# Homework Help: Stability of solutions to perturbations

1. Aug 28, 2006

### pivoxa15

How would you answer questions like 'is the solution stable to arbitary small perturbations in the intial values of x(0), x'(0) and x''(0)'?

2. Aug 28, 2006

### HallsofIvy

The first thing I would do is look up "stable"! What happens to the solution to the problem (third order differential equation?) if x(0), x'(0), x"(0) are slightly different? Is the new solution only slightly different from the old solution (stable) or can it become, as t increases, very different from the old solution (unstable).

3. Aug 28, 2006

### pivoxa15

When you say solution to the problem, do you mean x(t)?

I understand what Stability means but it is not well defined is it, more of a qualitative thing.

Yes, this is a third order problem. So you would approach it by changing the initial value (IV) of x''(0) and see the changes in x'(0) and x(0). And the consequences of these changes to x(t).

Then change IV of x'(0) (leave x''(0) as it is) and see changes in x(0) and how it affects x(t)

Changes in x(0) shouldn't have effects on the stability of x(t) as it only shifts the graph upward or downward

Last edited: Aug 28, 2006
4. Aug 29, 2006

### Benny

Not sure but the solution has a texp(t) factor in it which is independent of the initial conditions.

5. Aug 29, 2006

### HallsofIvy

Yes, by "solution to the problem" I mean x(t).

No, "stability" is perfectly well defined. According to the "Science and Technology Dictionary http://www.answers.com/topic/asymptotic-stability, stability is
"The property of a vector differential equation which satisfies the conditions that (1) whenever the magnitude of the initial condition is sufficiently small, small perturbations in the initial condition produce small perturbations in the solution; and (2) there is a domain of attraction such that whenever the initial condition belongs to this domain the solution approaches zero at large times. "
Yes, we are not given a specific size for the "sufficiently small perturbations" nor the "domain of attraction" but stability only requires that there exist such things no matter how small.

Benny, do you know something I don't? Since we weren't given the equation itself, how do you know there is a t exp(t) term in the solution?

6. Aug 29, 2006

### Benny

I believe that pivoxa is working on the same assignment as I am which is why I mentioned the texp(t) term. Although I think what I said in my previous reply is a little misleading so it's probably best if that comment is ignored. I have no idea as to how to do the problem anyway, it all seems quite vague. But I'll just go with what I think should be done. I hope I haven't caused too much confusion.

7. Aug 30, 2006

### pivoxa15

I ended up putting different IV (adding in an arbitary small change, epsilon) in the matrices when constructing the solutions for x, x', x'' and so got a new set of x, x', x'' which had an extra addition (epsilon)e^t. So clearly after the small purtabations in IV, near t=0, the solutions are still stable.

The te^t shouldn't be of any trouble because it is a particular solution and independent of IV.

8. Aug 30, 2006

### Benny

x(t) is the solution with x(0) = a, x'(0) = b, x''(0) = c. x_0(t) is the solution with x(0) = 1, x'(0) = x''(0) = 0. For a simple choice of a,b and c |x(t)-x_0(t)| 'blows up' at a rate proportional to exp(t) as t increases. This is assuming that I obtained the correct solution to the IVP though. I'd rather not think about the assignment anymore, finishing it caused quite a few problems for me.