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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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 p(x)[u'(x)v(x)  u(x)v'(x)] o = -p(1)[u'(1)v(1)  u(1)v (1)] + p(0)[u/(0)v(0)  u(0)v/(0)]
= p(1) 0.
 ^ u(1)v(1) + ^ u(1)v(1)
+ p(0)
 ^ u(0)v(0) + ^ u(0)v(0)
2For brevity we sometimes use the notation f dx rather than f (x) dx in this chapter.
0
a
a
2
11.2 Sturm-Liouville Boundary Value Problems
631
The same result holds if either a2 or b2 is zero; the proof in this case is even simpler, and is left for you. Thus, if the differential operator L is defined by Eq. (3), and if the functions u and v satisfy the boundary conditions (2), Lagranges identity reduces to
f {L[u]v  uL[v]} dx = 0. (6)
Jo
Let us now write Eq. (6) in a slightly different way. In Eq. (4) of Section 10.2 we introduced the inner product (u, v) of two real-valued functions u and v on a given interval; using the interval 0 < x < 1, we have
(u, v) = f u(x)v(x) dx. (7)
Jo
In this notation Eq. (6) becomes
(L[u],v)  (u, L[v]) = 0. (8)
In proving Theorem 11.2.1 below it is necessary to deal with complex-valued functions. By analogy with the definition in Section 7.2 for vectors, we define the inner
product of two complex-valued functions on 0 < x < 1 as
(u, v) = f u(x)v(x) dx, (9)
J0
where v is the complex conjugate of v. Clearly, Eq. (9) coincides with Eq. (7) if u(x) and v(x) are real. It is important to know that Eq. (8) remains valid under the stated conditions if u and v are complex-valued functions and if the inner product (9) is used.
To see this, one can start with the quantity L [u] v dx and retrace the steps leading to Eq. (6), making use of the fact that p(x), q(x), a^ a2, bv and b2 are all real quantities (see Problem 22).
We now consider some of the implications of Eq. (8) for the Sturm-Liouville boundary value problem (1), (2). We assume without proof3 that this problem actually has eigenvalues and eigenfunctions. In Theorems 11.2.1 to 11.2.4 below, we state several of their important, but relatively elementary, properties. Each of these properties is illustrated by the basic Sturm-Liouville problem
/ + ky = 0, y(0) = 0, y(1) = 0, (10)
whose eigenvalues are kn = n2n2, with the corresponding eigenfunctions 0n (x) = sin nn x.
Theorem 11.2.1 All the eigenvalues of the Sturm-Liouville problem (1), (2) are real.
To prove this theorem let us suppose that k is a (possibly complex) eigenvalue of the problem (1), (2) and that 0 is a corresponding eigenfunction, also possibly complexvalued. Let us write k = fi + iv and 0(x) = U(x) + iV(x), where p, v, U(x), and V(x) are real. Then, if we let u = 0 and also v = 0 in Eq. (8), we have
(L[0],0) = (0, L[0]). (11)
3The proof may be found, for example, in the references by Sagan (Chapter 5) or Birkhoff and Rota (Chapter 10).
632
Chapter 11. Boundary Value Problems and Sturm-Liouville Theory
Theorem 11.2.2
However, we know that L [0] = Xr0, so Eq. (11) becomes
(Xr0, 0) = (0, Xr0). (12)
Writing out Eq. (12) in full, using the definition (9) of the inner product, we obtain
/1
Jo
Since r(x) is real, Eq. (13) reduces to
(X  X) f r(x)0(x)0(x) dx = 0, o
or
(X  X) f r(x) [U2(x) + V2(x)] dx = 0. (14)
0
The integrand in Eq. (14) is nonnegative and not identically zero. Since the integrand is also continuous, it follows that the integral is positive. Therefore, the factor X  X = 2iv must be zero. Hence v = 0 and X is real, so the theorem is proved.
An important consequence of Theorem 11.2.1 is that in finding eigenvalues and eigenfunctions of a Sturm-Liouville boundary value problem, one need look only for real eigenvalues. Recall that this is what we did in Chapter 10. It is also possible to
show that the eigenfunctions of the boundary value problem (1), (2) are real. A proof
is sketched in Problem 23.
J
Jo
Xr(x)0(x)0(x) dx = I 0(x)Xr(x)0(x) dx. (13)
If 01 and 02 are two eigenfunctions of the Sturm-Liouville problem (1), (2) corresponding to eigenvalues X1 and X2, respectively, and if X1 = X2, then
f r(x)01(x)02(x) dx = 0. (15)
0
This theorem expresses the property of orthogonality of the eigenfunctions with
respect to the weight function r. To prove the theorem we note that 01 and 02 satisfy
the differential equations
L [01] = X1r01 (16)
and
L [02] = X2r 02, (17)
respectively. If we let u = 01, v = 02, and substitute for L[u] and L [v] in Eq. (8), we obtain
(X1r01, 02)  (01, X2r02) = 0,
or, using Eq. (9),
XH r(x)01(x)02(x) dx  X2 I 01 (x)r(x)02(x) dx = 0.
1 0 1 2 2 0 1 2
11.2 Sturm-Liouville Boundary Value Problems
633
Because k2, r(x), and are real, this equation becomes
(A.j  X2) f r(x)0j(x)02(x) dx = 0. (18)
Jo
Since by hypothesis ^ = k2, it follows that 0 and 0 must satisfy Eq. (15), and the theorem is proved.
Theorem 11.2.3 The eigenvalues of the Sturm-Liouville problem (1), (2) are all simple; that is, to each eigenvalue there corresponds only one linearly independent eigenfunction. Further, the eigenvalues form an infinite sequence, and can be ordered according to increasing magnitude so that
^1 < ^2 < k3 < ? ? ? < kn < ? ? ? .
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