2
$\begingroup$

This is probably a known problem but I was not able to find exactly what I am looking for.

I have the following linear heat equation with zero-flux boundary conditions:

\begin{equation} \begin{cases} \dot{u} - \Delta u = u \quad \text{in} \quad \Omega;\\ \nabla u \cdot \boldsymbol{n} = 0 \quad \text{on} \quad \partial \Omega, \end{cases} \end{equation}

on a NON-CONVEX polygonal domain $\Omega \subset \mathbb{R}^2$, with $\Delta$ being the Laplace operator. From basic energy arguments, we have the following estimate

\begin{equation} \tag{1} \label{true_estimate} \|u(t)\|_{H^1(\Omega)} \leq C \|u(0)\|_{H^1(\Omega)} \exp(Ct), \qquad t >0. \end{equation}

Question: if $\Gamma \subset \partial \Omega$ is an edge of $\Omega$ -or a portion of an edge to avoid corner singularities-, do we have a similar estimate for the trace of $u$ on $\Gamma$? That is:

\begin{equation} \tag{2} \label{desired_estimate} \|Tr(u(t))\|_{H^1(\Gamma)} \leq C \left(\|Tr(u(0))\|_{H^1(\Gamma)} + \|u(0)\|_{H^1(\Omega)}\right)\exp(Ct), \qquad t>0. \end{equation}

If \eqref{desired_estimate} does not hold true, could you please provide a counterexample.

Challenges

  • Estimate \eqref{desired_estimate} cannot be inferred from \eqref{true_estimate} via trace theorem because the trace of a $H^1(\Omega)$ function is only $H^{1/2}(\Gamma)$. So we try to exploit higher regularity in the interior.

  • But we don't have enough regularity in the interior due to non-convexity. The best we can expect is a $H^{1+\varepsilon}(\Omega)$ estimate, with $0< \varepsilon < 1/2$. So, by the trace theorem we could get a $H^{1/2 + \varepsilon}(\Gamma)$ estimate on the boundary, with $0< \varepsilon < 1/2$.

$\endgroup$
1
$\begingroup$

The estimate $(2)$ is false even in a half-plane. Indeed, let $w$ be any solution to the heat equation on $\mathbb{R}^2 \times [0,\,\infty)$ that is even in $y$ (so $w$ solves the Neumann problem for the heat equation in the upper half-plane), and vanishes on the $x$-axis at $t = 0$. Then $w_R(x,\,y,\,t) := w(Rx,\,Ry,\,R^2t)$ solves the Neumann problem for the heat equation in the upper half-plane, vanishes on the boundary at $t = 0$, and has initial $H^1$ norm independent of $R$ (by the two-dimensionality of the spatial domain). We have in addition that $$f_R(t) := \int_{\mathbb{R}} |\partial_x w_R|^2(s,\,0,\,t)\,ds = Rf_1(R^2t).$$ Thus, if we let $u_R = e^tw_R$, then $u_R$ solves the desired Neumann problem in the upper half-plane, but at $t = R^{-2}$ the left side of $(2)$ is larger than $\sqrt{f_R(R^{-2})}\sim R^{1/2}$ and the right side is $\sim 1$ (independent of $R$). Taking $R \rightarrow \infty$ we see that such an estimate cannot hold.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you very much. Your argument also implies that estimate (2) can fail regardless of the regularity class of the initial condition right? However, does (2) become true by accounting for the $H^2(\Omega)$ norm of the initial datum? For instance, in your specific example the $H^2$ norm of $w_R$ is not bounded uniformly in $R$. Is that correct? $\endgroup$ – Massimo Frittelli Feb 27 at 9:32
  • 1
    $\begingroup$ That is right, the initial condition in the example can be smooth, and if you replace the $H^1(\Omega)$ norm with the $H^2(\Omega)$ norm on the right side of $(2)$ then the estimate follows from trace inequalities. $\endgroup$ – Connor Mooney Feb 27 at 17:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.