5.3 A stability test for equilibrium solutions
Notice how when constructing the phase line diagram we relied on the behavior of solutions around the equilibrium solution to classify the stability. As an alternative we can also use the point at the equilibrium solution itself.
To do this we are going to consider the general differential equation \(\displaystyle \frac{dy}{dt}=f(y)\). We are going to assume that we have an equilibrium solution at \(y=y_{*}\).
We use local linearization to construct a locally linear approximation to \(L(y)\) to \(f(y)\) at \(y=y_{*}\):
\[\begin{equation} L(y) = f(y_{*}) + f'(y_{*}) \cdot (y-y_{*}) \tag{5.1} \end{equation}\]
Once we have the local linearization, there are two follow-on steps to simplify Equation (5.1). First, because we have an equilibrium solution, \(f(y_{*}) =0\). Second, Equation (5.1) can be written with a new variable \(P\), defined by variable \(P = y-y_{*}\). Then Equation (5.1) translates to
\[\begin{equation} \frac{dP}{dt} = f'(y_{*}) \cdot P \tag{5.2} \end{equation}\]
Does Equation (5.2) look familiar? It should! This equation is similar to the example where we classified the stability of \(\displaystyle \frac{dy}{dt} = k \cdot y\) – cool! So let’s use what we learned in Example 5.3 above to classify the stability:
Local linearization stability test for equilibrium solutions: For a differential equation \(\displaystyle \frac{dy}{dt} = f(y)\) with equilibrium solution \(y_{*}\), we can classify the stability of the equilibrium solution through the following:
- If \(f'(y_{*})’>0\) at an equilibrium solution, the equilibrium solution \(y=y_{*}\) will be unstable.
- If \(f'(y_{*}) <0\) at an equilibrium solution, the equilibrium solution \(y=y_{*}\) will be stable.
- If \(f'(y_{*}) = 0\), we cannot conclude anything about the stability of \(y=y_{*}\).