# Exponential decay of solution in $L^p$ with $p>2$

Consider the following evolution equation

$$u_t=\Delta u$$ in a bounded and regular open subset $$\Omega$$ of $$\mathbb{R}^N$$, with smooth initial conditions $$u_0\geq 0$$ and homogeneous Dirichlet boundary conditions.

It is known that this equation has a smooth global solution $$u$$. By differentiating $$E(t):=\int_\Omega u^2 dx$$ and using integration parts we get $$E'(t)=2\int_\Omega uu_t=2\int_\Omega u\Delta u=-2\int_\Omega |\nabla u|^2.$$ Now by using Poincaré's inequality we get $$E'(t)\leq (-\frac{2}{c^2})E(t)$$ where $$c$$ is the Poincaré constant. This implies that $$|u(t)|_{L^2}\leq e^{\left( -\frac{1}{c^2}\right)t}|u(0)|_{L^2}$$ which means that the solution decays exponentially.

My question is: can we prove the above property but in $$L^p(\Omega)$$ for $$p>2$$ or maybe even $$L^\infty(\Omega)$$? $$p=2$$ is very important in the proof I gave because after the integration by parts we get the term $$\int_\Omega |\nabla u|^2dx$$ which we know how to estimate in terms of $$\int_\Omega u^2 dx.$$

• I think you can, using Gronwalls inequality. – mattos Jul 29 '19 at 3:00
• @Mattos If I take $E(t)=\int u^p$, then $E'(t)=pd\int u^{p-1}\Delta u+pa\int u^p=-pd\int \nabla u^{p-1}.\nabla u+paE(t)$. The problem is I don't know any kind of inequality between $\int \nabla u^{p-1}.\nabla u dx$ and $\int u^p dx$ except when $p=2$ which is Poincaré's inequality. – David Lingard Jul 29 '19 at 14:55