# How to prove that the L2 norm is a non-increasing function of time?

1. Sep 13, 2014

### Su3liminal

1. The problem statement, all variables and given/known data

2. Relevant equations
How can I start the proof? Shall I use the Poincare inequality?

3. The attempt at a solution
Well, I know that this norm is defined by , but still I don't know how to start constructing the proof?

Last edited: Sep 13, 2014
2. Sep 13, 2014

### Ray Vickson

Start by omitting the square root. Take $\partial / \partial t$ inside the integral sign (justify!), then use the DE to eliminate, or at least, modify $\partial{u^2}/\partial t$.

3. Sep 13, 2014

### Su3liminal

Thanks! I have done what you said (note that I just made a change in variables so I stick to the symbol convention of integration by parts.):

$\\2\int_{0}^{L}\frac{\partial^2 s}{\partial x^2}s dx \\ \\ \\u=s, dv=\frac{\partial^2 s}{\partial x^2}dx \\du=\frac{\partial s}{\partial x}dx,v=\frac{\partial s}{\partial x} \\ \\ \\\therefore 2\int_{0}^{L}\frac{\partial^2 s}{\partial x^2}s dx=2s\frac{\partial s}{\partial x}\mid-2\int_{0}^{L}\frac{\partial s}{\partial x}\frac{\partial s}{\partial x}dx=-2\int_{0}^{L}(\frac{\partial s}{\partial x})^{2}dx$

Is that sufficient?

Last edited: Sep 13, 2014
4. Sep 14, 2014

Standard for me would be to start with $\frac{\partial u}{\partial t}= \frac{\partial^2 u}{\partial x^2}$ and multiply by $u$ then integrate over the spatial domain $[0,L]$.