Books in black and white
 Books Biology Business Chemistry Computers Culture Economics Fiction Games Guide History Management Mathematical Medicine Mental Fitnes Physics Psychology Scince Sport Technics

# Elementary differential equations 7th edition - Boyce W.E

Boyce W.E Elementary differential equations 7th edition - Wiley publishing , 2001. - 1310 p.
ISBN 0-471-31999-6
Previous << 1 .. 275 276 277 278 279 280 < 281 > 282 283 284 285 286 287 .. 486 >> Next

Next we substitute from Eq. (30) for u in Eq. (25). From the first two terms on the right side of Eq. (25) we formally obtain
d [p(x)uJx - q(x)u = dx
p(xbn(t)<p'n(x)
n
n=1
- q (x)^2 bn (t)\$n(x)
n=1
= ^2 bn(t){[p(x)K(x)]' - q(x)0n(x)}. (31)
n1
Since [p(x)0'n(x)]' — q(x)0n(x) = —knr(x)\$ (x), we obtain finally
TO
[p(x)ux]x - q(x)u = -r(x)? bn(t)kn0n(x).
n=1
(32)
11.3 Nonhomogeneous Boundary Value Problems
647
Now consider the term on the left side of Eq. (25). We have
n TO
d \ >
r(x)ut = r(x)—2_^ bn(t)0n(x)
n=1
= r (x )? b'n (t)0n (x). (33)
n=1
We must also express the nonhomogeneous term in Eq. (25) as a series of eigenfunctions. Once again, it is convenient to look at the ratio F(x, t)/r (x) and to write
^TT = ? (')*„ (x), (34)
r( x) n n
n=1
where the coefficients are given by
f1 F (x, t)
Yn(t) = r(x)———0n(x) dx
J0 r(x)
= f F(x, t)0n(x) dx, n = 1, 2,.... (35)
Jd
Since F(x, t) is given, we can consider the functions yn (t) to be known.
Gathering all these results together, we substitute from Eqs. (32), (33), and (34) in Eq. (25), and find that
TO TO TO
r (x )E bn (t)0n (x) = -r (x )? bn (t)kn 0n (x) + r (x )? Yn (t)0n (x). (36)
n=1 n=1 n=1
To simplify Eq. (36) we cancel the common nonzero factor r(x) from all terms, and write everything in one summation:
TO
E [b’n(t) + knbn(t) - Yn(t)]0n(x) = 0. (37)
n=1
Once again, if Eq. (37) is to hold for all x in 0 < x < 1, it is necessary that the quantity in square brackets be zero for each n (again see Problem 14). Hence bn(t) is a solution of the first order linear ordinary differential equation
bn (t) + knbn (t) = yn (t), n = 1, 2,..., (38)
where yn(t) is given by Eq. (35). To determine bn(t) completely we must have an initial condition
bn(0) = an, n = 1, 2,... (39)
for Eq. (38). This we obtain from the initial condition (27). Setting t = 0 in Eq. (30) and using Eq. (27), we have
TOTO
u(x, 0) = E bn(0)0n(x) = E an0n(x) = f (x). (40)
n=1 n=1
648
Chapter 11. Boundary Value Problems and Sturm Liouville Theory
Thus the initial values an are the coefficients in the eigenfunction expansion for f (x). Therefore,
an = f r (x) f (x )\$n (x) dx, n = 1, 2,.... (41)
Jo
Note that everything on the right side of Eq. (41) is known, so we can consider an as known.
The initial value problem (38), (39) is solved by the methods of Section 2.1. The integrating factor is fr(t) = exp(Xnt), and it follows that
bn(t) = ane~xnt + f e~kn(t~s)yn(s) ds, n = 1, 2,.... (42)
n n o n
The details of this calculation are left to you. Note that the first term on the right side of Eq. (42) depends on the function f through the coefficients an, while the second depends on the nonhomogeneous term F through the coefficients yn (s).
Thus an explicit solution of the boundary value problem (25) to (27) is given by Eq. (30),
TO
u(x, t) = ^2 bn(t)\$n(x),
n=1
where the coefficients bn(t) are determined from Eq. (42). The quantities an and yn(s) in Eq. (42) are found in turn from Eqs. (41) and (35), respectively.
Summarizing, to use this method to solve a boundary value problem such as that given by Eqs. (25) to (27) we must
1. Find the eigenvalues Xn and the normalized eigenfunctions of the homogeneous problem (28), (29).
2. Calculate the coefficients an and yn(t) from Eqs. (41) and (35), respectively.
3. Evaluate the integral in Eq. (42) to determine bn(t).
4. Sum the infinite series (30).
Since any or all of these steps may be difficult, the entire process can be quite formidable. One redeeming feature is that often the series (30) converges rapidly, in which case only a very few terms may be needed to obtain an adequate approximation to the solution.
Find the solution of the heat conduction problem
ut = uxx + xe~l, (43)
u(0, t) = 0, ux(1, t) + u(1, t) = 0, (44)
u(x, 0) = 0. (45)
Again we use the normalized eigenfunctions of the problem (19), and assume that
u is given by Eq. (30),
TO
u(x, t) = ^2, bn(t)\$n(x).
n=1
11.3 Nonhomogeneous Boundary Value Problems
649
The coefficients bn are determined from the differential equation
b'n + Kbn = Yn (t)> (46)
where Kn is the nth eigenvalue of the problem (19) and Yn (t) is the nth expansion
coefficient of the nonhomogeneous term xe~ t in terms of the eigenfunctions . Thus
we have
Yn(t) = xe~l0n(x) dx = e~l x0n(x) dx
Jo Jo
= cne~(, (47)
where cn = I x4>n(x) dx is given by Eq. (23). The initial condition for Eq. (46) is
non
bn (0) = o (48)
since the initial temperature distribution (45) is zero everywhere. The solution of the initial value problem (46), (48) is
f TO , e(xn-1)t _ 1
I I ‘ ^ ^l/1--
I
c
^nsc e s ds = c e ----------------------------
l'/ ua cn^ ^ — 1
bn (t) = e n I e ns Cn e s ds = Cn?
= -^T (e- l-e- xnt). (49)
kn - 1
Thus the solution of the heat conduction problem (43) to (45) is given by
TO sin Jx~ (e 1 - e xnl) sin Jx~x u(x, t) = 4 V V n ------------------- 2 n . (50)
n=1 Kn (Kn- 1)(1 + cosVkn)
The solution given by Eq. (50) is exact, but complicated. To judge whether a satisfactory approximation to the solution can be obtained by using only a few terms in this series, we must estimate its speed of convergence. First we split the right side of Eq. (50) into two parts:
Previous << 1 .. 275 276 277 278 279 280 < 281 > 282 283 284 285 286 287 .. 486 >> Next