# How do you solve a nonlinear recurrence relation?

1. Jul 8, 2011

### lugita15

While solving a problem involving equilibrium positions of charges on a line, I came up with a recurrence relation which is nonlinear, and moreover implicitly defined. Here it is: $x_{0}=0$ and $\sum^{n-1}_{i=0} \frac{1}{(x_{n}-x_{i})^{2}} = 1$. I should also mention that $0 \leq x_{n}< x_{n+1}$ for all $n$.

I can't even find an explicit expression for $x_{n}$ as a function of the previous terms, let alone as a function of $n$. I've dealt with linear recurrences before, but how would I go about solving a nonlinear recurrence like this? Is it even possible to find a closed-form expression using elementary functions? If an exact solution is impossible, is there some way to get a numerical approximation?

Any help would be greatly appreciated.

2. Jul 9, 2011

### hunt_mat

I think the only way forward is numerical methods.

3. Jul 9, 2011

### Pythagorean

Numerical and graphical solutions. "nonlinear dynamics and chaos" is an excellent book by Strogatz that outlines such methods.

4. Jul 9, 2011

### Antiphon

What you have is a statement about the field in integral (sum) form. Write it as a differential equation which will become a difference equation. Then you're home free linear or not.

5. Jul 9, 2011

### lugita15

How exactly would I "write it as a differential equation which will become a difference equation"?

6. Jul 17, 2011

### lugita15

OK, I have two additional facts: $x_{n}\geq n$ for all $n$ and $lim_{n\rightarrow\infty} (x_{n}-x_{n-1}) = \infty$

I don't know how much they'll help.

7. Jul 18, 2011

### Antiphon

The steady-state solution to many charge and current flow problems can be formulated as solution to Laplace's equation. That in turn is a differential operator that can be interpreted as an averaging operation. This is essentially a differencing operation between the function at a point and the average of it's neighboring values. Once you cast your problem in this form, iteration leads to the correct solution.