$u$ vanishes identically in unit ball Let $B_1$ be the unit open ball in $\Bbb R^n$ centered at zero and let $a_i(x_1,... x_n), i =1,... n$ be continuous functions on $\bar B_1$. Suppose$\sum_1^n a_i(x_1,... x_n)x_i <0$ on $\partial B_1$,the boundary of $B_1$. Assume that $u$ is a $C^1$ solution of
$\sum_1^n a_i(x_1,... x_n)u_{x_i} = u$ in $\bar B_1$; prove that $u$ vanishes identically.
could you anyone please help me to do this problem..
Thanks in Advance
 A: We should think of the functions $a_i$ as components of a vector field $ F = (a_1,\dots,a_n)$. The assumptions say that $ F$ points inward on the boundary of $B_1$, and that $u$ satisfies an ordinary differential equation on each trajectory of $ F$. To make the latter more precise, consider the initial value problem $\gamma'(t) =  F(\gamma(t))$ with $\gamma(0)\in B_1$. A solution exists (by the Peano existence theorem) as long as $\gamma$ stays in $B_1$. But it stays in $B_1$ forever, because exit is prohibited: $F$ points inward on $\partial B_1$.  So, the solution exists for all $t\ge 0$. 
The function $u$ satisfies 
$$\frac{d}{dt}u(\gamma(t)) = u(\gamma(t)) \tag{1}$$
which you can check with the chain rule. Solve (1) to obtain
$$u(\gamma(t))= e^t u(\gamma(0))\tag{2}$$  
Suppose $u(x_0)\ne 0$ for some $x_0\in B_1$. Let $\gamma$ be as above, with $\gamma(0) = x_0$. By (2), $u(\gamma(t))$ is an unbounded function of $t\in [0,\infty)$. But $\gamma(t)\in B_1$ for all $t$, and $u$ is bounded on $B_1$. Contradiction.
