José Carlos Santos <firstname.lastname@example.org> wrote in news:avkirmF3aj0U1 @mid.individual.net:
> On 15-05-2013 21:15, Bart Goddard wrote: > >> Pretty much out of my field, (I'll do anything to avoid grading >> final exams) but I notice that if you take the >> derivative of both side of the DE and substitute, you discover >> that f''(x)=2. As I understand, Runge-Kutta approximates the >> second derivative and self-adjusts itself to account. > > This works indeed for the ODE f'(x) = 2 sqrt(f(x)), but, more > generally, I am interested in ODEs of the type f'(x) = g(f(x)), under > the initial condition f(0) = 0 and where g(0) = 0. Then, all I get is: > > f''(x) = g'(f(x))*g(f(x)) > > and I don't see how to proceed from here.
Me either, really, but my thought was you could plug in 0 to get f''(0) = g'(0)*g(0) = a. (Where a might be a limit here.) If we're lucky and a is not 0, then we assume y = (a/2)x^2 is a reasonable approximation to f around x=0. Then run the numerical method, using stepsize h, and assume that the first three points on the solution are (0,0), (h, ah^2/2) and (2h, a4h^2/2), which should be enough to seed a RK method.
If we're unlucky and a = 0, then take another derivative, plug in 0 and if the 3rd derivative is not 0, approximate by bx^3/6 and so on. Some derivative has to be nonzero, and that should give us a bump off of the f(x)=0 solution.(?)