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UDC 517.93
MATHEMATICS
D. K. GVazava
ON THE THEORY OF BOUNDARY-VALUE PROBLEMS FOR THE EQUATION
\(y^m u_{xx}+u_{yy}=k(x,y)e^u\) IN THE LARGE
(Presented by Academician M. A. Lavrent’ev, 17 VI 1965)
Let \(D\) be a simply connected domain of the half-plane \(y>0\), bounded by the segment \(AB\) of the \(x\)-axis and by a curve \(\sigma\), Hölder-continuous in the sense of a curve, which approaches the \(x\)-axis orthogonally at the points \(A(-1,0)\), \(B(1,0)\). Consider the equation
\[ Eu(x,y)=y^m u_{xx}+u_{yy}=k(x,y)e^{u(x,y)}, \tag{1} \]
where \(m\) is a positive number, and \(k(x,y)\) is a function nonnegative everywhere in the domain \(D\), possessing continuous first derivatives in \(\overline D\). Let \(F(s)\) be a continuous function, prescribed on the boundary \(\Gamma=\sigma+AB\) of the domain \(D\), of the arc abscissa \(s\), measured in the positive direction.
Theorem 1. There exists a unique solution, twice continuously differentiable in the domain \(D\), of equation (1), continuous in the closed domain \(\overline D\) and satisfying the condition
\[ u(x,y)\big|_{\Gamma}=F(s). \tag{2} \]
In view of the fact that \(k(x,y)>0\) in the domain \(D\), the uniqueness of the solution of problem (1)—(2) follows from the extremum principle \((^1)\).
We shall carry out the proof of the existence of a solution according to the scheme proposed in \((^2)\).
The solution \(u_0(x,y)\) of the equation
\[ Eu(x,y)\equiv 0, \tag{3} \]
satisfying condition (2), exists and is unique. The existence of the Green function for problem (2)—(3) was proved in \((^3)\). In particular, in the case of the so-called normal domain, when the curve \(\sigma\) is given by the equation
\[ x^2+\frac{4}{(m+2)^2}y^{m+2}=1,\qquad y\geqslant 0, \]
the function \(G(x,y;x_1,y_1)\) is constructed explicitly \((^{3,4})\).
With the aid of the Green function, problem (1)—(2) is reduced to the equivalent integral equation
\[ u(x,y)=-\int_D G(x,y;x_1,y_1)\,k(x_1,y_1)e^{u(x_1,y_1)}\,dx_1\,dy_1+u_0(x,y). \tag{4} \]
Into equation (4) we introduce the parameter \(\lambda\):
\[ u(x,y)=-\lambda\int_D G(x,y;x_1,y_1)\,k(x_1,y_1)e^{u(x_1,y_1)}\,dx_1\,dy_1+u_0(x,y). \tag{5} \]
A solution \(u_\lambda(x,y)\) of equation (5) always exists for small values of the parameter \(\lambda>0\).
If we show the existence of a solution of equation (5) for \(\lambda=1\), then thereby the existence of a solution of problem (1)—(2) will be proved.
We note that, for any \(\lambda>0\), every solution of the integral equation (5) is bounded.
Indeed, by virtue of the inequality \(G(x,y;x_1,y_1)>0\), it follows from (5) that
\(u_\lambda(x,y)\leq u_0(x,y)\), i.e.
\[
0<e^{u_\lambda(x,y)}\leq e^{u_0(x,y)}\leq e^{\max u_0(x,y)}=e^{\max F},
\]
and, consequently,
\[
u_0(x,y)-\lambda\int_D G(x,y;x_1,y_1)k(x_1,y_1)e^{\max F}\,dx_1\,dy_1
\leq u_\lambda(x,y)\leq u_0(x,y).
\tag{6}
\]
Suppose that the solution \(u_\nu(x,y)\) of the integral equation (5) is known for
\(\lambda=\nu<1\). Starting from this, we shall prove the existence of a solution
\(u_{\nu+\alpha}(x,y)\) of equation (5) for \(\lambda=\nu+\alpha\), where
\(\alpha\) is some positive number.
It is not difficult to see that the function
\[
\xi(x,y)=u_{\nu+\alpha}(x,y)-u_\nu(x,y)
\]
satisfies the integral equation
\[
\begin{aligned}
\xi(x,y)&+(\nu+\alpha)\int_D G(x,y;x_1,y_1)k(x_1,y_1)e^{u_\nu(x_1,y_1)}
\xi(x_1,y_1)\,dx_1\,dy_1 \\
&=-\alpha\int_D G(x,y;x_1,y_1)k(x_1,y_1)e^{u_\nu(x_1,y_1)}\,dx_1\,dy_1 \\
&\quad-(\nu+\alpha)\sum_{n=2}^{\infty}\frac{1}{n!}
\int_D G(x,y;x_1,y_1)k(x_1,y_1)e^{u_\nu(x_1,y_1)}
\xi^n(x_1,y_1)\,dx_1\,dy_1 .
\end{aligned}
\tag{7}
\]
By virtue of (6) we have
\[
\max |\xi(x,y)|=\delta\leq
\max\int_D G(x,y;x_1,y_1)k(x_1,y_1)e^{\max F}\,dx_1\,dy_1=d.
\tag{8}
\]
Since problem (2) with zero right-hand side for the equation
\[
E\xi(x,y)=(\nu+\alpha)k(x,y)e^{u_\nu(x,y)}\xi(x,y)
\]
is equivalent to the integral equation
\[
\xi(x,y)+(\nu+\alpha)\int_D G(x,y;x_1,y_1)k(x_1,y_1)e^{u_\nu(x_1,y_1)}
\xi(x_1,y_1)\,dx_1\,dy_1=0,
\tag{9}
\]
the latter cannot have nontrivial solutions. Hence, in turn, it follows that equation (7) is equivalent to the integral equation
\[
\begin{aligned}
\xi(x,y)&=-\alpha\int_D R(x,y;x_1,y_1;\nu+\alpha)\,dx_1\,dy_1 \\
&\quad-(\nu+\alpha)\sum_{n=2}^{\infty}\frac{1}{n!}
\int_D R(x,y;x_1,y_1;\nu+\alpha)\xi^n(x_1,y_1)\,dx_1\,dy_1,
\end{aligned}
\tag{10}
\]
where \(R(x,y;x_1,y_1;\nu+\alpha)\) is the resolvent of the kernel
\[
(\nu+\alpha)G(x,y;x_1,y_1)k(x_1,y_1)e^{u_\nu(x_1,y_1)}.
\]
Since the function
\[
\int_D R(x,y;x_1,y_1;\nu+\alpha)\,dx_1\,dy_1,
\qquad 0\leq \nu+\alpha\leq 1,
\]
is continuous in the closed domain \((x,y)\in\overline D\), the estimates
\[
0\leq\int_D R(x,y;x_1,y_1;\nu+\alpha)\,dx_1\,dy_1\leq B,
\tag{11}
\]
hold, where \(B\) is a constant.
From (8) and (11) follows the validity of the inequalities
\[ (\nu+\alpha)\left|\sum_{n=2}^{\infty}\frac{1}{n!}\int_D R(x,y;x_1,y_1;\nu+\alpha)\zeta^n(x_1,y_1)\,dx_1\,dy_1\right| \leq B_1\delta^2, \]
\[ (\nu+\alpha)\left|\sum_{n=2}^{\infty}\frac{1}{n!}\int_D R(x,y;x_1,y_1;\nu+\alpha)\{\zeta^n(x_1,y_1)- \bar{\xi}^{\,n}(x_1,y_1)\}\,dx_1\,dy_1\right| \leq B_2\max|\zeta(x,y)-\bar{\xi}(x,y)|, \]
where \(\bar{\xi}(x,y)\) is an arbitrary function of the class \(C(\bar D)\), \(\max|\zeta(x,y)|<d\), and the positive constants \(B_1\) and \(B_2\) do not depend on \(\nu\) and \(\alpha\).
According to these inequalities, from (10) we obtain
\[ \delta\leq aB+B_1\delta^2. \]
Let
\[ \tau=1-\sqrt{1-\alpha\cdot 4BB_1}/2B_1 \]
be the smaller root of the quadratic equation
\[ \tau=aB+B_1\tau^2. \]
In the work \((^2)\) it is proved that if
\[ B_2\tau=q<1, \tag{12} \]
then there exists a unique solution of the integral equation (10), satisfying the condition \(|\zeta(x,y)|<\tau\).
Condition (12) will be satisfied for
\[ a<(4B_2-B_1)/4BB_2^2=a^*. \tag{13} \]
This proves the existence of the solution \(u_{\nu+\alpha}(x,y)\) for any positive \(a<a^*\).
Repeating the above argument step by step, we reach the value \(\lambda=1\). Otherwise this process would break off for some value of the parameter \(\lambda=\mu+\alpha\), i.e. the integral equation
\[ \zeta(x,y)+(\mu+\alpha)\int_D G(x,y;x_1,y_1)k(x_1,y_1)e^{u_\mu(x_1,y_1)} \zeta(x_1,y_1)\,dx_1\,dy_1=0 \]
would have a nontrivial solution, which is impossible.
Thus the existence of a solution of problem (1)—(2) has been proved.
Theorem 2. There exists a unique twice continuously differentiable in the domain \(D\) solution of equation (1), satisfying the conditions
\[ u(x,y)\big|_\sigma=\varphi(s), \tag{14} \]
\[ \partial u(x,y)/\partial y\big|_{y=0}=\nu(x), \tag{15} \]
where \(\varphi(s)\) is a prescribed continuous function on \(\sigma\), and \(\nu(x)\) is also a prescribed continuous function on the interval \(0<x<1\), which may tend to infinity of order less than \(2/(2+m)\) as \(x\to0\) or \(x\to1\).
The uniqueness of the solution of this problem again follows from the extremum principle.
Equation (3) has a unique solution satisfying conditions (14) and (15), which can be written with the help of the corresponding Green’s function \(G^*(x,y;x_1,y_1)\). For this problem the Green’s function exists, and it is written explicitly in the case of a normal domain \((^5,^6)\).
The problem (1)—(14)—(15), with the aid of the Green’s function, is reduced to a nonlinear integral equation, whose study is carried out analogously to the study of equation (5).
The boundary-value problems (2) and (14)—(15) for the equation
\[ Eu(x,y)=f(x,y;u) \]
are studied analogously, under the condition that the function \(f(x,y;u)\) can be expanded in a series in powers of \(u(x,y)\), is positive everywhere in the domain \(D\), is continuous in \(x,y\), and the derivative \(f_u'(x,y;u)\) is nonnegative in the domain \(D\).
In conclusion, the author expresses his deep gratitude to Corresponding Member of the Academy of Sciences of the USSR A. V. Bitsadze, who drew the author’s attention to these questions and supervised the execution of the present work.
Novosibirsk State University
Institute of Mathematics
Siberian Branch of the Academy of Sciences of the USSR
Received
15 VI 1965
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