Poisson Equation Inequality

Hi all. I’m new to the forums; I’m a PhD student in engineering with a Masters in mathematics, and I’m taking a course in PDEs right now. This statement below was claimed to be ‘clear' in our course notes by applying integration by parts, but I’m not seeing how. And, the professor’s out of town and unreachable (of course).

Let $\displaystyle u\in C_{0}^{\infty}(\Omega)$ (i.e. u is smooth with compact support on a region Omega) solve the Poisson equation (i.e. $\displaystyle \nabla(u)(x)=-f(x)$ in $\displaystyle \Omega$). Then, this inequality holds:

$\displaystyle \int_{\Omega}\sum_{k=1}^{n}|\frac{d^{2}u}{dx_{k}^{ 2}}|^{2} dx \leq \int_{\Omega}|f(x)|^{2} dx$

I’m getting really frustrated if it’s as ‘clear’ as the notes claim it is. Help is much appreciated in advance!

Re: Poisson Equation Inequality

Hey teonichelli.

What about considering the Triangle Inequality for Norms along with the definition of the total differential?

Total derivative - Wikipedia, the free encyclopedia

Triangle inequality - Wikipedia, the free encyclopedia

Re: Poisson Equation Inequality

I don’t think total derivatives should be necessary here. And anyway, wouldn’t applying the Triangle Inequality intuitively provide us with the wrong inequality?

Re: Poisson Equation Inequality

The triangle inequality can relate the terms inside the norm to the individual normed terms.

This relates the sum of the norms (the sigma of norm terms) to one norm and the idea was to highlight the link between the total differential (it will be with respect to the 2nd derivatives) and the one single norm containing all the 2nd derivatives.

You may also want to look at results for the Hessian matrix and norm identities relating to that.