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

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400 'V e-(2n-1) 2n2a2t/400 (2n-1) nx

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

400

u(x,t) = ---------- Ó

rr

n 2n—1 2 0

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 — 1 e-9n2 (.86) 30/400 }

, n 3

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

— e-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' - o1a 2T = 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 ^ ro we must have G1 < 0. Thus, setting o1 = -X2, we have T' + a2X2T = 0. For X" - a2X = 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 = -ö , then X + ö 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):

(7 0cosnn+5 0)/nn, and thus

u(x,t) = 3x + ^

70cosnn + 50

-0.86n2n2t/400-

nnx

e

since

20

n=1

a2 = .86 for aluminum.

9b.

9c.

M u

9d. Using just the first term of

of the sum, we have n

_n— = 15 ± .15. Thus

4

222

Section 10.6

---e °.86n2t/4°°sinn = .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 — ro we see that all terms in the series decay to zero except the constant term, 2/n. Hence 2/n.

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