Numerical method for functional equations similar to $2f(x) - f(x+2) - f(x-2) = 0$

Question

I have some functional equations, that are of the form similar to something like $$ 2f(x) - f(x+2) - f(x-2) = 0 \text, $$ which I need to find the solution to. I know that equations like these are generally hard to solve analytically (if even possible) so I was hoping to know if there were any numerical methods that could be used to solve functional equations of this form.

In general I'm looking to solve functional equations of the form $$ \sum_{i=1}^n a_i(x) f\bigl(g_i(x)\bigr) = 0 \text. $$

I know in some cases it is entirely possible to have there be multiple solutions, so that may throw a wrench in the works.

Answer 1

Set $g(x)=f(x)-f(x-2)$ to find $g(x+2)=g(x)$. So $g$ is any $2$-periodic function, not even continuity is required. Next set $h(x)=f(x)-\frac12xg(x)$ to find $$\begin{multline} h(x)-h(x-2)=f(x)-f(x-2)-\frac12(xg(x)-(x-2)g(x-2))\\=g(x)-\frac12(x-(x-2))g(x)=0. \end{multline}$$ So $h$ is another arbitrary $2$-periodic function.

In total there are many solutions $f(x)=h(x)+\frac12xg(x)$ to that functional equation. It is not clear what a numerical method should achieve.

Answer 2

What would the numerical method achieve in this case? If you try to get the solution on groups of points using the recursion, there is no guarantee that you will not be switching between different solutions.

If you restrict the domain to natural numbers and solve the corresponding difference equation, you'll see that every solution must satisfy $$ f(n) = (c_1 n +c_2) (-1)^n + c_3 n + c_4, \quad n \in \mathbb{N} $$

and the constants $c_i$ are computed from initial conditions.