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How to solve this nonlinear differential equation numerically?

  1. Jul 15, 2010 #1
    -f''-(2/x)f'+(2/x^2)f+f^3-f=0

    the boundary condition is f(x=0)=0 and f(x=\infty)=1

    how to solve f numerically?
     
  2. jcsd
  3. Jul 16, 2010 #2
    This is similar to the Riccati equation. I hope one of the methods for tackling the Riccati equation is of some use.
     
  4. Jul 16, 2010 #3

    phyzguy

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    Since you have two boundary conditions at two different points, this is usually solved by the shooting method, where you assume a value of f'(0), and then integrate out to see if it satisfies the boundary condition at infinity, then adjust your assumption to match the desired boundary condition. However, I tried this briefly using Mathematica's NDSolve function:

    NDSolve[{-D[D[f[x], x], x] - 2/x D[f[x], x] + 2/x^2 f[x] + f[x]^3 -
    f[x] == 0, f[.00001] == 0,
    f'[.00001] == 1.0}, f[x], {x, .00001, 20}]

    and I got oscillatory solutions that tended to zero as x->inf (see attachment). Perhaps there is a particluar value of f' at zero that will give the solution you want?
     

    Attached Files:

  5. Jul 16, 2010 #4
    thanks a lot

    but i am always concerned with the singularity at the origin

    the coefficients diverge there

    so i do not know how to deal with it
     
  6. Jul 16, 2010 #5
    the derivative f' at the origin has to be scanned

    but i am concerned with how to integrate the eq accurately for a given f'
     
  7. Jul 16, 2010 #6

    phyzguy

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    There is no singularity at the origin. Assume as x->0 that f looks like f = K x. Then f->0, f^3 ->0, f''->0, and f'->K, and the equation reduces to: -2K/x+2K/x=0. So everything is well behaved. In essence, the two "singular" terms cancel. You can then start your integration at some small distance away from the origin (I chose x=1E-5, but it won't really matter as long as you start where the other terms are negligible).
     
  8. Jul 19, 2010 #7
    You can simplify things making [tex]y(x)=u(x)/x[/tex]. Then you have:
    [tex]u''+(1-2/x^{2})u-u^{3}/x^{2}[/tex]
    this equation has two independent solutions as [tex]x\rightarrow \infty[/tex], namely:
    [tex]u\rightarrow C_{1}Sin(x)+C_{2}Cos(x)[/tex]
    You can integrate the equation backwards. I am not very sure about there being no singularity. If you keep all the terms, except for [tex]u^{3}/x^{2}[/tex], you still obtain a solution:
    [tex]u(x)\equiv C_{1}(Sin(x)-Cos(x)/x)+C_{2}(Cos(x)-Sin(x)/x)[/tex]
    which should be valid before the neglected term is noticeable. I see no way for any solution being bounded at x=0, unless the term with [tex]u^3[/tex] is dominant somehow
     
  9. Jul 19, 2010 #8
    i guess the substitution y=xu may be better?
     
  10. Jul 19, 2010 #9
    The change I suggest is standard. It removes the first order differentiation, allowing to study the decay and the oscillation separately
     
  11. Jul 19, 2010 #10
    thanks a lot.

    any reference´╝č
     
  12. Jul 19, 2010 #11
    how about guess a taylor expansion of f around x=0

    and integrate outside from x_0<<1?

    the boundary condition at x=0 indicates that

    f(x)=ax + b x^2 + c x^3+ d x^4+....

    we can determine the relations between a, b, c, d by the differential equation.
     
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