I'm trying to wrap my head around a certain situation involving a boundary value problem. Here's the scoop:
We've got a Lipschitz open and bounded subset $\Omega \subset \mathbb{R}$, with its boundary $\partial\Omega$ composed of two disjoint, open sets $\Gamma_1$ and $\Gamma_2$ (so $\partial\Omega = \Gamma_1 \cup \Gamma_2$).
I'm considering a function $f$ that lies in $L^{2}(\Gamma_2)$ and another function $u$ from $H^{1}(\Omega)$. $u$ happens to be the solution to:
$$ \begin{cases} -\Delta u=0 & \Omega\\ u=0 & \Gamma_1 \\ u=f & \Gamma_2 \end{cases} $$
Let's say we have a non-smooth subset $B$ of $\Omega$. Can we still say that $\Delta u = 0$ applies inside $B$? And if so, in what sense might that be true? The challenge I'm facing is that I don't have access to any regularity theorems.
Any insights or suggestions on how to proceed, given this limitation, would be greatly appreciated!
Yes, you can say that. Essentially, the what you are trying to show is an interior property, so the boundary information is actually irrelevant. Indeed, that fact that $u\in H^1(\Omega)$ satisfies $\Delta u=0$ in $\Omega$ means that $$ \int_\Omega \nabla u \cdot \nabla \varphi \, dx =0 $$ for all $\varphi \in C^\infty_0(\Omega)$.
Now if $B\subset \Omega$ is open then any $\varphi \in C^\infty_0(B)$ is also in $C^\infty_0(\Omega)$, so $$ \int_B \nabla u \cdot \nabla \varphi \, dx =\int_\Omega \nabla u \cdot \nabla \varphi \, dx =0. $$ Thus, $\Delta u=0$ in $B$ in the weak sense. In fact, this argument still holds true in the very weak sense, that is, we say $u\in L^1_{\mathrm{loc}}(\Omega)$ satisfies $\Delta u=0$ in $\Omega$ in the very weak sense if $$ \int_\Omega u \Delta \varphi \, dx =0 $$ for all $\varphi \in C^\infty_0(\Omega)$.
Also, you mention that 'you don't have access to regularity theorems', but you do! In particular, all the interior regularity theorem still remain valid in you situation, so $u$ is actually smooth (in fact, analytic) in $\Omega$ just not up to the boundary. Since this is the case, the statement also holds in the classical (strong) case too.