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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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an3 n3 L L
n = 1
226
Section 10.7
6c.
u(2.5,t)
u(5,t)
9. Assuming that u(x,t) = X(x)T(t) and substituting for u in Eq.(1) leads to the pair of O.D.E. X" + GX = 0,
T" + a2GT = 0. Applying the B.C. u(0,t) = 0 and
ux(L,t) = 0 to u(x,t) we see that we must have X(0) = 0
and X'(L) = 0. By considering the three cases G < 0,
G = 0, and G > 0 it can be shown that nontrivial solutions of the problem X" + GX = 0, X(0) = 0, X'(L) = 0 are possible if
and only if G = (2n-1) 2n 2/4L2, n = 1,2,... and the
corresponding solutions for X(x) are proportional to sin[(2n-1)nx/2L]. Using these values for G we find that T(t) is a linear combination of sin[(2n-1)nat/2L] and cos[(2n-1)nat/2L]. Now, the I.C. ut(x,0) implies that T (0) = 0 and thus functions of the form
un(x,t) = sin[(2n-1)nx/2L]cos[(2n-1)nat/2L], n = 1,2,... satsify the P.D.E. (1), the B.C. u(0,t) = 0, ux(L,t) = 0,
and the I.C. ut(x,0) = 0. We now seek a superposition of
these fundamental solutions un that also satisfies the
I.C. u(x,0) = f(x). Thus we assume that
u(x,t) = cnsin[(2n-1)nx/2L]cos[(2n-1)nat/2L]. The
n=1
I.C. now implies that we must have
f(x) = cnsin[(2n-1)nx/2L]. From Problem 39 of Section
n=1
10.4 we see that f(x) can be represented by such a series and that
cn = (2/L) [Lf(x)sin[(2n-1) nx/2L]dx, n = 1,2,... .
0
Substituting these values into the above series for u(x,t) yields the desired solution.
10a. From Problem 9 we have
-4
[cos(-------------------) - cos(
4L
(2n-1)n(L+2)
(2n-1)(L-2) )] 4L
(2n-1) _ (2n-1)
(2n-1)
sin
using the
4
2L
Section 10.7
227
trigonometric relations for cos(A B). Substituting this value of cn into u(x,t) in Problem 9 yields the desired solution.
10b.
10c.
13. Using the chain rule we obtain ux = u^x + u^Hx =
u? + u^ since ?x = = 1. Differentiating a second time
gives uxx = u?? + 2u^n + . In a similar way we obtain
ut = u^t + unnt = -au? + aun, since ?t = -a, nt = a. Thus
utt = a2 (u?? - 2u^n + ^nn). Substituting for uxx and utt
in the wave equation, we obtain u^ = 0. Integrating both sides of u^n = 0 with respect to n yields u?(?,n) = Y(?) where Y is an arbitrary function of ?.
Integrating both sides of u?(?,n) = Y(?) with respect to ?
yields u(?,n) = JY(?)d? + y(n) = (?) + V(n) where y(n) is an arbitrary function of n and JY(?)d? is some function of ? denoted by (?). Thus u(x,t) = u(? (x,t),n (x,t)) = (x - at) + y(x + at).
14. The graph of y = sin(x-at) for the various values of t is
indicated in the figure on the next page. Note that the
graph of y = sinx is displaced to the right by the
distance "at" for each value of t.
228
Section 10.7
Similarly, the graph of y = (x + at) would be displaced to the left by a distance "at" for each t. Thus (x + at) represents a wave moving to the left.
16. Write the equation as a2uxx = utt + a 2u and assume u(x,t) = X(x)T(t). This gives a2X"T = XT" + a 2XT,
X" 1 T" 2
or = ( + a2) = a. The separation constant a is X a2 T
X2 using the same arguments as in the text and earlier problems. Thus X" + X 2X = 0, X(0) = 0, X(L) = 0 and T" + (a2 + z2X2)T = 0, T'(0) = 0. If we let pn = X^a2+a2,
nnx nnx
we then have un(x,t) = cospntsin-, where X n = -.
LL
nnx
oo
Using superposition we obtain u(x,t) = cncospntsin
n = 1 J_1
^ nnx
and thus u(x,0) = cnsin---- = f(x). Hence cn are given
L
n = 1
by Eq. (22).
17a. We have u(x,t) = ^-at) + y(x+at) and thus ut(x,t) = -a/(x-at) + a%'(x+at). Hence u(x,0) = ^) + y(x) = f(x) and
ut(x,0) = -a/(x) + ay'(x) = 0. Dividing the last
equation by a yields the desired result.
17b. Using the hint and the first equation obtained in part (a) leads to ^) + y(x) = 2^) + c = f(x) so ^) = (1/2)f(x) - c/2 and y(x) = (1/2)f(x) + c/2. Hence
u(x,t) = (x - at) + y(x + at) = (1/2)[f(x - at) - c] + (1/2)[f(x + at) + c] = (1/2)[f(x - at) + f(x + at)].
f(x + at)
17c. Substituting x + at for x in f(x) yields
I 2 -1 < x + at < 1
[0 otherwise
Subtracting "at" from both sides of the inequality then yields
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