Proving the Non-negativity Property of a Diffusion Equation Solution

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Homework Statement



Let u(x,t) satisfy


Homework Equations




(\partialu/\partialt) = (\partial^{2}u/\partialx^{2})...(0<x<1,t>0)

u(0,t)=u(1,t)=0...(t\geq0)

u(x,0)=f(x)...(o\leqx\leq1),

where f\inC[0.1] show that for any T\geq0

\int from 0..1 (u(x,T))^{2}dx \leq \int from 0..1 (f(x))^{2}dx


The Attempt at a Solution



not sure
 
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That's 3 posts and no work shown on anyone of them. You *must* show an attempt at a solution, or we cannot be of help to you. Please click on the "Rules" link at the top of the page.
 
im sorry, I am obviously knew to this forum...

For this problem, I am trying to use the identity as follows

2u((\partialu/\partialt)-(\partial^{2}u/\partialx^{2})) = (\partialu^{2}/\partialt)-(\partial/\partialx)*(u*(\partialu/\partialx))+2*(\partialu/\partialx)^{2}
 
It's a diffusion equation, so you might expect this sort of behavior. Your 'identity' is a little messed up. Can you fix it? Once you've done that substitute the PDE in. You should be able to show that ((u^2),t)/2-(u*(u,x)),x=(-(u,x)^2)<=0. I'm using commas for partial derivatives, forgive my laziness. Now integrate dx between 0 and 1. Can you show the (u*(u,x)),x term vanishes? Once you have integal (u^2),t<=0 you are home free.
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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