# An inequality about linear PDE

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.

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