Finding the stationary points & stability of differential equations

Question

For the following system of differential equations, find two stationary points and determine the stability of one of them (your choice).

X' = XY - X

Y' = X - Y

Answer 1

In order to find the stationary points, we must solve the (non-linear) system:

XY - X=0

X-Y=0.

The first equation may be written as X(Y-1)=0, which implies that either X=0, or Y=1. But the second equation tells us that X=Y. Therefore, our stationary points are (0,0) and (1,1).

Now, in order to determine the stability of (0,0), we analyze the linear stability, so we compute the derivative matrix, or Hessian, of the vector-valued function F(X,Y) = (XY-X, X-Y). Recall that if F(X,Y) = (P(X,Y), Q(X,Y)), where P, Q are scalar functions, then the derivative matrix is

DF(X,Y) = [[P_X, P_Y],[Q_X, Q_Y]]. Here P_X denotes partial derivative of P w.r.t. X.

In our case we get DF(X,Y) = [[Y-1, X],[1,-1]]. Then we evaluate at the stationary point:

DF(0,0) = [[-1,0],[1,-1]], and compute the eitenvalues of it, which are the roots of the characteristic polynomial

p(t) = (-1-t)(-1-t). That is, we only have one eigenvalue t=-1, with multiplicity 2. Since the eigenvalues are both negative, we conclude that this point is stable.

A similar reasoning can be applied to (1,1). In this case,

DF(1,1) = [[0,1],[1,-1]]. The characteristic polynomial is p(t) = -t(-1-t) - 1 = t²+t-1, which has roots (-1±√5)/2. So we get a positive eigenvalue and a negative eigenvalue. This means that (1,1) is a saddle point, that is, it is unstable.

Note: the linear stability analysis we just performed only fails to describe the stability of the non-linear system in the case when there are eigenvalues of the derivative matrix at a stationary point that are complex with zero real part, that is, purely imaginary.