How can I solve this differential equation whit Runge-Kutta 4° order

$$ f''' + f f'' + 1- f'^2= 0$$

$f(0)=0$ $f'(0)=0$ $f'(\infty)=1$


You have to transform the equation into a system of first order, as most numerical methods are designed for systems of first order. Usually this is done by taking the lower derivatives as state vector $(u_0(x),u_1(x),u_2(x))=(f(x), f'(x), f''(x))$. Then \begin{align} u_0'&=f'&&=u_1\\ u_1'&=f''&&=u_2\\ u_2'&=f'''=-ff''-1+f'^2&&=-u_0u_2-1+u_1^2 \end{align} Now apply the steps of the Runge-Kutta method to this as vector-valued function.

With the added boundary conditions with one boundary at infinity you get a different problem which is not an ordinary boundary value problem (BVP). You would need to find a suitable reformulation to get a BVP on a finite interval and then employ a BVP solver. There the Runge-Kutta method or similar is the trivial part of the solver.

  • $\begingroup$ Thanks, The steps of Runge-Kutta for this ecuation are? $$u_1= f (u_0,u_1,u_2) $$ $$u_2= g (u_0,u_1,u_2) $$ $$u'_2= p(u_0,u_1,u_2) $$ and start whit $$k_0= h f((u_0,u_1,u_2)$$ $$k_1= h f((u_0+ \frac{1}{2}h,u_1+\frac{1}{2}k_0,u_2+\frac{1}{2}k_0)$$ $$k_2= h f((u_0+ \frac{1}{2}h,u_1+\frac{1}{2}k_1,u_2+\frac{1}{2}k_1)$$ . . . and $$l_0= h g((u_0,u_1,u_2)$$ $$l_1= h g((u_0+ \frac{1}{2}h,u_1+\frac{1}{2}l_0,u_2+\frac{1}{2}l_0)$$ . . . and $$m_0= h p((u_0,u_1,u_2)$$ $$m_1= h p((u_0+ \frac{1}{2}h,u_1+\frac{1}{2}m_0,u_2+\frac{1}{2}m_0)$$ . . . $\endgroup$ – Conan Oct 16 '18 at 19:28
  • $\begingroup$ No, you need to mix the steps as for instance $k_1=hf(u_0+0.5k_0,\, u_1+0.5l_0,\, u_2+0.5m_0)$ etc. This means that for the computation of $k_1$ you need already have computed $l_0$ and $m_0$. $\endgroup$ – LutzL Oct 16 '18 at 20:09
  • $\begingroup$ ok, I try, but initial conditions only give me zero. $\endgroup$ – Conan Oct 16 '18 at 20:42
  • $\begingroup$ Are you sure that it is $f(∞)=1$ and not $f'(∞)=1$? $\endgroup$ – LutzL Oct 16 '18 at 21:13

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