Approximate Solutions to Non-Linear Differential Equations

[nassboy]

I'm interested in coming up with an function that approximates the solution to a non-linear differential equation. (There is no known closed form solution)

The equation is Y''=(1/Y)(Y')^2-Y*A+Y^2*A where "A" is a constant, and Y' and Y'' are the first and second derivatives with respect to x.

How would I look for approximations to the answer in the literature? Is this equation a member of a certain class of differential equations?

Is there a technique known to work well?

Any help would be appreciated!

 

[HallsofIvy]

Since the independent variable does not appear explicitely in this equation, it looks like a candidate for "quadrature". Taking x to be the independent variable, let z= dy/dx. Then d²y/dx²= dz/dx= (dz/dy)(dy/dx)= zdz/dx. Now the equation becomes
z\frac{dz}{dy}= \frac{z^2}{y}- Ay+ Ay^2
a separable first order equation.

Another typical method for non-linear equations is "perturbations". The "WKB approximation" in quantum mechanics is a type of perburbation method.

 

[nassboy]

quadrature as in numerical integration?

 

[nassboy]

If I work the problem using this form z\frac{dz}{dy}= \frac{z^2}{y}- Ay+ Ay^2

, I'm still left with an integral that has no solution when I try to integrate a second time!

 

[arildno]

Well, set u=\frac{1}{2}z^{2}, rewriting the equation as:
\frac{du}{dy}-\frac{2}{y}u=-Ay+Ay^{2}
Multiplying this by the integrating factor y^-2, we get:
\frac{d}{dy}(\frac{1}{y^{2}}u)=-\frac{A}{y}+A,
or, by integrating:
u=Cy^{2}+Ay^{3}-Ay^{2}\ln(y)
Thus, we get:
\frac{dy}{dx}=\pm\sqrt{2(Cy^{2}+Ay^{3}-Ay^{2}\ln(y))}

which is separable, if not easy to solve..

 

[nassboy]

Thanks for the help!

\int \frac{1}{\sqrt{2\left(C Y^2+A Y^3-A Y^2 \text{Log}[Y]\right)}}

(I hope that is right, the latex preview isn't working for me)

After separating the variables, one is left with the integral above which has no closed form solution. What is the best way to approximate the integral, keeping in mind that I want to solve algebraically for y in the end.