\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2011 (2011), No. 36, pp. 1--13.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2011 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2011/36\hfil Hyperbolic equations with degenerate curve] {Oblique derivative problems for second-order hyperbolic equations with degenerate curve} \author[G.-C. Wen\hfil EJDE-2011/36\hfilneg] {Guo-Chun Wen} \address{Guo-Chun Wen \newline LMAM, School of Mathematical Sciences, Peking University, Beijing 100871, China} \email{Wengc@math.pku.edu.cn} \thanks{Submitted December 22, 2010. Published March 3, 2011.} \subjclass[2000]{35L20, 35L80} \keywords{Oblique derivative problem; hyperbolic equations; degenerate curve} \begin{abstract} The present article concerns the oblique derivative problem for second order hyperbolic equations with degenerate circle arc. Firstly the formulation of the oblique derivative problem for the equations is given, next the representation and estimates of solutions for the above problem are obtained, moreover the existence of solutions for the problem is proved by the successive iteration of solutions of the equations. In this article, we use the complex analytic method, namely the new partial derivative notations, hyperbolic complex functions are introduced, such that the second order hyperbolic equations with degenerate curve are reduced to the first order hyperbolic complex equations with singular coefficients, then the advantage of complex analytic method can be applied. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \allowdisplaybreaks \section{Formulation of the oblique derivative problem} In \cite{b1,b2,r1,s1,s2,w3,w4,w5}, the authors posed and discussed the Cauchy problem, Dirichlet problem and oblique derivative boundary value problem of second order hyperbolic equations and mixed equations with parabolic degenerate straight lines by using the methods of integral equations, functional analysis, energy integrals, complex analysis and so on, the obtained results possess the important applications. Here we generalize the above results to the oblique derivative problem of hyperbolic equations with degenerate circle arc. In this article, the used notations are the same as in \cite{w1,w2,w3,w4,w5}. Let $D$ be a simply connected bounded domain $D$ in the hyperbolic complex plane $\mathbb{C}$ with the boundary $\partial D=L\cup L_0$, where $L=L_1\cup L_2$. Herein and later on, denote $\hat y=y - \sqrt{R^2-x^2}$, and \begin{gather*} L_1 = \{x + G(\hat y) = R_*,x \in [R_*,0]\},\quad L_2 = \{x - G(\hat y) = R^*,x \in [0,R^*]\}, \\ L_0 = \{R_*\le x\le R^*, \hat y=0\}, \end{gather*} in which $K(\hat y)=-|\hat y|^m$, $m, R$ are positive numbers, $R_*=-R$, $R^*=R$, $z_0=z_1=jy_0=jy_1$ the intersection of $L_1, L_2$, $G(\hat y)=\int_0^{\hat y} \sqrt{|K(t)|}dt$, $H(\hat y)=|K(\hat y)|^{1/2}$. In this article we use the hyperbolic unit $j$ with the condition $j^2=1$ in $\overline{D}$, and $x+jy, w(z)=U(z)+jV(z)=[H(\hat y)u_x-ju_y]/2$ are called the hyperbolic number and hyperbolic complex function in $D$. Consider the second-order linear equation of hyperbolic type with degenerate circle arc $$Lu=K(\hat y)u_{xx}+u_{yy}+au_x+bu_y+cu=-d\quad\text{in }D,\label{e1}$$ where $a,b,c,d$ are real functions of $z$ ($z\in \overline D$), and suppose that the equation \eqref{e1} satisfies the following conditions: \subsection*{Condition C} The coefficients $a,b,c,d$ in $\overline{D}$ satisfy $$\begin{gathered} \tilde C[d,\overline{D}]=C[d,\overline{D}]+C[d_x,\overline{D}] \le k_1, \quad \tilde C[\eta,\overline{D}]\le k_0,\quad \eta=a,b,c, \\ |a(x,y)|{|\hat y|^{1-m/2}}=\varepsilon_1(\hat y)\quad\text{as } \hat y\to0,\;m\ge2, \;z\in\overline{D}, \end{gathered}\label{e2}$$ in which $\varepsilon_1(\hat y)$ is a non-negative function satisfying the condition: $\varepsilon_1(\hat y)\to 0$ as $\hat y\to0$. To write the complex form of the above equation, denote $Y=G(\hat y)$, $\hat y=y-\sqrt{R^2-x^2}$, $\hat x=x$, and \begin{gather*} W(z)=U+jV = \frac12[H(\hat y)u_x -ju_y]=\frac{H(\hat y)} 2[u_x - ju_Y] = H(\hat y)u_Z, \\ H(\hat y)W_{\overline Z}=\frac{H(\hat y)}2[W_x+jW_Y] =\frac12[H(\hat y)u_x+jW_y]=W_{\overline{\tilde z}}\quad\text{in }\overline{D}, \end{gather*} where $Z = Z(z) = x + jY = x + jG(\hat y)$ in $\overline{D}$, $G(\hat y)=\int_0^{\hat y}H(t)dt$, $H(\hat y)=\sqrt{|K(\hat y)|}$. Moreover, \begin{gathered} \begin{aligned} -K(\hat y)u_{xx}-u_{yy} &=H(\hat y)[H(\hat y)u_x-ju_y]_x +j[H(\hat y)u_x-ju_y]_y-[jH_y+HH_x]u_x \\ &=4H(\hat y)W_{\overline Z} - [jH_y/H + H_x]Hu_x \\ &= au_x + bu_y + cu+d, \end{aligned} \\ \begin{aligned} &H(\hat y)W_{\overline Z}\\ &= H[W_x + jW_Y]/2 \\ &= H[(U + jV)_x + j(U + jV)_Y]/2 \\ &=\frac14(e_1 - e_2)(H_{\hat y}/H)Hu_x + (e_1 + e_2) [(H_x + a/H)Hu_x + bu_y + cu + d], \end{aligned} \\ (U+V)_\mu=\frac1{4H}\{2[H_{\hat y}/H + H_x+a/H]U-2bV+cu+d\}, \quad \text{in }D,\\ (U-V)_\nu=\frac1{4H}\{-2[H_{\hat y}/H - H_x - a/H]U - 2bV +cu+d\}, \quad \text{in }D, \end{gathered} \label{e3} where $e_1 = (1+j)/2$, $e_2 = (1-j)/2$, $x=\mu+\nu$, $Y=\mu-\nu$, $\partial x/\partial\mu=1/2=\partial Y/\partial\mu$, $\partial x/\partial\nu=1/2=-\partial Y/\partial\nu$. Hence the complex form of \eqref{e1} can be written as $$\begin{gathered} W_{\overline{\tilde z}}\, =A_1W+A_2\overline W+A_3u+A_4\quad\text{in }\overline{D}, \\ u(z) = 2\operatorname{Re} \int_{z_0}^z [\frac{U(z)}{H(\hat y)} - jV(z)]dz + b_0 \quad\text{in }\overline{D}, \end{gathered}\label{e4}$$ where $b_0=u(z_0)$, $z_0=jy_0$, and the coefficients $A_l=A_l(z)$ ($l=1,2,3,4$) are as follows \begin{gather*} A_1 = \frac14[\frac a{H}+\frac{jH_{\hat y}}{H} + H_x -jb],\quad A_2 = \frac14[\frac a{H}+\frac{jH_{\hat y}}{H} + H_x +jb],\\ A_3=\frac c4,\quad A_4=\frac d4\quad\text{in }\overline{D}. \end{gather*} For convenience, sometimes the hyperbolic complex number $\hat z=\hat x+j\hat y=x+j(y-\sqrt{R^2-x^2})$ and the function $F[z(Z)]$ are simply written as $z=x+j\hat y$ and $F(Z)$ respectively. We mention that in this article, three domains; i.e., the original domain $D$, the characteristic domain $D_{\hat z}$ and the image domain $D_Z$ are used, and the corresponding characteristic domain $D_{\hat z}$ almost is written as the original domain $D$. The oblique derivative problem for \eqref{e1} may be formulated as follows. \subsection*{Problem O} Find a continuous solution $u(z)$ of \eqref{e1} in $\overline D\backslash L_0$, which satisfies the boundary conditions $$\begin{gathered} \frac12\frac{\partial u}{\partial l} = \frac1{H(y)}\operatorname{Re}[\overline{\lambda(z)}u_{\tilde z}] = \operatorname{Re}[\overline{\Lambda(z)}u_z] = r(z),\quad z \in L = L_1 \cup L_2,\\ u(z_0)=b_0,\quad \frac1{H(\hat y)}\operatorname{Im}[\overline{\lambda(z)}u_{\tilde z}]|_{z=z_0} =\operatorname{Im}[\overline{\Lambda(z)} u_z]|_{z=z_0}=b_{1}, \end{gathered}\label{e5}$$ in which $l$ is a given vector at every point $z\in L$, $u_{\tilde z}=[H(\hat y)u_x -ju_{y}]/2$, $u_{\bar{\tilde z}}=[H(\hat y)u_x+ju_{y}]/2, b_0,b_1$ are real constants, $\lambda(z)=\lambda_1(x)+j\lambda_2(x)$, $\Lambda(z) = \cos(l,x) + j\cos(l,y)$, $R(z) = H(\hat y)r(z)$, $z \in L$, $b'_1 = H(\hat y_1)b_1$, $\lambda_1(z)$ and $\lambda_2(x)$ are real functions, $\lambda(z),r(z),b_0,b_1$ satisfy the conditions $$\begin{gathered} C^1[\lambda(z),L] \le k_0, \quad C^1[r(z),L] \le k_2, \quad |b_0|,|b_1| \le k_2,\\ \max_{z\in L_1}\frac1{|\lambda_1(x) - \lambda_2(x)|},\quad \max_{z\in L_2}\frac1{|\lambda_1(x) + \lambda_2(x)|}\le k_0, \end{gathered}\label{e6}$$ in which $k_0, k_2$ are positive constants. For the Dirichlet problem (Problem D) with the boundary condition: $$u(z)=\phi(x)\quad\text{on }L=L_1\cup L_2,\label{e7}$$ where $L_1, L_2$ are as stated before, we find the derivative for \eqref{e7} according to the parameter $s=x$ on $L_1,L_2$, and obtain \begin{gather*} u_s=u_x+u_yy_x=u_x-\frac{u_y}{H(\hat y)}=\phi'(x)\quad\text{on }L_1, \\ u_s=u_x+u_yy_x=u_x+\frac{u_y}{H(\hat y)}=\phi'(x) \quad\text{on }L_2; \end{gather*} i. e., \begin{gather*} U(z)+V(z)=\frac12H(\hat y)\phi'(x)=R(z)\quad\text{on }L_1,\\ U(z)-V(z)=\frac12H(\hat y)\phi'(x)=R(z)\quad\text{on }L_2; \end{gather*} i. e., \begin{gather*} \operatorname{Re}[(1+j)(U+jV)]=U(z)+V(z)=R(z)\quad\text{on }L_1, \\ \operatorname{Im}[(1+j)(U+jV)]|_{z=z_0-0}=[U(z)+V(z)]|_{z=z_0-0}=R(z_0-0), \\ \operatorname{Re}[(1-j)(U+jV)]=U(z)-V(z)=R(z)\quad\text{on }L_2, \\ \operatorname{Im}[(1-j)(U +jV)]|_{z=z_0+0} = [-U(z) + V(z)]|_{z=z_0+0} = -R(z_0 + 0), \end{gather*} where \begin{gather*} U(z)=\frac12H(\hat y)u_x,\quad V(z)=-\frac{u_y}2, \\ \lambda_1+j\lambda_2=1-j,\quad \lambda_1=1\not=\lambda_2=-1\quad\text{on }L_1,\\ \lambda_1+j\lambda_2=1+j,\quad \lambda_1=1\not= -\lambda_2=-1\quad\text{on }L_2. \end{gather*} From the above formulas, we can write the complex forms of boundary conditions of $U+jV$: $$\begin{gathered} \operatorname{Re}[\overline{\lambda(z)}(U+jV)]=R(z)\quad\text{on }L, \\ \operatorname{Im}[\overline{\lambda(z)}(U+jV)]|_{z=z_0-0}=R(z_0-0)=b'_1, \\ \lambda(z) = \begin{cases} 1 - j = \lambda_1 + j\lambda_2, \\ 1 + j = \lambda_1 + j\lambda_2, \end{cases} \quad R(z) = \begin{cases} H(\hat y)\phi'(x)/2 &\text{on }L_1, \\ H(\hat y)\phi'(x)/2 &\text{on }L_2, \end{cases}\\ u(z)=2\operatorname{Re}\int_{z_0}^z[\frac{U(z)}{H(\hat y)}-jV(z)]dz+\phi(z_0)\quad\text{in } D. \end{gathered}\label{e8}$$ Hence Problem D is a special case of Problem O. Noting that the condition \eqref{e6}, we can find a twice continuously differentiable functions $u_0(z)$ in $\overline{D}$, for instance, which is a solution of the oblique derivative problem with the boundary condition in \eqref{e5} for harmonic equations in $D$ (see \cite{w1,w2}), thus the functions $v(z)=u(z)-u_0(z)$ in $D$ is the solution of the following boundary value problem in the form \begin{gather} K(\hat y)v_{xx}+v_{yy}+av_x+bv_y+cv =-\hat d\quad\text{in }D,\label{e9}\\ \begin{gathered} \operatorname{Re}[\overline{\lambda(z)}v_{\tilde z}(z)] = r(z)\quad\text{on }L,\\ v(z_0)=b_0, \quad \operatorname{Im}[\overline{\lambda(z_0)}v_{\tilde z}(z_0)]=b'_1, \end{gathered} \label{e10} \end{gather} where $W(z)=U+jV=v_{\tilde z}$ in $\overline{D}$, $r(z)=0$ on $L$, $b_0=b'_1=0$. Hence later on we only discuss the case of the homogeneous boundary condition. From $v(z)=u(z)-u_0(z)$ in $\overline{D}$, we have $u(z)=v(z)+u_0(z)$ in $\overline{D}$, and $v_y=2\tilde R_0(x)$ on $L_0=D_{\hat z}\cap\{\hat y=0\}$, in which $\tilde R_0(x)$ is an undermined real function. The boundary vale problem \eqref{e9}, \eqref{e10} is called Problem $\tilde O$. \section{Properties of solutions to the oblique derivative problem} In this section, we consider the special mixed equation $$\begin{gathered} u_{\tilde z\overline{\tilde z}}=W_{\overline{\tilde z}}=0,\quad\text{i.e.,}\\ (U+V)_\mu=0,\quad (U-V)_\nu=0\quad\text{in }\overline D, \end{gathered}\label{e11}$$ where $U(z)=\operatorname{Re} W(z)$, $V(z)=\operatorname{Im} W(z)$. \begin{theorem} \label{thm1} Any solution $u(z)$ of Problem O for the hyperbolic equation \eqref{e11} can be expressed as $$u(z) = u(x) - 2 \int_0^{\hat y} V(\hat y)d\hat y = 2\operatorname{Re} \int_{z_0}^z[\frac{\operatorname{Re} W(z)}{H(\hat y)} - j\operatorname{Im} W(z)]dz + b_0\quad\text{in }\overline{D},\label{e12}$$ where \begin{aligned} W(z)&= U + jV = f(x - Y)e_1 + g(x + Y)e_2\\ &= f(\nu)e_1 + g(\mu)e_2 \\ &= \frac12\{f(x-Y) + g(x+Y) + j[f(x-Y) - g(x+Y)]\}, \end{aligned}\label{e13} in which $Y=G(\hat y)$. For convenience denote by the functions $\lambda_1(x),\lambda_2(x),r(x)$ of $x$ the functions $\lambda_1(z),\lambda_2(z),r(z)$ of $z$ in \eqref{e10}, and $f(x-Y)=f(\nu)$, $g(x+Y)=g(\mu)$ possess the forms \begin{gathered} \begin{aligned} f(\nu) = f(x - Y) &= \frac{2r((x-Y+R_*)/2)}{\lambda_1((x-Y +R_*)/2)-\lambda_2((x-Y +R_*)/2)}\\ &\quad -\frac{[\lambda_1((x - Y+R_*)/2)+\lambda_2((x-Y+R_*)/2)] g(R^*)}{\lambda_1((x-Y+R_*)/2) - \lambda_2((x-Y+R_*)/2)}, \end{aligned}\\ R_*\le x-Y\le R^*, \\ (\lambda_1(0) + \lambda_2(0))g(R_*) = (\lambda_1(0) + \lambda_2(0))(U(z_1) - V(z_1)) = r(0) - b_1\quad\text{or }0, \\ \begin{aligned} g(\mu)&=g(x+Y)= \\ &=\frac{2r((x + Y + R^*)/2) - [\lambda_1((x + Y + R^*)/2) - \lambda_2((x + Y + R^*)/2)]f(R^*)}{\lambda_1((x+Y+R^*)/2)+\lambda_2((x+Y+R^*)/2)}, \end{aligned} \\ R_*\le x+Y\le R^*, \\ (\lambda_1(0) - \lambda_2(0))f(R^*) = (\lambda_1(0) - \lambda_2(0))(U(z_1) + V(z_1)) = r(0) + b_1\quad\text{or }0. \end{gathered}\label{e14} Moreover $u(z)$ satisfies the estimate $$C^1_\delta[u(z),\overline D]\le M_1,\quad C^1_\delta[u(z),\overline D]\le M_2k_1,\label{e15}$$ where $\delta = \delta(\alpha,k_0,k_1,D)<1$, $M_1 = M_1(\alpha,k_0,k_1,D)$, $M_2 = M_2 (\alpha,k_0,D)$ are positive constants. \end{theorem} \begin{proof} Let the general solution $$W(z)=u_{\tilde z}=\frac12\{f(x-Y)+g(x+Y)+j[f(x-Y)-g(x+Y)]\}$$ of \eqref{e11} be substituted in the boundary condition \eqref{e10}, thus \eqref{e10} can be rewritten as \begin{gather*} \lambda_1(x)U(z) -\lambda_2(x)V(z) = r(z)\quad\text{on }L, \\ \overline{\lambda(z_1)}W(z_1) = r(z_1)+jb_1; \end{gather*} i.e., \begin{gather*} {[\lambda_1(x)-\lambda_2(x)]f(2x- R_*)+[\lambda_1(x)+\lambda_2(x)]g(R_*) =2r(x)\quad\text{on }L_1,} \\ {[\lambda_1(x)-\lambda_2(x)]f(R^*)+[\lambda_1(x)+\lambda_2(x)]g(2x-R^*) =2r(x)\quad\text{on }L_2,} \end{gather*} the above formulas can be rewritten as \begin{gather*} \begin{aligned} &\Big[\lambda_1\Big(\frac{t+ R_*}2\Big) - \lambda_2\Big(\frac{t+ R_*}2 \Big)\Big]f(t)+ \Big[\lambda_1\Big(\frac{t+ R_*}2\Big) + \lambda_2\Big(\frac{t+ R_*}2\Big)\Big]g(R_*) \\ &=2r\Big(\frac{t+R_*}2\Big),\quad t\in[R_*,R^*], \end{aligned}\\ (\lambda_1(0) + \lambda_2(0))g(R_*) = (\lambda_1(0) + \lambda_2(0))(U(z_1) - V(z_1)) = r(0) - b_1\quad\text{or}\quad 0,\\ \begin{aligned} &\Big[\lambda_1\Big(\frac{t+ R^*}2\Big) - \lambda_2\Big(\frac{t+R^*}2\Big) \Big]f(R^*) + \Big[\lambda_1\Big(\frac{t+ R^*}2) +\lambda_2\Big(\frac{t+ R^*}2\Big)\Big]g(t) \\ &=2r(\frac{t+R^*}2),\quad t\in[R_*,R^*], \end{aligned} \\ (\lambda_1(0) - \lambda_2(0))f(R^*) = (\lambda_1(0) - \lambda_2(0))(U(z_1) + V(z_1)) = r(0) + b_1\quad\text{or} \quad 0, \end{gather*} thus the solution $W(z)$ can be expressed as \eqref{e13}. Here we mention that for the oblique derivative boundary condition, by \eqref{e10}, we have $(\lambda_1(0) + \lambda_2(0))g(R_*) = 0$, $(\lambda_1(0) - \lambda_2(0))f(R^*) = 0$. If $(\lambda_1(x) + \lambda_2(x))g(R_*)$ on $L_1$ and $(\lambda_1(x) - \lambda_2(x))f(R^*)$ on $L_2$ are known. From the condition \eqref{e6} and the relation \eqref{e12}, we see that the estimate \eqref{e15} of the solution $u(z)$ for \eqref{e11}, \eqref{e12} is obviously true. \end{proof} \section{Uniqueness of solutions to the oblique derivative problem} The representation of solutions of Problem O for equation \eqref{e1} is as follows. \begin{theorem} \label{thm2} Under Condition C, any solution $u(z)$ of Problem O for equation \eqref{e1} in $D$ can be expressed as $$\begin{gathered} u(z) = 2\operatorname{Re} \int_{z_0}^z[\frac{\operatorname{Re} W}{H(\hat y)}-j\operatorname{Im} W]{\rm d}z+b_0, \\ W(z) = w(z)+{\Phi}(z)+{\Psi}(z)\quad\text{in }D, \\ w(z) = f(\nu)e_1 + g(\mu)e_2, {\Phi}(z) = \tilde f(\nu)e_1 + \tilde g(\mu)e_2, \\ {\Psi}(z) = \int^\mu_{R_*} g_1(z)e_1{\rm d}\mu + \int_{R^*}^{\nu} g_2(z)e_2d\nu, \\ g_l(z) = A_l\xi+B_l\eta+Cu+D,\quad l=1,2.\end{gathered} \label{e16}$$ Here $$\begin{gathered} A_1 = \frac1{4H}[\frac aH + H_x + \frac{H_y}H - b],\quad B_1 = \frac1{4H}[\frac aH + H_x + \frac{H_y}H + b], \quad C = \frac c{4H},\\ A_2 = \frac1{4H}[\frac aH + H_x - \frac{H_y}H - b], \quad B_2 = \frac1{4H}[\frac aH + H_x - \frac{H_y}H + b],\quad D = \frac d{4H}, \end{gathered}\label{e17}$$ where $f(\nu), g(\mu)$ are as stated in \eqref{e14}, and $\tilde f(\nu), \tilde g(\mu)$ are similar to $f(\nu), g(\mu)$, and $\Phi(z)$ satisfy the boundary condition $$\begin{gathered} \operatorname{Re}[\overline{\lambda(z)}(\Phi(z) + {\Psi}(z))] = 0, \quad z\in L,\\ \operatorname{Im}[\overline{\lambda(z_0)}({\Phi}(z_0) + {\Psi}(z_0))] = 0. \end{gathered}\label{e18}$$ \end{theorem} \begin{proof} Since Problem O is equivalent to the Problem A for \eqref{e4}, from Theorem \ref{thm1} and \eqref{e3}, it is not difficult to see that the function ${\Psi}(z)$ satisfies the complex equation $$[{\Psi}]_{\bar{\tilde z}} = H\{[A_1\xi + B_1\eta + Cu + D]e_1 + [A_2\xi + B_2\eta + Cu + D]e_2\}\quad\text{in } D,\label{e19}$$ and ${\Phi}(z)=W(z)-w(z)-{\Psi}(z)$ satisfies \eqref{e11} and the boundary conditions $$\begin{gathered} \operatorname{Re}[\overline{\lambda(z)}{\Phi}(z)] = -\operatorname{Re}[\overline{\lambda(z)}{\Psi}(z)]\quad\text{on }L, \\ \operatorname{Im}[\overline{\lambda(z_0)}{\Phi}(z_0)]=-\operatorname{Im} [\overline{\lambda(z_0)}{\Psi}(z_0)]. \end{gathered} \label{e20}$$ By the representation of solutions of Problem A for \eqref{e4} as stated in the final four formulas of \eqref{e16}, we can obtain the representation of solutions of Problem O for \eqref{e1} as stated in the first formula of \eqref{e16}. \end{proof} Next, we prove the uniqueness of solutions of Problem O for equation \eqref{e1}. \begin{theorem} \label{thm3} Suppose that \eqref{e1} satisfies the Condition C. Then Problem O for \eqref{e1} in $D$ has a unique solution. \end{theorem} \begin{proof} Let $u_1(z), u_2(z)$ be two solutions of Problem O for \eqref{e1}. Then $u(z)=u_1(z)-u_2(z)$ is a solution of the homogeneous equation $$K(\hat y)u_{xx} + u_{yy} + au_x + bu_y + cu=0\quad\text{in }D, \label{e21}$$ satisfying the boundary conditions $$\begin{gathered} u(z) = 0 ;\quad\text{i.e., } \operatorname{Re}[\overline{\lambda(z)}u_{\tilde z}(z)] = 0 \quad\text{on }L, \\ u(z_0)=0,\quad \operatorname{Im}[\overline{\lambda(z_0)}u_{\tilde z}(z_0)]=0, \end{gathered}\label{e22}$$ where the function $W(z)=[H(\hat y)u_x-ju_y]/2$ is a solution of the homogeneous problem of Problem A; namely $W(z)$ satisfies the homogeneous equation and boundary conditions $$\begin{gathered} W_{\bar z}=A_1W+A_2\overline W+A_3u\quad\text{in }D, \\ u(z)=2\operatorname{Re} \int_{z_0}^z[\frac{\operatorname{Re} W}{H(\hat y)}-j\operatorname{Im} W]dz,\\ \operatorname{Re}[\overline{\lambda(z)}W(z)] = 0\quad \text{on }L, \quad \operatorname{Im}[\overline{\lambda(z_0)}W(z_0)] = 0. \end{gathered}\label{e23}$$ On the basis of Theorem \ref{thm2}, the function $W(z)$ can be expressed in the form \begin{gathered} W(z)={\Phi}(z)+{\Psi}(z),\\ \begin{aligned} {\Psi}(z) &= \int^{\mu}_{R_*} [A_1\xi + B_1\eta + Cu]e_1d\mu + \int^{\nu}_{R^*} [A_2\xi + B_2\eta + Cu]e_2d\nu\\ &= \int^{\hat y}_{y_1'} 2H(\hat y)[A_1\xi + B_1\eta + Cu]e_1dy - \int^{\hat y}_{y_1''} 2H(\hat y)[A_2\xi + B_2\eta + Cu]e_2dy \end{aligned} \end{gathered}\label{e24} in $D$, where $z_1'=x_1'+j\hat y'_1, z_1''=x_1''+j\hat y''_1$ are two intersection points of $L_1,L_2$ and two families of characteristics lines $$s_1 : \frac{{\rm d}x}{{\rm d}y}={\sqrt{|K(\hat y)|}}=H(\hat y),\quad s_2 :\frac{{\rm d}x}{{\rm d}y}= -{\sqrt{|K(\hat y)|}}=-H(\hat y)\label{e25}$$ passing through $z=x+\hat y\in\overline D$ respectively. Suppose $w(z)\not\equiv0$ in the neighborhood of the point $z_1$. We may choose a sufficiently small positive number $R_0$, such that $8M_2MR_0 < 1$, where $M_2 = \max\{C[A_1,Q_0]$, $C[B_1,Q_0],C[A_2,Q_0]$, $C[B_2,Q_0],C[C,Q_0]\}$, $M = 1 + 4k_0^2(1 + 2k_0^2)$ is a positive constant, and $M_0 = C[W(z),\overline{Q_0}]+C[u(z),\overline{Q_0}] > 0$. Herein $\|W(z)\|=\hat C[W(z),\overline{Q_0}]=C[\operatorname{Re} W(z)/H(\hat y) +j\operatorname{Im} W(z),\overline{Q_0}],$ $Q_0 = \{R_* \le \mu \le R_*+R_0\} \cap\{R^*-R_0 \le \nu \le R^*\}$. From \eqref{e14}--\eqref{e18}, \eqref{e24} and Condition C, we have $$\|{\Psi}(z)\|\le8M_2M_0R_0,\quad \|{\Phi}(z)\|\le32M_2k_0^2(1+2k_0^2)M_0R_0,$$ thus an absurd inequality $M_0\le 8M_2MM_0R_00$ on $\{00$, $00$, \$0