2

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

Community
  • 1
math110
  • 93,304

1 Answers1

1

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.

Paul
  • 19,140