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Let $u\in C^1$ in the unit closed disk $\Omega$ be a solution of the PDE $$a(x,y)u_x+b(x,y)u_y=-u $$ Suppose that $a(x,y)x+b(x,y)y>0$ in $\partial\Omega$. Show that $u=0$.

Hint: Show that $\max_{\Omega} u\leq 0 $ and $\min_{\Omega} u\geq 0 $.

I think that neither there is some mistake or i am missing something, since if $(x_0,y_0)$ is the value there $u$ arrives its maximum in $\omega$ then $u_x=u_y=0$ in $x_0$. But $$-u=a(x_0,y_0)u_x+b(x_0,y_0)u_y=0 $$ Then $u(x_0,y_0)=0$. This argument is also true for the minimum, then $u=0$. What am I missing?

Thanks!

EQJ
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1 Answers1

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Since $\Omega$ is compact and $u$ continuous over $\Omega$, $u$ achieve its absolute maximum at some $p \in \Omega$.

  1. If $p \in {\bf int}\,\Omega$, then $\nabla u(p) = 0\implies \max_\Omega u = u(p) = 0$.

  2. Otherwise, $p \in \partial\Omega$ and $p$ will be an absolute maximum for $u$ when restricted on $\partial \Omega$. There are two subcases:

    • If $\nabla u(p) = 0$, we have $\max_\Omega u = u(p) = 0$ again.

    • If not, then $p$ being a local maximum of $u$ along $\partial \Omega$, $\nabla u(p)$ will be pointing along the radial direction. Since $u$ achieves maximum at $p$, $\nabla u(p)$ will be pointing outwards. i.e. we can find a positive number $\lambda$ such that $\nabla u(p) = \lambda p$. Let $p = (x,y)$, the condition imposed on the boundary implies $$u(p) = -( a(x,y) u_x(p) + b(x,y) u_y(p) ) = -\lambda (a(x,y) x + b(x,y) y) < 0$$ We have $\max_\Omega u = u(p) < 0$ in this cases.

Combine all these case, we have $\max_\Omega u \le 0$ in general. By a similar argument, we have $\min_\Omega u \ge 0$. This force $u = 0$ over the whole $\Omega$.

achille hui
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  • Why we can not used in general the fact that $\text{ grad }u(p)=0$ – EQJ Jun 11 '15 at 07:21
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    @YotasTrejos the $\nabla u(p) = 0$ test only works in the interior. On the boundary $\partial \Omega$, we only have $\nabla u(p) \cdot t = 0$ where $t$ is any tangent vector of the $\partial \Omega$. As a simple example, consider the function $f(x,y) = x^2+y^2$, it achieves its maximum over unit disk on the boundary but $\nabla f \ne 0$ there. – achille hui Jun 11 '15 at 07:24
  • For the first case, why $\nabla u(p)=0$ implies $u(p)=0$? – Chee Han Jun 11 '15 at 07:44
  • @CheeHan $\nabla u(p) = 0 \iff u_x = u_y = 0 \implies u = -( a u_x + b u_y ) = 0$ – achille hui Jun 11 '15 at 08:33
  • Ahhhh right, I thought you obtained the conclusion without the PDE itself. Now I look stupid in front of you sorry ! – Chee Han Jun 11 '15 at 09:03