Time dependent delta-prime interactions in dimension one Текст научной статьи по специальности «Математика»

Time dependent delta-prime interactions in dimension one

C. Cacciapuoti1, A. Mantile2, A. Posilicano1

1 DiSAT, Sezione di Matematica, Universita dell'Insubria, via Valleggio 11, 22100 Como, Italy 2 Laboratoire de Mathematiques, Universite de Reims - FR3399 CNRS, Moulin de la Housse BP 1039, 51687 Reims, France claudio.cacciapuoti@uninsubria.it, andrea.mantile@univ-reims.fr, andrea.posilicano@uninsubria.it

PACS 02.30.Jr, 03.65.Db, 02.30.Rz DOI 10.17586/2220-8054-2016-7-2-303-314

We solve the Cauchy problem for the Schrodinger equation corresponding to the family of Hamiltonians HY(t) in L2(R) which describes a S'-interaction with time-dependent strength 1/y(t). We prove that the strong solution of such a Cauchy problem exists whenever the map t ^ y(t) belongs to the fractional Sobolev space H3/4(R), thus weakening the hypotheses which would be required by the known general abstract results. The solution is expressed in terms of the free evolution and the solution of a Volterra integral equation. Keywords: time dependent point interactions, delta-prime interaction, non-autonomous Hamiltonians. Received: 10 February 2016

1. Introduction

In this paper, we address the generation problem for the family of time-dependent Hamiltonians HY(t), where HY(t), for any fixed real t, denotes the self-adjoint operator in L2(R) describing a ^'-interaction of strength 1/y(t) (see [1,2], [3, Chapter I.4] and references therein).

Most of the literature on time dependent point interactions focuses on perturbations of the free dynamics of the form of a Dirac's delta time dependent potential. In three dimensions time dependent ^-interactions were studied in [4, 5] and in [6] in relation with the problem of ionization under periodic perturbations, see also [7]. In two dimensions, very recently, the problem of the well-posedness was studied in [8]. In one dimension, this kind of non-autonomouss Hamiltonians was analyzed in [9], see also [10].

It is well known that in one dimension, the family of point perturbations of the Laplacian is richer than in two and three dimensions, and includes 8 and 8' perturbations, as well as their combinations. In this paper, we focus attention on the topical case of a time dependent 8'-interaction.

We remark that time-dependent 8-interactions have a non-linear counterpart, see, e.g., [11-13] in three dimensions, and [14,15]. More recently, a systematic study of the blow-up in the one dimensional case was started in [16]. In one dimension, in particular, such models find applications to the propagation of optical waves in Kerr media, or one-dimensional many body systems, see, e.g., [17-20] and references therein. The problem of the derivation of non-linear S-interactions from scaled regular dynamics was recently studied in one- and three-dimensions [21-23].

Several results discussed in the present paper set the groundworks for defining the nonlinear point interactions of 8'-type and for the study of the problem of their derivation from scaled regular dynamics.

We recall that the definition of HY is given by the theory of self-adjoint extensions of the symmetric operator:

H° = -A = , D(H°) = Co00(r\{0}), dx2

and, for any real y, reads as follows:

H7^(x) = -(x), x = 0 , (1)

D(H7) = {^ G L2(r) : ^ = 0 + qn, 0 G X2, q G c, 0'(O) = 7q} , (2)

where n(x) := 1 sgn(x) and for any v > 0 we defined Xv as the space of tempered distributions with Fourier transform in L2(r, |k|2vdk).

We remark that if f G Xv, then its Fourier transform might be a distribution as well. Moreover, for v = m + a, with m integer and 1/2 < a < 1, if f G Xv then f G Cm(r), see Prop. 2.1 below. Hence, 0 in D(HY) is a C^r) function and 0'(O) in the boundary condition is well defined.

The action of the operator HY can be understood also by exploiting the decomposition = 0 + this leads to

H7-(x) = -0''(x), x G r. (3)

When 7(t) is assigned as a real valued function of time, the domain D(HY(t)) changes in time with the boundary condition 0'(O) = Y(t)q. In contrast, the quadratic form corresponding to HY is given by

Qy(-) = I0'!2 + Y|q|2 , D(Qy) = {- G L2(r) : - = 0 + qn, 0 G X1, q G c }, and so QY(t) has a time-independent domain. Thus, by the abstract results in [24] and [10], assuming that the map t m- y(t) is differentiable, there exists an unitary propagator Ut,s in L2(r), continuously mapping D(HY(s)) onto D(HY(t)), such that —(t) := Ut,0—0 is the (strong) solution of the Cauchy problem:

d

—(t) = HY(t) —(t) (4)

_ —(0) = —0 G D(Hy(0)) .

However, as the case of time-dependent self-adjoint extensions Ha(t) (corresponding to a i-in-teraction) studied in [9] suggests, the quite explicit knowledge of the action and operator domain of HY should allow one to solve the Cauchy problem (4) under weaker regularity conditions on Y(t). Indeed, as we show in this paper, this is the case and problem (4) has a strong, unique solution whenever the map t m Y(t) is in the fractional Sobolev space H3/4(r), a condition weaker than the differentiability hypotheses required in [24] and [10]. Such a H3/4 hypothesis is the same required in the paper [9] in order to guarantee that the Cauchy problem for the family Ha(t) has a strong solution. However, in contrast to [9], here we make use neither of sophisticated analytic tools (paraproducts) nor of abstract generation theorems (as the ones provided in [24] and [25]); instead, following the same strategy as in the paper [26], we apply a more direct approach which exploits definitions (1) and (2), providing a relatively explicit expression for the solution of (4) with initial datum —0 = 00 + q0n in D(HY(0)):

m = №) + q(t)n, (5)

with

t

$(t) = eitA0o - j ds q(s)ei(t-s)A'q , (6)

0

where t m q(t) solves the Volterra-type integral equation

_ t

« ds^SM (7)

n J yjt - S

0

and the source term f0 is defined as:

_ t

4i f (eisA^o)'(0) (8)

- ds -n- . (8)

n J yjt - S

0

We shall prove the following well-posedness result:

Theorem 1. Let T > 0, 7 e H3/4(0,T), and set 70 = y(0). Let ^ = <p0 + qon e D(Hl0). Then, for any t e [0,T], there exists a unique strong solution for the Cauchy problem (4) given by ip(t) = <fr(t) + q(t)n as in Eqs. (6) - (8). Moreover, the map t m HY(ty^(t) belongs to

C ([0,T ],L2'

We briefly discuss the heuristic derivation of the solution. The solution of the Schrodinger equation with HY as Hamiltonian satisfies the distributional equation:

d

iJtm = -r(t) + q(t)8'0, (9)

where 80 is the first derivative of the Dirac delta-distribution. Let us assume, in the first part of this discussion, that the source term q(t) is an assigned function. Since n" = 8'0, it is natural to seek for solutions of the form (5). Setting ^(t) = 4>(t) + q(t)n in Eq. (9) gives the following equation for 4>(t):

d

i-^(t) = -4>"(t) - iq(t)n.

Eq. (6) follows directly from the Duhamel's formula. Indeed by integration by parts, see Section 2.3 (in particular Eqs. (15) and (17)), one obtains the following equation for ^(t):

t

^(t) = eitA^0 - i J ds q(s)ei(t-s)A8l0 . (9a)

0

This can be understood as Duhamel's formula applied to Eq. (9).

The equation for q(t) is obtained by imposing the boundary condition 4>'(0) = jq, using Eq. (6) to compute the l.h.s. in the boundary condition. We postpone the details of the calculation to Section 2.3. Here we note that the boundary condition turns the flow associated

to Eq. (9) into a unitary flow. In fact, one can show that:

d

-U(t)\\2 = 2Im q(t)#(0,t).

d

Hence, if the boundary condition is satisfied, one has — ||-0(i)|| = 0.

We remark that a function ^ e D(HY) can be written as the sum of a regular and singular part with both functions in L2 by introducing a regularization parameter A. More precisely, define:

p-\/X|x|

Ga(X) := -A > 0.

The function GA is the solution of the distributional equation G'A = + AGa. The domain D(Hy ) can be rewritten as

D(H7) = {^ e L2(r): ^ = 0a + qG", 0a e H2(r), q e c, 0A(0) = (7 + ^)q}, and the action of can be understood by the identity:

(HY + A)^(x) = -0A(x) + A0A(x), x e r,

see, e.g., [3]. Eq. (3) is obtained by taking A ^ 0.

We note that the charge equation (7) does not depend on A, it is easy to see that:

_ t ,

« f ds ,

n J v t - s

0

see Eqs. (18) and (19) below. The equation for the regular part 0A, instead, does involve the regularization parameter, precisely

0a (t) = -itA0A,o -J ds (q(s) + Aq(s))-i(t-s)AGA. 0

We note that, even if the regularization would avoid few issues with convolutions and Fourier transforms, which must otherwise be interpreted in distributional sense, it makes formulae more involved and introduces an unnecessary parameter. For those reasons, we decided to avoid it.

The paper consists of one additional section in which we prove Theorem 1.

t

2. Proof of Theorem 1

2.1. Notation and preliminaries

In what follows, C denotes a generic positive constant whose value may change from line to line.

We denote by — the spatial Fourier transform of —:

-(k) = J dxe-ikx-(x) .

r

The time-Fourier transform of f is denoted by Ff and defined as:

FfM = J di--iWtf(t) .

r

With these definitions, the Fourier transform of the convolution is:

)(k) = ^(k)0(k),

and similarly for the time-Fourier transform.

In the following, we denote by U(t) the free unitary group -iAt, we recall that its explicit expression is given by:

i(x-y)

ye 4t

dy Tint

which in Fourier transform reads:

ttMk) = e-ik^(k).

Proposition 2.1. For v = m + a, with m integer and 1/2 < a < 1, it results Xv C Cm Proof. In Fourier transform:

f (m)(x) - f(m) (y) = 71 i dk (zk)m(eifcx - eiky)f (k).

2n

We note that:

dk (zk)m(eikx - eiky)/ (k)

k|<i

Moreover:

dk (ik)m(eikx - eiky)/(k)

k|>i

<C|x - dk |k|m+CT|f (k)|

|k|<i

<C|x - y|CT 11/||L2(R)|fc|2vdk).

<c / dk |kn/(k)|

iki>i i

<C

dk

J |k|2CT Vfci>i /

||f ||L2(R,|k|2v dk).

(10)

(11)

Then, the continuity of ffollows from the bounds (10) and (11), and the dominated convergence theorem. □

We will make use of fractional Sobolev spaces; for this reason we recall few definitions.

For any < a < b < and v e (0,1), we set:

[/]Hv(a,b) : =

I A1/2

i dSdS'|f - f<8'"2

|s - s'|

1+2v

/

which is sometimes referred to as Gagliardo (semi)norm of f. The space Hv(a,b), for < a < b < and v e (0,1), is the space of functions for which the norm

\\f \\hv (a,b) = \\f \\L2(a,b) + [f]uv (a,b)

is finite. To define the space Hv(a, b) for v > 1 not integer, one sets v = m + a, where m is an integer and a e (0,1). Then Hv(a,b) is the space of functions such that f e Hm(a,b) and f(m) e H°(a, b).

Remark 2.2. Note that, for v e (0,1) there exists a constant Cv such that:

[f]Hv (r) = Cv\\Ff \\l2 (r,M2v du),

for any f e Xv, this is a direct consequence of Plancherel's theorem (see [27], Proposition 1.37). This identity, together with Prop. 2.1 implies that, for all v > 1/2, and a and b finite, if f e Xv then f e Hv(a,b), and, consequently, it belongs to Ha, b) for all 0 < ^ < v. Also, if

f e L2(a, b) and f e Xv, then f e Hv (a, b), and, consequently, in H^ (a, b) for all 0 < ^ < v.

We recall that, for < a < b < the space L2(a,b) can be identified with

H0(a, b), and L2(r) can be identified with X0.

For the norms, we shall use the notation \\ ■ \\ = \\ ■ \\L2(R). We denote by I the operator:

t

If (t) = A=fds -f^L. (12)

vW V t - s

0

We shall use the following results which establish the regularization properties of the operator I.

Lemma 2.3. Let v > 0 and T > 0. Assume that f e Xv and has support in [0,T], then

If e XV+1/2.

Proof. The integral kernel:

AM = -T ^

yjn yjt

where 6 is the Heaviside function, is a tempered distribution and

FA(u) = -|= (^6(u) + -2 (6(-u) + i6(u))

Let f e Xv. The convolution of A and f, If = A * f, is a tempered distributions and FIf = FAFf, see, e.g., [28, Th. 14.25]. Then,

\\|-r+1/2FIf \\ < C\\|-rFf \\.

We recall the following technical lemma:

Lemma 2.4. Let -<x < a < b < and let f e Hv(a, b) with v > 0. Define

f (s) if s e [a,b];

□

f (s) =

1 0 otherwise.

i) If 0 < v < 1/2, then f e Hv

ii) If 1/2 < v < 3/2 and f (a) = f (b) = 0, then f e Hv (r) .

For the proof, see for example [29, Th. 11.4], see also [30, Th. III.3.2]. We shall also use the following:

Proposition 2.5. Let ^ > 1/2 and 0 < v < If g e HM(a,b) and f e Hv(a, b) then fg e Hv(a, b).

For the proof we refer to [30].

2.2. Well-posedness of the charge equation

In this section, we study the well-posedness of the charge equation (7). We start with the following lemma, which gives the regularity properties of the inhomo-geneous term in Eq. (7):

Lemma 2.6. Let 0O e X2, then (U(-)0O)'(0) e X3/4.

Proof. Since 00 e L2(r), one has that the distributional identity:

i(x-y)2 - 41

r

shows that (U(t)0O)' e L2(r). By using the Fourier transform, one has that:

(U (t)0o)'(0) = 2n / dfc--ifc2t0O(fc).

r

By splitting the integral in dk for k > 0 and k < 0, and by using the change of variables k = \fw for k > 0 and k = - Vw for k < 0, it follows that:

oo

(U(t)0o)' (0) = J V= --iWt (<0O(V^) + 0O(-VW)).

O

Hence:

F((U(•)0o)'(0)) (w) = ^6(-w) (0O(V=W) + 0O(-V=W)) ,

where 6 denotes the Heaviside Unction. To prove that F ((U(•)0O)'(0)) e L2(r, |w|3dw), it is enough to note that

|||-|4F((U(•)0o)'(0)) ||< C||H0OH = C||H20o||, where we used the change of variables k2 = w. □

We are now ready to prove the main result of this section.

Lemma 2.7. Let T > 0, y e H3/4(0,T), and set YO = 7(0). Let = 0O + qOn e D(HY0). Then, Eq. (7) admits a unique solution q e H5/4(0,T).

Proof. We split the proof in two steps: first, we prove that there exists a unique solution

q e L2(0,T), then, by

a bootstrap argument, we show that such solution belongs to H5/4(0,T). We start by step 1. We use several results from the monograph [31]. We set:

Y(s) Vi - s

(U (t)0O)'(x) = J

k(t,s)

and rewrite the equation as:

t

q(t) = f0(t) -J ds k(t, s)q(s). (13)

0

This is a linear nonconvolution Volterra equation to which we can apply the results in [31, Ch. 9]. We start by noticing that for any finite interval J c r+, k(t, s) is a Volterra kernel of type L2, more precisely:

|||k|||L2(j) := sup [ i dsdt lh(t)k(t,s)g(s)l< J1^\\l-jj).

\\h\\L2{j)<1J J

NIl2(j)<1 J J

Hence, the interval [0,T] can be divided into finitely many subintervals Ji such that |||k|||L2(Ji) < 1 on each Ji, and, as a consequence of Cor. 9.3.14 in [31], one has that k has a resolvent of type L2 on [0,T]. By applying Th. 9.3.6 of [31], we conclude that Eq. (13) has a unique solution in L2(0,T).

We can now proceed to the second step of the proof, which consists in showing that such a solution belongs to H5/4(0, T). By Lemma 2.6 and Rem. 2.2, one has (U(-)fo)'(0) e Hv(0, T) for all 0 < v < 3/4. We set:

Q(t) = q(t) - q0 and F(t) = V4i((U(t)<h)'(0) - l(t)q(t)) t e [0,T].

We denote by Q the function obtained by prolonging Q to zero outside [0, T] and remark that the claim Q e Xv implies Q e H^(0,T) for all 0 < p < v, see Rem. 2.2, therefore q e H^(0,T).

By the charge equation (7), the identity Q = IF holds true for a.a. t e [0,T], here I is the operator defined in (12). Since, by Prop. 2.5, F e L2(0,T) we can define F e L2(r) by extending it to zero. Then, by Lemma 2.3, Q = IF e X1/2, hence, Q e H 1/4(0,T) and q e H 1/4(0,T).

We can repeat the argument. We start with the observation that now we know that F e H1/4(0,T) and conclude that q e H3/4(0,T). Here, we use Lemma 2.4-i) to claim that F e H1/4(R) which in turn implies F e X1/4.

To conclude the proof, we must slightly adjust the argument above. So far, we have proved that F e H3/4(0, T), moreover we know that F(0) = 0, because the boundary condition 4>0(0) = Y0q0 holds true by assumption. Define Fs : [0, 2T] m c by reflection of F about t = T. We have that Fs (0) = Fs(2T) = 0. We define Fs : r m C by extending Fs to zero and use Lemma 2.4-ii) to claim that Fs e H3/4(R), and, consequently, F e X3/4. Reapplying Lemma 2.3, we conclude that q e H5/4(0, T). □

2.3. Proof of Theorem 1

The function 4>(t) defined by Eq. (6) exists and is unique for all t e [0,T]. Next we prove that 4>(t) e X2. Let us rewrite Eq. (6) as:

where we set:

m = U m0+$(t),

0(t) = — dsq(s)U(t - s)n.

(14)

One has that U(t)00 G X2, because ||U(t)00||L2(R,|fc|4dfc) = ||00||L2(R,|fc|4dfc).

We are left to prove that 0 G X2. We recall that the Fourier transform of n is the

distribution — i PV — (where PV stands for principal value). We have that:

k

l2(r,|fc4|dfc) =2n j dkk2

<x

-— dww22

2n J

o

dse-ifc2(i-s)q(s)

ds eiwsq(s)

(14a)

< C||(||h 1/4(0,T).

Here, the inequality follows from the same argument used in the proof of Prop. 3.3 in [23].

Next, we prove that ^(t) = 0(t) + q(t)n G L2(r). Since 0(t) G C^r), see Prop. 2.1, and n is bounded, ^(t) G L2oc(r). Hence, it is enough to prove that (1 — x)^(t) G L2(r), where X is the characteristic function of the interval [—1,1]. In the definition of 0(t), see Eq. (6), we use the identity:

dsq(s)U(t — s)n = q(t)n — q0U(t)n — i dsq(s) —U(t — s)n,

which gives:

f d

^(t) = U (t)^o + dsq(s) — U (t — s)n.

(15)

Since U(t)^0 G L2(r), we are left to prove that the second term at the r.h.s., times the function (1 — x), is also in L2(r). We note that:

■■(*-y)2

ye 4t

11

2 Vinit

dyei4t — dyei4t

From which, we get:

(16)

d ,rT, s s, s 1 1 x ,• x2 i ^/t d ,• x2

«(U (i)n)(x) = — 2 TS^e'" = —V ".

We remark that the first equality can be understood in distributional sense as:

d

- (U (t)n) = i(U (t)n)" = i(U (t)n") = i(U (t)i0), from which, one deduces that Eq. (15) is equivalent to Eq. (5).

t

2

t

2

t

t

t

t

2

2

This then gives:

i— t

' d i 1 f d ■ X2 ds q(s)w- (U(t - s)n) (x) = \ — ds q(s)y/t - s—ei4(t-s) dt v n x j ds

00

t t

x 2

- - ( -q0 -teix - J ds q(s) -1 - s ei4(x->) +1 j ds eJ4(x>)

00 We gained a factor 1/x which gives the bound:

t

f d

(1 - X) dsq(s)dtU(t - s)n

< C (\\q\\L~(0,T) + \\q\\L 1(0,T)) < C t e [0,T].

Next, we prove that the boundary condition $(0) = y(t)q holds true for all t e [0,T]. From Eq. (16), we obtain:

(U (t)n)'(0) = -4=t, (18)

hence

t

fa'(0,t) = (u(t)<h)'(0) - ids 1 q(s).

J J4-Ki(t - s) 0

We apply the operator I, defined in (12), and use the charge equation (7) to obtain

(Ifa'(0, ■)) (t) = (I(U(-)<h)'(0)) (t) - (q(t) - q0) = (Ijq)(t), which implies the boundary condition. Here, we used the identities:

I (n('))-1/2(t)

t

I ds -1 - s —ns 0

1 and 12f (t)= dsf (s). (19)

By Eq. (3), to prove the continuity of the map t m HY(t)^(t) in L2(r), it is enough to show the continuity of \\fa''(t)\\. As the continuity of U(t)fa0 is obvious, we just need to show that:

lim

0

fat + 8) - fa(t)

L2(R,lk4ldk)

By Eqs. (14) and (14a), this is reduced to show that:

lim dk k2

^0 /

t+S

dse-ik2sq(s)

0.

For the proof of this statement, we refer to the proof of Prop. 3.3 in [23].

□

t

2

0

2

Acknowledgements

The authors acknowledge the support of the FIR 2013 project "Condensed Matter in Mathematical Physics", Ministry of University and Research of Italian Republic (code RBFR13WAET) .

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