Problems in Cartesian Coordinates ⑦

3.1.1. Problems in Cartesian Coordinates ⑦

The three-dimensional sourceless heat equation in the rectangular Cartesian system of coordinates has the form

It governs three-dimensional thermal phenomena in quiescent media or solids with constant thermal diffusivity. A similar equation is used to study the corresponding three-dimensional unsteady mass-exchange processes with constant diffusivity.

3.1.1-1. Particular solutions:

2 ) cos( ➂ 3 ❹ + ❿ 3 ) exp ➁ −( ➂ 1 + ➂ 2 + ➂ 3 ) ❷ ➀ ➄ , ( ❻ , ❼ , , ➀

( , , ❹ , )= cos( ➂ 1 ❻ + ❿ 1 ) cos( ➂ 2 ❼ + ❿

)= cos( 1 + 1 ) cos( 2 + 2 ) sinh( 3 ❹ + 3 ) exp −( 1 + 2 − 3 ) ❷ ➀➅➄ , ( ❻ , ❼ , , ➀ )= ❽ cos( ➂ 1 ❻ + ❿ 1 ) cos( ➂ 2 ❼ + ❿ 2 ) cosh( ➂ 3 + ❿ 3 ) exp ➁ −( ➂ 2 + ➂ 1 2 2 − ➂ 2 3 ) ❷ ➀ ❶ ➄ ❹ ❹ ❹ , ( ❻ , ❼ , , ➀ )= ❽ exp(− ➂ 1 ❻ − ➂ 2 ❼ − ➂ 3 ) cos( ➂ ❻ −2 ❷ ➂ 2 1 1 ➀ ) cos( ➂ 2 ❼ −2 ❷ ➂ 2 2 ➀ ) cos( ➂ 3 −2 ❷ ➂ 2 ❹ ➀ ❹ 3 ),

where ❽ , ❾ , ❿ , ❿ 1 , ❿ 2 , ❿ 3 , ➂ 1 , ➂ 2 , ➂ 3 , ❻ 0 , ❼ 0 , 0 , and ➀ 0 are arbitrary constants. Fundamental solution: ➊ ➊

3.1.1-2. Formulas to construct particular solutions. Remarks on the Green’s functions.

1 ➌ . Apart from usual solutions with separated variables,

, ❼ , , ➀ )= ➎ 1 ( ❻ , ➀ ) ➎ 2 ( ❼ , ➀ ) ➎ 3 ( , ➀ ), where the functions ➎ 1 = ➎ 1 ( ❻ , ➀ ), ➎ 2 = ➎ 2 ( ❼ , ➀ ), and ➎ 3 = ➎ 3 ( ❼ , ➀ ) are solutions of the one-dimensional

heat equations

= ❷ ⑩ 2 ❹ 2 , treated in Subsection 1.1.1. ❶ ❶ ❹

2 ➌ . Suppose = ( ❻ , ❼ , , ➀ ) is a solution of the three-dimensional heat equation. Then the functions

where ❽ , ❿ 1 , ❿ 2 , ❿ 3 , ❿ 4 , ➐ , ➐ 1 , ➐ 2 , ➐ 3 , ➒ , and ❶ ➑ are arbitrary constants, are also solutions of this equation. The signs at ➐ in the formula for 1 can be taken independently of one another.

3 ➌ . For the three-dimensional boundary value problems considered in Subsection 3.1.1, the Green’s function can be represented in the product form

where ❹

1 ( ❻ , → , ➀ ), 2 ( ❼ , ➣ , ➀ ), 3 ( , ↔ , ➀ ) are the Green’s functions of the corresponding one- dimensional boundary value problems; these functions can be found in Subsections 1.1.1 and

Example 1. The Green’s function of the mixed boundary value problem for a semiinfinite layer (− ↕ < ➙ < ↕ , 0≤ ➛ < ↕ ,0≤ ➜ < ➝ ) presented in Paragraph 3.1.1-14 is the product of three one-dimensional Green’s functions from Paragraph 1.1.2-1 (Cauchy problem for − ↕

< ➙ < ↕ ), Paragraph 1.1.2-2 (first boundary value problem for 0 ≤ ➛ < ↕ ), and Paragraph 1.1.2-6 (second boundary value problem for 0 ≤ ➜ < ➝ ), in which one needs to carry out obvious renaming of variables.

< ❻ < ➞ ,− ➞ < ❼ < ➞ ,− ➞ < < ➞ . Cauchy problem. An initial condition is prescribed:

3.1.1-3. Domain: − ❹

Example 2. The initial temperature is constant and is equal to ➠ ➠ ➠ ➢ 1 in the domain | ➙ | < ➙ 0 ,| ➛ | < ➛ 0 ,| ➜ | < ➜ 0 and is equal

to ➢ 2 in the domain | ➙ | > ➙ 0 ,| ➛ | > ➛ 0 ,| ➜ | > ➜ 0 ; specifically,

Reference : H. S. Carslaw and J. C. Jaeger (1984).

3.1.1-4. Domain: 0 ≤ ➽ < ➞ ,− ➞ < ➾ < ➞ ,− ➞ < ➚ < ➞ . First boundary value problem.

A half-space is considered. The following conditions are prescribed:

= ➍ ( ➽ , ➾ , ➚ ) at ➀ =0 (initial condition),

= ➶ ( ➾ , ➚ , ➀ ) at ➽ =0 (boundary condition). Solution:

References : A. G. Butkovskiy (1979), H. S. Carslaw and J. C. Jaeger (1984).

3.1.1-5. Domain: 0 ≤ ➽ < ➞ ,− ➞ < ➾ < ➞ ,− ➞ < ➚ < ➞ . Second boundary value problem.

A half-space is considered. The following conditions are prescribed:

= ➍ ( ➽ , ➾ , ➚ ) at ➬ =0 (initial condition),

= ➶ ( ➾ , ➚ , ➬ ) at ➽ =0 (boundary condition). Solution:

Reference : A. G. Butkovskiy (1979).

3.1.1-6. Domain: 0 ≤ ➽ < ➞ ,− ➞ < ➾ < ➞ ,− ➞ < ➚ < ➞ . Third boundary value problem.

A half-space is considered. The following conditions are prescribed:

= ➍ ( ➽ , ➾ , ➚ ) at ➬ =0 (initial condition),

− Ï ➪ = ➶ ( ➾ , ➚ , ➬ ) at ➽ =0 (boundary condition). The solution ➪ ( ➽ , ➾ , ➚ , ➬ ) is determined by the formula in Paragraph 3.1.1-5 where

Reference : H. S. Carslaw and J. C. Jaeger (1984).

3.1.1-7. Domain: − Ù < Ú < Ù ,− Ù < Û < Ù ,0≤ Ü ≤ Ý . First boundary value problem. An infinite layer is considered. The following conditions are prescribed:

= ß ( Ú , Û , Ü ) at à =0 (initial condition),

= á 1 ( Ú , Û , à ) at Ü =0 (boundary condition),

(boundary condition). Solution:

Reference : H. S. Carslaw and J. C. Jaeger (1984). ä

3.1.1-8. Domain: − Ù < Ú < Ù ,− Ù < Û < Ù ,0≤ Ü ≤ Ý . Second boundary value problem. An infinite layer is considered. The following conditions are prescribed:

= ß ( Ú , Û , Ü ) at à =0 (initial condition),

= 1 ( , , ) at

=0 (boundary condition),

(boundary condition). Solution:

= á 2 ( Ú , Û , à ) at Ü = Ý

where ä

, Ü , , , æ , à )= î ➹ Ýïà exp é −

cos Ý cos

exp ó

Reference : H. S. Carslaw and J. C. Jaeger (1984). ä

3.1.1-9. Domain: − Ù < Ú < Ù ,− Ù < Û < Ù ,0≤ Ü ≤ Ý . Third boundary value problem. An infinite layer is considered. The following conditions are prescribed:

= ß ( Ú , Û , Ü ) at à =0 (initial condition),

− 1 = á 1 ( Ú , Û , à ) at Ü =0 (boundary condition),

(boundary condition). The solution Þ ( Ú , Û , Ü , à ) is determined by the formula in Paragraph 3.1.1-8 where ñ ñ

1 + 2 are positive roots of the transcendental equation

tan( )

Here, the Ý

3.1.1-10. Domain: − Ù < Ú < Ù ,− Ù < Û < Ù ,0≤ Ü ≤ Ý . Mixed boundary value problem. An infinite layer is considered. The following conditions are prescribed:

= ß ( Ú , Û , Ü ) at à =0 (initial condition),

= á 1 ( Ú , Û , à ) at Ü =0 (boundary condition),

(boundary condition). Solution:

Reference : A. G. Butkovskiy (1979). ä

3.1.1-11. Domain: − Ù < Ú < Ù ,0≤ Û < Ù ,0≤ Ü ≤ Ý . First boundary value problem.

A semiinfinite layer is considered. The following conditions are prescribed:

= ß ( Ú , Û , Ü ) at à =0 (initial condition),

= á 1 ( Ú , Ü , à ) at Û =0 (boundary condition),

= á 2 ( Ú , Û , à ) at Ü =0 (boundary condition),

(boundary condition). Solution:

3.1.1-12. Domain: − Ù < Ú < Ù ,0≤ Û < Ù ,0≤ Ü ≤ Ý . Second boundary value problem.

A semiinfinite layer is considered. The following conditions are prescribed:

= ß ( Ú , Û , Ü ) at à =0 (initial condition),

= á 1 ( Ú , ✆ Ü , ê à ) at Û =0 (boundary condition),

2 ( , , ) at

=0 (boundary condition),

(boundary condition). Solution:

) at

where

exp −

exp −

+ exp −

cos

cos Ý exp Ý ó − Ý 2 Ö ë .

3.1.1-13. Domain: − Ù < Ú < Ù ,0≤ Û < Ù ,0≤ Ü ≤ Ý . Third boundary value problem.

A semiinfinite layer is considered. The following conditions are prescribed:

= ß ( Ú , Û , Ü ) at à =0 (initial condition),

− 1 = á 1 ( Ú , Ü , à ) at Û =0 (boundary condition),

− 2 = á 2 ( Ú , Û , à ) at Ü =0 (boundary condition),

(boundary condition). The solution Þ ( Ú , Û , Ü , à ) is determined by the formula in Paragraph 3.1.1-12 where

are positive roots of the transcendental equation

3.1.1-14. Domain: − Ù < Ú < Ù ,0≤ Û < Ù ,0≤ Ü ≤ Ý . Mixed boundary value problems.

1 ✟ . A semiinfinite layer is considered. The following conditions are prescribed:

= ß ( Ú , Û , Ü ) at à =0 (initial condition),

= á 1 ( Ú , Ü , à ) at Û =0 (boundary condition),

= á 2 ( Ú , Û , à ) at Ü =0 (boundary condition),

(boundary condition). Solution:

, , ) at

where ä

exp é −

exp é −

− exp é −

cos

cos

exp ó

2 ✟ . A semiinfinite layer is considered. The following conditions are prescribed:

= ß ( Ú , Û , Ü ) at à =0 (initial condition),

= á 1 ( Ú , Ü , à ) at Û =0 (boundary condition),

= á 2 ( Ú , Û , à ) at Ü =0 (boundary condition),

(boundary condition). Solution:

3.1.1-15. Domain: 0 ≤ Ú < Ù ,0≤ Û < Ù ,0≤ Ü < Ù . First boundary value problem. An octant is considered. The following conditions are prescribed:

= ß ( Ú , Û , Ü ) at à =0 (initial condition),

= á 1 ( Û , Ü , à ) at Ú =0 (boundary condition),

= á 2 ( Ú , Ü , à ) at Û =0 (boundary condition),

= á 3 ( Ú , Û , à ) at Ü =0 (boundary condition). Solution:

where

( Ú , í , à ) = exp é

− exp −

Example 3. The initial temperature is uniform, ✎ ( ✏ , ✑ , ✒ )= ✓ ➘ þ 0 . The faces are maintained at zero temperature, 1 = ✓ 2 = ✓ 3 =0 . Solution:

Reference : H. S. Carslaw and J. C. Jaeger (1984).

3.1.1-16. Domain: 0 ≤ Ú < Ù ,0≤ Û < Ù ,0≤ Ü < Ù . Second boundary value problem. An octant is considered. The following conditions are prescribed:

= ß ( Ú , Û , Ü ) at à =0 (initial condition),

= á 1 ( Û , ê Ü ✙ , à ) at Ú =0 (boundary condition),

= á 2 ( Ú , Ü , à ) at Û =0 (boundary condition),

= á 3 ( Ú , Û , à ) at ê Ü ÷ =0 (boundary condition).

3.1.1-17. Domain: 0 ≤ Ú < Ù ,0≤ Û < Ù ,0≤ Ü < Ù . Third boundary value problem. An octant is considered. The following conditions are prescribed:

= ß ( Ú , Û , Ü ) at à =0 (initial condition),

− 1 = 1 ( , , ) at

=0 (boundary condition),

) at Û =0 (boundary condition),

− 3 = á 3 ( Ú , Û , à ) at Ü =0 (boundary condition). The solution Þ ( Ú , Û , Ü , à ) is determined by the formula in Paragraph 3.1.1-16 where

( , , ; ) = exp −

+ exp −

exp ✢✣✚ 2 à

+( Ú + í ) ✤ erfc ó

Example 4. The initial temperature is uniform, ✎ ( ✏ , ✑ , ✓ ✒ )= þ 0 . The temperature of the contacting media is zero, 1 = ✓ 2 = ✓ 3 =0 . Solution:

= þ 0 ✦ erf

+ exp( 1 + 1 ) erfc

+ exp( ✧ 2 ✑ + ✧ 2 2 ✕ ü✗✖ ✘ ) erfc ✔ ü✗✖

+ exp( ✧ 3 ✒ + ✧ 2 3 ü✗✖ ) erfc ✔

Reference : H. S. Carslaw and J. C. Jaeger (1984).

3.1.1-18. Domain: 0 ≤ ✫ < ✬ ,0≤ ✭ < ✬ ,0≤ ✮ < ✬ . Mixed boundary value problems.

1 ✟ . An octant is considered. The following conditions are prescribed:

= ✰ ( ✫ , ✭ , ✮ ) at ✱ =0 (initial condition),

= ✲ 1 ( ✭ , ✮ , ✱ ) at ✫ =0 (boundary condition),

= 2 ( , , ) at

=0 (boundary condition),

= ✲ 3 ( ✫ , ✭ , ✱ ) at ✮ =0 (boundary condition). Solution:

2 ● . An octant is considered. The following conditions are prescribed:

= ■ ( ❇ , ❊ , ❋ ) at ❄ =0 (initial condition),

= ❏ 1 ( ❊ , ❋ , ❄ ) at ❇ =0 (boundary condition),

= ❏ 2 ( ❇ , ❋ , ❄ ) at ❊ =0 (boundary condition),

= ❏ 3 ( ❇ , ❊ , ❄ ) at ❋ =0 (boundary condition). Solution:

3.1.1-19. Domain: 0 ≤ ❇ ≤ ❯ 1 ,0≤ ❊ ≤ ❯ 2 ,− ❱ < ❋ < ❱ . First boundary value problem. An infinite cylindrical domain of a rectangular cross-section is considered. The following conditions

are prescribed:

= ■ ( ❇ , ❊ , ❋ ) 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:

3.1.1-20. Domain: 0 ≤ ❇ ≤ ❯ 1 ,0≤ ❊ ≤ ❯ 2 ,− ❱ < ❋ < ❱ . Second boundary value problem. An infinite cylindrical domain of a rectangular cross-section is considered. The following conditions

are prescribed:

= ■ ( ❇ , ❊ , ❋ ) at ❄ =0

(initial condition),

= ❏ 1 ( ❊ , ❋ , ❄ ) at ❇ =0 (boundary condition),

= ❏ 2 ( ❊ , ❋ , ❄ ) at ❇ = ❯

1 (boundary condition),

= ❏ ( ❇ , ❋ , ❄ ) at ❊ =0 (boundary condition),

= ❏ 4 ( ❇ , ❋ , ❄ ) at ❊ = ❯ 2 (boundary condition).

3.1.1-21. Domain: 0 ≤ ❇ ≤ ❯ 1 ,0≤ ❊ ≤ ❯ 2 ,− ❱ < ❋ < ❱ . Third boundary value problem. An infinite cylindrical domain of a rectangular cross-section is considered. The following conditions

are prescribed:

= ■ ( ❇ , ❊ , ❋ ) at ❑ ❄ ❫ =0 (initial condition),

) at ❇ =0 (boundary condition),

+ ❍ ❵ 2 = ❏ 2 ( ❊ , ❋ , ❄ ) at ❇ = ❯

1 (boundary condition),

− ❍ ❵ 3 = ❏ 3 ( ❇ , ❋ , ❄ ) at ❊ =0 (boundary condition),

+ ❵ ❍ 4 = ❏ 4 ( ❇ , ❋ , ❄ ) at ❊ = ❯ 2 (boundary condition). The solution ❍ ( ❇ , ❊ , ❋ , ❄ ) is determined by the formula in Paragraph 3.1.1-20 where

and ❢ ❞ are positive roots of the transcendental equations

tan( ❝ ❯ 1 )

1 + ❵ 2 tan( ❢ ❯ 2 )

◆P◗

3.1.1-22. Domain: 0 ≤ ❇ ≤ ❯ 1 ,0≤ ❊ ≤ ❯ 2 ,− ❱ < ❋ < ❱ . Mixed boundary value problem. An infinite cylindrical domain of a rectangular cross-section is considered. The following conditions

are prescribed:

= ■ ( ❇ , ❊ , ❋ ) 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:

3.1.1-23. Domain: 0 ≤ ❇ ≤ ❯ 1 ,0≤ ❊ ≤ ❯ 2 ,0≤ ❋ < ❱ . First boundary value problem.

A semiinfinite cylindrical domain of a rectangular cross-section is considered. The following conditions are prescribed:

= ■ ( ❇ , ❊ , ❋ ) at ❄ =0

(initial condition),

= ❏ 1 ( ❊ , ❋ , ❄ ) at ❇ =0 (boundary condition),

= ❏ 2 ( ❊ , ❋ , ❄ ) at ❇ = ❯ 1 (boundary condition),

= ❏ 3 ( ❇ , ❋ , ❄ ) at ❊ =0 (boundary condition),

= ❏ 4 ( ❇ , ❋ , ❄ ) at ❊ = ❯ 2 (boundary condition),

= ❏ 5 ( ❇ , ❊ , ❄ ) at ❋ =0 (boundary condition).

3.1.1-24. Domain: 0 ≤ ② ≤ ③ 1 ,0≤ ④ ≤ ③ 2 ,0≤ ✇ < ⑤ . Second boundary value problem.

A semiinfinite cylindrical domain of a rectangular cross-section is considered. The following conditions are prescribed: ⑥

= ⑦ ( ② , ④ , ✇ ) at ✉ =0

(initial condition),

= ⑨ 1 ( ④ , ✇ , ✉ ) at ② =0 (boundary condition),

= ⑨ 2 ( ④ , ✇ , ✉ ) at ② = ③ 1 (boundary condition),

3 ( ② , ✇ , ✉ ) at ④ =0 (boundary condition),

= ⑨ 4 ( ② , ✇ , ✉ ) at ④ = ③ 2 (boundary condition),

= ⑨ 5 ( ② , ④ , ✉ ) at ✇ =0 (boundary condition). Solution: ⑥

3.1.1-25. Domain: 0 ≤ ② ≤ ③ 1 ,0≤ ④ ≤ ③ 2 ,0≤ ✇ < ⑤ . Third boundary value problem.

A semiinfinite cylindrical domain of a rectangular cross-section is considered. The following conditions are prescribed:

= ⑦ ( ② , ④ , ✇ ) at ➇ =0

(initial condition),

− ➉ 1 ⑥ = ⑨ 1 ( ④ , ✇ , ➇ ) at ② =0 (boundary condition),

+ ➉ 2 ⑥ = ⑨ 2 ( ④ , ✇ , ➇ ) at ② = ③ 1 (boundary condition),

− ➉ 3 ⑥ = ⑨ 3 ( ② , ✇ , ➇ ) at ④ =0 (boundary condition),

+ ➉ 4 ⑥ = ⑨ 4 ( ② , ✇ , ➇ ) at ④ = ③ 2 (boundary condition),

− ➉ 5 = ⑨ 5 ( ② , ④ , ➇ ) at ✇ =0 (boundary condition).

The solution ( ② , ④ , ✇ , ➇ ) is determined by the formula in Paragraph 3.1.1-24 where

and the functions ➎ 1 ( ↔ , ➌ , ➇ ) and ➎ 2 ( ↕ , ➄ , ➇ ) can be found in Paragraph 3.1.1-21.

3.1.1-26. Domain: 0 ≤ ↔ ≤ ➙ 1 ,0≤ ↕ ≤ ➙ 2 ,0≤ ➏ < ➛ . Mixed boundary value problems.

1 ➜ . A semiinfinite cylindrical domain of a rectangular cross-section is considered. The following conditions are prescribed: ➝

= ➞ ( ↔ , ↕ , ➝ ➏ ) at ➇ =0

(initial condition),

= ➟ 1 ( ↕ , ➏ , ➝ ➇ ) at ↔ =0 (boundary condition),

= 2 ( , , ) at

= 1 (boundary condition),

= ➟ 3 ( ↔ , ➏ , ➇ ) at ➝ ↕ =0 (boundary condition),

= ➟ 4 ( ↔ , ➏ , ➇ ) at ↕ = ➙ 2 (boundary condition),

= ➟ 5 ( ↔ , ↕ , ➇ ) at ➏ =0 (boundary condition).

2 ➜ . A semiinfinite cylindrical domain of a rectangular cross-section is considered. The following ➝

conditions are prescribed:

= ➞ ( ↔ , ↕ , ➏ ) at ➇ =0

(initial condition),

= ➟ 1 ( ↕ , ➏ , ➇ ) at ↔ =0 (boundary condition),

= ➟ 2 ( ↕ , ➏ , ➇ ) at ↔ = ➙ 1 (boundary condition),

= ➟ 3 ( ↔ , ➏ , ➇ ) at ↕ =0 (boundary condition),

= ➟ 4 ( ↔ , ➏ , ➇ ) at ↕ = ➙ 2 (boundary condition), = ➟ 5 ( ↔ , ↕ , ➇ ) at ➏ =0 (boundary condition).

Solution: ➝

3.1.1-27. Domain: 0 ≤ ✟ ≤ ✗ 1 ,0≤ ✠ ≤ ✗ 2 ,0≤ ✡ ≤ ✗ 3 . First boundary value problem.

A rectangular parallelepiped is considered. The following conditions are prescribed:

= ✥ ( ✟ , ✠ , ✡ ) at ✍ =0

(initial condition),

= ✦ 1 ( ✠ , ✡ , ✍ ) at ✟ =0 (boundary condition),

= ✦ 2 ( ✠ , ✡ , ✍ ) at ✟ = ✗ 1 (boundary condition),

= ✦ 3 ( ✟ , ✡ , ✍ ) at ✠ =0 (boundary condition),

= ✦ 4 ( ✟ , ✡ , ✍ ) at ✠ = ✗ 2 (boundary condition),

= ✦ 5 ( ✟ , ✠ , ✍ ) at ✡ =0 (boundary condition),

= ✦ 6 ( ✟ , ✠ , ✍ ) at ✡ = ✗ 3 (boundary condition). Solution:

Reference : H. S. Carslaw and J. C. Jaeger (1984).

3.1.1-28. Domain: 0 ≤ ✟ ≤ ✗ 1 ,0≤ ✠ ≤ ✗ 2 ,0≤ ✡ ≤ ✗ 3 . Second boundary value problem.

A rectangular parallelepiped is considered. The following conditions are prescribed:

= ✥ ( ✟ , ✠ , ✡ ) at ✍ =0

(initial condition),

= ✦ 1 ( ✠ , ✡ , ✍ ) at ✟ =0 (boundary condition),

= 2 ( , , ) at

= 1 (boundary condition),

= ✦ 3 ( ✟ , ✴ ✡ , ✍ ) at ✬ ✠ =0 (boundary condition),

= ✦ 4 ( ✟ , ✡ , ✬ ✍ ✴ ) at ✠ = ✗ 2 (boundary condition),

= ✦ 5 ( ✬ ✟ ✵ , ✠ , ✍ ) at ✡ =0 (boundary condition),

= ✦ 6 ( ✟ , ✠ , ✍ ) at ✡ = ✗ 3 (boundary condition). Solution:

3.1.1-29. Domain: 0 ≤ ✟ ≤ ✗ 1 ,0≤ ✠ ≤ ✗ 2 ,0≤ ✡ ≤ ✗ 3 . Third boundary value problem.

A rectangular parallelepiped is considered. The following conditions are prescribed:

= ✥ ( ✟ , ✠ , ✡ ) at ✍ =0

(initial condition),

, ✍ ) at ✬ ✟ =0 (boundary condition),

+ ✶ 2 = ✦ 2 ( ✠ , ✡ , ✍ ) at ✟ = ✗ 1 (boundary condition),

− ✶ 3 = ✦ 3 ( ✟ , ✡ , ✍ ) at ✠ =0 (boundary condition),

+ ✤ 4 = ✦ 4 ( ✟ , ✡ , ✍ ) at ✠ = ✗

2 (boundary condition),

5 ( ✟ , ✠ , ✍ ) at ✡ =0 (boundary condition),

+ ✶ 6 = ✦ 6 ( ✟ , ✠ , ✍ ) at ✡ = ✗ 3 (boundary condition).

The solution ✤ ( ✟ , ✠ , ✡ , ✍ ) is determined by the formula in Paragraph 3.1.1-28 where

The functions 1 ( ✟ , ☛ , ✍ ) and 2 ( ✠ , ☞ ✍

, ) can be found in Paragraph 3.1.1-21, and the function

3 ( , , ) is given by

5 + ✶ 6 where the ✹ are positive roots of the transcendental equation

tan( ✹ ✗ 3 )

3.1.1-30. Domain: 0 ≤ ✟ ≤ ✗ 1 ,0≤ ✠ ≤ ✗ 2 ,0≤ ✡ ≤ ✗ 3 . Mixed boundary value problems.

1 ✺ . A rectangular parallelepiped is considered. The following conditions are prescribed:

= ✥ ( ✟ , ✠ , ✡ ) at ✍ =0

(initial condition),

= ✦ 1 ( ✠ , ✡ , ✍ ) at ✟ =0 (boundary condition),

= ✦ 2 ( ✠ , ✡ , ✍ ) at ✟ = ✗ 1 (boundary condition),

= ✦ 3 ( ✟ , ✡ , ✍ ) at ✠ =0 (boundary condition),

= ✦ 4 ( ✟ , ✡ , ✍ ) at ✠ = ✗ 2 (boundary condition),

= ✦ 5 ( ✟ , ✠ , ✍ ) at ✡ =0 (boundary condition),

= ✦ 6 ( ✬ ✟ ✵ , ✠ , ✍ ) at ✡ = ✗ 3 (boundary condition).

Solution:

2 ✺ . A rectangular parallelepiped is considered. The following conditions are prescribed:

= ✥ ( ✟ , ✠ , ✡ ) at ✍ =0

(initial condition),

= ✦ 1 ( ✠ , ✡ , ✍ ) at ✟ =0 (boundary condition),

= ✦ 2 ( ✠ , ✡ , ✍ ) at ✟ = ✗ 1 (boundary condition),

= ✦ 3 ( ✟ , ✡ , ✬ ✍ ✴ ) at ✠ =0 (boundary condition),

= ✦ 4 ( ✟ , ✬ ✡ ✴ , ✍ ) at ✠ = ✗ 2 (boundary condition),

= ✦ 5 ( ✟ , ✠ , ✍ ) at ✡ =0 (boundary condition),

= ✦ 6 ( ✟ , ✠ , ✍ ) at ✡ = ✗ 3 (boundary condition). Solution:

where

exp ✗ ✜ −

sin ✜

sin ✗ ✜