Dear Mark, this ODE can be writen as a generalized Emden–Fowler equation. There is a classical reference

Habets P; Zanolin F: Upper and lower solutions for a generalized Emden–Fowler equation, Journal of Mathematical Analysis and Applications 1994 Vol: 181:684-700. DOI: 10.1006/jmaa.1994.1052

where the authors develop a method of upper and lower solutions to solve this type of problems. It can take some time to understand if your concrete example can be solved by this method, but there is a rich bibliography about it.

What I'm curious to know is: why do you think there is a unique c for which the problem is solvable? Have you done a numerical study?

Dear Pedro,

Thanks for your comment. We have done a numerical study and it indeed suggests that there is a solution for a single positive value of c.

Best regards,

Mark

Dear Pedro,

In researching this, I learned that the Emden–Fowler equation takes the form y'' = A f(x) y^m, where f(x) = x^n. Our problem is therefore a generalized Emden–Fowler equation in which m = -1/2 and f(x) is a cubic polynomial. (There is also a constant term on the right-hand side: c y'' = x(x^2 - 1)/\sqrt(y) - 1.) Unlike the problem considered in Habets and Zanolin, however, f(x) is nonsingular at the endpoints of the interval.

Best regards,

Mark

Right, if the equation is written as y''=g(x,y), then g is singular at y=0 but it is regular in the independent variable x. In a sense, this makes the problem easier, but it is necessary a careful checking. I can recommend the monography

I. Rachůnková, S. Staněk, M. Tvrdý: Solvability of Nonlinear Singular Problems for Ordinary Differential Equations. Hindawi Publishing Corporation, New York, USA, 2009, 268 pages.

http://phoenix.inf.upol.cz/~rachunl/mathair/rst.pdf

for this type of problems. But typically the mathematical methods we use are robust, that is, usually we get an open interval of available values for the parameters...

What would be really nice is if we could somehow get rid of the constant term on the right-hand side of c y'' = x(x^2 - 1)/\sqrt(y) - 1. There seem to be a lot of results on the problem y'' + f(x) y^m = 0 where m is negative, and they could then be applied to our problem.

Pedro --- The equation comes from a study of the dynamics of a solid surface. Since the problem has a physical basis, from a physicist's standpoint it is not necessary that there be an open interval of available values for the parameters. We think that there is some interesting physics to be extracted from the model.

Suppose t=sqrt(y), implies that dy=2tdt .