Really Struggling :(

• Oct 17th 2010, 06:03 AM
OliviaB
Really Struggling :(
Hi,
I am really struggling with a subject at my university which is all about partial differential equations. It takes me a long time to understand even the most basic concepts, and I really need some help understanding and completing the following questions.

Let $\mathcal{R}$ be a bounded region in $\mathbb{R}^3$, and suppose $p(x) > 0$ on $\mathcal{R}$.

(i) If $u$ is a solution of

$\bigtriangledown^2 u = p(x) u \ \ x \in \mathcal{R} \ \ \bigtriangledown \cdot n = 0 \ \x \in \partial \mathcal{R}$

show that $u \equiv 0$ on $\mathcal{R}$

(ii) If $u$ is a solution of

$\bigtriangledown^2 u = p(x) u \ \ x \in \mathcal{R} \ \ \bigtriangledown \cdot n = g(x) \ \x \in \partial \mathcal{R}$

show that $u$ is unique (It can be assumed that part (i) is true).
• Oct 18th 2010, 06:33 AM
OliviaB
I haven't made any progress on this one, but this is Poisson's equation, isn't it? And its inhomogenous? I think I need to use Greens identity to solve this. Is that right?
• Oct 18th 2010, 08:38 AM
Rebesques
Quote:

Originally Posted by OliviaB
Hi,
I am really struggling with a subject at my university which is all about partial differential equations. It takes me a long time to understand even the most basic concepts, and I really need some help understanding and completing the following questions.

Let $\mathcal{R}$ be a bounded region in $\mathbb{R}^3$, and suppose $p(x) > 0$ on $\mathcal{R}$.

(i) If $u$ is a solution of

$\bigtriangledown^2 u = p(x) u \ \ x \in \mathcal{R} \ \ \bigtriangledown \cdot n = 0 \ \x \in \partial \mathcal{R}$

show that $u \equiv 0$ on $\mathcal{R}$

(ii) If $u$ is a solution of

$\bigtriangledown^2 u = p(x) u \ \ x \in \mathcal{R} \ \ \bigtriangledown \cdot n = g(x) \ \x \in \partial \mathcal{R}$

show that $u$ is unique (It can be assumed that part (i) is true).

Well, it looks like a typical application of the Lax-Milgram theorem to me.

The weak formulation of this problem on $L^2(\mathcal{R})$ reads

$\int_{\mathcal{R}}(\bigtriangledown u)^2v=\int_{\mathcal{R}}puv,\ \forall v\in L^2(\mathcal{R})$

or, by integrating the left hand side by parts and using the boundary conditions,

$\int_{\mathcal{R}}[\langle \nabla u,\nabla v\rangle+puv]=0$.

Define the bilinear form on $L^2(\mathcal{R})\times L^2(\mathcal{R})$,

$\alpha(u,v)=\int_{\mathcal{R}}[\langle \nabla u,\nabla v\rangle+puv]$.

To show that $\alpha(u,v)=0$ is solvable for some $u$, we can invoke the Lax-Milgram theorem: We need just show that $\alpha$ is bounded and coercive.

You can show it is bounded, ie there exists a $B>0$ such that

$|\alpha(u,v)|\leq B||u||||v||$, where $||u||^2=\int_{\mathcal{R}}u^2+|\nabla u|^2.$

As for $\alpha$ being coercive, we must show there is $C>0$ with $\alpha(u,u)\geq C||u||^2$. We compute

$\alpha(u,u)=\int_{\mathcal{R}}|\nabla u|^2+pu^2\geq\min\{1,\inf_{\mathcal{R}} p\}||u||^2$.

Now the Lax-Milgram theorem applies, and there is a $u_0\in L^2(\mathcal{R})$ such that
$\alpha(u_0,v)=0 \ \forall v\in L^2(\mathcal{R})$.
And reflexivity implies $u_0=0$.

For part ii), consider two solutions $u_1,u_2\in L^2(\mathcal{R})$ of the problem at hand.
Then their difference $u_1-u_2$ is a solution for the problem given in part i), and therefore must be zero.

Ps. You cannot construct a Green's function here, as zero is an eigenvalue for $-\bigtriangledown^2+p$ on a region.