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