UDC 517.947.43

APPLICATION OF THE WIENER–HOPF METHOD TO THE SOLUTION OF A BOUNDARY VALUE PROBLEM FOR THE HEAT-CONDUCTION EQUATION

Ya. F. Rutner, L. P. Skryabina

The Wiener–Hopf method makes it possible to considerably extend the class of problems that can be solved by means of integral transforms. It makes it possible to solve boundary value problems when, on some piecewise-smooth portion of the surface bounding the domain, boundary conditions of different kinds are imposed. The method, which has found wide application in diffraction problems, for example in the works of L. A. Vainshtein, unfortunately has not become widespread in solving boundary value problems for the heat-conduction equation.

The present paper is one of the first attempts to fill this gap.

Consider the boundary value problem for the heat-conduction equation

\[ \frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} \qquad (t>0) \tag{1} \]

for the half-plane \(y \geq 0,\ -\infty < x < \infty\), with zero initial condition and boundary conditions

\[ u(x,0,t)=f(t)e^{-bx}\quad (b>0),\qquad x>0, \tag{2} \]

\[ \frac{\partial u(x,0,t)}{\partial y}=0,\qquad x<0. \tag{3} \]

Here \(|f(t)|<\mathrm{const}\, e^{\sigma_0 t}\) for \(t>0\), and the sought function \(u(x,y,t)\), together with its derivatives \(\frac{\partial u}{\partial x}\), \(\frac{\partial u}{\partial y}\), tends to zero as \(x \to \pm \infty\) and \(y \to \infty\), which permits the Laplace and Fourier transforms with respect to the variables \(t\) and \(x\), respectively, to be applied to this boundary value problem.

We apply to this boundary value problem successively the one-sided Laplace transform with respect to time \(t\):

\[ \bar u(x,y)=\int_0^\infty u(x,y,t)e^{-st}\,dt, \qquad \operatorname{Re}s>\sigma_0,\quad s=\sigma+i\tau, \]

and the Fourier transform with respect to the variable \(x\):

\[ U(y)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\bar u(x,y)e^{ipx}\,dx, \qquad p=\alpha+i\beta,\quad -b<\beta<b, \]

which gives

\[ \left(\frac{d^2}{dy^2}-\gamma^2\right)U(y)=0,\qquad -b<\beta<b, \tag{1'} \]

\[ U_+(0)=\frac{F(k^2)}{\sqrt{2\pi}(b-ip)},\qquad \beta>-b, \tag{2'} \]

\[ U'_-(0)=0. \tag{3'} \]

Here the following notation has been introduced and chosen:

\[ \operatorname{Re}s>0;\qquad k^2=s,\quad k=k_1+ik_2,\quad k_1>0;\qquad \gamma^2=k^2+p^2,\quad \operatorname{Re}\gamma>0; \]

\[ U_+(y)=\frac{1}{\sqrt{2\pi}}\int_0^\infty \bar u(x,y)e^{ipx}\,dx,\qquad \beta>-b; \]

\[ U_-(y)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^0 \bar u(x,y)e^{ipx}\,dx,\qquad \beta<b; \]

\[ F(s)=\int_0^\infty f(t)e^{-st}\,dt,\qquad \operatorname{Re}s>\sigma_0. \]

The general solution of (1′) is

\[ U(y)=A(p)e^{-\gamma y}+B(p)e^{\gamma y}. \tag{4} \]

Since as \(y\to+\infty\) the required solution \(u(x,y,t)\) tends to zero, from (4) we have \(B(p)=0\), and

\[ U(y)=A(p)e^{-\gamma y}. \tag{4'} \]

From (4′), using the notation introduced above, we obtain

\[ U_+(0)+U_-(0)=A(p),\qquad U'_+(0)+U'_-(0)=-\gamma A(p), \tag{5} \]

whence

\[ U'_+(0)+U'_-(0)=-\gamma[U_+(0)+U_-(0)]. \tag{5'} \]

Substituting in (5′), instead of \(U_+(0)\) and \(U'_-(0)\), their values (2′) and (3′), we arrive at the functional equation

\[ U'_+(0)+\sqrt{p^2+k^2}\,U_-(0)= \frac{\sqrt{p^2+k^2}\,F(k^2)}{\sqrt{2\pi}(ip-b)}. \tag{6} \]

This equation must be satisfied in the strip \(-m<\beta<m\) of the complex \(p\)-plane, where \(m=\min[b,k_1]\); moreover, \(U'_+(0)\) is regular in the half-plane \(\beta>-m\), and \(U_-(0)\) in the half-plane \(\beta<m\).

Since \(\gamma=\sqrt{p+ik}\sqrt{p-ik}\), the function \(\sqrt{p+ik}\) is regular and has no zeros in the half-plane \(\beta>-k_1\), while the function \(\sqrt{p-ik}\) is regular and has no zeros in the half-plane \(\beta<k_1\) (here the branches of each factor are chosen so that in the strip \(-k_1<\beta<k_1\), \(\operatorname{Re}\sqrt{p\pm ik}>0\) when \(\operatorname{Re}(p\pm ik)>0\)); hence, dividing (6) by \(\sqrt{p+ik}\), we obtain the equation

\[ \frac{U'_+(0)}{\sqrt{p+ik}}+\sqrt{p-ik}\,U_-(0)= \frac{\sqrt{p-ik}\,F(k^2)}{(ip-b)\sqrt{2\pi}}. \tag{7} \]

Here \(\dfrac{U_+'(0)}{\sqrt{p+ik}}\) is regular in the upper half-plane \(\beta>-m\), while the function \(\sqrt{p-ik}\,U_-(0)\) is regular in the lower half-plane \(\beta<m\).

Now both terms on the left-hand side of equation (7) are regular in the strip \(-m<\beta<m\), and the function standing on the right-hand side of equation (7),

\[ \Phi(p)=\frac{\sqrt{p-ik}\,F(k^2)}{\sqrt{2\pi}\,(ip-b)}, \]

as is easy to see, is also regular in the strip \(-m<\beta<m\).

According to Cauchy’s theorem, it may be represented in the form \(\Phi(p)=\Phi_+(p)+\Phi_-(p)\), where \(\Phi_+(p)\) is regular in the half-plane \(\beta>-m\), and the function \(\Phi_-(p)\) is regular in the half-plane \(\beta<m\), with

\[ \Phi_+(p)=\frac{1}{2\pi i}\int_{ci-\infty}^{ci+\infty}\frac{\Phi(z)}{z-p}\,dz, \]

\[ \Phi_-(p)=-\frac{1}{2\pi i}\int_{di-\infty}^{di+\infty}\frac{\Phi(z)}{z-p}\,dz,\qquad -m<c<\beta<d<m. \]

Closing the contours of integration in the lower half-plane, where the integrand has no branch points, and applying the residue theorem, we obtain

\[ \Phi_+(p)=\frac{F(k^2)}{2\pi i\sqrt{2\pi}} \int_{ci-\infty}^{ci+\infty} \frac{\sqrt{z-ik}}{(iz-b)(z-p)}\,dz = \frac{F(k^2)\sqrt{-ib-ik}}{\sqrt{2\pi}\,(ip-b)}, \]

\[ \Phi_-(p)= \frac{\sqrt{p-ik}-\sqrt{-ib-ik}}{ip-b}\, \frac{F(k^2)}{\sqrt{2\pi}}. \]

We regroup (7) and introduce a new function \(J(p)\) by the equalities

\[ J(p)=\frac{U_+'(0)}{\sqrt{p+ik}} -\frac{F(k^2)\sqrt{-ib-ik}}{\sqrt{2\pi}\,(ip-b)} = \]

\[ =-\sqrt{p-ik}\,U_-(0) +\frac{\left(\sqrt{p-ik}-\sqrt{-ib-ik}\right)F(k^2)} {\sqrt{2\pi}\,(ip-b)}. \tag{8} \]

The function \(J(p)\) is defined in the strip \(-m<\beta<m\); however, the left-hand side of equality (8) is defined and regular in the half-plane \(\beta>-m\), and the right-hand side in the half-plane \(\beta<m\).

Consequently, by analytic continuation one can define \(J(p)\) on the entire \(p\)-plane. According to the generalized Liouville theorem, \(J(p)\) is some polynomial [1,2].

Since \(U_+'(0)\) must tend to zero as \(p\to\infty\), it follows from (8) that \(J(p)\) also tends to zero as \(p\to\infty\). But a polynomial tends to zero as \(p\to\infty\) only if it is identically zero: \(J(p)=0\). Hence

\[ \frac{U_+'(0)}{\sqrt{p+ik}} -\frac{F(k^2)\sqrt{-ib-ik}}{\sqrt{2\pi}\,(ip-b)} = \]

\[ =-\sqrt{p-ik}\,U_-(0) +\frac{\left(\sqrt{p-ik}-\sqrt{-ib-ik}\right)F(k^2)} {\sqrt{2\pi}\,(ip-b)} =0. \]

or

\[ U_{+}'(0)=\frac{\sqrt{-ib-ik}\,F(k^2)\sqrt{p+ik}}{\sqrt{2\pi}(ip-b)} \]

and

\[ U_{-}(0)=\frac{(\sqrt{p-ik}-\sqrt{-ib-ik})F(k^2)} {\sqrt{2\pi}(ip-b)\sqrt{p-ik}} . \tag{9} \]

One can find the solution of the given problem by means of contour integrals, determining \(A(p)\) from equations (5). However, in this form the contour integrals are very difficult to compute; therefore we apply the following method.

In order to simplify the subsequent exposition, let us consider the case \(f(t)=1\), \(b=1\) (the general case is carried out analogously). Applying the inverse Fourier transform and using tables [3], we determine \(\dfrac{\partial}{\partial y}\bar u(x,0)\) for \(x>0\):

\[ -\frac{\partial}{\partial y}\bar u(x,0) = \frac{1}{\sqrt{\pi x}}\frac{1}{s}e^{-x\sqrt{s}}\sqrt{1+\sqrt{s}} +\frac{1}{s}\sqrt{s-1}\,e^{-x} - \operatorname{erfc}\sqrt{(\sqrt{s}-1)x}\,e^{-x}\frac{1}{s}\sqrt{s-1}. \tag{10} \]

Here, as also below, the branches of the roots are chosen so that when the real part of the radicand is greater than zero, the real part of the root is greater than zero.

Using the well-known relation of operational calculus

\[ \frac{F(\sqrt{s})}{\sqrt{s}} \doteq \frac{1}{\sqrt{\pi t}}\int_{0}^{\infty} f(\tau)e^{-\frac{\tau^2}{4t}}\,d\tau, \tag{11} \]

where \(f(t)\doteq F(s)\), which follows from Efros’s theorem, we pass to the original [3], finding in this way the original of each term of (10) separately. Then, relying on the relation

\[ \frac{e^{-sx}\sqrt{s+1}}{s} \doteq \begin{cases} 0, & \text{for } t<x,\\[6pt] \dfrac{1}{\sqrt{\pi(t-x)}}\exp(x-t)+\operatorname{erf}\sqrt{t-x}, & \text{for } t>x, \end{cases} \]

we obtain

\[ \frac{1}{\sqrt{\pi x}}\frac{1}{s}e^{-x\sqrt{s}}\sqrt{1+\sqrt{s}} \doteq \frac{1}{\pi\sqrt{tx}}\int_{x}^{\infty}e^{-\frac{\tau^2}{4t}} \times \]

\[ \times \left[ \frac{e^{-(\tau-x)}}{\sqrt{\pi(\tau-x)}}+\operatorname{erf}\sqrt{\tau-x} \right]\,d\tau = I_1. \tag{12} \]

The inverse of the second term in (10) is found directly:

\[ \frac{1}{s}\sqrt{s-1}\,e^{-x}\div e^{-x} \left[ \frac{1}{\sqrt{\pi t}}\,e^t - \frac{2}{\sqrt{\pi}}\int_0^{\sqrt{t}} e^{v^2}\,dv \right] = I_2 . \tag{13} \]

To find the inverse of the third term in (10), we represent it in the form

\[ e^{-x}\frac{F^*(\sqrt{s})}{\sqrt{s}}, \]

whence it follows that

\[ F^*(s)=\operatorname{erfc}\sqrt{(s-1)x}\,\frac{\sqrt{s^2-1}}{s}. \]

Since

\[ \operatorname{erfc}\sqrt{(s-1)x}\sqrt{s-1} - \frac{e^{-x(s-1)}}{\sqrt{\pi x}} \div \begin{cases} 0, & \text{for } t<x,\\[4pt] -\dfrac{e^t}{2\sqrt{\pi t^3}}, & \text{for } t>x, \end{cases} \]

and

\[ \frac{\sqrt{s+1}}{s}\div \frac{e^{-t}}{\sqrt{\pi t}}+\operatorname{erf}\sqrt{t}, \]

then, on the basis of the convolution theorem, we have

\[ \left[ \operatorname{erfc}\sqrt{(s-1)x}\sqrt{s-1} - \frac{e^{-x(s-1)}}{\sqrt{\pi x}} \right] \frac{\sqrt{s+1}}{s} \div \]

\[ \div \begin{cases} 0, & \text{for } t<x,\\[6pt] -\displaystyle\int_x^t \frac{e^\tau}{2\sqrt{\pi\tau^3}} \left[ \frac{e^{-(t-\tau)}}{\sqrt{\pi(t-\tau)}} + \operatorname{erf}\sqrt{t-\tau} \right]\,d\tau, & \text{for } t>x, \end{cases} \]

and, on the basis of the shift theorem, we obtain

\[ \frac{\sqrt{s+1}}{s}e^{-x(s-1)} \div \begin{cases} 0, & \text{for } t<x,\\[6pt] e^x\left[ \dfrac{e^{-(t-x)}}{\sqrt{\pi(t-x)}} + \operatorname{erf}\sqrt{t-x} \right], & \text{for } t>x. \end{cases} \]

And finally, on the basis of relation (11),

\[ e^{-x}\operatorname{erfc} \sqrt{(\sqrt{s}-1)x}\, \frac{\sqrt{s-1}}{s} \div \]

\[ \div \frac{e^{-x}}{\sqrt{\pi t}} \int_x^\infty e^{-\frac{\gamma^2}{4t}} \left\{ e^x\left[ \frac{e^{-\gamma+x}}{\sqrt{\pi(\gamma-x)}} + \operatorname{erf}\sqrt{\gamma-x} \right] - \int_x^\gamma \frac{e^\tau}{2\sqrt{\pi\tau^3}} \left[ \frac{e^{-\gamma+\tau}}{\sqrt{\pi(\gamma-\tau)}} + \operatorname{erf}\sqrt{\gamma-\tau} \right]\,d\tau \right\}\,d\gamma = -I_3 . \tag{14} \]

We have now obtained a boundary-value problem of the following form:

\[ \frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} \]

with boundary conditions

\[ \frac{\partial u}{\partial y} = \begin{cases} 0, & \text{for } x<0,\\ -(I_1+I_2+I_3), & \text{for } x>0, \end{cases} \]

whose solution can be determined with the aid of the known Green’s function

\[ G(x,y,t;\,x_0,y_0,t_0) = \frac{1}{4\pi(t-t_0)} \times \]

\[ \times \left[ e^{-\frac{(x-x_0)^2+(y-y_0)^2}{4(t-t_0)}} + e^{-\frac{(x-x_0)^2+(y+y_0)^2}{4(t-t_0)}} \right], \]

which gives

\[ u(x,y,t) = \int_0^t dt_0 \int_0^\infty G(x,y,t;\,x_0,0,t_0) \times \]

\[ \times \left[I_1(x_0,t_0)+I_2(x_0,t_0)+I_3(x_0,t_0)\right]\,dx_0 = \]

\[ = \int_0^t dt_0 \int_0^\infty \left[I_1(x_0,t_0)+I_2(x_0,t_0)+I_3(x_0,t_0)\right] \times \]

\[ \times \frac{ \exp\left[-\frac{(x-x_0)^2+y^2}{4(t-t_0)}\right] }{ 2\pi(t-t_0) }\,dx_0 . \]

REFERENCES

1. Wiener N., Paley R. Fourier Transforms in the Complex Domain. Moscow, Nauka Publishing House, 1964.

2. Noble B. Applications of the Wiener–Hopf Method for the Solution of Differential Equations in Partial Derivatives. Moscow, IL, 1962.

3. Ditkin V. A., Kuznetsov P. I. Handbook of Operational Calculus. Moscow–Leningrad, GITTL, 1951.

Received by the editors
February 6, 1965

Kuibyshev Polytechnic Institute
named after V. V. Kuibyshev