Stabilization of higher order Schrödinger equations on a finite interval: Part II
SSTABILIZATION OF HIGHER ORDER SCHR ¨ODINGEREQUATIONS ON A FINITE INTERVAL: PART II
T ¨URKER ¨OZSARI a, ∗ AND KEMAL CEM YILMAZ ba Department of Mathematics, Bilkent UniversityC¸ ankaya, Ankara, 06800 Turkey b Department of Mathematics, Izmir Institute of TechnologyUrla, Izmir, 35430 Turkey
Abstract.
Backstepping based controller and observer models were designed forhigher order linear and nonlinear Schr¨odinger equations on a finite interval in[2] where the controller was assumed to be acting from the left endpoint of themedium. In this companion paper, we further the analysis by considering bound-ary controller(s) acting at the right endpoint of the domain. It turns out thatthe problem is more challenging in this scenario as the associated boundary valueproblem for the backstepping kernel becomes overdetermined and lacks a smoothsolution. The latter is essential to switch back and forth between the original plantand the so called target system. To overcome this difficulty we rely on the strategyof using an imperfect kernel, namely one of the boundary conditions in kernel PDEmodel is disregarded. The drawback is that one loses rapid stabilization in com-parison with the left endpoint controllability. Nevertheless, the exponential decayof the L -norm with a certain rate still holds. The observer design is associatedwith new challenges from the point of view of wellposedness and one has to provesmoothing properties for an associated initial boundary value problem with inho-mogeneous boundary data. This problem is solved by using Laplace transform intime. However, the Bromwich integral that inverts the transformed solution is as-sociated with certain analyticity issues which are treated through a subtle analysis.Numerical algorithms and simulations verifying the theoretical results are given.2020 Mathematics Subject Classification.
Key words and phrases. higher order Schr¨odinger equation, backstepping, stabilization, observer,boundary controller, exponential stability.*Corresponding author: T¨urker ¨Ozsarı, [email protected]. a r X i v : . [ m a t h . O C ] F e b STABILIZATION OF HIGHER ORDER SCHR ¨ODINGER EQUATIONS
Contents
1. Introduction 21.1. Statements of problems and main results 21.2. Preliminaries 111.3. Outline 132. Auxiliary lemmas 133. Controller design 403.1. Backstepping kernel 403.2. Wellposedness 443.3. Stability 504. Observer design 514.1. Wellposedness 524.2. Stability 575. Numerical simulations 625.1. Controller design 625.2. Observer design 65Appendix A. Deduction of the kernel pde model (1.7) 68Appendix B. Deduction of the kernel pde model (1.18) 71Appendix C. Roots of the characteristic equation (2.32). 74Acknowledgements 76References 761.
Introduction
Statements of problems and main results.
Backstepping based controllerand observer models were designed for higher order linear and nonlinear Schr¨odingerequations on a finite interval in [2] where the controller was assumed to be actingfrom the left endpoint of the medium. In this companion paper, we further theanalysis by considering boundary controller(s) acting at the right endpoint of thedomain. We will consider only the linear higher order Schr¨odinger (HLS) equationin this paper: iu t + iβu xxx + αu xx + iδu x = 0 , x ∈ (0 , L ) , t ∈ (0 , T ) ,u (0 , t ) = 0 , u ( L, t ) = h ( t ) , u x ( L, t ) = h ( t ) ,u ( x,
0) = u ( x ) , (1.1)where β > α, δ ∈ R , h ( t ) = h ( u ( · , t )) and h ( t ) = h ( u ( · , t )) are feedbacks actingat the right endpoint of the domain. The control design results of this paper can be TABILIZATION OF HIGHER ORDER SCHR ¨ODINGER EQUATIONS 3 extended to associated higher-order nonlinear Schr¨odinger equations iu t + iβu xxx + αu xx + iδu x + f ( u ) = 0as in Part I (see [2]) with additional assumptions on the coefficients, but this topicis omitted here considering the volume of current text and postponed to a futurepaper. In addition, it is also possible to consider other sets of boundary conditionshere as in Part I that involves second order traces such as u (0 , t ) = 0 , u x ( L, t ) = h ( t ) , u xx ( L, t ) = h ( t ) , but this will also be discussed in another place.The higher-order nonlinear Schr¨odinger equation was originally given by iu t + 12 u xx + | u | u + (cid:15)i (cid:0) β u xxx + β ( | u | u ) x + β u | u | x (cid:1) = 0 , (1.2)which has been used to describe the evolution of femtosecond pulse propagationin a nonlinear optical fiber [15, 16]. In this equation the first term represents theevolution, second term is the group velocity dispersion, third term is self-phase mod-ulation, fourth term is the higher order linear dispersive term, fifth term is relatedto self-steepening and sixth term is related to self-frequency shift due to the stimu-lated Raman scattering. In the absence of the last three terms, the model becomesclassical nonlinear Schr¨odinger equation (NLS) which describes slowly varying waveenvelopes in a dispersive medium. It has applications in several fields of physicssuch as plasma physics, solid-state physics, nonlinear optics. It also describes thepropagation of picosecond optical pulse in a mono-mode fiber [33]. However, for thepulses in the femtosecond regime, the NLS equation becomes inadequate and higherorder nonlinear and dispersive terms become crucial. See [1] for a detailed discussionof the higher order effects upon the propagation of an optical pulse.Higher order linear and nonlinear Schr¨odinger equations were studied from thepoint of many different aspects. Regarding the wellposedness of solutions, we referthe reader to [5, 6, 7, 14, 19, 29, 30]. A numerical study of this problem was givenin [21]. From the controllability and stabilization perspective, we refer the reader to[8] for exact boundary controllability, [3] and [10] for internal feedback stabilizationand [2] for boundary feedback stabilization.From a practical point of view, stabilization of solutions is necessary in order toprevent the transmission of an undesirable pulse propagation. Our study offers apractical solution to this issue because: (i) the stabilization is fast, i.e. the absorp-tion effect is exponential and (ii) the control acts only from the boundary which isdesirable when access to medium is limited. In the absence of feedback controllers, L − norm of the solution satisfies ddt (cid:107) u ( · , t ) (cid:107) L (0 ,L ) = − β | u x (0 , t ) | ≤ . (1.3) STABILIZATION OF HIGHER ORDER SCHR ¨ODINGER EQUATIONS
This can be shown by taking L − inner product of the main equation in (1.1) with u and applying integration by parts. From the estimate (1.3), we infer that L − norm ofthe solution does not increase in time. Furthermore, it was shown that if α +3 βδ > L ∈ N . = (cid:40) πβ (cid:115) k + kl + l βδ + α : k, l ∈ Z + (cid:41) , (1.4)then the L − norm of the solution does not necessarily decay to zero. Here N is theset of critical lengths in the context of exact boundary controllability for the HLS(see [8, 12] for the derivation of this set of critical lengths). For instance choosingthe coefficients β = 1, α = 2 and δ = 8 with k = 1 and l = 2, we obtain L = π ∈ N .Moreover, choosing the initial state as u ( x ) = 3 − e ix − e − ix , we see that u ( x, t ) = u ( x ) solves (1.1). Therefore, we find a time–independentsolution with a constant energy if no control acts on the system.In this paper, we are interested in constructing suitable feedback controllers tomake sure that we can steer all solutions to zero with an exponential rate of decayon domains of both critical and uncritical lengths. More precisely, we consider theproblem below: Problem 1.1.
Given
L > , find λ > and feedback control laws h ( t ) = h ( u ( · , t )) and h ( t ) = h ( u ( · , t )) such that the solution of (1.1) satisfies (cid:107) u ( · , t ) (cid:107) L (0 ,L ) = O ( e − λt ) for some t > . In order to solve this problem, we use backstepping method (see [18] for a gen-eral discussion on the backstepping method), which is a well studied method for thesecond order evolutionary partial differential equations [17, 20, 26, 27, 28]. In recentyears, researchers studied backstepping stabilization of several higher order evolu-tionary equations that include third order dispersion term [2, 9, 22, 23, 31, 32]. Inthese studies on KdV type equations, a single boundary feedback control is locatedat one endpoint and the number of boundary conditions located at the opposite end-point are two. In particular, Part I of this study [2] assumes a control input actingfrom the left endpoint, and there are two homogeneous boundary conditions that areimposed from the right endpoint. Conversely, if there are two boundary controllersacting from the right endpoint and a single homogeneous boundary condition im-posed at the left endpoint, which is the subject of the present paper, the situationbecomes mathematically very different as we explain below.
TABILIZATION OF HIGHER ORDER SCHR ¨ODINGER EQUATIONS 5
To this end, we want to transform the original plant via the backstepping trans-formation w ( x, t ) = [( I − Υ k ) u ]( x, t ) . = u ( x, t ) − (cid:90) x k ( x, y ) u ( y, t ) dy (1.5)to a target system which already has the desired exponential stability. The classi-cal approach is to take the linearly damped version of the same type of pde withhomogeneous boundary conditions: iw t + iβw xxx + αw xx + iδw x + irw = 0 , x ∈ (0 , L ) , t ∈ (0 , T ) ,w (0 , t ) = w ( L, t ) = w x ( L, t ) = 0 ,w ( x,
0) = w ( x ) . = u ( x ) − (cid:82) x k ( x, y ) u ( y ) dy. (1.6)The key point here is to be able to show the existence of a sufficiently smooth kerneland that the transformation I − Υ k has a bounded inverse on a suitable space.Then, the wellposedness and stability properties for the target model will also betrue for the original plant. Figure 1 below summarizes the standard algorithm of thebackstepping method. Linear Plant
State variable u ( x, t ) Target
State variable w ( x, t ) w ( x, t ) = u ( x, t ) − (cid:82) x k ( x, y ) u ( y, t ) dy Inverse transformation
Figure 1.
BacksteppingTo prove the existence of a kernel, we differentiate (1.5) and use the original planttogether with the target system to see what conditions k must satisfy. After somecalculations (see Appendix A for details), we deduce that k must solve the followingboundary value problem β ( k xxx + k yyy ) − iα ( k xx − k yy ) + δ ( k x + k y ) + rk = 0 ,k ( x, x ) = k y ( x,
0) = k ( x,
0) = 0 ,k x ( x, x ) = rx β , (1.7)where ( x, y ) belongs to the triangular region ∆ x,y . = { ( x, y ) ∈ R | y ∈ (0 , x ) , x ∈ (0 , L ) } . To solve this problem, we change variables as s = x − y and t = y and write STABILIZATION OF HIGHER ORDER SCHR ¨ODINGER EQUATIONS G ( s, t ) ≡ k ( x, y ). Then (1.7) transforms into β (3 G sst − G tts + G ttt ) + iα ( G tt − G ts ) + δG t + rG = 0 ,G (0 , t ) = G t ( s,
0) = G ( s,
0) = 0 ,G s (0 , t ) = rt β , (1.8)where ( s, t ) belongs to ∆ s,t . = { ( s, t ) ∈ R | t ∈ (0 , L − s ) , s ∈ (0 , L ) } . See Figure 2for transformation of the triangular region under the above change of variables. y xLL Triangular region ∆ x,y s = x − yt = y t sLL Triangular region ∆ s,t
Figure 2.
Triangular regionsObserve that there is a mismatch between G t ( s,
0) and G s (0 , t ) in the sense that0 = G ts (0 , (cid:54) = G st (0 ,
0) = r β . This implies that (1.8) cannot have a smooth solutionand the standard algorithm of backstepping method fails. This issue was previouslyobserved in Korteweg de-Vries equation [9] and later treated in the case of uncriticaldomains in [11] and in the case of critical domains in [24]. The idea of the latterwork was to drop one of the boundary conditions from the kernel pde model andtake r sufficiently small. Note that if r is small, then the mismatch is also small andone can hope that the solution of the corrected pde model will yield a kernel whichis good enough for our purposes. Once we drop the boundary condition G t ( s,
0) = 0from (1.8), the corrected version of the pde model (1.8) becomes β (3 G ∗ sst − G ∗ tts + G ∗ ttt ) + iα ( G ∗ tt − G ∗ ts ) + δG ∗ t + rG ∗ = 0 ,G ∗ (0 , t ) = G ∗ ( s,
0) = 0 ,G ∗ s (0 , t ) = rt β , (1.9) TABILIZATION OF HIGHER ORDER SCHR ¨ODINGER EQUATIONS 7 where ( s, t ) ∈ ∆ s,t . Setting k ∗ ( x, y ) = G ∗ ( s, t ), we deduce that k ∗ is the sought aftersolution of β ( k ∗ xxx + k ∗ yyy ) − iα ( k ∗ xx − k ∗ yy ) + δ ( k ∗ x + k ∗ y ) + rk ∗ = 0 ,k ∗ ( x, x ) = k ∗ ( x,
0) = 0 ,k ∗ x ( x, x ) = rx β , (1.10)where ( x, y ) ∈ ∆ x,y . Existence of a smooth k ∗ is given in Lemma 3.1.If we use the backstepping transformation w ∗ ( x, t ) = u ( x, t ) − (cid:90) x k ∗ ( x, y ) u ( y, t ) dy, (1.11)then the corresponding target model becomes iw ∗ t + iβw ∗ xxx + αw ∗ xx + iδw ∗ x + irw ∗ = iβk ∗ y ( x, w ∗ x (0 , t ) , x ∈ (0 , L ) , t ∈ (0 , T ) ,w ∗ (0 , t ) = w ∗ ( L, t ) = w ∗ x ( L, t ) = 0 ,w ∗ ( x,
0) = w ∗ ( x ) . = u ( x ) − (cid:82) x k ∗ ( x, y ) u ( y ) dy. (1.12)See (A.5)-(A.9) in Appendix A for details. Notice that (1.12) is a modified versionof (1.6) in the sense that there is a trace term at the right hand side of the mainequation. This trace term is due to disregarding the condition k y ( x,
0) = 0 and usingthe relation u x (0 , t ) = w ∗ x (0 , t ). It is shown in Proposition 3.6 that the solution of(1.12) exponentially decays to zero for small r . Furhermore, it is important that thetransformation (1.11) has a bounded inverse (see Lemma 3.2). Graphical illustrationof the new scheme is shown in Figure 3. Linear Plant
State variable u ( x, t ) Modified targetwith a trace term
State variable w ∗ ( x, t ) w ∗ ( x, t ) = u ( x, t ) − (cid:82) x k ∗ ( x, y ) u ( y, t ) dy Inverse transformation
Figure 3.
Backstepping with an imperfect kernelBased on the above strategy feedback controllers take the following forms: h ( t ) = (cid:90) L k ∗ ( L, y ) u ( y, t ) dy, h ( t ) = (cid:90) L k ∗ x ( L, y ) u ( y, t ) dy. (1.13)Let us introduce the space X (cid:96)T . = C ([0 , T ]; H (cid:96) (0 , L )) ∩ L (0 , T ; H (cid:96) +1 (0 , L )) , (cid:96) ≥ . STABILIZATION OF HIGHER ORDER SCHR ¨ODINGER EQUATIONS
Now, regarding plant (1.1), we prove the following theorem.
Theorem 1.2.
Let
T, β > , α, δ ∈ R , u ∈ L (0 , L ) . Assume that the rightendpoint feedback controllers h ( t ) , h ( t ) are given by (1.13) and let k ∗ be a smoothbackstepping kernel solving (1.10) . Then, we have the following: (i) (Wellposedness) (1.1) has a unique solution u ∈ X T satisfying also u x ∈ C ([0 , L ]; L (0 , T )) . Moreover, if u ∈ H (0 , L ) and w ∗ = ( I − Υ k ∗ )[ u ] satisfiesthe compatibility conditions, then u ∈ X T . (ii) (Decay) Suppose u ∈ L (0 , L ) . Then, there is r > such that λ = β (cid:32) rβ − (cid:107) k ∗ y ( · , r ) (cid:107) L (0 ,L ) (cid:33) > . Moreover, the solution u of (1.1) with feedback controllers (1.13) satisfies (cid:107) u ( · , t ) (cid:107) L (0 ,L ) ≤ c k ∗ (cid:107) u (cid:107) L (0 ,L ) e − λt , t ≥ , where c k ∗ is a nonnegative constant which depends on k ∗ .Remark . It is shown in the proof of Proposition 3.6 that there exists r > λ >
0. See also Table 1 for different values of r, λ . Notealso from the same table that smallness of r is essential for λ to be positive.In the second part of the paper, we consider the case where the state of the systemis not fully measurable, in particular at time t = 0. However, we assume that thefirst order boundary trace y ( t ) = u x (0 , t ) and the second order boundary trace y ( t ) = u xx (0 , t ) are known, say detectable through boundary sensors. In order todeal with the robustness of the state, we construct an observer, which uses the givenboundary measurements. To this end, we propose the following observer model i ˆ u t + iβ ˆ u xxx + α ˆ u xx + iδ ˆ u x − p ( x )( y ( t ) − ˆ u x (0 , t )) − p ( x )( y ( t ) − ˆ u xx (0 , t )) = 0 , x ∈ (0 , L ) , t ∈ (0 , T ) , ˆ u (0 , t ) = 0 , ˆ u ( L, t ) = h ( t ) , ˆ u x ( L, t ) = h ( t ) , ˆ u ( x,
0) = ˆ u ( x ) . (1.14)In the above model, the feedback controllers h ( t ), h ( t ) depend on the state of theobserver model. The same controllers will also be applied to the original plant (1.1). p , p are set to be observer gains and they will be constructed in such a way thatthe so called error ˜ u = u − ˆ u must approach to zero as t gets larger. More precisely,we want to solve the following problem: Problem 1.4.
Given
L > , find observer gains p , p , and feedback laws h ( t ) = h (ˆ u ( · , t )) , h ( t ) = h (ˆ u ( · , t )) such that there exists λ > for which the solution u of (1.1) satisfies (cid:107) u ( · , t ) (cid:107) L (0 ,L ) = O ( e − λt ) . TABILIZATION OF HIGHER ORDER SCHR ¨ODINGER EQUATIONS 9
Note that the error function ˜ u satisfies i ˜ u t + iβ ˜ u xxx + α ˜ u xx + iδ ˜ u x + p ( x )˜ u x (0 , t ) + p ( x )˜ u xx (0 , t ) = 0 , x ∈ (0 , L ) , t ∈ (0 , T ) , ˜ u (0 , t ) = ˜ u ( L, t ) = ˜ u x ( L, t ) = 0 , ˜ u ( x,
0) = ˜ u ( x ) . (1.15)In order to guarantee the decay of solutions of (1.15) at an exponential rate, wetreat p and p as control inputs and suitably construct them via the backsteppingtechnique. To this end, we transform (1.15) using the transformation˜ u ( x, t ) = ˜ w ( x, t ) − (cid:90) x p ( x, y ) ˜ w ( y, t ) dy, (1.16)to a target error model i ˜ w t + iβ ˜ w xxx + α ˜ w xx + iδ ˜ w x + ir ˜ w = 0 , x ∈ (0 , L ) , t ∈ (0 , T ) , ˜ w (0 , t ) = ˜ w ( L, t ) = ˜ w x ( L, t ) = 0 , ˜ w ( x,
0) = ˜ w ( x ) , (1.17)which has exponential decay property. Differentiating (1.16) and using (1.15) with(1.17), one can see that p must satisfy the following boundary value problem β ( p xxx + p yyy ) − iα ( p xx − p yy ) + δ ( p x + p y ) − rp = 0 ,p ( x, x ) = p ( L, y ) = p x ( L, y )0 ,p x ( x, x ) = r β ( L − x ) , (1.18)on ∆ x,y (see Appendix B for detailed calculations). However, changing variables as s = L − x , t = x − y and defining H ( s, t ) ≡ p ( x, y ), one can see that the resultingboundary value problem is overdetermined in the sense that there is a mismatchbetween the boundary conditions: 0 = H st (0 , (cid:54) = H ts (0 ,
0) = r β . Therefore, therecannot exist a smooth kernel satisfying all boundary conditions. Following a similarapproach as in the earlier part of the paper, we could consider disregarding one ofthe boundary conditions, namely p x ( L, y ) = 0, and take r sufficiently small. Then,the corrected version of the pde model (1.18) becomes β ( p ∗ xxx + p ∗ yyy ) − iα ( p ∗ xx − p yy ) + δ ( p ∗ x + p ∗ y ) − rp ∗ = 0 ,p ∗ ( x, x ) = p ∗ ( L, y ) = 0 ,p ∗ x ( x, x ) = r β ( L − x ) . (1.19)Now if we use the backstepping transformation˜ u ( x, t ) = ˜ w ∗ ( x, t ) − (cid:90) x p ∗ ( x, y ) ˜ w ∗ ( y, t ) dy, (1.20) where p ∗ solves (1.19), then the corresponding target error model for (1.6) becomes i ˜ w ∗ t + iβ ˜ w ∗ xxx + α ˜ w ∗ xx + iδ ˜ w ∗ x + ir ˜ w ∗ = 0 , x ∈ (0 , L ) , t ∈ (0 , T ) , ˜ w ∗ (0 , t ) = ˜ w ∗ ( L, t ) = 0 , ˜ w ∗ x ( L, t ) = (cid:82) L p ∗ x ( L, y ) ˜ w ∗ ( y, t ) dy, ˜ w ∗ ( x,
0) = ˜ w ∗ ( x ) , (1.21)if the control gains are chosen such that p ( x ) = iβp ∗ y ( x, − αp ∗ ( x,
0) and p ( x ) = − iβp ∗ ( x,
0) (see Appendix B for details). Note that nonhomogeneous Neumanntype boundary condition in (1.21) is due to disregarding the condition p x ( L, y ) = 0.Nevertheless, we still have the exponential decay of solutions of (1.21) assuming that r is sufficiently small. This is given in Proposition 4.3. Also, it is not difficult to seethat p ∗ and k ∗ are related via p ∗ ( x, y ) ≡ k ∗ ( L − y, L − x ; − r ) , which immediately guarantees the existence of a smooth kernel p ∗ . This yields theexponential decay of the solution of the error model.Next, we apply the backstepping method to the observer model. To this end, wedifferentiate ˆ w ∗ ( x, t ) = ˆ u ( x, t ) − (cid:90) x k ∗ ( x, y )ˆ u ( y, t ) dy (1.22)and use (1.14) to deduce that ˆ w ∗ solves target observer model given by i ˆ w ∗ t + iβ ˆ w ∗ xxx + α ˆ w ∗ xx + iδ ˆ w ∗ x + ir ˆ w ∗ = iβk ∗ y ( x,
0) ˆ w ∗ x (0 , t )+[( I − Υ k ∗ ) p ]( x ) ˜ w x (0 , t ) + [( I − Υ k ∗ ) p ]( x ) ˜ w xx (0 , t ) , x ∈ (0 , L ) , t ∈ (0 , T ) , ˆ w ∗ (0 , t ) = ˆ w ∗ ( L, t ) = ˆ w ∗ x ( L, t ) = 0 , ˆ w ∗ ( x,
0) = ˆ w ∗ ( x ) . = ˆ u ( x ) − (cid:82) x k ∗ ( x, y )ˆ u ( y, t ) dy. (1.23)Notice that we still use the solution of the corrected kernel pde model in the back-stepping transformation given in (1.22). Therefore, as in the first part of this paper,an extra trace term shows up in the main equation of (1.23). We prove in Proposition4.4 that the solution of (1.23) exponentially decays to zero in time, again assumingthat r is sufficiently small.Thanks to the bounded invertibility of the transformations ( I − Υ k ∗ ), ( I − Υ p ∗ ),stability estimates with same decay rates also hold for observer and error models.We have the following theorem for the wellposedness and stabilization of the plant-observer-error system (1.1)-(1.14)-(1.15): Theorem 1.5.
Let
T, β > , α, δ ∈ R , and u , ˆ u ∈ H (0 , L ) . Assume that the rightendpoint feedback controllers are given by h ( t ) = (cid:90) L k ∗ ( L, y ; r )ˆ u ( y, t ) dy, h ( t ) = (cid:90) L k ∗ x ( L, y ; r )ˆ u ( y, t ) dy, TABILIZATION OF HIGHER ORDER SCHR ¨ODINGER EQUATIONS 11 where k ∗ and p ∗ are smooth solutions of (1.10) and (1.19) , respectively. Then, wehave the following: (i) (Wellposedness) Suppose that ˆ w ∗ = ( I − Υ k ∗ )[ˆ u ] satisfies the compatibilityconditions and the pair ( ˜ w ∗ , ψ ) , where ˜ w ∗ = ( I − Υ p ∗ ) − [˜ u ] , ψ = ψ ( ˜ w ) . = (cid:82) L p ∗ x ( L, y ) ˜ w ( y, t ) dy , satisfies the higher order compatibility conditions. Thenthe plant-observer-error system (1.1) - (1.14) - (1.15) has a unique solution ( u, ˆ u, ˜ u ) ∈ X T × X T × X T . (ii) (Decay) Moreoever, for sufficiently small r > , there exists µ > ν > suchthat, (cid:107) u ( · , t ) (cid:107) L (0 ,L ) (cid:46) c k,p (cid:0) (cid:107) ˆ u (cid:107) L (0 ,L ) + c p (cid:107) u − ˆ u (cid:107) H (0 ,L ) (cid:1) e − νt + (cid:107) u − ˆ u (cid:107) L (0 ,L ) e − µt , (cid:107) ˆ u ( · , t ) (cid:107) L (0 ,L ) (cid:46) c k,p (cid:0) (cid:107) ˆ u (cid:107) L (0 ,L ) + (cid:107) u − ˆ u (cid:107) H (0 ,L ) (cid:1) e − νt , (cid:107) ( u − ˆ u )( · , t ) (cid:107) L (0 ,L ) (cid:46) c p (cid:107) u − ˆ u (cid:107) L (0 ,L ) e − µt , (cid:107) ( u − ˆ u )( · , t ) (cid:107) H (0 ,L ) (cid:46) c p (cid:48) (cid:107) u − ˆ u (cid:107) H (0 ,L ) e − µt . for t ≥ , where c k , c p , c p (cid:48) , c k,p are nonnegative constants depending on theirsubindices. Preliminaries.
In this section, we state a few important inequalities and no-tation which will be useful in our proofs.1.2.1.
Notation. L p (0 , L ), 1 ≤ p < ∞ , is the usual Lebesgue space and given u ∈ L p (0 , L ), a Lebesgue measurable function, we will denote its L p − norm by (cid:107) u (cid:107) p , i.e. (cid:107) u (cid:107) p . = (cid:18)(cid:90) L | u | p dx (cid:19) p . If p = ∞ , then the corresponding norm is given by (cid:107) u (cid:107) ∞ . = ess sup x ∈ (0 ,L ) | u | . Given k >
0, we denote the L − based Sobolev space by H k (0 , L ). In particular, H (0 , L ) is the space of functions which belong to H (0 , L ) that vanish at the end-points in the sense of traces. If A is a linear and bounded operator on L (0 , L ),we will denote its operator norm on L (0 , L ) by (cid:107) A (cid:107) → . We will write a (cid:46) b todenote an inequality a ≤ cb where c > Definition 1.6 (Compatibility) . Let
T, L > φ ∈ H (0 , L ), ψ ∈ H (0 , T ) are such that φ (0) = 0 , φ ( L ) = 0 , φ (cid:48) ( L ) = ψ (0) (1.24)then we say ( φ, ψ ) satisfies compatibility conditions.(ii) If φ ∈ H (0 , L ), ψ ∈ H (0 , T ) and ˜ φ . = βφ (cid:48)(cid:48)(cid:48) − iαφ (cid:48)(cid:48) + δφ (cid:48) , then we say ( φ, ψ )satisfies higher order compatibility conditions provided it satisfies compati-bility conditions and˜ φ (0) = 0 , ˜ φ ( L ) = 0 , ˜ φ (cid:48) ( L ) = ψ (cid:48) (0) . (1.25)If ψ ≡
0, then we only refer to φ regarding compatibility conditions.Finally, starting from Section 2, we drop the superscript notation ∗ that ex-presses modified target models and modified backstepping kernels, and simply write k, p, w, ˆ w, ˜ w , etc.1.2.2. Some useful inequalities.
We will use the Cauchy–Schwarz inequality, for u, v ∈ L (0 , L ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:90) L u ( x ) v ( x ) dx (cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:107) u (cid:107) (cid:107) v (cid:107) , and (cid:15) − Young’s inequality | uv | ≤ (cid:15)p | u | p + 1 q(cid:15) / ( p − | v | q , where (cid:15) >
0, 1 < p < ∞ and p + q = 1. We will use Gagliardo–Nirenberg’sinterpolation inequality in our estimates: Let p ≥ α = 1 / − /p and u ∈ H (0 , L ).Then, (cid:107) u (cid:107) p ≤ c (cid:107) u (cid:48) (cid:107) α (cid:107) u (cid:107) − α . We will also use its higher order version: Let u ∈ H k (0 , L ) ∩ H (0 , L ) for α = j/k ≤ j, k ∈ N . Then (cid:107) u ( j ) (cid:107) p ≤ c (cid:107) u ( k ) (cid:107) α (cid:107) u (cid:107) − α . Special case of the Gronwall’s inequality reads: given f : [0 , t ] → R + and α, β > f ( t ) ≤ α + β (cid:90) t f ( s ) ds implies f ( t ) ≤ αe βt . TABILIZATION OF HIGHER ORDER SCHR ¨ODINGER EQUATIONS 13
Outline.
In Section 2, we prove smoothing properties for a nonhomogeneousinitial–boundary value problem with inhomogeneous boundary conditions. Thesewill be useful for the wellposedness analysis that will be carried out in Section 3and Section 4. The tools we mainly use are semigroup theory, multipliers and theLaplace transform. In Section 3, we first show the existence of an infinitely differen-tiable smooth backstepping kernel in Lemma 3.1 and then state the invertibility ofthe backstepping transformation with a bounded inverse in Lemma 3.2. Then, westudy the wellposedness and exponential decay properties of modified target model(1.12). Finally, thanks to the bounded invertibility of the backstepping transfor-mation, we obtain the wellposedness and exponential decay for the original plant.Section 4 is devoted to the observer design problem where we assume that the state ofthe system is not known and only some partial boundary measurements are available.We first make the wellposedness analysis for modified target error model (1.21) andmodified target observer model (1.23), respectively. Thanks to the bounded invert-ibility of the backstepping transformations, we show the wellposedness for error andobserver, which imply the wellposedness of the original plant. Next, we study thedecay properties of the target error and target observer models. Again, by using theinvertibility of backstepping transformations, we obtain the exponential stability ofplant–observer–error system. In Section 5, we introduce a numerical algorithm andthen provide two numerical simulations for controller and observer designs. Finally,in appendices, we give details of some calculations.2.
Auxiliary lemmas
In this section we prove some auxiliary lemmas which will be useful in order toshow wellposesness results in Section 3.2 and Section 4.1. Let us start by consideringthe following model iy t + iβy xxx + αy xx + iδy x = f ( x, t ) , x ∈ (0 , L ) , t ∈ (0 , T ) ,y (0 , t ) = ψ ( t ) , y ( L, t ) = ψ ( t ) , y x ( L, t ) = ψ ( t ) ,y ( x,
0) = φ ( x ) . (2.1)We will denote the solution of (2.1) by y [ φ, f, ψ , ψ , ψ ]. Let A be the linear operatordefined by Ay . = − βy (cid:48)(cid:48)(cid:48) + iαy (cid:48)(cid:48) − δy (cid:48) (2.2)with domain D ( A ) = { y ∈ H (0 , L ) : y (0) = y ( L ) = y (cid:48) ( L ) = 0 } . (2.3)It is shown in [8] that A generates a strongly continuous semigroup of contractionson L (0 , L ) denoted by S ( t ), t ≥
0. Thus, by standard semigroup theory, (2.1) with f = 0 and ψ i ≡ i = 1 , ,
3, admits a unique mild solution (see [25]) for φ ∈ L (0 , L ). In the presence of nonhomogeneous boundary conditions ψ i , i = 1 , ,
3, but with zeroforcing and zero initial state, i.e. φ ≡ f ≡
0, analysis of solutions of (2.1) will becarried out by obtaining a representation for the solution via the Laplace transformin t . Also, in the following lemmas below we obtain regularity estimates for solutionscorresponding to initial, interior and boundary data in (2.1). Lemma 2.1.
Let f ≡ ψ i ≡ , i = 1 , , . Then, for T > and φ ∈ L (0 , L ) , y ( · ) = S ( · ) φ = y [ φ, , , , satisfies space-time estimates (i) (cid:107) y (cid:107) C ([0 ,T ]; L (0 ,L )) + β (cid:107) y x (0 , · ) (cid:107) L (0 ,T ) = (cid:107) φ (cid:107) , (ii) (cid:107) y (cid:107) L (0 ,T ; H (0 ,L )) (cid:46) (1 + T ) (cid:107) φ (cid:107) and the time-space estimate (iii) sup x ∈ [0 ,L ] (cid:107) y x ( x, · ) (cid:107) L (0 ,T ) (cid:46) (1 + √ T ) (cid:107) φ (cid:107) .Proof. We first assume that φ ∈ D ( A ) and the solution is sufficiently smooth. Thegeneral case then can be shown by using the classical density argument.(i) We take L − inner product of the main equation in (2.1) by 2 y and get2 (cid:61) (cid:90) L iy t ¯ ydx + 2 (cid:61) (cid:90) L iβy xxx ¯ ydx + 2 (cid:61) (cid:90) L αy xx ¯ ydx + 2 (cid:61) (cid:90) L iδy x ¯ ydx = 0 . (2.4)The first term at the left hand side of (2.4) can be written as2 (cid:61) (cid:90) L iy t ¯ ydx = 2 (cid:60) (cid:90) L y t ¯ ydx = ddt | y ( · , t ) | . (2.5)The second term can be integrated by parts in x , and using boundary condi-tions we have2 (cid:61) (cid:90) L iβy xxx ¯ ydx = − (cid:60) (cid:90) L βy xx ¯ y x dx = β | y x (0 , t ) | . (2.6)The third term, again via integration by parts in x , gives2 (cid:61) (cid:90) L αy xx ¯ ydx = − (cid:61) (cid:90) L α | y x | dx = 0 . (2.7)The fourth term vanishes since2 (cid:61) (cid:90) L iδu x ¯ udx = δ | u ( x, t ) | (cid:12)(cid:12)(cid:12)(cid:12) L = 0 . (2.8)Combining (2.5)-(2.8) and integrating with respect to t , we arrive at (cid:107) y ( · , t ) (cid:107) + β (cid:107) y x (0 , · ) (cid:107) L (0 ,T ) = (cid:107) φ (cid:107) . (2.9)Passing to supremum on both sides over [0 , T ] yields the desired result. TABILIZATION OF HIGHER ORDER SCHR ¨ODINGER EQUATIONS 15 (ii) Now we take L − inner product of the main equation (2.1) by 2 xy and considerthe imaginary parts of both sides to get2 (cid:61) (cid:90) L ixy t ¯ ydx + 2 (cid:61) (cid:90) L iβxy xxx ¯ ydx + 2 (cid:61) (cid:90) L αxy xx ¯ ydx + 2 (cid:61) (cid:90) L iδxy x ¯ ydx = 0 . (2.10)The first term at the left hand side of (2.10) can be written as2 (cid:61) (cid:90) L ixy t ¯ ydx = ddt (cid:90) L x | y | dx. (2.11)The second term can be integrated by parts in x and due to the boundaryconditions we have2 (cid:61) (cid:90) L iβxy xxx ¯ ydx = 2 β (cid:18) − (cid:90) L y xx ¯ udx − (cid:90) L xy xx y x dx (cid:19) = 2 β (cid:18)(cid:90) L | y x | dx − (cid:90) L x ddx | y x | dx (cid:19) = 2 β (cid:18) (cid:107) y x ( · , t ) (cid:107) dx + 12 (cid:90) L | y x | dx (cid:19) = 3 β (cid:107) y x ( · , t ) (cid:107) . (2.12)Third and fourth terms, again via integration by parts in x give us2 (cid:61) (cid:90) L αxy xx ¯ ydx = 2 α (cid:18) −(cid:61) (cid:90) L y x ydx − (cid:61) (cid:90) L x | y x | dx (cid:19) (2.13)and 2 (cid:61) (cid:90) L iδxy x ¯ ydx = 2 δ (cid:90) L x ddx | y | dx = − δ (cid:107) y ( · , t ) (cid:107) . (2.14)Combining (2.11)-(2.14), we get ddt (cid:90) L x | y | dx + 3 β (cid:107) y x ( · , t ) (cid:107) = δ (cid:107) y ( · , t ) (cid:107) + 2 α (cid:61) (cid:90) L y x ydx which, by (cid:15) − Young’s inequality applied to the second term at the right handside, is equivalent to ddt (cid:90) L x | y | dx + (3 β − (cid:15) ) (cid:107) y x ( · , t ) (cid:107) ≤ c α,δ,(cid:15) (cid:107) y ( · , t ) (cid:107) . Integrating this result with respect to t over [0 , T ] yields (cid:90) L x | y | dx + (3 β − (cid:15) ) (cid:90) T (cid:107) y x ( · , t ) (cid:107) dt = (cid:90) L x | φ | dx + c α,δ,(cid:15) (cid:90) T (cid:107) y ( · , t ) (cid:107) dt. Combining this result with (2.9), using Poincare inequality and choosing (cid:15) > φ in L ( R ), denoted φ ∗ , with the property that (cid:107) φ ∗ (cid:107) L ( R ) (cid:46) (cid:107) φ (cid:107) L (0 ,L ) . Consider the Cauchy problem (cid:40) iv t + iβv xxx + αv xx + iδv x = 0 , x ∈ R , t ∈ (0 , T ) ,v ( x,
0) = φ ∗ ( x ) . (2.15)Using the Fourier transform ˆ ϕ ( ξ ) = (cid:82) ∞−∞ e − ixξ ϕ ( x ) dx , the solution of theabove model can be represented as v ( x, t ) = (cid:90) ∞−∞ e ixξ e iω ( ξ ) t ˆ φ ∗ ( ξ ) dξ, where ω ( ξ ) . = βξ − αξ − δξ . We pick a smooth cut-off function θ ( ξ ) = , a ≤ ξ ≤ b, smooth , a − (cid:15) < ξ < a or b < ξ < b + (cid:15) , ξ ≤ a − (cid:15) or ξ ≥ b + (cid:15), where (cid:15) > | θ | ≤
1, and a and b will be chosen below in asuitable manner. Now, we decompose v as v ( x, t ) = (cid:90) ∞−∞ e ixξ e iω ( ξ ) t θ ( ξ ) ˆ φ ∗ ( ξ ) dξ + (cid:90) ∞−∞ e ixξ e iω ( ξ ) t (1 − θ ( ξ )) ˆ φ ∗ ( ξ ) dξ. = v ( x, t ) + v ( x, t ) . (2.16)Using Cauchy-Schwarz inequality on v , Plancherel theorem and consideringthat θ is a compactly supported function, we get | ∂ x v ( x, t ) | = (cid:12)(cid:12)(cid:12)(cid:12)(cid:90) ∞−∞ iξe ixξ e iω ( ξ ) t θ ( ξ ) ˆ φ ∗ ( ξ ) dξ (cid:12)(cid:12)(cid:12)(cid:12) = (cid:18)(cid:90) b + (cid:15)a − (cid:15) | iξ | | θ ( ξ ) | dξ (cid:19) (cid:18)(cid:90) ∞−∞ | ˆ φ ∗ ( ξ ) | dξ (cid:19) (cid:46) (cid:107) φ ∗ (cid:107) L ( R ) . Taking square of both sides and integrating over [0 , T ] yields (cid:107) ∂ x v ( x, · ) (cid:107) L (0 ,T ) (cid:46) T (cid:107) φ ∗ (cid:107) L ( R ) . (2.17)By similar arguments, we also have (cid:107) ∂ jt v ( x, · ) (cid:107) L (0 ,T ) (cid:46) T (cid:107) φ ∗ (cid:107) L ( R ) (2.18) TABILIZATION OF HIGHER ORDER SCHR ¨ODINGER EQUATIONS 17 for j = 0 ,
1, where the constant of the inequality depends on j and θ . Inter-polating, we get (cid:107) v ( x, · ) (cid:107) H m (0 ,T ) (cid:46) T (cid:107) φ ∗ (cid:107) L ( R ) (2.19)for any m ∈ [0 , θ and m .In particular, for m = 1 /
3, we have (cid:107) v ( x, · ) (cid:107) H / (0 ,T ) (cid:46) √ T (cid:107) φ ∗ (cid:107) L ( R ) . (2.20)The last inequality is a smoothing property and will have a particular impor-tance in our wellposedness analysis later. Next, consider the second term in(2.16) and rewrite it as v = (cid:90) a −∞ · + (cid:90) ∞ b · . = v − + v . Consider the change of variable given by τ = ω − ( ξ ) = ω ( ξ ) : ( −∞ , a ] → ( −∞ , ω ( a )] ,τ = ω + ( ξ ) = ω ( ξ ) : [ b, ∞ ) → [ ω ( b ) , ∞ ) , (2.21)where we define their inverses as ξ = ω − − ( τ ) . = ξ − ( τ ) ,ξ = ω − ( τ ) . = ξ + ( τ ) (2.22)for each integral respectively. Indeed a suitable choice of the support of thecut-off function ensures that the transformations given by (2.21) are 1 − ω (cid:48) ( ξ ) stays away from zero in (2.24). Depending on the sign of α + 3 βδ , we have three different cases:(a) Let α + 3 βδ >
0. Then, any choice a < α − √ α +3 βδ β and b > α + √ α +3 βδ β provides that the mapping is 1 − ω (cid:48) ( ξ ) stays away from zero on( −∞ , a ] ∪ [ b, ∞ ).(b) Let α + 3 βδ = 0. Then the mapping is 1 − a < b . Onthe other hand, choosing a < α β < b will provide that ω (cid:48) ( ξ ) stays awayfrom zero on ( −∞ , a ] ∪ [ b, ∞ ).(c) Let α + 3 βδ <
0. Then the mapping is 1 − ω (cid:48) ( ξ ) stays away fromzero on ( −∞ , a ] ∪ [ b, ∞ ) for all choices of a < b .Assume that we choose appropriate a and b values for each case of α + 3 βδ described above. Following from (2.21)-(2.22), we have dξ = 13 βξ ∓ ( τ ) − αξ ∓ ( τ ) − δ dτ. (2.23) Hence v becomes v ( x, t ) = (cid:90) ω ( a ) −∞ e iξ − ( τ ) x e iτt (1 − θ ( ξ − ( τ ))) ˆ φ ∗ ( ξ − ( τ ))3 βξ − ( τ ) − αξ − ( τ ) − δ dτ + (cid:90) ∞ ω ( b ) e iξ + ( τ ) x e iτt (1 − θ ( ξ + ( τ ))) ˆ φ ∗ ( ξ + ( τ ))3 βξ ( τ ) − αξ + ( τ ) − δ dτ. Let us first consider v − and observe that the function (cid:40) e iξ − ( τ ) x (1 − θ ( ξ − ( τ ))) ˆ φ ∗ ( ξ − ( τ ))3 βξ − ( τ ) − αξ − ( τ ) − δ , τ ∈ ( −∞ , ω ( a )] , , elsewhere,is the Fourier transform of v − with respect to the second component. Thus, (cid:107) v − ( x, · ) (cid:107) H (0 ,T ) ≤ (cid:107) v − ( x, · ) (cid:107) H t ( R ) = (cid:90) ∞−∞ (1 + τ ) | ˆ v − ( x, τ ) | dτ = (cid:90) ω ( a ) −∞ (1 + τ ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) e iξ − ( τ ) x (1 − θ ( ξ − ( τ ))) ˆ φ ∗ ( ξ − ( τ ))3 βξ − ( τ ) − αξ − ( τ ) − δ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) dτ ≤ (cid:90) ω ( a ) −∞ (1 + τ ) | | ˆ φ ∗ ( ξ − ( τ )) | | βξ − ( τ ) − αξ − ( τ ) − δ | dτ. Changing variables back as τ = ω − ( ξ ), it follows from the above estimatethat (cid:107) v − ( x, · ) (cid:107) H (0 ,T ) ≤ (cid:90) a −∞ (1 + ω ( ξ )) | ˆ φ ∗ ( ξ ) | | βξ − αξ − δ | (3 βξ − αξ − δ ) dξ (2.24) (cid:46) (cid:90) a −∞ (1 + ξ ) | ˆ φ ∗ ( ξ ) | βξ − αξ − δ dξ (cid:39) (cid:90) a −∞ | ˆ φ ∗ ( ξ ) | dξ ≤ (cid:107) φ ∗ (cid:107) L ( R ) . (cid:107) v ( x, · ) (cid:107) H (0 ,T ) (cid:46) (cid:107) φ ∗ (cid:107) L ( R ) can be shown similarly. Hence (cid:107) v ( x, · ) (cid:107) H (0 ,T ) (cid:46) (cid:107) φ ∗ (cid:107) L ( R ) . Combining this with (2.17), we get (cid:107) v ( x, · ) (cid:107) H (0 ,T ) (cid:46) (1 + √ T ) (cid:107) φ ∗ (cid:107) L ( R ) . Differentiating in x and repeating the above arguments it also follows that (cid:107) ∂ x v ( x, · ) (cid:107) L (0 ,T ) (cid:46) (1 + √ T ) (cid:107) φ ∗ (cid:107) L ( R ) . TABILIZATION OF HIGHER ORDER SCHR ¨ODINGER EQUATIONS 19
We also have the continuity of the mappings x → (cid:107) v ( x, · ) (cid:107) H / (0 ,T ) and x →(cid:107) ∂ x v ( x, · ) (cid:107) L (0 ,T ) . To this end, one needs to show that, given { x n } ⊂ R converging to x ∈ R (cid:107) v ( x, · ) − v ( x n , · ) (cid:107) H / (0 ,T ) → , as n → ∞ and (cid:107) ∂ x v ( x, · ) − ∂ x v ( x n , · ) (cid:107) L (0 ,T ) → , as n → ∞ hold. These can be easily shown by using the dominated convergence theo-rem. Now, we can represent the solution of (2.1) as y [ φ, , , ,
0] = v | (0 ,L ) − y [0 , , v (0 , · ) , v ( L, · ) , v x ( L, · )] , where y [0 , , v (0 , · ) , v ( L, · ) , v x ( L, · )] is the solution of (2.1) with f ≡ φ ≡ ,y (0 , t ) . = v (0 , t ) ∈ H / (0 , T ) , y ( L, t ) . = v ( L, t ) ∈ H / (0 , T ) ,y x ( L, t ) . = v x ( L, t ) ∈ L (0 , T ) . Hence, part (iii) follows by combining the boundary smoothing property of v and the inhomogeneous boundary value problem given in Lemma 2.6 below. (cid:3) Lemma 2.2.
Let φ ≡ ψ i ≡ , i = 1 , , , T > , and f ∈ L (0 , T ; L (0 , L )) . Thenthe solution y = y [0 , f, , , of (2.1) satisfies space-time estimates (i) (cid:107) y (cid:107) C ([0 ,T ]; L (0 ,L )) + β (cid:107) y x (0 , · ) (cid:107) L (0 ,T ) (cid:46) (cid:107) f (cid:107) L (0 ,T ; L (0 ,L )) , (ii) (cid:107) y (cid:107) L (0 ,T ; H (0 ,L )) (cid:46) (1 + T ) (cid:107) f (cid:107) L (0 ,T ; L (0 ,L )) and the time-space estimate (iii) sup x ∈ [0 ,L ] (cid:107) y x ( x, · ) (cid:107) L (0 ,T ) (cid:46) (1 + √ T ) (cid:107) f (cid:107) L (0 ,T ; L (0 ,L )) .Proof. (i) Multiplying the main equation by 2¯ y , integrating over [0 , T ] × [0 , L ] and using(2.5)-(2.8), we get (cid:107) y ( · , t ) (cid:107) + β (cid:107) y x (0 , · ) (cid:107) L (0 ,T ) ≤ (cid:90) T (cid:90) L | f ( x, t ) || y ( x, t ) | dxdt. (2.25)We apply Cauchy–Schwarz and (cid:15) − Young’s inequalities to the right hand sideof (2.25) to obtain (cid:107) y ( · , t ) (cid:107) + β (cid:107) y x (0 , · ) (cid:107) L (0 ,T ) ≤ (cid:15) sup t ∈ [0 ,T ] (cid:107) y ( · , t ) (cid:107) + c (cid:15) (cid:107) f (cid:107) L (0 ,T ; L (0 ,L )) . Right hand side is independent of t . So passing to supremum on both sidesover [0 , T ] and choosing (cid:15) > (ii) Multiplying the main equation by 2 xy , integrating over [0 , T ] × [0 , L ] andusing the same arguments in (2.11)-(2.14), we get (cid:90) L x | y ( x, t ) | dx + 3 β (cid:90) T (cid:107) y x ( · , t ) (cid:107) dt = δ (cid:90) T (cid:107) y ( · , t ) (cid:107) dt +2 α (cid:61) (cid:90) T (cid:90) L y x ( x, t ) y ( x, t ) dxdt +2 (cid:61) (cid:90) T (cid:90) L xf ( x, t ) y ( x, t ) dxdt. (2.26)Second term at the right hand side of (2.26) can be estimated via (cid:15) − Young’sinequality as2 α (cid:61) (cid:90) T (cid:90) L y x ( x, t ) y ( x, t ) dxdt ≤ (cid:15) (cid:90) T (cid:107) y x ( · , t ) (cid:107) dt + c α,(cid:15) (cid:90) T (cid:107) y ( · , t ) (cid:107) dt. (2.27)Using Cauchy–Schwarz inequality and (cid:15) − Young’s inequality, and thanks to(i), the third term at the right hand side of (2.26) can be estimated as2 (cid:61) (cid:90) T (cid:90) L xf ( x, t ) y ( x, t ) dxdt ≤ sup t ∈ [0 ,T ] (cid:107) y ( · , t ) (cid:107) + L (cid:107) f (cid:107) L (0 ,T ; L (0 ,L )) ≤ c L (cid:107) f (cid:107) L (0 ,T ; L (0 ,L )) . (2.28)Using (2.27)-(2.28), it follows from (2.26) that (cid:90) L x | y ( x, t ) | dx + (3 β − (cid:15) ) (cid:90) T (cid:107) y x ( · , t ) (cid:107) dt ≤ c α,δ,(cid:15) (cid:90) T (cid:107) y ( · , t ) (cid:107) + c L (cid:107) f (cid:107) L (0 ,T ; L (0 ,L )) ≤ c α,δ,(cid:15) T sup t ∈ [0 ,T ] (cid:107) y ( · , t ) (cid:107) + c L (cid:107) f (cid:107) L (0 ,T ; L (0 ,L )) ≤ c L,α,δ,(cid:15) (1 + T ) (cid:107) f (cid:107) L (0 ,T ; L (0 ,L )) , (2.29)where we used (i) in the last line again. Finally, choosing (cid:15) ∈ (0 , β ), usingthe Poincare inequality and dropping the first term at the left hand side, weconclude with (ii).(iii) Using Duhamel’s principle, the solution is of the form y ( x, t ) = (cid:90) t S ( t − τ ) f ( x, τ ) dτ. TABILIZATION OF HIGHER ORDER SCHR ¨ODINGER EQUATIONS 21
By differentiating with respect to x | ∂ x y ( x, t ) | = (cid:12)(cid:12)(cid:12)(cid:12) ∂ x (cid:20)(cid:90) t S ( t − τ ) f ( x, τ ) dτ (cid:21)(cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:90) t | ∂ x [ S ( t − τ ) f ( x, τ )] | dτ ≤ (cid:90) T | ∂ x [ S ( t − τ ) f ( x, τ )] | dτ. Taking L − norm of both sides with respect to t on [0 , T ] and using the resultin Lemma 2.1-(iii) (cid:107) ∂ x y ( x, · ) (cid:107) L (0 ,T ) ≤ (cid:13)(cid:13)(cid:13)(cid:13)(cid:90) T | ∂ x [ S ( · − τ ) f ( x, τ )] | dτ (cid:13)(cid:13)(cid:13)(cid:13) L (0 ,T ) ≤ (cid:90) T (cid:107) ∂ x [ S ( · − τ ) f ( x, τ )] (cid:107) L (0 ,T ) dτ (cid:46) (cid:90) T (1 + √ T ) (cid:107) f ( · , τ ) (cid:107) L ( R ) dτ (cid:46) (1 + √ T ) (cid:107) f (cid:107) L (0 ,T ; L (0 ,L )) . Passing to supremum in x over [0 , L ] ends the proof of (iii). (cid:3) Now let us turn our attention to the nonhomogeneous boundary value problemwith φ ≡ f ≡ y [0 , , ψ , ψ , ψ ]in terms of the boundary data ψ m , m = 1 , ,
3, where we consider, ψ ∗ m , of ψ m ’s from(0 , T ) to R satisfying (cid:107) ψ ∗ j (cid:107) H / ( R ) (cid:46) (cid:107) ψ (cid:107) H / (0 ,T ) , j = 1 , (cid:107) ψ ∗ (cid:107) L ( R ) (cid:46) (cid:107) ψ (cid:107) L (0 ,T ) .We can further assume that ψ ∗ m ( t ) = 0 for t <
0. For simplicity, we denote theextended functions again by ψ m . Our approach for obtaining a representation forthe solution is to apply the Laplace transformation in time:˜ f ( s ) = (cid:90) ∞ e − st f ( t ) dt. This approach is motivated from [4] on the KdV equation. However, due to theparameters β, α, δ and assuming that L may be critical, the situation gets morecomplicated and the treatment of the problem is more subtle. To this end, we apply the Laplace transformation and transform (2.1) to thefollowing infinite family of third–order boundary value problems (cid:40) is ˜ y ( x, s ) + iβ ˜ y xxx ( x, s ) + α ˜ y xx ( x, s ) + iδ ˜ y x ( x, s ) = 0 , ( x, s ) ∈ (0 , L ) × C , ˜ y (0 , s ) = ˜ ψ ( s ) , ˜ y ( L, s ) = ˜ ψ ( s ) , ˜ y x ( L, s ) = ˜ ψ ( s ) , (2.30)where a suitable set for the complex valued independent variable s is specified below.Using the Bromwich integral, y can be represented as y ( x, t ) = 12 πi (cid:90) r + i ∞ r − i ∞ e st ˜ y ( x, s ) ds, (2.31)where the vertical integration path ( r − i ∞ , r + i ∞ ) in the complex plane is chosenso that, all possible singularities of ˜ y lie at the left of it. Note that for sufficientlylarge r the characteristic equation, s + βλ − iαλ + δλ = 0 , (2.32)for (2.30) has distinct roots. In fact, there exists only finitely many s for which (2.32)has double or possibly triple roots. We can classify these cases depending on the signof the quantity α + 3 βδ . To this end, let λ j , j = 1 , ,
3, denote the roots of (2.32)and assume that λ = λ for some s ∈ C . Then direct calculations (see Appendix Cfor details) yield the following cases.(i) If α + 3 βδ >
0, then there exists only two possible values of s and thesevalues belong to the imaginary axis.(ii) If α + 3 βδ = 0, then we have one and only one possible value of s and thisvalue belongs to the imaginary axis. Note that for this value of s , we have λ = λ = λ .(iii) If α + 3 βδ <
0, then there exists only two possible values of s and thesevalues are symmetric with respect to the imaginary axis.Consequently, for sufficiently large r (2.32) has distinct roots on the line (cid:60) ( s ) = r and solution of (2.30) is of the form˜ y ( x, s ) = (cid:88) j =1 c j ( s ) e λ j ( s ) x , (2.33)where the column vector ( c ( s ) , c ( s ) , c ( s )) T is the solution of the linear system e λ ( s ) L e λ ( s ) L e λ ( s ) L λ ( s ) e λ ( s ) L λ ( s ) e λ ( s ) L λ ( s ) e λ ( s ) L c ( s ) c ( s ) c ( s ) = ˜ ψ ( s )˜ ψ ( s )˜ ψ ( s ) . (2.34)Applying Cramer’s rule, these coefficients can be obtained as c j = ∆ j ∆ , where ∆ isthe determinant of the coefficient matrix and ∆ j ’s are determinants of the matrices TABILIZATION OF HIGHER ORDER SCHR ¨ODINGER EQUATIONS 23 formed by replacing the j − th column of the coefficient matrix by the column vector( ˜ ψ , ˜ ψ , ˜ ψ ) T . Thus y is of the form y ( x, t ) = 12 πi (cid:88) j =1 (cid:90) r + i ∞ r − i ∞ e st ∆ j ( s )∆( s ) e λ j ( s ) x ds. (2.35)We can rewrite y [0 , , ψ , ψ , ψ ] as y ≡ (cid:80) m =1 y m , where y m solves the same problemwith boundary data ψ j ≡ j (cid:54) = m , j = 1 , ,
3. Thus y m ’s can be expressed as y m ( x, t ) = 12 πi (cid:88) j =1 (cid:90) r + i ∞ r − i ∞ e st ∆ j,m ( s )∆( s ) e λ j ( s ) x ˜ ψ m ( s ) ds, m = 1 , , . (2.36)Here ∆ j,m ’s, m = 1 , ,
3, are obtained from ∆ j , where ψ m is replaced by 1 and ψ j ’s, j (cid:54) = m , are replaced by 0 for each j .To change the integration path in (2.36) by a more convenient one, one needs toinvestigate possible zeros of ∆( s ) in the complex plane. These points occur not onlydue to the double or possibly triple roots of (2.32) but may also occur due to theeigenvalues of the operator A defined in (2.2) with domain D ( A ) defined in (2.3).Note that A is a dissipative operator: (cid:60) ( Aϕ, ϕ ) = (cid:60) (cid:90) L ( − βϕ (cid:48)(cid:48)(cid:48) + iαϕ (cid:48)(cid:48) − δϕ (cid:48) )( x ) ϕ ( x ) dx = − β | ϕ (cid:48) (0) | ≤ . Thus, in particular, all eigenvalues of A lie on the left complex half plane or possiblyon the imaginary axis. The latter situation occurs only if the problem (cid:40) − βϕ (cid:48)(cid:48)(cid:48) + iαϕ (cid:48)(cid:48) − δϕ (cid:48) = λϕ, in (0 , L ) ,ϕ (0) = ϕ ( L ) = ϕ (cid:48) (0) = ϕ (cid:48) ( L ) , (2.37)has nontrivial solutions. Using the corresponding characteristic equation − βm + iαm − δm = λ for the main equation of (2.37) together with the boundary conditions, one can obtainthat the roots m j , j = 1 , ,
3, must be distinct, i.e. ϕ ( x ) = (cid:80) j =1 d e m j x . Togetherwith the boundary conditions, this implies that m j ’s must satisfy e m L = e m L = e m L (see [13, Proposition 2] for similar calculations). Therefore, we have m − m = 2 kπiL ,m − m = 2 lπiL , where without loss of generality, upon relabeling m j ’s we can assume that k, l ∈ Z + .Using m + m + m = iαβ , we get m = iα β + 2( − k − l ) πi L ,m = iα β + 2( k − l ) πi L ,m = iα β + 2( k + 2 l ) πi L .
Substituting these into m m + m m + m m = δβ , after some calculations, we obtain δβ = − α β + 4 π ( k + kl + l )3 L or equivalently α + 3 βδ = 4 π β ( k + kl + l ) L . (2.38)Consequently, depending on the sign of α + 3 βδ and the interval length L , it is pos-sible to obtain a nontrivial solution of (2.37), therefore there can be some eigenvalueson the imaginary axis.(i) Let α + 3 βδ >
0. Then, choosing L from the set of critical lengths givenin (1.4) imply the existence of some eigenvalues that are located on theimaginary axis. Now from the equation m m m = − λβ and using L =2 πβ (cid:113) k + kl + l α +3 βδ , one obtains after some calculations that λ = i β (cid:2) α − α ( α + 3 βδ ) + 2( α + 3 βδ ) / H ( k, l ) (cid:3) (2.39)where H ( k, l ) = ( − k − l )( k − l )( k + 2 l )2( k + kl + l ) . (2.40)It is not difficult to see that − < H ( k, l ) < k, l ∈ Z + . Thus, from(2.39), we deduce that (cid:61) ( λ ) ∈ ( (cid:61) s +2 , (cid:61) s +1 ), where s +1 . = i β (cid:2) α − α ( α + 3 βδ ) + 2( α + 3 βδ ) / (cid:3) and s +2 . = i β (cid:2) α − α ( α + 3 βδ ) − α + 3 βδ ) / (cid:3) . See Figure 4 for the graph of H ( k, l ). TABILIZATION OF HIGHER ORDER SCHR ¨ODINGER EQUATIONS 25
Figure 4. H ( k, l ) : [1 , ∞ ) × [1 , ∞ ) → ( − , s +1 and s +2 are the points where (2.30) assumes doubleroot (see Appendix C for detailed calculations), hence zeros of ∆( s ). Thisfact together with the location of the possible pure imaginary eigenvaluesimply that all possible singular points of (2.35) belonging to the imaginaryaxis lie in the closed interval [ (cid:61) ( s +2 ) , (cid:61) ( s +1 )]. Thus, we can deform the verticalintegration path of (2.35) by shifting the parts { s : (cid:61) ( s ) > (cid:61) ( s +1 ) + ρ } and { s : (cid:61) ( s ) < (cid:61) ( s +2 ) − ρ } to C +1 . = ( s +1 + iρ, i ∞ ) and C +3 . = ( − i ∞ , s +2 − iρ )respectively, ρ > ρ units from the imaginary axis, and avoid the points s +1 , s +2 by aquarter-circular arcs to the upper-right and lower-right respectively, denotedby C +2 . See Figure 5 below. (cid:60) ( s ) (cid:61) ( s ) s +2 s +1 C +3 C +2 C +1 Figure 5.
Integration path for the case α + 3 βδ > Thus we can express (2.36) as y m ( x, t ) = 12 πi (cid:88) j =1 (cid:90) C +1 e st ∆ j,m ( s )∆( s ) e λ j ( s ) x ˜ ψ m ( s ) ds + 12 πi (cid:88) j =1 (cid:90) C +2 e st ∆ j,m ( s )∆( s ) e λ j ( s ) x ˜ ψ m ( s ) ds + 12 πi (cid:88) j =1 (cid:90) C +3 e st ∆ j,m ( s )∆( s ) e λ j ( s ) x ˜ ψ m ( s ) ds. (2.41)Now we change the variable in the first and the third integral as s = iω ( ξ ) = i ( βξ − αξ − δξ ). For α + 3 βδ >
0, the function ω ( ξ ) has one local maximumand one local minimum. After some calculations one can find that the mostright inverse image of s +1 and the most left inverse image of s +2 are given by ξ +1 . = α + 2 (cid:112) α + 3 βδ β , ξ +2 . = α − (cid:112) α + 3 βδ β respectively (see Figure 6). Thus inverse images of the paths C +1 and C +3 under the transformation s = iω ( ξ ) become ( ξ +1 + η +1 , ∞ ) and ( −∞ , ξ +2 − η +2 )respectively for η +1 , η +2 >
0. Consequently, (2.41) becomes ω ( ξ ) ξ +2 ξ +1 (cid:61) ( s +2 ) (cid:61) ( s +1 ) ξ (cid:61) ( s ) Figure 6.
Plot of transformation (cid:61) ( s ) = ω ( ξ ) when α + 3 βδ > TABILIZATION OF HIGHER ORDER SCHR ¨ODINGER EQUATIONS 27 y m ( x, t ) = 12 πi (cid:88) j =1 (cid:90) ∞ ξ +1 + η +1 e iω ( ξ ) t ∆ ∗ j,m ( ξ )∆ ∗ ( ξ ) e λ ∗ j ( ξ ) x (3 βξ − αξ − δ ) ˜ ψ m ∗ ( ξ ) dξ + 12 πi (cid:88) j =1 (cid:90) C +2 e st ∆ j,m ( s )∆( s ) e λ j ( s ) x ˜ ψ m ( s ) ds + 12 πi (cid:88) j =1 (cid:90) ξ +2 − η +2 −∞ e iω ( ξ ) t ∆ ∗ j,m ( ξ )∆ ∗ ( ξ ) e λ ∗ j ( ξ ) x (3 βξ − αξ − δ ) ˜ ψ m ∗ ( ξ ) dξ. = y + m, ( x, t ) + y + m, ( x, t ) + y + m, ( x, t ) , (2.42)where the superscript ∗ stands for the transformed functions under the changeof variable given above. Note also that, with respect to the new variable, wehave the following explicit representation for the roots of the characteristicequation (2.32): λ ∗ ( ξ ) = iξ,λ ∗ ( ξ ) = − i ( βξ − α ) − (cid:112) β ξ − βαξ − α − βδ β ,λ ∗ ( ξ ) = − i ( βξ − α ) + (cid:112) β ξ − βαξ − α − βδ β . (2.43)(ii) Let α +3 βδ = 0. Then we see that (2.38) does not hold for any k, l ∈ Z + , thusthe eigenvalue problem (2.37) has only trivial solution. But this contradictswith the fact that ϕ is an eigenvalue. Thus (cid:60) ( Aϕ, ϕ ) < A are strictly negative. On the otherhand s . = iα β is the point where (2.32) assumes triple root (see Appendix C for details).Thus the integrand of (2.36) is continuous for all r > s bya half-circular arc to the right with a radius ρ > C . Definingalso C . = ( s + iρ, i ∞ ) and C . = ( − i ∞ , s − iρ ) (see Figure 7 below) (cid:60) ( s ) (cid:61) ( s ) C s C C Figure 7.
Integration path for the case α + 3 βδ = 0.we can express (2.36) as y m ( x, t ) = 12 πi (cid:88) j =1 (cid:90) C e st ∆ j,m ( s )∆( s ) e λ j ( s ) x ˜ ψ m ( s ) ds + 12 πi (cid:88) j =1 (cid:90) C e st ∆ j,m ( s )∆( s ) e λ j ( s ) x ˜ ψ m ( s ) ds + 12 πi (cid:88) j =1 (cid:90) C e st ∆ j,m ( s )∆( s ) e λ j ( s ) x ˜ ψ m ( s ) ds. (2.44)Now let us consider changing variables as s = iω ( ξ ) = i ( βξ − αξ − δξ )for the first and the third integrals. For α + 3 βδ = 0, note that ω ( ξ ) isnondecreasing and ( ξ , s ) is the inflection point of ω ( ξ ), where after somecalculations, ξ can be obtained as ξ . = α β . See Figure 8 for a graphical illustration. Hence we can find the unique inverseimages of C and C as ( ξ + η , ∞ ) and ( −∞ , ξ − η ) respectively for some TABILIZATION OF HIGHER ORDER SCHR ¨ODINGER EQUATIONS 29 ω ( ξ ) ξ (cid:61) ( s ) ξ (cid:61) ( s ) Figure 8.
Plot of transformation (cid:61) ( s ) = ω ( ξ ) when α + 3 βδ = 0. η , η >
0. Thus (2.44) becomes y m ( x, t ) = 12 πi (cid:88) j =1 (cid:90) ∞ ξ + η e iω ( ξ ) t ∆ ∗ j,m ( ξ )∆ ∗ ( ξ ) e λ ∗ j ( ξ ) x (3 βξ − αξ − δ ) ˜ ψ m ∗ ( ξ ) dξ + 12 πi (cid:88) j =1 (cid:90) C +2 e st ∆ j,m ( s )∆( s ) e λ j ( s ) x ˜ ψ m ( s ) ds + 12 πi (cid:88) j =1 (cid:90) ξ − η −∞ e iω ( ξ ) t ∆ ∗ j,m ( ξ )∆ ∗ ( ξ ) e λ ∗ j ( ξ ) x (3 βξ − αξ − δ ) ˜ ψ m ∗ ( ξ ) dξ. = y m, ( x, t ) + y m, ( x, t ) + y m, ( x, t ) , (2.45)where λ ∗ j ’s are given as in (2.43).(iii) Let α + 3 βδ <
0. Then (2.38) does not hold for any k, l ∈ Z + and alleigenvalues lie on the left half complex plane. On the other hand, there exitstwo values of s for which (2.32) assumes double root. These values, say s − and s − with (cid:60) ( s − ) > > (cid:60) ( s − ) which are symmetric with respect to theimaginary axis (see Appendix C), are also the branch points of the squareroot function (cid:113) ( s − s − )( s − s − )where we choose the branch cut as { s ∈ C : (cid:61) ( s ) = (cid:61) ( s − ) , (cid:60) ( s − ) < (cid:60) ( s ) < (cid:60) ( s − ) } . Indeed changing variables as s = i (cid:61) ( s − ) + r and than making somecalculations, roots of the characteristic equation (2.32) can be expressed as λ † j ( r ) = 13 β (cid:18) iα − α + 3 βδ Λ j ( r ) + Λ j ( r ) (cid:19) , whereΛ j ( r ) = − β e πij − (cid:113) r + α +3 βδ )729 β , j = 1 , , . Note that (cid:60) ( s − ) and (cid:60) ( s − ) are now the zeros of the square root above.In conclusion, what distinguishes this case from the previous cases is that,we have now a zero of ∆( s ) that lies on the right half complex plane whichis at the endpoint of the branch cut. Therefore, to deform the integrationpath, we first shift the vertical integration line to the left until we meet s − .Then we deform a part of the path near s − by a half-circular arc to the rightwith a radius ρ >
0. Next we deform the rest of the integration path as, firstby horizontal line segments to the left starting from the end points of the arcthrough the imaginary axis and second continuing from the imaginary axis inthe vertical direction towards + i ∞ and − i ∞ respectively. See Figure 9 forthe path deformation described here. (cid:60) ( s ) (cid:61) ( s ) (cid:60) ( s − ) s − C − C − C − C − C − Figure 9.
Integration path for the case α + 3 βδ < TABILIZATION OF HIGHER ORDER SCHR ¨ODINGER EQUATIONS 31
Consequently, we can write (2.36) as y m ( x, t ) = 12 πi (cid:88) j =1 (cid:90) C − e st ∆ j,m ( s )∆( s ) e λ j ( s ) x ˜ ψ m ( s ) ds + 12 πi (cid:88) j =1 (cid:90) C − ∪ C − ∪ C − e st ∆ j,m ( s )∆( s ) e λ j ( s ) x ˜ ψ m ( s ) ds + 12 πi (cid:88) j =1 (cid:90) C − e st ∆ j,m ( s )∆( s ) e λ j ( s ) x ˜ ψ m ( s ) ds. (2.46)Now let us apply change of variable s = iω ( ξ ) = i ( βξ − αξ − δξ ) for thefirst and third integrals. Note that for α + 3 βδ <
0, this mapping is strictlyincreasing and the inverse image of (cid:61) ( s − ) under the transformation ω ( ξ ) isthe point ξ − . = α β . Then C − and C − are mapped to ( ξ − + η − , ∞ ) and ( −∞ , ξ − − η − ) for some η − , η − >
0. See Figure 10. Thus (2.46) becomes ω ( ξ ) ξ − (cid:61) ( s − ) ξ (cid:61) ( s ) Figure 10.
Plot of transformation (cid:61) ( s ) = ω ( ξ ) when α + 3 βδ < y m ( x, t ) = 12 πi (cid:88) j =1 (cid:90) ∞ ξ − + η − e iω ( ξ ) t ∆ ∗ j,m ( ξ )∆ ∗ ( ξ ) e λ ∗ j ( ξ ) x (3 β ξ − αξ − δ ) ˜ ψ m dξ + 12 πi (cid:88) j =1 (cid:90) C − ∪ C − ∪ C − e st ∆ j,m ( s )∆( s ) e λ j ( s ) x ˜ ψ m ( s ) ds + 12 πi (cid:88) j =1 (cid:90) ξ − − η − −∞ e iω ( ξ ) t ∆ ∗ j,m ( ξ )∆ ∗ ( ξ ) e λ ∗ j ( ξ ) x (3 β ξ − αξ − δ ) ˜ ψ m ∗ ( ξ ) dξ. = y − m, ( x, t ) + y − m, ( x, t ) + y − m, ( x, t ) . (2.47)where λ ∗ j ’s are given by (2.43).In the following three lemmas we provide estimates for y m for each m . Note that foreach solution representation (2.42), (2.45) and (2.47) corresponding to the differentcases of α + 3 βδ , second integrals are bounded on the corresponding integrationpaths. However, these paths lie on the right half complex plane. Therefore, for agiven T >
0, we can find c > e cT . On the other hand, for each case of α + 3 βδ , we have to estimate thefollowing form of integrals I m ( x, t ) = 12 πi (cid:88) j =1 (cid:90) ∞ γ e iω ( ξ ) t ∆ ∗ j,m ( ξ )∆ ∗ ( ξ ) e λ ∗ j ( ξ ) x (3 βξ − αξ − δ ) ˜ ψ m ∗ ( ξ ) dξ (2.48)and J m ( x, t ) = 12 πi (cid:88) j =1 (cid:90) γ −∞ e iω ( ξ ) t ∆ ∗ j,m ( ξ )∆ ∗ ( ξ ) e λ ∗ j ( ξ ) x (3 βξ − αξ − δ ) ˜ ψ m ∗ ( ξ ) dξ, (2.49)where γ ∈ { ξ +1 + η +1 , ξ + η , ξ − + η − } and γ ∈ { ξ +2 + η +2 , ξ + η , ξ − + η − } . Thus,it is enough to study (2.48) and (2.49) in order to obtain desired norm estimates for y m , m = 1 , , Lemma 2.3.
Let
T, β > , α, δ ∈ R and ψ ∈ H / (0 , T ) . Then y = y [0 , , ψ , , belongs to the space C ([0 , T ]; L (0 , L )) ∩ L (0 , T ; H (0 , L )) and it also satisfies ∂ x y ∈ C ([0 , L ]; L (0 , T )) . Moreover, there exists a constant c > such that (cid:107) y (cid:107) C ([0 ,T ]; L (0 ,L )) + (cid:107) y (cid:107) L (0 ,T ; H (0 ,L )) (cid:46) e cT (cid:107) ψ (cid:107) H / (0 ,T ) (2.50) and sup x ∈ [0 ,L ] (cid:107) ∂ x y ( x, · ) (cid:107) L (0 ,T ) (cid:46) e cT (cid:107) ψ (cid:107) H / (0 ,T ) . (2.51) If α + 3 βδ < , then c > (cid:60) ( s − ) > where s − is the value for which (2.32) assumesdouble root. TABILIZATION OF HIGHER ORDER SCHR ¨ODINGER EQUATIONS 33
Proof.
Let us first obtain the asymptotic behaviours of the ratios (cid:12)(cid:12)(cid:12) ∆ ∗ j, ( ξ )∆ ∗ ( ξ ) (cid:12)(cid:12)(cid:12) for largevalues of ξ . Using the relation λ ∗ + λ ∗ + λ ∗ = iαβ , we have∆( ξ ) ∗ = e iαLβ (cid:0) e − λ ∗ ( ξ ) L ( λ ∗ ( s ) − λ ∗ ( ξ )) − e − λ ∗ ( ξ ) L ( λ ∗ ( ξ ) − λ ∗ ( ξ )) + e − λ ∗ ( ξ ) L ( λ ∗ ( ξ ) − λ ∗ ( ξ )) (cid:1) (2.52)and ∆ ∗ , ( ξ ) = e iαLβ e − λ ∗ ( ξ ) L ( λ ∗ ( ξ ) − λ ∗ ( ξ )) , (2.53)∆ ∗ , ( ξ ) = e iαLβ e − λ ∗ ( ξ ) L ( λ ∗ ( ξ ) − λ ∗ ( ξ )) , (2.54)∆ ∗ , ( ξ ) = e iαLβ e − λ ∗ ( ξ ) L ( λ ∗ ( ξ ) − λ ∗ ( ξ )) . (2.55)Using the roots of the characteristic equation (2.43) in the variable ξ , we obtain thefollowing large ξ asymptotics (cid:12)(cid:12)(cid:12)(cid:12) ∆ ∗ j, ( ξ )∆ ∗ ( ξ ) (cid:12)(cid:12)(cid:12)(cid:12) ∼ e − √ ξL , j = 1 , , j = 2 ,e −√ ξL , j = 3 . (2.56)Let us start by taking L − norm of I with respect to its first component and apply[4, Lemma 2.5] to get (cid:107) I ( · , t ) (cid:107) (cid:46) (cid:88) j =1 (cid:90) ∞ γ (cid:16) e L (cid:60) ( λ ∗ j ( ξ )) + 1 (cid:17) (cid:12)(cid:12)(cid:12)(cid:12) ∆ ∗ j, ( ξ )∆ ∗ ( ξ ) (cid:12)(cid:12)(cid:12)(cid:12) (cid:12)(cid:12) βξ − αξ − δ (cid:12)(cid:12) (cid:12)(cid:12)(cid:12) ˜ ψ ∗ ( ξ ) (cid:12)(cid:12)(cid:12) dξ. Using the asymptotic behaviours (2.56), we have (cid:16) e L (cid:60) ( λ ∗ j ( ξ )) + 1 (cid:17) (cid:12)(cid:12)(cid:12)(cid:12) ∆ ∗ j, ( ξ )∆ ∗ ( ξ ) (cid:12)(cid:12)(cid:12)(cid:12) ∼ e −√ ξL , j = 1 , , j = 2 ,e −√ ξL , j = 3 , (2.57)as ξ → ∞ . Thus, we can write (cid:107) I ( · , t ) (cid:107) (cid:46) (cid:90) ∞ γ (cid:12)(cid:12) βξ − αξ − δ (cid:12)(cid:12) (cid:12)(cid:12)(cid:12) ˜ ψ ∗ ( ξ ) (cid:12)(cid:12)(cid:12) dξ. Changing variables as µ = βξ − αξ − δξ , we get (cid:107) I ( · , t ) (cid:107) (cid:46) (cid:90) ∞ ω ( γ ) (1 + µ ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:90) ∞ e − iµτ ψ ( τ ) dτ (cid:12)(cid:12)(cid:12)(cid:12) dµ ≤ (cid:107) ψ (cid:107) H / (0 ,T ) . Passing to supremum over t ∈ [0 , T ], we obtain (cid:107) I (cid:107) C ([0 ,T ]; L (0 ,L )) (cid:46) (cid:107) ψ (cid:107) H / (0 ,T ) . (2.58)Next, we differentiate I with respect to its first component, take L − norm on(0 , T ) and change variables as µ = βξ − αξ − δξ to get (cid:107) ∂ x I ( x, · ) (cid:107) L (0 ,T ) = (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) (cid:88) j =1 π (cid:90) ∞ γ e iω ( ξ ) t λ ∗ j ( ξ ) e λ ∗ j ( ξ ) x ∆ ∗ j, ( ξ )∆ ∗ ( ξ ) (3 βξ − αξ − δ ) ˜ ψ ∗ ( ξ ) dξ (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) L (0 ,T ) (cid:46) (cid:88) j =1 (cid:13)(cid:13)(cid:13)(cid:13)(cid:90) ∞ ω ( γ ) e iµt λ ∗ j ( θ ( µ )) e λ ∗ j ( θ ( µ )) x ∆ ∗ j, ( θ ( µ ))∆ ∗ ( θ ( µ )) ˜ ψ ∗ ( θ ( µ )) dµ (cid:13)(cid:13)(cid:13)(cid:13) L (0 ,T ) , (2.59)where θ ( µ ) is the real solution of µ = βξ − αξ − δξ for γ < ξ < ∞ . Observe thatthe function (cid:40) λ ∗ j ( θ ( µ )) e λ ∗ j ( θ ( µ )) x ∆ ∗ j, ( θ ( µ ))∆ ∗ ( θ ( µ )) ˜ ψ ∗ ( θ ( µ )) , µ ∈ ( ω ( γ ) , ∞ ) , , elsewhere,is the Fourier transform of the function given by the integral. So, thanks to thePlancherel’s theorem, we can write (cid:107) ∂ x I ( x, · ) (cid:107) L (0 ,T ) (cid:46) (cid:88) j =1 (cid:90) ∞ ω ( γ ) (cid:12)(cid:12)(cid:12)(cid:12) λ ∗ j ( θ ( µ )) e λ ∗ j ( θ ( µ )) x ∆ ∗ j, ( θ ( µ ))∆ ∗ ( θ ( µ )) ˜ ψ ∗ ( θ ( µ )) (cid:12)(cid:12)(cid:12)(cid:12) dµ (2.60)for all x ∈ [0 , L ]. It follows that (cid:107) ∂ x I (cid:107) L (0 ,L ; L (0 ,T )) ≤ sup x ∈ [0 ,L ] (cid:107) ∂ x I ( x, · ) (cid:107) L (0 ,T ) (cid:46) (cid:88) j =1 (cid:90) ∞ γ | λ ∗ j ( ξ ) | sup x ∈ [0 ,L ] (cid:16) e (cid:60) ( λ ∗ j ( ξ )) x (cid:17) (cid:12)(cid:12)(cid:12)(cid:12) ∆ ∗ j, ( ξ )∆ ∗ ( ξ ) (cid:12)(cid:12)(cid:12)(cid:12) (3 βξ − αξ − δ ) (cid:12)(cid:12)(cid:12) ˜ ψ ∗ ( ξ ) (cid:12)(cid:12)(cid:12) dξ. (2.61)Using (2.43) and (2.56), one can obtain the following asymptoic behaviours in ξ | λ ∗ j ( ξ ) | sup x ∈ [0 ,L ] (cid:16) e (cid:60) ( λ ∗ j ( ξ )) x (cid:17) (cid:12)(cid:12)(cid:12)(cid:12) ∆ ∗ j, ( ξ )∆ ∗ ( ξ ) (cid:12)(cid:12)(cid:12)(cid:12) ∼ ξ e −√ ξL , j = 1 ,ξ , j = 2 ,ξ e −√ ξL , j = 3 . (2.62) TABILIZATION OF HIGHER ORDER SCHR ¨ODINGER EQUATIONS 35
Using (2.62) in (2.61), and then changing variables back as µ = βξ − αξ − δξ , weget (cid:107) ∂ x I (cid:107) L (0 ,L ; L (0 ,T )) ≤ sup x ∈ [0 ,L ] (cid:107) ∂ x I ( x, · ) (cid:107) L (0 ,T ) (cid:46) (cid:90) ∞ γ ξ (3 βξ − αξ − δ ) | ˜ ψ ∗ ( ξ ) | dξ (cid:46) (cid:90) ∞ ω ( γ ) (1 + µ ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:90) ∞ e − iµτ ψ ( τ ) dτ (cid:12)(cid:12)(cid:12)(cid:12) dµ (cid:46) (cid:107) ψ (cid:107) H / (0 ,T ) (2.63)Changing the integration order on (cid:107) ∂ x I (cid:107) L (0 ,L ; L (0 ,T )) and using Poincare inequality,we conclude that (2.50) and (2.51) holds for I .To show that the mapping x ∈ [0 , L ] → (cid:107) ∂ x I ( x, · ) (cid:107) L (0 ,T ) is continuous, let { x n } n ∈ N ⊂ [0 , L ] be such that x n → x as n → ∞ and let us write ∂ x I ( x, t ) − ∂ x I ( x n , t )= 12 πi (cid:88) j =1 (cid:90) ∞ γ e iω ( ξ ) t λ ∗ j ( ξ ) (cid:16) e λ ∗ j ( ξ ) x − e λ ∗ j ( ξ ) x n (cid:17) ∆ ∗ j, ( ξ )∆ ∗ ( ξ ) ˜ ψ ∗ ( ξ ) dξ (2.64)Applying the arguments above in (2.59)-(2.63), one can obtain that (cid:107) ∂ x I ( x, · ) − ∂ x I ( x n , · ) (cid:107) L (0 ,T ) (cid:46) (cid:88) j =1 (cid:90) ∞ ω ( γ ) (cid:12)(cid:12)(cid:12)(cid:12) λ ∗ j ( θ ( µ )) (cid:16) e λ ∗ j ( θ ( µ )) x − e λ ∗ j ( θ ( µ )) x n (cid:17) ∆ ∗ j, ( θ ( µ ))∆ ∗ ( θ ( µ )) ˜ ψ ∗ ( θ ( µ )) (cid:12)(cid:12)(cid:12)(cid:12) dµ (cid:46) (cid:107) ψ (cid:107) H / (0 ,T ) , for all n ∈ N . Hence, by the dominated convergence theorem, we see thatlim n →∞ (cid:107) ∂ x I ( x, · ) − ∂ x I ( x n , · ) (cid:107) L (0 ,T ) → . Applying a similar procedure yields the same results for J . (cid:3) Lemma 2.4.
Let
T, β > , α, δ ∈ R and ψ ∈ H / (0 , T ) . Then y = y [0 , , , ψ , belongs to the space C ([0 , T ]; L (0 , L )) ∩ L (0 , T ; H (0 , L )) and also satisfies ∂ x y ∈ C ([0 , L ]; L (0 , T )) . Moreover, there exists a constant c > such that (cid:107) y (cid:107) C ([0 ,T ]; L (0 ,L )) + (cid:107) y (cid:107) L (0 ,T ; H (0 ,L )) (cid:46) e cT (cid:107) ψ (cid:107) H / (0 ,T ) (2.65) and sup x ∈ [0 ,L ] (cid:107) ∂ x y ( x, · ) (cid:107) L (0 ,T ) (cid:46) e cT (cid:107) ψ (cid:107) H / (0 ,T ) . (2.66) If α + 3 βδ < , then c > (cid:60) ( s − ) > where s − is the value for which (2.32) assumesdouble root.Proof. We start by obtaining large ξ asymptotics for (cid:12)(cid:12)(cid:12) ∆ ∗ j, ( ξ )∆ ∗ ( ξ ) (cid:12)(cid:12)(cid:12) . Let us write∆ ∗ , ( ξ ) = λ ∗ ( ξ ) e λ ∗ ( ξ ) L − λ ∗ ( ξ ) e λ ∗ ( ξ ) L , (2.67)∆ ∗ , ( ξ ) = λ ∗ ( ξ ) e λ ∗ ( ξ ) L − λ ∗ ( ξ ) e λ ∗ ( ξ ) L , (2.68)∆ ∗ , ( ξ ) = λ ∗ ( ξ ) e λ ∗ ( ξ ) L − λ ∗ ( ξ ) e λ ∗ ( ξ ) L , (2.69)and then, use ∆ ∗ given in (2.52) and characteristic roots given in (2.43) to obtain (cid:12)(cid:12)(cid:12)(cid:12) ∆ ∗ j, ( ξ )∆ ∗ ( ξ ) (cid:12)(cid:12)(cid:12)(cid:12) ∼ , j = 1 , , j = 2 ,e − √ ξL , j = 3 . (2.70)The rest of the proof is as in the proof of previous lemma. (cid:3) Lemma 2.5.
Let
T, β > , α, δ ∈ R and ψ ∈ L (0 , T ) . Then y = y [0 , , ψ , , belongs to the space C ([0 , T ]; L (0 , L )) ∩ L (0 , T ; H (0 , L )) and also satisfies ∂ x y ∈ C ([0 , L ]; L (0 , T )) . Moreover, there exists a constant c > such that (cid:107) y (cid:107) C ([0 ,T ]; L (0 ,L )) + (cid:107) y (cid:107) L (0 ,T ; H (0 ,L )) (cid:46) e cT (cid:107) ψ (cid:107) L (0 ,T ) (2.71) and sup x ∈ [0 ,L ] (cid:107) ∂ x y ( x, · ) (cid:107) L (0 ,T ) (cid:46) e cT (cid:107) ψ (cid:107) L (0 ,T ) (2.72) If α + 3 βδ < , then c > (cid:60) ( s − ) > where s − is the value for which (2.32) assumesdouble root.Proof. As in the previous proofs, let us obtain large ξ asymptotics of the ratios (cid:12)(cid:12)(cid:12) ∆ ∗ j, ( ξ )∆ ∗ ( ξ ) (cid:12)(cid:12)(cid:12) . To this end, we write∆ ∗ , ( ξ ) = e λ ∗ ( ξ ) L − e λ ∗ ( ξ ) L , (2.73)∆ ∗ , ( ξ ) = e λ ∗ ( ξ ) L − e λ ∗ ( ξ ) L , (2.74)∆ ∗ , ( ξ ) = e λ ∗ ( ξ ) L − e λ ∗ ( ξ ) L , (2.75)and then use ∆ ∗ given in (2.52) and characteristic roots given in (2.43) to obtain (cid:12)(cid:12)(cid:12)(cid:12) ∆ ∗ j, ( ξ )∆ ∗ ( ξ ) (cid:12)(cid:12)(cid:12)(cid:12) ∼ ξ − , j = 1 ,ξ − , j = 2 ,ξ − e − √ ξL , j = 3 . (2.76)Proceeding as in the proof of Lemma 2.3, we can obtain the desired result. (cid:3) TABILIZATION OF HIGHER ORDER SCHR ¨ODINGER EQUATIONS 37
Combining Lemmas 2.3, 2.4 and 2.5, we obtain the following result for y [0 , , ψ , ψ , ψ ]. Lemma 2.6.
Let φ ≡ f ≡ . For T > and ( ψ , ψ , ψ ) ∈ H / (0 , T ) × H / (0 , T ) × L (0 , T ) , (2.1) admits a unique solution which belongs to the space C ([0 , T ]; L (0 , L )) ∩ L (0 , T ; H (0 , L )) with y x ∈ C ([0 , L ]; L (0 , T )) . Moreover there exists a constant c > such that (cid:107) y (cid:107) C ([0 ,T ]; L (0 ,L )) + (cid:107) y (cid:107) L (0 ,T ; H (0 ,L )) (cid:46) e cT (cid:107) ψ (cid:107) L (0 ,T ) (2.77) and sup x ∈ [0 ,L ] (cid:107) ∂ x y ( x, · ) (cid:107) L (0 ,T ) (cid:46) e cT (cid:107) ψ (cid:107) L (0 ,T ) (2.78) If α + 3 βδ < , then c > (cid:60) ( s − ) > where s − is the value for which (2.32) assumesdouble root. Now for the sake of our study, let us consider the problem iy t + iβy xxx + αy xx + iδy x = 0 , x ∈ (0 , L ) , t ∈ (0 , T ) ,y (0 , t ) = 0 , y ( L, t ) = 0 , y x ( L, t ) = ψ ( t ) ,y ( x,
0) = φ ( x ) , (2.79)From Lemma 2.1 and Lemma 2.5, we know that solution y of (2.79) belongs to thespace X T and satisfies (cid:107) y (cid:107) X T (cid:46) (cid:107) φ (cid:107) + e cT (cid:107) ψ (cid:107) L (0 ,T ) . Let v = y t . Then v solves the linear model below iv t + iβv xxx + αv xx + iδv x = 0 , x ∈ (0 , L ) , t ∈ (0 , T ) ,v (0 , t ) = 0 , v ( L, t ) = 0 , v x ( L, t ) = ψ (cid:48) ( t ) ,v ( x,
0) = ˜ φ ( x ) , (2.80)where ˜ φ . = − βφ (cid:48)(cid:48)(cid:48) + iαφ (cid:48)(cid:48) − δφ (cid:48) . Assume that ˜ φ ∈ L (0 , L ). From Lemma 2.1 andLemma 2.5, v satisfies (cid:107) v (cid:107) X T (cid:46) (cid:107) ˜ φ (cid:107) + e cT (cid:107) ψ (cid:48) (cid:107) L (0 ,T ) . Set y ( x, t ) = φ ( x ) + (cid:82) t v ( x, τ ) dτ . Then due to compatibility conditions and φ (cid:48) ( L ) = ψ (0), y satisfies the initial and boundary conditions y ( x,
0) = φ ( x ) ,y (0 , t ) = φ (0) + (cid:90) t v (0 , τ ) dτ = φ (0) + y (0 , t ) − y (0 ,
0) = 0 ,y ( L, t ) = φ ( L ) + (cid:90) t v ( L, τ ) dτ = φ ( L ) + y ( L, t ) − y ( L,
0) = 0 ,y x ( L, t ) = φ (cid:48) ( L ) + (cid:90) t v x ( L, τ ) dτ = φ (cid:48) ( L ) + y x ( L, t ) − y x ( L,
0) = ψ ( t ) . Moreover,( iy t + iβy xxx + αy xx + iδy x )( x, t )= iv ( x, t ) + (cid:90) t ( iβv xxx + αv xx + iδv x )( x, τ ) dτ + iβφ (cid:48)(cid:48)(cid:48) ( x ) + αφ (cid:48)(cid:48) ( x ) + iδφ (cid:48) ( x )= iv ( x,
0) + (cid:90) t iv t ( x, τ ) dτ + (cid:90) t ( iβv xxx + αv xx + iδv x )( x, τ ) dτ + iβφ (cid:48)(cid:48)(cid:48) ( x ) + αφ (cid:48)(cid:48) ( x ) + iδφ (cid:48) ( x )= iv ( x,
0) + (cid:90) t ( − iβv xxx − αv xx − iδv x )( x, τ ) dτ + (cid:90) t ( iβv xxx + αv xx + iδv x )( x, τ ) dτ + iβφ (cid:48)(cid:48)(cid:48) ( x ) + αφ (cid:48)(cid:48) ( x ) + iδφ (cid:48) ( x )=0 . Thus, y solves (2.79). Now, from the main equation of (2.79), we have β (cid:107) y xxx ( · , t ) (cid:107) ≤ (cid:107) v ( · , t ) (cid:107) + α (cid:107) y xx ( · , t ) (cid:107) + δ (cid:107) y x ( · , t ) (cid:107) (2.81)Applying Gagliardo–Nirenberg interpolation inequality and then (cid:15) − Young’s inequal-ity to the second term at the right hand side of (2.81), we get α (cid:107) y xx ( · , t ) (cid:107) ≤ c α (cid:107) y xxx ( · , t ) (cid:107) (cid:107) y ( · , t ) (cid:107) ≤ (cid:15) (cid:107) y xxx ( · , t ) (cid:107) + c α,(cid:15) (cid:107) y ( · , t ) (cid:107) . (2.82)Similarly, for the third term at the right hand side of (2.81), we have δ (cid:107) y x ( · , t ) (cid:107) ≤ c δ (cid:107) y xxx ( · , t ) (cid:107) (cid:107) y ( · , t ) (cid:107) ≤ (cid:15) (cid:107) y xxx ( · , t ) (cid:107) + c δ,(cid:15) (cid:107) y ( · , t ) (cid:107) . (2.83) TABILIZATION OF HIGHER ORDER SCHR ¨ODINGER EQUATIONS 39
Using (2.82)-(2.83) on (2.81), we obtain( β − (cid:15) ) (cid:107) y xxx ( · , t ) (cid:107) ≤ (cid:107) v ( · , t ) (cid:107) + c α,δ,(cid:15) (cid:107) y ( · , t ) (cid:107) . Therefore, for sufficiently small (cid:15) >
0, we get (cid:107) y xxx ( · , t ) (cid:107) (cid:46) (cid:107) v ( · , t ) (cid:107) + (cid:107) y ( · , t ) (cid:107) . (2.84)Passing to supremum over t ∈ [0 , T ] and using the fact that the right hand sidebelongs to C ([0 , T ] , L (0 , L )), we have y ∈ C ([0 , T ]; H (0 , L )). Next, differentiatingthe main equation of (2.79) with respect to x and taking L − norms of each term,we get β (cid:107) y xxxx ( · , t ) (cid:107) ≤ (cid:107) v x ( · , t ) (cid:107) + α (cid:107) y xxx ( · , t ) (cid:107) + δ (cid:107) y xx ( · , t ) (cid:107) . (2.85)Thanks to Gagliardo–Nirenberg’s interpolation inequality, (cid:15) − Young’s inequality andPoincare inequality, the second term at the right hand side of (2.85) can be estimatedas α (cid:107) y xxx ( · , t ) (cid:107) ≤ c α (cid:107) y xxxx ( · , t ) (cid:107) (cid:107) y ( · , t ) (cid:107) , ≤ (cid:15) (cid:107) y xxxx ( · , t ) (cid:107) + c α,(cid:15) (cid:107) y ( · , t ) (cid:107) ≤ (cid:15) (cid:107) y xxxx ( · , t ) (cid:107) + c α,(cid:15) (cid:107) y x ( · , t ) (cid:107) . (2.86)Using the same inequalities, the third term in (2.85) is estimated as δ (cid:107) y xx ( · , t ) (cid:107) ≤ c δ (cid:107) y xxxx ( · , t ) (cid:107) (cid:107) y ( · , t ) (cid:107) ≤ (cid:15) (cid:107) y xxxx ( · , t ) (cid:107) + c δ,(cid:15) (cid:107) y ( · , t ) (cid:107) ≤ (cid:15) (cid:107) y xxxx ( · , t ) (cid:107) + c δ,(cid:15) (cid:107) y x ( · , t ) (cid:107) . (2.87)Using (2.86)-(2.87) on (2.85) and choosing (cid:15) > (cid:107) y xxxx ( · , t ) (cid:107) (cid:46) (cid:107) v x ( · , t ) (cid:107) + (cid:107) y x ( · , t ) (cid:107) . (2.88)Right hand side belongs to L (0 , T ), so the left hand side does too. This implies y ∈ L (0 , T ; H (0 , L )). Combining this result with the previous one, we proved thefollowing lemma. Lemma 2.7.
Let
T > , ( φ, ψ ) ∈ H (0 , L ) × H (0 , T ) satisfy the compatibilityconditions. Then (2.79) has a unique solution y ∈ X T with y t ∈ X T and it satisfiesthe following estimate (cid:107) y (cid:107) X T + (cid:107) y t (cid:107) X T (cid:46) (cid:107) φ (cid:107) H (0 ,L ) + e cT (cid:107) ψ (cid:107) H (0 ,T ) . Now letting z = v t , one can see that z solves the following model iz t + iβz xxx + αz xx + iδz x = 0 , x ∈ (0 , L ) , t ∈ (0 , T ) ,z (0 , t ) = 0 , z ( L, t ) = 0 , z x ( L, t ) = ψ (cid:48)(cid:48) ( t ) ,z ( x,
0) = ˜˜ φ ( x ) , (2.89) where ˜˜ φ . = − β ˜ φ (cid:48)(cid:48)(cid:48) + iα ˜ φ (cid:48)(cid:48) − δ ˜ φ (cid:48) . Let ψ (cid:48)(cid:48) ( t ) ∈ L (0 , T ) and ˜˜ φ ∈ L (0 , L ). Then fromLemma 2.1 and Lemma 2.5, z satisfies (cid:107) z (cid:107) X T (cid:46) (cid:107) ˜˜ φ (cid:107) + e cT (cid:107) ψ (cid:48)(cid:48) (cid:107) L (0 ,T ) . (2.90)Define v ( x, t ) . = ˜ φ ( x ) + (cid:82) t z ( x, τ ) dτ . If ( ˜ φ, ψ (cid:48) ) satisfies the compatibility conditions,one can show that v satisfies the following initial–boundary conditions: v ( x,
0) = ˜ φ ( x )and v (0 , t ) = 0 , v ( L, t ) = 0 , v x ( L, t ) = φ (cid:48) ( t ) . Then one can also show that v solves (2.80). Now defining v = y t and repeating thesame analysis as we did through (2.80)-(2.88), one concludes the following lemma. Lemma 2.8.
Let
T > , ( φ, ψ ) ∈ H (0 , L ) × H (0 , T ) satisfy the higher ordercompatibility conditions. Then (2.79) has a unique solution in X T with y tt ∈ X T andit satisfies the following estimate, (cid:107) y (cid:107) X T + (cid:107) y tt (cid:107) X T (cid:46) (cid:107) φ (cid:107) H (0 ,L ) + e cT (cid:107) ψ (cid:107) H (0 ,T ) . Controller design
In this section, first we prove the existence of a smooth backstepping kernel. Thenwe state the result of the invertibility of the backstepping transformation with abounded inverse. Next, we prove the global wellposedness and exponential stabilityresults.3.1.
Backstepping kernel.
Let us express the main equation in (1.9) as G sst = DG . = 13 β [ β (3 G tts − G ttt ) − iα ( G tt − G ts ) − δG t − rG ] . Integrating the above expression in the first variable and using G s (0 , t ) = rt β weobtain G st ( s, t ) = r β + (cid:90) s [ DG ]( ξ, t ) dξ. Integrating once again in the first variable and using G (0 , t ) = 0 we get G t ( s, t ) = r β s + (cid:90) s (cid:90) ω [ DG ]( ξ, t ) dξdω. Finally, integrating in the second variable and using G ( s,
0) = 0 we obtain that G solves G ( s, t ) = r β st + (cid:90) t (cid:90) s (cid:90) ω [ DG ]( ξ, η ) dξdωdη. (3.1) TABILIZATION OF HIGHER ORDER SCHR ¨ODINGER EQUATIONS 41
So the solution of the boundary value problem (1.9) can be constructed by applyinga successive approximation method to the integral equation (3.1).
Lemma 3.1.
There exists a C ∞ − function G such that G solves the integral equation (3.1) .Proof. Let P be defined by( P f )( s, t ) . = (cid:90) t (cid:90) s (cid:90) ω [ Df ]( ξ, η ) dξdωdη. (3.2)Then we express (3.1) as G ( s, t ) = r β st + P G ( s, t ) . (3.3)Define G ≡ , G ( s, t ) = − r β st, and G n +1 = G + P G n , n ≥
1. Then we have G n +1 − G n = P ( G n − G n − ) , n ≥ . (3.4)To prove the existence of a solution of (3.3), it is enough to show that the sequence( G n ) and its partial derivatives are Cauchy with respect to the supremum norm (cid:107)·(cid:107) ∞ .To this end, define H ( s, t ) = st , H n = βr ( G n +1 − G n ). Then by (3.4), H n +1 = P H n and for j > i , G j − G i = j − (cid:88) n = i ( G n +1 − G n ) = r β j − (cid:88) n = i H n . (3.5)We see from (3.5) that the sequence ( G n ) (and its partial derivatives) is Cauchywith respect to the norm (cid:107) · (cid:107) ∞ , which implies that ( G n ) is convergent and its limitsolves (3.1) if and only if the sequence ( H n ) (and its partial derivatives) is absolutelysummable sequence with respect to the same norm.To show that H n ’s are absolutely summable, let us express P as sum of six oper-ators P = P , − + P , − + P , − + P , + P , + P , , where P , − f . = (cid:90) t (cid:90) s (cid:90) ω f tts ( ξ, η ) dξdωdη, P , − f . = − (cid:90) t (cid:90) s (cid:90) ω f ttt ( ξ, η ) dξdωdη,P , − f . = − iα β (cid:90) t (cid:90) s (cid:90) ω f tt ( ξ, η ) dξdωdη, P , f . = 2 iα β (cid:90) t (cid:90) s (cid:90) ω f ts ( ξ, η ) dξdωdη,P , f . = − δ β (cid:90) t (cid:90) s (cid:90) ω f t ( ξ, η ) dξdωdη, P , f . = − r β (cid:90) t (cid:90) s (cid:90) ω f ( ξ, η ) dξdωdη. Then H n = P n H = ( P , − + P , − + P , − + P , + P , + P , ) n st = n (cid:88) r =1 R r,n st, (3.6)where R r,n . = P i r,n ,j r,n P i r,n − ,j r,n − · · · P i r, ,j r, , i r,q ∈ { , } , j r,q ∈ {− , − , , } , for 1 ≤ q ≤ n . Observe that for positive integers m and nonnegative integers k , P , − s m t k = c , − s m +1 t k − , c , − = (cid:40) , k ≤ km +1 , else, (3.7) P , − s m t k = c , − s m +2 t k − , c , − = (cid:40) , k ≤ − k ( k − m +1)( m +2) , else, (3.8) P , − s m t k = c , − s m +2 t k − , c , − = (cid:40) , k ≤ − iαk β ( m +1)( m +2) , else, (3.9) P , s m t k = c , s m +1 t k , c , = 2 iα β ( m + 1) , (3.10) P , s m t k = c , s m +2 t k , c , = − δ β ( m + 1)( m + 2) , (3.11) P , s m t k = c , s m +2 t k +1 , c , = − r β ( m + 1)( m + 2)( k + 1) . (3.12)Let σ = σ ( r ) ≡ (cid:80) nq =1 j r,q . Then from (3.7)-(3.12), for eeach n and r , R r,n st = (cid:40) , if σ ≤ − ,C r,n s γ t σ +1 , if σ > − , (3.13)where n + 1 ≤ γ ≤ n + 1 and C r,n is a constant which only depends on r and n . Let M = max { , αβ , δβ , rβ } . We claim that for each r and n , | C r,n | ≤ M n ( n + 1)!( σ + 1)! . (3.14)Taking m = 1, k = 1 in (3.7)-(3.12) we see that (3.14) holds for n = 1. Supposeit holds for n = (cid:96) − r ∈ { , , . . . , (cid:96) − } . Then for n = (cid:96) and r ∗ ∈{ , , . . . , (cid:96) } , using (3.7) and (3.13), we get R r ∗ ,(cid:96) st = P i,j R r,(cid:96) − st = C r,(cid:96) − P i,j s γ t σ +1 = C r,(cid:96) − c i,j s γ ∗ t σ ∗ +1 TABILIZATION OF HIGHER ORDER SCHR ¨ODINGER EQUATIONS 43 for some i ∈ { , } , j ∈ {− , − , , } and r ∈ { , , .., (cid:96) − } , where γ ∗ is either γ + 1or γ + 2, σ ∗ = σ + j . By the induction assumption (3.14), C r,(cid:96) − ≤ M (cid:96) − (cid:96) !( σ + 1)! . Moreover using (3.7)-(3.12) and the fact that γ ≥ (cid:96) we see that | c i,j | ≤ M σ +1 (cid:96) +1 for j = − , − | c i, | < M(cid:96) +1 , and | c i, | < M ( σ +2)( (cid:96) +1) . Hence for each i ∈ { , } and j ∈ {− , − , , } we obtain | C r ∗ ,(cid:96) | = | C r, ( (cid:96) − c i,j | ≤ M (cid:96) ( (cid:96) + 1)!( σ + j + 1)! = M (cid:96) ( (cid:96) + 1)!( σ ∗ + 1)! , which proves that the claim holds for n = (cid:96) as well.Using (3.6), (3.13), (3.14) and the fact that 0 ≤ s, t ≤ L in the triangle ∆ s,t weobtain (cid:107) H n (cid:107) ∞ ≤ n M n L n +2 ( n + 1)! . (3.15)This shows H n is absolutely summable. On the other hand since H n is a linearcombination of 6 n monomials of the form s γ t σ +1 with γ ≤ n + 1 and σ ≤ n , anypartial derivative ∂ as ∂ bt H n of H n will be absolutely less than(2 n + 1) a ( n + 1) b n M n L n +2 − a − b ( n + 1)! , (3.16)which is a summable sequence. (cid:3) See Figure 11 for a graph and a contour plot of the kernel and Figure 12 for thecorresponding control gains for L = π , β = 1, α = 2, δ = 8 and r = 0 . Figure 11.
Backstepping kernel on ∆ x,y for L = π , β = 1, α = 2, δ = 8 and r = 0 . Figure 12.
Control gains for the Dirichlet (left) and Neumann (right)boundary conditions. L = π , β = 1, α = 2, δ = 8 and r = 0 . η = η ( x, y ) be a C ∞ − function defined on ∆ x,y and Υ η : H l (0 , L ) → H l (0 , L ), l ≥ η ϕ ]( x ) . = (cid:90) x η ( x, y ) ϕ ( y ) dy. Then, we have the following lemma for the operator I − Υ η . Lemma 3.2. I − Υ η is invertible with a bounded inverse from H l (0 , L ) → H l (0 , L ) ( l ≥ ). Moreover, ( I − Υ η ) − can be written as I + Φ , where Φ is a bounded operatorfrom L (0 , L ) into H l (0 , L ) for l = 0 , , and from H l − (0 , L ) into H l (0 , L ) for l > . We omit the proof since it can be done as in [20, 24].3.2.
Wellposedness.
We first investigate the local and global wellposedness of thetarget model. Then, using Lemma 3.1 and Lemma 3.2, we deduce the wellposednessof the original plant (1.1). To this end, let us consider the modified target model iw t + iβw xxx + αw xx + iδw x + irw = iβk y ( x, w x (0 , t ) , x ∈ (0 , L ) , t ∈ (0 , T ) ,w (0 , t ) = w ( L, t ) = w x ( L, t ) = 0 ,w ( x,
0) = w ( x ) . = u ( x ) − (cid:82) x k ( x, y ) u ( y, t ) dy. (3.17)Consider the operator A defined in (2.2) with domain D ( A ) defined in (2.3). Let usexpress (3.17) in the abstract operator theoretic form as (cid:40) ˙ y = Ay + F y,y (0) = y , TABILIZATION OF HIGHER ORDER SCHR ¨ODINGER EQUATIONS 45 where
F ϕ . = − irϕ + iβa ( · )Γ ϕ and Γ is the first order trace operator at the leftend point. Operator A generates strongly continuous semigroup of contractions, { S ( t ) } t ≥ , in L (0 , L ) [8]. Define the operator w = [Ψ z ]( t ) . = S ( t ) w + (cid:90) t S ( t − s ) F z ( s ) ds (3.18)and the space Y T . = { z ∈ X T : z x ∈ C ([0 , L ]; L (0 , T )) } (3.19)endowed with the norm (cid:107) z (cid:107) Y T . = (cid:16) (cid:107) z (cid:107) C ([0 ,T ]; L (0 ,L )) + (cid:107) z (cid:107) L (0 ,T ; H (0 ,L )) + (cid:107) z x (cid:107) C ([0 ,L ]; L (0 ,T )) (cid:17) . (3.20)We prove the following result. Proposition 3.3 (Local wellposedness) . Let T (cid:48) > and w ∈ L (0 , L ) . Then, thereexists T ∈ (0 , T (cid:48) ) which is independent of size of w such that (3.17) possesses aunique local solution w ∈ Y T .Proof. We first show that Ψ, defined by (3.18) maps Y T into itself. To see this, firstof all, we obtain from (3.18) that (cid:107) w (cid:107) Y T = (cid:107) Ψ z (cid:107) Y T ≤ (cid:107) S ( t ) w (cid:107) Y T + (cid:13)(cid:13)(cid:13)(cid:13)(cid:90) t S ( t − s )[ F z ]( s ) ds (cid:13)(cid:13)(cid:13)(cid:13) Y T . By using Lemma 2.1, the first term at the right hand side of the above inequalitycan be estimated as (cid:107) S ( t ) w (cid:107) Y T (cid:46) √ T (cid:107) w (cid:107) . (3.21)Using Lemma 2.2 and then applying Cauchy–Schwarz inequality, the second termcan be estimated as (cid:13)(cid:13)(cid:13)(cid:13)(cid:90) t S ( t − s ) F z ( s ) ds (cid:13)(cid:13)(cid:13)(cid:13) Y T ≤ C √ T (cid:107) − rz + βk y ( · , z x (0 , · ) (cid:107) L (0 ,T ; L (0 ,L )) ≤ c β,r (cid:112) T (1 + T )(1 + (cid:107) k y ( · , (cid:107) ) (cid:107) z (cid:107) Y T . (3.22)Combining (3.21) and (3.22), we see that Ψ maps Y T into itself. To see that Ψ iscontraction on Y T , let z , z ∈ Y T and w = Ψ z , w = Ψ z . Using the similararguments as above, we get (cid:107) w − w (cid:107) Y T = (cid:107) Ψ z − Ψ z (cid:107) Y T ≤ c β,r (cid:112) T (1 + T )(1 + (cid:107) k ( · , (cid:107) ) (cid:107) z − z (cid:107) Y T . In order for the map Ψ to be a contraction, we choose T ∈ (0 , T (cid:48) ) such that 0 < (cid:112) T (1 + T ) ≤ ( c β,r (1 + (cid:107) k ( · , (cid:107) )) − which is independent of the size of the initialdatum. This guarantees the existence of a unique local solution w ∈ Y T . (cid:3) This proposition shows the existence of a maximal time, T max , of the existence ofthe solution w ∈ Y T for all T < T max . To prove that w is global, it is enough to showthat lim T → T − max (cid:107) w (cid:107) Y T < ∞ . Proposition 3.4 (Global wellposedness) . Let w ∈ L (0 , L ) . Then w extends as aglobal solution in Y T .Proof. Taking L − inner product of the main equation of (3.17) by 2 w , taking theimaginary parts of both sides and applying several integration by parts together withimposing the boundary conditions, we derive ddt (cid:107) w ( · , t ) (cid:107) + β | w x (0 , t ) | + 2 r (cid:107) w ( · , t ) (cid:107) = 2 β (cid:60) (cid:90) L k y ( x, w x (0 , t ) w ( x, t ) dx. (3.23)By using (cid:15) − Young’s inequality, right hand side of (3.23) can be estimated as2 β (cid:60) (cid:90) L k y ( x, w x (0 , t ) w ( x, t ) ≤ β (cid:15) (cid:90) L | k y ( x, | | w ( x, t ) | dx + 2 (cid:15)βL | w x (0 , t ) | Choosing (cid:15) = L , (3.23) becomes ddt (cid:107) w ( · , t ) (cid:107) + β | w x (0 , t ) | ≤ βL (cid:107) k y ( · , (cid:107) ∞ − r ) (cid:107) w ( · , t ) (cid:107) . Now integrating the above inequality over (0 , t ) yields2 (cid:107) w ( · , t ) (cid:107) + β (cid:90) t | w x (0 , τ ) | dτ ≤ (cid:107) w (cid:107) + 4( βL (cid:107) k y ( · , (cid:107) ∞ − r ) (cid:90) t (cid:107) w ( · , τ ) (cid:107) dτ. (3.24)Define E ( t ) . = 2 (cid:107) w ( · , t ) (cid:107) + β (cid:82) t | w x (0 , τ ) | dτ . Then, from (3.24) E ( t ) ≤ (cid:107) w (cid:107) + 4 (cid:12)(cid:12) βL (cid:107) k y ( · , (cid:107) ∞ − r (cid:12)(cid:12) (cid:90) t E ( τ ) dτ. Thanks to Gronwall’s inequality, E ( t ) = 2 (cid:107) w ( · , t ) (cid:107) + β (cid:90) t | w x (0 , τ ) | dτ ≤ (cid:107) w (cid:107) e ( βL (cid:107) k y ( · , (cid:107) ∞ − r ) t , (3.25)for all t ∈ [0 , T ]. Passing to supremum on [0 , T ] and then letting T → T − max , we getlim T → T max (cid:107) w (cid:107) C ([0 ,T ]; L (0 ,L )) ≤ (cid:107) w (cid:107) e ( βL (cid:107) k ( · , (cid:107) ∞ − r ) T max < ∞ . (3.26)Using sup ≤ t ≤ T (cid:107) w ( · , t ) (cid:107) = 1 T (cid:107) w (cid:107) L (0 ,T ; L (0 ,L )) (3.27) TABILIZATION OF HIGHER ORDER SCHR ¨ODINGER EQUATIONS 47 and then letting T → T − max , we also getlim T → T − max (cid:107) w (cid:107) L (0 ,T ; L (0 ,L )) ≤ (cid:112) T max (cid:107) w (cid:107) e ( βL (cid:107) k ( · , (cid:107) ∞ − r ) T max . (3.28)Next, we multiply the main equation of (3.17) by 2 xw , integrate over (0 , t ) × (0 , L ),consider the imaginary parts and apply several integration by parts to get (cid:90) L x | w ( x, t ) | dx + 3 β (cid:90) t (cid:90) L | w x ( x, τ ) | dxdτ + 2 r (cid:90) L x | w ( x, t ) | dx = (cid:90) L x | w ( x ) | dx + δ (cid:90) t (cid:90) L | w ( x, τ ) | dxdτ + 2 β (cid:90) t (cid:90) L xk y ( x, w x (0 , τ ) w ( x, τ ) dxdτ (3.29)Thanks to Cauchy–Schwarz inequality, the last term at the right hand side of (3.29)can be estimated as2 β (cid:90) t (cid:90) L xk y ( x, x ) w x (0 , τ ) y ( x, τ ) dxdτ ≤ βL (cid:90) t (cid:90) L | k y ( x, || w x (0 , τ ) || w ( x, τ ) | dxdτ ≤ βL (cid:107) k y ( · , (cid:107) ∞ (cid:90) t | w x (0 , τ ) | dτ + βL (cid:90) t (cid:90) L | w ( x, τ ) | dxdτ. Dropping the first and third terms at the left hand side of (3.29), and using theabove estimate, it follows that (cid:107) w x (cid:107) L (0 ,t ; L (0 ,L )) ≤ L β (cid:107) w (cid:107) + βL + δ β (cid:90) t (cid:90) L | w ( x, τ ) | dxdτ + L (cid:107) k y ( · , (cid:107) ∞ (cid:90) t | y x (0 , τ ) | dτ ≤ L β (cid:107) w (cid:107) + (cid:18) βL + δ β + L (cid:107) k y ( · , (cid:107) ∞ β (cid:19) (cid:90) t E ( τ ) dτ. Using (3.25) we get,lim T → T max (cid:107) w x (cid:107) L (0 ,T ; L (0 ,L )) ≤ (cid:115) L β (cid:107) w (cid:107) + (cid:32)(cid:115) βL + δ β + L (cid:107) k x ( · , (cid:107) ∞ √ β (cid:33) (cid:112) T max (cid:107) w (cid:107) e ( βL (cid:107) k y ( · , (cid:107) ∞ − r ) T max . (3.30) Combining (3.28) and (3.30), we deduce thatlim T → T − max (cid:107) w (cid:107) L (0 ,T ; H (0 ,L )) ≤ (cid:115) L β (cid:107) w (cid:107) + (cid:32) (cid:115) βL + δ β + L (cid:107) k y ( · , (cid:107) ∞ √ β (cid:33) (cid:112) T max (cid:107) w (cid:107) e ( βL (cid:107) k y ( · , (cid:107) ∞ − r ) T max < ∞ . (3.31)From Proposition 3.3, w is the fixed point of (3.17), so it satisfies w = S ( t ) w + (cid:90) t S ( t − τ )[ F w ]( τ ) dτ for some t ∈ (0 , T (cid:48) ). From Lemma 2.1-(iii) and 2.6-(iii), we now thatsup x ∈ [0 ,L ] (cid:107) ∂ x [ S ( t ) w ]( x ) (cid:107) L (0 ,T ) (cid:46) √ T (cid:107) w (cid:107) (3.32)and sup x ∈ [0 ,L ] (cid:13)(cid:13)(cid:13)(cid:13) ∂ x (cid:20)(cid:90) t S ( t − τ )[ F w ]( τ ) dτ (cid:21) ( x ) (cid:13)(cid:13)(cid:13)(cid:13) L (0 ,T ) (cid:46) √ T (cid:90) T (cid:107) [ F w ]( · , t ) (cid:107) dt (3.33)holds. Using the definition of F w , right hand side of (3.33) can be estimated as √ T (cid:90) T (cid:107) [ F w ]( · , t ) (cid:107) L (0 ,L ) dt ≤ √ T r (cid:90) T (cid:107) w ( · , t ) (cid:107) dt + √ T β (cid:107) k ( · , (cid:107) (cid:90) T | w x (0 , t ) | dt ≤ √ T r (cid:90) T (cid:112) E ( t ) dt + √ T (cid:107) k ( · , (cid:107) (cid:112) E ( t ) ≤ (cid:32) T √ T r √ T β (cid:107) k ( · , (cid:107) (cid:33) √ (cid:107) w (cid:107) e ( βL (cid:107) k y ( · , (cid:107) ∞ − r ) T . (3.34)Finally using (3.32)-(3.34)lim T → T − max (cid:107) w x (cid:107) C ([0 ,L ]; L (0 ,T )) (cid:46) (cid:112) T max (cid:107) w (cid:107) + (cid:32) T √ T r √ T β (cid:107) k ( · , (cid:107) (cid:33) √ (cid:107) w (cid:107) e ( βL (cid:107) k y ( · , (cid:107) ∞ − r ) T < ∞ . This completes the proof. (cid:3)
Choosing w ∈ H (0 , L ) that satisfies compatibility conditions, the global solutionenjoys higher order regularity given by the following proposition. TABILIZATION OF HIGHER ORDER SCHR ¨ODINGER EQUATIONS 49
Proposition 3.5 (Regularity) . Let w ∈ H (0 , L ) satisfy the compatibility condi-tions. Then w ∈ Y T .Proof. Let v = w t and consider the following problem iv t + iβv xxx + αv xx + iδv x + irv = iβk y ( x, v x (0 , t ) , x ∈ (0 , L ) , t ∈ (0 , T ) ,v (0 , t ) = v ( L, t ) = v x ( L, t ) = 0 ,v ( x,
0) = v ( x ) , where v ( x ) . = − βw (cid:48)(cid:48)(cid:48) ( x ) + iαw (cid:48)(cid:48) ( x ) − δw (cid:48) ( x ) − rw ( x ) + βk y ( x, w (cid:48) (0). For a given v ∈ L (0 , L ), we know from Proposition 3.3 that v ∈ Y T . Set w ( x, t ) = w ( x ) + (cid:82) t v ( x, τ ) dτ . Under the compatibility conditions, one can show that w solves (3.17).From the main equation of (3.17), we have iβw xxx ( x, t ) = ( − iv − αw xx − iδw x − irw )( x, t ) + iβk y ( x, w x (0 , t ) . Observe that w x (0 , t ) = − (cid:82) L w xx ( x, t ) dx . Using this in the above expression andthen taking L − norms of both sides with respect to x , we get β (cid:107) w xxx ( · , t ) (cid:107) ≤ (cid:107) v ( · , t ) (cid:107) + (cid:0) α + β (cid:107) k y ( · , (cid:107) (cid:1) (cid:107) w xx ( · , t ) (cid:107) + δ (cid:107) w x ( · , t ) (cid:107) + r (cid:107) w ( · , t ) (cid:107) . Similar work as we did on (2.83)-(2.84) yields (cid:107) w xxx ( · , t ) (cid:107) (cid:46) (cid:107) v ( · , t ) (cid:107) + (cid:107) w ( · , t ) (cid:107) . Taking supremum on both sides, we obtain w ∈ C ([0 , T ]; H (0 , L )). Next, we dif-ferentiate the main equation of (3.17) with respect to x and take L − norm of bothsides with respect to x to get β (cid:107) w xxxx ( · , t ) (cid:107) ≤(cid:107) v x ( · , t ) (cid:107) + α (cid:107) w xxx ( · , t ) (cid:107) + (cid:0) δ + β (cid:107) k yx ( · , (cid:107) (cid:1) (cid:107) w xx ( · , t ) (cid:107) + r (cid:107) w x ( · , t ) (cid:107) . Proceeding as in (2.86)-(2.87), we get (cid:107) w xxxx ( · , t ) (cid:107) (cid:46) (cid:107) v x ( · , t ) (cid:107) + (cid:107) w x ( · , t ) (cid:107) . Now the right hand side belongs to L (0 , T ), so w belongs to L (0 , T ; H (0 , L )).Combining with the previous result, we deduce that w ∈ X T if w ∈ H (0 , L ). (cid:3) Now the first part of Theorem 1.2 follows from the fact that backstepping kernel isa smooth function over a compact set and backstepping transformation is invertibleon L (0 , L ) and H (0 , L ). Stability.
In this part, we obtain exponential stability for the original plant.This will be done by first obtaining the exponential stability result for the modifiedtarget model (3.17). Thanks to the invertibility of the backstepping transformation,this result will imply the exponential decay of solutions of the original plant.
Proposition 3.6.
Let β > , α, δ ∈ R , k be a smooth backstepping kernel solving (1.10) . Then for sufficiently small r > , there exists λ = β (cid:16) rβ − (cid:107) k y ( · , r ) (cid:107) (cid:17) > such that solution, w , of (1.12) satisfies the following decay estimate (cid:107) w ( · , t ) (cid:107) ≤ (cid:107) w (cid:107) e − λt for t ≥ .Proof. Taking the L − inner product of the main equation of (1.12) by 2 w and pro-ceeding as in (2.5)-(2.8), we get ddt (cid:107) w ( · , t ) (cid:107) + β | w x (0 , t ) | + 2 r (cid:107) w ( · , t ) (cid:107) = 2 β (cid:60) (cid:18) w x (0 , t ) (cid:90) L k y ( x, w ( x, t ) dx (cid:19) . (3.35)Using (cid:15) − Young’s inequality and then the Cauchy–Schwarz inequality, the term atthe right hand side can be estimated as2 β (cid:60) (cid:90) L k y ( x, w x (0 , t ) w ( x, t ) dx ≤ β (cid:32) (cid:15) | w x (0 , t ) | + (cid:15) (cid:18)(cid:90) L | k y ( x, || w ( x, t ) | dx (cid:19) (cid:33) ≤ β (cid:15) | w x (0 , t ) | + 2 β(cid:15) (cid:107) k y ( · , (cid:107) (cid:107) w ( · , t ) (cid:107) . (3.36)Combining this estimate with (3.35) and choosing (cid:15) = , we get ddt (cid:107) w ( · , t ) (cid:107) + 2 β (cid:18) rβ − (cid:107) k y ( · , (cid:107) (cid:19) (cid:107) w ( · , t ) (cid:107) ≤ , which implies (cid:107) w ( · , t ) (cid:107) (cid:46) (cid:107) w (cid:107) e − λt , λ . = β (cid:18) rβ − (cid:107) k y ( · , (cid:107) (cid:19) . (3.37)Next we prove that λ > α = αβ , ˜ δ = δβ , ˜ r = rβ and let M = max { , ˜ α, ˜ δ, ˜ r } . Differentiating (3.5) with respectto t , taking i = 0 and passing to limit as n → ∞ , we obtain G t ( s, t ) = ˜ r ∞ (cid:88) n =0 H nt ( s, t ) . (3.38) TABILIZATION OF HIGHER ORDER SCHR ¨ODINGER EQUATIONS 51
Then, considering ˜ r <
1, we first see that M is independent of ˜ r . Using this, we getfrom (3.16) that the term inside the summation (3.38) is absolutely less than someconstant c L, ˜ α, ˜ δ which is independent of ˜ r . So from (3.38), we get (cid:107) G t (cid:107) L ∞ (∆ s,t ) ≤ ˜ rc L, ˜ α, ˜ δ , and therefore we have (cid:107) k y ( · , (cid:107) ≤ L (cid:107) k y ( · , (cid:107) ∞ ≤ L (cid:107) G t ( · , (cid:107) ∞ ≤ L (cid:107) G t (cid:107) L ∞ (∆ s,t ) ≤ L ˜ r c L, ˜ α, ˜ δ . Using this estimate, we get λ = 2 β (cid:18) ˜ r − (cid:107) k y ( · , (cid:107) (cid:19) = 2 β ˜ r (cid:18) r − (cid:107) k y ( · , (cid:107) r (cid:19) ≥ β ˜ r (cid:32) r − Lc L, ˜ α, ˜ δ (cid:33) , which remains positive for sufficiently small r . (cid:3) Now using (1.11) and the fact that k = k ( x, y ) is a smooth function on a compactset ∆ x,y , we have (cid:107) w (cid:107) ≤ (cid:0) (cid:107) k ( · , · ) (cid:107) L (∆ x,y ) (cid:1) (cid:107) u (cid:107) . (3.39)Moreover, using the invertibility of the backstepping transformation given by Lemma3.2, we have (cid:107) u ( · , t ) (cid:107) ≤ (cid:107) I − Υ k (cid:107) → (cid:107) w ( · , t ) (cid:107) . (3.40)Combining (3.39) and (3.40), we deduce (cid:107) u ( · , t ) (cid:107) ≤ (cid:107) I − Υ k (cid:107) → (cid:0) (cid:107) k ( · , · ) (cid:107) L (∆ x,y ) (cid:1) (cid:107) u (cid:107) e − λt . So we conclude the proof of the second part of Therem 1.2.Table 1 below shows some values of r and corresponding decay rates λ . Resultsare obtained by choosing β = 1, α = 2, δ = 8, domain length L = π and N = 1001spatial node points. 4. Observer design
In this section, our aim is to prove the wellposedness and exponential stability ofthe plant–observer–error system. r λ = β (cid:18) rβ − (cid:107) k y ( · , r ) (cid:107) (cid:19) .
001 0 . .
01 0 . .
02 0 . .
03 0 . .
04 0 . .
05 0 . . . . − . . − . − . Table 1.
Some numerical values for the decay rate λ correspondingto various values of r .4.1. Wellposedness.
We start by the wellposedness analysis of the error model(1.15). To this end, we first study the target error model given by (1.21) and thenuse the invertibility of the transformation (1.20) and deduce that same results alsohold for (1.15). To see that (1.20) is invertible with a bounded inverse, we changevariables as s = L − y and t = L − x on (1.19), and obtain that p = p ( x, y ) solves(1.19) if and only if p ( x, y ) ≡ H ( s, t ) solves β ( H sss + H ttt ) − iα ( H ss − H tt ) + δ ( H s + H t ) − rH = 0 ,H ( s, s ) = H ( s,
0) = 0 ,H s ( s, s ) = − rs β . Observe that this model is exactly the same model given in (1.10) except that r isreplaced by − r . Therefore, we obtain the following relation p ( x, y ) = H ( s, t ) = k ( s, t ; − r ) = k ( L − y, L − x ; − r ) , (4.1)where k solves (1.10). Consequently existence of smooth kernel p = p ( x, y ) is guar-anteed and (1.20) is invertible with a bounded inverse. See Figure 13 for a graphand a contour plot of p ( x, y ) for L = π , β = 1, α = 2, δ = 8 and r = 0 . TABILIZATION OF HIGHER ORDER SCHR ¨ODINGER EQUATIONS 53
Figure 13. p ( x, y ) defined on ∆ x,y for L = π , β = 1, α = 2, δ = 8and r = 0 . Error model.
Let us prove the wellposedness of the target error model. Tothis end, let us first consider the following model i ˜ w t + iβ ˜ w xxx + α ˜ w xx + iδ ˜ w x + ir ˜ w = 0 , x ∈ (0 , L ) , t ∈ (0 , T ) , ˜ w (0 , t ) = ˜ w ( L, t ) = 0 , ˜ w x ( L, t ) = ˜ ψ ( t ) , ˜ w ( x,
0) = ˜ w ( x ) . (4.2)Note that the function y , defined by y ( · , t ) . = e rt ˜ w ( · , t ) together with the initial andboundary conditions ˜ w and ψ ( t ) . = e rt ˜ ψ ( t ), satisfies the results obtained in Lemma2.1 and Lemma 2.6. Thus, for given ˜ w ∈ H (0 , L ), ψ ∈ H (0 , T ) satisfying thehigher order compability conditions, Lemma 2.8 implies that (cid:107) y (cid:107) X T + (cid:107) y tt (cid:107) X T (cid:46) (cid:107) ˜ w (cid:107) H (0 ,L ) + e cT (cid:107) ψ (cid:107) H (0 ,T ) . (4.3)Notice that the original boundary condition of the problem (1.21) is of feedback type,given by ψ ( t ) = ψ ( ˜ w )( t ) = (cid:90) L p x ( L, y ) ˜ w ( y, t ) dy. We will treat the wellposedness of the target error model by using a fixed pointargument. To this end, let us define the Banach space Q T ≡ { ˜ w ∈ X T : ˜ w tt ∈ X T } and its complete metric subspace ˜ Q T ≡ { ˜ w ∈ Q T : ˜ w ( · ,
0) = ˜ w ( · ) } equipped withthe metric induced by the norm associated with Q T . Since p = p ( x, y ) is a smooth solution of (1.19), for a given ˜ w ∗ ∈ ˜ Q T , we have (cid:107) ψ ( ˜ w ∗ ) (cid:107) H (0 ,T ) = (cid:13)(cid:13)(cid:13)(cid:13)(cid:90) L p x ( L, y ) ˜ w ∗ ( y, · ) dy (cid:13)(cid:13)(cid:13)(cid:13) H (0 ,T ) ≤ √ T (cid:107) p x ( L, · ) (cid:107) (cid:88) j =0 (cid:107) ∂ jt ˜ w ∗ (cid:107) X T < ∞ . Thus by the Lemma 2.8, for ψ ( ˜ w ∗ )( t ) ∈ H (0 , T ), the problem (2.1) with f ≡ Q T → ˜ Q T , Ψ ˜ w ∗ = ˜ w . Now let˜ w , ˜ w ∈ ˜ Q T . Using the estimates (4.3), we get (cid:107) Ψ ˜ w − Ψ ˜ w (cid:107) ˜ Q T ≤ C (cid:107) ψ ( ˜ w ) − ψ ( ˜ w ) (cid:107) H (0 ,T ) ≤ C √ T (cid:107) ˜ w − ˜ w (cid:107) ˜ Q T . For sufficiently small T , we can guarantee that the mapping Ψ : ˜ Q T → ˜ Q T is con-traction. Thanks to the Banach fixed point theorem, this yields the existence of aunique local solution of the target error system. As we show in Proposition 4.3 inthe following section, the local solution remains uniformly bounded in time. Thisyields the unique global solution and we have the following proposition. Proposition 4.1.
Let
T, β > , α, δ ∈ R , p be a smooth backstepping kernel solving (1.19) and ( ˜ w , ψ ) ∈ H (0 , L ) × H (0 , T ) satisfies higher order compatibility condi-tions where ψ = ψ ( ˜ w ) = (cid:82) L p x ( L, y ) ˜ w ( y, t ) dy . Then (1.21) admits a unique globalsolution ˜ w ∈ X T . Thanks to the bounded invertibility of the backstepping transformation (1.20), weobtain under the same assumptions that ˜ u ∈ X T .4.1.2. Observer model.
Consider the target observer model 1.23 i ˆ w t + iβ ˆ w xxx + α ˆ w xx + iδ ˆ w x + ir ˆ w = iβk y ( x,
0) ˆ w x (0 , t )+ f ( x, t ) , x ∈ (0 , L ) , t ∈ (0 , T ) , ˆ w (0 , t ) = ˆ w ( L, t ) = ˆ w x ( L, t ) = 0 , ˆ w ( x,
0) = ˆ w ( x ) . (4.4)where f ( x, t ) . = [( I − Υ k ) p ]( x ) ˜ w x (0 , t ) + [( I − Υ k ) p ]( x ) ˜ w xx (0 , t ). Recall that thebackstepping transformation (1.20) transforms (1.14) to (4.4) if p , p are chosen suchthat p ( x ) iβp y ( x, − αp ( x,
0) and p ( x ) = − iβp ( x,
0) where p is the backsteppingkernel that solves (1.19). An example for the real and imaginary parts of the observergains for a problem defined on [0 , π ] and the coefficients β = 1 , α = 2 , δ = 8 , r = 0 . TABILIZATION OF HIGHER ORDER SCHR ¨ODINGER EQUATIONS 55
Figure 14.
Observer gains for L = π , β = 1, α = 2, δ = 8 and r = 0 . w ∈ L (0 , L ), let us first show that ˆ w ∈ X T . Thanks to Proposition3.4, given ˆ w ∈ L (0 , L ), we know that solution, ˆ w , of 4.4 with f ≡ X T . Let us express it as ˆ w i ( x, t ) . = W ( t ) ˆ w ( x ) and now consider the problemwhere ˆ w ≡
0. Let us express its solution asˆ w f ( x, t ) . = (cid:90) t W ( t − τ ) f ( x, τ ) dτ. Recall that k = k ( x, y ), p ( x ) = − iβp y ( x,
0) + αp ( x,
0) and p ( x ) = iβp ( x,
0) aresmooth functions. Also, we will see in Proposition 4.3-(ii) below that, if ˜ w ∈ H (0 , L ), then ˜ w x (0 , t ) , ˜ w xx (0 , t ) ∈ L (0 , T ). This implies that f ∈ L (0 , T ; H ∞ (0 , L )).Now to see that ˆ w f ∈ X T , first observe that (cid:107) ˆ w f ( · , t ) (cid:107) ≤ (cid:90) t (cid:107) W ( t − τ ) f ( · , τ ) (cid:107) dτ ≤ (cid:90) t (cid:107) f ( · , τ ) (cid:107) dτ = (cid:107) f (cid:107) L (0 ,T ; L (0 ,L )) . (4.5)Taking supremum in t ∈ [0 , T ] yields (cid:107) ˆ w f (cid:107) C ([0 ,T ]; L (0 ,L )) ≤ (cid:107) f (cid:107) L (0 ,T ; L (0 ,L )) ≤ (cid:107) f (cid:107) L (0 ,T ; H ∞ (0 ,L )) . Following from (4.5), we also have (cid:107) ˆ w f (cid:107) L (0 ,T ; L (0 ,L )) ≤ √ T (cid:107) f (cid:107) L (0 ,T ; L (0 ,L )) ≤ √ T (cid:107) f (cid:107) L (0 ,T ; H ∞ (0 ,L )) . Using similar arguments, one can get (cid:107) ∂ x ˆ w f (cid:107) L (0 ,T ; L (0 ,L )) ≤ √ T (cid:107) f (cid:107) L (0 ,T ; H ∞ (0 ,L )) . Consequently, ˆ w f ∈ X T . As a conclusion given ˆ w ∈ L (0 , L ), ˜ w ∈ H (0 , L ) and f ∈ L (0 , T ; H ∞ (0 , L )), we obtain that ˆ w ∈ X T . Next, we show that ˆ w ∈ X T . To this end, we set ˜ z = ˜ w t . Then ˜ z satisfies i ˜ z t + iβ ˜ z xxx + α ˜ z xx + iδ ˜ z x + ir ˜ z = 0 , x ∈ (0 , L ) , t ∈ (0 , T ) , ˜ z (0 , t ) = ˜ z ( L, t ) = 0 , ˜ z x ( L, t ) = ψ (cid:48) (˜ z ) , ˜ z ( x,
0) = ˜ z ( x ) , where ˜ z . = − β ˜ w (cid:48)(cid:48)(cid:48) + iα ˜ w (cid:48)(cid:48) − δ ˜ w (cid:48) − r ˜ w . Applying the arguments in Section 4.1.1, wesee that if (˜ z , ψ (cid:48) ) ∈ H (0 , L ) × H (0 , T ) satisfies compatibility conditions, then z ∈ X T . Moreover, thanks to the Proposition 4.3-(ii) below, we have ˜ z x (0 , t ) , ˜ z xx (0 , t ) ∈ L (0 , T ). This implies by using ˜ z = ˜ w t that, ˜ w xt (0 , t ) , ˜ w xxt (0 , t ) ∈ L (0 , T ) where˜ w solves target error model satisfying ( ˜ w , ψ ) ∈ H (0 , L ) × H (0 , T ) higher ordercompatibility. Thus f ∈ W , (0 , T ; H ∞ (0 , L )).Now let us set ˆ v = ˆ w t . Then ˆ v solves i ˆ v t + iβ ˆ v xxx + α ˆ v xx + iδ ˆ v x + ir ˆ v = iβk y ( x, v x (0 , t )+ f t ( x, t ) , x ∈ (0 , L ) , t ∈ (0 , T ) , ˆ v (0 , t ) = ˆ v ( L, t ) = ˆ v x ( L, t ) = 0 , ˆ v ( x,
0) = ˆ v ( x ) , whereˆ v ( x ) . = − β ˆ w (cid:48)(cid:48)(cid:48) ( x ) + iα ˆ w (cid:48)(cid:48) ( x ) − δ ˆ w (cid:48) ( x ) − r ˆ w ( x ) + βk y ( x,
0) ˆ w (cid:48) (0) − if t ( x, . Assume that ˆ v ∈ L (0 , L ). Then, from the above study, we deduce that ˆ v ∈ X T . Ifˆ w satisfies the compatibility conditions, then we can also show that ˆ w , defined byˆ w ( x, t ) = ˆ w ( x ) + (cid:82) t ˆ v ( x, τ ) dτ solves (4.4). Now from the main equation of (4.4), wehave iβ ˆ w xxx ( x, t ) = ( − i ˆ v − α ˆ w xx − iδ ˆ w x − ir ˆ w )( x, t ) + iβk y ( x,
0) ˆ w x (0 , t ) + f ( x, t ) . Using ˆ w x (0 , t ) = − (cid:82) L ˆ w xx ( x, t ) dx and taking L − norms of both side we get β (cid:107) ˆ w xxx ( · , t ) (cid:107) ≤ (cid:107) ˆ v ( · , t ) (cid:107) + (cid:0) α + β (cid:107) k y ( · , (cid:107) (cid:1) (cid:107) ˆ w xx ( · , t ) (cid:107) + δ (cid:107) ˆ w x ( · , t ) (cid:107) + r (cid:107) ˆ w ( · , t ) (cid:107) + (cid:107) f ( · , t ) (cid:107) , Using Gagliardo–Nirenberg’s interpolation inequality and (cid:15) − Young’s inequality, sec-ond and third terms at the right hand side can be estimated as( α + β (cid:107) k y ( · , (cid:107) ) (cid:107) ˆ w xx ( · , t ) (cid:107) ≤ (cid:15) (cid:107) ˆ w xxx (cid:107) + c β,α,k,(cid:15) (cid:107) ˆ w ( · , t ) (cid:107) and δ (cid:107) ˆ w x ( · , t ) (cid:107) ≤ (cid:15) (cid:107) ˆ w xxx (cid:107) + c δ,(cid:15) (cid:107) ˆ w ( · , t ) (cid:107) respectively. Choosing (cid:15) > (cid:107) ˆ w xxx ( · , t ) (cid:107) (cid:46) (cid:107) ˆ v ( · , t ) (cid:107) + (cid:107) ˆ w ( · , t ) (cid:107) + (cid:107) f ( · , t ) (cid:107) . TABILIZATION OF HIGHER ORDER SCHR ¨ODINGER EQUATIONS 57
Notice from Proposition 4.3-(ii) that supremum of the trace terms ˜ w x (0 , t ) , ˜ w xx (0 , t )exist. Therefore, taking supremum on both sides, we deduce that ˆ w ∈ C ([0 , T ]; H (0 , L )).Next, again from the main equation, we have iβ ˆ w xxxx ( x, t ) = ( − i ˆ v x − α ˆ w xxx − iδ ˆ w xx − ir ˆ y x )( x, t ) + iβk yx ( x,
0) ˆ w x (0 , t ) + f x ( x, t ) , and therefore we get β (cid:107) ˆ w xxxx ( · , t ) (cid:107) ≤ (cid:107) ˆ v x ( · , t ) (cid:107) + α (cid:107) ˆ w xxx ( · , t ) (cid:107) + (cid:0) δ + β (cid:107) k xy ( · , (cid:107) (cid:1) (cid:107) ˆ w xx ( · , t ) (cid:107) + r (cid:107) ˆ w x ( · , t ) (cid:107) + (cid:107) f x ( · , t ) (cid:107) . Similarly, by Gagliardo–Nirenberg’s inequality and then (cid:15) − Young’s inequality, weget α (cid:107) ˆ w xxx ( · , t ) (cid:107) ≤ (cid:15) (cid:107) ˆ w xxxx ( · , t ) (cid:107) + c α,(cid:15) (cid:107) ˆ w ( · , t ) (cid:107) , (cid:0) δ + β (cid:107) k y ( · , (cid:107) (cid:1) (cid:107) ˆ w xx ( · , t ) (cid:107) ≤ (cid:15) (cid:107) ˆ w xxxx ( · , t ) (cid:107) + c β,δ,k,(cid:15) (cid:107) ˆ w ( · , t ) (cid:107) ,r (cid:107) ˆ w x ( · , t ) (cid:107) ≤ (cid:15) (cid:107) ˆ w xxxx ( · , t ) (cid:107) + c r,(cid:15) (cid:107) ˆ w ( · , t ) (cid:107) . Using these estimates, we obtain that (cid:107) ˆ w xxxx ( · , t ) (cid:107) (cid:46) (cid:107) ˆ v x ( · , t ) (cid:107) + (cid:107) ˆ w ( · , t ) (cid:107) + (cid:107) f x ( · , t ) (cid:107) . We see that right hand side belongs to L (0 , T ), so ˆ w belongs to L (0 , T ; H (0 , L )).Combining with the previous result, we obtain that ˆ w ∈ X T if ˆ w ∈ H (0 , L ). Thisfinishes the proof of the following proposition. Proposition 4.2.
Let
T, β > , α, δ ∈ R , k and p be smooth backstepping kernelssolving (1.10) and (1.19) respectively, and p ( x ) = − iβp y ( x,
0) + αp ( x, , p ( x ) = iβp ( x, . Assume that ˆ w ∈ H (0 , L ) satisfies the compatibility conditions and theinitial–boundary pair of the target error model ( ˜ w , ψ ) ∈ H (0 , L ) × H (0 , T ) satisfiesthe higher order compatibility conditions. Then (4.4) admits a unique solution ˆ w ∈ X T . Finally, thanks to bounded invertibility of the backstepping transformation (1.11),we obtain under the same assumptions that ˆ u ∈ X T . Combining the wellposedness ofˆ u and ˜ u , we obtained the wellposedness of (1.1) and proved the first part of Theorem1.5.4.2. Stability.
In this part, we obtain exponential stability estimates for the plant–observer–error system. This will be done by first considering the target error andtarget observer models. Then we use the bounded invertibility of the backsteppingtransformations, which will yield the exponential stability for the error and observermodels, consequently for the original plant.
Error model.
Proposition 4.3.
Let β > , α, δ ∈ R and p is the smooth backstepping kernel thatsolves (1.19) . Then for sufficiently small r > , it is true that µ = β (cid:18) rβ − (cid:107) p x ( L, · ; r ) (cid:107) (cid:19) > . Moreover, the solution ˜ w of (1.21) satisfies the following estimates (i) (cid:107) ˜ w ( · , t ) (cid:107) ≤ (cid:107) ˜ w (cid:107) e − µt , (ii) | ˜ w xx (0 , t ) | + | ˜ w x (0 , t ) | + (cid:107) ˜ w ( · , t ) (cid:107) H (0 ,L ) (cid:46) (cid:107) ˜ w (cid:107) H (0 ,L ) e − µt ,for t ≥ .Proof. (i) Taking the L − inner product of the main equation of (1.21) with 2 ˜ w , fol-lowing (2.5)-(2.8) and applying Cauchy–Schwarz inequality at the right handside, we get ddt (cid:107) ˜ w ( · , t ) (cid:107) + 2 r (cid:107) ˜ w ( · , t ) (cid:107) + β | ˜ w x (0 , t ) | = β | ˜ w x ( L, t ) | = β (cid:12)(cid:12)(cid:12)(cid:12)(cid:90) L p x ( L, y ) ˜ w ( y, t ) dy (cid:12)(cid:12)(cid:12)(cid:12) ≤ β (cid:107) p x ( L, · ) (cid:107) (cid:107) ˜ w ( · , t ) (cid:107) . It follows from the last expression that ddt (cid:107) ˜ w ( · , t ) (cid:107) + 2 β (cid:18) rβ − (cid:107) p x ( L, · ) (cid:107) (cid:19) (cid:107) ˜ w ( · , t ) (cid:107) ≤ . Denoting µ . = β (cid:16) rβ − (cid:107) p x ( L, · ) (cid:107) (cid:17) and integrating the above estimate yields (i).Recall that that p x ( L, y ) = − k y ( x, µ > r > t , take L − inner product by 2 ˜ w t andfollowing similar steps as in part (i), we obtain ddt (cid:107) ˜ w t ( · , t ) (cid:107) + 2 β (cid:18) rβ − (cid:107) p x ( L, · ) (cid:107) (cid:19) (cid:107) ˜ w t ( · , t ) (cid:107) ≤ , which implies (cid:107) ˜ w t ( · , t ) (cid:107) ≤ (cid:107) ˜ w t ( · , (cid:107) e − µt . (4.6) TABILIZATION OF HIGHER ORDER SCHR ¨ODINGER EQUATIONS 59
In particular, from the main equation of (1.21) together with (4.6), we get (cid:107) ˜ w t ( · , t ) (cid:107) ≤ (cid:107) ˜ w t ( · , (cid:107) e − µt = (cid:107) − β ˜ w (cid:48)(cid:48)(cid:48) + iα ˜ w (cid:48)(cid:48) − δ ˜ w (cid:48) − r ˜ w (cid:107) e − µt (cid:46) (cid:107) ˜ w (cid:107) H (0 ,L ) e − µt . (4.7)On the other hand, again from (1.21), we also have β (cid:107) ˜ w xxx ( · , t ) (cid:107) ≤ α (cid:107) ˜ w xx ( · , t ) (cid:107) + δ (cid:107) ˜ w x ( · , t ) (cid:107) + r (cid:107) ˜ w ( · , t ) (cid:107) + (cid:107) ˜ w t ( · , t ) (cid:107) . (4.8)Applying Gagliardo–Nirenberg interpolation inequality and then (cid:15) − Young’sinequality, the first term at the right hand side can be estimated as α (cid:107) ˜ w xx ( · , t ) (cid:107) ≤ c α (cid:107) ˜ w xxx ( · , t ) (cid:107) (cid:107) ˜ w ( · , t ) (cid:107) ≤ (cid:15) (cid:107) ˜ w xxx ( · , t ) (cid:107) + c α,(cid:15) (cid:107) ˜ w ( · , t ) (cid:107) . (4.9)Similarly, the second term can be estimated as δ (cid:107) ˜ w x ( · , t ) (cid:107) ≤ (cid:15) (cid:107) ˜ w xxx ( · , t ) (cid:107) + c δ,(cid:15) (cid:107) w ( · , t ) (cid:107) . (4.10)Combining (4.9)-(4.10) with (4.8), and then choosing (cid:15) > (cid:107) ˜ w ( · , t ) (cid:107) H (0 ,L ) (cid:46) (cid:107) ˜ w ( · , t ) (cid:107) + (cid:107) ˜ w t ( · , t ) (cid:107) , Using (4.7) and (i) , it follows that (cid:107) ˜ w ( · , t ) (cid:107) H (0 ,L ) (cid:46) (cid:107) ˜ w (cid:107) H (0 ,L ) e − µt . (4.11)To estimate the trace terms in (ii), we take L − inner product of (1.21) by2( L − x ) ˜ w xx and consider only the imaginary terms to get2 (cid:60) (cid:90) L ˜ w t ˜ w xx ( L − x ) dx + 2 β (cid:60) (cid:90) L ˜ w xxx ˜ w xx ( L − x ) dx + 2 α (cid:61) (cid:90) L ˜ w xx ˜ w xx ( L − x ) dx + 2 δ (cid:60) (cid:90) L ˜ w x ˜ w xx ( L − x ) dx + 2 r (cid:60) (cid:90) L ˜ w ˜ w xx ( L − x ) dx = 0 . (4.12)Integrating by parts, the second term is equivalent to2 β (cid:60) (cid:90) L ˜ w xxx ˜ w xx ( L − x ) dx = β (cid:60) (cid:90) L ddx | ˜ w xx | ( L − x ) dx = β (cid:0) − L | ˜ w xx (0 , t ) | + (cid:107) ˜ w xx ( · , t ) (cid:107) (cid:1) . The third term vanishes since it is pure real. The fourth term, again byintegration by parts, can be expressed as2 δ (cid:60) (cid:90) L ˜ w x ˜ w xx ( L − x ) dx = δ (cid:60) (cid:90) L ddx | ˜ w x | ( L − x ) dx = δ (cid:0) − L | ˜ w x (0 , t ) | + (cid:107) ˜ w x ( · , t ) (cid:107) (cid:1) . Using these estimates in (4.12), we obtain that L ( β | ˜ w xx (0 , t ) | + δ | ˜ w x (0 , t ) | ) =2 (cid:60) (cid:90) L ˜ w t ˜ w xx ( L − x ) dx + β (cid:107) ˜ w xx ( · , t ) (cid:107) + δ (cid:107) ˜ w x ( · , t ) (cid:107) + 2 r (cid:60) (cid:90) L ˜ w ˜ w xx ( L − x ) dx. Applying Cauchy–Schwarz inequality and then Young’s inequality on the firstand last terms at the right hand side, using (4.7) and (4.11), we get | ˜ w xx (0 , t ) | + | ˜ w x (0 , t ) | (cid:46) (cid:107) ˜ w t ( · , t ) (cid:107) + (cid:107) ˜ w ( · , t ) (cid:107) H (0 ,L ) (cid:46) e − µt (cid:107) ˜ w (cid:107) H (0 ,L ) . Combining this result with (4.11) yields (ii). (cid:3)
Since p is a smooth function on a compact set ∆ x,y and the backstepping trans-formation (1.19) is invertible on L (0 , L ) and H (0 , L ) with a bounded inverse, weobtain that (cid:107) ˜ u ( · , t ) (cid:107) ≤ c p (cid:107) ˜ u (cid:107) , c p = (cid:0) (cid:107) p (cid:107) L (∆ x,y ) (cid:1) (cid:107) ( I − Υ p ) − (cid:107) → (4.13)and (cid:107) ˜ u ( · , t ) (cid:107) H (0 ,L ) ≤ c p (cid:48) (cid:107) ˜ u (cid:107) H (0 ,L ) , c p (cid:48) = (cid:0) (cid:107) p (cid:107) H (∆ x,y ) (cid:1) (cid:107) ( I − Υ p ) − (cid:107) H (0 ,L ) → H (0 ,L ) . (4.14)4.2.2. Observer model.
Proposition 4.4.
Let β > , α, δ ∈ R and k, p be the smooth backstepping kernelssolving (1.10) , (1.19) respectively. Then for sufficiently small r, (cid:15) > , it is true that ν . = β (cid:18) rβ − (cid:107) k y ( · , r ) (cid:107) − (cid:15) (cid:0) (cid:107) Π (cid:107) + (cid:107) Π (cid:107) (cid:1)(cid:19) > , where (cid:107) Π j (cid:107) = (cid:107) ( I − Υ k ) p j (cid:107) , j = 1 , with p ( x ) = − iβp y ( x,
0) + αp ( x, , p ( x ) = iβp ( x, . Moreover, the solution ˆ w of (1.23) satisfies the following estimate (cid:107) ˆ w ( · , t ) (cid:107) (cid:46) e − νt (cid:0) (cid:107) ˆ w (cid:107) + (cid:107) ˜ w (cid:107) H (0 ,L ) (cid:1) (4.15) for t ≥ . TABILIZATION OF HIGHER ORDER SCHR ¨ODINGER EQUATIONS 61
Proof.
We take L − inner product of the main equation of (1.23) by 2 ˆ w and followingthe steps (2.5)-(2.8), we get ddt (cid:107) ˆ w ( · , t ) (cid:107) + β | ˆ w x (0 , t ) | + 2 r (cid:107) ˆ w ( · , t ) (cid:107) =2 β (cid:60) (cid:90) L k y ( x,
0) ˆ w x (0 , t ) ˆ w ( x, t ) dx + 2 (cid:61) (cid:90) L Π ( x ) ˜ w x (0 , t ) ˆ w ( x, t ) dx + 2 (cid:61) (cid:90) L Π ( x ) ˜ w xx (0 , t ) ˆ w ( x, t ) dx. (4.16)Using Young’s inequality and then Cauchy–Schwarz inequality, the first term at theright hand side can be estimated as2 β (cid:60) (cid:90) L k y ( x,
0) ˆ w x (0 , t ) ˆ w ( x, t ) dx ≤ β | ˆ w x (0 , t ) | + β (cid:107) k y ( · , (cid:107) (cid:107) ˆ w ( · , t ) (cid:107) . Applying Cauchy–Schwarz inequality and (cid:15) − Young’s inequality to the second andthird terms at the right hand side of (4.16), we get (cid:12)(cid:12)(cid:12)(cid:12) (cid:61) (cid:90) L Π j ( x ) ˜ w x (0 , t ) ˆ w ( x, t ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:15)β (cid:13)(cid:13) Π j (cid:13)(cid:13) (cid:107) ˆ w ( · , t ) (cid:107) + 12 (cid:15)β | ˜ w x (0 , t ) | , j = 1 , . Using these estimates in (4.16), we obtain that ddt (cid:107) ˆ w ( · , t ) (cid:107) + 2 β (cid:18) rβ − (cid:107) k y ( · , (cid:107) − (cid:15) (cid:0) (cid:107) Π (cid:107) + (cid:107) Π (cid:107) (cid:1)(cid:19) (cid:107) ˆ w ( · , t ) (cid:107) ≤ (cid:15)β (cid:0) | ˜ w x (0 , t ) | + | ˜ w xx (0 , t ) | (cid:1) . (4.17)From Proposition 3.6 we know that, there exists a sufficiently small r > (cid:16) rβ − (cid:107) k y ( · , (cid:107) (cid:17) remains positive. So choosing (cid:15) sufficiently small, we areable to guarantee that the term ν . = β (cid:18) rβ − (cid:107) k y ( · , (cid:107) − (cid:15) (cid:0) (cid:107) Π (cid:107) + (cid:107) Π (cid:107) (cid:1)(cid:19) remains positive. Now applying Proposition 4.3-(ii) to the right hand side of (4.17),we get ddt (cid:107) ˆ w ( · , t ) (cid:107) + 2 ν (cid:107) ˆ w ( · , t ) (cid:107) (cid:46) (cid:107) ˜ w (cid:107) H (0 ,L ) e − µt . Also, observing (cid:107) p x ( L, · ) (cid:107) = (cid:107) k y ( · , (cid:107) and comparing ν with µ , we observe that µ > ν . Thus, integrating the above inequality from 0 to t , we finally obtain (cid:107) ˆ w ( · , t ) (cid:107) (cid:46) e − νt (cid:0) (cid:107) ˆ w (cid:107) + (cid:107) ˜ w (cid:107) H (0 ,L ) (cid:1) . (cid:3) Since k , p are smooth backstepping kernels on the triangular domain ∆ x,y andthanks to the invertibility of the corresponding backstepping transformations (1.10),(1.19) on L (0 , L ) and H (0 , L ) respectively, with a bounded inverse, we deduce that (cid:107) ˆ u ( · , t ) (cid:107) (cid:46) c k,p e − νt (cid:0) (cid:107) ˆ u (cid:107) + (cid:107) ˜ u (cid:107) H (0 ,L ) (cid:1) , (4.18)where c k,p is the maximum of c k = (cid:0) (cid:107) k (cid:107) L (∆ x,y ) (cid:1) (cid:107) ( I − Υ k ) − (cid:107) → and c p = (cid:0) (cid:107) p (cid:107) H (∆ x,y ) (cid:1) (cid:107) ( I − Υ p ) − (cid:107) H (0 ,L ) → H (0 ,L ) . Finally, combining (4.13) and (4.18) (cid:107) u ( · , t ) (cid:107) = (cid:107) (ˆ u + ˜ u )( · , t ) (cid:107) ≤ (cid:107) ˆ u ( · , t ) (cid:107) + (cid:107) ( u − ˆ u )( · , t ) (cid:107) (cid:46) c k,p (cid:0) (cid:107) ˆ u (cid:107) + (cid:107) u − ˆ u (cid:107) H (0 ,L ) (cid:1) e − νt + c p (cid:107) u − ˆ u (cid:107) e − µt . This gives us the second part of Theorem 1.5.5.
Numerical simulations
In this part, we present our numerical algorithm and numerical simulations forcontroller and observer designs.5.1.
Controller design.
Our algorithm consists of three steps. We first obtain anapproximation for the backstepping kernel k by solving the integral equation (3.1).Then we solve the modified target equation (1.12) numerically. As a third and finalstep, we use the invertibility of the backstepping transformation and end up withthe numerical solution to the original plant. Details are given in the below. Step i.
We solve the integral equation G j +1 ( s, t ) = r β st + (cid:90) t (cid:90) s (cid:90) ω [ DG j ]( ξ, η ) dξdωdη, j = 1 , , . . . iteratively, where the iteration is initialized with G ( s, t ) = r β st. As the initial function is a polynomial, the result of the each iteration yieldsagain a polynomial. Thus, here, we use the advantage of the fact that sum-mation and multiplication with a scalar of polynomials, their differentiationand integration can be carried out easily by simple algebraic operations. To
TABILIZATION OF HIGHER ORDER SCHR ¨ODINGER EQUATIONS 63 perform these operations computationally, we express a given n − th degreepolynomial with complex coefficients, say P ( s, t ) = α , + α , s + α , t + α , s + α , st + α , t + · · · + α n, s n + α n − , s n − t + α n − , s n − t + · · · + α ,n t n , (5.1)in a more convenient form as[P] = α , α , · · · α ,n − α ,n α , α , · · · α ,n − ... ... ... α n − , α n − , α n, . (5.2)Once we introduce this matrix representation (5.2) of P in our algorithm, thenit is easy to perform summation and scalar multiplication. Moreover, usingthe elementary row and column operations, one can perform differentiationand integration. For instance multiplying the j − th row of [ P ] by j − j − − th row and repeating this process for each j , j = 2 , , . . . , n + 1 yields the matrix representation of P s ( s, t ), [ P s ]. Similarly,multiplying the j − th row of [ P ] by 1 /j , writing the result to the ( j + 1) − throw and repeating this process for each j , j = 1 , , . . . , n + 1 yields [ (cid:82) s P ds ].Differentiation and integration with respect to t can be done by performinganalogous column operations. Step ii.
Let us consider the uniform discretization of [0 , L ] with the set of
M > { x m } Mm =1 where x m = ( m − h x and h x = LM − is the the uniformspatial grid spacing. Let us introduce the following finite dimensional vectorspace X M . = (cid:8) w = [ w · · · w M ] T ∈ C M (cid:9) where each w ∈ X M satisfies w ( t ) = w M ( t ) = 0 , (5.3) w M − ( t ) − w M − ( t ) + 3 w M ( t )2 h x = 0 , (5.4)for t >
0. Note that w m ( t ) is an approximation to w ( x, t ) at the point x = x m and (5.3) and (5.4) correspond to Dirichlet and Neumann type boundary con-ditions respectively. Consider the standard forward and backward differenceoperators ∆ + : X M → X M and ∆ − : X M → X M , respectively and let us introduce the following finite difference operators on X M : ∆ . = 12 ( ∆ + + ∆ − ) ∆ . = ∆ + ∆ − ∆ . = ∆ + ∆ + ∆ − . (5.5)Next assume N to be a positive integer, T be the final time and consider thenodal points in time axis t n = ( n − k , where n = 1 , . . . , N is time index and h t = TN − is the time step size. Let w n = [ w n · · · w nm ] T be an approximation ofthe solution at the n -th time step where w nm is an approximation to w ( x, t ) atthe point ( x m , t n ). Discretizing (1.12) in space by using the finite differenceoperators (5.5) and in time by using Crank–Nicolson time stepping, we endup with the discrete problem: Given w n ∈ X M , find w n +1 ∈ X M such that (cid:18) I M + h t A − βh t K My ( · , Γ ,M (cid:19) w n +1 = Fw n , n = 1 , , . . . , N, (5.6)where I M is the identity matrix on X M , A is defined as A . = β ∆ − iα ∆ + δ ∆ + r I M , (5.7) K My ( · ,
0) is an M × M diagonal matrix, where each element on the diagonalconsists of the elements of the form k y ( x m , m = 1 , . . . , M and k y ( x,
0) isobtained exactly in the previous step, Γ ,M is a discrete counterpart of thetrace operator Γ and given by an M × M matrix Γ ,M = 12 h x − − · · · − − · · · − − · · · , (5.8)and F . = I M − h t A + βh t K My ( · , Γ ,M . Note that the nonzero elements in the matrix Γ ,M given in (5.8) are due tothe one–sided second order finite difference approximation to the first orderderivative at the point x = 0. Step iii.
Now we find the inverse image, u , of w under the backstepping transforma-tion: Given w , we find u by using succession method. More precisely, we set v . = Υ k u , therefore we obtain u = v + w and substitute u by v + w on (1.11)to get v ( x, t ) = (cid:90) x k ( x, y ) w ( y, t ) dy + (cid:90) x k ( x, y ) v ( y, t ) dy. TABILIZATION OF HIGHER ORDER SCHR ¨ODINGER EQUATIONS 65
Now given w obtained numerically in the previous step, we solve this equationsuccessively for v . Using the numerical results for w and v on u = v + w , weobtain a numerical solution for u .Now, let us present a numerical simulation that verifies our stability results. Wetake M = 1001 spatial nodes, N = 5001 time steps. The iteration for the backstep-ping kernel is performed j = 27 times so that the error is aroundmax ( s,t ) ∈ ∆ s,t | G j +1 − G j | ∼ − . We consider the following model iu t + iu xxx + 2 u xx + 8 iu x = 0 , x ∈ (0 , π ) , t ∈ (0 , T ) ,u (0 , t ) = 0 , u ( π, t ) = h ( t ) , u x ( π, t ) = h ( t ) ,u ( x,
0) = 3 − e ix − e − ix . (5.9)In the absence of controllers, i.e. h ( t ) ≡ h ( t ) ≡
0, we have a stationary solution u ( x, t ) = 3 − e ix − e − ix . Let us choose r = 0 .
05. This choice yields a positiveexponent value λ , defined in Proposition 3.6, i.e. solution is decaying exponentiallyin time (see Figure 1). Contour plot of the corresponding solution and time evolutionof its L − norm are given by Figures 15. Figure 15.
Numerical results in the presence of controllers. Left:Time evolution of | u ( x, t ) | . Right: Time evolution of (cid:107) u ( · , t ) (cid:107) .5.2. Observer design.
Our algorithm consists of five steps. First we obtain anapproximation for the backstepping kernel p . In the second and third steps, weobtain a numerical solution for the error model (1.15) and modified target observermodel (1.23), respectively. As a fourth step, we get a numerical solution for theobserver model by using the invertibility of the backstepping transformation (1.10). At the fifth and the last step, we deduce numerical solution of the original plant via u = ˆ u + ˜ u . Step i.
Following the same procedure we introduced in the first step of Section 5.1and then changing the variables first as ˜ s = L − t , ˜ t = L − s − t then as s = x − y , t = y , we get G (˜ s, ˜ t ; − r ) = G ( L − t, L − s − t ; − r ) = k ( L − y, L − x ; − r ) = p ( x, y ) . Note that using p , we also derive p ( x ) = − iβp y ( x,
0) + αp ( x,
0) and p ( x ) = iβp ( x, Step ii.
To solve (1.15), we apply the same discretization procedure as we introducedin the second step of Section 5.1. The trace terms included in the mainequation of (1.15) are approximated by the following one sided second orderfinite differences˜ u x (0 , t ) ≈ − u ( t ) + 4˜ u ( t ) − ˜ u ( t )2 h x , ˜ u xx (0 , t ) ≈ u ( t ) − u ( t ) + 4˜ u ( t ) − ˜ u ( t ) h x . (5.10) Step iii.
Applying the similar discretization procedure, now we solve (1.23) numer-ically. Note that using ˜ w (0 , t ) = 0 and p ( x, x ) = 0, one can show byusing the backstepping transformation (1.20) that ˜ w x (0 , t ) = ˜ u x (0 , t ) and˜ w xx (0 , t ) = ˜ u xx (0 , t ). Therefore, instead of approximating the first order andsecond order traces of ˜ w at the left end point, we can use (5.10). Note alsothat a discrete counterpart, Υ Mk , of Υ k can be obtained by applying a suitablenumerical integration technique. For instance applying composite trapezoidalrule yields the following representation Υ Mk = h x · · · k ( x , x ) k ( x , x ) · · · k ( x M − , x ) k ( x M − , x ) · · · k ( x M − , x M − ) 0 k ( x M , x ) k ( x M , x ) · · · k ( x M , x M − ) k ( x M , x M ) . (5.11) Step iv.
Using the invertibility of the backstepping transformation (1.11), we obtaininverse image ˆ u of ˆ w . This will be done by applying a similar procedure aswe introduced in the first step of Section 5.1. Step v.
Using the numerical results for the observer and error models and setting u = ˆ u + ˜ u , we deduce an approximation for the solution of the original plant.Now let us go on with the numerical simulations. We obtain our results by taking M = 1001 spatial nodes, N = 5001 time steps. We performed the iteration for p ( x, y ) TABILIZATION OF HIGHER ORDER SCHR ¨ODINGER EQUATIONS 67 and k ( x, y ) several times so that the error is around max ( s,t ) ∈ ∆ s,t | G j +1 − G j | ∼ − . Weconsider the same model (cid:40) iu t + iu xxx + 2 u xx + 8 iu x = 0 , x ∈ (0 , π ) , t ∈ (0 , T ) ,u (0 , t ) = 0 , u ( π, t ) = h ( t ) , u x ( π, t ) = h ( t ) , where, unlike the controller design case, the feedback controllers use the state of theobserver model. We initialize the error model as ˜ u ( x,
0) = 3 − e ix − e − ix andobserver model ˆ u ( x, ≡
0. We take r = 0 .
05. Since the problem parameters aresame as the previous numerical example, this choice will yield positive exponent vales µ > ν > µ, ν are defined in Proposition 4.3 and Proposition 4.4.Contour plot of the numerical solution of original plant is given at the left sideof Figure 16. At the right, we show time evolution of the L − norms of solutions ofplant-observer-error system. Figure 16.
Numerical results. Left: Time evolution of | u ( x, t ) | .Right: Time evolution of | u ( · , t ) | . Appendix A. Deduction of the kernel pde model (1.7)In this section, we present the details of the calculations for obtaining the kernelmodel given in (1.7). Differentiating both sides of (1.5) with respect to t we get iw t ( x, t ) = iu t ( x, t ) − (cid:90) x ik ( x, y ) u t ( y, t ) dy = iu t ( x, t ) + (cid:90) x k ( x, y )( iβu yyy ( y, t ) + αu yy ( y, t ) + iδu y ( y, t )) dy = iu t ( x, t )+ iβ (cid:32) k ( x, y ) u yy ( y, t ) − k y ( x, y ) u y ( y, t ) + k yy ( x, y ) u ( y, t ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) x − (cid:90) x k yyy ( x, y ) u ( y, t ) dy (cid:19) + α (cid:32) k ( x, y ) u y ( y, t ) − k y ( x, y ) u ( y, t ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) x + (cid:90) x k yy ( x, y ) u ( y, t ) dy (cid:33) + iδ (cid:32) k ( x, y ) u ( y, t ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) x − (cid:90) x k y ( x, y ) u ( y, t ) dy (cid:33) . Using the boundary condition u (0 , t ) = 0, rearrenging the last expression in termsof u ( x, t ), u x ( x, t ), u xx ( x, t ), u x (0 , t ) and u xx (0 , t ), we obtain iw t ( x, t ) = iu t ( x, t ) + (cid:90) x ( − iβk yyy + αk yy − iδk y ) ( x, y ) u ( y, t ) dy + ( iβk yy ( x, x ) − αk y ( x, x ) + iδk ( x, x )) u ( x, t )+ ( − iβk y ( x, x ) + αk ( x, x )) u x ( x, t ) + iβk ( x, x ) u xx ( x, t )+ ( iβk y ( x, − αk ( x, u x (0 , t ) − iβk ( x, u xx (0 , t ) . (A.1)Next we differentiate both sides of (1.11) with respect to x up to the order three andmultiply the results by iδ , α and iβ respectively to obtain iδw x ( x, t ) = iδu x ( x, t ) − iδ ∂∂x (cid:90) x k ( x, y ) u ( y, t ) dy = iδu x ( x, t ) − (cid:90) x iδk x ( x, y ) u ( y, t ) dy − iδk ( x, x ) u ( x, t ) , (A.2) TABILIZATION OF HIGHER ORDER SCHR ¨ODINGER EQUATIONS 69 αw xx ( x, t ) = αu xx ( x, t ) − α ∂∂x (cid:90) x k x ( x, y ) u ( y, t ) dy − α ∂∂x ( k ( x, x ) u ( x, t ))= αu xx ( x, t ) − (cid:90) x αk xx ( x, y ) u ( y, t ) dy + α (cid:18) − k x ( x, x ) − ddx k ( x, x ) (cid:19) u ( x, t ) − αk ( x, x ) u x ( x, t ) , (A.3)and iβw xxx ( x, t ) = iβu xxx ( x, t ) − iβ ∂∂x (cid:90) x k xx ( x, y ) u ( y, t ) dy − iβ ∂∂x (cid:18)(cid:18) k x ( x, x ) + ddx k ( x, x ) (cid:19) u ( x, t ) + k ( x, x ) u x ( x, t ) (cid:19) = iβu xxx ( x, t ) − (cid:90) x iβk xxx ( x, y ) u ( y, t ) dy + iβ (cid:18) − k xx ( x, x ) − ddx k x ( x, x ) − d dx k ( x, x ) (cid:19) u ( x, t )+ iβ (cid:18) − k x ( x, x ) − ddx k ( x, x ) (cid:19) u x ( x, t ) − iβk ( x, x ) u xx ( x, t ) . (A.4)Adding (A.1)-(A.4) side by side together with irw ( x, t ) = iru ( x, t ) − ir (cid:82) x k ( x, y ) u ( y, t ) dy and using the main equation of the linear plant, we obtain iw t + iβw xxx + αw xx + iδw x + irw = (cid:90) x ( − iβ ( k xxx + k yyy ) − α ( k xx − k yy ) − iδ ( k x + k y ) − irk ) ( x, y ) u ( y, t ) dy (A.5)= (cid:18) iβ (cid:18) k yy ( x, x ) − k xx ( x, x ) − ddx k x ( x, x ) − d dx k ( x, x ) (cid:19) + α (cid:18) − k y ( x, x ) − k x ( x, x ) − ddx k ( x, x ) (cid:19) + ir (cid:19) u ( x, t ) (A.6) − iβ (cid:18) k y ( x, x ) + k x ( x, x ) + 2 ddx k ( x, x ) (cid:19) u x ( x, t ) (A.7)+ ( iβk y ( x, − αk ( x, u x (0 , t ) (A.8) − iβk ( x, u xx (0 , t ) . (A.9)From (A.9) we have k ( x,
0) = 0 and therefore, from (A.8) we get k y ( x,
0) = 0. Usingthe relation ddx k ( x, x ) = k x ( x, x ) + k y ( x, x ), we obtain from (A.7) that ddx k ( x, x ) = 0 and thanks to k ( x,
0) = 0, this implies k ( x, x ) = 0. Next, we differentiate ddx k ( x, x ) = k x ( x, x ) + k y ( x, x ) with respect to x and use ddx k ( x, x ) = 0 to obtain k yy ( x, x ) = − k xy ( x, x ) − k xx ( x, x ). Using this result on (A.6), we deduce that ddx k x ( x, x ) = r β which, by the implications k y ( x,
0) = 0 ⇒ k x ( x,
0) = 0 ⇒ k x (0 ,
0) = 0, is equivalentto k x ( x, x ) = rx β . Also note that taking x = 0 in the backstepping transformation implies w (0 , t ) = u (0 , t ) = 0 and, taking x = L implies w ( L, t ) = u ( L, t ) − (cid:90) L k ( L, y ) u ( y, t ) dy = 0and w x ( L, t ) = u x ( L, t ) − (cid:90) L k x ( L, y ) u ( y, t ) dy − k (0 , u (0 , t ) = 0 . So the boundary conditions are being satisfied without any extra conditions on k .As a conclusion, linear plant is mapped to target model (not modified one) if k ( x, y ) satisfies the following boundary value problem β ( k xxx + k yyy ) − iα ( k xx − k yy ) + δ ( k x + k y ) + rk = 0 ,k ( x, x ) = k y ( x,
0) = k ( x,
0) = 0 ,k x ( x, x ) = rx β , on ∆ x,y . TABILIZATION OF HIGHER ORDER SCHR ¨ODINGER EQUATIONS 71
Appendix B. Deduction of the kernel pde model (1.18)In this section, we present the details of the calculations for obtaining the kernelmodel given in (1.19). Differentiating (1.16) with respect to t , we get i ˜ u t ( x, t ) = i ˜ w t ( x, t ) − (cid:90) x ip ( x, y ) ˜ w t ( y, t ) dy = i ˜ w t ( x, t ) + (cid:90) x p ( x, y )( iβ ˜ w yyy ( y, t ) + α ˜ w yy ( y, t ) + iδ ˜ w y ( y, t ) + ir ˜ w ) dy = i ˜ w t ( x, t )+ iβ (cid:32) p ( x, y ) ˜ w yy ( y, t ) − p y ( x, y ) ˜ w y ( y, t ) + p yy ( x, y ) ˜ w ( y, t ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) x − (cid:90) x p yyy ( x, y ) ˜ w ( y, t ) dy (cid:19) + α (cid:32) p ( x, y ) ˜ w y ( y, t ) − p y ( x, y ) ˜ w ( y, t ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) x + (cid:90) x p yy ( x, y ) ˜ w ( y, t ) dy (cid:33) + iδ (cid:32) p ( x, y ) ˜ w ( y, t ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) x − (cid:90) x p y ( x, y ) ˜ w ( y, t ) dy (cid:33) + ir (cid:90) L p ( x, y ) ˜ w ( y, t ) dy. Using the boundary conditions w (0 , t ) = 0, rearranging the last expression in termsof ˜ w ( x, t ), ˜ w x ( x, t ), ˜ w xx ( x, t ), ˜ w x (0 , t ) and ˜ w xx (0 , t ), we obtain i ˜ u t ( x, t ) = i ˜ w t ( x, t ) + (cid:90) x ( − iβp yyy + αp yy − iδp y + irp ) ( x, y ) ˜ w ( y, t ) dy + ( iβp yy ( x, x ) − αp y ( x, x ) + iδp ( x, x )) ˜ w ( x, t )+ ( − iβp y ( x, x ) + αp ( x, x )) ˜ w x ( x, t ) + iβp ( x, x ) ˜ w xx ( x, t )+ ( iβp y ( x, − αp ( x, w x (0 , t ) − iβp ( x,
0) ˜ w xx (0 , t ) . (B.1)Next we differentiate (1.16) up to order three and multiply the results by iδ , α and iβ respectively to obtain iδ ˜ u x ( x, t ) = iδ ˜ w x ( x, t ) − iδ ∂∂x (cid:90) x p ( x, y ) ˜ w ( y, t ) dy = iδ ˜ w x ( x, t ) − (cid:90) x iδp x ( x, y ) ˜ w ( y, t ) dy − iδp ( x, x ) ˜ w ( x, t ) , (B.2) α ˜ u xx ( x, t ) = α ˜ w xx ( x, t ) − α ∂∂x (cid:90) x p x ( x, y ) ˜ w ( y, t ) dy − α ∂∂x ( p ( x, x ) ˜ w ( x, t ))= α ˜ w xx ( x, t ) − (cid:90) x αp xx ( x, y ) ˜ w ( y, t ) dy + α (cid:18) − p x ( x, x ) − ddx p ( x, x ) (cid:19) ˜ w ( x, t ) − αp ( x, x ) ˜ w x ( x, t ) , (B.3)and iβ ˜ u xxx ( x, t ) = iβ ˜ w xxx ( x, t ) − iβ ∂∂x (cid:90) x p xx ( x, y ) ˜ w ( y, t ) dy − iβ ∂∂x (cid:18)(cid:18) p x ( x, x ) + ddx p ( x, x ) (cid:19) ˜ w ( x, t ) + p ( x, x ) ˜ w x ( x, t ) (cid:19) = iβ ˜ w xxx ( x, t ) − (cid:90) x iβp xxx ( x, y ) ˜ w ( y, t ) dy + iβ (cid:18) − p xx ( x, x ) − ddx p x ( x, x ) − d dx p ( x, x ) (cid:19) ˜ w ( x, t )+ iβ (cid:18) − p x ( x, x ) − ddx p ( x, x ) (cid:19) ˜ w x ( x, t ) − iβp ( x, x ) ˜ w xx ( x, t ) . (B.4)From (B.2) and (B.3) we also have p ( x )˜ u x (0 , t ) = p ( x ) ˜ w x (0 , t ) (B.5)and p ( x )˜ u xx (0 , t ) = p ( x )( ˜ w xx (0 , t ) − p (0 ,
0) ˜ w x (0 , t )) . (B.6) TABILIZATION OF HIGHER ORDER SCHR ¨ODINGER EQUATIONS 73
Adding (B.1)-(B.6) side by side we obtain i ˜ u t + iβ ˜ u xxx + α ˜ u xx + iδ ˜ u x + p ( x )˜ u x (0 , t ) + p ( x )˜ u xx (0 , t )= i ˜ w t + iβ ˜ w xxx + α ˜ w xx + iδ ˜ w x (B.7)+ (cid:90) x ( − iβ ( p xxx + p yyy ) − α ( p xx − p yy ) − iδ ( p x + p y ) + irp ) ( x, y ) ˜ w ( y, t ) dy (B.8)= (cid:18) iβ (cid:18) p yy ( x, x ) − p xx ( x, x ) − ddx p x ( x, x ) − d dx p ( x, x ) (cid:19) + α (cid:18) − p y ( x, x ) − p x ( x, x ) − ddx p ( x, x ) (cid:19)(cid:19) ˜ w ( x, t ) (B.9) − iβ (cid:18) p y ( x, x ) + p x ( x, x ) + 2 ddx p ( x, x ) (cid:19) ˜ w x ( x, t ) (B.10)+ ( iβp y ( x, − αp ( x,
0) + p ( x ) − p (0 , p ( x )) ˜ w x (0 , t ) (B.11)+ ( − iβp ( x,
0) + p ( x )) ˜ w xx (0 , t ) . (B.12)Note that, taking x = L on (1.20) and using the boundary condition ˜ u ( L, t ) = 0,we must have p ( L, y ) = 0 in order to get ˜ w ( L, t ) = 0. On the other hand, using therelation ddx p ( x, x ) = p x ( x, x ) + p y ( x, x ), we get from (B.10) that ddx p ( x, x ) = 0 , (B.13)which, thanks to p ( L, y ) = 0 implies p ( x, x ) = 0 . Next, we differentiate ddx p ( x, x ) = p x ( x, x ) + p y ( x, x ) with respect to x and use ddx p ( x, x ) = 0 to obtain p yy ( x, x ) = − p xy ( x, x ) − p xx ( x, x ). Using this result in(B.9), we deduce that ddx p x ( x, x ) = − r β . which, due to the implications p ( L, y ) = 0 ⇒ p y ( L, y ) = 0 ⇒ p y ( L, L ) = 0 ⇒ p x ( L, L ) = 0, is equivalent to p x ( x, x ) = r β ( L − x ) . On the other hand we obtain from (B.11)-(B.12) that p ( x ) = − iβp y ( x,
0) + αp ( x, p ( x ) = iβp ( x, . Note that for x = 0 in (1.20), we have ˜ u (0 , t ) = ˜ w (0 , t ) = 0. For x = L and thanksto p ( L, y ) = 0, we have ˜ u ( L, t ) = ˜ w ( L, t ) = 0. Also for x = L on (B.2), we see that˜ w x ( L, t ) = 0 holds if p x ( L, y ) = 0.As a conclusion, the error model is mapped to the target error model (not modifiedone), if p satisfies the following boundary value problem β ( p xxx + p yyy ) − iα ( p xx − p yy ) + δ ( p x + p y ) − rp = 0 ,p ( x, x ) = p ( L, y ) = p x ( L, y ) = 0 ,p x ( x, x ) = r β ( L − x ) , on ∆ x,y . Appendix C. Roots of the characteristic equation (2.32) . In this part, we investigate the roots λ j = λ j ( s ), j = 1 , ,
3, of the characteristicequation s + βλ + iαλ + δλ = 0that is obtained by the one parameter family of boudary value problems (2.30). Moreprecisely, we show that (2.32) has double or possibly triple roots only for finitelymany exceptional cases of the values of s in the complex plane. The location of s inthe complex plane is directly related with the sign of the quantity α + 3 βδ . Alsoin the following calculations, we drop notation for s dependence and simply write λ j ( s ) = λ j , j = 1 , , λ and λ , are equal for some s ∈ C .Then λ and λ satisfy λ + 2 λ = iαβ , (C.1)2 λ λ + λ = δβ , (C.2) λ λ = − sβ , (C.3)where s ∈ ( r − i ∞ , r + i ∞ ) for some r ∈ R + . Let λ = a + ib , λ = λ = c + id , a, b, c, d are real functions of s . From the real and imaginary parts of (C.1)-(C.2), TABILIZATION OF HIGHER ORDER SCHR ¨ODINGER EQUATIONS 75 we have following equations: a + 2 c = 0 , (C.4) b + 2 d = αβ , (C.5)2 ac − bd + c − d = δβ , (C.6) ad + bc + cd = 0 . (C.7)Using (C.4) in (C.7), we get c ( b − d ) = 0.(i) Let b = d . By (C.5), b = d = α β . Substituting these into (C.6) yields c (2 a + c ) = α + 3 βδ β . Using (C.4), we get a = − α + 3 βδ )9 β , c = − α + 3 βδ β . Assuming α + 3 βδ > α + 3 βδ = 0 yields λ = λ = λ = iα β . For this case we see from (C.3) that, the only value for s is pure imaginary and given by s = iα β . Now let α + 3 βδ <
0. Then we have a , = ∓ (cid:112) − ( α + 3 βδ )3 β , c , = ± (cid:112) − ( α + 3 βδ )3 β . Note that using (C.4) and b = d , we obtain from (C.3) there are two possiblevalues of s , denoted by s − and s − , which are given by (cid:60) ( s − ) = 227 β ( − α − βδ ) / , (cid:60) ( s − ) = − β ( − α − βδ ) / and (cid:61) ( s − ) = (cid:61) ( s − ) = − α β (2 α + 9 βδ ) . (ii) Let c = 0. Then, by (C.4), a = 0. Using this, direct calculation from (C.5)-(C.6) yields b , = α ∓ (cid:112) α + 3 βδ β , d , = α ± (cid:112) α + 3 βδ β . Observe that α + 3 βδ < α + 3 βδ = 0 ends upwith the triple root case investigated in (i). Now let α + 3 βδ >
0. From(C.3), we see that there are two possible values of s , denoted by s +1 and s +2 ,given by s +1 = i β (cid:0) α − α ( α + 3 βδ ) + 2( α + 3 βδ ) / (cid:1) and s +2 = i β (cid:0) α − α ( α + 3 βδ ) − α + 3 βδ ) / (cid:1) . Acknowledgements . We would like to thank Professor Bing-Yu Zhang (University of Cincinnati) andProfessor Shu Ming Sun (Virginia Tech) for their fruitful comments regarding theapplication of the Laplace transform method that we used in Section 2 to prove theboundary smoothing properties.
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