I am trying to solve the following problem:

Let $\Omega$ be a bounded smooth domain in $\mathbb{R}^n, \ n\geq 2$. Let $u\in C^2(\overline{\Omega})$ be a solution of $$\left\{\begin{array}{ll}u_t-\Delta u=f(x)& \text{in } \Omega\times(0,\infty)\\ u=0&\text{on }\partial \Omega\times (0,\infty) \\ u=g(x) & \text{on } \Omega\times\{0\}\end{array}\right.$$ Show that $$\max_{0\leq t\leq T} \int_\Omega u^2(x,t)dx+\int_0^T\int_\Omega|\nabla u(x,t)|^2dx\, dt\leq C\left(\int_\Omega g^2(x)dx+\int_0^T|f(x)|^2 dx\,dt\right)$$ for some constant $C$ independent of $f,\ g$ and $u$.

My attempt:

Multiplying the first equation by $u$ and taking integration on $\Omega$, we have $$\frac{1}{2}\frac{d}{dt}||u||^2_{L^2}+\int_\Omega \nabla u\cdot \nabla u=\int_\Omega fu.$$ (Here we also use the Green's identity and the boundary condition of $u$)

Taking integration on $[0,s]$ where $s\leq T$. Then we have $$\frac{1}{2}||u(x,s)||^2_{L^2}+\int_0^s\int_\Omega|\nabla u(x,t)|^2dx\, dt= \frac{1}{2} \int_\Omega g^2(x)dx+\int_0^s\int_\Omega fu\ dt$$ where $$\int_\Omega fu\leq ||f||_{L^2} ||u||_{L^2}\leq \epsilon ||u||_{L^2}^2 + C(\epsilon ) ||f||_{L^2}$$

Then we have $$(\frac{1}{2}-s\epsilon )||u(x,s)||^2_{L^2}+\int_0^s\int_\Omega|\nabla u(x,t)|^2dx\, dt \leq \frac{1}{2} \int_\Omega g^2(x)dx+\int_0^s C(\epsilon ) ||f||_{L^2} dt$$

Then I was stuck here. I don't know where can I derive the part $$\max_{0\leq t\leq T}||u(x,t)||.$$ Also, I am struggling how to make the two coefficients before the two terms on the left the same so that we can get the desired inequality.

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    $\begingroup$ Solved it. Tell me if someone needs the answer. The idea is right. Just need some subtle modification about $\epsilon$. $\endgroup$ – Aolong Li Dec 14 '18 at 7:29
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    $\begingroup$ I for one would like to see your solution. Cheers! $\endgroup$ – Robert Lewis Dec 14 '18 at 7:38
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    $\begingroup$ the usual convention is to post your own answer and accept it. $\endgroup$ – dezdichado Dec 14 '18 at 15:56
  • $\begingroup$ @dezdichado OK thanks! Will do! $\endgroup$ – Aolong Li Dec 14 '18 at 16:00

What I did above tells us that $$||u(x,s)||^2_{L^2}+\frac{1}{\frac{1}{2}-s\epsilon}\int_0^s\int_\Omega|\nabla u(x,t)|^2dx\, dt \leq \frac{1}{\frac{1}{2}-s\epsilon}\left(\frac{1}{2} \int_\Omega g^2(x)dx+\int_0^s C(\epsilon ) ||f||_{L^2} dt \right) $$ for any $s>0$.

Now, pick up an appropriate $\epsilon>0$ such that $$\frac{1}{2}-\epsilon>0$$ and $$\int_0^T\int_\Omega |\nabla u|^2 \ dx\ dt\leq \frac{1}{\frac{1}{2}-\epsilon}\int_0^{t_0}\int_\Omega |\nabla u|^2 \ dx\ dt$$ where $t_0$ is the point where $||u(x,t)||^2_{L^2}$ obtain the maximum. Then we have $$\begin{eqnarray}||u(x,t_0)||^2_{L^2}+\int_0^T\int_\Omega |\nabla u|^2 \ dx\ dt &\leq& ||u(x,t_0)||^2_{L^2} + \frac{1}{\frac{1}{2}-\epsilon}\int_0^{t_0}\int_\Omega |\nabla u|^2 \ dx\ dt\\ &\leq& \frac{1}{\frac{1}{2}-\epsilon}\left(\frac{1}{2} \int_\Omega g^2(x)dx+\int_0^T C(\epsilon ) ||f||_{L^2} dt\right)\\ &\leq& C'\left( \int_\Omega g^2(x)dx+\int_0^T C(\epsilon ) ||f||_{L^2} dt\right) \end{eqnarray}$$ for some constant $C'$.


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