Numerical solution ODE

1. Nov 21, 2014

Vrbic

Hello,
can someone advise me how to solve numerically ODE which consist of function with "critical point" (Im not sure if it is good definition)? I mean for example this one:
$y'(x)=\frac{\sin{x}}{x}$, where in x=0 has function a "problem". I know that limit ->1 but in numerical solutions it blows up.
I know that for example Mathematica can do that analytically but I would like to know general procedure for this issue.

2. Nov 21, 2014

zoki85

Writting polynomial expansion of sin x , than integrating term by term is one way.

3. Nov 21, 2014

Vrbic

Ou, nice and easy idea :) Can you think of another way?
I have system of two eq. and problem is coming from zero in denominator subtracting these unknowns... $x'(r)=\frac{f(x(r),y(r),r)}{x(r)-y(r)}, y'(r)=\frac{g(x(r),y(r),r)}{x(r)-y(r)}$. Any idea? :)
And thank you for reply ;)

4. Nov 21, 2014

zoki85

I'm affraid I don't understand meaning of notation you used here. Why don't you write instead of f(x(r),y(r),r) and g(x(r),y(r),r) just f(r) and g(r)?

5. Nov 21, 2014

Vrbic

I wanted to emphasize that there are unknown functions and Im not able to carry any trick like expansion. Exact form of equation is here https://www.physicsforums.com/threa...umerical-solution-with-critical-point.783090/ but no comments. I wrote it probably in wrong way so Im trying step by step find how to solve it.

6. Nov 21, 2014

the_wolfman

The word you're are looking for is "singular point."

The integrand $\frac{\sin{x}}{x}$ has a removable singularity. We can get around it by defining $\frac{\sin{x}}{x} =1$ for x=0 . This insight comes from observing the limiting behavior for small $x$ . Integrating the Taylor series term by term works for small x but fails miserably for x>1.

What are the functions $f\left(x,y,r\right)$ and $g\left(x,y,r\right)$. There are a number of "tricks" that allow you to treat different kinds of singularities. But different tricks work for different kinds of singular points.

7. Nov 21, 2014

Vrbic

Ah, yes it is clear. Thank you for a comment.

$u'=\frac{D_1}{D}$
$\rho'=\frac{D_2}{D}$, where $D_1=\frac{2a^2/r-\alpha/r^2}{\rho}$, $D_2=\frac{2u^2/r-\alpha/r^2}{u}$ and $D=\frac{u^2-a^2}{\rho u}$, where $a(r)=a_0\big(\frac{\rho(r)}{\rho_0}\big)^{(\Gamma-1)/2}$, $\rho(r)$ and $u(r)$ are function of r and $\Gamma, a_0, \rho_0, \alpha$ are constant.

This is the system. Here is described all this problem: https://www.physicsforums.com/threa...umerical-solution-with-critical-point.783090/
Thank you again.

8. Nov 22, 2014

zoki85

Ok then. Without getting into mess of other thread, here is what I can say. This is a system of differential equation (generally, a nonlinear one):

dx/dr = G(x,y,r)/z(x,y)
dy/dr = F(x,y,r)/z(x,y)
where z(x,y)= x-y

For behaviour near singularity you must specify or explore what's going on when z(x,y) → 0. Also explore if there is a stability issue for solutions to the system ( for instance, check Lyapunov's criteria etc). And for z(x,y) ≠ 0 the system yields equation F(x,y,r)⋅dx - G(x,y,r)⋅dy = 0 which is a special case of well known Pfaff's PDE in 3D: P(x,y,z)⋅dx + Q(x,y,z)⋅dy +T(x,y,z)⋅dz =0

Regards

Last edited: Nov 22, 2014
9. Nov 22, 2014

Vrbic

Ou thank you very much. And what about z(x,y)=0. What possibilities are? And could you give me a reference to literature which concern in this problems or something like that. Thank you very very much ;)

10. Nov 22, 2014

zoki85

For z(x,y)=0 you have to explore at line y=x : If G≠0,F≠0 than "asymptotic" cases. If G=0 or/and F=0 than "0/0" cases to check .
Since you can't exactly solve your nonlinear system you should linearize it. Here is instructional video how to do it.
And for probing system stability see this paper . Treating all the variables like they are independent (via Pfaffian) sometimes may help to explore behaviour of system too . Many things depend on form of the functions F, G.

11. Nov 22, 2014

Vrbic

Thank you again, understand.
Im looking for solution which is in "0/0". In my def. of function Im looking specially for case when "my" D=D1=D2=0 and (I mean) it is "smooth" solution.