TITLE: Fritz John PDE book chapter 1 exercise: Prove that $u$ vanishes identically if $au_x+bu_y=-u$
QUESTION [9 upvotes]: I was trying out this question:
Let $u$ be a solution in $C^1$ of the PDE
$$ a(x,y)u_x + b(x,y)u_y = -u $$
on the closed unit disc $\Omega$ in the xy-plane. Let $a(x,y)x + b(x,y)y > 0$ on the boundary of $\Omega$. Prove that $u$ vanishes identically.
According to the hint, we are supposed to show that $\max_\Omega u \leq 0 $ and $\min_\Omega u \geq 0$. If the maxima/minima occurs in the interior of $\Omega$, since $u_x = u_y = 0$ we get $\max u = \min u = 0 $, and so $u$ vanishes identically. Now, if, say the maxima occurs on the boundary, we take $f(\theta) = u(\cos\theta,\sin\theta)$ and differentiate to get a condition on maxima:
$$ f'(\theta) = -u_xy + u_yx = 0 \implies u_x = \lambda x; \, u_y = \lambda y $$
This is where I am stuck. To specify that this is a maxima and not a minima, we need to differentiate $f$ further. But we aren't allowed to do that since $u \in C^1$. Even if I disregard that constraint and differentiate anyway (and use $u_x,\,u_y$ from above), I am stuck in terms containing $\lambda_x,\,\lambda_y$. 
What am I missing? Or am I doing something completely wrong?

REPLY [7 votes]: A key relevant concept is directional derivative. The expression $au_x+bu_y$ is the directional derivative of $u$ along vector $(a,b)$. And this is the direction you should differentiate in, not along the boundary. 
You want to rule out two scenarios: (a) $u$ has a strictly positive maximum; (b) $u$ has a strictly negative minumum. They are similar (and one reduces to the other by considering $-u$), so let's suppose (a) holds.  
You correctly observed that having $u(x_0,y_0)\ne 0$ at an interior stationary point is impossible, so this positive maximum $(x_0,y_0)$ has to be on the boundary. At this point, the directional derivative $au_x+bu_y$ is strictly negative. Crucially, the vector $(a,b)$ points outside of the disk. Hence,  we can move in the opposite direction and stay in the domain. This yields higher values of $u$:  $$u(x_0-\epsilon a, y_0-\epsilon b)>u(x_0,y_0)$$
A contradiction.