Verifying the Minimum Principle for $u(x,y)$

[evinda]

Hello! (Wave)

Let $u(x,y), x^2+y^2 \leq 1$ a solution of $u_{xx}(x,y)+u_{yy}(x,y)+(1+x^2) e^{-u(x,y)}=0, x^2+y^2 \leq 1$.
Show that $\min_{x^2+y^2 \leq 1} u(x,y)= \min_{x^2+y^2=1} u(x,y)$.

Is the following right?

$u_{xx}(x,y)+u_{yy}(x,y)=-(1+x^2) e^{-u(x,y)}$

$(1+x^2) e^{-u(x,y)}>0 \ \ \ \ \ \ \forall x,y \text{ with } x^2+y^2 \leq 1$.

So $u_{xx}(x,y)+u_{yy}(x,y) \leq 0$.

Thus the Minimum principe is satisfied for $u$ and so we have: $\min_{x^2+y^2 \leq 1} u(x,y)= \min_{x^2+y^2=1} u(x,y)$.

 

[I like Serena]

Hello! (Wave)

Let $u(x,y), x^2+y^2 \leq 1$ a solution of $u_{xx}(x,y)+u_{yy}(x,y)+(1+x^2) e^{-u(x,y)}=0, x^2+y^2 \leq 1$.
Show that $\min_{x^2+y^2 \leq 1} u(x,y)= \min_{x^2+y^2=1} u(x,y)$.

Is the following right?

$u_{xx}(x,y)+u_{yy}(x,y)=-(1+x^2) e^{-u(x,y)}$

$(1+x^2) e^{-u(x,y)}>0 \ \ \ \ \ \ \forall x,y \text{ with } x^2+y^2 \leq 1$.

So $u_{xx}(x,y)+u_{yy}(x,y) \leq 0$.

Thus the Minimum principe is satisfied for $u$ and so we have: $\min_{x^2+y^2 \leq 1} u(x,y)= \min_{x^2+y^2=1} u(x,y)$.

Sounds good to me! (Nod)