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

Let $\Omega\in \mathbb{R}^d$ 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. Thanks so much for any direction you can give me. Thanks.

Addendum: Following @WolfgangBangerth's advice we have the following.

Step One, convert non-homogeneous boundary conditions to a right hand side.

Let $u_{BC}\in H^1(\Omega) \times \mathbb{R}^+$ be the weak solution to the following elliptic problem

$$ \left\{ \begin{align*} -\nabla \cdot \nabla u_{BC} &= 0 & \mathrm{in}& (\Omega \times \mathbb{R}^+)\\ u_{BC} &=u_d &\mathrm{on} &(\partial\Omega \times \mathbb{R}^+) \end{align*} \right. $$

So $u_{BC}$ is the weak solution of an elliptic equation at each instant in time.

Now define $w = u-u_{BC}$. Then we have that $w$ is the weak solution of the following homogeneous initial boundary value problem. $$ \left\{ \begin{align*} w_t - \nabla \cdot \nabla w &= -(u_{BC})_t& \mathrm{in}& (\Omega\times \mathbb{R}^+)\\ w &= 0& \mathrm{on}& (\partial\Omega \times \mathbb{R}^+)\\ w &= -u_{BC}\big|_{t=0} &\mathrm{at} &(\Omega\times \{0\}) \end{align*} \right. $$

Step Two Provide bounds for each eigenfunction.

Let $\left\{(\lambda_i^2, V_i)\right\}_{i=1}^\infty\in (\mathbb{R}^+ \times H_0^1)$ be the the eigenvalue and eigenfunction pairs ordered by increasing eigenvalue. Expand $w$ as $w = \sum_{i=1}^\infty \omega_i(t) V_i$ where $\omega_i(t)$ is the weight of the $i$'th eigenfunction at time $t$.

Then after testing the PDE with a test function $V_i$ we get

$$ \frac{d}{dt} \omega_i(t) + \lambda_i^2 \omega_i(t) = -(V_i, \frac{d}{dt} u_{BC}) $$

This ODE has solution

$$\omega_i(t) = \exp(-\lambda^2_i t) \omega_i(0) - \int_{\tau =0 }^t \exp(-\lambda_i^2(t-\tau))(V_i, \frac{\mathrm{d}}{\mathrm{d}\tau}u_{BC})\, \mathrm{d}\tau $$

Then we integrate by parts on the integral term to get

$$ \begin{align*}\omega_i(t) =& \exp(-\lambda^2_i t) \omega_i(0) - (V_i, u_{BC}(t)) + (V_i, u_{BC}(0)) \exp(-\lambda^2_i t) \\ &+ \int_{\tau=0}^t \lambda_i^2 \exp(-\lambda_i^2 (t-\tau)) ( V_i, u_{BC}(\tau))\, \mathrm{d}\tau \end{align*} $$

Now, the sum of the first and third term vanishes because of the intial conditions for $u$ in the original problem.

So we are left with $$ \begin{align*}\omega_i(t) =& - (V_i, u_{BC}(t)) + \int_{\tau=0}^t \lambda_i^2 \exp(-\lambda_i^2 (t-\tau)) ( V_i, u_{BC}(\tau))\, \mathrm{d}\tau \end{align*} $$

Now expanding $u$ as $u = \sum_{i=1}^\infty \mathcal{u}_{i}(t) V_i$ and using the previous equation we get

$$\mathcal{u}_i(t) = \int_{\tau=0}^t \lambda_i^2 \exp(-\lambda_i^2 (t-\tau)) ( V_i, u_{BC}(\tau))\, \mathrm{d}\tau$$

Step Three Estimate.

Now $$ \begin{align*} \|u\|_{L^\infty(\mathbb{R}^+, L^2(\Omega)}^2\\ &= \sup_{t\in\mathbb{R}^+} \sum_{i=1}^{\infty} \left(\mathcal{u}_i(t)\right)^2\\ &= \sup_{t\in\mathbb{R}^ +} \sum_{i=1}^{\infty} \left(\int_{\tau=0}^t \lambda_i^2\exp(-\lambda_i^2 (t-\tau)) (V_i,u_{BC}(\tau))\, \mathrm{d} \tau \right)^2\\ \end{align*} $$

This is where I am stuck. I have attempted estimates like the following but I run into problems because of the countably infinite quantity of eigenpairs.

$$ \begin{align*} & \sup_{t\in\mathbb{R}} \sum_{i=1}^{\infty} \left(\int_{\tau=0}^t \lambda_i^2\exp(-\lambda_i^2 (t-\tau)) (V_i,u_{BC}(\tau))\, \mathrm{d} \tau \right)^2\\ &\le \sup_{t\in\mathbb{R}} \sum_{i=1}^{\infty} \left(\sup_{\tau\in (0,t)} (V_i, u_{BC}(\tau)) \right)^2 \left(\int_{\tau=0}^t \lambda_i^2 \exp(-\lambda_i^2 \tau)\, \mathrm{d}\tau \right)^2\\ &\le \sup_{t\in\mathbb{R}} \sum_{i=1}^\infty \left(\sup_{\tau\in(0,t)}(V_i, u_{BC}(\tau) ) \right)^2 \end{align*} $$

Any further estimation I perform I lose the boundedness. I am guessing I just need a different inequality I am not thinking of. Any ideas? Thanks.

Another Addendum:

I don't think it is legal to expand $u$ as $u= \sum_{i=1}^\infty \mathcal{u}_i V_i$ as all the $V_i$'s are zero on the boundary condition. So a new strategy would be to bound $w$ and use Lax-Milgram and to $u_{BC}$ and then use the triangle inequality to bound $u$.

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    $\begingroup$ Isn't this straightforward application of maximum principle ? Time varying boundary condition is not an issue since for applying max. principle to heat equation, time domain is also part of the boundary, i.e. the boundary is $\Omega\times [0,T]$ $\endgroup$ – Piyush Grover Jul 3 '15 at 22:49
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    $\begingroup$ I'm voting to close this question as off-topic because it belongs on Math.SE. $\endgroup$ – David Ketcheson Jul 4 '15 at 8:29
  • $\begingroup$ @PiyushGrover I don't think the maximum principle applies as I stated the problem. The problem is that $ H^{1/2}(\partial\Omega) \not\subseteq L^{\infty}(\partial\Omega)$ at least that inclusion is not implied by the Sobolev Embedding theorem. My intuition isn't as strong as it could be when dealing with fractional spaces. Please correct me if I am wrong. Thanks. $\endgroup$ – fred Jul 4 '15 at 18:19
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You immediately get $L^\infty$ stability by using the eigenfunction decomposition of the solution. The coefficient of each mode is strictly decreasing exponentially if you don't have a right hand side (or boundary) values, and consequently the $L_2$ norm of the solution decays. In your case, you need to write down the differential equation for each mode and prove stability for it; the stability for the entire solution the follows immediately by summation.

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  • $\begingroup$ I have attempted your program and have run into a problem. I am sending you an email with my approach. Feel free to ignore it if you lack the time to look at it. Thanks, $\endgroup$ – fred Jul 4 '15 at 19:39
  • $\begingroup$ Post it as an addendum to your question above -- this way others get to learn from your obstruction, and can help out with it. $\endgroup$ – Wolfgang Bangerth Jul 4 '15 at 22:36

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