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