Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

From a bank of past master's exams:

Let $u(x,y)$ satisfy $$ -\left( \frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}\right) = \lambda u $$ in a bounded region $\mathcal{D} \subset \mathbb{R}^2$ with smooth boundary $\mathcal B$. Assume $u=0$ on $\mathcal B$ (but $u \not \equiv 0$ on $\mathcal D$). Show that $$ \lambda = \frac{\iint_{\mathcal D} |\nabla u|^2dxdy}{\iint_{\mathcal D} u^2dxdy}. $$ (Suggestion: Compute the divergence $\operatorname{div}(u\nabla u)$.)

So taking the suggestion, I get the following: $$ \begin{align} \nabla \cdot (u \nabla u) &= \nabla \cdot \left( u \frac{\partial u}{\partial x}, u \frac{\partial u}{\partial y} \right) \\ &= u \left( \frac{ \partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} \right) + \left( \frac{\partial u}{\partial x} \right)^2 + \left( \frac{\partial u}{\partial y} \right)^2 \\ &= -\lambda u^2 + |\nabla u|^2 \end{align} $$

If I get $$ \iint_D \nabla \cdot (u \nabla u) dA = 0,$$ then I'll have $$\begin{align} \iint_D (|\nabla u|^2 - \lambda u^2) dA &= 0 \\ \iint_D |\nabla u|^2 dA = \iint_D \lambda u^2 dA &= \lambda \iint_D u^2 dA \\ \frac{\iint_D |\nabla u|^2 dA}{\iint_D u^2 dA} &= \lambda \end{align}$$

But I don't know how to show that (I was thinking Stokes's Theorem, but I don't think the integrand is the curl of a vector field). It's probably something very simple (my vector calculus is quite rusty) -- what am I missing here?

share|improve this question
4  
Well, you're almost there and browsing through wikipedia would have answered your question: see here –  t.b. Apr 5 '11 at 18:22
2  
@Theo: Ah, I see. So then $\iint_D \nabla \cdot (u \nabla u) dA = \oint_B (u \nabla u) \cdot \vec{n} ds = \oint_B (0) \nabla u \cdot \vec{n} ds = 0$. Thank you. Now I'm not sure about the protocol here. Do I ask you to make that an official answer that I accept? Do I close the question? Do I answer my own question? –  Michael Chen Apr 5 '11 at 18:34
    
Exactly, that's what I was getting at! I only added a comment because I have to run now. I suggest that you add your own comment as an answer and accept it. Maybe you wait a while before accepting because it is more likely that people look at your question when it is still unanswered and maybe somebody has something interesting to add. –  t.b. Apr 5 '11 at 18:40

1 Answer 1

up vote 1 down vote accepted

With help from @Theo Buehler:

The Wikipedia article on Green's Theorem refers to the two-dimensional version of the divergence theorem, by which $$ \iint_A \operatorname{div}(F) dA = \oint_C F \cdot \vec{n} ds, $$ Then $$\iint_D \nabla \cdot (u \nabla u) dA = \oint_B (u \nabla u) \cdot \vec{n} ds = \oint_B (0) \nabla u \cdot \vec{n} ds = 0, $$ which is exactly what is needed to complete the proof.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.