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Consider the following elliptic PDE boundary value problem, \begin{eqnarray} & a u_x + b u_y + \frac{\alpha}{2} u_{xx} + \beta u_{xy} + \frac{\gamma}{2} u_{yy} = 0 \;, \quad {\rm ~for~} -\varepsilon < x < \varepsilon \;, -1 < y < 1 \;, & \\ & u(-\varepsilon,y) = 0 \;, \quad u(\varepsilon,y) = 0 \;, \quad u(x,-1) = 0 \;, \quad u(x,1) = 1 \;, & \end{eqnarray} where $\varepsilon > 0$ is small. We can think of $u(x,y)$ as representing the probability that a certain generalized random walk starting from $(x,y)$ exits the rectangle $[-\varepsilon,\varepsilon] \times [-1,1]$ through the edge $y=1$. The PDE is elliptic, so $\alpha \gamma > \beta^2$, and we can assume $\alpha, \gamma > 0$.

I would like to obtain a reasonable upper bound for $u(0,0)$, for small $\varepsilon$. How can I do this?

At the very least the bound should limit to zero as $\varepsilon \to 0$. Given that when $a = b = \beta = 0$, the exact solution is \begin{equation} u(x,y) = \cos \left( \frac{\pi x}{2 \varepsilon} \right) \frac{\sinh \left( \frac{\pi (y+1) \sqrt{\alpha}}{2 \varepsilon \sqrt{\gamma}} \right)} {\sinh \left( \frac{\pi \sqrt{\alpha}}{\varepsilon \sqrt{\gamma}} \right)} \;, \end{equation} and so in this case, $u(0,0) \approx {\rm exp} \left( \frac{-\pi \sqrt{\alpha}}{2 \varepsilon \sqrt{\gamma}} \right)$, which is very small, I would guess that we could prove $u(0,0) < M {\rm exp} \left( \frac{-C}{\varepsilon} \right)$, for some constants $M$ and $C$. I've been unsuccessful in trying to derive a bound by using the maximum principle somehow.

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Note that this PDE is separable when $\beta=0$ . So you can first try to find the variable transformation to eliminate the mix-derivative term. –  doraemonpaul Nov 24 '12 at 11:24
    
It is not suggested to accommodate $\varepsilon$ so that it appears in the denominator as it will increase the difficulties of analyzing. –  doraemonpaul Nov 24 '12 at 11:34

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