INTRODUCTION

In this section we will investigate the existence of periodic solutions of nonlinear plane autonomous systems and introduce special periodic solutions called limit cycles.

We saw in Sections 11.3 and 11.4 that an analysis of critical points using linearization can provide valuable information on the behavior of solutions near critical points and insight into a variety of biological and physical phenomena. There are, however, some inherent limitations to this approach. When the eigenvalues of the Jacobian matrix are pure imaginary, we cannot conclude that there are periodic solutions near the critical point. In some cases we were able to solve dy/dx = Q(x, y)/P(x, y), obtain an implicit representation f(x, y) = c of the solution curves, and investigate whether any of these solutions formed closed curves. More often than not, this differential equation will not possess closed-form solutions. For example, the Lotka–Volterra competition model cannot be handled by this procedure. The first goal of this section is to determine conditions under which we can either exclude the possibility of periodic solutions or assert their existence.

We encountered an additional problem in studying the models in Section 11.3. FIGURE 11.5.1 illustrates the common situation in which a region R contains a single asymptotically stable critical point X₁. We can assert that lim_t→∞ X(t) = X₁ when the initial position X(0) = X₀ is “near” X₁, but under what conditions is lim_t→∞ X(t) = X₁ for all initial positions in R? Such a critical point is called globally stable in R. A second goal is to determine conditions under which an asymptotically stable critical point is globally stable.

In motivating and discussing the methods in this section, we will use the fact that the vector field V(x, y) = (P(x, y), Q(x, y)) can be interpreted as defining a fluid flow in a region of the plane, and a solution to the autonomous system may be interpreted as the resulting path of a particle as it moves through the region.

Negative Criteria

A number of results can sometimes be used to establish that there are no periodic solutions in a given region R of the plane. We will assume that P(x, y) and Q(x, y) have continuous first partial derivatives in R and that R is simply connected. Recall that in a simply connected region, any simple closed curve C in R encloses only points in R. Therefore, if there is a periodic solution X = X(t) in R, then R will contain all points in the interior of the resulting curve.

EXAMPLE 1 No Periodic Solutions

EXAMPLE 2 Lotka–Volterra Competition Model

Another sometimes useful result can be formulated in terms of the divergence of the vector field V(x, y) = (P(x, y), Q(x, y)):

PROOF:

Suppose, to the contrary, that there is a periodic solution X = X(t) lying in R, and let C be the resulting simple closed curve and R₁ the region bounded by C. Green’s theorem states that

whenever M(x, y) and N(x, y) have continuous first partials in R. If we let N = P and M = –Q, we obtain

Since X = X(t) is a solution with period p, we have x′(t) = P(x(t), y(t)) and y′(t) = Q(x(t), y(t)), and so

Since div V = ∂P/∂x + ∂Q/∂y is continuous and does not change sign in R, it follows that either div V ≥ 0 in R or div V ≤ 0 in R, and so

This contradiction establishes that there are no periodic solutions in R.≡

EXAMPLE 3 Bendixson Negative Criterion

EXAMPLE 4 Sliding Bead and Periodic Solutions

The sliding bead discussed in Section 11.4 satisfies the differential equation

Show that there are no periodic solutions.

SOLUTION

The corresponding plane autonomous system is

and so div V = By Theorem 11.5.2 we conclude the system has no periodic solutions.≡

The following theorem is a generalization of the Bendixson negative criterion, which leaves it to the reader to construct an appropriate function δ(x, y).

There are no general techniques for constructing an appropriate function δ(x, y). Instead, we experiment with simple functions of the form ax² + by², e^ax+by, x^ay^b, and so on, and try to determine constants for which ∂(δP)/∂x + (δQ)/∂y is nonzero in a given region.

EXAMPLE 5 Dulac Negative Criterion

EXAMPLE 6 Dulac Negative Criterion

Use δ(x, y) = 1/(xy) to show that the Lotka–Volterra competition equations

have no periodic solutions in the first quadrant.

SOLUTION

If δ(x, y) = 1/(xy), then

and so

For (x, y) in the first quadrant, the latter expression is always negative. Therefore, there are no periodic solutions.≡

Positive Criteria: Poincaré–Bendixson Theory

The Poincaré–Bendixson theorem is an advanced result that describes the long-range behavior of a bounded solution to a plane autonomous system. Rather than present the result in its full generality, we will concentrate on a number of special cases that occur frequently in applications. One of these cases will lead to a new type of periodic solution called a limit cycle.

FIGURE 11.5.2 shows two standard types of invariant regions. A Type I invariant region R is bounded by a simple closed curve C, and the flow at the boundary defined by the vector field V(x, y) = (P(x, y), Q(x, y)) is always directed into the region. This prevents a particle from crossing the boundary. A Type II invariant region is an annular region bounded by simple closed curves C₁ and C₂, and the flow at the boundary is again directed toward the interior of R. The following theorem provides a method for verifying that a given region is invariant.

PROOF:

If θ is the angle between V(x, y) and n(x, y), then from V ⋅ n = ∥V∥ ∥n∥ cos θ, we may conclude that cos θ ≥ 0 and so θ is between 0° and 90°. The flow is therefore directed into the region (or at worst along the boundary) for any boundary point (x, y). This prevents a solution that starts in R from leaving R. Therefore, R is an invariant region for the plane autonomous system.≡

The problem of finding an invariant region for a given nonlinear system is an extremely difficult one. An excellent first step is to use software that plots the vector field V(x, y) = (P(x, y), Q(x, y)) together with the curves P(x, y) = 0 (along which the vectors are vertical) and Q(x, y) = 0 (along which the vectors are horizontal). This can lead to choices for R. In the following examples, we will construct invariant regions bounded by lines and circles. In more complicated cases, we will be content to offer empirical evidence that an invariant region exists.

EXAMPLE 7 Circular Invariant Region

EXAMPLE 8 Annular Invariant Region

EXAMPLE 9 The Van der Pol Equation

The Van der Pol equation is a nonlinear second-order differential equation that arises in electronics, and as a plane autonomous system it takes the form

FIGURE 11.5.3 shows the corresponding vector field for µ = 1, together with the curves y = 0 and (x² – 1)y = –x along which the vectors are vertical and horizontal, respectively. (For convenience we have sketched the normalized vector field V/∥V∥.) It is not possible to find a simple invariant region whose boundary consists of lines or circles. The figure does offer empirical evidence that an invariant region R, with (0, 0) in its interior, does exist. Advanced methods are required to demonstrate this mathematically.*≡

We next present two important special cases of the Poincaré–Bendixson theorem that guarantee the existence of periodic solutions.

The flow interpretation portrayed in Figure 11.5.2 can be used to make the result plausible. If a particle is released at a point X₀ in a Type II invariant region R, then, with no escape from the region and with no resting points, the particle will begin to rotate around the boundary C₂ and settle into a periodic orbit. It is not possible for the particle to return to an earlier position unless the solution is itself periodic.

EXAMPLE 10 Existence of a Periodic Solution

EXAMPLE 11 Limit Cycle in the Van der Pol Equation

Show that the Van der Pol differential equation

x″ + µ(x² – 1)x′ + x = 0

has a periodic solution when µ > 0.

SOLUTION

We will assume that there is a Type I invariant region R for the corresponding plane autonomous system and that this region contains (0, 0) in its interior (see Example 9 and Figure 11.5.3). The only critical point is (0, 0), and the Jacobian matrix is given by

Therefore, τ = µ, Δ = 1, and τ² – 4Δ = µ² – 4. Since µ > 0, the critical point is either an unstable spiral point or an unstable node. By part (a) of Theorem 11.5.5, the system has at least one periodic solution in R. FIGURE 11.5.4 shows solutions corresponding to X(0) = (0.5, 0.5) and X(0) = (3, 3) for µ = 1. Each of these solutions spirals around the origin and approaches a limit cycle. It can be shown that the Van der Pol differential equation has a unique limit cycle for all values of the parameter µ.≡

Global Stability

Another version of the Poincaré–Bendixson theorem can be used to show that a locally stable critical point is globally stable:

In Theorem 11.5.6 the particle cannot escape from R, cannot return to any of its prior positions, and therefore, in the absence of cycles, must be attracted to some stable critical point X₁.

EXAMPLE 12 A Globally Stable Critical Point

Investigate global stability for the system from Example 7:

x′ = –y – x³

y′ = x – y³.

SOLUTION

In Example 7 we showed that the circular region defined by x² + y² ≤ r² serves as an invariant region for the system for any r > 0. Since ∂P/∂x + ∂Q/∂y = –3x² – 3y² does not change sign, there are no periodic solutions by the Bendixson negative criterion. It is not hard to show that (0, 0) is the only critical point and that the Jacobian matrix is

Since τ = 0 and Δ = 1, (0, 0) may be either a stable or an unstable spiral (it cannot be a center). Theorem 11.5.6, however, guarantees that lim_t→∞ X(t) = X₁ for some critical point X₁. Since (0, 0) is the only critical point, we must have lim_t→∞ X(t) = (0, 0) for any initial position X₀ in the plane. The critical point is therefore a globally stable spiral point. FIGURE 11.5.5 shows two views of the solution satisfying X(0) = (4, 4). Figure 11.5.5(b) is an enlarged view of the curve around (0, 0). Notice how slowly the solution spirals toward (0, 0).

11.5 Exercises Answers to selected odd-numbered problems begin on page ANS-31.

In Problems 1–8, show that the given plane autonomous system (or second-order differential equation) has no periodic solutions.

1. x′ = 2 + xy
   y′ = x – y
2. x′ = 2x – xy
   y′ = –1 – x² + 2x – y²
3. x′ = –x + y²
   y′ = x – y
4. x′ = xy² – x²y
   y′ = x²y – 1
5. x′ = –µx – y
   y′ = x + y³
   for µ < 0
6. x′ = 2x + y²
   y′ = xy – y
7. x″ – 2x + (x′)⁴ = 0
8. x″ + x = [ + 3(x′)²]x′ – x²

In Problems 9 and 10, use the Dulac negative criterion to show that the given plane autonomous system has no periodic solutions. Experiment with simple functions of the form δ(x, y) = ax² + by², e^ax+by, or x^ay^b.

9. x′ = –2x + xy
   y′ = 2y – x²
10. x′ = –x³ + 4xy
    y′ = –5x² – y²
11. Show that the plane autonomous system

    x′ = x(1 – x² – 3y²)

    y′ = y(3 – x² – 3y²)

    has no periodic solutions in an elliptical region about the origin.

12. If ∂g/∂x′ ≠ 0 in a region R, prove that x″ = g(x, x′) has no periodic solutions in R.
13. Show that the predator–prey model

    

    in Problem 20 of Exercises 11.4 has no periodic solutions in the first quadrant.

In Problems 14 and 15, find a circular invariant region for the given plane autonomous system.

14. x′ = –y – xe^x+y
    y′ = x – ye^x+y
15. x′ = –x + y + xy
    y′ = x – y – x² – y³
16. Verify that the region bounded by the closed curve x⁶ + 3y² = 1 is an invariant region for the second-order nonlinear differential equation x″ + x′ = –(x′)³ – x⁵. See FIGURE 11.5.6.

    

17. The plane autonomous system in Example 8 has only one critical point. Can we conclude that this system has at least one periodic solution?
18. Use the Poincaré–Bendixson theorem to show that the second-order nonlinear differential equation

    x″ = x′[1 – 3x² – 2(x′)²] – x

    has at least one periodic solution. [Hint: Find an invariant annular region for the corresponding plane autonomous system.]

19. Let X = X(t) be the solution of the plane autonomous system

    

    that satisfies X(0) = (x₀, y₀). Show that if + < 1, then lim_t→∞ X(t) = (0, 0). [Hint: Select r < 1 with + < r² and first show that the circular region R defined by x² + y² ≤ r² is an invariant region.]

20. Investigate global stability for the system

    x′ = y – x

    y′ = –x – y³.

21. Empirical evidence suggests that the plane autonomous system

    

    has a Type I invariant region R that lies inside the rectangle defined by 0 ≤ x ≤ 2, 0 ≤ y ≤ 1.

    1. Use the Bendixson negative criterion to show that there are no periodic solutions in R.
    2. If X₀ is in R and X = X(t) is the solution satisfying X(t) = X₀, use Theorem 11.5.6 to find lim_t→∞ X(t).
22. 1. Find and classify all critical points of the plane autonomous system

       

    2. FIGURE 11.5.7 shows the vector field V/∥V∥ and offers empirical evidence that there is an invariant region R in the first quadrant with a critical point in its interior. Assuming that such a region exists, prove that there is at least one periodic solution.

       

*See M. Hirsch and S. Smale, Differential Equations, Dynamical Systems, and Linear Algebra (New York: Academic Press, 1974).