A boundary value problem on the unit disk

Question

I'm working through some problems in PDE by myself and came across one that I can't seem to figure out, I'm betting it's an easy observation I'm just not making.

Problem Let $B$ be the unit disc in $\mathbb{R}^2$ and $a$ and $b$ be continuous functions in $B$ with $$a(x, y)x + b(x, y)y > 0$$ on $\partial{B}$. Assume $u$ is a $C^1$ solution of $$a(x, y)u_x + b(x, y)u_y = -u$$ in $B$. Prove that $u$ vanishes identically.

I tried solving this as a straightforward Cauchy problem and also by making the change to polar coordinates, but when I set up the system of ODEs I'm not clear how to solve them, nor where the condition comes into play.

Answer 1

This is a maximum principle problem. First of all, $u=0$ at any point of interior extremum, since $\nabla u=0$ there. The condition that the vector field $F = ( a,b)$ satisfies $F(x,y)\cdot (x,y)>0$ on the boundary should help analyzing the possible boundary extrema.

Suppose $u$ attains maximum at a boundary point $(x,y)$. Since $(x,y)$ is a point of maximum, $\nabla u$ must point in the direction of outward normal: $\nabla u(x,y)=\lambda (x,y)$ for some $\lambda\ge 0$. Then $$-u(x,y) = F(x,y) \cdot \nabla u(x,y) = \lambda F(x,y) \cdot (x,y) \ge 0$$ so $u(x,y)\le 0$. Similarly, $u(x,y)\ge 0$ at any point of boundary minimum, completing the proof.

It seems that nonstrict inequality $F(x,y)\cdot (x,y)\ge 0$ would be enough for the conclusion.