How to solve this nonlinear differential equation numerically?

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Discussion Overview

The discussion revolves around the numerical solution of a nonlinear differential equation characterized by boundary conditions at two different points. Participants explore various methods for solving the equation, including the shooting method and transformations to simplify the problem.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents the nonlinear differential equation and its boundary conditions, seeking numerical solutions.
  • Another participant suggests that the problem resembles a Riccati equation, proposing that methods for Riccati equations might be applicable.
  • Several participants discuss the shooting method as a potential approach, noting the need to adjust the initial derivative to meet the boundary condition at infinity.
  • Concerns are raised about the singularity at the origin, with one participant arguing that the singular terms cancel out, allowing for a well-behaved solution near zero.
  • Another participant proposes a transformation of variables to simplify the equation, suggesting that this standard change can help analyze the behavior of solutions.
  • There is a suggestion to use a Taylor expansion around x=0 to derive relationships between coefficients in the series expansion of the solution.

Areas of Agreement / Disagreement

Participants express differing views on the presence and implications of singularities in the equation, with some asserting that there is no singularity while others remain uncertain. The discussion does not reach a consensus on the best method for solving the equation or the nature of the singularity.

Contextual Notes

Participants note limitations related to the singular behavior at the origin and the dependence on the choice of initial conditions for numerical methods. The discussion highlights unresolved mathematical steps and the need for careful consideration of boundary conditions.

wdlang
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-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?
 
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This is similar to the Riccati equation. I hope one of the methods for tackling the Riccati equation is of some use.
 
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?
 

Attachments

phyzguy said:
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?

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
 
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'
 
wdlang said:
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

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).
 
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
 
gato_ said:
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

i guess the substitution y=xu may be better?
 
The change I suggest is standard. It removes the first order differentiation, allowing to study the decay and the oscillation separately
 
  • #10
gato_ said:
The change I suggest is standard. It removes the first order differentiation, allowing to study the decay and the oscillation separately

thanks a lot.

any reference?
 
  • #11
gato_ said:
The change I suggest is standard. It removes the first order differentiation, allowing to study the decay and the oscillation separately

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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