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

ISBN 0-471-31999-6

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X(x) = sin(nnx/2), n = 1,2,... . Finally, the solution of the

D.E. for T yields T(t) = exp(-X2t/4) = exp(-n2n2t/16). Thus we have found un(x,t) = exp(-n2n2t/16)sin(nnx/2). Setting t = 0 in this last expression indicates that un(x,0) has, for the correct choices of n, the same form as the terms in u(x,0), the initial condition. Using the principle of superposition we know that

u(x,t) = c1u1(x,t) + c2u2(x,t) + c4u4(x,t) satisfies the

P.D.E. and the B.C. and hence we let t = 0 to obtain u(x,0) = c1u1(x,0) + c2u2(x,0) + c4u4(x,0) =

c1sinnx/2 + c2sinnx + c4sin2nx. If we choose c1 = 2, c2 = -1 and c4 = 4 then u(x,0) here will match the given initial condition, and hence substituting these values in u(x,t) above then gives the desired solution.

10. Since the B.C. for this heat conduction problem are

u(0,t) = u(40,t) = 0, t > 0, the solution u(x,t) is given

by Eq.(19) with a2 = 1 cm2/sec, L = 40 cm, and the coefficients cn determined by Eq.(21) with the I.C. u(x,0) = f(x) = x, 0 < x < 20; = 40 - x, 20 < x < 40.

1 f 20 nnx f 40 nnx

Thus cn = [ xsin-----dx + (40-x)sin----dx]

20 0 40 20 40

n2n2 2

160 V-1

u(x,t) = V

16 0 nn

?sin. It follows that

= i60V sin(nn/2) e-n2n2t/1600sinnnx n2 ^ n2 40

n=1

Section 10.5

219

15a.

15b.

15c.

15d. As in Example 1, the maximum temperature will be at the midpoint, x = 20, and we use just the first term, since the others will be negligible for this temperature, since t is so large. Thus

u(20,t) = 1

160

- sin(n/2)e'

-n2t/1600

sin(20n/40). Solving

for t, we obtain e 1600 160

t =

2

ln

2

?n2t/l600 = n 2/160, or = 451.60 sec.

18a. Since the B.C. for this heat conduction problem are

u(0,t) = u(20,t) = 0, t > 0, the solution u(x,t) is given by Eq.(19) with L = 20 cm, and the coefficients cn determined by Eq.(21) with the I.C. u(x,0) = f(x) = 100oC. Thus cn = (1/10) 120 (10 0)sin(nnx/2 0)dx = -200 [(-1)n-1]/nn and hence

220

Section 10.5

c2n = 0 and c2n_i = 400/(2n-1)n. Substituting these values into Eq.(19) yields

400 'V e"(2n_1) 2n22t/400 (2n-1)nx

----------------sin-----------

400

u(x,t) = ---------- >

rr

n 2n1 20

n=1

18b. For aluminum, we have a2 = .86 cm2/sec (from Table 10.5.1) and thus the first two terms give

U(10,30) = 400 {e-n2 (.86) 30/400 - 1e-9rc2 (.86) 30/400 j

, n 3

= 67.228oC. If an additional term is used, the temperature is increased by

80e-25n2 (.86) 30/400 = 3 x 10-6 degrees C. n

19b. Using only one term in the series for u(x,t), we must solve the equation 5 = (400/n)exp[-n2(.86)t/400] for t. Taking the logarithm of both sides and solving for t yields t = 400ln(80/n)/n2(.86) = 152.56 sec.

20. Applying the chain rule to partial differentiation of u with respect to x we see that ux = u^x = u^(l/L) and uxx = u^(l/L)2. Substituting u^/L2 for uxx in the heat

equation gives a 2u^/L2 = ut or u^ = (L2/a 2)ut. In a

L2

2 2 L similar manner, ut = uTTt = uT(a /L ) and hence -ut = uT

a2

and thus u^ = uT.

22. Substituting u(x,y,t) = X(x)Y(y)T(t) in the P.D.E. yields a2 (X"YT + XY"T) = XYT', which is equivalent to X" Y" t'

+ = ----------. By keeping the independent variables x

X Y a 2T

and y fixed and varying t we see that T'/a2T must equal some constant o1 since the left side of the equation is fixed. Hence, X"/X + y"/y = T'/a 2T = o1, or X"/X = o1 - Y"/Y and T' - o1a2T = 0. By keeping x fixed and varying y in the equation involving X and Y we see that o1 - Y"/Y must equal some constant o2 since the left side of the equation is fixed. Hence,

X"/X = o1 - Y"/Y = o2 so X" - o2X = 0 and Y" -(o1 - o2)Y = 0. For T' - o1a2T = 0 to have solutions that remain bounded as t ^ ^ we must have G1 < 0. Thus, setting o1 = -^2, we have T' + a2^2T = 0. For X" - o2X = 0 and homogeneous B.C., we conclude, as in Sect. 10.1, that o2 < 0 and, if we let

Section 10.6

221

2 //2

02 = , then X + ^ X = 0. With these choices for a1 and 02

we then have Y" + (X 2-^2)Y = 0.

Section 10.6, Page 588

3. The steady-state temperature distribution v(x) must satisfy Eq.(9) and also satsify the B.C. vx(0) = 0, v(L) = 0. The general solution of v" = 0 is v(x) = Ax + B. The B.C. vx(0) = 0 implies A = 0 and then

v(L) = 0 implies B = 0, so the steady state solution is

v(x) = 0.

7. Again, v(x) must satisfy v" = 0, v'(0) -v(0) = 0 and v(L) = T.

The general solution of v" = 0 is v(x) = ax + b, so v(0) = b,

v'(0) = a and v(L) = T. Thus a - b = 0 and aL + b = T, which give a = b = T/(1+L). Hence v(x) = T(x+1)/(L+1).

9a. Since the B.C. are not homogeneous, we must first find

the steady state solution. Using Eqs.(9) and (10) we have v" = 0 with v(0) = 0 and v(20) = 60, which has the

solution v(x) = 3x. Thus the transient solution w(x,t) satisfies the equations a 2wxx = wt, w(0,t) = 0, w(20,t) = 0 and w(x,0) = 25 - 3x, which are obtained from Eqs.(13) to (15). The solution of this problem is given by Eq.(4) with the cn given by Eq.(6):

(70cosnn+50)/nn, and thus

70cosnn + 50

e

?0.86n2n2t/400si

nnx

20

since

n=1

a2 = .86 for aluminum.

9b.

9c.

M u

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