Heat Equation ✁

2.1. Heat Equation ✁

2.1.1. Boundary Value Problems in Cartesian Coordinates

In rectangular Cartesian coordinates, the two-dimensional sourceless heat equation has the form ✆ ✝ ✠ ✆

It governs two-dimensional unsteady heat transfer processes in quiescent media or solid bodies with constant thermal diffusivity ✟ . A similar equation is used to study analogous two-dimensional unsteady mass transfer phenomena with constant diffusivity; in this case the equation is called a diffusion equation.

2.1.1-1. Particular solutions:

( , , )= ✌ cos( ✑ 1 + ✍ 1 ) cos( ✑ 2 + ✍ 2 ) exp ✏ −( ✑ 2 1 + ✑ 2 2 ) ✟ ,

cos( ✑ 1 + ✍ 1 ) sinh( ✑ 2 + ✍ 2 ) exp ✏ −( ✑ 2 1 − ✑ 2 2 ) ✟ ,

cos( ✑ 1 + ✍ 1 ) cosh( ✑ 2 + ✍ ) exp ✏ −( ✑ 2 − ✑ 2 ) ✟ ,

exp(− ✕ − ✖ ) cos( ✕ −2 ✟ ✕ 2 + ✍ 1 ) cos( ✖ −2 ✟ ✖ 2 + ✡ ✍ ✡ ☛ ☛ 2 ),

2 where ✌ , ✎ , ✍ 1 , ✍ 2 , ✍ 3 , ✑ 1 , ✑ 2 , 0 , 0 , and 0 are arbitrary constants.

Fundamental solution:

exp −

2.1.1-2. Formulas to construct particular solutions. Remarks on the Green’s functions.

. Apart from usual separable solutions ( , , )= ✢ 1 ( ) ✢ 2 ( ) ✢ 3 ( ), the equation in question has more sophisticated solutions in the product form

= ✣ ( , ) and ✤ = ✤ ( , ) are solutions of the one-dimensional heat equations

considered in Subsection 1.1.1.

. Suppose = ( , , ) is a solution of the heat equation. Then the functions

, ✍ 1 , ✍ 2 , ✍ 3 , ✦ , , ✖ , ✖ 1 and ✝ ✖ 2 are arbitrary constants, are also solutions of this equation. ✪✒✫ The signs at

’s in the formula for 1 are taken arbitrarily, independently of each other.

Reference : W. Miller, Jr. (1977).

3 ✜ . For all two-dimensional boundary value problems discussed in Subsection 2.1.1, the Green’s function can be represented in the product form

, ) and 2 ( , ✮ , ) are the Green’s functions of the corresponding one-dimensional boundary value problems (these functions are specified in Subsections 1.1.1 and 1.1.2).

where ✞ 1 ( ,

Example 1. The Green’s function of the first boundary value problem for a semiinfinite strip (0 ≤ ✯ ≤ ✰ ,0≤ ✱ < ✲ ), considered in Subsection 2.1.1-12, is the product of two one-dimensional Green’s functions. The first Green’s function is that of the first boundary value problem on a closed interval (0 ≤ ✯ ≤ ✰ ) presented in Subsection 1.1.2-5. The second Green’s function is that of the first boundary value problem on a semiinfinite interval (0 ≤ ✱ < ✲ ) presented in Subsection 1.1.2-2, where ✯ and ✳ must be renamed ✱ and ✴ , respectively.

2.1.1-3. Transformations that allow separation of variables. Table 18 lists possible transformations that allow reduction of the two-dimensional heat equation to ✡ ☛ ✞

a separable equation. All transformations of the independent variables have the form ( ✞ , , ✡ ) ✸ ✵✶ ✷ ☛

( ✭ , ✮ , ). The transformations that can be obtained by interchange of independent variables, , are omitted. The anharmonic oscillator functions are solutions of the second-order ordinary differential equation ✹ ✼✽✼ ✺✻✺ +( ✟ ✾ 4 + ✿✧✾ 2 + ❀ ) ✹ = 0. The Ince polynomials are the 2 ✛ -periodic solutions of the Whittaker–Hill equation ✹ ✼✽✼ ✺✻✺ + ✑ sin 2 ✾ ✹ ✼ ✺ +( ✟ − ✿❁✑ cos 2 ✾ ) ✹ ✪✒✫ = 0; see Arscott (1964, 1967).

Reference : W. Miller, Jr. (1977).

< < ❂ ,− ❂ < < ❂ . Cauchy problem. An initial condition is prescribed:

2.1.1-4. Domain: − ☛

Example 2. The initial temperature is piecewise-constant and equal to ❄

1 in the domain | ✯ | < ✯ 0 ,| ✱ | < ✱ 0 and ❆ 2 in the domain | ✯ | > ✯ 0 ,| ✱ | > ✱ 0 , specifically, ❇

( ✯ , ✱ )= ❈ ❆ 1 for | ✯ | < ✯ 0 ,| ✱ | < ✱ 0 ❆ , 2 for | ✯ | > ✯ 0 ,| ✱ | > ✱ 0 .

TABLE 18 ✞

Transformations ( ✝ , , ) ✵✶ ✷ ✞ ( ✭ , ✮ , ) that allow solutions with ✞ ❏ -separated

variables, ✆ ▼✔✝ = exp[ ✆ ◆ ◆ ✝ ❏ ( ✭ ✆ , ❖❁❖ ✮ , ✝ )] ✢ ( ✭ ) ❑ ( ✮ ) ▲ ( ), for the two-dimensional heat ✞

. Everywhere, the function ▲ ( ) is exponential No

equation =

Transformations

Factor exp ❏

Function ✢ ( ✭ ) Function ❑ ( ✮ )

| ❏ | =0 Hermite function

function

Hermite function Hermite function | |

Parabolic cylinder Parabolic cylinder = ✭ ✮

Bessel function

function = cosh ✭ cos ✮ ,

Modified Mathieu Mathieu

= sinh ❏ =0 sin

Laguerre function Exponential = ✙

| | sin function

= ✙ | ✞ | cosh ✭ cos ✮ ,

Ince polynomial Ince polynomial

= ✙ | | sinh ✭ sin ✮

Exponential Airy function = +

Airy function

Parabolic cylinder = ✮ ✙

2 Hermite function

=− 1 4 ( ✭ 2 + ✮ ✮ 2 ) Exponential Exponential =

Airy function Airy function = ✮ + ✿ 2

15 = ✞ ✮ + ✿❋P + 1 Airy function

Airy function

2 ( ✟ ✭ + ✿✽✮ ) P

=− 1 16 ( ✭ 2 ✮ 2 2 Parabolic cylinder Parabolic cylinder = ✭ ✮

2 Parabolic cylinder

Parabolic cylinder

2 Hermite function =

Hermite function

= sign(1 − )

TABLE 18 (continued)

No Transformations

Factor exp ❏

Function ✢ ( ✭ ) Function ❑ ( ✮ )

Anharmonic Anharmonic = ✭ ✮

oscillator function oscillator function

Anharmonic Anharmonic

oscillator function oscillator function

= ✭ cos ✮ ,

Exponential = ✭ sin ✮

Bessel function function

= ✞ cosh ✭ cos ✮ ,

Modified = sinh

4 (sinh ✭ ✭ sin ✮ + cos 2 ✮ )

Mathieu function Mathieu function

1+ 2 ✞ ✞ ✭ cos ✮ , ❏ 1 ✭ 2 Whittaker Exponential = ✙ 1+ 2 ✭ sin ✮

function function

|1 − 2 | ✭ cos ✮ ,

Laguerre function Exponential

Ince polynomial Ince polynomial sinh ✭ sin ✮

cos ✮ , ❏

=− 4 (sinh + cos )

|1 − 2 | cosh ✭ cos ✮ , ❏ =− 1 (sinh 2 ✭ + cos ✞ 2 ❘ ✞ ☛ ✮ 4 ) ,

26 Ince polynomial Ince polynomial = ◗ |1 − 2 | sinh ✭ sin ✮

If the initial temperature distribution ✢ ( ❨ , ❫ ❩ ) is an infinitely differentiable function in both arguments, then the solution can be represented in the series form ❫ ❝❡❞

❤✒✐ Such a representation is useful for small ❭ .

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

2.1.1-5. Domain: 0 ≤ ❨ < ❥

,− < < . First boundary value problem.

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

= ( ❨ , ❩ ) at ❭ =0 (initial condition),

= ❦ ( ❩ , ❭ ) at ❨ =0 (boundary condition).

Solution: ❞

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

2.1.1-6. Domain: 0 ≤ ❨ < ❥ ,− ❥ < ❩ < ❥ . Second boundary value problem.

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

= ( ❨ , ❩ ) at ❭ =0 (initial condition),

= ❦ ( ❩ , ❭ ) at ❨ =0 (boundary condition). Solution:

2.1.1-7. Domain: 0 ≤ ❨ < ❥ ,− ❥ < ❩ < ❥ . Third boundary value problem.

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

= ( ❨ , ❩ ) at ❭ =0 (initial condition),

− ⑦ ❬ = ❦ ( ❣ ❩ ⑥ , ❭ ) at ❨ =0 (boundary condition). The solution ❬ ( ❨ , ❩ , ❭ ) is determined by the formula in Paragraph 2.1.1-6 where

2.1.1-8. Domain: 0 ≤ ❨ < ❥ ,0≤ ❩ < ❥ . First boundary value problem.

A quadrant of the plane is considered. The following conditions are prescribed: ❞

= ( ❨ , ❩ ) at ❭ =0 (initial condition),

= ❦ 1 ( ❩ , ❭ ) at ❨ =0 (boundary condition),

= ❦ 2 ( ❨ , ❭ ) at ❩ =0 (boundary condition). Solution:

Example 3. The initial temperature is uniform, ❷ ⑩ ( ✯ , ❶ )= ② 0 . The boundary is maintained at zero temperature, 1 ( ❶ , ❸ )= 2 ( ✯ , ❸ ) = 0. Solution:

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

2.1.1-9. Domain: 0 ≤ ❼ < ❥ ,0≤ ❞ ❽ < ❥ . Second boundary value problem.

A quadrant of the plane is considered. The following conditions are prescribed:

= ( ❼ , ❽ ) at ❿ =0 (initial condition),

= ❦ 1 ( ❽ , ❿ ) at ❼ =0 (boundary condition),

) at ❣ ❽ =0 (boundary condition).

2.1.1-10. Domain: 0 ≤ ❼ < ❥ ,0≤ ❞ ❽ < ❥ . Third boundary value problem.

A quadrant of the plane is considered. The following conditions are prescribed:

at ❿ =0

(initial condition),

at ❣ ❼ ⑥ =0 (boundary condition),

(boundary condition). The solution ❾ ( ❼ , ❽ , ❿ ) is determined by the formula in Paragraph 2.1.1-9 where

Example 4. The initial temperature is constant, ⑩ ( ✯ , ❶ )= ② 0 . The temperature of the environment is zero, ❷ ❷ 1 ( ❶ , ❸ )= 2 ( ✯ , ❸ ) = 0. Solution:

= ② 0 ➆ erf ❹

+ exp( ➇ 1 ✯ + ✇❋➇ 2 1 ❸ ) erfc ❹

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

t❋✈

2.1.1-11. Domain: 0 ≤ ❼ < ❥ ,0≤ ❽ < ❥ . Mixed boundary value problems.

1 ➉ . A quadrant of the plane is considered. The following conditions are prescribed: ❞

= ( ❼ , ❽ ) at ❿ =0 (initial condition),

= ❦ 1 ( ❽ , ❿ ) at ❼ =0 (boundary condition),

= ❦ 2 ( ❼ , ❿ ) at ❽ =0 (boundary condition). Solution:

−exp q −

exp q

+exp −

2 ➉ . A quadrant of the plane is considered. The following conditions are prescribed: ❞

= ( ❼ , ❽ ) at ❿ =0 (initial condition),

− ⑦ ❾ = ❦ 1 ( ❣ ❽ ⑥ , ❿ ) at ❼ =0 (boundary condition),

= ❦ 2 ( ❼ , ❿ ) at ❽ =0 (boundary condition). Solution:

Example 5. The initial temperature is uniform, ⑩ ( ✯ , ❶ )= ② 0 . Heat exchange with the environment of zero temperature occurs at one side and the other side is maintained at zero temperature: ❷ 1 ( ❶ , ❸ )= ❷ 2 ( ✯ , ❸ ) = 0. Solution:

= ② 0 ➆ erf ❹

+ exp(

) erfc

erf

2.1.1-12. Domain: 0 ≤ ❼ ≤ ➍ ,0≤ ❞ ❽ < ❥ . First boundary value problem.

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

= ( ❼ , ❽ ) at ❿ =0 (initial condition),

= ❦ 1 ( ❽ , ❿ ) at ❼ =0 (boundary condition),

= ❦ 2 ( ❽ , ❿ ) at ❼ = ➍

(boundary condition),

= ❦ 3 ( ❼ , ❿ ) at ❽ =0 (boundary condition). Solution: ❞

Example 6. The initial temperature is uniform, ❷ ⑩ ( ✯ , ❶ )= ② 0 . The boundary is maintained at zero temperature, 1 ( ❶ , ❸ )= 2 ( ❶ , ❸ )= ❷ 3 ( ✯ , ❸ ) = 0. Solution:

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

2.1.1-13. Domain: 0 ≤ ➛ ≤ ➍ ,0≤ ➜ < ➝ . Second boundary value problem.

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

= ➟ ( ➛ , ➜ ) at ➡ ➠ ➢ =0 (initial condition),

= ➤ 1 ( ➜ , ➡ ➠ ) at ➛ =0 (boundary condition),

= ➤ 2 ( ➜ , ➡ ➠ ➥ ) at ➛ = ➍ (boundary condition),

= ➤ 3 ( ➛ , ➠ ) at ➜ =0 (boundary condition). Solution:

2.1.1-14. Domain: 0 ≤ ➛ ≤ ➸ ,0≤ ➜ < ➝ . Third boundary value problem.

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

= ➡ ➟ ( ➛ , ➜ ➢ ) at ➠ =0 (initial condition),

1 1 ( , ) at

(boundary condition),

+ ➞ 2 = ➤ 2 ( ➜ , ➠ ) at ➛ = ➸

(boundary condition),

− ➞ ➱ 3 = ➤ 3 ( ➛ , ➠ ) at ➜ =0 (boundary condition). The solution ➞ ( ➛ , ➜ , ➠ ) is determined by the formula in Paragraph 2.1.1-13 where the Green’s function ➭ ( ➛ , ➜ , ➩ , ➫ , ➠ ) is the product of the Green’s function of Subsection 1.1.1-11 and that of

Subsection 1.1.1-8; one should replace ➛ , ➩ , and ➱ by ➜ , ➫ and ➱ 3 , respectively, in the last Green’s function.

2.1.1-15. Domain: 0 ≤ ➛ ≤ ➸ ,0≤ ➜ < ➝ . Mixed boundary value problems.

1 ✃ . A semiinfinite strip is considered. The following conditions are prescribed:

= ➟ ( ➛ , ➜ ) at ➠ =0 (initial condition),

= ➤ 1 ( ➜ , ➠ ) at ➛ =0 (boundary condition),

= ➤ 2 ( ➜ , ➠ ) at ➛ = ➸

(boundary condition),

= ➤ 3 ( ➛ , ➠ ) at ➜ =0 (boundary condition). Solution:

where

sin ➶

sin ➶

exp − ➸ 2 ➘ ,

exp q −

+ exp q −

2 ✃ . A semiinfinite strip is considered. The following conditions are prescribed:

= ➟ ( ➛ , ➜ ) at ➠ =0 (initial condition),

= ➤ 1 ( ➜ , ➠ ) at ➛ =0 (boundary condition),

= ➤ 2 ( ➜ , ➠ ) at ➛ = ➸

(boundary condition),

= ➤ 3 ( ➛ , ➠ ) at ➜ =0 (boundary condition). Solution:

2.1.1-16. Domain: 0 ≤ ➛ ≤ ➸ 1 ,0≤ ➜ ≤ ➸ 2 . First boundary value problem.

A rectangle 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:

Example 7. The initial temperature is uniform, ⑩ ✯ ↔ Ð ↔ Ð ↔ Ð ↔ ( , Ï )= ➾ 0 . The boundary is maintained at zero temperature,

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

2.1.1-17. Domain: 0 ≤ à ≤ á 1 ,0≤ â ≤ á 2 . Second boundary value problem.

A rectangle is considered. The following conditions are prescribed:

at å =0

(initial condition),

at à =0

(boundary condition),

at à = á 1 (boundary condition),

at â =0

(boundary condition),

at â = á 2 (boundary condition). Solution:

2 2 Reference : H. S. Carslaw and J. C. Jaeger (1984).

2.1.1-18. Domain: 0 ≤ ü

≤ 1 ,0≤ ≤ 2 . Third boundary value problem.

A rectangle is considered. The following conditions are prescribed:

(initial condition), − ✠ 1 = è ✁ 1 ( ,) at û =0 (boundary condition),

, ) at

2 = è ✁ 2 ( ,) at û ✞ = é 1 (boundary condition),

− ✁ ✠ ✆ = è ( ,) at =0 (boundary condition),

4 = è 4 ( û ,) at

= 2 (boundary condition).

The solution ✁ ( , , ) is determined by the formula in Paragraph 2.1.1-17 where ö ö

Here, the ✎

and ✏ þ are positive roots of the transcendental equations

tan( ü

1 + ✠ 2 tan( ✏ 2 )

2.1.1-19. Domain: 0 ≤ ü

≤ 1 ,0≤ ≤ 2 . Mixed boundary value problems.

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

, ) at

(initial condition),

= ✁ è 1 ( ,) at

=0 (boundary condition),

( ,) at

= 1 (boundary condition),

= è 3 ( û ,) at

=0 (boundary condition),

= 2 (boundary condition). Solution:

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

, ) at

(initial condition),

= ✁ è 1 ( ,) at

=0 (boundary condition),

2 ✁ ( ,) at

= 1 (boundary condition),

3 ( û ,) at

=0 (boundary condition),

4 ( û ,) at

= 2 (boundary condition).

2.1.2. Problems in Polar Coordinates

The sourceless heat equation with two space variables in the polar coordinate system ✛ , has the form

= ( ✛ , ) are considered in Subsection 1.2.1.

One-dimensional problems with axial symmetry that have solutions of the form ✆

2.1.2-1. Domain: 0 ≤ ✛ < ✢ ,0≤ ≤2 ø . Cauchy problem. An initial condition is prescribed: ☛

2.1.2-2. Domain: 0 ≤ ✛ ≤ ✪ ,0≤ ≤2 ø . First boundary value problem.

A circle is considered. The following conditions are prescribed: ☛

= ✝ ( ✛ , ) at

(initial condition),

(boundary condition). Solution:

= è ( ,) ☛ at ✛ = ✪

Here,

) cos[ ( − )] exp(− ),

where ö ( ) are the Bessel functions (the prime denotes the derivative with respect to the argument), and ✌ þ are positive roots of the transcendental equation ✥ ✱ ( ✌ ✪ ) = 0.

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

2.1.2-3. Domain: 0 ≤ ✛ ≤ ✪ ,0≤ ≤2 ✵ . Second boundary value problem.

A circle is considered. The following conditions are prescribed: ☛

= ✝ ( ✛ , ) at

(initial condition),

(boundary condition). Solution: ☛

2 cos[ ( − )] exp(− ),

( ✰ ) are the Bessel functions, and ✌ þ are positive roots of the transcendental equation

2.1.2-4. Domain: 0 ≤ ✛ ≤ ✪ ,0≤ ≤2 ✵ . Third boundary value problem.

A circle is considered. The following conditions are prescribed: ☛

= ✝ ( ✛ , ) at

(initial condition),

+ ✠ ✆ = è ( ,) ☛ at ✛ = ✪

(boundary condition).

The solution ☛ ( ✛ , , ) is determined by the formula in Paragraph 2.1.2-3 where

cos[ ( − )] exp(− ),

( ✰ ) are the Bessel functions, and ✌ þ are positive roots of the transcendental equation

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

2.1.2-5. Domain: ✪ 1 ≤ ✛ ≤ ✪ 2 ,0≤ ≤2 ✵ . First boundary value problem. An annular domain is considered. The following conditions are prescribed: ☛

= ✝ ( ✛ , ) at

(initial condition),

= è 1 ( ,) ☛ at ✛ = ✪ 1 (boundary condition),

= è 2 ( ☛ ,) at ✛ = ✪ 2 (boundary condition). Solution:

( ö 1 ) ✺ ( ✌ þ ✛ )− ✺ ( ✌ þ ✪ 1 ) ✱ ( ✌ þ ✛ ), where ✱ ( ✛ ) and ✺ ( ✛ ) are the Bessel functions, and ö ö ö ✌ þ ö are positive roots of the transcendental

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

2.1.2-6. Domain: ✪ 1 ≤ ✛ ≤ ✪ 2 ,0≤ ≤2 ✵ . Second boundary value problem. An annular domain is considered. The following conditions are prescribed: ☛

= ✝ ( ✛ , ) at

(initial condition),

= ✻ ( ,) at ✛ = ✪

(boundary condition),

= ✻ 2 ( ☛ ,) at ✛ = ✪ 2 (boundary condition). Solution:

( ✌ þ ✛ ) ✸ ( ✌ þ ) cos[ ✳ ( − ✧ )] exp(− ✌ 2 þ ★ ) ( , , , ,)=

( ö ✌ þ ✛ )= ✱ ✲ ( ✌ þ ✪ 1 ) ✺ ( ✌ þ ✛ )− ✺ ✲ ( ✌ þ ✪ 1 ) ✱ ( ✌ þ ✛ ), where 0 = 1 and

= 2 for ✳ = 1, 2, ✴✕✴✕✴ ; ✱ ( ✛ ) and ✺ ( ✛ ) are the Bessel functions, and ✌ þ are positive roots of the transcendental equation ö ö

2.1.2-7. Domain: ✪ 1 ≤ ✛ ≤ ✪ 2 ,0≤ ≤2 ✵ . Third boundary value problem. An annular domain is considered. The following conditions are prescribed: ☛

= ✝ ( ✛ , ) at

(initial condition),

− ✠ ✆ 1 = ✻ 1 ( ,) at ✛ = ✪ 1 (boundary condition),

,) at ✛ = ✪ 2 (boundary condition). Solution:

) cos[ ✛ ✳ ( − ✬ ✧ )] exp(− ✌ 2 þ ★ ) ( , , , ,)=

where 0 = 1 and = 2 for ✳ = 1, 2, ✴✕✴✕✴ ; ✱ ( ✛ ) and ✺ ( ✛ ) are the Bessel functions, and ✌ are positive roots of the transcendental equation ö ö ö ö ✰ ✰

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

2.1.2-8. Domain: 0 ≤ ✛ < ✢ ,0≤ ≤ 0 . First boundary value problem.

A wedge domain is considered. The following conditions are prescribed: ☛ ☛

= ❅ ( ✛ , ❆ ) at ❇ =0

(initial condition),

= ✻ 1 ( ✛ , ❇ ) at ❆ =0

(boundary condition),

= ✻ 2 ( ✛ , ❇ ) at ❆ = ❆ 0 (boundary condition). Solution:

) are the modified Bessel functions.

Reference ❏❖◆ : B. M. Budak, A. A. Samarskii, and A. N. Tikhonov (1980).

P✕❘

2.1.2-9. Domain: 0 ≤ ❊ < ❱ ,0≤ ❆ ≤ ❆ 0 . Second boundary value problem.

A wedge domain is considered. The following conditions are prescribed:

= ❅ ( ❊ , ❆ ) at ❇ =0

(initial condition),

= ❉ ✻ 1 ( ❊ , ❇ ) at ❆ =0

(boundary condition),

= ✻ 2 ( ❊ , ❇ ) at ❆ = ❆ 0 (boundary condition). Solution:

) are the modified Bessel functions.

Reference ❏❖◆ : B. M. Budak, A. A. Samarskii, and A. N. Tikhonov (1980).

2.1.2-10. Domain: 0 ≤ ❣ ≤ ❥ ,0≤ ≤ 0 . First boundary value problem.

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

= ❧ ( ❣ , ) at ❴ =0

(initial condition),

= ❢ 1 ( , ❴ ) at ❣ = ❥

(boundary condition),

= ❢ 2 ( , ❴ ) at

(boundary condition),

, ❴ ) at

= 0 (boundary condition).

sin ❞ ❡ exp(− ),

where the ✉ ❜

0 ( ❣ ) are the Bessel functions, and the t q are positive roots of the transcendental equation

0 ( t ❥ ) = 0.

Example. The initial temperature is uniform, ① ( ② , ③ )= ④ 0 . The boundary is maintained at zero temperature, ⑤ 1 ( ③ , ⑥ )=

❤☎✐ where the ❺

are positive roots of the transcendental equation

References : B. M. Budak, A. A. Samarskii, and A. N. Tikhonov (1980), H. S. Carslaw and J. C. Jaeger (1984).

2.1.2-11. Domain: 0 ≤ ❣ ≤ ❥ ,0≤ ≤ 0 . Second boundary value problem.

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

= ❧ ( ❣ , ) at ❴ =0

(initial condition),

= ❢ 1 ( , ❴ ) at ❣ = ❥

(boundary condition),

(boundary condition),

= 3 ( ❣ , ❴ ) at

= 0 (boundary condition).

exp(− t 2 q ❭ ❴ ),

where the ❢

0 ( ❣ ) are the Bessel functions, and the t q are positive roots of the transcendental equation

( t ❥ ) = 0.

❤☎✐ r ❜

Reference r : B. M. Budak, A. A. Samarskii, and A. N. Tikhonov (1980). ✉

2.1.2-12. Domain: 0 ≤ ❣ ≤ ❥ ,0≤ ≤ 0 . Mixed boundary value problem.

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

= ❧ ( ❣ , ) at ❴ =0

(initial condition),

= ( , ❴ ) at ❣ = ❥

(boundary condition),

(boundary condition),

at

= 0 (boundary condition).

Solution: ❢

( t q ❾ ❨ ) cos( ➔

) cos( ➔ ❾ ❩ ➋ ) exp(− t 2 q ❭ ❴ ),

=0 q =1 ➒

where are positive roots of the transcendental equation ❾ ➓ ( ❣ ) are the Bessel functions, and t q r

Reference : B. M. Budak, A. A. Samarskii, and A. N. Tikhonov (1980). r ➓ ✉ r ➓

2.1.2-13. Domain: ❥ 1 ≤ ❣ ≤ ❥ 2 ,0≤ ≤ 0 . Different boundary value problems.

Some problems for this domain were studied in Budak, Samarskii, and Tikhonov (1980). ❢

2.1.3. Axisymmetric Problems

In the case of angular symmetry, the two-dimensional sourceless heat equation in the cylindrical coordinate system has the form

This equation governs two-dimensional unsteady thermal processes in quiescent media or solid ↕ bodies (bounded by coordinate surfaces of the cylindrical system) in the case where the initial and

boundary conditions are independent of the angular coordinate. A similar equation is used to study analogous two-dimensional unsteady mass transfer phenomena.

2.1.3-1. Particular solutions. Remarks on the Green’s functions.

. Apart from usual separable solutions ❦ ( ❣ , , ➟ )= ❧ 1 ( ❣ ) ❧ 2 ( ) ❧ 3 ( ➟ ), the equation in question has more sophisticated solutions in the product form ↕

where ↕

= ➠ ( ❣ , ➟ ) and ➡ = ➡ ( , ➟ ) are solutions of the simpler one-dimensional equations

(see Subsection 1.2.1-1 for particular solutions of this equation),

2 (see Subsection 1.1.1-1 for particular solutions of this equation).

2 ↕ ➞ . For all two-dimensional boundary value problems considered in Subsection 2.1.3, the Green’s function can be represented in the product form

where ↕

1 ( , ❨ , ➟ ) and ➈ 2 ( , ➥ , ➟ ) are the Green’s functions of appropriate one-dimensional boundary value problems. ➤

2.1.3-2. Domain: 0 ≤ ≤ ❥ ,0≤ < ➦ . First boundary value problem.

A semiinfinite circular cylinder is considered. The following conditions are prescribed: ↕

= ❧ ( , ) at ➟ =0

(initial condition),

= ➧ 1 ( ➤ , ↕ ➟ ) at

(boundary condition),

= ➧ 2 ( ↕ , ➟ ) at ➤ =0 (boundary condition).

, where the ➪

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

Example 1. The initial temperature is the same at every point of the cylinder, ➮ ( ➱ , ✃ )= ❐ 0 . The lateral surface and the

end face are maintained at zero temperature, ❒ 1 ( ➱ , ⑥ )= ❒ 2 ( ✃ , ⑥ ) = 0. Ó

Example 2. The initial temperature of the cylinder is everywhere zero, ➮ ( ➱ , ✃ ) = 0. The lateral surface ➱ = ⑧ is

maintained at a constant temperature ❐ 0 , and the end face ✃ =0 at zero temperature.

where the Õ

are positive zeros of the Bessel function, 0 ( Õ ⑧ ) = 0.

References : B. M. Budak, A. A. Samarskii, and A. N. Tikhonov (1980), H. S. Carslaw and J. C. Jaeger (1984). Ô

2.1.3-3. Domain: 0 ≤ ➸ ≤ ➲ ,0≤ ➭ < Û . Second boundary value problem.

A semiinfinite circular cylinder is considered. The following conditions are prescribed:

= ➫ ( ➸ , ➭ ) at ➺ =0

(initial condition),

= ➽ 1 ( ➭ , ➺ ) at ➸ = ➲

(boundary condition),

= ➽ 2 ( ➸ , ➺ ) at ➭ =0 (boundary condition).

where the ✜

are positive zeros of the first-order Bessel function, 1 ( ✢ ) = 0.

2.1.3-4. Domain: 0 ≤ ✟ ≤ ✕ ,0≤ ✠ < ✪ . Third boundary value problem.

A semiinfinite circular cylinder is considered. The following conditions are prescribed:

( ✟ , ✠ ) at ✡ ✫ =0 (initial condition),

1 ✏ = 1 ( ✠ , ✡ ) at ✟ = ✕

(boundary condition),

− ✗ ✞ 2 = 2 ( ✟ , ✡ ) at ✠ =0 (boundary condition).

The solution ✞ ( ✟ , ✠ , ✡ ) is determined by the formula in Paragraph 2.1.3-3 where

) is the zeroth Bessel function and the ✢ are positive roots of the transcendental equation

Example 3. The initial temperature is the same at every point of the cylinder, ✰ ( ✱ , ✲ )= ✝ 0 . At the lateral surface and the end face, heat exchange of the cylinder with the zero temperature environment occurs, ✳ 1 ( ✲ , ✴ )= ✳ 2 ( ✱ , ✴ ) = 0.

+ exp( ✵ 2 ✲ + ✵ 2 ☎✺✴ ) erfc ✸

where the ❀

are positive roots of the transcendental equation ❁ 1 ( ❁

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

2.1.3-5. Domain: 0 ≤ ✟ ≤ ✕ ,0≤ ✠ < ✪ . Mixed boundary value problems.

1 ❅ . A semiinfinite circular cylinder is considered. The following conditions are prescribed:

= ( ✟ , ✠ ) at ✡ =0

(initial condition),

, ✡ ) at ✟ = ✕

(boundary condition),

, ✡ ) at ✠ =0 (boundary condition). Solution: ✗

where the ✜

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

2 ❅ . A semiinfinite circular cylinder is considered. The following conditions are prescribed:

= ( ✟ , ✠ ) at ✡ =0

(initial condition),

, ✡ ) at ✟ = ✕

(boundary condition),

, ✡ ) at ✠ =0 (boundary condition). Solution: ✗

where the ✜

are positive zeros of the first-order Bessel function, 1 ( ✢ ) = 0.

3 ❅ . A semiinfinite circular cylinder is considered. The following conditions are prescribed:

= ( ✟ , ✠ ) at ✡ =0

(initial condition),

= 1 ( ✠ , ✡ ) at ✟ = ✕

(boundary condition),

, ✡ ) at ✠ =0 (boundary condition).

The solution ✗ (

, ✠ , ✡ ) is determined by the formula in Paragraph 2.1.3-5, Item 2 ❅ where

where the ✢ are positive roots of the transcendental equation

Example 4. The initial temperature is the same at every point of the cylinder, ✰ ( ✱ , ✲ )= ✝ 0 . Heat exchange of the cylinder with the zero temperature environment occurs at the lateral surface, ✳ 1 ( ✲ , ✴ ) = 0. The end face is maintained at zero temperature, ✳ 2 ( ✱ , ✴ ) = 0. Solution:

Example 5. The initial temperature of the cylinder is everywhere zero, ❀

( ✱ , ✲ ) = 0. Heat exchange of the cylinder with the zero temperature environment occurs at the lateral surface, ✳ 1 ( ✲ , ✴ ) = 0. The end face is maintained at a constant temperature, ✳ 2 ( ✱ , ✴ )= ✝ 0 . Solution:

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

2.1.3-6. Domain: 0 ≤ ✟ ≤ ✕ ,0≤ ✠ ≤ ❊ . First boundary value problem.

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

= ( ✟ , ✠ ) at ✡ =0

(initial condition),

, ✡ ) at ✟ = ✕

(boundary condition),

, ✡ ) at ✠ =0 (boundary condition),

, ✡ ) at ✠ = ❊

(boundary condition).

Solution: ✗

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

Example 6. The initial temperature is the same at every point of the cylinder, ✰ ( ✱ , ✲ )= ✝ 0 . The lateral surface and the

end faces are maintained at zero temperature, ✳ 1 ( ✲ , ✴ )= ✳ 2 ( ✱ , ✴ )= ✳ 3 ( ✱ , ✴ ) = 0.

are positive zeros of the Bessel function, 0 ( ❉

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

2.1.3-7. Domain: 0 ≤ ✟ ≤ ✕ ,0≤ ✠ ≤ ❊ . Second boundary value problem.

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

= ( ✟ , ✠ ) at ✡ =0

(initial condition),

, ✡ ) at ✟ = ✕

(boundary condition),

, ✡ ) at ✠ =0 (boundary condition),

, ✡ ) at ✠ = ❊

(boundary condition).

where the ✜

are positive zeros of the first-order Bessel function, 1 ( ✢ ) = 0.

2.1.3-8. Domain: 0 ≤ ✟ ≤ ✕ ,0≤ ✠ ≤ ❊ . Third boundary value problem.

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

= ( ✟ , ✠ ) at ✡ =0

(initial condition),

, ✡ ) at ✟ = ✕

(boundary condition),

( ✟ , ✡ ) at ✠ ✮ =0 (boundary condition),

3 ✞ = 3 ( ✟ , ✡ ) at ✠ = ❊

(boundary condition).

The solution ✗ (

, ✠ , ✡ ) is determined by the formula in Paragraph 2.1.3-7 where

and ◆ ◗ ▼ are positive roots of the transcendental equations

tan( )

2.1.3-9. Domain: 0 ≤ ❙ ≤ ✕ ,0≤ ❖ ≤ ❊ . Mixed boundary value problems.

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

= ( ❙ , ❖ ) at ❘ =0

(initial condition),

, ❘ ) at ❙ = ✕

(boundary condition),

, ❘ ) at ❖ =0 (boundary condition),

(boundary condition). Solution: ✗

, where the ❡

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

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

( ❤ , ❦ ) at =0

(initial condition),

, ) at

(boundary condition),

= ❫ ❤ ❥ 2 ( , ) at

=0 ♠ (boundary condition),

(boundary condition). Solution: ❫

where the ❡

are positive zeros of the first-order Bessel function, 1 ( ❢ ) = 0.

2.1.3-10. Domain: 1 ≤ ≤ 2 ,0≤ ❦ ≤ . First boundary value problem.

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

( ❤ , ❦ ) at =0

(initial condition),

, ❤ ) at = 1 (boundary condition),

2 ② ( , ) at = 2 (boundary condition),

= ② ❤ 3 ( ❥ , ) at

(boundary condition),

(boundary condition). Solution: ④

= ④ 4 ( , ) at

) and ❿ 0 ( ❢ ) are the Bessel functions, the ❢ are positive roots of the transcendental equation

where ② 0 (

2.1.3-11. 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),

, ) at

= 1 (boundary condition),

( , ) at

= 2 (boundary condition),

( ② , at

(boundary condition),

= ④ 4 ( , ) at

(boundary condition).

) and ❿ ( ❢ ) are the Bessel functions of order ➄ =0 , 1 and the ❢ are positive roots of the transcendental equation

2.1.3-12. Domain: 1 ≤ ≤ 2 ,0≤ ❦ ≤ . Third boundary value problem.

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

( ❤ , ) at =0

(initial condition),

1 ❤ 1 ( , ) at = 1 (boundary condition),

, ) at

= 2 (boundary condition),

3 ② ( , ) at

(boundary condition),

(boundary condition). For the solution of this problem, see Subsection 2.2.3-4 with ④

4 q = 4 ( , ) at ❦ =

2.2. Heat Equation with a Source ➆ ➇ = ➉ ➊ 2 ➇ + ➋ ( ➌ , ➍ , ➈ )

2.2.1. Problems in Cartesian Coordinates ➈

In the rectangular Cartesian coordinate system, the heat equation has the form

It governs two-dimensional unsteady thermal processes in quiescent media or solids with constant thermal diffusivity in the cases where there are volume thermal sources or sinks.

< < → ,− → < < → . Cauchy problem. An initial condition is prescribed:

2.2.1-1. Domain: − ➓

Reference ① : A. G. Butkovskiy (1979).

2.2.1-2. Domain: 0 ≤ ➓ <

,− → < < → . First boundary value problem.

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

= ➐ ( , ) at =0 (initial condition),

= ➒ ( , ) at =0 (boundary condition).

Solution: ④

References ① : A. G. Butkovskiy (1979), H. S. Carslaw and J. C. Jaeger (1984).

2.2.1-3. Domain: 0 ≤ ➓ <

,− → < < → . Second boundary value problem.

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

( ➓ , ➐ ) at

(initial condition),

=0 (boundary condition). Solution:

2.2.1-4. Domain: 0 ≤ ➓ <

,− → < < → . Third boundary value problem.

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

( , ) at

=0 (initial condition),

= ➧ ( , ) at

=0 (boundary condition).

The solution ➐ ( , , ) is determined by the formula in Paragraph 2.2.1-3 where

2.2.1-5. Domain: 0 ≤ ➓ <

,0≤ < → . First boundary value problem.

A quadrant of the plane is considered. The following conditions are prescribed:

( ➓ , ➐ ) at

(initial condition),

( ➒ , ) at

(boundary condition),

=0 (boundary condition). Solution:

= ➧ 2 ( , ) at

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

2.2.1-6. Domain: 0 ≤ ➓ <

,0≤ < → . Second boundary value problem.

A quadrant of the plane is considered. The following conditions are prescribed:

( , ) at

(initial condition),

, ➐ ) at ➓

(boundary condition),

=0 (boundary condition). Solution:

2.2.1-7. Domain: 0 ≤ ➓ <

,0≤ < → . Third boundary value problem.

A quadrant of the plane is considered. The following conditions are prescribed:

( , ) at

=0 (initial condition),

1 = ➧ 1 ( , ) at

=0 (boundary condition),

2 = ➧ 2 ( , ) at

=0 (boundary condition).

The solution ➐ ( , , ) is determined by the formula in Paragraph 2.2.1-6 where

exp ➪❄➯ ➺ 1 + ➺ 1 ( + ➫ ) ➶ erfc ➹ ➘

exp ➪❄➯ ➺ 2 2 + ➺ 2 ( ➬ + ➮ ) ➶ erfc ➹

2.2.1-8. Domain: 0 ≤ ➘ < Ô ,0≤ ➬ < Ô . Mixed boundary value problem.

A quadrant of the plane is considered. The following conditions are prescribed:

( , ➬ ) at

=0 (initial condition),

1 ( ➬ , ) at

(boundary condition),

2 ( ➘ , ) at ➬ =0 (boundary condition). Solution:

2.2.1-9. Domain: 0 ≤ ≤ Ú ➘ ,0≤ ➬ < Ô . First boundary value problem.

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

( ➬ ➘ , ➐ ) at

=0 (initial condition),

= ➧ 1 ( ➬ , ) at

(boundary condition),

= ➧ 2 ( ➬ , ) at

(boundary condition),

, ) at ➬ =0 (boundary condition). The solution is given by the formula of Subsection 2.1.1-12 with the additional term

which takes into account the equation’s nonhomogeneity.

2.2.1-10. Domain: 0 ≤ ≤ Ú ,0≤ ➬ < Ô ➘ . Second boundary value problem.

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

( ➘ , ➬ ) at

=0 (initial condition),

1 ( ➬ , ) at

=0 (boundary condition),

2 ( ➬ , ) at

(boundary condition),

3 ( ➘ , ) at ➬ =0 (boundary condition). Solution:

2.2.1-11. Domain: 0 ≤ ≤ Ú ➘ ,0≤ ➬ < Ô . Third boundary value problem.

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

( , ➬ ) at

=0 (initial condition),

) at ➘

(boundary condition),

, ➐ ) at

(boundary condition),

− ➺ Õ 3 = ➧ 3 ( , ) at ➬ =0 (boundary condition).

, ➐ ➬ , ) is determined by the formula in Paragraph 2.2.1-10 where the Green’s function Ø ( ➘ , ➬ , ➫ , ➮ , ) is the product of the Green’s function of Subsection 1.1.1-11 and that of Subsection 1.1.1-8; ➘ , ➫ , and ➺ in the last Green’s function must be replaced by ➬ , ➮ , and ➺ 3 , respectively.

The solution ➘ (

2.2.1-12. Domain: 0 ≤ ≤ Ú ➘ ,0≤ ➬ < Ô . Mixed boundary value problem.

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

( , ➬ ) at

=0 (initial condition),

1 ( ➬ , ) at

(boundary condition),

2 ( ➬ , ) at

(boundary condition),

(boundary condition). The solution is given by the formula of Subsection 2.1.1-15 (Item 1 á ) with the additional term

, ) at

which takes into account the equation’s nonhomogeneity.

2.2.1-13. Domain: 0 ≤ ≤ Ú 1 ,0≤ ➬ ≤ Ú ➘ 2 . First boundary value problem.

A rectangle is considered. The following conditions are prescribed:

, ➬ ) at æ =0

(initial condition),

= ➧ 1 ( ➬ , æ ) at

(boundary condition),

= ➧ 2 ( ➬ , æ ) at

1 (boundary condition),

, æ ) at ➬ =0 (boundary condition),

, æ ) at ➬ = Ú 2 (boundary condition). The solution is given by the formula of Subsection 2.1.1-16 with the additional term

which takes into account the equation’s nonhomogeneity. è❄é

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

2.2.1-14. Domain: 0 ≤ ≤ Ú ➘ 1 ,0≤ ➬ ≤ Ú 2 . Second boundary value problem.

A rectangle is considered. The following conditions are prescribed:

, ➬ ) at æ × =0 ➦ (initial condition),

= ➧ 1 ( ➬ , æ ) at

(boundary condition),

= ➧ 2 ( ➬ , æ ) at

1 (boundary condition),

, æ ) at ➬ =0 (boundary condition),

, æ ) at ➬ = Ú 2 (boundary condition). Solution:

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

2.2.1-15. Domain: 0 ≤ ≤ Ú ➘ 1 ,0≤ ➬ ≤ Ú 2 . Third boundary value problem.

A rectangle is considered. The following conditions are prescribed:

( ➘ , ) at

(initial condition),

( ➬ , æ ) at

(boundary condition),

( ➬ , æ ) at

1 (boundary condition),

, ) at

(boundary condition),

, æ ) at ➬ = Ú 2 (boundary condition). The solution Õ (

, ➬ , æ ) is determined by the formula in Paragraph 2.2.1-14 where Ý Ý

Here, the ô

and õ î are positive roots of the transcendental equations

tan( ó Ú 1 )

1 2 tan( 2 ) =

2.2.1-16. Domain: 0 ≤ ø ≤ Ú 1 ,0≤ ➬ ≤ Ú 2 . Mixed boundary value problem.

A rectangle is considered. The following conditions are prescribed:

= Ö ( ø , ➬ ) at æ =0

(initial condition),

= ➧ 1 ( ➬ , æ ) at ø =0 (boundary condition),

= ➧ 2 ( ➬ , æ ) at ø = Ú 1 (boundary condition),

= ➧ 3 ( ø , æ ) at ➬ =0 (boundary condition),

= ➧ 4 ( ø , æ ) at ➬ = Ú 2 (boundary condition). The solution is given by the formula of Subsection 2.1.1-19 (Item 1 á ) with the additional term

which takes into account the equation’s nonhomogeneity.

2.2.2. Problems in Polar Coordinates

The heat equation with a volume source in the polar coordinate system ù , is written as

and govern plane thermal processes with central symmetry, are presented in Subsection 1.2.2.

Solutions of the form Õ = Õ ( ñ ù , æ ) that are independent of the angular coordinate

2.2.2-1. Domain: 0 ≤ ù < Ô ,0≤ ≤2 ➱ . Cauchy problem. An initial condition is prescribed: ñ

2.2.2-2. Domain: 0 ≤ ù ≤ û ,0≤ ≤2 ➱ . Different boundary value problems.

1 á . The solution of the first boundary value problem for a circle of radius ñ û is given by the formula from Subsection 2.1.2-2 with the additional term

which allows for the equation’s nonhomogeneity. ñ

2 á . The solution of the second boundary value problem for a circle is given by the formula in Paragraph 2.1.2-3 with the additional term (1).

3 á . The solution of the third boundary value problem for a circle is given by the formula in Paragraph 2.1.2-4 with the additional term (1).

2.2.2-3. Domain: û 1 ≤ ù ≤ û 2 ,0≤ ≤2 ➱ . Different boundary value problems.

1 á . The solution of the first boundary value problem for an annular domain is given by the formula ñ

in Paragraph 2.1.2-5 with the additional term

which allows for the equation’s nonhomogeneity. ñ

2 á . The solution of the third boundary value problem for an annular domain is given by the formula in Paragraph 2.1.2-7 with the additional term (2).

2.2.2-4. Domain: 0 ≤ ù < Ô ,0≤ ≤ 0 . Different boundary value problems.

1 á . The solution of the first boundary value problem for a wedge domain is given by the formula in ñ ñ

Paragraph 2.1.2-8 with the additional term

which allows for the equation’s nonhomogeneity. ñ

2 á . The solution of the second boundary value problem for a wedge domain is given by the formula in Paragraph 2.1.2-9 with the additional term (3).

2.2.2-5. Domain: 0 ≤ ù ≤ û ,0≤ ≤ 0 . Different boundary value problems.

1 á . The solution of the first boundary value problem for a sector of a circle is given by the formula ñ ñ

of Paragraph 2.1.2-10 with the additional term

which allows for the equation’s nonhomogeneity. ñ

2 á . The solution of the mixed boundary value problem for a sector of a circle is given by the formula of Paragraph 2.1.2-11 with the additional term (4).

2.2.3. Axisymmetric Problems

In the case of axial symmetry, the heat equation in the cylindrical coordinate system is written as

provided there are heat sources or sinks. One-dimensional axisymmetric problems that have solutions of the form Õ = Õ ( ù , æ ) can be

found in Subsection 1.2.2. 2.2.3-1. Domain: 0 ≤ ÿ

≤ û ,0≤ < Ô . Different boundary value problems.

1 á . The solution to the first boundary value problem for a semiinfinite circular cylinder of radius û

is given by the formula of Subsection 2.1.3-2 with the term

added; this term takes into account the nonhomogeneity of the equation.

2 á . The solution to the second boundary value problem for a semiinfinite circular cylinder is given by the formula of Subsection 2.1.3-3 with the additional term (1).

3 á . The solution to the third boundary value problem for a semiinfinite circular cylinder is given by the formula of Subsection 2.1.3-4 with the additional term (1).

4 á . The solutions to various mixed boundary value problems for a semiinfinite circular cylinder are defined by formulas of Subsection 2.1.3-5 with additional terms of the form (1).

2.2.3-2. Domain: 0 ≤ ÿ

≤ û ,0≤ ≤ Ú . Different boundary value problems.

1 á . The solution to the first boundary value problem for a circular cylinder of radius û and length Ú

is given by the formula of Subsection 2.1.3-6 with the term

added; this term takes into account the nonhomogeneity of the equation.

2 á . The solution to the second boundary value problem for a finite circular cylinder is given by the formula of Subsection 2.1.3-7 with the additional term (2).

3 á . The solution to the third boundary value problem for a finite circular cylinder is given by the formula of Subsection 2.1.3-8 with the additional term (2).

4 á . The solutions to various mixed boundary value problems for a finite circular cylinder are defined by formulas of Subsection 2.1.3-9 with additional terms of the form (2).

2.2.3-3. Domain: ÿ

1 ≤ ù ≤ û 2 ,0≤ ≤ Ú . First and second boundary value problems.

1 á . The solution to the first boundary value problem for a hollow circular cylinder of interior radius û 1 , exterior radius û 2 , and length Ú is given by the formula of Subsection 2.1.3-10 with the

term

added; this term takes into account the equation’s nonhomogeneity.

2 á . The solution to the second boundary value problem for a finite hollow circular cylinder is given by the formula of Subsection 2.1.3-11 with the additional term (3).

2.2.3-4. Domain: ÿ

1 ≤ ù ≤ û 2 ,0≤ ≤ Ú . Third boundary value problem.

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

( ù , ) at æ =0

(initial condition),

− ð Õ 1 = ✁ ÿ 1 ( , æ ) at ù = û 1 (boundary condition),

2 Õ = ✁ 2 ( , æ ) at ù = û 2 (boundary condition),

, æ ) at

(boundary condition),

4 Õ = ✁ 4 ( ù , æ ) at

(boundary condition).

Here, the Green’s function is given by

( )= ó î cos( ó î )+ ð 3 sin( ó î ),

0 ( ), ✞ 1 ( õ ), ✏ 0 ( õ ), and ✏ 1 ( õ ) are the Bessel functions, the õ are positive roots of the transcendental equation ✝

− ð 2 ✞ 0 ( õ û 2 )− õ ✞ 1 ( õ û 2 ) ✟ ð 1 ✏ 0 ( õ û 1 )+ õ ✏ 1 ( õ û 1 ) ✟ =0 , and the ó î are positive roots of the transcendental equation

tan ó

Reference : A. G. Butkovskiy (1979).

2.2.3-5. Domain: ÿ

1 ≤ ù ≤ û 2 ,0≤ ≤ ✑ . Mixed boundary value problem.

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

( ù , ) at æ =0

(initial condition),

− ð ✒ 1 = ✁ ÿ 1 ( , æ ) at ù = û 1 (boundary condition),

2 ✁ = 2 ( , æ ) at ù = û 2 (boundary condition),

, æ ) at

(boundary condition),

, æ ) at

(boundary condition).

where the expression of ✘ 1 ( ù , ★ , ✚ ) is specified in Subsection 2.2.3-4.

Subsection 3.2.2 presents solutions of other boundary value problems; a more general, three- dimensional equation is discussed there.