Inverse spectral problem for singular AKNS operator on [0,1]

0.34 MB·

by  Frédéric Serier

Inverse spectral problem for singular AKNS
operator on [0,1].
6
0
0
2 Fr´ed´eric SERIER
‡
n LMJL - Laboratoire de Math´ematiques Jean Leray
a
UMR CNRS 6629-UFR Sciences et Techniques
J
0 2 rue de la Houssini`ere - BP 92208
2 F-44322 Nantes Cedex 3

] E-mail: [email protected]
P
S
. Abstract. We consider an inverse spectral problem for a class of singular AKNS
h
t operators Ha,a N with an explicit singularity. We construct for each a N, a
a standard map λa∈ κa with spectral data λa and some norming constant κ∈a. For
m a=0,λa κa was×knowntobealocalcoordinatesystemonL2(0,1) L2(0,1). Using
R R
[ × ×
adapted transformation operators, we extend this result to any non-negative integer

1 a, give a description of isospectral sets and we obtain a Borg-Levinsontype theorem.
v
5
9
4 AMS classification scheme numbers: 35R30; 65L09
1
0
6
0 Submitted to: Inverse Problems
/
h
t
a
m
:
v
i
X
r
a
Present address: Institut fu¨r Mathematik, Universit¨at Zu¨rich, Winterthurerstrasse 190 CH-8057
‡
Zu¨rich
Inverse spectral problem for singular AKNS operator on [0,1]. 2
1. Introduction
The Schr¨odinger operator = ∆+q( x ) with a radial potential q, acting on the unit
H − k k
ball of R3, through a decomposition via spherical harmonics (see [19], p. 160 161), is
−
unitary equivalent to a collection of singular differential operators (q), a N acting
a
H ∈
on L2(0,1), with Dirichlet boundary conditions, defined by
R
d2 a(a+1)
(y)(x) := + +q(x) y(x) = λy(x), x [0,1],λ C.
Ha −dx2 x2 ∈ ∈
(cid:18) (cid:19)
With this splitting, it makes sense to study inverse spectral problems not for itself
H
but for each .
a
H
The inverse spectral problem for these operator is the construction for each a N,
∈
of a regular coordinate system λa κa for potentials q L2(0,1) where λa represent the
R
× ∈
spectrum of and κa are convenient complementary data (regularity means stability
a
H
of the inverse spectral problem).
This question is not new and has been answered: Borg [6] and Levinson [15]
first, proved that λ0 κ0 was one-to-one on L2(0,1); then P¨oschel and Trubowitz [18]
R
×
completed this result obtaining λ0 κ0 as a global real-analytic coordinate system on
×
L2(0,1). Guillot and Ralston [13] extended their results to λ1 κ1, passing through
R
×
the singularity inside the equation. Next Zhornitskaya and Serov [24], and Carlson [7],
proved that for all real a 1/2, λa κa is one-to-one on L2(0,1). Finally, the author
R
≥ − ×
[21] completed theses works proving that for all a N the map λa κa was a local
∈ ×
(hence global) diffeomorphism on L2(0,1).
R
Then, it is natural and interesting to wonder if these kind of results can be found
for an other physical equation: the Dirac equation. Hence, as the radial Schr¨odinger
operator, the Dirac operator with a radial electric potential acting on the unit ball of R3
is decomposed (see for instance [23]) into a collection of operators H defined on [0,1]
a
by
0 1 d 0 a
H (V)Y(x) := − + −x +V(x) Y(x) = λY(x), (1)
a 1 0 dx a 0
(cid:18)(cid:20) (cid:21) (cid:20) −x (cid:21) (cid:19)
where Y = (Y ,Y ), λ C and
1 2
∈
q(x)+m 0
V(x) = ,m R;
0 q(x) m ∈
(cid:20) − (cid:21)
with general boundary conditions
sinβ
Y (0) = 0; Y(1) u = 0 u = , β R. (2)
2 β β
· cosβ ∈
(cid:20) (cid:21)
Written this way, the Dirac operator seems to be unadapted in view of inverse
spectral problems. Indeed for a = 0, as raised by Levitan and Sargsjan in [16](Chap. 7)
and pointed out more generally by Clark and Gesztesy in [8] (section 6), the existence
of a gauge transformation on the potential V leaving the spectrum invariant leads to
Inverse spectral problem for singular AKNS operator on [0,1]. 3
choose a normal form for the problem, namely, the AKNS system, obtained from (1)
considering potentials V of the following shape:
q(x) p(x)
V(x) = − , (p,q) L2(0,1) L2(0,1). (3)
R R
p(x) q(x) ∈ ×
(cid:20) (cid:21)
Moreover, there are some clues showing that the inverse spectral problem is kind
of degenerated: for instance, the Ambarzumian type theorem obtained by Kiss [14] who
proves that forallm = 0andq ([0,1];R), if H (V) has thesame eigenvalues asH (0)
0 0
6 ∈ C
then q = 0. An other reason, to turn to the AKNS operator, is it similarity with the
Schr¨odinger operator as figured out the papers of Gr´ebert and Guillot [11] and Amour
and Guillot [3]. And finally, technical difficulties arise when computing asymptotics for
solutions of the Dirac equation, see remark page 9.
Our purpose is the stability of the inverse spectral problem for H ((1)-(2)-(3)).
a
For this, we construct for each a N, a spectral map λa κa for potentials V with
∈ ×
spectral data λa and some norming constant κa. The framework is the work of Gr´ebert
and Guillot [11] for the regular operator (a = 0). They constructed a local coordinate
system λ0 κ0 on L2(0,1) L2(0,1) and proved it is global on Hj(0,1) Hj(0,1)
R R R R
× × ×
for j = 1,2. With the singularity, interesting problem arise and add supplementary
difficulties, especially when we study theinvertibility of theFr´echet derivative ofλa κa.
×
For this, we use some transformation operators who, roughly speaking, reduce the
singularity.
Our result is that for all a N , λa κa is a local diffeomorphism on L2(0,1)
R
∈ × ×
L2(0,1) and one-to-one on H1(0,1) H1(0,1). Moreover, we locally describe sets
R R R
×
of isospectral potentials as smooth submanifolds of L2(0,1) L2(0,1) with explicitly
R R
×
tangent and normal spaces.
2. The direct spectral problem
We will omit proofs which are nearly repetitions of the regular case (for details see [22]).
2.1. Solutions Properties
In this section, V is any 2 2 matrix with L2(0,1) coefficients. A fundamental system
C
×
of solutions for (1) when V = 0 is given by
1 j (λx) η (λx)
R(x,λ) = a−1 , S(x,λ) = λa − a−1 ,
λa j (λx) η (λx)
(cid:20) − a (cid:21) (cid:20) a (cid:21)
where j and η are spherical Bessel functions (see section 4.1). These functions are
a a
called fundamental since their wronskian is equal to 1. From their behavior near x = 0,
R(x,λ) is called the regular solution, it is analytic on [0,1] C; S(x,λ) is called the
×
singular solution, it is analytic on (0,1] C.
×
Following Blancarte, Gr´ebert and Weder [5], we construct solutions for (1) by a
Picard’s iteration method from R and S.
Inverse spectral problem for singular AKNS operator on [0,1]. 4
˜
Let and ß be defined by
R
˜
(x,λ,V) = R (x,λ,V), ß(x,λ,V) = S (x,λ,V)
k k
R
k≥0 k≥0
X X
with
R (x,λ,V) = R(x,λ),
0
x (4)
 R (x,λ,V) = (x,t,λ)V(t)R (t,λ,V)dt, k N;
k+1 k
 G ∈
Z0
S (x,λ,V) = S(x,λ),
 0
1 (5)
 S (x,λ,V) = (x,t,λ)V(t)S (t,λ,V)dt, k N.
k+1 k
 − G ∈
Zx
is called Green function and is given by (see [4])

G
(x,t,λ) = S(x,λ)R(t,λ)⊤ R(x,λ)S(t,λ)⊤. (6)
G −
This construction is justified with the following
Lemma 2.1 Series defined by (4), respectively by (5), uniformly converge on bounded
sets of [0,1] C (L2(0,1))4, respectively of (0,1] C (L2(0,1))4, towards solutions
C C
× × × ×
of (1). Moreover, they satisfy the integral equations
x
(x,λ,V) = R(x,λ)+ (x,t,λ)V(t) (t,λ,V)dt,
R G R
Z0
1
˜ ˜
ß(x,λ,V) = S(x,λ) (x,t,λ)V(t)ß(t,λ,V)dt,
− G
Zx
and the estimates
a
x
(x,λ,V) Ce|Imλ|x ,
|R | ≤ 1+ λ x
(cid:18) | | (cid:19)
a
1+ λ x
˜ß(x,λ,V) Ce|Imλ|(1−x) | | ,
≤ x
(cid:18) (cid:19)
(cid:12) (cid:12)
with C uniform(cid:12) on bound(cid:12)ed sets of (L2(0,1))4.
(cid:12) (cid:12) C
˜
Proof. We give it for , it is similar for ß. Estimate (A.2) for Bessel functions gives
R
a
x
R(x,λ) Ce|Imλ|x . (7)
| | ≤ 1+ λ x
(cid:18) | | (cid:19)
Iterative relation (4) leads to
x
R (x,λ,V) = (x,t,λ)V(t)R(t,λ)dt, (8)
1
G
Z0
which, combining (7) and the Green function estimates (A.4), is bounded by
x a x
R (x,λ,V) C2e|Imλ|x V(t) dt,
1
| | ≤ 1+ λ x | |
(cid:18) | | (cid:19) Z0
By successive iterations and recurrence, for all positive integer n, we get
Cn+1 x a x n
R (x,λ,V) e|Imλ|x V(t) dt .
n
| | ≤ n! 1+ λ x | |
(cid:18) | | (cid:19) (cid:18)Z0 (cid:19)
Inverse spectral problem for singular AKNS operator on [0,1]. 5
This proves uniform convergence on bounded sets of [0,1] C (L2(0,1))4 for and
C
× × R
(cid:3)
the estimate. Integral equation follows from (4).
This uniform convergence gives us the following
Proposition 2.1 (Analyticity of solutions)
(a) For all x [0,1], (x,λ,V) is analytic on C (L2(0,1))4. Moreover, it is real
C
∈ R ×
valued on R (L2(0,1))4.
R
×
(b) The map : (λ,V) ( ,λ,V) is analytic from C (L2(0,1))4 to H1([0,1],C2).
C
R 7→ R · ×
(c) For all x (0,1], ˜ß(x,λ,V) is analytic on C (L2(0,1))4 and real valued on
C
∈ ×
R (L2(0,1))4.
R
×
˜
Let (λ,V) be the wronskian of and ß, defined by:
W R
˜ ˜
(λ,V) := (x,λ,V),ß(x,λ,V) = det (x,λ,V),ß(x,λ,V) .
W W R R
(cid:16) (cid:17) (cid:16) (cid:17)
Recallthat (λ,V)isindependent ofx. Wefollowtheconstruction ofasimilar solution
W
byGuillotandRalstonin[13]: (λ,V)isnotequalto1. However, aswewillseefurther,
W
for λ large enough, doesn’t vanishes (see Theorem 3.2). Thus we may define the
| | W
so-called singular solution by
˜
ß(x,λ,V)
ß(x,λ,V) = , x (0,1].
(λ,V) ∈
W
Regularity of leads to existence of derivatives, obtained following [18]:
R
Proposition 2.2 For all v (L2(0,1))4, we have
C
∈
x
˜
[d (x,λ,V)](v) = (x,t,λ,V)v(t) (t,λ,V)dt, (9)
V
R G R
Z0
∂
R(x,λ,V) = [d (x,λ,V)](Id), (10)
V
∂λ − R
where
˜(x,t,λ,V) = ß(x,λ,V) (t,λ,V)⊤ (x,λ,V)ß(t,λ,V)⊤.
G R −R
Notations 1 For simplicity, we name the components of solutions by
Y (x,λ,p,q) Y (x,λ,p,q)
(x,λ,p,q) = 1 , ß(x,λ,p,q) = 2
R Z (x,λ,p,q) Z (x,λ,p,q)
(cid:20) 1 (cid:21) (cid:20) 2 (cid:21)
and we introduce the following quantities
a(x,λ,p,q) = [Y (x,λ,p,q)Z (x,λ,p,q)+Z (x,λ,p,q)Y (x,λ,p,q)],
1 2 1 2
−
b(x,λ,p,q) = [Y (x,λ,p,q)Y (x,λ,p,q) Z (x,λ,p,q)Z (x,λ,p,q)].
1 2 1 2
−
Now precise derivative expressions for AKNS potentials defined by (3). First, we
define L2(0,1)-gradients for multiple variable functions.
C
Inverse spectral problem for singular AKNS operator on [0,1]. 6
Definition 2.1 Let H be an Hilbert space. For a continuously differentiable complex
valued map f : (p,q) f(p,q), the L2(0,1)-gradient with respect to (p,q) is the vector
C
7→
valued function
∂f ∂f
f = ,
p,q
∇ ∂p ∂q
(cid:18) (cid:19)
∂f ∂f
where , resp. is the Riesz representant of the partial differential D f, resp. D f
p q
∂p ∂q
defined by
d f(v ,v ) = D f(v )+D f(v ), (v ,v ) H H.
p,q 1 2 p 1 q 2 1 2
∈ ×
Remark. If f is valued in Cn, this notation is understood component by component.
Corollary 2.1 (AKNS Gradients) For all (p,q) L2(0,1), we have
C
∈
∂
R(x,λ,p,q) (t) = ll (t) ß(x,λ,p,q)[2Y (t,λ,p,q)Z (t,λ,p,q)]
[0,x] 1 1
∂p
(cid:20) (cid:21)
h
+ (x,λ,p,q)a(t,λ,p,q) , (11)
R
∂ i
R(x,λ,p,q) (t) = ll (t) ß(x,λ,p,q) Z (t,λ,p,q)2 Y (t,λ,p,q)2
[0,x] 1 1
∂q −
(cid:20) (cid:21)
h
(cid:2) (cid:3)
+ (x,λ,p,q)b(t,λ,p,q) , (12)
R
∂ x i
R(x,λ,p,q) = ß(x,λ,p,q) Y (t,λ,p,q)2 +Z (t,λ,p,q)2
1 1
∂λ −
(cid:20) (cid:21) Z0
h
(cid:2) (cid:3)
+ (x,λ,p,q)[Y (t,λ,p,q)Y (t,λ,p,q)+Z (t,λ,p,q)Z (t,λ,p,q)] dt.(13)
1 2 1 2
R
i
2.2. Spectra
Conditionatx = 0 selects a solutioncollinear to , condition atx = 1reduces spectrum
R
to an eigenvalues-sequence. To this end, we set:
Notations 2 Let D(λ,V) be defined by:
D(λ,V) = (1,λ,V) u . (14)
β
R ·
Moreover, for all u = (a,b) C2, we define u⊥ by
∈
(a,b)⊥ = (b, a). (15)
−
Proposition 2.3 D is analytic in λ and V. The roots of λ D(λ,V) are exactly the
7→
eigenvalues for (1)-(2). Moreover, if V is real-valued, they are all simple.
Proof. Analyticity of D comes from . Since ,ß is a basis for the solutions of (1),
R {R }
the identification between eigenvalues and roots of λ D(λ,V) follows.
7→
Now suppose V is real-valued et let λ be an eigenvalue of the problem. Simplicity lies
0
on
∂D
( ,λ ,V) 2 = ( (1,λ ,V) u ⊥) (λ ,V). (16)
kR · 0 kL2R(0,1) − R 0 · β ∂λ 0
Inverse spectral problem for singular AKNS operator on [0,1]. 7
Indeed, from (9) and (10) we have
∂D
(λ ,V) = (ß(1,λ ,V) u ) ( ,λ ,V) 2 .
∂λ 0 − 0 · β kR · 0 kL2R(0,1)
Then, rewriting the wronskian of (1,λ ,V) and ß(1,λ ,V) in the orthonormal basis
0 0
R
u ,u ⊥ , we obtain (1,λ ,V) u ⊥ (ß(1,λ ,V) u ) = 1. (cid:3)
β β 0 β 0 β
R · ·
From now, V is defined by (3), corresponding to an AKNS operator.
(cid:8) (cid:9) (cid:0) (cid:1)
2.3. H1(0,1)-estimates
C
In order to obtain accurate asymptotics, we add some regularity on potentials. We use
this roundabout method not because of the singularity a/x in the equation, but because
of the AKNS operator itself. Indeed, contrary to the Schr¨odinger operator, there is no
explicit decreasing for the Green function with respect to λ; so we have to force it
G
allowing some derivation. For the regular case (a = 0), see for instance [11].
Theorem 2.1 For (p,q) (H1(0,1))2, we have
C
∈
a+1
x
(x,λ,p,q) R(x,λ) C V ln[2+ λx ]e|Imλ|x+CkVk2, (17)
R − ≤ k kHC1(0,1) 1+ λx | |
(cid:12) (cid:12) (cid:20) | |(cid:21)
un(cid:12)iformly on [0,1] C (H(cid:12) 1(0,1) H1(0,1)), where V 2 = p 2 + q 2 .
(cid:12) × × (cid:12) C × C k kHC1(0,1) k kHC1(0,1) k kHC1(0,1)
Proof. From relation (4) at k = 1 and (6), we have
x
R (x,λ,p,q) = S (x,λ) R (t,λ)⊤V(t)R (t,λ)dt
1 0 0 0
Z0
x
R (x,λ) S (t,λ)⊤V(t)R (t,λ)dt
0 0 0
−
Z0
x
= S (x,λ) q(t) R2(t,λ)2 R1(t,λ)2 +2p(t)R1(t,λ)R2(t,λ) dt
0 0 − 0 0 0
Z0
(cid:2)x (cid:0) (cid:1) (cid:3)
R (x,λ) q(t) S2(t,λ)R2(t,λ) S1(t,λ)R1(t,λ)
− 0 0 0 − 0 0
Z0
(cid:2) (cid:0) (cid:1)
+p(t) S1(t,λ)R2(t,λ)+S2(t,λ)R1(t,λ) dt.
0 0 0 0
We can write R (x,λ,p,q) = λ−a[X(q(cid:0))+Y(p)], where (cid:1)(cid:3)
1
η (λx) x
X(q) = − a−1 [j (λt)]2 [j (λt)]2 q(t)dt
a a−1
η (λx) −
(cid:20) a (cid:21)Z0
n o
j (λx) x
a−1
+ η (λt)j (λt) η (λt)j (λt) q(t)dt,
a a a−1 a−1
j (λx) −
(cid:20) − a (cid:21)Z0 h i
η (λx) x
a−1
Y(p) = 2j (λt)j (λt) p(t)dt
a−1 a
η (λx)
(cid:20) − a (cid:21)Z0 h i
j (λx) x
a−1
η (λt)j (λt)+η (λt)j (λt) p(t)dt.
a−1 a a a−1
− j (λx)
(cid:20) − a (cid:21)Z0 h i
Inverse spectral problem for singular AKNS operator on [0,1]. 8
Estimation for X(q):
Integrating by parts, we get
1 0
X(q) = q(x)
λ j (λx)
(cid:20) − a (cid:21)
1 x η (λx)j (λt)+j (λx)η (λt)
+ − a−1 a−1 a−1 a−1 j (λt)q′(t)dt.
a
λ η (λx)j (λt) j (λx)η (λt)
Z0 (cid:20) a a−1 − a a−1 (cid:21)
Estimates (A.2),(A.4) and Sobolev inequality q C q give
k k∞ ≤ k kHC1(0,1)
a+1
C λx
X(q) | | e|Imλ|x q . (18)
| | ≤ λ 1+ λx k kHC1(0,1)
| | (cid:18) | |(cid:19)
Estimation for Y(p):
Withnotationsfromlemmas Appendix A.1andAppendix A.2, integrationbyparts
gives :
η (λx) 1 x 1 x
Y(p) = a−1 F (λt)p(t) F (λt)p′(t)dt
1 1
η (λx) λ − λ
(cid:20) − a (cid:21)(cid:18)(cid:20) (cid:21)0 Z0 (cid:19)
j (λx) 1 x 1 x
a−1 F (λt)p(t) F (λt)p′(t)dt .
2 2
− j (λx) λ − λ
(cid:20) − a (cid:21)(cid:18)(cid:20) (cid:21)0 Z0 (cid:19)
When λx 1. Estimations A.2, A.3 and part (i) from lemmas Appendix A.1 and
| | ≤
Appendix A.2 lead to
a+1
C λx
Y(p) | | p e|Imλ|x.
| | ≤ λ 1+ λx k kHC1(0,1)
| | (cid:18) | |(cid:19)
When λx 1. Now, we only consider Y(p) second component, the proof is similar
| | ≥
for the first one. Terms to estimate contain :
g(x,t) := η (λx)F (λt) j (λx)F (λt), 0 t x.
a 1 a 2
− ≤ ≤
a+1
λx
If λt 1. As for λx 1, we get g(x,t) 2C | | e|Imλ|x.
| | ≤ | | ≤ | | ≤ 1+ λx
(cid:18) | |(cid:19)
If λt 1. Using points (ii) from lemmas Appendix A.1 and Appendix A.2,
| | ≥
expressions (A.6) and (A.7), it follows :
g(x,t) = η (λx)r (λt)
a a
aπ aπ
a cos λx ci(2λt)+sin λx Si(2λt) P (λx)
a
− − 2 − 2
h (cid:16) aπ (cid:17) (cid:16) aπ(cid:17) i
+a sin λx ci(2λt) cos λx Si(2λt) I (λx)
a
− 2 − − 2
h (cid:16) (cid:17) (cid:16) (cid:17) aiπ
+(P (λx)p (λt) I (λx)q (λt))cos λ(x 2t)
a a a a
− − − 2
h aπi
(P (λx)q (λt)+I (λx)p (λt))sin λ(x 2t) .
a a a a
− − − 2
h i
(To lighten, the polynomial variable X is replaced by 1/X.) First term is bounded
by Ce|Imλ|x thanks to (A.3). The last two terms are uniformly bounded by
Ce|Imλ|(x−2t) on the considered area. Now remains the following expression
aπ aπ
h(x,t) := cos λx ci(2λt)+sin λx Si(2λt).
− 2 − 2
(cid:16) (cid:17) (cid:16) (cid:17)
Inverse spectral problem for singular AKNS operator on [0,1]. 9
According to [1] and [2], we have
log(z2) sinz 1 cosz 1
ci(z) = γ + 1+ 1+ ,
− − 2 z O1 z2 − z2 O2 z2
(cid:18) (cid:18) (cid:19)(cid:19) (cid:18) (cid:18) (cid:19)(cid:19)
π√z2 cosz 1 sinz 1
Si(z) = 1+ 1+ .
2z − z O1 z2 − z2 O2 z2
(cid:18) (cid:18) (cid:19)(cid:19) (cid:18) (cid:18) (cid:19)(cid:19)
Thus, we get
log(2λt)2 aπ π (2λt)2 aπ
h(x,t) = γ + cos λx + sin λx
− 2 − 2 2λt − 2
(cid:20) (cid:21) p
(cid:16) (cid:17) (cid:16) (cid:17)
1 1 aπ
1+ sin λ(x 2t)
1
−2λt O (2λt)2 − − 2
(cid:18) (cid:18) (cid:19)(cid:19)
h i
1 1 aπ
1+ cos λ(x 2t) .
−(2λt)2 O2 (2λt)2 − − 2
(cid:18) (cid:18) (cid:19)(cid:19)
h i
The last three terms are also uniformly controlled by Ce|Imλ|(x−2t); the first one is
boundedbyCln λt e|Imλ|x. Combiningtheaboveestimates, weobtainthefollowing
| |
uniform estimate
a+1
C λx
Y(p) | | ln[2+ λ x] p e|Imλ|x. (19)
| | ≤ λ 1+ λx | | k kHC1(0,1)
| | (cid:18) | |(cid:19)
Relations (18)-(19) and the concavity rule
x + y x 2 + y 2
(x,y) R2, | | | | | | | | ,
∀ ∈ 2 ≤ 2
r
imply
a+1
C λx
R (x,λ,p,q) | | ln[2+ λ x] V e|Imλ|x. (20)
| 1 | ≤ λ a+1 1+ λx | | k kHC1(0,1)
| | (cid:18) | |(cid:19)
(cid:3)
From this estimate, as in the proof of lemma 2.1, we deduce estimate (17).
Remark. A similar computation for the Dirac operator is not easy, even if a = 0.
Indeed, when we compute the term R , we do not only get a term, loosely speaking,
1
in (1/λ) but also in (1). And when iterating this, we get at each time a new
O O
term (1) and (1/λ). A way through this problem is given in [22] using the latter
O O
gauge transformation to deduce, for any a N, some partial results from AKNS to
∈
Dirac operator: spectrum, asymptotic expansion for eigenvalues and eigenvectors, Borg-
Levinson theorem type...
2.4. L2(0,1)-Estimates
C
To transform H1(0,1)-estimates into L2(0,1)-estimates, we need an auxiliary lemma
C C
(for the regular case, see [3], [17] and [10]).
Lemma 2.2 Let V L2(0,1) L2(0,1), r 0, ε 0 and let V H1(0,1) H1(0,1)
0 C C 0 ǫ C C
∈ × ≥ ≥ ∈ ×
such that V V < ε. Then, for all V L2(0,1) L2(0,1) such that V V < r
k 0 − εk2 ∈ C × C k − 0k2 0
Inverse spectral problem for singular AKNS operator on [0,1]. 10
and for all (x,λ) [0,1] C∗, we have
∈ ×
ln λ
|R(x,λ,p,q)−R(x,λ)| ≤ C r0 +ε+ λ| |kVεkHC1(0,1)
(cid:18) | | (cid:19)
a
x
e|Imλ|x+CkVk2. (21)
× 1+ λx
(cid:18) | |(cid:19)
Proof. Since V H1(0,1) H1(0,1), estimate (20) obtained during the proof of
ǫ C C
∈ ×
Theorem 2.1 becomes
a+1
C λx
R (x,λ,V ) | | ln[2+ λ x] V e|Imλ|x.
| 1 ε | ≤ λ a+1 1+ λx | | k εkHC1(0,1)
| | (cid:18) | |(cid:19)
Using V V < ε and V V < r in (8), estimations (A.4) and (A.2) lead to
k 0 − εk2 k − 0k2 0
a
C λx
R (x,λ,p,q) R (x,λ,V ) | | (r +ε)e|Imλ|x.
| 1 − 1 ε | ≤ λ a 1+ λx 0
| | (cid:18) | |(cid:19)
Combining these two inequalities, we get
a
x ln[2+ λ x]
R (x,λ,p,q) C r +ε+ | | V e|Imλ|x.
| 1 | ≤ 1+ λx 0 λ k εkHC1(0,1)
(cid:18) | |(cid:19) (cid:18) | | (cid:19)
Iterating this with (4), we deduce for every n N
∈
Cn+1 ln[2+ λ x]
R (x,λ,p,q) r +ε+ | | V
| n+1 | ≤ n! 0 λ k εkHC1(0,1)
(cid:18) | | (cid:19)
x a 1 n
e|Imλ|x V(t) dt .
× 1+ λx | |
(cid:18) | |(cid:19) (cid:18)Z0 (cid:19)
(cid:3)
Then, summing up, estimation (21) follows.
We now deduce the following
Proposition 2.4 Let (p,q) (L2(0,1))2, we have uniformly on [0,1],
C
∈
a
x
(x,λ,p,q) = R(x,λ)+o e|Imλ|x , λ . (22)
R 1+ λx | | → ∞
(cid:20)(cid:18) | |(cid:19) (cid:21)
Proof. From Lemma 2.2 with r = 0, given δ > 0 there exists λ > 0 such that
0 δ
a
x
(x,λ,p,q) R(x,λ) δ e|Imλ|x+CkVk2,
|R − | ≤ 1+ λx
(cid:18) | |(cid:19)
for all λ such that λ > λ . (cid:3)
δ
| |
2.5. Spectrum localization
Theorem 2.2 (Counting Lemma)
Let (p ,q ) L2(0,1) L2(0,1), there exist ε > 0 and an integer N > 0 such that for
0 0 C C 0
∈ ×
all (p,q) L2(0,1) L2(0,1) with (p,q) (p ,q ) < ε, the following statements
∈ C × C k − 0 0 kL2C(0,1)
hold:
For all n > N , λ D(λ,p,q) has exactly one root in λ nπ + aπ +β < π,
• | | 0 7→ − 2 2
(cid:12) (cid:0) (cid:1)(cid:12)
(cid:12) (cid:12)