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My problem is the following.

Let $x_{0}\in\partial\Omega$ and $\Omega\subseteq\mathbb{R}^{2}$ open and connected domain. Suppose there exists $R\in\mathbb{R}$ such that $\Omega\subseteq B_{x_{0},R}$. Let $u$ being a harmonic function on $\Omega$ that is continuous on $\overline{\Omega}=\Omega\cup\partial\Omega$ except on $x_{0}$. If \begin{align*} \frac{u(x)}{\log\frac{2R^{2}}{\left\vert x-x_{0}\right\vert^{2}}}\rightarrow 0 \end{align*} when $x\rightarrow x_{0}$ then $u\leq M$ in $\Omega$.

Obviously what I'm trying to prove is that, unless $u$ is discontinuous on $x_{0}$, the maximum principle holds. Therefore I define the following

\begin{align*} u(x)=\log\frac{2R^{2}}{\left\vert x-x_{0}\right\vert^{2}}\varphi(x)+M \end{align*} note that $\log\frac{2R^{2}}{\left\vert x-x_{0}\right\vert^{2}}\geq 0$ since $\left\vert x-x_{0}\right\vert^{2}\leq R^{2}<2R^{2}$ for all $x\in\Omega$. Also, note that since $\Omega\subseteq B_{x_{0},R}$ then $\Omega$ is bounded.

I must find the PDE for $\varphi$ \begin{align*} \nabla u &=\nabla \log\frac{2R^{2}}{\left\vert x-x_{0}\right\vert^{2}}\varphi(x)\\ &=\log\frac{2R^{2}}{\left\vert x-x_{0}\right\vert^{2}}\nabla\varphi+\varphi\frac{\left\vert x-x_{0}\right\vert^{2}}{2R^{2}}\nabla\frac{2R^{2}}{\left\vert x-x_{0}\right\vert^{2}}\\ &=\log\frac{2R^{2}}{\left\vert x-x_{0}\right\vert^{2}}\nabla\varphi+\varphi\left\vert x-x_{0}\right\vert^{2}\nabla\frac{1}{\left\vert x-x_{0}\right\vert^{2}}\\ &=\log\frac{2R^{2}}{\left\vert x-x_{0}\right\vert^{2}}\nabla\varphi+\varphi\left\vert x-x_{0}\right\vert^{2}(-2\left\vert x-x_{0}\right\vert^{-3})\nabla\left\vert x-x_{0}\right\vert\\ &=\log\frac{2R^{2}}{\left\vert x-x_{0}\right\vert^{2}}\nabla\varphi+\varphi\left\vert x-x_{0}\right\vert^{2}(-2\left\vert x-x_{0}\right\vert^{-3})\frac{\left\vert x-x_{0}\right\vert^{-1}}{2}2(x-x_{0})\\ &=\log\frac{2R^{2}}{\left\vert x-x_{0}\right\vert^{2}}\nabla\varphi-2\varphi\frac{x-x_{0}}{\left\vert x-x_{0}\right\vert^{2}} \end{align*} and therefore \begin{align*} \nabla^{2} u &=\nabla\cdot\log\frac{2R^{2}}{\left\vert x-x_{0}\right\vert^{2}}\nabla\varphi-2\nabla\cdot\varphi\frac{x-x_{0}}{\left\vert x-x_{0}\right\vert^{2}}\\ &=\log\frac{2R^{2}}{\left\vert x-x_{0}\right\vert^{2}}\nabla^{2}\varphi-2\frac{x-x_{0}}{\left\vert x-x_{0}\right\vert^{2}}\cdot\nabla\varphi-2\varphi\nabla\cdot\frac{x-x_{0}}{\left\vert x-x_{0}\right\vert^{2}}-2\nabla\varphi\cdot\frac{x-x_{0}}{\left\vert x-x_{0}\right\vert^{2}} \end{align*} but \begin{align*} \nabla\cdot\frac{x-x_{0}}{\left\vert x-x_{0}\right\vert^{2}}&=\frac{n}{\left\vert x-x_{0}\right\vert^{2}}-\frac{2}{\left\vert x-x_{0}\right\vert^{2}} \end{align*} which in the case of $n=2$, since we are in $\mathbb{R}^{2}$, vanishes. Then \begin{align*} \nabla^{2} u &=\log\frac{2R^{2}}{\left\vert x-x_{0}\right\vert^{2}}\nabla^{2}\varphi-4\frac{x-x_{0}}{\left\vert x-x_{0}\right\vert^{2}}\cdot\nabla\varphi \end{align*} and since $u$ is harmonic the equation for $\varphi$ reads \begin{align*} \log\frac{2R^{2}}{\left\vert x-x_{0}\right\vert^{2}}\nabla^{2}\varphi-4\frac{x-x_{0}}{\left\vert x-x_{0}\right\vert^{2}}\cdot\nabla\varphi=0 \end{align*} this equation is also elliptic but with a term of first-order derivatives and non-constant coefficients. What I would expect is that it has some maximum principle. If it has one, I would prove that $\varphi$ attains its maximum over $\partial U$ and therefore probe that $u$ attains its maximum at $\partial\Omega$.

However, how can I prove that PDE for $\phi$ has a maximum principle and how can I conclude the proof. I have run out of ideas. Please help.

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The equation is elliptic and I have been reading about generalizations of maximum principle for general elliptic linear equations, then I can adapt one of the proofs to obtain maximum principle for this equation. However, the main claim of the problem is such that I can't grasp it. I can't see how to use that PDE for $\varphi$ can aid me to conclude that for $u$ is true that $u\leq M$ in $\Omega$. – elessartelkontar Nov 30 '12 at 3:21

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