Equilibrium Solutions and Stability

We have seen how a lot of information about the solutions of a differential equation is available without solving it explicitly. In particular, there are oftentimes constant solutions which we called equilibrium solutions or fixed points of the differential equation. These are important because they represent behaviour which persists in time.


The Logistic Equation Revisited

When we looked at the logistic equation

\[  \frac{dy}{dt} = ry(1-y) 
 \]

we saw that there were two equilibrium solutions---namely, $  \bar{y} = 0  $ and $  \bar{y} = 1  $ . The first equilibrium solution, $  \bar{y} = 0  $ represents the situation when there is no population. This condition persists because there are no members of the population to die or reproduce.

The second equilibrium solution $  \bar{y} = 1  $ also represents persistent behaviour. In this case, however, the rate of reproduction exactly balances the mortality and so the population remains constant at this level.

If we think a bit harder about this example though, we see that the behaviour of the solutions near the two equilbrium solutions is decidedly different.

Notice how solutions which start out near $  y = 0  $ eventually move far away from this value. We call this an unstable equilibrium. However, solutions which start out near $  y = 1  $ only get closer to that equilibrium solution and so we call that a stable equilibrium.

Here is an analogy: consider a rigid rod which is firmly attached to the wall on one end and allowed to pivot as a pendulum. If we work very hard, we could balance it so that it points straight up. However, just the slightest touch will cause it to move and eventually point downwards. This is an unstable equilibrium. Another equilibrium is when the pendulum points downwards. If we move it to some nearby place, it will swing until it eventually returns to the downward position. This is a stable equilibrium.

More generally, near a stable equilibrium, solutions look like:

while solutions near an unstable equilibrium look like:

In practice, a stable equilibrium is important because it represents behaviour which cannot easily be changed---it represents a fundamental feature of the system. For example, stable equilibria can be useful for making predictions because lots of solutions eventually settle down near the stable equilibria (like in the pendulum).

What we would like to do is devise a simple test to determine whether an equilibrium solution to a differential equation is stable or not.


The Linear Stability Test

Let's first represent the behaviour of the logistic equation in a slightly different way. Here we are thinking about the equation

\[  \frac{dy}{dt} = 2y(1-y) 
 \]

If we plot $  \frac{dy}{dt}  $ versus y , we see the equilibrium solutions as given by $  \frac{dy}{dt} = 0  $ ---that is, where the graph passes through the y axis. Remember also that when $  \frac{dy}{dt} > 0  $ , y is increasing which means we are moving to the right on this graph. Similarly, when $  \frac{dy}{dt} < 0  $ , then y is decreasing and we are moving to the left of the graph. This is the meaning of the arrows on the y axis in the figure below.

From this picture, we are able to determine the nature of the equilibrium solutions.

Equilibrium $  \bar{y} = 0  $ Equilibrium $  \bar{y} = 1  $
unstable stable

What we are trying to point out in this case is that the stability is determined by what happens right around the equilibrium solution. In particular, when the derivative is negative to the left and positive to the right of the equilibrium solution, we have an unstable equilibrium. However, when the derivative is positive to the left and negative to the right, it is stable.

More generally, let's consider the differential equation

\[ 
\frac{dy}{dt} = f(y) 
 \]

We know that an equilibrium solution corresponds to a value $  \bar{y}  $ such that $  f(\bar{y}) = 0  $ . Now to determine the stability, we can look at the graph $  \frac{dy}{dt} = f(y)  $ .

Equilibrium $  \bar{y}   $ Equilibrium $  \bar{y}   $
$  f^\prime(\bar{y}) > 0  $ $  f^\prime(\bar{y}) < 0  $
unstable stable

This more general observation is known as the Linear Stability Test. It tells us simply that if we know an equilibrium solution $  \bar{y}  $ to the differential equation $  \frac{dy}{dt} = f(y)  $ and if $  f^\prime(\bar{y}) > 0  $ , then the equilibrium solution is unstable. However, if $  \frac{dy}{dt} < 0  $ , then the equilibrium solution is stable.

You may wonder why this is a useful thing to know. If we can draw a plot of the differential equation, then stability is pretty easy to determine. However, when the differential equation becomes more complicated, it can be more difficult to sketch its graph. Computing its derivative is much easier though. In addition, this test can be applied in many other diverse contexts, such as systems of differential equations, so that it really is a useful thing to be familiar with. Shortly, we will see an application.

As an example, let's consider the differential equation

\[ 
\frac{dy}{dt} = f(y) = y - y^3 
 \]

We have seen that there are equilibrium solutions $  \bar{y} = 0,-1, 1  $ . To determine the stability of these solutions, all we need is to do is calculate $  f^\prime(y) = 1 - 3y^2  $ . Then we see that

$  f^\prime(-1) = -2 < 0  $ $  \bar{y} = -1  $ is stable
$  f^\prime(0) = 1 > 0  $ $  \bar{y} = 0  $ is unstable
$  f^\prime(1) = -2 < 0  $ $  \bar{y} = 1  $ is stable