31.3. STABILITY OF EQUILIBRIUM POINTS 605

point is unstable. Think f (y) = (y−a)3. Then f has a 0 derivative at a but is increasing.Similarly, you could have f decreasing through (a,0) in which case, the equilibrium pointwould be stable. The point is, anything can happen when f ′ (a) = 0.

You should regard f ′ (a) as an eigenvalue for the linear map x→ f ′ (a)x. The eigen-value is negative implies stability. The eigenvalue is positive implies not stable. The eigen-value is 0 means anything can happen.

The situation is completely similar for nonlinear systems of equations.

Definition 31.3.1 Consider y′ = f (y) , where we always assume f is C1. A point a iscalled an equilibrium point when f (a) = 0.

From the notion of differentiability,

f (a+y) = 0+Df (a)y+o(y)

The situation which generalizes what happens with functions of one variable is as follows.Let a be an equilibrium point for the differential equation y′ = f (y). Thus f (a) = 0.

eigenvalues of Df (a) have negative real parts equilibrium is stablesome eigenvalue of Df (a) has positive real part equilibrium is unstablesome eigenvalue of Df (a) has zero real part you have no idea

So what exactly is meant by stable? It is the same as in the case of scalar valuedequations.

Definition 31.3.2 An equilibrium point a for y′ = f (y) is stable if whenever the initialcondition y0 is sufficiently close to a, it follows that the solution to the initial value problemy′ = f (y) ,y (0) = y0 will stay close to a. Also, a is asymptotically stable if whenever y0is close enough to a, then the solution of the initial value problem just described convergesto a as t→ ∞.

In fact, one has a little more in case all eigenvalues are negative.

eigenvalues of Df (a) have negative real parts equilibrium is asymptotically stablesome eigenvalue of Df (a) has positive real part equilibrium is unstablesome eigenvalue of Df (a) has zero real part you have no idea

Example 31.3.3 Consider the following system of equations for which(

0 0 0)T

isan equilibrium point. Determine whether this is a stable equilibrium. x

yz

′

=

 xy−12z−5xy2− y−2x−6z

xy2 + x+2z

You need to get the derivative of the right side. The matrix of this is y−5 x −12

−2 2y−1 −6y2 +1 2xy 2

