\documentclass[reqno]{amsart} \usepackage{hyperref} \usepackage{graphicx} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2012 (2012), No. 202, pp. 1--7.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2012 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2012/202\hfil Existence and uniqueness] {Existence and uniqueness for boundary-value problem with additional single point conditions of the Stokes-Bitsadze system} \author[M. Tahir \hfil EJDE-2012/202\hfilneg] {Muhammad Tahir} \address{Muhammad Tahir \newline Mathematics Department\\ HITEC University, Taxila Cantonment, Pakistan} \email{mtahir@hitecuni.edu.pk} \thanks{Submitted July 24, 2012. Published November 15, 2012.} \subjclass[2000]{35J57} \keywords{Bitsadze system; boundary value problem; single point conditions} \begin{abstract} This article shows the uniqueness of a solution to a Bitsadze system of equations, with a boundary-value problem that has four additional single point conditions. It also shows how to construct the solution. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \allowdisplaybreaks \section{Introduction} The planar Stokes flow based on stream function $\psi(x,y)$ and stress function $\phi(x,y)$, is expressed as $$\label{e1} \begin{gathered} \phi_{xx}-\phi_{yy}=-4\eta\psi_{xy},\\ -\phi_{xy}=\eta(\psi_{yy}-\psi_{xx}), \end{gathered}$$ where $\eta$ is a material constant, see for the details \cite{c1,d1,o2}. The re-scaling (2$\eta\psi\to \psi$) reduces the system \eqref{e1} to $$\label{e2} \begin{gathered} \phi_{xx}-\phi_{yy}+2\psi_{xy}=0,\\ \psi_{xx}-\psi_{yy}-2\phi_{xy}=0, \end{gathered}$$ which is the famous second order elliptic system called the Bitsadze system of equations and is identified as Stokes-Bitsadze system \cite{t1}. In the literature Bitsadze appears to have been the first to question the uniqueness and existence or even the well-posedness of \eqref{e2} subject to certain boundary conditions, see for reference \cite{b2,b3,k1}. Oshorov \cite{o1} finds well-posed problems for the Cauchy-Riemann system and extends those to the Bitsadze system \eqref{e2}. Vaitekhovich \cite{v1} discusses Dirichlet and Schwarz problems for the inhomogeneous Bitsadze equation for a circular ring domain. In the interior of unit disc a boundary value problem for the Bitsadze equation is considered by Babayan \cite{b1} and is proved to be Noetherian. In his paper Babayan also proposes solvability conditions for the inhomogeneous Bitsadze equation. The unique solvability in a unit disc for the inhomogeneous Bitsadze system is discussed in \cite{h1}. The Stokes-Bitsadze system \eqref{e2} can be expressed in the matrix form as $$\label{e3} A\mathbf{U}_{xx}+2B\mathbf{U}_{xy}+C\mathbf{U}_{yy}=\mathbf{0},$$ where $A = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix},\quad B = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix},\quad C=-A, \quad \mathbf{U}(x,y)= \begin{pmatrix} \phi \\ \psi\end{pmatrix}.$ In a domain $\Omega\subset\mathbb{R}^2$ with boundary $\Gamma$ a linear boundary value problem of Poincar\'{e} for the system \eqref{e3} can be formulated as $$\label{e4} p_1\mathbf{U}_{x}+p_2\mathbf{U}_{y}+q\mathbf{U}=\boldsymbol{\alpha}(x,y),\quad (x,y)\in \Gamma$$ where $p_1,p_2, q$ are real 2$\times$2 matrices and $\boldsymbol{\alpha}(x,y)$ a real vector given on the boundary $\Gamma$. The boundary-value problems of Poincar\'e for the Stokes-Bitsadze system will be discussed elsewhere. In this paper we are interested in a boundary value problem with four additional single point conditions. \section{A boundary value problem with additional single point conditions} We consider the Stokes-Bitsadze system \eqref{e2} in domain $\Omega\subset\mathbb{R}^2$ with boundary $\Gamma$ subject to the following boundary conditions. $$\label{e5} \psi=f, \quad \psi_{n}=g \quad \text{on } \Gamma,$$ and $$\label{e6} \phi=\phi^P,\quad \nabla\phi=(\nabla\phi)^P,\quad \Delta\phi=(\Delta\phi)^P, \quad \text{at a single point P \in \bar{\Omega}}.$$ \begin{theorem} \label{thm2.1} For $f,g\in C(\Gamma)$, the boundary value problem \eqref{e5}--\eqref{e6} for the Stokes-Bitsadze system \eqref{e2} has a unique solution $(\phi,\psi)\in C^4(\Omega)\times C^4(\Omega)$. \end{theorem} \begin{proof} Suppose $\phi,\psi\in C^4(\Omega)$. If $(\phi,\psi)$ satisfies \eqref{e2}, then $\phi$ and $\psi$ are biharmonic in $\Omega$, and for $f,g\in C(\Gamma)$ the problem $$\label{e7} \begin{gathered} \Delta^2 \psi=0 \quad \text{in } \Omega\\ \psi=f \quad \text{on } \Gamma \\ \psi_n=g \quad \text{on } \Gamma \end{gathered}$$ has a unique solution $\psi\in C^4(\Omega)$, \cite{t2}, that satisfies \eqref{e2} and \eqref{e5}. Let the unique solution be denoted by $\widetilde{\psi}$. Now we show that for the unique $\widetilde{\psi}$ if there exists $\phi$ satisfying \eqref{e2} and \eqref{e5}--\eqref{e6} then that $\phi$ is unique. Assume that the pairs $(\phi_1,\widetilde{\psi})$ and $(\phi_2,\widetilde{\psi})$ with $\phi_1\neq \phi_2$ satisfy \eqref{e2} and \eqref{e5}--\eqref{e6} and that $\delta = \phi_1-\phi_2$. Then from \eqref{e2} it immediately follows that $$\label{e8} \delta_{xx}-\delta_{yy}=0, \quad \delta_{xy}=0\quad \text{on }\Omega.$$ But \eqref{e6} then yields $$\label{e9} \delta=0,\quad \nabla\delta=0,\quad \Delta\delta=0 \quad\text{at }P,$$ and the general solution of the system \eqref{e8} becomes, $$\label{e10} \delta=ax+by+c(x^{2}+y^{2})+d,$$ which on imposing the conditions \eqref{e9} gives $\delta\equiv 0$ in $\bar{\Omega}$ and uniqueness of $\phi$ thus follows. Hence there exists at most one pair $(\phi,\psi)\in C^4(\Omega)\times C^4(\Omega)$ that can satisfy \eqref{e2} and \eqref{e5}--\eqref{e6}. We are now in a position to assume (without proof) that $(\widetilde{\phi},\widetilde{\psi})$ is a solution of \eqref{e2} and \eqref{e5}--\eqref{e6}. Next, we suppose that $P(x_{P},y_{P})$ and $Q(x,y_{P})$ are the points in $\bar{\Omega}$, refer to the Figure \ref{fig1}. \begin{figure}[hbtp] \begin{center} \includegraphics[width=0.7\textwidth]{fig1} %\includegraphics[scale=0.5]{img} \end{center} \caption{Boundary conditions and additional single point conditions} \label{fig1} \end{figure} At point $P$ the expressions \eqref{e2}(a) and \eqref{e6}(c) respectively take the form $$\label{e11} \begin{gathered} \phi_{xx}^P-\phi_{yy}^P=-2\psi_{xy}^P,\\ \phi_{xx}^P+\phi_{yy}^P=\Delta\phi^P, \end{gathered}$$ from which it is obvious that $\phi_{xx}^P$ and $\phi_{yy}^P$ are known at $P$. Since $(\widetilde{\phi},\widetilde{\psi})$ satisfies \eqref{e2}(b), therefore $$\label{e12} \widetilde{\phi}_{xyy}=\frac{1}{2}[\widetilde{\psi}_{xxy}-\widetilde{\psi}_{yyy}],\/extract_tex] and on integration along PQ we have \begin{gather} \label{e13} \widetilde{\phi}_{yy}(x,y_{P})=\phi _{yy}^P+\frac{1}{2}\int_{x_P}^x [\widetilde{\psi}_{xxy}(\lambda,y_P)-\widetilde{\psi}_{yyy}(\lambda,y_P)]d\lambda,\\ \label{e14} \widetilde{\phi}_y(x,y_P)=\phi_{y}^P+\frac{1}{2}\int_{x_P}^x [\widetilde{\psi}_{xx}(\lambda,y_P)-\widetilde{\psi}_{yy}(\lambda,y_P)]d\lambda. \end{gather} Since all the terms on right hand sides of \eqref{e13} and \eqref{e14} are known therefore \widetilde{\phi}_{yy} and \widetilde{\phi}_{y} are known along PQ. Since (\widetilde{\phi},\widetilde{\psi}) satisfies \eqref{e2}(a), we have $$\label{e15} \widetilde{\phi}_{xx}=\widetilde{\phi}_{yy}-2\widetilde{\psi}_{xy},$$ and using \eqref{e13}, can further be expressed as $$\label{e16} \widetilde{\phi}_{xx}(x,y_P) =\phi_{yy}^P +\frac{1}{2}\int_{x_P}^x[\widetilde{\psi}_{xxy}(\lambda,y_P) -\widetilde{\psi}_{yyy}(\lambda,y_P)]d\lambda-2 \widetilde{\psi}_{xy}(\lambda,y_P).$$ Further on integration along PQ, we have $$\label{e17} \begin{split} \widetilde{\phi}_{x}(x,y_P) &=\phi_{x}^P+ \int_{x_P}^x\bigg[\phi_{yy}^P +\frac{1}{2}\int_{x_P}^\mu [ \widetilde{\psi}_{xxy}(\lambda,y_P) -\widetilde{\psi}_{yyy}(\lambda,y_P)]\bigg ]\,d\lambda \,d\mu \\ &\quad - 2\int _{x_P}^x\widetilde{\psi}_{xy}(\lambda,y_P)\,d\lambda, \end{split}$$ whence $$\label{e18} \begin{split} &\widetilde{\phi}(x,y_P)\\ &=\phi^P+(x-x_P)\phi_x^ P +\frac{1}{2}(x-x_P)^2\phi_{yy}^P - 2 \int_{x_P}^x \int_{x_P}^\mu \widetilde{\psi}_{xy}(\lambda,y_P)d\lambda\,\,d\mu\\ &\quad +\frac{1}{2}\int_{x_P}^x \int_{x_P}^\nu\int_{x_P}^\mu \big [\widetilde{\psi}_{xxy}(\lambda,y_P)-\widetilde{\psi}_{yyy}(\lambda,y_P) \big ] \,d\lambda\,\,d\mu\,\,d\nu. \end{split}$$ Since all the terms on right hand sides of \eqref{e15}, \eqref{e16}, \eqref{e17} are known therefore \widetilde{\phi}_{xx}, \widetilde{\phi}_x  and \widetilde{\phi} are known along PQ and hence we know \widetilde{\phi}, \nabla\widetilde{\phi} and \Delta\widetilde{\phi} at Q(x,y_{P}). Now from the point Q we draw the line QR where R(x,y)\in\bar{\Omega} is an arbitrary point. Again, since (\widetilde{\phi},\widetilde{\psi}) satisfies \eqref{e2}(b); therefore $$\label{e19} \widetilde{\phi}_{xxy}=\frac{1}{2}[\widetilde{\psi}_{xxx}-\widetilde{\psi}_{xyy}],$$ which on integration, along QR, gives \begin{gather} \label{e20} \widetilde{\phi}_{xx}(x,y) =\widetilde{\phi}_{xx}(x,y_P)+\frac{1}{2}\int_{y_P}^y [\widetilde{\psi}_{xxx}(x,\lambda)-\widetilde{\psi}_{xyy}(x,\lambda)]d\lambda,\\ \label{e21} \widetilde{\phi}_{x}(x,y)=\widetilde{\phi}_{x}(x,y_P) +\frac{1}{2}\int_{y_P}^y[\widetilde{\psi}_{xx}(x,\lambda) -\widetilde{\psi}_{yy}(x,\lambda)]d\lambda. \end{gather} But the following expression from \eqref{e2}(a) $$\label{e22} \widetilde{\phi}_{yy}=\widetilde{\phi}_{xx}+2\widetilde{\psi}_{xy},$$ on integration along QR gives $$\label{e23} \widetilde{\phi}_y(x,y)=\widetilde{\phi}_{y}(x,y_P) +\int_{y_P}^y[\widetilde{\phi}_{xx}(x,\lambda) +2\widetilde{\psi}_{xy}(x,\lambda)]\,d\lambda.$$ Using \eqref{e14} and \eqref{e20} the expression \eqref{e23} takes the form $$\label{e24} \begin{split} \widetilde{\phi}_y(x,y) &=\phi_y^P+\frac{1}{2}\int_{x_P}^x[\widetilde{\psi}_{xx}(\lambda,y_P) -\widetilde{\psi}_{yy}(\lambda,y_P)]d\lambda+(y-y_P)\widetilde{\phi}_{xx}(x,y_P)\\ &\quad +\frac{1}{2}\int_{y_P}^y\int_{y_P}^\mu[\widetilde{\psi}_{xxx}(x,\lambda) -\widetilde{\psi}_{xyy}(x,\lambda)]d\lambda\,\,d\mu +2 \int_{y_P}^y \widetilde{\psi}_{xy}(x,\lambda)d\lambda. \end{split}$$ Integrating along QR we obtain from \eqref{e24} as follows. $$\label{e25} \begin{split} \widetilde{\phi}(x,y) &= \widetilde{\phi}(x,y_P)+(y-y_P)\phi _y^P + \frac{1}{2}(y-y_P)^2 \widetilde{\phi}_{xx}(x,y_P)\\ &\quad +\frac{1}{2}(y-y_P)\int_{x_P}^x[\widetilde{\psi}_{xx}(\lambda,y_P) -\widetilde{\psi}_{yy}(\lambda,y_P)]d\lambda\\ &\quad +\frac{1}{2}\int_{y_P}^y \int_{y_P}^\nu \int _{y_P}^\mu [\widetilde{\psi}_{xxx}(x,\lambda)-\widetilde{\psi}_{xyy}(x,\lambda)] d\lambda \,d\mu d\nu\\ &\quad + 2\int_{y_P}^y \int_{y_P}^\mu\widetilde{\psi}_{xy}(x,\lambda) d\lambda \,d\mu. \end{split}$$ Using \eqref{e16} and \eqref{e18} we finally obtain the following expression for \widetilde{\phi}(x,y) at an arbitrary point (x,y)\in \bar{\Omega}. $$\label{e26} \begin{split} &\widetilde{\phi}(x,y)\\ &=\phi^P+(x-x_P)\phi_x^P+(y-y_P)\phi_y^P +\frac{1}{2}[(x-x_P)^2+(y-y_P)^2]\phi_{yy}^P\\ &\quad -(y-y_P)^2\widetilde{\psi}_{xy}(x,y_P) +\frac{1}{2}(y-y_P)\int_{x_P}^x[\widetilde{\psi}_{xx}(\lambda,y_P) -\widetilde{\psi}_{yy}(\lambda,y_P)]\,d\lambda\\ &\quad + \frac{1}{4}(y-y_P)^2 \int_{x_P}^x[\widetilde{\psi}_{xxy}(\lambda,y_P) -\widetilde{\psi}_{yyy}(\lambda,y_P)]d\lambda \\ &\quad -2\int_{x_P}^x\int_{x_P}^\mu \widetilde{\psi}_{xy}(\lambda,y_P)\,d\lambda \,d\mu +2\int_{y_P} ^y \int_{y_P}^\mu \widetilde{\psi}_{xy}(x,\lambda)d\lambda \,d\mu \\ &\quad +\frac{1}{2}\int_{x_P}^x\int_{x_P}^\nu \int_{x_P}^\mu [\widetilde{\psi}_{xxy}(\lambda,y_P)-\widetilde{\psi}_{yyy}(\lambda,y_P)] d\lambda\,\,d\mu\,\,d\nu\\ &\quad +\frac{1}{2}\int_{y_P}^y\int_{y_P}^\nu\int_{y_P}^\mu [\widetilde{\psi}_{xxx}(x,\lambda)-\widetilde{\psi}_{xyy}(x,\lambda)] d\lambda\,\,d\mu\,d\nu. \end{split}$$ Obviously we have obtained an explicit representation for \widetilde{\phi} in terms of the point conditions and \widetilde{\psi}, on the assumption that (\widetilde{\phi},\widetilde{\psi}) satisfies \eqref{e2} and \eqref{e5}--\eqref{e6}. Next we show that (\widetilde{\phi},\widetilde{\psi}) actually satisfies the Bitsadze system \eqref{e2} and the conditions \eqref{e6}. From expression \eqref{e26} it is easy to verify that \widetilde{\phi}(x_P,y_P )=\phi^P. We use \eqref{e17} in \eqref{e21} to obtain \begin{align*} \widetilde{\phi}_x(x,y) &=\phi_x ^P+\int_{x_P}^x[\phi_{yy}^P+\frac{1}{2} \int_{x_P}^\mu [ \widetilde{\psi}_{xxy}(\lambda,y_P) -\widetilde{\psi}_{yyy}(\lambda,y_P)] d\lambda] \,d\mu \\ &\quad - 2\int_{x_P}^x\widetilde{\psi}_{xy}(\lambda,y_P)\,d\lambda +\frac{1}{2}\int _{y_P}^y[\widetilde{\psi}_{xx}(x,\lambda) -\widetilde{\psi}_{yy}(x,\lambda)]d\lambda, \end{align*} and it can be easily verified that \widetilde{\phi}_x(x_P,y_P)=\phi_x^P. Similarly from \eqref{e14} and \eqref{e24} we have \[ \widetilde{\phi}_y(x,y)=\phi_y ^P+\frac{1}{2}\int_{x_P}^x [\widetilde{\psi}_{xx}(\lambda,y_P)-\widetilde{\psi}_{yy}(\lambda,y_P)] \,d\lambda+\int_{y_P}^y[\widetilde{\phi}_{xx}(x,\lambda) +2\widetilde{\psi}_{xy}(x,\lambda)]\,d\lambda, and it follows that $\widetilde{\phi}_y(x_P,y_P)=\phi_y ^P$. Again, from \eqref{e16}and \eqref{e20} we obtain \begin{align*} \widetilde{\phi}_{xx}(x,y) &=\phi_{yy}^P+\frac{1}{2}\int_{x_P}^x[\widetilde{\psi}_{xxy}(\lambda,y_P) -\widetilde{\psi}_{yyy}(\lambda,y_P)]\,d\lambda -2\widetilde{\psi}_{xy}(x,y_P)\\ &\quad +\frac{1}{2}\int_{y_P}^y[\widetilde{\psi}_{xxx}(x,\lambda) -\widetilde{\psi}_{xyy}(x,\lambda)]\,d\lambda, \end{align*} which at $P$ yields $$\label{e27} \widetilde{\phi}_{xx}(x_P,y_P)=\phi_{yy}^P-2\widetilde{\psi}_{xy}(x_P,y_P),$$ and from \eqref{e11}(a) we obtain $\widetilde{\phi}_{xx}(x_P,y_P)=\phi_{xx}^P$. Also from \eqref{e22} it is obvious that $$\label{e28} \widetilde{\phi}_{yy}(x_P,y_P) =\widetilde{\phi}_{xx}(x_P,y_P)+2\widetilde{\psi}_{xy}(x_P,y_P),$$ and \eqref{e27}--\eqref{e28} yield $\widetilde{\phi}_{yy}(x_P,y_P)={\phi}_{yy}^P$. Now we verify that $\widetilde{\phi}(x,y)$ satisfies \eqref{e2}(a). Using \eqref{e14} in \eqref{e24} and then differentiating with respect to $x$ we obtain \begin{align*} &\widetilde{\phi}_{xy}(x,y) =\frac{1}{2}[\widetilde{\psi}_{xx}(x,y_P)-\widetilde{\psi}_{yy} (x,y_P)]+\frac{1}{2}(y-y_P)[\widetilde{\psi}_{xxy}(x,y_P) -\widetilde{\psi}_{yyy}(x,y_P)]\\ &\quad -2(y-y_P)\widetilde{\psi}_{xxy}(x,y_P) + \frac{1}{2}\int_{y_p} ^y \int_{y_P}^\mu[\widetilde{\psi}_{xxxx}(x,\lambda) - \widetilde{\psi}_{xxyy}(x,\lambda)] \,d\lambda \,d\mu \\ &\quad +2\widetilde{\psi}_{xx}(x,y)-2\widetilde{\psi}_{xx}(x,y_P), \end{align*} which, since $\Delta^2\widetilde{\psi}=0$, can be simplified as $$\label{e29} \begin{split} &\widetilde{\phi}_{xy}(x,y)\\ &=- \frac{1}{2}[3\widetilde{\psi}_{xx}(x,y_P)+\widetilde{\psi}_{yy}(x,y_P)] -\frac{1}{2}(y-y_P)[3\widetilde{\psi}_{xxy}(x,y_P) +\widetilde{\psi} _{yyy}(x,y_P)]\\ &\quad - \frac{1}{2}[3\widetilde{\psi}_{xx}(x,y_P)+\widetilde{\psi}_{yy}(x,y)] + \frac{1}{2}[3\widetilde{\psi}_{xx}(x,y_P)+\widetilde{\psi}_{yy}(x,y_P)]\\ &\quad +\frac{1}{2}(y-y_P)[3\widetilde{\psi}_{xxy}(x,y_P) +\widetilde{\psi} _{yyy}(x,y_P)]+2\widetilde{\psi}_{xx}(x,y), \end{split}$$ and we obtain $$\label{e30} \begin{split} \widetilde{\phi}_{xy}(x,y)=\frac{1}{2}[\widetilde{\psi}_{xx}(x,y)-\widetilde{\psi}_{yy}(x,y)]. \end{split}$$ Then, to verify that $\widetilde{\phi}(x,y)$ satisfies \eqref{e2}(b), we use \eqref{e26} to obtain \begin{align*} &\widetilde{\phi}_{xx}(x,y)-\widetilde{\phi}_{yy}(x,y)\\ &=-(y-y_P)^2\widetilde{\psi}_{xxxy}(x,y_P) +\frac{1}{2}(y-y_P)[\widetilde{\psi}_{xxx}(x,y_P)-\widetilde{\psi}_{xyy}(x,y_P)] \\ &\quad +\frac{1}{4}(y-y_P)^2[\widetilde{\psi}_{xxxy}(x,y_P) -\widetilde{\psi}_{xyyy}(x,y_P)]\\ &\quad +2\int_{y_P}^y \int_{y_P}^\mu\widetilde{\psi}_{xxxy}(x,\lambda)d\lambda \,d\mu +\frac{1}{2}\int_{x_P}^x[\widetilde{\psi}_{xxy}(\lambda,y_P) -\widetilde{\psi}_{yyy}(\lambda,y_P)]\,d\lambda\\ &\quad +\frac{1}{2}\int_{y_P}^y\int_{y_P}^\nu\int_{y_P}^\mu [\widetilde{\psi}_{xxxxx}(x,\lambda)-\widetilde{\psi}_{xxxyy}(x,\lambda)] \,d\lambda \,d\mu \,d\nu \\ &\quad -\frac{1}{2}\int_{x_P}^x[\widetilde{\psi}_{xxy}(\lambda,y_P) -\widetilde{\psi}_{yyy}(\lambda,y_P)]d\lambda-2\widetilde{\psi}_{xy}(x,y)\\ &\quad -\frac{1}{2}\int_{y_P}^y[\widetilde{\psi}_{xxx}(x,\lambda) -\widetilde{\psi}_{xyy}(x,\lambda)]\,d\lambda, \end{align*} which can further be simplified to obtain \begin{align*} &\widetilde{\phi}_{xx}(x,y)-\widetilde{\phi}_{yy}(x,y)\\ &=-\frac{1}{4}(y-y_P)^2[3\widetilde{\psi}_{xxxy}(x,y_P) +\widetilde{\psi}_{xyyy}(x,y_P)]\\ &\quad -\frac{1}{2}(y-y_P)[3\widetilde{\psi}_{xxx}(x,y_P) +\widetilde{\psi}_{xyy}(x,y_P)]\\ &\quad -\frac{1}{2}\int_{y_P}^y[3\widetilde{\psi}_{xxx}(x,\lambda) +\widetilde{\psi}_{xyy}(x,\lambda)]\,d\lambda +\frac{1}{2}(y-y_P)[3\widetilde{\psi}_{xxx}(x,y_P) +\widetilde{\psi}_{xyy}(x,y_P)]\\ &\quad +\frac{1}{4}(y-y_P)^2[3\widetilde{\psi}_{xxxy}(x,y_P) +\widetilde{\psi}_{xyyy}(x,y_P)]\\ &\quad -2\widetilde{\psi}_{xy}(x,y) +\frac{1}{2}\int_{y_P}^y[3\widetilde{\psi}_{xxx}(x,\lambda) +\widetilde{\psi}_{xyy}(x,\lambda)]d\lambda, \end{align*} and finally we have $\widetilde{\phi}_{xx}(x,y)-\widetilde{\phi}_{yy}(x,y) =-2 \widetilde{\psi}_{xy}(x,y),$ which completes the proof. \end{proof} \subsection*{Conclusion} It has been proved by construction that there exists a unique solution $(\widetilde{\phi},\widetilde{\psi})$ in $C^4(\Omega)\times C^4(\Omega)$ to the Stokes-Bitsadze system \eqref{e2} subject to the boundary conditions \eqref{e5} along with additional single point conditions \eqref{e6}. \subsection*{Acknowledgements} The author is grateful to Professor A. 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