7.3: Autonomous First Order Equations

In this section we will study the stability of nonlinear first order autonomous equations. We will then extend this study in the next section to looking at families of first order equations which are connected through a parameter.

Recall that a first order autonomous equation is given in the form

\[\dfrac{d y}{d t}=f(y) \label{7.8} \]

We will assume that \(f\) and \(\dfrac{\partial f}{\partial y}\) are continuous functions of \(y\), so that we know that solutions of initial value problems exist and are unique.

A solution \(y(t)\) of Equation \ref{7.8} is called an equilibrium solution, or a fixed point solution, if it is a constant solution satisfying \(y^{\prime}(t)=0\). Such solutions are the roots of the right-hand side of the differential equation, \(f(y)=0\).

Example \(\PageIndex{1}\)

Find the equilibrium solutions of \(y^{\prime}=1-y^{2}\).

Solution

The equilibrium solutions are the roots of \(f(y)=1-y^{2}=0\). The equilibria are found to be \(y=\pm 1\).

Once we have determined the equilibrium solutions, we would like to classify them. Are they stable or unstable? As we had seen previously, we are interested in the behavior of solutions near the equilibria. This classification can be determined using a linearization of the given equation. This will provide an analytic criteria to establish the stability of equilibrium solutions without geometrically drawing the phase lines as we had done previously.

Definition: Linearization of first order equations

Let \(y^{*}\) be an equilibrium solution of Equation \ref{7.8}. Then, any solution can be written in the form

\[y(t)=y^{*}+\xi(t) \label{Lin}\]

where \(\xi(t)\) measures how far the solution is from the equilibrium at any given time.

Inserting Equation \ref{Lin} form into Equation \ref{7.8}, we have

\[\dfrac{d \xi}{d t}=f\left(y^{*}+\xi\right) \nonumber \]

We now consider small \(\xi(t)\) in order to study solutions near the equilibrium solution. For such solutions, we can expand \(f(y)\) about the equilibrium solution,

\[f\left(y^{*}+\tilde{\xi}\right)=f\left(y^{*}\right)+f^{\prime}\left(y^{*}\right) \xi+\dfrac{1}{2 !} f^{\prime \prime}\left(y^{*}\right) \xi^{2}+\cdots\nonumber \]

Since \(y^{*}\) is an equilibrium solution, \(f\left(y^{*}\right)=0\), the first term in the Taylor series vanishes. If the first derivative does not vanish, then for solutions close to equilibrium, we can neglect higher order terms in the expansion. Then, \(\xi(t)\) approximately satisfies the differential equation

\[\dfrac{d \xi}{d t}=f^{\prime}\left(y^{*}\right) \xi \label{7.9} \]

This is called a linearization of the original nonlinear equation about the equilibrium point. This equation has exponential solutions for \(f^{\prime}\left(y^{*}\right) \neq 0\),

\[\xi(t)=\xi_{0} e^{f^{\prime}\left(y^{*}\right) t}.\nonumber \]

Now we see how the stability criteria arise. If \(f^{\prime}\left(y^{*}\right)>0, \xi(t)\) grows in time. Therefore, nearby solutions stray from the equilibrium solution for large times. On the other hand, if \(f^{\prime}\left(y^{*}\right)<0, \xi(t)\) decays in time and nearby solutions approach the equilibrium solution for large \(t\). Thus, we have the results:

 

\[ \begin{aligned} & \begin{aligned}&f^{\prime}\left(y^{*}\right)<0, \quad y^{*} \text { is stable. } \\&f^{\prime}\left(y^{*}\right)>0, \quad y^{*} \text { is unstable. }\end{aligned} \end{aligned}\label{7.10} \]

The stability criteria for equilibrium solutions of a first order differential equation.

Example \(\PageIndex{2}\)

Determine the stability of the equilibrium solutions of \(y^{\prime}=1-y^{2}\).

In the last example we found the equilibrium solutions, \(y^{*}=\pm 1 .\) The stability criteria require computing

\[f^{\prime}\left(y^{*}\right)=-2 y^{*}\nonumber \]

For this problem we have \(f^{\prime}(\pm 1)=\mp 2\). Therefore, \(y^{*}=1\) is a stable equilibrium and \(y^{*}=-1\) is an unstable equilibrium.

Example \(\PageIndex{3}\)

Find and classify the equilibria for the logistic equation \(y^{\prime}=y-y^{2} .\)

Solution

We had already investigated this problem using phase lines. There are two equilibria, \(y=0\) and \(y=1\).

We next apply the stability criteria. Noting that \(f^{\prime}(y)=1-2 y\), the first equilibrium solution gives \(f^{\prime}(0)=1 .\) So, \(y=0\) is an unstable equilibrium. Since \(f^{\prime}(1)=-1<0\), we see that \(y=1\) is a stable equilibrium. These results are the same as we hade determined earlier using phase lines.