Given linear PDE $$x \cdot \nabla u +u=0$$ on a unit open ball, $u$ is $C^1$ on the ball and $C^0$ on the boundary, prove $u$ is $0$ on the unit ball.
I solved that $u$ is constant on the boundary, $u(0)=0$. I am thinking about using proof by contradiction and applying for the mean value theorem.
This PDE can be solved using the method of characteristics. We first solve the PDE, and then deduce the claimed result.
Solution of PDE
The characteristics $X$ are the solution to the parameterised ODE $$ \partial_tX(x_0,t) = X(x_0,t), \\ X(x_0,0) = x_0$$ where $x_0$ ranges over the points in $\partial B_1(0)$ i.e. $|x_0|=1$. (we remark that $\partial B_1(0)$ is non-characteristic, i.e. none of the characteristics are tangent to it.) The solution is $$X(x_0,t) = x_0 e^t.$$
Lets say that on $\partial B_1(0)$, $u=g$ for some continuous function $g:\partial B_1(0)\to\mathbb R$. Setting $U(x_0,t):=u(X(x_0,t))$, we note that $$ \partial_t U = ((\nabla u)\circ X)\cdot\partial_t X = X\cdot u(X) = -U$$ which gives $$U(x_0,t) = g(x_0)e^{-t}.$$ $X$ has the inverse $(x_0(x),t(x)) =(\frac x{|x|} , \log |x|)$. so we can use this to write the solution as $$u(x) = U(x_0(x),t(x)) =g\left(\frac x{|x|}\right) \frac1{|x|}$$ We can easily verify this factor of $\frac1{|x|}$ is correct: say $g\equiv 1$, i.e. $u(x)=\frac1{|x|}$ then $$ \nabla u(x) = \frac{-x}{|x|^3} \implies x\cdot\nabla u(x) = \frac{-1}{|x|} = -u(x).$$
Deducing that $u$ vanishes
Once we have the above form, it is easy to conclude. If the boundary data $g(x)=g_0\neq 0$ at any point $x$, then along the line joining $0$ to $x$, the solution takes the form $\frac{g_0}{|x|}$, which explodes as $x\to 0$. Since we assumed the solution to be continuous, this means $g$ must be identically zero.
However, from the form of $u$, this means that $u\equiv 0$.