如何从拉普拉斯方程的弱解中恢复出经典解？

Alright, let's walk through how to recover a classical solution from a weak solution of Laplace's equation, using the setup you've outlined. We'll break this down into clear, logical steps:

First, let's restate the weak form of the Dirichlet Laplace problem you provided:
$$
\begin{cases}
\int_\Omega \nabla u \cdot \nabla \phi = 0, & \forall \phi \in C^1_c(\Omega), \
u = 0, & x\in \partial\Omega.
\end{cases}
$$
Using the Riesz Representation Theorem and Lax-Milgram Theorem, we can prove there's a unique weak solution $u \in H^1_0(\Omega)$ to this problem. For context, $H^1_0(\Omega)$ is the Sobolev space of square-integrable functions on $\Omega$ with square-integrable first derivatives, and zero trace (boundary value) on $\partial\Omega$.

You mentioned that $C^{\infty_c(\mathbb{R}}n)$ is a dense subspace of $H^1(\mathbb{R}n)$. A direct consequence of this is that $C^\infty_c(\Omega)$ (smooth functions with compact support inside $\Omega$) is dense in $H^1_0(\Omega)$. This density result is critical—it lets us bridge weak (Sobolev space) solutions to classical (smooth) solutions.

First, we need to link the weak form to distributional derivatives. Recall that a distribution $T$ satisfies $\Delta T = 0$ if for every test function $\phi \in C^\infty_c(\Omega)$,
$$
\langle T, \Delta \phi \rangle = 0.
$$
For our weak solution $u \in H^1_0(\Omega)$, take any $\phi \in C^\infty_c(\Omega)$ (which is also in $C^1_c(\Omega)$, so it's a valid test function for the weak form). Apply integration by parts to the weak form equation:
$$
\int_\Omega \nabla u \cdot \nabla \phi dx = -\int_\Omega u \Delta \phi dx + \int_{\partial\Omega} u \frac{\partial \phi}{\partial n} dS.
$$
Since $u \in H^1_0(\Omega)$, its trace on $\partial\Omega$ is zero, so the boundary integral vanishes. The left-hand side is zero by the weak form, so we get:
$$
\int_\Omega u \Delta \phi dx = 0 \quad \forall \phi \in C^\infty_c(\Omega).
$$
This means $u$ satisfies Laplace's equation in the distributional sense: $\Delta u = 0$ (as a distribution).

Now we use elliptic regularity to upgrade the weak solution to a smooth (classical) solution. Here's the core result for Laplace's equation:

• If $u \in H^1_{loc}(\Omega)$ is a distributional solution to $\Delta u = 0$, then $u$ is real-analytic in $\Omega$. For our case, since $u \in H^1_0(\Omega) \subset H^1_{loc}(\Omega)$, this immediately tells us $u$ is $C^\infty(\Omega)$ (and even real-analytic).

For the boundary regularity: if $\Omega$ has a $C^2$ smooth boundary, we can use Sobolev regularity estimates to show $u \in H^2(\Omega)$, then bootstrap this to higher Sobolev spaces ($H^k(\Omega)$ for any $k \geq 1$). Using the Sobolev Embedding Theorem, when $k > \frac{n}{2} + 2$, $H^k(\Omega)$ embeds continuously into $C^{2,\alpha}(\overline{\Omega})$ (Hölder continuous second derivatives on the closure of $\Omega$). This means $u$ is $C^2(\overline{\Omega})$, so it has continuous second derivatives on all of $\overline{\Omega}$.

Since $u \in H^1_0(\Omega)$, its trace (boundary value) is zero in the Sobolev sense. Once we've shown $u \in C(\overline{\Omega})$ (via Sobolev embedding for $k > \frac{n}{2}$), the Sobolev trace coincides with the classical pointwise boundary value. So we have $u(x) = 0$ for all $x \in \partial\Omega$, which is the classical Dirichlet boundary condition.

Putting it all together:

1. The weak solution $u \in H^1_0(\Omega)$ satisfies Laplace's equation in the distributional sense.
2. Elliptic regularity theory tells us $u$ is $C^\infty(\Omega)$ (and $C^2(\overline{\Omega})$ for smooth $\Omega$).
3. The Sobolev trace condition translates to the classical zero boundary condition.

Thus, $u$ is a classical solution to the Dirichlet Laplace problem:
$$
\begin{cases}
\Delta u = 0, & x \in \Omega, \
u = 0, & x \in \partial\Omega.
\end{cases}
$$

内容的提问来源于stack exchange，提问作者gdlm