# Thread: 2-Dimensional Boundary Value Problem

1. ## 2-Dimensional Boundary Value Problem

Let $\displaystyle D$ be the unit disk in $\displaystyle \mathbb{R}^2$ and let $\displaystyle \partial D$ denote the boundary. Show that the only solution of the boundary value problem:

$\displaystyle \Delta u - u = 0$ in $\displaystyle D$
$\displaystyle u = 0$ on $\displaystyle \partial D$

is $\displaystyle u=0$. The hint given is to apply the 2-D divergence theorem on the vector field $\displaystyle u\nabla u$, but I don't see how that even enters in.

2. Originally Posted by mathematicalbagpiper
Let $\displaystyle D$ be the unit disk in $\displaystyle \mathbb{R}^2$ and let $\displaystyle \partial D$ denote the boundary. Show that the only solution of the boundary value problem:

$\displaystyle \Delta u - u = 0$ in $\displaystyle D$
$\displaystyle u = 0$ on $\displaystyle \partial D$

is $\displaystyle u=0$. The hint given is to apply the 2-D divergence theorem on the vector field $\displaystyle u\nabla u$, but I don't see how that even enters in.
You are going to want to calculate an integral two different ways

First note that

$\displaystyle \Delta u=u \implies u\Delta u=u^2$

The right hand side of the above equation is non negative so we get that

$\displaystyle \iint_{D}u\Delta udA= \iint_{D}u^2dA \ge 0$

Now consider the flux integral out of the disk

$\displaystyle \oint_{\partial D} (u \nabla u)\cdot d\mathbf{s}=0$

This is zero because u=0 on the boundary.

Now if we use the divergence theorem we get

$\displaystyle 0=\oint_{\partial D} (u \nabla u)\cdot d\mathbf{s}=\iint_{D}\nabla \cdot (u \nabla u)dA=\iint_{D}|\nabla u|^2dA+\iint_{D}u\Delta u dA$

This gives

$\displaystyle -\iint_{D}u\Delta u dA = \iint_{D}|\nabla u|^2dA \ge 0 \iff \iint_{D}u\Delta u dA \le 0$

Combining these two equations gives

$\displaystyle 0 \le \iint_{D}u\Delta u dA \le 0 \implies \iint_{D}u\Delta u dA=0$

So the integrand must be equal to zero so

$\displaystyle u\Delta u=0 \iff u=0 \text{ or } \Delta u=0$

If it is the first cases we are done if it is the plug that into the pde and conclue.

3. I am now doubting the above the very last line does not hold!

$\displaystyle 0 \le \iint_{D}u\Delta u dA \le 0 \implies \iint_{D}u\Delta u dA=0$

So the integrand must be equal to zero
This is only true if the integrand does not change signs and I have not shown that!

There are many counter examples. Sorry

4. Ahh okay here is the fix. It is true that

$\displaystyle 0 \le \iint_{D}u\Delta u dA \le 0 \implies \iint_{D}u\Delta u dA=0$

Now if we sub this back into the above equation we get

$\displaystyle 0=\iint_{D}|\nabla u|^2dA+\iint_{D}u\Delta u dA \iff 0=\iint_{D}|\nabla u|^2dA$

Now this integrand is non negative so it must be zero this gives

$\displaystyle |\nabla u|^2=0 \iff \nabla u=0 \implies u=c$

is a constant but u on the boundary is zero so $\displaystyle u \equiv 0$