Show that $u = 0$ in $B_1 (0)$ if we only assume that $u \in C^1 (B_1 (0))$.

Question

Let $n \geq 2$ and $B_1 (0)$ be the open unit ball in $\Bbb R^n$. Suppose $u \in C^1 (B_1(0))\cap C(\overline{B_1 (0)})$ is a solution of linear PDE $$x \cdot Du=-u; x \in B_1 (0)$$

(a) Show that $u = 0$ in $B_1 (0)$.

(b) Show that $u = 0$ in $B_1 (0)$ if we only assume that $u \in C^1 (B_1 (0))$.

Here I can clearly see that $u=0$ is a solution as it satisfies the PDE. I was also trying to solve it using characteristic line method e.g. taking $z=u(x(s))$ and $p=Du$ then we have $F(p,z,x)=x \cdot p+z$ and hence $\dot x(s)=x$ and $\dot z(s)=-z$. I don't think this method would be useful as there is no boundary condition.

How to attempt part a and part b then? Please help.

Answer 1

For each $x\in B_1(0)$, consider

$$f(t) = u(tx).$$

Then

$$f'(t) = \frac{d}{dt} u(tx) = x\cdot Du(tx) = \frac{1}{t} (tx)\cdot Du(tx) = -\frac{1}{t} f(t).$$

So $(tf)' = tf' + f= 0$ and so $t u(tx)$ is constant in $t$. Or, that

$$ u(x) = u \left( \| 2x\| \frac{x}{\|2x\|}\right) =\frac{1}{\|2x\|} \|2x\|u\left( \| 2x\| \frac{x}{\|2x\|}\right) = \frac{1}{2\|x\|}c\left(\frac{x}{2\|x\|}\right)$$

The only way that this $u$ is continuous at $x=0$ is that $c = u = 0$.