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# 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
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ax j(0) + a2/(0) = 0, y(1) + b2/(1) = 0.
(2)
642
Chapter 11. Boundary Value Problems and Sturm Liouville Theory
As in Section 11.2 we assume that p, p', q, and r are continuous on 0 < x < 1 and that p(x) > 0 and r(x) > 0 there. We will solve the problem (1), (2) by making use of the eigenfunctions of the corresponding homogeneous problem consisting of the differential equation
and the boundary conditions (2). Let ^ < ^2 < <^n < '' be the eigenvalues
of this problem, and let 01,02,... ,0n, • • • be the corresponding normalized eigenfunctions.
We now assume that the solution y = 0(x) of the nonhomogeneous problem (1), (2) can be expressed as a series of the form
However, since we do not know 0 (x), we cannot use Eq. (5) to calculate bn. Instead, we will try to determine bn so that the problem (1), (2) is satisfied, and then use Eq. (4) to find 0 (x). Note first that 0 as given by Eq. (4) always satisfies the boundary conditions
(2) since each 0n does.
Now consider the differential equation that 0 must satisfy. This is just Eq. (1) with y replaced by 0:
We substitute the series (4) into the differential equation (6) and attempt to determine bn so that the differential equation is satisfied. The term on the left side of Eq. (6)
becomes
where we have assumed that we can interchange the operations of summation and differentiation.
Note that the function r appears in Eq. (7) and also in the term pr(x)0(x) in Eq. (6). This suggests that we rewrite the nonhomogeneous term in Eq. (6) as r(x) [ f (x)/r(x)] so that r(x) also appears as a multiplier in this term. If the function f/r satisfies the conditions of Theorem 11.2.4, then
L [y] = Xr (x)y
(3)
TO
(4)
n=1
From Eq. (34) of Section 11.2 we know that
n = 1, 2,....
(5)
L [0](x) = pr (x)0(x) + f (x).
(6)
TO
TO
L[0](x) = L ^2bn0n (x) = ^2 b„L[0n](x)
TO
(7)
(8)
where, using Eq. (5) with 0 replaced by f/r,
n = 1, 2,.... (9)
11.3 Nonhomogeneous Boundary Value Problems
643
Upon substituting for \$(x), L [tp](x), and f (x) in Eq. (6) from Eqs. (4), (7), and (8), respectively, we find that
CO CO O
Z2 bnknr(x)<Pn(x) = (x)Z2 bn\$n(x) + r(x)? cn(x).
n=1 n=l n=l
After collecting terms and canceling the common nonzero factor r(x) we have
E[(An - i)bn — CJ\$n(x) = 0 (10)
n=1
If Eq. (10) is to hold for each x in the interval 0 < x < 1, then the coefficient of (x) must be zero for each n; see Problem 14 for a proof of this fact. Hence
(Xn — ^)bn — cn = °, n = 1, 2,.... (11)
We must now distinguish two main cases, one of which also has two subcases.
First suppose that i = Xn for n = 1, 2, 3,...; that is, fi is not equal to any eigenvalue of the corresponding homogeneous problem. Then
bn = —, n = 1, 2, 3,..., (12)
Xn - I1
and
Oc
y = <P(x) = Z2 X n \$n(x)- (13)
n=1 Xn 1
Equation (13), with cn given by Eq. (9), is a formal solution of the nonhomogeneous boundary value problem (1), (2). Our argument does not prove that the series (13) converges. However, any solution of the boundary value problem (1), (2) clearly satisfies the conditions of Theorem 11.2.4; indeed, it satisfies the more stringent conditions given in the paragraph following that theorem. Thus it is reasonable to expect that the series (13) does converge at each point, and this fact can be established provided, for example, that f is continuous.
Now suppose that i is equal to one of the eigenvalues of the corresponding homogeneous problem, say, i = Xm; then the situation is quite different. In this event, for n = m Eq. (11) has the form 0 ? bm — cm = 0. Again we must consider two cases.
If i = Xm and cm = 0, then it is impossible to solve Eq. (11) for bm, and the nonhomogeneous problem (1), (2) has no solution.
If i = Xm and cm = 0, then Eq. (11) is satisfied regardless of the value of bm; in other words, bm remains arbitrary. In this case the boundary value problem (1), (2) does have a solution, but it is not unique, since it contains an arbitrary multiple of the eigenfunction \$m.
Since cm is given by Eq. (9), the condition cm = 0 means that
f f(x)\$m(x) dx = 0. (14)
J0
Thus, if i = Xm, the nonhomogeneous boundary value problem (1), (2) can be solved only if f is orthogonal to the eigenfunction corresponding to the eigenvalue Xm.
The results we have formally obtained are summarized in the following theorem.
644
Chapter 11. Boundary Value Problems and Sturm Liouville Theory
Theorem 11.3.1
Theorem 11.3.2
EXAMPLE
1
The nonhomogeneous boundary value problem (1), (2) has a unique solution for each continuous f whenever fi is different from all the eigenvalues of the corresponding homogeneous problem; the solution is given by Eq. (13), and the series converges for each x in 0 < x < 1. If i is equal to an eigenvalue Xm of the corresponding homogeneous problem, then the nonhomogeneous boundary value problem has no solution unless f is orthogonal to \$m ; that is, unless the condition (14) holds. In that case, the solution is not unique and contains an arbitrary multiple of \$m (x).
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