What Is Stability in the Context of the Hill Equation for a Mass-Spring System?

Drokz
Messages
3
Reaction score
0
When trying to solve a problem I arrive at the following equation of motion / Hill equation:

\frac{d^{2}y}{dx^2} + \frac{4 k_0}{m w^2} cos(2x)y = 0

There exists a value x_0 such that for all x>x_0 the motion is stable.

I actually don't know what is meant by this 'stability'. Can someone help, please?

Thanks, Drokz
 
Physics news on Phys.org
I've usually seen Hill's equation with omega^2, not omega^(-2). Is this a typo?
 
I don't think it is a typo. Omega is just a constant here, I think.
 
Ok. Solutions are products of exponential terms and periodic functions, so your stability condition is needed to keep the exponential parts of the solutions bounded. Do you have access to a reference on Mathieu functions like Whittaker and Watson, or Erdelyi's Higher Transcendental Functions? The latter one has a clear discussion of the stable and unstable solution regions (p. 101 vol. 3).
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top