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.