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 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