I have a $C^1$ function $u$ on the closed unit ball centered in $0$ (actually the exercise doesn't specify where the ball is, I just assume it being centered in $0$ just for simplicity), called $\Omega$, such that $$B(x) \cdot \nabla u(x)=-u$$ on the ball and with $B$ a vector field such that $B(x) \cdot x>0$ if $x \in \partial \Omega$. I should prove that $u=0$ in $\Omega$, and there's a hint to show that $\max (u) \leq 0$ and $\min (u) \geq 0$ in $\Omega$.
I have tried writing the usual system for first-order PDE, and I got $$\dot x(s)=B(x)\text{ and } \dot z(s)=-z(s)$$
So $z(s)=ke^{-s}$ with $k$ that must be determined. Then I noticed that $\frac{d}{ds} ||x(s)||^2=2x(s) \dot x(s)= 2 xB(x)>0 $ if $x \in \partial \Omega$. So if a curve $x(s)$ arrives to $\partial \Omega$, it gets out.
Now, I am pretty much stuck and I donn't know how to go on. I tried writing the expression for the partial derivatives of $u$, but I didn't see how it could be useful.
The hint looks like it is suggesting a maximum principle argument, not the method of characteristics. For a maximum principle argument, just examine the point $x$ where $u$ attains its maximum value. If $x$ is in the interior of the ball, then $\nabla u(x) =0$ and hence $-u(x) = \nabla u(x)\cdot B(x) =0$ at the max, i.e., $u \leq 0$ in the ball. If $x$ is on the boundary, then
$$\nabla u(x) = \lambda x \ \ \ \ \ \ (*)$$
where $\lambda$ is a Lagrange multiplier (obtained from maximizing $u$ over the boundary $|x|^2=1$ of the ball). The outward normal to the ball is $x$, hence we must also have $\nabla u(x) \cdot x \geq 0$ at the max. Dotting both sides of (*) with $x$ we have
$$ \lambda |x|^2 = \nabla u(x) \cdot x \geq 0,$$
and so $\lambda \geq 0$. Since $B(x)\cdot x > 0$ (since $x\in \partial \Omega$) and $\nabla u(x) = \lambda x$ with $\lambda \geq 0$ we have
$$-u(x) = \nabla u(x) \cdot B(x) = \lambda x\cdot B(x) \geq 0.$$
Again, we obtain $u(x) \leq 0$ at the max. Therefore $\max u \leq 0$. The argument for $\min u \geq 0$ is similar. Notice we actually only needed $B(x)\cdot x \geq 0$ on $\partial \Omega$.