Wave Equation ✂ 2 = ✄ 2

5.1. Wave Equation ✂ 2 = ✄ 2

5.1.1. Problems in Cartesian Coordinates

The wave equation with two space variables in the rectangular Cartesian system of coordinates has the form

5.1.1-1. Particular solutions and some relations.

1 ✌ . Particular solutions:

( , , )= ✍ exp ✎ ✏ 1 + ✏ 2 ✟

( , , )= ✍ sin( ✏ 1 + ✕ 1 ) sin( ✏ 2 + ✕ 2 ) sin ✎ ✟

( , , )= ✍ sin( ✏ 1 + ✕ 1 ) sin( ✏ 2 + ✕ 2 ) cos ✎✖✟

( , , )= ✍ sinh( ✏ 1 + ✕ 1 ) sinh( ✏ 2 + ✕ 2 ) sinh ✎ ✟

( , , )= ✍ sinh( ✏ 1 + ✕ 1 ) sinh( ✏ 2 + ✕ 2 ) cosh ✎ ✟

( , , )= ✗ ( sin ✘ + cos ✘ + ✟ )+ ✙ ( sin ✘ + cos ✘ − ✟ ),

where ✍ , ✕ 1 , ✕ 2 , ✏ 1 , ✏ 2 , and ✘ are arbitrary constants, and ✗ ( ✚ ) and ✙ ( ✚ ) are arbitrary functions.

2 ✌ . Particular solutions that are expressed in terms of solutions to simpler equations:

cos( ✏ )+ ✢ sin( ✏ ☛ ✣✥✤ ✞ )= ☛ ) ☛ ( ,

( ✡ , ☛ , ✞ )= ✛ ✍ cosh( ✏ )+ ✢ sinh( ✏ ) ✣ ✤ ( , ),

For particular solutions of equations (1) and (2) for the function ✞

( , ), see the Klein–Gordon ✡ ☛

equation 4.1.3. For particular solutions of equations (3) and (4) for the function ✤ ( ✡ , ), see Subsection 7.3.2. For particular solutions of the heat equation (5) for the function ✤ ( , ✭ ), see Subsection 1.1.1.

3 ✌ . Fundamental solution: ✰ ✰

4 ✌ . Infinite series solutions that contain arbitrary functions of the space variables: ✹ ✹

( , ) and ( , ✡ ) are any infinitely differentiable functions. The first solution satisfies ☛ ✡ ☛ ✆ ✦ ✡ ☛

the initial conditions ✝ ( , , 0) =

( , ), ✽ ✡ ☛ ( , , 0) = 0 and the second solution to the initial ✡ ☛ ✽ ✡ ☛

conditions ( , , 0) = 0, ( , , 0) = ( , ). The sums are finite if ✶ ( , ) and ( , ) are bivariate polynomials. ✿❁❀

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

5 ✌ . A wide class of solutions to the wave equation with two space variables are described by the formulas

( ❃ ) is an arbitrary analytic function of the complex argument ❃ related to the variables ( , , ) by the implicit relation

( ❃ ) is any analytic function and 0 , 0 are arbitrary constants. Solutions of the forms (6), (7) find wide application in the theory of diffraction. If the argument ❃ obtained by solving (7) with

a prescribed ❄ ( ❃ ) is real in some domain ❅ , then one should set Re ❂ ( ❃ )= ❂ ( ❃ ) in relation (6) everywhere in ❅ ✿❁❀ .

Reference : V. I. Smirnov (1974, Vol. 3, Pt. 2).

6 ☛ . Suppose = ( , , ) is a solution of the wave equation. Then the functions

, ✕ 1 , ❆ ✕ 2 , ✕ 3 , ✏ 1 , ✏ 2 , ✏ 3 , ❇ , and are arbitrary constants, are also solutions of the equation. The signs at in the expression of ✝ 1 can be taken independently of one another. The function ✝ 2 results from the invariance of the wave equation under the Lorentz transformation.

More detailed information about particular solutions and transformations of the wave equation with two space variables can be found in the references cited below. ✿❁❀

References : E. Kalnins and W. Miller, Jr. (1975, 1976), W. Miller, Jr. (1977).

< < ❏ ,− ❏ < < ❏ . Cauchy problem. Initial conditions are prescribed:

5.1.1-2. Domain: − ☛

Solution (Poisson’s formula):

where the integration is performed over the interior of the circle of radius ☛

with center at ( , ).

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

5.1.1-3. 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),

= ❙ 2 (boundary condition). Solution:

) sin( ) sin( ),

The problem of vibration of a rectangular membrane with sides ❙ 1 and ❙ 2 rigidly fixed in its contour is characterized by homogeneous boundary conditions, ❞ ❡ ❣❁❤ ≡0( ❢ = 1, 2, 3, 4).

References : M. M. Smirnov (1964), B. M. Budak, A. A. Samarskii, and A. N. Tikhonov (1980).

5.1.1-4. 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). Solution:

) cos( q

) cos( ❝ ④ r ) sin( ❳ ④ ✈

References : A. G. Butkovskiy (1979), B. M. Budak, A. A. Samarskii, and A. N. Tikhonov (1980).

5.1.1-5. Domain: 0 ≤ t ≤ ✐ 1 ,0≤ ✉ ≤ ✐ 2 . Third boundary value problem.

A rectangle is considered. The following conditions are prescribed:

= ❦ 0 ( t , ✉ ) at ✈ =0 (initial condition),

= ❦ 1 ( t , ✉ ) at ✈ =0 (initial condition),

− 1 = 1 ( , ) at

=0 (boundary condition),

+ 2 = 2 ( , ) at

= 1 (boundary condition),

3 ( t , ✈ ) at ✉ =0 (boundary condition),

+ ❸ 4 = ❞ 4 ( t , ✈ ) at ✉ = ✐ 2 (boundary condition). The solution ❥ ( t , ✉ , ✈ ) is determined by the formula in Paragraph 5.1.1-4 where

( t , ✉ , q , r , ✈ )=

sin( ❝ ❻ t

× sin( ❻ ❝ q + ❽ ❝ ) sin( ❼ ④ r + ❾ ④ ) sin ❿➀❳ ✈✓❜ ❻ 2 ❝ + ❼ 2 ④ ➁ ,

= arctan , ❾ ④

= arctan ❝ , ④ = ➂✧✐

where the ❻ ❝ and ❼ ④ are positive roots of the transcendental equations

2 − ❸ 1 ❸ 2 =( ❸ 1 + ❸ 2 ) ❻ cot( ✐ 1 ❻ ), ❼ 2 − ❸ 3 ❸ 4 =( ❸ 3 + ❸ 4 ) ❼ cot( ✐ 2 ❼ ).

References : B. M. Budak, A. A. Samarskii, and A. N. Tikhonov (1980).

5.1.1-6. Domain: 0 ≤ t ≤ ✐ 1 ,0≤ ✉ ≤ ✐ 2 . Mixed boundary value problems.

1 ➋ . A rectangle is considered. The following conditions are prescribed:

= ❦ 0 ( t , ✉ ) at ✈ =0

(initial condition),

= ❦ 1 ( t , ✉ ) at ✈ =0

(initial condition),

= ❞ 1 ( ✉ , ✈ ) at t =0 (boundary condition),

= ❞ 2 ( ✉ , ✈ ) at t = ✐ 1 (boundary condition),

= 3 ( , ) at

=0 (boundary condition),

= ❞ 4 ( t , ✈ ) at ✉ = ✐ 2 (boundary condition). Solution:

( ❝ , , , , )= sin( ) cos( ) sin( q ) cos( ⑧ ④

2 ➋ . A rectangle is considered. The following conditions are prescribed:

= ❦ 0 ( t , ✉ ) at ✈ =0

(initial condition),

= ❦ 1 ( t , ✉ ) at ✈ =0

(initial condition),

= ❞ 1 ( ✉ , ✈ ) at t =0 (boundary condition),

= ❞ 2 ( ✉ , ✈ ) at t = ✐ 1 (boundary condition),

= ❞ 3 ( t , ✈ ) at ✉ =0 (boundary condition),

= ❞ 4 ( t , ✈ ) at ✉ = ✐ 2 (boundary condition).

) sin( ) sin( ),

5.1.2. Problems in Polar Coordinates

The wave equation with two space variables in the polar coordinate system has the form

= ➒ ( ➏ , ✈ ) that are independent of the angular coordinate ➐ are considered in Subsection 4.2.1.

One-dimensional solutions ➐

5.1.2-1. Domain: 0 ≤ ➏ ≤ ➓ ,0≤ ➐ ≤2 ⑩ . First boundary value problem.

A circle is considered. The following conditions are prescribed:

= ➔ 0 ( ➏ , ➐ ) at ✈ =0

(initial condition),

= 1 ( ➏ , ➐ ) at ✈ =0

(initial condition),

(boundary condition). Solution:

, )= ↕ ➓ 2 ❻ ➞ ④ ➞ ➟ ❻ ➜ ④ ➓ 2 ( ➏ ) ( ❻ ④ q ) cos[ ⑨ ( ➐ − r )] sin( ❻ ④ ↕ ✈ ),

where the ➜

( q ) are the Bessel functions (the prime denotes the derivative with respect to the ⑤ ⑤

argument) and the ❻ ④ are positive roots of the transcendental equation ➞ ( ❻ ➓ ) = 0. The problem of vibration of a circular membrane of radius ➓ rigidly fixed in its contour is ➡❁➢ characterized by the homogeneous boundary condition, ❞ ( , ✈ ➐ ) ≡ 0.

References : N. S. Koshlyakov, E. B. Gliner, and M. M. Smirnov (1970), A. G. Butkovskiy (1979), B. M. Budak, A. A. Samarskii, and A. N. Tikhonov (1980).

5.1.2-2. Domain: 0 ≤ ➏ ≤ ➓ ,0≤ ➐ ≤2 ⑩ . Second boundary value problem.

A circle is considered. The following conditions are prescribed:

= ➔ 0 ( ➏ , ➐ ) at ✈ =0

(initial condition),

= 1 ( ➏ , ➐ ) at

(initial condition),

= ➙ ➓ ➤ ➐ ➏ (boundary condition). Solution:

2 cos[ ( − )] sin( ),

where the ➜

( q ) are the Bessel functions and the ⑤ ⑤ ❻ ④ are positive roots of the transcendental equation

References : A. G. Butkovskiy (1979), B. M. Budak, A. A. Samarskii, and A. N. Tikhonov (1980).

5.1.2-3. Domain: 0 ≤ ➏ ≤ ➓ ,0≤ ➐ ≤2 ⑩ . Third boundary value problem.

A circle is considered. The following conditions are prescribed:

, ➐ ) at

(initial condition),

= ➙ ➔ 1 ( ➏ , ➐ ) at

(initial condition),

(boundary condition). The solution ➒ (

at

, ➐ , ) is determined by the formula in Paragraph 5.1.2-2 where ➜ ➜ ➜ ➜ ➜ ➜

2 cos[ ( − )] sin( ),

Here, the ➞ ( q ) are the Bessel functions and the ⑤ ➜ ⑤ ❻ ④ ➜ are positive roots of the transcendental equation

5.1.2-4. Domain: ➓ 1 ≤ ➏ ≤ ➓ 2 ,0≤ ➐ ≤2 ⑩ . First boundary value problem. An annular domain is considered. The following conditions are prescribed:

= ➔ 0 ( ➏ , ➐ ) at ✈ =0

(initial condition),

= ➔ 1 ( ➏ , ➐ ) at ✈ =0

(initial condition),

= ❞ 1 ( ➐ , ✈ ) at ➏ = ➓ 1 (boundary condition),

= ❞ 2 ( ➐ , ✈ ) at ➏ = ➓ 2 (boundary condition).

) ➦ ( ❻ ➜ q ) cos[ ➜ ➜ ( − r )] sin( ❻ ④ ↕ ✈ ),

( ➜ ❻ ④ )= ➞ ( ❻ ④ ➓ 1 ) ➩ ( ❻ ④ )− ➩ ( ❻ ➜ ④ ➏ ➓ ➏ 1 ) ➞ ( ❻ ④ ➏ ), where the ➞ ( ) and ➩ ( ) are the Bessel functions, and the ❻ ④ ➏ ➏ ➜ ➜ ➜ ➜ are positive roots of the transcen-

dental equation

≤ 2 ,0≤ ➐ ≤2 ⑩ . Second boundary value problem. An annular domain is considered. The following conditions are prescribed:

5.1.2-5. Domain: ➓ 1 ≤

= ➔ 0 ( ➏ , ➐ ) at ➫ =0

(initial condition),

= ➔ 1 ( , ) at ➙ ➫ → ➏ ➐ =0

(initial condition),

= 1 ( ➐ , ) at ➏ = 1 (boundary condition),

= ➏ ➓ 2 (boundary condition). Solution:

) cos[ ( − )] sin( )

where ➶ 0 = 1 and = 2 for = 1, 2, ➱✖➱✖➱ ; the ➬ ( ➵ ⑨ ) and ➩ ( ➵ ) are the Bessel functions; and the

are positive roots of the transcendental equation ➘ ➘ ➶ ➴ ➶ ➴ ➶ ➴

≤2 ⑩ . Third boundary value problem. An annular domain is considered. The following conditions are prescribed:

5.1.2-6. Domain: ➺ 1 ≤ ➵ ≤ 2 ,0≤

) at Ò =0

(initial condition),

= Ñ 1 ( ➵ , ) at Ò =0

(initial condition),

= ❞ 1 ( ➐ , Ò ) at ➵ = 1 (boundary condition),

, Ò ) at ➵ = 2 (boundary condition). The solution Ð ( ➵ , , Ò

) is determined by the formula in Paragraph 5.1.2-5 where ➶ ➴ ➶ ➶ ➴ ➶ ➶ ➴ ➶ ➴ ➶

) cos[ ( − ➲ )] sin(

Here, ➶ 0 = 1 and = 2 for Ù = 1, 2, ➱✖➱✖➱ ; the ➬ ( ➵ ) and Ø ( ➵ ) are the Bessel functions; and the

are positive roots of the transcendental equation ➴ ➶ ➴ ➶ ➴ ➴ ➴ ➶ ➴ ➘ ➘ ➶

5.1.2-7. Domain: 0 ≤ ➵

≤ ➐ 0 . First boundary value problem.

A circular sector is considered. The following conditions are prescribed:

= Ñ 0 ( ➵ , ) at Ò ➐ =0

(initial condition),

) at

(initial condition),

= ❞ 1 ( , Ò ➐ ) at ➵

(boundary condition),

= ❞ 2 ( ➵ , Ò ) at ➐ =0

(boundary condition),

= ❞ 3 ( ➵ , Ò ) at ➐ = ➐ 0 (boundary condition). Solution:

where the ➴

are positive roots of the transcendental equation ➬

0 ( ➵ ) are the Bessel functions and the Ú

5.1.2-8. Domain: 0 ≤ ➵

≤ ➐ 0 . Second boundary value problem.

A circular sector is considered. The following conditions are prescribed:

0 ( ➵ , ➐ ) at Ò =0

(initial condition),

= Ñ 1 ( ➵ , ) at Ò =0

(initial condition),

= ❞ 1 ( ➐ , Ò ) at ➵ =

(boundary condition),

= ❞ 2 ( ➵ , Ò ) at ➐ =0

(boundary condition),

= ❞ 3 ( ➵ , Ò ) at ➐ = ➐ 0 (boundary condition). Solution: Ó

are positive roots of the transcendental equation ➺ ➬ ➮

where the ➬ ➼ à 0 ( ➵ ) are the Bessel functions and the

5.1.2-9. Domain: 0 ≤ ➺ ➵ ≤ ,0≤

≤ ➐ 0 . Mixed boundary value problem.

A circular sector is considered. The following conditions are prescribed:

= Ñ 0 ( ➵ , ) at Ò =0

(initial condition),

) at

(initial condition),

at ➵

(boundary condition),

at ➐ =0

(boundary condition),

at ➐ = ➐ 0 (boundary condition). Solution: Ó

) cos( ➴ ➶ ➐ ) cos( ) sin( ),

where the ➬ ❡äå ( ➵ ) are the Bessel functions and the

are positive roots of the transcendental equation

5.1.3. Axisymmetric Problems

In the axisymmetric case the wave equation in the cylindrical system of coordinates has the form

One-dimensional problems with axial symmetry that have solutions Ð = ( ➵ , Ò ) are considered in Subsection 4.2.1.

In the solution of the problems considered below, the modified Green’s function æ

2 â ➯ ➳ ( ➵ , , ➯ , ➲ , Ò ) is used for convenience. 5.1.3-1. Domain: 0 ≤ æ

≤ ➺ ,0≤ ≤ ë . First boundary value problem.

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

0 ( ➵ , ) at Ò =0

(initial condition),

1 ( ➵ , ) at Ò =0

(initial condition),

= ❞ 1 ( , Ò ) at ➵ =

(boundary condition),

2 ( ➵ , Ò ) at

=0 (boundary condition),

(boundary condition). Solution:

where the

are positive zeros of the Bessel function, ➬ 0 ( ) = 0.

5.1.3-2. Domain: 0 ≤ æ

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

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

0 ( ñ , ) at Ò =0

(initial condition),

1 ( ñ , ) at Ò =0

(initial condition),

) at ñ =

(boundary condition),

, Ò ) at

=0 (boundary condition),

, Ò ) at

(boundary condition).

where the ✆ ✁

are zeros of the first-order Bessel function, ☎ 1 (

5.1.3-3. Domain: 0 ≤ ✡

≤ ,0≤ ✠ ≤ . Third 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 = ✔ 1 ( , ☛ ) at ✏ = ✕ ✠

(boundary condition),

− ✍ ✓ 2 = ✔ 2 ( ✏ , ☛ ) at

=0 (boundary condition),

(boundary condition). The solution ✍ ( ✏ ,

3 ( ✏ , ☛ ) at ✠ =

, ☛ ) is determined by the formula in Paragraph 5.1.3-2 where

= ✕ 2 + ★ ✂ 2 , 2 ✂ ✚ ( ) = cos( ★ ✂ )+ ✂ sin( ★ ✂ ✠ ✠ ★ ✠ ),

Here, the ✆ ✁ and ★ ✂ are positive roots of the transcendental equations

tan( ★ )

5.1.3-4. Domain: 0 ≤ ✡

≤ ✕ ,0≤ ✠ ≤ . Mixed boundary value problems.

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

) at

(initial condition),

= ✎ 1 ( ✏ , ✠ ) at ☛ =0

(initial condition),

= ✔ 1 ( , ☛ ) at ✏ = ✕

(boundary condition),

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

(boundary condition). Solution:

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:

3 ( ✏ , ☛ ) at ✠ =

are zeros of the first-order Bessel function, 1 ( ✼ )=0( ✼ 0 = 0). 5.1.3-5. 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:

where 0 ( ✼ ) and ❃ 0 ( ✼ ) are the Bessel functions, and the ✼ ✁ are positive roots of the transcendental equation

) ❃ 0 ( ❀❁✼ )− 0 ( ❀❁✼ ) ❃ 0 ( ✼ ) = 0.

5.1.3-6. Domain: ✡

1 ≤ ✏ ≤ ✕ 2 ,0≤ ✠ ≤ . Second boundary value problem.

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

) 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).

) and ❃ ( ✼ ) are the Bessel functions ( ✓ = 0, 1); and the ✼ ✁ are positive roots of the transcendental equation