# Prove that function in two dimensions with some conditions are zero.

Let $\Omega=\{(x,y) \in \mathbb{R}^2: x^2+y^2<1\}$, and let $u\in C^1(\bar{\Omega})$ satisfy

$$\alpha(x,y)u_x+\beta(x,y)u_y =-u \hspace{0.2in} \forall (x,y)\in\bar{\Omega},$$

where $\alpha$ and $\beta$ are continuous functions on $\bar{\Omega}$ such that

$$\alpha(x,y)x+\beta(x,y)y >0 \hspace{0.2in} \forall (x,y)\in\bar{\Omega},$$

Prove that $u\equiv 0$.

Let $X_0$ be a maximum of $u$.

If $X_0$ is interior, then $u(X_0)=-\alpha(X_0) u_x(X_0) -\beta(X_0) u_y(X_0) =0$ since $\nabla u(X_0)=0$, and therefore $u\le 0$.

If $X_0$ is in the boundary, then since $v=(\alpha(X_0),\beta(X_0))$ is an exterior vector to the ball at $X_0$ (i.e. $v\cdot X_0>0$), we have $\nabla u(X_0)\cdot v \ge 0$. That is $u(X_0)\le 0$ and thus again $u\le0$.

Similarly you can see that $u\ge 0$.

• Why $\nabla u(X_0)\cdot v =\alpha(x,y)u_x+\beta(x,y)u_y \geq 0$? – Worawit Tepsan Oct 9 '13 at 17:49
• $\nabla u(X_0)\cdot v=\lim_{t\rightarrow 0^-}\frac{u(X_0+tv)-u(X_0)}{t}\ge 0$. – Pocho la pantera Oct 9 '13 at 18:23
• I get it now. Thank you so much. Next step, we also do the same with minimum. We can conclude from the minimum that $u\geq0$. Therefore, $u\equiv 0$ – Worawit Tepsan Oct 9 '13 at 19:16
• You are welcome. – Pocho la pantera Oct 9 '13 at 19:37
• I have a question. Is $t\to 0^+$? (Because $X_0+tv$ shoud goes inside the ball.) – Worawit Tepsan Oct 9 '13 at 19:58

Suppose not. Then either $\max_{\overline{\Omega}} u>0$ or $\min_{\overline{\Omega}} u<0$. Assume it's the first one. The PDE implies that $u$ cannot have an interior critical point with nonzero value. Therefore, the maximum is at some boundary point $(x,y)$. But then the derivative along the (inward-pointing) vector $-(\alpha,\beta)$ is positive at that point, which is impossible.

• If $u\equiv -1$, then $u \not \equiv 0$ and $\max u<0$. – Pocho la pantera Oct 9 '13 at 14:06
• Thank you so much for help me solve my problem. – Worawit Tepsan Oct 9 '13 at 19:18