. 
2. 4uxx - 4uxy - 2uyz + uy + Uz = 0. 
3. uxy - Uxz + Ux + uy ,- Uz = 0. 
3* 
36 Statement of Boundary Valuc Problems 
4. UXX + 2uay - 2uxz + 2uyV + 2uzz = O. 
5. UXX + 2uxy - 4uxz - 6uI/z - Uzz = o. 
6. Uxx + 2uxy + 2uyy + 2uI/z + 2uyt + 2uzz + 3utt = O. 
7. u xy - Uxt + Uzz - 2uzt + 2utt = O. 
8. u xy + u xz + Uxt + Uzt = O. 
9. Uxx + 2uxy - 2uxz - 4ugz + 2ug t + Uzz = O. 
10. Uxx + 2uxz -2uxt + Uyy + 2ugz + 2ugt + 2uzz+2utt = O. 
n n-1 
11. UX1X1 + 2 ~ uxkxk - 2 ~ uxkx = O. 
k=! k=1 k+l 
n 
12. UX1X1 - 2 ~ (-1)kUXk Xk = O. k=! -1 
n 
13. ~ kUxkxk + 2 ~ lUxZxk = O. 
k=1 l<k 
n 
14. ~ uxkxk + ~ uxZxk = O. 
k=1 Z<k 
15. ~ uXZxk =0. 
Z<k 
The equation 
a (x, y) Uxx + 2b (x, y) uxy + c (x, y) Ugy = <P (.r, y, u, ux , ug), (2.1) 
where I a I + I b I + I c I =1= 0, belongs (at a point or in a region) 
to the 
hyperbolic type if b2 - ac > 0, 
parabolic type if h2 - ac = 0, 
elliptic type if b2 - ac < O. 
The characteristic equation 
a (x, y) (dy)2 - 2b (x, y) dx dy + c (x, y) (dx)2 = 0 
of Eq. (2.1) splits into two equations, 
ady-(b+ Vb2 -ac) dx= 0, 
a dy-(b- Vb2 - ac) dx= O. 
(2.2) 
(2.3) 
Equations of the hyperbolic type: b2 - ac > O. The general solutions 
(r (x, y) = Cl and 'IjJ (x, y) = c 2 of Eqs. (2.2) and (2.3) are real and 
d istincL They determine two different families of real characteristics 
ofEq. (2.1). By a change of variables, ; = cf (x, y) and Yj = 'IjJ (x, y), 
Eq. (1. 1) is red uced to canonical form 
u sTJ = cI\ (~, Yj, u, 1I~, UTJ). 
Equations oj the parabolic type: b2 - ac = O. Equations (2.2) and 
(2.3) coincide. The general solution cy (x, y) = c of Eq. (2.2) deter-
mines the family of real characteristics of Eq. (2.1). A change of 
variables ~ = cp (x, y) and Yj = 'IjJ (x, V), where 'IjJ (x, y) is any smooth 
function such that the change of variables is uniqlle in the region 
considered, reduces Eq. (2.1) to canonical form 
UTJTJ = <PI (~, Yj, u, u;, UTJ)· 
Deriving Equations of Mathematical Physics 37 
Equations 0/ the elliptic type: b2 - ac< O. Let q> (x, y) + i'I/J (x, y) = 
= c be the general solution of Eq. (2.2), where q> (x, y) and", (x, y) 
are real-valued functions. (If a, b, and c are analytic functions, the 
existence of a general solution for Eq. (2.2) follows from Kovalev-
skaya's theorem.) Then the change of variables 6 = q> (x, y) and 
1) = '" (x, y) reduces Eq. (2.1) to canonical form 
u;; + U'l1'11 = <PI (6, 1), u, u;, U'l1). 
2.2. Reduce to canonical form the equations given below in regions 
that preserve the type of the equation. 
1. Uxx - 2uxy - 3uyy + uy = O. 
2. Uxx - 6uxv + 10uyy + Ux - 3uy = O. 
3. 4uxx + 4uxy + Uyy - 2uy = O. 4. Uxx - XU yy = O. 
5. Uxx - YUyy = O. 6. xUxx - YUyy = O. 7. YUxx - XU yy = O. 
8. x2uxx + y2Uyg = O. 9. y2uxx + X2Ugy = O. 
10. y2uxx - X2Uyy = 0.11. (1 + x2) u.u+ (1+ y2) Uyy + YU y = O. 
12. 4y2uxx - e2Xuyy = O. 
13. Uxx - 2 sin xUxy.+ (2 - cos2 x) Uyy = O. 
14. y2uxx + 2yuxy + Uyy = O. 
15. x2uxx - 2xuxy + Uyy = O. 
Suppose the coefficients of Eq. (2.1) are continuous in a region D. 
The function U (x, y) is said to be a solution of Eq. (2.1) if it belongs 
to the dass C2 (D) and satisfies Eq. (2.1) in D. The collection of all 
the solutions of Eq. (2.1) is said to be the general solution of Eq. (2.1). 
2.3. Find the general solution of each of the equations with constant 
coefficients given below: 
1. uxy = O. 2. Uxx - a2uyy = O. 
3. Uxx - 2uxy - 3uyy = O. 4. uxy + aux = O. 
5. 3uxx - 5uxy - 2uyy + 3ux + uy = 2. 
6. uxy + aux + buy + abu = O. 
7. uxy - 2ux - 3uy + 6u = 2eX+Y• 
8. Uxx + 2auxy + a2uyy + Ux + auy = O. 
2.4. Prove that the equation with constant coefficients 
uxy + aux+ buy+ cu = 0 
is reduced to the form vxy + (c - ab) v = 0 by the change of vari-
ables u (x, y) = v (x, y) e-bx - ay• 
2.5. Prove that the general solution of the equation uxy = u has 
the form 
x 
u(x, y) = \ !(t)Jo(2iVy(x-t»dt 
" o 
Y 
+ j' g (t) Jo (2i V x (y-t») dt+ [f (0) +g(O)] Jo(2i V xy), 
o 
38 Statement of Boundary Value Proalems 
where J 0 (z) is the Bessel function of the zeroth order, and fand g 
are arhitrary functions of the Ci dass. 
2.6. Prove that the general solution of the equation u xy = F (x, y), 
where FE C (I x - Xo / < a, / y - Yo / < b), has the form 
x y 
u (x, y) = f (x) + g (y) + ~ ~ F (~, lj) dlj d~, 
Xo Yo 
where fand gare arbitrary functions of the e2 dass. 
2.7. Prove that the general solution of the equation u xy + 
+ A (x, y) U x = 0, where A (x, y) E Ci (/ x - Xo / < a, / y - Yo /< 
< b), has the form 
x y 
u (x, y) = t (!I) + ~ g (~) exp { - ~ A(~, lj) dlj} d{;, 
xo Yo 
where fand gare arbitrary functions of dasses C2 and Ci, respecti-
vely. 
2.8. Prove that the general solution of the equation 
u __ 1_ u +_1_ u =0 
xy x-.y x x-y y 
has the form u (x, y) = f (x;~: (Y), where fand gare arbitrary 
functions of the C2 class. 
2.9. Prove that the general solution of the equation 
n m 0 UXy---ux+-- u y = , x-y x-y 
where n and m are positive integers, has the form 
on+m-2 [f (x) + g (y) ] 
u (x, y) = ßxm 1 ßyn 1 x-y , 
where fand gare arbitrary functions of classes cm+1 and cn+1, re-
spectively. 
2.10. Prove that the general solution of the equation 
n m 0 UXy+--ux---uy = , x-y x-y 
where n and mare nonnegative integers, has the form 
u (x y) = (x - y)n+m+1 ßn+m [I (x) + g (Y)J 
' ßxn ßym x-y , 
where fand gare arbitrary functions of classes cn+2 and cm+2, re-
~pectively. 
2.11. Find the general solution of each of the equations below ia 
a region that preserves the type of the equation. 
Derivill~ Equations of Mathematical Physics 39 
1. YUxx + (x - y) uxy - xuYU = O. 2. x 2uxx - y2uuu = O. 
3. :r;2uxx + 2xyu:!..u - 3y2Uyy - 2xux = O. 
4. x 2uxx + 2xyuxy + y2Uyy = O. 
5. uxy - XU x + U = O. 6. U;y + 2xyuy - 2xu = O. 
7. uxy + Ux + yuu + (y - 1) U = O. 
8. uxy + XU x + 2yuy + 2xyu = O. 
Answers to Problems of Sec. 2 
1 1 
2.1. 1. u~s+unn+u~c=O; S=x, 'YJ=y-x, ~=x-2Y+T z. 
1 1 1 
2. uGS-UTlTI+u~~+UTj=O; S=T x , 'YJ= TX+Y' ~= -2 x-
-y+z. 3. UH-unTj+2u~=0; S=x+y, 'YJ=y-x, I;=y+z. 
4. U;, +uTITI = 0; S = x, 'YJ = y-x, ~= 2x- y + z. 5. ue~-uTlTI-
-U~t=O; S=x, 'YJ=Y-X, ~= ; x--}·y+-}z. 6. Utt+unn+ 
+ ucc + Un = 0; s = x, 'YJ = y - x, I; = x - 11 + z, 't = 2x -
- 2y + z + t. 7. uss - uTlTj + uc?: + un = 0; S = x+ y, 'YJ = 
= y - x, 1; = z, 't = Y + z + t. 8. Ust - UT]T] + ucc - un = 0; 
S = x + y. 'YJ = x - y, ~ = -2y + z + t, 't = Z - t. 9. Ust-
- UT]n + ut:c = 0; s = x, 'YJ = y - x, I; = 2x - y+ z, T = x+ 
+ z+ t. 
10. uss+unTl=O; s=x, 'Yj=y, 1;= -x-y+z, 't=x-y+t. 
n k n 
11. 2J USkSk = 0, Sk = 2J Xl; k = 1, 2, ... , n. 12. 2J (-1)k+1U SkSk = 
k=1 1=1 k=1 
k n 
=0, Sk= ~ XI, k= 1,2, ... , n. 13. 2J USk~k=O, St=Xt, Sh= 
1=1 k=1 n ___ 
=Xk- Xh-ll k =2,3, ... ,n. 14. ~ USkSh=O, Sh=1/k~1X 
k=1 
x (Xk-+ ~ XI): k=1, 2, ... , n. 
l<h 
n 
15. US1S1 - ~ U ShSh = 0, 
h=2 
k=2, 
40 Statement of Boundary Value Problems 
1 1 y- y--TU;+1jUTI=0'6= lxi, fJ= IYI (x>O, y>O or x<O, 
1 1 y- y-Y<O); Un+UTlTI-T U;-""1] uTI=o, 6= lxi, YI= IYI (x>o, 
y<o or x<o, y>O). 7. u,a-uTlTI+ i~ ue- iY) UTI =0, G= Ix I3/ 2 , 
3/2 ( 0' 1" 'I') = IYI x>O, y> or x<O, 'Y<Ü); U;;+ UTITI + 3~ U,+ 
+ 3~ UTI=O, G= Ix I3/ 2, fJ= IYI 3/ 2 (X>0, y<O or x<O, y>Ü). 
8. uss+UTlTI-U;-UTI=O, G=lnlxl, fJ=ln IYI (in each quadrant). 
9. u;; + uTITI + ;~ u, + 2~ UT) = 0, s = y2, fJ = x2 (in each quadrant). 
10. u'TI + 2 (1/- ~2) (fJU; - GUTI) = 0, 6 = y2 - x2, lJ = y2 + x2 (in each 
quadrant). 11. U;,+uflTl-tanhsu;=O, 6=ln(x+ Y1+x2 ), fJ= 
=ln(y+Y1+y2). 12. U;TI- 2(~~T]) (U;-UT)) + 4(;~T]) (u,+ 
+UTI)=O, 6=y2+ ell , fJ=y2_ eX(y>0 or y<O) 13. u,,+uTlTI + 
+ COS SUTI = 0, S = x, fJ = y - cos x. 14. UTlT] - 2u" S = 2x -
- y2, fJ = y. 15. UTITI - SU; = 0; S = xell, fJ = y. 
2.3. 1. I (x) + g (y). 2. I (y + ax) + g (y - ax). 3. I (x - y) + 
+ g (3x + y). 4. I (y) + g (x) e-UII• 5. x - y + I (x - 3y) + 
+ g (2x + y) e(3Y-X)/7. 6. I! (x) + g (y)1 e-bx -ay. 7. eX+Y + 
+ rt (x) + g (y)1 e3x+2Y. 8. I (y - ax) + g (y - ax) e-X. 
2.11. 1. I (x,+ y) + (x - y) g (x2 - y2)(X> -l/ or x< -y). 
2. I (xy) + Y I xy I g (x/y) (in eacft quadrant). 3. I (xy) + 
+ I xy 1314 g (x3/y) (in each quadrant). 4. f (qJ) + rg (qJ), x = 
= r cos qJ, y = r sin qJ, (x2 + y2 =1= 0). 5. xl (y) - f' (y) + 
x 
+ J (x - s) g (s) esY ds. Hint. Introducing the notation Ux = v, 
o 
find the relationships U = xv - vy, vxy - XVx = 0. 6. 2yg (x) + 
Y 
+ ! g' (x) + .\ (y - s) I (6) e-xlsdS. Hint. Introducing the nota-
o 
tion Uy = v, find the relationships u = ;x Vx + Yl~, vxy + 2xyvy = 
Y 
= 0. 7. e-Y [yl (x) + f' (x) + ~ (y - 11) g (fJ) e-XTj dl']l. Hint. In-
o 
troducing the notation, uy + U = v, find the relationships U = 
= Vx + yv, vxy + Vx + YVy+ yv = Ü. 8. e-XY [yl (x) + f' (x) + 
Y 
+ ~ (y - fJ) g (fJ) e-XTj dl']J. Hint. Introducing the notation uy + 
o + xu = v, find the relationships U = Vx + 2yv, (vy + xv)x + 
+ 2y (vl/+ xv) = 0. 
Chapter 11 
Function Spaces and Integral Equations 
3 Measurable Functions. The Lebesgue Integral 
The set E c Rn is called a set 0/ (n-dimensional) measure zero if for 
each positive e there exists a countable set of open (n-dimensional) 
cubes that cover E and whose total volume is less than e. 
Suppose Q c Rn is a region. If a certain property is valid every-
where in Q except, perhaps, on a set of measure zero, we say that 
this property is valid almost everywhere in Q. A function / (x) definedin Q is said to be measurable in Q if it is the limit of a sequence of 
functions from C (0 that converges almost everywhere in Q. If 
/ (x) = g (x) almost everywhere in Q, the two functions are said to be 
equivalent in Q. 
3.1. See whether the following sets are sets of measure zero: 
(1) a finite set of points, 
(2) a countable set of points, 
(3) the intersection of a countable set of sets of measure zero, 
(4) the union of a countahle set of sets of measure zero, 
(5) a smooth (n - l)-climensional surface, 
(6) a smooth k-dimensional surface (k ~ n - 1). 
In Problems 3.2-3.9 prove the propositions. 
3.2. Prove that the Dirichlet function X (x) (wh ich takes the value 1 
at rational points and 0 at irrational points) is zero almost every-
where. 
3.3. The function / (x) = 1- ~ x I is continuous in Rn almost 
everywhere 
3.4. The sequence of the functions In (x) = I x In converges to 
zero in the ball I x I ~ 1 alm ost everywhere. 
3.5. Theorem A set E is a set 01 measure zero i/ and only il there 
exists a cover 01 E consisting 01 a countable system 0/ open cubes with 
a finite sum 01 their volumes and such that each point 0/ E proL'es to be 
covered by an infinite set 01 cubes. 
3.6. A function that belongs to C (Q) is measurable. 
42 Function SpaC8! and Integral Equations 
3.7. If I (x) and g (x) are equivalent and g (x) is measurable in Q, 
then I (x) is measurable in Q, too. 
3.8. The limit of a sequence of measurable functions that cOBverges 
almost everywhere is a measurable function. 
3.9. A function that is continuous on Q everywhere except on a sub-
set consisting of a finite or countable number of smooth k-dimen-
sional surfaces (k::::;;; n - 1) is measurable in Q. 
3.10. Establish whether the following functions defined on [-1, 1] 
are measurable: 
{ 
sin _1_ if x =f= 0, 
(a) y=signx; (b)y= x ° if x=o; 
{ 
sign (sin~ ) if x =f= 0, 
(c) y= ° if x=o; 
fl.. if x = mln, with m and n relatively prime, 
(d) y= t n ° if x is irrational. 
3.11. Let the functions I (x) and g (x) be measurable in Q. Establish 
wh ether the functions given below are measurable. 
(a) I (x) g (x), (b) I (x)/g (x) (provided g (x) =f= 0, x E Q), 
(c) 1 I (x) I; (d) (f (x»f(x) if I (x) > 0. 
3.12. Suppose I (x) E C (Q) and at each point x of Q the function 
I (x) has a derivative, Ix,. Prove that lXi is measurable in Q. 
3.13. (a) Suppose I (x) and g (x) are measurable in Q. Prove that 
max {f (x), g (x)} and min {f (x), g (x)} are measurable in Q. 
(b) Prove that every measurable function f (x) is the differenee 
of two nonnegative measurable functions, 1+ (x) = max {t (x), O}, 
and 1- (x) = max {O, -I (x)}. 
3.14. Prove that a function that is nondecreasing (nonincreasing) 
in [a, b] is measurahle. 
3.15. Prove that if I (a:) is measurable in Q, then there is a sequence 
of polynomials that converges to I (x) almost everywhere in Q. 
We will assume that a function I (x) defined in a region Q belongs 
to the class L+ (Q) if there is a nondecreasing sequence of finite 
functions In (x), n = 1, 2, ... , that are continuous in Q such that 
it converges to I (x) alm ost everywhere in Q and such that the sequence 
of the (Riemann) integrals J In (x) da: is bounded from above. The 
Q 
Measurable Functions. The Lebesgue Integral 
Lebesgue integral 01 I (x) E L+ (Q) is then defined thus: 
(L) ~ fdx=sup ~ Indx= lim ~ Indx. 
Q n Q n-+oo Q 
43 
A funetion f (x) is said to be Lebesgue integrable over a region Q 
if it ean be represented in the form of the differenee I (x) = fl (x) -
- 12 (x) of two funetions fl (x) and /2 (x) that belong to L+ (Q). The 
Lebesgue integral of f (x) is then defined as follows: 
(L) \ fdx=(L) \ Itdx-(L) \ fzdx. 
Q Q Q 
A eomplex-valued funetion f (x) = Re f (x) + i Im f (x) is said 
to be Lebesgue integrable over a region Q if the funetions Re f (x) 
and Im f (x) are Lebesgue integrable. By definition we assume that 
(L) ~ f dx = (L) ~ Re I dx + i (L) ~ Im f dx. 
Q Q Q 
We denote the set of eomplex-valued funetions that are Lebesgue 
integrable over a domain Q and are identified if they are equivalent 
by LI (Q). 
Funetions that belong to LI (Q) are finite almost everywhere in Q. 
If a funetion is Riemann integrable, it is Lebesgue integrable, too, 
and the Riemann and Lebesgue integrals eoineide. Beeause of this 
we will drop the (L) in front of the integral sign; an integral will be 
always understood as a Lebesgue integral and the integrand as a 
Lebesgue integrable funetion. Moreover, if a funetion is an absolutely 
improperly Riemann integrable, it is Lebesgue integrable, too, and 
the Riemann and Lebesgue integrals eoineide. 
The following theorems play an important role in the theory of 
Lebesgue integrable funetions: 
(a) if a lunetion / (x) is measurable in a region Q and I / (x) I ~ 
~ g (x), where g (x) E LI (Q), then I (x) belongs to LI (Q). In partieu-
lar, a measurable bounded funetion belongs to LI (Q) in a bounded 
region Q. 
(b) Lebesgue' s theorem 1/ a sequenee 0/ funetions that are measur-
able in Q, say 11 (x), /2 (x), ... , fn (x), ... , n = 1, 2, ... , eon-
verges to a lunetion f (x) almost everywhere in Q and if I In (x) I ~ g (x), 
where g belongs to LI (Q), then / also belongs to LI (Q) and ~ fn (x) dx--+ 
--+ .\ I dx as n --+ 00. 
Q 
Q 
(e) Fubini's theorem 1/ f (x, y) E LI (Q X P), x = (Xl' ... , xn ) E 
E Q and y = (YI' ... , Ym) E P, where Q and P are regions, then 
j / (x, y) dx E LdP ), ~ f (x, y) dy E 1 (Q) 
V P 
44 Function Spaces and Integral Equations 
and 
J f (x, y) dx dy = ~ dx .\ f (x, y) dy = ~ dy ~ f (x, y) dx. 
QxP Q P P Q 
If f (x, y) is measurable in Q X P and for almost all x E Q the 
function I f (x, y) I E LI (P) and ~ I f (x, y) I dy E LI (Q), then 
P 
f (x, y) E LI (Q X P). 
In Problems 3.16-3.20 prove the validity of the propositions. 
3.16. If f (x) ;;:. 0 and ~ f (x) dx = 0, then f (x) = 0 almost every-
Q 
where in Q. 
3.17. If f (x) = 0 almost everywhere in Q, then ~ t dx = O. 
Q 
3.18. If f, gELl (Q), then af + ßg E LI (Q) for all constant a 
and ß. 
3.19. If fE" LI (Q), then I f I E LI (Q) and 
l~/dxl~i I/Idx. 
Q Q 
3.20. If fELl (Q), then for every positive e there exists a finite 
function ge E C (Q) such that .I 1/- ge I dx< e. 
Q 
3.21. Prove that the Dirichlet function 
{ 1 for x rational, 
f (x) = 0 for x irrational 
is Lebesgue integrable on [0, 1] but is not Riemann integrable. 
What is its Lebesgue integral equal to? 
3.22. Find the integral over the segment [0, 1] of each of the functions 
given below (first prove the finiteness of the integral): 
{ 
X2 for x irrational, 
(a) f (x) = 0 for x rational; 
{ 
X2 for x irrational and greater than 113, 
(b) fex) = x3 for x irrational and less than 1/3, 
o for x rational; 
{ 
sin:rtx for x irrational and less than 1/2, 
(c) f (x) = x2 for x irrational and greater than 1/2, 
o if x is rational; 
Measurablc Functions. The Lebesgue Integral 45 
(d) I(X)={ 
(e) 1 (x) = { 
11n for x = m/n, where m and n are relatively prime, 
o for x irrational; 
x- 1/ 3 for x irrational, 
x' for x rational; 
(f) 1 (x) = sign (sin ;). 
3.23. At which values of aare the following functions integrable 
over the ball 1 xl< 1: 
(a) I(x)= Ix1l a ; (b) I(X)=(1_11XI)a; (e) l(x)=XIX~~·I~Xn. 
3.24. Let g (x) be a measurable and bounded function in a region Q. 
Show that the function 1 (x) = \ r< (~) ds belongs to eh (Rn) at 
J I x-~ la 
Q 
k<n-[a]. 
3.25. Suppose 1 belongs to LI (Q). Show thl1t the function 
_ { 1 (x) if 1I (x) 1 < N at point x, 
IN(X)- N if I/(x)I~N at point x 
is integrable over Q and that 
lim ) IN (x) dx = ) 1 (x) dx. 
f'tT-jo-OO Q Q 
3.26. Suppose Q = (0 < Xl < 1, 0 < X 2 < 1) and let 1 (x) be 
defined in Q in the following manner: 
x _ { Xl X2/1xl 4 for (XI' x2 ) =1= (0, 0), 
(a) 1 ( ) - 0 for XI = X2 = 0; 
{ 
Xli-I:~ for (x,x2) =1= (0, 0), 
(b) 1 (x) = x ° for Xi = X 2 = 0; 
f 
I 
(c) 1 (x) = 1 
I 
l 
1 
-2 for 0 < x, < X2 < 1, 
X 2 
1 
- -2 for ° < X2 < XI < 1, 
Xl 
° at other points. 
(1) Do these