11.3 Linearization and Local Stability

INTRODUCTION

The key idea in this section is that of linearization. Recall from calculus and Section 2.6, a linearization of a differentiable function f(x) at a number x1 is the equation of the tangent line to the graph of f at the point:

For x close to x1 the points on the graph of f are close to the points on the tangent line so values y(x) obtained from the equation of the tangent line are said to be local linear approximations to the corresponding function values f(x). Similarly, a linearization of a function of two variables f(x, y) that is differentiable at a point (x1, y1) is the equation of the tangent plane to the graph of f at the point:

where fx and fy are partial derivatives with respect to x and y, respectively. In this section we will use linearization as a means of analyzing nonlinear DEs and nonlinear systems; the idea is to replace them by linear DEs and linear systems.

Sliding Bead

We start this section by refining the stability concepts introduced in Section 11.2 in such a way that they will apply to nonlinear autonomous systems as well. Although the linear system X′ = AX had only one critical point when det A ≠ 0, we saw in Section 11.1 that a nonlinear system may have many critical points. We therefore cannot expect that a particle placed initially at X0 will remain near a given critical point X1 unless X0 has been placed sufficiently close to X1 to begin with. The particle might well be driven to a second critical point X2. To emphasize this idea we will consider the physical system shown in FIGURE 11.3.1 in which a bead slides along the curve z = f(x) under the influence of gravity alone. We will show in Section 11.4 that the x-coordinate of the bead satisfies a nonlinear second-order differential equation x″ = g(x, x′), and therefore letting y = x′ satisfies the nonlinear autonomous system

A bead slides on a curve graphed on an x z plane. The curve labeled z equals f(x) enters the top left of the second quadrant, goes down and to the right through a black dot, and reaches a point in the second quadrant that corresponds to the x value of x subscript 1. Then, it goes up and to the right through a black dot, and reaches a point in the second quadrant that corresponds to the x value of x subscript 2. Again, it goes down and to the right through the positive z axis and a black dot, reaches a low point in the first quadrant that corresponds to the x value of x subscript 3, then goes up and to the right, and exits the top right of the first quadrant. An arrowhead on the curve between the first black dot and the point that corresponds to the x value of x subscript 1 points toward the point. An arrowhead on the curve between the second black dot and the point that corresponds to the x value of x subscript 1 points toward the point. Another arrowhead on the curve between the third black dot and the point that corresponds to the x value of x subscript 3 points toward the point.

FIGURE 11.3.1 Bead sliding on graph of z = f(x)

If the bead is positioned at P = (x, f(x)) and given zero initial velocity, the bead will remain at P provided f′(x) = 0. If the bead is placed near the critical point located at x = x1, it will remain near x = x1 only if its initial velocity does not drive it over the “hump” at x = x2 toward the critical point located at x = x3. Therefore X(0) = (x(0), x′(0)) must be near (x1, 0).

In the next definition we will denote the distance between two points X and Y by |X – Y|. Recall that if X = (x1, x2, … , xn) and Y = (y1, y2 … , yn), then

|X – Y| =

DEFINITION 11.3.1 Stable Critical Points

Let X1 be a critical point of an autonomous system, and let X = X(t) denote the solution that satisfies the initial condition X(0) = X0, where X0 ≠ X1. We say that X1 is a stable critical point when, given any radius ρ > 0, there is a corresponding radius r > 0 such that if the initial position X0 satisfies |X0 – X1| < r, then the corresponding solution X(t) satisfies |X(t) – X1| < ρ for all t > 0. If, in addition, limt→∞ X(t) = X1 whenever |X0 – X1| < r, we call X1 an asymptotically stable critical point.

This definition is illustrated in FIGURE 11.3.2(a). Given any disk of radius ρ about the critical point X1, a solution will remain inside this disk provided X(0) = X0 is selected sufficiently close to X1. It is not necessary that a solution approach the critical point in order for X1 to be stable. Stable nodes, stable spiral points, and centers are all examples of stable critical points for linear systems. To emphasize that X0 must be selected close to X1 the terminology locally stable critical point is also used.

Two images. A small, shaded circle with radius r. Another big circle with radius rho. A point X subscript 0 on the small circle goes in a circular and anticlockwise direction. Caption. Stable. A small, shaded circle. Another big circle with radius rho. A point X subscript 0 on the small circle goes in an anticlockwise direction but not in a regular circular motion. Caption. Unstable.

FIGURE 11.3.2 Critical points

By negating Definition 11.3.1 we obtain the definition of an unstable critical point.

DEFINITION 11.3.2 Unstable Critical Point

Let X1 be a critical point of an autonomous system, and let X = X(t) denote the solution that satisfies the initial condition X(0) = X0, where X0 ≠ X1. We say that X1 is an unstable critical point if there is a disk of radius ρ > 0 with the property that, for any r > 0, there is at least one initial position X0 that satisfies |X0 – X1| < r, yet the corresponding solution X(t) satisfies |X(t) – X1| ≥ ρ for at least one t > 0.

If a critical point X1 is unstable, no matter how small the neighborhood about X1, an initial position X0 can always be found that results in the solution leaving some disk of radius ρ at some future time t. See Figure 11.3.2(b). Therefore unstable nodes, unstable spiral points, and saddle points are all examples of unstable critical points for linear systems. In Figure 11.3.1 the critical point (x2, 0) is unstable. The slightest displacement or initial velocity results in the bead sliding away from the point (x2, f(x2)).

EXAMPLE 1 A Stable Critical Point

Show that (0, 0) is a stable critical point of the nonlinear plane autonomous system

considered in Example 6 of Section 11.1.

SOLUTION

In Example 6 of Section 11.1 we showed that in polar coordinates

is the solution of the system. If X(0) = (r0, θ0) is the initial condition in polar coordinates, then

Note that r ≤ r0 for t ≥ 0, and r approaches (0, 0) as t increases. Therefore, given ρ > 0, a solution that starts less than ρ units from (0, 0) remains within ρ units of the origin for all t ≥ 0. Hence the critical point (0, 0) is stable and is in fact asymptotically stable. A typical solution is shown in FIGURE 11.3.3. ≡

A curve is graphed on an x y plane. It lies inside a shaded circle whose center is at the origin. The curve starts at a black dot on the circumference of the circle in the first quadrant, spirals in through the positive y axis, negative x axis, and negative y axis, and ends at a black dot at the origin (0, 0).

FIGURE 11.3.3 Asymptotically stable critical point in Example 1

EXAMPLE 2 An Unstable Critical Point

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

Show that (x, y) = (0, 0) is an unstable critical point.

SOLUTION

Since x = r cos θ and y = r sin θ, we have

From dr/dt = 0.05r(3 – r) we see that dr/dt = 0 when r = 0 and can conclude that (x, y) = (0, 0) is a critical point by substituting r = 0 into the new system.

The differential equation dr/dt = 0.05r(3 – r) is a logistic equation that can be solved using separation of variables. If r(0) = r0 and r0 ≠ 0, then

,

where c0 = (3 – r0)/r0. Since

it follows that no matter how close to (0, 0) a solution starts, the solution will leave a disk of radius 1 about the origin. Therefore (0, 0) is an unstable critical point. A typical solution that starts near (0, 0) is shown in FIGURE 11.3.4.≡

A solution curve is graphed on an x y plane. A shaded circle representing a disk of radius 1 is centered at the origin. The curve starts at a point on the positive x axis inside the circle, spirals out in counter clockwise direction leaving the disk, and ends at the point (3, 0) forming an outer circle of radius 3.

FIGURE 11.3.4 Unstable critical point in Example 2

Linearization

It is rarely possible to determine the stability of a critical point of a nonlinear system by finding explicit solutions, as in Examples 1 and 2. Instead we replace the term g(X) in the original autonomous system X′ = g(X) by a linear term A(X – X1) that most closely approximates g(X) in a neighborhood of X1. This replacement process, called linearization, will be illustrated first for the first-order differential equation x′ = g(x).

An equation of the tangent line to the curve y = g(x) at x = x1 is y = g(x1) + g′(x1)(x – x1), and if x1 is a critical point of x′ = g(x), we have

x′ = g(x) ≈ g′(x1)(x – x1).

The general solution to the linear differential equation

x′ = g′(x1)(x – x1)

is x = x1 +, where λ1 = g′(x1). Thus if g′(x1) < 0, then x(t) approaches x1. Theorem 11.3.1 asserts that the same behavior occurs in the original differential equation provided x(0) = x0 is selected close enough to x1.

Autonomous first-order DEs were discussed in Section 2.1.

THEOREM 11.3.1 Stability Criteria for x′ = g(x)

Let x1 be a critical point of the autonomous first-order differential equation x′ = g(x), where g is differentiable at x1.

  1. If g′(x1) < 0, then x1 is an asymptotically stable critical point.
  2. If g′(x1) > 0, then x1 is an unstable critical point.

EXAMPLE 3 Stability in a Nonlinear First-Order DE

Both x = π/4 and x = 5π/4 are critical points of the autonomous differential equation x′ = cos x – sin x. This differential equation is difficult to solve explicitly, but we can use Theorem 11.3.1 to predict the behavior of solutions near these two critical points.

Since g′(x) = –sin x – cos x, g′(π/4) = – < 0 and g′(5π/4) = > 0. Therefore x = π/4 is an asymptotically stable critical point, but x = 5π/4 is unstable. In FIGURE 11.3.5 we used a numerical solver to investigate solutions that start near (0, π/4) and (0, 5π/4). Observe that solution curves that start close to (0, 5π/4) quickly move away from the line x = 5π/4, as predicted.≡

Four solution curves are graphed on the t x plane. Two dashed lines x equals pi over 4 and x equals 5 pi over 4 are drawn in the first quadrant. The first curve starts at the origin, goes up and to the right, remains flat and lies along the line x equals pi over 4, and exits the right of the first quadrant. The second curve starts approximately at the point (0, pi), goes down and to the right, remains flat at the end lying along the line x equals pi over 4, and exits the right of the first quadrant. The third curve starts approximately below the point (0, 5pi over 4), goes down and to the right, remains flat at the end lying along the line x equals pi over 4, and exits the right of the first quadrant. The fourth curve starts approximately above the point (0, 5 pi over 4), goes up and to the right, and exits the top of the first quadrant. The region above and below the line x equals 5pi over 4 is shaded. It covers the starting point of the third and fourth curves indicating that the curves quickly move away from the line.

FIGURE 11.3.5 In Example 3, π/4 is asymptotically stable; 5π/4 is unstable

EXAMPLE 4 Stability Analysis of the Logistic DE

Without solving explicitly, analyze the critical points of the logistic differential equation (see Section 2.8) x′ = x(K – x), where r and K are positive constants.

SOLUTION

The two critical points are x = 0 and x = K. Since

g′(x) = (K – 2x),

g′(0) = r and g′(K) = –r and it follows from Theorem 11.3.1 that x = 0 is an unstable critical point and x = K is an asymptotically stable critical point.≡

Jacobian Matrix

A similar analysis may be carried out for a plane autonomous system. An equation of the tangent plane to the surface z = g(x, y) at X1 = (x1, y1) is

and g(x, y) may be approximated by its tangent plane in a neighborhood of X1.

When X1 is a critical point of a plane autonomous system, P(x1, y1) = Q(x1, y1) = 0 and we have

The original system X′ = g(X) may be approximated in a neighborhood of the critical point X1 by the linear system X′ = A(X – X1), where

This matrix is called the Jacobian matrix at after the German mathematician Carl Gustav Jacob Jacobi (1804–1851). The Jacobian matrix at is also denoted by If we let H = X – X1, then the linear system X′ = A(X – X1) becomes H′ = AH, which is the form of the linear system analyzed in Section 11.2. The critical point X = X1 for X′ = A(X – X1) now corresponds to the critical point H = 0 for H′ = AH. If the eigenvalues of A have negative real parts, then, by Theorem 11.2.1, 0 is an asymptotically stable critical point for H′ = AH. If there is an eigenvalue with a positive real part, H = 0 is an unstable critical point. Theorem 11.3.2 asserts that the same conclusions can be made for the critical point X1 of the original system.

THEOREM 11.3.2 Stability Criteria for Plane Autonomous Systems

Let X1 be a crtical point of the plane autonomous sytem X′ = g(X), where P(x, y) and Q(x, y) have continuous first partial derivatives in a neighborhood of X1.

  1. If the eigenvalues of A = g′(X1) have negative real part, then X1 is an asymptotically stable critical point.
  2. If A = g′(X1) has an eigenvalue with positive real part, then X1 is an unstable critical point.

EXAMPLE 5 Stability Analysis of Nonlinear Systems

Classify (if possible) the critical points of each of the following plane autonomous systems as stable or unstable.

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

The critical points of each system were determined in Example 4 of Section 11.1.

  1. The critical points are (, 2) and (–, 2), the Jacobian matrix is

and so

and

Since the determinant of A1 is negative, A1 has a positive real eigenvalue. Therefore (, 2) is an unstable critical point. Matrix A2 has a positive determinant and a negative trace, and so both eigenvalues have negative real parts. It follows that (–, 2) is a stable critical point.

  1. The critical points are (0, 0), (0, 60), (100, 0), and (50, 50), the Jacobian matrix is

and so

Since the matrix A1 has a positive determinant and a positive trace, both eigenvalues have positive real parts. Therefore (0, 0) is an unstable critical point. The determinants of matrices A2 and A3 are negative, and so in each case one of the eigenvalues is positive. Therefore both (0, 60) and (100, 0) are unstable critical points. Since the matrix A4 has a positive determinant and a negative trace, (50, 50) is a stable critical point.≡

In Example 5 we did not compute τ2 – 4Δ (as in Section 11.2) and attempt to further classify the critical points as stable nodes, stable spiral points, saddle points, and so on. For example, for X1 = (–, 2) in Example 4(a), τ2 – 4Δ < 0, and if the system were linear, we would be able to conclude that X1 was a stable spiral point. FIGURE 11.3.6 shows several solution curves near X1 that were obtained with a numerical solver, and each solution does appear to spiral in toward the critical point.

Four curves are graphed on an x y plane. The curves from all directions in the second quadrant spiral in toward a common central point. The spirals follow clockwise direction.

FIGURE 11.3.6 Point appears to be a stable spiral point

Classifying Critical Points

It is natural to ask whether we can infer more geometric information about the solutions near a critical point X1 of a nonlinear autonomous system from an analysis of the critical point of the corresponding linear system. The answer is summarized in FIGURE 11.3.7, but you should note the comments following the figure.

A parabola is graphed on a tau delta coordinate plane. The curve is labeled tau^2 = 4 delta. The area above the parabola is shaded and the region is labeled tau^2 minus 4 delta < 0. Portions of the curve are marked stable and other portions are marked unstable. There are other graphs: an unstable spiral that goes outward. A stable spiral that goes inward. A stable node and an unstable node with two curves and two lines. A sixth graph represents saddle with curves with and two lines.

FIGURE 11.3.7 Geometric summary

  1. In five separate cases (stable node, stable spiral point, unstable spiral point, unstable node, and saddle) the critical point may be categorized like the critical point in the corresponding linear system. The solutions have the same general geometric features as the solutions to the linear system, and the smaller the neighborhood about X1, the closer the resemblance.
  2. If τ2 = 4Δ and τ > 0, the critical point X1 is unstable, but in this borderline case we are not yet able to decide whether X1 is an unstable spiral, unstable node, or degenerate unstable node. Likewise, if τ2 = 4Δ and τ < 0, the critical point X1 is stable but may be either a stable spiral, a stable node, or a degenerate stable node.
  3. If τ = 0 and Δ > 0, the eigenvalues of A = g′(X) are pure imaginary, and in this borderline case X1 may be either a stable spiral, an unstable spiral, or a center. It is therefore not yet possible to determine whether X1 is stable or unstable.

EXAMPLE 6 Classifying Critical Points of a Nonlinear System

Classify each critical point of the plane autonomous system in Example 5(b) as a stable node, a stable spiral point, an unstable spiral point, an unstable node, or a saddle point.

SOLUTION

For the matrix A1 corresponding to (0, 0), Δ = 3, τ = 4, and so τ2 – 4Δ = 4. Therefore (0, 0) is an unstable node. The critical points (0, 60) and (100, 0) are saddles since Δ < 0 in both cases. For matrix A4, Δ > 0, τ < 0, and τ2 – 4Δ > 0. It follows that (50, 50) is a stable node. Experiment with a numerical solver to verify these conclusions.≡

EXAMPLE 7 Stability Analysis for a Soft Spring

Recall from Section 3.11 that the second-order differential equation mx″ + kx + k1x3 = 0, for k > 0, represents a general model for the free, undamped oscillations of a mass m attached to a nonlinear spring. If k = 1 and k1 = –1, the spring is called soft and the plane autonomous system corresponding to the nonlinear second-order differential equation x″ + x – x3 = 0 is

x′ = y

        y′ = x3 – x.

Find and classify (if possible) the critical points.

SOLUTION

Since x3 – x = x(x2 – 1), the critical points are (0, 0), (1, 0), and (–1, 0). The corresponding Jacobian matrices are

Since det A2 < 0, critical points (1, 0) and (–1, 0) are both saddle points. The eigenvalues of matrix A1 are ±i, and according to comment (iii) on page 657, the status of the critical point at (0, 0) remains in doubt. It may be either a stable spiral, an unstable spiral, or a center.≡

The Phase-Plane Method

The linearization method, when successful, can provide useful information on the local behavior of solutions near critical points. It is of little help if we are interested in solutions whose initial position X(0) = X0 is not close to a critical point or if we wish to obtain a global view of the family of solution curves. The phase-plane method is based on the fact that

and it attempts to find y as a function of x using one of the methods available for solving first-order differential equations (Chapter 2). As we show in Examples 8 and 9, the method can sometimes be used to decide whether a critical point such as (0, 0) in Example 7 is a stable spiral, an unstable spiral, or a center.

EXAMPLE 8 Phase-Plane Method

Use the phase-plane method to classify the sole critical point (0, 0) of the plane autonomous system

x′ = y2

y′ = x2.

SOLUTION

The determinant of the Jacobian matrix

is 0 at (0, 0), and so the nature of the critical point (0, 0) remains in doubt. Using the phase-plane method, we obtain the first-order differential equation

which can be easily solved by separation of variables:

y2 dy = x2 dx    or    y3 = x3 + c.

If X(0) = (0, y0), it follows that y3 = x3 + or y = . FIGURE 11.3.8 shows a collection of solution curves corresponding to various choices for y0, and the nature of the critical point is clear. No matter how close to (0, 0) the solution starts, X(t) moves away from the origin as t increases. The critical point at (0, 0) is therefore unstable.≡

Six curves are graphed on an x y plane. Three curves start from the bottom left of the third quadrant, go up and to the right, bulge around the left of the origin in the form of concentric semicircles, then go up and to the right, and exits the top right of the first quadrant. The other three curves start from the bottom left of the third quadrant, go up and to the right, bulge around the right of the origin in the form of concentric semicircles, then go up and to the right, and exits the top right of the first quadrant.

FIGURE 11.3.8 Phase portrait of nonlinear system in Example 8

EXAMPLE 9 Phase-Plane Analysis of a Soft Spring

Use the phase-plane method to determine the nature of the solutions to x″ + x – x3 = 0 in a neighborhood of (0, 0).

SOLUTION

If we let dx/dt = y, then dy/dt = x3 – x. From this we obtain the first-order differential equation

which can be solved by separation of variables. Integrating

gives

After completing the square, we can write the solution as y2 = (x2 – 1)2/2 + c0. If X(0) = (x0, 0), where 0 < x0 < 1, then , and so

Note that y = 0 when x = –x0. In addition, the right-hand side is positive when –x0 < x < x0, and so each x has two corresponding values of y. The solution X = X(t) that satisfies X(0) = (x0, 0) is therefore periodic, and so (0, 0) is a center.

FIGURE 11.3.9 shows a family of solution curves or phase portrait of the original system. We used the original plane autonomous system to determine the directions indicated on each trajectory.≡

Two sets of five curves each and two circles are graphed on an x y plane. The first set of curves start from the top of the second quadrant, goes down and to the right, then goes up and to the right, and exits the top of the first quadrant. The curves are concentric and the bottommost curve is a parabola that has its vertex at a point on the negative y axis. The second set of curves start from the bottom of the fourth quadrant, goes up and to the left, then goes down and to the left, and exits the bottom of the third quadrant. The curves are concentric and the topmost curve is a parabola that has its vertex at a point on the positive y axis. Two concentric circles that follow clockwise direction are centered at the origin. They are enclosed between the two parabolas.

FIGURE 11.3.9 Phase portrait of nonlinear system in Example 9

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

  1. Show that (0, 0) is an asymptotically stable critical point of the nonlinear autonomous system

    x′ = αx – βy + y2

    y′ = βx + αy – xy

    when α < 0 and an unstable critical point when α > 0.
    [Hint: Switch to polar coordinates.]

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

    Show that (0, 0) is an asymptotically stable critical point if and only if α < 0.

In Problems 3–10, without solving explicitly, classify the critical points of the given first-order autonomous differential equation as either asymptotically stable or unstable. All constants are assumed to be positive.

  1. = kx(n + 1 – x)
  2. = –kx ln , x > 0
  3. = k(T – T0)
  4. m = mg – kv
  5. = k(α – x)(β – x), α > β
  6. = k(α – x)(β – x)(γ – x), α > β > γ
  7. = P(a – bP)(1 – cP–1), P > 0, a < bc
  8. A > 0

In Problems 11–20, classify (if possible) each critical point of the given plane autonomous system as a stable node, a stable spiral point, an unstable spiral point, an unstable node, or a saddle point.

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

In Problems 21–26, classify (if possible) each critical point of the given second-order differential equation as a stable node, a stable spiral point, an unstable spiral point, an unstable node, or a saddle point.

  1. θ″ = (cos θ – 0.5) sin θ, |θ| < π
  2. x″ + x = ( – 3(x′)2)x′ – x2
  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 [Hint: x|x| = 2|x|.]
  7. Show that the nonlinear second-order differential equation

    (1 + α2x2)x″ + (β + α2(x′)2)x = 0

    has a saddle point at (0, 0) when β < 0.

  8. Show that the dynamical system

    x′ = –αx + xy

        y′ = 1 – βy – x2

    has a unique critical point when αβ > 1 and that this critical point is stable when β > 0.

    1. Show that the plane autonomous system

      x′ = –x + y – x3

      y′ = –x – y + y2

      has two critical points by sketching the graphs of –x + y – x3 = 0 and –x – y + y2 = 0. Classify the critical point at (0, 0).

    2. Show that the second critical point X1 = (0.88054, 1.56327) is a saddle point.
    1. The equation

      is called Rayleigh’s differential equation after the English physicist John William Strutt, 3rd Baron Rayleigh (1842–1919). Strutt was the co-discoverer of argon gas and was awarded the Nobel Prize in physics in 1904. Show that is the only critical point of this differential equation.

    2. Show that (0, 0) is unstable when ϵ > 0. When is (0, 0) an unstable spiral point?
    3. Show that (0, 0) is stable when ϵ < 0. When is (0, 0) a stable spiral point?
    4. Show that (0, 0) is a center when ϵ = 0.
  11. Use the phase-plane method to show that (0, 0) is a center of the nonlinear second-order differential equation x″ + 2x3 = 0.
  12. Use the phase-plane method to show that the solution to the nonlinear second-order differential equation x″ + 2x – x2 = 0 that satisfies x(0) = 1 and x′(0) = 0 is periodic.
    1. Find the critical points of the plane autonomous system

      x′ = 2xy

                  y′ = 1 – x2 + y2,

      and show that linearization gives no information about the nature of these critical points.

    2. Use the phase-plane method to show that the critical points in part (a) are both centers. [Hint: Let u = y2/x, and show that (x – c)2 + y2 = c2 – 1.]
  14. The origin is the only critical point of the nonlinear second-order differential equation x″ + (x′)2 + x = 0.
    1. Show that the phase-plane method leads to the Bernoulli differential equation dy/dx = –y – xy–1.
    2. Show that the solution satisfying x(0) = and x′(0) = 0 is not periodic.
  15. A solution of the nonlinear second-order differential equation x″ + x – x3 = 0 satisfies x(0) = 0 and x′(0) = v0. Use the phase-plane method to determine when the resulting solution is periodic. [Hint: See Example 9.]
  16. The nonlinear differential equation x″ + x = 1 + ϵx2 arises in the analysis of planetary motion using relativity theory. Classify (if possible) all critical points of the corresponding plane autonomous system.
  17. When a nonlinear capacitor is present in an LRC-series circuit, the voltage drop is no longer given by q/C but is more accurately described by αq + βq3, where α and β are constants and α > 0. Differential equation (34) of Section 3.8 for the free circuit is then replaced by

    Find and classify all critical points of this nonlinear differential equation. [Hint: Divide into the two cases β > 0 and β < 0.]

  18. The nonlinear second-order differential equation

    mx″ + kx + k1x3 = 0,

    for k > 0, represents a general model for the free, undamped oscillations of a mass m attached to a spring. If k1 > 0, the spring is called hard (see Example 1 in Section 3.11). Determine the nature of the solutions to x″ + x + x3 = 0 in a neighborhood of (0, 0).

  19. The nonlinear differential equation

    can be interpreted as a model for a certain pendulum with a constant driving function.

    1. Show that (π/6, 0) and (5π/6, 0) are critical points of the corresponding plane autonomous system.
    2. Classify the critical point (5π/6, 0) using linearization.
    3. Use the phase-plane method to classify the critical point (π/6, 0).
    1. Show that (0, 0) is an isolated critical point of the plane autonomous system

      x′ = x4 – 2xy3

      y′ = 2x3y – y4

      but that linearization gives no useful information about the nature of this critical point.

    2. Use the phase-plane method to show that a family of solutions of the system in part (a) is

      The graphs of the members of this one-parameter family are called folia of Descartes. [Hint: The differential equation in x and y is homogeneous.]

    3. Parametric equations for the folia in part (b) are

      Use graphing software or a numerical solver to graph various solution curves defined by these equations. Based on your graphs, would you classify the critical point as stable or unstable? Would you classify the critical point as a node, saddle point, center, or spiral point?