Equilibrium Solutions and Their Stability

Now consider the direction field for the differential equation [latex]y^{\prime} = (x - 3)(y^2 - 4)[/latex], shown below. This direction field has several interesting properties that reveal important information about the behavior of solutions.

First, notice the horizontal dashes that appear all the way across the graph at [latex]y = -2[/latex] and [latex]y = 2[/latex]. This means that when [latex]y = -2[/latex], we have [latex]y^{\prime} = 0[/latex]. Let’s verify this by substituting into the differential equation:

[latex](x - 3)(y^2 - 4) = (x - 3)((-2)^2 - 4) = (x - 3)(0) = 0 = y^{\prime}[/latex]

Therefore [latex]y = -2[/latex] is a solution to the differential equation. Similarly, [latex]y = 2[/latex] is also a solution.

These are the only constant-valued solutions to this differential equation. Here’s why: If [latex]y = k[/latex] is a constant solution, then [latex]y^{\prime} = 0[/latex]. Substituting into our equation gives:

[latex]0 = (x - 3)(k^2 - 4)[/latex]

Since this must be true for all values of [latex]x[/latex], we need [latex]k^2 - 4 = 0[/latex], which gives us [latex]k = -2[/latex] and [latex]k = 2[/latex].

These constant solutions are called equilibrium solutions to the differential equation.

equilibrium solutions

Consider the differential equation [latex]y^{\prime} = f(x,y)[/latex]. An equilibrium solution is any solution of the form [latex]y = c[/latex], where [latex]c[/latex] is a constant.

[latex]\\[/latex]

To find equilibrium solutions: Set the right-hand side equal to zero and solve for constant values. An equilibrium solution [latex]y = k[/latex] satisfies [latex]f(x,k) = 0[/latex] for all values of [latex]x[/latex] in the domain of [latex]f[/latex].

Stability of Equilibrium Solutions

An important characteristic of equilibrium solutions is whether other solutions approach them as asymptotes for large values of [latex]x[/latex]. This leads us to classify equilibrium solutions by their stability.

stability classifications

Consider the differential equation [latex]y^{\prime} = f(x,y)[/latex] with equilibrium solution [latex]y = k[/latex]:

• Asymptotically Stable: If there exists [latex]\epsilon > 0[/latex] such that solutions starting near [latex]k[/latex] (within [latex]\epsilon[/latex]) approach [latex]k[/latex] as [latex]x \to \infty[/latex].
• Asymptotically Unstable: If there exists [latex]\epsilon > 0[/latex] such that solutions starting near [latex]k[/latex] never approach [latex]k[/latex] as [latex]x \to \infty[/latex].
• Semi-stable: If the solution is neither stable nor unstable.

Let’s examine the behavior of solutions to [latex]y^{\prime} = (x - 3)(y^2 - 4)[/latex] with initial condition [latex]y(0) = 0.5[/latex]. The figure below shows the direction field and corresponding solution.

Notice that the solution values stay between [latex]y = -2[/latex] and [latex]y = 2[/latex] (our equilibrium solutions). As [latex]x[/latex] approaches infinity, [latex]y[/latex] approaches [latex]2[/latex].

The behavior is similar if the initial value is higher than [latex]2[/latex], say [latex]y(0) = 2.3[/latex]. In this case, solutions decrease and approach [latex]y = 2[/latex] as [latex]x[/latex] approaches infinity. Therefore [latex]y = 2[/latex] is asymptotically stable.

What happens when the initial value is below [latex]y = -2[/latex]? The figure below illustrates this scenario with initial value [latex]y(0) = -3[/latex].

The solution decreases rapidly toward negative infinity as [latex]x[/latex] approaches infinity. If the initial value is slightly higher than [latex]-2[/latex], the solution approaches [latex]2[/latex] instead. Since solutions don’t approach [latex]y = -2[/latex] in either case, [latex]y = -2[/latex] is asymptotically unstable.

Create a direction field for the differential equation [latex]y^{\prime} ={\left(y - 3\right)}^{2}\left({y}^{2}+y - 2\right)[/latex] and identify any equilibrium solutions. Classify each of the equilibrium solutions as stable, unstable, or semi-stable.

Show Solution

The direction field is shown below.

The equilibrium solutions are [latex]y=-2,y=1[/latex], and [latex]y=3[/latex]. To classify each of the solutions, look at an arrow directly above or below each of these values. For example, at [latex]y=-2[/latex] the arrows directly below this solution point up, and the arrows directly above the solution point down. Therefore all initial conditions close to [latex]y=-2[/latex] approach [latex]y=-2[/latex], and the solution is stable. For the solution [latex]y=1[/latex], all initial conditions above and below [latex]y=1[/latex] are repelled (pushed away) from [latex]y=1[/latex], so this solution is unstable. The solution [latex]y=3[/latex] is semi-stable, because for initial conditions slightly greater than [latex]3[/latex], the solution approaches infinity, and for initial conditions slightly less than [latex]3[/latex], the solution approaches [latex]y=3[/latex].

Analysis

It is possible to find the equilibrium solutions to the differential equation by setting the right-hand side equal to zero and solving for [latex]y[/latex]. This approach gives the same equilibrium solutions as those we saw in the direction field.