How to solve a second order nonlinear ordinary differential equation?

Questions
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1

Hollo, I am trying to solve a second order nonlinear ODE with a given boundary conditions at infinity. My equation looks like this

\frac{d^2 u}{dx^2} = - \frac{dV}{du}

where V=\frac{1}{2}u^2 - \frac{1}{4}u^4.

I tried to use this line
sol = DSolve[u''\[x\] == u\[x\]^3 - u\[x\], u[-Infinity] == 1 , u[Infinity] == -1, u\[x\], x]
but it resulted in nothing that I can comprehend
DSolve::dsfun: "u[-\[Infinity]]==1 cannot be used as a function."
The solution for this boundary condition problem is

u = \mp \tanh \left( \frac{x}{\sqrt{2}} \right)
2

Using the substitution u'(x)=v(u(x)) which results in u''(x)=v'(u)u'(x) or u''(x)=v'(u)v(u). Plugging these in the ode result in a first order nonlinear ode in v(u) which is v'(u)+\frac{u}{v(u)}-\frac{u^3}{v(u)}=0 which can now be solved for v(u). Now use this solution to solve for the first oder order ode u'(x)=v(u(x)).
The problem is with the boundary conditions at \pm \infty. btw, you had them reversed.

I had to sneek in a simpler boundary conditions at zero.

The solution is in terms of inverse function with logs, which you can try to convert to trig. But I plotted it and it does match the \frac{\tanh(x)}{\sqrt{2}}. I only used one of the 2 solutions here.

Clear[u, v, x];
eq = v'[u] + u/v[u] - u^3/v[u] == 0;
sol = v[u] /. First@DSolve[{eq, v[0] == 1/Sqrt[2]}, v[u], u];
sol = u[x] /. First@DSolve[{u'[x] == (sol /. u -> u[x]), u[0] == 0}, u[x], x]

8794Untitled.png

Plotting it side-by-side to the proposed solution shows it is the same

Row[{
  Plot[sol /. x -> t, {t, -Pi, Pi}, ImageSize -> Medium, PlotLabel -> "DSolve"],
  Plot[Tanh[x/Sqrt[2]], {x, -Pi, Pi}, ImageSize -> Medium, PlotLabel -> "Given solution"]}
 ]

10259Untitled.png
10259Untitled.png835×271 3.61 KB

3

Any tip on how can I solve v'(u) + \frac{u}{v(u)} - \frac{u^3}{v(u)} =0 analytically?

4

Any tip on how can I solve v?(u)+\frac{u}{v(u)}?\frac{u^3}{v(u)}=0
analytically?

This is separable, so direct integration solves it. Since v'(u)=f(v,u) can be written as v'(u)= \frac{-1}{v} (u-u^3) so it is separable. Rearranging gives v v'(u) = u^3-u or v \frac{dv}{du}=u^3-u now integrating both sides gives \int vdv=\int (u^3-u)du or \frac{v^2}{2}=\frac{u^4}{4}-\frac{u^2}{2}+c. Using the initial condition that v(0)=\frac{1}{\sqrt{2}} gives c=\frac{1}{4}, plugging this back and simplifying gives v(u)=\frac{\sqrt{(u^2-1)^2}}{\sqrt{2}} which is what Mathematica gives.