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.