Equations of the Form ê
4.4.1. Equations of the Form ê
For Ø õ ( ö , Ü ) ≡ 0, this equation governs free transverse vibration of a string, and also longitudinal vibration of a rod in a resisting medium with a velocity-proportional resistance coefficient.
1 ä ÷ . The substitution ( ö , Ü ) = exp ø − 1
2 ù Ü■ú➺û ( ö , Ü ) leads to the equation
2 + 1 4 2 ù û + exp ø 1 2 ù Ü■ú õ ( ö , Ü ),
which is considered in Subsection 4.1.3.
2 ÷ . Fundamental solution: ý ý
, ✆✞✝ where þ ( ☎ ) is the Heaviside unit step function and 0 ( ☎ ) is the modified Bessel function.
−| ö | ú exp ø − 1 2 ù Ü■ú ✁ 0 ø 1 Ü 2 ö 2 ✄ è 2 ú
Reference : V. S. Vladimirov, V. P. Mikhailov, A. A. Vasharin, et al. (1974).
3 ÷ . Domain: − Û < ö < Û . Cauchy problem. Initial conditions are prescribed:
where 0 ( ★ ) and 1 ( ★ ) are the modified Bessel functions of the first kind. ☞
. Domain: 0 ≤ ✒ ≤ ✪ . First boundary value problem. The following conditions are prescribed:
= ✬ ✟ 0 ( ✒ ) at ✑ =0 (initial condition),
= ✟ 1 ( ✒ ) at ✑ =0 (initial condition),
= ✢ 1 ( ✑ ) at ✒ =0 (boundary condition),
(boundary condition). Solution:
= ✢ 2 ( ✑ ) at ✒ = ✪
Example. Consider the homogeneous equation ( ✑ ≡ 0). The initial shape of the string is a triangle with base 0 ≤ ✒ ≤ ✓
and height ✔ at ✒ = ✕ , that is,
The initial velocities of the string points are zero, ✛ ( ✒ ) = 0. Solution:
References : M. M. Smirnov (1975), B. M. Budak, A. N. Tikhonov, and A. A. Samarskii (1980). ☎
5 ✶ . For the second and third boundary value problems on the interval 0 ≤ ✁ ≤ , see equation 4.4.1.2 (Items 5 ✶ and 6 ✶ with ✷ = 0).
Telegraph equation ✽ (with
> 0, ✷ < 0, and ❀ ( ✁ , ✄ ) ≡ 0).
1 ✶ . The substitution ❁ ( ✁ , ✄ ) = exp ❂ − 1 2 ✝ ✄❄❃❆❅ ( ✁ , ✄ ) leads to the equation
which is considered in Subsection 4.1.3.
2 ✶ . Fundamental solutions:
where ( ❑ ) is the Heaviside unit step function, ❏ 0 ( ❑ ) and ❏ 1 ( ❑ ) are the Bessel functions, and ❊ 0 ( ❑ ) and ❊ 1 ( ❉ ❑ ) are the modified Bessel functions.
3 ✶ . Domain: − ▲ < ✁ < ▲ . Cauchy problem. Initial conditions are prescribed:
= ▼ ( ✁ ) at ✄ = 0,
= ❖ ( ✁ ) at ✄ = 0.
Solution for ✷
Solution for ❲
4 ♣ . Domain: 0 ≤ ❭ ≤ q . First boundary value problem. ❙ ♥
The following conditions are prescribed:
= ❵ ❇ 0 ( ❭ ) at ❬ =0 (initial condition),
= ❵ 1 ( ❭ ) at ❬ =0 (initial condition),
= ❡ 1 ( ❬ ) at ❭ =0 (boundary condition),
(boundary condition). Solution:
4 Let ❴ 2 ✇ 2 ⑤ 2 − ❦♠q 2 − 1 2 q 2 ≤ 0 for ⑤ = 1, ⑨⑩⑨⑩⑨ , ❶ and ❴ 2 ✇ 2 ⑤ 2 − ❦♠q 2 − 1 4 2 ❜ 4 ❜ q 2 > 0 for ⑤ = ❶ + 1, ❶ + 2, ⑨⑩⑨⑩⑨ ✡
Then
sinh ⑥ ❸
( ❭ , ❪ , ❬ )= exp ①
sin ①
sin
sin ❳ ❬⑦⑥ ⑧ ❃
exp − ②
sin ①
sin ①
5 ♣ . Domain: 0 ≤ ❭ ≤ q . Second boundary value problem. The following conditions are prescribed:
0 ( ❭ ) at ❬ =0 (initial condition),
1 ( ❭ ) at ❬ =0 (initial condition), = ❡ 1 ( ❬ ) at ❭ =0 (boundary condition),
(boundary condition). Solution: ❙
. If the inequality ✡ ❴ 2
2 − ✡ ❹ < 0 holds for several first values ⑤ = 1, ⑨⑩⑨⑩⑨ , ❶ , then the expressions
2 ❻ 2 − ❹ should be replaced by ❩ | ❴ 2 ❻ 2 − ❹ | and the sines by the hyperbolic sines in the corre- sponding terms of the series.
6 ♣ . Domain: 0 ≤ ❭ ≤ q . Third boundary value problem. The following conditions are prescribed:
0 ( ❭ ) at ❬ =0 (initial condition),
= ❵ 1 ( ❭ ) at ❇ ❬ =0 (initial condition),
− ❼ ❧ 1 = ❡ 1 ( ❇ ❬ ) at ❭ =0 (boundary condition),
+ ❼ ❧ 2 = ❡ 2 ( ❬ ) at ❭ = q
(boundary condition).
The solution ❙ (
, ❬ ) is determined by the formula in Item 5 ✡ ✡ ♣ ✡ with
1 + ❼ 2 Here, the
tan( ❻ )
are positive roots of the transcendental equation
− ⑤ < 0 holds for several first values = 1, ⑨⑩⑨⑩⑨ , ❶ , then the expres- sions ❩ ❴ 2 ❻ 2 − ❹ should be replaced by ❩ | ❴ 2 ❻ 2 − ❹ | and the sines by the hyperbolic sines in the
If the inequality ❴ 2 ❻ 2 ❹ ✡
corresponding terms of the series.
1 1 . The substitution ( ,
) = exp ❳ − 2 ❦❱❭ − 2 ❜ ❬❄➈❆➉ ( ❭ , ❬ ) leads to the equation
2 + ❳●❨ + 4 ❜ − 4 ❦ ➈ ❭ ➉ + exp ❳ 2 ❦❱❭ + 2 ❜ ❬ ➈ ❥ ( ❭ , ❬ ), which is discussed in Subsection 4.1.3.
2 ♣ . Fundamental solutions:
( ➏ ) is the Heaviside unit step function, 0 ( ➏ ) and 1 ( ➏ ) are the Bessel functions, and 0 ( ➏ ) and 1 ( ➏ ) are the modified Bessel functions.
3 ♣ . Domain: − ❲ ➐ < ❭ < ➐ . Cauchy problem. Initial conditions are prescribed:
= ( ❭ ) at ❬ = 0,
= ❡ ( ❭ ) at ❬ = 0.
Solution for ➒
Solution for ➒
Reference ➓ : A. N. Tikhonov and A. A. Samarskii (1990).
4 ✃ . Domain: 0 ≤ ➭ ≤ ❐ . First boundary value problem. The following conditions are prescribed:
= ➼ 0 ( ➭ ) at ❒ ➫ ➑ =0 (initial condition),
= ➼ 1 ( ➭ ) at ➫ =0 (initial condition),
= ➺ 1 ( ➫ ) at ➭ =0 (boundary condition),
(boundary condition). Solution:
Reference : A. G. Butkovskiy (1979).
5 ✃ . Domain: 0 ≤ ➭ ≤ ❐ . Second boundary value problem. The following conditions are prescribed:
0 ( ➭ ) at ➫ =0 (initial condition),
= 1 ( ➭ ) at ➫ =0 (initial condition),
= ➺ 1 ( ➫ ) at ➭ =0 (boundary condition),
(boundary condition). Solution:
= ➺ 2 ( ➫ ) at ➭ = ❐
, Ö , Ü and the functions ( ➭ ) remain as before. If the inequality Ö <0 holds for several first values Ô = 1, ×⑩×⑩× , ❶ , then the expressions
where the coefficient Ó
must be replaced by Õ | Ö | and the sines by the hyperbolic sines in the corresponding terms of the series.
6 ✃ . Domain: 0 ≤ ➭ ≤ ❐ . Third boundary value problem. The following conditions are prescribed:
= ➼ 0 ( ➭ ) at ➫ =0 (initial condition),
= ➼ 1 ( ➭ ) at ➫ =0 (initial condition),
− ➷ ä 1 = ➺ 1 ( ➫ ) at ➭ =0 (boundary condition),
(boundary condition). The solution ➷ ( ➭ ,
+ ä ➷ 2 = ➺ 2 ( ➫ ) at ➭ = ❐
) is determined by the formula in Item 5 Ó ✃ Ó with
2 8 ➳ 4 Ü 2 where the Ü are positive roots of the transcendental equation