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

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bn = (2/an ah(x)sin(nnx/a)dx and u(x,b) = g(x) gives 0

ansinh(nnb/a) + bncosh(nnb/a) = (2/anag(x)sin(nnx/a)dx, which

0

can be solved for an since bn is known.

5. Using Eq.(20) and following the same arguments as

presented in the text, we find that R(r) = k^n + k2r-n and 0(9) = c^inn9 + c2cosn9, for n a positive integer, and u0(r,9) = 1 for n = 0. Since we require that u(r,9) be bounded as r ^ ^, we conclude that k1 = 0. The fundamental solutions are therefore un(r,9) = r-ncosn9, vn(r,9) = r-nsinn9, n = 1,2,... together with u0(r,9) = 1. Assuming that u can be expressed as a linear combination of the fundamental solutions we have

u(r,9) = c0/2 + r-n(cncosn9 + knsinn9). The B.C.

n=1

requires that

u(a,9) = c0/2 + a-n(cncosn9 + knsinn9) = f(9) for

n=1

0 < 9 < 2ï. This is precisely the Fourier series representation for f(9) of period 2ï and thus

a-ncn = (1/ï) |2f(9)cosn9d9, n = 0,1,2,... and 0

a-nkn = (1/ï) |2f(9)sinn9d9, n = 1,2... .

0

7. Again we let u(r,9) = R(r)0(9) and thus we have

r2R" + rR' - OR = 0 and 0" + O0 = 0, with R(0) bounded and

the B.C. 0(0) = 0(a) = 0. For î < 0 we find that

0(0) = 0, so we let î = X2 (X2 real) and thus

0(9) = cicosX9 + c2sinX9. The B.C. 0(0) = 0 ^ ci = 0 and

the B.C. 0(a) = 0 ^ X = nn/a, n = 1,2,... .

Substituting these values into Eq.(31) we obtain R(r) = k1rnn/a + k2r-nn/a. However k2 = 0 since R(0) must be bounded, and thus the fundamental solutions are un(r,9) = rnn/asin(nn9/a). The desired solution may now be formed using previously discussed procedures.

Section 10.8

223

8a. Separating variables, as before, we obtain

X" + X2X = 0, X(0) = 0, X(a) = 0 and Y" _ X2Y = 0, Y(y) bounded

as y ^ ^. Thus X(x) = sin(nïx/a), and X2 = ^ï/a) 2.

However, since neither sinhy nor coshy are bounded as y ^ ^, we must write the solution to Y" _ ^ï/a)2Y = 0 as

Y(y) = c1exp[nïy/a] + c2exp[_nïy/a]. Thus we must choose

c1 = 0 so that u(x,y) = X(x)Y(y) ^ 0 as y ^ ro. The

fundamental solutions are then un(x,t) = e_nÏY/asin(nïx/a).

u(x,y) = cnun(x,y) then gives

n=1

V"1 2 ("1 nïx

u(x,0) = Ó cnsin(nïx/a) = f(x) so that cn = — f(x)sin--------------dx.

^ a J0 a

n=1

2

2 fa nïx 4 a

8b. cn = — x(a-x)sin-dx = ——- (1-cosn^

a^0 a ï3ï3

8c. Using just the first term and letting a = 5, we have

2 0 0 _ ïó/5 ïx

u(x,y) = --------e Y/5 sin--, which, for a fixed y, has a maximum

ï3 5

at x = 5/2 and thus we need to find y such that

2 0 0 _ ïÓ/5

u(5/2,y) = ——e ïó/5 = .1. Taking the logarithm of both

ï3

sides and solving for y yields y0 = 6.6315. With an equation solver, more terms can be used. However, to four decimal places, three terms yield the same result as above.

13a. Assuming that u(x,y) = X(x)Y(y) and substituting into

Eq.(1) leads to the two O.D.E. X" _ OX = 0, Y" + OY = 0.

The B.C. u(x,0) = 0, uy(x,b) = 0 imply that Y(0) = 0 and Y'(b) = 0. For nontrivial solutions to exist for Y" + OY = 0 with these B.C. we find that O must take the values (2n_1)2ï2/4b2, n = 1,2,...; the corresponding solutions for Y(y) are proportional to si^^n^^y^^. Solutions to X" _ [(2n_1)2ï2/4b2]X = 0 are of the form

X(x) = Asinh[(2n_1)ïx/2b] + Bcosh[(2n_1)ïx/2b]. However,

the boundary condition u(0,y) = 0 implies that X(0) = B = 0. It follows that the fundamental solutions are un(x,y) = cïsinh[(2n_1)ïx/2b]sin[(2n_1)ïy/2b], n = 1,2,... . To satisfy the remaining B.C. at x = a we

assume that we can represent the solution u(x,y) in the

form u(x,y) = cïsinh[(2n_1)ïx/2b]sin[(2n_1)ïy/2b].

n=1

The coefficients cn are determined by the B.C.

234

Section 10.2

u(a,y) = I cnsinh[(2n-1)na/2b]sin[(2n-1)ny/2b] = f(y).

n=1

By properly extending f as a periodic function of period 4b as in Problem 39, Section 10.4, we find that the coefficients cn are given by

cnsinh[(2n-1)na/2b] = (2/b)[bf(y)sin[(2n-1)ny/2b]dy,

J0

n = 1,2,... .

235

CHAPTER 11

Section 11.1, Page 626

2. Since the B.C. at x = 1 is nonhomogeneous, the B.V.P. is nonhomogeneous.

4. The D.E. may be written y" + (X-x2)y = 0 and is thus homogeneous, as are both B.C.

5. Since the D.E. contains the nonhomogeneous term 1, the B.V.P. is nonhomogeneous.

9. If X = 0, then y(x) = cix + c2 and thus y(0) = c2,

y(1) = c1 + c2, y'(0) = c1 and y'(1) = c1. Hence the B.C. yield the two equations c2 - c1 = 0 and c1 + c2 + c1 = 0

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