$L^{\infty}$ (uniform) decay of Dirichlet heat equation $u_t=\Delta u$

Question

Let $\Omega$ be a smooth bounded open subset of $\mathbb{R}^N$. Consider the following initial-boundary value problem for the heat equation:

\begin{equation} \begin{cases} u_t=\Delta u\quad\quad\quad\;\; \text{in}\;\Omega\times(0,\infty) , \\ u=0 \quad\quad\quad\;\;\;\;\; \text{on}\;\partial\Omega\times(0,\infty), \\ u(x,0)=u_0(x),\quad x\in\Omega. \end{cases} \end{equation}

It is known that this problem has a smooth solution $u(t,x)$. My goal is to explore its behaviour as $t$ goes to infinity.

Multiply by $u$: $$uu_t -u\Delta u = 0$$ Integrate $$\int uu_t -\int u\Delta u = 0$$ Using Green's theorem $$\frac{1}{2}\frac{d}{dt}\int u^2+\int |\nabla u|^2= 0$$

Move the second term onto the RHS and using Poincaré's inequality $$\frac{1}{2}\frac{d}{dt}\int u^2=-\int |\nabla u|^2 \leq -\lambda_1\int u^2$$ where $\lambda_1>0$ is the smallest eigenvalue of the (negative) Laplacian. This can also be written as

$$\frac{d}{dt}|u(t)|^2_{L^2(\Omega)} \leq -2\lambda_1|u(t)|^2_{L^2(\Omega)}$$ By solving this differential inequality using Gronwall's Lemma, we get $$|u(t)|^2_{L^2(\Omega)} \leq |u_0|^2_{L^2(\Omega)}e^{-2\lambda_1 t}$$ which means $$|u(t)|_{L^2(\Omega)} \leq |u_0|_{L^2(\Omega)}e^{-\lambda_1 t},$$ this implies that $$\lim_{t\to \infty}|u(t)|_{L^2(\Omega)}=0$$ which means the heat goes to zero in the ${L^2(\Omega)}$ norm.

My question is how do we prove that uniform convergence to $0$ also holds, more specifically: $$\lim_{t\to \infty}\sup_{x\in \Omega}|u(t,x)|=0$$or by another notation $$\lim_{t\to \infty}|u(t)|_{L^{\infty}(\Omega)}=0$$ I know that convergence to $0$ in $L^{\infty}$ implies convergence to $0$ in $L^2$, because $$|u(t)|_{L^{2}(\Omega)}^2\leq meas(\Omega)\sup_{x\in \Omega}|u^2(t,x)|=meas(\Omega)\left(\sup_{x\in \Omega}|u(t,x)|\right)^2=meas(\Omega)|u(t)|_{L^{\infty}(\Omega)}^2$$ The converse is not true in general. Does something special here makes it true? like compactness of trajectories in $L^{2}$ or something?

Answer 1

Let $u_1, u_2, \dots \in C^{\infty}(\Omega)$ be an orthonormal basis of $L^2$ consisting of eigenfunctions of $-\Delta$ on $\Omega$. We have $$u(t) = e^{-t\Delta}u_0,$$ that is, $$u(t, x) = \sum_{j = 1}^{\infty}e^{-t\lambda_j}\hat{u_0}(j)u_j(x).$$ Let $s > d/2$. Then \begin{align} \|u(t)\|_{H^s}^2 = \sum_{j = 1}^{\infty}\lambda_j^se^{-2t\lambda_j}|\hat{u_0}(j)|^2 \end{align} The function $f(x) = x^se^{-2tx}$ on $[0, \infty)$ has it's maximum at $x = \frac{s}{2t}$ and decreases after that, so when $t$ is large enough that $s/2t \leq \lambda_1$, $$\|u(t)\|_{H^s}^2 \leq \lambda_1^s e^{-2t\lambda_1}\sum_{j = 1}^{\infty}|\hat{u_0}(j)|^2 = \lambda_1^s e^{-2t\lambda_1}\|u_0\|_{L^2}^2.$$ By Sobolev embeddding, $$\|u(t)\|_{L^{\infty}} \leq C_s \|u(t)\|_{H^s} \leq C_s \lambda_1^{s/2} e^{-t\lambda_1}\|u_0\|_{L^2}.$$

Answer 2

WLOG we may assume that $0\in\Omega$ and consider the function $v(x,t)=e^{-bt}\displaystyle\prod\limits_{k=1}^N\cos cx_k$, where $b=c^2N>0$. It's easy to prove that $v_t=\Delta v$. Next, consider a cube $I=\left\{x\in\mathbb{R}^N:\;|x_k|<\dfrac{\pi}{4c},\;k=1,2,...,N\right\}$ and let $c$ be so small that $\overline{\Omega}\subset I$. Then clearly $v(x,0)=\displaystyle\prod\limits_{k=1}^N\cos cx_k\ge\prod\limits_{k=1}^N\cos \frac{\pi}{4}=\frac{1}{2^{N/2}}$. Therefore $|u(x,0)|\le M\le\dfrac{K}{2^{N/2}}\le Kv(x,0),$ for sufficiently large $K$. On $\partial\Omega\times[0,+\infty)$ we have $|u|=0\le Kv$. By comparsion principle, $|u(x,t)|\le Kv(x,t)\le Ke^{-bt}$ in $\Omega\times(0,+\infty)$ and result follows.