Equations of the Form ✦ 2 = ✩
5.3. Equations of the Form ✦ 2 = ✩
5.3.1. Problems in Cartesian Coordinates ★
The two-dimensional nonhomogeneous Klein–Gordon equation with two space variables in the rectangular Cartesian coordinate system is written as
5.3.1-1. Fundamental solutions.
where ✷ ( ✌ ) is the Heaviside unit step function.
References : V. S. Vladimirov, V. P. Mikhailov, A. A. Vasharin, et al. (1974), B. M. Budak, A. A. Samarskii, and A. N. Tikhonov (1980).
< < ❂ ,− ❂ < < ❂ . Cauchy problem. Initial conditions are prescribed:
5.3.1-2. Domain: − ✲
( ✱ , ✲ ) at ✒ = 0,
= ✕ ( , ) at ✒ = 0.
2 1 ❃ ✎ 2 . Solution for ✴ =− ✯ < 0:
2 ✎ . Solution for ✴ = ✯ 2 ❃ 2 > 0:
Reference : B. M. Budak, A. A. Samarskii, and A. N. Tikhonov (1980).
5.3.1-3. Domain: 0 ≤ ✲ ≤
1 ,0≤ ≤ ✍ 2 . First boundary value problem.
A rectangle is considered. The following conditions are prescribed:
0 ( , ) at ✒ =0
(initial condition),
1 ( , ) at ✒ =0
(initial condition),
1 ( , ✒ ) at
=0 (boundary condition),
2 ( , ✒ ) at
= ✍ 1 (boundary condition),
3 ( , ✒ ) at
=0 (boundary condition),
4 ( , ✒ ) at
= ✍ 2 (boundary condition).
) sin( P ▼ ) sin(
) sin( P ▼ ✢ ) sin( ▼ ✒ ),
5.3.1-4. Domain: 0 ≤ ✲ ≤
1 ,0≤ ≤ ❚ 2 . Second boundary value problem.
A rectangle is considered. The following conditions are prescribed:
0 ( , ) at ❲ =0
(initial condition),
1 ( , ) at ❲ =0
(initial condition),
1 ( , ❲ ) at
=0 (boundary condition),
2 ( , ❲ ) at
= ❚ 1 (boundary condition),
3 ( , ❲ ) at
=0 (boundary condition),
4 ( , ❲ ) at
= ❚ 2 (boundary condition).
) cos( P ▼ ) cos( ❖ ❴ ) cos( P ▼ ❵ ) sin( ▼ q ),
5.3.1-5. Domain: 0 ≤ ❧
≤ 1 ,0≤ ≤ 2 . Third boundary value problem.
A rectangle is considered. The following conditions are prescribed:
) at
=0 (initial condition),
at ❳ q ❩ ) =0 (initial condition),
− ✇ ② 1 = ❬ 1 ( , q ) at ✈ =0 (boundary condition),
2 = ❬ 2 ( , q ) at ✈ = 1 (boundary condition),
3 = ❬ 3 ( ✈ , q ) at
=0 (boundary condition),
4 ( ✈ , q ) at
= 2 (boundary condition).
The solution ✲ (
, , q ) is determined by the formula in Paragraph 5.3.1-3 where
sin( ④ ✈ + ⑥ ) sin( ⑤ ▼ ✲
+ ⑥ ) sin( ▼ ❵ + ⑦ ▼ ) sin ❢ q t ❝
= arctan ❧ , ⑦ = arctan ❧
Here, the ④ and ⑤ ▼ are positive roots of the transcendental equations
References : A. G. Butkovskiy (1979), B. M. Budak, A. A. Samarskii, and A. N. Tikhonov (1980).
5.3.1-6. Domain: 0 ≤ ❧
≤ 1 ,0≤ ≤ 2 . Mixed boundary value problems.
1 ➂ . A rectangle is considered. The following conditions are prescribed:
0 ( ✈ , ) at q =0
(initial condition),
1 ( ✈ , ) at q =0
(initial condition),
1 ( , q ) at ✈ =0 (boundary condition),
2 ( , q ) at ✈ = 1 (boundary condition),
3 ( ✈ , q ) at
=0 (boundary condition),
4 ( ✈ , q ) at
= 2 (boundary condition).
sin( ❖ ✈ ) cos( P ▼ ) sin( ❖ ❴ ) cos( P ▼ ❵ ) sin( ▼ q
2 ➂ . A rectangle is considered. The following conditions are prescribed:
0 ( ✈ , ) at q =0
(initial condition),
1 ( ✈ , ) at q =0
(initial condition),
1 ( , q ) at ✈ =0 (boundary condition),
2 ( , ) at ✈ = 1 (boundary condition),
3 ( ✈ , q ) at
=0 (boundary condition),
) ❧ at = 2 (boundary condition).
sin( ➈ ✈ ) sin( ➉ ➆ ) sin( ➈ ❴ ) sin( ➉ ➆ ❵ ) sin( ➆ q ),
5.3.2. Problems in Polar Coordinates
A nonhomogeneous Klein–Gordon equation with two space variables in the polar coordinate system has the form
, ➒ ) independent of the angular coordinate ➎ are considered in Subsection 4.2.5.
One-dimensional solutions ✇ = (
5.3.2-1. Domain: 0 ≤ ➍ ≤ ➔ ,0≤ ➎ ≤2 → . First boundary value problem.
A circle is considered. The following conditions are prescribed:
= ↔ 0 ( ➍ , ➎ ) at ➒ =0
(initial condition),
= ↔ 1 ( ➍ , ➎ ) at ➒ =0
(initial condition),
(boundary condition). Solution:
at ➍ = ➔
( ➜ ) are the Bessel functions (the prime denotes the derivative with respect to the ➾ ➾
argument) and the ➶ ➽ are positive roots of the transcendental equation ➚ ( ➶ ➔ ) = 0.
5.3.2-2. Domain: 0 ≤ ➹ ≤ ➔ ,0≤ ➴ ≤2 → . Second boundary value problem.
A circle is considered. The following conditions are prescribed:
= ↔ 0 ( ➹ , ➴ ) at ➬ =0
(initial condition),
= ↔ 1 ( ➹ , ➴ ) at ➬ =0
(initial condition),
(boundary condition). Solution:
= 1, 2, ❒✄❒✄❒ ; the ➚ ( ➜ ) are the Bessel functions; and the ➶ ➽ are positive roots of the transcendental equation ➚ ➪ ➾ ➾ ( ➶ ➔ ) = 0.
where ➼ 0 = 1 and = 2 for
5.3.2-3. Domain: 0 ≤ ➹ ≤ ➔ ,0≤ ➴ ≤2 → . Third boundary value problem.
A circle is considered. The following conditions are prescribed:
= ↔ 0 ( ➹ , ➴ ) at ➬ =0
(initial condition),
= ↔ 1 ( ➹ , ➴ ) at ❮ ➬ ❰ =0 (initial condition),
(boundary condition). The solution Ó ( ➹ , ➴ , ➬ ) is determined by the formula in Paragraph 5.3.2-2 where ➼ ➼ ➼ ➼ ➼ ➼
Here, the ➚ ( ➜ ) are the Bessel functions and the ➾ ➼ ➾ ➶ ➽ ➼ are positive roots of the transcendental equation
* In the expressions of the Green’s functions specified in Subsection 5.3.2, the ratios sin ÕÖ➸Ø× ➩ 2 Ù 2 Ú Û + ➯ Ü✐Ý × ➩ 2 Ù 2 Ú Û + ➯
must be replaced by sinh ÕÞ➸Ø× | ➩ 2 Ù 2 Ú Û + ➯ | Ü✐Ý × | ➩ 2 Ù 2 Ú Û + ➯ | if ➩ 2 Ù 2 Ú Û + ➯ < 0.
2 Ô ,0≤ ➴ ≤2 . First boundary value problem. An annular domain is considered. The following conditions are prescribed:
5.3.2-4. Domain: Ò 1 ≤ ➹ ≤ Ò
= ß 0 ( ➹ , ➴ ) at ➬ =0
(initial condition),
= ß 1 ( ➹ , ➴ ) at ➬ =0
(initial condition),
= ❬ 1 ( ➴ , ➬ ) at ➹ = Ò 1 (boundary condition),
= ❬ 2 ( ➴ , ➬ ) at ➹ = Ò 2 (boundary condition). Solution:
( ➼ ➶ ➽ ➹ )= ➚ ( ➶ ➽ Ò 1 ) ä ( ➶ ➽ ➹ )− ä ( ➶ ➼ ➽ Ò 1 ) ➚ ( ➶ ➽ ➹ ), where the ➚ ( ➹ ) and ä ( ➹ ) are the Bessel functions, and the ➶ ➽ ➼ ➼ ➼ ➼ are positive roots of the transcen-
dental equation
5.3.2-5. Domain: Ô 1 ≤ ≤ 2 ,0≤ ≤2 . Second boundary value problem. An annular domain is considered. The following conditions are prescribed:
= ß 0 ( ➹ , ➴ ) at ❨ ➬ =0
(initial condition),
= ß 1 ( ➹ , ➴ ) at ➬ =0
(initial condition),
= ❬ 1 ( ➴ , ➬ ) at ➹ = Ò 1 (boundary condition),
= ❬ 2 ( ➴ , ➬ ) at ➹ = Ò 2 (boundary condition). Solution:
Here, sin ➷❣➬✐Ð ✃ ❐
) cos[ ( − )] sin ➷✺➬ ➮ ➠ 2 ➶ 2 ➽ + ➼ ✃ ❐
( ➹ ) and ä ( ➹ ) are the Bessel functions, and the ➶ ➽ ➼ ➼ ➼ ➼ are positive roots of the transcendental equation
1 Ô ≤ ➹ ≤ Ò 2 ,0≤ ➴ ≤2 . Third boundary value problem. An annular domain is considered. The following conditions are prescribed:
5.3.2-6. Domain: Ò
= ß 0 ( ➹ , ➴ ) at ➬ =0
(initial condition),
= ß 1 ( ➹ , ➴ ) at ➬ =0
(initial condition),
= ( , ) at
(boundary condition),
+ Ñ 2 Ó = ❬ 2 ( ➴ , ➬ ) at ➹ = Ò 2 (boundary condition). The solution Ó ( ➹ , ➴ , ➬ ) is determined by the formula in Paragraph 5.3.2-5 where ➼ ➼ ➼ ➼ ➼
( ➜ ) cos[ ➘ ( ➴ − ➝ )] sin( ç , ➬
= 1, 2, ❒✄❒✄❒ ; ç ➽ = ➮ ➠ 2 ➶ 2 ➽ + ✃ ; the ➚ ( ➹ ) and ä ( ➹ ) are the Bessel functions; and the ➼ ➶ ➽ ➾ ➾ ➼ are positive roots of the transcendental equation ➼ ➼
Here, ➼ 0 = 1 and = 2 for
5.3.2-7. Domain: 0 ≤ ➹ ≤ Ò ,0≤ ➴ ≤ ➴ 0 . First boundary value problem.
A circular sector is considered. The following conditions are prescribed:
= ß 0 ( ➹ , ➴ ) at ➬ =0
(initial condition),
= ß 1 ( ➹ , ➴ ) at ➬ =0
(initial condition),
= ❬ 1 ( ➴ , ➬ ) at ➹ = Ò
(boundary condition),
= ❬ 2 ( ➹ , ➬ ) at ➴ =0
(boundary condition),
= ❬ 3 ( ➹ , ➬ ) at ➴ = ➴ 0 (boundary condition).
where the ➼ ➚
0 ( ➹ ) are the Bessel functions and the è ➶ ➽ are positive roots of the transcendental equation ➚ ➙ ê 0 ( ➶ Ò ) = 0, and ç ➽ = ➮ ➠ 2 ➶ 2 è ➽ + ✃ .
5.3.2-8. Domain: 0 ≤ è
≤ Ò ,0≤ ➴ ≤ ➴ 0 . Second boundary value problem.
A circular sector is considered. The following conditions are prescribed:
= ß 0 ( ➹ , ➴ ) at ➬ =0
(initial condition),
( ➹ , ➴ ) at ➬ =0
(initial condition),
= ❬ 1 ( ➴ , ➬ ) at ➹ = Ò
(boundary condition),
= ❬ 2 ( ➹ , ➬ ) at ➴ =0
(boundary condition),
= ❬ 3 ( ➹ , ➬ ) at ➴ = ➴ 0 (boundary condition). Solution: ❮
where the ✄
0 ( ) are the Bessel functions and the ✁ ✂ are positive roots of the transcendental equation ☎ ✁ ✌ ✆ ✝
5.3.2-9. Domain: 0 ≤ õ
≤ ✡ ,0≤ ≤ 0 . Mixed boundary value problem.
A circular sector is considered. The following conditions are prescribed:
= ✡ 0 ( õ , ) at ö =0
(initial condition),
= ✡ 1 ( õ , ) at ö =0
(initial condition),
at õ
(boundary condition),
=0 ✡ at =0
(boundary condition),
=0 ✡ at = 0 (boundary condition). Solution:
) ☎ ✜✣✢ ➺ ✄ ➺ ☎ ✜✣✢ ✄ ✁ ✂ ( ✁ ✂ ò ) cos( ✤ ✁ ✡ ) cos( ✤ ✁ ó ) sin ü ö☞✥ ø 2 ✄ 2 ✁ ✂ + þ ÿ ,
where the ☎ ✜✣✢
( ✄ õ ) are the Bessel functions and the ✁ ✂ are positive roots of the transcendental equation
5.3.3. Axisymmetric Problems
In the axisymmetric case, a nonhomogeneous Klein–Gordon equation in the cylindrical system of coordinates has the form
In the solutions of the problems considered below, the modified Green’s function ★
2 ✦ ò ô ( õ , , ò , ó , ö ) is used for convenience.
5.3.3-1. Domain: 0 ≤ ★
≤ ,0≤ ≤ ✬ . First boundary value problem.
A circular cylinder of finite length is considered. The following conditions are prescribed:
, ) at ö =0
(initial condition),
, ) at ö =0
(initial condition),
= ù ❬ 1 ( , ö ) at õ =
(boundary condition),
2 ( õ , ö ) at
=0 (boundary condition),
3 ( õ , ö ) at
(boundary condition).
+ þ , where the ✄ ✁ are positive zeros of the Bessel function, ☎ ✄ 0 ( ) = 0.
5.3.3-2. Domain: 0 ≤ ★
≤ ,0≤ ≤ ✬ . Second boundary value problem.
A circular cylinder of finite length is considered. The following conditions are prescribed:
, ) at ö =0
(initial condition),
, ) at ö =0
(initial condition),
= ù ❬ 1 ( , ö ) at õ =
(boundary condition),
2 ( õ , ö ) at
=0 (boundary condition),
(boundary condition). Solution:
3 ( õ , ö ) at
5.3.3-3. Domain: 0 ≤ õ
. Third boundary value problem.
A circular cylinder of finite length is considered. The following conditions are prescribed:
, ) at ö =0
(initial condition),
, ) at ö =0
(initial condition),
+ ù 1 = ❬ 1 ( , ö ) at õ =
(boundary condition),
2 = ❬ 2 ( õ , ö ) at
=0 (boundary condition),
3 = ❬ 3 ( õ , ö ) at
(boundary condition).
The solution ★ (
, , ö ) is determined by the formula in Paragraph 5.3.3-2 where
Here, the ✶ ✁ and ✂ are positive roots of the transcendental equations
5.3.3-4. Domain: 0 ≤ õ
. Mixed boundary value problems.
1 ✷ . A circular cylinder of finite length is considered. The following conditions are prescribed:
0 ( õ , ) at ö =0
(initial condition),
1 ( õ , ) at ö =0
(initial condition),
) ù at õ =
(boundary condition),
2 ( õ , ö ) at
=0 (boundary condition),
3 ( õ , ö ) at
(boundary condition).
sin ❑▲✾◆▼ ❖ ❇ ✫ P
where the ●
are positive zeros of the Bessel function, 0 ( ❉ ) = 0.
2 ❙ . A circular cylinder of finite length is considered. The following conditions are prescribed:
= ❯ 0 ( ✽ , ) at ✾ =0
(initial condition),
1 ( ✽ , ) at ✾ =0
(initial condition),
1 ( , ✾ ) at ✽ =
(boundary condition),
2 ( ✽ , ✾ ) at
=0 (boundary condition),
(boundary condition). Solution: ■
where the ❈
are zeros of the first-order Bessel function, 1 ( ❉ )=0( ❉ 0 = 0).
❡✑❣
5.3.3-5. Domain: ❏ 1 ≤
≤ 2 ,0≤ ≤ . First boundary value problem.
A hollow circular cylinder of finite length is considered. The following conditions are prescribed: ■
= ❯ ❨ 0 ( ✽ , ) at ❴ =0
(initial condition),
1 ( ✽ , ) at ❴ =0
(initial condition),
1 ( , ❴ ) at ✽ = 1 (boundary condition),
2 ( , ❴ ) at ✽ = 2 (boundary condition),
3 ( ✽ , ❴ ) at
=0
(boundary condition),
(boundary condition). Solution: ■
) and t 0 ( ❉ ) are the Bessel functions, and the ❉ are positive roots of the transcendental equation
where ❈ 0 (
) t 0 ( qr❉ )− 0 ( qr❉ ) t 0 ( ❉ ) = 0.
5.3.3-6. Domain: ❏ 1 ≤
≤ 2 ,0≤ ≤ . Second boundary value problem.
A hollow circular cylinder of finite length is considered. The following conditions are prescribed: ■
= ❯ 0 ( ✽ , ) at ❴ =0
(initial condition),
1 ( ✽ , ) at ❴ =0
(initial condition),
1 ( , ❴ ) at ✽ = 1 (boundary condition),
2 ( , ❴ ) at ✽ = 2 (boundary condition),
3 ( ✽ , ❴ ) at
=0
(boundary condition),
4 ( ✽ , ❴ ) at
(boundary condition).
( ③ ) and t ② ( ③ ) are the Bessel functions ( ⑥ = 0, 1); and the ③ are positive roots of the transcendental equation
) t 1 ( q ③ )− 1 ( q ③ ) t 1 ( ③ ) = 0.