Chapter 5
Hyperbolic Equations with Two Space Variables 2
5.1. Wave Equation
=
2
2
2
5.1.1. Problems in Cartesian Coordinates The wave equation with two space variables in the rectangular Cartesian system of coordinates has the form 2
2
2
2
=
2
+
2
2
.
5.1.1-1. Particular solutions and some relations. 1 . Particular solutions:
( , , )=
( , , )=
( , , )=
( , , )=
( , , )=
exp
sin(
1
sin(
1
sinh( sinh(
1,
2
+
1 ) sin( 2
+
1 ) sin( 2
+
1
12 + 22 ,
+
2 ) sin
+
2 ) cos
+
1 ) sinh( 2
+
1
( , , ) = ( sin +
where ,
+
1
1 ) sinh( 2
+
12 + 22 ,
2 ) sinh
2 ) cosh
cos + ) + ( sin +
12 + 22 , 12 + 22 , 12 + 22 , cos − ),
2 , 1 , 2 , and are arbitrary constants, and ( ) and ( ) are arbitrary functions.
2 . Particular solutions that are expressed in terms of solutions to simpler equations:
( , , ) =
( , , )=
( , , ) =
( , , )=
cosh(
) + sin(
cos(
) + sinh(
) ( , ),
where
) ( , ),
where
cos( ) + sin( ) $ ( , ),
where
cosh( ) + sinh( ) ( , ),
( , , ) = exp
2'
( , ( ), ( =
where
)
2
,
where
!"! !"! # # # # *
= 2 =
2
# # # #
+
%%
+
%% # #
='
− 2 2 , +
2 2
(1)
,
(2)
2
(3)
= −( & ) , 2
= ( & ) ,
(4)
.
(5)
For particular solutions of equations (1) and (2) for the function ( , ), see the Klein–Gordon
equation 4.1.3. For particular solutions of equations (3) and (4) for the function ( , ), see Subsection 7.3.2. For particular solutions of the heat equation (5) for the function ( , ( ), see Subsection 1.1.1.
© 2002 by Chapman & Hall/CRC Page 341
3 . Fundamental solution: +
( , , )= where - =
2
+
,
(
−- )
2. /
2 2
−-
2
, ,
1 for ≥ 0, 0 for < 0,
( ) = 0
2.
4 . Infinite series solutions that contain arbitrary4 functions of the space variables: 4
354
( , , ) = 1 ( , )+ 2 =1
9 8 53 4 ( , , ) = ( , )+ 2
8
( )2 (26 + 1)! 7
=1
( )2 4 1 ( , ), (26 )! 7 4 8
7
≡
2 2
+
2 2
,
( , ),
where 1 ( , ) and ( , ) are any infinitely functions. The first solution satisfies
! differentiable
( , , 0) = 1 ( , ), the initial conditions to8 the
!
8 ( , , 0) = 0 and the second solution
initial conditions ( , , 0) = 0, ( , , 0) = ( , ). The sums are finite if 1 ( , ) and ( , ) are bivariate polynomials.
: ) ( , , ) = Im = (> ). and (6)
Here, = (> ) is an arbitrary analytic function of the complex argument > related to the variables ( , , ) by the implicit relation
− ( − 0 )> + ( − 0 ) / 1 − > 2 = ? (> ), (7)
where ? (> ) is any analytic function and 0 , 0 are arbitrary constants. Solutions of the forms (6), (7) find wide application in the theory of diffraction. If the argument > obtained by solving (7) with a prescribed ? (> ) is real in some domain @ , then one should set Re = (> ) = = (> ) in relation (6) everywhere in @ .
: 0, > 0, 0
= 0).
# 5.3.3-3. Domain: 0 ≤ õ ≤ ù , 0 ≤ ≤ ' . Third boundary value problem.
A circular cylinder of finite length is considered. The following conditions are prescribed: " Y "
= ñ
= " ñ " + 1 =[ . " − 2" = [ . " + 3" = [
#
0 (õ
, ) at
ö =0
(initial condition),
1 (õ # 1( 2 (õ 3 (õ
, ) ,ö ) ,ö ) ,ö )
ö =0 õ =ù #
(initial condition), (boundary condition), (boundary condition), (boundary condition).
#
at at at at
#
=0 ='
#
The solution " (õ , , ö ) is determined by the formula in Paragraph 5.3.3-2 where
# 2 # õ ò ( 2ò 0
& (õ , , ò , ó , ö ) = ù 2 » 0 ù ù 2 ù 2 + 2 ) 2( ) 0 » =1 =1 ( 1 0 2 2 # # ø 2 ç = ù 2 + ø 21 2 + þ , ( ) = cos(1 ) + 1 sin(1 1 2 + 2 0
0 2 ' 2 2 2= 1 3 + 1 2 + 1 + 1 22 . 1 2 2 2 + 3 2 2 2
) (ó ) sin ügö ý ç ÿ 0
#
2
ý ç
,
),
Here, the and 1 are positive roots of the transcendental equations ù 1 ( ) − 1 0 ( ) = 0,
tan(1 ' ) 1
= 1
2+ 3 . 2− 2 3
# 5.3.3-4. Domain: 0 ≤ õ ≤ ù , 0 ≤ ≤ ' . Mixed boundary value problems.
1 2 . A circular cylinder of finite length is considered. The following conditions are prescribed: #
"
= 3 0 (õ , ) at
ö =0
(initial condition),
= 3 1 (õ , ) at " = [ ( # , ö ) at 1
ö =0 õ =ù #
(initial condition), (boundary condition),
Y " . " . "
#
= [ 2 (õ , ö ) = [ 3 (õ , ö )
at at
#
=0 ='
(boundary condition), (boundary condition).
© 2002 by Chapman & Hall/CRC Page 373
Solution:
" (õ , # , ö ) =
+
2
−;
2
+;
2
Here, # 26 A? @ & (8 , , 6 , 7 , 9 ) = > 2 » '
? =1 @
»
B =0
# 3 1 (6 , 7 ) & (8 , , 6 , 7 , 9 ) : 6 : 7
4 0 ( 4 0Y 5
−;
+
# 3 0 (6 , 7 ) & (8 , , 6 , 7 , 9 ) : 6 : 7
ö 4 0( 4 05
4 0 4 0( Y
[ 1 (7 , < ) )
# & ( 8 , , 6 , 7 , 9 − < )* + 6
# [ 2 (6 , < ) & (8 , , 6 , 0, 9 − < ) : 6 :
0,
for 2 for
= ( G 22 - 22; + = 2 - 22 − 2 ) I 2 (= - 2 ) − ( G 12 - 12 + = 2 - 12 − 2 ) I 2 (= ( ( I (= ) = = < > (= - 1 ) − G 1 < (= - 1 )6LK (= )
H
5
(
− = K > (= 5
(
1)
− G 1 K (= -
< (=
1 )6
(
1 ),
),
where the < ( ) and K ( ) are the Bessel functions, and the = are positive roots of the transcendental equation 5
= < > (= -
1)
− G 1 < (= -
1 )6
5
= K > (= -
+ G 2 K (= -
2)
= = K > (= 5
5.4.2-7. Domain: 0 ≤
(
≤- ,0≤
)
≤
)
0.
1)
2 )6
− G 1 K (= -
1 )6
5
= < > (= -
2)
+ G 2 < (= -
2 )6
.
First boundary value problem.
A circular sector is considered. The following conditions are prescribed: &
(
)
(
)
= . 0 ( , ) at
=0
(initial condition),
= . 1 ( , ) at & = 0 ( ) , ) at 1
=0
=-
(initial condition), (boundary condition),
=0 ) = 0
(boundary condition), (boundary condition).
% /&
(
& &
= 0 2( , ) ( = 0 3( , )
at at
(
)
)
© 2002 by Chapman & Hall/CRC Page 384
Solution: & (( , ) , ) = %
%
03
1
0M 0
+ 1
( ) . 0 ( , ) ( , , , , ) 4 4
0
1
0
2
−# +#
2
−#
2
+
1 1
0
/ 1
1
1
0
0M
(
4
)
( , , , , ) =
) 2 -
0
1
% %
)
(
4 4 7
=
)
( , , , , − 7 )9 N
8 %
(
( , , , , − 7 )9 : (
8 %
)
( , , , , ) 4 4
34 4 7
=0
)
( , , , , − 7 )9 N
=
4 4 7 0
( ) + ( , , 7 ) ( , , , , − 7 ) 4 4 M 4 7 .
03
1
exp − 12
0 3 ( , 7 )
0
Here,
1
0 2 ( , 7 )
03
1
0
/
0M
03
0/
1
1
%
0 1 ( , 7 ) 8 %
(
. 1 ( , ) + . 0 ( , )6
03/ 5
0M 1
< 2 O
=1
(
) < 2 O 0 (= ) [ < > 2 O 0 (= - )]2 M M ) sin ? , # 2 = 2 + $ − 2E 4 M × sin ' ) * sin ' ) * , 0 0 , # 2 = 2 + $ − 2E 4
=1
(
(=
0
where the < 2 O 0 ( ) are the Bessel functions and the = equation < 2 O 0 (= - ) = 0.
are positive roots of the transcendental
M
M
(
5.4.2-8. Domain: 0 ≤
)
≤- ,0≤
)
≤
0.
Second boundary value problem.
A circular sector is considered. The following conditions are prescribed: &
% /& % C & (
−1 %
&
(
−1 %
&
Solution:
M
)
(
)
=0
(initial condition),
= . 1 ( , ) at ) = 0 1 ( , ) at
=0
(initial condition), (boundary condition),
(
= 0 2( , )
at
= 0 3( , )
at
(
M (
(
= . 0 ( , ) at
)
& ( , , ) = %
%
+ 1
0M
−#
2
+#
2
03/
1
5
1/ 1
0
0/ 1
1/
0
1
0M
0
1
0M
03
1
=
)
(boundary condition), 0
(boundary condition).
(
)
( , , , , ) 4 4
( ) 0 1 ( , 7 ) ( , , - , , − 7 ) 4 4 7
)
( ) 0 3 ( , 7 ) ( , , ,
0
1
=0
( ) 0 2 ( , 7 ) ( , , , 0, − 7 ) 4 4 7
03
1
)
. 1 ( , ) + . 0 ( , )6 0
+ # 2-
+
03
1
0M 0
=)
( ) . 0 ( , ) ( , , , , ) 4 4
0
1
(
03
0,
−7 )4 4 7
( ) + ( , , 7 ) ( , , , , − 7 ) 4 4 4 7 .
© 2002 by Chapman & Hall/CRC Page 385
Here,
(
)
( , , , , ) = exp
2 sin ? , $ − 2E 4 ) D +4 ) 2 2 E 4 0, $ −
− 21
× cos '
)
0
)
=0 =1
' * cos
0
(
< 2 O ( ) are the Bessel functions and the = 0 > 2 O 0 (= - ) = 0. M
where the equation
0,
Í
, 0 ≤ ≤ . Third boundary value problem.
A circular cylinder of finite length is considered. The following conditions are prescribed: Ö Ù ÚÖ Ù Û Ö Ù ß Ö
+ Ü 1Ö
= × 0 (Õ , Ø ) at = × 1 (Õ , Ø ) at
= Ý 1(Ø , Ï ) − Ü 2 Ö = Ý 2 (Õ , Ï ) Ù ß Ö + Ü 3 Ö = Ý 3 (Õ , Ï )
Ï =0 Ï =0
(initial condition), (initial condition),
at
Õ =Þ
(boundary condition),
at at
Ø =0 Í
(boundary condition), (boundary condition).
Ø = Ö Ã Ã Ã Ê Ê The solution (Õ , Ø , Ï ) is determined by the à formula in Paragraph 5.4.3-2 where Ê ÂÃ Â Ä ( Ø ) Ä ( ) sin ¼?Ï ¿ Ã Ð Ä ½ à Æ Æ Ç Õ Ç 2 ã â ã Î Á É É â (Õ , Ø , , , Ï ) = 2 exp ¼ − 12 á Ï ½ Á . 0È 0È Ä 2 ¿ Ð Ä Þ Þ Þ Î =1 Ä =1 â Å Ã Ã Here, à à à à 2 2 2 Ç 2 2 å Ä2 á , Ð Ä = ä Ç = 2 2 + + Ò − , Æ ä Þ 2 4 ( Ü 1 Þ + Ç 2 ) 02 (Ç ) Í å Å 2 2 ã ã Ä Ü 3 +Ü 2 Ü 2 Ü 22 å Ä Ø ) + Ü 2 sin(å Ä Ø ), 2 Ä Ä ( Ø ) = cos( = + + È 1+ å 2 É ; à â å Ä å å å 2 2 2 2 Ä Ä +Ü 3 â 2 Ä 2 Ä 2 å Ä the Ç and are positive roots of the transcendental equations Í Æ Æ Ü 2+Ü 3 tan(å ) Ç 1 (Ç ) − Ü 1 Þ 0 (Ç ) = 0, = å 2 . å −Ü Ü 2 3
© 2002 by Chapman & Hall/CRC Page 388
ç =ê æ è
æè
−í ì
Í
5.4.3-4. Domain: 0 ≤ Õ ≤ Þ , 0 ≤ Ø ≤ . Mixed boundary value problems. 1 ò . A circular cylinder of finite length is considered. The following conditions are prescribed: Ö Ù ÚÖ
= × 0 (Õ , Ø ) at
=× =Ý Ù ß Ö =Ý Ù ß Ö =Ý
1 (Õ
,Ø ) 1(Ø , Ï ) 2 (Õ , Ï ) 3 (Õ , Ï )
Ö
Solution:
Ê Ù
(initial condition),
Ï =0
(initial condition), (boundary condition), (boundary condition), (boundary condition).
at at at at
Ï =0 Õ =Þ Ø =0 Í Ø = Ê
Ê
à × 0 ( , ) (Õ , Ø , , , Ï ) ö ö õ Î Î Ê 0 Ê Ê Ê Î à × 1 ( , ) + á × 0 ( , )ø ( Õ , Ø , , , Ï ) ö ö + ó 0 ô ó 0Ú õ ÷ Î Î Î Î Ê ÙÊ à ö ö ù − 2 Ý 1( , ù ) ú Ù ( Õ , Ø , , , Ï − ù )û ü ä ó 0 ó 0ô Î Ú ÎÊ Î = Ê Ê Ê Ê Ê Ú à à Í − 2 Ý 2 ( , ù ) (Õ , Ø , , 0, Ï − ù ) ö ö ù +õ 2 Ý 3 ( , ù ) (Õ , Ø , , , Ï − ù ) ö ö ù ä óÚ 0 ó 0 õ ä ó 0 ó 0õ Ê Ê Ê à + ( , , ù ) (Õ , Ø , , , Ï − ù ) ö ö ö ù . ó 0 ó 0ô ó 0õ ý Î Î Î Ê Ê þ Here, Ê Ú à Æ Ç Õ Æ Ç Ø sin( Ð Ï ) 2 −ÿ 2 Í É 0 É cos Ë Í Ì É cos Ë ÍÌ Î É (Õ , Ø , , , Ï ) = , Æ 2 0 2 Ð Þ Þ Þ ) Î =1 =0 1Å ( Ç Ö (Õ , Ø , Ï ) = Ù
0ô ó
Ï ó
Ð
where the
=
2 Ç 2
2 2
+ ä Ì 2Ë
2
2
+ −
, 4 are zeros of the Bessel function, 0 ( ) = 0.
ä
2
=
1 for 2 for Ë Ë
= 0, > 0,
2 . A circular cylinder of finite length is considered. The following conditions are prescribed:
= × 0 ( , ) at
= × 1 ( , ) at = 1 ( , ) at
= 2 ( , ) = 3 ( , )
at at
=0
(initial condition),
=0 =
(initial condition), (boundary condition),
=0 =
(boundary condition), (boundary condition).
Solution:
( , , ) = +
+
0 0
2
ä
× 1( , ) + × 0( , )!" ( , , , , )
2
ä
× 0( , ) ( , , , , )
0 0
0 0
+ −
0 0
0 0 0
+
,# )
#
3(
,# ) $
#
#
.
2(
$
( , , , , − # )% &
( , , , , − # )% &
( , , # ) ( , , , , − # )
2
ä
0 0
1(
, # ) ( , , , , − # )
#
=0
'
=
© 2002 by Chapman & Hall/CRC Page 389
Here, ( , , , , ) =
4
−ÿ (
*
2
2 )+*
)
*
1
2 0 ( =0 , =1*
)
= ,
0 *
0-
1
2
*
*
2 2
.
2 2
+
2
0-
Ì 2
*
2
/ 2
sin .
+3 −
2 4
5.4.3-5. Domain:
1
≤ ≤
2,
0≤
sin .
Ì /
.
sin( 0
,
)
, 0
,
,
4
are zeros of the first-order Bessel function, 1 ( ) = 0 (
where the
Ì /
*
= 0).
0
≤ . First boundary value problem.
A hollow circular cylinder of finite length is considered. The following conditions are prescribed:
= =
, ) at 1 ( , ) at
5
=0 =0
5 0 (
= 1( , ) = 2( , )
at at
= 3 ( , ) = 4 ( , )
at at
= =
(initial condition), (initial condition), (boundary condition), (boundary condition),
1
2
=0 =
(boundary condition), (boundary condition).
Solution:
( , , ) =
2
2
2
2 0
2
, )+
0
,# ) $
( , , , , ) = Ì *
( ) =
> 0( ?
2
2 * 1
)@
(
0-
−ÿ
2
)*
* A ?
1
.
=1 , =1
− @ 0( ?
=
,# )
4 (
,# ) $
'
( , , , , − # )%
&
1
*
2
0-
( , , , , − # )% &
( , , # ) ( , , , , − # )
?
*
1
.
#
#
#
2
=0
=
.
#
*
*
#
*
) 9 ( ) 9 ( ) sin *) − A 2( 78 ) 0 *
2 0 (
)>
6
=
$
1
0 ( 78
*
( , , , , − # )%
3 (
*
2
6
2
0 0
)
, )!" ( , , , , )
( , , , , − # )%
1
,# ) $
*
5 0 ( 4
1 2
2
Here,
9
2 (
0 0
2
2
+
1 (
0 0
−
, ) ( , , , , )
2
5 0 (
1
5 ( 1
0 1
+
+
0
2
+
−
2
, 7
=
2 *1
, 0
Ì /
,
=
sin .
*
2 2 2
?
2 1
+
Ì /
.
2 2 2
Ì
/ 2
*
sin : 0 (? ) are the Bessel functions, and the ? are positive roots of the transcendental equation @ 0 ( ? ) > 0 ( 78? ) − @ 0 ( 78? ) > 0 ( ? ) = 0.
© 2002 by Chapman & Hall/CRC Page 390
5.5. Other Equations with Two Space Variables 1.
2C B
2 B
+
C E
=
B
D
B
D
The transformation
2 F
2C B -
2 B
K
2C
+ B
G
2 B
C
+ .
H
B
C
+
I 1 B G
B
1 24
(L , M , N ) = O (L , M , # ) exp - −
J
C
.
H
+ 3 2M 2 2
3 1L
− N
+
I 2 B
.
,
= #
2
2
N
leads to the equation from Subsection 5.1.3: P
2
2C
P O
P Q
=
2
2 P
P O
L 2
2
+ P
O
M 2
+å
R
2
2
+
4
4
2
2C
C
å =
, O
2
−
2
1 2 2 ( 3 + 3 ). 4 4 1 2 2
2C
–1 B = B 2 + B 2. 2 DVS B D B D B G B H Domain: − W < L < W , − W < M < W . Cauchy problem. K Initial conditions are prescribed:
2.
DTS
B
2
+
U
K , N K Solution for 1≤
N
,
X 2
c 4
where f
Y
= {d
2
≤
2c 4
= 5 (L , M ) at N
= 0,
= Z (L , M )
N
= 0.
at
< 2: /
1 (L , M , N ) = 2[
ikj
X 2P Y
N
2−
c
=
P
P N
b
\ ] \^
2 2−e
(_ , ` ) a 5 4
2c N 2−
, d
=
b
c
_ a `
−
d 2
+
1 2[
\ ] \^
b
4
2c N 2−
c
_ a `
−d
2
,
( L − _ )2 + ( M − ` ) 2 ,
} is the circle with center at (L , M ) and radius
Reference: M. M. Smirnov (1975).
(_ , ` ) a Z
c 4
N
1 XVgVh
.
© 2002 by Chapman & Hall/CRC Page 391
