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THE CAUCHY PROBLEM FOR THE LAPLACE EQUATION IN AN INFINITE STRIP
V. K. IVANOV
- The Cauchy problem for the Laplace equation, as was shown by J. Hadamard [1], is unstable with respect to small changes in the initial data. However, if some additional information about the solution is known, then one can construct approximations converging to the desired solution. At present a whole series of methods for such approximate solution is known (for the bibliography see [3] and [6]). Usually in these methods an upper bound for the modulus of the solution is assumed known in advance, which corresponds to specifying, in the corresponding function space, a compact set to which the solution belongs (stability in the sense of A. N. Tikhonov; see [7], [6]). Recently A. N. Tikhonov has shown [8], [9] that in some cases one can construct approximations to the solution without knowing in advance the size of the compact set to which the solution belongs. In this connection it is of interest to consider the Cauchy problem for the Laplace equation under the weakest possible requirements imposed on the sought solution.
The distinctive features of the formulation of the problem considered in the present work are as follows:
1) the domain in which the solution is sought is infinite and closed (the strip \(-\infty < x < +\infty,\ 0 \le y \le b\));
2) regarding the solution \(u(x,y)\) it is assumed only that it exists, and that on the boundary of the strip the function \(u(x,b)\) is bounded and continuous.
These assumptions make it possible to approximate the solution arbitrarily closely, provided only that the initial data on the line \(y=0\) are given with sufficiently small error. If, in addition, the modulus of continuity of the function \(u(x,b)\) is known, then one can also estimate the error of the approximate solution.
- The problem that we shall consider is posed as follows.
Find in the strip \(-\infty < x < +\infty,\ 0 \le y \le b\) a solution of the Laplace equation
\[ \frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=0 \tag{1} \]
under the conditions \(u(x,0)=f(x),\ u'_y(x,0)=g(x)\), where \(f(x)\) and \(g(x)\) are bounded real analytic functions of the real variable \(x\). We shall suppose that the functions \(f(x)\) and \(g(x)\) can be continued analytically into the complex domain as functions of the complex argument \(z=x+yi\) in such a way that the continued functions \(f(x+yi)\) and \(g(x+yi)\) will be analytic in the strip \(-\infty < x < +\infty,\ 0 \le y < b\)
and continuous in its closure. Under these conditions our problem has a unique solution. Assuming at first that \(f(x)\) and \(g(x)\) are known to us exactly, we shall solve the problem by applying the Fourier transform with respect to \(x\). If the original function \(f(x)\) is absolutely integrable on \((-\infty, +\infty)\), then its Fourier transform
\[ \mathbf{F}[f]=\int_{-\infty}^{\infty} f(x)e^{ixs}\,dx=F(s) \]
is a continuous function of \(s\).
In the general case, when \(f(x)\) is only bounded, we shall regard it as a functional on the space \(K\) of infinitely differentiable finite functions of I. M. Gelfand and G. E. Shilov [2]. Then \(F(s)\) is a generalized function from the space \(Z'\) (see p. 190 in [2]). Analogous considerations apply also to \(g(x)\), for which we put
\[ \mathbf{F}[g]=G(s). \]
For \(U(s,y)=\mathbf{F}[u(x,y)]\), from (1) we obtain an ordinary differential equation, solving which with the prescribed initial conditions, we find
\[ U(s,y)=F(s)\operatorname{Ch}sy+G(s)\frac{\operatorname{sh}sy}{s}. \tag{2} \]
For every \(y\) this is a generalized function from \(Z'\).
- Following the idea of A. N. Tikhonov [8, 9] on the regularization of ill-posed problems, we replace the generalized function \(U(s,y)\) by
\[ U_{\alpha}(s,y)=e^{-\alpha^{2}s^{2}}U(s,y). \tag{3} \]
The inverse Fourier transform of \(e^{-\alpha^{2}s^{2}}\) is
\[ W_{\alpha}(x)=\mathbf{F}[e^{-\alpha^{2}s^{2}}]=\frac{1}{2\alpha\sqrt{\pi}}e^{-\frac{x^{2}}{4\alpha^{2}}}. \tag{4} \]
The inverse Fourier transform of \(U_{\alpha}(s,y)\)
\[ \mathbf{F}^{-1}[U_{\alpha}(s,y)]=u_{\alpha}(x,y) \tag{5} \]
is an ordinary function, expressible in the form of a convolution
\[ u_{\alpha}(x,y)=\int_{-\infty}^{\infty} W_{\alpha}(x-t)u(t,y)\,dt. \tag{6} \]
Denote
\[ R_{\alpha}(x,y)=\frac{1}{2\pi}\int_{-\infty}^{\infty} e^{-\alpha^{2}s^{2}}\operatorname{Ch}sy\,ds, \tag{7} \]
\[ T_{\alpha}(x,y)=\frac{1}{2\pi}\int_{-\infty}^{\infty} e^{-\alpha^{2}s^{2}-ixs}\frac{\operatorname{Sh}sy}{s}\,ds. \tag{7} \]
Then from (2), (3), (5), and (6) we obtain
\[ u_{\alpha}(x,y)=\frac{1}{2\pi}\int_{-\infty}^{\infty} R_{\alpha}(x-t,y)f(t)\,dt+ \frac{1}{2\pi}\int_{-\infty}^{\infty} T_{\alpha}(x-t,y)g(t)\,dt. \tag{8} \]
Computing the integrals (7), we find:
\[ R_\alpha(x,y)=\frac{1}{2\alpha\sqrt{\pi}}e^{\frac{y^2-x^2}{4\alpha^2}}\cos\frac{xy}{2\alpha^2}, \]
\[ T_\alpha(x,y)=\frac{1}{2\alpha\sqrt{\pi}}e^{-\frac{x^2}{4\alpha^2}}\int_0^{\frac{y}{2\alpha}} e^{t^2}\cos\frac{tx}{\alpha}\,dt, \tag{9} \]
- The function \(u_\alpha(x,y)\), defined by (8), may be regarded as an approximate solution of the problem. Denote \(u(x,b)=\varphi(x)\), \(u_\alpha(x,b)=\varphi_\alpha(x)\). Under the assumptions adopted, \(\varphi(x)\) exists and is continuous.
Theorem 1. If the modulus of continuity of the function \(\varphi(x)\) is equal to \(\omega(h)\), then
\[ |\varphi(x)-\varphi_\alpha(x)|\leq \omega(\alpha)\left(1+\frac{2}{\sqrt{\pi}}\right). \tag{10} \]
Proof. According to (5),
\[ \varphi_\alpha(x)=\int_{-\infty}^{\infty} W_\alpha(x-t)\varphi(t)\,dt =\int_{-\infty}^{\infty} W_\alpha(t)\varphi(x-t)\,dt. \tag{11} \]
Multiplying both sides of the identity
\[ 1=\int_{-\infty}^{\infty} W_\alpha(t)\,dt, \]
which follows from (4), by \(\varphi(x)\) and subtracting from (9), we find
\[ |\varphi(x)-\varphi_\alpha(x)| \leq \int_{-\infty}^{\infty}|\varphi(x)-\varphi(x-t)|W_\alpha(t)\,dt. \]
But \(|\varphi(x)-\varphi(x-t)|\leq \omega(|t|)\). If \(\mu>0\), then by the property of the modulus of continuity (see [5], p. 69)
\[ \omega(|t|)=\omega\left(\frac{1}{\mu}\mu|t|\right) \leq \omega\left(\frac{1}{\mu}\right)(\mu|t|+1). \]
Thus,
\[ |\varphi(x)-\varphi_\alpha(x)| \leq \omega\left(\frac{1}{\mu}\right) \int_{-\infty}^{\infty}(\mu|t|+1)W_\alpha(t)\,dt. \]
Substituting \(W_\alpha(t)\) from (4) and computing the integral, we find
\[ |\varphi(x)-\varphi_\alpha(x)| \leq \omega\left(\frac{1}{\mu}\right) \left(\frac{2\alpha\mu}{\sqrt{\pi}}+1\right). \]
Putting \(\mu=\frac{1}{\alpha}\), we obtain (10).
Corollary 1. The equality
\[ \lim_{\alpha\to 0}\varphi_\alpha(x)=\varphi(x), \]
holds uniformly with respect to \(x\) on \((-\infty,+\infty)\). Similarly, one obtains
Corollary 2. For every \(y \in [0,b]\) the equality
\[ \lim_{\alpha \to 0} u_\alpha(x,y)=u(x,y), \]
holds uniformly with respect to \(x\) on \((-\infty,+\infty)\).
- Suppose now that, instead of \(f(x)\) and \(g(x)\), their approximations \(f_\delta(x)\) and \(g_\delta(x)\) are given, about which we know that they are continuous, bounded (but in general nonanalytic), and differ from \(f(x)\) and, respectively, from \(g(x)\) by no more than a prescribed \(\delta>0\):
\[ |f(x)-f_\delta(x)| \leq \delta,\quad |g(x)-g_\delta(x)| \leq \delta. \tag{12} \]
If (8) is applied to \(f_\delta(x)\) and \(g_\delta(x)\), then instead of \(u_\alpha(x,y)\) we obtain its approximation
\[ u_{\alpha\delta}(x,y)=\frac{1}{2\pi}\int_{-\infty}^{\infty} R_\alpha(x-t,y)f_\delta(t)\,dt +\frac{1}{2\pi}\int_{-\infty}^{\infty} T_\alpha(x-t,y)g_\delta(t)\,dt . \tag{13} \]
Accordingly, instead of \(\varphi_\alpha(x)\) we shall have \(\varphi_{\alpha\delta}(x)\). The function \(\varphi_{\alpha\delta}(x)\) is an approximate solution of the problem which can actually be found. Its deviation from the true solution \(\varphi(x)\) can be estimated according to the scheme of M. M. Lavrent'ev [6] and F. John [4]:
\[ |\varphi(x)-\varphi_{\alpha\delta}(x)| \leq |\varphi(x)-\varphi_\alpha(x)|+ \]
\[ +|\varphi_\alpha(x)-\varphi_{\alpha\delta}(x)| =\varepsilon_1(\alpha)+\varepsilon_2(\alpha,\delta). \tag{14} \]
The estimate of the first term \(\varepsilon_1(\alpha)\) is given by formula (10). To estimate \(\varepsilon_2(\alpha,\delta)\) we use (8) and (12), putting \(y=b\) in them. Then
\[ |\varphi_\alpha(x)-\varphi_{\alpha\delta}(x)| \leq \frac{1}{2\pi}\int_{-\infty}^{\infty} |R_\alpha(x-t,b)|\cdot |f(t)-f_\delta(t)|\,dt+ \]
\[ +\frac{1}{2\pi}\int_{-\infty}^{\infty} |T_\alpha(x-t,b)|\cdot |g(t)-g_\delta(t)|\,dt \leq \]
\[ \leq \frac{\delta}{2\pi}\int_{-\infty}^{\infty}|R_\alpha(x,b)|\,dx +\frac{\delta}{2\pi}\int_{-\infty}^{\infty}|T_\alpha(x,b)|\,dx . \]
Estimating the integrals with the aid of (9), we find
\[ |\varphi_\alpha(x)-\varphi_{\alpha\delta}(x)| \leq \frac{\delta}{2\pi} \left( e^{\frac{b^2}{4\alpha^2}}+ \int_{0}^{\frac{b}{2\alpha}} e^{t^2}\,dt \right). \tag{15} \]
Estimate (15) is sharp in order. Let, for example,
\[ f(x)-f_\delta(x)=\delta\cos\frac{bx}{4\alpha^2},\quad g(x)-g_\delta(x)=0. \]
THE CAUCHY PROBLEM FOR THE LAPLACE EQUATION IN AN INFINITE STRIP
Then, on the basis of (8), (13), and (9),
\[ \varphi_a(0)-\varphi_{a\delta}(0)= \frac{\delta}{4a\pi\sqrt{\pi}} e^{\frac{b^2}{4a^2}} \times \]
\[ \times \int_{-\infty}^{\infty} e^{-\frac{t^2}{4a^2}}\cos^2\frac{bt}{2a^2}\,dt = \frac{\delta}{4\pi}\left(e^{\frac{b^2}{4a^2}}+1\right). \]
6. Theorem 2. Let the Cauchy problem for the initial functions \(f(x)\) and \(g(x)\) have a continuous solution in the closed strip \(-\infty<x<+\infty,\ 0\le y\le b\), and let \(f_\delta(x)\) and \(g_\delta(x)\) be continuous approximations to \(f(x)\) and \(g(x)\), satisfying, for a given \(\delta>0\), the inequalities (12). If \(a=a(\delta)\) is a root of the equation
\[ \frac{1}{2\pi}\left(e^{\frac{b^2}{4a^2}}+\int_0^{\frac{b}{2a}} e^{t^2}\,dt\right)=\frac{1}{\sqrt{\delta}} \tag{16} \]
and \(u_{a\delta}(x,y)\) is defined by formula (13), then
\[ \lim_{\delta\to 0} u_{a\delta}(x,y)=u(x,y) \]
uniformly in the entire strip.
Proof. It is enough to prove that
\[ \lim_{\delta\to 0}\varphi_{a\delta}(x)=\varphi(x)\quad (a=a(\delta)). \]
When (16) is satisfied, we have \(\varepsilon_2(a,\delta)=\dfrac{1}{\sqrt{\delta}}\), therefore
\[ \lim_{\delta\to 0}\varepsilon_2(a(\delta),\delta)=0. \]
Moreover, as is seen from (16), \(\lim_{\delta\to 0} a(\delta)=0\), hence \(\lim_{\delta\to 0}\varepsilon_1(a(\delta))=0\). Then, as (14) shows,
\[ \lim_{\delta\to 0}\varphi_{a\delta}(x)=\varphi(x)\quad (a=a(\delta)). \]
Theorem 2 makes it possible to approach the solution without bound if \(\lim_{\delta\to 0} f_\delta(x)=f(x)\) and \(\lim_{\delta\to 0} g_\delta(x)=g(x)\), knowing only that it exists and is continuous on the line \(y=b\). It should be noted that the deviation \(|\varphi(x)-\varphi_{a\delta}(x)|\) will be several orders of magnitude larger than \(\delta\); therefore the accuracy in specifying the initial data must be very high. This is connected with the fact that we use very little information about the solution: only the fact of its existence. If, in addition, the degree of smoothness of the solution is known to us, then with the aid of (14) one can also give a quantitative estimate of the deviation of \(\varphi_{a\delta}(x)\) from \(\varphi(x)\).
References
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- Korovkin P. P. Linear operators and approximation theory. Fizmatgiz, Moscow, 1959.
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Received by the editors
September 5, 1964
Ural State University
named after A. M. Gorky