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International Journal of Control

ISSN: 0020-7179 (Print) 1366-5820 (Online) Journal homepage: https://www.tandfonline.com/loi/tcon20

Stabilisation of Schrödinger equation in dynamicboundary feedback with a memory-typed heatequation

Lu Lu & Jun-Min Wang

To cite this article: Lu Lu & Jun-Min Wang (2019) Stabilisation of Schrödinger equation in dynamicboundary feedback with a memory-typed heat equation, International Journal of Control, 92:2,416-430, DOI: 10.1080/00207179.2017.1358826

To link to this article: https://doi.org/10.1080/00207179.2017.1358826

Accepted author version posted online: 26Jul 2017.Published online: 08 Aug 2017.

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Stabilisation of Schrödinger equation in dynamic boundary feedback with amemory-typed heat equation

Lu Lua and Jun-Min Wangb

aElementary College, Beijing Polytechnic, Beijing, People’s Republic of China; bDepartment of Mathematics and Statistics, Beijing Institute ofTechnology, Beijing, People’s Republic of China

ARTICLE HISTORYReceived March Accepted July

KEYWORDSSchrödinger equation; heatequation with memory;spectrum; asymptoticanalysis; Riesz basis;exponential stability

ABSTRACTIn this work, we study the dynamic behaviour for a heat equationwith exponential polynomial kernelmemory to be a controller for a Schrödinger system. By introducing some new variables, the time-variant system is transformed into a time-invariant one. Remarkably, the resolvent of the closed-loopsystem operator is not compact anymore. The residual spectrum is shown to be empty and the con-tinuous spectrum consisting of finite isolated points are obtained. It is shown that the sequence ofgeneralised eigenfunctions forms a Riesz basis for theHilbert state space. This deduces the spectrum-determined growth condition for the C0-semigroup, and the exponential stability is then established.

1. Introduction

There has been extensive literature on control of theSchrödinger equation (Guo & Shao, 2005; Krstic, Guo, &Smyshlyaev, 2011; Machtyngier, 1994; Phung, 2001). Fora single Schrödinger equation, in Krstic et al. (2011), thecollocated boundary control is designed to exponentiallystabilise the system

⎧⎪⎪⎨⎪⎪⎩

wt (x, t ) + iwxx(x, t ) = 0, x ∈ (0, 1), t > 0,wx(1, t ) = 0, t ≥ 0,w(0, t ) = U (t ), t ≥ 0,Y (t ) = wx(0, t ), t ≥ 0,

(1.1)

where U(t) is the control input and Y(t) is the out-put observation. When U(t) = −icY(t), where c > 0is a positive constant, Krstic et al. (2011) showed thatthe system operator of the closed-loop system generatesan exponentially stable semigroup in the energy space;and the eigenvalues approach a vertical line parallel tothe imaginary axis. Recently, in Wang, Ren, and Krstic(2012), an alternative design method to system (1.1) ispresented

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

wt (x, t ) + iwxx(x, t ) = 0, x ∈ (0, 1), t > 0,ut (x, t ) − uxx(x, t ) = 0, x ∈ (0, 1), t > 0,w(1, t ) = ux(1, t ) = 0, t ≥ 0,w(0, t ) = ku(0, t ), t ≥ 0,ux(0, t ) = ikwx(0, t ), t ≥ 0,

(1.2)

CONTACT Lu Lu lulu@.com, @dky.bjedu.cn

where k � 0. The output Y(t) of the Schrödinger equa-tion is fed into the boundary heat flux of the heatequation while the boundary temperature of the heatequation is fed into the Schrödinger equation. Such adesign replaces the static feedback by dynamic feed-back governed by a heat equation with a collocatedinput/output pair and shows the exponential stability andGevrey regularity for the closed-loop system.Moreover, itmoves the eigenvalues of the Schrödinger and heat equa-tions into the second quadrant, which are approachingtwo asymptotically symmetric parabolas relative to theline Reλ = −Imλ.

However, it is indicated that the coupling system doesnot take the memory effect into account, which may existin some materials particularly in low temperature. Fur-thermore, the fact that the thermal disturbance at onepoint affects the whole elastic body instantly, which isimplied by (1.2), is not physically acceptable (Fatori &Muñoz Rivera, 2001). Gurtin and Pipkin (1968) indicatedthat by a heat equation, which is parabolic, the speedof propagation of thermal disturbance is infinite, and inthe same paper, they proposed a general theory for heatconduction with finite propagation speed. The linearisedGurtin–Pipkin heat equation is described by

θt (x, t ) =∫ t

∞k(t − s)θxx(x, s)ds,

where the kernel k is supposed to be a positive non-increasing function of its variable. Instead of the memory

© Informa UK Limited, trading as Taylor & Francis Group

INTERNATIONAL JOURNAL OF CONTROL2019, VOL. 92, NO. 2, 416–430

θt(x, t) − (g ∗ θxx)(x, t) = 0

yt(x, t) + iyxx(x, t) = 0

k(g ∗ θx)(1, t) ikyx(1, t)

y0(x) y(x, t)

θ0(x) θ(x, t)

Figure . Interconnected system of Schrödinger heat equation with memory.

from infinity, Pandolfi (2005) considered this type of heatequation with memory starting from the starting point:

θt (x, t ) =∫ t

0k(t − s)θxx(x, s)ds.

The cosine operator approach was used to study its wellposedness. In Wang, Guo, and Fu (2009), the same heatequation under the Dirichlet boundary condition wasconsidered, with a special type of kernel

k(t ) = N�j=1

a2j e−b jt , 0 < a j, b j ∈ R, 1 ≤ N ∈ N.

It is shown that this system achieves strongly exponentialstability, though the resolvent of the system operator isnot compact.

Hence, with this inspiration, it is natural to raise aquestion of whether a heat equation with memory can beregarded as a compensator to stabilise the Schrödingerequation, so we study the following interconnected sys-tem (see Figure 1):

⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

yt (x, t ) + iyxx(x, t ) = 0, t > 0,θt (x, t ) − (g ∗ θxx)(x, t ) = 0, x ∈ (0, 1),y(0, t ) = θ (0, t ) = 0, t ≥ 0,y(1, t ) = k(g ∗ θx)(1, t ), t ≥ 0,θ (1, t ) = ikyx(1, t ), t ≥ 0,y(x, 0) = y0(x), θ (x, 0) = θ0(x), x ∈ [0, 1],

(1.3)

where k � 0. The sign ‘*’ denotes the convolutionproduct:

(g ∗ l)(x, t ) =∫ t

0g(t − s)l(x, s)ds.

The kernel being taken as an exponential type

g(t ) = N�j=1

a2j e−b jt , 0 < a j, b j ∈ R, 1 ≤ N ∈ N.

For simplicity, we assume that

0 < b1 < b2 < · · · < bN . (1.4)

This paper is organised as follows. In Section 2, weintroduce some new variables so that system (1.3) isreduced to a time-invariant system. And then, we formu-late the system into an abstract evolution equation. Sec-tion 3 is devoted to the spectral analysis of the system.Riesz basis property and exponential stability of the sys-tem are given in Section 4.

2. System operator set-up

Let

φ j(x, t ) =∫ t

0a je−b j(t−s)θx(x, s)ds, j = 1, 2, . . . ,N,

we have (φj)t = ajθx − bjφj. Then, Equation (1.3) can bechanged into

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

yt (x, t ) + iyxx(x, t ) = 0, t > 0,

θt (x, t ) −[

N�j=1

a jφ j(x, t )]x= 0, x ∈ (0, 1),

(φ j)t (x, t ) = a jθx(x, t ) − b jφ j(x, t ), t > 0,y(0, t ) = θ (0, t ) = φ j(x, 0) = 0,

t ≥ 0, j = 1, 2, . . . ,N,

y(1, t ) = kN�j=1

a jφ j(1, t ), t ≥ 0,

θ (1, t ) = ikyx(1, t ), t ≥ 0.(2.1)

The energy function for (1.3) is given by

E(t ) = 12

∫ 1

0|y(x, t )|2 + θ2(x, t ) + N

�j=1

φ2j (x, t )dx. (2.2)

We consider system (1.3) in the energy state space

H = (L2(0, 1))N+2,

INTERNATIONAL JOURNAL OF CONTROL 417

with the norm induced by the following inner product:

〈X,Y 〉 =∫ 1

0ueudx +

∫ 1

0vevdx + N

�j=1

∫ 1

0h jeh jdx

∀ X = (u, v, h1, . . . hN ) ∈ H,

Y = (eu,ev,eh1, . . .ehN ) ∈ H. (2.3)

Now, define the system operator A : D(A)(⊂ H) → Hby

A

⎛⎜⎜⎜⎜⎜⎜⎜⎝

uv

h1...hN

⎞⎟⎟⎟⎟⎟⎟⎟⎠

�

=

⎛⎜⎜⎜⎜⎜⎜⎜⎝

−iu′′(x)[N�j=1

a jh j(x)]′

a1v ′(x) − b1h1(x)...

aNv ′(x) − bNhN (x)

⎞⎟⎟⎟⎟⎟⎟⎟⎠

�

, (2.4)

D(A) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

⎛⎜⎜⎜⎜⎜⎜⎜⎝

uv

h1...hN

⎞⎟⎟⎟⎟⎟⎟⎟⎠

�∣∣∣∣∣∣∣∣∣∣∣∣∣

u(0) = v(0) = 0, u ∈ H2(0, 1),

u(1) = kN�j=1

a jh j(1),

v(1) = iku′(1),

h j ∈ L2(0, 1), j = 1, . . . ,N,

N�j=1

a jh j(x) ∈ H1(0, 1).

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

.

Then, (1.3) can be formulated as an abstract evolutionequation inH:

⎧⎨⎩

ddt

Z(t ) = AZ(t ),Z(0) = Z0,

(2.5)

where Z(t ) = (y(·, t ), θ (·, t ), φ1(·, t ), . . . , φN (·, t )) isthe state variable and Z0 = (y0(x), θ0(x), 0, …, 0) is theinitial value.Theorem 2.1: LetA be defined by (2.4). ThenA−1 existsand hence 0 ∈ ρ(A), the resolvent set of A. Moreover, Ais dissipative and thusA generates a C0-semigroup of con-tractions eAt onH.Proof: Let Z = (u, v, h1, . . . , hN ) ∈ H. Solve AZ = Zfor Z = (u, v, h1, . . . , hN ) ∈ D(A), that is⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

−iu′′(x) = u(x),[ N�j=1

a jh j(x)]′

= v(x),

a jv′(x) − b jh j(x) = h j(x), j = 1, 2, . . . ,N,

u(0) = v(0) = 0, v(1) = iku′(1),

u(1) = kN�j=1

a jh j(1).

(2.6)

From (2.6), we have

h j(x) = a j

b jv ′(x) − 1

b jh j(x), j = 1, 2, . . . ,N, (2.7)

and

N�j=1

a jh j(x)∣∣∣1x

=∫ 1

xv(s)ds,

i.e.N�j=1

a jh j(1) − N�j=1

a jh j(x) =∫ 1

xv(s)ds. (2.8)

Multiplying k to both sides of the second equation of (2.8)and using the last boundary condition, we get

N�j=1

a jh j(x) = 1ku(1) −

∫ 1

xv(s)ds. (2.9)

Substituting (2.7) into (2.9), we get

v ′(x) = 1T

N�j=1

a j

b jh j(x) + 1

kTu(1) − 1

T

∫ 1

xv(s)ds, (2.10)

where T =∑ j=1N

a2jb j

>0. Integrating both sides of (2.10) in xfrom 0 to x, and using v(0) = 0, we obtain

v(x) = 1T

∫ x

0

N�j=1

a j

b jh j(τ )dτ + 1

kTu(1)x

− 1T

∫ x

0

∫ 1

τ

v(s)dsdτ. (2.11)

Integrating both sides of the first equation of (2.6) twice,first from x to 1 and then from 0 to x, with u(0) = 0, toachieve

u(x) = u′(1)x − i∫ x

0

∫ 1

τ

u(s)dsdτ. (2.12)

Setting x = 1 in both (2.11) and (2.12) respectively, wehave

v(1) = 1kT

u(1) + A, u(1) = u′(1) + B, (2.13)

where

A = 1T

∫ 1

0

N�j=1

a j

b jh j(τ )dτ − 1

T

∫ 1

0

∫ 1

τ

v(s)dsdτ,

and

B = −i∫ 1

0

∫ 1

τ

u(s)dsdτ.

418 L. LU AND J.M. WANG

Together with the boundary condition v(1) = iku′(1), toachieve

u′(1) = kAT + Bik2T − 1

, u(1) = kAT + Bik2T − 1

+ B. (2.14)

Collecting (2.7), (2.10), (2.11) and (2.12), we get theunique solution Z to equation (2.6). Hence, A−1 exists,or 0 ∈ ρ(A).

Now, we show thatA is dissipative inH. Given Z= (u,v , h1, . . . , hN ) ∈ D(A), we have

〈AZ,Z〉 =

�⎛⎜⎜⎜⎜⎜⎜⎜⎝

−iu′′(x)[N�j=1

a jh j(x)]′

a1v ′(x) − b1h1(x)...

aNv ′(x) − bNhN (x)

⎞⎟⎟⎟⎟⎟⎟⎟⎠

�

,

⎛⎜⎜⎜⎜⎜⎝

u(x)v(x)h1(x)...

hN (x)

⎞⎟⎟⎟⎟⎟⎠

��

=∫ 1

0(−iu′′)udx +

∫ 1

0

[N�j=1

a jh j

]′vdx

+∫ 1

0

N�j=1

(a jv′ − b jh j)h jdx

= −iuu′∣∣∣∣1

0+ i∫ 1

0|u′|2dx + v

[N�j=1

a jh j

] ∣∣∣∣1

0

−∫ 1

0

[N�j=1

a jh j

]v ′dx +

∫ 1

0

[N�j=1

a jh j

]v ′dx

− N�j=1

b j

∫ 1

0|h j(x)|2dx,

and hence

Re〈AZ,Z〉 = − N�j=1

b j

∫ 1

0|h j(x)|2dx ≤ 0. (2.15)

So, A is dissipative and generates a C0-semigroup eAt ofcontractions on H by the Lumer–Philips theorem (Pazy,1983, p. 14). �

3. Spectral analysis of the system

In this section, we analyse the spectrum of A. First,we consider the eigenvalue problem. Suppose AZ = λZfor λ ∈ C and 0 �= Z = (u, v, w, h1, . . . , hN ) ∈ D(A).Then,⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

−iu′′(x) = λu(x),[ N�j=1

a jh j(x)]′

= λv(x),

a jv′(x) − b jh j(x) = λh j(x), j = 1, 2, . . . ,N,

u(0) = v(0) = 0, v(1) = iku′(1),

u(1) = kN�j=1

a jh j(1).

(3.1)

Lemma 3.1: LetA be defined by (2.4). Then, for each λ ∈σp(A), we have Reλ < 0.

Proof: By Theorem 2.1, since A is dissipative, wehave for each λ ∈ σ (A), Reλ � 0. So we only needto show that there is no eigenvalue on the imaginaryaxis. Let λ = ±iμ2 ∈ σp(A) with μ ∈ R

+ and 0 �= Z =(u, v, h1, . . . , hN ) ∈ D(A) be its associated eigenfunc-tion ofA. Then by (2.15), we have

Re〈AZ,Z〉 = − N�j=1

b j

∫ 1

0|h j(x)|2dx = 0.

Hence, hj(x)= 0, j= 1, 2, …,N. Then, from (3.1), we havev(x)= 0. Furthermore, the first equation of (3.1) satisfies

{−iu′′(x) = λu(x),u(0) = u(1) = u′(1) = 0. (3.2)

A direct computation shows that the above equationadmits only zero solution. Therefore, Z = (u, v , h1, …,hN) = 0. There is no eigenvalue on the imaginary axis.The proof is complete. �

Proposition 3.1: Let A be defined by (2.4). Then, λ =−b j ∈ ρ(A), j = 1, 2, . . . ,N, where ρ(A) is the resol-vent set ofA.

Proof: We only give the proof for λ = −b1because other cases can be treated similarly. Letλ = −b1 and Z = (u, v, h1, . . . , hN ) ∈ D(A),G = ( f , g, ϕ1, . . . , ϕN ) ∈ H. Solve (−b1I − A)Z = G,that is,

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

−b1u(x) + iu′′(x) = f (x),

−b1v(x) −[

N�j=1

a jh j(x)]′

= g(x),

−a1v ′(x) = ϕ1(x),

−b1h j(x) −[a jv

′(x) − b jh j(x)]

= ϕ j(x),

j = 2, . . . ,N,

u(0) = v(0) = 0, v(1) = iku′(1),

u(1) = kN�j=1

a jh j(1).

(3.3)

From the third equation of (3.3) and the boundary con-dition v(0) = 0, we obtain

v(x) = − 1a1

∫ x

0ϕ1(τ )dτ. (3.4)

INTERNATIONAL JOURNAL OF CONTROL 419

Substituting (3.4) into the fourth equation of (3.3) to get

h j(x) = 1b j − b1

[ϕ j(x) + a jv′(x)]

= 1b j − b1

[ϕ j(x) − a j

a1ϕ1(x)], j = 2, . . . ,N.

From the first equation of (3.3) and the boundary condi-tion u(0) = 0 and u′(1) = v(1)/ik, we obtain

u(x) = c1e√−ib1x + c2e−

√−ib1x

− 1√ib1

∫ x

0sinh[

√−ib1(x − s)] f (s)ds, (3.5)

where

c1 = 12√ib1 cos(

√ib1)

[1ka1

∫ 1

0ϕ1(τ )dτ

+∫ 1

0cos[

√ib1(1 − s)] f (s)ds

],

c2 = − 12√ib1 cos(

√ib1)

[1ka1

∫ 1

0ϕ1(τ )dτ

+∫ 1

0cos[

√ib1(1 − s)] f (s)ds

].

Now, we calculate h1(x). Noting the last boundary con-dition, we integrate both sides of the second equation of(3.3) from x to 1, then

h1(x) = 1a1k

u(1) + 1a1

∫ 1

xb1v(τ ) + g(τ )dτ

− 1a1

N�j=2

a jh j(x)

= i tan√ib1

a1k√ib1

[1ka1

∫ 1

0ϕ1(τ )dτ

+∫ 1

0cos

√ib1(1 − s) f (s)ds

]

− ia1k

√ib1

∫ 1

0sin√ib1(1 − s) f (s)ds

−b1a21

∫ 1

x

∫ τ

0ϕ1(s)dsdτ + 1

a1

∫ 1

xg(τ )dτ

− 1a1

N�j=2

[a j

b j − b1

(ϕ j(x) − a j

a1ϕ1(x)

)].

So, Z = (u, v , h1, …, hN) is uniquely determined. Hence,(−b1I − A)−1 exists and is bounded. Similarmanner canbe applied to λ = −bj, j = 2, 3, …, N. Therefore, −b j ∈ρ(A), j = 1, 2, . . . ,N, we complete the proof. �

When λ � −bj, j= 1, …,N, it follows from (3.1) that

h j(x) = a j

λ + b jv ′(x), j = 1, 2, . . . ,N, (3.6)

u and v satisfies⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

u′′(x) = iλu(x),N�j=1

a2jλ + b j

v ′′(x) = λv(x),

u(0) = v(0) = 0, v(1) = iku′(1),

u(1) = kN�j=1

a2jλ + b j

v ′(1).

(3.7)

Lemma 3.2: LetA be defined by (2.4) and

={

λ ∈ C

∣∣∣ N�j=1

a2jλ + b j

= 0

}. (3.8)

Then

∩ σp(A) = ∅,

where σp(A) stands for the point spectrum ofA.

Proof: Obviously, 0 �. Suppose λ0 � is an eigen-value of A, and 0 �= Z = (u, v, h1, . . . , hN ) ∈ D(A) sat-isfiesAZ = λ0Z. Then, (3.7) becomes

v(x) ≡ 0.

Then, the equation of u with its boundary conditionschanges to be

{u′′(x) = iλu(x),u(1) = u(0) = u′(1) = 0, (3.9)

which yield u(x)= 0. We can get hj(x)= 0, j= 1, 2, …,Ndirectly from (3.6). These contradict Z � 0, which com-pletes the proof. �Lemma3.3: LetA be defined as in (2.4) and let be givenby (3.8). Then

= {λc,k ∈ (−bk+1, −bk) ⊂ R, k = 1, 2, . . . ,N − 1}.(3.10)

Proof: Since −b j �∈ , j = 1, 2, . . . ,N, p(λ) = 0 isequivalent to q(λ) = 0, where

p(λ) = N�j=1

a2jλ + b j

, q(λ) = p(λ)N�j=1

(λ + b j). (3.11)

420 L. LU AND J.M. WANG

Thus, the elements of are zeros of q(λ). However, q(λ)is an (N − 1)th order polynomial, and hence there areat most N − 1 number of zeros of p(λ). Now, we find allthese zeros. Notice that for any j = 1, 2, …, N − 1, wheni is even, q(− bj)> 0, q(− bj + 1)< 0. By Rolle’s theorem,there exists a solution q(λ) = 0 in ( − bj + 1, −bj). Thiscompletes the proof by (1.4). �

We need the Lemma 3.3 of Wang et al. (2009).

Lemma 3.4: Suppose that λ � 0 and λ�. Then,

λN�j=1

(λ + b j) = a(λ2 + bλ + c)q(λ) + h(λ) (3.12)

where q(λ) is given by (3.11)

a =( N

�j=1

a2j)−1

, b = aN�j=1

a2jb j, c = aN�j=1

a2jb2j + b2, (3.13)

and h(λ) is a residual polynomial in λ with order N − 2.

By Lemma 3.2 and 3.4, the eigenvalue problem (3.7) isequivalent to the following problem:

⎧⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎩

u′′(x) = iλu(x),v ′′(x) − a[λ2 + bλ + c + a−1h(λ)q−1(λ)]v(x) = 0,

u(0) = v(0) = 0, v(1) = iku′(1),

P(λ)u(1) := a[λ2 + bλ + c

+ a−1h(λ)q−1(λ)]u(1) = kλv ′(1),

(3.14)

where a, b and c are constants given by (3.13), q(λ) is apolynomial in λ with order N − 1 given by (3.11) andh(λ) is given by (3.12):

h(λ) = λN�j=1

(λ + b j) − a(λ2 + bλ + c)q(λ).

Therefore,{q(λ) �= 0, for any λ �= 0 and λ /∈ ,

h(λ)q−1(λ) = O(λ−1) as |λ| → ∞. (3.15)

Lemma 3.5: Suppose that λ � 0 and λ�. Then, for x �[0, 1],

e√aλx, e−

√aλx (3.16)

are linearly independent fundamental solutions of v ′′(x)−aλ2v(x) = 0, and

v ′′(x) − a[λ2 + bλ + c + a−1h(λ)q−1(λ)]v(x) = 0,as |λ| → ∞,

has the following asymptotic fundamental solutions:

{v1(x) = e

√aλx[v10(x) + v11(x)λ−1] + O(λ−2),

v2(x) = e−√aλx[v20(x) + v21(x)λ−1] + O(λ−2),

(3.17)

where

⎧⎪⎨⎪⎩

v10(x) = e(1/2)√abx, v11(x) = −1

8√a(b2 − 4c)xe(1/2)

√abx,

v20(x) = e(−1/2)√abx, v21(x) = 1

8√a(b2 − 4c)xe(−1/2)

√abx.

(3.18)

Proof: The first claim is trivial.We only need to show that(3.17) is the asymptotic fundamental solutions of v ′′(x)− a[λ2 + bλ + c + a−1h(λ)q−1(λ)]v(x) = 0. This can bedone along the same way of Birkhoff (1908) and Naimark(1967). Here, we present briefly a simple calculation to(3.17):

Let

ev1(x, λ) := e√aλx[v10(x) + v11(x)

λ

],

ev2(x, λ) := e−√aλx[v20(x) + v21(x)

λ

], (3.19)

where v i0(x) and v i1(x) (i=1,2) are some functions to bedetermined, and

D(v ) = v ′′(x) − a[λ2 + bλ + c + a−1h(λ)q−1(λ)]v(x),(3.20)

where h(λ)q−1(λ) = O(λ−1). Substitute ev1(x, λ) andev2(x, λ) into D(v), respectively, to yield

e−√aλxD(ev1(x, λ))

= aλ2[v10(x) + v11(x)

λ

]+ 2

√aλ[v ′10(x) + v ′

11(x)λ

]

+[v ′′10(x) + v ′′

11(x)λ

]

− a[v10(x) + v11(x)

λ

][λ2 + bλ + c + a−1h(λ)q−1(λ)]

= λ[2√av ′

10(x) − abv10(x)] + [2√av ′

11(x) − abv11(x)

+ v ′′10(x) − acv10(x)] + λ−1F1(x, λ),

INTERNATIONAL JOURNAL OF CONTROL 421

and

e√aλxD(ev2(x, λ))

= aλ2[v20(x) + v21(x)

λ

]− 2

√aλ[v ′20(x) + v ′

21(x)λ

]

+[v ′′20(x) + v ′′

21(x)λ

]

− a[v20(x) + v21(x)

λ

][λ2 + bλ + c + a−1h(λ)q−1(λ)]

= −λ[2√av ′

20(x) + abv20(x)] − [2√av ′

21(x) + abv21(x)− v ′′

20(x) + acv20(x)] + λ−1F2(x, λ),

where

Fi(x, λ) = v ′′i1(x) − acvi1(x)

− [λvi0(x) + vi1(x)]h(λ)q−1(λ), i = 1, 2,

in which

|[λvi0(x) + vi1(x)]h(λ)q−1(λ)| ≤ M,

|Fi(x, λ)| ≤ M, ∀x ∈ [0, 1],

for some positive constant M. Thus, letting the coeffi-cients of λ1 and λ0 be zero gives

2√av ′

10(x) − abv10(x) = 0, 2√av ′

20(x) + abv20(x) = 0,and

2√av ′

11(x) − abv11(x) + v ′′10(x) − acv10(x) = 0,

2√av ′

21(x) + abv21(x) − v ′′20(x) + acv20(x) = 0.

Now, use the conditions v i0(0)= 1, v i1(0)= 0, i= 1, 2, toobtain

v10(x) = e(1/2)√abx, v11(x) = −1

8√a(b2 − 4c)xe(1/2)

√abx,

v20(x) = e(−1/2)√abx, v21(x) = 1

8√a(b2 − 4c)xe(−1/2)

√abx.

These are (3.18). When |λ| is large enough, we obtain thelinearly independent asymptotic fundamental solutionsof v ′′(x) − a[λ2 + bλ + c + a−1h(λ)q−1(λ)]v(x) = 0 givenby (3.17) (see Birkhoff, 1908):

v1(x) = e√aλx[v10(x) + v11(x)λ−1] + O(λ−2),

and

v2(x) = e−√aλx[v20(x) + v21(x)λ−1] + O(λ−2).

The proof is complete. �

Let λ � 0 and λ�, and let

u(x) = d1e√iλx + d2e−

√iλx, v(x) = d3v1(x) + d4v2(x),

where v i(x), i = 1, 2 are given by (3.17). Then, substi-tute the above equalities into the boundary conditions of(3.14) to obtain

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

d1 + d2 = 0,d3v1(0) + d4v2(0) = 0,

d3v1(1) + d4v2(1) = ik√iλ(d1e

√iλ − d2e−

√iλ),

P(λ)(d1e√iλ + d2e−

√iλ) = kλ[d3v ′

1(1) + d4v ′2(1)].

Hence, (3.14) has non-trivial solution if and only if thecharacteristic equation

det((λ)) = 0, (3.21)

where

(λ) =

⎡⎢⎢⎢⎢⎢⎣

1 1 0 00 0 v1(0) v2(0)

ik√iλe

√iλ −ik

√iλe−

√iλ −v1(1) −v2(1)

P(λ)e√iλ P(λ)e−

√iλ −kλv ′

1(1) −kλv ′2(1)

⎤⎥⎥⎥⎥⎥⎦ .

(3.22)

and the eigenvalues of (3.14) are the zeros of (3.21).Notice that

det((λ))

= −e−√iλv1(0)[ik2λ

√iλv ′

2(1) + P(λ)v2(1)]

+ e−√iλv2(0)[ik2λ

√iλv ′

1(1) + P(λ)v1(1)]

+ e√iλv1(0)[−ik2λ

√iλv ′

2(1) + P(λ)v2(1)]

− e√iλv2(0)[−ik2λ

√iλv ′

1(1) + P(λ)v1(1)],

that is

12det((λ)) = ik2λ

√iλ cosh

√iλ[v ′

1(1)v2(0)

− v1(0)v ′2(1)] + P(λ) sinh

√iλ

× [v1(0)v2(1) − v1(1)v2(0)]. (3.23)

422 L. LU AND J.M. WANG

Simple calculation gives,⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

v ′1(1)v2(0) − v1(0)v ′

2(1)

= e√a(λ+ b

2 )

[√aλ − 1

8a(b2 − 4c) + 1

2√ab+ O(λ−1)

]

+ e−√a(λ+ b

2 )

[√aλ + 1

8a(b2 − 4c) + 1

2√ab+ O(λ−1)

],

v1(0)v2(1) − v1(1)v2(0)

= e−√a(λ+ b

2 )

[1 + 1

8√a(b2 − 4c)λ−1

]

− e√a(λ+ b

2 )

[1 − 1

8√a(b2 − 4c)λ−1

]+ O(λ−2).

When λ � 0, the coefficient of the highest order of λ goesto

√ae

√a(λ+ b

2 ) + √ae−

√a(λ+ b

2 ) + O(λ−1) = 0. (3.24)

Finally, since the solutions of e2√a(λ+ b

2 ) + 1 = 0 are

λn = −b2

+√aa

(n − 1

2

)π i, n ∈ Z.

Applying Rouché’s theorem, we get the solutions of(3.24)

λn = −b2

+√aa

(n − 1

2

)π i + O(n−1), n ∈ Z.

Theorem 3.1: The eigenvalue of (3.14) has the followingasymptotic expressions:

λn = −b2

+√aa

(n − 1

2

)π i + O(n−1), n ∈ Z, (3.25)

in particular,

Reλn → −12b = −1

2�N

i=1a2i bi�N

i=1a2i< 0, as |n| → ∞, (3.26)

that is, Reλ = − 12b is the asymptote of the eigenvalues λn

given by (3.25). Here, b is given by (3.13). Furthermore, thecorresponding eigenfunctions vn(x), n ∈ Z have the asymp-totic expressions

un(x) = o(n−1), vn(x) = sin(n − 1

2

)πx + O(n−1),

(3.27)

and

λ−1n v ′

n(x) = −i√a cos

(n − 1

2

)πx + O(n−1). (3.28)

Moreover, {vn(x), n ∈ Z} and {λ−1n v ′

n(x), n ∈ Z} areapproximately normalised in L2(0, 1), in the sense that

there exist positive constants c1 and c2 independent of nsuch that for n ∈ Z,

c1 ≤ ||vn||L2, ||λ−1n v ′

n||L2 ≤ c2. (3.29)

Proof: Equation (3.25) has been proved. Only proofs for(3.27)–(3.29) are needed. Since λ � 0, λ�, in view of(3.22), (3.25), Lemma 3.5 and some facts in linear algebra,the eigenfunctions u and v corresponding to the eigen-value λ are given by

u(x) =

∣∣∣∣∣∣∣∣∣∣∣

1 1 0 00 0 v1(0) v2(0)

ik√iλe

√iλ −ik

√iλe−

√iλ −v1(1) −v2(1)

e√iλx e−

√iλx 0 0

∣∣∣∣∣∣∣∣∣∣∣=∣∣∣∣ 1 1e√iλx e−

√iλx

∣∣∣∣∣∣∣∣ v1(0) v2(0)−v1(1) −v2(1)

∣∣∣∣= 2 sinh(

√iλx)[v1(0)v2(1) − v1(1)v2(0)],

v(x) =

∣∣∣∣∣∣∣∣∣∣∣

1 1 0 00 0 v1(0) v2(0)

ik√iλe

√iλ −ik

√iλe−

√iλ −v1(1) −v2(1)

0 0 v1(x) v2(x)

∣∣∣∣∣∣∣∣∣∣∣= −ik

√iλ∣∣∣∣ 1 1e√iλ −e−

√iλ

∣∣∣∣∣∣∣∣ v1(0) v2(0)v1(x) v2(x)

∣∣∣∣= −ik

√iλ(−e−

√iλ − e

√iλ)[v1(0)v2(x)

− v2(0)v1(x)]

= 2ik√iλ cosh(

√iλ)[v1(0)v2(x)

− v2(0)v1(x)].

Owing to the fact of (3.25) that

12√ab+ √

aλn = √a(

λn + b2

)

=(n − 1

2

)π i + O(n−1), n ∈ Z.

Equation (3.27) and (3.28) are thus proved by taking

un(x) = 14k

√iλn cosh(

√iλn)

u(λn, x)

= v1(0)v2(1) − v1(1)v2(0)2k

√iλn

sinh[√iλnx]

cosh(√iλn)

= O(n− 12 )O(e−c

√n) = o(n−1), (c > 0),

INTERNATIONAL JOURNAL OF CONTROL 423

and

vn(x) = 14k

√iλn cosh(

√iλn)

v(λn, x)

= − 12i[v1(0)v2(x) − v2(0)v1(x)]

= − 12i[1 + O(λ−2

n )] {e−√

aλnx[v20(x)

+ v21(x)λ−1n ] + O(λ−2

n )} + 12i[1 + O(λ−2

n )]

×{e√aλnx[v10(x) + v11(x)λ−1

n ] + O(λ−2n )}

= 12i

[e√a(λn+ 1

2 b)x − e−√a(λn+ 1

2 b)x]

+ O(λ−1n )

= sin(n − 1

2

)πx + O(n−1).

Finally,

||vn||L2 =∫ 1

0sin2

(n − 1

2

)πxdx + O(n−1)

= 12

+ O(n−1),

and

||λ−1n v ′

n||L2

=∫ 1

0

(−i

√a cos

(n − 1

2

)πx)

×(

−i√a cos

(n − 1

2

)πx)dx

+O(n−1)

= a2

+ O(n−1).

These give (3.29). The proof is complete. �

By Lemma 3.2, the eigenvalue problem (3.7) is equiv-alent to the following problem:

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

u′′(x) = iλu(x),v ′′(x) = s2(λ)v(x),

u(0) = v(0) = 0, v(1) = iku′(1),

P(λ)u(1) = kλv ′(1),

(3.30)

where p(λ) is given by (3.11), and s(λ) =√

λp(λ)

. Hence,

u(x) = e√iλx − e−

√iλx, v(x) = es(λ)x − e−s(λ)x.(3.31)

By boundary condition v(1) = iku′(1), and P(λ)u(1) =kλv ′(1), we have

⎧⎨⎩es(λ) − e−s(λ) = ik

√iλ(e

√iλ + e−

√iλ),

P(λ)(e√iλ − e−

√iλ) = kλs(λ)(es(λ) + e−s(λ)),

Further calculation shows that

P(λ)(es(λ) − e−s(λ))(e√iλ − e−

√iλ)

= ik2λs(λ)(es(λ) + e−s(λ))(e√iλ + e−

√iλ),

then

e2s(λ) = P(λ)(e√iλ − e−

√iλ) + ik2λs(λ)(e

√iλ + e−

√iλ)

P(λ)(e√iλ − e−

√iλ) − ik2λs(λ)(e

√iλ + e−

√iλ)

.

So for any λc � , when λ → λc, μ = λ − λc → 0, wehave

e2s(λ) − 1 − 2ik2λs(λ)(e2√iλ + 1)

P(λ)(e2√iλ − 1) − ik2λs(λ)(e2

√iλ + 1)

= e2s(μ) + 1 + O(μ) = 0. (3.32)

The above equation is achieved from the fact that whenλ → λc,

p(λ) =N∑j=1

a2jλ + b j

=N∑j=1

a2jλc + b j

11 + λ−λc

λc+b j

= −μ

N∑j=1

a2j

[1

(λc + b j)2− μ

(λc + b j)3+ O(μ2)

],

and

2ik2λs(λ)(e2√iλ + 1)

P(λ)(e2√iλ − 1) − ik2λs(λ)(e2

√iλ + 1)

= 2ik2λ√

λ(e2√iλ + 1)

P(λ)√p(λ)(e2

√iλ − 1) − ik2λ

√λ(e2

√iλ + 1)

= 2P(λ)

√p(λ)(e2

√iλ−1)

ik2λ√

λ(e2√iλ+1)

− 1= 2

O(μ) − 1

= −2[1 + O(μ)].

From (3.32), e2s(μ) + 1 = 0 yields

s(μ) =(n − 1

2

)π i, i.e.

λ

p(λ)= −

(n − 1

2

)2

π2,

n = 1, 2, . . . . (3.33)

424 L. LU AND J.M. WANG

We also have

λ

p(λ)= λc + μ

p(λ)

= − 1μ

λc(1 + μ

λc)

N∑i=1

a2i(λc + bi)2

⎡⎣1 −

∑Ni=1

a2i(λc+bi)3∑N

i=1a2i

(λc+bi)2

μ + O(μ2)

⎤⎦

= −λc(1 + μ

λc)

μ

[1 +

μ

]+ O(μ)

= − 1μ

λc

[1 +

(1λ c

+e

)μ

]+ O(μ), (3.34)

where

=N∑j=1

a2j(λc + b j)2

, =N∑j=1

a2j(λc + b j)3

.

This together with (3.33), we have

− 1μ

λc

[1 +

(1λ c

+e

)μ

]+ O(μ)

= −(n − 1

2

)2

π2, n → ∞.

Thus

μn = 1(n − 1

2 )2π2

λc

+ O(n−3), n → ∞,

or

λn = λc + μn = λc + 1(n − 1

2 )2π2

λc

+ O(n−3),

n → ∞.

We summarise these results as Proposition 3.2.Proposition 3.2: Let A be defined by (2.4) and λ be aneigenvalue ofA, satisfying λ � −bj, j= 1, …, N. Then, theeigenfunction corresponding to λ is of the form

(u(x), v(x),

a1λ + b1

v ′(x), . . . ,aN

λ + bNv ′(x)

),

where u(x) = o(n−1), v(x) = sin(n − 12 )πx + O(n−1),

for some n ∈ N+. Moreover,

(i) When |λ|→ �, the eigenvalues {λn0, λn0} ofA havethe following asymptotic expressions:

λn0 = −b2

+√aa

(n − 1

2

)π i + O(n−1), n ∈ Z.

(3.35)

The corresponding eigenfunctions (u(x), v(x),a1

λ+b1v ′(x), . . . , aN

λ+bNv ′(x)), satisfy

un(x) = o(n−1), vn(x) = sin(n − 1

2

)πx

+O(n−1). (3.36)

(ii) For any 1 � k � N − 1, there is a sequence of eigen-values {λn, k} of A, which has the following asymp-totic expressions:

λn,k = λc,k + 1(n − 1

2 )2π2

λc,k

k+ O(n−3),

n → ∞, (3.37)

where

k =N∑j=1

a2j(λc,k + b j)2

.

Furthermore, the corresponding eigenfunctions(u(x), v(x), a1

λ+b1v ′(x), . . . , aN

λ+bNv ′(x)), satisfy

(3.36).

The following result is a direct consequence of theabove analysis.

Theorem 3.2: LetA be defined as in (2.4). Then

(i) A has the eigenvalues

{λn0, λn0, λn,1, λn,2, . . . , λn,N−1, n ∈ N+},

where λn0 and λn, k, k = 1, 2, …, N − 1has the asymptotic expressions (3.35) and (3.37),respectively.

(ii) When |λ|→ �, the eigenfunctions corresponding toλn0 and λn0 are given by

Wn0(x) =(0, sin

(n − 1

2

)πx,

− i√aa1 cos

(n − 1

2

)πx, . . . ,

− i√aaN cos

(n − 1

2

)πx)

+ (o(n−1),O(n−1),O(n−1),O(n−1),

. . . ,O(n−1)),

INTERNATIONAL JOURNAL OF CONTROL 425

and

Wn0(x) =(0, sin

(n − 1

2

)πx,

i√aa1 cos

(n − 1

2

)πx, . . . ,

i√aaN cos

(n − 1

2

)πx)

+ (o(n−1),O(n−1),O(n−1),O(n−1)

, . . . ,O(n−1)), (3.38)

for n → �.(iii) When λn, k → λc, k, n → �, the eigenfunctions cor-

responding to λn, k are given by

Wn,k(x) =(0, 0,

a1λn,k + b1

cos(n − 1

2

)πx, . . . ,

× aNλn,k + bN

cos(n − 1

2

)πx)

+ (o(n−2),O(n−1),O(n−1),O(n−1)

, . . . ,O(n−1)) (3.39)

for k = 1, 2, …, N − 1.

In order to investigate the residual and continuousspectrum ofA, we need the adjoint operatorA∗.

Lemma 3.6: LetA be defined by (2.4). Then

A∗

⎛⎜⎜⎜⎜⎜⎝

uv

h1...hN

⎞⎟⎟⎟⎟⎟⎠

�

= −

⎛⎜⎜⎜⎜⎜⎜⎝

−iu′′(�N

j=1a jh j

)′

a1v ′ + b1h1...

aNv ′ + bNhN

⎞⎟⎟⎟⎟⎟⎟⎠

�

, (3.40)

with

D(A∗) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

⎛⎜⎜⎜⎜⎜⎜⎜⎝

uvwh1...hN

⎞⎟⎟⎟⎟⎟⎟⎟⎠

�∣∣∣∣∣∣∣∣∣∣∣∣∣∣

u(0) = v(0) = 0, u ∈ H2(0, 1),

u(1) = kN�j=1

a jh j(1),

v(1) = iku′(1),

h j ∈ L2(0, 1), j = 1, . . . ,N,

N�j=1

a jh j(x) ∈ H1(0, 1).

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

.

(3.41)

Theorem3.3: LetA be defined by (2.4). Then σr(A) = ∅,where σr(A) denotes the set of residual spectrum ofA.

Proof: Since λ ∈ σr(A), λ ∈ σp(A∗), the proof will beaccomplished if we can show that σp(A) = σp(A∗). This

is because obviously the eigenvalues of A are symmet-ric on the real axis. From (3.40), the eigenvalue problemA∗Z = λZ for λ ∈ C and 0 �= Z = (u, v, h1, . . . , hN ) ∈D(A∗), we have

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

iu′′(x) = λu(x),−(�N

i=1a jh j(x))′ = λv(x),

−[a jv′(x) + b jh j(x)] = λh j(x), j = 1, . . . ,N,

u(0) = v(0) = 0, v(1) = iku′(1),

u(1) = kN�j=1

a jh j(1).

(3.42)Equation (3.42) is the same as (3.1) by settingeu = −u andev = −v . Hence,A∗ has the same eigenvalues withA. Theproof is complete. �Theorem 3.4: Let A be defined as in (2.4) and let begiven by (3.8). Then

σc(A) = , (3.43)

where σc(A) is the set of the continuous spectrum ofA.

Proof: Let λ �∈ σp(A) and λ � −bj. For any G =( f , g, ϕ1, . . . ϕn) ∈ H, solve (λI − A)[u, v, h1, . . . , hN]= G, that is⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

λu(x) + iu′′(x) = f (x),

λv(x) −[

N�j=1

a jh j(x)]′

= g(x),

λh j(x) − [a jv′(x) − b jh j(x)] = ϕ j(x),

j = 1, 2, . . . ,N,

u(0) = v(0) = 0, v(1) = iku′(1),

u(1) = kN�j=1

a jh j(1).

(3.44)

Step I: We first show ⊆ σc(A). From (3.44), we get

h j = a j

λ + b jv ′(x) + 1

λ + b jϕ j(x),

N�j=1

a jh j(x)

= N�j=1

a2jλ + b j

v ′(x) + N�j=1

a j

λ + b jϕ j(x). (3.45)

Let

H(x) = N�j=1

a jh j(x), �(λ, x) = N�j=1

a j

λ + b jϕ j(x). (3.46)

We claim that ⊆ σc(A). In fact, when λ � , it has

N�j=1

a2jλ + b j

= 0.

426 L. LU AND J.M. WANG

By (3.45) and (3.46), we have

H(x) = �(λ, x).

SinceH(x)�H1(0, 1), the above identity holds true unless�(λ, x) � H1(0, 1). This shows that λ /∈ ρ(A), or

⊆ σc(A),

by Lemma 3.2 and Theorem 3.3.Step II: Now, we show that σc(A) ⊆ , or equiva-

lently, any λ /∈ σp(A) ∪ belongs to ρ(A). To do this,suppose that λ /∈ σp(A) ∪ . We write (3.44) as

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

u′′(x) − iλu(x) = −i f (x),H ′(x) = λv(x) − g(x),

v ′(x) = μ(λ)H(x) − μ(λ)�(λ, x),

u(0) = v(0) = 0, v(1) = iku′(1),

u(1) = kH(1),

(3.47)

where

μ(λ) =(

N�j=1

a2jλ + b j

)−1

, for any λ /∈ σp(A) ∪ .

Hence, the equations in (3.47) can be rewritten as the fol-lowing system of first-order differential equations:

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

ddx

⎡⎢⎢⎣

uu′

Hv

⎤⎥⎥⎦ = M(λ)

⎡⎢⎢⎣

uu′

Hv

⎤⎥⎥⎦−

⎡⎢⎢⎣

0i fg

μ(λ)�

⎤⎥⎥⎦ ,

u(0) = v(0) = 0, v(1) = iku′(1),

u(1) = kH(1),

(3.48)

where

M(λ) =[A(λ) 00 B(λ)

],A(λ) =

[0 1iλ 0

],

B(λ) =[

0 λ

μ(λ) 0

].

Note that

eM(λ)x =[eA(λ)x 00 eB(λ)x

],

eA(λ)x =[a11(λ, x) a12(λ, x)a21(λ, x) a22(λ, x)

],

eB(λ)x =[b11(λ, x) b12(λ, x)b21(λ, x) b22(λ, x)

],

where

⎧⎨⎩a11(λ, x) = cosh(

√iλx), a12(λ, x) = 1√

iλsinh(

√iλx),

a21(λ, x) =√iλ sinh(

√iλx), a22(λ, x) = cosh(

√iλx);⎧⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎩

b11(λ, x) = cosh(√

λμ(λ)x),

b12(λ, x) =√

λ√μ(λ)

sinh(√

λμ(λ)x),

b21(λ, x) =√

μ(λ)√λ

sinh(√

λμ(λ)x),

b22(λ, x) = cosh(√

λμ(λ)x).

By u(0) = v(0) = 0, the general solution of (3.48) isgiven by

⎡⎢⎢⎣

u(x)u′(x)H(x)v(x)

⎤⎥⎥⎦ = eM(λ)x

⎡⎢⎢⎣

u(0)u′(0)H(0)v(0)

⎤⎥⎥⎦

−∫ x

0eM(λ)(x−τ )

⎡⎢⎢⎣

0i f (τ )

g(τ )

μ(λ)�(λ, τ )

⎤⎥⎥⎦ dτ

=

⎡⎢⎢⎣a12(λ, x)u′(0)a22(λ, x)u′(0)b11(λ, x)H(0)b21(λ, x)H(0)

⎤⎥⎥⎦

−∫ x

0

⎡⎢⎢⎣

ia12(λ, x − τ ) f (τ )

ia22(λ, x − τ ) f (τ )

b11(λ, x − τ )g(τ ) + b12(λ, x − τ )μ(λ)�(λ, τ )

b21(λ, x − τ )g(τ ) + b22(λ, x − τ )μ(λ)�(λ, τ )

⎤⎥⎥⎦ dτ,

that is

⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

u(x) = a12(λ, x)u′(0) − i∫ x

0a12(λ, x − τ ) f (τ )dτ,

u′(x) = a22(λ, x)u′(0) − i∫ x

0a22(λ, x − τ ) f (τ )dτ,

H(x) = b11(λ, x)H(0) − ξ1(λ, x),v(x) = b21(λ, x)H(0) − ξ2(λ, x),

(3.49)where

ξ j(λ, x) =∫ x

0[b j1(λ, x − τ )g(τ )

+ b j2(λ, x − τ )μ(λ)�(λ, τ )]dτ, j = 1, 2.

INTERNATIONAL JOURNAL OF CONTROL 427

Putting the boundary conditions of v(1) = iku′(1), u(1)= kH(1) into (3.49) to eliminate u′(0), we have

ik cosh√iλ[i∫ 1

0

a12(λ, 1 − τ )

a12(λ, 1)f (τ )dτ

+ kb11(λ, 1)a12(λ, 1)

H(0) − kξ1(λ, 1)a12(λ, 1)

]

=√

μ(λ)√λ

sinh√

λμ(λ)H(0) − ξ2(λ, 1)

− k∫ 1

0cosh[

√iλ(1 − τ )] f (τ )dτ.

The coefficient of H(0) equals

ik2 cosh√iλb11(λ, 1)a12(λ, 1)

−√

μ(λ)√λ

sinh√

λμ(λ)

= 1λ sinh

√iλ[ik2λ

√iλ cosh

√iλ cosh

√λμ(λ)

−√

λμ(λ) sinh√iλ sinh

√λμ(λ)] �= 0,

since λ /∈ σp(A) and by (3.23). So H(0) and hence v canbe uniquely determined by (3.49) and v ′ � L2(0, 1). Oncev ′(x) is known, the hj(x) and u are also uniquely deter-mined by (3.45) and (3.49), with

u′(0) = i∫ 1

0

a12(λ, 1 − τ )

a12(λ, 1)f (τ )dτ + kb11(λ, 1)

a12(λ, 1)H(0)

− kξ1(λ, 1)a12(λ, 1)

.

Hence, (λI − A)−1 exists and is bounded. Therefore, λ ∈ρ(A). The proof is complete. �

4. Riesz basis property and exponential stability

In this section, we study the Riesz basis property for sys-tem (2.5). To do this, we need the following result men-tioned in Guo and Zhang (2012) (see also Guo & Zwart,2001).Theorem 4.1: Let A be a densely closed linear operatorin a Hilbert space H with isolated eigenvalues {λi}∞1 andσ r(A)= �. Let {φn}∞1 be a Riesz basis for H. Suppose thatthere are N0 � 1 and a sequence of generalised eigenvectors{ψn}∞N0

of A such that

∞∑n=N0

‖ψn − φn‖2H < ∞. (4.1)

Then, there exists M( � N0) number of generalised eigen-vectors {ψn0}M1 such that {ψn0}M1

⋃{ψn}∞M+1 forms a Rieszbasis for H.

Theorem 4.2: LetA be defined by (2.4). Then

(i) There is a sequence of generalised eigenfunctions ofA, which forms a Riesz basis for the state spaceH.

(ii) All eigenvalues with large modulus are algebraicallysimple.

Therefore, for the semigroup eAt , the spectrum-determined growth condition holds true: s(A) = ω(A),where s(A) = sup{Reλ | λ ∈ σ (A)} is the spectral boundof A and ω(A) = limt→∞ 1

t ln ‖eAt‖ is the growth orderof eAt .

Proof: For any n ∈ N+, set

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

e1 = (1, 0, 0, . . . , 0),

ψn0(x) =(0, sin

(n − 1

2

)πx, −i

√aa1 cos

(n − 1

2

)πx

, . . . , −i√aaN cos

(n − 1

2

)πx)

,

ψn0(x) =(0, sin

(n − 1

2

)πx, i

√aa1 cos

(n − 1

2

)πx

, . . . , i√aaN cos

(n − 1

2

)πx)

,

ψn,k(x) =(0, 0,

a1λn,k + b1

, . . . ,aN

λn,k + bN

)

× cos(n − 1

2

)πx, k = 1, 2, . . . ,N − 1.

(4.2)We need to prove

{e1, ψn0, ψn0, ψn,1, ψn,2, . . . , ψn,N−1

}∞

1(4.3)

forms a Riesz basis inH. Since we know that

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

e1 = (1, 0, 0, . . . , 0),

φn0(x) =(0,

√2 sin

(n − 1

2

)πx, 0, . . . , 0

),

φn,1(x) =(0, 0,

√2 cos

(n − 1

2

)πx, 0, . . . , 0

),

φn,2(x) =(0, 0, 0,

√2 cos

(n − 1

2

)πx, . . . , 0

),

...

φn,N (x) =(0, 0, 0, 0, . . . , 0,

√2 cos

(n − 1

2

)πx)

,

(4.4)forms an orthonormal basis in H. Then, there exists aninvertible matrix Tn such that

{e1, ψn0, ψn0, ψn,1, ψn,2, . . . , ψn,N−1}∞1 Tn= {e1, φn0, φn,1, φn,2, . . . , φn,N}∞1 ,

428 L. LU AND J.M. WANG

where

Tn = 1√2

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

√2 0 0 0 0 · · · 00

√2

√2 0 0 · · · 0

0 i√aa1 −i

√aa1

a1λn,1 + b1

a1λn,2 + b1

· · · a1λn,N−1 + b1

0 i√aa2 −i

√aa2

a2λn,1 + b2

a2λn,2 + b2

· · · a2λn,N−1 + b2

· · · · · · · · · · · ·0 i

√aaN −i

√aaN

aNλn,1 + bN

aNλn,2 + bN

· · · aNλn,N−1 + bN

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

.

Since bj � bk, λn, j � λn, k, 1� j< k�N, a direct compu-tation shows

detTn = −2i√2a(

N�i=1

ai)

�

1≤i< j≤N(bi − b j) �

1≤i< j≤N(λn,i − λn, j)

�N−1i=1 [�N

j=1(λn,i + b j)]�= 0,

where a is given by (3.13). Then, the two branchesof vectors {e1, ψn0, ψn0, ψn,1, ψn,2, . . . , ψn,N−1}∞1 and{e1, φn0, φn,1, φn,2, . . . , φn,N}∞1 are equivalent. Hence,

{e1, ψn0, ψn0, ψn,1, ψn,2, . . . , ψn,N−1

}∞

1

forms a Riesz basis forH. By (4.2) and Theorem 3.2, thereexists an N0 ∈ N

+, such that

∞∑n=N0

(‖Wn0 − ψn0‖2H + ∥∥Wn0 − ψn0

∥∥2H

+N−1∑k=1

∥∥Wn,k − ψn,k∥∥2H

)= O(n−2) < ∞. (4.5)

So by Theorem 4.1, we conclude that the generalisedeigenfunctions ofA form a Riesz basis inH, then (i) andhence (ii) hold true. The proof is complete. �

Now, we establish the exponential stability of system(2.4).

Theorem 4.3: Let A be defined by (2.4). Then, thespectrum-determined growth condition ω(A) = s(A)

holds true for the C0-semigroup eAt generated byA. More-over, the system (2.4) is exponentially stable, i.e. there existtwo positive constantsM andω such that the C0-semigroupeAt satisfies

‖eAt‖ ≤ Me−ωt , (4.6)

for some M, ω > 0.

Proof: The spectrum-determined growth condition fol-lows from Theorem 4.2. By Lemma 3.1, Lemma 3.3,Theorem 3.3 and Theorem 3.4 for each λ ∈ σ (A),we have Reλ < 0. This, together with (3.25) and thespectrum-determined growth condition, shows that eAt

is exponentially stable. �

Acknowledgments

The authors would like to thank anonymous reviewers andeditors for their careful reading and valuable suggestions toimprove the paper.

Disclosure statement

No potential conflict of interest was reported by the authors.

Funding

This work is supported by the National Natural Science Foun-dation of China [grant number 61673061].

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430 L. LU AND J.M. WANG