APPLICATION OF SOME SUBSTITUTIONS TO THE SOLUTION OF BOUNDARY VALUE PROBLEMS OF HEAT CONDUCTION

V. V. Ivanov, V. V. Salomatov

(Presented by Prof. G. I. Fuks, Doctor of Technical Sciences)

In the theory of heat transfer it is sometimes necessary to solve the heat-conduction equation \(\nabla^2 \vartheta = 0\) with mixed boundary conditions of the first and third kinds. In a number of cases, by means of appropriate analytical techniques, it is possible to reduce a problem with mixed boundary conditions to a simpler problem with homogeneous boundary conditions of the first kind. Below we give several substitutions that make it possible to avoid the difficulties associated with boundary conditions reflecting convective heat exchange on the surface of a body.

Let it be required to solve the equation

\[ \frac{\partial^2 \vartheta}{\partial x^2}+\frac{\partial^2 \vartheta}{\partial y^2}=0 \]

with boundary conditions:

for \(x=x_i\),

\[ a\vartheta=-\lambda \frac{\partial \vartheta}{\partial x} \]

or

\[ \vartheta+\frac{\lambda}{a}\frac{\partial \vartheta}{\partial x}=0, \]

while on the remaining part of the boundary of the body constant temperatures \(\vartheta=\vartheta_i\) are prescribed.

Introduce the substitution

\[ \varepsilon(x,y)=\vartheta(x,y)+\frac{\lambda}{a}\frac{\partial \vartheta(x,y)}{\partial x}. \tag{1} \]

Then

\[ \frac{\partial^2 \vartheta}{\partial x^2} = \frac{\partial^2 \varepsilon}{\partial x^2} - \frac{\lambda}{a}\frac{\partial}{\partial x} \left( \frac{\partial^2 \vartheta}{\partial x^2} \right), \]

\[ \frac{\partial^2 \vartheta}{\partial y^2} = \frac{\partial^2 \varepsilon}{\partial y^2} - \frac{\lambda}{a}\frac{\partial}{\partial x} \left( \frac{\partial^2 \vartheta}{\partial y^2} \right) = \frac{\partial^2 \varepsilon}{\partial y^2} + \frac{\lambda}{a}\frac{\partial}{\partial x} \left( \frac{\partial^2 \vartheta}{\partial x^2} \right). \]

Adding these equalities, we obtain

\[ \frac{\partial^2 \varepsilon}{\partial x^2} + \frac{\partial^2 \varepsilon}{\partial y^2} =0. \tag{2} \]

The boundary conditions for equation (2) will have the form

\[ \varepsilon = 0 \quad \text{for } x = x_i, \]

\(\varepsilon=\vartheta_i\), on the remaining part of the boundary of the body.

Let the solution of equation (2) be \(\varepsilon=\psi(x,y)\). Then the sought temperature field for \(\vartheta\) is found by integrating the differential equation

\[ \frac{\partial \vartheta}{\partial x}+\frac{\alpha}{\lambda}\vartheta = \frac{\alpha}{\lambda}\psi . \]

If it is desired to find the solution \(\vartheta=f(\rho,\varphi)\) of the heat-conduction equation in cylindrical coordinates

\[ \rho^2\frac{\partial^2\vartheta}{\partial \rho^2} + \rho\frac{\partial\vartheta}{\partial \rho} + \frac{\partial^2\vartheta}{\partial \varphi^2} =0 \]

with boundary conditions:

\[ \text{for } \rho=R,\quad \varphi_i<\varphi<\varphi_k,\quad \alpha\vartheta=-\lambda\frac{\partial\vartheta}{\partial \rho}, \]

and on the remaining part of the boundary of the body constant temperatures \(\vartheta=\vartheta_i\) are prescribed, then the substitution used is

\[ \varepsilon(\rho,\varphi) = Bi\vartheta(\rho,\varphi) + \rho\frac{\partial\vartheta(\rho,\varphi)}{\partial \rho}, \qquad \left(Bi=\frac{\alpha R}{\lambda}\right). \tag{3} \]

Since

\[ Bi\frac{\partial\vartheta}{\partial \rho} = \frac{\partial\varepsilon}{\partial \rho} - \rho\frac{\partial^2\vartheta}{\partial \rho^2} - \frac{\partial\vartheta}{\partial \rho}, \tag{4} \]

\[ Bi\frac{\partial^2\vartheta}{\partial \rho^2} = \frac{\partial^2\varepsilon}{\partial \rho^2} - 2\frac{\partial^2\vartheta}{\partial \rho^2} - \rho\frac{\partial^3\vartheta}{\partial \rho^3}, \tag{5} \]

\[ Bi\frac{\partial^2\vartheta}{\partial \varphi^2} = \frac{\partial^2\varepsilon}{\partial \varphi^2} - \rho\frac{\partial}{\partial \rho} \left( \frac{\partial^2\vartheta}{\partial \varphi^2} \right) = \frac{\partial^2\varepsilon}{\partial \varphi^2} + \rho\frac{\partial}{\partial \rho} \left( \rho^2\frac{\partial^2\vartheta}{\partial \rho^2} + \rho\frac{\partial\vartheta}{\partial \rho} \right), \]

then, multiplying (4) and (5) respectively by \(\rho\) and \(\rho^2\), we obtain

\[ \rho^2\frac{\partial^2\varepsilon}{\partial \rho^2} + \rho\frac{\partial\varepsilon}{\partial \rho} + \frac{\partial^2\varepsilon}{\partial \varphi^2} =0. \tag{6} \]

The boundary conditions for equation (6) take the form

\[ \varepsilon=0 \quad \text{for } \rho=R,\quad \varphi_i<\varphi<\varphi_k, \]

The sought temperature field \(\vartheta=f(\rho,\varphi)\) is found by integrating the equation

\[ \frac{\partial\vartheta}{\partial \rho} + Bi\,\frac{\vartheta}{\rho} = \frac{\varepsilon}{\rho}. \]

Substitutions (1) and (3) make it possible to solve certain stationary heat-conduction problems with mixed boundary conditions (especially in bodies of complex configuration), using the effective method of conformal mappings. It is known [1] that with the aid of conformal mappings one can solve only boundary-value problems of heat conduction with boundary conditions of the 1st and 2nd kinds.

Suppose it is required to solve Laplace’s equation in spherical coordinates

\[ r^2\frac{\partial^2\vartheta}{\partial r^2} + 2r\frac{\partial\vartheta}{\partial r} + \operatorname{ctg}\Theta\frac{\partial\vartheta}{\partial \Theta} + \frac{\partial^2\vartheta}{\partial \Theta^2} =0 \]

with boundary conditions: for \(r=R,\ \Theta_i<\Theta<\Theta_k,\ x\vartheta=-\lambda\,\dfrac{\partial \vartheta}{\partial r}\); on the remaining part of the surface of the body constant temperatures \(\vartheta=\vartheta_i\) are prescribed.

Introducing the substitution

\[ \varepsilon(r,\Theta)=\vartheta(r,\Theta)+\frac{r}{Bi}\,\frac{\partial \vartheta}{\partial r}, \tag{7} \]

we obtain

\[ \frac{\partial \vartheta}{\partial r} = \frac{\partial \varepsilon}{\partial r} -\frac{1}{Bi}\,r\,\frac{\partial^2 \vartheta}{\partial r^2} -\frac{1}{Bi}\,\frac{\partial \vartheta}{\partial r}, \tag{8} \]

\[ \frac{\partial^2\vartheta}{\partial r^2} = \frac{\partial^2\varepsilon}{\partial r^2} -\frac{2}{Bi}\,\frac{\partial^2\vartheta}{\partial r^2} -\frac{1}{Bi}\,r\,\frac{\partial^3\vartheta}{\partial r^3}, \tag{9} \]

\[ \frac{\partial \vartheta}{\partial \Theta} = \frac{\partial \varepsilon}{\partial \Theta} -\frac{1}{Bi}\,r\,\frac{\partial^2\vartheta}{\partial r\,\partial \Theta}, \tag{10} \]

\[ \begin{aligned} \frac{\partial^2\vartheta}{\partial \Theta^2} &= \frac{\partial^2\varepsilon}{\partial \Theta^2} -\frac{1}{Bi}\,r\,\frac{\partial}{\partial r} \left(\frac{\partial^2\vartheta}{\partial \Theta^2}\right) \\ &= \frac{\partial^2\varepsilon}{\partial \Theta^2} +\frac{1}{Bi}\,r \left( r^2\frac{\partial^2\vartheta}{\partial r^2} -2r\frac{\partial\vartheta}{\partial r} +\operatorname{ctg}\Theta\,\frac{\partial\vartheta}{\partial \Theta} \right). \end{aligned} \]

Multiplying equations (8), (9), and (10), respectively, by \(2r\), \(r^2\), and \(\operatorname{ctg}\Theta\), and summing term by term, we shall have

\[ r^2\frac{\partial^2\varepsilon}{\partial r^2} +2r\frac{\partial\varepsilon}{\partial r} +\operatorname{ctg}\Theta\,\frac{\partial\varepsilon}{\partial \Theta} +\frac{\partial^2\varepsilon}{\partial \Theta^2} =0. \tag{11} \]

For equation (11) the boundary conditions will have the form, for

\[ r=R,\quad \Theta_i<\Theta<\Theta_k,\quad \varepsilon=0, \]

and on the remaining part \(\varepsilon=\vartheta_i\).

Substituting the found \(\varepsilon=f(r,\Theta)\) into (7), we obtain an equation from which the required \(\vartheta\) is easily found.

As an example, let us solve the following problem [2]. A sphere of radius \(R\) is heated by a plane-parallel heat flux of density \(q\), incident on its surface, and gives up heat to the surrounding medium in accordance with Newton’s law. Find the temperature field in the sphere.

The required temperature distribution must satisfy Laplace’s equation \(\nabla^2 t(r,\Theta)=0\) and the boundary conditions

\[ \left. \frac{\partial t}{\partial r}+\frac{\alpha}{\lambda}t \right|_{r=R} = \begin{cases} \dfrac{q}{\lambda}\cos\Theta, & 0\leq \Theta\leq \pi/2,\\[6pt] 0, & \dfrac{\pi}{2}\leq \Theta\leq \pi. \end{cases} \]

We introduce substitution (7). Then the boundary conditions take the form

\[ \text{for } r=R,\ 0\leq \Theta\leq \frac{\pi}{2},\quad \varepsilon=\frac{q}{\lambda}\cos\Theta,\qquad \frac{\pi}{2}\leq \Theta\leq \pi,\quad \varepsilon=0. \]

The solution for \(\varepsilon\) is constructed in the form of a series

\[ \varepsilon(r,\Theta) = \sum_{n=0}^{\infty} a_n \left(\frac{r}{R}\right)^n P_n(\cos\Theta), \]

where

\[ a_n=\frac{2n+1}{2}\int_{-\frac{\pi}{2}}^{0}\frac{q}{\lambda}\cos\Theta\,P_n(\cos\Theta)\,d(\cos\Theta). \]

On the basis of [3] we have

\[ a_0=\frac{q}{4\lambda},\qquad a_1=\frac{q}{2\lambda},\qquad a_{2n+1}=0, \]

\[ a_{2n}=\frac{q}{\lambda}(-1)^n \frac{(2n-2)!(4n+1)}{2^{2n+1}(n-1)!(n+1)!}. \]

Therefore

\[ \varepsilon(r,\Theta)=\frac{q}{4\lambda}+\frac{q}{2\lambda}\cdot\frac{r}{R}\cos\Theta+ \sum_{n=1}^{\infty}(-1)^{2n}\frac{q}{\lambda} \frac{(2n-2)!(4n+1)}{2^{2n+1}(n-1)!(n+1)!} \cdot\left(\frac{r}{R}\right)^{2n}P_{2n}(\cos\Theta). \tag{12} \]

Substituting (12) into (7) and solving the resulting equation, we obtain

\[ t(r,\Theta)=\frac{q}{4\lambda} +\frac{1}{(Bi+1)}\cdot\frac{q}{2\lambda}\left(\frac{r}{R}\right)\cos\Theta+ \]

\[ +\frac{q}{\lambda}\sum_{n=1}^{\infty}(-1)^{2n} \frac{(2n-2)!(4n+1)}{2^{2n+1}(n-1)!(n+1)!} \cdot\frac{1}{Bi+2n}\left(\frac{r}{R}\right)^{2n}P_{2n}(\cos\Theta), \]

which corresponds to the solution given in [3].

REFERENCES

1. Lavrent'ev M. A. and Shabat B. V. Methods of the Theory of Functions of a Complex Variable. Gostekhizdat, 1951.

2. Lebedev N. N., Skalskaya I. P., Uflyand Ya. S. Collection of Problems in Mathematical Physics. Gostekhizdat, 1955.

3. Koshlyakov N. S., Gliner E. B., Smirnov M. M. Basic Differential Equations of Mathematical Physics. Fizmatgiz, 1962.