# Decompose solution of pde in harmonic and non-harmonic part

Let $w$ be the solution of the pde $$\begin{cases} \Delta w = f(w) & \mbox{on } \Omega \\ w=g & \mbox{on } \partial \Omega\end{cases} \tag{1}$$ where $\Omega \subset \mathbb{R}^N$ open, $w: \Omega \to \mathbb{C}$, $f: \mathbb{C} \to \mathbb{C}$ and $g:\Omega \to \mathbb{C}$.

Recently, I read that for some elliptic arguments it is convenient to decompose the solution of some pde as: $$w=w_1+w_2$$ where $$\begin{cases} \Delta w_1 =0 & \mbox{on } \Omega \\ w_1=w & \mbox{on } \partial \Omega \end{cases}$$ and $$\begin{cases} \Delta w_2 = f(w) & \mbox{on } \Omega \\ w_2=0 & \mbox{on } \partial \Omega \end{cases}.$$

I do understand that $w_1+w_2$ solves the original equation if there are such functions $w_1$ and $w_2$. But my question is:

Why can every solution of (1) be decomposed that way?

• Is there a typo? For $w_1+w_2$ to solve (1) it seems you need $f(w_2) = f(w_1 + w_2)$. – Anthony Carapetis Aug 5 '13 at 8:42
• @AnthonyCarapetis Yes, thank you! I corrected it. – mjb Aug 5 '13 at 9:03

The equation for $w_1$ is simply Laplace's equation with Dirichlet boundary conditions $w$, and likewise the equation for $w_2$ is Poisson's equation with source term $f(w)$ and zero boundary condition. Since we have already fixed a solution $w$ to (1), there exists such a decomposition so long as $w$ and $f(w)$ are regular enough for the usual elliptic existence theory to apply; e.g. $w \in H^1(\Omega)$, $f(w) \in H^{-1}(\Omega)$.
In fact, all you really need is one of the above conditions: if for example $w \in H^1(\Omega)$ then we get a $w_1$ solving one equation, and $w_2 := w - w_1$ immediately solves the second. Since the boundary condition for (1) is $w=g$ on $\partial\Omega$, this reduces to just needing $g \in H^1(\Omega)$; hopefully this is required to solve (1) in the first place? If not then you may need some facts about the regularity of solutions to (1).
• This is very useful for analyzing a solution when it is known to exist, and you have some pre-existing regularity for it. However, unlike the linear theory, this will not give you existence or uniqueness of a solution, and without some a priori bound on the solution (in some $L^p$ space, for example) you cannot use this method alone to get refinements. – Ray Yang Aug 5 '13 at 16:31