Telegraph Equation ⑦ ⑦
5.4. Telegraph Equation ⑦ ⑦
5.4.1. Problems in Cartesian Coordinates
A two-dimensional nonhomogeneous telegraph equation in the rectangular Cartesian coordinate system is written as
5.4.1-1. Reduction to the two-dimensional Klein–Gordon equation. ➀ ➁ ❾ ❾➇➆☞➈ ➀ ➁ ❾
The substitution ❽ ( ❼ , ➈ , ) = exp ❼ ➅ ➈ − 1 2 ⑥ ❼ ➈ ( , , ) leads to the equation
( , , ), which is discussed in Subsection 5.3.1.
+ exp ➅ 2 ⑥
5.4.1-2. Fundamental solutions.
2 + 2 and ➛ ( ➠ ) is the Heaviside unit step function.
1 2 . Case 2 ➃ −
1 cosh ➅➉↕ ➝ ❾ 2 − ✽ 2 ➞ ❿ ( 2 , , )= ( − ) exp −
Reference : V. S. Vladimirov, V. P. Mikhailov, A. A. Vasharin, et al. (1974).
5.4.1-3. Domain: − ➥ < < ➥ ,− ➥ < < ➥ . Cauchy problem. Initial conditions are prescribed:
5.4.1-4. Domain: 0 ≤ ≤ ➹ 1 ,0≤ ≤ ➹ 2 . First boundary value problem.
A rectangle is considered. The following conditions are prescribed: ➀ ➁ ❾
= ➦ 0 ( ➀ , ➁ ) at ❾ =0
(initial condition),
= ➦ 1 ( ➁ , ❾ ) at ➀ =0
(initial condition),
= ➨ 1 ( ➁ , ❾ ) at ➀ =0 (boundary condition),
= ➨ 2 ( ➀ , ❾ ) at ➁ = ➹ 1 (boundary condition),
= ➨ 3 ( ➀ , ❾ ) at ➁ =0 (boundary condition),
= ➨ 4 ( , ) at
= ➹ 2 (boundary condition).
) sin( Ò ❰ Ó ) sin( Ð ➪ ) sin( Ò ❰ ➶ ) sin( ❰ Ô ),
5.4.1-5. Domain: 0 ≤ ≤ ➹ 1 ,0≤ Ó ≤ ➹ 2 . Second boundary value problem.
A rectangle is considered. The following conditions are prescribed: Ñ
= Ñ Ü 0 ( , Ó ) at Ô =0
(initial condition),
1 Ñ ( , Ó ) at Ô =0
(initial condition),
= 1 ( , ) at
=0 (boundary condition),
= ß 2 ( Ó , Ô ) at
= ➹ 1 (boundary condition),
= ß ➱ Ñ 3 ( , Ô ) at Ó =0 (boundary condition),
= ß 4 ( , Ô ) at Ó = ➹ 2 (boundary condition). Solution: Ñ
5.4.1-6. Domain: 0 ≤ æ ≤ í 1 ,0≤ ç ≤ í 2 . Third boundary value problem.
A rectangle is considered. The following conditions are prescribed:
( æ , ç ) at è =0 (initial condition),
= Ü 1 ( æ , Ý ç ) at è =0 (initial condition),
− ✌ Û 1 = ✍ 1 ( ç , è ) at æ =0 (boundary condition),
+ Û ✌ 2 = ✍ 2 ( ç , è ) at æ = í 1 (boundary condition),
− Û ✌ 3 = ✍ 3 ( æ , è ) at ç =0 (boundary condition),
+ Û ✌ 4 = ✍ 4 ( æ , è ) at ç = í 2 (boundary condition). The solution Û ( æ , ç , è ) is determined by the formula in Paragraph 5.4.1-5 where
( ÿ , , ) = 4 exp −
sin( ✑ æ + ✓ ) sin( ✒ ❰ ç + ✔ ❰ )
× sin( ✑ ã + ✓ ) sin( ✒ ❰ ä + ✔ ❰ ) sin ✕☞è ✖✝ ✞ 2 ✑ 2 + ✞ 2 ✒ 2 ✗ + ✟ − 1 2 4 ✘ ✠ . Here,
= arctan , ✔ = arctan
3 )( ✌ + ✒ 2 4 ✗ ) the ✑ and ✒ ✗ are positive roots of the transcendental equations
2 − ✌ 1 ✌ 2 =( ✌ 1 + 2 ✌ 2 ) ✑ cot( í 1 ✑ ), ✒ − ✌ 3 ✌ 4 =( ✌ 3 + ✌ 4 ) ✒ cot( í 2 ✒ ).
5.4.1-7. Domain: 0 ≤ æ ≤ í 1 ,0≤ ç ≤ í 2 . Mixed boundary value problems.
1 ➣ . A rectangle is considered. The following conditions are prescribed:
= Ü 0 ( æ , ç ) at è =0
(initial condition),
= Ü 1 ( æ , ç ) at è =0
(initial condition),
= ✍ 1 ( ç , è ) at æ =0 (boundary condition),
= ✍ 2 ( ç , è ) at æ = í 1 (boundary condition),
= ✍ 3 ( æ , è ) at ç =0 (boundary condition),
= ✍ 4 ( æ , è ) at ç = í 2 (boundary condition).
) sin( ) cos( ) sin( ),
2 ➣ . A rectangle is considered. The following conditions are prescribed:
= Ü 0 ( æ , ç ) at ☛ è =0 (initial condition),
= Ü 1 ( æ , ç ) at è =0
(initial condition),
= ( , ) at
=0 (boundary condition),
= ✍ 2 ( ç , è ) at æ = í 1 (boundary condition),
( æ , è ) at ç =0 (boundary condition),
= ✍ 4 ( æ , è ) at ç = í 2 (boundary condition). Solution:
5.4.2. Problems in Polar Coordinates
A two-dimensional nonhomogeneous telegraph equation in the polar coordinate system has the form
For one-dimensional solutions ✭ = ( ,
), see equation 4.4.2.2.
5.4.2-1. Domain: 0 ≤ ✮ ≤
,0≤ ≤2 ✤ . First 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:
=2 ( ✥ = 1, 2, ❇✄❇✄❇ ), 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.
5.4.2-2. 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),
at
(boundary condition).
=2 ( ✥ = 1, 2, ❇✄❇✄❇ ), where the ❁ ✚ ( ✏ ) are the Bessel functions and the ❀ ❂ ✛ are positive roots of the transcendental equation ❀
5.4.2-3. 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),
at
(boundary condition).
The solution ✮ ( , ,
) is determined by the formula in Paragraph 5.4.2-2 where
( ❂ ✚ ✛ ) ❁ ✚ ( ❂ ✚ ✛ ✏ ) cos[ ✥ ( − ✑ )] sin ✔❄✒❆❅
=2 ( ✥ = 1, 2, ❇✄❇✄❇ ). Here, the ❁ ✚
( ❀ ✏ ) are the Bessel functions and the ❂ ✛ are positive roots of the transcendental equation
1 ≤ ≤ ✲ 2 ,0≤ ≤2 ✤ . First boundary value problem. An annular domain is considered. The following conditions are prescribed:
5.4.2-4. Domain: ✮
0 ( , ) at ✒ =0
(initial condition),
1 ( , ) at ✒ =0
(initial condition),
1 ( , ✒ ) at
= ✲ 1 (boundary condition),
2 ( , ✒ ) at
= ✲ 2 (boundary condition).
sin ) cos[ ✥ ( − ✑ )]
= ★ ❂ ✚ ✛ + ✩ − 1 4 ✕ , where the ❁ ✚ ✭
( ) are the Bessel functions, and the ❂ ✚ ✛ are positive roots of the transcen- dental equation
( ✭ ) and
1 ≤ ≤ ✲ 2 ,0≤ ≤2 ✤ . Second boundary value problem. An annular domain is considered. The following conditions are prescribed:
5.4.2-5. Domain: ✮
0 ( , ) at ✒ =0
(initial condition),
1 ( , ) at ✒ =0
(initial condition),
1 ( , ✒ ) at
= ✲ 1 (boundary condition),
= ✲ 2 (boundary condition). Solution:
2 ( , ✒ ) at
( ❂ ✚ ✛ ) ■ ✚ ( ❂ ✚ ✛ ✏ ) cos[ ✥ ( − ✑ )] sin ✔❄✒ ✱ ★ 2 ❂ ✚ 2 ✛ + ✩ − 2 ✕ ❊ 4 ✖
( ❀ ) are the Bessel functions, and the ❂ ✚ ✛ are positive roots of the transcendental equation
1 ≤ ≤ ✲ 2 ,0≤ ≤2 ✤ . Third boundary value problem. An annular domain is considered. The following conditions are prescribed:
5.4.2-6. Domain: ✮
= ✳ 0 ( ✭ , ✮ ) at ✒ =0
(initial condition),
= ( , ) at
(initial condition),
= ✵ 1 ( ✮ , ✒ ) at
= ✲ ✭ 1 (boundary condition),
+ ● 2 ✫ = ✵ 2 ( , ✒ ) at
= ✲ 2 (boundary condition).
The solution ✮ ( , ,
) is determined by the formula in Paragraph 5.4.2-5 where
exp ✔ − 2 ✕ ✒✗✖
( ❂ ✚ ✛ ) ■ ✚ ( ❂ ✚ ✛ ✏ ) cos[ ✥ ( − ✑ )] sin( ✚ ✛ ✒ ).
where the ✺
( ) and ❑ ✚ ( ) are the Bessel functions, and the ❂ ✚ ✛ are positive roots of the transcen- dental equation
5.4.2-7. Domain: 0 ≤ ✮ ≤
,0≤ ≤ 0 . First boundary value problem.
A circular sector is considered. The following conditions are prescribed:
0 ( , ) at ✒ =0
(initial condition),
1 ( , ) at ✒ =0
(initial condition),
1 ( , ✒ ) at
(boundary condition),
2 ( , ✒ ) at
(boundary condition),
3 ( , ✒ ) at
= 0 (boundary condition).
+ ✩ − 2 ✕ ❊ 4 where the ✭ ❁
0 ( ) are the Bessel functions and the ❂ ✚ ✛ are positive roots of the transcendental equation ❁ ✚ ✷ ❖ 0 ( ❂ ✲ ) = 0.
5.4.2-8. Domain: 0 ≤ ✮ ≤
,0≤ ≤ 0 . Second boundary value problem.
A circular sector is considered. The following conditions are prescribed:
0 ( , ) at ✒
(initial condition),
= ( ✮ , ) at ✒ ✭ =0
(initial condition),
= ✵ 1 ( , ✒ ) at
(boundary condition),
(boundary condition),
= 0 (boundary condition). Solution:
= ✵ 3 ( , ✒ ) at
0 ( ) are the Bessel functions and the ❂ ✚ ✛ are positive roots of the transcendental equation ❁ ✚ ❃ ✷ ❖ ▼ 0 ( ❂ ✲ ) = 0.
5.4.2-9. Domain: 0 ≤ ✮ ≤
,0≤ ≤ 0 . Mixed boundary value problem.
A circular sector is considered. The following conditions are prescribed:
0 ( , ) at ✒ =0
(initial condition),
1 ( , ) at ✒ =0
(initial condition),
at ✮ = ✲
(boundary condition),
at ✮ =0 ✮
(boundary condition),
= 0 (boundary condition). Solution:
) cos( ● ❙ ) sin ❱❝❜ ❭ ❩ ◗✗❲ ,
where the ❥ ❪❴❫ ( ❵ ) are the Bessel functions and the ❣ ❭ are positive roots of the transcendental equation
( ❣ ❤ )+ P ❥ ❪❴❫ ( ❣ ❤ ) = 0.
5.4.3. Axisymmetric Problems
In the axisymmetric case, a nonhomogeneous telegraph equation in the cylindrical coordinate system has the form
2 + ❵ ✪ ❵ + ✪ ♣ 2 q − ✩ + r ( ❵ , , ◗ ),
✈✄①
✈✄①
In the solutions of the problems considered below, the modified Green’s function ♣
( ❵ , , ❷ , ❸ , ◗ )=
2 ❷ ❹ ( ❡ ❵ , , ❷ , ❸ , ◗ ) is used for convenience. 5.4.3-1. Domain: 0 ≤ ♣
≤ ❤ ,0≤ ≤ ❺ . First boundary value problem.
A circular cylinder of finite length is considered. The following conditions are prescribed:
0 ( ❵ , ) at ◗
=0
(initial condition),
= ❻ 1 ( ❵ ♣ , ) at ◗ =0
(initial condition),
= ❽ 1 ( , ◗ ) at ❵ ♣ = ❤
(boundary condition),
= ❽ 2 ( ❵ , ◗ ) at ♣ =0 (boundary condition),
(boundary condition). Solution:
where the →
are positive zeros of the Bessel function, 0 ( ➣ ) = 0.
5.4.3-2. Domain: 0 ≤ ➜
≤ ,0≤ ➋ ≤ . Second boundary value problem.
A circular cylinder of finite length is considered. The following conditions are prescribed:
= ➢ 0 ( ➁ , ➋ ) at ➟ =0
(initial condition),
= ➢ 1 ( ➁ , ➋ ) at ➟ =0
(initial condition),
1 ( ➋ , ➟ ) at ➁ =
(boundary condition),
= ➦ 2 ( ➁ , ➟ ) at ➋ =0 (boundary condition),
3 ( ➁ , ➟ ) at ➋ =
(boundary condition).
2 for > 0, and the ❐ ➬ are zeros of the first-order Bessel function, 1 ( ➬ )=0( ➬ 0 = 0).
5.4.3-3. Domain: 0 ≤ ❮
≤ ,0≤ ➋ ≤ . Third boundary value problem.
A circular cylinder of finite length is considered. The following conditions are prescribed:
= × 0 ( Õ , Ø ) at Ï =0
(initial condition),
= × 1 ( Õ , Ø ) at Ï =0
(initial condition),
+ Ö Ü 1 = Ý 1 ( Ø , Ï ) at Õ = Þ
(boundary condition),
− Ö Ü 2 = Ý 2 ( Õ , Ï ) at Ø =0 (boundary condition),
(boundary condition). The solution Ö ( Õ ✃ , Ø , Ï ) is determined by the formula in Paragraph 5.4.3-2 where ➹ ➹ ✃
the å ➬ and ➘ are positive roots of the transcendental equations
tan( ❮
5.4.3-4. Domain: 0 ≤ ❮
≤ Þ ,0≤ Ø ≤ . Mixed boundary value problems.
1 ò . A circular cylinder of finite length is considered. The following conditions are prescribed:
= × 0 ( Õ , Ø ) at Ï =0
(initial condition),
= × 1 ( Õ , Ø ) at Ï =0
(initial condition),
= Ý 1 ( Ø , Ï ) at Õ = Þ
(boundary condition),
= Ý 2 ( Õ , Ï ) at Ø =0 (boundary condition),
(boundary condition). Solution:
> 0, where the ❐
4 2 for
are 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: ✕
3 ( ✒ , ✔ ) at ✓ =
5.4.3-5. 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:
1 1 1 1 4 where ❅ 0 ( ❄ ) and ❃ 0 ( ❄ ) are the Bessel functions, and the ❄
are positive roots of the transcendental equation