# the comparison principle for the Laplace operator

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.

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.