Question:
Let $U \subset\mathbb{R}^2$ be an open and bounded set and $u\in C^2(U)\cap C(U)$ satisfy $\Delta u(x)=\frac{-1}{1+|x|}$ in $U$. Show that $u$ attains it's minimum on the boundary of $U$.
My attempt:
$$\text{Assume u attains a minimum on the interior of}\space U. \text{For some}\space(x_0,y_0)\in U.$$
$$\Rightarrow \quad \frac{\partial u}{\partial x}(x_0,y_0)=\frac{\partial u}{\partial y} (x_0,y_0)=0 \space\text{ and }\space \frac{\partial^2 u}{\partial x^2}(x_0,y_0),\frac{\partial^2 u}{\partial y^2}(x_0,y_0)\ge0.\quad\text{So we have...}$$
$$\Rightarrow \quad\Delta u(x_0,y_0)=\frac{\partial^2 u}{\partial x^2}(x_0,y_0)+\frac{\partial^2 u}{\partial y^2}(x_0,y_0)\ge0.$$
$$\text{However we know}\space \Delta u(x)=\frac{-1}{1+|x|}\lt0\quad \forall x\in U.\text{Hence a contradiction}\therefore \text{the minimum is attained on the boundary}$$
Help!
Here is an attempt at a question I'm doing on my partial differential equations course. I am not happy with my solution as it's definitely missing bits. Any alternative solutions/hints would be greatly appreciated! :)
$$\text{Assume u attains a minimum on the interior of}\space U. \text{For some}\space(x_0,y_0)\in U.$$
$$\Rightarrow \quad \frac{\partial u}{\partial x}(x_0,y_0)=\frac{\partial u}{\partial y} (x_0,y_0)=0 \space\text{ and }\space \frac{\partial^2 u}{\partial x^2}(x_0,y_0),\frac{\partial^2 u}{\partial y^2}(x_0,y_0)\ge0.\quad\text{So we have...}$$
$$\Rightarrow \quad\Delta u(x_0,y_0)=\frac{\partial^2 u}{\partial x^2}(x_0,y_0)+\frac{\partial^2 u}{\partial y^2}(x_0,y_0)\ge0.$$
$\text{However we know}\space \Delta u(x)=\frac{-1}{1+|x|}\lt0\quad \forall x\in U.\text{Hence a contradiction}\therefore \text{by the maximum principle u must attain a minimum somewhere in }\overline{U}.$ $\text{Since it is not in U it must be on } \partial U$