THE COMPLEX AIRY OPERATOR ON THE LINE
WITH A SEMI-PERMEABLE BARRIER
Denis Grebenkov, Bernard Helffer, Raphael Henry
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Denis Grebenkov, Bernard Helffer, Raphael Henry. THE COMPLEX AIRY OPERATOR ON
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1
2
THE COMPLEX AIRY OPERATOR ON THE LINE WITH A
SEMI-PERMEABLE BARRIER
3
DENIS S. GREBENKOV∗ , BERNARD HELFFER† , AND RAPHAEL HENRY‡
4
5
6
7
8
9
Abstract. We consider a suitable extension of the complex Airy operator, −d2 /dx2 + ix, on the
real line with a transmission boundary condition at the origin. We provide a rigorous definition of
this operator and study its spectral properties. In particular, we show that the spectrum is discrete,
the space generated by the generalized eigenfunctions is dense in L2 (completeness), and we analyze
the decay of the associated semi-group. We also present explicit formulas for the integral kernel of
the resolvent in terms of Airy functions, investigate its poles, and derive the resolvent estimates.
10
11
Key words. Airy operator, transmission boundary condition, spectral theory, Bloch-Torrey
equation
12
AMS subject classifications. 35P10, 47A10, 47A75
18
19
20
21
1. Introduction. The transmission boundary condition which is considered in
this article appears in various exchange problems such as molecular diffusion across
semi-permeable membranes [37, 34, 33], heat transfer in composite materials [11,
18, 8], or transverse magnetization evolution in nuclear magnetic resonance (NMR)
experiments [20]. In the simplest setting of the latter case, one considers the local
transverse magnetization G(x, y ; t) produced by the nuclei that started from a fixed
initial point y and diffused in a constant magnetic field gradient g up to time t. This
magnetization is also called the propagator or the Green function of the Bloch-Torrey
equation [39]:
22
(1.1)
23
with the initial condition
24
(1.2)
25
26
where δ(x) is the Dirac distribution, D the intrinsic diffusion coefficient, ∆ = ∂ 2 /∂x21 +
. . . + ∂ 2 /∂x2d the Laplace operator in Rd , γ the gyromagnetic ratio, and x1 the coordinate in a prescribed direction. From an experimental point of view, the local
transverse magnetization is too weak to be detected but the macroscopic signal produced by all the nuclei in a sample can be measured. In other words, one can access
the double integral of G(x, y ; t) ρ(y) over the starting and arrival points y and x,
where ρ(y) is the initial density of the nuclei in the sample. The microstructure
of the sample, which can eventually be introduced through boundary conditions to
(1.1), affects the motion of the nuclei, the magnetization G(x, y ; t), and the resulting
macroscopic signal. Measuring the signal at various times t and magnetic field gradients g, one aims at inferring structural properties of the sample [19]. Although this
13
14
15
16
17
27
28
29
30
31
32
33
34
35
∂
G(x, y ; t) = (D∆ − iγgx1 ) G(x, y ; t) ,
∂t
G(x, y ; t = 0) = δ(x − y),
∗ Laboratoire de Physique de la Matière Condensée, CNRS–Ecole Polytechnique, University ParisSaclay, 91128 Palaiseau, France ([email protected], http://pmc.polytechnique.fr/
pagesperso/dg/).
† Laboratoire de Mathématiques Jean Leray, Université de Nantes 2 rue de la Houssinière 44322
Nantes, France, and Laboratoire de Mathématiques, Université Paris-Sud, CNRS, Univ. Paris Saclay,
France ([email protected]).
‡ Laboratoire de Mathématiques, Université Paris-Sud, CNRS, Univ.
Paris Saclay, France
([email protected]).
1
This manuscript is for review purposes only.
2
36
37
38
39
40
41
42
43
44
D. S. GREBENKOV, B. HELFFER, AND R. HENRY
non-invasive experimental technique has found numerous applications in material sciences and medicine, mathematical aspects of this formidable inverse problem remain
poorly understood. Even the forward problem of relating a given microstructure to
the macroscopic signal is challenging because of the non-selfadjoint character of the
Bloch-Torrey operator D∆ − iγgx1 in (1.1). In particular, the spectral properties of
this operator were rigorously established only on the line R (no boundary condition)
and on the half-axis R+ with Dirichlet or Neumann boundary conditions (see Sec. 3).
Throughout this paper, we focus on the one-dimensional situation
(d = 1), in which the operator
Dx2 + ix = −
45
d2
+ ix
dx2
is called the complex Airy operator and appears in many contexts: mathematical
physics, fluid dynamics, time dependent Ginzburg-Landau problems and also as an
interesting toy model in spectral theory (see [3]). We will consider a suitable exten+
+
49 sion A1 of this differential operator and its associated evolution operator e−tA1 . The
+
50 Green function G(x, y ; t) is the distribution kernel of e−tA1 . A separate article will
51 address this operator in higher dimensions [24].
46
47
48
52
53
54
For the problem on the line R, an intriguing property is that this non self-adjoint
operator, which has compact resolvent, has empty spectrum (see Section 3.1). How55 ever, the situation is completely different on the half-line R+ . The eigenvalue problem
56
(Dx2 + ix)u = λu,
57
58
59
60
61
for a spectral pair (u, λ) with u ∈ H 2 (R+ ) and xu ∈ L2 (R+ ) has been thoroughly
analyzed for both Dirichlet (u(0) = 0) and Neumann (u0 (0) = 0) boundary conditions. The spectrum consists of an infinite sequence of eigenvalues of multiplicity one
explicitly related to the zeros of the Airy function (see [36, 26]). The space generated by the eigenfunctions is dense in L2 (R+ ) (completeness property) but there is
no Riesz basis of eigenfunctions.1 Finally, the decay of the associated semi-group has
been analyzed in detail. The physical consequences of these spectral properties for
NMR experiments have been first revealed by Stoller, Happer and Dyson [36] and
then thoroughly discussed in [15, 19, 22].
62
63
64
65
66
67
In this article, we consider another problem for the complex Airy operator on the
line but with a transmission condition at 0 which reads [22]:
0
u (0+ ) = u0 (0− ) ,
69 (1.3)
u0 (0)
= κ u(0+ ) − u(0− ) ,
68
70
71
72
73
74
75
where κ ≥ 0 is a real parameter. In physical terms, the transmission condition accounts for the diffusive exchange between two media R− and R+ across the barrier
at 0, while κ is defined as the ratio between the barrier permeability and the bulk
diffusion coefficient. This situation is particularly relevant for biological samples and
applications [19, 21, 22]. The case κ = 0 corresponds to two independent Neumann
problems on R− and R+ for the complex Airy operator. When κ tends to +∞, the
1
We recall that a collection of vectors (xk ) in a Hilbert space H is called Riesz basis if it is an
image of an orthonormal basis in H under some isomorphism.
This manuscript is for review purposes only.
THE COMPLEX AIRY OPERATOR ON THE LINE WITH A SEMI-PERMEABLE BARRIER3
97
second relation in (1.3) becomes the continuity condition, u(0+ ) = u(0− ), and the
barrier disappears. As a consequence, the problem tends (at least formally) to the
standard problem for the complex Airy operator on the line.
The main purpose of this paper is to define the complex Airy operator with transmission (Section 4) and then to analyze its spectral properties. Before starting the
analysis of the complex Airy operator with transmission, we first recall in Section 2
the spectral properties of the one-dimensional Laplacian with the transmission condition, and summarize in Section 3 the known properties of the complex Airy operator.
New properties are also established concerning the Robin boundary condition and
the behavior of the resolvent for real λ going to +∞. In Section 4 we will show that
2
the complex Airy operator A+
1 = Dx + ix on the line R with a transmission property (1.3) is well defined by an appropriate sesquilinear form and an extension of the
Lax-Milgram theorem. Section 5 focuses on the exponential decay of the associated
semi-group. In Section 6, we present explicit formulas for the integral kernel of the
resolvent and investigate its poles. In Section 7, the resolvent estimates as |Im λ| → 0
are discussed. Finally, the proof of completeness is reported in Section 8. In five
Appendices, we recall the basic properties of Airy functions (Appendix A), determine
the asymptotic behavior of the resolvent as λ → +∞ for extensions of the complex
Airy operator on the line (Appendix B) and in the semi-axis (Appendix C), give
the statement of the needed Phragmen-Lindelöf theorem (Appendix D) and finally
describe the numerical method for computing the eigenvalues (Appendix E).
We summarize our main results in the following:
98
99
100
+
Theorem 1. The semigroup exp(−tA+
1 ) is contracting. The operator A1 has a
discrete spectrum {λn (κ)}. The eigenvalues λn (κ) are determined as (complex-valued)
solutions of the equation
101
(1.4)
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
2πAi0 (e2πi/3 λ)Ai0 (e−2πi/3 λ) + κ = 0,
where Ai0 (z) is the derivative of the Airy function.
For all κ ≥ 0, there exists N such that, for all n ≥ N , there exists a unique eigenvalue
±2πi/3 0
± −1
±
an , and a0n are the zeros of
), where λ±
of A+
n = e
1 in the ball B(λn , 2κ|λn |
Ai0 (z).
Finally, for any κ ≥ 0 the space generated by the generalized eigenfunctions of the
107 complex Airy operator with transmission is dense in L2 (R− ) × L2 (R+ ).
102
103
104
105
106
108
109
110
111
112
Remark 2. Numerical computations suggest that all the spectral projections have
rank one (no Jordan block) but we shall only prove in Proposition 36 that there are
at most a finite number of eigenvalues with nontrivial Jordan blocks. It will be shown
in [6] that the eigenvalues are actually simple. Hence one can replace “generalized
eigenfunctions” by “eigenfunctions” in the Theorem 1.
2. The free Laplacian with a semi-permeable barrier. As an enlighting
exercise, let us consider in this section the case of the free one-dimensional Laplacian
d2
115 − dx
2 on R \ {0} with the transmission condition (1.3) at x = 0. We work in the
116 Hilbert space
113
114
117
H := L2− × L2+ ,
118
119
120
where L2− := L2 (R− ) and L2+ := L2 (R+ ) .
An element u ∈ L2− × L2+ will be denoted by u = (u− , u+ ) and we shall use the
s
s
notation H−
= H s (R− ) , H+
= H s (R+ ) for s ≥ 0 .
This manuscript is for review purposes only.
4
D. S. GREBENKOV, B. HELFFER, AND R. HENRY
121
So (1.3) reads
122
(2.1)
123
124
In order to define appropriately the corresponding operator, we start by considering a sesquilinear form defined on the domain
125
1
1
V = H−
× H+
.
126
127
128
The space V is endowed with the Hilbert norm k · kV defined for all u = (u− , u+ ) in
V by
kuk2V = ku− k2H 1 + ku+ k2H 1 .
129
We then define a Hermitian sesquilinear form aν acting on V × V by the formula
Z 0 0
aν (u, v) =
u0− (x)v̄−
(x) + ν u− (x)v̄− (x) dx
u0+ (0)
u0+ (0)
= u0− (0) ,
= κ u+ (0) − u− (0) .
−
130
−∞
Z +∞
0
u0+ (x)v̄+
(x) + ν u+ (x)v̄+ (x) dx
0
+ κ u+ (0) − u− (0) v+ (0) − v− (0) ,
+
131
132
+
(2.2)
for all pairs u = (u− , u+ ) and v = (v− , v+ ) in V . For z ∈ C, z̄ denotes the complex
conjugate of z. The parameter ν ≥ 0 will be determined later to ensure the coercivity
135 of aν .
133
134
136
Lemma 3. The sesquilinear form aν is continuous on V .
137
138
139
Proof
We want to show that, for any ν ≥ 0 , there exists a positive constant c such that, for
all (u, v) ∈ V × V ,
140
(2.3)
|aν (u, v)| ≤ c kukV kvkV .
We have, for some c0 > 0 ,
Z 0 0
0
142
u
(x)v̄
(x)
+
ν
u
(x)v̄
(x)
dx
−
−
−
−
−∞
Z +∞ 0
143
+
u0+ (x)v̄+
(x) + ν u+ (x)v̄+ (x) dx ≤ c0 kukV kvkV .
141
0
144
On the other hand,
145
(2.4)
|u+ (0)|2 = −
Z
+∞
(u+ ū+ )0 (x) dx ≤ 2 kukL2 ku0 kL2 ,
0
and similarly for |u− (0)| , |v+ (0)|2 and |v− (0)|2 . Thus there exists c1 > 0 such that,
for all (u, v) ∈ V × V ,
148
κ u− (0) − u+ (0) v− (0) − v+ (0) ≤ c1 kukV kvkV ,
146
147
149
150
151
152
153
2
and (2.3) follows with c = c0 + c1 .
The coercivity of the sesquilinear form aν for ν large enough is proved in the
following lemma. It allows us to define a closed operator associated with aν by using
the Lax-Milgram theorem.
This manuscript is for review purposes only.
THE COMPLEX AIRY OPERATOR ON THE LINE WITH A SEMI-PERMEABLE BARRIER5
Lemma 4. There exist ν0 > 0 and α > 0 such that, for all ν ≥ ν0 ,
154
155
(2.5)
aν (u, u) ≥ α kuk2V .
∀u ∈ V ,
Proof
The proof is elementary for κ ≥ 0. For completeness, we provide below the proof for
the case κ < 0 (in which an additional difficulty occurs), but we will keep considering
the physically relevant case κ ≥ 0 throughout the paper.
Using the estimate (2.4) as well as the Young inequality
1 2
161
δe + δ −1 f 2 ,
∀e, f, δ > 0 , ef ≤
2
156
157
158
159
160
we get that, for all ε > 0 , there exists C(ε) > 0 such that, for all u ∈ V ,
Z 0
Z +∞
2
0
2
0
2
163 (2.6)
u− (0) − u+ (0) ≤ ε
|u− (x)| dx +
|u+ (x)| dx + C(ε)kuk2L2 .
162
−∞
164
0
Thus for all u ∈ V we have
Z
0
|u0− (x)|2
aν (u, u) ≥ (1 − |κ|ε)
165
Z
dx +
−∞
166
(2.7)
+∞
|u0+ (x)|2
dx
0
+ ν − |κ|C(ε) kuk2L2 .
167
168
169
170
171
Choosing ε < |κ|−1 and ν > |κ|C(ε) , we get (2.5).
175
and the operator T0 satisfies, for all (u, v) ∈ D(T0 ) × V ,
The sesquilinear form aν being symmetric, continuous and coercive in the sense of
(2.5) on V × V , we can use the Lax-Milgram theorem [26] to define a closed, densely
defined selfadjoint operator Sν associated with aν . Then we set T0 = Sν − ν . By
172 construction, the domain of Sν and T0 is
173
D(T0 ) = u ∈ V : v 7→ aν (u, v) can be extended continuously
174 (2.8)
on L2− × L2+ ,
176
aν (u, v) = hT0 u, vi + νhu, vi .
177
Now we look for an explicit description of the domain (2.8). The antilinear form
a(u, ·) can be extended continuously on L2− × L2+ if and only if there exists wu =
(wu− , wu+ ) ∈ L2− × L2+ such that
178
179
180
∀v ∈ V ,
aν (u, v) = hwu , vi .
According to the expression (2.2), we have necessarily
182
wu = − u00− + ν u− , −u00+ + ν u+ ∈ L2− × L2+ ,
181
where u00− and u00+ are a priori defined in the sense of distributions respectively in
D0 (R− ) and D0 (R+ ). Moreover (u− , u+ ) has to satisfy conditions (1.3). Consequently
we have
n
1
1
186
D(T0 ) = u = (u− , u+ ) ∈ H−
× H+
: (u00− , u00+ ) ∈ L2− × L2+
o
187
and u satisfies conditions (1.3) .
183
184
185
This manuscript is for review purposes only.
6
D. S. GREBENKOV, B. HELFFER, AND R. HENRY
188
Finally we have introduced a closed, densely defined selfadjoint operator T0 acting by
189
T0 u = −u00
on (−∞, 0) ∪ (0, +∞) , with domain
2
2
191
D(T0 ) = u ∈ H−
× H+
: u satisfies conditions (1.3) .
190
192
193
194
195
196
Note that at the end T0 is independent of the ν chosen for its construction.
We observe also that because of the transmission conditions (1.3), the operator T0
might not be positive when κ < 0, hence there can be a negative spectrum, as can be
seen in the following statement.
Proposition 5. For all κ ∈ R , the essential spectrum of T0 is
197
(2.9)
σess (T0 ) = [0, +∞) .
198
Moreover, if κ ≥ 0 the operator T0 has empty discrete spectrum and
199
(2.10)
σ(T0 ) = σess (T0 ) = [0, +∞) .
On the other hand, if κ < 0 there exists a unique negative eigenvalue −4κ2 , which is
simple, and
202 (2.11)
σ(T0 ) = − 4κ2 ∪ [0, +∞) .
200
201
203
204
205
206
207
Proof
Let us first prove that [0, +∞) ⊂ σess (T0 ) . This can be achieved by a standard
singular sequence construction.
Let (aj )j∈N be a positive increasing sequence such that, for all j ∈ N , aj+1 − aj >
2j + 1 . Let χj ∈ C0∞ (R) (j ∈ N) such that
(p)
Supp χj ⊂ (aj − j, aj + j) , ||χj ||L2+ = 1 and sup |χj | ≤
208
C
, p = 1, 2 ,
jp
for some C independent of j. Then, for all r ≥ 0 , the sequence urj (x) = 0, χj (x)eirx
210 is a singular sequence for T0 corresponding to z = r 2 in the sense of [17, Definition
211 IX.1.2]. Hence according to [17, Theorem IX.1.3], we have [0, +∞) ⊂ σess (T0 ) .
209
212
213
214
215
216
Now let us prove that (T0 − µ) is invertible for all µ ∈ (−∞, 0) if κ ≥ 0 , and for
all µ ∈ (−∞, 0) \ {−4κ2 } if κ < 0 .
Let µ < 0 and f = (f− , f+ ) ∈ L2− × L2+ . We are going to determine explicitly the
solutions u = (u− , u+ ) to the equation
217
(2.12)
T0 u = µ u + f .
Any solution of the equation −u00± = µ u± + f± has the form
(2.13)
Z x
√
√
√
√
1
219
u± (x) = √
f± (y) e− −µ (x−y) − e −µ (x−y) dy + A± e −µ x + B± e− −µ x ,
2 −µ 0
218
for some A± , B± ∈ R .
We shall now determine A+ , A− , B+ and B− such that (u− , u+ ) belongs to the
domain D(T0 ) . The conditions (1.3) yield
A+ − B+ = A− − B− ,
√
223
−µ A+ − B+
= −κ A− + B− − A+ − B+ .
220
221
222
This manuscript is for review purposes only.
THE COMPLEX AIRY OPERATOR ON THE LINE WITH A SEMI-PERMEABLE BARRIER7
224
225
2
Moreover, the decay conditions at ±∞ imposed by u± ∈ H±
lead to the following
values for A+ and B− :
226
1
(2.14) A+ = √
2 −µ
Z
+∞
√
f+ (y)e−
−µ y
1
B− = √
2 −µ
dy ,
0
Z
0
√
f− (y)e
−µ y
dy .
−∞
The remaining constants A− and B+ have to satisfy the system
A− +B+ = A+ + B− ,
√
√
228 (2.15)
−µ + κ B+ =
−µ − κ A+ + κB− .
−κA− +
227
229
230
231
232
We then notice that the equation (2.12) has a unique solution u = (u− , u+ ) if and
only if κ ≥ 0 or µ 6= −4κ2 .
Finally in the case κ < 0 and µ = −4κ2 , the homogeneous equation associated with
(2.12) (i.e with f ≡ 0) has a one-dimensional space of solutions, namely
u(x) = − Ke−2κx , Ke2κx
233
234
235
236
237
238
with K ∈ R . Consequently if κ < 0 , the eigenvalue µ = −4κ2 is simple, and the
desired statement is proved.
The expression (2.14) along with the system (2.15) yield the values of A− and
B+ when µ ∈
/ σ(T0 ) :
Z +∞
√
2κ
f+ (y)e− −µ y dy
A− = √
√
2 −µ −µ + 2κ 0
Z 0
√
1
+ √
f− (y)e −µ y dy
2 −µ + 2κ −∞
239
240
and
B+ =
241
242
243
2
√
1
−µ + 2κ
+∞
Z
f+ (y)e−
√
−µy
dy
0
2κ
+ √
√
2 −µ −µ + 2κ
Z
0
√
f− (y)e
−µ y
dy .
−∞
Using (2.13), we are then able to obtain the expression of the integral kernel of (T0 −
µ)−1 . More precisely we have, for all f = (f− , f+ ) ∈ L2− × L2+ ,
−1
(T0 − µ)
244
=
R−−
µ
R+−
µ
R−+
µ
R++
µ
,
245
246
where for ε, σ ∈ {−, +} the operator Rεµσ : Rσ → Rε is an integral operator whose
kernel (still denoted Rεµσ ) is given for all (x, y) ∈ Rε × Rσ by
247
(2.16)
248
249
Noticing that the first term in the right-hand side of (2.16) is the integral kernel of
the Laplacian on R , and that the second term is the kernel of a rank one operator,
√
√
1
1
e− −µ(|x|+|y|) .
Rε,σ
e− −µ |x−y| + εσ √
µ (x, y) = √
2 −µ
2 −µ + 2κ
This manuscript is for review purposes only.
8
250
251
D. S. GREBENKOV, B. HELFFER, AND R. HENRY
we finally get the following expression of (T0 − µ)−1 as a rank one perturbation of the
Laplacian:
(T0 − µ)−1 = (−∆ − µ)−1
1
+ √
2 −µ + 2κ
252
h · , `µ i− (`µ )−
−h · , `µ i− (`µ )+
−h · , `µ i+ (`µ )−
h · , `µ i+ (`µ )+
,
√
253
254
255
where `µ (x) = e− −µ |x| and h · , · i± denotes the L2 scalar product on R± .
Here the operator (−∆ − µ)−1 denotes the operator acting on L2− × L2+ like the
resolvent of the Laplacian on L2 (R):
256
(−∆ − µ)−1 (u− , u+ ) := (−∆ − µ)−1 (u− 1(−∞,0) + u+ 1(0,+∞) ) ,
257
258
composed with the map L2 (R) 3 v 7→ (v|R− , v|R+ ) ∈ L2− × L2+ .
3. Reminder on the complex Airy operator. Here we recall relatively basic
facts coming from [32, 3, 10, 26, 25, 28, 29] and discuss new questions concerning
261 estimates on the resolvent and the Robin boundary condition. Complements will also
262 be given in Appendices A, B and C.
259
260
263
264
265
266
3.1. The complex Airy operator on the line. The complex Airy operator
on the line can be defined as the closed extension A+ of the differential operator
− ∗
−
∞
+
2
2
A+
0 := Dx + i x on C0 (R). We observe that A = (A0 ) with A0 := Dx − i x and
that its domain is
267
D(A+ ) = {u ∈ H 2 (R) , x u ∈ L2 (R)} .
268
269
In particular, A+ has a compact resolvent. It is also easy to see that −A+ is the
generator of a semi-group St of contraction,
270
(3.1)
271
272
Hence the results of the theory of semi-groups can be applied (see for example [12]).
In particular, we have, for Re λ < 0 ,
273
(3.2)
274
A very special property of this operator is that, for any a ∈ R,
275
(3.3)
276
277
278
where Ta is the translation operator: (Ta u)(x) = u(x − a) .
As an immediate consequence, we obtain that the spectrum is empty and that the
resolvent of A+ ,
279
G0+ (λ) = (A+ − λ)−1 ,
St = exp(−tA+ ) .
||(A+ − λ)−1 || ≤
1
.
|Re λ|
Ta A+ = (A+ − ia) Ta ,
280
which is defined for any λ ∈ C, satisfies
281
(3.4)
282
The most interesting property is the control of the resolvent for Re λ ≥ 0.
||(A+ − λ)−1 || = ||(A+ − Re λ)−1 || .
This manuscript is for review purposes only.
THE COMPLEX AIRY OPERATOR ON THE LINE WITH A SEMI-PERMEABLE BARRIER9
Proposition 6 (W. Bordeaux-Montrieux [10]).
As Re λ → +∞, we have
r
1
3
π
4
285 (3.5)
||G0+ (λ)|| ∼
(Re λ)− 4 exp
(Re λ) 2 ,
2
3
283
284
286
where f (λ) ∼ g(λ) means that the ratio f (λ)/g(λ) tends to 1 in the limit λ → +∞.
287
288
This improves a previous result (see Appendix B) by J. Martinet [32] (see also in
[26, 25]) who also proved2
Proposition 7.
kG0+ (λ)kHS = kG0+ (Re λ)kHS ,
289
(3.6)
290
and
291
(3.7)
292
293
294
295
where k · kHS is the Hilbert-Schmidt norm. This is consistent with the well-known
translation invariance properties of the operator A+ , see [26]. The comparison between the HS-norm and the norm in L(L2 (R)) immediately implies that Proposition
7 gives the upper bound in Proposition 6.
kG0+ (λ)kHS ∼
r
1
π
(Re λ)− 4 exp
2
3
4
(Re λ) 2
3
as Re λ → +∞ ,
3.2. The complex Airy operator on the half-line: Dirichlet case. It is
not difficult to define the Dirichlet realization A±,D of Dx2 ± ix on R+ (the analysis on
298 the negative semi-axis is similar). One can use for example the Lax-Milgram theorem
299 and take as form domain
296
297
1
V D := {u ∈ H01 (R+ ) , x 2 u ∈ L2+ } .
300
301
It can also be shown that the domain is
2
DD := {u ∈ V D , u ∈ H+
}.
302
303
This implies
304
305
306
Proposition 8. The resolvent G ±,D (λ) := (A±,D − λ)−1 is in the Schatten class
C for any p > 32 (see [16] for definition), where A±,D is the Dirichlet realization of
Dx2 ± ix, as emphasized by the superscript D.
p
More precisely we provide the distribution kernel G −,D (x, y ; λ) of the resolvent for
the complex Airy operator Dx2 − ix on the positive semi-axis with Dirichlet bound309 ary condition at the origin (the results for G +,D (x, y ; λ) are similar). Matching the
310 boundary conditions, one gets
(3.8)

Ai(e−iα w ) 
2π Ai(e−iα wy0 ) Ai(eiα wx )Ai(e−iα w0 )




−Ai(e−iα wx )Ai(eiα w0 ) (0 < x < y) ,
311
G −,D (x, y ; λ) =
−iα
wx ) iα
−iα

2π Ai(e
w0 )

Ai(e−iα w0 ) Ai(e wy )Ai(e



−iα
−Ai(e wy )Ai(eiα w0 ) (x > y) ,
307
308
2 The
coefficient was wrong in [32] and is corrected here, see Appendix B.
This manuscript is for review purposes only.
10
312
D. S. GREBENKOV, B. HELFFER, AND R. HENRY
where Ai(z) is the Airy function,
wx = ix + λ,
313
314
and
α = 2π/3 .
315
316
The above expression can also be written as
317
(3.9)
G −,D (x, y ; λ) = G0− (x, y ; λ) + G1−,D (x, y ; λ),
where G0− (x, y ; λ) is the resolvent for the complex Airy operator Dx2 − ix on the whole
line,
(
2πAi(eiα wx )Ai(e−iα wy ) (x < y),
−
320 (3.10)
G0 (x, y ; λ) =
2πAi(e−iα wx )Ai(eiα wy ) (x > y),
318
319
321
and
322
(3.11)
323
324
325
326
327
The resolvent is compact. The poles of the resolvent are determined by the zeros of
Ai(e−iα λ), i.e., λn = eiα an , where the an are zeros of the Airy function: Ai(an ) = 0 .
The eigenvalues have multiplicity 1 (no Jordan block). See Appendix A.
G1−,D (x, y ; λ) = −2π
Ai(eiα λ)
Ai e−iα (ix + λ) Ai e−iα (iy + λ) .
Ai(e−iα λ)
As a consequence of the analysis of the numerical range of the operator, we have
Proposition 9.
||G ±,D (λ)|| ≤
1
, if Re λ < 0 ;
|Re λ|
328
(3.12)
329
and
330
(3.13)
331
332
This proposition together with the Phragmen-Lindelöf principle (Theorem 54) and
Proposition 8 implies (see [2] or [16])
333
334
Proposition 10. The space generated by the eigenfunctions of the Dirichlet realization A±,D of Dx2 ± ix is dense in L2+ .
335
336
337
It is proven in [28] that there is no Riesz basis of eigenfunctions.
At the boundary of the numerical range of the operator, it is interesting to analyze
the behavior of the resolvent. Numerical computations lead to the observation that
338
(3.14)
339
340
341
||G ±,D (λ)|| ≤
1
, if ∓ Im λ > 0 .
|Im λ|
lim ||G ±,D (λ)||L(L2+ ) = 0 .
λ→+∞
As a new result, we will prove
Proposition 11. When λ tends to +∞, we have
(3.15)
1
1
||G ±,D (λ)||HS ≈ λ− 4 (log λ) 2 .
This manuscript is for review purposes only.
THE COMPLEX AIRY OPERATOR ON THE LINE WITH A SEMI-PERMEABLE BARRIER
11
342
343
The convention “A(λ) ≈ B(λ) as λ → +∞” means that there exist C and λ0 such
that
|A(λ)|
1
≤
≤C,
C
|B(λ)|
344
345
346
347
348
∀λ ≥ λ0 ,
or, in other words, A = O(|B|) and B = O(|A|).
The proof of this proposition will be given in Appendix C.
Note that, as ||G ±,D (λ)||L(L2 ) ≤ ||G ±,D (λ)||HS , the estimate (3.15) implies (3.14).
3.3. The complex Airy operator on the half-line: Neumann case. Similarly, we can look at the Neumann realization A±,N of Dx2 ± ix on R+ (the analysis
351 on the negative semi-axis is similar).
352
One can use for example the Lax-Milgram theorem and take as form domain
349
350
1
353
1
V N = {u ∈ H+
, x 2 u ∈ L2+ }.
354
355
356
We recall that the Neumann condition appears when writing the domain of the operator A±,N .
As in the Dirichlet case (Proposition 8), this implies
357
358
Proposition 12. The resolvent G ±,N (λ) := (A±,N −λ)−1 is in the Schatten class
C for any p > 23 .
359
More explicitly, the resolvent of A−,N is obtained as
p
G −,N (x, y ; λ) = G0− (x, y ; λ) + G1−,N (x, y ; λ)
360
361
where G0− (x, y ; λ) is given by (3.10) and G1−,N (x, y ; λ) is
362
(3.16)
G1−,N (x, y ; λ) = −2π
for (x, y) ∈ R2+ ,
eiα Ai0 (eiα λ)
Ai e−iα (ix + λ) Ai e−iα (iy + λ) .
0 −iα
−iα
e Ai (e λ)
The poles of the resolvent are determined by zeros of Ai0 (e−iα λ), i.e., λn = eiα a0n ,
0
364 where a0n are zeros of the derivative of the Airy function: Ai (a0n ) = 0. The eigenvalues
365 have multiplicity 1 (no Jordan block). See Appendix A.
366 As a consequence of the analysis of the numerical range of the operator, we have
363
Proposition 13.
||G ±,N (λ)|| ≤
1
, if Re λ < 0 ;
|Re λ|
367
(3.17)
368
and
369
(3.18)
370
This proposition together with Proposition 12 and the Phragmen-Lindelöf principle
implies the completeness of the eigenfunctions:
371
372
373
374
||G ±,N (λ)|| ≤
1
, if ∓ Im λ > 0 .
|Im λ|
Proposition 14. The space generated by the eigenfunctions of the Neumann realization A±,N of Dx2 ± ix is dense in L2+ .
This manuscript is for review purposes only.
12
D. S. GREBENKOV, B. HELFFER, AND R. HENRY
375
At the boundary of the numerical range of the operator, we have
376
Proposition 15. When λ tends to +∞, we have
1
1
||G ±,N (λ)||HS ≈ λ− 4 (log λ) 2 .
377
(3.19)
378
379
Proof
Using the Wronskian (A.3) for Airy functions, we have
380
(3.20)
381
Hence
G −,D (x, y ; λ) − G −,N (x, y ; λ) = −ieiα
||G
382
−,D
(x, y ; λ) − G
−,N
(x, y ; λ)||2HS
Ai(e−iα wx )Ai(e−iα wy )
.
Ai(e−iα λ)Ai0 (e−iα λ)
R +∞
( 0 |Ai(e−iα wx )|2 dx)2
=
.
|Ai(e−iα λ)|2 |Ai0 (e−iα λ)|2
We will show in (8.10) that there exists C > 0 such that
Z +∞
4 3
−iα
2
− 12
2
|Ai(e wx )| dx ≤ Cλ exp
λ
384
.
3
0
383
On the other hand, using (A.5) and (A.6), we obtain, for λ ≥ λ0
1
4 3
0 −iα
−iα
386
|Ai(e λ)Ai (e λ)| ≥
exp
λ2
4π
3
385
387
388
(this argument will also be used in the proof of (8.7)). We have consequently obtained
that there exist C > 0 and λ0 > 0 such that, for λ ≥ λ0 ,
389
(3.21)
390
The proof of the proposition follows from Proposition 11.
1
||G −,D (λ) − G −,N (λ)||HS ≤ C |λ|− 4 .
3.4. The complex Airy operator on the half-line: Robin case. For completeness, we provide new results for the complex Airy operator on the half-line with
the Robin boundary condition that naturally extends both Dirichlet and Neumann
cases:
∂ −,R
395 (3.22)
G
(x, y ; λ, κ) − κ G −,R (x, y ; λ, κ)
= 0,
∂x
x=0
391
392
393
394
with a positive parameter κ. The operator is associated with the sesquilinear form
1
1
defined on H+
× H+
by
Z +∞
Z +∞
398 (3.23)
a−,R (u, v) =
u0 (x)v̄ 0 (x) dx − i
xu(x)v̄(x) dx + κ u(0)v̄(0) .
396
397
0
399
The distribution kernel of the resolvent is obtained as
G −,R (x, y ; λ) = G0− (x, y ; λ) + G1−,R (x, y ; λ, κ)
400
401
402
0
for (x, y) ∈ R2+ ,
where
(3.24)
ieiα Ai0 (eiα λ) − κAi(eiα λ)
− κAi(e−iα λ)
× Ai e−iα (ix + λ) Ai e−iα (iy + λ) .
G1−,R (x, y ; λ, κ) = −2π
ie−iα Ai0 (e−iα λ)
This manuscript is for review purposes only.
THE COMPLEX AIRY OPERATOR ON THE LINE WITH A SEMI-PERMEABLE BARRIER
13
Setting κ = 0, one retrieves (3.16) for the Neumann case, while the limit κ → +∞
yields (3.11) for the Dirichlet case, as expected. As previously, the resolvent is compact
and actually in the Schatten class C p for any p > 32 (see Proposition 8). Its poles are
406 determined as (complex-valued) solutions of the equation
403
404
405
407
f R (κ, λ) := ie−iα Ai0 (e−iα λ) − κAi(e−iα λ) = 0 .
(3.25)
Except for the case of small κ, in which the eigenvalues might be localized close to
the eigenvalues of the Neumann problem (see Section 4 for an analogous case), it does
not seem easy to localize all the solutions of (3.25) in general. Nevertheless one can
prove that the zeros of f R (κ, ·) are simple. If indeed λ is a common zero of f R and
(f R )0 , then either λ + κ2 = 0, or e−iα λ is a common zero of Ai and Ai0 . The second
option is excluded by the properties of the Airy function, whereas the first option is
414 excluded for κ ≥ 0 because the spectrum is contained in the positive half-plane.
415
As a consequence of the analysis of the numerical range of the operator, we have
408
409
410
411
412
413
Proposition 16.
||G ±,R (λ, κ)|| ≤
1
, if Re λ < 0 ;
|Re λ|
416
(3.26)
417
and
418
(3.27)
419
420
This proposition together with the Phragmen-Lindelöf principle (Theorem 54) and
the fact that the resolvent is in the Schatten class C p , for any p > 32 , implies
421
422
Proposition 17. For any κ ≥ 0, the space generated by the eigenfunctions of
the Robin realization A±,R of Dx2 ± ix is dense in L2+ .
423
424
At the boundary of the numerical range of the operator, it is interesting to analyze
the behavior of the resolvent. Equivalently to Propositions 11 or 15, we have
425
||G ±,R (λ, κ)|| ≤
1
, if ∓ Im λ > 0 .
|Im λ|
Proposition 18. When λ tends to +∞, we have
1
1
||G ±,R (λ, κ)||HS ≈ λ− 4 (log λ) 2 .
426
(3.28)
427
428
Proof
The proof is obtained by using Proposition 15 and computing, using (A.3),
429
||G
−,N
(λ) − G
−,R
(κ, λ)||2HS
Z
+∞
|Ai(e
=
−iα
2
2
wx )| dx
0
430
×
1
κ
.
2π |ie−iα Ai0 (e−iα λ) − κAi(e−iα λ)|2 |Ai0 (e−iα λ)|2
431
432
As in the proof of Proposition 15, we show that for any κ0 > 0, there exist C > 0 and
λ0 such that, for λ ≥ λ0 and κ ∈ [0, κ0 ],
433
||G −,N (λ) − G −,R (λ, κ)||HS ≤ C|κ|λ− 4 .
3
4. The complex Airy operator on the line with a semi-permeable barrier: definition and properties. In comparison with Section 2, we now replace
+
d2
d2
436 the differential operator − dx
2 by A1 = − dx2 + ix but keep the same transmission
434
435
This manuscript is for review purposes only.
14
D. S. GREBENKOV, B. HELFFER, AND R. HENRY
condition. To give a precise mathematical definition of the associated closed operator,
we consider the sesquilinear form aν defined for u = (u− , u+ ) and v = (v− , v+ ) by
Z 0 0
439
u0− (x)v̄−
(x) + i xu− (x)v̄− (x) + ν u− (x)v̄− (x) dx
aν (u, v) =
437
438
−∞
Z +∞
0
u0+ (x)v̄+
(x) + i xu+ (x)v̄+ (x) + ν u+ (x)v̄+ (x) dx
0
+κ u+ (0) − u− (0) v+ (0) − v− (0) ,
+
440
441
(4.1)
where the form domain V is
o
n
1
1
1
443
V := u = (u− , u+ ) ∈ H−
× H+
: |x| 2 u ∈ L2− × L2+ .
442
The space V is endowed with the Hilbert norm
12
1
kukV := ku− k2H 1 + ku+ k2H 1 + k|x| 2 uk2L2
445
.
444
−
446
447
+
We first observe
Lemma 19. For any ν ≥ 0, the sesquilinear form aν is continuous on V .
Proof
The proof is similar to that of Lemma 3, the additional term
Z 0
Z +∞
i
x u− (x) v̄− (x) dx +
450
x u+ (x) v̄+ (x) dx
448
449
−∞
451
452
453
454
0
being obviously bounded by kukV kvkV .
2
2
and satisfy the boundary
× H+
Let us notice that, if u and v belong to H−
conditions (1.3), then an integration by parts yields
Z 0
455
aν (u, v) =
− u00− (x) + ixu− (x) + ν u− (x) v̄− (x) dx
−∞
Z +∞
− u00+ (x) + ixu+ (x) + ν u+ (x) v̄+ (x) dx
0
+ u0+ (0) + κ(u− (0) − u+ (0)) v− (0) − v+ (0)
d2
= h − 2 + ix + ν u, viL2− ×L2+ .
dx
+
456
457
458
Hence the operator associated with the form aν , once defined appropriately, will act
d2
∞
460 as − dx
2 + ix + ν on C0 (R \ {0}) .
459
461
462
463
464
As the imaginary part of the potential ix changes sign, it is not straightforward
to determine whether the sesquilinear form aν is coercive, i.e., whether there exists
ν0 such that for ν ≥ ν0 the following estimate holds:
465
(4.2)
466
Let us show that it is indeed not true. Consider for instance the sequence
467
un (x) = (χ(x + n), χ(x − n)) ,
∃α > 0 , ∀u ∈ V ,
|aν (u, u)| ≥ α kuk2V .
n ≥ 1,
This manuscript is for review purposes only.
THE COMPLEX AIRY OPERATOR ON THE LINE WITH A SEMI-PERMEABLE BARRIER
15
where χ ∈ C0∞ (−1, 1) is an even function such that χ(x) = 1 for x ∈ [−1/2, 1/2] .
Then ku0n kL2 (−∞,0) and ku0n kL2 (0,+∞) are bounded, and
Z
470
x |un (x)|2 dx = 0 ,
468
469
R
1
471
since x 7→ x|un (x)|2 is odd, whereas k|x| 2 un kL2 −→ +∞ as n → +∞ . Consequently
472
|aν (un , un )|
−→ 0 as n → +∞ ,
kun k2V
473
474
475
476
and (4.2) does not hold.
Due to the lack of coercivity, the standard version of the Lax-Milgram theorem
does not apply. We shall instead use the following generalization introduced in [4].
Theorem 20. Let V ⊂ H be two Hilbert spaces such that V is continuously embedded in H and V is dense in H . Let a be a continuous sesquilinear form on V × V ,
and assume that there exist α > 0 and two bounded linear operators Φ1 and Φ2 on V
480 such that, for all u ∈ V ,
|a(u, u)| + |a(u, Φ1 u)| ≥ α kuk2V ,
481 (4.3)
|a(u, u)| + |a(Φ2 u, u)| ≥ α kuk2V .
477
478
479
Assume further that Φ1 extends to a bounded linear operator on H .
Then there exists a closed, densely-defined operator S on H with domain
484
D(S) = u ∈ V : v 7→ a(u, v) can be extended continuously on H ,
482
483
485
such that, for all u ∈ D(S) and v ∈ V ,
a(u, v) = hSu, viH .
486
Now we want to find two operators Φ1 and Φ2 on V such that the estimates (4.3)
hold for the form aν defined by (4.1).
First we have, as in (2.7),
Z 0
Z +∞
0
2
0
2
Re aν (u, u) ≥ (1 − |κ|ε)
|u− (x)| dx +
490
|u+ (x)| dx
−∞
0
491
+ ν − |κ|C(ε) kuk2L2 .
487
488
489
Thus by choosing ε and ν appropriately we get, for some α1 > 0 ,
Z 0
Z +∞
493 (4.4)
|aν (u, u)| ≥ α1
|u0− (x)|2 dx +
|u0+ (x)|2 dx + kuk2L2 .
492
−∞
0
1
494
495
It remains to estimate the term k|x| 2 ukL2 appearing in the norm kukV . For this
purpose, we introduce the operator
496
ρ : (u− , u+ ) 7−→ (−u− , u+ ) ,
498
which corresponds to the multiplication operator by the function sign x .
It is clear that ρ maps H onto H and V onto V. Then we have
499
(4.5)
497
1
Im aν (u, ρu) = k|x| 2 uk2L2 .
This manuscript is for review purposes only.
16
500
D. S. GREBENKOV, B. HELFFER, AND R. HENRY
Thus using (4.4), there exists α0 such that, for all u ∈ V ,
|aν (u, u)| + |aν (u, ρu)| ≥ αkuk2V .
501
502
Similarly, for all u ∈ V ,
503
|aν (u, u)| + |aν (ρu, u)| ≥ αkuk2V .
504
505
506
In other words, the estimate (4.3) holds, with Φ1 = Φ2 = ρ . Hence the assumptions
of Theorem 20 are satisfied, and we can define a closed operator A+
1 := S − ν, which
is given by the identity
507
508
∀u ∈ D(A+
1 ) , ∀v ∈ V ,
aν (u, v) = hA+
1 u + νu, viL2− ×L2+
on the domain
510
D(A+
1 ) = D(S) = u ∈ V : v 7→ aν (u, v) can be extended continuously
on L2− × L2+ .
511
512
513
514
Now we shall determine explicitly the domain D(A+
1 ).
Let u ∈ V . The map v 7→ aν (u, v) can be extended continuously on L2− × L2+ if
and only if there exists some wu = (wu− , wu+ ) ∈ L2− × L2+ such that, for all v ∈ V ,
aν (u, v) = hwu , viL2 . Then due to the definition of aν (u, v) , we have necessarily
509
515
wu− = −u00− + ixu− + νu−
and
wu+ = −u00+ + ixu+ + νu+
in the sense of distributions respectively in R− and R+ , and u satisfies the conditions
(1.3). Consequently, the domain of A+
1 can be rewritten as
n
00
00
2
2
518
D(A+
1 ) = u ∈ V : (−u− + ixu− , −u+ + ixu+ ) ∈ L− × L+
519
and u satisfies conditions (1.3) .
516
517
b
We now prove that D(A+
1 ) = D where
n
2
2
b = u ∈ V : (u− , u+ ) ∈ H−
521
D
× H+
, (xu− , xu+ ) ∈ L2− × L2+
522
and u satisfies conditions (1.3) .
520
2
2
. The only problem is at
× H+
It remains to check that this implies (u− , u+ ) ∈ H−
+∞. Let u+ be as above and let χ be a nonnegative function equal to 1 on [1, +∞)
and with support in ( 12 , +∞). It is clear that the natural extension by 0 of χu+ to R
belongs to L2 (R) and satisfies
d2
527
− 2 + ix (χu+ ) ∈ L2 (R) .
dx
523
524
525
526
528
529
530
531
532
533
534
535
536
One can apply for χu+ a standard result for the domain of the accretive maximal
extension of the complex Airy operator on R (see for example [26]).
Finally, let us notice that the continuous embedding
1
1
V ,→ L2 (R; |x|dx) ∩ H−
× H+
implies that A+
1 has a compact resolvent; hence its spectrum is discrete.
b we
Moreover, from the characterization of the domain and its inclusion in D,
deduce the stronger
This manuscript is for review purposes only.
THE COMPLEX AIRY OPERATOR ON THE LINE WITH A SEMI-PERMEABLE BARRIER
17
537
538
−1
Proposition 21. There exists λ0 (λ0 = 0 for κ > 0) such that (A+
1 − λ0 )
3
p
belongs to the Schatten class C for any p > 2 .
539
Note that if it is true for some λ0 it is true for any λ in the resolvent set.
540
541
542
Remark 22. The adjoint of A+
1 is the operator associated by the same construction with Dx2 − ix. A−
+
λ
being
injective,
this implies by a general criterion [26] that
1
A+
+
λ
is
maximal
accretive,
hence
generates
a contraction semigroup.
1
543
The following statement summarizes the previous discussion.
544
Proposition 23. The operator A+
1 acting as
d2
d2
+
u 7→ A1 u = − 2 u− + ixu− , − 2 u+ + ixu+
dx
dx
545
546
on the domain
2
2
2
2
D(A+
1 ) = u ∈ H− × H+ : xu ∈ L− × L+
547
548
549
550
551
(4.6)
and u satisfies conditions (1.3)
is a closed operator with compact resolvent.
There exists some positive λ such that the operator A+
1 + λ is maximal accretive.
Remark 24. We have
−
ΓA+
1 = A1 Γ ,
552
(4.7)
553
where Γ denotes the complex conjugation:
Γ(u− , u+ ) = (ū− , ū+ ) .
554
555
This implies that the distribution kernel of the resolvent satisfies:
556
(4.8)
557
for any λ in the resolvent set.
558
559
560
Remark 25 (PT-Symmetry). If (λ, u) is an eigenpair, then (λ̄, ū(−x)) is also an
eigenpair. Let indeed v(x) = ū(−x). This means v− (x) = ū+ (−x) and v+ (x) =
ū− (−x). Hence we get that v satisfies (2.1) if u satisfies the same condition:
561
0
v−
(0) = −ū0+ (0) = κ(ū− (0) − ū+ (0)) = +κ (v+ (0) − v− (0)) .
G(x, y ; λ) = G(y, x; λ) ,
Similarly one can verify that
d2
d2
− 2 + ix v+ (x) = − 2 − ix u− (−x)
dx
dx
d2
563
− 2 + ix u− (−x)
=
dx
562
= λ̄ v+ (x) .
564
565
566
567
5. Exponential decay of the associated semi-group. In order to control the
decay of the associated semi-group, we follow what has been done for the Neumann
or Dirichlet realization of the complex Airy operator on the half-line (see for example
[26] or [28, 29]).
This manuscript is for review purposes only.
18
D. S. GREBENKOV, B. HELFFER, AND R. HENRY
568
569
Theorem 26. Assume κ > 0, then for any ω < inf{Re σ(A+
1 )}, there exists Mω
such that, for all t ≥ 0,
570
|| exp(−tA+
1 )||L(L2− ×L2+ ) ≤ Mω exp(−ωt) ,
571
+
where σ(A+
1 ) is the spectrum of A1 .
572
573
574
To apply the quantitative Gearhart-Prüss theorem (see [26]) to the operator A+
1 , we
should prove that
−1
sup k(A+
k ≤ Cω ,
1 − z)
Re z≤ω
Re σ(A+
1)
576
for all ω < inf
:= ω1 .
First we have by accretivity (remember that κ > 0), for Re λ < 0 ,
577
(5.1)
578
So it remains to analyze the resolvent in the set
575
−1
||(A+
|| ≤
1 − λ)
0 ≤ Re λ ≤ ω1 − ,
579
1
.
|Re λ|
|Im λ| ≥ C > 0 ,
580
where C > 0 is sufficiently large. Let us prove the following lemma.
581
582
Lemma 27. For any α > 0, there exist Cα > 0 and Dα > 0 such that for any
λ ∈ {ω ∈ C : Re ω ∈ [−α, +α] and |Im ω| > Dα } ,
583
(5.2)
584
585
Proof
Without loss of generality, we treat the case when Im λ > 0 . As in [9], the main idea
−1
of the proof is to approximate (A+
by a sum of two operators: one of them
1 − λ)
is a good approximation when applied to functions supported near the transmission
point, while the other one takes care of functions whose support lies far away from
this point.
The first operator Ǎ is associated with the sesquilinear form ǎ defined for u =
(u− , u+ ) and v = (v− , v+ ) by
Z 0
0
ǎ(u, v) =
u0− (x)v̄−
(x) + i xu− (x)v̄− (x) + λ u− (x)v̄− (x) dx
586
587
588
589
590
591
592
−1
||(A±
|| ≤ Cα .
1 − λ)
−Im λ/2
Z
594
595
(5.3)
Im λ/2
0
u0+ (x)v̄+
(x) + i xu+ (x)v̄+ (x) + λ u+ (x)v̄+ (x) dx
0
+ κ u+ (0) − u− (0) v+ (0) − v− (0) ,
+
593
where u and v belong to the following space:
596
H10 (Sλ , C) := H 1 (Sλ− ) × H 1 (Sλ+ ) ∩ {u− (−Im λ/2) = 0 , u+ (Im λ/2) = 0} ,
597
598
599
600
601
with Sλ− := (−Im λ/2, 0) and Sλ+ := (0, +Im λ/2).
The domain D(Ǎ) of Ǎ is the set of u ∈ H 2 (Sλ− ) × H 2 (Sλ+ ) such that u− (−Im λ/2) =
0 , u+ (Im λ/2) = 0 and u satisfies conditions (1.3). Denote the resolvent of Ǎ by
R1 (λ) in L(L2 (Sλ− , C) × L2 (Sλ+ , C)) and observe also that R1 (λ) ∈ L(L2 (Sλ− , C) ×
L2 (Sλ+ , C), H10 (Sλ , C)).
This manuscript is for review purposes only.
THE COMPLEX AIRY OPERATOR ON THE LINE WITH A SEMI-PERMEABLE BARRIER
19
602
We easily obtain (looking at the imaginary part of the sesquilinear form) that
kR1 (λ)k ≤
2
.
Im λ
603
(5.4)
604
Furthermore, we have, for u = R1 (λ) f (with u = (u− , u+ ), f = (f− , f+ ))
kDx R1 (λ)f k2 = kDx uk2
2
≤ k(A+
1 − λ)ukkuk + Re λkuk
≤ kf kkR1 (λ)f k + |α|kR1 (λ)f k2
2
4|α|
≤
kf k2 .
+
|Im λ| |Im λ|2
605
606
Hence there exists C0 (α) such that, for Im λ ≥ 1 and Re λ ∈ [−α, +α],
607
(5.5)
1
kDx R1 (λ)k ≤ C0 (α) |Im λ|− 2 .
Far from the transmission point 0 , we approximate by the resolvent G0+ of the complex
Airy operator A+ on the line. Denote this resolvent by R2 (λ) when considered as
an operator in L(L2− × L2+ ). We recall from Section 3 that the norm kR2 (λ)k is
independent of Im λ. Since R2 (λ) is an entire function of λ, we easily obtain a uniform
612 bound on kR2 (λ)k for Re λ ∈ [−α, +α]. Hence,
608
609
610
611
kR2 (λ)k ≤ C1 (α) .
613
(5.6)
614
As for the proof of (5.5), we then show
615
(5.7)
kDx R2 (λ)k ≤ C(α) .
We now use a partition of unity in the x variable in order to construct an approximate inverse Rapp (λ) for A+
1 − λ. We shall then prove that the difference between
the approximation and the exact resolvent is well controlled as Im λ → +∞. For this
619 purpose, we define the following triple (φ− , ψ, φ+ ) of cutoff functions in C ∞ (R, [0, 1])
620 satisfying
616
617
618
φ− (t) = 1 on (−∞, −1/2] ,
621
φ− (t) = 0 on [−1/4, +∞)
ψ(t) = 1 on [−1/4, 1/4] ,
ψ(t) = 0 on (−∞, −1/2] ∪ [1/2, +∞) ,
φ+ (t) = 1 on [1/2, +∞) ,
φ+ (t) = 0 on (−∞, 1/4],
2
2
2
φ− (t) + ψ(t) + φ+ (t) = 1 on R ,
622
623
624
and then set
x x , ψλ (x) = ψ
.
Im λ
Im λ
The approximate inverse Rapp (λ) is then constructed as
φ±,λ (x) = φ±
Rapp (λ) = φ−,λ R2 (λ)φ−,λ + ψλ R1 (λ)ψλ + φ+,λ R2 (λ)φ+,λ ,
625
(5.8)
626
627
where φ±,λ and ψλ denote the operators of multiplication by the functions φ±,λ and
ψλ . Note that ψλ maps L2− × L2+ into L2 (Sλ− ) × L2 (Sλ+ ). In addition,
628
ψλ : D(Ǎ) → D(A+
1 ),
φλ : D(A+ ) → D(A+
1 ),
This manuscript is for review purposes only.
20
D. S. GREBENKOV, B. HELFFER, AND R. HENRY
629
630
where we have defined φλ (u− , u+ ) as (φ−,λ u− , φ+,λ u+ ).
From (5.4) and (5.6) we get, for sufficiently large Im λ,
631
(5.9)
632
Note that
633
(5.10)
634
app
Next, we apply A+
to obtain that
1 − λ to R
635
(5.11)
636
where I is the identity operator on L2− × L2+ , and
kRapp (λ)k ≤ C3 (α).
|φ0λ (x)| + |ψλ0 (x)| ≤
641
C
.
|Im λ|2
+
+
[A+
1 , φλ ] := A1 φλ − φλ A1
= [Dx2 , φλ ]
x 1
2i 0 x φ
Dx −
φ00
.
= −
2
Im λ
Im λ
(Im λ)
Im λ
638
640
|φ00λ (x)| + |ψλ00 (x)| ≤
+
app
(A+
(λ) = I + [A+
1 − λ)R
1 , ψλ ]R1 (λ)ψλ + [A1 , φλ ]R2 (λ)φλ ,
637
639
C
,
|Im λ|
(5.12)
A similar relation holds for [A+
1 , ψλ ]. Here we have used (5.8), and the fact that
(A+
1 − λ)R1 (λ)ψλ u = ψλ u ,
(A+
1 − λ)R2 (λ)φλ u = φλ u ,
∀u ∈ L2− × L2+ .
642
Using (5.4), (5.5), (5.7), and (5.12) we then easily obtain, for sufficiently large Im λ,
643
(5.13)
644
645
+
Hence, if |Im λ| is large enough then I + [A+
1 , ψλ ]R1 (λ)ψλ + [A1 , φλ ]R2 (λ)φλ is invertible in L(L2− × L2+ ), and
646
(5.14)
647
Finally, since
648
649
650
+
k[A+
1 , ψλ ] R1 (λ)k + k[A1 , φλ ] R2 (λ)k ≤
C4 (α)
.
|Im λ|
−1 +
,
ψ
]R
(λ)ψ
+
[A
,
φ
]R
(λ)φ
I + [A+
≤ C5 (α) .
λ
1
λ
λ
2
λ
1
1
+
−1
(A+
= Rapp (λ) ◦ I + [A+
1 − λ)
1 , ψλ ]R1 (λ)ψλ + [A1 , φλ ]R2 (λ)φλ
−1
,
we have
−1 +
−1
.
k(A+
k ≤ kRapp (λ)k I + [A+
1 − λ)
1 , ψλ ]R1 (λ)ψλ + [A1 , φλ ]R2 (λ)φλ
651
Using (5.9) and (5.14) we conclude that (5.2) is true.
652
653
654
655
Remark 28. One could alternatively use more directly the expression of the ker−1
nel G + (x, y ; λ) of (A+
in terms of Ai and Ai0 , together with the asymptotic
1 − λ)
expansions of the Airy function, see Appendix A and the discussion at the beginning
of Section 7.
This manuscript is for review purposes only.
THE COMPLEX AIRY OPERATOR ON THE LINE WITH A SEMI-PERMEABLE BARRIER
21
6. Integral kernel of the resolvent and its poles. Here we revisit some
of the computations of [22, 23] with the aim to complete some formal proofs. We
−1
are looking for the distribution kernel G − (x, y; λ) of the resolvent (A−
which
1 − λ)
659 satisfies in the sense of distribution
∂2
660 (6.1)
G − (x, y ; λ) = δ(x − y),
−λ − ix −
∂x2
656
657
658
as well as the boundary conditions
∂ −
∂ −
G (x, y ; λ)
=
G (x, y ; λ)
∂x
∂x
662 (6.2)
x=0+
x=0−
− +
= κ G (0 , y; λ) − G − (0− , y; λ) .
661
663
664
Sometimes, we will write G − (x, y ; λ, κ), in order to stress the dependence on κ.
Note that one can easily come back to the kernel of the resolvent of A+
1 by using
665
(6.3)
666
Using (4.8), we also get
667
(6.4)
G + (x, y ; λ) = G − (y, x; λ̄) .
G + (x, y ; λ) = G − (x, y ; λ̄) .
We search for the solution G − (x, y ; λ) in three subdomains: the negative semi-axis
(−∞, 0), the interval (0, y), and the positive semi-axis (y, +∞) (here we assumed
670 that y > 0; the opposite case is similar). For each subdomain, the solution is a linear
671 combination of two Airy functions:

−
−iα
−
iα

(x < 0) ,
A Ai(e wx ) + B Ai(e wx )
−
672 (6.5)
G (x, y ; λ) = A+ Ai(e−iα wx ) + B + Ai(eiα wx ) (0 < x < y) ,

 +
C Ai(e−iα wx ) + D+ Ai(eiα wx )
(x > y) ,
668
669
673
with six unknown coefficients (which are functions of y > 0). We recall that
α=
674
675
676
and
677
The boundary conditions (6.2) read as
2π
3
wx = ix + λ .
B − ieiα Ai0 (eiα w0 )
678
(6.6)
= A+ ie−iα Ai0 (e−iα w0 ) + B + ieiα Ai0 (eiα w0 )
= κ A+ Ai(e−iα w0 ) + B + Ai(eiα w0 ) − B − Ai(eiα w0 ) ,
where w0 = λ and we set A− = 0 and D+ = 0 to ensure the decay of G − (x, y ; λ) as
680 x → −∞ and as x → +∞ , respectively.
681 We now look at the condition at x = y in order to have (6.1) satisfied in the distribu682 tion sense. We write the continuity condition,
679
683
A+ Ai(e−iα wy ) + B + Ai(eiα wy ) = C + Ai(e−iα wy ) ,
This manuscript is for review purposes only.
22
684
D. S. GREBENKOV, B. HELFFER, AND R. HENRY
and the discontinuity jump of the derivative,
685
A+ ie−iα Ai0 (e−iα wy ) + B + ieiα Ai0 (eiα wy ) = C + ie−iα Ai0 (e−iα wy ) + 1 .
686
687
This can be considered as a linear system for A+ and B + . Using the Wronskian
(A.3), one expresses A+ and B + in terms of C + :
688
(6.7)
689
We can rewrite (6.6) in the form
690
(6.8)
691
and
A+ = C + − 2πAi(eiα wy ) ,
B − = e−2iα
Ai0 (e−iα w0 ) +
A + B+ ,
Ai0 (eiα w0 )
A+ ie−iα Ai0 (e−iα w0 ) + B + ieiα Ai0 (eiα w0 )
Ai0 (e−iα w0 )
.
= κ A+ Ai(e−iα w0 ) − e−2iα Ai(eiα w0 ) 0 iα
Ai (e w0 )
692
(6.9)
693
Using again the Wronskian (A.3), we obtain
A+ Ai0 (e−iα w0 ) + B + e2iα Ai0 (eiα w0 ) = −κA+
694
695
B + = 2πAi(e−iα wy ) .
1
,
2πAi (eiα w0 )
0
that is
2
A+ (f (λ) + κ) + B + (2π)e2iα Ai0 (eiα w0 ) = 0 ,
696
697
where
698
(6.10)
699
So we now get
700
(6.11)
A+ = −
702
(6.12)
B − = 2πAi(e−iα wy ) − 2π
703
and
704
(6.13)
705
Combining these expressions, one finally gets
706
(6.14)
f (λ) := 2πAi0 (e−iα λ)Ai0 (eiα λ) .
2
1
(2π)2 e2iα Ai0 (eiα w0 ) Ai(e−iα wy ) ,
f (λ) + κ
701
C + = 2πAi(eiα wy ) − 4π 2
f (λ)
Ai(e−iα wy ) ,
f (λ) + κ
e2iα [Ai0 (eiα λ)]2
Ai(e−iα wy ) .
f (λ) + κ
G − (x, y ; λ, κ) = G0− (x, y ; λ) + G1 (x, y ; λ, κ) ,
where G0− (x, y ; λ) is the distribution kernel of the resolvent of the operator A∗0 :=
d2
708 − dx
2 − ix on the line (given by Eq. (3.10)), whereas G1 (x, y ; λ, κ) is given by the
709 following expressions

e2iα [Ai0 (eiα λ)]2


 −4π 2
Ai(e−iα wx )Ai(e−iα wy ) , (x > 0) ,
f (λ) + κ
710 (6.15) G1 (x, y ; λ, κ) =
f (λ)


Ai(eiα wx )Ai(e−iα wy ) ,
(x < 0) ,
 −2π
f (λ) + κ
707
This manuscript is for review purposes only.
THE COMPLEX AIRY OPERATOR ON THE LINE WITH A SEMI-PERMEABLE BARRIER
23
711
for y > 0, and



 −2π
f (λ)
Ai(e−iα wx )Ai(eiα wy ) ,
(x > 0) ,
f (λ) + κ
712 (6.16) G1 (x, y ; λ, κ) =
e−2iα [Ai0 (e−iα λ)]2


Ai(eiα wx )Ai(eiα wy ) , (x < 0) ,
 −4π 2
f (λ) + κ
713
for y < 0. Hence the poles are determined by the equation
714
(6.17)
715
with f defined in (6.10).
716
717
Remark 29. For κ = 0, one recovers the conjugated pairs associated with the
zeros a0n of Ai0 . We have indeed as poles
718
(6.18)
719
720
721
722
where a0n is the n-th zero (starting from the right) of Ai0 . Note that a0n < 0 so that
Re λ±
n > 0, as expected.
In this case, the restriction of G1 (x, y ; λ, 0) to R2+ is the kernel of the resolvent of the
Neumann problem in R+ .
723
724
725
726
f (λ) = −κ ,
iα 0
λ+
n = e an ,
−iα 0
λ−
an ,
n =e
We also know that the eigenvalues for the Neumann problem are simple. Hence
by the local inversion theorem we get the existence of a solution close to each λ±
n for
+
κ small enough (possibly depending on n) if we show that f 0 (λ±
)
=
6
0.
For
λ
n
n , we
)
=
0
,
have, using the Wronskian relation (A.3) and Ai0 (e−iα λ+
n
0 iα +
−iα
f 0 (λ+
Ai00 (e−iα λ+
n ) = 2π e
n )Ai (e λn )
727
−iα +
= 2πe−2iα λ+
λn )Ai0 (eiα λ+
n Ai(e
n)
(6.19)
= −iλ+
n .
728
Similar computations hold for λ−
n . We recall that
−
λ+
n = λn .
729
−
The above argument shows that f 0 (λn ) 6= 0, with λn = λ+
n or λn = λn . Hence by
∗
∗
731 the holomorphic inversion theorem we get that, for any n ∈ N (with N = N∗ \ {0}),
732 and any , there exists hn () such that for |κ| ≤ hn (), we have a unique solution
733 λn (κ) of (6.17) such that |λn (κ) − λn | ≤ .
730
734
735
736
We would like to have a control of hn () with respect to n. What we should do
is inspired by the Taylor expansion given in [23] (Formula (33)) of λ±
n (κ) for fixed n :
737
(6.20)
π
±
±i 6
λ±
n (κ) = λn + e
1
κ + On (κ2 ) .
a0n
2
±
Since |λn | behaves as n 3 (see Appendix A), the guess is that λ±
n+1 (κ) − λn (κ) behaves
1
739 as n− 3 .
740 To justify this guess, one needs to control the derivative in a suitable neighborhood
741 of λn .
738
742
743
744
Proposition 30. There exist η > 0 and h∞ > 0, such that, for all n ∈ N∗ , for
any κ such that |κ| ≤ h∞ there exists a unique solution of (6.17) in B(λn , η|λn |−1 )
with λn = λ±
n.
This manuscript is for review purposes only.
24
D. S. GREBENKOV, B. HELFFER, AND R. HENRY
Proof of the proposition
Using the previous arguments, it is enough to establish the proposition for n large
enough. Hence it remains to establish a local inversion theorem uniform with respect
748 to n for n ≥ N . For this purpose, we consider the holomorphic function
745
746
747
749
B(0, η) 3 t 7→ φn (t) = f (λn + tλ−1
n ).
750
751
752
To have a local inversion theorem uniform with respect to n, we need to control |φ0n (t)|
from below.
Lemma 31. For any η > 0, there exists N such that, ∀n ≥ N ,
753
|φ0n (t)| ≥
754
(6.21)
755
Proof of the lemma
We have
756
and
φ0n (0) = −i .
759
760
761
762
∀t ∈ B(0, η) .
0
−1
φ0n (t) = λ−1
n f (λn + tλn ) ,
757
758
1
,
2
Hence it remains to control φ0n (t) − φ0n (0) in B(0, η). We treat the case λn = λ+
n.
We recall that
(6.22)
f 0 (λ) = 2πe−iα Ai00 (e−iα λ)Ai0 (eiα λ) + 2πeiα Ai0 (e−iα λ)Ai00 (eiα λ)
= 2πλ e−2iα Ai(e−iα λ)Ai0 (eiα λ) + e2iα Ai0 (e−iα λ)Ai(eiα λ)
= −iλ + 4πλe2iα Ai0 (e−iα λ)Ai(eiα λ).
763
Hence we have
764
(6.23)
765
766
767
768
769
with λ = λn + tλ−1
n .
770
771
772
773
774
775
776
777
2iα
φ0n (t) − φ0n (0) = 4πλλ−1
Ai0 (e−iα λ)Ai(eiα λ) − itλ−2
n e
n ,
The last term in (6.22) tends to zero. It remains to control Ai0 (e−iα λ)Ai(eiα λ) in
B(λn , η|λn |−1 ) and to show that this expression tends to zero as n → +∞.
We have
Ai0 (e−iα λ) = e−iα (λ − λn )Ai00 (e−iα λ̃) = e−2iα (λ − λn ) λ̃ Ai(e−iα λ̃) ,
with λ̃ ∈ B(λn , η|λn |−1 ).
Hence it remains to show that the product |Ai(e−iα λ̃)Ai(eiα λ)| for λ and λ̃
in B(λn , η|λn |−1 ) tends to 0. For this purpose, we will use the known expansion
for the Airy function (recalled in Appendix A) in the balls B(e−iα λn , η|λn |−1 ) and
B(eiα λn , η|λn |−1 ).
(i) For the factor |Ai(e−iα λ̃)|, we need the expansion of Ai(z) for z in a neighbor778 hood of a0n of size C|λn |−1 . Using the asymptotic relation (A.7), we observe that
3
2 3
2
exp ±i z 2 = exp ±i
779
(−a0n ) 2 (1 + O(1/|a0n |2 ))
= O(1) .
3
3
This manuscript is for review purposes only.
THE COMPLEX AIRY OPERATOR ON THE LINE WITH A SEMI-PERMEABLE BARRIER
25
780
Hence we get
1
∀ λ̃ ∈ B(λn , η|λn |−1 ) .
|Ai(e−iα̃ λ)| ≤ C |a0n |− 4
781
(ii) For the factor |Ai(eiα λ)|, we use (A.5) to observe that
3
3
2
2
exp − (eiα λ) 2 = exp −i (−a0n ) 2 (1 + O((−a0n )−2 )) ,
783
3
3
782
784
and we get, for any λ ∈ B(λn , η|λn |−1 )
785
(6.24)
786
787
This completes the proof of the lemma and of the proposition.
Actually, we have proved on the way the more precise
788
789
Proposition 32. For all η > 0 and 0 ≤ κ < η2 , there exists N such that, for all
n ≥ N , there exists a unique solution of (6.17) in B(λn , η |λn |−1 ).
1
|Ai(eiα λ)| ≤ C |a0n |− 4 .
Figure 1 illustrates Proposition 30. Solving Eq. (6.17) numerically, we find the
first 100 zeros λn (κ) with Im λn (κ) > 0. According to Proposition 30, these zeros are
792 within distance 1/|λn | from the zeros λn = λn (0) = eiα a0n which are given explicitly
793 through the zeros a0n . Moreover, the second order term in (6.20) that was computed
794 in [23], suggests that the rescaled distance
790
791
δn (κ) = |λn (κ) − λn ||λn |/κ,
795
(6.25)
796
behaves as
797
(6.26)
798
799
800
801
with a nonzero constant c. Figure 1(top) shows that the distance δn (κ) remains below
1 for three values of κ: 0.1, 1, and 10. The expected asymptotic behavior given in
(6.26) is confirmed by Figure 1(bottom), from which the constant c is estimated to
be around 0.31.
802
803
Remark 33. The local inversion theorem with control with respect to n permits
to have the asymptotic behavior of the λn (κ) uniformly for κ small:
804
(6.27)
805
806
An improvement of (6.27) (as formulated by (6.26)) results from a good estimate on
1
φ00n (t). Observing that |φ00n (t)| ≤ C|a0n |− 2 in the ball B(0, η), we obtain
807
(6.28)
1
1
δn (κ) = 1 − cκn− 3 + o(n− 3 ),
π
±
±i 6
λ±
n (κ) = λn + e
π
±
±i 6
λ±
n (κ) = λn + e
1
1
κ + 0 O(κ2 ) .
a0n
an
1
1
2
κ+
3 O(κ ) .
a0n
|a0n | 2
If one needs finer estimates, one can compute φ00n (0) and estimate φ000
n , and so on.
It would also be interesting to analyze the case κ → +∞. The limiting problem in
810 this case is the realization of the complex Airy operator on the line which has empty
811 spectrum. See [23] for a preliminary non rigorous analysis.
808
809
812
813
In the remaining part of this section, we describe the distribution kernel of the
±
projector Π±
n associated with λn (κ).
This manuscript is for review purposes only.
26
D. S. GREBENKOV, B. HELFFER, AND R. HENRY
|(λn(κ) − λn)λn|/κ
1
0.8
0.6
0.4
κ = 0.1
κ=1
κ = 10
0.2
0
0
20
40
60
80
100
n
0
(δn − 1) n1/3/κ
−0.2
−0.4
−0.6
κ = 0.1
κ=1
κ = 10
−0.8
−1
0
20
40
60
80
100
n
Fig. 1. Illustration of Proposition 30 by the numerical computation of the first 100 zeros λ+
n (κ)
+
+
of (6.17). At the top, the rescaled distance δn (κ) from (6.25) between λ+
n (κ) and λn = λn (0). At
the bottom, the asymptotic behavior of this distance.
Proposition 34. There exists κ0 > 0 such that, for any κ ∈ [0, κ0 ] and any
±
n ∈ N∗ , the rank of Π±
n is equal to one. Moreover, if ψn is an eigenfunction, then
Z +∞
816 (6.29)
ψn± (x)2 dx 6= 0 .
814
815
−∞
Proof
±
To write the projector Π±
n associated with an eigenvalue λn we integrate the resolvent
±
±
along a small contour γn around λn :
Z
1
(A± − λ)−1 dλ .
820 (6.30)
Π±
=
n
2iπ γn± 1
817
818
819
This manuscript is for review purposes only.
THE COMPLEX AIRY OPERATOR ON THE LINE WITH A SEMI-PERMEABLE BARRIER
27
If we consider the associated kernels, we get, using (6.14) and the fact that G0− is
holomorphic in λ:
Z
1
±
823 (6.31)
G1 (x, y ; λ, κ) dλ .
Πn (x, y ; κ) =
2iπ γn±
821
822
The projector is given by the following expression (with wx±,n = ix + λ±
n ) for
y>0
(6.32)

2
e2iα [Ai0 (eiα λ±
n )]

 −4π 2
Ai(e−iα wx±,n )Ai(e−iα wy±,n ) (x > 0) ,
0 (λ± )
±
f
n
826
Πn (x, y ; κ) =
κ

 2π
Ai(eiα wx±,n )Ai(e−iα wy±,n )
(x < 0) ,
f 0 (λ±
n)
824
825
827
828
and for y < 0
(6.33)

κ
−iα ±,n
iα ±,n

 2π f 0 (λ± ) Ai(e wx )Ai(e wy )
n
Π±
2
n (x, y ; κ) =
e−2iα [Ai0 (e−iα λ±
n )]

 −4π 2
Ai(eiα wx±,n )Ai(eiα wy±,n )
±
0
f (λn )
(x > 0) ,
(x < 0) .
Here we recall that we have established that for |κ| small enough f 0 (λ±
n ) 6= 0. It
remains to show that the rank of Π±
n is one that will yield an expression for the eigenfunction. It is clear from (6.32) and (6.33) that the rank of Π±
n is at most two and that
−iα ±,n
iα ±,n
wx )) ,
every function in the range of Π±
n has the form (c− Ai(e wx ) , c+ Ai(e
where c− , c+ ∈ R . It remains to establish the existence of a relation between c− and
834 c+ . This is directly obtained by using the first part of the transmission condition. If
835 κ 6= 0, the functions in the range of Π±
n have the form
829
830
831
832
833
836
iα ±,n
2iα
−iα ±,n
cn Ai0 (e−iα λ±
Ai0 (eiα λ±
wx ) ,
n )Ai(e wx ), e
n )Ai(e
837
838
with cn ∈ C. Inequality (6.29) results from an abstract lemma in [7] once we have
proved that the rank of the projector is one. We have indeed
839
(6.34)
840
841
842
1
||Π±
.
n || = R +∞ ±
| −∞ ψn (x)2 dx|
More generally, what we have proven can be formulated in this way:
Proposition 35. If f (λ) + κ = 0 and f 0 (λ) 6= 0 , then the associated projector
has rank 1 (no Jordan block).
The condition of κ being small in Proposition 34 is only used for proving the property
f 0 (λ) 6= 0. For the case of the Dirichlet or Neumann realization of the complex
Airy operator in R+ , we refer to Section 3. The nonemptiness was obtained directly
by using the properties of the Airy function. Note that our numerical solutions did
not reveal projectors of rank higher than 1. We conjecture that the rank of these
848 projectors is 1 for any 0 ≤ κ < +∞ but we could only prove the weaker
843
844
845
846
847
849
850
Proposition 36. For any κ ≥ 0, there is at most a finite number of eigenvalues
with nontrivial Jordan blocks.
851
852
Proof
We start from
853
f (λ) := 2π Ai0 (eiα λ) Ai0 (e−iα λ) ,
This manuscript is for review purposes only.
28
D. S. GREBENKOV, B. HELFFER, AND R. HENRY
854
and get by derivation
855
(6.35)
1 0
f (λ) = eiα Ai00 (eiα λ)Ai0 (e−iα λ) + e−iα Ai0 (eiα λ)Ai00 (e−iα λ) .
2π
What we have to prove is that f 0 (λ) is different from 0 for a large solution λ of
f (λ) = −κ. We know already that Re λ ≥ 0. We note that f (0) > 0. Hence 0 is not
a pole for κ ≥ 0. More generally f is real and strictly positive on the real axis. Hence
f (λ) + κ > 0 on the real axis.
860 We can assume that Im λ > 0 (the other case can be treated similarly). Using the
861 equation satisfied by the Airy function, we get
856
857
858
859
1 0
f (λ) = e−iα Ai(eiα λ)Ai0 (e−iα λ) + eiα Ai0 (eiα λ)Ai(e−iα λ) ,
2πλ
862
(6.36)
863
and by the Wronskian relation (A.3):
864
(6.37)
865
866
Suppose that f (λ) = −κ and that f 0 (λ) = 0 .
We have
i
.
4π
and get
κ=−
869
870
871
i
.
2π
−eiα Ai0 (eiα λ)Ai(e−iα λ) = e−iα Ai0 (e−iα λ)Ai(eiα λ) =
867
868
e−iα Ai0 (e−iα λ)Ai(eiα λ) − eiα Ai0 (eiα λ)Ai(e−iα λ) =
ieiα Ai0 (eiα λ)
ie−iα Ai0 (e−iα λ)
=
.
iα
2 Ai(e λ)
2 Ai(e−iα λ)
Using the last equality and the asymptotics (A.5), (A.6) for Ai and Ai0 , we get as
|λ| → +∞ satisfying the previous condition
1 1
|λ| 2 ,
2
872
κ∼
873
which cannot be true for λ large. This completes the proof of the proposition.
874
875
876
877
878
7. Resolvent estimates as |Im λ| → +∞. The resolvent estimates have been
already proved in Section 5 and were used in the analysis of the decay of the associated
semigroup. We propose here another approach which leads to more precise results.
We keep in mind (6.14) and the discussion in Section 5.
For λ = λ0 + iη , we have
879
kG0− ( · , · ; λ)kL2 (R2 ) = kG0− ( · , · ; λ0 )kL2 (R2 ) .
880
881
882
883
Hence the Hilbert-Schmidt norm of the resolvent (A+ − λ)−1 does not depend on the
imaginary part of λ .
As a consequence, to recover Lemma 27 by this approach, it only remains to check
the following lemma
884
885
Lemma 37. For any λ0 , there exist C > 0 and η0 > 0 such that
(7.1)
sup kG1 (· , ·; λ0 + iη)kL2 (R2 ) ≤ C .
|η|>η0
This manuscript is for review purposes only.
THE COMPLEX AIRY OPERATOR ON THE LINE WITH A SEMI-PERMEABLE BARRIER
29
10
5
0
−5
−10
0
1
2
3
4
5
6
1
2
3
4
5
6
10
5
0
−5
−10
0
Fig. 2. Numerically computed pseudospectrum in the complex plane of the complex Airy operator with Neumann boundary conditions (top) and with the transmission boundary condition at
the origin with κ = 1 (bottom). The red points show the poles λ±
n (κ) found by solving numerically
Eq. (6.17) that corresponds to the original problem on R. The presented picture corresponds to a
zoom (eliminating numerical artefacts) in a computation done for a large interval [−L, +L] with the
transmission condition at the origin and Dirichlet boundary conditions at ±L. The pseudospectrum
was computed for L3 = 104 by projecting the complex Airy operator onto the orthogonal basis of
eigenfunctions of the corresponding Laplace operator and then diagonalizing the obtained truncated
matrix representation (see Appendix E for details). We only keep a few lines of pseudospectra for the
clarity of the picture. As predicted by the theory, the vertical lines are related to the pseudospectrum
of the free complex Airy operator on the line.
886
887
888
889
The proof is included in the proof of the following improvement which is the main
result of this section and is confirmed by the numerical computations. One indeed
observes that the lines of the pseudospectrum are asymptotically vertical as Im λ →
±∞ when Re λ > 0 , see Figure 2.
This manuscript is for review purposes only.
30
D. S. GREBENKOV, B. HELFFER, AND R. HENRY
Proposition 38. For any λ0 > 0,
890
lim kG1 (· , ·; λ0 + iη)kL2 (R2 ) = 0 .
891
η→±∞
892
Moreover, this convergence is uniform for λ0 in a compact set.
893
894
895
Proof
We have
896
897
and
eiα λ = eiα λ0 − eiπ/6 η
e−iα λ = e−iα λ0 + e−iπ/6 η .
Then according to (A.6), one can easily check that the term Ai0 (e±iα λ) decays exponentially as η → ∓∞ and grows exponentially as η → ±∞ . On the other hand, the
term Ai0 (eiα λ) decays exponentially as η → ±∞ .
More precisely, we have
√ 3
1
|Ai0 (eiα (λ0 + iη))|2
∼ |c|2 η 2 exp 2 3 2 η 2 , as η → +∞ ;
√ 3
1
∼ |c|2 (−η) 2 exp − 2 3 2 η 2 , as η → −∞ ;
√ 3
902 (7.2)
1
|Ai0 (e−iα (λ0 + iη))|2 ∼ |c|2 η 2 exp − 2 3 2 η 2 , as η → +∞ ;
√ 3
1
∼ |c|2 (−η) 2 exp 2 3 2 η 2 , as η → −∞ .
898
899
900
901
903
As a consequence, the function f (λ), which was defined in (6.10) by
f (λ) := 2πAi0 (e−iα λ)Ai0 (eiα λ) ,
904
905
has the following asymptotic behavior as η → ∓∞ :
906
(7.3)
1
f (λ0 + iη) = 2π|c|2 |η| 2 1 + o(1) .
We treat the case η > 0 (the other case can be deduced by considering the
complex conjugate).
909 Coming back to the two formulas giving G1 in (6.15) and (6.16) and starting with the
910 first one, we have to analyze the L2 norm over R+ × R+ of
907
908
911
912
(x, y) 7→ −4π 2
e2iα [Ai0 (eiα λ)]2
Ai(e−iα wx )Ai(e−iα wy ) .
f (λ) + κ
This norm N1 is given by
N1 := 4π 2 |Ai0 (eiα λ)|2 |f (λ) + κ|−1 ||Ai(e−iα wx )||2L2 (R+ ) .
R +∞
914 Hence we have to estimate 0
|Ai(e−iα wx )|2 dx. We observe that
913
π
915
e−iα wx = e−i 6 (x + η) + e−iα λ0 ,
and that the argument of e−iα wx is very close to − π6 as η → +∞ (uniformly for
x > 0). This is rather simple for η > 0 because x and η have the same sign. We can
use the asymptotics (A.5) (with z = e−iα wx ) in order to get
!
√
Z +∞
1
3
2
2
919 (7.4)
|η| 2 .
|Ai(e−iα wx )|2 dx ≤ C (|η|2 + 1)− 2 exp −
3
0
916
917
918
This manuscript is for review purposes only.
THE COMPLEX AIRY OPERATOR ON THE LINE WITH A SEMI-PERMEABLE BARRIER
31
Here we have used that, for β > 0,
Z +∞
3
3
1
2
921
exp −βy 2 dy =
exp −βη 2 (1 + O(|η|− 2 )) .
3β
η
920
922
The control of |Ai0 (eiα (λ0 + iη))|2 given in (7.2) and (7.3) finally yields
923
(7.5)
924
By the notation . , we mean that there exists a constant C such that
1
N1 . (|η|2 + 1)− 2 .
1
N1 ≤ C(|η|2 + 1)− 2 .
925
For the L2 -norm of the second term (see (6.16)),
f (λ)
927
,
N2 := −2π
Ai(eiα wx )Ai(e−iα wy )
+
f (λ) + κ
L2 (R−
x ×Ry )
926
928
we observe that
N2 . ||Ai(eiα wx )||L2 (R− ) ||Ai(e−iα wx )||L2 (R+ ) ,
R0
930 and having in mind (7.4), we have only to bound −∞ |Ai(eiα wx )|2 dx . We can no
931 more use the asymptotic for the Airy function as (x + η) is small. We have indeed
929
π
eiα wx = −ei 6 (x + η) + eiα λ0 .
932
933
We rewrite the integral as the sum
Z 0
Z −η−C
iα
2
|Ai(e wx )| dx =
|Ai(eiα wx )|2 dx
−∞
934
−∞
Z −η+C
+
|Ai(eiα wx )|2 dx +
−η−C
Z
0
|Ai(eiα wx )|2 dx .
−η+C
The integral in the middle of the r.h.s. is bounded. The first one is also bounded
according to the behavior of the Airy function. So the dominant term is the third one
Z 0
Z η
π
|Ai(eiα wx )|2 dx =
|Ai(−ei 6 x + eiα λ0 )|2 dx
−η+C
C
!
√
937
1
2 2 3
2
|η| 2 .
≤ C̃(|η| + 1) 4 exp +
3
935
936
938
Combining with (7.4), the L2 -norm of the second term decays as η → +∞:
939
(7.6)
940
This achieves the proof of the proposition, the uniformity for λ0 in a compact being
controlled at each step of the proof.
941
942
943
944
945
946
947
1
N2 . (|η|2 + 1)− 8 .
8. Proof of the completeness. We have already recalled or established in Section 3 (Propositions 10, 14, and 17) the results for the Dirichlet, Neumann or Robin
realization of the complex Airy operator in R+ . The aim of this section is to establish
the same result in the case with transmission. The new difficulty is that the operator
is no longer sectorial.
This manuscript is for review purposes only.
32
948
949
D. S. GREBENKOV, B. HELFFER, AND R. HENRY
8.1. Reduction to the case κ = 0. We first reduce the analysis to the case
κ = 0 by comparison of the two kernels. We have indeed
950 (8.1)
G − (x, y ; λ, κ) − G − (x, y ; λ, 0) = G1 (x, y ; λ, κ) − G1 (x, y ; λ, 0)
= −κ(f (λ) + κ)−1 G1 (x, y ; λ, 0) ,
951
952
953
where G − (x, y ; λ, κ) denotes the kernel of the resolvent for the transmission problem
associated to κ ≥ 0 and Dx2 − ix.
We will also use the alternative equivalent relation:
954
(8.2) G − (x, y ; λ, κ) = G − (x, y ; λ, 0)f (λ) (f (λ) + κ)−1 + κ(f (λ) + κ)−1 G0− (x, y ; λ, 0) .
955
956
957
Remark 39. This formula gives another way for proving that the operator with
kernel G ± (x, y ; λ, κ) is in a suitable Schatten class (see Proposition 21). It is indeed
enough to have the result for κ = 0, that is to treat the Neumann case on the half line.
958
Another application of this formula is
959
Proposition 40. There exists M > 0 such that for all λ > 0 ,
1
1
−1
k(A±
kHS ≤ M (1 + λ)− 4 (log λ) 2 .
1 − λ)
960
(8.3)
961
962
Proof
Proposition 40 is a consequence of Proposition 15, and Formula (8.1).
963
964
965
Remark 41. Similar estimates are obtained in the case without boundary (typically for a model like the Davies operator Dx2 + ix2 ) by Dencker-Sjöstrand-Zworski
[14] or more recently by Sjöstrand [35].
8.2. Estimate for f (λ). We recall that f (λ) was defined in (6.10) by
966
f (λ) := 2πAi0 (e−iα λ)Ai0 (eiα λ) .
967
968
969
Recalling the asymptotic expansions (A.6) and (A.8) of Ai0 , it is immediate to
get
Lemma 42. The function λ 7→ f (λ) is an entire function of type
exists D > 0 such that
3
972 (8.4)
|f (λ)| ≤ D exp D|λ| 2 , ∀λ ∈ C .
970
971
3
2,
i.e. there
Focusing now on the main purpose of this section, we get from (A.6) that for any
> 0 there exists λ1 > 0 such that, for λ ≥ λ1 ,
1− 1
4 3
975 (8.5)
|Ai0 (eiα λ)|2 = |Ai0 (e−iα λ)|2 ≥
λ 2 exp
λ2 .
4π
3
973
974
976
Here we have also used that
977
(8.6)
Ai(z) = Ai(z̄)
and
Ai0 (z) = Ai0 (z̄)
(note that Ai(x) is real for x real). Thus there exists C1 > 0 such that, for λ ≥ 1,
1
C1
4 3
979 (8.7)
≤ 1 exp − λ 2 .
|f (λ)|
3
λ2
978
This manuscript is for review purposes only.
THE COMPLEX AIRY OPERATOR ON THE LINE WITH A SEMI-PERMEABLE BARRIER
33
980
981
8.3. Estimate of the L2 norm of G1 (· , · ; λ, 0). Having in mind (6.15)-(6.16)
and noting that, for λ > 0 ,
982
|Ai0 (eiα λ)|2
1
=
,
|f (λ)|
2π
it is enough to estimate
Z +∞
Z
−iα
2
984 (8.8)
|Ai(e (ix + λ))| dx = I0 (λ) =
983
0
|Ai(eiα (ix + λ))|2 dx .
−∞
0
985
986
It is enough to observe from (3.10), (8.6) and the comparison of the domain of integration in R2 , that
987
(8.9)
988
Applying (3.7), we get
2I0 (λ)2 ≤ ||G0− (· , · ; λ)||2 .
exp
4 3
λ2
3
.
989
(8.10)
990
Hence, coming back to (8.1), we have obtained
991
992
Proposition 43. There exist κ0 , C and λ0 > 0 such that, for all κ ∈ [0, κ0 ], for
all λ ≥ λ0 ,
993
(8.11)
994
995
996
Hence we are reduced to the case κ = 0 which can be decoupled (see Remark 29) in
two Neumann problems on R− and R+ .
997
998
Using (8.2) and the estimates established for G0− (· , · ; λ, 0) (which depends only
on Re λ) (see (3.7) or (3.5)), we have
999
1000
Proposition 44. For all κ0 , there exist C and λ0 > 0 such that, for all κ ∈
[0, κ0 ], for all real λ ≥ λ0 , one has
1001
(8.12)
1002
This immediately implies
1003
I0 (λ) . λ
− 14
3
||G − (· , · ; λ, κ) − G − (· , · ; λ, 0)||L2 (R2 ) ≤ Cκ |λ|− 4 .
3
||G − (· , · ; λ, κ) − (f (λ)(f (λ) + κ)−1 )G − (· , · ; λ, 0)||L2 (R2 ) ≤ Cκ |λ|− 4 .
Proposition 45. For any g = (g− , g+ ), h = (h− , h+ ) in L2− × L2+ , we have
3
|hG − (λ, κ)g , hi − (f (λ)(f (λ) + κ)−1 )hG − (λ, 0)g , hi| ≤ C(g, h)κ |λ|− 4 ,
1004
(8.13)
1005
where h·, ·i denotes the scalar product in the Hilbert space L2− × L2+ .
1006
1007
1008
We now adapt the proof of the completeness from [2].
If we denote by E the closed space generated by the generalized eigenfunctions of A−
1,
the proof of [2] in the presentation of [28] consists in introducing
1009
F (λ) = hG − (λ, κ)g , hi,
1010
where
1011
(8.14)
h ∈ E⊥
and g ∈ L2− × L2+ .
This manuscript is for review purposes only.
34
1012
1013
1014
1015
1016
1017
1018
1019
1020
1021
1022
1023
1024
1025
1026
1027
1028
D. S. GREBENKOV, B. HELFFER, AND R. HENRY
As a consequence of the assumption on h, one observes that F (λ) is an entire function
and the problem is to show that F is identically 0. The completeness is obtained if
we prove this statement for any g and h satisfying the condition (8.14).
Outside the numerical range of A−
1 , i.e. in the negative half-plane, it is immediate to
see that F (λ) tends to zero as Re λ → −∞. If we show that |F (λ)| ≤ C(1 + |λ|)M for
some M > 0 in the whole complex plane, we will get by Liouville’s theorem that F
is a polynomial and, with the control in the left half-plane, we should get that F is
identically 0.
Hence it remains to control F (λ) in a neighborhood of the positive half-plane {λ ∈
C , Re λ ≥ 0}.
As in [2], we apply Phragmen-Lindelöf principle (see Appendix D). The natural
idea (suggested by the numerical picture) is to control the resolvent on the positive
real axis. We first recall some additional material present in Chapter 16 in [2].
Theorem 46. Let φ(λ) be an entire complex-valued function of finite order ρ
(and φ(λ) is not identically equal to 0). Then for any > 0 there exists an infinite
increasing sequence (rk )k∈N in R+ and tending to +∞ such that
min |φ(λ)| > exp(−rkρ+ ) .
1029
|λ|=rk
1031
1032
1033
1034
1035
For this theorem (Theorem 6.2 in [2]), S. Agmon refers to the book of E.C. Titchmarsh
[38] (p. 273).
This theorem is used for proving an inequality of the type ρ with ρ = 2 in the
Hilbert-Schmidt case. We avoid an abstract lemma [2] (Lemma 16.3) but follow the
scheme of its proof for controlling directly the Hilbert-Schmidt norm of the resolvent
along an increasing sequence of circles.
1036
1037
Proposition 47. For > 0, there exists an infinite increasing sequence (rk )k∈N
in R+ tending to +∞ such that
1030
3
max ||G ± (· , · ; λ, κ)||HS ≤ exp rk2
1038
+ |λ|=rk
.
1039
1040
Proof
We start from
1041
(8.15) G − (x, y ; λ, κ) = G − (x, y ; λ, 0)f (λ) (f (λ)+κ)−1 +κ(f (λ)+κ)−1 G0− (x, y ; λ, 0) .
We apply Theorem 46 with φ(λ) = f (λ) + κ. It is proven in Lemma 42 that f is
of type 32 . Hence we get for > 0 (arbitrary small) the existence of a sequence
r1 < r2 < · · · < rk < · · · such that
3
1
≤ exp r 2 + .
1045
max k
|λ|=rk f (λ) + κ
1042
1043
1044
1046
In view of (8.15), it remains to control the Hilbert-Schmidt norm of
1047
G − (x, y ; λ, 0)f (λ) + κ G0− (x, y ; λ, 0) .
1048
1049
1050
Hence the remaining needed estimates only concern the case κ = 0. The estimate on
the Hilbert-Schmidt norm of G0− is recalled in (3.7). It remains to get an estimate for
the entire function G − (x, y ; λ, 0)f (λ).
This manuscript is for review purposes only.
THE COMPLEX AIRY OPERATOR ON THE LINE WITH A SEMI-PERMEABLE BARRIER
35
1051
1052
1053
Because κ = 0 , this is reduced to the Neumann problem on the half-line for the
complex Airy operator Dx2 − ix. For y > 0 and x > 0 , f (λ) G1N (x, y ; λ) is given by
the following expression
1054
(8.16)
1055
1056
1057
We only need the estimate for λ in a sector containing R+ × R+ . This is done in [28]
but we will give a direct proof below. In the other region, we can first control the
resolvent in L(L2 ) and then use the resolvent identity
1058
G ±,N (λ) − G ±,N (λ0 ) = (λ − λ0 )G ±,N (λ)G ±,N (λ0 ) .
1059
1060
1061
1062
This shows that in order to control the Hilbert-Schmidt norm of G ±,N (λ) for any λ,
it is enough to control the Hilbert-Schmidt norm of G ±,N (λ0 ) for some λ0 , as well as
the L(L2 ) norm of G ±,N (λ), the latter being easier to estimate.
f (λ) G1N (x, y ; λ) = −4π 2 [e2iα Ai0 (eiα λ)]2 Ai(e−iα wx )Ai(e−iα wy ) .
More directly the control of the Hilbert-Schmidt norm is reduced to the existence
of a constant C > 0 such that
Z +∞
3
1065
|Ai(e−iα (ix + λ))|2 dx ≤ C exp(C|λ| 2 ) .
1063
1064
0
In this case, we have to control the resolvent in a neighborhood of the sector
Im λ ≤ 0 , Re λ ≥ 0, which corresponds to the numerical range of the operator.
π
As x → +∞, the dominant term in the argument of the Airy function is ei(−α+ 2 ) x =
−i π
e 6 x. As expected we arrive in a zone of the complex plane where the Airy function
is exponentially decreasing. It remains to estimate for which x we enter in this zone.
We claim that there exists C > 0 such that if x ≥ C|λ| and |λ| ≥ 1, then
3
|Ai(e−iα (ix + λ))| ≤ C exp(−C(x + |λ|) 2 ) .
1066
3 In the remaining zone, we obtain an upper bound of the integral by O exp(C|λ| 2 ) .
1067
1068
1069
1070
1071
1072
We will then use the Phragmen-Lindelöf principle (Theorem 54). For this purpose,
it remains to control the resolvent on the positive real line. It is enough to prove the
theorem for g + = (0, g+ ) and g − = (g− , 0). In other words, it is enough to consider
F+ (resp. F− ) associated with g + (resp. g − ).
Let us treat the case of F+ and use Formula (8.2) and Proposition 45:
1073
(8.17) |hG − (λ, κ)g + , hi − (f (λ)(f (λ) + κ)−1 )hG − (λ, 0)g+ , h+ i| ≤ C(g, h)κ |λ|− 4 .
1074
1075
1076
1077
1078
1079
This estimate is true on the positive real axis. It remains to control the term
|hG − (λ, 0)g + , hi|. Along this positive real axis, we have by Proposition 15 the decay
of F+ (λ). Using Phragmen-Lindelöf principle completes the proof.
Note that for F− (λ), we have to use the symmetric (with respect to the real axis)
curve in Im λ > 0.
In summary, we have obtained the following
1080
1081
1082
Proposition 48. For any κ ≥ 0, the space generated by the generalized eigenfunctions of the complex Airy operator on the line with transmission is dense in
L2− × L2+ .
3
This manuscript is for review purposes only.
36
D. S. GREBENKOV, B. HELFFER, AND R. HENRY
Appendices.
1083
1084
1085
1086
1087
Appendix A. Basic properties of the Airy function.
In this Appendix, we summarize the basic properties of the Airy function Ai(z)
and its derivative Ai0 (z) that we used (see [1] for details).
We recall that the Airy function is the unique solution of
1088
(Dx2 + x)u = 0
1089
on the line such that u(x) tends to 0 as x → +∞ and
Ai(0) =
1090
1091
1092
1093
1094
1095
1
2
3
3 Γ
2
3
.
This Airy function extends into a holomorphic function in C .
The Airy function is positive decreasing on R+ but has an infinite number of zeros
in R− . We denote by an (n ∈ N) the decreasing sequence of zeros of Ai. Similarly we
denote by a0n the sequence of zeros of Ai0 . They have the following asymptotics (see
for example [1]), as n → +∞,
1096
(A.1)
1097
and
1098
(A.2)
an
a0n
∼
−
∼
−
n→+∞
n→+∞
3π
(n − 1/4)
2
23
3π
(n − 3/4)
2
23
,
.
The functions Ai(eiα z) and Ai(e−iα z) (with α = 2π/3) are two independent
1100 solutions of the differential equation
d2
1101
− 2 + z w(z) = 0 .
dz
1099
1102
Their Wronskian reads
1103
(A.3)
1104
Note that these two functions are related to Ai(z) by the identity
1105
(A.4)
e−iα Ai0 (e−iα z)Ai(eiα z) − eiα Ai0 (eiα z)Ai(e−iα z) =
Ai(z) + e−iα Ai(e−iα z) + eiα Ai(eiα z) = 0
i
2π
∀ z ∈ C.
∀ z ∈ C.
The Airy function and its derivative satisfy different asymptotic expansions depending on their argument:
(i) For | arg z| < π,
3 1
2 3
− 14
2
1109 (A.5)
Ai(z) = √ z
exp − z
1 + O(|z|− 2 ) ,
3
2 π
1
3 2 3
1
0
4
2
1110 (A.6)
1 + O(|z|− 2 )
Ai (z) = − √ z exp − z
3
2 π
1106
1107
1108
1111
1112
(moreover O is, for any > 0, uniform when | arg z| ≤ π − ) .
This manuscript is for review purposes only.
THE COMPLEX AIRY OPERATOR ON THE LINE WITH A SEMI-PERMEABLE BARRIER
37
1113
1114
(ii) For | arg z| < 23 π ,
(A.7)
1115
1116
(A.8)
1117
1118
1
3
1
π
2 3
Ai(−z) = √ z − 4 sin
z2 +
(1 + O(|z|− 2 )
3
4
π
!
−1
3
5 2 3
2 3
π
−
−
z2
cos
z2 +
(1 + O(|z| 2 )
72 3
3
4
3
1 1
π
2 3
0
4
2
Ai (−z) = − √ z
z +
(1 + O(|z|− 2 ))
cos
3
4
π
!
−1
3
7 2 3
2 3
π
−
+
z2
sin
z2 +
(1 + O(|z| 2 ))
72 3
3
4
(moreover O is for any > 0, uniform in the sector {| arg z| ≤
2π
3
− }).
Appendix B. Analysis of the resolvent of A+ on the line for λ > 0 (after
[32]).
∞
On the line R, A+ is the closure of the operator A+
0 defined on C0 (R) by
+
2
A0 = Dx + ix. A detailed description of its properties can be found in [26]. In this
1123 appendix, we give the asymptotic control of the resolvent (A+ − λ)−1 as λ → +∞.
1124 We successively discuss the control in L(L2 (R)) and in the Hilbert-Schmidt space
1125 C 2 (L2 (R)). These two spaces are equipped with their canonical norms.
1119
1120
1121
1122
1126
1127
B.1. Control in L(L2 (R)). Here we follow an idea present in the book of E. B.
Davies [13] and used in J. Martinet’s PHD [32] (see also [26]).
1128
Proposition 49.
For all λ > λ0 ,
1129
1130
(B.1)
+
k(A − λ)
−1
kL(L2 (R)) ≤
√
− 41
2π λ
exp
4 3
λ2
3
1 + o(1) .
1131
1132
Proof
The proof is obtained by considering A+ in the Fourier space, i.e.
1133
(B.2)
d
Ab+ = ξ 2 −
.
dξ
The associated semi-group Tt := exp(−Ab+ t) is given by
t3
2
2
1135 (B.3)
Tt u(ξ) = exp −ξ t − ξt −
u(ξ − t) ,
3
1134
∀ u ∈ S(R) .
Tt appears as the composition of a multiplication by exp(−ξ 2 t − ξt2 −
3
1137 translation by t. Computing supξ {exp(−ξ 2 t − ξt2 − t3 )} leads to
3
t
1138 (B.4)
kTt kL(L2 (R)) ≤ exp −
.
12
1136
t3
3)
1139
It is then easy to get an upper bound for the resolvent. For λ > 0, we have
1140
(B.5)
1141
1142
k(A+ − λ)−1 kL(L2 (R)) = k(Ab+ − λ)−1 kL(L2 (R))
Z +∞
≤
exp(tλ) kTt kL(L2 (R)) dt
0
Z +∞
t3
≤
exp tλ −
dt .
12
0
This manuscript is for review purposes only.
and a
38
D. S. GREBENKOV, B. HELFFER, AND R. HENRY
The right hand side can be estimated by using the Laplace integral method. Setting
1
t = λ 2 s, we have
Z +∞
Z +∞
1
3
t3
s3
1145 (B.6)
exp tλ −
dt = λ 2
exp λ 2 s −
ds .
12
12
0
0
1143
1144
3
s
We observe that φ̂(s) = s − 12
admits a global non-degenerate maximum on [0, +∞)
4
1147 at s = 2 with φ̂(2) = 3 and φ̂00 (2) = −1. The Laplace integral method gives the
1148 following equivalence as λ → +∞ :
Z +∞
√
3
3
s3
4 3
1149 (B.7)
exp λ 2 s −
ds ∼ 2π λ− 4 exp
λ2 .
12
3
0
1146
1150
1151
This completes the proof of the proposition. We note that this upper bound is not
optimal in comparison with Bordeaux-Montrieux’s formula (3.5).
1152
1153
1154
B.2. Control in Hilbert-Schmidt norm. In this part, we give a proof of
Proposition 7. As in the previous subsection, we use the Fourier representation and
analyze Ab+ . Note that
1155
(B.8)
k(Ab+ − λ)−1 k2HS = k(A+ − λ)−1 k2HS .
We have then an explicit description of the resolvent by
Z ξ
1 3
1157
(Ab+ − λ)−1 u(ξ) =
u(η) exp
(η − ξ 3 ) + λ(ξ − η) dη .
3
−∞
1156
1158
Hence, we have to compute
k(Ab+ − λ)−1 k2HS =
1159
Z Z
exp
η<ξ
2 3
(η − ξ 3 ) + 2λ(ξ − η) dηdξ .
3
1
1
After the change of variable (ξ1 , η1 ) = (λ− 2 ξ, λ− 2 η), we get
Z Z
3
2 3
+
−1 2
3
b
2
1161
k(A − λ) kHS = λ
exp λ
(η − ξ1 ) + 2(ξ1 − η1 )
dξ1 dη1 .
3 1
η1 <ξ1
1160
1162
With
1163
(B.9)
1164
we can write
1165
(B.10)
1166
where
3
h = λ− 2 ,
2
k(Ab+ − λ)−1 k2HS = h− 3 Φ(h),
Z Z
1167
(B.11)
Φ(h) =
exp
y<x
1168
with
1169
(B.12)
2
[φ(x) − φ(y)] dxdy ,
h
φ(x) = x −
x3
.
3
This manuscript is for review purposes only.
THE COMPLEX AIRY OPERATOR ON THE LINE WITH A SEMI-PERMEABLE BARRIER
39
1170
Φ(h) can now be split in three terms
1171
(B.13)
1172
with
Φ(h) = I1 (h) + I2 (h) + I3 (h) ,
2
[φ(x) − φ(y)] dxdy ,
y<x
h
y>0
Z Z
2
[φ(x) − φ(y)] dxdy ,
I2 (h) =
exp
y<x
h
x<0
Z Z
2
I3 (h) =
[φ(x) − φ(y)] dxdy .
exp
x∈R+
h
−
Z Z
I1 (h) =
1173
1174
1175
exp
y∈R
1176
We observe now that by the change of variable (x, y) 7→ (−y, −x), we get
I1 (h) = I2 (h) ,
1177
1178
and that
I3 (h) = I4 (h)2 ,
1179
1180
with
Z
I4 (h) =
1181
exp
R+
1182
1183
2
φ(x) dx .
h
Hence, it remains to estimate, as h → 0, the integrals I1 (h) and I4 (h).
B.2.1. Control of I1 (h).
√
√
The function
√ φ(x) is positive on (0, 3) and negative decreasing on ( 3, +∞), with
φ(0) = φ( 3) = 0. It admits a unique non-degenerate maximum at x = 1 with
φ(1) = 32 .
We first observe the trivial estimates
√
2
exp − φ(y) ≤ 1 , ∀ y ∈ [0, 3] ,
1189
h
1184
1185
1186
1187
1188
1190
and
√
2
1191
exp
φ(x) ≤ 1 , ∀ x ∈ [ 3, +∞[ .
h
√
1192 We will also have to estimate, for x ≥ 3,
Z x
2
1193
J(h, x) := √ exp − φ(y) dy .
h
3
For this purpose, we integrate by parts, observing that
2
h 1 d
2
1195
exp − φ(y) = −
exp
−
φ(y)
.
h
2 φ0 (y) dy
h
1194
This manuscript is for review purposes only.
40
D. S. GREBENKOV, B. HELFFER, AND R. HENRY
1196
We get
1197
J(h, x) = −
1198
This implies
1199
Z 0
h 1
h x 1
2
2
h 1
√
+
(y)
exp
−
exp
−
φ(x)
−
φ(y)
dy .
2 φ0 (x)
h
2 φ0 ( 3) 2 √3 φ0
h
J(h, x) ≤
2
h 1
h 1
√ + ChJ(h, x) ,
exp
−
φ(x)
−
2 x2 − 1
h
2 φ0 ( 3)
and finally, for h small enough and another constant C > 0
√
2
h 1 + Ch
J(h, x) ≤
1201 (B.14)
exp
−
φ(x)
+ Ch , ∀x ∈ [ 3, +∞) .
2
2 x −1
h
1200
Similarly, one can show that
Z +∞
2
h
exp
1203 (B.15)
φ(x)
dx ≤ .
√
h
4
3
1202
1204
1205
1206
1207
1208
1209
1210
1211
1212
1213
1214
With these estimates, we can bound I1 (h) from above in the following way
Z x
Z √3
2
2
φ(x)
I1 (h) =
exp
exp − φ(y) dy dx
h
h
0
0
Z √3
!
Z +∞
2
2
+ √ exp
φ(x)
exp − φ(y) dy dx
h
h
3
0
Z +∞
2
+ √ exp
φ(x) J(h, x)
h
3
√
Z √3
!
Z 3
2
2
≤
exp
exp − φ(y) dy dx
φ(x)
h
h
0
0
Z
+∞
√
2
φ(x) dx
+( 3 + Ch) √ exp
h
3
Z
h +∞ 1
+(1 + Ch)
dx
2 √3 x2 − 1
2
φ(x)
≤ 3 sup
exp
√
h
[0, 3]
√
( 3 + Ch)h
+
4
Z
h +∞ 1
dx
+(1 + Ch)
2 √3 x2 − 1
√
Z +∞
4
3h(1 + Ch) h
1
≤ 3 exp
+
+ (1 + Ch) √
dx .
2
3h
4
2
x −1
3
Hence we have shown the existence of Ĉ > 0 and h0 > 0 such that, for h ∈ (0, h0 ),
4
1216 (B.16)
I1 (h) ≤ Ĉ exp
.
3h
1215
1217
1218
Consequently, I1 (h) and I2 (h) appear as remainder terms.
This manuscript is for review purposes only.
THE COMPLEX AIRY OPERATOR ON THE LINE WITH A SEMI-PERMEABLE BARRIER
41
1219
1220
1221
1222
B.2.2. Asymptotic of I4 (h). Here, using the properties of φ, we get by the
standard Laplace integral method
I4 (h) ∼
(B.17)
p
√
π/2
h exp
4
3h
.
Hence, putting altogether the estimates, we get, as h → 0,
πh
Φ(h) ∼
exp
2
8
3h
1223
(B.18)
1224
Coming back to (B.8), (B.9) and (B.10), this achieves the proof of Proposition 7.
1225
1226
Appendix C. Analysis of the resolvent for the Dirichlet realization in
the half-line.
C.1. Main statement. The aim of this appendix is to give the proof of Proposition 11. Although it is not used in our main text, it is interesting to get the main
1229 asymptotic for the Hilbert-Schmidt norm of the resolvent in Proposition 11.
1227
1228
1230
Proposition 50. As λ → +∞, we have
||G
−,D
1231
(C.1)
1232
1233
C.2. The Hilbert-Schmidt norm of the resolvent for real λ . The HilbertSchmidt norm of the resolvent can be written as
1234
||G −,D ||2HS
(C.2)
(λ)||HS
√
1
1
3
∼ √ λ− 4 (log λ) 2 .
2 2
Z
|G
=
−,D
(x, y ; λ)| dxdy = 8π
2
Z∞
Q(x; λ)dx,
0
R2+
1235
2
where
|Ai(e−iα (ix + λ))|2
×
|Ai(e−iα λ)|2
Zx
2
× Ai(eiα (iy + λ))Ai(e−iα λ) − Ai(e−iα (iy + λ))Ai(eiα λ) dy.
Q(x; λ) =
1236
(C.3)
0
1237
Using the identity (A.4), we observe that
1238
(C.4)
1239
Hence we get
(C.5)
1240
Ai(eiα (iy + λ))Ai(e−iα λ) − Ai(e−iα (iy + λ))Ai(eiα λ)
= e−iα Ai(e−iα (iy + λ))Ai(λ) − Ai(iy + λ)Ai(e−iα λ) .
Q(x; λ) = |Ai(e
−iα
2
Zx Ai(λ)
−iα
(ix + λ))|
Ai(e (iy + λ)) Ai(e−iα λ) − Ai(iy + λ) dy .
2
0
This manuscript is for review purposes only.
42
D. S. GREBENKOV, B. HELFFER, AND R. HENRY
C.3. More facts on Airy expansions. As a consequence of (A.5), we can
write for λ > 0 and x > 0
3
exp − 23 λ 2 u(x/λ)
3 −iα
1243 (C.6)
1 + O(λ− 2 ) ,
|Ai(e (ix + λ))| =
√
1
2
2
2 π(λ + x ) 8
1241
1242
1244
1245
1246
1247
1248
1249
1250
1251
1252
1253
1254
1255
1256
1257
1258
where
√
p√
3
1 + s2 + 1 ( 1 + s2 − 2)
−1
(C.7)
√
.
u(s) = −(1 + s ) cos
tan (s) =
2
2
√
We note indeed that |e−iα (ix + λ)| = x2 + λ2 ≥ λ ≥ λ0 and that we have a control
π
of the argument arg(e−iα (ix + λ)) ∈ [− 2π
3 , − 6 ] which permits to apply (A.5).
Similarly, we obtain
3
exp 2 λ 2 u(x/λ)
3 (C.8)
|Ai(ix + λ)| = √ 3
1 + O(λ− 2 ) .
1
2 π(λ2 + x2 ) 8
√
We note indeed that |ix + λ| = x2 + λ2 and arg((ix + λ)) ∈ [0, + π2 ] so that one
can then again apply (A.5). In particular the function |Ai(ix + λ)| grows superexponentially as x → +∞.
Figure 3 illustrates that, for large λ, both equations (C.6) and (C.8) are very accurate
approximations for |Ai(e−iα (ix + λ))| and |Ai(ix + λ)|, respectively.
The control of the next order term (as given in (A.5)) implies that there exist C > 0
2
and 0 > 0, such that, for any ∈ (0, 0 ], any λ > ε− 3 and any x ≥ 0, one has
3
3
exp − 23 λ 2 u(x/λ)
exp − 23 λ 2 u(x/λ)
−iα
(C.9) (1−C) √
≤ |Ai(e (ix+λ))| ≤ (1+C) √
1
1
2 π (λ2 + x2 ) 8
2 π (λ2 + x2 ) 8
2
3
4
and
3
3
exp 32 λ 2 u(x/λ)
exp 23 λ 2 u(x/λ)
(1 − C) √
≤ |Ai(ix + λ)| ≤ (1 + C) √
1
1 ,
2 π (λ2 + x2 ) 8
2 π (λ2 + x2 ) 8
1259
(C.10)
1260
1261
1262
1263
where the function u is explicitly defined in Eq. (C.7).
1264
(C.11)
1265
and u has the following expansion at the origin
1266
(C.12)
1267
For large s, one has
1268
(C.13)
1269
1270
One concludes that the function u is monotonously increasing on [0, +∞) with
u(0) = −1 and lims→+∞ u(s) = +∞ .
Basic properties of u.
Note that
3
s
u0 (s) = √ p
≥0
√
2 2
1 + 1 + s2
(s ≥ 0),
3
u(s) = −1 + s2 + O(s4 ) .
8
3
s2
u(s) ∼ √ ,
2
1
3s 2
u0 (s) ∼ √ .
2 2
This manuscript is for review purposes only.
THE COMPLEX AIRY OPERATOR ON THE LINE WITH A SEMI-PERMEABLE BARRIER
43
C.4. Upper bound. We start from the simple upper bound (for any > 0)
1
1272 (C.14)
Q1 (x, λ) + (1 + )Q2 (x, λ) ,
Q(x, λ) ≤ 1 +
1271
1273
with
Q1 (x, λ) := |Ai(e
1274
−iα
|Ai(λ)|2
(ix + λ))|
|Ai(e−iα λ)|2
2
Zx
|Ai(e−iα (iy + λ))|2 dy
0
1275
and
Q2 (x, λ) := |Ai(e
1276
−iα
2
Zx
(ix + λ))|
|Ai(iy + λ)|2 dy .
0
1277
We then write
Q1 (x, λ) ≤ |Ai(e
1278
−iα
|Ai(λ)|2
(ix + λ))|
|Ai(e−iα λ)|2
2
+∞
Z
|Ai(e−iα (iy + λ))|2 dy
0
1279
and integrating over x
Z
+∞
1280
Q1 (x, λ)dx ≤ I0 (λ)2
0
1281
1282
1283
|Ai(λ)|2
,
|Ai(e−iα λ)|2
where I0 (λ) is given by (8.8).
Using (8.10) and (A.5), we obtain
Z +∞
1
1284 (C.15)
Q1 (x, λ)dx ≤ Cλ− 2 .
0
Hence at this stage, we have proven the existence of C > 0, 0 > 0 and λ0 such
that for any ∈ (0, 0 ] and any λ ≥ λ0


Z∞
1
1287 (C.16)
||G −,D ||2HS ≤ (1 + ) 8π 2
Q2 (x; λ)dx + Cλ− 2 −1 .
1285
1286
0
1288
It remains to estimate
Z
1289
(C.17)
Z
Q2 (x, λ)dx =
0
1290
+∞
Zx
+∞
dx
0
|Ai(e−iα (ix + λ))Ai(iy + λ)|2 dy .
0
Using the estimates (C.6) and (C.8), we obtain
2
Lemma 51. There exist C and 0 , such that, for any ∈ (0, 0 ), for λ > − 3 , the
integral of Q2 (x; λ) can be bounded as
Z +∞
1
1
2
Q2 (x, λ)dx ≤ (1 + C) I(λ) ,
1293 (C.18)
(1 − C) I(λ) ≤ 8π
2
2
0
1291
1292
This manuscript is for review purposes only.
44
1294
D. S. GREBENKOV, B. HELFFER, AND R. HENRY
where
Z∞
1295
(C.19)
Zx
3
exp − 43 λ 2 u(x/λ)
dx
I(λ) =
1
(λ2 + x2 ) 4
0
1296
1297
dy
0
(C.20)
I(λ) = λ
dx
Zx
3
exp − 34 λ 2 u(x)
1
(1 + x2 ) 4
0
dy
0
1299
Hence, introducing
1300
(C.21)
1301
ˆ defined for t ≥ t0 by
we reduce the analysis to I(t)
1302
(C.22)
t=
ˆ :=
I(t)
Z∞
dx
0
1303
.
Control of I(λ).
It remains to control I(λ) as λ → +∞ . Using a change of variables, we get
Z∞
1298
4 32
3 λ u(y/λ)
1
(λ2 + y 2 ) 4
exp
dy
1
(1 + x2 ) 4
.
4 3
λ2 ,
3
Zx
1
4 32
3 λ u(y)
1
(1 + y 2 ) 4
exp
exp t(u(y) − u(x))
1
(1 + y 2 ) 4
0
,
with
4 3
b
2
I(λ) = λ I
λ
.
3
1304
(C.23)
1305
1306
1307
1308
The following analysis is close to that of the asymptotic behavior of a Laplace integral.
1309
1310
1311
1312
1313
ˆ
Asymptotic upper bound of I(t).
Although u(y) ≤ u(x) in the domain of integration in Eq. (C.22), a direct use of this
upper bound will lead to an upper bound by +∞.
Let us start by a heuristic discussion. The maximum of u(y) − u(x) should be
on x = y. For x − y small, we have u(y) − u(x) ∼ (y − x)u0 (x). This suggests a
concentration near x = y = 0, whereas a contribution for large x is of smaller order.
More rigorously, we write
1314
(C.24)
1315
with, for 0 < < ξ,
b = Ib1 (t, ) + Ib2 (t, , ξ) + Ib3 (t, ),
I(t)
Z
Ib1 (t, ) =
dx
0
Z
1316
(C.25)
Ib2 (t, , ξ) =
dx
Z
Ib3 (t, ξ) =
+∞
dy
1
(1 + x2 ) 4
exp t(u(y) − u(x))
1
(1 + y 2 ) 4
0
Zx
1
dx
ξ
dy
1
(1 + x2 ) 4
ξ
Zx
1
exp t(u(y) − u(x))
1
(1 + y 2 ) 4
0
Zx
1
dy
1
(1 + x2 ) 4
0
,
,
exp t(u(y) − u(x))
1
(1 + y 2 ) 4
This manuscript is for review purposes only.
.
THE COMPLEX AIRY OPERATOR ON THE LINE WITH A SEMI-PERMEABLE BARRIER
45
We now observe that u(s) has the form u(s) = v(s2 ) where v 0 > 0 , so that
(C.26)
∀x, ys.t. 0 ≤ y ≤ x ≤τ0 ,
1318
supτ ∈[0,τ0 ] v 0 (τ ) (x2 − y 2 ) ≥ u(x) − u(y) ≥ inf τ ∈[0,τ0 ] v 0 (τ ) (x2 − y 2 ) .
1317
Analysis of Ib1 (t, ).
Using the right hand side of inequality (C.26) with τ0 = , we show the existence of
constants C and 0 > 0, such that, ∀ ∈ (0, 0 )
3
3
b
1322 (C.27)
(1 − C)J (1 + C) t ≤ I1 (t, ) ≤ (1 + C)J (1 − C) t ,
8
8
1319
1320
1321
1323
with
Z
J (σ) :=
1324
0
1325
1326
0
J (σ) = J1 (σ) + J2 (σ) ,
with
J1 (σ) :=
J2 (σ)
√1
σ
Z
Z0 1329
1330
exp σ(y 2 − x2 ) dy ,
which has now to be estimated for large σ.
For √1σ ≤ , we write
1327
1328
x
Z
dx
:=
Z x
dx
exp σ(y 2 − x2 ) dy ,
Z 0x
dx
exp σ(y 2 − x2 ) dy .
√1
σ
0
Using the trivial estimate
Z
1331
x
exp σ(y 2 − x2 ) dy ≤ x,
0
1332
we get
1333
(C.28)
J1 (σ) ≤
1
.
2σ
We have now to analyze J2 (σ).
The formula giving J2 (σ) can be expressed by using the Dawson function (cf [1], p.
295 and 319)
Z s
1337
s 7→ D(s) :=
exp(y 2 − s2 ) dy
1334
1335
1336
0
1338
1339
and its asymptotics as s → +∞ ,
1340
(C.29)
1341
1342
where the function δ(s) satisfies δ(s) = O(s−1 ) .
We get indeed
D(s) =
1
(1 + δ(s)) ,
2s
1
1343
J2 (σ)
1
=
σ
Z
σ 2
1
D(s)ds .
− 2
This manuscript is for review purposes only.
46
1344
1345
D. S. GREBENKOV, B. HELFFER, AND R. HENRY
By taking small enough to use the asymptotics of D(·) , we get
(C.30)


Z σ 12 Z σ 12 1
δ(s)
1

ds +
ds
J2 (σ) =
1
1
2σ
s
− 2 s
− 2
=
1 log σ C
+ (log + O(1)) .
4 σ
σ
1346
1347
Hence we have shown the existence of constants C > 0 and 0 such that if t ≥ C−3
and ∈ (0, 0 ), then
1348
(C.31)
log t 1 1
2 log t
b
+C +
.
I1 (t, ) ≤
3 t
t
t
1349
1350
Analysis of Ib3 (t, ξ)
We start from
+∞
Z
Ib3 (t, ξ) =
1351
dx
ξ
Zx
1
dy
1
(1 + x2 ) 4
exp t(u(y) − u(x))
1
(1 + y 2 ) 4
0
.
1352
1353
Having in mind the properties of u, we can choose ξ large enough in order to have for
some cξ > 0 the property that for x ≥ ξ and x2 ≤ y ≤ x,
1354
(C.32)
u(x) ≥ cξ x 2
3
u(x) − u(x/2) ≥ cξ x 2 ,
1
u(x) − u(y) ≥ cξ x 2 (x − y) .
1355
1356
This determines our choice of ξ. Using these inequalities, we rewrite Ib3 (t, ξ) as the
sum
1357
Ib3 (t, ξ) = Ib31 (t) + Ib32 (t) ,
3
1358
with
x
Z
+∞
Ib31 (t) =
dx
ξ
Z2
1
1359
Z
Ib32 (t) =
+∞
dx
ξ
dy
1
(1 + x2 ) 4
0
Zx
1
dy
1
(1 + x2 ) 4
x
2
exp t(u(y) − u(x))
,
1
(1 + y 2 ) 4
exp t(u(y) − u(x))
1
(1 + y 2 ) 4
This manuscript is for review purposes only.
.
THE COMPLEX AIRY OPERATOR ON THE LINE WITH A SEMI-PERMEABLE BARRIER
47
1360
Using the monotonicity of u, we obtain the upper bound
x
+∞
Z
Ib31 (t) ≤
dx
ξ
1
x2 ) 4
(1 +
1
2
Z
1
≤
2
Z
≤
1
3
Z
≤
3 1
exp −cξ ξ 2 t .
3cξ t
≤
1361
+∞
Z2
1
dy exp t(u(y) − u(x))
0
1
x 2 exp t(u(x/2) − u(x)) dx
ξ
+∞
1
3
x 2 exp −cξ t x 2 dx
ξ
+∞
exp −cξ ts ds
3
ξ2
1362
Hence, there exists ξ > 0 such that as t → +∞ ,
1363
(C.33)
1364
The last term to control is Ib32 (t). Using (C.32) and
1365
(1 + y 2 )− 4 ≤ (1 + (x/2)2 )− 4 ≤
Ib31 (t) = O exp(−ξ t) .
1
1366
1
√
1
2 (4 + x2 )− 4 ≤
√
1
2 (1 + x2 )− 4
for x/2 ≤ y ≤ x, we get
Ib32 (t) ≤
√ Z
2
+∞
dx
ξ
√ Z
≤ 2
1367 (C.34)
ξ
dy exp(t(u(y) − u(x))
1
(1 + x2 ) 2
+∞
dx
Zx
1
x
2
Zx
1
(1 + x2 )
1
2
1
dy exp −cξ tx 2 (x − y)
x
2
√ Z +∞
√
2
2 2
− 32
≤
.
x dx = √
cξ t ξ
ξcξ t
1368
1369
Hence putting together (C.33) and (C.34) we have, for this choice of ξ, the existence
of Ĉξ > 0 and tξ > 0 such that
1370
(C.35)
1371
1372
Analysis of Ib2 (t, , ξ).
We recall that
∀t ≥ tξ ,
Z
1373
Ib2 (t, , ξ) =
dx
1374
Zx
1
dy
1
(1 + x2 ) 4
exp t(u(y) − u(x))
1
(1 + y 2 ) 4
0
.
We first observe that
Z
1375
ξ
Ib3 (t) ≤ Ĉξ /t .
Ib2 (t, , ξ) ≤
Zx
ξ
Z
dy exp t(u(y) − u(x)) ≤
dx
0
Zx
ξ
dx
dy exp cξ t(y 2 − x2 ) ,
0
This manuscript is for review purposes only.
48
1376
D. S. GREBENKOV, B. HELFFER, AND R. HENRY
with
cξ = inf v 0 > 0 .
1377
[0,ξ]
Using now
1378
Z
1379
x
exp cξ t(y 2 − x2 ) dy ≤
0
Z
x
exp cξ tx(y − x) dy =
0
1380
we get
1381
(C.36)
1
1
1 − exp(−cξ tx2 ) ≤
,
cξ tx
cξ tx
1
Ib2 (t, , ξ) ≤
(log ξ − log ) .
cξ t
Putting together (C.24), (C.31), (C.35) and (C.36), we have shown the existence of
C > 0 and 0 such that if t ≥ C−3 and ∈ (0, 0 ), then
b ≤ 2 log t + C log t + 1 1 .
1384 (C.37)
I(t)
3 t
t
t
1382
1383
Coming back to (C.23) and using (C.16), we show the existence of C > 0 and 0 such
that if λ ≥ C−2 , then
1 −1
3 −1
−,D
2
− 21
2
2
1387
||G
(λ)||HS ≤ λ log λ + C λ log λ + λ
.
8
1385
1386
1388
1389
1
Taking = (log λ)− 2 , we obtain
Lemma 52. There exist C > 0 and λ0 such that for λ ≥ λ0
||G −,D (λ)||2HS ≤
1390
1
3 −1
λ 2 (1 + C (log λ)− 2 ) log λ .
8
1391
1392
C.5. Lower bound. Once the upper bounds are established, the proof of the
lower bound is easy. We start from the simple lower bound (for any > 0)
1393
(C.38)
1
Q(x, λ) ≥ − Q1 (x, λ) + (1 − )Q2 (x, λ) ,
and consequently
Z +∞
Z
1395 (C.39)
Q(x, λ) dx ≥ (1 − )
1394
0
0
+∞
1
Q2 (x, λ) dx −
Z
+∞
Q1 (x, λ) dx .
0
1
Taking = (log λ)− 2 and using the upper bound (C.15), it remains to find a lower
R +∞
1397 bound for 0
Q2 (x, λ)dx, which can be worked out in the same way as for the upper
1398 bound. We can use (C.18), (C.27), (C.30) and
b ≥ Ib1 (t, ) ≥ 2 log t − C log t + 1 1 .
1399 (C.40)
I(t)
3 t
t
t
1396
1400
1401
1402
This gives the proof of
Lemma 53. There exist C > 0 and λ0 such that for λ ≥ λ0
||G −,D (λ)||2HS ≥
1
3 −1
λ 2 (1 − C (log λ)− 2 ) log λ .
8
This manuscript is for review purposes only.
THE COMPLEX AIRY OPERATOR ON THE LINE WITH A SEMI-PERMEABLE BARRIER
49
10
30
10
10
10
10
|Ai(λ+ix)|
(λ+ix))|
−iα
|Ai(e
λ=5
λ = 10
λ = 5 (asymp)
λ = 10 (asymp)
20
0
−10
10
−20
10
−30
10
λ=5
λ = 10
λ = 5 (asymp)
λ = 10 (asymp)
0
5
10
10
10
0
10
−10
15
20
25
10
30
0
5
10
15
20
25
30
x
x
1.015
1.005
λ=5
λ = 10
λ=5
λ = 10
1.01
ratio
ratio
1
1.005
0.995
1
0.995
0
5
10
15
20
25
0.99
0
30
5
10
15
20
25
30
x
x
Fig. 3.
(Top) Asymptotic behavior of |Ai(e−iα (ix + λ))| (left) and
|Ai(ix + λ)| (right) for large λ. (Bottom) The ratio between these functions and their asymptotics
given by (C.6) and (C.8).
1403
1404
Appendix D. Phragmen-Lindelöf theorem.
The Phragmen-Lindelöf Theorem (see Theorem 16.1 in [2]) reads
1405
Theorem 54. (Phragmen-Lindelöf ) Let us assume that there exist two rays
1406
R1 = {reiθ1 : r ≥ 0} and R2 = {reiθ2 : r ≥ 0}
1407
1408
1409
π
with (θ1 , θ2 ) such that |θ1 − θ2 | = α
, and a continuous function F in the closed sector
delimited by the two rays, holomorphic in the open sector, satisfying the properties
(i)
∃C > 0,
1410
1411
1412
1413
1414
1415
1416
∃N ∈ R, s. t. ∀λ ∈ R1 ∪ R2 ,
|F (λ)| ≤ C |λ|2 + 1
N/2
.
(ii) There exist an increasing sequence (rk ) tending to +∞, and C such that
(D.1)
∀k,
max |F (λ)| ≤ C exp(rkβ ) ,
|λ|=rk
with β < α.
Then we have
|F (λ)| ≤ C |λ|2 + 1
N/2
for all λ between the two rays R1 and R2 .
1417
1418
Appendix E. Numerical computation of eigenvalues.
In order to compute numerically the eigenvalues of the realization A+,D
1,L of the
1419
1420
d
2
complex Airy operator A+
0 := Dx +ix = − dx2 +ix on the real line with a transmission
condition, we impose auxiliary Dirichlet boundary conditions at x = ±L, i.e., we
2
This manuscript is for review purposes only.
50
D. S. GREBENKOV, B. HELFFER, AND R. HENRY
search for eigenpairs {λL , uL (·)} of the following problem:
d2
− 2 + ix uL (x) = λL uL (x),
(−L < x < L),
dx
1422 (E.1)
uL (±L) = 0 ,
u0L (0+ ) = u0L (0− ) = κ uL (0+ ) − uL (0− ) ,
1421
1423
1424
1425
1426
1427
1428
1429
1430
1431
1432
1433
1434
with a positive parameter κ.
Since the interval [−L, L] is bounded, the spectrum of the above differential operator is discrete. To compute its eigenvalues, one can either discretize the second
derivative, or represent this operator in an appropriate basis in the form of an infinitedimensional matrix. Following [21], we choose the second option and use the basis
d2
formed by the eigenfunctions of the Laplace operator − dx
2 with the above boundary
conditions. Once the matrix representation is found, it can be truncated to compute
the eigenvalues numerically. Finally, one considers the limit L → +∞ to remove the
auxiliary boundary conditions at x = ±L.
There are two sets of Laplacian eigenfunctions in this domain:
(i) symmetric eigenfunctions
p
(E.2)
vn,1 (x) = 1/L cos(π(n + 1/2)x/L), µn,1 = π 2 (n + 1/2)2 /L2 ,
enumerated by the index n ∈ N.
(ii) antisymmetric eigenfunctions
(
√
+(βn / L) sin(αn (1 − x/L)) (x > 0),
√
1437 (E.3)
vn,2 (x) =
−(βn / L) sin(αn (1 + x/L)) (x < 0),
1435
1436
1438
with µn,2 = αn2 /L2 , where αn (n = 0, 1, 2, . . .) satisfy the equation
1439
(E.4)
αn ctan(αn ) = −2κL ,
while the normalization constant βn is
1441 (E.5)
βn = 1 +
1440
2κL
αn2 + 4κ2 L2
− 12
.
The solutions αn of Eq. (E.4) lie in the intervals (πn + π/2, πn + π), with n ∈ N .
In what follows, we use the double index (n, j) to distinguish symmetric and antisymmetric eigenfunctions and to enumerate eigenvalues, eigenfunctions, as well as the
elements of governing matrices and vectors. We introduce two (infinite-dimensional)
1446 matrices Λ and B to represent the Laplace operator and the position operator in the
1447 Laplacian eigenbasis:
1442
1443
1444
1445
1448
(E.6)
1449
and
Λn,j;n0 ,j 0 = δn,n0 δj,j 0 µn,j ,
ZL
1450
(E.7)
Bn,j;n0 ,j 0 =
dx vn,j (x) x vn0 ,j 0 (x) .
−L
1451
The symmetry of eigenfunctions vn,j implies Bn,1;n0 ,1 = Bn,2;n0 ,2 = 0, while
Bn,1;n0 ,2 = Bn0 ,2;n,1
1452 (E.8)
= −2Lβn0
sin(αn0 )(αn2 0 + π 2 (n + 1/2)2 ) − (−1)n (2n + 1)παn0
.
(αn2 0 − π 2 (n + 1/2)2 )2
This manuscript is for review purposes only.
κ=1
κ=0
THE COMPLEX AIRY OPERATOR ON THE LINE WITH A SEMI-PERMEABLE BARRIER
51
L
4
6
8
10
∞
4
6
8
10
∞
λ1,L
0.5161 - 0.8918i
0.5094 - 0.8823i
0.5094 - 0.8823i
0.5094 - 0.8823i
0.5094 - 0.8823i
1.0516 - 1.0591i
1.0032 - 1.0364i
1.0029 - 1.0363i
1.0029 - 1.0363i
1.0029 - 1.0363i
λ3,L
1.2938 - 2.1938i
1.1755 - 3.9759i
1.1691 - 5.9752i
1.1691 - 7.9751i
1.3441
1.1725
1.1691
1.1691
-
2.0460i
3.9739i
5.9751i
7.9751i
λ5,L
3.7675 - 1.9790i
1.6066 - 2.7134i
1.6233 - 2.8122i
1.6241 - 2.8130i
1.6241 - 2.8130i
4.1035 - 1.7639i
1.7783 - 2.7043i
1.8364 - 2.8672i
1.8390 - 2.8685i
1.8390 - 2.8685i
Table 1
The convergence of the eigenvalues λn,L computed by diagonalization of the matrix Λ + iB
truncated to the size 100×100. Due to the reflection symmetry of the interval, all eigenvalues appear
in complex conjugate pairs: λ2n,L = λ̄2n−1,L . The last line presents the poles of the resolvent of the
complex Airy operator A+
1 obtained by solving numerically the equation (6.17). The intermediate
column shows the eigenvalue λ3,L coming from the auxiliary boundary conditions at x = ±L (as
a consequence, it does not depend on the transmission coefficient κ). Since the imaginary part of
these eigenvalues diverges as L → +∞, they can be easily identified and discarded.
The infinite-dimensional matrix Λ+iB represents the complex Airy operator A+,D
1,L
on the interval [−L, L] in the Laplacian eigenbasis. As a consequence, the eigenvalues
and eigenfunctions can be numerically obtained by truncating and diagonalizing this
1456 matrix. The obtained eigenvalues are ordered according to their increasing real part:
1453
1454
1455
1457
Re λ1,L ≤ Re λ2,L ≤ . . .
1458
1459
1460
Table 1 illustrates the rapid convergence of these eigenvalues to the eigenvalues of the
complex Airy operator A+
1 on the whole line with transmission, as L increases. The
same matrix representation was used for plotting the pseudospectrum of A+
1 (Fig. 2).
This manuscript is for review purposes only.
52
1461
1462
1463
1464
1465
1466
1467
1468
1469
1470
1471
1472
1473
1474
1475
1476
1477
1478
1479
1480
1481
1482
1483
1484
1485
1486
1487
1488
1489
1490
1491
1492
1493
1494
1495
1496
1497
1498
1499
1500
1501
1502
1503
1504
1505
1506
1507
1508
1509
1510
1511
1512
1513
1514
1515
1516
1517
1518
1519
1520
1521
D. S. GREBENKOV, B. HELFFER, AND R. HENRY
REFERENCES
[1] M. Abramowitz and I. A. Stegun, Handbook of Mathematical Functions, Dover Publisher,
New York, 1965.
[2] S. Agmon, Elliptic boundary value problems, D. Van Nostrand Company, 1965.
[3] Y. Almog, The stability of the normal state of superconductors in the presence of electric
currents, SIAM J. Math. Anal. 40 (2) (2008), pp. 824-850.
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