11.1 Autonomous Systems

INTRODUCTION

We introduced the notions of autonomous first-order DEs, critical points of an autonomous DE, and the stability of a critical point in Section 2.1. This earlier consideration of stability was purposely kept at a fairly intuitive level; it is now time to give the precise definition of this concept. To do this we need to examine autonomous systems of first-order DEs. In this section we define critical points of autonomous systems of two first-order DEs; the autonomous systems can be linear or nonlinear.

Autonomous Systems

A system of first-order differential equations is called autonomous when the system can be written in the form

(1)

Notice that the independent variable t does not appear explicitly on the right-hand side of each differential equation. Compare (1) with the general system given in (2) of Section 10.1.

EXAMPLE 1 A Nonautonomous System

The system of nonlinear first-order differential equations

is not autonomous because of the presence of t on the right-hand side of both DEs. ≡

When n = 1 in (1) a single first-order differential equation takes on the form dx/dt = g(x). This last equation is equivalent to (1) of Section 2.1 with the symbols x and t playing the parts of y and x, respectively. Explicit solutions can be constructed since the differential equation dx/dt = g(x) is separable, and we will make use of this fact to give illustrations of the concepts in this chapter.

See pages 154 and 155.

Second-Order DE as a System

Any autonomous second-order differential equation x″ = g(x, x′) can be written as an autonomous system. As we did in Section 3.7, if we let y = x′, then x″ = g(x, x′) becomes y′ = g(x, y). Thus the second-order differential equation becomes the system of two first-order equations

x′ = y

y′ = g(x, y).

EXAMPLE 2 The Pendulum DE as an Autonomous System

In (6) of Section 3.11 we showed that the displacement angle θ for a pendulum satisfies the nonlinear second-order differential equation

If we let x = θ and y = θ′, this second-order differential equation can be rewritten as the autonomous system

≡

Matrix Forms of (1)

If X(t) and g(X) denote the respective column vectors

then the autonomous system (1) may be written in the compact column vector form X′ = g(X). The homogeneous linear system X′ = AX studied in Section 10.2 is an important special case.

In this chapter it is also convenient to write (1) using row vectors. If we let

X(t) = (x1(t), x2(t), … , xn(t))

and g(X) = (g1(x1, x2, … , xn), g2(x1, x2, … , xn), … , gn(x1, x2, … , xn)),

then the autonomous system (1) can also be written in the compact row vector form X′ = g(X). It should be clear from the context whether we are using column or row vector form, and therefore we will not distinguish between X and XT, the transpose of X. In particular, when n = 2 it is convenient to use row vector form and write an initial condition as X(0) = (x0, y0).

When the variable t is interpreted as time, we can refer to the system of differential equations in (1) as a dynamical system and a solution X(t) as the state of the system or the response of the system at time t. With this terminology, a dynamical system is autonomous when the rate X′(t) at which the system changes depends only on the system’s present state X(t). The linear system X′ = AX + F(t) studied in Chapter 10 is then autonomous when F(t) is constant. In the case n = 2 or 3 we call a solution a path or trajectory since we may think of x = x1(t), y = x2(t), z = x3(t) as parametric equations of a curve.

Vector Field Interpretation

When n = 2 the system in (1) is called a plane autonomous system, and we write the system as

The vector V(x, y) = (P(x, y), Q(x, y)) defines a vector field in a region of the plane, and a solution to the system may be interpreted as the resulting path of a particle as it moves through the region. To be more specific, let V(x, y) = (P(x, y), Q(x, y)) denote the velocity of a stream at position (x, y), and suppose that a small particle (such as a cork) is released at a position (x0, y0) in the stream. If X(t) = (x(t), y(t)) denotes the position of the particle at time t, then X′(t) = (x′(t), y′(t)) is the velocity vector v. When external forces are not present and frictional forces are neglected, the velocity of the particle at time t is the velocity of the stream at position X(t); that is, X′(t) = V(x(t), y(t)), or

Thus the path of the particle is a solution to the system that satisfies the initial condition X(0) = (x0, y0). We will frequently call on this simple interpretation of a plane autonomous system to illustrate new concepts.

EXAMPLE 3 Plane Autonomous System of a Vector Field

A vector field for the steady-state flow of a fluid around a cylinder of radius 1 is given by

where V0 is the speed of the fluid far from the cylinder. If a small cork is released at (–3, 1), the path X(t) = (x(t), y(t)) of the cork satisfies the plane autonomous system

subject to the initial condition X(0) = (–3, 1). See FIGURE 11.1.1.≡

A graph. A circle representing a cylinder and a curve representing a fluid flow around the cylinder are graphed on an x y plane. The circle is centered at the origin and shaded. The curve with two right arrows on it is placed above the circle. It starts at a point (negative 3, 1) marked with a dot, goes slightly up and to the right through the positive y axis, then goes slightly down and to the right, and exits the right of the first quadrant. Lines of right arrows placed below and above the circle from left to right represent the vector field.

FIGURE 11.1.1 Vector field of a fluid flow around a cylinder in Example 3

Types of Solutions

If P(x, y), Q(x, y), and the first-order partial derivatives ∂P/∂x, ∂P/∂y, ∂Q/∂x, and ∂Q/∂y are continuous in a region R of the plane, then a solution to the plane autonomous system

that satisfies X(0) = X0 is unique and one of three basic types:

  1. A constant solution x(t) = x0, y(t) = y0 (or X(t) = X0 for all t). A constant solution is called a critical or stationary point. When the particle is placed at a critical point X0 (that is, X(0) = X0), it remains there indefinitely. For this reason a constant solution is also called an equilibrium solution. Note that since X′(t) = 0, a critical point is a solution of the system of algebraic equations

P(x, y) = 0

Q(x, y) = 0.

  1. A solution x = x(t), y = y(t) defines an arc—a plane curve that does not cross itself. Thus the curve in FIGURE 11.1.2(a) can be a solution to a plane autonomous system, whereas the curve in Figure 11.1.2(b) cannot be a solution. There would be two solutions that start from the point P of intersection.
  2. A periodic solution x = x(t), y = y(t). A periodic solution is called a cycle. If p is the period of the solution, then X(t + p) = X(t) and a particle placed on the curve at X0 will cycle around the curve and return to X0 in p units of time. See FIGURE 11.1.3.
The first curve shown in (a) starts at a dot labeled X(0), follows a spiral pattern, spirals in and ends at the center. Two arrows on the curve represent the direction of the spiral in counter clockwise direction. The second curve shown in (b) starts at a dot labeled X(0), goes up and to the right through a dot labeled P, then curves back forming a closed curve, and reaches P. From P, it then goes down to the right, and exits the bottom. Four arrows on the curve represent the counter clockwise direction. The part of the curve that starts from point P and goes up is labeled 1. The part of the curve that starts from point P and goes down is labeled 2.

FIGURE 11.1.2 Curve in (a) is called an arc

A closed curve passes through a dot labeled X(0). Two arrows on the curve represent the cycle in counter-clockwise direction.

FIGURE 11.1.3 Periodic solution or a cycle

EXAMPLE 4 Finding Critical Points

Find all critical points of each of the following plane autonomous systems.

  1. x′ = –x + y
      y′ = x – y
  2. x′ = x2 + y2 – 6
      y′ = x2 – y
  3. x′ = 0.01x(100 – x – y)
      y′ = 0.05y(60 – y – 0.2x)
SOLUTION

We find the critical points by setting the right-hand side of the differential equations equal to zero.

  1. The solution to the system

–x + y = 0

  x – y = 0

consists of all points on the line y = x. Thus there are infinitely many critical points.

  1. To solve the system

x2 + y2 – 6 = 0

       x2 – y = 0

we substitute the second equation, x2 = y, into the first equation to obtain y2 + y – 6 = (y + 3)(y – 2) = 0. If y = –3, then x2 = –3, and so there are no real solutions. If y = 2, then x = ±, and so the critical points are (, 2) and (–, 2).

  1. Finding the critical points requires a careful consideration of cases. The equation 0.01x(100 – x – y) = 0 implies x = 0 or x + y = 100.

If x = 0, then substituting in 0.05y(60 – y – 0.2x) = 0 we have y(60 – y) = 0. Thus y = 0 or 60, and so (0, 0) and (0, 60) are critical points.

If x + y = 100, then 0 = y(60 – y – 0.2(100 – y)) = y(40 – 0.8y). It follows that y = 0 or 50, and so (100, 0) and (50, 50) are critical points.≡

When the plane autonomous system is linear, we can use the methods in Chapter 10 to investigate solutions.

EXAMPLE 5 Discovering Periodic Solutions

Determine whether the given linear dynamical system possesses a periodic solution.

  1. x′ = 2x + 8y
      y′ = –x – 2y
  2. x′ = x + 2y
      y′ = – x + y

In each case sketch the graph of the solution that satisfies X(0) = (2, 0).

SOLUTION

(a) In Example 6 of Section 10.2 we used the eigenvalue–eigenvector method to show that

Thus every solution is periodic with period p = π. The solution satisfying X(0) = (2, 0) is x = 2 cos 2t + 2 sin 2t, y = –sin 2t. This solution generates the ellipse shown in FIGURE 11.1.4(a).

Two graphs. The graph (a) has a solution curve graphed on an x y plane. The curve is an ellipse that starts at the point (negative 2, 0), goes up and to the left, then goes down and to the right through the positive y axis, and reaches the point (2, 0) marked with a dot. Then, it goes down and to the right, goes up and to the left through the negative y axis, and ends at the starting point (negative 2, 0). Two arrows on the curve represent the clockwise direction. The graph (b) has a solution curve graphed on an x y plane. The curve starts at the point (2, 0) marked with a dot, goes down and to the left through the negative y axis, and exits the bottom left of the third quadrant. Two arrows on the curve represent the clockwise direction.

FIGURE 11.1.4 Solution curves in Example 5

  1. Using the eigenvalue–eigenvector method, we can show that

x = c1(2et cos t) + c2(2et sin t), y = c1(–et sin t) + c2(et cos t).

Because of the presence of et in the general solution, there are no periodic solutions (that is, cycles). The solution satisfying X(0) = (2, 0) is x = 2et cos t, y = –et sin t, and the resulting curve is shown in Figure 11.1.4(b).≡

Changing to Polar Coordinates

Except for the case of constant solutions, it is usually not possible to find explicit expressions for the solutions of a nonlinear autonomous system. We can solve some nonlinear systems, however, by changing to polar coordinates. From the formulas r2 = x2 + y2 and θ = tan–1(y/x) we obtain

(2)

We can sometimes use (2) to convert a plane autonomous system in rectangular coordinates to a simpler system in polar coordinates.

EXAMPLE 6 Changing to Polar Coordinates

Find the solution of the nonlinear plane autonomous system

satisfying the initial condition X(0) = (3, 3).

SOLUTION

Substituting for dx/dt and dy/dt in the expressions for dr/dt and dθ/dt in (2), we obtain

Since (3, 3) is (3, π/4) in polar coordinates, the initial condition X(0) = (3, 3) becomes r(0) = and θ(0) = π/4. Using separation of variables, we see that the solution of the system is

for r ≠ 0. (Check this!) Applying the initial conditions then gives

The spiral r = is sketched in FIGURE 11.1.5.≡

A graph of a solution curve graphed on an x y plane. The curve is a spiral that starts at the point (3, 3) marked with a dot, goes down and to the left through (0, 1), then goes down and to the right, spirals in, and ends at the origin (0, 0). Two arrows on the curve represent the counter clockwise direction.

FIGURE 11.1.5 Solution curve in Example 6

EXAMPLE 7 Solutions in Polar Coordinates

When expressed in polar coordinates, a plane autonomous system takes the form

Find and sketch the solutions satisfying X(0) = (0, 1) and X(0) = (3, 0) in rectangular coordinates.

SOLUTION

Applying separation of variables to dr/dt = 0.5(3 – r) and integrating dθ/dt leads to the solution

r = 3 + c1e–0.5t,   θ = t + c2.

If X(0) = (0, 1), then r(0) = 1 and θ(0) = π/2, and so c1 = –2 and c2 = π/2. The solution curve is the spiral r = 3 – 2e–0.5(θ–π/2). Note that as t → ∞, θ increases without bound and r approaches 3.

If X(0) = (3, 0), then r(0) = 3 and θ(0) = 0. It follows that c1 = c2 = 0, and so r = 3 and θ = t. Hence x = r cos θ = 3 cos t and y = r sin θ = 3 sin t, and so the solution is periodic. The solution generates a circle of radius 3 about (0, 0). Both solutions are shown in FIGURE11.1.6.≡

Two solution curves are graphed on an x y plane. The first curve is a spiral that starts at the point (0, 1) marked with a dot, goes down and to the left through (negative 2, 0), then goes down and to the right, spirals out, and ends at a point in the third quadrant. Four arrows on the curve represent the counter clockwise direction. The second curve is a circle that is centered at the origin (0, 0) and its radius is 3. It starts at the point (3, 0) marked with a dot, and ends at the same point. Four arrows on the circle represent the counter clockwise direction. The first curve, spiral is placed inside the circle. The end point of the spiral is on the circumference of the circle.

FIGURE 11.1.6 Solution curves in Example 7

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

In Problems 1–6, write the given nonlinear second-order differential equation as a plane autonomous system. Find all critical points of the resulting system.

  1. x″ + 9 sin x = 0
  2. x″ + (x′)2 + 2x = 0
  3. x″ + x′(1 – x3) – x2 = 0
  4. x″ + 4 + 2x′ = 0
  5. x″ + x = ϵx3 for ϵ > 0
  6. x″ + x – ϵx|x| = 0 for ϵ > 0

In Problems 7–16, find all critical points of the given plane autonomous system.

  1. x′ = x + xy
    y′ = –y – xy
  2. x′ = y2 – x
    y′ = x2 – y
  3. x′ = 3x2 – 4y
    y′ = x – y
  4. x′ = x3 – y
    y′ = x – y3
  5. x′ = x(10 – x – y)
    y′ = y(16 – y – x)
  6. x′ = –2x + y + 10
    y′ = 2x – y – 15
  7. x′ = x2ey
    y′ = y(ex – 1)
  8. x′ = sin y
    y′ = ex–y – 1
  9. x′ = x(1 – x2 – 3y2)
    y′ = y(3 – x2 – 3y2)
  10. x′ = –x(4 – y2)
    y′ = 4y(1 – x2)

In Problems 17–22, for the given linear dynamical system (taken from Exercises 10.2)

  1. find the general solution and determine whether there are periodic solutions,
  2. find the solution satisfying the given initial condition, and,
  3. with the aid of a graphing utility, plot the solution in part (b) and indicate the direction in which the curve is traversed.
  1. x′ = x + 2y
    y′ = 4x + 3y, X(0) = (2, –2) (Problem 1, Exercises 10.2)
  2. x′ = –6x + 2y
    y′ = –3x + y, X(0) = (3, 4) (Problem 6, Exercises 10.2)
  3. x′ = 4x – 5y
    y′ = 5x – 4y, X(0) = (4, 5) (Problem 39, Exercises 10.2)
  4. x′ = x + y
    y′ = –2x – y, X(0) = (–2, 2) (Problem 36, Exercises 10.2)
  5. x′ = 5x + y
    y′ = –2x + 3y, X(0) = (–1, 2) (Problem 37, Exercises 10.2)
  6. x′ = x – 8y
    y′ = x – 3y, X(0) = (2, 1) (Problem 40, Exercises 10.2)

In Problems 23–26, solve the given nonlinear plane autonomous system by changing to polar coordinates. Describe the geometric behavior of the solution that satisfies the given initial condition(s).

  1. x′ = –y – x(x2 + y2)2
    y′ = x – y(x2 + y2)2, X(0) = (4, 0)
  2. x′ = y + x(x2 + y2)
    y′ = –x + y(x2 + y2), X(0) = (4, 0)
  3. x′ = –y + x(1 – x2 – y2)
    y′ = x + y(1 – x2 – y2), X(0) = (1, 0); X(0) = (2, 0)
    [Hint: The resulting differential equation for r is a Bernoulli differential equation. See Section 2.5.]
  4. x′ = y – (4 – x2 – y2)
    y′ = –x – (4 – x2 – y2),
    X(0) = (1, 0); X(0) = (2, 0)
    [Hint: See Example 4 in Section 2.2.]

If a plane autonomous system has a periodic solution, then there must be at least one critical point inside the curve generated by the solution. In Problems 27–30, use this fact together with a numerical solver to investigate the possibility of periodic solutions.

  1. x′ = –x + 6y
    y′ = xy + 12
  2. x′ = –x + 6xy
    y′ = –8xy + 2y
  3. x′ = y
    y′ = y(1 – 3x2 – 2y2) – x
  4. x′ = xy
    y′ = –1 – x2 – y2
  5. If z = f(x, y) is a function with continuous first partial derivatives in a region R, then a flow V(x, y) = (P(x, y), Q(x, y)) in R may be defined by letting P(x, y) = (x, y) and Q(x, y) = (x, y). Show that if X(t) = (x(t), y(t)) is a solution of the plane autonomous system

    x′ = P(x, y)

     y′ = Q(x, y),

    then f(x(t), y(t)) = c for some constant c. Thus a solution curve lies on the level curves of f. [Hint: Use the Chain Rule to compute f(x(t), y(t)).]