7
$\begingroup$

This is a cross-post from Computational Science.

I am interested in proving or obtaining a counterexample to the following conjecture.

Let $\Omega\subset\mathbb{R}^d$ be a bounded open domain. Let $u_d\in H^{1/2}(\partial\Omega) \times \mathbb{R}^+$. Suppose $u\in L^2((0,\infty); H^1(\Omega))$ is the weak solution of $$ \left\{ \begin{align*} u_t- \nabla \cdot \nabla u &= 0&\, \mathrm{in}\,& (\Omega\times \mathbb{R}^+) \\ u &= u_d&\, \mathrm{on}\,& (\partial\Omega \times \mathbb{R}^+)\\ u &= 0&\, \mathrm{at}&\, (\Omega \times \{0\}) \end{align*} \right. $$ Then there exists a constant $C$ dependent only on the domain $\Omega$ such that $ \left\| u \right\|_{L^\infty((0,\infty); L^2(\Omega))} \le C \left\| u_d \right\|_{L^{\infty}((0,\infty); H^{1/2}(\partial\Omega))}. $

I haven't been able to find many texts that treat the heat equation with time varying boundary conditions. Also, feel free to change the Sobolev spaces if you need to.

Addendum #1 Response to Denis Serre. Thanks for the response. I would like to check my understanding. To simplify things I am going to consider a 2-dimensional problem.

We modify the problem as follows.

Let $\Omega \subset \mathbb{R}^2$ be defined as $\Omega = \{(x,y)\in \mathbb{R}^2: x>0\}$

Suppose $u\in L^2(\mathbb{R}^+; H^1(\Omega)) $ is the weak solution of the above initial boundary value problem.

Step #1, Laplace Transform in $t$.

$0=\tau U(\tau,x,\eta) - \frac{\mathrm{d}^2}{\mathrm{d}y^2} U(\tau,x,\eta) - \frac{\mathrm{d}^2}{\mathrm{d}x^2}U(t,x,\eta)$

Step #2, Fourier transform in $y$.

$0 = \tau\, \mathcal{U}(\tau, x,\eta) + |\eta|^2 \mathcal{U}(\tau, x,\eta) - \frac{\mathrm{d}^2}{\mathrm{d}x^2} \mathcal{U}(\tau,x,\eta)$

Step #3

Solve the resulting ODE.

$\mathcal{U}(\tau, x,\eta) = \mathcal{U}_d(\tau, \eta) \exp(-x\sqrt{\tau+|\eta|^2)}$

Step #4

Estimate $L^2$ norm.

$\int_{x\in \mathbb{R}^+}|\mathcal{U}|^2\, \mathrm{d}x = \cdots = |\mathcal{U}_d(\tau, \eta)|^2 \frac{1}{2\sqrt{\tau+|\eta|^2}} $

Step #5

Go back by Plancherel

$$ \begin{align*} \int_{(x,y)\in \Omega} |U(\tau, x,y)|^2 \mathrm{d}x \mathrm{d}y &= \int_{\eta\in \mathbb{R}} \int_{x\in \mathbb{R}^+} |\mathcal{U} (\tau,x,\eta)|^2 \mathrm{d}x\, \mathrm{d}\eta \\ &= \int_{\eta\in \mathbb{R}} |\mathcal{U}_d(\tau, \eta)|^2 \frac{1}{2\sqrt{\tau+|\eta|^2}}\mathrm{d}\eta\\ &= \int_{\eta\in \mathbb{R}} |\mathcal{U}_d(\tau,\eta)|^2 \frac{1}{2 \sqrt{\tau+ |\eta|^2}} \frac{(1+|\eta|^2)^{1/4}}{(1+|\eta|^2)^{1/4}}\, \mathrm{d}\eta \\ &\le \frac{1}{2\sqrt{\tau}} \left\|U_d(\tau,\cdot)\right\|^2_{H^{1/2}(\partial\Omega)} \end{align*} $$

I honestly don't know how to apply Paley-Wiener here. Any hints would be appreciated. If I am totally off track, please let me know too. Thanks.

$\endgroup$
  • $\begingroup$ Do you have access to L.C.Evans' very large book? $\endgroup$ – Jesse C. McKeown Aug 5 '15 at 17:10
  • $\begingroup$ Yes, I have it. The AMS book that is titled "Partial Differential Equations". $\endgroup$ – fred Aug 5 '15 at 20:42
2
+100
$\begingroup$

The theory of linear Intial-boundary value problem proceeds the following way. In your case, because of the zero initial condition, you may replace ${\mathbb R}^+$ by $\mathbb R$ and define $u\equiv0$, $u_d\equiv0$ for $t<0$. Then what matters is the special case where $\Omega$ is a half-space, say $\{x_d>0\}$. In this simpler situation, you can (and you must) perform a Fourier transform in the tangential variables $x_1,\ldots,x_{d-1}$, and a Laplace transform in time. You end up with an ODE, parametrized by the frequency variables $\eta\in{\mathbb R}^{d-1}$ and $\Re\tau>0$ : $$\tau v+|\eta|^2v-\frac{d^2v}{dx_d^2}=0,\qquad v:=\hat u$$ with an ''initial'' data $$v(\tau,\eta,0)=\hat u_d(\tau,\eta).$$ Here, $x_d\mapsto v$ is the unique square integrable solution over ${\mathbb R}^+$.

Write the explicit form of $v(\tau,\eta,\cdot)$, estimate its $L^2$-norm (or every $H^s$-norm you wish). Then go back by Plancherel and Paley-Wiener. You obtain an accurate estimate in weighted spaces $L^2(e^{-\gamma t}dt;L^2(\Omega))$.

The method is detailed in my book with S. Benzoni-Gavage, in the case of hyperbolic equations (like the wave equation), but it can be applied to any well-posed IBVP. It is also used to detect the well- or ill-posedness (key word : Lopatinskii condition).

$\endgroup$
1
$\begingroup$

In the paper by Weidermann, Maximal regularity for parabolic equations with inhomogeneous boundary conditions in Sobolev spaces with mixed $L_p$−norm, 2002, you can find in particular the estimate $$ \Vert u \Vert_{L^2 ([0, T]; H^2 (\Omega))} + \Vert u \Vert_{H^1 ([0, T]; L^2 (\Omega))} \le C \Bigl(\Vert u_d \Vert_{L^2 ([0, T]; H^{3/2} (\partial \Omega))} + \Vert u_d \Vert_{H^{3/4} ([0, T]; L^2 (\partial \Omega))} \Bigr), $$ provided that $u_d (0) = 0$ on $\{0\} \times \partial \Omega$.

This estimate implies an estimate in $L^\infty ([0, T]; L^2 (\Omega))$. Compared to the question, remark that there we need some regularity in time on the Dirichlet boundary condition.

One way to obtain such estimates for linear problems with nonhomogeneous boundary conditions is to first construct an extension $v$ satisfying your initial and boundary condition with suitable estimate (this is called trace theory) and then apply the theory for homogeneous boundary conditions to the new function $u - v$.

$\endgroup$
  • $\begingroup$ I appreciate the answer. I am going to go through the paper this weekend. I haven't looked at it yet. Do you know if the result can be extended from $L^{\infty}([0,T]; L^2(\Omega))$ to $L^{\infty}(\mathbb{R}^+; L^2(\Omega))$. I ask because a lot of similar estimates use Gronwall's lemma/ inequality in such a way that the hidden constant goes to infinity exponentially as $T\to \infty$, $\endgroup$ – fred Aug 7 '15 at 20:09

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.