0387986995Basic TheoryC

Step 1. To prove (i), apply Lemma VI-3-8 to pi(x) = 12(x) = p(x), 91(x) = g(x, A1), and 92(x) = g(x, A2). If A2 < A1, then g2(x) > 91(x) on Z(a, b). Hence,

9(x, A2) > 9(x, A1) on a < x < b, since 9(a, A2) = 9(a, Al) = a.

Step 2. To prove (ii), choosing a real number m and a positive number P so that u(x) > m and p(x) < P on I(a, b), determine 0o(x, A) by the initial-value problem

P = 1 + (P(m - A) - 1) sin2(8o), 9o(a, A) = 0.

Then, since g(x, A) > m -A and P > p(x) > 0 for x E I(a, b), -oo < A < +oo and a > 0, it follows from Lemma VI-3-8 that

9(c, A) > 9o(c, A) for a < c < b and - oo < A </+oo.

Observe that v = P tan(Oo) satisfies the differential equation = 1 + (MP A ) v2.

Hence, tan(9o(x, A)) = P(m - A) tan ( mP A (x - a) f for A < m. Note that

tan(9o(x, A)) = 0 if and only if tan I VmP A (x - a)) = 0, i.e.,

m - A

and that 9o(x, A) is strictly decreasing with respect to A if x > a (cf. Step 1). Furthermore, since tan(9o(x, m)) = x P a, we must have 0 < 9o(x, m) < 2 for

Thus, we conclude that acclim 9(c, A) = +oa if c > a.

Step 3. To prove (iii, first notice that 0(c, A) > 0 for -oo < A < +oo, since

Remark VI-3-15. Lemma VI-3-8, Lemma VI-10, and Theorem VI-3-11 are also found in (CL, §§1 and 2 of Chapter 8] and [Hart, Chapter XI].

VI-4. Eigenfunction expansions

Let us define a differential operator L[y] = dx (p(z))!LV + u(x)y and the vector

space V(a, b) over the real number field R in the same way as in §VI-2. Also, as in §VI-2, define the inner product (f, g) for two real-valued continuous functions f

(cf. Theorem VI-2-5). Define a norm of a continuous function on X(a, b) by 11f 11 =

(f, f). Then, 1(f, g)f < Ilf II llgll. The following theorem is a basic result in the theory of self-adjoint boundary-value problems.

Theorem VI-4-1. Let Al and A2 be two distinct eigenvalues of the problem

Let r11(x) and r}2(x) be eigenfunctions corresponding to Al and A2, respectively.

Then, (rll,rh) = 0.

Proof.

Note that (r71,L[r12]) = A2(rh,g2) and (Ljrll1,rh) = Al(rl1,rn). This implies that

A2(r11, r12) = AI (r11, r12). Therefore, (r1i, r12) = 0 since Al 0 A2 0

It is known that problem (EP) has real eigenvalues

such that 11,11 = 1 (n = 1, 2, ... ). Theorem VI-4-1 implies that (' h,''m) = 0 if h # k. From these properties of the eigenfunctions rjn, we obtain

for any continuous function f on Z(a, b).

In this section, we explain the generalized Fourier expansion of a function f (x) in terms of the orthonormal sequence {rl (x) : n = 1,2,... }. As a preparation, we

prove the following theorem.

164 VI. BOUNDARY-VALUE PROBLEMS

Theorem VI-4-3. For any continuous function f (x) on Z(a, b), the sum

Proof

Set 91 (f ] = 9 If, - 1: (f, TO 17h . Then, it suffices to show

h=1=1 Ah - PO

for every continuous function f (x) on Z(a, b). We prove (VIA.1) in four steps. Without any loss of generality, we assume that Ah - uo < 0 for h > t . Also, note that

for any real-valued continuous functions f and 9 on Z(a, b). Step 1. It can be shown easily that Gr[>jh] = chrjh, where

Let y be an eigenvalue of 9t and let ¢(r) be an eigenfunction of 9t associated with

implies that (0,9471h]) = (91'101,170. Hence, ch(¢, nh) = y(¢, nh) Consequently,

(¢, rjh) = 0 if y 0 ch. Therefore, (¢, rjh) = 0 for h = 1, ... , I and y¢ = G[¢] if y 96 0. Hence, either y = O or y = ch (h > P).

Step 2. In this step, we prove that

sup 1(91[f],f)I Also, since (Ge[f ±9],f ±9) = (911f],f)+(9t[9],9)±2(9t[f],9),

Of H=1

we obtain

4(91[f],9) = (91 [f+9) ,f+9) - (9e[f-9],f-9)

< ( sup l(Ge[w], w)I (IIf + 9II2 + If - 9112) lH'°t=1

= 2 sup I(9e[wj,w)l (11f112+119112).

N++ N=1

Step 3. In this step, we prove that - sup I (Gt If], f) I is an eigenvalue of Gt. To

IlfI1=1

do this, set c = sup I(Ge[f], f)I. Note first that c > I(Gt[ne+I],ne+i)I = Ice+II >

c = sup (9t If], f ). Then, there exists a sequence f (j = 1, 2.... ) of continuous

11111=1

functions on T(a, b) such that lim (Gt[ fi], fi) = c and IIf, II = 1(j = 1, 2, . . . )

Since {Gt[ fj] : j = 1,2.... } is bounded and equicontinuous on T(a, b), assume without any loss of generality that lim Gt[ f)] = g E G(a, b) uniformly on T(a, b).

Gt[g] = cg. Also, from (VI.4.4), it follows that IIgII2 = c2 > 0. Therefore, c must be an eigenvalue of Gt. However, since all eigenvalues of Gt are nonpositive, this is a

contradiction. Therefore, -c = inf (91 [fit f ). We can now prove in a similar way

Illil=l

that

1

-c = ct+1 =

),ttl - µ0

and the proof of (VI.4.1) is completed.

The following theorem is the basic result concerning eigenfunction expansion.

+00

Theorem VI-4-4. For every f E V(a, b), the series f = 1: (f, nh)nh converges to

h=1

f uniformly on T(a, b).

Proof.

Set g = .Co[f], then f = g[g]. Therefore,

Then, fe satisfies the conditions of Example VI-4-7, while fo(x) satisfies the conditions of Example VI-4-8. Thus, we obtain the Fourier series

series converges uniformly to f (x) on Z(-7r, 7r).

Remark VI-4-10. The Fourier series of Example VI-4-9 can be constructed also

from the eigenvalue problem L2 = Ay, y(-7r) = y(7r), y'(-7r) = y'(ir). Including

this eigenvalue problem, more general cases are systematically explained in [CL, Chapters 7 and 11]. For the uniform convergence of the Fourier series, it is not necessary to assume that f (x) be twice continuously differentiable. For example, it suffices to assume that f (x) is continuous and f'(x) is piecewise continuous on

1(-7r, 7r), and f (7r) = f(-7r). For those informations, see [Z].

Remark VI-4-11. The results in §§V1-2, VI-3, and VI-4 can be extended to the case when p(x) is continuous on 1(a, b), p(x) > 0 on a < x < b, and p(a)p(b) =

0 under some suitable assumptions and with some suitable boundary conditions.

Although we do not go into these cases in this book, the following example illustrates such a case.

Example VI-4-12. Let us consider the boundary-value problem

d f_dy . _

(VI.4.5)

y(x) is bounded in a neighborhood of x = 0, and y(l) = 0,

where u(x) and f (x) are real-valued and continuous on the interval 1(0, 1).

Step 1. Consider the linear homogeneous equation

on the interval 1(0,1). Since 1 and logs form a fundamental set of solutions of

the differential equation ± (xdx = 0, a general solution of (VI.4.6) can be

constructed by solving the intergral)equation

where c1 and c2 are arbitrary constants. In this way, we find two solutions O(x) and

-i(x) of (VI.4.6) such that O(x) and 4'(x) are continuous on 1(0,1), lim O(x) = 1,

and lien x¢'(x) = 0, whereas tI'(z) and tY(x) are continuous for 0 < x < 1, r_0+

lim (O(z) - logs) = 0, and urn (xo'(x) - 1) = 0. In general, this step of analy-

sis is very important. The behavior of solutions at x = 0 determines the nature of eigenvalues of the given problem. Denote also by p(x) the unique soltion of (VI.4.6)

such that p(1)=0andp'(1)=1.