\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2014 (2014), No. 193, pp. 1--9.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2014 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2014/193\hfil Solutions to Neumann-Tricomi problems] {Properties of solutions to Neumann-Tricomi problems for Lavrent'ev-Bitsadze equations \\ at corner points} \author[M. A. Sadybekov, N. A. Yessirkegenov \hfil EJDE-2014/193\hfilneg] {Makhmud A. Sadybekov, Nurgissa A. Yessirkegenov} % in alphabetical order \address{Makhmud A. Sadybekov \newline Institute of Mathematics and Mathematical Modeling, 125 Pushkin str., 050010 Almaty, Kazakhstan} \email{makhmud-s@mail.ru} \address{Nurgissa A. Yessirkegenov \newline Institute of Mathematics and Mathematical Modeling, 125 Pushkin str., 050010 Almaty, Kazakhstan. \newline Al-Farabi Kazakh national university, 71 ave. Al-Farabi, 050040 Almaty, Kazakhstan} \email{nurgisa@hotmail.com} \thanks{Submitted July 9, 2014. Published September 16, 2014.} \subjclass[2000]{35M10} \keywords{Neumann-Tricomi problem; n-regular solution; \hfill\break\indent Lavrent'ev-Bitsadze equation} \begin{abstract} We consider the Neumann-Tricomi problem for the Lavrent'ev-Bitsadze equation for the case in which the elliptic part of the boundary is part of a circle. For the homogeneous equation, we introduce a new class of solutions that are not continuous at the corner points of the domain and construct nontrivial solutions in this class in closed form. For the nonhomogeneous equation, we introduce the notion of an n-regular solution and prove a criterion for the existence of such a solution. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{remark}[theorem]{Remark} \allowdisplaybreaks \section{Introduction} Let $\Omega \subset R^{2}$ be a finite domain bounded for $y<0$ by the characteristics $AC: x+y=0$ and $BC:x-y=1$ of the Lavrent'ev-Bitsadze equation \begin{equation} \operatorname{sgn}(y)u_{xx}+u_{yy}=f(x,y) \label{e1.1} \end{equation} and for $y>0$ by the circular arc $\sigma_{\delta}=\{(x,y):(x-1/2)^{2}+(y+\delta)^{2}=1/4+\delta^{2},\ y>0\}$. \subsection*{Neumann-Tricomi problem (problem N-T)} Find a solution of \eqref{e1.1} with the boundary condition \begin{gather} u|_{AC}=0, \label{e1.2} \\ \frac{\partial u}{\partial n}\big|_{\sigma_{\delta}}=0, \label{e1.3} \end{gather} where $\frac{\partial }{\partial n}=(x-1/2)\partial_{x}+(y+\delta)\partial_{y}$. We assume that the classical transmission conditions \begin{equation} u(x,+0)=u(x,-0),\quad u_{y}(x,+0)=u_{y}(x,-0), \quad 00\}$ and $\Omega_2=\Omega\cap\{y<0\}$. A solution of the problem is understood as a function in the class $C^{2}(\Omega_1)\cap C^{2}(\Omega_2)\cap C^1(\Omega)\cap C^1(\sigma_{\delta})$. We denote the angle at which the curve $\sigma_{\delta}$ approaches the line of change of type satisfies \begin{equation} \gamma_{\delta}=\operatorname{arccot} (2\delta), \quad 0<\gamma_{\delta}<\pi.\label{e2.1} \end{equation} \begin{theorem} \label{thm2.1} There are infinitely many solutions $u_{k}(x,y)\in C^{-\alpha_{k},\alpha_{k}}_{A,B}(\overline{\Omega})\cap C^1(\sigma_{\delta})$ to the homogeneous problem N-T $(f\equiv 0)$. These solutions are given by the relations \begin{equation} u_{k}(x,y)=\begin{cases} \operatorname{Re}\big(\frac{1-x+iy}{(1-x)^{2}+y^{2}}-1 \big)^{\alpha_{k}} +\operatorname{Im}\big(\frac{1-x+iy}{(1-x)^{2}+y^{2}}-1\big)^{\alpha_{k}} & \text{for }y>0, \\ \big(\frac{1}{1-x-y}-1 \big)^{\alpha_{k}}& \text{for } y<0; \end{cases} \label{e2.2} \end{equation} where \begin{equation} \alpha_{k}=\pi (1+4k)/(4\gamma_{\delta}),\quad k=0,1,\dots \label{e2.3} \end{equation} In addition, $u_{k}(x,y)\notin L_2(\Omega)$ for $k\geq 1$, and $u_{0}(x,y)\in L_2(\Omega)$ only if \begin{equation} \gamma_{\delta}>\pi/4.\label{e2.4} \end{equation} \end{theorem} \begin{theorem} \label{thm2.2} There are infinitely many solutions $\upsilon_{k}(x,y)\in C^{\alpha_{k},-\alpha_{k}}_{A,B}(\overline{\Omega})\cap C^1(\sigma_{\delta})$ to the homogeneous problem N-T* $(g\equiv 0)$, where $\alpha_{k}$ is given by relation \eqref{e2.3}. These solutions are given by the formulas \begin{equation} \upsilon_{k}(x,y)=\begin{cases} \operatorname{Re}\big(\frac{x+iy}{x^{2}+y^{2}}-1\big)^{\alpha_{k}} +\operatorname{Im}\big(\frac{x+iy}{x^{2}+y^{2}}-1\big)^{\alpha_{k}} &\text{for } y>0,\\ \big(\frac{1}{x-y}-1 \big)^{\alpha_{k}}& \text{for } y<0; \end{cases} \label{e2.5} \end{equation} in addition, $\upsilon_{k}(x,y)\notin L_2(\Omega)$ for $k\geq 1$, and $\upsilon_{0}(x,y)\in L_2(\Omega)$ only under condition \eqref{e2.4}. \end{theorem} \begin{proof}[Proof of Theorem \ref{thm2.1}] \textbf{1.} We denote $$ w(x,y)=\Big(\frac{1-x+iy}{(1-x)^2+y^2}-1\Big) ^{\alpha _k}, $$ then $u(x,y)=\operatorname{Re}w(x,y)+\operatorname{Im}w(x,y)$, $y>0$. For $y>0$ and $f=0$ equation \eqref{e1.1} can be written as $$ u_{xx}+u_{yy}=0. $$ By a direct calculation, we have \begin{align*} w_{xx}&=\alpha _k(\alpha _k-1)\Big(\frac{1-x+iy}{(1-x)^2+y^2}-1\Big) ^{\alpha_k-2}\\ &\quad \times \frac{(1-x)^4+4(1-x)^3yi-6(1-x)^2y^2-4(1-x)y^3i+y^4} {\left((1-x)^2+y^2\right)^4} \\ &\quad +\alpha _k\Big(\frac{1-x+iy}{(1-x)^2+y^2}-1\Big) ^{\alpha _k-1} \frac{2(1-x)^3+6(1-x)^2yi-6(1-x)y^2-2y^3i}{\left((1-x)^2+y^2\right)^3}, \end{align*} \begin{align*} w_{yy}&=\alpha _k(\alpha _k-1)\Big(\frac{1-x+iy}{(1-x)^2+y^2}-1\Big)^{\alpha _k-2}\\ &\quad \times \frac{-(1-x)^4-4(1-x)^3yi+6(1-x)^2y^2+4(1-x)y^3i-y^4}{\left((1-x)^2+y^2\right)^4} \\ &\quad +\alpha _k\Big(\frac{1-x+iy}{(1-x)^2+y^2}-1\Big) ^{\alpha-1} \frac{-2(1-x)^3-6(1-x)^2yi+6(1-x)y^2+2y^3i}{\left((1-x)^2+y^2\right)^3}. \end{align*} Thus, $w_{xx}+w_{yy}=0$, since $(1-x)^2+y^2\neq0$ in $\Omega_1$, hence, $$ \operatorname{Re}(w_{xx}+w_{yy})+\operatorname{Im}(w_{xx}+w_{yy})=0 \Rightarrow u_{xx}+u_{yy}=0. $$ For $y<0$ and $f=0$ equation \eqref{e1.1} can be written as $$ u_{xx}-u_{yy}=0. $$ By a direct calculation, from \eqref{e2.2}, for $y<0$ we have \begin{align*} u_{xx}(x,y)&=\alpha _k(\alpha _k-1)\Big(\frac{1}{1-x-y}-1\Big)^{\alpha _k-2}\frac{1}{(1-x-y)^4}\\ &\quad +\alpha _k\Big(\frac{1}{1-x-y}-1\Big)^{\alpha _k-1}\frac{2}{(1-x-y)^3}, \end{align*} \begin{align*} u_{yy}(x,y)&=\alpha _k(\alpha _k-1)\Big(\frac{1}{1-x-y}-1\Big)^{\alpha _k-2}\frac{1}{(1-x-y)^4} \\ &\quad +\alpha _k\Big(\frac{1}{1-x-y}-1\Big)^{\alpha_k-1}\frac{2}{(1-x-y)^3}. \end{align*} Thus, $u_{xx}-u_{yy}=0$, since $x+y\neq1$ in $\Omega_2$. The function in \eqref{e2.2} satisfies equation \eqref{e1.1} for both $y>0$ and $y<0$. \textbf{2.} By \eqref{e2.2}, for $y<0$, $$ u(x,y)=\Big(\frac{1}{1-x-y}-1\Big)^{\alpha_k}=\Big(\frac{x+y}{1-x-y}\Big)^{\alpha _k}, $$ we have $$ u|_{AC}=u|_{x+y=0}= \Big(\frac{x+y}{1-x-y}\Big)^{\alpha _k}\Big|_{x+y=0}=0, $$ since $\alpha_k>0$. Thus, function in \eqref{e2.2} satisfies the boundary condition \eqref{e1.2} in the hyperbolic part of the domain. The contour $\sigma _{\delta}$ has the form $$ 2y\delta =(1-x)-(1-x)^2-y^2. $$ Thus, boundary condition \eqref{e1.3} can be written as $$ \frac{\partial u}{\partial n}|_{(1-x)-(1-x)^2-y^2=2y\delta}=0. $$ By definition \eqref{e2.1} for the number $\gamma _{\delta}$, we have $$ \frac{\partial u}{\partial n}|_{\sigma_{\delta}} =\Big(\frac{\alpha_{k} y^{\alpha_{k}-1}}{2(\sin\gamma _{\delta})^{\alpha _{k}}((1-x)^{2}+y^{2})^{\alpha _{k}}}\Big)(\cos(\alpha_{k} \gamma _{\delta})-\sin(\alpha_{k} \gamma _{\delta})). $$ By the definition of $\alpha_{k}$ in \eqref{e2.3}, we obtain $$ \alpha _k=\frac{\frac{\pi}{4}+\pi k}{\gamma_{\delta}}\;\Rightarrow \; \cos(\alpha_{k} \gamma _{\delta})-\sin(\alpha_{k} \gamma_{\delta})=0. $$ Thus, $$ \frac{\partial u}{\partial n}\big|_{\sigma _{\delta}}=0. $$ The function in \eqref{e2.2} satisfies the boundary condition \eqref{e1.3}. \textbf{3.} To check conditions \eqref{e1.4}, from the representation of \eqref{e2.2}, we obtain \begin{gather*} u(x,-0)=u(x,+0)=\Big(\frac{1}{1-x}-1\Big)^{\alpha _k}, \\ u_y(x,-0)=u_y(x,+0)=\alpha _k\Big(\frac{1}{1-x}-1\Big)^{\alpha _k-1} \frac{1}{(1-x)^2}, \end{gather*} and conditions \eqref{e1.4} are satisfied. As a result, function in \eqref{e2.2} is solution of the homogeneous Problem N-T. It is easy to see that function in \eqref{e2.2} belongs to the class $C^{-\alpha_{k},\alpha_{k}}_{A,B}(\overline{\Omega})\cap C^1(\sigma_{\delta})$. Next, we prove the final statement of theorem \ref{thm2.1}. \begin{equation} \|u_k\|^2_{L_2(\Omega)}=\iint_{\Omega}|u_k(x,y)|^2\,dx\,dy<\infty . \label{e2.6} \end{equation} Note that \begin{gather*} \iint_{\Omega}|u_k(x,y)|^2\,dx\,dy= \iint_{\Omega _2}|u_k(x,y)|^2\,dx\,dy+ \iint_{\Omega _1}|u_k(x,y)|^2\,dx\,dy, \\ \begin{aligned} \iint_{\Omega _2}|u_k(x,y)|^2\,dx\,dy &=\iint_{\Omega _2}\Big|\Big(\frac{x+y}{1-(x+y)}\Big)^{2\alpha _k}\Big|\,dx\,dy\\ &< \iint_{\Omega_2}\big(1-(x+y)\big)^{-2\alpha _k}\,dx\,dy. \end{aligned} \end{gather*} Using the change of variables $x+y=\xi$, $x-y=\eta$, we obtain \begin{align*} \iint_{\Omega_2}|u_k(x,y)|^2\,dx\,dy &<\frac{1}{2}\iint_{\Omega_2}(1-\xi)^{-2\alpha _k}d\xi d\eta =\frac{1}{2}\int_0^1d\eta\int_0^{\eta}(1-\xi)^{-2\alpha_k}d\xi \\ &=\frac{1}{2(2\alpha _k-1)}\int_0^1 \big((1-\eta)^{1-2\alpha _k}-1\big)d\eta <\infty \end{align*} for $1-2\alpha _k>-1$, $\Leftrightarrow \alpha _k<1$. By using the change of variables $x=\frac{r^2+r\cos\varphi}{1+2r\cos\varphi+r^2}$, $y=\frac{r\sin\varphi}{1+2r\cos\varphi+r^2}$ in $\Omega_1$, we obtain \begin{align*} \iint_{\Omega _1}|u_k(x,y)|^2\,dx\,dy &=\iint_{\Omega _1} \frac{r^{2\alpha _k}(1+\sin 2\alpha _k\varphi) r} {(1+2r\cos\varphi+r^2)^2}\,dr\,d\varphi \\ &=\int_0^{\gamma _\delta}(1+\sin 2\alpha _k\varphi)d\varphi \int_0^{\infty}\frac{r^{2\alpha _k+1}dr} {(1+2r\cos\varphi+r^2)^2}<\infty \end{align*} for $2\alpha _k+1+1-4<0\;\Leftrightarrow \;\alpha _k<1$. Thus, ratio \eqref{e2.6} is satisfied for $\alpha _k<1$. Taking into account the definition of $\alpha _k$ in \eqref{e2.3}, it is easy to see that ratio \eqref{e2.6} is satisfied only for $k=0$, and $$ \frac{\pi}{4\gamma _{\delta}}<1\;\Leftrightarrow \;\gamma _{\delta}>\frac{\pi}{4}. $$ \end{proof} Theorem \ref{thm2.2} can be proved in a similar way; so we omit its proof. Let us proceed to the analysis of the nonhomogeneous problem N-T and N-T*. An \emph{n-regular solution} of Problem N-T (N-T*) is defined as a solution, \begin{gather*} u(x,y)\in C^{2}(\Omega_1)\cap C^{2}(\Omega_2)\cap C^1(\Omega)\cap C^1(\sigma_{\delta})\cap C^{-n,0}_{A,B}(\overline{\Omega})\\ (\upsilon(x,y)\in C^{2}(\Omega_1)\cap C^{2}(\Omega_2)\cap C^1(\Omega)\cap C^1(\sigma_{\delta})\cap C^{0,-n}_{A,B}(\overline{\Omega})). \end{gather*} The following theorems hold for the nonhomogeneous Problems N-T and N-T*. \begin{theorem} \label{thm2.3} The solution of Problem N-T is n-regular for any right-hand side $f(x,y)\in C(\overline{\Omega})$ if and only if \begin{equation} \gamma_{\delta}<\pi/(4n),\quad n=1,2,\dots \label{e2.7} \end{equation} The solution is n-regular for arbitrary approach angles $\gamma_{\delta}$ only if the right-hand side of \eqref{e1.1} satisfies the conditions \begin{equation} \iint_\Omega \upsilon_{k}(x,y)f(x,y)\,dx\,dy=0,\quad k=0,\dots ,j_{0}, \label{e2.8} \end{equation} where $\upsilon_{k}$ are the functions given by \eqref{e2.5}, $j_{0}=[n\gamma_{\delta}/\pi - 1/4]$, and $[z]$ is the integer part of $z$. In this case, the number of conditions \eqref{e2.8} depends on the angle $\gamma_{\delta}$, and their maximum number is equal to n (as $\gamma_{\delta}\rightarrow \pi$). \end{theorem} \begin{theorem} \label{thm2.4} Condition \eqref{e2.7} is necessary and sufficient for the n-regularity of the solution of Problem N-T* for any right-hand side $g(x,y)\in C(\overline{\Omega})$; for arbitrary approach angles $\gamma_{\delta}$, the right-hand side of \eqref{e1.5} satisfies the relations \begin{equation} \iint _\Omega u_{k}(x,y)g(x,y)\,dx\,dy=0,\quad k=0,\dots ,j_{0}, \label{e2.9} \end{equation} where the $u_{k}$ are the functions given by \eqref{e2.2} and $j_{0}=[n\gamma_{\delta}/\pi-1/4]$. In this case, the number of conditions \eqref{e2.9} depends on the angle $\gamma _{\delta}$, and their maximum number is equal to n (as $\gamma_{\delta}\rightarrow \pi$). \end{theorem} \begin{remark} \rm Conditions \eqref{e2.8} and \eqref{e2.9} with $k\geq 1$ are not orthogonality conditions in $L_2(\Omega)$, and for $k\geq0$ they are orthogonality conditions only if inequality \eqref{e2.4} holds. This immediately follows from theorems \ref{thm2.1} and \ref{thm2.2}. \end{remark} \begin{proof}[Proof of Theorem \ref{thm2.3}] Set $u(x,y)=\tau (x)$ and $u_{y}(x,0)=\nu (x)$. In the hyperbolic part of the domain $\Omega_2$, we consider the Cauchy-Goursat problem $$ -u_{xx}+u_{yy}=f(x,y),\quad u|_{AC}=0,\quad u_y(x,0)=\nu (x). $$ The solution of this problem has the form \cite[p.121]{sm1}: $$ u(x,y)=\int_0^{x+y}\nu (t)dt-\frac{1}{2} \int_0^{x+y}d\xi _1\int_{\xi _1}^{x-y} f\big( \frac{\xi _1+\eta _1}{2}, \frac{\xi _1-\eta _1}{2}\big)\, d\eta _1. $$ Then we obtain the main relation $$ \tau(x)=\int_{0}^{x}\nu(t)dt -\frac{1}{2}\int _{0}^{x} d\xi_1 \int_{\xi_1}^{x}f \big(\frac{\xi_1+\eta_1}{2}, \frac{\xi_1-\eta_1}{2}\big)d\eta_1,\quad 01, \end{cases} \] from \eqref{e2.10}, we obtain the relation \begin{equation} \overline{\tau}(s)=-\frac{1}{s}\overline{\nu}(s)+\overline{F_1}(s).\label{e2.11} \end{equation} Here \begin{gather} \overline{\tau}(s)=\int_0^{\infty} x^{s-1}\tau \big(\frac{x}{1+x}\big)dx,\quad \overline{\nu}(s)=\int _0^{\infty} x^{s-1}x\frac{\nu(x/(1+x))}{(1+x)^{2}}dx, \label{e2.12} \\ \overline{F_1}(s)=\int_0^{\infty}x^{s-1}F_1\big(\frac{x}{1+x}\big)dx. \label{e2.13} \end{gather} In the elliptic part $\Omega_1$, we consider the problem $$ u_{xx}+u_{yy}=f(x,y),\quad \frac{\partial u}{\partial n}\big|_{\sigma_{\delta}}=0,\quad u(x,0)=\tau (x). $$ By making the change of variables \begin{equation} x=\frac{r^{2}+r \cos \varphi}{1+2r \cos \varphi +r^{2}},\quad y=\frac{r \sin \varphi}{1+2r \cos \varphi +r^{2}}, \label{e2.14} \end{equation} by using the Mellin transform, and by solving the resulting problem, we obtain \begin{equation} \overline{\nu}(s)=\tan(s\gamma_{\delta})s\overline{\tau}(s)-\int _0^{\gamma_{\delta}}\overline{u_2}(s,t)\overline{f}(s,t)dt, \label{e2.15} \end{equation} where \begin{gather} \overline{f}(s,\varphi) =\int_0^{\infty}r^{s-1}\frac{r^{2}}{(1+2r \cos \varphi +r^{2})^{2}}f\big(\frac{r^{2}+r \cos \varphi}{1+2r \cos \varphi+r^{2}}, \frac{r \sin \varphi}{1+2r\ cos \varphi+r^{2}}\big)dr, \label{e2.16} \\ \overline{u_2}(s,\varphi)=\cos s \varphi+\sin s \varphi\frac{\sin s\gamma_{\delta}}{\cos s\gamma_{\delta}}, \label{e2.17} \end{gather} and the functions $\overline{\tau}(s)$ and $\overline{\nu}(s)$ are defined in \eqref{e2.12}. Now from relations \eqref{e2.11} and \eqref{e2.15}, we obtain \begin{gather} \overline{\nu}(s)= \Big[\tan (s\gamma_{\delta})s\overline{F_1}(s)-\int _0^{\gamma_{\delta}}\overline{u_2}(s,t)\overline{f}(s,t)dt\Big] [1+\tan (s\gamma_{\delta})]^{-1},\label{e2.18} \\ \overline{\tau}(s)= \Big[s\overline{F_1}(s)+\int _0^{\gamma_{\delta}}\overline{u_2}(s,t)\overline{f}(s,t)dt\Big] [s(1+\tan (s\gamma_{\delta}))]^{-1}. \label{e2.19} \end{gather} First, let us analyze definitions \eqref{e2.12} of the functions $\overline{\tau}(s)$ and $\overline{\nu}(s)$ and their relationship with the original functions $\tau(x)$ and $\nu(x)$. By making the obvious change of variables $t=x/(1+x)$, we reduce relation \eqref{e2.12} to the form $$ \overline{\tau}(s) =\int _0^1 t^{s-1}(1-t)^{-s-1}\tau(t)dt,\quad \overline{\nu}(s)=\int _0^1t^{s}(1-t)^{-s}\nu(t)dt. $$ Hence, it follows that if the function $\overline{\tau}(s)$ is continuous on the interval $(-1, 0)$, then the function $\tau(t)$ is continuous at the point $t=0$ and has a zero of order $\geq1$ there. As a result, by taking into account the definitions of the functions $\tau(x)$ and $\nu(x)$, for the $n$-regularity of the solution, the right-hand sides in relations \eqref{e2.18} and \eqref{e2.19} should be continuous for $-n