INTRODUCTION

We have seen that the plane autonomous system

gives rise to a vector field V(x, y) = (P(x, y), Q(x, y)), and a solution X = X(t) of the system may be interpreted as the resulting path of a particle that is initially placed at position X(0) = X₀. If X₀ is a critical point of the system, then the particle remains stationary. In this section we examine the behavior of solutions when X₀ is chosen close to a critical point of the system.

Some Fundamental Questions

Suppose that X₁ is a critical point of a plane autonomous system and X = X(t) is a solution of the system that satisfies X(0) = X₀. If the solution is interpreted as a path of a moving particle, we are interested in the answers to the following questions when X₀ is placed near X₁:

1. Will the particle return to the critical point? More precisely, if X = X(t) is the solution that satisfies X(0) = X₀, is lim_t→∞ X(t) = X₁?
2. If the particle does not return to the critical point, does it remain close to the critical point or move away from the critical point? It is conceivable, for example, that the particle may simply circle the critical point, or it may even return to a different critical point or to no critical point at all. See FIGURE 11.2.1.

If in some neighborhood of the critical point case (a) or (b) in Figure 11.2.1 always occurs, we call the critical point locally stable. If, however, an initial value X₀ that results in behavior similar to (c) can be found in any given neighborhood, we call the critical point unstable. These concepts will be made more precise in Section 11.3, where questions (i) and (ii) will be investigated for nonlinear systems.

Stability Analysis

We will first investigate these two stability questions for linear plane autonomous systems and lay the foundation for Section 11.3. The solution methods of Chapter 10 enable us to give a careful geometric analysis of the solutions to

(1)

in terms of the eigenvalues and eigenvectors of the coefficient matrix

Here a, b, c, and d are constants. To ensure that X₀ = (0, 0) is the only critical point, we will assume that the determinant Δ = ad – bc ≠ 0. If τ = a + d is the trace* of matrix A, then the characteristic equation det(A – λI) = 0 can be rewritten as

λ² – τλ + Δ = 0.

Therefore the eigenvalues of A are λ = (τ ±)/2, and the usual three cases for these roots occur according to whether τ² – 4Δ is positive, negative, or zero. In the next example we use a numerical solver to discover the nature of the solutions corresponding to these cases.

EXAMPLE 1 Eigenvalues and the Shape of Solutions

Find the eigenvalues of the linear system

x′ = –x + y

y′ = cx – y

in terms of c, and use a numerical solver to discover the shapes of solutions corresponding to the cases c = , 4, 0, and –9.

SOLUTION

The coefficient matrix has trace τ = –2 and determinant Δ = 1 – c, and so the eigenvalues are

The nature of the eigenvalues is therefore determined by the sign of c.

If c = , then the eigenvalues are negative and distinct: λ = – and –. In FIGURE 11.2.2(a) we have used a numerical solver to generate solution curves or trajectories that correspond to various initial conditions. Note that, except for the trajectories drawn in red in the figure, the trajectories all appear to approach 0 from a fixed direction. Recall from Chapter 10 that a collection of trajectories in the xy-plane or phase plane is called a phase portrait of the system.

When c = 4 the eigenvalues have opposite signs, λ = 1 and –3, and an interesting phenomenon occurs. All trajectories move away from the origin in a fixed direction except for solutions that start along the single line drawn in red in Figure 11.2.2(b). We have already seen behavior like this in the phase portrait given in Figure 10.2.2. Experiment with your numerical solver and verify these observations.

The selection c = 0 leads to a single real eigenvalue λ = –1. This case is very similar to the case c = with one notable exception. All solution curves in Figure 11.2.2(c) appear to approach 0 from a fixed direction as t increases.

Finally, when c = –9, λ = –1 ± = –1 ± 3i. Thus the eigenvalues are conjugate complex numbers with negative real part –1. Figure 11.2.2(d) shows that solution curves spiral in toward the origin 0 as t increases.≡

The behaviors of the trajectories observed in the four phase portraits in Figure 11.2.2 in Example 1 can be explained using the eigenvalue–eigenvector solution results from Chapter 10.

Case I: Real Distinct Eigenvalues (τ² – 4Δ > 0)

According to Theorem 10.2.1 in Section 10.2, the general solution of (1) is given by

(2)

where λ₁ and λ₂ are the eigenvalues and K₁ and K₂ are the corresponding eigenvectors. Note that X(t) can also be written as

(3)

(a) Both eigenvalues negative (τ² – 4Δ > 0, τ < 0, and Δ > 0)

Stable Node (λ₂ < λ₁ < 0): Since both eigenvalues are negative, it follows from (2) that lim_t→∞ X(t) = 0. If we assume that λ₂ < λ₁, then λ₂ – λ₁ < 0 and so is an exponential decay function. We may therefore conclude from (3) that X(t) ≈ c₁K₁ for large values of t. When c₁ ≠ 0, X(t) will approach 0 from one of the two directions determined by the eigenvector K₁ corresponding to λ₁. If c₁ = 0, X(t) = c₂K₂ and X(t) approaches 0 along the line determined by the eigenvector K₂. FIGURE 11.2.3 shows a collection of solution curves around the origin. A critical point is called a stable node when both eigenvalues are negative.

(b) Both eigenvalues positive (τ² – 4Δ > 0, τ > 0, and Δ > 0)

Unstable Node (0 < λ₂ < λ₁): The analysis for this case is similar to (a). Again from (2), X(t) becomes unbounded as t increases. Moreover, again assuming λ₂ < λ₁ and using (3), we see that X(t) becomes unbounded in one of the directions determined by the eigenvector K₁ (when c₁ ≠ 0) or along the line determined by the eigenvector K₂ (when c₁ = 0). FIGURE 11.2.4 shows a typical collection of solution curves. This type of critical point, corresponding to the case when both eigenvalues are positive, is called an unstable node.

(c) Eigenvalues have opposite signs (τ² – 4Δ > 0 and Δ < 0)

Saddle Point (λ₂ < 0 < λ₁): The analysis of the solutions is identical to (b) with one exception. When c₁ = 0, X(t) = c₂K₂ and, since λ₂ < 0, X(t) will approach 0 along the line determined by the eigenvector K₂. If X(0) does not lie on the line determined by K₂, the line determined by K₁ serves as an asymptote for X(t). Thus the critical point is unstable even though some solutions approach 0 as t increases. This unstable critical point is called a saddle point. See FIGURE 11.2.5.

EXAMPLE 2 Real Distinct Eigenvalues

Classify the critical point (0, 0) of each of the following linear systems X′ = AX as either a stable node, an unstable node, or a saddle point.

1. 
2. 

In each case discuss the nature of the solutions in a neighborhood of (0, 0).

SOLUTION

(a) Since the trace τ = 3 and the determinant Δ = –4, the eigenvalues are

The eigenvalues have opposite signs, and so (0, 0) is a saddle point. It is not hard to show (see Example 1, Section 10.2) that eigenvectors corresponding to λ₁ = 4 and λ₂ = –1 are

respectively. If X(0) = X₀ lies on the line y = –x, then X(t) approaches 0. For any other initial condition X(t) will become unbounded in the directions determined by K₁. In other words, the line y = x serves as an asymptote for all these solution curves. See FIGURE 11.2.6.

(b) From τ = –29 and Δ = 100 it follows that the eigenvalues of A are λ₁ = –4 and λ₂ = –25. Both eigenvalues are negative, and so (0, 0) is in this case a stable node. Since eigenvectors corresponding to λ₁ = –4 and λ₂ = –25 are

respectively, it follows that all solutions approach 0 from the direction defined by K₁ except those solutions for which X(0) = X₀ lies on the line y = – x determined by K₂. These solutions approach 0 along y = – x. See FIGURE 11.2.7.≡

Case II: A Repeated Real Eigenvalue (τ² – 4Δ = 0)

Degenerate Nodes: Recall from Section 10.2 that the general solution takes on one of two different forms depending on whether one or two linearly independent eigenvectors can be found for the repeated eigenvalue λ₁.

(a) Two linearly independent eigenvectors

If K₁ and K₂ are two linearly independent eigenvectors corresponding to λ₁, then the general solution is given by

If λ₁ < 0, then X(t) approaches 0 along the line determined by the vector c₁K₁ + c₂K₂ and the critical point is called a degenerate stable node. See FIGURE 11.2.8(a). The arrows in Figure 11.2.8(a) are reversed when λ₁ > 0, and we have a degenerate unstable node.

(b) A single linearly independent eigenvector

When only a single linearly independent eigenvector K₁ exists, the general solution is given by

where (A – λ₁I)P = K₁ (see Section 10.2, (12)–(14)), and the solution may be rewritten as

If λ₁ < 0, then lim_t→∞ t= 0 and it follows that X(t) approaches 0 in one of the directions determined by the vector K₁ (see Figure 11.2.8(b)). The critical point is again called a degenerate stable node. When λ₁ > 0, the solutions look like those in Figure 11.2.8(b) with the arrows reversed. The line determined by K₁ is an asymptote for all solutions. The critical point is again called a degenerate unstable node.

Case III: Complex Eigenvalues (τ² – 4Δ < 0)

If λ₁ = α + iβ and ₁ = α – iβ are the complex eigenvalues and K₁ = B₁ + iB₂ is a complex eigenvector corresponding to λ₁, the general solution can be written as X(t) = c₁X₁(t) + c₂X₂(t), where

X₁(t) = (B₁ cos βt – B₂ sin βt)e^αt and X₂(t) = (B₂ cos βt + B₁ sin βt)e^αt.

See equations (23) and (24) in Section 10.2. A solution can therefore be written in the form

x(t) = e^αt(c₁₁ cos βt + c₁₂ sin βt), y(t) = e^αt(c₂₁ cos βt + c₂₂ sin βt),(4)

and when α = 0 we have

x(t) = c₁₁ cos βt + c₁₂ sin βt, y(t) = c₂₁ cos βt + c₂₂ sin βt.(5)

(a) Pure imaginary roots (τ² – 4Δ < 0, τ = 0)

Center: When α = 0, the eigenvalues are pure imaginary and, from (5), all solutions are periodic with period p = 2π/β. Notice that if both c₁₂ and c₂₁ happened to be 0, then (5) would reduce to

x(t) = c₁₁ cos βt,   y(t) = c₂₂ sin βt,

which is a standard parametric representation for the ellipse x²/ + y²/ = 1. By solving the system of equations in (4) for cos βt and sin βt and using the identity sin² βt + cos² βt = 1, it is possible to show that all solutions are ellipses with center at the origin. The critical point (0, 0) is called a center, and FIGURE 11.2.9 shows a typical collection of solution curves. The ellipses are either all traversed in the clockwise direction or all traversed in the counterclockwise direction.

(b) Nonzero real part (τ² – 4Δ < 0, τ ≠ 0)

Spiral Points: When α ≠ 0, the effect of the term e^αt in (4) is similar to the effect of the exponential term in the analysis of damped motion given in Section 3.8. When α < 0, e^αt → 0, and the elliptical-like solution spirals closer and closer to the origin. The critical point is called a stable spiral point. When α > 0, the effect is the opposite. An elliptical-like solution is driven farther and farther from the origin, and the critical point is now called an unstable spiral point. See FIGURE 11.2.10.

EXAMPLE 3 Repeated and Complex Eigenvalues

Classify the critical point (0, 0) of each of the following linear systems X′ = AX.

(a)

(b)

In each case discuss the nature of the solution that satisfies X(0) = (1, 0). Determine parametric equations for each solution.

SOLUTION

(a) Since τ = –6 and Δ = 9, the characteristic polynomial is λ² + 6λ + 9 = (λ + 3)², and so (0, 0) is a degenerate stable node. For the repeated eigenvalue λ = –3 we find a single eigenvector K₁ = , and so the solution X(t) that satisfies X(0) = (1, 0) approaches (0, 0) from the direction specified by the line y = x/3.

(b) Since τ = 0 and Δ = 1, the eigenvalues are λ = ±i, and so (0, 0) is a center. The solution X(t) that satisfies X(0) = (1, 0) is an ellipse that circles the origin every 2π units of time.

From Example 4 of Section 10.2 the general solution of the system in part (a) is

The initial condition gives c₁ = 0 and c₂ = 2, and so x = (6t + 1)e^–3t, y = 2te^–3t are parametric equations for the solution.

The general solution of the system in part (b) is

.

The initial condition gives c₁ = 0 and c₂ = 1, and so x = cos t – sin t, y = –sin t are parametric equations for the ellipse. Note that y < 0 for small positive values of t, and therefore the ellipse is traversed in the clockwise direction.

The solutions of parts (a) and (b) are shown in FIGURES 11.2.11(a) and 11.2.11(b), respectively.≡

Classifying Critical Points

FIGURE 11.2.12 conveniently summarizes the results of this section. The general geometric nature of the solutions can be determined by computing the trace and determinant of A. In practice, graphs of the solutions are most easily obtained not by constructing explicit eigenvalue–eigenvector solutions but rather by generating the solutions numerically using a numerical solver and a method such as the Runge–Kutta method for first-order systems (Section 6.4).

EXAMPLE 4 Classifying Critical Points

We can now give answers to each of the questions posed at the beginning of this section for the linear plane autonomous system

x′ = ax + by

y′ = cx + dy

with ad – bc ≠ 0. The answers are summarized in the theorem below.

11.2 Exercises Answers to selected odd-numbered problems begin on page ANS-30.

In Problems 1–8, the general solution of the linear system X′ = AX is given.

1. In each case discuss the nature of the solution in a neighborhood of (0, 0).
2. With the aid of a graphing utility plot the solution that satisfies X(0) = (1, 1).

In Problems 9–16, classify the critical point (0, 0) of the given linear system by computing the trace τ and determinant Δ and using Figure 11.2.12.

9. x′ = –5x + 3y
   y′ = 2x + 7y
10. x′ = –5x + 3y
    y′ = 2x – 7y
11. x′ = –5x + 3y
    y′ = –2x + 5y
12. x′ = –5x + 3y
    y′ = –7x + 4y
13. x′ = – x + y
    y′ = –x – y
14. x′ = x + y
    y′ = –x + y
15. x′ = 0.02x – 0.11y
    y′ = 0.10x – 0.05y
16. x′ = 0.03x + 0.01y
    y′ = –0.01x + 0.05y
17. Determine conditions on the real constant µ so that (0, 0) is a center for the linear system

    x′ = –µx + y

    y′ = –x + µy.

18. Determine a condition on the real constant µ so that (0, 0) is a stable spiral point of the linear system

    

19. Show that (0, 0) is always an unstable critical point of the linear system

    x′ = µx + y

    y′ = –x + y,

    where µ is a real constant and µ ≠ –1. When is (0, 0) an unstable saddle point? When is (0, 0) an unstable spiral point?

20. Let X = X(t) be the response of the linear dynamical system

    x′ = αx – βy

    y′ = βx + αy

    that satisfies the initial condition X(0) = X₀. Determine conditions on the real constants α and β that will ensure lim_t→∞ X(t) = (0, 0). Can (0, 0) be a node or saddle point?

21. Show that the nonhomogeneous linear system X′ = AX + F has a unique critical point X₁ when Δ = det A ≠ 0. Conclude that if X = X(t) is a solution to the nonhomogeneous system, τ < 0 and Δ > 0, then lim_t→∞ X(t) = X₁.
    [Hint: X(t) = X_c(t) + X₁.]
22. In Example 4(b) show that (0, 0) is a stable node when bc < 1.

In Problems 23–26, a nonhomogeneous linear system X′ = AX + F is given.

1. In each case determine the unique critical point X₁.
2. Use a numerical solver to determine the nature of the critical point in part (a).
3. Investigate the relationship between X₁ and the critical point (0, 0) of the homogeneous linear system X′ = AX.

23. x′ = 2x + 3y – 6
    y′ = –x – 2y + 5
24. x′ = –5x + 9y + 13
    y′ = –x – 11y – 23
25. x′ = 0.1x – 0.2y + 0.35
    y′ = 0.1x + 0.1y – 0.25
26. x′ = 3x – 2y – 1
    y′ = 5x – 3y – 2

*In general, if A is an n × n matrix, the trace of A is the sum of the main diagonal entries.