A trivial solution of a PDE 
Let $u\in C^1$  in the unit closed disk $\Omega$ be a solution of the PDE
  $$a(x,y)u_x+b(x,y)u_y=-u $$
  Suppose that $a(x,y)x+b(x,y)y>0$ in $\partial\Omega$. Show that $u=0$.
Hint: Show that $\max_{\Omega} u\leq 0 $ and $\min_{\Omega} u\geq 0 $.

I think that neither there is some mistake or i am missing something, since if $(x_0,y_0)$ is the value there $u$ arrives its maximum in $\omega$ then $u_x=u_y=0$ in $x_0$. But 
$$-u=a(x_0,y_0)u_x+b(x_0,y_0)u_y=0 $$
Then $u(x_0,y_0)=0$. This argument is also true for the minimum, then $u=0$. What am I missing? 
Thanks!
 A: Since $\Omega$ is compact and $u$ continuous over $\Omega$, $u$ achieve its absolute maximum at some $p \in \Omega$. 


*

*If $p \in {\bf int}\,\Omega$, then $\nabla u(p) = 0\implies \max_\Omega u = u(p) = 0$.

*Otherwise, $p \in \partial\Omega$ and $p$ will be an absolute maximum for $u$ when restricted on $\partial \Omega$. There are two subcases:


*

*If $\nabla u(p) = 0$, we have $\max_\Omega u = u(p) = 0$ again.

*If not, then $p$ being a local maximum of $u$ along $\partial \Omega$,
$\nabla u(p)$ will be pointing along the radial direction. Since $u$ achieves maximum at $p$, $\nabla u(p)$ will be pointing outwards. i.e. we can find a positive
number $\lambda$ such that $\nabla u(p) = \lambda p$. Let $p = (x,y)$, the condition imposed on the boundary implies
$$u(p) = -( a(x,y) u_x(p) + b(x,y) u_y(p) ) = -\lambda (a(x,y) x + b(x,y) y) < 0$$
We have $\max_\Omega u = u(p) < 0$ in this cases.
Combine all these case, we have $\max_\Omega u \le 0$ in general. By a similar argument, we have $\min_\Omega u \ge 0$. This force $u = 0$ over the whole $\Omega$.
