# Neumann/Robin Laplacian semigroup well-known estimate

Let $$\Delta_R:D(\Delta_R)\to L^2(\Omega)$$ the Robin Laplacian defined on:

$$D(\Delta_R)=\left\{u\in H^1(\Omega)\ \big |\ \Delta u\in L^2(\Omega),\ \dfrac{\partial u}{\partial\nu}+bu=0 \ \text{on}\ \partial\Omega\right\}$$,

where $$b\in L^{\infty}(\partial\Omega)$$ (can be taken positive if needed). Denote by $$T(t)_{t\geq 0}$$ the semigroup generated by $$\Delta_R$$. Here $$\Omega\subseteq\mathbb{R}^N$$ is an open and bounded set with uniform Lipschitz boundary.

I read in some article that it can be shown that for any $$1\leq q\leq p\leq +\infty$$ there is a constant $$C=C(\Omega,p,q)>0$$ (depending only on $$\Omega,p,q$$) such that following estimate hold:

$$\Vert T(t)\phi\Vert_{L^p(\Omega)}\leq C t^{-\frac{N}{2}\left (\frac{1}{q}-\frac{1}{p}\right )}\Vert\phi\Vert_{L^q(\Omega)},\ \forall\ \phi\in L^q(\Omega).$$

How can we prove that inequality?

I read the proof for Dirichlet boundary conditions in T. Cazenave & A. Haraux - An introduction to Semilinear Evolution Equations,1998, page 44. But how can it be done for Neumann or Robin boundary conditions?

• The exponent $-N/2\times (1/p-1/q)$ should be replaced by $-N/2 \times (1/q-1/p)$? Feb 21, 2021 at 12:37
• Of course. Sorry. Feb 21, 2021 at 13:46
• At least, if $p=\infty, q=1$, and $\Omega$ is a bounded Lipschitz domain, your inequality should follow from a Sobolev type inequality. Feb 21, 2021 at 14:13

Since the Robin Laplacian is associated to a bilinear form with form domain $$H^1(\Omega)$$, part (v) of the theorem is useful to answer the question: it tells us - in the case $$N > 2$$ - that we have ultra contractivity of the semigroup if and only if $$H^1(\Omega)$$ embeds into $$L^{2N/(N-2)}(\Omega)$$, i.e., the question reduces to a Sobolev embedding theorem. Such an embedding theorem is satisfied if the domain $$\Omega$$ isn't too rough; more precisely:
Assume that the domain $$\Omega$$ has the extension property (which is, for instance, satisfied if $$\Omega$$ is bounded and has Lipschitz boundary). If $$N > 2$$, then the embedding $$H^1(\Omega) \hookrightarrow L^{2N/(N-2)}(\Omega)$$ is true, so we get the desired ultracontractivity [op. cit., Subsection 7.3.6].
• Thanks for your answer! I read the theorem that you have mentioned above. I work mainly in spaces of dimension $N=2$. But from what I see on page 69 (Neumann Laplacean) in the survey it seems to work even if $N=2$ (if we have the extension property satisfied). So similarly we can deduce that it holds for Robin bc. Am I right? Feb 21, 2021 at 15:03
• @Bogdan: Good question; right now I'm somewhat confused by the case $N=2$, but this is most likely just due to my ignorance. I'll let you know in case that my confusion fades away... Feb 21, 2021 at 17:04
• @Bogdan: A few days ago, I briefly discussed the case $N=2$ with a colleague. Could you contact me via email about this (since the information I got is somewhat vague, I'd prefer not to discuss this publicly)? (Please see my profile for a link to my webpage, where my email address can be found.) Mar 21, 2021 at 16:52