CERTAIN PROBLEMS FOR THE HEAT-CONDUCTION EQUATION WITH MIXED BOUNDARY CONDITIONS

E. F. AFANAS'EV

A solution is given for certain boundary-value problems for the heat-conduction equation in a half-plane with one or two points of discontinuity in the boundary conditions. The solutions are represented in the form of a simple-layer potential. The boundary conditions lead to integral equations of the first kind with a finite or semi-infinite interval of variation of the variables. The solutions of the equations are found in the form of quadratures or a series of quadratures.

The solutions obtained may be used in problems of heat conduction, diffusion, and nonstationary filtration.

1. Consider the two-dimensional heat-conduction equation

\[ \frac{\partial^2 T}{\partial x^2}+\frac{\partial^2 T}{\partial y^2} = \frac{\partial T}{\partial t}. \tag{1.1} \]

The solution of the Neumann problem for equation (1.1), with zero initial condition in the half-plane \(y>0\), is represented in the form of a simple-layer potential [1]

\[ T(x,y,t) = -\frac{1}{2\pi} \int_0^t d\tau \int_{-\infty}^{+\infty} \frac{ e^{-\frac{(x-\xi)^2+y^2}{4(t-\tau)}} }{t-\tau} \times \]

\[ \times \left[ \frac{\partial T(\xi,y,\tau)}{\partial y} \right]_{y=0} d\xi \quad (y>0). \tag{1.2} \]

The solution of the analogous Dirichlet problem is represented in the form of a double-layer potential [1]

\[ T(x,y,t) = \frac{1}{4\pi} \int_0^t d\tau \int_{-\infty}^{+\infty} \frac{ y e^{-\frac{(x-\xi)^2+y^2}{4(t-\tau)}} }{(t-\tau)^2} T(\xi,0,\tau)\,d\xi \quad (y \ge 0). \tag{1.3} \]

It follows from (1.3) that the normal derivative of the function \(T(x,y,t)\) on the boundary is equal to

\[ \left. \frac{\partial T}{\partial y} \right|_{y=0} = -\left\{ \frac{T(x,0,t)}{\sqrt{\pi t}} + \right. \]

\[ \left. + \frac{1}{4\pi} \int_0^t d\tau \int_{-\infty}^{+\infty} \frac{ e^{-\frac{(x-\xi)^2}{4(t-\tau)}} }{(t-\tau)^2} \,[T(x,0,t)-T(\xi,0,\tau)]\,d\xi \right\}. \tag{1.4} \]

We note that, in Laplace transforms with respect to \(t\), formula (1.2) for \(y=0\) takes the form

\[ T^*(x,0,\lambda)=-\frac{1}{\pi}\int_{-\infty}^{+\infty} K_0\bigl(\sqrt{\lambda}\,|x-\xi|\bigr)\times \]

\[ \times\left[\frac{\partial T^*(\xi,y,\lambda)}{\partial y}\right]_{y=0}\,d\xi . \tag{1.5} \]

Here \(K_0\) is the Macdonald function, \(\sqrt{\lambda}>0\) for \(\lambda>0\),

\[ T^*(x,y,\lambda)=\int_0^\infty e^{-\lambda t}T(x,y,t)\,dt . \]

Formula (1.5) establishes an integral relation between the function \(T^*(x,y,\lambda)\) and its normal derivative on the boundary \(y=0\); in what follows this formula will be used as the basis for solving all the problems.

2. Let us find the solution of equation (1.1) in the half-plane \(y>0\) under a zero initial condition and the boundary conditions:

\[ T\big|_{y=0}=p_0(x,t)\quad \text{for } x>0, \tag{2.1} \]

\[ -\frac{\partial T}{\partial y}\bigg|_{y=0}=-q_0(x,t)\quad \text{for } x<0. \tag{2.2} \]

Taking into account (1.5), (2.1), and (2.2), after simple transformations, in order to determine the function

\[ q^*(x,\lambda)=-\frac{\partial T^*}{\partial y}\bigg|_{y=0}\quad (x>0) \]

we obtain an integral equation of the first kind

\[ \frac{1}{\pi}\int_0^\infty K_0\bigl(\sqrt{\lambda}\,|x-\xi|\bigr)q^*(\xi,\lambda)\,d\xi = p^*(x,\lambda)\quad (x>0), \tag{2.3} \]

where the right-hand side is equal to

\[ p^*(x,\lambda)=p_0^*(x,\lambda) -\frac{1}{\pi}\int_0^\infty K_0\bigl(\sqrt{\lambda}\,|x+\xi|\bigr)q_0^*(-\xi,\lambda)\,d\xi . \tag{2.4} \]

The integral equation (2.3) is an equation of Wiener–Hopf type with a kernel depending on the absolute value of the difference of the arguments, and with a semi-infinite interval of variation of the variables. Following the theory developed by G. A. Grinberg [2], we represent the solution of this equation in the form

\[ q^*(x,\lambda)=u^*(x,\lambda)+\int_0^\infty u^*(-\xi,\lambda)v(\xi,x,\lambda)\,d\xi \quad (x>0), \tag{2.5} \]

where the functions \(u^*(x,\lambda)\) and \(v(\eta,x,\lambda)\) satisfy the integral equations:

\[ \frac{1}{\pi}\int_{-\infty}^{+\infty} K_0\bigl(\sqrt{\lambda}\,|x-\xi|\bigr)u^*(\xi,\lambda)\,d\xi = p^*(x,\lambda) \quad (-\infty<x<\infty), \tag{2.6} \]

\[ \int_0^\infty K_0\bigl(\sqrt{\lambda}\,|x-\xi|\bigr)\,v(\eta,\xi,\lambda)\,d\xi =K_0\bigl[\sqrt{\lambda}(x+\eta)\bigr]\quad (x,\eta>0). \tag{2.7} \]

The function \(p_0^*(x,\lambda)\), which enters the right-hand side of equation (2.6), is defined for positive values of the argument \(x\). According to the method being used, in solving equation (2.6) the function \(p_0^*(x,\lambda)\) may be continued into the region of negative values of \(x\) in an arbitrary continuous manner.

The integral equation (2.6) is equivalent to formula (1.5), if in the latter one puts

\[ T^*(x,0,\lambda)=p^*(x,\lambda),\qquad \left.\frac{\partial T^*}{\partial y}\right|_{y=0}=-u^*(x,\lambda). \tag{2.8} \]

Taking into account (1.4), (1.5), and (2.8), we find the expression for the original \(u(x,t)\) in the form

\[ u(x,t)=\frac{p(x,t)}{\sqrt{\pi t}}+ \]

\[ +\frac{1}{4\pi}\int_0^t d\tau\int_{-\infty}^{+\infty} \frac{e^{-\frac{(x-\xi)^2}{4(t-\tau)}}}{(t-\tau)^2} [p(x,t)-p(\xi,\tau)]\,d\xi. \tag{2.9} \]

To solve equation (2.7) we shall apply the Wiener–Hopf method [3]. Multiplying both sides of equation (2.7) by \(e^{-\sqrt{\lambda}\,zx}\), where \(\operatorname{Re} z>0\), and integrating with respect to \(x\) from zero to infinity, we obtain, for the function

\[ V(z)=\int_0^\infty e^{-\sqrt{\lambda}\,zx}v(\eta,x,\lambda)\,dx, \]

the functional equation

\[ \frac{V(z)}{\sqrt{1-z^2}}+\frac{1}{\pi}\int_1^\infty \frac{V(\xi)\,d\xi}{(z-\xi)\sqrt{\xi^2-1}} = \]

\[ =\frac{1}{\pi}\int_1^\infty \frac{e^{-\sqrt{\lambda}\eta\xi}\,d\xi}{(z+\xi)\sqrt{\xi^2-1}} \quad (0<\operatorname{Re}z<1). \tag{2.10} \]

Omitting the reasoning and transformations carried out when using the method indicated above, we find from (2.10) the function \(V(z)\), regular in the half-plane \(\operatorname{Re}z>0\) and tending to zero as \(z\to\infty\):

\[ V(z)=e^{-\sqrt{\lambda}\eta z}\operatorname{Erfc}\{[\sqrt{\lambda}\eta(z+1)]^{1/2}\}\quad (\operatorname{Re}z>0). \tag{2.11} \]

Applying the inversion theorem with respect to \(z\) to (2.11), we obtain the solution of the integral equation (2.7) in the form

\[ v(\eta,x,\lambda)=\frac{\sqrt{\eta}\,e^{-\sqrt{\lambda}(\eta+x)}}{\pi\sqrt{x}(\eta+x)}. \tag{2.12} \]

Substituting (2.12) into formula (2.5) and carrying out in the latter the inverse transform with respect to \(\lambda\), we find the function \(q(x,t)\):

\[ q(x,t)=u(x,t)+\frac{1}{2\pi^{3/2}\sqrt{x}}\int_0^t d\tau \int_0^\infty \frac{\sqrt{\xi}\,e^{-\frac{(x+\xi)^2}{4(t-\tau)}}}{(t-\tau)^{3/2}}\,u(-\xi,\tau)\,d\xi \qquad (x>0). \]

Knowing the normal derivative

\[ \left.\frac{\partial T}{\partial y}\right|_{y=0} = \begin{cases} -q_0(x,t), & (x<0),\\ -q(x,t), & (x>0), \end{cases} \]

we obtain the solution of equation (1.1) in the form (1.2).

1. Let us find the solution of equation (1.1) in the half-plane \(y>0\) with zero initial condition and boundary conditions:

\[ \left. T\right|_{y=0}=p_0(x,t)\qquad (0<x<1); \tag{3.1} \]

\[ \left.\frac{\partial T}{\partial y}\right|_{y=0} = \begin{cases} -q_1(x,t), & (x<0),\\ -q_2(x,t), & (x>1). \end{cases} \tag{3.2} \]

Taking into account (1.5), (3.1), and (3.2), we shall have

\[ \frac{1}{\pi}\int_0^1 K_0\!\left(\sqrt{\lambda}\,|x-\xi|\right)q^*(\xi,\lambda)\,d\xi = p^*(x,\lambda) \qquad (0<x<1), \tag{3.3} \]

where

\[ q^*(x,\lambda)=-\left.\frac{\partial T^*}{\partial y}\right|_{y=0} \qquad (0<x<1) \]

is the unknown function, and the right-hand side is equal to

\[ p^*(x,\lambda)=p_0^*(x,\lambda)-\frac{1}{\pi}\int_0^\infty \left[ K_0\!\left(\sqrt{\lambda}\,|x+\xi|\right)q_1^*(-\xi,\lambda)+ \right. \]

\[ \left. +\,K_0\!\left(\sqrt{\lambda}\,|\xi+1-x|\right)q_2^*(\xi+1,\lambda) \right]d\xi . \tag{3.4} \]

Following [3], we represent the solution of equation (3.3) in the form

\[ q^*(x,\lambda)=u^*(x,\lambda)+\int_0^\infty \left[u^*(-\xi,\lambda)\varphi(\xi,x,\lambda)+ \right. \]

\[ \left. +\,u^*(\xi+1,\lambda)\varphi(\xi,1-x,\lambda)\right]d\xi \tag{3.5} \]

\[ (0<x<1). \]

Here the function \(u^*(x,\lambda)\) satisfies the integral equation (2.6), whose solution in the originals has the form (2.9), where, by virtue of (3.4), the function \(p(x,t)\) is equal to

\[ p(x,t)=p_0(x,t)- \]

\[ -\frac{1}{2\pi}\int_0^t d\tau \int_0^\infty \frac{ q_1(-\xi,\tau)e^{-\frac{(x+\xi)^2}{4(t-\tau)}}+ q_2(\xi+1,\tau)e^{-\frac{(x-\xi-1)^2}{4(t-\tau)}} }{t-\tau}\,d\xi . \]

The function \(\varphi(\eta,x,\lambda)\) satisfies the Fredholm integral equation of the second kind

\[ \varphi(\eta,x,\lambda)=\nu(\eta,x,\lambda)+\int_0^\infty \nu(\eta,\xi+1,\lambda)\varphi(\xi,1-x,\lambda)\,d\xi \tag{3.6} \]

\[ (0<x<1), \]

in this case \(v(\eta, x, \lambda)\) is a solution of equation (2.7), which has the form (2.12).

Solving the Fredholm equation (3.6) by the method of successive approximations and substituting \(\varphi(\eta, x, \lambda)\) into (3.5), we shall have

\[ q^{*}(x,\lambda)=u^{*}(x,\lambda)+\sum_{n=0}^{\infty}\left[U_{2n}^{(1)}(x,\lambda)+U_{2n+1}^{(1)}(1-x,\lambda)+\right. \]
\[ \left.+\,U_{2n}^{(2)}(1-x,\lambda)+U_{2n+1}^{(2)}(x,\lambda)\right]\quad (0<x<1). \tag{3.7} \]

Here the functions

\[ U_{n}^{(i)}(x,\lambda)=-\frac{1}{\pi\sqrt{x}} \int_{0}^{\infty}\frac{\sqrt{\xi}\,e^{-\sqrt{\lambda}(x+\xi)}}{x+\xi}\, U_{n-1}^{(i)}(\xi+1,\lambda)\,d\xi \tag{3.8} \]

\[ (n=0;\ 1;\ \ldots;\ i=1;\ 2), \]

where

\[ U_{-1}^{(1)}(x,\lambda)=u^{*}(1-x,\lambda),\qquad U_{-1}^{(2)}(x,\lambda)=u^{*}(x,\lambda). \]

The series (3.7) converges uniformly under the condition \(\operatorname{Re}\lambda^{1/2}>0\). Performing the inverse transform in (3.7), we obtain the function

\[ q(x,t)=u(x,t)+\sum_{n=0}^{\infty}\left[u_{2n}^{(1)}(x,t)+u_{2n+1}^{(1)}(1-x,t)+\right. \]
\[ \left.+\,u_{2n}^{(2)}(1-x,t)+u_{2n+1}^{(2)}(x,t)\right]\quad (0<x<1), \]

where

\[ u_{n}^{(i)}(x,t)=\frac{1}{2\pi^{3/2}\sqrt{x}} \int_{0}^{t}d\tau\int_{0}^{\infty} \frac{\sqrt{\xi}\,e^{-\frac{(x+\xi)^2}{4(t-\tau)}}}{(t-\tau)^{3/2}}\, u_{n-1}^{(i)}(\xi+1,\tau)\,d\xi \]

\[ (n=0;\ 1;\ \ldots;\ i=1;\ 2), \]

\[ u_{-1}^{(1)}(x,t)=u(1-x,t),\qquad u_{-1}^{(2)}(x,t)=u(x,t). \]

Knowing the derivative

\[ \left.\frac{\partial T}{\partial y}\right|_{y=0} = \begin{cases} -q_{1}(x,t), & (x<0),\\ -q(x,t), & (0<x<1),\\ -q_{2}(x,t), & (x>0), \end{cases} \]

we find the function \(T(x,y,t)\) in the form (1.2).

1. Let us find the solution of equation (1.1) in the half-plane \(y>0\) with zero initial condition and boundary conditions:

\[ \left.T\right|_{y=0}= \begin{cases} p_{1}(x,t), & (x<0),\\ p_{2}(x,t), & (x>1); \end{cases} \tag{4.1} \]

\[ \left.\frac{\partial T}{\partial y}\right|_{y=0} =-q_{0}(x,t)\qquad (0<x<1). \tag{4.2} \]

According to (1.5) and (4.2), we have

\[ \begin{aligned} &\frac{1}{\pi}\int_{0}^{\infty}\bigl[K_0(\sqrt{\lambda}|x+\xi|)q_1^*(-\xi,\lambda)+{}\\ &\qquad\qquad{}+K_0(\sqrt{\lambda}|x-1-\xi|)q_2^*(\xi+1,\lambda)\bigr]\,d\xi+{}\\ &\qquad{}+\frac{1}{\pi}\int_{0}^{1}K_0(\sqrt{\lambda}|x-\xi|)q_0^*(\xi,\lambda)\,d\xi={}\\ &\qquad{}=T^*(x,0,\lambda)\qquad (-\infty<x<\infty), \end{aligned} \tag{4.3} \]

where

\[ q_1^*(x,\lambda)=-\left.\frac{\partial T^*}{\partial y}\right|_{y=0}\quad (x<0),\qquad q_2^*(x,\lambda)=-\left.\frac{\partial T^*}{\partial y}\right|_{y=0}\quad (x>1) \]

are unknown functions.

Taking (4.1) into account, from (4.3), after simple transformations, we obtain

\[ \begin{aligned} &\frac{1}{\pi}\int_{0}^{\infty}\{K_0(\sqrt{\lambda}|x-\xi|)q_1^*(-\xi,\lambda)+{}\\ &\qquad\qquad{}+K_0[\sqrt{\lambda}(x+1+\xi)]q_2^*(\xi+1,\lambda)\}\,d\xi={}\\ &\qquad{}=p_1^*(-x,\lambda)-\frac{1}{\pi}\int_{0}^{1}K_0[\sqrt{\lambda}(x+\xi)]q_0^*(\xi,\lambda)\,d\xi \qquad (x>0); \end{aligned} \tag{4.4} \]

\[ \begin{aligned} &\frac{1}{\pi}\int_{0}^{\infty}\{K_0(\sqrt{\lambda}|x-\xi|)q_2^*(\xi+1,\lambda)+{}\\ &\qquad\qquad{}+K_0[\sqrt{\lambda}(x+1+\xi)]q_1^*(-\xi,\lambda)\}\,d\xi={}\\ &\qquad{}=p_2^*(x+1,\lambda)-\frac{1}{\pi}\int_{0}^{1}K_0[\sqrt{\lambda}(x+\xi)]q_0^*(1-\xi,\lambda)\,d\xi \qquad (x>0). \end{aligned} \tag{4.5} \]

From equations (4.4) and (4.5), by addition and subtraction, we obtain that the combinations of functions

\[ q_{\pm}^*(x,\lambda)=q_1^*(-x,\lambda)\pm q_2^*(x+1,\lambda)\qquad (x>0) \tag{4.6} \]

satisfy equations of one and the same type,

\[ \frac{1}{\pi}\int_{0}^{\infty}\{K_0(\sqrt{\lambda}|x-\xi|)\pm K_0[\sqrt{\lambda}(x+1+\xi)]\}\times \]

\[ \times q_{\pm}^*(\xi,\lambda)\,d\xi=p_{\pm}^*(x,\lambda)\qquad (x>0), \tag{4.7} \]

where the right-hand side is equal to

\[ p_{\pm}^*(x,\lambda)=p_1^*(-x,\lambda)\pm p_2^*(x+1,\lambda)- \]

\[ -\frac{1}{\pi}\int_0^1 K_0\bigl[\sqrt{\lambda}(x+\xi)\bigr]\,[q_0^*(-\xi,\lambda)\pm q_0^*(1-\xi,\lambda)]\,d\xi . \tag{4.8} \]

We seek the solution of equation (4.7) in the form of the series

\[ q_\pm^*(x,\lambda)=\sum_{n=0}^{\infty} q_{\pm n}^*(x,\lambda), \tag{4.9} \]

where the functions \(q_{\pm n}^*(x,\lambda)\) satisfy the equations:

\[ \frac{1}{\pi}\int_0^\infty K_0(\sqrt{\lambda}|x-\xi|)\,q_{\pm 0}^*(\xi,\lambda)\,d\xi = p_\pm^*(x,\lambda)\quad (x>0); \tag{4.10} \]

\[ \int_0^\infty K_0(\sqrt{\lambda}|x-\xi|)\,q_{\pm n}^*(\xi,\lambda)\,d\xi = \mp\int_0^\infty K_0\bigl[\sqrt{\lambda}(x+1+\xi)\bigr]\,q_{\pm(n-1)}^*(\xi,\lambda)\,d\xi \tag{4.11} \]

\[ (n=1;\,2;\,\ldots;\ x>0). \]

Equation (4.10) is of the same type as equation (2.3). Its solution in the originals has the form

\[ q_{\pm 0}(x,t)=\frac{p_\pm(x,t)}{\sqrt{\pi t}}+ \]

\[ +\frac{1}{4\pi}\int_0^t d\tau\int_{-\infty}^{+\infty} \frac{e^{-\frac{(x-\xi)^2}{4(t-\tau)}}}{(t-\tau)^2} \,[p_\pm(x,t)-p_\pm(\xi,\tau)]\,d\xi, \tag{4.12} \]

where, according to (4.8), the function \(p_\pm(x,t)\) is equal to

\[ p_\pm(x,t)=p_1(-x,t)\pm p_2(x+1,t)- \]

\[ -\frac{1}{2\pi}\int_0^t d\tau\int_0^1 \frac{e^{-\frac{(x+\xi)^2}{4(t-\tau)}}}{t-\tau} \,[q_0(\xi,\tau)\pm q_0(1-\xi,\tau)]\,d\xi. \]

Applying to equations (4.11) the Wiener—Hopf technique, we obtain

\[ q_{\pm n}^*(x,\lambda)=\mp\frac{1}{\pi\sqrt{x}} \int_0^\infty \frac{\sqrt{\xi+1}\,e^{-\sqrt{\lambda}(x+\xi+1)}}{x+\xi+1}\, q_{\pm(n-1)}^*(\xi,\lambda)\,d\xi \tag{4.13} \]

\[ (n=1;\,2;\,\ldots;\ x>0). \]

Carrying out the inverse transformation in (4.13), we obtain

\[ q_{\pm n}(x,t)=\mp\frac{1}{2\pi^{3/2}\sqrt{x}} \int_0^t d\tau\int_0^\infty \frac{\sqrt{\xi+1}\,e^{-\frac{(x+\xi+1)^2}{4(t-\tau)}}}{(t-\tau)^{3/2}}\, q_{\pm(n-1)}(\xi,\tau)\,d\xi \]

\[ (n=1;\,2;\,\ldots;\ x>0). \]

As a result, the solution of equation (4.7), according to (4.9), in the originals will have the form

\[ q_{\pm}(x,t)=\sum_{n=0}^{\infty} q_{\pm n}(x,t), \]

where the series converges uniformly in \(x\) and \(t\) \((x,t>0)\).

Taking (4.6) into account, we find the functions \(q_1(-x,t)\) and \(q_2(x+1,t)\) in the form:

\[ q_1(-x,t)=\frac{1}{2}\,[q_+(x,t)+q_-(x,t)], \]

\[ q_2(x+1,t)=\frac{1}{2}\,[q_+(x,t)-q_-(x,t)]. \]

Knowing the derivative

\[ \left.\frac{\partial T}{\partial y}\right|_{y=0} = \begin{cases} -q_1(x,t) & (x<0),\\ -q_0(x,t) & (0<x<1),\\ -q_2(x,t) & (x>1), \end{cases} \]

we obtain the function \(T(x,y,t)\) in the form (1.2).

We note that, under the most general restrictions on the functions entering the boundary conditions, one can prove the existence and uniqueness of the solution of the integral equations (2.3), (3.3), and (4.7), using for this the path employed, for example, in [4].

In an analogous way, problems for the heat-conduction equation in the half-plane \(y>0\) with one or two points of discontinuity in the boundary conditions can be solved, when on part of the boundary a condition of the third kind is prescribed,

\[ a\,\frac{\partial T}{\partial y}+bT=f(x,t)\quad (y=0), \]

where \(a,b\) are constants, and \(f(x,t)\) is a known function.

References

1. Montz G. Integral equations. GTTI, 1, 1934.
2. Grinberg G. A. DAN SSSR, 128, No. 3, 1959.
3. Noble B. Application of the Wiener–Hopf method to the solution of partial differential equations. IL, 1962.
4. Afanas’ev E. F. PMM, 28, issue 5, 1964.

Received by the editors
October 30, 1964

Institute of Mechanics
Academy of Sciences of the USSR