# Elementary differential equations 7th edition - Boyce W.E

ISBN 0-471-31999-6

**Download**(direct link)

**:**

**258**> 259 260 261 262 263 264 .. 486 >> Next

u(x, t) = X(x)T(t) (10)

and substituting for u in Eq. (1), we obtain

— = ~2 — = — (11)

X a2 T ’ V ;

where k is a separation constant. Thus we find that X(x) and T (t) satisfy the ordinary differential equations

X" + = 0, (12)

T" + a2kT = 0. (13)

Further, by substituting from Eq. (10) for u(x, t) in the boundary conditions (3) we find that X(x) must satisfy the boundary conditions

X (0) = 0, X (L ) = 0. (14)

Finally, by substituting from Eq. (10) into the second of the initial conditions (9), we also find that T (t) must satisfy the initial condition

T'(0) = 0. (15)

Our next task is to determine X(x), T(t), and k by solving Eq. (12) subject to the boundary conditions (14) and Eq. (13) subject to the initial condition (15).

The problem of solving the differential equation (12) subject to the boundary conditions (14) is precisely the same problem that arose in Section 10.5 in connection with

the heat conduction equation. Thus we can use the results obtained there and at the end of Section 10.1: The problem (12) and (14) has nontrivial solutions if and only if k is an eigenvalue,

k = n2n2/L2, n = 1, 2,..., (16)

and X(x) is proportional to the corresponding eigenfunction sin(nnx/L ).

Using the values of k given by Eq. (16) in Eq. (13), we obtain

2 2 2

T" + n n2a T = 0. (17)

L2

594

Chapter 10. Partial Differential Equations and Fourier Series

Therefore

nn at nn at

T(t) = k, cos + k2sin----------------, (18)

1 L 2 L

where k1 and k2 are arbitrary constants. The initial condition (15) requires that k2 = 0, so T(t) must be proportional to cos(nnat/L).

Thus the functions

. nn x nn at

un (x, t ) = sin ——— cos ———, n = 1, 2,..., (19)

satisfy the partial differential equation (1), the boundary conditions (3), and the second initial condition (9). These functions are the fundamental solutions for the given problem.

To satisfy the remaining (nonhomogeneous) initial condition (9) we will consider a superposition of the fundamental solutions (19) with properly chosen coefficients. Thus we assume that u(x, t) has the form

CO CO -

nnx nn at

u(x, t) = 2^ cnun(x, t) = 2^ cn sin l c°s——, (20)

n=1 n=1

where the constants cn remain to be chosen. The initial condition u(x, 0) = f (x) requires that

00

nn x

u(x, 0) = ^] cn sin l = f (x). (21)

n = l

Consequently, the coefficients cn must be the coefficients in the Fourier sine series of period 2L for f; hence

2 fL „ . nnx , , N

c = — I f (x) sin dx, n = 1, 2,.... (22)

n L Jo L ’ ’ ’ v ;

Thus the formal solution of the problem of Eqs. (1), (3), and (9) is given by Eq. (20) with the coefficients calculated from Eq. (22).

For afixed value of n the expression sin(nnx/L) cos(nnat/L) inEq. (19) is periodic in time t with the period 2L /na; it therefore represents a vibratory motion of the string having this period, or having the frequency nn a/L. The quantities Xa = nn a/L for n = 1, 2,... are the natural frequencies of the string, that is, the frequencies at which the string will freely vibrate. The factor sin (nn x / L) represents the displacement pattern occurring in the string when it is executing vibrations of the given frequency. Each displacement pattern is called a natural mode of vibration and is periodic in the space variable x; the spatial period 2L /n is called the wavelength of the mode of frequency nn a/L. Thus the eigenvalues n2n2/L2 of the problem (12), (14) are proportional to the squares of the natural frequencies, and the eigenfunctions sin (nn x / L) give the natural modes. The first three natural modes are sketched in Figure 10.7.3. The total motion of the string, given by the function u (x, t) of Eq. (20), is thus a combination of the natural modes of vibration, and is also a periodic function of time with period 2L /a.

10.7 The Wave Equation: Vibrations of an Elastic String

595

EXAMPLE

1

FIGURE 10.7.3 First three fundamental modes of vibration of an elastic string. (a) Frequency = na/L, wavelength = 2L; (b) frequency = 2na/L, wavelength = L; (c) frequency = 3na/L, wavelength = 2 L /3.

Consider a vibrating string of length L = 30 that satisfies the wave equation

0 < x < 30, t > 0

4uxx = utt,

(23)

Assume that the ends of the string are fixed, and that the string is set in motion with no initial velocity from the initial position

u(x, 0) = f (x) =

x/10,

(30 - x)/20,

0 < x < 10, 10 < x < 30.

(24)

Find the displacement u(x, t) of the string and describe its motion through one period. The solution is given by Eq. (20) with a = 2 and L = 30, that is,

(x, t) =

n=1

c sin-----------

n 30

nn x 2nn t

cos

30

(25)

where cn is calculated from Eq. (22). Substituting from Eq. (24) into Eq. (22), we obtain

2 f10 x . nnx 2 f

— I — sin dx -I I

30 Jq 10 30 30 A,

30 3 0 — x . nnx

sin dx.

10 20 30

By evaluating the integrals in Eq. (26), we find that

9

nn

2l2

n = 1, 2,....

(26)

(27)

The solution (25), (27) gives the displacement of the string at any point x at any time t. The motion is periodic in time with period 30, so it is sufficient to analyze the solution for 0 < t < 30.

**258**> 259 260 261 262 263 264 .. 486 >> Next