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.

In a textbook of functional analysis I found this equation derived from Green's first identity

$$\int _{ \Omega }^{ }{ u{ \nabla }^{ 2 }ud\tau } =\int _{ \partial \Omega }^{ }{ u\frac { \partial u }{ \partial n } ds } -\int _{ \Omega }^{ }{ \left| \nabla u \right| ^{2}d\tau }$$

Then it goes on saying that if the boundary conditions on u are such that the integral over the boundary vanishes then the operator $ -\nabla^{2}$ is positive definite. Why ?

What I can see is that $$\int _{ \Omega }^{ }{ u{ \nabla }^{ 2 }u+{ \left| \nabla u \right| }^{ 2 }d\tau } =0$$ and what I'd need to declare that the operator is positive definite is : $$\left< -{ \nabla }^{ 2 }u,u \right> >0\Leftrightarrow \int _{ \Omega }^{ }{ -{ \nabla }^{ 2 }u\bar { u } d\tau } >0$$

So far I don't see how to prove that the operator is positive definite... Thanks for any kind of help.

share|improve this question
    
The form of Green's first identity your wrote down only applies to real valued functions, because for complex numbers $z^2 \neq |z|^2$ in general. –  Willie Wong Jul 6 '12 at 7:58
add comment

2 Answers 2

up vote 2 down vote accepted

Green's identity reads:

$$\int_U \left( \psi \nabla^{2} \varphi + \nabla \varphi \cdot \nabla \psi\right)\, dV = \oint_{\partial U} \psi \left( \nabla \varphi \cdot \mathbf{n} \right)\, dS$$

Select $\psi=\bar{u}$ and $\varphi=u$ and negate:

$$-\int_U\bar{u}\nabla^2u+\nabla \bar{u}\cdot\nabla u\; dV=-\oint_{\partial U}\bar{u}(\nabla u\cdot\mathbf{n})dS.$$

Of course $\nabla u\cdot\mathbf{n}=0$ by hypothesis on the boundary conditions, so we may rearrange this to

$$\left\langle -\nabla^2 u,u\right\rangle=\int_U \overline{\nabla u}\cdot \nabla u\;dV.$$

The integrand on the right is $\sum_i|\partial u/\partial x_i|^2$, so of course it is nonnegative, and is in fact only zero when $\nabla u=0$. In fact the integral on the right displays a means to defining an inner product for complex-valued vector functions, hence $\langle\nabla u,\nabla u\rangle $ in Andrew's answer, and knowing this a priori would provide a very direct means to seeing positive definiteness. The inner product is

$$\langle \mathbf v,\mathbf w\rangle =\int_U \mathbf v\cdot \overline{\mathbf w}\; dV.$$

It is somewhat unclear to me if the statement about $-\Delta^2$'s positive definiteness is in the context of real-valued or complex-valued functions $u$. In the former situation $u=\bar{u}$ so you already had all you needed, and in the latter situation the identity it had was slightly off (didn't involve a complex conjugate) for the purpose at hand, albeit a slight modification was all that was necessary.

share|improve this answer
    
Thanks to all the comments, enlightening :) I was slightly lost to not see a complex conjugate. With that , I can go on ! –  zebullon Jul 6 '12 at 8:47
add comment

According to last but one equation $$\left< -{ \nabla }^{ 2 }u,u \right>= \left< { \nabla }u,\nabla u \right>>0 $$ for $u\not\equiv 0\;$.

share|improve this answer
    
By "last but one" do you mean second-to-last? Do note that the middle equation in the OP does not include a complex conjugate, which is what is tripping the OP up I believe. –  anon Jul 6 '12 at 7:29
add comment

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.