# Bound the probability of unlikely escape through one end of a thin rectangle

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 $$u(x,y) = \cos \left( \frac{\pi x}{2 \varepsilon} \right) \frac{\sinh \left( \frac{\pi (y+1) \sqrt{\alpha}}{2 \varepsilon \sqrt{\gamma}} \right)} {\sinh \left( \frac{\pi \sqrt{\alpha}}{\varepsilon \sqrt{\gamma}} \right)} \;,$$ 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

Let $u=v+\dfrac{y+1}{2}$ ,

Then $u_x=v_x$

$u_{xx}=v_{xx}$

$u_{xy}=v_{xy}$

$u_y=v_y+\dfrac{1}{2}$

$u_{yy}=v_{yy}$

$\therefore av_x+b\left(v_y+\dfrac{1}{2}\right)+\dfrac{\alpha}{2}v_{xx}+\beta v_{xy}+\dfrac{\gamma}{2}v_{yy}=0$

$\alpha v_{xx}+2\beta v_{xy}+\gamma v_{yy}+2av_x+2bv_y=-b$ with $v(x,-1)=0$ , $v(x,1)=0$

When $\beta=0$ , the homogeneous part of the PDE is separable.

Let $v(x,y)=X(x)Y(y)$ ,

Then $aX'(x)Y(y)+bX(x)Y'(y)+\dfrac{\alpha}{2}X''(x)Y(y)+\dfrac{\gamma}{2}X(x)Y''(y)=0$

$\dfrac{\alpha}{2}X''(x)Y(y)-aX'(x)Y(y)=-\dfrac{\gamma}{2}X(x)Y''(y)+bX(x)Y'(y)$

$\dfrac{(\alpha X''(x)+2aX'(x))Y(y)}{2}=-\dfrac{X(x)(\gamma Y''(y)+2bY'(y))}{2}$

$\dfrac{\alpha X''(x)+2aX'(x)}{X(x)}=-\dfrac{\gamma Y''(y)+2bY'(y)}{Y(y)}$

$\therefore$ Let $v(x,y)=\sum\limits_{n=0}^\infty C(x,n)e^{-\frac{by}{\gamma}}\cos\dfrac{(2n+1)\pi y}{2}$ so that it automatically satisfies $v(x,-1)=0$ and $v(x,1)=0$ ,

Then $\sum\limits_{n=0}^\infty\alpha C_{xx}(x,n)e^{-\frac{by}{\gamma}}\cos\dfrac{(2n+1)\pi y}{2}+\sum\limits_{n=0}^\infty\dfrac{b^2}{\gamma}C(x,n)e^{-\frac{by}{\gamma}}\cos\dfrac{(2n+1)\pi y}{2}+\sum\limits_{n=0}^\infty b(2n+1)\pi C(x,n)e^{-\frac{by}{\gamma}}\sin\dfrac{(2n+1)\pi y}{2}-\sum\limits_{n=0}^\infty\dfrac{\gamma(2n+1)^2\pi^2}{4}C(x,n)e^{-\frac{by}{\gamma}}\cos\dfrac{(2n+1)\pi y}{2}+\sum\limits_{n=0}^\infty2aC_x(x,n)e^{-\frac{by}{\gamma}}\cos\dfrac{(2n+1)\pi y}{2}-\sum\limits_{n=0}^\infty\dfrac{2b^2}{\gamma}C(x,n)e^{-\frac{by}{\gamma}}\cos\dfrac{(2n+1)\pi y}{2}-\sum\limits_{n=0}^\infty b(2n+1)\pi C(x,n)e^{-\frac{by}{\gamma}}\sin\dfrac{(2n+1)\pi y}{2}=-b$

$\sum\limits_{n=0}^\infty\left(\alpha C_{xx}(x,n)+2aC_x(x,n)-\dfrac{\gamma^2(2n+1)^2\pi^2+4b^2}{4\gamma}C(x,n)\right)e^{-\frac{by}{\gamma}}\cos\dfrac{(2n+1)\pi y}{2}=-b$

$\sum\limits_{n=0}^\infty\left(\alpha C_{xx}(x,n)+2aC_x(x,n)-\dfrac{\gamma^2(2n+1)^2\pi^2+4b^2}{4\gamma}C(x,n)\right)\cos\dfrac{(2n+1)\pi y}{2}=-be^\frac{by}{\gamma}$

For $-1<y<1$ , with reference to Wave equation with initial and boundary conditions - is this function right? ,

$\sum\limits_{n=0}^\infty\left(\alpha C_{xx}(x,n)+2aC_x(x,n)-\dfrac{\gamma^2(2n+1)^2\pi^2+4b^2}{4\gamma}C(x,n)\right)\cos\dfrac{(2n+1)\pi y}{2}=-\sum\limits_{n=0}^\infty\int_{-1}^1be^\frac{by}{\gamma}\cos\dfrac{(2n+1)\pi y}{2}dy~\cos\dfrac{(2n+1)\pi y}{2}$

$\sum\limits_{n=0}^\infty\left(\alpha C_{xx}(x,n)+2aC_x(x,n)-\dfrac{\gamma^2(2n+1)^2\pi^2+4b^2}{4\gamma}C(x,n)\right)\cos\dfrac{(2n+1)\pi y}{2}=-\sum\limits_{n=0}^\infty\dfrac{4b\gamma^2(-1)^n(2n+1)\pi\cosh\dfrac{b}{\gamma}}{\gamma^2(2n+1)^2\pi^2+4b^2}\cos\dfrac{(2n+1)\pi y}{2}$

$\therefore\alpha C_{xx}(x,n)+2aC_x(x,n)-\dfrac{\gamma^2(2n+1)^2\pi^2+4b^2}{4\gamma}C(x,n)=-\dfrac{4b\gamma^2(-1)^n(2n+1)\pi\cosh\dfrac{b}{\gamma}}{\gamma^2(2n+1)^2\pi^2+4b^2}$

$4\alpha\gamma C_{xx}(x,n)+8a\gamma C_x(x,n)-(\gamma^2(2n+1)^2\pi^2+4b^2)C(x,n)=-\dfrac{16b\gamma^3(-1)^n(2n+1)\pi\cosh\dfrac{b}{\gamma}}{\gamma^2(2n+1)^2\pi^2+4b^2}$

$C(x,n)=A(n)e^{-\frac{ax}{\alpha}}\cosh\dfrac{\sqrt{\alpha\gamma^2(2n+1)^2\pi^2+4(a^2+b^2\alpha)}x}{2\alpha}+B(n)e^{-\frac{ax}{\alpha}}\sinh\dfrac{\sqrt{\alpha\gamma^2(2n+1)^2\pi^2+4(a^2+b^2\alpha)}x}{2\alpha}+\dfrac{16b\gamma^3(-1)^n(2n+1)\pi\cosh\dfrac{b}{\gamma}}{(\gamma^2(2n+1)^2\pi^2+4b^2)^2}$

$\therefore u(x,y)=\sum\limits_{n=0}^\infty A(n)e^{-\frac{ax}{\alpha}-\frac{by}{\gamma}}\cosh\dfrac{\sqrt{\alpha\gamma^2(2n+1)^2\pi^2+4(a^2+b^2\alpha)}x}{2\alpha}\cos\dfrac{(2n+1)\pi y}{2}+\sum\limits_{n=0}^\infty B(n)e^{-\frac{ax}{\alpha}-\frac{by}{\gamma}}\sinh\dfrac{\sqrt{\alpha\gamma^2(2n+1)^2\pi^2+4(a^2+b^2\alpha)}x}{2\alpha}\cos\dfrac{(2n+1)\pi y}{2}+\sum\limits_{n=0}^\infty\dfrac{16b\gamma^3(-1)^n(2n+1)\pi\cosh\dfrac{b}{\gamma}}{(\gamma^2(2n+1)^2\pi^2+4b^2)^2}e^{-\frac{by}{\gamma}}\cos\dfrac{(2n+1)\pi y}{2}+\dfrac{y+1}{2}$

$u(-\varepsilon,y)=0$ , $u(\varepsilon,y)=0$ :

$\begin{cases}\sum\limits_{n=0}^\infty A(n)e^{\frac{a\varepsilon}{\alpha}-\frac{by}{\gamma}}\cosh\dfrac{\sqrt{\alpha\gamma^2(2n+1)^2\pi^2+4(a^2+b^2\alpha)}\varepsilon}{2\alpha}\cos\dfrac{(2n+1)\pi y}{2}-\sum\limits_{n=0}^\infty B(n)e^{\frac{a\varepsilon}{\alpha}-\frac{by}{\gamma}}\sinh\dfrac{\sqrt{\alpha\gamma^2(2n+1)^2\pi^2+4(a^2+b^2\alpha)}\varepsilon}{2\alpha}\cos\dfrac{(2n+1)\pi y}{2}+\sum\limits_{n=0}^\infty\dfrac{16b\gamma^3(-1)^n(2n+1)\pi\cosh\dfrac{b}{\gamma}}{(\gamma^2(2n+1)^2\pi^2+4b^2)^2}e^{-\frac{by}{\gamma}}\cos\dfrac{(2n+1)\pi y}{2}+\dfrac{y+1}{2}=0\\\sum\limits_{n=0}^\infty A(n)e^{-\frac{a\varepsilon}{\alpha}-\frac{by}{\gamma}}\cosh\dfrac{\sqrt{\alpha\gamma^2(2n+1)^2\pi^2+4(a^2+b^2\alpha)}\varepsilon}{2\alpha}\cos\dfrac{(2n+1)\pi y}{2}+\sum\limits_{n=0}^\infty B(n)e^{-\frac{a\varepsilon}{\alpha}-\frac{by}{\gamma}}\sinh\dfrac{\sqrt{\alpha\gamma^2(2n+1)^2\pi^2+4(a^2+b^2\alpha)}\varepsilon}{2\alpha}\cos\dfrac{(2n+1)\pi y}{2}+\sum\limits_{n=0}^\infty\dfrac{16b\gamma^3(-1)^n(2n+1)\pi\cosh\dfrac{b}{\gamma}}{(\gamma^2(2n+1)^2\pi^2+4b^2)^2}e^{-\frac{by}{\gamma}}\cos\dfrac{(2n+1)\pi y}{2}+\dfrac{y+1}{2}=0\end{cases}$

$\begin{cases}\sum\limits_{n=0}^\infty A(n)e^{\frac{a\varepsilon}{\alpha}}\cosh\dfrac{\sqrt{\alpha\gamma^2(2n+1)^2\pi^2+4(a^2+b^2\alpha)}\varepsilon}{2\alpha}\cos\dfrac{(2n+1)\pi y}{2}-\sum\limits_{n=0}^\infty B(n)e^{\frac{a\varepsilon}{\alpha}}\sinh\dfrac{\sqrt{\alpha\gamma^2(2n+1)^2\pi^2+4(a^2+b^2\alpha)}\varepsilon}{2\alpha}\cos\dfrac{(2n+1)\pi y}{2}=-\sum\limits_{n=0}^\infty\dfrac{16b\gamma^3(-1)^n(2n+1)\pi\cosh\dfrac{b}{\gamma}}{(\gamma^2(2n+1)^2\pi^2+4b^2)^2}\cos\dfrac{(2n+1)\pi y}{2}-\dfrac{y+1}{2}e^\frac{by}{\gamma}\\\sum\limits_{n=0}^\infty A(n)e^{-\frac{a\varepsilon}{\alpha}}\cosh\dfrac{\sqrt{\alpha\gamma^2(2n+1)^2\pi^2+4(a^2+b^2\alpha)}\varepsilon}{2\alpha}\cos\dfrac{(2n+1)\pi y}{2}+\sum\limits_{n=0}^\infty B(n)e^{-\frac{a\varepsilon}{\alpha}}\sinh\dfrac{\sqrt{\alpha\gamma^2(2n+1)^2\pi^2+4(a^2+b^2\alpha)}\varepsilon}{2\alpha}\cos\dfrac{(2n+1)\pi y}{2}=-\sum\limits_{n=0}^\infty\dfrac{16b\gamma^3(-1)^n(2n+1)\pi\cosh\dfrac{b}{\gamma}}{(\gamma^2(2n+1)^2\pi^2+4b^2)^2}\cos\dfrac{(2n+1)\pi y}{2}-\dfrac{y+1}{2}e^\frac{by}{\gamma}\end{cases}$

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