General Solution to Nonlinear ODE

Question

The functions $f(x) = \tanh(x/2)$ and $f(x) = \coth(x/2)$ satisfy the nonlinear differential equation:

$$-f'' + \textstyle\frac{1}{2}(f^2 - 1)f = 0$$

Is there any way I can find the more general solution to this differential equation knowing the example tanh and coth solutions?

Answer 1

First integral

As usual with a kinetic equation in a potential field, multiply with $2f'$ and integrate $$ -f'(x)^2+\frac14(f(x)^2-1)^2=C $$

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Reparametrization of the level curve

This has the general form $A^2-B^2=C$ and can thus be parametrized by hyperbolic functions.

• Case $C=R^2>0$, $R\ne0$: $f'(x)=R\sinh(g(x))$, $f(x)^2-1=2R\cosh(g(x))$, so that also $$ f(x)f'(x)= R\sinh(g(x))g'(x)= f'(x)g'(x). $$ Consequently $g'(x)= f(x)$, so $$g''(x) = R\sinh(g(x)),$$ which has no simple solution, WA gives solutions in terms of the Jacobi amplitude function.

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• Case $C=-R^2<0$, $R\ne0$: $f'(x)=R\cosh(g(x))$, $f(x)^2-1=2R\sinh(g(x))$, again $g'(x)=f(x)$, $$g''(x)=R\cosh(g(x))$$ follows. Again WA gives solutions in terms of the Jacobi amplitude function.

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• Case $C=0$: Now $f'(x)=\pm\frac12(f(x)^2-1)$. This results in the two functions that were the point of departure for this question, the form $$ f(x)=\pm \frac{e^x-K}{e^x+K} $$ contains all the variants.