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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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OO
u(x, 0) = Y2 bn(0)(x) = ^2 an(x) = f (x). (40)
n= 1 n= 1
648
Chapter 11. Boundary Value Problems and Sturm-Liouville Theory
EXAMPLE
2
Thus the initial values an are the coefficients in the eigenfunction expansion for f (x). Therefore,
an = f r (x) f () (x) dx, n = 1, 2,.... (41)
J0
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 fz(t) = exp(Xnt), and it follows that
bn(t) = ane~l'nt + f e~xn(t~s)yn(s) ds, n = 1, 2,.... (42)
n n 0 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) = Y2 bn({)(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). n n
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-t, (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()(x).
n= 1
11.3 Nonhomogeneous Boundary Value Problems
649
The coefficients bn are determined from the differential equation
b'n + Xnbn = Yn (t), (46)
where Xn 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~t(x) dx = e-t x(x) dx
Jo Jo
= cne~t, (47)
where cn = xn(x) dx is given by Eq. (23). The initial condition for Eq. (46) is
non
bn (0) = 0 (48)
since the initial temperature distribution (45) is zero everywhere. The solution of the initial value problem (46), (48) is
't e(Xn-1)t _ 1
"Xnsc e s ds = C e Xnt
bn(t) = e Xnt ^ eXnscne s ds = Cne X - 1
c
= -^~(e-t - e~Xnt). (49)
Xn - 1
Thus the solution of the heat conduction problem (43) to (45) is given by
sin JX~ (et - e Xnt) sin JX~x u(x, t) = 4 V n ---------------------- 2 n . (50)
n=1 Xn (Xn-1)(1 + cosv^n)
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:
sin, sin, x e~X'nt sin, sin, x
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