Nonhomogeneous Wave Equation ❋
5.2. Nonhomogeneous Wave Equation ❋
5.2.1. Problems in Cartesian Coordinates
5.2.1-1. Domain: − ◆ < ❖ < ◆ ,− ◆ < P < ◆ . Cauchy problem. Initial conditions are prescribed:
Reference : N. S. Koshlyakov, E. B. Gliner, and M. M. Smirnov (1970).
5.2.1-2. Domain: 0 ≤ ❖ ≤ ❣ 1 ,0≤ P ≤ ❣ 2 . First boundary value problem.
A rectangle is considered. The following conditions are prescribed:
= ◗ ( ❖ , P ) at ❘ =0
(initial condition),
= ◗ 1 ( ❖ , P ) at ❘ =0
(initial condition),
= ❚ 1 ( P , ❘ ) at ❖ =0 (boundary condition),
= ❚ 2 ( P , ❘ ) at ❖ = ❣ 1 (boundary condition),
= ❚ 3 ( ❖ , ❘ ) at P =0 (boundary condition),
= ❚ 4 ( ❖ , ❘ ) at P = ❣ 2 (boundary condition).
The solution ❤ ( ❖ , P , ❘ ) is given by the formula in Paragraph 5.1.1-3 with the additional term
which allows for the equation’s nonhomogeneity; this term is the solution of the nonhomogeneous ❛
equation with homogeneous initial and boundary conditions. ❞❢❡
Reference : B. M. Budak, A. A. Samarskii, and A. N. Tikhonov (1980).
5.2.1-3. Domain: 0 ≤ ❖ ≤ ❣ 1 ,0≤ P ≤ ❣ 2 . Second boundary value problem.
A rectangle is considered. The following conditions are prescribed:
= ◗ ( ❖ , P ) at ❘ =0
(initial condition),
= ◗ ( ❖ , P ) at ❘ =0
(initial condition),
= ❚ 1 ( P , ❘ ) at ❖ =0 (boundary condition),
= ❚ 2 ( P , ❘ ) at ❖ = ❣ 1 (boundary condition),
= ❚ 3 ( ❖ , ❘ ) at P =0 (boundary condition),
= ❚ 4 ( ❖ , ❘ ) at P = ❣ 2 (boundary condition). The solution ❤ ( ❖ , P , ❘ ) is given by the formula in Paragraph 5.1.1-4 with the additional term specified in Paragraph 5.2.1-2 (the Green’s function is taken from Paragraph 5.1.1-4). ❞❢❡
References : A. G. Butkovskiy (1979), B. M. Budak, A. A. Samarskii, and A. N. Tikhonov (1980).
5.2.1-4. Domain: 0 ≤ ❖ ≤ ❣ 1 ,0≤ P ≤ ❣ 2 . Third boundary value problem.
A rectangle is considered. The following conditions are prescribed:
= ◗ 1 ( ❖ , P ) at ❘ =0
(initial condition),
2 ( ❖ , P ) at ❘ =0
(initial condition),
− ❤ ❧ 1 = ❚ 1 ( P , ❘ ) at ❖ =0
(boundary condition),
+ ❧ ❤ 2 = ❚ 2 ( P , ❘ ) at ❖ = ❣ 1 (boundary condition),
− ❤ ❧ 3 = ❚ 3 ( ❖ , ❘ ) at P =0 (boundary condition),
+ ❧ ❤ 4 = ❚ 4 ( ❖ , ❘ ) at P = ❣ 2 (boundary condition). The solution ❤ ( ❖ , P , ❘ ) is the sum of the solution to the homogeneous equation with non- homogeneous initial and boundary conditions (see Paragraph 5.1.1-5) and the solution to the nonhomogeneous equation with homogeneous initial and boundary conditions. This solution is given by the formula in Paragraph 5.2.1-2 in which one should substitute the Green’s function of Paragraph 5.1.1-5). ❞❢❡
References : A. G. Butkovskiy (1979), B. M. Budak, A. A. Samarskii, and A. N. Tikhonov (1980).
5.2.1-5. Domain: 0 ≤ ❖ ≤ ❣ 1 ,0≤ P ≤ ❣ 2 . Mixed boundary value problems.
1 s . A rectangle is considered. The following conditions are prescribed:
= ◗ 1 ( ❖ , P ) at ❘ =0
(initial condition),
= ◗ 2 ( ❖ , P ) at ❘ =0
(initial condition),
= ❚ 1 ( P , ❘ ) at ❖ =0 (boundary condition),
= ❚ 2 ( P , ❘ ) at ❖ = ❣ 1 (boundary condition),
= ❚ 3 ( ❖ , ❘ ) at P =0 (boundary condition),
= ❚ 4 ( ❖ , ❘ ) at P = ❣ 2 (boundary condition). The solution ❤ ( ❖ , P , ❘ ) is given by the formula in Paragraph 5.1.1-6, Item 1 s , with the additional term specified in Paragraph 5.2.1-2.
2 s . A rectangle is considered. The following conditions are prescribed:
= ◗ ( ❖ , P ) at ❘ =0
(initial condition),
= ◗ 2 ( ❖ , P ) at ❘ =0
(initial condition),
= ❚ 1 ( P , ❘ ) at ❖ =0 (boundary condition),
= ❚ 2 ( P , ❘ ) at ❖ = ❣ 1 (boundary condition),
= ❚ 3 ( ❖ , ❘ ) at P =0 (boundary condition),
= ❚ 4 ( ❖ , ❘ ) at P = ❣ 2 (boundary condition). The solution ❤ ( ❖ , P , ❘ ) is given by the formula in Paragraph 5.1.1-6, Item 2 s , with the additional term specified in Paragraph 5.2.1-2.
5.2.2. Problems in Polar Coordinates
A nonhomogeneous wave equation in the polar coordinate system has the form
are consid- ered in Subsection 4.2.2.
One-dimensional boundary value problems independent of the angular coordinate ❛
5.2.2-1. Domain: 0 ≤ ✇ ≤ ⑥ ,0≤ ① ≤2 ⑦ . First boundary value problem.
A circle is considered. The following conditions are prescribed:
= t ⑧ 0 ( ✇ , ① ) at ③ =0
(initial condition),
= ⑧ 1 ( ✇ , ① ) at ③ =0
(initial condition),
(boundary condition). The solution ❤ (
at ✇ = ⑥
, ① , ③ ) is given by the formula in Paragraph 5.1.2-1 with the additional term
which allows for the equation’s nonhomogeneity; this term is the solution of the nonhomogeneous ❛ equation with homogeneous initial and boundary conditions. ❞❢❡
References : N. S. Koshlyakov, E. B. Gliner, and M. M. Smirnov (1970), B. M. Budak, A. A. Samarskii, and A. N. Tikhonov (1980).
5.2.2-2. Domain: 0 ≤ ✇ ≤ ⑥ ,0≤ ① ≤2 ⑦ . Second boundary value problem.
A circle is considered. The following conditions are prescribed:
= ⑧ 0 ( ✇ , ① ) at ③ =0
(initial condition),
= ⑧ 1 ( ✇ , ① ) at ③ =0
(initial condition),
at ✇ = ⑥
(boundary condition).
The solution ( ✇ , ① , ③ ) is given by the formula in Paragraph 5.1.2-2 with the additional term (1).
References : A. G. Butkovskiy (1979), B. M. Budak, A. A. Samarskii, and A. N. Tikhonov (1980).
5.2.2-3. Domain: 0 ≤ ✇ ≤ ⑥ ,0≤ ① ≤2 ⑦ . Third boundary value problem.
A circle is considered. The following conditions are prescribed:
= t ⑧ 0 ( ✇ , ① ) at ③ =0
(initial condition),
= ⑧ 1 ( ✇ , ① ) at ③ =0
(initial condition),
(boundary condition). The solution ❤ (
at ✇ = ⑥
, ① , ③ ) is the sum of the solution to the homogeneous equation with nonho- mogeneous initial and boundary conditions (see Paragraph 5.1.2-3) and the solution to the nonho- mogeneous equation with homogeneous initial and boundary conditions [this solution is given by formula (1) in which one should substitute the Green’s function in Paragraph 5.1.2-3]. ❞❢❡
References : A. G. Butkovskiy (1979), B. M. Budak, A. A. Samarskii, and A. N. Tikhonov (1980).
5.2.2-4. Domain: ⑥ 1 ≤ ✇ ≤ ⑥ 2 ,0≤ ① ≤2 ⑦ . First boundary value problem. An annular domain is considered. The following conditions are prescribed:
0 ( ✇ , ① ) at ③ =0
(initial condition),
= ⑧ 1 ( ✇ , ① ) at ③ =0
(initial condition),
= ⑨ 1 ( ① , ③ ) at ✇ = ⑥ 1 (boundary condition),
= ⑨ 2 ( ① , ③ ) at ✇ = ⑥ 2 (boundary condition). The solution ❤ (
, ① , ③ ) is given by the formula in Paragraph 5.1.2-4 with the additional term
which allows for the equation’s nonhomogeneity; this term is the solution of the nonhomogeneous ❛ equation with homogeneous initial and boundary conditions.
5.2.2-5. Domain: ⑥ 1 ≤ ✇ ≤ ⑥ 2 ,0≤ ① ≤2 ⑦ . Second boundary value problem. An annular domain is considered. The following conditions are prescribed:
= ⑧ 0 ( ✇ , ① ) at ③ =0
(initial condition),
= ⑧ 1 ( ✇ , ① ) at ③ =0
(initial condition),
= ⑨ 1 ( ① , ③ ) at ✇ = ⑥ 1 (boundary condition),
= ⑨ 2 ( ① , ③ ) at ✇ = ⑥ 2 (boundary condition). The solution ❤ (
, ① , ③ ) is given by the formula in Paragraph 5.1.2-5 with the additional term (2).
5.2.2-6. Domain: ⑥ 1 ≤ ✇ ≤ ⑥ 2 ,0≤ ① ≤2 ⑦ . Third boundary value problem. An annular domain is considered. The following conditions are prescribed:
= ⑧ 0 ( ✇ , ① ) at ③ =0
(initial condition),
= ⑧ 1 ( ✇ , ① ) at ③ =0
(initial condition),
− 1 = ⑨ 1 ( ① , ③ ) at ✇ = ⑥ 1 (boundary condition),
+ ❧ 2 = ⑨ 2 ( ① , ③ ) at ✇ = ⑥ 2 (boundary condition). The solution ❤ (
, ① , ③ ) is the sum of the solution to the homogeneous equation with nonho- mogeneous initial and boundary conditions (see Paragraph 5.1.2-6) and the solution to the nonho- mogeneous equation with homogeneous initial and boundary conditions [this solution is given by formula (2) in which one should substitute the Green’s function in Paragraph 5.1.2-6].
5.2.2-7. Domain: 0 ≤ ✇ ≤ ⑥ ,0≤ ① ≤ ① 0 . First boundary value problem.
A circular sector is considered. The following conditions are prescribed:
0 ( ✇ , ① ) at ③ =0
(initial condition),
= ⑧ 1 ( ✇ , ① ) at ③ =0
(initial condition),
= ⑨ 1 ( ① , ③ ) at ✇ = ⑥
(boundary condition),
= ⑨ 2 ( ✇ , ③ ) at ① =0
(boundary condition),
= ⑨ 3 ( ✇ , ③ ) at ① = ① 0 (boundary condition). The solution ❤ (
, ① , ③ ) is given by the formula in Paragraph 5.1.2-7 with the additional term t
which allows for the equation’s nonhomogeneity. ❛
5.2.2-8. Domain: 0 ≤ ✇ ≤ ⑥ ,0≤ ① ≤ ① 0 . Second boundary value problem.
A circular sector is considered. The following conditions are prescribed:
= ⑧ 0 ( ✇ , ① ) at ③ =0
(initial condition),
= ⑧ 1 ( ✇ , ① ) at ③ =0
(initial condition),
= ⑨ 1 ( ① , ③ ) at ✇ = ⑥
(boundary condition),
= ⑨ 2 ( ✇ , ③ ) at ① =0
(boundary condition),
= ⑨ 3 ( ✇ , ③ ) at ① = ① 0 (boundary condition). The solution ❤ (
, ① , ③ ) is given by the formula in Paragraph 5.1.2-8 with the additional term (3). ❼
5.2.2-9. Domain: 0 ≤ ✇ ≤ ⑥ ,0≤ ① ≤ ① 0 . Mixed boundary value problem.
A circular sector is considered. The following conditions are prescribed:
= ⑧ 0 ( ✇ , ① ) at ③ =0
(initial condition),
= ⑧ 1 ( ✇ , ① ) at ③ =0
(initial condition),
at ✇ = ⑥
(boundary condition),
at ① =0
(boundary condition),
at ① = ① 0 (boundary condition). The solution ❤ (
, ① , ③ ) is given by the formula in Paragraph 5.1.2-9 with the additional term (3). ❼
5.2.3. Axisymmetric Problems
In the axisymmetric case, a nonhomogeneous wave equation in the cylindrical system of coordinates has the form
5.2.3-1. Domain: 0 ≤ ✇ ≤ ⑥ ,0≤ ❾ ≤ ❣ . First boundary value problem.
A circular cylinder of finite length is considered. The following conditions are prescribed:
= ⑧ 0 ( ✇ , ❾ ) at ③ =0
(initial condition),
= ⑧ 1 ( ✇ , ❾ ) at ③ =0
(initial condition),
= ⑨ 1 ( ❾ , ③ ) at ✇ = ⑥
(boundary condition),
= ⑨ 2 ( ✇ , ③ ) at ❾ =0 (boundary condition),
(boundary condition). The solution ❤ (
= ⑨ 3 ( ✇ , ③ ) at ❾ = ❣
, ❾ , ③ ) is given by the formula in Paragraph 5.1.3-1 with the additional term t
which allows for the equation’s nonhomogeneity; this term is the solution of the nonhomogeneous ❛
equation with homogeneous initial and boundary conditions.
5.2.3-2. Domain: 0 ≤ ✇ ≤ ⑥ ,0≤ ❾ ≤ ❣ . Second boundary value problem.
A circular cylinder of finite length is considered. The following conditions are prescribed:
= ⑧ 0 ( ✇ , ❾ ) at ③ =0
(initial condition),
= ⑧ 1 ( ✇ , ❾ ) at ③ =0
(initial condition),
= ⑨ 1 ( ❾ , ③ ) at ✇ = ⑥
(boundary condition),
= ⑨ 2 ( ✇ , ③ ) at ❾ =0 (boundary condition),
(boundary condition). The solution ❤ (
= ⑨ 3 ( ✇ , ③ ) at ❾ = ❣
, ❾ , ③ ) is given by the formula in Paragraph 5.1.3-2 with the additional term (1).
5.2.3-3. Domain: 0 ≤ ✇ ≤ ⑥ ,0≤ ❾ ≤ ❣ . Third boundary value problem.
A circular cylinder of finite length is considered. The following conditions are prescribed:
= ⑧ 0 ( ✇ , ❾ ) at ③ =0
(initial condition),
= ⑧ 1 ( ✇ , ❾ ) at ③ =0
(initial condition),
+ ❧ 1 = ⑨ 1 ( ❾ , ③ ) at ✇ = ⑥
(boundary condition),
− ❧ ❤ 2 = ⑨ 2 ( ✇ , ③ ) at ❾ =0 (boundary condition),
(boundary condition). The solution ❤ (
+ ❧ 3 = ⑨ 3 ( ✇ , ③ ) at ❾ = ❣
, ❾ , ③ ) is the sum of the solution to the homogeneous equation with nonho- mogeneous initial and boundary conditions (see Paragraph 5.1.3-3) and the solution to the nonho- mogeneous equation with homogeneous initial and boundary conditions [this solution is given by formula (1) in which one should substitute the Green’s function in Paragraph 5.1.3-3].
5.2.3-4. Domain: 0 ≤ ☛ ≤ ☞ ,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),
= ( , ) at
=0 (boundary condition),
(boundary condition). The solution ✏ ( ☛ , ✌ , ✒ ) is given by the formula in Paragraph 5.1.3-4, Item 1 ✎ , with the additional
= ✕ 3 ( ☛ , ✒ ) at ✌ = ✍
term (1).
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). The solution ✏ ( ☛ , ✌ , ✒ ) is given by the formula in Paragraph 5.1.3-4, Item 2 ✎ , with the additional
= ✕ 3 ( ☛ , ✒ ) at ✌ = ✍
term (1). 5.2.3-5. 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),
= ✕ 4 ( ☛ , ✒ ) at ✌ = ✍
(boundary condition).
The solution ✏ ( ☛ , ✌ , ✒ ) is given by the formula in Paragraph 5.1.3-5 with the additional term
which allows for the equation’s nonhomogeneity; this term is the solution of the nonhomogeneous equation with homogeneous initial and boundary conditions.
5.2.3-6. 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