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Topic:
Numerical ODEs
Replies:
10
Last Post:
May 16, 2013 4:22 PM




Re: Numerical ODEs
Posted:
May 14, 2013 7:07 AM


José Carlos Santos wrote:
> 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?
There is a standard trick for dealing with this type of problem for ODEs of the form dy/dx = g(y). Note that y = f(x) is a solution of this equation, then so is y = f(C + x) with an arbitrary constant C. Also, y = f(C  x) is a solution of dy/dx = g(y). You can use this to work backward and forward from an arbitrary point where the equation is not singular and then translate the solution along the xaxis to satisfy the boundary condition.
For the case you mention, you might start by solving the IVP dy/dx = 2 sqrt y y(0) = 1 (instead of 1 any positive number will do) This should give you a solution that looks like f_1(x) = (1x)^2 on [0, 1]. (What happens past 1 probably depends on the algorithm used). Because f_1(1) = 0, you know that there is a solution to your original problem that looks like (1  (1  x))^2 on [(11), (10)]. You can then complete the solution by solving the IVP dy/dx = 2 sqrt y y(1) = 1
 Niels Diepeveen



