# Equilibrium stable or unstable - differential euqation

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• July 23rd 2010, 06:21 AM
mathfilip
Equilibrium stable or unstable - differential euqation
So this is actually a physics problem but im having problem with the math again.

My function is: y'' + y*sin(a)*k^2 = 0 , where a and k are constants.

from this I get the following solution;

y = y(max)*cos(w*t + a)

where w = sin(a)*k.

This far I understand everything. However after this my teacher has written "thus it is stable"

My question is how can he tell that it is stable? I know there is some way to do like secondary derivative tests? Can I do that or can it be seen someother way wheter its stable or not?

• July 23rd 2010, 06:37 AM
Ackbeet
I don't quite agree with your solution. You should have w = k*sqrt(sin(a)). Unless you have a typo and the original DE said (sin(a)*k)^2.

An equilibrium point is a point where the potential energy function has a local extremum. It's exactly like the top or bottom of a hill and a wheel. If you're on top of a hill (local max), you have an unstable equilibrium point. If you're at the bottom of a hill (local min), you have a stable equilibrium point. So yes, it is related to a second derivative test:

1. Form the potential energy function as a function of position y.
2. Then find its extrema. At each extremum, compute the second derivative of potential energy with respect to position y.
3. If the second derivative is positive, you've got a stable equilibrium position. If negative, then it's unstable.

Make sense?
• July 23rd 2010, 08:05 AM
mathfilip
Thanks for answering. Yes I forgot to square the sine. I dont think I can go that way. I am sorry for the notation but the y represents a small angular displacement in this exercise and it will be hard to relate it to the potential energy even though it might be possible. Since my teacher havent written that he makes any tests it should probably be seen quite easily but I dont know how.

There is also a second equation where he makes the same statement:

y'' + y(g/a - k^2) = 0 stable if g/a > k^2; unstable if g/a < k^2

Can you see how he gets whether they are stable or not from this? Otherwise I will have to try go into by defining some expression for the potential energy.
• July 23rd 2010, 08:11 AM
Ackbeet
Ah, I remember now how to do this in your context. The form of your solution changes depending on whether y'' and y have the same sign. If they have the same sign, then think about a positive function that has to be concave up. There's no way that function is going to settle down to any stable value. In addition, your solution is going to look like exponentials, one of which will blow up at positive infinity, and the other at negative infinity. On the other hand, if y'' and y have the opposite sign, then your solution is a sum of sines and cosines. You'll get oscillation, but y is not going to go off to infinity. So. In your problem, do y'' and y have the same or opposite signs?
• July 23rd 2010, 08:20 AM
lvleph
EDIT: Ignore the following, DEQ had typo.

A very minor point on the first problem
$w = \pm \sqrt{\sin a}\cdot |k|$

EDIT: The other problem with the first problem is that the form of the solution depends on the sign of $\sin a$. If $\sin a < 0$ we have one form, if $\sin a > 0$ there is another form, and finally if $\sin a = 0$ there is yet another form.
• July 23rd 2010, 08:22 AM
Ackbeet
lvleph: See post # 3. The original DE, as I understand it, should have been the following:

$y'' + y\,k^{2}\,\sin^{2}(a) = 0.$
• July 23rd 2010, 08:29 AM
lvleph
Ah, then ignore my edit. I was thinking that the problem was more difficult than I would expect.
$w = |k|\cdot |\sin a|$
• July 23rd 2010, 08:32 AM
Ackbeet
I don't think it really matters what the sign of k and sin(a) are. After taking two derivatives, they'll get squared anyway.
• July 23rd 2010, 08:40 AM
lvleph
It matters because of the first derivative. You want to make sure it is increasing when it is suppose to and decreasing when it is suppose to.
• July 23rd 2010, 08:44 AM
Ackbeet
What if you don't know that information a priori? Then you would have ruled out those options.
• July 23rd 2010, 08:51 AM
lvleph
Are you referring to my edit? If so the fact that we have the square makes the difference why we don't care about the sign. However, we need the absolute value for the reason I stated in my last post. If we didn't have the square it matters because of the eigenvalues. If we have imaginary eigenvalues then the solution is a combination of sines and cosines, if we have real eigenvalues we get a combination of exponentials, etc.
• July 23rd 2010, 09:01 AM
Ackbeet
I wasn't referring to your edit, but to the comment about first derivatives. I think this DE comes from classical mechanics. That means, most likely, that k is real, and sin(a) is real. The form of the solutions is determined no matter what sign k and sin(a) have, and you should be able to apply the initial conditions to determine the correct values of the constants of integration. I don't see the need to ensure that the first derivative have a particular sign in any region of interest. Why would you need to be able to do that? k and a are either known a priori or determined experimentally.
• July 23rd 2010, 09:08 AM
lvleph
It comes from the eigenvalues.
$y'' + k^2 \sin^2 a y = 0$
Which gives eigenvalues $\lambda = \pm \sqrt{-k^2 \sin^2 a} = |k| |\sin a| i$.
Now let $\omega = |k| |\sin a|$ and then the general solution is of the form
$y = A \cos \omega t + B \sin \omega t$.
If we then look at the first derivative we have
$y' = -A \omega \sin \omega t + B \omega \cos \omega t$.
This is why the absolute value is important. Because, the sign of $k$ and $\sin a$ would change the behavior of the first derivative.
• July 23rd 2010, 09:15 AM
mathfilip
Comment on your little discussion first: Yes this is a physical problem so we dont care about absolute signs since its a positive physical quantity.

Back to the real question. You seem to be almost right Ackbeet but it should be the otherway around. If they have opposite signs then it is unstable. If we have y = y(max) * cos (w*t + a), then taking the derivative twive gives y'' = - y(max)w^2*cos(w*t + a)

But then your explanation is correct since we get y'' - y which will then always be negative.

Thanks again. Really helped!
• July 23rd 2010, 09:20 AM
Ackbeet
Reply to lvleph at Post # 13:

You're not really listening to me. I'm saying that controlling what happens to the first derivative may not be necessary, desirable, or even possible!

Reply to mathfilip at Post # 14:

I think it matters whether you think of y'' and y as being on the same side of the equals sign or not. Here's what I mean:

Strip away the constants, and just look at this DE:

y'' = -y. Now here, y'' and y have the opposite sign. The solutions are sines and cosines: stable.

y'' = y. Now here, y'' and y have the same sign. The solutions are exponentials. Unstable.

Does this help?
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