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 $u$.

  • 1
    $\begingroup$ 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 $\Omega\times [0,T]$ $\endgroup$ Jul 3, 2015 at 22:49
  • 1
    $\begingroup$ I'm voting to close this question as off-topic because it belongs on Math.SE. $\endgroup$ Jul 4, 2015 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, 2015 at 18:19

1 Answer 1


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.

  • $\begingroup$ 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, 2015 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. $\endgroup$ Jul 4, 2015 at 22:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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