2. Autonomous scalar problemsΒΆ
Our general form for a first-order scalar ODE is \(x'=f(t,x)\). In the special case where \(f\) does not depend explicitly on time, i.e., when
we say the equation is autonomous. These equations are the simplest place to start.
Even before attempting a full solution of the problem, we can see that a value \(\hat{x}\) that satisfies \(f(\hat{x})=0\) is special: the constant function \(x(t)\equiv \hat{x}\) is a solution. Such a value is called a steady state, equilibrium solution, or fixed point. One of the most important characteristics of each steady state is its stability, as we first encountered in the phase plane. Here we use some definitions to formalize the idea a bit more.
We will use a shorthand to help pin down these properties. Let \(x(t,X)\) be the solution of the initial-value problem
The steady state \(\hat{x}\) is said to be asymptotically stable if there is a positive number \(\delta\) such that
whenever \(|X-\hat{x}|< \delta\).
In words, asymptotic stability means that if you start close enough to \(\hat{x}\), then you end up there as \(t \to \infty\). There is a weaker form of the property that is a bit more difficult to parse.
The steady state \(\hat{x}\) is said to be stable if, for any \(\epsilon>0\), you can ensure that \(|x(t,X)-\hat{x}|<\epsilon\) for all time whenever \(|X-\hat{x}|<\delta\), where \(\delta\) depends on \(\epsilon\).
Stability means that you can stay as close as you like to \(\hat{x}\) by starting off close enough to it. Any point that is asymptotically stable is stable by this definition as well. We will call this weak stability to highlight the contrast with asymptotic stability. Finally, a steady state that is not at least weakly stable is called unstable.
One tool for discriminating between stable and unstable equilibria is a phase line diagram.
Example
Consider the ODE \(x'=x-x^3\). We will start by graphing \(f(x)=x-x^3\).
The equilibrium solutions occur at the crossings of the \(x\)-axis in the plot, where \(dx/dt=0\).
Now we highlight those parts of the plot where the graph is above the \(x\)-axis, i.e., where \(dx/dt > 0\).
Imagine that the solution is represented by a point \(x(t)\) sliding along the \(x\)-axis. Under the highlighted regions, this point has to be sliding rightward, because \(x'>0\) there.
Similarly, on the remaining segments of the \(x\)-axis, the solution must be sliding leftward.
Graphically it is now clear that the steady states at \(x=\pm 1\) are stable, while the one at \(x=0\) is unstable.
An analytical test for stability is:
If \(f'(\hat{x})<0\), then \(\hat{x}\) is asymptotically stable.
If \(f'(\hat{x})>0\), then \(\hat{x}\) is unstable.
As you might surmise, the case \(f'(\hat{x})=0\) is ambiguous, and a phase line or other analysis is needed to reveal the stability.