Message Boards Message Boards

Get an analytical solution of a non-linear 3rd order ode?

GROUPS:

The equation that I need an analytical solution of, is

$f'''= \frac{1}{3}f'^2+\frac{2}{3}ff''$

The boundary conditions are:

at $x=0; f=0, f'=1$

at $x=\infty; f'=0$

If I use Dsolve, it doesn't return a solution

sol = Dsolve[y'''[x] == (1/3)*(y'[x])^2 - (2/3)*(y[x])*(y''[x]), y[0] == 0, y'[0] = 1, y'[inf] == 0, y[x], x]
POSTED BY: Pragyan Sarma
Answer
1 month ago

Likely, that means Mathematica doesn't know a closed-form solution, if it exists.

 DSolve[{f'''[x] == 1/3*f'[x]^2 - 2/3*f[x]*f''[x], f[0] == 0, f'[0] == 1, f'[Infinity] == 0}, f[x], x]

 (* Returns unevaluated  *)

Numeric method:

sol = NDSolve[{f'''[x] == 1/3*f'[x]^2 - 2/3*f[x]*f''[x], f[0] == 0, f'[0] == 1, f'[20] == 0}, f, {x, 0, 20}]
Plot[{f[x] /. sol, Evaluate@D[f[x] /. sol, x]}, {x, 0, 20}, PlotLegends -> {"f[x]", "f'[x]"}]

enter image description here

POSTED BY: Mariusz Iwaniuk
Answer
1 month ago

Thanks, the numerical solution is what I expected it to be.

POSTED BY: Pragyan Sarma
Answer
1 month ago

It can be reduced to second-order ODE with substitution:

$f'''(x)=u''(f) u(f)^2+u'(f)^2 u(f)$

$f''(x)=u(f) u'(f)$

$f'(x)=u(f)$ where $f(x)=f$

Putting to equation:

  u''[f]*u[f]^2 + u'[f]^2*u[f] == 1/3*u[f]^2 - 2/3*f*u[f]*u'[f]
  u''[f]*u[f] + u'[f]^2 == 1/3*u[f] - 2/3*f*u'[f]

  eq2 = u''[f]*u[f] + u'[f]^2 + 2/3*f*u'[f] - 1/3*u[f] == 0

  DSolve[eq2, u[f], f]
  (* Returns unevaluated  *)

With help of Maple it can be reduced to Abel second kind equation,but DSolve returns unevaluated.

enter image description here

  DSolve[y'[x]*y[x] - x^2/3 + 2/3*y[x] == 0, y[x], x]
  ( * Input ?*)

I can only say Mathematica 11.3 and Maple 2018 doesn't know a closed-form solution.

POSTED BY: Mariusz Iwaniuk
Answer
1 month ago

For analytic solution, you may try from here and use Mathematica to find a solution from lower degree.

sketch

odeclass4

POSTED BY: Shenghui Yang
Answer
1 month ago

Group Abstract Group Abstract