0387986995Basic TheoryC

188	VI. BOUNDARY-VALUE PROBLEMS

Step 4. Let us prove that

dzf(a) J< 0

dal

> 0

if	f (a) = 2 and	d(\) = 0,
if	f (A) _ -2 and	df(A)	= 0.
		d,\

First observe that Q is a perfect square if (f(A)l = 2. Hence,

71K (A)	= ± ft [c101(t, A) + c202(t, \)12 dt = 0,
	0

where cl and c2 are some real numbers. Therefore, cl = 0 and c2 = 0, since 01 and

02 are linearly independent. This means that

(IV)	02(f,,\)=0,		' (e, A)= 0,		ax
			UX
Since det	1, we conclude that
	fi(A) =	12	if f (a) = 2 and		d (a) = 0,
(V)		- I2	if f (A) = -2 and		df
					-(A) = 0,
where 12 is the 2 x 2 identity matrix. Furthermore,
z	e	-	(t, A)0 (t, A)2 + da (		')(e,
2 (a) =						a)2
	Jo			d
	+		d	) (e, A) } m (t, A)m2(t, A)1 dt.
Also, if f (A) = 2 and d		(A) = 0, it follows from (VI.10.11) and (IV) that
			I
	d(e, a) =
		z {e, ) = j42(s,A)2ds,
	d			t
				0i (s, A)2ds,
				0
	da (			f t 0r (s, A)02(s, A)ds.
				0
Hence,
		t(0f(t,
d(A) _ - f t f			A)02{s, A) - 0e(s, A)02(t,A)j2dtds < 0

(3) and (4) are verified.

Step 5. The functions

(IV), and (V) that 01(x, A) and 02(x, A) are two linearly independent solutions for problem (B). Similarly, if w2r_1 = v2,,, it follows from (VI.10.2), (IV), and (V) that 01(x, A) and 62(x, A) are two linearly independent solutions for problem (C). Thus,

10. PERIODIC POTENTIALS

189

if f (A) = 2 and I(A) = 0. Similarly,

d2f	t t
2 (A) = f f [Ol (t, A)02(s, A) - 41(s, A)4 2(t, A)J2dtds > 0

if f (A) = -2 and df (A) = 0. Note that, if A2,,-1 = A2,,, it follows from (VI. 10.2),

(respectively -yn(x)) have an even (respectively odd)

number of zeros on the interval 0 < x < 1, since /3, (0) = Qn(1) and 7n(0)	Yn(1)
As can be seen in Figure 1,
112n-i < A2n-1 A2n < 12n+l-
Therefore, by virtue of Theorems VI-3-11 and VI-1-1, we conclude that fit,	1 and

per have more than 2n - 1, and less than 2n + 2, zeros on the interval 0 < x < 1.

Hence, they have exactly 2n zeros there. Similarly, since

(VII)

/12n-2 < V2n-1 V2n < 112n,

and -t2n have exactly 2n - I zeros on the interval 0 < x < e. The function

,8a does not have any zero on the interval 0 < x < t since Ao < 111. Thus, (5), (6), and (7) are verified. Finally, (2) follows from (VI.10.9), (VI), and (VII).

Definition VI-10-7.

(I)The set {A : f(A)2 < 4} is called the stability region of the differential equation (VI.10.1).

(11)The set {A : f(A)2 > 4} is called the instability region of the differential equation (6210.1).

Example VI-10-8. A periodic function u(x) is called a finite-zone potential if the

function f (A)' - 4 of the differential equation d2y + (A - u(x))y = 0 has a finite dx2

number of simple zeros (i.e., all other zeros are double). For example, consider the case when u(x) = a, where a is a constant. The differential equation becomes

d? 2

dx+ (A - a)y = 0 and, hence,

cos(v1'"T-_ax)	if	A > a,
01 (X, A) = 1	if	A =a,
cosh( a --Ax)	if	A < a

190	VI. BOUNDARY-VALUE PROBLEMS
and		( sin(v1_T_-ax)
				if A>a,
		A-a
					A =a,
	02(x, A) =	x		if
		sinh(x)		if	A < a.
Therefore,
			12 cos(\/-), --at)
AA) =	A) +	(e, A) _	2

2 cosh( a --At)

if A >a,

if A=a,

if A < a.

Thus, we conclude that in this case, f (A)2 -4 has only one simple zero a (cf. Figure

2).

FicURE 2.

The materials in this section are also found in [CL, Chapter 8j.

EXERCISES VI

VI-1. Assume that u(x) is a real-valued continuous function on the interval 20 = {x : 0 < x < +oo} such that u(x) > rno for x >_ xo for some positive numbers "to and x0. Show that

(1) every nontrivial solution of the differential equation

(E)	day	- u(x)y = 0
	&2

has at most a finite number of zeros on Z0,

(2) the differential equation (E) has a nontrivial solution 1)(x) such that hm q(x) = 0.

Hint.

(1) Note that if y(xo) > 0, then y"(xo) > 0. Hence, y(x) > 0 for x > x0 if

1/(xo) > 0.

(2) It is sufficient to find a solution 0(x) such that 0(x) > 0 and 0'(x) < 0 for

x>xo.

	EXERCISES VI		191
VI-2. For the eigenvalue-problem
(EP)	L2 + u(x)y = Ay,	Y(()) = y(1),	y(0) = y'(1),

where u(x) is real-valued and continuous on the interval 0 < x < 1,

(1)construct Green's function,

(2)show that (EP) is self-adjoint,

(3)show that (EP) has infinitely many eigenvalues.

VI-3. Let A, > A2 > - > An > - be eigenvalues of the boundary-value problem

d2y
+ u(x)y = Ay,	y(a) = 0, y '(b) = 0,

where u(x) is real-valued and continuous on the interval a < x < b. Show that there exists a positive number K such that

2

	n +	ir	K	for n = 1, 2, 3, ... .
		b - a
	n2		n

VI-4. Assuming that u(x) is real-valued and continuous on the interval 0 < x <

+oo and that lim u(x) = +oo, consider the eigenvalue-problem r+oo

dx2 - u(x)y = Ay,	y(0) cos a - y'(0) sin a = 0,	z lim y(x) = 0.
where a is a non-negative constant. Show that
(a) there exist infinitely many real eigenvalues AI > A2 > . . . such that			lim An =
-00,			n-.ioo

(b)eigenfunctions corresponding to the eigenvalue An have exactly n -1 zeros on the interval 0 < x < +oo.

Hint. Let A1(b) > A2(b) > . . . be the eigenvalues of

dx2 - u(x)y = Ay,

y(0) cos a - y'(0) sin a = 0, y(b) = 0,

where b > 0. Define Am by lim A,(b). See JCL, Problem 1 on p. 2541.

6 +oo

VI-5. Show that if a function O(x) is real-valued, twice continuously differen- tiable, and ¢"(x) + e-'O(x) = 0 on the interval 10 = {x : 0 < x < +oo} and

if JJ0

O(x)2dx < +oo, then O(x) is identically equal to zero on I .

192	VI. BOUNDARY-VALUE PROBLEMS

VI-6. Using the notations and definitions of §VI-4, show that

+oo

(f,C(f)) = j>n(f,17n)2

n=1

b

= j {u(x)f(x)- P(x)f'(x)2}dx + P(b)f(b)f'(b) - P(a)f(a)f'(a),

if f E V(a, b).

VI-7. Assume that u(x) is real-valued and continuous and u(x) < 0 on the interval

I(a, b), where a < b. Denote by yh(x, A) the unique solution of the initial-value

problem 2 + u(x)y = Ay, y(a) = 0, y(a) = 1, where A is a comlex parameter.

Show that

(i)O(b, A) is an entire function of A,

(ii)¢(b, A) 54 0 if A is a positive real number,

(iii) ¢(b, A) has infinitely many zeros A,, such that 0 > A0 > Al > A2 >								and
llm		A" - -	\	r l 2			-
					/) .
n +oo n2			\b - a
VI-8. Find the unique solution O(x) of the differential equation
I x ) = y such that						is analytic at .x = 0 and 0(0) = 1. Also, show that
(i)\\\ m(x)	is an entire function of x,
(ii) ¢(x) A 0 if x is a positive real number,
(iii) ¢(x) has infinitely many zeros an such that 0 > \o > a1 > A2 > .								and
lim		1n = -oc,
n-+00
(iv) J0			nx)dx = 0 if n 96 m.
VI-9. Show that the Legendre polynomials
		Pn(x)			2"n1	d"	(n = 0,1,2,...)
				=		!dx"((x2 - 1)'J

satisfy the following conditions:

(i)deg P,, (x) = n (n = 0,1,2,...),

(ii)J 1 Pn(x)Pm(x)dx = 0 if n 0 m,

(iii) j P n (x)2dx =	1 (n = 0,1, 2, ... ),
n	z

(iv) J xkPn(x)dx = 0 for k = 0,... , n - 1,

(v) Pn(x) (n > 1) has n simple zeros in the interval Ixl < 1,

: x E Zo, A E It} such that

EXERCISES VI

193

(vi) if f (x) is real-valued and continuous on the interval jxj < 1, then

lim f

N-+oo

1

1

N	2

f (x) - E (n+ Z) (f, PP)Pn(x)) dx = 0,

n=O

where (f, Pn) = f1 f (x)PP(x)dx,

(vii) the series +00F, (n + 21 (f, PP)Pn(x) converges to f uniformly on the interval

n=0

jxj < 1 if f, f', and f" are continuous on the interval jxj < 1.

Hint. See Exercise V-13. Also, note that if f (x) is continuous on the interval jxj < 1, then f (x) can be approximated on this interval uniformly by a polynomial in x. To prove (vii), construct the Green function G(x,t) for the boundary-value problem

((I -x2)d/ +aoy=f(W),

y(x) is bounded in the neighborhood of x = ±1,

1 1

where ao is not a non-negative integer. Show that f f G(x, )2dx< < +oo.

J! 1J 1

Then, we can use a method similar to that of §VI-4.

VI-10. Assume that (1) p(x) and p'(x) are continuous on an interval Zo(a,b) _

{x : a < x < b}, (2) p(x) > 0 on Zo, and (3) u(x, A) is a real-valued and continuous function of (x, A) on the region Zo x ll = {(x, A)

lim u(x, A) = boo uniformly for x E Zo. Assume also that u(x, A) is strictly

A-too

decreasing in A E R for each fixed x on I. Denote by O(x, A) the unique solution of the intial-value problem

(P(x)L ) + u(x,.\)y = 0 y(a) = 0, y(a) = 1.

Show that there exists a sequence {pn : n = 0, 1, 2, ... } of real numbers such that

(i) pn < 1An-1 (n = 1, 2, ... ), and	lim pn = -oo,
	n .+oo

(ii) 4(b, pn) = 0 (n = 0,1, 2, ... ),

(iii) ¢(x, .\) ,- 0 on a < x < b for A > po, and q(x, A) (n > 1) has n simple zeros

on a<x<bforpn<\<,un-1,

(iv)	strictly decreases from +oo to -oo as A decreases from pn_1

to An-

VI-11. Assume that p(x) and u(x) are real-valued and continuous on the interval

1(0,1) = {x : 0 <- x < 1} and that p(x) is also continuous on 7(0,1), p(x) > 0

for 0 < x < 1, p(0) = 0, and p'(0) < 0. Show that the differential equation

194			VI. BOUNDARY-VALUE PROBLEMS
(px) d		+u(x)y = 0 has a fundamental set {0, ii} of solutions on the interval
dx	dx)
0 < x < 1 such that
(i) slim 4(x) = 1 and Zli o p(x)O'(x) = 0,
(ii)	slo (i(z) - 11		F )
			0 and =li m- p(x)ii (x) = 1.
VI-12. Set				0<x<
			for
			for	2 < x 1.

Denote by 01(x, A) and 42(x, A) the two unique solutions of the differential equation y + (A - u(x))y = 0 satisfying the initial conditions 4i(0, A) = 1, X1(0, A) = 0,

42(0, A) = 0, and (0, A) = 1, where A is a real parameter. Sketch the graph of

the function f (A) = 4(1, A) + (1, A).

VI-13. For the scattering data {r({) = 0, (1,2,3), (1,1,1)}, find the potential u(x) and the Jost solutions ft(x,().

VI-14. Calculate the scattering data for each of u(x + 1) and u(-x), assuming that {r({), (271, ... ,T)N), (cl, ... , cN)} are the scattering data for u(x) and that u(x) satisfies a condition Eu(x)I < Ae-k1=I for some positive numbers A and k.

Hint. Let f f(x, () be the Jost solutions for u(x). Two quantities a(() and are given by

a(S) =	and	1
2C I f+(x,C) f_(x,<) I		(X,0 f-' (x:0

respectively.

The Jost solutions f o r u(x + 1 ) are e ' ft (x + 1, (). Therefore, the scattering data for u(x + 1) are

(171,... ,nN), (e-2"'c1,... ,e-2"r'cN)}.

The Jost solutions for u(-x) are ff(-x, (). Hence, the scattering data for u(-x) are

b(l)'	(171,...,17N)I	(	1	,...,-	1	l}
a(C)			cia<(iv71)2		CNa<(LT,N)21 J

Note that

f-(x,iil3) = i a<(iul,)f+(x,iu,)

EXERCISES VI

195

VI-15. Let u(x) be real-valued, continuous, and periodic of period t > 0. Also, for every real , let O(x, {, A) be the solution of the differential equation

(Eq)	2 + (A - u(x))y = 0

satisfying the initial conditions 0({, {,A) = 0 and {G'(£, t, A) = 1. Let

A = Al (0 < A2W < µ3(f) < ...

be all roots of ii({ + t, , A) = 0 with respect to A. Show that

( + [', , A) = Q+00	Am(t) - A
	2
m=1	e

Hint.

Step 1. Let us construct O(x, , A) for negative A. To do this, change (Eq) to the integral equation

sinh(p(x - )) + 1 r sinh(µ(x -

p

p f

where it = v > 0. Since e-;`(=-() sinh(µ(x - )) is bounded as p - +oo, it can be shown that

e-v(x-E)+G(x, , A) =

e-al=-E) such{µ x- f )) + O( A 1

on the interval 0 < x - < t as A -e -oo. Thus, we derive

lim t'(4+f,.,A)= 1.

(sinh(e))

Step 2. There exists an entire function R(A) of A such that

	sin(Pf)	for	A > 0,
	R( A ) = J a
	sinh(CX)	for	A <0.

This function R(A) has the following factorization:

-_,

R(A) = t

m=1 \1

196	VI. BOUNDARY-VALUE PROBLEMS
where Cm =	mr
	t

Step 3. If the function V)((+1, 1, A) is entire in .\ and ''(+t, 1,,\) = O(exp(1 JaJ))
as A -+ oo, we can write 1'(( + 1, t,,\) in the following form:
lp(f + t, (, a) = c		[i(A]
where c is independent of A. Here, we used the fact that				cn,J is bounded
as m +oc (cf. Theorem VI-3-14).
Step 4. Note that	'c' + [0-W
_((+t,t,X) -
(1)		) H	_- a	]
R(A)	(		cam, - A

Note also that

c -a

cm -A

This implies the uniform convergence of the infinite product on the right-hand side

	of (1) for -oo < A < 0. Thus, lim	Z'((+ 1, t, _ = 1 = c
		R(a)	l
	a-.-OO

Remark. For expressions of analytic functions in infinite products, see, for exam- ple, (Pa, pp. 490-5041.

VI-16. The functions are defined in §VI-9. Calculate G4(u) and G5(u).

I

VI-17. Use the same notations as in Theorem VI-10-6. Show that if J u(t)dt = 0,

0

ft

then Ao < 0. Also show that if / u(t)dt = 0 and = 0, then u(x) = 0 identically

on-oc<x<+00. Jo

Hint. Since j3o(z) # 0 on -oc < x < +oo, set w(z) = !Lx). Then, w'(x) +

w(x)Z + A - u(x) = 0 (cf. 1A'IW, Theorem 4.4, p. 621).

V1-18. Set F(() = log f (() for `a( > 0, ( # 0, where

(i) f (() is continuous for QJ( > 0, ( 0, (ii) f (() is analytic for `£( > 0,

(iii)((f (() - 1) is bounded for !a( > 0,

(iv)f (() # 0 for !'( > 0, (0 0.

Denote by SR the semicircle {(: 1(I = R,0 < arg(< r) which is oriented counter- clockwise, where R is a positive number. Show that

(a) 2ri IR		1	1 F(z) dz = f F(()		(`1( > 0,1(1 < R),
		2Ri	0		(`'(<0,1(1<R),
(b)	F(() = tai j	F'(() d{ + 7 f-		g	for a(> 0,

where fl() is the complex conjugate of F().

In order to measure the

CHAPTER VII

ASYMPTOTIC BEHAVIOR OF

SOLUTIONS OF LINEAR SYSTEMS

In this chapter, we explain the behavior of solutions of a homogeneous linear

system

= A(t)y as t

+oo in the case when the coefficient matrix A(t)

has a limit A0 as t -+ +oo. The purpose is to show how much information we can glean from the limit matrix Ao. We are interested in the exponential growth of solutions and the asymptotic behavior of solutions.

exponential growth of a function, we use Liapounoff's type numbers which was originally introduced by A. Liapounoff in [Lia]. Liapounoff's type numbers are explained in §VII-1. Also, we explain Liapounoff's type numbers of solutions of a homogeneous linear system in §VII-2. (Liapounoff's type numbers are also found

in L. Cesari [Ce, pp.	50-551.) In §VII-3, assuming that	lim A(t) = Ao, we
		t-too

calculate Liapotimoff's type numbers of solutions in terms of the eigenvalues of A0 (cf. [Huk3j, [Hart, Chapter X1, and [Si9]). In §VII-4. we explain how to derive the asymptotic behavior of solutions by diagonalizing the given system. A theorem of

M. Hukuhara and N. Nagumo gives the original motivation (cf. Theorem VII-4-1). The main result is Theorem of N. Levinson (cf. Theorem VII-4-2). (Generalizations and refinements of Theorem of Levinson are found, for example, in [Bell], [Bel2],

[CK], [Cop1]. [Dev], [DK], [El], [E2], [Gi], [GHS], [HarL1], [HarL2], [HarL3], [HW1],

[HW2], [HX1], [HX2], and [HX3].) In §VII-5, the Theorem of Levinson is applied to a system whose matrix has a limit as t -+ +oc and its derivative is absolutely integrable on the interval 0 _< t < oo. The topics of §§VII-4 and VII-5 are also found in [CL, §8 of Chapter 3, pp. 91-97]. In §VII-6, we explain how we can reduce

some problems such as the differential equation dt2 + {X + h(t) sin(at)}rj = 0 to

the Theorem of Levinson, even if the derivative of h(t)sin(at) is small but not absolutely integrable on the interval 0 < t < oo. The main idea is to apply the

Floquet theorem (Theorem IV-4-1) to z + {1 + e sin(ot)}rt = 0 to eliminate

the periodic parts of coefficients so that we can use the Theorem of Levinson (cf.

[HaS3]; see also [HarLl], [HarL2], [HarL3]).

VII-1. Liapounoff's type numbers

In order to measure the exponential growth of a function, let us introduce Liapounof''s type numbers.

Definition VII-1-1. Let f (t) be a C"-valued function whose entries are continuous on an interval Z = {t : to < t < +oo}. Let us denote by A the set of all real

197