Wave Equation ❧ 2 = ♦ 2

6.1. Wave Equation ❧ 2 = ♦ 2

6.1.1. Problems in Cartesian Coordinates ♥

The wave equation with three space variables in the rectangular Cartesian coordinate system has the ❑

form

This equation is of fundamental importance in sound propagation theory, the propagation of elec- tromagnetic fields theory, and a number of other areas of physics and mechanics.

6.1.1-1. Particular solutions and their properties.

1 ② . Particular solutions:

( ✉ , ✈ , ✇ , ③ )= ④ exp ⑤ ⑥ 1 ✉ + ⑥ 2 ✈ + ⑥ 3 ✇ ⑦ q ③

( ✉ , ✈ , ✇ , ③ )= ④ sin( ⑥ 1 ✉ + ❢ 1 ) sin( ⑥ 2 ✈ + ❢ 2 ) sin( ⑥ 3 2 2 ✇ 2 + ❢ 3 ) sin ⑤✧q ③ ⑥ 1 + ⑥ 2 + ⑥ 3 ⑧ ,

( ✉ , ✈ , ✇ , ③ )= ④ sin( ⑥ 1 ✉ + ❢ 1 ) sin( ⑥ 2 ✈ + ❢ 2 ) sin( ⑥ 3 ✇ + ❢ 3 ⑤ q ③

( ✉ , ✈ , ✇ , ③ )= ④ sinh( ⑥ 1 ✉ + ❢ 1 ) sinh( ⑥ 2 ✈ + ❢ 2 ) sinh( ⑥ 3 ✇ + ❢ 3 ) cosh ⑤✧q ③

where ④ , ❢ 1 , ❢ 2 , ❢ 3 , ⑥ 1 , ⑥ 2 , and ⑥ 3 are arbitrary constants.

2 ② . Fundamental solution: ⑨ ⑨

( ✉ , ✈ , ✇ , ③ )= ⑩ q ❶ ( q ③ − ❷ ),

where ❶ ( ❹ ❺❦❻ ) is the Dirac delta function.

Reference : V. S. Vladimirov (1988).

3 ❿ ② . Infinite series solutions containing arbitrary functions of space variables:

( ➀ ③ t 2 ) ( ✉ , ✈ , ✇ , ③ )= ③➄➃ ( ✉ , ✈ , ✇ )+ ③ ❽

the initial conditions t (

, ✈ , ✇ , 0) = ❼ ( ✉ , ✈ , ✇ ), s ➅ ( ✉ , ✈ , ✇ , 0) = 0, and the second solution initial

conditions t (

, ✈ , ✇ , 0) = 0, s ➅ ( ✉ , ✈ , ✇ , 0) = ➃ ( ✉ , ✈ , ✇ ). The sums are finite if ❼ ( ✉ , ✈ , ✇ ) and ➃ ( ✉ , ✈ , ✇ ) are polynomials in ✉ , ✈ , ✇ .

❺❦❻

Reference : A. V. Bitsadze and D. F. Kalinichenko (1985).

4 ② . Suppose t = t ( ✉ , ✈ , ✇ , ③ ) is a solution of the wave equation. Then the functions

2 − ➀ 2 ③ 2 ❷ 2 − ➀ 2 ③ 2 , ❷ 2 ➀ 2 ③ 2 , ❷ 2 ➀ 2 ③ 2 − , − ❷ 2 − ➀ 2 ③ 2 ① , where ④ , ➇ , ➈ , and ➆ are arbitrary constants, are also solutions of the equation. The signs at ➆ in

the expression of t 1 can be taken independently of one another. The function 2 is a consequence of the invariance of the wave equation under the Lorentz transformation.

❺❦❻

References : G. N. Polozhii (1964), W. Miller, Jr. (1977), A. V. Bitsadze and D. F. Kalinichenko (1985).

6.1.1-2. Domain: − ➊ < ✉ < ➊ ,− ➊ < ✈ < ➊ ,− ➊ < ✇ < ➊ . Cauchy problem. Initial conditions are prescribed:

= ❼ ( ✉ , ✈ , ✇ ) at ③ = 0,

= ➃ ( ✉ , ✈ , ✇ ) at ③ = 0.

Solution (Kirchhoff’s formula):

where the integration is performed over the surface of the sphere of radius ➀ ③ with center at ( ✉ , ✈ , ✇ ).

❺❦❻

References : N. S. Koshlyakov, E. B. Glizer, and M. M. Smirnov (1970), A. N. Tikhonov and A. A. Samarskii (1990).

6.1.1-3. Domain: 0 ≤ ✉ ≤ ➔ 1 ,0≤ ✈ ≤ ➔ 2 ,0≤ ✇ ≤ ➔ 3 . First boundary value problem.

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

= ❼ 0 ( ✉ , ✈ , ✇ ) at ③ =0

(initial condition),

= ❼ 1 ( ✉ , ✈ , ✇ ) at ③ =0

(initial condition),

= ➃ 1 ( ✈ , ✇ , ③ ) at ✉ =0 (boundary condition),

= ➃ 2 ( ✈ , ✇ , ③ ) at ✉ = ➔ 1 (boundary condition),

= ➃ 3 ( ✉ , ✇ , ③ ) at ✈ =0 (boundary condition),

= ➃ 4 ( ✉ , ✇ , ③ ) at ✈ = ➔ 2 (boundary condition),

= ➃ 5 ( ✉ , ✈ , ③ ) at ✇ =0 (boundary condition),

= ➃ 6 ( ✉ , ✈ , ③ ) at ✇ = ➔ 3 (boundary condition).

) sin( ) sin( )

× sin( ➲ ❹ ) sin( ➳ ➫ ➐ ) sin( ➵ ➑ ) sin( ➀ ➆ ➫ ➥ ), where

6.1.1-4. Domain: 0 ≤ ➡ ≤ ➔ 1 ,0≤ ➢ ≤ ➔ 2 ,0≤ ➤ ≤ ➔ 3 . Second boundary value problem.

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

= ❼ 0 ( ➡ , ➢ , ➤ ) at ➥ =0

(initial condition),

= ❼ 1 ( ➡ , ➢ , ➠ ➤ ) at ➥ =0 (initial condition),

= 1 ( , , ) at

=0 (boundary condition),

= ➃ 2 ( ➢ , ➤ , ➥ ) at ➡ = ➔ 1 (boundary condition),

= ➃ 3 ( ➡ , ➤ , ➥ ) at ➢ =0 (boundary condition),

= ➃ 4 ( ➡ , ➤ , ➥ ) at ➢ = ➔ 2 (boundary condition),

= ➃ 5 ( ➡ , ➢ , ➥ ) at ➤ =0 (boundary condition),

= ➃ 6 ( ➡ , ➢ , ➥ ) at ➤ = ➔ 3 (boundary condition).

) cos( ➳ ➢ ) cos( ➭ ➵ ➤ ) ❿ ➭

× cos( ❿ ➲ ❹ ) cos( ➭ ➳ ➫ ➐ ❿ ) cos( ➵ ➑ ) sin( ➀ ➆ ➫ ➥ ),

2 1 for ➲ ➁

3 2 for ➁ > 0. The summation here is performed over the indices satisfying the condition ➶ ➁ +

+ ➻ > 0; the term corresponding to ➁ = ➺ = ➻ = 0 is singled out.

6.1.1-5. Domain: 0 ≤ ➡ ≤ ➔ 1 ,0≤ ➢ ≤ ➔ 2 ,0≤ ➤ ≤ ➔ 3 . Third boundary value problem.

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

= ➘ 0 ( ➡ , ➢ , ➤ ) at ➥ =0

(initial condition),

= ➘ 1 ( ➡ , ➢ , ➤ ) at ➥ =0

(initial condition),

− 1 = 1 ( , , ) at ➡ =0 (boundary condition),

) at ➡ = ➔ 1 (boundary condition),

− ➴ ➠ ➽ ➚ 3 = ➃ 3 ( ➡ , ➤ , ➥ ) at ➢ =0 (boundary condition),

+ ➴ 4 = ➃ 4 ( ➡ , ➤ , ➥ ) at ➢ = ➔ 2 (boundary condition),

− 5 = 5 ( , , ) at

=0 (boundary condition),

+ ➴ ➠ ➽ ➪ 6 = ➃ 6 ( ➡ , ➢ , ➥ ) at ➤ = ➔ 3 (boundary condition). The solution ➽ ( ➡ , ➢ , ➤ , ➥ ) is determined by the formula in Paragraph 6.1.1-4 where

sin( ➲ ➡ + ➱ ) sin( ➳ ➫ ➢

× sin( ➲ ➷ + ➱ ) sin( ➳ ➫ ➐ + ✃ ➫ ) sin( ➵ ➑ + ❐ ) sin ❒✧➀ ➥ ➼ ➲ 2 + ➳ 2 ➫ + ➵ 2

6 ) Here, the ➲ , ➳ ➫ , and ➵ are positive roots of the transcendental equations

4 ) ➳ cot( ➔ 2 ➳ ), ➵ − ➴ 5 ➴ 6 =( ➴ 5 + ➴ 6 ) ➵ cot( ➔ 3 ➵ ).

6.1.1-6. Domain: 0 ≤ ➡ ≤ ➔ 1 ,0≤ ➢ ≤ ➔ 2 ,0≤ ➤ ≤ ➔ 3 . Mixed boundary value problems.

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

= ➘ 0 ( ➡ , ➢ , ➤ ) at ➥ =0

(initial condition),

= ➘ 1 ( ➡ , ➢ , ➤ ) at ➥ =0

(initial condition),

= ➃ 1 ( ➢ , ➤ , ➥ ) at ➡ =0 (boundary condition),

= ➃ 2 ( ➢ , ➤ , ➥ ) at ➡ = ➔ 1 (boundary condition),

= ➃ 3 ( ➡ , ➤ , ➠ ➥ ➚ ) at ➢ =0 (boundary condition),

= ➃ 4 ( ➡ , ➤ , ➥ ) at ➢ = ➔ 2 (boundary condition),

= ➃ 5 ( ➡ , ➢ , ➥ ) at ➤ =0 (boundary condition),

= ➃ 6 ( ➡ , ➢ , ➥ ) at ➤ = ➔ 3 (boundary condition). Solution:

) cos( ➳ ➫ Ó ) cos( ➵ ) sin( ➀

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

= ➘ 0 ( ➡ , ➢ , ➤ ) at ➥ =0

(initial condition),

= 1 ( , , ) at

(initial condition),

= × 1 ( ➢ , ➤ , ➥ ) at ➡ =0 (boundary condition),

= ( ➢ , ➤ , ➥ ) at ➡ = 1 (boundary condition),

= × 3 ( ➡ , ➤ , ➥ ) at ➢ =0 (boundary condition),

= × 4 ( ➡ , ➚ ➤ , ➥ ) at ➢ = 2 (boundary condition),

= × 5 ( ➡ , ➢ , ➥ ) at ➤ =0 (boundary condition),

6 ( ➡ , ➢ ➠ ➥ ) at ➤ ➪ , = 3 (boundary condition). Solution:

sin( é ➡ ) sin( ê ç å ➢ ) sin( ➵ ➤ )

) sin( ê ç Ó ) sin( ➵ ) sin( å

6.1.2. Problems in Cylindrical Coordinates

The three-dimensional wave equation in the cylindrical coordinate system is written as

One-dimensional problems with axial symmetry that have solutions ô = ( , Û ) are considered õ

in Subsection 4.2.1. Two-dimensional problems whose solutions have the form ô = ( ,

, Û ) or

= ô ( , ù , Û ) are discussed in Subsections 5.1.2 and 5.1.3.

6.1.2-1. Domain: 0 ≤ ≤ ý ,0≤ ø ≤2 þ ,0≤ ù ≤ ÿ . First boundary value problem.

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

= 0 ( õ , ø , ) at Û ù =0

(initial condition),

= 1 ( , ø , ù ) at õ Û =0

(initial condition),

= ✂ 1 ( õ ø , ù , Û ) at

(boundary condition),

= ✂ 2 ( õ , ø , Û ) at ù =0 (boundary condition),

(boundary condition). Solution: õ

where the ✕ ✦ (

) are the Bessel functions (the prime denotes the derivative with respect to the argument) and the ✗ ✦ ç are positive roots of the transcendental equation ✕ ✦ ( ✗ ý ) = 0.

6.1.2-2. Domain: 0 ≤ ≤ ý ,0≤ ✧ ≤2 þ ,0≤ ★ ≤ ÿ . Second boundary value problem.

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

= 0 ( õ , ✧ , ★ ) at ✠ =0

(initial condition),

= 1 ( , ✧ , ★ ) at õ ✠ =0

(initial condition),

= ☛ ✂ ✪ 1 ( ✧ õ , ★ , ✠ ) at

(boundary condition),

= ✭ 2 ( õ , ✧ , ✬ ✠ ) at ☛ ★ =0 (boundary condition),

= 3 ( , , ) at

(boundary condition).

cos[ ✚ ( ✧ − ✛ )] cos ✶

2 for ✽ > 0, where the ✔

( ✿ ) are the Bessel functions and the ✻ ✦ ç are positive roots of the transcendental equation

6.1.2-3. Domain: 0 ≤ ✴ ≤ ,0≤ ≤2 ,0≤ ★ ≤ ✮ . Third boundary value problem.

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

= ✹ 0 ( ❁ , ✧ , ★ ) at =0

(initial condition),

) ✹ at =0

(initial condition),

at ❁ = ✫

(boundary condition),

) at ★ =0 (boundary condition),

(boundary condition). The solution ✩ ( ❁ , ✧ , ★ ✹ , ) is determined by the formula in Paragraph 6.1.2-2 where

+ ✹ 3 = ✭ 3 ( ❁ , ✧ , ) at ★ = ✮

1 ( ç ❁ ) ✾ ( ✻ ç ✿ ) cos[ ✽ ( ✧ − ❄ )] ❊ ( ★ ) ❊ ( ✲ ) sin( ❋ ç ) ( , , , , , , )= ✴

= ■ ❏ ✻ 2 ç + ❑ 2 , ❊ ( ★ ) = cos( ❑ ★ )+ ✼ 2 sin( ❑ ★ ), ❍ ❍ 2 ❊ ❑ 3 = + ✼ 2 2 2 2 ✼ 2 + ✼ 2 + ▲

Here, 0 = 1 and = 2 for ✽ = 1, 2, ❖P❖P❖ ; the ✾ ( ✿ ) are the Bessel functions; and the ✻ ç and ❑

are positive roots of the transcendental equations ✔ ✔

tan( ▲ )

6.1.2-4. 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),

, ✏ ) at ✞ = ✟

(boundary condition),

2 ( ✔ ✞ , ✠ , ✏ ) at ☛ =0 (boundary condition),

(boundary condition). Solution:

× cos[ ✵ ( ✠ − ✚ )] cos ✶ ✷

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:

, ✠ , ☛ ) at ✏ =0

(initial condition),

= ✎ 1 ( ✞ , ✠ , ☛ ) at ✏ =0

(initial condition),

= ✓ 1 ( ✠ , ☛ , ✏ ) at ✞ = ✟

(boundary condition),

= ✓ 2 ( ✞ , ✠ , ✏ ) at ☛ =0 (boundary condition),

(boundary condition). Solution:

= ✓ 3 ( ✞ , ✠ , ✏ ) at ☛ = ☞

( ) are the Bessel functions and the ✲ ✬ ✯ are positive roots of the transcendental equation

6.1.2-5. 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),

(boundary condition). Solution:

× cos[ ✵ ( ✠ − ✚ )] sin ✶ ✷

sin ✶ ✷

for

2 ✪ for

where the ✰ ( ✞ ) and ❃ ( ✞ ) are the Bessel functions, and the

are positive roots of the transcen- dental equation

6.1.2-6. 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),

(boundary condition). Solution:

( ✯ ✞ )= ✰ ✱ ( ✲ ✬ ✟ 1 ) ❃ ( ✲ ✬ ✞ )− ❃ ✪ ✱ ( ✲ ✬ ✟ 1 ) ✰ ( ✲ ✬ ✞ ); the ✰ ( ✞ ) and ❃ ( ✞ ) are the Bessel functions, and the

are positive roots of the transcendental equation

6.1.2-7. 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 ( ✠ , ☛ , ✏ ) at ✞ = ✟ 1 ✑ (boundary condition), ✿

2 ✑ (boundary condition), ✔

2 2 ( , , ) at

− ✍ 3 = ✓ 3 ( ✞ , ✠ , ✏ ) at ☛ =0

(boundary condition),

(boundary condition). The solution ✍ ( ✞ , ✠ , ☛ , ✏ ) is determined by the formula in Paragraph 6.1.2-6 where

( ) ✬ ❂ ( ) cos[ ✵ ( ✠ − ✚ )] ( ☛ ) ( ✛ ) sin ✹✺✣ ✏ ✻ ✲ 2 ✬ + ✴ 2 ( 2 2 ✟ 2 2 ✙ + ✲ 2 ✬ ✟ 2 2 − ✵ 2 ) 2 ✬ ( ✟ 2 )−( 2 1 2 ✟ 2 1 + ✲ ✬ ✟ 2 ✷ . 1 − ✵ ✷ 2 ) 2 ❂ ✬ ❂ ( ✟ 1 )

( ☛ ) = cos( ✴ ☛ )+ ✷ sin( ✴ ✴ ☛ ),

where the ✰ ( ✞ ) and ❃ ( ✞ ) are the Bessel functions; the ✲ ✬ are positive roots of the transcendental equation

are positive roots of the transcendental equation

6.1.2-8. 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),

= ✓ 2 ( ✠ , ☛ , ✏ ) at ✞ = ✟ 2 (boundary condition),

= ✓ 3 ( ✞ , ✠ , ✏ ) at ☛ =0

(boundary condition),

= ✓ 4 ( ✞ , ✠ , ✏ ) at ☛ = ☞

(boundary condition).

× 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),

) at ✏ =0

(initial condition),

= ✓ 1 ( ✠ , ☛ , ✏ ) at ✞ = ✟

1 (boundary condition),

= ✓ 2 ( ✠ , ☛ , ✏ ) at ✞ = ✟ 2 (boundary condition),

= ✓ 3 ( ✞ , ✠ , ✏ ) at ☛ =0

(boundary condition),

(boundary condition). Solution:

= ✓ 4 ( ✞ , ✠ , ✏ ) at ☛ = ☞

( ✯ ✞ )= ✰ ✱ ( ✲ ✬ ✟ 1 ) ❃ ( ✲ ✬ ✞ )− ❃ ✪ ❂ ✱ ( ✲ ✬ ✟ 1 ) ✰ ( ✲ ✬ ✞ ); the ✰ ( ✞ ) and ❃ ( ✞ ) are the Bessel functions, and the

are positive roots of the transcendental equation

6.1.2-9. Domain: 0 ≤ ✞ ≤ ✟ ,0≤ ✠ ≤ ✠ 0 ,0≤ ☛ ≤ ☞ . First boundary value problem.

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

= ✎ 0 ( ✞ , ✠ , ☛ ) at ✏ =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:

= ✓ 5 ( ✞ , ✠ , ✏ ) at ☛ = ☞

0 ( ✞ ) are the Bessel functions and the ✲ ✬ are positive roots of the transcendental equation ✰ ✗ ■ ❋ 0 ( ✲ ✟ ) = 0.

6.1.2-10. 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),

= ✓ ( ✞ , ☛ , ✏ ) at ✠ = ✠

(boundary condition),

= ✓ ( ✞ , ✠ , ✏ ) at ☛ =0

(boundary condition),

(boundary condition). Solution:

0 = 1 and = 2 for ✷ ≥ 1; the ✰ ✗ ■ ❋ 0 ( ✞ ) are the Bessel functions; and the ✲ ✬ are positive roots of the transcendental equation ✰ ✯ ✯ ✗ ■ ❋ 0 ( ✲ ✟ ) = 0.

6.1.3. Problems in Spherical Coordinates

The three-dimensional wave equation in the spherical coordinate system is represented as

One-dimensional problems with central symmetry that have solutions ✍ = ✍ ( ✞ , ✏ ) are considered in Subsection 4.2.3.

6.1.3-1. 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 ✭ ✭ ❖ ( ✲ ) as

− 1) , and the ✴ ✬ are positive roots of the transcendental equation ✰

6.1.3-2. Domain: 0 ≤ ✞ ≤ ✟ ,0≤ ❑ ≤ ✡ ,0≤ ✠ ≤2 ✡ . Second boundary value problem.

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

= ✎ ( ✞ , ❑ , ✠ ) at ✏ =0

(initial condition),

= ✎ ( ✞ , ❑ , ✠ ) at ✏ =0

(initial condition),

at ✞ = ✟

(boundary condition).

× ❖ (cos ❑ ) ❖ (cos ✚ ) cos[ ( ✠ − ✛ )] sin( ✴ ✬ ✣ ✏

Here, the ✪

+1 ■ 2 ( ✞ ) are the Bessel functions, the ❖ ( ✲ ) are the associated Legendre functions (see Paragraph 6.1.3-1), and the ✴ ✬ are positive roots of the transcendental equation ✪ ✪

Reference : M. M. Smirnov (1975).

6.1.3-3. 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),

at ✞ = ✷ ✟

(boundary condition).

The solution ✪ (

, ❑ , ✠ , ✏ ) is determined by the formula in Paragraph 6.1.3-2 where ❄ ◆ ✪ ❄ ✪ ✪ ✪ ✪

× ❖ (cos ❑ ❄ ✪ ) ❖ (cos ✚ ) cos[ ❯ ( ✠ − ✛ )] sin( ✴ ✬ ✣ ✏ ),

Here, the ✷

+1 ■ 2 ( ✞ ) are the Bessel functions, the ❖ ( ✲ ) are the associated Legendre functions (see Paragraph 6.1.3-1), and the ✴ ✬ ✪ are positive roots of the transcendental equation ✽ ✪

6.1.3-4. 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).

( )] sin( ). Here,

× ❖ (cos ❑ ) ❖ (cos ✚ ) cos[ ✠ − ✛

where the ✪

+1 ■ 2 ( ✞ ) are the Bessel functions, the ❖ ✭ ( ✲ ) are the associated Legendre functions ✪

expressed in terms of the Legendre polynomials ✭ ✭ ❖

and the ✴ ✬ are positive roots of the transcendental equation ❂ +1 ■ 2 ( ✴ ✟ 2 ) = 0.

6.1.3-5. 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:

0 ( , , ) ( , , , , , , ) sin

, , ✏ , 2 , ✠ , , ✚ ✛ ) sin ✚ ✢ ✢ ✚ ✢ ✛

1 ✓ 1 ( ✚ , ✛ , ✤ ) ✜ ( ✞ , ❑ , ✠ , ✟ 1 , ✚ , ✛ , ✏ − ✤ ) sin ✚ ✢ ✚ ✢ ✛ ✢ ✤

2 ✓ 2 ( ✚ , ✛ , ✤ ) ✜ ( ✞ , ❑ , ✠ , ✟ 2 , ✚ , ✛ , ✏ − ✤ ) sin ✚ ✢ ✚ ✢ ✛ ✢ ✤ ,

× ❖ (cos ❑ ) ❖ (cos ✚ ) cos[ ✠ − ✛

Here, the ✪

+1 ■ 2 ( ✞ ) and ❃ +1 ■ 2 ( ✞ ) are the Bessel functions, the ❖ ( ✲ ) are the associated Legendre functions (see Paragraph 6.1.3-4), and the ✴ ✬ are positive roots of the transcendental equation

6.1.3-6. 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),

) at ✏ =0

(initial condition),

1 (boundary condition), + ✍

) at

2 = ✓ 2 ( ❑ , ✠ , ✏ ) at ✞ = ✟ 2 (boundary condition). The solution ✍ ( ✞ , ❑ , ✠ , ✏ ) is determined by the formula in Paragraph 6.1.3-5 where ✪

× ❖ (cos ❑ ) ❖ (cos ✚ ) cos[ ❯ ( ✠ − ✛ )] sin( ✴ ✬ ✣ ✏ ). Here,

where the ✪

+1 ■ 2 ( ✞ ) and ❃ +1 ■ 2 ( ✞ ) are the Bessel functions, the ❖ ( ✲ ) are the associated Legendre functions (see Paragraph 6.1.3-4), and the ✴ ✬ are positive roots of the transcendental equation