# Pullback-Forward Dynamics for Damped Schrödinger Equations with Time-Dependent Forcing

Authors: Lianbing She; Yangrong Li; Renhai Wang
Journal: Discrete Dynamics in Nature and Society
 (2018)

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## Abstract

This paper deals with pullback dynamics for the weakly damped Schrödinger equation with time-dependent forcing. An increasing, bounded, and pullback absorbing set is obtained if the forcing and its time-derivative are backward uniformly integrable. Also, we obtain the forward absorption, which is only used to deduce the backward compact-decay decomposition according to high and low frequencies. Based on a new existence theorem of a backward compact pullback attractor, we show that the nonautonomous Schrödinger equation has a pullback attractor which is compact in the past. The method of energy, high-low frequency decomposition, Sobolev embedding, and interpolation are quite involved in calculating a priori pullback or forward bound.


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## Body

## 1. Introduction

This paper is concerned with the backward compact dynamics of space-periodic solutions for the nonautonomous complex-valued Schrödinger equation inR:(1)ut+iuxx+iu2u+αu-iu=ft,x.The above equation for α,f=0 was introduced in [1] as a model for the propagation of solitons and laser beams. In such a case (without damping and forcing), it is easy to prove the energy conservation law; that is, u(t)=u(0) (see, e.g., [2]). So no attractors exist. To obtain an attractor, we have to assume that the equation has a positive damping parameter α>0.The dynamical behavior of the damped Schrödinger equation was widely investigated by many physicists and mathematicians (see, e.g., [3–11]) but restricted in the autonomous case; that is, the force f is time-independent (only space dependent).This paper deals with dynamics for the nonautonomous Schrödinger equation; that is, the forcef is time-dependent. To the best of our knowledge, there is no literature treating nonautonomous dynamics (including random dynamics) for the Schrödinger equation, even in the simple case for the existence of a pullback attractor, although the theory and application of pullback attractors had been widely developed for many other PDEs (see [12–16]), and for pullback random attractors, see, for example, [17–20].When one tries to look for a bounded pullback absorbing set for (1) (see Lemma 7), it seems to be assumed that f is backward bounded in L2; that is,(2)sups⩽t⁡fs<∞,∀t∈R,rather than the ordinary tempered integrable condition. Another difficulty is that we must treat the time-derivative ft, which is assumed to be backward tempered. All those assumptions are different from dealing with the pullback dynamics for other nonautonomous dissipative equations (see, e.g., [17, 21–28]).On the other hand, the backward condition (2) permits us to consider further properties of the pullback attractor, for example, backward compactness, as considered in [29–31], where a pullback attractor is called backward compact if the union over the past is precompact.So, in Section2, we establish a new abstract theorem on a backward compact pullback attractor for adecomposable evolution process; that is, it has a backward compact-decay decomposition. For such a decomposable process, we show that the existence of a backward compact attractor is equivalent to the existence of anincreasing, bounded, and pullback absorbing set (see Theorem 4).We then apply the abstract result to the nonautonomous Schrödinger equation. In Section3, the increasingly pullback absorption is verified if f is assumed to be backward bounded and the time-derivative ft is backward tempered.The difficulty arises from verifying the compact-decay decomposition according to the high and low frequency of the Fourier series. In this case, the pullback absorption may not be suitable for verifying such a decomposable property. So, in Section4, we have to give an auxiliary result on theforward absorption. It may be possible to deduce a forward attractor (cf. [32, 33]), but we do not pursue this forward attractor in the present paper.In Section5, we present some techniques of splitting the solutions of (1) into high and low frequency parts and establish a new equation with initial value zero in the high-frequency part. Then the forward absorption obtained in Section 4 can be applied to prove that the new equation has a forward uniformly bounded solution in H2, which further proves that the component system is backward asymptotically compact in H1. Also, we prove that the difference of solutions from both equations in the high-frequency part is backward exponential decay and so obtain the compact-decay decomposition as required.The final existence result of a backward compact attractor is given in Theorem14. It is worth pointing out that the pullback-forward method (involving the high-low frequency decomposition) may be special for the nonautonomous Schrödinger equation, which is different from treating the pullback dynamics for other nonautonomous dissipative equations.


## 2. Backward Compact Dynamics for Decomposable Systems

Let(X,·X) be a Banach space equipped with the class B of all bounded subset in X. We consider anonautonomous processS on X, which means S(t,s):X→X is a continuous nonlinear mapping such that S(s,s)=idX and S(t,s)=S(t,r)S(r,s) for all t⩾r⩾s with s∈R.We assume that the process isdecomposable in the following sense.Definition 1.
A nonautonomous processS is said to have abackward compact-decay decompositionS(t,s)=S1(t,s)+S2(t,s)  (t⩾s) if
(i)S1 isbackward asymptotically compact, that is, the sequence {S1(sn,sn-τn)xn}n=1∞ is precompact whenever sn⩽t with fixed t∈R, τn→+∞ and {xn}n=1∞ is bounded in X;
(ii)S2 isbackward exponential decay, that is, for each t∈R and B∈B, there exist two positive constants c1=c1(t,B) and c2=c2(t,B) such that, for some τ0⩾0,(3)sups⩽t⁡supx∈B⁡S2s,s-τxX⩽c1e-c2τ-τ0,∀τ⩾τ0.A nonautonomous setD={D(t)}t∈R in X means a set-valued mapping D(·):R→2X∖∅, which is said to have some topological properties (such as compactness, boundedness, and closedness) if the component set D(t) has the corresponding properties.Definition 2.
A nonautonomous setD={D(t)}t∈R is called
(i)backward (resp.,forward or globally)compact if D(t) and ∪s⩽tD(s)¯ (resp., ∪r⩾tD(r)¯ or ∪r∈RD(r)¯) are compact for each t∈R;
(ii)increasing (resp.,decreasing) if D(s)⊂D(t) (resp., D(s)⊃D(t)) for all s⩽t.Definition 3.
A backward compact setA={A(t)}t∈R in X is called abackward compact attractor for a nonautonomous process S if A isinvariant, that is, S(t,s)A(s)=A(t) for t⩾s, andpullback attracting, that is, for any t∈R and B∈B, (4)limτ→+∞⁡distXSt,t-τB,At=0,where the Hausdorff semi-distance distX(A,B)=supa∈Ainfb∈Ba-bX.A backward compact attractor must be unique and minimal, where the minimality meansA⊂K for any closed attracting set K (see [30]).For the purpose of applying to the Schrödinger equation, we need to establish a new existence theorem of a backward compact attractor for a backward compact-decay process, although other existence criteria were established in [29, 30].Recall that a nonautonomous setK is calledpullback absorbing for S if for each t∈R and B∈B there is a τ0=τ0(t,B) such that (5)St,t-τB⊂Kt,∀τ⩾τ0.Theorem 4.
Suppose a nonautonomous processS has a backward compact-decay decomposition S(t,s)=S1(t,s)+S2(t,s) for t⩾s. Then the following statements are equivalent.
(i) There exists an increasing, bounded, and pullback absorbing setK={K(t)}t∈R.
(ii) There is a backward compact attractorA={A(t)}t∈R given by(6)At=ωKt,t≔⋂τ0>0⋃τ⩾τ0St,t-τKt¯,t∈R.Proof.
The necessity is easily proved by settingK(t)=N1(∪s⩽tA(s)), the 1-neighborhood of ∪s⩽tA(s), for each t∈R. Since ∪s⩽tA(s) is obviously an increasing family in t∈R, it follows that K is increasing. Since ∪s⩽tA(s) is precompact, it follows that K(t) is bounded (not necessarily precompact). Finally, it is easy to deduce the pullback absorption of K from the pullback attraction of A.
Conversely, suppose (i) is true; we show thatA(t)≔ω(K(t),t) is a backward compact attractor in three steps.
Step  1. We show the invariance. As usual (cf. [12]), it is easy to prove the positive invariance S(t,s)A(s)⊂A(t). On the other hand, let y∈A(t)=ω(K(t),t). We choose yn∈K(t) and τn→+∞ such that(7)limn→∞⁡St,t-τnyn=y.Let s⩽t and τ~n=s-t+τn→+∞. Then τ~n>0 if n is large enough. By the compact-decay decomposition, we know that {S1(s,s-τ~n)yn} is precompact and {S2(s,s-τ~n)yn} is exponential decay. Then, passing to subsequences, there is an z∈X such that (8)S1s,s-τ~nkynk⟶z,S2s,s-τ~nkynk⟶0,where τ~nk>0 for all k∈N. Hence, it further implies(9)Ss,s-τ~nkynk=S1s,s-τ~nkynk+S2s,s-τ~nkynk⟶z.Since K is pullback absorbing and increasing, for each i∈N, we can choose a ki in {nk} such that τ~ki-i is large enough; it follows that(10)zi≔Ss-i,s-i-τ~ki-iyki∈Ks-i⊂Ks.By the process property, as i→∞,(11)Ss,s-izi=Ss,s-τ~kiyki⟶z.By (10)-(11), we have z∈ω(K(s),s)=A(s). By the continuity of S, it follows from (7) and (9) that(12)y=limi→∞⁡St,t-τkiyki=limk→∞⁡St,sSs,s-τ~kiyki=St,sz∈St,sAs,which proves the negative invariance.
Step  2. We show the attraction of A. If it is not true, then there are δ>0, t∈R, τn→+∞ as n→∞ and xn∈B with B∈B such that(13)distXSt,t-τnxn,At⩾δ,foreachn∈N.By using the compact-decay decomposition, we can take subsequences such that (14)limk→∞⁡S1t,t-τnkxnk=x,limk→∞⁡S2t,t-τnkxnk=0,where x∈X. Therefore,(15)limk→∞⁡St,t-τnkxnk=limk→∞⁡S1t,t-τnkxnk+limk→∞⁡S2t,t-τnkxnk=x.Like we did in Step  1, we choose a subsequence {ki} of {nk} such that (16)yi≔St-i,t-i-τki-ixki∈Kt-i⊂Kt,and by (15),(17)limk→∞⁡St,t-iyi=limk→∞⁡St,t-τkixki=x.The above two formulations imply x∈A(t), which contradicts with (13).
Step  3. It remains to show the precompact of Bt≔∪s⩽tA(s) with fixed t∈R. Let s⩽t. Since K is pullback absorbing and increasing, there is a τ0(s) such that (18)⋃τ⩾τ0Ss,s-τKs¯⊂Ks¯⊂Kt¯,and so ω(K(s),s)⊂K(t)¯, which further implies (19)Bt=⋃s⩽tAs=⋃s⩽tωKs,s⊂Kt¯.Hence Bt is at least bounded. To prove the precompactness of Bt, we take a sequence zn∈Bt=∪s⩽tA(s) and then zn∈A(sn) with sn⩽t. Let 0⩽τn→∞. Then invariance of A implies (20)zn∈Asn=Ssn,sn-τnAsn-τn⊂Ssn,sn-τnBt.Since Bt is proved to be bounded, there is a bounded sequence {xn} such that zn=S(sn,sn-τn)xn. By the backward compact-decay decomposition, we know S1 is backward asymptotically compact on the bounded set Bt. Then, passing to a subsequence, there is a x∈X such that (21)S1snk,snk-τnkxnk⟶x.By using the decay property of S2, we know (22)S2snk,snk-τnkxnk⩽c1e-c2τnk-τ0⟶0.The above limits imply (23)znk=Ssnk,snk-τnkxnk=S1snk,snk-τnkxnk+S2snk,snk-τnkxnk⟶x.Hence, Bt is precompact. In particular, A(t) is precompact. But A(t) is obviously closed and so it is compact. The proof is complete.


## 3. Pullback Absorption in Schrödinger Equations

We come back to the Schrödinger equation with time-dependent force as follows:(24)ut+iuxx+iu2u+αu-iu=ft,x,ut,0=ut,1,t⩾s,us,x=u0x,s∈R,x∈Ω=0,1,where α>0, u(t,x) is an unknown complex-valued function.


### 3.1. Hypotheses and Existence of Solutions

LetL2(Ω) be the space of complex-valued L2-functions whose norm is denoted by ·. Let H1 be the space of all one-periodic H1-function with the norm uH12=u2+ux2. The norm in Lp(Ω) is denoted by ·p. We then give some hypotheses on the time-space dependent force f=f(t,x).Assumption
A1. f is continuous andbackward bounded: (25)F1t≔sups⩽tfs,·2<+∞,∀t∈R.Assumption
A2. The time-derivative ft isbackward tempered: (26)F2t≔sups⩽t⁡∫-∞seαr-sfrr,·2dr<+∞,∀t∈R,where α>0 as given in (24).AssumptionA1 implies f∈Lloc2(R,L2(Ω)) and the finiteness of F3(t), where (27)F3t≔sups⩽t⁡∫-∞seαr-sfr,·2dr⩽F1t1-e-α.The last inequality can be proved as follows. (28)F3t=sups⩽t⁡∑m=0∞∫s-m+1s-meαr-sfr,·2dr⩽sups⩽t⁡∑m=0∞e-mα∫s-m+1s-mfr,·2dr⩽∑m=0∞e-mαsupr⩽t⁡fr,·2=F1t1-e-α.Note that all functions Fi(·) (i=1,2,3) are finite, nonnegative, andincreasing.We will repeatedly use the following two energy inequalities.Lemma 5.
Letu(t)=u(t,s,u0), t∈[s,+∞), be the approximation solution of (24). Then(29)ddtu2+αu2⩽1αft,·2,(30)ddtΦ1u+αΦ1u⩽9αu6+3αu4+1αftt,·2,where(31)Φ1u=ux2+u2-12u44+2Im⁡∫Ωfu¯dx.Proof.
Equation (29) is easily obtained by multiplying (24) with u¯ (the complex conjugate of u) and taking the real part of the final equation.
To prove (30), we multiply (24) by -(u¯t+αu¯) and take the imaginary part; after some complex-valued calculations, we obtain(32)ddtΦ1u+2αΦ2u-2Re∫Ωftu¯dx=0,where Φ1(u) is given by (31), and (33)Φ2u=ux2+u2-u44+Im∫Ωfu¯dx.By the Agmon inequality u∞2⩽2u(ux+u), we have u44⩽2u3(ux+u). Hence, (34)Φ1u-2Φ2u=-ux2-u2+32u44⩽-ux2-u2+3u3ux+u⩽9u6+3u4-u2.We then rewrite the energy equation (32) as follows: (35)ddtΦ1u+αΦ1u=αΦ1u-2Φ2u+2Im⁡∫Ωftu¯dx⩽9αu6+3αu4+1αft2,which is just the needed energy inequality.Based on the above energy inequalities, one can obtain a priori estimate (for absorption, see the next subsection). Then it is similar as the autonomous case (see [5, 8]) to prove that (24) is well posed in H1. Namely, for each u0∈H1 and s∈R, (24) has a unique solution u(·,s,u0)∈C([s,+∞),H1) and the solution u(t,s,·):H1→H1 is continuous. This well-posed property permits us to define an evolution process S on H1 by(36)St,su0=ut,s,u0fort⩾s,u0∈H1.


### 3.2. Increasing, Bounded, and Pullback Absorbing Sets

In order to use the results of Theorem4, we need to look for an increasing, bounded, and pullback absorbing set.Lemma 6.
LetA1 be satisfied. Suppose BR be a ball in H1 (R is the radius) and t∈R, then there are τ0=τ0(R) such that, for all u0∈BR and τ⩾τ0,(37)sups⩽t⁡us,s-τ,u02⩽1+F3tα⩽1+F1tα1-e-α,(38)sups⩽t⁡∫s-τseαr-sur,s-τ,u04dr⩽c11+F12t,(39)sups⩽t⁡∫s-τseαr-sur,s-τ,u06dr⩽c21+F13t,where c1,c2 are positive constants and F1,F3 are functions given inA1.Proof.
Lett∈R and τ⩾0. We apply the Gronwall inequality on (29) over the interval [s-τ,r] with s-τ⩽r⩽s⩽t and then find(40)ur,s-τ,u02⩽e-αr-s+τu02+1α∫s-τreαr^-rfr^,·2dr^.Letting r=s, we have (41)us,s-τ,u02⩽e-ατu0H12+1αF3t,which yields (37) if we take τ0=2ln⁡R+1/α. On the other hand, by (40) again, (42)ur,s-τ,u02⩽e-αr-s+τu02+1α2F1t,where we use the fact that f(r^,·)2⩽F1(t) for r^⩽r⩽t. We then consider the sixth power to obtain (43)ur,s-τ,u06⩽8e-3αr-s+τu06+8α6F13t,which yields (44)∫s-τseαr-sur,s-τ,u06dr⩽8u06∫s-τseαr-se-3αr-s+τdr+cF13t∫s-τseαr-sdr⩽8e-3ατu06∫-τ0e-2αrdr+cF13t∫-∞seαr-sdr⩽4αe-ατu06+cF13t.From this, it is easy to deduce (39) with τ0(R)=2ln⁡2+6ln⁡R-ln⁡α/α and similarly obtain (38).We then consider the backward bound of solutions inH1 as follows.Lemma 7.
LetA1,A2 be satisfied. Then for each ball BR in H1 and t∈R, there exists τ1=τ1(R,t), such that, for all τ⩾τ1 and u0∈BR,(45)sups⩽t⁡us,s-τ,u0H12⩽M1t≔c31+F13t+F2t,where c3≔c3(α) is a constant and thus M1(·) is finite and increasing.Proof.
Lets⩽t and τ⩾0. Applying the Gronwall inequality on (30) over [s-τ,s], we get (46)Φ1us,s-τ,u0⩽e-ατΦ1u0+9α∫s-τseαr-sur,s-τ6dr+3α∫s-τseαr-sur,s-τ4dr+1α∫s-τseαr-sfrr,·2dr.By Lemma 6 andA2, there is a τ1=τ1(R) such that, for all τ⩾τ1 and u0∈BR,(47)sups⩽t⁡Φ1us,s-τ,u0⩽e-ατΦ1u0+c1+F13t+F12t+F2t.On the other hand, by the definition of Φ1 given in (31), we have (48)Φ1us-τ,s-τ,u0⩽2u0H12+fs-τ,·2⩽2u0H12+F1t.In particular, at the initial value, we have, if τ is large enough, then(49)e-ατΦ1u0⩽2e-ατu0H12+F1t⩽1.Both (47) and (49) imply that, for all u0∈BR and τ⩾τ0 (with a larger τ0),(50)sups⩽t⁡Φ1us,s-τ,u0⩽c1+F13t+F2t. By the Agmon inequality again, it follows from (31) that(51)Φ1u⩾ux2+u2-u3ux+u+2Im⁡∫Ωfu¯dx⩾12uH12-2u6-u4-2fs,·2,which together with (50) and Lemma 6 imply that, for all u0∈BR and τ⩾τ0 (with a larger τ0), (52)sups⩽t⁡us,s-τ,u0H12⩽2sups⩽t⁡Φ1us,s-τ,u0+4sups⩽t⁡us,s-τ,u06+2sups⩽t⁡us,s-τ,u04+4sups⩽t⁡fs,·2⩽c1+F13t+F2t+41+cF1t3+21+cF1t2+4F1t⩽c1+F13t+F2t,which shows the needed result.Under the light of Lemma7, we have the following increasing absorption.Theorem 8.
LetA1, A2 be true. Then the nonautonomous Schrödinger equation possesses an increasing, bounded, and pullback absorbing set K={K(t)}t∈R in H1 given by(53)Kt=w∈H1:wH12⩽M1t≔c31+F13t+F2t,t∈R.Proof.
ByA1,A2, we can see that M1(t) is an increasing and finite function with respect to t; this fact along with Lemma 7 shows that the nonautonomous set K is increasing, bounded, and pullback absorbing in H1. In fact, by (45), the absorption is backward uniform.Remark 9.
It seems not to prove that the nonautonomous Schrödinger equation has a bounded absorbing set inH1 if the assumptionA1 is replaced by the weaker assumption that F3(t)<∞, although this weak assumption is enough for reaction-diffusion systems (see [29]), BBM equations (see [30]), and Navier-Stokes equations (see [31]).


## 3.1. Hypotheses and Existence of Solutions

LetL2(Ω) be the space of complex-valued L2-functions whose norm is denoted by ·. Let H1 be the space of all one-periodic H1-function with the norm uH12=u2+ux2. The norm in Lp(Ω) is denoted by ·p. We then give some hypotheses on the time-space dependent force f=f(t,x).Assumption
A1. f is continuous andbackward bounded: (25)F1t≔sups⩽tfs,·2<+∞,∀t∈R.Assumption
A2. The time-derivative ft isbackward tempered: (26)F2t≔sups⩽t⁡∫-∞seαr-sfrr,·2dr<+∞,∀t∈R,where α>0 as given in (24).AssumptionA1 implies f∈Lloc2(R,L2(Ω)) and the finiteness of F3(t), where (27)F3t≔sups⩽t⁡∫-∞seαr-sfr,·2dr⩽F1t1-e-α.The last inequality can be proved as follows. (28)F3t=sups⩽t⁡∑m=0∞∫s-m+1s-meαr-sfr,·2dr⩽sups⩽t⁡∑m=0∞e-mα∫s-m+1s-mfr,·2dr⩽∑m=0∞e-mαsupr⩽t⁡fr,·2=F1t1-e-α.Note that all functions Fi(·) (i=1,2,3) are finite, nonnegative, andincreasing.We will repeatedly use the following two energy inequalities.Lemma 5.
Letu(t)=u(t,s,u0), t∈[s,+∞), be the approximation solution of (24). Then(29)ddtu2+αu2⩽1αft,·2,(30)ddtΦ1u+αΦ1u⩽9αu6+3αu4+1αftt,·2,where(31)Φ1u=ux2+u2-12u44+2Im⁡∫Ωfu¯dx.Proof.
Equation (29) is easily obtained by multiplying (24) with u¯ (the complex conjugate of u) and taking the real part of the final equation.
To prove (30), we multiply (24) by -(u¯t+αu¯) and take the imaginary part; after some complex-valued calculations, we obtain(32)ddtΦ1u+2αΦ2u-2Re∫Ωftu¯dx=0,where Φ1(u) is given by (31), and (33)Φ2u=ux2+u2-u44+Im∫Ωfu¯dx.By the Agmon inequality u∞2⩽2u(ux+u), we have u44⩽2u3(ux+u). Hence, (34)Φ1u-2Φ2u=-ux2-u2+32u44⩽-ux2-u2+3u3ux+u⩽9u6+3u4-u2.We then rewrite the energy equation (32) as follows: (35)ddtΦ1u+αΦ1u=αΦ1u-2Φ2u+2Im⁡∫Ωftu¯dx⩽9αu6+3αu4+1αft2,which is just the needed energy inequality.Based on the above energy inequalities, one can obtain a priori estimate (for absorption, see the next subsection). Then it is similar as the autonomous case (see [5, 8]) to prove that (24) is well posed in H1. Namely, for each u0∈H1 and s∈R, (24) has a unique solution u(·,s,u0)∈C([s,+∞),H1) and the solution u(t,s,·):H1→H1 is continuous. This well-posed property permits us to define an evolution process S on H1 by(36)St,su0=ut,s,u0fort⩾s,u0∈H1.


## 3.2. Increasing, Bounded, and Pullback Absorbing Sets

In order to use the results of Theorem4, we need to look for an increasing, bounded, and pullback absorbing set.Lemma 6.
LetA1 be satisfied. Suppose BR be a ball in H1 (R is the radius) and t∈R, then there are τ0=τ0(R) such that, for all u0∈BR and τ⩾τ0,(37)sups⩽t⁡us,s-τ,u02⩽1+F3tα⩽1+F1tα1-e-α,(38)sups⩽t⁡∫s-τseαr-sur,s-τ,u04dr⩽c11+F12t,(39)sups⩽t⁡∫s-τseαr-sur,s-τ,u06dr⩽c21+F13t,where c1,c2 are positive constants and F1,F3 are functions given inA1.Proof.
Lett∈R and τ⩾0. We apply the Gronwall inequality on (29) over the interval [s-τ,r] with s-τ⩽r⩽s⩽t and then find(40)ur,s-τ,u02⩽e-αr-s+τu02+1α∫s-τreαr^-rfr^,·2dr^.Letting r=s, we have (41)us,s-τ,u02⩽e-ατu0H12+1αF3t,which yields (37) if we take τ0=2ln⁡R+1/α. On the other hand, by (40) again, (42)ur,s-τ,u02⩽e-αr-s+τu02+1α2F1t,where we use the fact that f(r^,·)2⩽F1(t) for r^⩽r⩽t. We then consider the sixth power to obtain (43)ur,s-τ,u06⩽8e-3αr-s+τu06+8α6F13t,which yields (44)∫s-τseαr-sur,s-τ,u06dr⩽8u06∫s-τseαr-se-3αr-s+τdr+cF13t∫s-τseαr-sdr⩽8e-3ατu06∫-τ0e-2αrdr+cF13t∫-∞seαr-sdr⩽4αe-ατu06+cF13t.From this, it is easy to deduce (39) with τ0(R)=2ln⁡2+6ln⁡R-ln⁡α/α and similarly obtain (38).We then consider the backward bound of solutions inH1 as follows.Lemma 7.
LetA1,A2 be satisfied. Then for each ball BR in H1 and t∈R, there exists τ1=τ1(R,t), such that, for all τ⩾τ1 and u0∈BR,(45)sups⩽t⁡us,s-τ,u0H12⩽M1t≔c31+F13t+F2t,where c3≔c3(α) is a constant and thus M1(·) is finite and increasing.Proof.
Lets⩽t and τ⩾0. Applying the Gronwall inequality on (30) over [s-τ,s], we get (46)Φ1us,s-τ,u0⩽e-ατΦ1u0+9α∫s-τseαr-sur,s-τ6dr+3α∫s-τseαr-sur,s-τ4dr+1α∫s-τseαr-sfrr,·2dr.By Lemma 6 andA2, there is a τ1=τ1(R) such that, for all τ⩾τ1 and u0∈BR,(47)sups⩽t⁡Φ1us,s-τ,u0⩽e-ατΦ1u0+c1+F13t+F12t+F2t.On the other hand, by the definition of Φ1 given in (31), we have (48)Φ1us-τ,s-τ,u0⩽2u0H12+fs-τ,·2⩽2u0H12+F1t.In particular, at the initial value, we have, if τ is large enough, then(49)e-ατΦ1u0⩽2e-ατu0H12+F1t⩽1.Both (47) and (49) imply that, for all u0∈BR and τ⩾τ0 (with a larger τ0),(50)sups⩽t⁡Φ1us,s-τ,u0⩽c1+F13t+F2t. By the Agmon inequality again, it follows from (31) that(51)Φ1u⩾ux2+u2-u3ux+u+2Im⁡∫Ωfu¯dx⩾12uH12-2u6-u4-2fs,·2,which together with (50) and Lemma 6 imply that, for all u0∈BR and τ⩾τ0 (with a larger τ0), (52)sups⩽t⁡us,s-τ,u0H12⩽2sups⩽t⁡Φ1us,s-τ,u0+4sups⩽t⁡us,s-τ,u06+2sups⩽t⁡us,s-τ,u04+4sups⩽t⁡fs,·2⩽c1+F13t+F2t+41+cF1t3+21+cF1t2+4F1t⩽c1+F13t+F2t,which shows the needed result.Under the light of Lemma7, we have the following increasing absorption.Theorem 8.
LetA1, A2 be true. Then the nonautonomous Schrödinger equation possesses an increasing, bounded, and pullback absorbing set K={K(t)}t∈R in H1 given by(53)Kt=w∈H1:wH12⩽M1t≔c31+F13t+F2t,t∈R.Proof.
ByA1,A2, we can see that M1(t) is an increasing and finite function with respect to t; this fact along with Lemma 7 shows that the nonautonomous set K is increasing, bounded, and pullback absorbing in H1. In fact, by (45), the absorption is backward uniform.Remark 9.
It seems not to prove that the nonautonomous Schrödinger equation has a bounded absorbing set inH1 if the assumptionA1 is replaced by the weaker assumption that F3(t)<∞, although this weak assumption is enough for reaction-diffusion systems (see [29]), BBM equations (see [30]), and Navier-Stokes equations (see [31]).


## 4. Forward Absorption in Schrödinger Equations

This section establishes theforwardabsorption, which will be useful to deduce the compact-decay decomposition in the next section.In this case, we need to strengthen AssumptionsA1 andA2 as follows.Assumption
A1
′. f is continuous such that ρ1≔limt→+∞F1(t)<∞.Assumption
A2
′. ft exists such that ρ2≔limt→+∞F2(t)<∞.Assumption
A3. ftx exists such that ftx∈Lloc2(R,H-1(Ω)).Lemma 10.
LetA1′, A2′ be satisfied. Then for each ball BR in H1, there exists t1=t1(R)>0 such that(54)supt⩾t1⁡sups⩽0⁡supu0∈BR⁡us+t,s,u0H1⩽ρ3,(55)supt⩾t1⁡sups⩽0⁡supu0∈BR⁡uts+t,s,u0H-1⩽ρ4,where ρ3,ρ4 are constants depending on α and f.Proof.
Lets⩽0 and t⩾0. We apply the Gronwall inequality on (29) over [s,r] with r∈[s,s+t] and then find(56)ur,s,u02⩽e-αr-su02+1αF3r.Taking r=s+t in (56), we have (57)us+t,s,u02⩽e-αtu0H12+1αF3s+t⩽e-αtu0H12+cF1t.Let t1=2ln⁡R+1/α>0. By A1′, we know, for all t⩾t1 and u0∈BR,(58)us+t,s,u02⩽1+cρ1.We then consider the third power of (56) to obtain (59)ur,s,u06⩽8e-3αr-su06+cρ13,s⩽r⩽s+t,which yields (60)∫ss+teαr-s-tur,s,u06dr⩽8e-αtu06∫ss+te-2αr-sdr+cρ13∫ss+teαr-s-tdr⩽4αe-αtu061-e-2αt+cρ13⩽4αe-αtu06+cρ13.Rewriting t1 by t1(R)=(2ln⁡2+6ln⁡R+1-ln⁡α)/α, we deduce for all t⩾t1,(61)∫ss+teαr-s-tur,s,u06dr⩽1+cρ13,and similarly we have(62)∫ss+teαr-s-tur,s,u04dr⩽1+cρ12.To prove (54), we apply the Gronwall inequality on (30) over [s,s+t] to get (63)Φ1us+t,s,u0⩽e-αtΦ1u0+9α∫ss+teαr-s-tur,s6dr+3α∫ss+teαr-s-tur,s4+1α∫ss+teαr-s-tfrr,·2dr.By (61)-(62) and AssumptionA2, we easily deduce that, for t⩾t1,(64)Φ1us+t,s,u0⩽e-αtΦ1u0+cρ13+ρ12+cF2s+t.By the definition of Φ1 given in (31), we have (65)Φ1ur,s,u0⩽2uH12+fr,·2⩽2uH12+F1r,r⩾s.In particular, if t is large enough, then(66)e-αtΦ1us,s,u0⩽e-αt2u0H12+F1s⩽e-αt2u0H12+F10⩽1.By (61)-(62) and Assumption A2′, for t⩾t1 (with a larger t1),(67)Φ1us+t,s,u0⩽1+cρ13+ρ12+cρ2⩽C.By the Agmon inequality again, it is similar as (51) to prove that (68)Φ1us+t,s,u0⩾12uH12-2u6-u4-2fs+t,·2,which together with (67) and (58) imply that, for all u0∈BR, s⩽0, and t⩾t1 (with a larger τ0), (69)us+t,s,u0H12⩽2C+4us+t,s,u06+2us+t,s,u04+4fs+t,·2⩽c1+ρ13+ρ12+F1t⩽c1+ρ13+ρ12+ρ1,which shows the needed result (54).
To prove (55), we multiply (24) by a test function φ with φH1=1, then by the Agmon inequality, (70)ut,φ=-i∫uxxφ-i∫u2uφ+i-α∫uφ+∫fφ⩽ux2+φx2+u44+φ44+cu2+φ2+fs+t2⩽cux2+C+F1t⩽cux2+C.Then (55) follows from (54).


## 5. Compact-Decay Decomposition

### 5.1. High-Low Frequency Decomposition

We expandu(t) (the solution of (24)) into its Fourier series (71)ut=∑k∈Zukte2kπixand split u(t) into low frequency part and high-frequency part u(t)=PNu(t)+QNu(t) with (72)PNut=∑k⩽Nukte2kπix,QNut=∑k>Nukte2kπix.Lets⩽0 be arbitrary but fixed and take the initial value u0 in a ball BR of H1. Then we are concerned with two functions of t⩾0: (73)y=yt=yt;s=PNus+t,s,u0,z=zt=zt;s=QNus+t,s,u0.By the forward absorption given in Lemma 10, we know there is a t1>0 (depending on the radius of initial ball) such that(74)supt⩾t1⁡sups⩽0⁡yt;sH12+zt;sH12⩽C.The high-frequency part z(t) satisfies the following equation.(75)zt+αz+izxx-iz+iQNy+z2y+z=QNfs+t,t⩾t1,zt1=QNus+t1,s,u0.Then we consider the following equation on QNH1 with the initial value zero.(76)Zt+αZ+iZxx-iZ+iQNy+Z2y+Z=QNfs+t,t⩾t1,Zt1=Zt1;s=0,where Z≔Z(t) (if exists) is actually a function w.r.t. t⩾t1, s⩽0, u0∈BR, and N∈N. Sometimes, we write it as Z=Z(t;s). We can decompose the evolution process S by(77)us+t,s,u0=vt+wt,wherev≔y+Z,w≔z-Zfor all t⩾t1, s⩽0, and u0∈H1. In the sequel, the main task is to prove the asymptotic compactness of v and the exponential decay of w in QNH1 for large N.


### 5.2. The Uniformly Bounded Estimate forZ

To prove the existence of solutions for (76), we need to consider the approximation Zm, which is the solution for the projection of (76) on the subspace(78)PmQNH1=Zmt:Zmt=∑N<k⩽mukte2kπix,m>N.Then, by the standard Galerkin method and a priori estimate (see the next proposition), one can prove the existence of Z in QNH1 if N is large enough.In the following proposition, we actually prove the result forZm. However, for the sake of simplicity, we omit the subscript m and also omit the proof of convergence as m→∞.Proposition 11.
SupposeA1′,A2′ are satisfied. Then there exists N0=N0(α,f) such that, for N⩾N0, (76) has a uniformly bounded solution Z in QNH1 such that(79)supt⩾t1⁡sups⩽0⁡Zt;sH12⩽C,where C is a constant and t1 is the forward absorbing entering time.Proof.
Multiplying (76) by -Z¯t-αZ¯, taking the imaginary part, after some computations, we obtain an energy equation:(80)12ddtΨ1Z+αΨ1Z=Ψ2Z,t⩾t1with(81)Ψ1Z=ZH12+2Im⁡∫ΩfZ¯dx-12Z44-2Re⁡∫Ωy2yZ¯dx-2Re⁡∫ΩZ2Zy¯dx-2∫Ωy2Z2dx-2∫ΩRe⁡y¯Z2dx,Ψ2Z=αIm⁡∫ΩfZ¯dx+Im⁡∫ΩftZ¯dx+α2Z44-Re∫Ωy2ytZ¯dx-αRe⁡∫Ωy2yZ¯dx-∫ΩRe⁡y¯tZZ2dx+α∫ΩRe⁡y¯ZZ2dx-Re∫Ωy¯ytZ2dx-2∫ΩRe⁡y¯tZRe⁡y¯Zdx.We now consider the upper bound of Ψ2(Z); by AssumptionA1′, we have (82)I1≔αIm⁡∫ΩfZ¯dx+Im⁡∫ΩftZ¯dx⩽α16Z2+cfs+t,·2+cfts+t,·2⩽α16ZH12+cρ1+cfts+t,·2.By the classical interpolation and the Poincaré inequality on QNH1(83)Z4⩽cZ3/4ZH11/4,Z⩽cNZH1,the second term of Ψ2(Z) can be bounded by(84)I2≔α2Z44⩽cZ3ZH1⩽cN3ZH14.Note that H1 forms a Banach multiplicative algebra in one-dimension; that is,(85)uvwH1⩽cuH1vH1wH1.Then, by Lemma 10 and (74), the third term of Ψ2(Z) can be bounded by(86)I3≔-Re⁡∫Ωy2ytZ¯dx⩽ytH-1y2ZH1+2ytH-1yRe⁡y¯ZH1⩽3ytH-1yH12ZH1⩽CZH1⩽α32ZH12+C.By (74) and the embedding L6↪H1, the fourth term is bounded by (87)I4≔-αRe⁡∫Ωy2yZ¯dx⩽α32Z2+cy66⩽α32Z2+C.By the Agmon inequality and (83), we have(88)Z∞⩽cZ1/2ZH11/2⩽cNZH1.Then the fifth term of Ψ2(Z) is bounded by(89)I5≔-∫ΩRe⁡y¯tZZ2dx⩽cytH-1Z∞2ZH1⩽cNZH13⩽α32ZH12+cN2ZH14.Similarly, by (83) and the Hölder inequality, the sixth term of Ψ2(Z) is bounded by(90)I6≔α∫ΩRe⁡y¯ZZ2dx⩽αy4Z34/3=αy4Z43⩽cyH1Z9/4ZH13/4⩽cN9/4ZH13⩽α32ZH12+cN9/2ZH14.By (88), the rest terms can be bounded by(91)I7≔-Re⁡∫Ωy¯ytZ2dx-2∫ΩRe⁡y¯tZRe⁡y¯Zdx⩽3Z∞∫ΩytyZdx⩽3Z∞ytH-1yZH1⩽cZ∞ytH-1yH1ZH1⩽c1NZH12.By the above estimates, we know that Ψ2(Z) can be bounded by(92)Ψ2Z=∑i=17Ii⩽3α16+c1NZH12+cN2ZH14+cρ1+cfts+t,·2.Letting N0′=[256c1/α2]+1, where [·] denotes the integer-valued function, we have, for N⩾N0′,(93)Ψ2Z⩽α4ZH12+cN2ZH14+cρ1+cfts+t,·2.On the other hand, we similarly obtain a lower bound of Ψ1(Z).(94)Ψ1Z⩾ZH12-cfs+t2-12ZH12-12Z44-cy43Z4+y4Z43+y42Z42⩾12ZH12-C-Z44-gy4⩾12ZH12-cN3ZH14-C⩾12ZH12-cN2ZH14-C,where g(y4) is uniformly bounded in view of (74). By (93)-(94), we have (95)2Ψ2Z-αΨ1Z⩽cN2ZH14+C+cfts+t,·2.Then it follows from (80) that(96)ddtΨ1Z+αΨ1Z⩽cN2ZH14+C+cfts+t,·2.Since Z(t1)=0 and Ψ1(Z(t1))=Ψ1(0)=0, it follows from the Gronwall inequality on [t1,t] that(97)Ψ1Zt⩽cN2∫t1tZrH14eαr-tdr+C∫t1teαr-tdr+c∫t1teαr-tfts+r,·2dr⩽cN2∫t1tZrH14eαr-tdr+Cα+cF2s+t,which along with (94) and AssumptionA2′ imply that, for all t⩾t1,(98)ZtH12⩽cN2ZtH14+cN2∫t1tZrH14eαr-tdr+C.Hence, ξ(t)≔supr∈[t1,t]Z(r)H12 satisfies the following twice inequality:(99)ξt⩽c2N2ξ2t+C2,i.e.c2ξ2t-N2ξt+C2N2⩾0,where c2, C2 are positive constants. Let N0′′=[2c2C2]+1 and N0=max⁡(N0′,N0′′). Then for all N⩾N0 the discriminant of twice inequality is positive: (100)Δ=N4-4c2C2N2=N2N2-4c2C2>0.In this case, the equation c2ξ2-N2ξ+C2N2=0 has two positive roots. (101)α1=N2-NN2-4c2C22c2,α2=N2+NN2-4c2C22c2.We show that α1⩽2C2. Indeed, (102)α1⩽2C2⟺N2-4c2C2⩽NN2-4c2C2⟺N2-4c2C2⩽N,where the last inequality is obviously true. Now the twice inequality (99) has the solution (103)ξt⩽α1orξt⩾α2,t⩾t1.Since ξ(t1)=0 and the mapping t→ξ(t) is continuous on [t1,+∞), it follows that (104)ξt⩽α1⩽2C2,t⩾t1,which proves the required bound.


### 5.3. Further Regularity Estimates

We further show a regularity result forZ in H2.Proposition 12.
SupposeA1′,A2′, andA3 are satisfied. Then there exists N1=N1(α,f)⩾N0 such that, for N⩾N1, the solution Z of (76) is uniformly bounded in QNH2:(105)supt⩾t1⁡sups⩽0⁡Zt;sH22⩽CN+1,where C is a constant and t1 is the forward absorbing entering time.Proof.
Just like we did in the above proposition, we need to firstly show the bound ofZxm in H1 and then let m→+∞ to obtain the bound of Zx in H1. Thus, for the sake of convenience, we omit the detail of letting m→+∞ and also drop the superscript m of Zm and write Z=Zm, v=vm=y+Zm, QN=PmQN in this proof.
The first thing we shall do is to obtain the following equation by differentiating (76):(106)Zxt+αZx+QNA-F′vZx=G,t⩾t1,with initial value Zx(t1)=0, where A, F′, and G are given by(107)AZx=iZxxx-iZx,F′vZx=-iv2Zx-2iRe⁡v¯Zxv,G=QNfxs+t,·+F′vyx.
In order to show the needed result (105), we divide it into three steps.
Step  1. We show G∈Cb([t1,+∞);H-1). Let φ be a test function taken from H1 such that φH1=1. Then by AssumptionA1′, (108)φ,fxs+t=∫φfxs+t⩽cφx2+fs+t2⩽cφH12+F1s+t⩽c1+ρ1,which implies that fx(s+t)H-1⩽C. Since QN is a bounded operator from H-1 to H-1, it follows that QNfx(s+t)H-1⩽C for all t⩾t1 and s⩽0.
The rest term ofG can be rewritten as(109)QNF′vyx=-iQNv2yx-2iQNRe⁡v¯yxv.Note that we have the inverse inequality yxH1⩽2πNyH1 on PNH1. By Lemma 10 and Proposition 11, we know y,v∈Cb([t1,+∞);H1), where v=y+Z. Then, by the multiplicative algebra (see (85)), we have, for all t⩾t1,(110)-iQNvt2yxtH1⩽cvv¯yxH1⩽cvH12yxH1⩽cNyH1⩽CN,-2iQNRe⁡v¯tyxtvtH1⩽cNvH12yH1⩽CN.Therefore, QNF′(v)yx∈Cb([t1,+∞);H1) and so G∈Cb([t1,+∞);H-1) with G(t)H-1⩽C(N+1).
Step  2. We give the estimates of GtH-1, where it is easy to obtain (111)Gt=QNfxts+t,·-2iQNRe⁡v¯vtyx-iQNv2yxt-2iQNRev¯yxvt.By AssumptionA3, for the test function φH1=1, we have (112)φ,fxts+t=∫φfxts+t⩽cφx2+fts+t2.Hence, QNfxts+tH-1⩽c(1+fts+t,·2) for all t⩾t1 and s⩽0.
To estimate the rest terms, we note thatsupt⩾t1Zt(t)H-1⩽C, which can be obtained from (76) and the uniform bound of Z. By Lemma 10, ytH-1⩽C and so vtH-1⩽C. For the second term of Gt, by the multiplicative algebra (85) again, we have (113)-2iRe⁡v¯vtyx,φ⩽∫vtvyxφdx⩽vtH-1vyxφH1⩽CvH1φH1yxH1⩽CyxH1⩽CNyH1⩽CN,where φH1=1. Hence Re(v¯vt)yxH-1⩽CN and so QNRe⁡(v¯vt)yxH-1⩽CN in view of the boundedness of the operator QN from H-1 to H-1. It is similar to obtain the estimates for the rest terms in Gt and so(114)Gtt;sH-1⩽c1+N+fts+t,·2,t⩾t1,s⩽0.
Step  3. Finally, we bound ZxH1. Multiplying (106) by -Z¯xt-αZ¯x, taking the imaginary part of the resulting equation and then integrating over Ω, we obtain an energy equation:(115)ddtΛ1Zx+2αΛ1Zx=2Λ2Zxwhere (116)Λ1Zx=ZxH12-∫Ωv2Zx2dx-2∫ΩRe⁡v¯Zx2dx+2∫ΩIm⁡GZ¯xdx,(117)Λ2Zx=-∫ΩRe⁡v¯vtZx2dx-2∫ΩRe⁡v¯tZx·Re⁡v¯Zxdx+∫ΩIm⁡GtZ¯xdx+α∫ΩIm⁡GZ¯xdx.By the bound of v in H1 and vt in H-1, and by the inequality v∞⩽(c/N)vH1 on the space QNH1 (see (88)), we have, for N⩾N0, (118)Λ2Zx⩽3vtH-1v∞ZxH12+GtH-1ZxH1+GH-1ZxH1⩽cNvtH-1vH1ZxH12+GtH-1+GH-1ZxH1⩽c0N+α16ZxH12+cGtH-12+GH-12.Letting N1′=max⁡([256c0/9α2]+1,N0), then we have(119)Λ2Zx⩽α4ZxH12+cGtH-12+GH-12,forN⩾N1′.On the other hand, we have the lower bound of Λ1(Zx): (120)Λ1Zx⩾ZxH12-3v∞2ZxL22-2GH-1ZxH1⩾ZxH12-cN2ZxH12-2GH-1ZxH1⩾34-c1N2ZxH12-cGH-12.Letting N1=max([4c1]+1,N1′), we see, that for N⩾N1,(121)Λ1Zx⩾12ZxH12-cGH-12fort⩾t1.Therefore, we substitute (119)-(121) into (115) and use (114) to find(122)ddtΛ1Zx+αΛ1Zx⩽cGtH-12+GH-12⩽c1+N+fts+t,·2.Applying the Gronwall inequality on [t1,t] and observing Λ1(Zx(t1))=Λ1(0)=0, we get (123)Λ1Zxt⩽∫t1tceαr-t1+N+fts+r,·2dr⩽c1+N+F2t,which along withA2′ and (121) imply the needed result (105).


### 5.4. Exponential Decay for Large Times

Letz=z(t)=zt;s,u0=QNus+t,s,u0, which is a solution of (75). Let Z=Z(t)=Zt;s,u0 be the corresponding solution of (76). We need to prove the exponential decay of z-Z.Proposition 13.
SupposeA1′,A2′ are satisfied. Then there exists N2⩾N1 such that, for N⩾N2,(124)sups⩽0⁡zt;s-Zt;sH1⩽ce-αt-t1,fort⩾t1.Proof.
Recall thatv=y+Z and w=z-Z=u-v. Then it follows from (75) and (76) that w is the solution of(125)wt+αw+iwxx-iw=iQNu2+v2w+uvw¯,t⩾t1.Multiplying (125) by -w¯t-αw¯, taking the imaginary part of the resulting equation and then integrating over Ω, we obtain an energy equation:(126)12ddtΥ1w+αΥ1w=Υ2wwith(127)Υ1w=wH12-∫Ωu2+v2w2dx-Re∫Ωuvw¯2dx,Υ2w=-12∫Ωu2+v2tw2dx-12Re∫Ωuvtw¯2dx.By the boundedness of z, Z in QNH1 and ut,vt in H-1, we have the upper bound of Υ2(w):(128)Υ2w⩽cut+vtH-1u+vH1wH1wL∞⩽cNwH12,and also the lower bound of Υ1(w):(129)Υ1w⩾1-cN2wH12⩾12wH12if N is large enough. By (126)–(129), we have(130)ddtΥ1w+αΥ1w⩽c2N-αwH12⩽0if N⩾N2≔max⁡([4c22/α2]+1,N1) and t⩾t1. By the Gronwall inequality on [t1,t], we obtain(131)Υ1wt⩽e-αt-t1Υ1wt1.By Lemma 10, z(t1)H1=QNu(s+t1,s,u0)H1⩽C. Note that Z(t1)=0 and so w(t1)H1⩽C, which easily implies that Υ1(w(t1)) is bounded. Therefore, by (129), we have(132)wtH12⩽Ce-αt-t1,∀t⩾t1,which completes the proof.


## 5.1. High-Low Frequency Decomposition

We expandu(t) (the solution of (24)) into its Fourier series (71)ut=∑k∈Zukte2kπixand split u(t) into low frequency part and high-frequency part u(t)=PNu(t)+QNu(t) with (72)PNut=∑k⩽Nukte2kπix,QNut=∑k>Nukte2kπix.Lets⩽0 be arbitrary but fixed and take the initial value u0 in a ball BR of H1. Then we are concerned with two functions of t⩾0: (73)y=yt=yt;s=PNus+t,s,u0,z=zt=zt;s=QNus+t,s,u0.By the forward absorption given in Lemma 10, we know there is a t1>0 (depending on the radius of initial ball) such that(74)supt⩾t1⁡sups⩽0⁡yt;sH12+zt;sH12⩽C.The high-frequency part z(t) satisfies the following equation.(75)zt+αz+izxx-iz+iQNy+z2y+z=QNfs+t,t⩾t1,zt1=QNus+t1,s,u0.Then we consider the following equation on QNH1 with the initial value zero.(76)Zt+αZ+iZxx-iZ+iQNy+Z2y+Z=QNfs+t,t⩾t1,Zt1=Zt1;s=0,where Z≔Z(t) (if exists) is actually a function w.r.t. t⩾t1, s⩽0, u0∈BR, and N∈N. Sometimes, we write it as Z=Z(t;s). We can decompose the evolution process S by(77)us+t,s,u0=vt+wt,wherev≔y+Z,w≔z-Zfor all t⩾t1, s⩽0, and u0∈H1. In the sequel, the main task is to prove the asymptotic compactness of v and the exponential decay of w in QNH1 for large N.


## 5.2. The Uniformly Bounded Estimate forZ

To prove the existence of solutions for (76), we need to consider the approximation Zm, which is the solution for the projection of (76) on the subspace(78)PmQNH1=Zmt:Zmt=∑N<k⩽mukte2kπix,m>N.Then, by the standard Galerkin method and a priori estimate (see the next proposition), one can prove the existence of Z in QNH1 if N is large enough.In the following proposition, we actually prove the result forZm. However, for the sake of simplicity, we omit the subscript m and also omit the proof of convergence as m→∞.Proposition 11.
SupposeA1′,A2′ are satisfied. Then there exists N0=N0(α,f) such that, for N⩾N0, (76) has a uniformly bounded solution Z in QNH1 such that(79)supt⩾t1⁡sups⩽0⁡Zt;sH12⩽C,where C is a constant and t1 is the forward absorbing entering time.Proof.
Multiplying (76) by -Z¯t-αZ¯, taking the imaginary part, after some computations, we obtain an energy equation:(80)12ddtΨ1Z+αΨ1Z=Ψ2Z,t⩾t1with(81)Ψ1Z=ZH12+2Im⁡∫ΩfZ¯dx-12Z44-2Re⁡∫Ωy2yZ¯dx-2Re⁡∫ΩZ2Zy¯dx-2∫Ωy2Z2dx-2∫ΩRe⁡y¯Z2dx,Ψ2Z=αIm⁡∫ΩfZ¯dx+Im⁡∫ΩftZ¯dx+α2Z44-Re∫Ωy2ytZ¯dx-αRe⁡∫Ωy2yZ¯dx-∫ΩRe⁡y¯tZZ2dx+α∫ΩRe⁡y¯ZZ2dx-Re∫Ωy¯ytZ2dx-2∫ΩRe⁡y¯tZRe⁡y¯Zdx.We now consider the upper bound of Ψ2(Z); by AssumptionA1′, we have (82)I1≔αIm⁡∫ΩfZ¯dx+Im⁡∫ΩftZ¯dx⩽α16Z2+cfs+t,·2+cfts+t,·2⩽α16ZH12+cρ1+cfts+t,·2.By the classical interpolation and the Poincaré inequality on QNH1(83)Z4⩽cZ3/4ZH11/4,Z⩽cNZH1,the second term of Ψ2(Z) can be bounded by(84)I2≔α2Z44⩽cZ3ZH1⩽cN3ZH14.Note that H1 forms a Banach multiplicative algebra in one-dimension; that is,(85)uvwH1⩽cuH1vH1wH1.Then, by Lemma 10 and (74), the third term of Ψ2(Z) can be bounded by(86)I3≔-Re⁡∫Ωy2ytZ¯dx⩽ytH-1y2ZH1+2ytH-1yRe⁡y¯ZH1⩽3ytH-1yH12ZH1⩽CZH1⩽α32ZH12+C.By (74) and the embedding L6↪H1, the fourth term is bounded by (87)I4≔-αRe⁡∫Ωy2yZ¯dx⩽α32Z2+cy66⩽α32Z2+C.By the Agmon inequality and (83), we have(88)Z∞⩽cZ1/2ZH11/2⩽cNZH1.Then the fifth term of Ψ2(Z) is bounded by(89)I5≔-∫ΩRe⁡y¯tZZ2dx⩽cytH-1Z∞2ZH1⩽cNZH13⩽α32ZH12+cN2ZH14.Similarly, by (83) and the Hölder inequality, the sixth term of Ψ2(Z) is bounded by(90)I6≔α∫ΩRe⁡y¯ZZ2dx⩽αy4Z34/3=αy4Z43⩽cyH1Z9/4ZH13/4⩽cN9/4ZH13⩽α32ZH12+cN9/2ZH14.By (88), the rest terms can be bounded by(91)I7≔-Re⁡∫Ωy¯ytZ2dx-2∫ΩRe⁡y¯tZRe⁡y¯Zdx⩽3Z∞∫ΩytyZdx⩽3Z∞ytH-1yZH1⩽cZ∞ytH-1yH1ZH1⩽c1NZH12.By the above estimates, we know that Ψ2(Z) can be bounded by(92)Ψ2Z=∑i=17Ii⩽3α16+c1NZH12+cN2ZH14+cρ1+cfts+t,·2.Letting N0′=[256c1/α2]+1, where [·] denotes the integer-valued function, we have, for N⩾N0′,(93)Ψ2Z⩽α4ZH12+cN2ZH14+cρ1+cfts+t,·2.On the other hand, we similarly obtain a lower bound of Ψ1(Z).(94)Ψ1Z⩾ZH12-cfs+t2-12ZH12-12Z44-cy43Z4+y4Z43+y42Z42⩾12ZH12-C-Z44-gy4⩾12ZH12-cN3ZH14-C⩾12ZH12-cN2ZH14-C,where g(y4) is uniformly bounded in view of (74). By (93)-(94), we have (95)2Ψ2Z-αΨ1Z⩽cN2ZH14+C+cfts+t,·2.Then it follows from (80) that(96)ddtΨ1Z+αΨ1Z⩽cN2ZH14+C+cfts+t,·2.Since Z(t1)=0 and Ψ1(Z(t1))=Ψ1(0)=0, it follows from the Gronwall inequality on [t1,t] that(97)Ψ1Zt⩽cN2∫t1tZrH14eαr-tdr+C∫t1teαr-tdr+c∫t1teαr-tfts+r,·2dr⩽cN2∫t1tZrH14eαr-tdr+Cα+cF2s+t,which along with (94) and AssumptionA2′ imply that, for all t⩾t1,(98)ZtH12⩽cN2ZtH14+cN2∫t1tZrH14eαr-tdr+C.Hence, ξ(t)≔supr∈[t1,t]Z(r)H12 satisfies the following twice inequality:(99)ξt⩽c2N2ξ2t+C2,i.e.c2ξ2t-N2ξt+C2N2⩾0,where c2, C2 are positive constants. Let N0′′=[2c2C2]+1 and N0=max⁡(N0′,N0′′). Then for all N⩾N0 the discriminant of twice inequality is positive: (100)Δ=N4-4c2C2N2=N2N2-4c2C2>0.In this case, the equation c2ξ2-N2ξ+C2N2=0 has two positive roots. (101)α1=N2-NN2-4c2C22c2,α2=N2+NN2-4c2C22c2.We show that α1⩽2C2. Indeed, (102)α1⩽2C2⟺N2-4c2C2⩽NN2-4c2C2⟺N2-4c2C2⩽N,where the last inequality is obviously true. Now the twice inequality (99) has the solution (103)ξt⩽α1orξt⩾α2,t⩾t1.Since ξ(t1)=0 and the mapping t→ξ(t) is continuous on [t1,+∞), it follows that (104)ξt⩽α1⩽2C2,t⩾t1,which proves the required bound.


## 5.3. Further Regularity Estimates

We further show a regularity result forZ in H2.Proposition 12.
SupposeA1′,A2′, andA3 are satisfied. Then there exists N1=N1(α,f)⩾N0 such that, for N⩾N1, the solution Z of (76) is uniformly bounded in QNH2:(105)supt⩾t1⁡sups⩽0⁡Zt;sH22⩽CN+1,where C is a constant and t1 is the forward absorbing entering time.Proof.
Just like we did in the above proposition, we need to firstly show the bound ofZxm in H1 and then let m→+∞ to obtain the bound of Zx in H1. Thus, for the sake of convenience, we omit the detail of letting m→+∞ and also drop the superscript m of Zm and write Z=Zm, v=vm=y+Zm, QN=PmQN in this proof.
The first thing we shall do is to obtain the following equation by differentiating (76):(106)Zxt+αZx+QNA-F′vZx=G,t⩾t1,with initial value Zx(t1)=0, where A, F′, and G are given by(107)AZx=iZxxx-iZx,F′vZx=-iv2Zx-2iRe⁡v¯Zxv,G=QNfxs+t,·+F′vyx.
In order to show the needed result (105), we divide it into three steps.
Step  1. We show G∈Cb([t1,+∞);H-1). Let φ be a test function taken from H1 such that φH1=1. Then by AssumptionA1′, (108)φ,fxs+t=∫φfxs+t⩽cφx2+fs+t2⩽cφH12+F1s+t⩽c1+ρ1,which implies that fx(s+t)H-1⩽C. Since QN is a bounded operator from H-1 to H-1, it follows that QNfx(s+t)H-1⩽C for all t⩾t1 and s⩽0.
The rest term ofG can be rewritten as(109)QNF′vyx=-iQNv2yx-2iQNRe⁡v¯yxv.Note that we have the inverse inequality yxH1⩽2πNyH1 on PNH1. By Lemma 10 and Proposition 11, we know y,v∈Cb([t1,+∞);H1), where v=y+Z. Then, by the multiplicative algebra (see (85)), we have, for all t⩾t1,(110)-iQNvt2yxtH1⩽cvv¯yxH1⩽cvH12yxH1⩽cNyH1⩽CN,-2iQNRe⁡v¯tyxtvtH1⩽cNvH12yH1⩽CN.Therefore, QNF′(v)yx∈Cb([t1,+∞);H1) and so G∈Cb([t1,+∞);H-1) with G(t)H-1⩽C(N+1).
Step  2. We give the estimates of GtH-1, where it is easy to obtain (111)Gt=QNfxts+t,·-2iQNRe⁡v¯vtyx-iQNv2yxt-2iQNRev¯yxvt.By AssumptionA3, for the test function φH1=1, we have (112)φ,fxts+t=∫φfxts+t⩽cφx2+fts+t2.Hence, QNfxts+tH-1⩽c(1+fts+t,·2) for all t⩾t1 and s⩽0.
To estimate the rest terms, we note thatsupt⩾t1Zt(t)H-1⩽C, which can be obtained from (76) and the uniform bound of Z. By Lemma 10, ytH-1⩽C and so vtH-1⩽C. For the second term of Gt, by the multiplicative algebra (85) again, we have (113)-2iRe⁡v¯vtyx,φ⩽∫vtvyxφdx⩽vtH-1vyxφH1⩽CvH1φH1yxH1⩽CyxH1⩽CNyH1⩽CN,where φH1=1. Hence Re(v¯vt)yxH-1⩽CN and so QNRe⁡(v¯vt)yxH-1⩽CN in view of the boundedness of the operator QN from H-1 to H-1. It is similar to obtain the estimates for the rest terms in Gt and so(114)Gtt;sH-1⩽c1+N+fts+t,·2,t⩾t1,s⩽0.
Step  3. Finally, we bound ZxH1. Multiplying (106) by -Z¯xt-αZ¯x, taking the imaginary part of the resulting equation and then integrating over Ω, we obtain an energy equation:(115)ddtΛ1Zx+2αΛ1Zx=2Λ2Zxwhere (116)Λ1Zx=ZxH12-∫Ωv2Zx2dx-2∫ΩRe⁡v¯Zx2dx+2∫ΩIm⁡GZ¯xdx,(117)Λ2Zx=-∫ΩRe⁡v¯vtZx2dx-2∫ΩRe⁡v¯tZx·Re⁡v¯Zxdx+∫ΩIm⁡GtZ¯xdx+α∫ΩIm⁡GZ¯xdx.By the bound of v in H1 and vt in H-1, and by the inequality v∞⩽(c/N)vH1 on the space QNH1 (see (88)), we have, for N⩾N0, (118)Λ2Zx⩽3vtH-1v∞ZxH12+GtH-1ZxH1+GH-1ZxH1⩽cNvtH-1vH1ZxH12+GtH-1+GH-1ZxH1⩽c0N+α16ZxH12+cGtH-12+GH-12.Letting N1′=max⁡([256c0/9α2]+1,N0), then we have(119)Λ2Zx⩽α4ZxH12+cGtH-12+GH-12,forN⩾N1′.On the other hand, we have the lower bound of Λ1(Zx): (120)Λ1Zx⩾ZxH12-3v∞2ZxL22-2GH-1ZxH1⩾ZxH12-cN2ZxH12-2GH-1ZxH1⩾34-c1N2ZxH12-cGH-12.Letting N1=max([4c1]+1,N1′), we see, that for N⩾N1,(121)Λ1Zx⩾12ZxH12-cGH-12fort⩾t1.Therefore, we substitute (119)-(121) into (115) and use (114) to find(122)ddtΛ1Zx+αΛ1Zx⩽cGtH-12+GH-12⩽c1+N+fts+t,·2.Applying the Gronwall inequality on [t1,t] and observing Λ1(Zx(t1))=Λ1(0)=0, we get (123)Λ1Zxt⩽∫t1tceαr-t1+N+fts+r,·2dr⩽c1+N+F2t,which along withA2′ and (121) imply the needed result (105).


## 5.4. Exponential Decay for Large Times

Letz=z(t)=zt;s,u0=QNus+t,s,u0, which is a solution of (75). Let Z=Z(t)=Zt;s,u0 be the corresponding solution of (76). We need to prove the exponential decay of z-Z.Proposition 13.
SupposeA1′,A2′ are satisfied. Then there exists N2⩾N1 such that, for N⩾N2,(124)sups⩽0⁡zt;s-Zt;sH1⩽ce-αt-t1,fort⩾t1.Proof.
Recall thatv=y+Z and w=z-Z=u-v. Then it follows from (75) and (76) that w is the solution of(125)wt+αw+iwxx-iw=iQNu2+v2w+uvw¯,t⩾t1.Multiplying (125) by -w¯t-αw¯, taking the imaginary part of the resulting equation and then integrating over Ω, we obtain an energy equation:(126)12ddtΥ1w+αΥ1w=Υ2wwith(127)Υ1w=wH12-∫Ωu2+v2w2dx-Re∫Ωuvw¯2dx,Υ2w=-12∫Ωu2+v2tw2dx-12Re∫Ωuvtw¯2dx.By the boundedness of z, Z in QNH1 and ut,vt in H-1, we have the upper bound of Υ2(w):(128)Υ2w⩽cut+vtH-1u+vH1wH1wL∞⩽cNwH12,and also the lower bound of Υ1(w):(129)Υ1w⩾1-cN2wH12⩾12wH12if N is large enough. By (126)–(129), we have(130)ddtΥ1w+αΥ1w⩽c2N-αwH12⩽0if N⩾N2≔max⁡([4c22/α2]+1,N1) and t⩾t1. By the Gronwall inequality on [t1,t], we obtain(131)Υ1wt⩽e-αt-t1Υ1wt1.By Lemma 10, z(t1)H1=QNu(s+t1,s,u0)H1⩽C. Note that Z(t1)=0 and so w(t1)H1⩽C, which easily implies that Υ1(w(t1)) is bounded. Therefore, by (129), we have(132)wtH12⩽Ce-αt-t1,∀t⩾t1,which completes the proof.


## 6. Backward Compact Attractors

Finally, we state and prove the main result. We need to spilt the evolution processS. Let N⩾N2 be fixed, where N2 is the level given in Proposition 13. Then for all t⩾0, s∈R, and u0∈H1, we write S=S1+S2 with(133)S1s+t,su0=yt;s=PNus+t,s,u0if0⩽t⩽t1ors>0,vt;s=yt;s+Zt;sfort⩾t1,s⩽0,(134)S2s+t,su0=zt;s=QNus+t,s,u0for0⩽t⩽t1ors>0,wt;s=zt;s-Zt;sfort⩾t1,s⩽0,where t1=t1(u0H1) is the forward entering time.Theorem 14.
(a) SupposeA1,A2. Then the nonautonomous Schrödinger equation has an increasing, bounded, and pullback absorbing set K={K(s)}s∈R.
(b) SupposeA1′,A2′,A3. Then the nonautonomous Schrödinger equation has a backward compact pullback attractor A={A(s)}s∈R given by (135)Ar=ωKr,r=⋂t0>0⋃t⩾t0Sr,r-tKr¯=ω1Kr,r=⋂t0>0⋃t⩾t0S1r,r-tKr¯,r∈R.Proof.
The assertion (a) follows from Theorem8. To prove (b), by the abstract result (i.e., Theorem 4), it suffices to prove that the evolution process S has a backward compact-decay decomposition.
We need to prove thatS1 is backward asymptotically compact. Indeed, let r∈R and take some sequences rn⩽r, τn→+∞, and u0,n∈BR⊂H1. Without loss of generality, we assume τn⩾max⁡{t1,r}, where t1 is the forward entering time. Then by (133) we have (136)S1rn,rn-τnu0,n=S1rn-τn+τn,rn-τnu0,n=yτn;rn-τn,u0,n+Zτn;rn-τn,u0,n.By Lemma 10, we know that (137)yτn;rn-τn,u0,nH1⩽supt⩾t1⁡sups⩽0⁡supu0∈BR⁡yt;s,u0H1⩽C,which means that {y(τn;rn-τn,u0,n)} is a bounded sequence in the finite-dimensional space PNH1 and so, passing to a subsequence, it is convergent. By Proposition 12, we know (138)Zτn;rn-τn,u0,nH2⩽supt⩾t1⁡sups⩽0⁡supu0∈BR⁡Zt;s,u0H2⩽CN+1,which means that {Z(τn;rn-τn,u0,n)} is bounded in H2. By the Sobolev compact embedding, it is precompact in H1 and so is {S1(rn,rn-τn)u0,n}.
On the other hand, by Proposition13, we know, for r∈R, τ⩾max{t1,r} and u0∈BR, (139)supr^⩽r⁡S2r^,r^-τu0H1⩽sups⩽0⁡zτ;s,u0-Zτ;s,u0H1⩽Ce-ατ-t1,which implies S2 is backward decay.Remark 15.
(I) In fact, under AssumptionsA1′-A2′, it is possible to prove the existence of a forward attractor (see [32, 33]), which may be different from the pullback attractor. In the present paper, we only impose the forward absorption to deduce a backward compact-decay composition, which seems not to be deduced from the pullback absorption for the nonautonomous Schrödinger equation.
(II) Supposef,ft∈C(R,L2) and they are time-periodic, then f satisfies both conditionsA1′ andA2′. In this case, by using the theoretical result as given in [17], one can obtain a periodic pullback attractor and maybe a periodic forward attractor.



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