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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.

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  • $\begingroup$ 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. $\endgroup$ Jul 6, 2012 at 7:58

2 Answers 2

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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.

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  • $\begingroup$ Thanks to all the comments, enlightening :) I was slightly lost to not see a complex conjugate. With that , I can go on ! $\endgroup$
    – zebullon
    Jul 6, 2012 at 8:47
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According to last but one equation $$\left< -{ \nabla }^{ 2 }u,u \right>= \left< { \nabla }u,\nabla u \right>>0 $$ for $u\not\equiv 0\;$.

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  • $\begingroup$ 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. $\endgroup$
    – anon
    Jul 6, 2012 at 7:29

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