Is the Laplacian $\triangle:C^\infty(\Omega) \to C^\infty(\Omega)$, $\Omega\subseteq \mathbb{R}^n$, invertible? I have read Invertibility of laplacian operator which talks about thinking of the Laplacian as an operator on distributions. I do not know much about distributions, so I am looking for an answer which sticks to honest functions.
I know that (if $\overline{\Omega}$ is compact) for harmonic functions $u_1,u_2$ which agree on $\partial \Omega$ we must have $u_1=u_2$. (Folland PDE, 2.15) So let us assume $\Omega$ satisfies that condition.
Now if $\triangle u = \triangle v$ then $\triangle(u-v)=0$. But we can't say $u=v$ because we don't know they agree on the boundary. Indeed, adding a constant to $u$ preserves all the other equalities so we are dead if we don't make another assumption. Let us assume then that we have a fixed boundary condition $u = f$ on $\partial \Omega$. Then we definitely have $u=v$.
What about surjectivity? If I prescribe a $w \in C^\infty$, is there a $u$ satisfying the boundary condition such that $\triangle u = w$?
Best Answer
What Anthony Carapetis wrote in his answer, the weak solution, is still in distributional sense, namely, the Poisson equation $-\Delta u = f$ holds in $H^{-1}$.
For the smooth case, we don't have to go into the realm of weak solution.
The uniqueness of the solution, given that a solution exists, is obtained by maximum principle (see Evans 2.2.3):
Replacing $u$ with $-u$ we will see that $$ \min_{x\in \partial\Omega} u(x)\leq u(x) \leq \max_{x\in \partial\Omega} u(x). $$ Thus $\Delta w = 0$ and $w|_{\partial\Omega}=0$ give us $w=0$. Where this $w$ here is the difference between two smooth functions satisfying the same boundary value problem.
For existence, the construction is actually explicit! This is called Green's representation formula, please refer to Evans 2.2.4 and Theorem 12.
But the thing is the construction relies on the existence of $\phi(x,y)$ such that for a fixed $x\in \Omega$ $$ \begin{cases}\Delta_y \phi(x,y) = 0 & \text{in }\Omega, \\[4pt] \phi(x,y) = \Phi(y-x) &\text{on }\partial \Omega, \end{cases} \tag{$\star$} $$ where the $\Phi(y-x)$ is the Green function solving the $-\Delta u = \delta_x$ on the whole $\mathbb{R}^n$. If we can prove the existence of a solution for above boundary value problem of Laplace equation $(\star)$, we are done.
Moreover $u$ can be represented as: $$ u(x) = \int_{\Omega} f(y)G(x,y)dy - \int_{\partial \Omega} g(y) \frac{\partial G}{\partial n}(x,y)dS(y), $$ where $\partial G/\partial n = \nabla G\cdot n$ is the directional derivative, and $G(x,y) = \Phi(x-y)-\phi(x,y)$. This $u$ solves $$ \begin{cases}-\Delta u = f & \text{in }\Omega, \\[4pt] u = g &\text{on }\partial \Omega, \end{cases} $$ for $f$ and $g$ satisfying certain smoothness.
The existence of a solution for $(\star)$ is a difficult analysis problem, when $\partial \Omega$ is relatively "simple", we can transform the domain into a ball which has an explicit expression Green function. For the existence in general bounded, smooth domain, we can use Perron's method of subharmonic functions (See Gilbarg-Trudinger 2.8, exercise 2.10).