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.
So it seems like we could cook up a numerical method which took into account the information from the second derivative. This would get you off the 0 solution. In this case, the second derivative is constant, but in a more general case, we can always find the value of the second derivative at x=0. Then assume that the solution is the corresponding parabola to generate two more points and then use a Runge-Kutta method from there.
José Carlos Santos <firstname.lastname@example.org> wrote in news:av9p7dFl84qU1 @mid.individual.net:
> Hi all, > > This question is perhaps too vague to have a meaningful answer, but > here it goes. > > In what follows, I am only interested in functions defined in some > interval of the type [0,a], with a > 0. > > Suppose that I want to solve numerically the ODE f'(x) = 2*sqrt(f(x)), > under the condition f(0) = 0. Of course, the null function is a > solution of this ODE. The problem is that I am not interested in that > solution; the solution that I am after is f(x) = x^2. > > For my purposes, numerical solutions are enough, but if I try to solve > numerically an ODE of the type f'(x) = g(f(x)) (with g(0) = 0) and > f(0) = 0, what I get is the null function. So far, my way of dealing > with this has been to solve numerically the ODE f'(x) = g(f(x)) and > f(0) = k, where _k_ is positive but very small and to hope that the > solution that I get is very close to the solution of the ODE that I am > interested in (that is, the one with k = 0). Do you know a better way > of dealing with this problem? > > Best regards, > > Jose Carlos Santos