# ODE: limit as parameters tend to 0

1. Mar 4, 2006

### Vey2000

Maybe there's something I'm missing, but I just realized that if we take a certain general ODE such as y''+k^2 y=0 for example, and assume that in a specific case k=0 the solution is vastly different depending on whether k is set to 0 before or after solving the ODE.

The general solution to y''+k^2 y = 0 (with k>0) is y=A cos(kx)+B sin(kx).
If we then set k=0 the solution reduces to y=A.
Yet, if we set k=0 in the original ODE we get [1] y''=0, whose solution is [2] y=Bx+C.
Where A, B, and C are constants.

I get analogous results for y''-k^2 y = 0.

On the other hand, if I start with y''+2b y'+k^2 y = 0, then it seems that setting either b or k to 0, but not both, can be done either in the ODE itself or in the general solution.

Hrm, I just had another insight while writing this. The form of solution [2] is due to the characteristic equation having a repeated real root of r=0. Is this the reason why my initial attempt failed? Because I was setting the parameter to the same value as the eigenvalue of the ODE?

2. Mar 4, 2006

### HallsofIvy

Basically, that's correct. In your first two examples, as k goes to 0, the characteristic equation goes from having two distinct solutions to one solution. For your third example, the characteristic equation is r2+ 2br+ k2= 0 which, in general has two distinct solutions. Taking either parameter 0 gives r2+ k2= 0 which still has 2 (imaginary) solutions or r2+ 2br= 0 which also still has two solutions (one r= 0).

In technical terms, for your first two examples, k= 0 is a "bifurcation point".

Generally, solutions will have distinct qualitative changes at a bifurcation point.

Last edited by a moderator: Mar 4, 2006
3. Mar 4, 2006

### Vey2000

Thank you, now I have a starting point to look more into this.