Books
in black and white
Main menu
Share a book About us Home
Books
Biology Business Chemistry Computers Culture Economics Fiction Games Guide History Management Mathematical Medicine Mental Fitnes Physics Psychology Scince Sport Technics
Ads

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
Previous << 1 .. 498 499 500 501 502 503 < 504 > 505 506 507 508 509 510 .. 609 >> Next

Multiplying and integrating as suggested yields
J1[/ c (x)]r(x^ (x)dx = 0 or 0 nTn m
n=1
I cnt1 r(x).(x^m(x)dx = 0. The integral that multiplies c.
n=1
is just 5 [Eq.(22) of Section 11.2]. Thus the infinite sum
nm
becomes c and the last equation yields c = 0.
mm
A twice differentiable function v satisfying the boundary conditions can be found by assuming that v = ax + b.
Thus v(0) = b = 1 and v(1) = a + 1 while v'(1) = a.
Hence 2a + 1 = -2 or a = -3/2 and v(x) = 1 - 3x/2.
Section 11.3
245
19.
22.
Assuming y = u + v we have (u+v)" + 2(u+v) = u" + 2u + 2(1-3x/2) = 2 - 4x or u" + 2u = -x, u(0) = 0, u(1) + u'(1) = 0 which is the same as Example 1 of the text.
From Eq.(30) we assume u(x,t) = bn(t^ n(x), where the
n=1
n are the eigeefuectioes of the related eigenvalue problem y" + Xy = 0, y(0) = 0, y'(1) = 0 and the bn(t) are given by Eq.(42). From Problem 2, we have = V~2sin[(2n-1) nx/2] and X = (2n-1) 2n 2/4. To
nn
evaluate Eq.(42) we need to calculate
a = 11sin(nx/2) "\/~2sin[(2n-1)TCx/2dx [Eq.(41) with n0
r(x) = 1 and f(x) = sin(nx/2)], which is zero except for n = 1 in which case a1 = /2, and
Yn = J1(-x) "\/~2sin[(2n-1)TCx/2]dx [Eq.(35) with
F(x,t) = -x]. This integral is the negative of the cn in Problem 2 and thus
Yn = -4^/2 (-1)n+1/(2n-1)2n2 = -cn, n = 1,2,... . Setting
Y = -c in Eq.(42) we then have
nn
b = ^e-*2t/4 - c fVn2(t-s)/4ds
0
1 2 ^0
\/~2 4ci
V 2 e~n 2t/4 - ________1e_n 2(t-s)/4
2
\2 4c 4c
^e'n 2t/4 - + 1 e-n 2t/4 and
2
ft -X (t-s) -X (t-s)
similarly b = -c e n ds = -(c /X )e n n 0 n n
0
= -(c /X ) (1-e n ), where X = (2n-1)2n2/4,
n n n
n = 2,3,... . Substituting these values for bn along
with n = "\/"2sin[(2n-1)TCx/2] into the series for u(x,t) yields the solution to the given problem.
In this case a = 0 for all n and Y is given by = |1 " "
n
1
Yn = Je b(1-x) V~2sin[(2n-1)nx/2]dx
= e_ti 1(1-x) '\/~2sin[(2n-1) nx/2]dx. This last integral
0
can be written as the sum of two integrals, each of which has been evaluated in either Problem 6 or 7 of Section
t
0
t
246
Section 11.3
11.2. Letting c denote the value obtained, we then have
n
*t -X (t-s)
' n n n nJo
-t ft -X (t-s)
Y = c e t and thus b = c e n e sds =
n n n nJo
-X tft (X -1)s -X t
c e n e n ds = [c /(X -1)](e t-e n ), where n Jo n n
X. = (2n-1) 2n 2/4. Substituting these values into Eq.(30) yields the desired solution.
24. Using the approach of Problem 23 we find that v(x) satisfies v" = 2, v(0) = 1, v(1) = 0. Thus v(x) = x2 + c1x + c2 and the B.C. yield v(0) = c2 = 1 and
v(1) = 1 + c1 + 1 = 0 or c1 = -2. Hence
v(x) = x2 - 2x + 1 and w(x,t) = u(x,t) - v(x) where, from
Problem 23, we have w = w , w(0,t) = 0, w(1,t) = 0 and
t xx
w(x,0) = x2 - 2x + 2 - v(x) = 1. This last problem can be solved by methods of this section or by methods of Chapter 10. Using the approach of this section we have
w(x,t) = > b (t)6 (x) where (x) = V2 sinnnx
n n n
n=1
[which are the normalized eigenfunctions of the associated eigenvalue problem y" + Xy = 0, y(0) = 0, y(1) = 0] and the b. are given by Eq.(42). Since the
P.D.E. for w(x,t) is homogeneous Eq.(42) reduces to
b = a e n (X = n2n2 from the above eigenvalue problem),
n n n
where
1
=
n
111^V~2sinnnxdx = \p2 [1-(-1)n]/nn. Thus
0
^ / n
u(x,t) = x2-2x + 1 + ------[--(--)] e-n2n2^V"2sinnnx,
^ nn
n=1
which simplifies to the desired solution.
28a. Since yc = ci+c2x, we assume that
Y(x) = u1(x) + xu2 (x). Then Y' = u2 since we require
u1 + xu^ = 0. Differentiating again yields Y" = u^ and
thus u"2 = -f(x) by substitution into the D.E. Hence
u (x) = - xf(s)ds, u' = xf(x), and u (x) = xsf(s)ds.
2 J0 1 1 J0
Therefore Y = xsf(s)ds - ^xf(s)ds = - x(x-s)f(s)ds and J0 J0 J0
(x) = c + c x - Ix(x-s)f(s)ds.
Previous << 1 .. 498 499 500 501 502 503 < 504 > 505 506 507 508 509 510 .. 609 >> Next