3.11 Nonlinear Models

INTRODUCTION

In this section we examine some nonlinear higher-order mathematical models. We are able to solve some of these models using the substitution method introduced on page 153. In some cases where the model cannot be solved, we show how a nonlinear DE can be replaced by a linear DE through a process called linearization.

Nonlinear Springs

The mathematical model in (1) of Section 3.8 has the form

(1)

where F(x) = kx. Since x denotes the displacement of the mass from its equilibrium position, F(x) = kx is Hooke’s law, that is, the force exerted by the spring that tends to restore the mass to the equilibrium position. A spring acting under a linear restoring force F(x) = kx is naturally referred to as a linear spring. But springs are seldom perfectly linear. Depending on how it is constructed and the material used, a spring can range from “mushy” or soft to “stiff” or hard, so that its restorative force may vary from something below to something above that given by the linear law. In the case of free motion, if we assume that a nonaging spring possesses some nonlinear characteristics, then it might be reasonable to assume that the restorative force F(x) of a spring is proportional to, say, the cube of the displacement x of the mass beyond its equilibrium position or that F(x) is a linear combination of powers of the displacement such as that given by the nonlinear function F(x) = kx + k1x3. A spring whose mathematical model incorporates a nonlinear restorative force, such as

(2)

is called a nonlinear spring. In addition, we examined mathematical models in which damping imparted to the motion was proportional to the instantaneous velocity dx/dt, and the restoring force of a spring was given by the linear function F(x) = kx. But these were simply assumptions; in more realistic situations damping could be proportional to some power of the instantaneous velocity dx/dt. The nonlinear differential equation

(3)

is one model of a free spring/mass system with damping proportional to the square of the velocity. One can then envision other kinds of models: linear damping and nonlinear restoring force, nonlinear damping and nonlinear restoring force, and so on. The point is, nonlinear characteristics of a physical system lead to a mathematical model that is nonlinear.

Notice in (2) that both F(x) = kx3 and F(x) = kx + k1x3 are odd functions of x. To see why a polynomial function containing only odd powers of x provides a reasonable model for the restoring force, let us express F as a power series centered at the equilibrium position x = 0:

F(x) = c0 + c1x + c2x2 + c3x3 + ….

When the displacements x are small, the values of xn are negligible for n sufficiently large. If we truncate the power series with, say, the fourth term, then

F(x) = c0 + c1x + c2x2 + c3x3.

In order for the force at x > 0 (F(x) = c0 + c1x + c2x2 + c3x3) and the force at −x < 0 (F(−x) = c0 − c1x + c2x2 − c3x3) to have the same magnitude but act in the opposite directions, we must have F(−x) = −F(x). Since this means F is an odd function, we must have c0 = 0 and c2 = 0, and so F(x) = c1x + c3x3. Had we used only the first two terms in the series, the same argument yields the linear function F(x) = c1x. For discussion purposes we shall write c1 = k and c2 = k1. A restoring force with mixed powers such as F(x) = kx + k1x2, and the corresponding vibrations, are said to be unsymmetrical.

Hard and Soft Springs

Let us take a closer look at the equation in (1) in the case where the restoring force is given by F(x) = kx + k1x3, k > 0. The spring is said to be hard if k1 > 0 and soft if k1 < 0. Graphs of three types of restoring forces are illustrated in FIGURE 3.11.1. The next example illustrates these two special cases of the differential equation m d2x/dt2 + kx + k1x3 = 0, m > 0, k > 0.

Two curves and a line are graphed on an x F plane. The first curve labeled soft spring starts from the left of the third quadrant, goes up and to the right through the origin, and ends at the right of the first quadrant. The second curve labeled hard spring starts from the bottom of the third quadrant, goes up and to the right through the origin, and ends at the top of the first quadrant. The line labeled linear spring starts from the bottom left of the third quadrant, goes up and to the right through the origin, and ends at the top right of the first quadrant. The line is tangential to both curves.

FIGURE 3.11.1 Hard and soft springs

EXAMPLE 1 Comparison of Hard and Soft Springs

The differential equations

+ x + x3 = 0 (4)

and

+ x − x3 = 0 (5)

are special cases of (2) and are models of a hard spring and soft spring, respectively. FIGURE 3.11.2(a) shows two solutions of (4) and Figure 3.11.2(b) shows two solutions of (5) obtained from a numerical solver. The curves shown in red are solutions satisfying the initial conditions x(0) = 2, x′(0) = −3; the two curves in blue are solutions satisfying x(0) = 2, x′(0) = 0. These solution curves certainly suggest that the motion of a mass on the hard spring is oscillatory, whereas motion of a mass on the soft spring is not oscillatory. But we must be careful about drawing conclusions based on a couple of solution curves. A more complete picture of the nature of the solutions of both equations can be obtained from the qualitative analysis discussed in Chapter 11. ≡

The first graph is shown in part (a). It is the graph of two solutions for a hard spring. Two curves are graphed on the t x coordinate plane. The first curve is labeled x(0) = 2, x prime(0) = negative 3. It follows an oscillatory pattern. It starts at the point (0, 2), goes down and to the right through the positive t axis, and reaches a low point in the fourth quadrant. Then it goes up and to the right through the positive t axis, reaches a high point in the first quadrant, goes down and to the right through the positive t axis, and reaches a low point in the fourth quadrant. Again it goes up and to the right through the positive t axis, reaches a point in the first quadrant, and ends at the right of the first quadrant. The second curve is labeled x(0) = 2, x prime(0) = 0. It follows an oscillatory pattern. It starts at the point (0, 2), goes down and to the right through the positive t axis, and reaches a low point in the fourth quadrant. Then it goes up and to the right through the positive t axis, reaches a high point in the first quadrant, goes down and to the right through the positive t axis, and reaches a low point in the fourth quadrant. Again it goes up and to the right, and ends at the right of the fourth quadrant. The amplitude of the second curve is less than the amplitude of the first curve. The second graph is shown in part (b). It is the graph of two solutions for a soft spring. Two curves are graphed on the t x coordinate plane. The first curve is labeled x(0) = 2, x prime(0) = negative 3. It starts at the point (0, 2), goes down and to the right through the positive t axis, and exits the bottom of the fourth quadrant. The second curve is labeled x(0) = 2, x prime(0) = 0. It starts at the point (0, 2), goes up and to the right, and exits the top of the first quadrant.

FIGURE 3.11.2 Numerical solution curves

Nonlinear Pendulum

Any object that swings back and forth is called a physical pendulum. The simple pendulum is a special case of the physical pendulum and consists of a rod of length l to which a mass m is attached at one end. In describing the motion of a simple pendulum in a vertical plane, we make the simplifying assumptions that the mass of the rod is negligible and that no external damping or driving forces act on the system. The displacement angle θ of the pendulum, measured from the vertical as shown in FIGURE 3.11.3, is considered positive when measured to the right of OP and negative to the left of OP. Now recall that the arc s of a circle of radius l is related to the central angle θ by the formula s = lθ. Hence angular acceleration is

The simple pendulum consists of a thin rod of length l. The rod is fixed vertically at the top. A mass m is attached to the free end of the pendulum. The fixed end is labeled O. A dashed vertical line from O points downward and labeled P. The pendulum is displaced to the right by an angle theta from the vertical line. A dashed arc from the center of the mass goes to the left crossing the vertical line. The arc is labeled mg sin theta. A vertical downward arrow from the center of the mass is labeled W = mg. An arrow tangential to the arc goes to the left from the center of the mass. An arrow parallel to the displaced pendulum starts from the end of the tangential arrow and ends at the end of the vertical arrow. It is labeled mg cos theta.

FIGURE 3.11.3 Simple pendulum

From Newton’s second law we then have

From Figure 3.11.3 we see that the magnitude of the tangential component of the force due to the weight W is mg sin θ. In direction this force is −mg sin θ, since it points to the left for θ > 0 and to the right for θ < 0. We equate the two different versions of the tangential force to obtain ml d2θ/dt2 = −mg sin θ or

. (6)

Linearization

Because of the presence of sin θ, the model in (6) is nonlinear. In an attempt to understand the behavior of the solutions of nonlinear higher-order differential equations, one sometimes tries to simplify the problem by replacing nonlinear terms by certain approximations. For example, the Maclaurin series for sin θ is given by

and so if we use the approximation sin θ ≈ θ − θ3/6, equation (6) becomes d2θ/dt2 + (g/l)θ − (g/6l)θ3 = 0. Observe that this last equation is the same as the second nonlinear equation in (2) with m = 1, k = g/l, and k1 = −g/6l. However, if we assume that the displacements θ are small enough to justify using the replacement sin θ ≈ θ, then (6) becomes

. (7)

See Problem 24 in Exercises 3.11. If we set ω2 = g/l, we recognize (7) as the differential equation (2) of Section 3.8 that is a model for the free undamped vibrations of a linear spring/mass system. In other words, (7) is again the basic linear equation y″ + λy = 0 discussed on pages 175–176 of Section 3.9. As a consequence, we say that equation (7) is a linearization of equation (6). Since the general solution of (7) is θ(t) = c1 cos ωt + c2 sin ωt, this linearization suggests that for initial conditions amenable to small oscillations the motion of the pendulum described by (6) will be periodic.

EXAMPLE 2 Two Initial-Value Problems

The graphs in FIGURE 3.11.4(a) were obtained with the aid of a numerical solver and represent solution curves of equation (6) when ω2 = 1. The blue curve depicts the solution of (6) that satisfies the initial conditions θ(0) = , θ′(0) = whereas the red curve is the solution of (6) that satisfies θ(0) = , θ′(0) = 2. The blue curve represents a periodic solution—the pendulum oscillating back and forth as shown in Figure 3.11.4(b) with an apparent amplitude A ≤ 1. The red curve shows that θ increases without bound as time increases—the pendulum, starting from the same initial displacement, is given an initial velocity of magnitude great enough to send it over the top; in other words, the pendulum is whirling about its pivot as shown in Figure 3.11.4(c). In the absence of damping the motion in each case is continued indefinitely.

Graph of numerical solution curves is shown in part (a). Two curves are graphed on the t theta plane. The first curve is labeled theta(0) = 1 over 2, theta prime(0) = 1 over 2. It follows a constant-amplitude oscillatory pattern. It starts at a point on the positive theta axis just above the origin, goes up and to the right, reaches a high point, goes down and to the right through the positive t axis, and reaches a low point in the fourth quadrant. Then it goes up and to the right through the positive t axis, and reaches a high point in the first quadrant. It completes about two more cycles following the same oscillatory pattern and ends at the right of the fourth quadrant. The second curve is labeled theta(0) = 1 over 2, theta prime(0) = 2. It starts at the same starting point of the first curve, goes up and to the right, goes down and to the right, and exits the top of the first quadrant. An illustration of an oscillating pendulum is shown in part (b). The pendulum is displaced to the right of the pivot. The image of the pendulum displaced to the left is shown in dotted lines. A dotted arc connecting the center of the mass of the pendulum at the two positions has a right arrow from the left position and a left arrow from the right position. The illustration is labeled theta(0) = 1 over 2, theta prime(0) = 1 over 2. An illustration of a whirling pendulum is shown in part (c). The pendulum is displaced to the right of the pivot. The images of the pendulum displaced to the left and over the top are shown in dotted lines. A dotted circle connecting the center of the mass of the pendulum at the three positions has an upward arrow from the left position and a downward arrow from the top position. The illustration is labeled theta(0) = 1 over 2, theta prime(0) = 2.

FIGURE 3.11.4 Numerical solution curves in (a); oscillating pendulum in (b); whirling pendulum in (c) in Example 2

≡

Telephone Wires

The first-order differential equation

is equation (16) of Section 1.3. This differential equation, established with the aid of Figure 1.3.9 on page 24, serves as a mathematical model for the shape of a flexible cable suspended between two vertical supports when the cable is carrying a vertical load. In Exercises 2.2, you may have solved this simple DE under the assumption that the vertical load carried by the cables of a suspension bridge was the weight of a horizontal roadbed distributed evenly along the x-axis. With W = ρw, ρ the weight per unit length of the roadbed, the shape of each cable between the vertical supports turned out to be parabolic. We are now in a position to determine the shape of a uniform flexible cable hanging under its own weight, such as a wire strung between two telephone posts. The vertical load is now the wire itself, and so if ρ is the linear density of the wire (measured, say, in lb/ft) and s is the length of the segment P1P2 in Figure 1.3.9, then W = ρs. Hence,

(8)

Since the arc length between points P1 and P2 is given by

(9)

it follows from the Fundamental Theorem of Calculus that the derivative of (9) is

(10)

Differentiating (8) with respect to x and using (10) leads to the nonlinear second-order equation

. (11)

In the example that follows, we solve (11) and show that the curve assumed by the suspended cable is a catenary. Before proceeding, observe that the nonlinear second-order differential equation (11) is one of those equations having the form F(x,y′,y″) = 0 discussed in Section 3.7. Recall, we have a chance of solving an equation of this type by reducing the order of the equation by means of the substitution u = y′.

A photo of Telephones poles with suspended cables.

© Tad Denson/Shutterstock

Shape of hanging telephone wires is a catenary

EXAMPLE 3 An Initial-Value Problem

From the position of the y-axis in Figure 1.3.9 on page 24 it is apparent that initial conditions associated with the second differential equation in (11) are y(0) = a and y′(0) = 0. If we substitute u = y′, the last equation in (11) becomes Separating variables,

Now, y′(0) = 0 is equivalent to u(0) = 0. Since sinh−1 0 = 0, we find c1 = 0 and so u = sinh (ρx/T1). Finally, by integrating both sides of

Using y(0) = a, cosh 0 = 1, the last equation implies that c2 = a − T1/ρ. Thus we see that the shape of the hanging wire is given by y = (T1/ρ) cosh(ρx/T1) + a − T1/ρ. ≡

In Example 3, had we been clever enough at the start to choose a = T1/ρ, then the solution of the problem would have been simply the hyperbolic cosine y = (T1/ρ) cosh (ρx/T1).

Rocket Motion

In Section 1.3 we saw that the differential equation of a free-falling body of mass m near the surface of the Earth is given by

,

where s represents the distance from the surface of the Earth to the object and the positive direction is considered to be upward. In other words, the underlying assumption here is that the distance s to the object is small when compared with the radius R of the Earth; put yet another way, the distance y from the center of the Earth to the object is approximately the same as R. If, on the other hand, the distance y to an object, such as a rocket or a space probe, is large compared to R, then we combine Newton’s second law of motion and his universal law of gravitation to derive a differential equation in the variable y.

Suppose a rocket is launched vertically upward from the ground as shown in FIGURE 3.11.5. If the positive direction is upward and air resistance is ignored, then the differential equation of motion after fuel burnout is

, (12)

where k is a constant of proportionality, y is the distance from the center of the Earth to the rocket, M is the mass of the Earth, and m is the mass of the rocket. To determine the constant k, we use the fact that when y = R, kMm/R2 = mg or k = gR2/M. Thus the last equation in (12) becomes

. (13)

A circular two dimensional view of the Earth is shown. The center of Earth is marked with a large dot. The radius of the Earth is labeled R. A rocket is launched vertically upward from the ground with a velocity of v subscript 0. Distance from the center of the Earth to the rocket is represented by a vertical dashed line labeled y.

FIGURE 3.11.5 Distance to rocket is large compared to R

See Problem 14 in Exercises 3.11.

Variable Mass

Notice in the preceding discussion that we described the motion of the rocket after it has burned all its fuel, when presumably its mass m is constant. Of course during its powered ascent, the total mass of the rocket varies as its fuel is being expended. The second law of motion, as originally advanced by Newton, states that when a body of mass m moves through a force field with velocity v, the time rate of change of the momentum mv of the body is equal to applied or net force F acting on the body:

. (14)

See Problems 21 and 22 in Exercises 1.3. If m is constant, then (14) yields the more familiar form F = m dv/dt = ma, where a is acceleration. We use the form of Newton’s second law given in (14) in the next example, in which the mass m of the body is variable.

EXAMPLE 4 Rope Pulled Upward by a Constant Force

A uniform 10-foot-long heavy rope is coiled loosely on the ground. One end of the rope is pulled vertically upward by means of a constant force of 5 lb. The rope weighs 1 lb per foot. Determine the height x(t) of the end above ground level at time t. See FIGURE 1.R.2 and Problem 47 in Chapter 1 in Review.

SOLUTION

Let us suppose that x = x(t) denotes the height of the end of the rope in the air at time t, v = dx/dt, and that the positive direction is upward. For that portion of the rope in the air at time t we have the following variable quantities:

            weight: W = (x ft) · (1 lb/ft) = x,

mass: m = W/g = x/32,

        net force: F = 5 − W = 5 − x.

Thus from (14) we have

(15)

Since v = dx/dt the last equation becomes

(16)

The nonlinear second-order differential equation (16) has the form F(x,x′,x″) = 0, which is the second of the two forms considered in Section 3.7 that can possibly be solved by reduction of order. In order to solve (16), we revert back to (15) and use v = x′ along with the Chain Rule. From the second equation in (15) can be rewritten as

(17)

On inspection (17) might appear intractable, since it cannot be characterized as any of the first-order equations that were solved in Chapter 2. However, by rewriting (17) in differential form M(x,v) dx + N(x,v) dv = 0, we observe that the nonexact equation

(v2 + 32x − 160)dx + xv dv = 0 (18)

can be transformed into an exact equation by multiplying it by an integrating factor.* When (18) is multiplied by µ(x) = x, the resulting equation is exact (verify). If we identify = xv2 + 32x2 − 160x, = x2v, and then proceed as in Section 2.4, we arrive at

(19)

From the initial condition x(0) = 0 it follows that c1 = 0. Now by solving − 80x2 = 0 for v = dx/dt > 0 we get another differential equation,

which can be solved by separation of variables. You should verify that

(20)

This time the initial condition x(0) = 0 implies c2 = −3/8. Finally, by squaring both sides of (20) and solving for x we arrive at the desired result,

(21) ≡

The graph of (21) given in FIGURE 3.11.6 should not, on physical grounds, be taken at face value. See Problems 15 and 16 in Exercises 3.11.

A curve is graphed on the t x axis. The curve starts at the point (0, 0), goes up and to the right through the approximate points (0.5, 4.99) and (1, 7.32), and reaches a high point. Then it goes down and to the right through the approximate points (1.5, 6.97) and (2, 3.96), and ends at the approximate point (2.4, 0).

FIGURE 3.11.6 Graph of (21) in Example 4

3.11 Exercises Answers to selected odd-numbered problems begin on page ANS-8.

To the Instructor

In addition to Problems 24 and 25, all or portions of Problems 1−6, 8−13, and 23 could serve as Computer Lab Assignments.

Nonlinear Springs

In Problems 1−4, the given differential equation is a model of an undamped spring/mass system in which the restoring force F(x) in (1) is nonlinear. For each equation use a numerical solver to plot the solution curves satisfying the given initial conditions. If the solutions appear to be periodic, use the solution curve to estimate the period T of oscillations.

  1. + x3 = 0,

    x(0) = 1, x′(0) = 1; x(0) = , x′(0) = −1

  2. + 4x − 16x3 = 0,

    x(0) = 1, x′(0) = 1; x(0) = −2, x′(0) = 2

  3. + 2x − x2 = 0,

    x(0) = 1, x′(0) = 1; x(0) = , x′(0) = −1

  4. + xe0.01x = 0,

    x(0) = 1, x′(0) = 1; x(0) = 3, x′(0) = −1

  5. In Problem 3, suppose the mass is released from the initial position x(0) = 1 with an initial velocity x′(0) = x1. Use a numerical solver to estimate the smallest value of |x1| at which the motion of the mass is nonperiodic.
  6. In Problem 3, suppose the mass is released from an initial position x(0) = x0 with the initial velocity x′(0) = 1. Use a numerical solver to estimate an interval a ≤ x0 ≤ b for which the motion is oscillatory.
  7. Find a linearization of the differential equation in Problem 4.
  8. Consider the model of an undamped nonlinear spring/mass system given by x″ + 8x − 6x3 + x5 = 0. Use a numerical solver to discuss the nature of the oscillations of the system corresponding to the initial conditions:

    x(0) = 1, x′(0) = 1; x(0) = −2, x′(0) = ;

    x(0) = , x′(0) = 1; x(0) = 2, x′(0) = ;

    x(0) = 2, x′(0) = 0; x(0) = −, x′(0) = −1.

In Problems 9 and 10, the given differential equation is a model of a damped nonlinear spring/mass system. Predict the behavior of each system as t → . For each equation use a numerical solver to obtain the solution curves satisfying the given initial conditions.

  1. + x + x3 = 0,

    x(0) = −3, x′(0) = 4; x(0) = 0, x′(0) = −8

  2. + x − x3 = 0,

    x(0) = 0, x′(0) = ; x(0) = −1, x′(0) = 1

  3. The model mx″ + kx + k1x3 = F0 cos ωt of an undamped periodically driven spring/mass system is called Duffing’s differential equation after the German engineer Georg Duffing (1861–1944). Consider the initial-value problem x″ + x + k1x3 = 5 cos t, x(0) = 1, x′(0) = 0. Use a numerical solver to investigate the behavior of the system for values of k1 > 0 ranging from k1 = 0.01 to k1 = 100. State your conclusions.
    1. Find values of k1 < 0 for which the system in Problem 11 is oscillatory.
    2. Consider the initial-value problem

      x″ + x + k1x3 = cos t, x(0) = 0, x′(0) = 0.

      Find values for k1 < 0 for which the system is oscillatory.

Nonlinear Pendulum

  1. Consider the model of the free damped nonlinear pendulum given by

    Use a numerical solver to investigate whether the motion in the two cases λ2 − ω2 > 0 and λ2 − ω2 < 0 corresponds, respectively, to the overdamped and underdamped cases discussed in Section 3.8 for spring/mass systems. Choose appropriate initial conditions and values of λ and ω.

Rocket Motion

    1. Use the substitution v =dy/dt to solve (13) for v in terms of y. Assume that the velocity of the rocket at burnout is v = v0 and that y ≈ R at that instant; show that the approximate value of the constant c of integration is c = −gR + v02.
    2. Use the solution for v in part (a) to show that the escape velocity of the rocket is given by v0 = [Hint: Take y → and assume v > 0 for all time t.]
    3. The result in part (b) holds for any body in the solar system. Use the values g = 32 ft/s2 and R = 4000 mi to show that the escape velocity from the Earth is (approximately) v0 = 25,000 mi/h.
    4. Find the escape velocity from the Moon if the acceleration of gravity is 0.165g and R = 1080 mi.

Variable Mass

    1. In Example 4, show that equation (16) possesses a constant solution x(t) = k > 0.
    2. Why does the value of k in part (a) make intuitive sense?
    3. What is the initial velocity of the rope?
    1. In Example 4, what does (21) predict to be the maximum amount of rope lifted by the constant force?
    2. Explain any difference between your answer to part (a) in this problem and your answer to part (a) of Problem 15.
    3. Why would you expect the solution x(t) of the problem in Example 4 to be oscillatory?

Additional Mathematical Models

  1. Ballistic Pendulum Historically, in order to maintain quality control over munitions (bullets) produced by an assembly line, the manufacturer would use a ballistic pendulum to determine the muzzle velocity of a gun, that is, the speed of a bullet as it leaves the barrel. The ballistic pendulum, invented in 1742 by the British mathematician and military engineer Benjamin Robins (1707–1751), is simply a plane pendulum consisting of a rod of negligible mass to which a block of wood of mass mw is attached. The system is set in motion by the impact of a bullet that is moving horizontally at the unknown muzzle velocity vb; at the time of the impact, t = 0, the combined mass is mw + mb, where mb is the mass of the bullet embedded in the wood. We have seen in (7) of this section, that in the case of small oscillations, the angular displacement θ(t) of a plane pendulum shown in Figure 3.11.3 is given by the linear DE where corresponds to motion to the right of vertical. The velocity vb can be found by measuring the height h of the mass at the maximum displacement angle shown in FIGURE 3.11.7.
    A ballistic pendulum consists of a thin rod of length l fixed at the top. A block of wood of mass m subscript w is attached to the free end of the rod. A right arrow at the left of the mass represents a mass m subscript b and is labeled v subscript b. Another right arrow at the right of the mass is labeled uppercase V. The pendulum is displaced to the right by an angle theta subscript max. The displaced pendulum is represented by a dotted line and the mass of the displaced pendulum is labeled m subscript b + m subscript w. The height of the mass at the maximum displacement is labeled h.

    FIGURE 3.11.7 Ballistic pendulum in Problem 17

    Intuitively, the horizontal velocity V of the combined mass after impact is only a fraction of the velocity vb of the bullet, that is,

    Now recall, a distance s traveled by a particle moving along a circular path is related to the radius l and central angle θ by the formula By differentiating the last formula with respect to time t, it follows that the angular velocity of the mass and its linear velocity v are related by Thus the initial angular velocity at the time t at which the bullet impacts the wood block is related to V by or

    1. Solve the initial-value problem

    2. Use the result from part (a) to show that

    3. Use Figure 3.11.7 to express in terms of l and h. Then use the first two terms of the Maclaurin series for to express in terms of l and h. Finally, show that vb is given (approximately) by

  2. Ballistic Pendulum—Continued Use the result in Problem 17 to find the muzzle velocity vb when and h = 6 cm.
  3. Pursuit Curve In a naval exercise, a ship S1 is pursued by a submarine S2 as shown in FIGURE 3.11.8. Ship S1 departs point (0, 0) at t = 0 and proceeds along a straight-line course (the y-axis) at a constant speed v1. The submarine S2 keeps ship S1 in visual contact, indicated by the straight dashed line L in the figure, while traveling at a constant speed v2 along a curve C. Assume that S2 starts at the point at t = 0 and that L is tangent to C.
    1. Determine a mathematical model that describes the curve C. [Hint: where s is arc length measured along C.].
    2. Find an explicit solution of the differential equation. For convenience define
    3. Determine whether the paths of S1 and S2 will ever intersect by considering the cases and
      The pursuit curve is graphed on an x y coordinate plane. A ship labeled S subscript 1 is almost on the top of the y axis facing up. A submarine labeled S subscript 2 is in the first quadrant, slanted and facing the ship. A straight dashed line L connects the ship and submarine. A curve is drawn from the submarine and labeled C. It goes up and to the left initially, and then goes up to the top of the first quadrant.

      FIGURE 3.11.8 Pursuit curve in Problem 19

  4. Pursuit Curve In another naval exercise, a destroyer S1 pursues a submerged submarine S2. Suppose that S1 at (9, 0) on the x-axis detects S2 at (0, 0) and that S2 simultaneously detects S1. The captain of the destroyer S1 assumes that the submarine will take immediate evasive action and conjectures that its likely new course is the straight line indicated in FIGURE 3.11.9. When S1 is at (3, 0) it changes from its straight-line course toward the origin to a pursuit curve C. Assume that the speed of the destroyer is, at all times, a constant 30 mi/h and the submarine’s speed is a constant 15 mi/h.
    1. Explain why the captain waits until S1 reaches (3, 0) before ordering a course change to C.
    2. Using polar coordinates, find an equation r = f(θ) for the curve C.
    3. Let T denote the time, measured from the initial detection, at which the destroyer intercepts the submarine. Find an upper bound for T.
      The pursuit curve is graphed on an x y coordinate plane. A destroyer labeled S subscript 1 is in the first quadrant, slanted and facing top left. A dashed line connects the origin to the destroyer. The angle between the dashed line and the positive x axis is labeled theta. A submarine labeled S subscript 2 is in the second quadrant, slanted and facing top left. A straight line L connects the origin and the submarine. The points (9, 0) and (3, 0) are marked with heavy dot. A line connects the point (9, 0) to (3, 0). It has a left arrow on it. Another line connects the point (3, 0) to the destroyer. A curve from the destroyer goes up and to the left through the positive y axis and connects to the submarine. It has a left arrow on it.

      FIGURE 3.11.9 Pursuit curve in Problem 20

Discussion Problems

  1. Discuss why the damping term in equation (3) is written as

    1. Experiment with a calculator to find an interval 0 ≤ θ < θ1, where θ is measured in radians, for which you think sin θ ≈ θ is a fairly good estimate. Then use a graphing utility to plot the graphs of y = x and y = sin x on the same coordinate axes for 0 ≤ x ≤ π/2. Do the graphs confirm your observations with the calculator?
    2. Use a numerical solver to plot the solution curves of the initial-value problems

      + sin θ = 0, θ(0) = θ0, θ′(0) = 0

      and

      + θ = 0, θ(0) = θ0, θ′(0) = 0

      for several values of θ0 in the interval 0 ≤ θ < θ1 found in part (a). Then plot solution curves of the initial-value problems for several values of θ0 for which θ0 > θ1.

    1. Consider the nonlinear pendulum whose oscillations are defined by (6). Use a numerical solver as an aid to determine whether a pendulum of length l will oscillate faster on the Earth or on the Moon. Use the same initial conditions, but choose these initial conditions so that the pendulum oscillates back and forth.
    2. For which location in part (a) does the pendulum have greater amplitude?
    3. Are the conclusions in parts (a) and (b) the same when the linear model (7) is used?

Computer Lab Assignments

  1. Consider the initial-value problem

    for the nonlinear pendulum. Since we cannot solve the differential equation, we can find no explicit solution of this problem. But suppose we wish to determine the first time t1 > 0 for which the pendulum in Figure 3.11.3, starting from its initial position to the right, reaches the position OP—that is, find the first positive root of θ(t) = 0. In this problem and the next we examine several ways to proceed.

    1. Approximate t1 by solving the linear problem

      .

    2. Use the method illustrated in Example 3 of Section 3.7 to find the first four nonzero terms of a Taylor series solution θ(t) centered at 0 for the nonlinear initial-value problem. Give the exact values of all coefficients.
    3. Use the first two terms of the Taylor series in part (b) to approximate t1.
    4. Use the first three terms of the Taylor series in part (b) to approximate t1.
    5. Use a root-finding application of a CAS (or a graphing calculator) and the first four terms of the Taylor series in part (b) to approximate t1.
    6. In this part of the problem you are led through the commands in Mathematica that enable you to approximate the root t1. The procedure is easily modified so that any root of θ(t) = 0 can be approximated. (If you do not have Mathematica, adapt the given procedure by finding the corresponding syntax for the CAS you have on hand.) Reproduce and then, in turn, execute each line in the given sequence of commands.

    7. Appropriately modify the syntax in part (f) and find the next two positive roots of θ(t) = 0.
  2. Consider a pendulum that is released from rest from an initial displacement of θ0 radians. Solving the linear model (7) subject to the initial conditions θ(0) = θ0, θ′(0) = 0 gives θ(t) = θ0 cos t. The period of oscillations predicted by this model is given by the familiar formula T = 2π/ The interesting thing about this formula for T is that it does not depend on the magnitude of the initial displacement θ0. In other words, the linear model predicts that the time that it would take the pendulum to swing from an initial displacement of, say, θ0 = π/2 (= 90°) to −π/2 and back again would be exactly the same time to cycle from, say, θ0 = π/360 (= 0.5°) to −π/360. This is intuitively unreasonable; the actual period must depend on θ0.

    If we assume that g = 32 ft/s2 and l = 32 ft, then the period of oscillation of the linear model is T = 2π s. Let us compare this last number with the period predicted by the nonlinear model when θ0 = π/4. Using a numerical solver that is capable of generating hard data, approximate the solution of

    for 0 ≤ t ≤ 2. As in Problem 24, if t1 denotes the first time the pendulum reaches the position OP in Figure 3.11.3, then the period of the nonlinear pendulum is 4t1. Here is another way of solving the equation θ(t) = 0. Experiment with small step sizes and advance the time starting at t = 0 and ending at t = 2. From your hard data, observe the time t1 when θ(t) changes, for the first time, from positive to negative. Use the value t1 to determine the true value of the period of the nonlinear pendulum. Compute the percentage relative error in the period estimated by T = 2π.

 

*See pages 65–66 in Section 2.4.