Telegraph Equation ➄ ➄

6.4. Telegraph Equation ➄ ➄

6.4.1. Problems in Cartesian Coordinates

A three-dimensional nonhomogeneous telegraph equation in the rectangular Cartesian system of coordinates has the form

6.4.1-1. Reduction to the three-dimensional Klein–Gordon equation.

The substitution ➏ ( , , , ) = exp

( , , , ) leads to the equation

which is discussed in Subsection 6.3.1.

6.4.1-2. Domain: 0 ≤ ➔ ≤

1 ,0≤ ≤ ➜ 2 ,0≤ ≤ ➜ 3 . First boundary value problem.

A rectangular parallelepiped is considered. The following conditions are prescribed:

( , , ) at

(initial condition),

, ➔ , ➏ ) at ➒

(initial condition),

, ➔ , ➏ ) at ➒

(boundary condition),

= ➡ 2 ( , , ) at

= ➜ 1 (boundary condition),

3 ( , , ) at

=0 (boundary condition),

4 ( , , ) at

= ➜ 2 (boundary condition),

5 ( , , ) at

=0 (boundary condition),

= ➜ 3 (boundary condition). Solution:

6 ( , , ) at

) sin( Ñ ❒ ➦ ) sin( Ò ➧ ) sin ↔

6.4.1-3. Domain: 0 ≤ ×

≤ 1 ,0≤ ß ≤ 2 ,0≤ à ≤ 3 . Second boundary value problem.

A rectangular parallelepiped is considered. The following conditions are prescribed:

= â 0 ( Þ , ß , à ) at ã =0

(initial condition),

= â 1 ( Þ , ß , à ) at ã =0

(initial condition),

= ç 1 ( ß , à , ã ) at Þ =0 (boundary condition),

2 ( ß , à , ã ) at Þ = 1 (boundary condition),

= ç 3 ( Þ , à , ã ) at ß =0 (boundary condition),

4 ( Þ , à , ã ) at ß = 2 (boundary condition),

= ç 5 ( Þ , ß , ã ) at à =0 (boundary condition),

6 ( Þ , ß , ã ) at à = 3 (boundary condition). Solution:

) cos( Ñ þ í ) cos( Ò î ) sin ù✆☎

1 for ☛ = 0,

2 for > 0,

− 1 2 2 2 2 2 1 = 2 4 Ý , ✁ þ = Û ( Ð + Ñ þ + Ò )+ Ü − 4 Ý . The summation is performed over the indices satisfying the condition ☛ + ✍ + ✎ > 0; the term

corresponding to ☛ = ✍ = ✎ = 0 is singled out.

6.4.1-4. Domain: 0 ≤ ✌

≤ ✄ 1 ,0≤ ≤ 2 ,0≤ ≤ 3 . Third boundary value problem.

A rectangular parallelepiped is considered. The following conditions are prescribed:

= ✄ ✑ 0 ( Þ , ✂ , ) at ☎ =0

(initial condition),

, ✂ , ) at ☎ =0

(initial condition),

, , ) at

=0 (boundary condition),

2 ( ✂ ✄ , , ✎ ☎ ) at Þ = 1 (boundary condition),

− ✏ 3 = ✕ 3 ( Þ ✄ , , ☎ ) at ✂ =0 (boundary condition),

4 = 4 ( Þ , , ☎ ) at ✂ = 2 (boundary condition),

5 = ✕ 5 ( Þ , ✂ , ☎ ) at

=0 (boundary condition),

6 = 6 ( , , ) at

= 3 (boundary condition).

The solution ✄ ( Þ , ✂ , , ☎ ) is determined by the formula in Paragraph 6.4.1-3 where

sin( Ð Þ + ✣ ) sin( Ñ ✂ + ✤ ) sin( Ò + ✥ )

+ ✣ ) sin( Ñ í + ✤ ) sin( Ò î + ✥ ) sin ù✦☎

2 2 = arctan 2 , = arctan

( ✎ 1 + ✛ Ð 2 )( 2 Ð 2 2 Ñ 2 2 Ñ 2 2 Ò 2 2 Ò ✎ 2 2 + ) ø ( ✎ 3 + )( ✎ 4 + ) ø ( ✎ 5 + )( ✎ 6 + ) where the Ð , Ñ , and Ò are positive roots of the transcendental equations

1 ✎ 2 =( ✎ 1 + ✎ 2 ) Ð cot( 1 Ð ),

3 + ✎ 4 ) cot( 2 ),

5 ✎ 6 =( ✎ 5 + ✎ 6 ) Ò cot( 3 Ò ).

6.4.1-5. Domain: 0 ≤ ✌ ≤ 1 ,0≤ ≤ 2 ,0≤ ≤ 3 . Mixed boundary value problems.

1 ✵ . A rectangular parallelepiped is considered. The following conditions are prescribed:

= ✄ ✑ 0 ( Þ , ✂ , ) at ☎ =0

(initial condition),

= ✄ ✑ 1 ( Þ , ✂ , ) at ☎ =0

(initial condition),

1 ✄ ( ✂ , , ☎ ) at Þ =0 (boundary condition),

2 ( ✂ ✄ , , ☎ ) at Þ = 1 (boundary condition),

3 ✄ ( Þ , , ☎ ) at ✂ =0 (boundary condition),

4 ( Þ , , ☎ ) at ✂ = 2 (boundary condition),

= ✄ ✕ 5 ( Þ , ✂ , ☎ ) at =0 (boundary condition),

= 3 (boundary condition). Solution:

sin( ✷ Þ ý ) cos( ✸ þ ✂ ) cos( ✹

) cos( ✸ þ í ) cos( ✹ î ) sin ù✆☎

1 for

2 for ✍ > 0, ý

2 ✵ . A rectangular parallelepiped is considered. The following conditions are prescribed:

= ✄ ✑ 0 ( Þ , ✂ , ) at ☎ =0

(initial condition),

1 ✄ ( Þ , ✂ , ) at ☎ =0

(initial condition),

1 ✄ ( ✂ , , ☎ ) at Þ =0 (boundary condition),

2 ( , , ☎ ) at Þ = 1 (boundary condition),

3 ✄ ( Þ , , ☎ ) at ✂ =0 (boundary condition),

4 ( Þ , , ☎ ) at

= 2 (boundary condition),

) ✄ at =0 (boundary condition),

) ✄ at = 3 (boundary condition). Solution:

sin( ✷ ❀ ) sin( ▼

× sin( ✷ ● ✼ ) sin( ❏ ▼ ■ ✽ ● ) sin( ◆ ✾ ) sin ❏ ❊✆❖◗P ✁ ■ ,

6.4.2. Problems in Cylindrical Coordinates

A three-dimensional nonhomogeneous telegraph equation in the cylindrical coordinate system is written as ❨

( ✄ , , , ❖ ), = P ❛ ❖ 2 ❖ + ✂ 2 ❬ .

One-dimensional problems with axial symmetry that have solutions ❩ = ( , ❖ ) are treated in ❬

Subsection 4.4.2. Two-dimensional problems whose solutions have the form ❫ = ( , ,

) or

= ❩ ( , , ❖ ) are considered in Subsections 5.4.2 and 5.4.3.

6.4.2-1. Domain: 0 ≤ ✄ ≤ ♠ ,0≤ ≤2

,0≤ ≤ ❯ . First boundary value problem.

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),

, , ❖ ) at

=0 (boundary condition),

(boundary condition). Solution:

( ) are the Bessel functions (the prime denotes the derivative with respect to the argument), and the ✉ ➂ ❽ are positive roots of the transcendental equation ➀ ( ➂ ♠ ) = 0.

6.4.2-2. Domain: 0 ≤ ➇ ➃ ≤

,0≤ ✄ ,0≤ ≤ . Second boundary value problem.

A circular cylinder of finite length is considered. The following conditions are prescribed:

= ✄ ♥ 0 ( ➃ , , ) at ⑨ =0

(initial condition),

1 ( ➃ , , ➑ ) at ⑨ =0

(initial condition),

1 ( , ➑ , ⑨ ) at ➃ = ♠

(boundary condition),

2 ( ➃ , , ⑨ ) at ➑ =0 (boundary condition),

3 ( , , ⑨ ) at ➑ =

(boundary condition).

cos ➣ ➇ ↕ sin ➈✆⑨◗➛ →

2 cos[ ➄ ( ➝ − ➞ )] cos ➣ ➇ ↕ cos ➣ ➇ ↕

4 4 4 2 for ➄ > 0, where the ❼

( ➙ ) are the Bessel functions and the ➂ ❽

are positive roots of the transcendental equation ❿

6.4.2-3. Domain: 0 ≤ ➃

,0≤ ➝ ≤2 ➆ ,0≤ ➑ ≤ . Third boundary value problem.

A circular cylinder of finite length is considered. The following conditions are prescribed:

= ➤ 0 ( ➃ , ➝ , ➑ ) at ➥ =0

(initial condition),

= ➤ 1 ( ➃ , ➝ , ➑ ) at ➥ =0

(initial condition),

at ➃ = ➜

(boundary condition),

− ➅ ➏ 2 = ➧ ➐ 2 ( ➃ ➓ , ➝ , ➥ ) at ➑ =0 (boundary condition),

(boundary condition). The solution ➏ ( ➃ , ➝ , ➑ , ➥ ) is determined by the formula in Paragraph 6.4.2-2 where ➯ ➸ ➯ ➯ ➸ ➯ ➯ ➸ ➯

. Here, the ➾

× cos[ ➽ ( ➝ − ➞ )] ➾

( ➙ ) are the Bessel functions, ➯

1 for ➽ = 0, ➹ ➲

2 for ➽

) = cos(

sin( ➚

✄✂

✄✂

the ✍ ✎ ✏ and ✑ ✒ are positive roots of the transcendental equations

6.4.2-4. Domain: 0 ≤ ✘ ≤ ✕ ,0≤ ✙ ≤2 ✚ ,0≤ ✛ ≤ ✗ . Mixed boundary value problems.

1 ✜ . 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:

)] cos ❅ cos sin ✼❊✤✵❋ ✎ ✏ ✽ ,

✖❈✚

✖❈✚

2 + ● − 1 4 ✳ 2 for , = 0, ✎ = ❍ 2 for ❆ > 0, where the ❃

( ) are the Bessel functions (the prime denotes the derivative with respect to the argument) and the ✭ ✍ ✎ ✏ are positive roots of the transcendental equation ✓ ✎ ( ✍ ✕ ) = 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),

= ✧ 3 ( ✘ , ✙ , ✤ ) at ✛ = ✗

(boundary condition).

× cos[ ❆ ( ✙ − ✮ )] sin ❇

where the ✓ ✎ ( ) are the Bessel functions and the ✍ ✎ ✏ are positive roots of the transcendental equation

( ✍ ✕ ) = 0. ✭

6.4.2-5. Domain: ✕ 1 ≤ ✘ ≤ ✕ 2 ,0≤ ✙ ≤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),

= ✧ 4 ( ✘ , ✙ , ✤ ) at ✛ = ✗

(boundary condition).

), where the ✓ ✎ ( ✘ ) and ▲ ✎ ( ✘ ) are the Bessel functions, and the ✍ ✎ ✏ are positive roots of the transcen-

( ✘ )= ✓ ✎ ( ✍ ✎ ✏ ✕

)−

dental equation

( ✍ ✕ 1 ) ▲ ✎ ( ✍ ✕ 2 )− ▲ ✎ ( ✍ ✕ 1 ) ✓ ✎ ( ✍ ✕ 2 ) = 0.

6.4.2-6. Domain: ✕ 1 ≤ ✘ ≤ ✕ 2 ,0≤ ✙ ≤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).

); the ✓ ✎ ( ✘ ) and ▲ ✎ ( ✘ ) are the Bessel functions, and the ✍ ✎ ✏ are positive roots of the transcendental

6.4.2-7. Domain: ✕ 1 ≤ ✘ ≤ ✕ 2 ,0≤ ✙ ≤2 ✚ ,0≤ ✛ ≤ ✗ . Third 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 = ✧ 1 ( ✙ , ✛ , ✤ ) at ✘ = ✕

1 (boundary condition),

+ ✢ ✖ 2 = ✧ 2 ( ✙ , ✛ , ✤ ) at ✘ = ✕ 2 (boundary condition),

− ✢ ✖ 3 = ✧ 3 ( ✘ , ✙ , ✤ ) at ✛ =0

(boundary condition),

+ ✢ ✖ 4 = ✧ 4 ( ✘ , ✙ , ✤ ) at ✛ = ✗

(boundary condition).

The solution ✢ ( ✘ , ✙ , ✛ , ✤ ) is determined by the formula in Paragraph 6.4.2-6 where

( ✘ ) ✎ ✏ ( ) cos[ ❆ ( ✙ − ✮ )] ✒ ( ✛ ) ✒ ( ✯ ) sin ✼❊✤ ❋ ✶ 2 ✍ ✎ 2 ✏ + ✶ 2 ❅ 2 ✒ + ● − ✳ 2 ❚ 4 ✽

( ✛ ) = cos( ❅ ✒ ✛ )+ ❅ sin( ❅ ✒ ✛

where the ✓ ✎ ( ✘ ) and ▲ ✎ ( ✘ ) are the Bessel functions, the ✍ ✎ ✏ are positive roots of the transcendental equation

are positive roots of the transcendental equation

6.4.2-8. Domain: ✕ 1 ≤ ✘ ≤ ✕ 2 ,0≤ ✙ ≤2 ✚ ,0≤ ✛ ≤ ✗ . Mixed boundary value problems.

1 ✜ . 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),

= ✧ ( ✙ , ✛ , ✤ ) at ✘ = ✕

2 2 (boundary condition),

= ✧ 3 ( ✘ , ✙ , ✤ ) at ✛ =0

(boundary condition),

(boundary condition). Solution:

= ✧ 4 ( ✘ , ✙ , ✤ ) at ✛ = ✗

× cos[ ❆ ( ✙ − ✮ )] cos ❇

( ✘ )= ✓ ✎ ( ✍ ✎ ✏ ✕ 1 ) ▲ ✎ ( ✍ ✎ ✏ ✘ )− ▲ ✎ ( ✍ ✎ ✏ ✕ 1 ) ✓ ✎ ( ✍ ✎ ✏ ✘ ), where the ✓ ✎ ( ✘ ) and ▲ ✎ ( ✘ ) are the Bessel functions, and the ✍ ✎ ✏ are positive roots of the transcen-

dental equation

2 ✜ . 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:

= ✧ 4 ( ✘ , ✙ , ✤ ) at ✛ = ✗

Here,

sin ✼ ✤ ❄ ❁ ✑ ✰ ✽

sin ❇

sin ❇

sin

× cos[ ✽

( ✙ − ✮ )] sin ❇

sin sin

6.4.2-9. Domain: 0 ≤ ✘ ≤ ✕ ,0≤ ✙ ≤ ✙ 0 ,0≤ ✛ ≤ ✗ . First boundary value problem.

A cylindrical sector of finite thickness is considered. The following conditions are prescribed:

= 0 ( , , ) at

(initial condition),

= ✣ 1 ( ✘ , ✙ , ✛ ) at ✤ =0

(initial condition),

= ✧ 1 ( ✙ , ✛ , ✤ ) at ✘ = ✕

(boundary condition),

= ✧ 2 ( ✘ , ✛ , ✤ ) at ✙ =0

(boundary condition),

= ✧ 3 ( ✘ , ✛ , ✤ ) at ✙ = ✙ 0 (boundary condition),

= ✧ 4 ( ✘ , ✙ , ✤ ) at ✛ =0

(boundary condition),

(boundary condition). Solution:

+ ✶ ✖ ✚ ✗ + ● − ✳ 2 4 where the ❖

0 ( ✘ ) are the Bessel functions and the ✍ ✎ ✏ are positive roots of the transcendental equation ✓ ✎ ✫ ❯ 0 ( ✍ ✕ ) = 0.

6.4.2-10. Domain: 0 ≤ ✘ ≤ ✕ ,0≤ ✙ ≤ ✙ 0 ,0≤ ✛ ≤ ✗ . Mixed boundary value problem.

A cylindrical sector of finite thickness 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),

= ✧ 3 ( ✘ , ✛ , ✤ ) at ✙ = ✙ 0 (boundary condition),

= ✧ 4 ( ✘ , ✙ , ✤ ) at ✛ =0

(boundary condition),

(boundary condition). Solution:

+ ✶ ✖ ✚ ✗ + ● − ✳ 4 where ❖ 0 = 1 and = 2 for

≥ 1; the ✓ ✎ ❖ ✫ ❯ 0 ( ✘ ) are the Bessel functions; and the ✍ ✎ ✏ are positive roots of the transcendental equation ✓ ✎ ✫ ❯ 0 ( ✍ ❃ ✕ ❃ ) = 0.

6.4.3. Problems in Spherical Coordinates

A three-dimensional nonhomogeneous telegraph equation in the spherical coordinate system is written as

+ ✘ 2 sin

sin ❳

2 sin 2 ❳

6.4.3-1. Domain: 0 ≤ ✘ ≤ ✕ ,0≤ ❳ ≤ ✚ ,0≤ ✙ ≤2 ✚ . First boundary value problem.

A spherical domain is considered. The following conditions are prescribed:

= ✥ ✣ ✦ 0 ( ✘ , ❳ , ✙ ) at ✤ =0 (initial condition),

= ✣ 1 ( ✘ , ❳ , ✙ ) at ✤ =0 (initial condition),

(boundary condition). Solution:

(cos ❳ ) ❬ ✎ (cos ✮ ) cos[ ✖ ( ✙

Here, the ✓ ✎ +1 2 ( ✘ ) are the Bessel functions, the ❁ ❬ ✎ ( ✍ ) are the associated Legendre functions

expressed in terms of the Legendre polynomials ❁

are positive roots of the transcendental equation ✓ ✎ +1 2 ( ❅ ✕ ) = 0.

6.4.3-2. Domain: 0 ≤ ✘ ≤ ✕ ,0≤ ❳ ≤ ✚ ,0≤ ✙ ≤2 ✚ . Second boundary value problem.

A spherical domain is considered. The following conditions are prescribed:

= ✣ 0 ( ✘ , ❳ , ✙ ) at ✤ =0

(initial condition),

= ✣ 1 ( ✘ , ❳ , ✙ ) at ✤ =0

(initial condition),

(boundary condition). Solution:

at ✘ = ✕

0 ( , ✮ , ✯ ) ✰ ( ✘ , ❳ , ✙ , , ✮ , ✯ , ✤ ) 2 sin ✮ ✱ ✱ ✮ ✱ ✤ ✯

1 ( , ✮ , ✯ )+ ✳ ✣ 0 ( , ✮ , ✯ ) ✴ ✰ ( ✘ , ❳ , ✙ , , ✮ , ✯ , ✤ ) 2 sin ✮ ✱ ✱ ✮ ✱ ✯

2 ✕ 2 ✧ ( ✮ , ✯ , ✷ ) ✰ ( ✘ , ❳ , ✙ , ✕ , ✮ , ✯ , ✤ − ✷ ) sin ✮ ✱ ✮ ✱ ✯ ✱ ✷

) 2 sin ✮ ✱ ✱ ✮ ✱ ✯ ✱ ✷ , ) 2 sin ✮ ✱ ✱ ✮ ✱ ✯ ✱ ✷ ,

+1 2 ( ✎ ✏ ) ❬ ✎ (cos ❳ ) ❬ ✎ (cos ✮

Here, the ❖

+1 2 ( ✘ ) are the Bessel functions, the ❬ ✎ ( ✍ ) are the associated Legendre functions (see Paragraph 6.4.3-1), and the ❅ ✎ ✏ are positive roots of the transcendental equation

6.4.3-3. Domain: 0 ≤ ✘ ≤ ✕ ,0≤ ❳ ≤ ✚ ,0≤ ✙ ≤2 ✚ . Third boundary value problem.

A spherical domain is considered. The following conditions are prescribed:

= ✣ 0 ( ✘ , ❳ , ✙ ) at ✤ =0

(initial condition),

= ✣ 1 ( ✘ , ❳ , ✙ ) at ✤ =0

(initial condition),

(boundary condition). The solution ✢ ( ✘ , ❳ , ✙ , ✤ ) is determined by the formula in Paragraph 6.4.3-2 where

× ✎ (cos ) ❬ ✎ (cos ✮ ) cos[ ✗ ( ✙ − ✯ )]

Here, the ✪

+1 2 ( ✘ ) are the Bessel functions, the ❬ ✎ ( ✍ ) are the associated Legendre functions (see Paragraph 6.4.3-1), and the ❅ ✎ ✏ are positive roots of the transcendental equation

6.4.3-4. Domain: ✕ 1 ≤ ✘ ≤ ✕ 2 ,0≤ ❳ ≤ ✚ ,0≤ ✙ ≤2 ✚ . First boundary value problem.

A spherical layer 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).

× ✎ (cos ) ✎ (cos ) cos[ ( − )]

Here, the ✲

+1 2 ( ✘ ) are the Bessel functions, the ❁ ❬ ✎ ✍ ❁ ❁ ( ) are the associated Legendre functions expressed in terms of the Legendre polynomials ❁ ❬ ✎ ( ✍ ) as

and the ❅ ✎ ✏ are positive roots of the transcendental equation ✎ ❑ +1 2 ( ❅ ✕ 2 ) = 0.

6.4.3-5. Domain: ✕ 1 ≤ ✘ ≤ ✕ 2 ,0≤ ❳ ≤ ✚ ,0≤ ✙ ≤2 ✚ . Second boundary value problem.

A spherical layer 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). Solution:

, , , , , ) sin ✮ ✱ ✱ ✮ ✱ ✯

1 ( , ✮ , ✯ )+ ✳ 0 ( , ✮ , ✯ ) ✴ ✰ ( ✘ , ❳ , ✙ , , ✮ , ✯ , ✤ ) sin ✮ ✱ ✱ ✮ ✱ ✯

1 ✧ 1 ( ✮ , ✯ , ✷ ) ✰ ( ✘ , ❳ , ✙ , ✕ 1 , ✮ , ✯ , ✤ − ✷ ) sin ✮ ✱ ✮ ✱ ✯ ✱ ✷

2 ✧ 2 ( ✮ , ✯ , ✷ ) ✰ ( ✘ , ❳ , ✙ , ✕ 2 , ✮ , ✯ , ✤ − ✷ ) sin ✮ ✱ ✮ ✱ ✯ ✱ ✷

+ ( , ✮ , ✯ , ✷ ) ✰ ( ✘ , ❳ , ✙ , , ✮ , ✯ , ✤ − ✷ ) 2 sin ✮ ✱ ✱ ✮ ✱ ✯ ✱ ✷ ,

× ❬ ✎ (cos ❳ ) ❬ ✎ (cos ✮ ) cos[ ✖ ( ✙ − ✯ )]

where the ❖

+1 2 ( ✘ ) and ▲ ✎ +1 2 ( ✘ ) are the Bessel functions, the ❬ ✎ ( ✍ ) are the associated Legendre functions (see Paragraph 6.4.3-4), and the ❅ ✎ ✏ are positive roots of the transcendental equation

6.4.3-6. Domain: ✕ 1 ≤ ✘ ≤ ✕ 2 ,0≤ ❳ ≤ ✚ ,0≤ ✙ ≤2 ✚ . Third boundary value problem.

A spherical layer is considered. The following conditions are prescribed:

= 0 ( ✘ , ❳ , ✙ ) at ✤ =0

(initial condition),

= ✣ 1 ( ✘ , ❳ , ✙ ) at ✤ =0

(initial condition),

− ✢ ✖ 1 = ✧ 1 ( ❳ , ✙ , ✤ ) at ✘ = ✕

1 (boundary condition),

+ ✖ ✢ 2 = ✧ 2 ( ❳ , ✙ , ✤ ) at ✘ = ✕ 2 (boundary condition). The solution ✢ ( ✘ , ❳ , ✙ , ✤

) is determined by the formula in Paragraph 6.4.3-5 where

) ❬ ✎ (cos ✮ ) cos[ ✗ ( ✙ − ✯ )]

where the ✪

+1 2 ( ✘ ) and ▲ ✎ +1 2 ( ✘ ) are the Bessel functions, the ❬ ✎ ( ✍ ) are the associated Legendre functions (see Paragraph 6.4.3-4), and the ❅ ✎ ✏ are positive roots of the transcendental equation