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Consider the problem of Cauchy $$\left\{\begin{array}-\Delta u=1 \ \ \ on \ \Omega= (0,1)\times(0,1)\subset R^2\ \ \ \ (*)\\ u|_{\partial\Omega}=0 \end{array}\right.$$ Find upper and lower estimates on $u(1/2,1/2)$ using the comparison principle with $v(x,y):= A(x^2+ y^2) + B$ ($A$, $B$ are constants).

Remember: "The comparison principle for the Laplace operator":

if $\Omega\subset R^d$ is open, connected, and bounded, and for $u,v\in C^2 (\Omega)\cap C(\Omega)$ we have that $$-\Delta u\leq -\Delta v\ \ on\ \Omega\ \; \ \ u\leq v\ \ on\ \partial\Omega$$ then $$u\leq v\ \ on\ \Omega$$

My attempt is:

We have $\Delta v=4A$. Let $u(x,y)=-\frac{1}{4}(x^2+y^2)$. Thus $\Delta u=-1$. So $u(x,y)=-\frac{1}{4}(x^2+y^2)$ is one solution for our problem (*).

To use the "The comparison principle for the Laplace operator", we should have $-\Delta u\leq -\Delta v$ on $(0,1)\times(0,1)$. So $A\leq -\frac{1}{4}$, and $u\leq v$ on $\partial \Omega$; hence from $u(0,0)\leq v(0,0)$ we conclude $0\leq B$.

Therefore If $A\leq -\frac{1}{4}$, then we have $u\leq v$ on $\Omega= (0,1)\times(0,1)$. So \begin{eqnarray*}u(1/2,1/2)&\leq& v(1/2,1/2)\\ &=& \frac{1}{2}A+B\\ &\leq& -\frac{1}{8}+B\end{eqnarray*} Hence the estimate for sup $u(1/2,1/2)$ is $-\frac{1}{8}$.

My question is about $B$, and the inf $u(1/2,1/2)$. Please help me.

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You need to apply all the boundary conditions to identify both A and B. Namely $$\begin{cases}v(0,y) \geq 0\ (y \in [0,1])\\ v(x,0) \geq 0 \ (x \in [0,1])\\ v(1,y) \geq 0\ (y \in [0,1])\\ v(x,1) \geq 0\ (x \in [0,1])\end{cases}$$. Once you get a fully identified upper estimate function $v(x,y)$ then $u(\frac{1}{2}, \frac{1}{2}) \leq v(\frac{1}{2},\frac{1}{2})$. Conversely by taking $-\Delta v \leq -\Delta u$ and $v(x,y)|_{\partial \Omega} \leq u|_{\partial \Omega} = 0 $, you'll get a different $v(x,y)$ which is your lower estimate function.

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