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Is there a way to show in a relatively simple manner that for a Lipschitz bounded, connected, open domain $\Omega\subset\mathbb{R}^N$ for any $f\in L^2(\Omega)$ the solution of the problem:

$$\begin{cases}\dfrac{\partial u}{\partial t}=\Delta u, & (t,x)\in (0,T)\times \Omega\\ \dfrac{\partial u}{\partial\nu}=0, & (t,x)\in (0,T)\times\partial\Omega \\ u(0,x)=f(x), & x\in\Omega\end{cases}$$

is bounded and $\Vert u(t,\cdot)\Vert_{L^{\infty}(\Omega)}\leq c\cdot t^{-N/4}\cdot \Vert f\Vert_{L^2(\Omega)}$ for any $t\in (0,T)$, for some $c$ depending only on $\Omega$?

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  • $\begingroup$ Use the integral representation in terms of the fundamental solution? $\endgroup$
    – leo monsaingeon
    Commented Jan 12 at 20:35
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    $\begingroup$ In Theorem 6.3 and 6.4 this inequality in Ouhabaz (page 158) this inequality is equivalent with proving that $\Vert u\Vert^2_{\frac{2d}{d-2}}\leq c\cdot a(u,u),\ \forall u\in D(a)$. But this is not true in our case since $a(u,u)=\int_{\Omega}|\nabla u |^2\ dx$ and $D(a)=H^1(\Omega)$. For example $u=$constant do not satisfy such an inequality. $\endgroup$
    – Bogdan
    Commented Jan 13 at 4:40
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    $\begingroup$ One has to be a bit careful here, there a two (non-equivalent) forms of ultracontractivity. The one holds for the heat semigroup on full space and requires the $L^1$-$L^\infty$ bound from the question for all $t\geq 0$. As noted in Denis' answer, this cannot possible hold for the Neumann Laplacian on bounded domains. The second form of ultracontractivity requires the $L^1$-$L^\infty$ bound from the question only on bounded intervals with a constant $c$ depending on $T$. This form of ultracontractivity is true for the Neumann Laplacian on bounded domains with Lipschitz boundary. $\endgroup$
    – MaoWao
    Commented Jan 13 at 9:53
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    $\begingroup$ The equivalent characterization in terms of a Sobolev inequality takes the form $\lVert u\rVert_{2d/(d-1)}^2\leq c(a(u,u)+\lVert u \rVert_2^2)$, which is true for bounded domains with Lipschitz boundary. $\endgroup$
    – MaoWao
    Commented Jan 13 at 9:54
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    $\begingroup$ As explained by @MaoWao one has to apply Ouhabaz results to the form $a+1$ which gives global ultracontractivity to for the scaled semigroup and local for the original one. $\endgroup$
    – Giorgio Metafune
    Commented Jan 13 at 12:05

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The inequality is obviously false, because if $f\equiv c$ is a constant, then $u\equiv c$ remains constant. Since $\Omega$ is bounded, this $f$ is $L^2(\Omega)$, and yet $\|u(t)\|_\infty=|c|$ does not decay as $t\to+\infty$.

I suppose that you make a confusion with the heat equation in the hole space ($\Omega={\mathbb R}^N$, of course the boundary condition disappears). Then $$u(t)=H_t\star f$$ where the heat kernel is given by $$H_t(x)=\frac1{(2\pi t)^{N/2}}K\left(\frac x{\sqrt t}\right),\qquad K(y):=\exp\frac{-|y|^2}4.$$ From Young inequality (or just Hölder) $$\|u(t)\|_\infty\le\|H_t\|_2\|f\|_2=\frac c{T^{N/4}}\|f\|_2.$$

Turning back to the case of a bounded domain, the general rule is an exponential decay as $t\to+\infty$. For instance, the spectrum of $\Delta$ under the Dirichlet boundary condition $u|_{\partial\Omega}=0$ is real, bounded above by $-\mu_D<0$, and we get $\|u(t)\|_p=O(e^{-\mu_D t})$ for every $p\ge2$. In the case of the Neumann boundary condition, we must subtract the mean of the initial data: $$\|u(t)-\frac1{|\Omega|}\int_\Omega f(x)dx\|_p=O(e^{-\mu_N t}),\qquad\forall p\ge2.$$ Here $\mu_N$ is the least non-zero eigenvalue of $-\Delta$.

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