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Im currently working on an implementation of the laplace dirichlet problem for the purpose of image inpainting (see image below). My implementation works fine but I now need to give an explanation as to why.

The code basically follows the flow of the heat equation $\frac{\partial u}{\partial t} = \Delta u$ on the corrupted part until the time step is $\approx0$. Obviously this outcome is then (approximately) a stationary point to the laplace dirichlet problem.

What i now need help with is a way to rigorously explain that following the flow of the heat equation actually converges to this stationary point (for all we know it could explode to infinity). Any help with this would be appreciated, thanks!

enter image description here

Arctic Char
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Jord van Eldik
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  • You can do it concretely with separation of variables. I also think there's a maximum modulus principle approach for showing there's a stationary point. Check out Strauss' PDE text in the early heat equation sections. I think it's in there. – Cameron L. Williams May 25 '25 at 12:12
  • This should be true in bounded domains. In unbounded domains, in order to ensure a unique solution, you need some type of growth bound, or, you can use a stronger condition like an integrability condition. For instance, Tychonoff showed that without this condition, you can get non-unique solutions which grow rapidly. – K.defaoite May 25 '25 at 16:28

1 Answers1

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I will go purely into the theoretical aspects of your question, completely aside from your numerical algorithm. The question is: Given a domain $U\subset \mathbb R^n$ which is open and bounded, is it true that, given $u_0:U\to\Bbb R$ and $h:\partial U\to \Bbb R$ the solution to the initial-boundary value problem $$\begin{cases}\partial_tu=\Delta u & \text{in}~\Bbb R_+\times U \\ u=u_0 & \text{on}~\{0\}\times U \\ u=h & \text{on}~\Bbb R_+\times \partial U\end{cases}\tag{1}$$

Necessarily obeys $u(t,x)\to v (x)$ as $t\to\infty$ in some type of convergence (e.g $L^p$, pointwise a.e, or perhaps even uniformly) for suitably nice functions $u_0,h$, for some function $v:U\to \Bbb R$ ?

To make things as simple as possible, we will presume that $u_0\in C^2(\bar U)\cap L^2(U)$ and $h\in C^0(\partial U)\cap L^2(\partial U)$. Note that this ensures that $u_0|_{\partial U}=h$, i.e, the initial and boundary conditions are compatible with one another. We now define $u_\infty :U\to \Bbb R$ as the solution of the associated Laplace equation , $$\begin{cases}\Delta u=0 & \text{in}~U \\ u=h & \text{on}~\partial U\end{cases}$$ It is a well known consequence of standard PDE theory (e.g Lax-Milgram theorem) that $u_\infty$ exists, is unique, and moreover is in $C^\infty (\bar U)$.

We first establish a critical result that I have shown in another post concerning independence of initial conditions for the heat equation: if $u_1,u_2$ are solutions of $(1)$ with the same boundary condition $h$ but arbitrary initial conditions $u_{0,1},u_{0,2}$, then, it can be shown (with a great deal of effort) that $\Vert u_1(t,\cdot)-u_2(t,\cdot) \Vert_{\infty }\to 0$ as $t\to\infty$, no matter the choice of $u_{0,1},u_{0,2}$.

Armed with this, it is now clear that $u(t,\cdot)\to u_\infty$ pointwise a.e as $t\to\infty$, for arbitrary $u_0,h$. If we simply consider a function which is constant in time, $u_\infty (t,x):=u_\infty (x)$ then clearly $u_\infty(\cdot,\cdot)$ satisfies $(1)$ with the initial condition $u_\infty(\cdot)$. Therefore, any other solution $u$ of $(1)$ with the initial condition $u_0$ will satisfy $$\Vert u(t,\cdot)-u_\infty(\cdot)\Vert_\infty \to 0 \\ \text{as}~t\to\infty$$

Since $L^\infty$ convergence implies pointwise a.e convergence, we see that $u(t,x)\to u_\infty(x)$ as $t\to\infty$ pointwise, for $\mu^n$-almost every $x\in U$, which actually implies the result for every $x\in U$ when considering the smoothness properties of $u_\infty$.

K.defaoite
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