# Thread: 2-Dimensional Boundary Value Problem

1. ## 2-Dimensional Boundary Value Problem

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

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

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

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

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

is $u=0$. The hint given is to apply the 2-D divergence theorem on the vector field $u\nabla u$, 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

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

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

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

Now consider the flux integral out of the disk

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

This is zero because u=0 on the boundary.

Now if we use the divergence theorem we get

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

This gives

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

Combining these two equations gives

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

So the integrand must be equal to zero so

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

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!

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

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

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

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

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

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

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

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