Solving linear 2nd order IVP non-constant coefficient

[member 428835]

Hi PF!

Generally speaking, how would one solve $$f''-a(x) f = 0 : f(0)=0,f'(0)=1$$
Or if you could point me to a source that would be awesome too!

 

[Dr.D]

Assuming that a(x) is known, I'd solve it numerically with Runge-Kutta or something similar.

 

[member 428835]

Quick question (I hope it's quick). Variable $x$ is arc-length of a given in parametric form. If it's okay with you, I'll switch variables so that arc-length is $s$: $$f''(s) + a(s) f(s) = 0$$ where $a(s)$ is the square of the curvature of a line in the $x-z$ plane given in parametric form here $$x(s) = \frac{1}{c} \sin(c s)\\ z(s) = \frac{1}{c}(1- \cos(c s))$$
where $c$ is a constant. One formula of curvature is $|\hat T'(s)|/|\hat r'(s)|$ where $\hat T$ is the unit tangent vector to a curve given in parametric for as $\hat r(t)$. Computing this in Mathematica gives $a(s) = c/(2\sin(cs/2))$. Then the IVP is $$f''(s) + \frac{c}{2\sin(cs/2)} f(s) = 0$$ Any ideas for an analytic solution?

 

[Dr.D]

Now that you have changed notations, what do the primes mean? Is (') =d()/dx or is (') = d()/ds?

 

[member 428835]

Now that you have changed notations, what do the primes mean? Is (') =d()/dx or is (') = d()/ds?

Sorry, prime denotes $d/ds$. I tried letting $f = \exp(u(s))$ but that gives a non-linear 1st order IVP...

 

[Dr.D]

I still suggest obtaining a numerical solution first. This will tell you general shape of the solution, and from there educated guesses may give you a closed-form solution.

 

[member 428835]

Thanks!

 

[Nguyen Son]

What about using numerical method like Runge-Kutta 4?

 

[member 428835]

What about using numerical method like Runge-Kutta 4?

Yea that's what Dr.D suggested. I was just wondering if an analytic solution existed.

 

[member 428835]

I still suggest obtaining a numerical solution first. This will tell you general shape of the solution, and from there educated guesses may give you a closed-form solution.

I wrote an RK4 technique, but I get no good response (I check the code on IVPs with analytic solutions and it works). The non-constant coefficient has a singularity at $s=0$; how would you subvert this?

 

[Jone Oldman]

i offer a textbook of ode https://drive.google.com/open?id=1JZXHplbM7tIBWYOVxUPx5Oa0Q-L0VC55
you can use "reduction of order method" page 242.

 

[member 428835]

i offer a textbook of ode https://drive.google.com/open?id=1JZXHplbM7tIBWYOVxUPx5Oa0Q-L0VC55
you can use "reduction of order method" page 242.

Hi Jone Oldman, and welcome to PF! Reduction of order requires knowing one solution though, which I don't know.

 

[Dr.D]

The form you give in post #3 should tell you right away that there will be difficulties due to a singularity at the beginning. This is not "just a simple" IVP at all, but rather a very complex differential equation that will require special handling.

 

[member 428835]

The form you give in post #3 should tell you right away that there will be difficulties due to a singularity at the beginning. This is not "just a simple" IVP at all, but rather a very complex differential equation that will require special handling.

Any ideas on how to begin?

 

[Dr.D]

One possibility for a numerical approach is to propose a terminal value, and integrate backwards toward the beginning to see where that takes you. Obviously where it takes you depends in most cases on the terminal value, but you could learn something about the nature of the solution this way.

 

[member 428835]

One possibility for a numerical approach is to propose a terminal value, and integrate backwards toward the beginning to see where that takes you. Obviously where it takes you depends in most cases on the terminal value, but you could learn something about the nature of the solution this way.

Thanks, I'll play around with it!