# Elementary Differential Equations and Boundary Value Problems - Boyce W.E.

**Download**(direct link)

**:**

**497**> 498 499 500 501 502 503 .. 609 >> Next

an3 n3 L L

n = 1

226

Section 10.7

6c.

u(2.5,t)

u(5,t)

9. Assuming that u(x,t) = X(x)T(t) and substituting for u in Eq.(1) leads to the pair of O.D.E. X" + GX = 0,

T" + a2GT = 0. Applying the B.C. u(0,t) = 0 and

ux(L,t) = 0 to u(x,t) we see that we must have X(0) = 0

and X'(L) = 0. By considering the three cases G < 0,

G = 0, and G > 0 it can be shown that nontrivial solutions of the problem X" + GX = 0, X(0) = 0, X'(L) = 0 are possible if

and only if G = (2n-1) 2n 2/4L2, n = 1,2,... and the

corresponding solutions for X(x) are proportional to sin[(2n-1)nx/2L]. Using these values for G we find that T(t) is a linear combination of sin[(2n-1)nat/2L] and cos[(2n-1)nat/2L]. Now, the I.C. ut(x,0) implies that T (0) = 0 and thus functions of the form

un(x,t) = sin[(2n-1)nx/2L]cos[(2n-1)nat/2L], n = 1,2,... satsify the P.D.E. (1), the B.C. u(0,t) = 0, ux(L,t) = 0,

and the I.C. ut(x,0) = 0. We now seek a superposition of

these fundamental solutions un that also satisfies the

I.C. u(x,0) = f(x). Thus we assume that

u(x,t) = cnsin[(2n-1)nx/2L]cos[(2n-1)nat/2L]. The

n=1

I.C. now implies that we must have

f(x) = Ó cnsin[(2n-1)nx/2L]. From Problem 39 of Section

n=1

10.4 we see that f(x) can be represented by such a series and that

cn = (2/L) [Lf(x)sin[(2n-1) nx/2L]dx, n = 1,2,... .

0

Substituting these values into the above series for u(x,t) yields the desired solution.

10a. From Problem 9 we have

-4

[cos(-------------------) - cos(

4L

(2n-1)n(L+2)

(2n-1)Ï(L-2) )] 4L

(2n-1)Ï _ (2n-1)Ï

(2n-1)Ï

sin

using the

4

2L

Section 10.7

227

trigonometric relations for cos(A ± B). Substituting this value of cn into u(x,t) in Problem 9 yields the desired solution.

10b.

10c.

13. Using the chain rule we obtain ux = u^x + u^Hx =

u? + u^ since ?x = = 1. Differentiating a second time

gives uxx = u?? + 2u^n + . In a similar way we obtain

ut = u^t + unnt = -au? + aun, since ?t = -a, nt = a. Thus

utt = a2 (u?? - 2u^n + ^nn). Substituting for uxx and utt

in the wave equation, we obtain u^ = 0. Integrating both sides of u^n = 0 with respect to n yields u?(?,n) = Y(?) where Y is an arbitrary function of ?.

Integrating both sides of u?(?,n) = Y(?) with respect to ?

yields u(?,n) = JY(?)d? + y(n) = ô(?) + V(n) where y(n) is an arbitrary function of n and JY(?)d? is some function of ? denoted by ô(?). Thus u(x,t) = u(? (x,t),n (x,t)) = ô (x - at) + y(x + at).

14. The graph of y = sin(x-at) for the various values of t is

indicated in the figure on the next page. Note that the

graph of y = sinx is displaced to the right by the

distance "at" for each value of t.

228

Section 10.7

Similarly, the graph of y = ô(x + at) would be displaced to the left by a distance "at" for each t. Thus ô(x + at) represents a wave moving to the left.

16. Write the equation as a2uxx = utt + a 2u and assume u(x,t) = X(x)T(t). This gives a2X"T = XT" + a 2XT,

X" 1 T" 2

or — = — (— + a2) = a. The separation constant a is X a2 T

—X2 using the same arguments as in the text and earlier problems. Thus X" + X 2X = 0, X(0) = 0, X(L) = 0 and T" + (a2 + z2X2)T = 0, T'(0) = 0. If we let pn = X^a2+a2,

nnx nnx

we then have un(x,t) = cospntsin-, where X n = -.

LL

nnx

oo

Using superposition we obtain u(x,t) = cncospntsin

n = 1 J_1

^ nnx

and thus u(x,0) = Ó cnsin---- = f(x). Hence cn are given

L

n = 1

by Eq. (22).

17a. We have u(x,t) = ô^-at) + y(x+at) and thus ut(x,t) = -aô/(x-at) + a%'(x+at). Hence u(x,0) = ô^) + y(x) = f(x) and

ut(x,0) = -aô/(x) + ay'(x) = 0. Dividing the last

equation by a yields the desired result.

17b. Using the hint and the first equation obtained in part (a) leads to ô^) + y(x) = 2ô^) + c = f(x) so ô^) = (1/2)f(x) - c/2 and y(x) = (1/2)f(x) + c/2. Hence

u(x,t) = ô (x - at) + y(x + at) = (1/2)[f(x - at) - c] + (1/2)[f(x + at) + c] = (1/2)[f(x - at) + f(x + at)].

f(x + at)

17c. Substituting x + at for x in f(x) yields

I 2 -1 < x + at < 1

[0 otherwise

Subtracting "at" from both sides of the inequality then yields

**497**> 498 499 500 501 502 503 .. 609 >> Next