Let $u(x,y), x^2+y^2 \leq 1$, a solution of
$$u_{xx}(x,y)+2u_{yy}(x,y)+e^{u(x,y)}=0, x^2+y^2\leq 1$$
Show that $\min_{x^2+y^2 \leq 1} u(x,y)= \min_{x^2+y^2=1} u(x,y) $.
We suppose that $\min_{x^2+y^2 \leq 1} u(x,y) \neq \min_{x^2+y^2=1} u(x,y) $.
How can we find a function $U(X,Y)$ so that from $u_{xx}(x,y)+2u_{yy}(x,y)+e^{u(x,y)}=0$ we get the known Laplace's equation ?
After a change of variables $y=\sqrt2\,z$, we get $$ \underbrace{u_{xx}+u_{zz}}_{\text{Laplacian in $x,z$}}=-e^u\lt0 $$ which makes $-u$ subharmonic on $$ x^2+2z^2\le1 $$ and thus $-u$ obeys the Maximum Principle.