INTRODUCTION

We are now in a position to solve the boundary-value problem (11) discussed in Section 13.2. The vertical displacement u(x, t) of a string of length L that is freely vibrating in the vertical plane shown in Figure 13.2.2(a) is determined from

(1)

(2)

(3)

Solution of the BVP

With the usual assumption that u(x, t) = X(x)T(t), separating variables in (1) gives

so that (4)

(5)

As in Section 13.3, the boundary conditions (2) translate into X(0) = 0 and X(L) = 0. The ODE in (4) along with these boundary conditions is the regular Sturm–Liouville problem

(6)

Of the usual three possibilities for the parameter λ: λ = 0, λ = –α² < 0, and λ = α² > 0, only the last choice leads to nontrivial solutions. Corresponding to λ = α², α > 0, the general solution of (4) is

X(0) = 0 and X(L) = 0 indicate that c₁ = 0 and c₂ sin αL = 0. The last equation again implies that αL = nπ or α = nπ/L. The eigenvalues and corresponding eigenfunctions of (6) are λ_n = n²π²/L² and X(x) = c₂ sin x, n = 1, 2, 3, … . The general solution of the second-order equation (5) is then

By rewriting c₂c₃ as A_n and c₂c₄ as B_n, solutions that satisfy both the wave equation (1) and boundary conditions (2) are

(7)

and (8)

Setting t = 0 in (8) and using the initial condition u(x, 0) = f(x) gives

Since the last series is a half-range expansion for f in a sine series, we can write A_n = b_n:

(9)

To determine B_n we differentiate (8) with respect to t and then set t = 0:

In order for this last series to be the half-range sine expansion of the initial velocity g on the interval, the total coefficient B_nnπa/L must be given by the form b_n in (5) of Section 12.3—that is,

,

from which we obtain

(10)

The solution of the boundary-value problem (1)–(3) consists of the series (8) with coefficients A_n and B_n defined by (9) and (10), respectively.

We note that when the string is released from rest, then g(x) = 0 for every x in the interval [0, L] and consequently B_n = 0.

Plucked String

A special case of the boundary-value problem in (1)–(3) is the model of a plucked string. When a string is plucked, a point on the string at is raised to a height h above the x-axis so that its initial shape is given by the graph of a function of the form

The point is released from rest at See Figure 13.2.4 and Problems 7 and 8 in Exercises 13.4. We can see the motion of the string by plotting the solution or displacement for increasing values of time and using the animation feature of a CAS. Some frames of a “movie” generated in this manner are given in FIGURE 13.4.1; the initial shape of the string is given in Figure 13.4.1(a). You are asked to emulate the results given in the figure by plotting a sequence of partial sums of (8). See Problem 27 in Exercises 13.4.

Standing Waves

Recall from the derivation of the wave equation in Section 13.2 that the constant a appearing in the solution of the boundary-value problem in (1) – (3) is given by , where ρ is mass per unit length and T is the magnitude of the tension in the string. When T is large enough, the vibrating string produces a musical sound. This sound is the result of standing waves. The solution (8) is a superposition of product solutions called standing waves or normal modes:

u(x, t) = u₁(x, t) + u₂(x, t) + u₃(x, t) + .

In view of (6) and (7) of Section 3.8, the product solutions (7) can be written as

(11)

where C_n = and ø_n is defined by sin ø_n = A_n/C_n and cos ø_n = B_n/C_n. For n = 1, 2, 3, … the standing waves are essentially the graphs of sin(nπx/L), with a time-varying amplitude given by

Alternatively, we see from (11) that at a fixed value of x each product function u_n(x, t) represents simple harmonic motion with amplitude C_nsin(nπx/L) and frequency f_n = na/2L. In other words, each point on a standing wave vibrates with a different amplitude but with the same frequency. When n = 1,

is called the first standing wave, the first normal mode, or the fundamental mode of vibration. The first three standing waves, or normal modes, are shown in FIGURE 13.4.2. The dashed graphs represent the standing waves at various values of time. The points in the interval (0, L), for which sin(nπ/L)x = 0, correspond to points on a standing wave where there is no motion. These points are called nodes. For example, in Figures 13.4.2(b) and (c) we see that the second standing wave has one node at L/2 and the third standing wave has two nodes at L/3 and 2L/3. In general, the nth normal mode of vibration has n – 1 nodes.

The frequency

of the first normal mode is called the fundamental frequency or first harmonic and is directly related to the pitch produced by a stringed instrument. It is apparent that the greater the tension on the string, the higher the pitch of the sound. The frequencies f_n of the other normal modes, which are integer multiples of the fundamental frequency, are called overtones. The second harmonic is the first overtone, and so on.

Superposition Principle

The superposition principle, Theorem 13.1.1, is the key in making the method of separation of variables an effective means of solving certain kinds of boundary-value problems involving linear partial differential equations. Sometimes a problem can also be solved by using a superposition of solutions of two easier problems. If we can solve each of the problems,

(12)

then a solution of (1)–(3) is given by u(x, t) = u₁(x, t) + u₂(x, t). To see this we know that u(x, t) = u₁(x, t) + u₂(x, t) is a solution of the homogeneous equation in (1) because of Theorem 13.1.1. Moreover, u(x, t) satisfies the boundary condition (2) and the initial conditions (3) because, in turn,

and

You are encouraged to try this method to obtain (8), (9), and (10). See Problems 5 and 14 in Exercises 13.4.

13.4 ExercisesAnswers to selected odd-numbered problems begin on page ANS-34.

In Problems 1–6, solve the wave equation (1) subject to the given conditions.

In Problems 7–10, a string is secured to the x-axis at x = 0 and at x = L and its initial displacement u(x, 0) = f(x), 0 < x < L, is shown in the figure. Find u(x, t) if the string is released from rest.

7. 

8. 

9. 

10. 

11. The longitudinal displacement of a vibrating elastic bar shown in FIGURE 13.4.7 satisfies the wave equation (1) and the conditions

    

    The boundary conditions at x = 0 and x = 0 are called free-end conditions. Find the displacement u(x, t).

    

12. A model for the motion of a vibrating string whose ends are allowed to slide on frictionless sleeves attached to the vertical axes x = 0 and x = 0 is given by the wave equation (1) and the conditions

    

    

    See FIGURE 13.4.8. The boundary conditions indicate that the motion is such that the slope of the curve is zero at its ends for t > 0. Find the displacement u(x, t).

    

13. In Problem 10, determine the value of u(L/2, t) for t ≥ 0.
14. Rederive the results given in (8), (9), and (10), but this time use the superposition principle discussed on page 731.
15. A string stretched along the x-axis and secured at and for is given an initial displacement and released. If the transverse vibrations take place in a medium that imparts a linear damping proportional to the instantaneous velocity, then the wave equation takes on the form

    

    where is a constant satisfying Find the displacement if the boundary and initial conditions are

    

16. Show that a solution of the boundary-value problem

    

    is

    

17. Consider the boundary-value problem given in (1) – (3) of this section. If g(x) = 0 on 0 < x < L, show that the solution of the problem can be written as

    u(x, t) = [f(x + at) + f(x – at)].

    [Hint: Use the identity

    2 sin θ₁ cos θ₂ = sin(θ₁ + θ₂) + sin(θ₁ – θ₂).]

18. The vertical displacement u(x, t) of an infinitely long string is determined from the initial-value problem

    (13)

    This problem can be solved without separating variables.

    1. Show that the wave equation can be put into the form ∂²u/∂η∂ξ = 0 by means of the substitutions ξ = x + at and η = x – at.
    2. Integrate the partial differential equation in part (a), first with respect to η and then with respect to ξ, to show that u(x, t) = F(x + at) + G(x – at), where F and G are arbitrary twice differentiable functions, is a solution of the wave equation. Use this solution and the given initial conditions to show that

       

       and

       where x₀ is arbitrary and c is a constant of integration.

    3. Use the results in part (b) to show that

       u(x, t) = [f(x + at) + f(x – at)] + (s) ds.(14)

       Note that when the initial velocity g(x) = 0 we obtain

       u(x, t) = [f(x + at) + f(x – at)], –∞ < x < ∞.

       The last solution can be interpreted as a superposition of two traveling waves, one moving to the right (that is, f(x – at)) and one moving to the left ( f(x + at)). Both waves travel with speed a and have the same basic shape as the initial displacement f(x). The form of u(x, t) given in (14) is called d’Alembert’s solution after the French mathematician, physicist, philosopher, and music theorist Jean-Baptiste le Rond d’Alembert (1717–1783).

In Problems 19–21, use d’Alembert’s solution (14) to solve the initial-value problem in Problem 18 subject to the given initial conditions.

19. f(x) = sin x, g(x) = 1
20. f(x) = sin x, g(x) = cos x
21. f(x) = 0, g(x) = sin 2x
22. Suppose f(x) = 1/(1 + x²), g(x) = 0, and a = 1 for the initial-value problem given in Problem 18. Graph d’Alembert’s solution in this case at the time t = 0, t = 1, and t = 3.
23. The transverse displacement u(x, t) of a vibrating beam of length L is determined from a fourth-order partial differential equation

    

    If the beam is simply supported, as shown in FIGURE 13.4.9, the boundary and initial conditions are

    

    Solve for u(x, t). [Hint: For convenience use λ = α⁴ when separating variables.]

    

Computer Lab Assignments

24. If the ends of the beam in Problem 23 are embedded at x = 0 and x = L, the boundary conditions become, for t > 0,

    

    

    1. Show that the eigenvalues of the problem are λ = /L² where x_n, n = 1, 2, 3, … , are the positive roots of the equation cosh x cos x = 1.
    2. Show graphically that the equation in part (a) has an infinite number of roots.
    3. Use a CAS to find approximations to the first four eigenvalues. Use four decimal places.
25. A model for an infinitely long string that is initially held at the three points (–1, 0), (1, 0), and (0, 1) and then simultaneously released at all three points at time t = 0 is given by (13) with

    

    1. Plot the initial position of the string on the interval [–6, 6].
    2. Use a CAS to plot d’Alembert’s solution (14) on [–6, 6] for t = 0.2k, k = 0, 1, 2, … , 25. Assume that a = 1.
    3. Use the animation feature of your computer algebra system to make a movie of the solution. Describe the motion of the string over time.
26. An infinitely long string coinciding with the x-axis is struck at the origin with a hammer whose head is 0.2 inch in diameter. A model for the motion of the string is given by (13) with

    

    1. Use a CAS to plot d’Alembert’s solution (14) on [–6, 6] for t = 0.2k, k = 0, 1, 2, … , 25. Assume that a = 1.
    2. Use the animation feature of your computer algebra system to make a movie of the solution. Describe the motion of the string over time.
27. The model of the vibrating string in Problem 7 is a plucked string.
    1. Use a CAS to plot the partial sum S₆(x, t); that is, the first six nonzero terms of your solution u(x, t), for t = 0.1k, k = 0, 1, 2, … , 20. Assume that a = 1, h = 1, and L = π.
    2. Use the animation feature of your computer algebra system to make a movie of the solution to Problem 7.
28. Consider the vibrating string in Problem 10. Use a CAS to plot the partial sum S₆(x, t); that is, the first six nonzero terms of your solution u(x, t) for t = 0.25k, k = 0, 2, 3, 4, 6, 8, 10, 14. Assume that a = 1, h = 1, and L = π. Then superimpose the eight graphs on the same coordinate system.