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MATHEMATICS
B. V. MELENT'EV
ON THE EXISTENCE THEOREM FOR THE SOLUTION OF A BOUNDARY-VALUE PROBLEM FOR AN EQUATION OF MIXED TYPE
(Presented by Academician M. A. Lavrent'ev on 24 X 1961)
In the present note an existence theorem is proved for a boundary-value problem for the M. A. Lavrent'ev—A. V. Bitsadze equation
\[ \frac{\partial^2 z}{\partial x^2}+\operatorname{sgn} y\,\frac{\partial^2 z}{\partial y^2}=0. \tag{1} \]
§ 1. Consider a domain \(\Delta\), bounded in the half-plane \(y>0\) by the half-lines \(OA\), \(x=0\), and \(DE\), \(x=1\), and by a certain smooth curve \(DC\) joining the points \(C(m,0)\) and \(D(1,y)\), and by two characteristics \(OB\), \(y=-x\), and \(CB\), \(y=x-m\) \((0<m<1)\), in the half-plane \(y<0\).
Statement of the F. I. Frankl problem for the M. A. Lavrent'ev—A. V. Bitsadze equation. Find a solution of equation (1) in the domain \(\Delta\) \((y\ne 0)\), continuous up to the boundary, bounded at infinity, having continuous partial derivatives \(\partial z/\partial x\) and \(\partial z/\partial y\) in \(\Delta\); moreover, near the points \(O\) and \(C\) they may tend to infinity of order less than one, and satisfying on the boundary of the domain the conditions:
\[ \begin{aligned} &\text{on the characteristic } OB \qquad z=0;\\ &\text{on the half-lines } OA \text{ and } DE \qquad z=0;\\ &\text{on } CD \qquad a\frac{\partial z}{\partial x}+b\frac{\partial z}{\partial y}=0. \end{aligned} \tag{2} \]
§ 2. Theorem. In the domain \(\Delta\) there exists a solution of equation (1) satisfying the conditions listed above.
Proof. We construct the Green function of the Frankl problem for the Lavrent'ev—Bitsadze equation. In the lower half-plane the solution, as is known, has the form
\[ u(x,y)=f(x+y)+f_1(x-y), \tag{3} \]
where \(f\) and \(f_1\) are arbitrary twice differentiable functions.
On the characteristic \(y=-x\), \(u=0\); therefore \(f(0)+f_1(2x)=0\), whence
\[ u(x,y)=f(x+y)-f(0). \tag{4} \]
Differentiating (4) with respect to \(x\) and to \(y\) and letting \(y\) tend to zero, we obtain
\[ u'_x(x,0)=\tau'(x), \qquad u'_y(x,0)=\nu(x), \]
whence
\[ \tau'(x)=\nu(x). \tag{5} \]
The obtained condition (5) is equivalent to the condition prescribed on the characteristic \((z=0)\).
By direct verification it is easy to see that the function
\[ u(x,y)=\ln \sqrt{(x-x_0)^2+(y-y_0)^2}-\operatorname{arc\,tg}\frac{x-x_0}{y+y_0} \tag{6} \]
is a solution of equation (1) satisfying condition (5).
To construct a solution satisfying the zero conditions on \(OA\) and \(DE\), we use the method of reflection. Taking two terms of the series constructed in this way, we obtain the function
\[ \begin{aligned} g^*={}&\ln \sqrt{(x-x_0)^2+(y-y_0)^2} -\operatorname{arc\,tg}\frac{x-x_0}{y+y_0} -\ln \sqrt{(x-x_0)^2+(y-y_0)^2} \\ &+\operatorname{arc\,tg}\frac{x+x_0}{y+y_0} +\ln \sqrt{(x-2-x_0)^2+(y-y_0)^2} -\operatorname{arc\,tg}\frac{x-2-x_0}{y+y_0} \\ &-\ln \sqrt{(x-2+x_0)^2+(y-y_0)^2} +\operatorname{arc\,tg}\frac{x-2+x_0}{y+y_0}, \end{aligned} \tag{7} \]
which on the half-lines \(OA\), \(x=0\), and \(DE\), \(x=1\), \((y>0)\), takes, respectively, the values:
\[ \begin{aligned} g^*(0,y,x_0,y_0)={}& \ln \sqrt{(2+x_0)^2+(y-y_0)^2} -\ln \sqrt{(2-x_0)^2+(y-y_0)^2} \\ &+2\operatorname{arc\,tg}\frac{x_0}{y+y_0} +\operatorname{arc\,tg}\frac{2+x_0}{y+y_0} -\operatorname{arc\,tg}\frac{2-x_0}{y+y_0}, \end{aligned} \]
\[ g^*(1,y,x_0,y_0)= 2\left( \operatorname{arc\,tg}\frac{1+x_0}{y+y_0} -\operatorname{arc\,tg}\frac{1-x_0}{y+y_0} \right). \]
The functions obtained by us, \(g^*(0,y,x_0,y_0)\) and \(g^*(1,y,x_0,y_0)\), are analytic for all \(y,y_0\) \((y,y_0>0)\) and \(0\le x\le 1\).
Let \(\varphi(x,y,x_0,y_0)\) be the solution of Tricomi’s problem for the domain considered in the problem, vanishing on the characteristic \(y=-x\) and taking the values:
\[ \begin{array}{ll} \text{on } OA & \varphi(0,y,x_0,y_0)=g^*(0,y,x_0,y_0),\\ \text{on } DE & \varphi(1,y,x_0,y_0)=g^*(1,y,x_0,y_0). \end{array} \]
Then the function
\[ g(x,y,x_0,y_0)=g^*(x,y,x_0,y_0)-\varphi(x,y,x_0,y_0) \tag{8} \]
will be the Green’s function of the Frankl problem for the Lavrent’ev–Bitsadze equation. The existence of the function \(\varphi(x,y,x_0,y_0)\) follows from work \((^1)\).
§ 3. We shall seek the solution of the posed problem in the form
\[ z=\int_L \mu(s)g(x,y,x_0,y_0)\,ds. \tag{9} \]
Therefore
\[ \frac{\partial z}{\partial x} =\int_L \mu(s)\left(\frac{\partial g^*}{\partial x}-\frac{\partial\varphi}{\partial x}\right)ds, \qquad \frac{\partial z}{\partial y} =\int_L \mu(s)\left(\frac{\partial g^*}{\partial y}-\frac{\partial\varphi}{\partial y}\right)ds, \]
where \(L\) denotes the arc \(CD\). More explicitly:
\[ \frac{\partial z}{\partial x} =\int_L \mu(s)\left\{ \frac{x-x_0}{(x-x_0)^2+(y-y_0)^2} -\frac{x+x_0-2}{(x+x_0-2)^2+(y-y_0)^2} +P(x,y,x_0,y_0) \right\}, \]
\[ \frac{\partial z}{\partial y} =\int_L \mu(s)\left\{ \frac{y-y_0}{(x-x_0)^2+(y-y_0)^2} -\frac{y-y_0}{(x+x_0-2)^2+(y-y_0)^2} +Q(x,y,x_0,y_0) \right\}. \tag{10} \]
Here \(P(x,y,x_0,y_0)\) and \(Q(x,y,x_0,y_0)\) denote sums of regular terms.
Let the point \((x_0,y_0)\) tend to the boundary \(L\). In doing so we use the limiting-transition formulas
\[ \frac{\partial z_i}{\partial x} = \frac{\partial z_0}{\partial x} +\pi\mu_0\cos(nx), \qquad \frac{\partial z_i}{\partial y} = \frac{\partial z_0}{\partial y} -\pi\mu_0\cos(ny), \tag{11} \]
where the subscript zero indicates that the value is taken on the boundary \(L\).
Introduce new notation
\[ x_0=x(s);\qquad y_0=y(s);\qquad x=x(\sigma);\qquad y=y(\sigma) \]
and, applying formulas (11) to (10), substitute the obtained values of the partial derivatives into the last of the boundary conditions (2). We arrive at the singular integral equation:
\[ \begin{aligned} &\pi\{a(\sigma)\cos(nx)+b(\sigma)\cos(ny)\}\mu(\sigma) +a(\sigma)\int_L \mu(s)\frac{[x(\sigma)-x(s)]\,ds} {[x(\sigma)-x(s)]^2+[y(\sigma)-y(s)]^2} \\ &\quad +b(\sigma)\int_L \mu(s)\frac{[y(\sigma)-y(s)]\,ds} {[x(\sigma)-x(s)]^2+[y(\sigma)-y(s)]^2} \\ &\quad -a(\sigma)\int_L \mu(\sigma)\frac{[x(\sigma)+x(s)-2]\,ds} {[x(\sigma)+x(s)-2]^2+[y(\sigma)-y(s)]^2} \\ &\quad -b(\sigma)\int_L \mu(s)\frac{[y(\sigma)-y(s)]\,ds} {[x(\sigma)x(s)-2]^2+[y(\sigma)-y(s)]^2} +a(\sigma)\int_L \mu(s)P(\sigma,s)\,ds \\ &\quad +b(\sigma)\int_L \mu(s)Q(\sigma,s)\,ds=0. \end{aligned} \tag{12} \]
Transform the singular integrals in the left-hand side of equation (12), separating out the principal part:
\[ \begin{aligned} \int_L \mu(s)\frac{[x(\sigma)-x(s)]\,ds} {[x(\sigma)-x(s)]^2+[y(\sigma)-y(s)]^2} &= \int_L \mu(s)\frac{K_1(\sigma,s)}{\sigma-s}\,ds \\ &= \int_L \mu(s)\frac{K_1(\sigma,\sigma)}{\sigma-s}\,ds +\int_L \mu(s)\frac{K_1(\sigma,s)-K_1(\sigma,\sigma)}{\sigma-s}\,ds \\ &= -x'(\sigma)\int_L \frac{\mu(s)}{s-\sigma}\,ds -\int_L \mu(s)\frac{K_1(\sigma,s)-x'(\sigma)}{s-\sigma}\,ds, \end{aligned} \]
where
\[ K_1(\sigma,s)= \frac{(\sigma-s)[x(\sigma)-x(s)]} {[x(\sigma)-x(s)]^2+[y(\sigma)-y(s)]^2}, \]
\[ K_1(\sigma,\sigma)=\lim_{s\to\sigma}K_1(\sigma,s) = \lim_{s\to\sigma} \frac{\dfrac{x(\sigma)-x(s)}{\sigma-s}} {\left[\dfrac{x(\sigma)-x(s)}{\sigma-s}\right]^2+ \left[\dfrac{y(\sigma)-y(s)}{\sigma-s}\right]^2} =x'(\sigma). \]
Similarly we transform the other four integrals entering into the left-hand side of (12). Substituting these values into (12) and taking into account that
\[ a(\sigma)\cos(nx)+b(\sigma)\cos(ny) = a(\sigma)y'(\sigma)+b(\sigma)x'(\sigma), \]
we obtain
\[ \begin{aligned} &\pi\{a(\sigma)y'(\sigma)+b(\sigma)x'(\sigma)\}\mu(\sigma) -\{a(\sigma)x'(\sigma)+b(\sigma)y'(\sigma)\}\int_L\frac{\mu(s)}{s-\sigma}\,ds \\ &\quad -\{a(\sigma)x'(l)+b(\sigma)y'(l)\}\int_L\frac{\mu(s)\,ds}{\sigma-s-2l} -a(\sigma)\int_L\frac{K_1(\sigma,s)-x'(\sigma)}{s-\sigma}\mu(s)\,ds \\ &\quad -b(\sigma)\int_L\frac{K_2(\sigma,s)-y'(\sigma)}{s-\sigma}\mu(s)\,ds -a(\sigma)\int_L\frac{K_3(\sigma,s)-x'(l)}{\sigma+s-2l}\mu(s)\,ds \\ &\quad -b(\sigma)\int_L\frac{K_4(\sigma,s)-y'(l)}{\sigma+s-2l}\mu(s)\,ds +a(\sigma)\int_L P(\sigma,s)\mu(s)\,ds \\ &\quad +b(\sigma)\int_L Q(\sigma,s)\mu(s)\,ds=0. \end{aligned} \]
Introduce the notation: \(A(\sigma)=a(\sigma)y'(\sigma)+b(\sigma)x'(\sigma)\), \(B(\sigma)=a(\sigma)x'(\sigma)+b(\sigma)y'(\sigma)\), \(B(\sigma,l)=a(\sigma)x'(l)+b(\sigma)y'(l)\). Then the equation takes the form
\[ \begin{aligned} &A(\sigma)\mu(\sigma) -\frac{B(\sigma)}{\pi}\int_L\left\{\frac{1}{s-\sigma}+\frac{1}{\sigma+s-2l}\right\}\mu(s)\,ds -\frac{B(\sigma,l)-B(\sigma)}{\pi}\int_L\frac{\mu(s)\,ds}{\sigma+s-2l} \\ &\quad -\frac{a(\sigma)}{\pi}\int_L \frac{K_1(\sigma,s)-x'(\sigma)}{s-\sigma}\,\mu(s)\,ds -\frac{b(\sigma)}{\pi}\int_L \frac{K_2(\sigma,s)-y'(\sigma)}{s-\sigma}\,\mu(s)\,ds \\ &\quad -\frac{a(\sigma)}{\pi}\int_L \frac{K_3(\sigma,s)-x'(l)}{\sigma+s-2l}\,\mu(s)\,ds -\frac{b(\sigma)}{\pi}\int_L \frac{K_4(\sigma,s)-y'(l)}{\sigma+s-2l}\,\mu(s)\,ds \\ &\quad +\frac{a(\sigma)}{\pi}\int_L P(\sigma,s)\mu(s)\,ds +\frac{b(\sigma)}{\pi}\int_L Q(\sigma,s)\mu(s)\,ds =0. \end{aligned} \tag{13} \]
§ 4. Let us compute the index of the singular equation obtained, using the known formulas \((^2)\). As the investigation shows, if the solution of the problem is sought in the class of functions that tend to infinity at both ends of the arc \(L\) with order less than one, then the index \(x\) will be equal to one. Therefore, proceeding from the general theory \((^2)\), the singular integral equation (13) obtained by us can be regularized and has no more than one independent solution. Thus, the existence theorem is proved.
Kharkov Polytechnic Institute
named after V. I. Lenin
Received
20 X 1961
CITED LITERATURE
- A. V. Bitsadze, Trudy Mat. Inst. im. V. A. Steklova AN SSSR, 41 (1953).
- N. I. Muskhelishvili, Singular Integral Equations, 1946.