0387986995Basic TheoryC

178 VI. BOUNDARY-VALUE PROBLEMS

Remark VI-7-7. If the reflection coefficient r(t) = 0 and if there is no eigenvalue for the problem

- d2y + u(x)y = Ay, y E L2(-oo, +oo),

2

then u(x) = 0 identically for -oo < x < +oo. In fact, f+(x,() is analytic in everywhere in the (-plane. Furthermore, l e-`<= f+(x, () -11 -+ 0 as - oo. Hence, f+(x,() = e'(=. This shows that u(x) = 0.

Remark VI-7-8. If u(x) 0 but u(x) = 0 for }xI _> M, where M is a positive number, then the reflection coefficient r(S) # 0. In fact, if r(C) = 0, then u(x) is analytic for -oo < x < +oo. Hence, u(x) = 0. This is a contradiction. This remark reveals that the problem posed at the beginning of §VI-5 is not as simple as it looks.

VI-8. Construction of a potential for given data

If scattering data {0, (71,... 7x), (cl,... ,cN)} are given, formula (VI.7.3) gives the corresponding reflectionless potential u(x) as it is shown in Examples VI-

7-5 and VI-7-6. However, in these two examples, we assumed implicitly that such a potential exist. Since the existence of u(x) has not been shown yet, it must be proved. To do this, for given data

7i>0, 72>0, ..., 7N>0 and c1>0, c2>0, ..., cN>0,

define g, (j = 1, 2,... , N) by (VI.7.2) and define f+ and u by (VI.7.1) and (VI.7.3).

Hereafter, the first thing that we have to do is to show that y = f+ is one of the

Jost solutions of Ly = (2y.

Observation VI-8-1. Let gi (t = 1, 2,... , N) be determined by (VI.7.2) and u(z) be defined by (VI.7.3). Then,

(VI.8.1)	gi + 27r9t - ugt = 0		(=1,2,... , N).
Proof
Write (VI.7.2) in the form
	N
(VI.8.2)	E ae, (x)9, (x) = 1		(t = 1, 2,... , N)
	3=1
and differentiate both sides with reaspect to z. Then,
(VI.8.3)	Eat, (x)9f (x) + 2t,		9j(x) = 0	(t = 1, 2,... , N)
	}=1
and, hence,
N	N		g, (x) = 0	(t = 1,2,... ,N).
Eata(x)9j'(x) + 27t 11 -		1
1=1	i=1 7e+7)

8. CONSTRUCTION OF A POTENTIAL FOR A GIVEN DATA

179

From (VI.8.2), it follows that

	N			N
	E ajj(x){g,(x) + 2gegj(x)} - 237tE				g, (x) = 0
	j=1			j=1 ge +T7j
					(e = 1, 2, ... , N)
or
N			N		N	g, 92(x) = 0
Eatj(x){gf(x) + 2gtgj(x)} - 2>9,(x)+ 2E
3=1			j=1		j=1 711+17.,
						(e = 1, 2, ... , N).
Differentiating again, we obtain
N				2o[s
E ata (x){gj"(x) + 2r7eg, (x) } + 2ge				(9e(x) + 2gtgt(x)) - u(x)
,=1
	N
	+ 2E gj 9'(x) = 0			(e = 1,2,... N)
	3=1 ge+g,
or
N		N
Eat,j(x){g,"(x) + 217eg,(x)) - Eaea(x)u(x)9J(x)
,=1		f=1
	N		e2nas			(e = 1,2,N).
+	aea(x)2%g,'(x) + 2171				= 0
		C27jt		)
						... ,9e(x)
	,=1		ct
Thus, we derive
N
F'ae,,(x){9,(x) + 217,g,(x) - u(x)g,(x)}
j=1
	N				= 0
	2171 E atj (x)9, (x) + 217t		ct	91(x)		(t = 1, 2, ... , N),
	,=1
and (VI.8.3) implies that
N
F'at,1(x){9;'(x) + 2s7,g,,(x) - u(x)g,(x)} = 0						(e= 1,2,... ,N).
,=1

Therefore, (VI.8.1) follows.

180	VI. BOUNDARY-VALUE PROBLEMS

Observation VI-8-2. If we further define f+ by (VI.7.1), then £f+ = (2f+.

Proof.

(e-"=f+)11 +			(		u(e_" f+)
	J=1					,=1
	-i			+ i				N
		g1'	+ 2i(g - ttgJ -				2		9J
	-i N	g;,	+ 2rligl' - ugJ = 0.
	E
			(+ tnJ
	j=1
Observation VI-8-3. Since
	N		x			en,"
f+(x,it7t) = e-fez 1 -		g., (x)
					_ -gt(x) E L2(-oo,+oo),
	J=1 711+17.;					ct

N numbers -1l,2 (j = 1, 2,... , N) are eigenvalues.

N
Observation VI-8-4. Set a	(cf. (VI.6.9) with r			0) and
J=1
		N
f-(x,() = a(()f+(x, -() =	a(S)e-Cz	1 + iE	g3 (x)
			J=1(-ig,
(cf. Observation VI-7-1). Then, £f_ _ (2f_ . Furthermore, since
E gJ (-oo) = 1	(e = 1, 2, ... , N)
J=1171+17J

(cf. (VI.7.2)), we obtain

(VI.8.4)

a(() = 1 - i1 ____ ) j=1(+nJ

In fact, this follows from the fact that both sides of (VI.8.4) are rational functions in ( with the same zeros, the same poles, and the same limits as ( , oo.

Observation VI-8-5. The functions f±(.x, () are the Jost solutions of Ly = (2y.

Proof.

Note first that lim f+(x. ()e-'t= = 1 (cf. (VI.7.1)). Also, we have s+oo

lim f+(x,()e-`<x = 1 -	gg(-o) = a(()	(cf. (VI-7.1) and (VI.8.4)).
	i=1
This implies that

lim f-(x,()e'S= = a(()lim f+(x,-()e"x = a(()a(-() = 1. O
t--oo	x---oo

9. DIFF. EQS. SATISFIED BY REFLECTIONLESS POTENTIALS

181

Observation VI-8-6. Note that f_ (x, ) = a(() f+(x, -{) =		f+(x, ) for ( =
C real. This means that	0. Hence, r(C) = 0. Thus, we conclude that u(x) is

reflectionless.

Remark VI-8-7. For the general r((), the potential u(x) can be constructed by solving the integral equation of Gel'fand-Levitan:

K(x, r) + F(x + r) +		+00
		+ F(x + r + s)K(x, s)ds = 0,
		0
where	1J +
F(x) =		2cL` cle-2n,=
	R

oo	J=1

Find K(x, r) by this equation. Then, the potential is given by

u(x) 8x (x,0).

If we set r({) = 0 in this integral equation, we can derive (VI.7.2) and (VI.7.3).

Details are left to the reader as exercises (cf. (Ge1LJ ).

V1-9. Differential equations satisfied by reflectionless potentials

Suppose that u(x) is a reflectionless potential whose associated scattering data

are given by	0, (r11, r12, ... flN ),			(cl, c2,... , cN) },where 0 < ril < ]2 <
< rily. It was proven that one of the Jost solutions is given by
	f+(T,() = e`t=		1 -Ni 9j (T)
						j_1 (+
Furthermore,						N
	f+(x, -() = e`= 1 -
						g1 (r)
						-( + trj7
is also a solution of
	dX2 - (u(x) - A)y = 0, where J = (2.
Therefore,
		N
(VI.9.1)	P(x, A) = 11 ! 1+11-				f+(T,C)f+(x, -()
		3=1	A )
satisfies the differential equation
(E)	3	+ 2(u(x) - A) dP + ddx)P = 0.
	2 dz3

182	VI. BOUNDARY-VALUE PROBLEMS
Let
	+oo
(S)	P = E an	(PO 3 0)
	n=0

be a formal solution in powers of A-1 of differential equation (E). Then,

0 = P [-- + 2(u(x) - A)P + u'(x)P

P2	, +	(4)2 + (u(x)
		- A)P2

J

and, hence,

A(P2)' = I - PZ it + (4)2 + u(x)P2 }, .

Therefore,

(i) the coefficients pn can be determined successively by

po = ca, where co is a constant,

	n	n	n	nPtPn+1-t
2C Pn+l = 2	E PtFri-1 + 4E Ptpn-t + uE pip.-t -			r
	t=o	t=o	c=o	1=1
			(n = 0,1,2,...),

(ii) the coefficients pn are polynomials in u and its derivatives with constant coefficients,

(iii)the formal power series (S) is uniquely determined by the condition P = 1 for u = 0,

() i		unique	Y	(	) =	+00 G"( u)	(	Go = 1), then the
lv	f we denote the		P of (iii) by G u, A

n=0	An

general formal solution of (S) is given by P(u, A) = P(0, A)G(u, A).

Example VI-9-1. A straight forward calculation yields

Gl =	u	G2 =	3u2 - u"	10u3 - lOuu" - 5(u')2 + u(4)
	Z'		8	G3 =	32
Observe that P(x, A) of (VIA 1) is a polynomial in					of degree N. Therefore,

the potential u(x) satisfies a differential equation

(VI.9.2) GN+1(u) + a1GN(u) + a2GN-1(u) + ... + QNG1(u) = 0

for some suitable constants a,, a2, ... , 0N. More precisely speaking, it is shown above that P(u, A) = P(0, A)G(u, A). Compute the coefficients of A-(N+1) on both sides of this identity. In fact,

	N	N
u(x) = 2	g}(x)	e2,7ixgj(x)2
		4 1` 171
		C.,

i=1

1=1

10. PERIODIC POTENTIALS

183

implies that u = 0 if and only if gl = 0 (1 = 1, 2,			- , N). This, in turn, implies
that
P(0, A) = 11 C1 +		A
and		where		A=C2.
G(u, A) = f+(x, C)f+(x,
Hence, ao = 1 and	N	+
N			2
1 + E	=
3=1	1 =1

Example VI-9-2. The function u(x) = -2q2 sech2(17(x /+ p)) is a reflectionless potential corresponding to the data {r7, c}, where p = 2- In 1c) (cf. Example VI-

7-5). In this case, G2(u)+172G1(u) = 0. Also, G3(u)+(n1+7)2 2(u)+rI14Gi(u) _

0 in the case when N = 2 and {r(C) = 0, (i71, n2), (cl, c2)} are the scattering data.

In particular, G3 (U) + 13G2(u) +36G1(u) = 0 if i1 = 2 and '12 = 3.

Remark VI-9-3. The materials of §§VI-5-VI-9 are also found in (TDJ.

VI-10. Periodic potentials

In this section, we consider the differential equation

(VI.10.1)

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

under the assumption that

(I)u(x) is continuous for -co < x < +oo,

(II)u(x) is periodic of period 1, i.e., u(x + e) = u(x) for -oo < x < +oo.

The period a is a positive number and A is a real parameter. Denote by 01(x, A) and 02(x, A) the two linearly independent solutions of (VI.10.1) such that

(VI.10.2) 01(0,A) = 1,		1(o	A) = 0,	02 (0, A) = 0,		d2(0,A) = 1.
Set			01(e, A)	02(t, A)
.D(A)		=	O1 (t"\)	(e A)
The two eigenvalues Z1(A) and Z2 (A) of the matrix					are the multipliers of the
periodic system
d
(VI.10.3)	[Y2J = [u(--O-A 0, [y21
dx

184	VI. BOUNDARY-VALUE PROBLEMS

(cf. Definition IV-4-5). It is easy to show that det [1(A)] = 1 for -oo < A <

+oo. Hence, the two multipliers Z1(A) and Z2(A) are determined by the equation Z2 - f (,\)Z + 1 = 0, where

(VI.10.4)	f(A) =	2(t,A)

Note that f (A) is continuous for -oo < A < oo (cf. Theorem 11-1-2). We derive first the following conclusion.

Lemma VI-10-1. The two multipliers Z1(A) and Z2(A) of system (VI.10.3) are

2 l	f(,\)	+	f (A)2 -	4, ,
2 l	f (A) -		f (A)2 -	4, ,

where f (A) is given by M. 10.4). Therefore,

(i) if I f (A) I > 2, two multipliers are real and distinct, i.e..

ZI(A) > 1,	0 < Z2(A) < 1	if	f(A) > 2,
-1 <Z1(A) < 0,	Z2(A) < -1	if	f(A) < -2,

(ii) if I f (A) I < 2, two multipliers are complex and distinct, and

I Zt (A) I = 1

and

I Z2(A) 1 = 1,

(iii)if f (A) = 2, then Z1(A) = Z2(A) = 1.

(iv)if f(A) _ -2, then Z1(A) = Z2(A) = -1.

Observation VI-10-2. If f (A) = 2, let ICCl2 be an eigenvector of 4;(A) associated

J

with the eigenvalue 1. This means that cl and c2 are two real numbers not both zero and that

c141(t, A) + C202(1, A) = c1, Cl d ' (t, A) + c2d-i (t, A) = C2 -

Set 6(x,,\) = cl o1(x, A) + c202(x, A). Then, d(x, A) is a nontrivial solution of

(VI.10.1) such that

.0(t, A) = 0(0,A) and LO (t, A) = (0, A).

This implies that O(x, A) is a nontrivial periodic solution of (VI.10.1) of period t.

Observation VI-10-3. If f (A) = -2, it can be shown that equation (VI.10.1) has a nontrivial solution O(x, A) such that

0(t, A) = -0(0,A)

and

(t, A) = - (O, A)

	10. PERIODIC POTENTIALS			185
This implies that ¢(x, A) satisfies the condition
	¢(x + e, A) = -&, A)	for - oo < x < +oo.
Observation VI-10-4. Cconsider the eigenvalue problem
(VI.10.5)	2 + (A - tL(x) y = 0,		y(o) = 0,	y(e) = 0.

Applying Theorem VI-3-11 (with -A instead of A) to problem (VI.10.5), it can be shown that (VI.10.5) has infinitely many eigenvalues

(VI.10.6)	µI < i2 <	< µ <
which are determined by the equation
(VI.10.7)	t2(£, A) = 0.

The eigenfunction 02 (x, has exactly n -1 zeros on the open interval 0 < x < e.

Since .2 (0, A) = 1 > 0, it follows that

		dO2		< 0	for n is odd,
			dx (e'µ") I > 0		for n is even.
	Furthermore, condition (VI.10.7) implies that det t I (A)] =					A) = 1.
	Thus, f (A) = 01(e, A) +		1	follows from (VI.10.4). Therefore,
			Z V. A)
	f(A) > 2 if		(e, A) >0 and		f(A) :5-2 if	(e, A) < 0,
				> 2	if n is even,
			f (µ") {	< -2	if n is odd

(cf. (VI.10.7)). Here, use was made of the fact that jaj+ 1a1,

ifa76 0.

Observation VI-10-5. If

(VI.10.8)	A < min{u(x) : -oo < x < +oo},
then f (A) > 2. In fact,	dX2	1 (x, A)	= (u(x) - A)01 (x, A) > 0 as long as 01(x, A) > 0

and, hence, dol (x, A) > 0. Thus, we obtain s1(£, A) > 1. Similarly, 2 (P, A) > 1

if (VI.10.8) is satisfied. In this way, a rough picture of the graph of the function f (A) is obtained (cf. Figure 1).

	At	A2	3	44	A3=	f=2
3	T			-l4F
	I
	W				lA	f= -2
	Ul U2	U3=N3U4		u5	U6

FIGURE 1.

Actually, we can prove the following theorem.

186	VI. BOUNDARY-VALUE PROBLEMS

Theorem VI-10-6. Assuming that u(x) is continuous and periodic of period t > 0 on the entire real line R, consider three boundary-value problems:

(A)	2 + (A - u(x)) y = 0, y(0) = 0,		y(t) = 0,
($)2 + (A - u(x)) y = 0,		W) = y(0),	dx (t) _	y (0),
(C)	d2 + (A - u(x)) y = 0,	y(t) = -y(0),	dY(t) _	L(0).

Then, each of these three problems has infinitely many eigenvalues:

	µI	<µ2</A3...<An<...,
(VI.10.9)	A0<\1<A2<A3<a4 <...		<A2n_I<\2n <...,
	V1 <v2 <v3 :5 L14 <... <V2n-1 <V2n <... ,
respectively. Furthermore, if we denote by an(x),			and 7n(x) the eigenfunc-

tions associated with the eigenvalues µn, An, and vn, respectively, then

(1)A0 < v1,

(2)v2n-1 :5 92n-I 1'2n < A2n-1 !5 µ2n < A2n < V2n+1 < µ2n+1 < i/2(n+l) for

n = 1,2,...,

(3)02n_1(x) and /32n(x) are linearly independent if A2,,-1 = A2n,

(4)72n-1(x) and 72n (x) are linearly independent if v2n-1 = v2n,

(5)3D(x) does not have any zero on the interval 0 < x < t,

(6)02._1(x) and /32,,(x) have exactly 2n zeros on the interval 0< x < t,

(7)72n-1(x) and 72n (x) have exactly 2n - 1 zeros on the interval 0 < x < t.

Proof.

We prove this theorem in five steps. Note first that (VI.10.9) is obtained from

Figure 1.

Step 1. If ¢(x, A) is a solution of differential equation (VI.10.1), then	(x, A) is
a solution of the initial-value problem
(VI.10.10)

d2w + (A - u(x)) w + O(x, A) = 0,	w(0, A) =	(0, )L),
	i i(0,,\) =
dx2		TA_ (dx

(cf. §lI-2). Therefore, using the variation of parameters method, we obtain

(VI.10.11)		_
-1(x, A) _ (x, A) fT 01(t, A)02(t, A)dt	¢2(x, A)	f01(t"\)2dt'

d z (x,,\) = 01(x"\) 102(t, A)2dt - 02(X,,\) f -01(t, A) 02(t, A)dt

dA\) ¢2(e, A) < 0. Also, using Lemma VI-3-11 (with

		10. PERIODIC POTENTIALS					187
and, hence,
da (dd-x (x, A)) =		1(x, A) f ; m2(t, A)2dt -				(_, A) la	mi (t, A)yi2(t, A)dt.
Therefore, df	= rr	Q(01 (t, A), ¢2(t, A))dt follows from (VI. 10.4), where
dA	0
Q(Y1,Y2) = -02(e,A)Y2 + dTl(e,A)Y2 + [1(t1A) -							(e,A)Jr1Y2.
Step 2. The discriminant of the quadratic form Q is f(,\)2 - 4. In fact,
		2
1(t"\)		+ 4xd		l (e, A)o2(t, A)
		A)]
			2	+ 401(e, A) dO2 (e, A) - 4 = f(,\)2 - 4.
		2 (e, A)]
Note that
41(e, )		(e, A) -	' (e,		A) = det [ I (A)] = 1.

Thus, Q(01 (X, A), ¢2(x, A)) does not change sign for 0 < x < e if If (A)f < 2. It follows that

(I)	df(A)	0	if	If(A)f < 2.
	dA
Step 3. Also, it can be proven that

df(A)	< 0	if	A < pl and f(,\)2 = 4,
d,\
(-1)1 d('\) < 0		if	pl < A < pi+1 and f(A)2 = 4.

To prove this, notice that Q is a perfect square if f(,\)2 = 4 and that 02(1,.\) # 0 if A 0 pj for every j. Hence,

-A instead of A), we obtain

02 (1, A) > 0	if	A < p1,
A) > 0	if	p., < A < pj+1.

Thus, (1) is verified.