How does the heat equation work?

Question

Assume I have the heat equation on a well-behaved domain $\Omega$ (let's say a disc).

But the Laplacian I consider is the Dirichlet Laplacian

$$(\partial_t-\Delta_D)u=0.$$

This equation is well-posed for general $L^2$ initial data $u(0)=u_0 \in L^2(\Omega).$

My question is now the following: Will the Dirichlet Laplacian enforce now $0$ boundary data for any time $t>0?$ How does this follow?

Answer 1

NOTE. All functions here are real-valued.

Since you spoke of "functional calculus" in the comments, you question can be reformulated as follows. Let $\phi_0, \phi_1, \ldots $ be an $L^2$-orthonormal set of Dirichlet eigenfunctions, that is, $\phi_j\in C^\infty(\Omega)$ for all $j$ and $$ \begin{cases} -\Delta \phi_j=\lambda_j \phi_j, & \Omega,\\ \phi_j=0, & \partial\Omega.\end{cases}$$ Notice that $\lambda_j>0$ and that $\lambda_j\to\infty$.

The solution to the problem you wrote in the question is given by the following operator; $$\tag{1} e^{t\Delta_D}f(x):=\sum_{j=0}^\infty e^{-t\lambda_j}\hat{f}(j)\phi_j(x), $$ where $$ \hat{f}(j):=\int_{\Omega} f(x)\phi_j(x)\, dx.$$ As such, (1) makes sense for all $f\in L^2(\Omega)$.

Now, notice that $$\tag{2} g\in H^1_0(\Omega)\quad \iff\quad \sum_{j=0}^\infty \lambda_j\lvert\hat{g}(j)\rvert^2<\infty;$$ this is another standard proposition of the functional calculus (indeed, $H^1_0$ is the so-called form domain of $-\Delta_D$), but you can see it directly as follows; $$ \int_{\Omega}|\nabla g(x)|^2\, dx=\int_{\Omega}(-\Delta_D g) g\, dx=\sum_{j=0}^\infty \lambda_j \lvert \hat g(j)\rvert^2.$$ We can then give the following result.

Let $f\in L^2(\Omega)$. For all $t>0$, $e^{t\Delta_D}f\in H^1_0(\Omega)$.

The proof is an immediate application of the criterion (2) to the formula (1).