# Invertibility of Laplace operator for smooth functions

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 $\triangle u = w$?

• What is $\Omega$. Do you want us to add hypothesis on it or do you have some conditions on it? Commented Sep 3, 2013 at 15:32
• Assume $\Omega$ is open and bounded with $C^1$ or smooth boundary if it helps. Commented Sep 3, 2013 at 15:33

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):

If $\Omega$ is open and bounded (equivalent to your assumption that $\overline{\Omega}$ is comapct), if $\Delta u = 0$ in $\Omega$, for some $u$ is at least twice continuously differentiable in $\Omega$, and continuous up to the boundary of $\Omega$. Then $$\max_{x\in \Omega} u(x) = \max_{x\in \partial\Omega} u(x).$$

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).

• What is your opinion on the relative difficulties of learning these two approaches to the smooth problem? The Green's function representation is of course very useful, but as you note in general it just reduces the problem to a different elliptic existence problem on $\Omega$. Thus it is perhaps possible that the weak existence theory isn't actually all that more material to comprehend, and it gives a more generally applicable result. I guess you do have to establish a lot of technical properties of the the Sobolev spaces, however - there's a lot we take for granted once we have learnt it. Commented Sep 4, 2013 at 0:36
• @AnthonyCarapetis Weak formulation requires some backgrounds in functional analysis, in which the open mapping theorem plays a big role (bounded inverse iff the operator is bijective). Classical Newtonian potential approach requires multivariate+real only, yet judging its form, the formulae look complicated, however, it uses much lower tech that weak. I was just trying to address OP's concern from a standard PDE course point of view. To myself, my favorite approach to existence and uniqueness theory is still calculus of variations, like looking for the minimal electric potential energy. Commented Sep 4, 2013 at 1:16

The surjectivity question is just asking whether or not Poisson's equation $\Delta u = w$ has solutions for all source terms $w$. The answer is yes, but this is not straightforward to prove. The most common method is via a variational formulation and some theorems on Hilbert spaces, which give a "weak solution" which is not (a priori) $C^\infty$, but lives in the so-called Sobolev space $H^1 (\Omega)$. The "elliptic regularity theory" then gives a proof that (so long as $\Omega$ is nice enough, open and bounded suffices) this solution is in fact $C^\infty$ when the source term $w$ is.

You can read about this theory in detail in any book covering Elliptic PDE theory - it looks like Folland covers this in chapter 7.