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

Boyce W.E. Elementary Differential Equations and Boundary Value Problems - John Wiley & Sons, 2001. - 1310 p.
Download (direct link): elementarydifferentialequations2001.pdf
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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 < 2. This is precisely the Fourier series representation for f(9) of period 2 and thus
a-ncn = (1/) |2f(9)cosn9d9, n = 0,1,2,... and 0
a-nkn = (1/) |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 < 0 we find that
0(0) = 0, so we let = 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(nx/a), and X2 = ^/a) 2.
However, since neither sinhy nor coshy are bounded as y ^ ^, we must write the solution to Y" _ ^/a)2Y = 0 as
Y(y) = c1exp[ny/a] + c2exp[_ny/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_nY/asin(nx/a).
u(x,y) = cnun(x,y) then gives
n=1
V"1 2 ("1 nx
u(x,0) = cnsin(nx/a) = f(x) so that cn = f(x)sin--------------dx.
^ a J0 a
n=1
2
2 fa nx 4 a
8b. cn = x(a-x)sin-dx = - (1-cosn^
a^0 a 33
8c. Using just the first term and letting a = 5, we have
2 0 0 _ /5 x
u(x,y) = --------e Y/5 sin--, which, for a fixed y, has a maximum
3 5
at x = 5/2 and thus we need to find y such that
2 0 0 _ /5
u(5/2,y) = e /5 = .1. Taking the logarithm of both
3
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)22/4b2, n = 1,2,...; the corresponding solutions for Y(y) are proportional to si^^n^^y^^. Solutions to X" _ [(2n_1)22/4b2]X = 0 are of the form
X(x) = Asinh[(2n_1)x/2b] + Bcosh[(2n_1)x/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) = csinh[(2n_1)x/2b]sin[(2n_1)y/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) = csinh[(2n_1)x/2b]sin[(2n_1)y/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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