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Consider $u \in C_1^2(\Omega \times [0,T]), \Omega\subset\mathbb{R}^n$ as a solution of the problem

$ u_t - \Delta u = f, \text{ in } \Omega \times (0, T]$,

$u = 0, \text{ on } \partial\Omega \times [0,T]$,

$u(x,0) = g(x) , x \in \Omega$.

Then the so called "energy inequality"

$\int_\Omega |u(x,t)|^2 dx + \int_0^t \int_\Omega |\nabla u(x,\tau)|^2 dxd\tau \leq c \int_0^t \int_\Omega |f(x,\tau)|^2 dxd\tau + \int_\Omega |g(x)|^2 dx$

holds. I have never heard of such an energy inequality and I´m looking for a proof but I have not found one yet.

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This is nothing more than integration by parts. Multiply your equation by $u$, integrate and take the derivative w.r.t. to t out (I let the formal verification to you): $$ \frac{1}{2}\frac{d}{dt}\int_\Omega |u(x,t)|^2 dx = \int_\Omega ( u\Delta u )(x,t)dx + \int_\Omega f(x,t) dx. $$ Using integration by parts on the form $u\Delta u$ and integrating in time will give you what you want.

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  • $\begingroup$ I got to the point you wrote in your answer but integrating by parts won´t work for me. I also don´t see where the term with $g$ comes in. $\endgroup$ – user348978 Jun 20 '16 at 13:11
  • $\begingroup$ when you integrate the integral of the derivative you get the initial condition. $\endgroup$ – Sergio Almada Jun 21 '16 at 2:33
  • $\begingroup$ Could you write the solution down here? I can't reproduce what you mean because firstly I got $\int_\Omega f(x,t)u(x,t)$ instead of $\int_\Omega f(x,t)$, and the second point is that it's somehow unclear to me how to integrate the laplace operator. $\endgroup$ – user348978 Jun 21 '16 at 9:26

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