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$\newcommand{\R}{\mathbb R}$I am looking for any possible references (modelling, mathematical and numerical analysis, well-posedness, regularity, anything really!) for heat equation-like problems with nonlocal Dirichlet boundary conditions. The typical problem I might write down is for example $$ \begin{cases} \partial_t u=\Delta u & \mbox{in }(0,T)\times\Omega\\ u\rvert_{\partial\Omega}=\int_\Omega u\, \mathrm d x & \mbox{on }(0,T)\times\partial\Omega\\ u\rvert_{t=0}=u_0 & \mbox{in }\Omega. \end{cases} $$ Here $\Omega\subset \R^d$ is as usual your favourite smooth, bounded, open set. This looks like a standard Dirichlet boundary condition $u\rvert_{\partial\Omega}=g$ except that $g=\int_\Omega u$ depends nonlocally on the values of $u$ inside the domain at any given time $t>0$. Has anyone seen anything vaguely related to this kind of models, and if so can anyone provide references? Any help is much appreciated!

I am sorry I cannot really provide more details or a more focussed problem. This popped up in my research as a very secondary question, and as usual I would rather avoid reinventing the wheel if possible. And a quick bibliographical search did not provide much useful information (except for a few exotic papers in dimension $d=1$ only).

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  • $\begingroup$ Can you say (or link) what is known in 1d? $\endgroup$ Commented Oct 30, 2023 at 7:11
  • $\begingroup$ TeX note: \mbox should be dispreferred in comparison to \text for text in math mode because (besides being less semantically precise) it sizes poorly. Compare, for example, $x^\text{I am an exponent}$ x^\text{I am an exponent} to $x^\mbox{I am an exponent}$ x^\mbox{I am an exponent}. It doesn't affect your use, but maybe it is worth knowing in general. $\endgroup$
    – LSpice
    Commented Oct 30, 2023 at 17:22

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I keep the reduction of @Andrè Schlicting, assume $|\Omega|=1$ and consider the operator $(I-P)\Delta$ in $L^2_0=\{u \in L^2(\Omega),\ \int_\Omega u=0\}$, where $Pu=\int_{\Omega} u$.

Assuming regularity for $\Omega$, then $\Delta$ has domain $H^2 \cap H^1_0$ so that the domain of $(I-P)\Delta$ in $L^2_0$ is $H^2\cap H^1_0 \cap L^2_0$.

Let me also consider the usual inner product $a(u,v)=\int_{\Omega} \nabla u \cdot \nabla v$ which is equivalent to the inner product of $H^1_0$, by Poincare inequality.

It is immediate to see that, if $u \in H^2\cap H^1_0 \cap L^2_0$ and $(I-P)u=-f\in L^2_0$, then $a(u,v)=\int_\Omega f v$ for every $v \in H^1_0 \cap L^2_0$.

Conversely, let $u \in H^1_0 \cap L^2_0$ satisfy $a(u,v)=\int_\Omega f v$ for every $v \in H^1_0 \cap L^2_0$, with $f \in L^2_0$. Let me show that $u \in H^2$ and $(I-P) \Delta u=0$. Let $w \in H^1_0$ and $\phi$ smooth with a compact support and mass 1. Then $v=w-c\phi \in H^1_0 \cap L^2_0\ $, $c=\int_\Omega w$. Therefore $$ \int_\Omega \nabla u \cdot (\nabla w-c\nabla\phi)= \int_\Omega f(w-c\phi) $$ or $$ \int_\Omega \nabla u \cdot \nabla w= \int_\Omega w(f-a) $$ with $a=\int_\Omega (\nabla u \nabla \phi-f \phi)$. By elliptic regularity, since $w \in H^1_0$ is arbitrary, $u \in H^2$. Integrating by parts we then obtain $$ -\int_\Omega \Delta u ( w-c\phi)= \int_\Omega f(w-c\phi) $$ Take now a sequence $\phi_k \to 1$ each with mass 1 (and all dominated by a fixed constant). Letting $k \to \infty$ above we get, since $f$ has zero mean,
$$ -\int_\Omega \Delta u (w-c)= \int_\Omega f(w-c)=\int_\Omega fw $$ which says that $(I-P)\Delta u=-f$.

Therefore the operator $(I-P)\Delta$ is associated to the form $a$ in $L^2_0$ and then is self-adjoint in $L^2_0$ and the solvability of the parabolic problem follows.

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  • $\begingroup$ Thank you @Giorgio Metafune, this works indeed and gives a clean, concise answer. I'm now wondering what to do with nonlinear boundary conditions, but this is for me to work out. Grazie! $\endgroup$ Commented Oct 31, 2023 at 9:25
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A short observation, which is too long for a comment. Let's assume that $\Omega$ has unit measure, i.e. $|\Omega|=1$. We define $$ w = u - \int u \,\mathrm{d}x . $$ By doing so, $w$ solves the classical Dirichlet problem with a non-local source related to the mass exiting the domain $$ \begin{cases} \partial_t w = \Delta w - \int_{\partial\Omega} \partial_n w \, \mathrm{d}\sigma &\text{ in } (0,T) \times \Omega, \\ w|_{\partial \Omega} = 0 &\text{ on } (0,T) \times \partial\Omega, \\ w|_{t=0} = w_0 & \text{ in } \Omega . \end{cases} $$ In particular, we also get that $w$ has conserved mean-value $$ \frac{\mathrm{d}}{\mathrm{d}t} \int w \,\mathrm{d} x = 0. $$ The mass, which leaves $\Omega$ is distributed uniformly in the domain.

There is for sure a connection to mean hitting times of Brownian motion on the domain $\Omega$, which is related to the stationary equation for $w$. The mean hitting time can be written as $$ \begin{cases} \Delta T_1 = -1 & \text{ in } \Omega , \\ T_1 =0 & \text{ on } \partial \Omega. \end{cases} $$ Then $w* = \lambda T_1$ with $\lambda = - \bigl( \int_{\partial\Omega} \partial_n T_1 \, \mathrm{d}\sigma\bigr)^{-1}$ is a specific stationary solution. In particular, it has some specific mean-value. Also the connection to the dynamic problem is not obvious.

Note, that the weak-formulation in the $w$ variable is very similar to the one for the usual Dirichlet homogeneous problem except for an additional mass parameter fixing the mean-value of $w$. Hence, a weak solution theory seems feasible, however I'm not aware, where this was studied before.

There might be also some more distant relation to Dirichlet-to-Neumann maps and to the electrical impedance tomography problem, where one wants to recover from measurements of the Dirichlet-to-Neumann problem the variable conductivity within the domain.

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  • $\begingroup$ Thank you André for your input, once again I'm bumping into you! There is indeed a connection with Brownian hitting times, actually I came up with this problem when studying sticky Brownian motion. I think we'll meet soon, I'll be in CIRM in January 2024 for your conference "PDE & Probability in interaction". Perhaps we can finally get to talk a bit on this occasion? $\endgroup$ Commented Oct 31, 2023 at 9:16
  • $\begingroup$ Looking forward to see you eventually in person and I'm sure that there will be some time to talk. $\endgroup$ Commented Oct 31, 2023 at 11:18

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