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How do you solve a nonlinear recurrence relation?

  1. Jul 8, 2011 #1
    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: [itex]x_{0}=0[/itex] and [itex]\sum^{n-1}_{i=0} \frac{1}{(x_{n}-x_{i})^{2}} = 1[/itex]. I should also mention that [itex]0 \leq x_{n}< x_{n+1}[/itex] for all [itex]n[/itex].

    I can't even find an explicit expression for [itex]x_{n}[/itex] as a function of the previous terms, let alone as a function of [itex]n[/itex]. 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.

    Thank You in Advance.
  2. jcsd
  3. Jul 9, 2011 #2


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    I think the only way forward is numerical methods.
  4. Jul 9, 2011 #3


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    Numerical and graphical solutions. "nonlinear dynamics and chaos" is an excellent book by Strogatz that outlines such methods.
  5. Jul 9, 2011 #4
    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.
  6. Jul 9, 2011 #5
    How exactly would I "write it as a differential equation which will become a difference equation"?
  7. Jul 17, 2011 #6
    OK, I have two additional facts: [itex]x_{n}\geq n[/itex] for all [itex]n[/itex] and [itex]lim_{n\rightarrow\infty} (x_{n}-x_{n-1}) = \infty[/itex]

    I don't know how much they'll help.
  8. Jul 18, 2011 #7
    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.
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