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

ISBN 0-471-31999-6

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9d. Using just the first term of

of the sum, we have

0sin — = 15 ± .15. Thus

4

u(5,t) = 15 - l°e-0-86n2t/400si

222

Section 10.6

e °.86n2t/4°°sin— = .15, which yields t = 160.30 sec.

n

4

To obtain the answer in the text, the first two terms of the sum must be used, which requires an equation solver to solve for t. Note that this reduces t by only .01 seconds.

12a. Since the B.C. are ux(0,t) = ux(L,t) = 0, t > 0, the

solution u(x,t) is given by Eq.(35) with the coefficients cn determined by Eq.(37). Substituting the I.C. u(x,0) = f(x) = sin(nx/L) into Eq.(37) yields

c0 = (2/L) cn = (2/L)

= (1/L)

Jsin(nx/L)dx = 4/n and Jsin(nx/L)cos(nnx/L)dx

{sin[(n+1)nx/L] - sin[(n-1)nx/L]}dx

(1/n){[1 - cos(n+1) n]/(n+1) - [1 - cos(n-1) n]/(n-1)}

0, n odd; = -4/(n2-1) n, n even. Thus

u(x,t)

2/n-(4/n) exp[-4n2n2a2t/L2]cos(2nnx/L)/(4n2-1)

n=1

where we are now summing over even terms by setting n = 2n.

12b. As t ^ ^ we see that all terms in the series decay to zero except the constant term, 2/n. Hence 2/n.

lim u(x,t)

t——^

12c.

12d. The original heat in the rod is redistributed to give the final temperature distribution, since no heat is lost.

Section 10.6

223

14a. Since the ends are insulated, the solution to this

problem is given by Eq.(35), with a2 = 1 and L = 30, and

c0 ^ 2 2

Eq.(37). Thus u(x,t) = — + V cnexp(-n2n2t/900)cos(nnx/3 0),

n=1

where

2 f30 1 (*10 25

c0 = -------- f(x)dx =------ 25dx = ------ and

30 0 15 5 3

2 f3 nnx 1 f10 nnx 5 0 nn nn

cn = ------- f(x)cos----dx = ------------------- 25cos-------------dx = - [sin- - sin-].

3 0 J0 3 0 15 J5 3 0 nn 3 6

14b.

steady state

3

0 * ® J n 80 too

Although x = 4 and x = 1 are symmetrical to the initial temperature pulse, they are not symmetrical to the insulated end points.

15a. Substituting u(x,t) = X(x)T(t) into Eq.(1) leads to the two O.D.E. X" - OX = 0 and T' - a2OT = 0. An argument similar to the one in the text implies that we must have X(0) = 0 and X'(L) = 0. Also, by assuming O is real and

considering the three cases O < 0, O = 0, and O > 0 we can show that only the case O < 0 leads to nontrivial solutions of X" - OX = 0 with X(0) = 0 and X'(L) = 0.

Setting O = -X2, we obtain X(x) = k1sinXx + k2cosXx.

Now, X(0) = 0 ^ k2 = 0 and thus X(x) = k1sinXx.

224

Section 10.7

Differentiating and setting x = L yields Xk^osXL = 0. Since X = 0 and k1 = 0 lead to u(x,t) = 0, we must choose X so that cosXL = 0, or X = (2n-1)n/2L, n = 1,2,3,... . These values for X imply that o = -(2n-1)2n2/4L2 so the solutions T(t) of T' - a2oT = 0 are proportional to exp[-(2n-1)2n2a2t/4L2]. Combining the above results leads to the desired set of fundamental solutions.

15b. In order to satisfy the I.C. u(x,0) = f(x) we assume that u(x,t) has the form

u(x,t) = cnexp[-(2n-1) 2n 2a 2t/4L2]sin[(2n-1) nx/2L]. The

n=1

coefficients cn are determined by the requirement that u(x,0) = cnsin[(2n-1)nx/2L] = f(x). Referring to

n=1

Problem 39 of Section 10.4 reveals that such a representation for f(x) is possible if we choose the

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

J0

19. We must solve v{(x) = 0, 0 < x < a and v2(x) = 0,

a < x < L subject to the B.C. v^0) = 0, v2(L) = T and the continuity conditions at x = a. For the temperature to be continuous at x = a we must have v^a) = v2(a) and for the rate of heat flow to be continuous we must have -K 1A1v/1(a) = -K 2A2v'2(a), from Eq.(2) of Appendix A. The general solutions to the two O.D.E. are v^x) = ^x + D}_ and v2 (x) = C2x + D2. By applying the boundary and continuity conditions we may solve for C1, D1, and C2 and D2 to obtain the desired solution.

Section 10.7, Page 600

1a. Since the initial velocity is zero, the solution is given by Eq.(20) with the coefficients cn given by Eq.(22). Substituting f(x) into Eq.(22) yields

2 fl/2 2x nnx |"l 2(L-x) nnx

= — [ —sin dx + sin-----dx]

L 0 L L L/2 L L

|"l/2 2x . nnx fi

[ —sin------dx +

J0 l l Jl

8 nn

sin—. Thus Eq. (20) becomes

n2n2 2

8 V-1 1 nn ]

u(x,t) = — > — sin—sin-222

8 V 1 nn nnx nnat

____ -cos---------

rc2^n2 2 L L

n=1

c

n

Section 10.7

225

1b.

1e. The graphs in part b can best be understood using Eq.(28) (or the results of Problems 13 and 14). The original triangular shape is composed of two similar triangles of 1/2 the height, one of which moves to the right, h(x-at), and the other to the left, h(x+at). Recalling that the series are periodic then gives the results shown. The graphs in part c can then be visualized from those in part b.

6a. The motion is governed by Eqs.(1), (3) and (31), and thus

the solution is given by Eq.(34) where the kn are given by Eq.(36):

2 Pl/4 4x nnx f3L/4 nnx Cl 4(L-x) nnx

kn = ------ [ —sin dx + sin-dx +------ sin---dx]

nna J0 L L Jl/4 l J3L L L

8L nn 3nn

= ------- (sin--- + sin-----). Substituting this in Eq.(34)

n3n3a 4 4

nn 3nn

^ sin + sin---------

8L ^ 4 4 nnx nnat

yields u(x,t) = ------- 7 ---------------------sin----sin-------.

an3 ^ n3 L L

n = 1

226

Section 10.7

6c.

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