From: rusin@vesuvius.math.niu.edu (Dave Rusin)
Subject: Re: stability query
Date: 8 Apr 1999 02:03:59 GMT
Newsgroups: sci.math
Keywords: some elementary autonomous systems of differential equations

In article <37079E24.F13AA222@uk.com>, Mumsy  <stuart.levene@uk.com> wrote:
>I am trying to determine the stability of the equilibrium points of a
>dynamical system given by
>
>**x = f(x,y,*x,*y)
>**y = g(x,y,*x,*y)
>
>(*x := dx/dt, **x := d^2(x)/dt^2 etc; for `*' read `dot')
>
>I have decoupled the equations to put them into the form
>
>*x = u
>*y = v
>*u = f(x,y,u,v)
>*v = g(x,y,u,v)
>
>so that, writing X = (x,y,u,v) I have *X = F(X).  I then linearise F by
>Taylor expanding about points X0 where F(X0) = 0 to get
>
>*X = A X
>
>where X is a 4x4 matrix.  Then I find eigenvalues of A, and for my
>particular problem it turns out that these are P, -P, I*Q, -I*Q for some
>P,Q real (I^2 = -1); the motivation behind this is to try for a solution
[...]

Good. More precisely, the local behaviour of  X  will be that of the
solution to  *X = A X. With the information you gave, there are matrices
Q1  and  Q2, inverses of each other, with  A = Q1 D Q2  where  D  is the 
block sum of matrices  
        [ P    0 ]       [  0  Q ]
        [ 0   -P ]  and  [ -Q  0 ].
Then  Y = Q2 X   satisfies   *Y = D Y,  which is easily solved: 
y1 = c1 exp(P t), y2 = c2 exp(-P t), y3 = c3 cos(Q t) + c4 sin(Q t),
y4 = -c3 sin(Q t) + c4 cos(Q t). Then you can recover  X = Q1 Y  as a
linear combination of these four functions.

Of course in the original setting it is only the first two coordinates
(x,y)  of  X = (x,y,u,v)  which interest you. These are, still, just
linear combinations of  y1  through  y4.

>My question (eventually!) is: if I had a 2-D system and I got say I*Q,
>-I*Q for my eigenvalues I'd know that I had a centre (circular orbits)
>and if I got +P, -P I'd have a saddle point.  What can I conclude about
>the stability of X0 given the four eigenvalues I have found?  I can also
>find the eigenvectors easily enough but am unsure what their
>significance is.

You've hit the nail on the head here. What's of critical importance to
answer your question is to know just _which_ linear combinations of the
four functions applies to give (x,y). That is, you need to know the
first two rows of  Q1. If those two rows end in [0,0], you've got a 
saddle, for example.

The first two columns of  Q1  are the  +P and -P eigenvectors of  A,
respectively the other two columns are simple (real) linear combinations
of the  +QI and -QI  eigenvectors. So if you really do know the eigenvectors
of  A, you can compute  P1  and thus get an explicit description of your
solution, if you like.

dave