How prove this $u(x,y)\ge 0,(x,y)\in D$ let $$D=\{(x,y)|x^2+y^2<1\}$$
and $u(x,y)$ be  second order continuous partial derivatives on $\overline{D}$,
and $$\dfrac{\partial^2 u}{\partial x^2}+\dfrac{\partial^2 u}{\partial y^2}+\dfrac{\partial u}{\partial x}+\dfrac{\partial u}{\partial y}=2u,(x,y\in D)$$
$$u(x,y)\ge 0,(x,y)\in\partial D$$
show that
$$u(x,y)\ge 0,(x,y)\in D$$
I find this similar problem How prove $f=0$,if $\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}+\frac{\partial f}{\partial z}=f$
But I can't prove this problem,Thank you for you help
 A: Since $\overline{D}$ is compact, there exists $(x_0,y_0)$ such that 
$$u(x_0,y_0)=\min_{(x,y)\in\overline{D}}u(x,y).$$
There are two cases to be considered: $(x_0,y_0)\in\partial D$ or $(x_0,y_0)\in D$. 
If $(x_0,y_0)\in\partial D$, then  $u(x_0,y_0)=0$ since $u|_{\partial D}=0$ by assumption. This implies that 
$$u(x,y)\geq  \min_{(x,y)\in\overline{D}}u(x,y)=u(x_0,y_0)=0\mbox{ for all }(x,y)\in\overline{D},$$
as required. 
If $(x_0,y_0)\in D$, then we have 
$$ \nabla u(x_0,y_0)=0\mbox{ and }\Delta  u(x_0,y_0)\geq 0$$
since $(x_0,y_0)$ is a minimum point, that is, 
$$\tag{1}\frac{\partial u}{\partial x}(x_0,y_0)=\frac{\partial u}{\partial y}(x_0,y_0)=0\mbox{ and }
 \frac{\partial^2u}{\partial x}(x_0,y_0)+\frac{\partial^2u}{\partial x}(x_0,y_0)\geq 0.$$
Using $(1)$ we have
$$0\leq \dfrac{\partial^2 u}{\partial x^2}(x_0,y_0)+\dfrac{\partial^2 u}{\partial y^2}(x_0,y_0)+\dfrac{\partial u}{\partial x}(x_0,y_0)+\dfrac{\partial u}{\partial y}(x_0,y_0)=2u(x_0,y_0)$$
which implies again 
$$u(x,y)\geq  \min_{(x,y)\in\overline{D}}u(x,y)=u(x_0,y_0)\geq 0\mbox{ for all }(x,y)\in\overline{D},$$
as required.
