Freiling G., Yurko V. Lectures on Differential Equations of Mathematical Physics: A First Course
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222 G.
Freiling and V. Yurko
Step 2:
Proof of the convergence of the series (5.10). We use Lemma 5.1. Choose
M > 0 and a > 0 such that a function F of the form (5.5) is simultaneously a majorant for
all coefficients a
i jk
, b
i j
, c
i
of system (5.7). We consider the so-called majorizing system
∂v
i
∂t
= F(t,x
1
,
.
.. ,x
n
)
Ã
N
∑
j=1
n
∑
k=1
∂v
j
∂x
k
+
N
∑
j=1
v
j
+ 1
!
, (5.11)
which
is
obtained from (5.7) by replacing all coefficients a
i jk
, b
i j
, c
i
by their majorant F.
We will seek a solution of system (5.11) in the form
v
1
= . .. = v
N
= v(z), where z =
t
α
+ x
1
+ .
.
. + x
n
. (5.12)
Substituting (5.12) into (5.11) we obtain for v(z) the following ordinary differential equa-
tion
1
α
d
v
d
z
= F(
z)
µ
Nn
dv
d
z
+ N
v + 1
¶
,
where
F(z) :=
M
1 −z/a
.
Therefore,
d
v
d
z
= B
(z)(Nv + 1), (5.13)
where
B(z) :=
F(z)
1/α −N
nF(z
)
.
Choose α ∈ (0,1] such that 1/α −NnF(z) > 0 in a neighbourhood of the point z = 0.
Then in this neighbourhood B(z) is analytic. A particular solution of equation (5.13) has
the form
v(z) =
1
N
³
e
D(z)
−1
´
, D(z) := N
z
0
B(ξ)dξ. (5.14)
The
function v
(z) of the form (5.14) is analytic at the point z = 0, and all its Taylor
coefficients are positive. Indeed, the Taylor coefficients of F(z) are positive:
F(z) = M
∞
∑
k=0
z
k
a
k
.
Furthermore,
B(z)
= α
F(z)
∞
∑
s=0
(αNnF(z))
s
,
and consequently, the Taylor coefficients of the function B(z) are positive as well. Ob-
viously, the functions D(z) and e
D(z)
−1 possess the same property. Therefore, all the
Taylor coefficients of the function v(z) are positive.
Thus, system (5.11) has in a neighbourhood of the origin an analytic solution of the
form
v
i
(t, x
1
,. .. ,x
n
) =
∞
∑
k
0
,...,k
n
˜
α
i
k
0
,...,k
n
t
k
0
x
k
1
1
.. .x
k
n
n
, (5.15)
The Cauch
y-Kowalevsky Theorem 223
where
˜
α
i
k
0
,..
.,k
n
> 0, and the series (5.15) converges absolutely in a neighbourhood of the
origin.
On the other hand, the coefficients
˜
α
i
k
0
,...,k
n
can be calculated by the A-procedure, using
system (5.11) and the initial conditions v
i|t=0
= v(x
1
+...+ x
n
), in the same way as the co-
efficients α
i
k
0
,...,k
n
were calculated by the A-procedure, using (5.7) and the initial conditions
(5.9). We note that for calculating the coefficients α
i
k
0
,...,k
n
and
˜
α
i
k
0
,...,k
n
in the A-procedure
we use only the operations of addition and multiplication. Since F is a majorant for the
coefficients of system (5.7), and the initial data are majorants for the initial data (5.9), we
conclude that
|α
i
k
0
,...,k
n
| ≤
˜
α
i
k
0
,...,k
n
.
Therefore, the series (5.10) converges absolutely in a neighbourhood of the origin and gives
us the solution of the Cauchy problem (5.7), (5.9). Theorem 5.1 is proved. 2
Remark 5.1. The Cauchy-Kowalevsky theorem is also valid for a wide class of non-
linear systems of the form
∂
n
1
u
1
∂t
n
1
= F
1
³
t,x
1
,
.
.. ,x
n
,u
1
,. .. ,u
N
,. .. ,
∂
k
u
1
∂t
k
0
∂x
k
1
1
.
.
.∂x
k
n
n
,. ..
´
,
.. . . .. . ..
∂
n
N
u
N
∂t
n
N
= F
N
³
t,x
1
,
.
.. ,x
n
,u
1
,. .. ,u
N
,. .. ,
∂
k
u
1
∂t
k
0
∂x
k
1
1
.
.
.∂x
k
n
n
,. ..
´
,
(5.16)
where k
0
+ . .. + k
n
≤ n
j
, k
0
< n
j
. Such systems are called normal systems or the
Kowalevsky systems. For example, the equation of a vibrating string
∂
2
u
∂t
2
=
∂
2
u
∂x
2
has
the
form (5.16), but the heat conduction equation
∂u
∂t
=
∂
2
u
∂x
2
has
not
the form (5.16).
Remark 5.2. For systems, which have not the form (5.16), the Cauchy-Kowalevsky
theorem, generally speaking, is not valid. Example:
u
t
= u
xx
,
u
|t=0
=
1
1 −x
, |x| < 1.
(5.17)
If
an
analytic solution of the Cauchy problem (5.17) exists, then it must have the form
∞
∑
n=0
(2n)!t
n
n!(1 −x)
2n+1
.
Ho
we
ver, for t 6= 0 this series is divergent.
Chapter
6.
Exer
cises
This chapter contains exercises for the course ”Differential Equations of Mathematical
Physics”. The material here reflects all main types of equations of mathematical physics
and represents the main methods for the solution of these equations.
6.1. Classification of Second-Order Partial Differential Equa-
tions
In order to solve the exercises in this section use the theory from Section 1.2.
6.1.1. Reduce the following equations to the canonical form in each domain where the
equation preserves its type:
1. u
xx
−4u
xy
+ 5u
yy
+ u
x
−3u
y
+ 6u + 2x = 0 ;
2. u
xx
−6u
xy
+ 9u
yy
+ 7u
x
−3 = 0 ;
3. 2u
xx
−4u
yy
+ 3u
x
−5u
y
+ x + 1 = 0;
4. 3u
xx
+ u
xy
+ 3u
y
−5u + y = 0;
5. 5u
xx
+ 16u
xy
+ 16u
yy
−u
x
+ 3u
y
+ 9u + x + y −2 = 0;
6. u
xx
−yu
yy
= 0;
7. u
xx
−xu
yy
= 0;
8. u
xx
signy + 2u
xy
+ u
yy
= 0;
9. xu
xx
−yu
yy
= 0;
10. yu
xx
−xu
yy
= 0;
11. x
2
u
xx
−y
2
u
yy
= 0;
12. x
2
u
xx
+ y
2
u
yy
= 0;
226 G.
Freiling and V. Yurko
13. y
2
u
xx
+ x
2
u
yy
= 0;
14. y
2
u
x
x
−x
2
u
yy
= 0;
15. (1 + x
2
)
2
u
xx
+ u
yy
+ 2x(1 + x
2
)u
x
= 0;
16. (1 + x
2
)u
xx
+ (1 + y
2
)u
yy
+ xu
x
+ yu
y
−2u = 0;
17. x
2
u
xx
+ 2xyu
xy
+ y
2
u
yy
−2yu
x
+ ye
y/x
= 0;
18. u
xx
sin
2
x −2yu
xy
sinx + y
2
u
yy
= 0;
19. 4y
2
u
xx
−e
2x
u
yy
= 0;
20. x
2
u
xx
+ 2xyu
xy
−3y
2
u
yy
−2xu
x
+ 4yu
y
+ 16x
4
u = 0;
21. y
2
u
xx
−
2
yu
xy
+
u
yy
=
0;
22. xy
2
u
xx
−2x
2
yu
xy
+ x
3
u
yy
−y
2
u
x
= 0;
23. u
xx
−2 sinx u
xy
−cos
2
x u
yy
−cos xu
y
= 0;
24. xu
xx
+ 2xu
xy
+ (x −1)u
yy
= 0.
Solution of Problem 1. The equation is elliptic in the whole plane since
a
2
12
−a
11
a
22
= 2
2
−1 ·5 = −1 < 0.
The characteristic equation has the form
dy
d
x
= −2 ±i
,
and its general integral is
y + 2x ±ix = C.
The change of variables
ξ = 2x + y, η = x
yields
u
x
= 2u
ξ
+ u
η
, u
y
= u
ξ
,
u
xx
= 4u
ξξ
+ 4u
ξη
+ u
ηη
,
u
xy
= 2u
ξξ
+ u
ξη
, u
yy
= u
ξξ
.
Therefore, we reduce the equation to the canonical form
u
ξξ
+ u
ηη
−u
ξ
+ u
η
+ 6u + 2η = 0.
Solution of Problem 6. One has a
2
12
−a
11
a
22
= y ; hence the equation is hyperbolic for
y > 0 , it is elliptic for y < 0 , and y = 0 is the line where the equation is parabolic. In the
Exercises 227
domain y > 0, general inte
grals of the characteristic equation have the form x ±2
√
y = C .
The change
of variables
ξ = x + 2
√
y, η = x −2
√
y
yields
u
x
= u
ξ
+ u
η
, u
y
=
(u
ξ
−u
η
)
1
√
y
,
u
x
x
= u
ξξ
+ 2
u
ξη
+ u
ηη
,
u
yy
= (u
ξξ
−2u
ξη
+ u
ηη
)
1
y
−(u
ξ
−u
η
)
1
2y
√
y
,
ξ −η = 4
√
y.
Hence,
we
get the following canonical form:
u
ξη
+
1
2(ξ −η)
(u
ξ
−u
η
)
= 0
.
In the domain y < 0, the change of variables
ξ = x, η = 2
√
−y
yields
u
x
= u
ξ
, u
y
= −
1
√
−y
u
η
= −
2
η
u
η
,
u
x
x
= u
ξξ
, u
yy
= −
1
y
u
ηη
+
1
2y
√
−y
u
η
.
Hence,
we
get the following canonical form:
u
ξξ
+ u
ηη
−
1
η
u
η
= 0.
6.1.2. Reduce
the
following equations with constant coefficients to the canonical form,
and carry out further simplifications:
1. 2u
xx
+ 2u
xy
+ u
yy
−4u
x
+ u
y
+ u = 0;
2. 3u
xx
−5u
xy
−2u
yy
+ 3u
x
+ u
y
= 2;
3. u
xx
+ 2u
xy
+ u
yy
+ u
x
+ 2u
y
+ u = 0;
4. u
xx
+ u
xy
−2u
yy
+ 9u
x
−3u + x = 0;
5. u
xx
+ 4u
xy
+ u
yy
−3u
y
+ u = 0;
6. 4u
xx
+ 4u
xy
+ u
yy
−5u
x
+ 3u
y
−6u = 0;
7. u
xx
−6u
xy
+ 10u
yy
+ 7u
x
−5u
y
+ 2u = 0;
8. u
xx
−2u
xy
−3u
yy
+ 2u
x
−u
y
−9u = 0;
228 G.
Freiling and V. Yurko
9. u
xx
+ 2
u
xy
−3u
yy
−5u
x
+ u
y
−2u = 0;
10. u
xx
+ 4u
xy
+ 5u
yy
−u
y
+ 3u + 2x = 0;
11. u
xx
+ 4u
xy
+ 13u
yy
+ 3u
x
+ 24u
y
−9u + 9(x + y) = 0;
12. u
xx
−2u
xy
+ u
yy
+ u
x
−7u
y
−9u = 0;
13. u
xx
−u
yy
+ u
x
+ u
y
−4u = 0;
14. 9u
xx
−6u
xy
+ u
yy
−8u
x
+ u
y
+ x −3y = 0;
15. u
xy
+ 2u
yy
−u
x
+ 4u
y
+ u = 0;
16. u
xy
+ u
xx
−u
y
−10u + 4x = 0;
17. u
xx
−2u
xy
+ u
yy
−3u
x
+ 12u
y
+ 27u = 0.
Hint. After the reduction to the canonical form go on to a new unknown function
v(ξ,η) by the formula u(ξ, η) = e
λξ+µη
v(ξ,η) . Choose the parameters λ and µ such that
the terms with first derivatives or with the unknown function are absent.
Solution of Problem 1. The equation is elliptic in the whole plane since
a
2
12
−a
11
a
22
= 1
2
−2 ·1 = −1 < 0.
The characteristic equation has the form
dy
d
x
=
1
2
±
1
2
i
and
its
general integral is
y −
1
2
x ±
1
2
ix = C.
The
change
of variables
ξ = y −
x
2
, η =
x
2
reduces
the
equation to the canonical form
u
ξξ
+ u
ηη
+ 6u
ξ
−4u
η
+ 2u = 0.
Using the replacement u(ξ, η) = e
λξ+µη
v(ξ,η) we obtain the following equation with re-
spect to v(ξ,η) :
v
ξξ
+ v
ηη
+ (2λ + 6)v
ξ
+ (2µ −4)v + (λ
2
+ µ
2
+ 6λ −4µ + 2)v = 0.
Taking λ = −3 , µ = 2 , we get
v
ξξ
+ v
ηη
−11v = 0.
Exercises 229
6.1.3. Reduce the
following equations to the canonical form:
1. 2u
xx
+ u
yy
+ 6u
zz
+ 2u
xy
−2u
yz
= 0;
2. 5u
xx
+ 2u
yy
+ u
zz
+ 4u
xy
+ 2u
yz
= 0;
3. 3u
yy
−2u
xy
−2u
yz
+ 7u = 0;
4. 4u
xx
+ u
yy
+ u
zz
+ 4u
xy
+ 4u
xz
+ 2u
yz
+ 5u = 0;
5. 4u
xx
−2u
yz
−4u
xy
+ u
y
+ u
z
= 0;
6. 9u
xx
+ 4u
yy
+ u
zz
+ 12u
xy
+ 6u
xz
+ 4u
yz
−6u
x
−4u
y
−2u
z
= 0;
7. u
tt
+ 3u
xx
+ 2u
yy
+ 2u
zz
+ 2u
ty
+ 2u
yz
+ 2u
xy
= 0;
8. u
tt
+ 2u
zz
+ u
xy
−2u
zt
−u
xz
= 0;
9. u
xx
+ u
zz
+ 2u
xy
−2u
xz
−4u
yz
+ 2u
yt
= 0.
6.1.4. Find the general solutions for the following equations:
1. 2u
xx
−5u
xy
+ 3u
yy
= 0;
2. 2u
xx
+ 6u
xy
+ 4u
yy
+ u
x
+ u
y
= 0;
3. 3u
xx
−10u
xy
+ 3u
yy
−2u
x
+ 4u
y
+
5
16
u = 0;
4. 3u
x
x
+ 10u
x
y
+ 3u
yy
+ u
x
+ u
y
+
1
16
u −16xe
xp
µ
−
x + y
16
¶
= 0;
5. u
yy
−2u
x
y
+ 2
u
x
−u
y
= 4e
x
;
6. u
xx
−6u
xy
+ 8u
yy
+ u
x
−2u
y
+ 4 exp
µ
5x +
3y
2
¶
= 0;
7. u
x
x
−2
cosxu
xy
−(3 +sin
2
x)u
yy
+ u
x
+ (sin x −cosx −2)u
y
= 0;
8. e
−2x
u
xx
−e
−2y
u
yy
−e
−2x
u
x
+ e
−2y
u
y
+ 8e
y
= 0;
9. u
xy
+ yu
y
−u = 0.
6.2. Hyperbolic Partial Differential Equations
Hyperbolic partial differential equations usually describe oscillation processes and give a
mathematical description of wave propagation. The prototype of the class of hyperbolic
equations and one of the most important differential equations of mathematical physics is
the wave equation. Hyperbolic equations occur in such diverse fields of study as electro-
magnetic theory, hydrodynamics, acoustics, elasticity and quantum theory. In order to solve
exercises from this section use the theory from Chapter 2.
230 G.
Freiling and V. Yurko
1. Method
of travelling waves
6.2.1. Solve the following problems of oscillations of a homogeneous infinite string
(−∞ < x < ∞) if a = 1 in (2.1.1), and the initial displacements and the initial velocities
have the form:
1. u(x,0) = sin x, u
t
(x,0) = 0;
2. u(x,0) = 0, u
t
(x,0) = A sinx, A −const;
3. u(x,t
0
) = ϕ(x), u
t
(x,t
0
) = ψ(x), t > t
0
;
4. u(x,0) = x
2
, u
t
(x,0) = 4x.
6.2.2. Find the solution of the equation
u
xx
+ 2u
xy
−3u
yy
= 0, y > 0, −∞ < x < ∞,
satisfying the initial conditions
u(x,0) = 3x
2
, u
y
(x,0) = 0.
Solution. The equation is hyperbolic in the whole plane since
a
2
12
−a
11
a
22
= 1 + 3 = 4 > 0,
The characteristic equations have the form
dy
d
x
= −1
,
dy
d
x
= 3
.
The change of variables
ξ = x + y, η = 3x −y
reduces the equation to the canonical form
u
ξη
= 0.
The general solution of this equation is u = f (ξ) + g(η) , and consequently,
u(x,y) = f (x + y) + g(3x −y),
where f and g are arbitrary twice continuously differentiable functions. Using the initial
conditions we calculate
f (x) + g(3x) = 3x
2
, f
0
(x) + g
0
(3x) = 0,
hence
f (x) =
3x
2
4
, g(x)
=
x
2
4
.
Exercises 231
This yields
u
(x,y) = 3x
2
+ y
2
.
6.2.3. Find the solution of the equation
4y
2
u
xx
+ 2(1 −y
2
)u
xy
−u
yy
−
2y
1 + y
2
(2u
x
−u
y
)
= 0
,
satisfying the initial conditions
u(x,0) = ϕ
0
(x), u
y
(x,0) = ϕ
1
(x).
6.2.4. Find the solution of the equation
u
xx
+ 2 cosx u
xy
−sin
2
x u
yy
−sin xu
y
= 0, y > 0, −∞ < x < ∞,
satisfying the initial conditions
u
|y=sinx
= ϕ
0
(x), u
y|y=sinx
= ϕ
1
(x).
6.2.5. Find the solution of the equation
u
xx
+ yu
yy
+
1
2
u
y
= 0, y > 0, −∞ < x < ∞,
satisfying
the
conditions
u(x,0) = f (x), u
y
(x,0) = 0.
6.2.6. Solve the following Cauchy problems in the domain t > 0, −∞ < x < ∞ :
1. u
tt
= u
xx
+ 6, u(x,0) = x
2
, u
t
(x,0) = 4x;
2. u
tt
= u
xx
+ xt, u(x,0) = x, u
t
(x,0) = sin x;
3. u
tt
= u
xx
+ sin x, u(x, 0) = sin x, u
t
(x,0) = 0;
4. u
tt
= u
xx
+ sin x, u(x, 0) = 0, u
t
(x,0) = 0;
5. u
tt
= u
xx
+ x sint, u(x,0) = sin x, u
t
(x,0) = cos x;
6. u
tt
= u
xx
+ e
x
, u(x, 0) = sin x, u
t
(x,0) = x +cosx;
7. u
tt
= u
xx
+ e
−t
, u(x, 0) = sin x, u
t
(x,0) = cos x.
6.2.7. Using the reflection method (see Section 2.1) solve the following problems on
the half-line:
1. u
tt
= a
2
u
xx
, u
x
(0,t) = 0, u(x, 0) = ϕ(x), u
t
(x,0) = ψ(x);
2. u
tt
= a
2
u
xx
, u
x
(0,t) = u(x,0) = 0, u
t
(x,0) =
0, 0 < x < l,
v
0
, l < x < 2l,
0, 2l < x < ∞;