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It is well known that the non-linear equation $f'' + 2f(1-f^2) = 0$ admits a soliton solution $f = \tanh(x)$.

Is it possible to solve this equation numerically? For example on a finite interval $[-L,L]$, with boundary conditions $f(-L) = -1$, and $f(L) = 1$? (I guess this might not be possible since solitons-antisoliton pair can be inserted. So may be more conditions are needed.)

Are there any numerical stable iterative algorithm, so that the initial guess like $f_0(x) = x/L$ would converges to the solution $\tanh(x)$?

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  • $\begingroup$ Do you mean $f'' + 2 f (1-f^2) = 0$? $\endgroup$
    – Robert Israel
    Commented Jul 30, 2017 at 18:50
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    $\begingroup$ Hi, welcome to MO. What do you mean by soliton solutions? This is an ODE, and the solution changes with the $x$ variable. Is this ODE derived from some other PDE? $\endgroup$
    – Amir Sagiv
    Commented Jul 30, 2017 at 19:03
  • $\begingroup$ yes, it was a type, it should be $(1-f^2)$ in stead of $(1-f)$. thx~~ $\endgroup$
    – Einstein_is_Coding
    Commented Jul 30, 2017 at 20:50

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If you mean $f'' + 2 f (1-f^2)$, that can be solved either numerically (by standard techniques, available e.g. using Maple's dsolve(..., numeric)) or symbolically. The solutions of the differential equation with $f(0)=0$ can be written using the Jacobi elliptic function $\text{sn}$.

EDIT: Specifically, we have the solutions $$ f(x) = c \sqrt{\frac{2}{c^2+1}} \text{sn}\left(\sqrt{\frac{2}{c^2+1}} x, c\right) $$ which satisfy $f(0)=0$ and $f'(0) = 2c/(1+c^2)$. Note that for real $c$, $2c/(1+c^2)$ takes on all values in $[-1,1]$; for $c=1$ this solution is $\tanh(x)$. For $0 < c < 1$ the solution is periodic. Complex values of $c$ (still giving real solutions) will handle $f'(0) $ outside this interval. These solutions reach the value $1$ at a finite $x$ and eventually go to $\infty$.

Here is a plot of the solutions with $f(0)=0$ and $f'(0) = 12/13, 1$ and $14/13$.

enter image description here

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  • $\begingroup$ Mathematica or Maple can get some solution, but it may not be that simple. For $x\in ]-\infty,+\infty[$, and boundary condition $f(-\infty) = -1$, $f'(-\infty) = 0$, this eq admits at least two solutions $f=-1$ and $f = \tanh(x)$. Due to its non-linearity, the uniqueness of the solution is no longer valid. $\endgroup$
    – Einstein_is_Coding
    Commented Jul 30, 2017 at 21:00
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    $\begingroup$ What in the world does uniqueness have to do with nonlinearity? $\endgroup$
    – Robert Israel
    Commented Jul 31, 2017 at 1:04
  • $\begingroup$ To reinforce what Robert Israel said: the linear ODE $f'' = f$ has multiple distinct solutions satisfying $f(-\infty)= 0 = f'(-\infty)$. The "failure" of uniqueness is mostly an issue of you trying to prescribe initial data at infinity. $\endgroup$
    – Willie Wong
    Commented Jul 31, 2017 at 2:51

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