Gen. Math. Notes, Vol. 16, No. 2, June, 2013, pp.1-13
c
ISSN 2219-7184; Copyright ICSRS
Publication, 2013
www.i-csrs.org
Available free online at http://www.geman.in
Coupled Linear Feedback and Sliding
Model Control for a Serially
Connected Euler-Bernoulli Beam
Xuezhang Hou
Mathematics Department, Towson University
Baltimore, Maryland- 21252, USA
E-mail: xhou@towson.edu
(Received: 15-7-11 / Accepted: 2-4-13)
Abstract
A system of serially connected string and Euler-Bernoulli beam with coupled
linear feedback control and sliding model control is studied in the present paper.
The system is formulated by partial differential equations with the boundary
conditions. The eigenvalues and eigenfunctions of the system operator are
discussed in the appropriate Hilbert spaces. A sliding model control is applied to
the serially connected beam and it has been shown that the actual sliding mode
of the system can be approximated by ideal sliding modes in any accuracy under
certain conditions. In this paper, a significant semigroup property of restriction
of the system operator is derived.
Keywords: Serially Connected Euler-Bernoulli Beam, Feedback Control,
Sliding Model Control, Semigroup of Linear Operators.
1
Introduction
The vibration and control of serially connected strings and Euler-Bernoulli
beams with linear feedback controls at joins have been studied extensively
in the last two decades (see, e.g., [2-4,7,10,12,13,16,17]). In addition to the
analysis of the distribution of eigenvalues, one also needs to establish the socalled spectrum-determined growth condition in order to conclude exponential
stability for these infinite-dimensional systems form spectral analysis. In the
2
Xuezhang Hou
case of serially connected strings, the first results on exponential stability were
obtained in [12] for a 2-connected strings with linear feedbacks at the middle
of the span. The stability of N-connected strings under joints feedback was
studied in [13].
In this paper, we consider the following serially connected beams with linear
feedback control
ytt (x, t) + yxxxx (x, t) = 0,
Lj−1 < x < Lj ,
j = 1, 2, · · · , n.
The boundary condition are
(
y(0) = yxx (0) = 0,
yx (Ln ) = yxxx (Ln ) = 0.
(1)
(2)
The linear feedback control at the joint points Lj , j = 1, · · · , n − 1, take the
form

+
y(L−

j , t) = y(Lj , t),


y (L− , t) = y (L+ , t)
xx
xx
j
j
(3)
+
−
−
−
j
2

y
tx (Lj , t) − ytx (Lj , t) = (−1) rj yt (Lj , t) + pj yxx (Lj , t),



−
−
−
2
j
yxxx (L+
j , t) − yxxx (Lj , t) = −qj yt (Lj , t) + (−1) sj yxx (Lj , t),
Where 0 = L0 < L1 < · · · < Ln and
p2j ≥ 0,
qj2 ≥ 0,
p2j + qj2 > 0,
rj , sj ∈ R
p2j α2 + qj2 β 2 + (rj − sj )αβ ≥ 0,
∀α, β ∈ R
(4)
Let us defines the energy of system (1.1)-(1.4)as
n Z
1 X Lj 2
2
[y (x, t) + yxx
(x, t)]dx.
E(t) =
2 j=1 Lj−1 t
Then a simple computation shows that Ė(t) ≤ 0 and hence the system is
dissipative.
Without loss of generality, we may assume that n is odd. For j = 1, 2, · · · , n,
we set

1
(−1)j − 1


u
(x,
t)
=
y
(L
+
lj + (−1)j+1 lj x, t)

j
t
j

2
2




(−1)j+1
(−1)j − 1


+
y
(L
+
lj + (−1)j+1 lj x, t) ,

xx
j

2
lj
2
(5)
j

 v (x, t) = 1 [y (L + (−1) − 1 l + (−1)j+1 l x, t)

j
t
j
j
j


2
2




(−1)j+1
(−1)j − 1


−
y
(L
+
lj + (−1)j+1 lj x, t)],

xx
j
lj2
2
3
Coupled Linear Feedback and Sliding...
where lj = Lj − Lj−1 , j = 1, 2, · · · , n, 0 ≤ x ≤ 1.Then system (1.1)-(1.4) can
be transformed into the form of
"
#
#
 "
u(x,
t)
u(x,
t)

2
∂
∂


= K ∂x
2

 ∂t v(x, t)
v(x, t)
(6)
[A, B][ux (0), vx (0), u(0), v(0)]T = 0,




[E, F ][ux (1), vx (1), u(1), v(1)]T = 0,
where
u(x, t) = [u1 (x, t), u2 (x, t), ..., un (x, t)]T , v(x, t) = [v1 (x, t), v2 (x, t), ..., vn (x, t)]T
and
(2n × 2n)-matrices


0 0
0 ...
0
0 0
0 ...
0
 0 P21 0 . . .

0
0 P22 0 . . .
0




0
0 0 P42 . . .
0
A =  0 0 P41 . . .
(6a)
,
 .. ..

..
..
.. ..
..
..
..
 . .

. ...
.
. .
.
.
.
0




B=


0
0
. . . P(n−1)1 0
Pn1 0
0
0 P̃21 0
0
0 P̃41
..
..
..
.
.
.
0
0
0




E=






F =


P11 0
0 P31
..
..
.
.
0
0
0
0
P̃11 0
0 P̃31
..
..
.
.
0
0
0
0
...
...
...
...
0
0
0
..
.
. . . P̃(n−1)1
...
...
...
0
0
..
.
. . . P(n−2)1
...
0
...
...
..
.
0
0
..
.
. . . P̃(n−2)1
...
0
0
0
. . . P(n−1)2
Pn2 0
0
0 P̃22 0
0
0 P̃42
..
..
..
.
.
.
0
0
0
0 P12 0
0 0 P32
.. ..
..
. .
.
0 0
0
0 0
0
0 P̃12 0
0 0 P̃32
.. ..
..
. .
.
0 0
0
0 0
0
Pn1 =
1
1
, Pn2 =
1
−1

0
0
0
..
.



,



. . . P(n−2)2
...
0



,

0 
0
...
...
..
.
0
0
..
.

0
0
..
.
0
0
..
.
. . . P̃(n−2)2
...
0
,
(6b)
. . . P̃(n−1)2
0
0
..
.
...
...
..
.
where for j = 1, 2, · · · , n,
...
...
...
..
.



,

0 
0
(6c)
(6d)
4
Xuezhang Hou
0
 0
Pj 1 = 
 l1j


0
0
 0
0 
 , Pj 2  1
1

 lj
lj +1

−1
lj
1
lj
1
lj +1

0
0
1
lj +1
1
−lj +1


,




1
1
−1

1 
−1
−1 
 , P̃j 2 = 

2
 −pj − rj 0  .
0 
0
qj2 − sj
0
1

1
P̃j 1 = 
 p2j − rj
qj2 + sj
Now we confine ourselves to system (1.1)-(1.4) with A, B, E, F specified by
(1.6). Divide by ρω1 both sides of those equations which contain nonzero
factors ρ in the system M̃ C = 0; then we have becomes
M̃ C = 0,
(7)
where
M̃ =
M1 M2 M3 M4
(8)
and for 1 ≤ k ≤ 4.

Q0 k
0
..
.
0
Q2 k
..
.





0
0

Mk = 
Q
R
ω
ρl
 1k k 1
1kωk ρl1

.
..
..

.


0
0
0
0
0 ...
R2 k . . .
..
...
.
0 ...
0 ...
..
..
.
.
0
0
0
0
..
.
0
0
..
.
0
0
..
.
Qn−k
0
..
.
. . . Q(n−2)kωk ρln−2 R(n−2)kωk ρln−1
...
0
0
0
0
..
.




R( n − 1)k 

,
0


..

.


0
Qn 1ωk ρln
(9)
with
T
T
Q01 = 1 − i 1 + i
, Q02 = 1 + i 1 − i
,
Q03 = Q01 , Q04 = Q02 , Qn1 = Q01 ,
Qn2 = Q02 · i, Qn3 = −Qn1 , Qn4 = −Qn2 .

(10)
5
Coupled Linear Feedback and Sliding...
For j = 1, 3, · · · , n − 2, l = 2, 4, · · · , n − 1,
Qj1 =
h
1−i+
(1+i)p2j −(1−i)rj
,
ρω1
−(1 + i) +
(1−i)qj2 +(1+i)sj
,
ρω1
1 − i, 1 + i
iT
,
(11a)
iT
(1−i)p2j −(1+i)rj
(1+i)qj2 +(1−i)sj
Qj2 = −(1 − i) +
,
, −(1 + i) +
, 1 + i, 1 − i
ρω1
ρω1
(11b)
i
h
T
(1−i)qj2 +(1+i)sj
(1+i)p2j −(1−i)rj
,
Qj3 = −(1 − i) +
,
1
+
i
+
,
1
−
i,
1
+
i
ρω1
ρω1
(11c)
h
i
T
(1−i)p2j −(1+i)rj
(1+i)qj2 +(1−i)sj
Qj4 = 1 − i +
,
, 1+i+
, 1 + i, 1 − i
ρω
ρω
h
1
1
(11d)
iT
Ql1 =
h
Ql2 =
h
Ql3 =
h
(12b)
i
T
(1−i)p2l −(1+i)rl
(1+i)ql2 +(1−i)sl
,
−(1 + i) +
, 1−i+
, 1 + i, 1 − i
ρω1
ρω1
Ql4 =
h
(12d)
iT
(1−i)ql2 +(1+i)sl
(1+i)p2l −(1−i)rl
;
, −(1 − i) +
, 1 − i, 1 + i
−(1 + i) +
ρω1
ρω1
1+i+
(1−i)p2l −(1+i)rl
,
ρω1
−(1 − i) +
(1+i)ql2 +(1−i)sl
,
ρω1
1 + i, 1 − i
,
(12a)
1+i+
(1+i)p2l −(1−i)rl
,
ρω1
1−i+
(1−i)ql2 +(1+i)sl
,
ρω1
1 − i, 1 + i
iT
,
(12d)
T
Rj1 =
1 + i, 1 − i, −(1 + i), 1 − i
Rj2 =
1 + i, −(1 − i), −(1 − i), 1 + i
Rj3 =
Rl1 =
1 − i, 1 + i, −(1 − i), 1 + i
Rl2 =
−(1 − i), 1 + i, −(1 + i), 1 − i
T
−(1 + i), −(1 − i), −(1 + i), 1 − i
T
Rj4 = −(1 + i), 1 − i, −(1 − i), 1 + i
(13)
T
(14)
T
T
−(1 − i), −(1 + i), −(1 − i), 1 + i
T
Rl4 = 1 − i, −(1 + i), −(1 + i), 1 − i
Rl3 =
T
6
Xuezhang Hou
2
Spectral Analysis and Semigroup Generation
In this section, we derive the characteristic equation satisfied by eigenvalues of system (1.1-1.4). To begin with, we put system (1.1-1.4) into the
framework of evolutionary equations in an underlying Hilbert space H. Take
H = (L2 (0, 1))2n and define A : D(A)(⊂ H) → H by
∂2 u
u
A
=K 2
,
(15)
v
∂x v
where
T
T
2
2n [A, B][ux (0), vx (0), u(0), v(0)] = 0,
D(A) = [u, v] ∈ (H (0, 1)) [E, F ][ux (1), vx (1), u(1), v(1)]T = 0
and H 2 (0, 1) denotes the usual Sobolev space. With this setting, system
(1.1-1.4) can be considered as an abstract equation in H:
d u
u
=A
.
(16)
v
dt v
Obviously, A is densely defined in H. Next, we consider the eigenvalue problem
for A. For any given Φ = [f, g]T ∈ H, solve the following equation:
u
f
(λ − A)
=
,
(17)
v
g
i.e.,
" #
" #

2

u
u
∂


− K −1 Φ,
= λK −1

2

v
 ∂x v
T
[A, B][ux (0), vx (0), u(0), v(0)] = 0,




[E, F ][ux (1), vx (1), u(1), v(1)]T = 0,
(18)
which can be further written as a first-order ordinary differential equation of
the following form:
  
 

ux

"
# ux
"
#




v 
−1
−1


v
0
λK
K
Φ
∂




x
2n
x


,
 =
 −


I2n 02n  u 
0
 ∂x  u 
(19)
v
v




[A, B][ux (0), vx (0), u(0), v(0)]T = 0,






[E, F ][ux (1), vx (1), u(1), v(1)]T = 0,
7
Coupled Linear Feedback and Sliding...
where I2n denotes the 2n × 2n identity matrix. Set
02n λK −1
Kλ =
I2n 02n
Then the solution to the governing equation of 19 is




ux (x)
ux (0)
−1 Z x
 vx (x) 


Φ
Kλ (x−s) K

 = eKλ x  vx (0)  −
ds.
e
 u(x) 
 u(0) 
0
0
v(x)
v(0)
(20)
(21)
In order for 21 to satisfy 19, the last two boundary conditions should be fulfilled, i.e.,

T

[A, B][ux (0), vx (0), u(0), v(0)] = 0, Z
1
Kλ

[E, F ]e
[E, F ]eKλ (1−s) [K −1 Φ, 0]T ds,
T
[ux (0), vx (0), u(0), v(0)] =
0
(22)
Define
[A, B]
H(λ) =
.
[E, F ]eKλ
(23)
h(λ) = detH(λ) 6= 0,
(24)
Then for
it has


ux (0)
−1 Z x


Φ
Kλ x  vx (0) 
Kλ (x−s) K
R(λ, A)Φ = [02n , I2n ]e
[02n , I2n ]e
ds,
 u(0)  −
0
0
v(0)
where




ux (0)
0 
Z 1
 vx (0) 
−1

 = H −1 (λ) 
Φ
Kλ (1−s) K

 u(0) 
[E, F ]e
ds .
0
0
v(0)
(25)
(26)
Therefore, in this case, λ ∈ ρ(A) and R(λ, A) is compact.
On the order hand, if h(λ) = 0, for any 4n × 1 nonzero column vector
Z = (ux (0), vx (0), u(0), v(0))T satisfying H(λ)Z = 0, by setting Φ = 0 in 21,
we have


ux (0)
 vx (0) 
Kλ x


 u(0)  = e Z 6= 0
v(0)
8
Xuezhang Hou
and hence (ux (0), vx (0), u(0), v(0))T 6= 0. Therefore,
 
ux
 vx 
u
Kλ x

= [02n , I2n ] 
 u  = [02n , I2n ]e Z 6= 0
v
v
(27)
and satisfies
u
u
A
=λ
v
v
In other words, λ ∈ σ(A) = σp (A).
To sum up, we have obtained the following Theorem.
Theorem 1: Let h(λ) = detH(λ) be defined by 24. Then h(λ) is an entire
function of λ, and the following statements hold:
1. λ ∈ σ(A) if and only if h(λ) = 0, i.e.,
σ(A) = {λ|h(λ) = 0}.
(28)
The eigenvalues are symmetric with respect to the real axis.
2. For each λ ∈ σ(mathcalA), the corresponding eigenfunction [u, v]T is
given by 27, where Z is any nonzero solution of the algebraic equation
H(λ)Z = 0.
3. A is a densely defined discrete operator in H, i.e., A is densely defined
in H and R(λ, A) = (λ − A)−1 is compact for any λ ∈ ρ(A).
4. A is an infinitesimal generator of a C0 -semigroup in H.
3
Sliding Model Control
Let us establish a sliding model control for the system (15)

 ∂z = Az + Bw(z, t)
∂t
z(0) = z0
(1)
where B is a bounded linear operator from H to H, w(z, t) is the control of
the system (1) that is not continuous on the manifold S = Cz = 0, and C is
a bounded linear operator with S = S(z) = Cz ∈ Rn .
9
Coupled Linear Feedback and Sliding...
Now, we consider the δ-neighborhood of sliding mode S = Cz = 0, where
δ > 0 is an arbitrary given positive number. Using a continuous control w̃(z, t)
to replace w(z, t) in the system 1 yields

ż = Az + B w̃(z, t)
z(0) = z0
(2)
where ż = ∂z/∂t, and the solution of (2) belongs to the boundary layer
kS(z)k ≤ δ
Let Ṡ(z) = cż = 0. Applying C to the first equation of (1) leads to the
following the equivalent control:
weq (z, t) = −(CB)−1 C(Az)
with assumption that (CB)−1 exists. Substitute weq (z, t) into 1 to find
ż = [I − B(CB)−1 C]Az.
(3)
Denote P = B(CB)−1 C and A0 = (I − P )A, then 1 becomes
ż = A0 z,
z(0) = z0
(4)
In the rest part of this paper, we are going to show that the actual sliding
mode Z(t) will approach uniformly to the ideal sliding mode Z(t) under certain
conditions.
If (CB)−1 is a compact operator and P A = AP , then A0 = (I − P )A
generates a C0 -semigroup T2 (t) in H and T2 (t) = (I − p)T1 (t), where T1 (t) is
the C0 -semigroup generated by A.
Since (CB)−1 is a compact operator, B and C are bounded linear operators,
we see from the definition of P that P is compact, and therefor the range of
I − P is a closed subspace of H. Since P 2 = P and (1 − P )2 = I − P , I − P
can be viewed as the identity operator on (I − P )H. It can be easily seen that
T2 (t) = (I − P )T1 (t) is a C0 -semigroup in (I − P )H.
Next, we shall prove that the infinitesimal generator of T2 (t) is (I − P )A
and D((I − P )A) = (I − P )D(A).
In fact, for every x ∈ (I − P )D(A), there is a x1 ∈ D(A) such that
x = (I − P )x1 . It should be noted that T1 (t) and I − P are commutative
10
Xuezhang Hou
because A and P are commutative. We see that
lim+
t→0
T2 (t)x − x
(I − P )T1 (t)(I − P )x1 − (I − P )x1
= lim+
t→0
t
t
(I − P )2 T1 (t)x1 − (I − P )x1
= lim+
t→0
t
(I − P )T1 (t)x1 − (I − P )x1
= lim+
t→0
t
T1 (t)x1 − x1
= (I − P ) lim+
t→0
t
= (I − P )Ax1 .
Let à be the infinitesimal generator of T2 (t). Since the limit on the left
exists, we can assert that x ∈ D(Ã) and (I − P )D(A) ⊆ D(Ã).
On the other hand, for any x ∈ D(Ã), since D(Ã) ⊆ (I − P )H, there exists
x̃ ∈ H, such that x = (I − P )x̃, and
lim+
t→0
T2 (t)x − x
T2 (t)(I − P )˜(x) − (I − P )˜(x)
= lim+
t→0
t
t
(I − P )T1 (t)x̃ − (I − P )x̃
= lim+
t→0
t
T1 (t)x̃ − x̃
= (I − P ) lim+
t→0
t
= (I − P )Ax̃.
Since the limit of the left hand side exists, and so the limit of the right hand
side exists, and x̃ ∈ D(A) which implies that D(Ã) ⊆ (I − P )D(A). Thus,
D(Ã) = (I − P )D(A) and Ã, the infinitesimal generator of T2 (t), is (I − P )A.
The proof of the lemma is complete.
Suppose that in the system 1,
1. (CB)−1 exists and it is compact,
2. P A = AP , where P = B(CB)−1 C.
Then for any solution z(t) of the system 4 satisfying S(z 0 ) = 0, z 0 ∈ D(A0 )
and kz0 − z 0 k ≤ δ, z0 ∈ D(A), we have
limkz(t) − z(t)k = 0
δ→0
uniformly on [0, T ] for any positive number T .
We see from the Theorem 2 and Lemma 3 that A and A0 = (I − P )A
are infinitesimal generators of C0 -semigroups T1 (t) and T2 (t) respectively. It
11
Coupled Linear Feedback and Sliding...
follows from theory of semigroup of linear operators that there are positive
constants M1 , M2 , ω1 and ω2 such that
kT1 (t)k ≤ M1 eω1 t ,
kT2 (t)k ≤ M2 eω2 t .
(0 ≤ t ≤ T )
(5)
In the boundary layer kT1 (t)k ≤ δ, the equivalent control is
weq (z, t) = −(CB)−1 CAz + (CB)−1 C ż
(6)
Substitute (6) into (1) to find
ż = (I − P )Az + P ż
Hence, the solution of (7) can be expressed as follows:
Z t
z(t) = T2 (t)z0 +
T2 (t − s)P ż(s)ds,
(7)
(8)
0
and the solution of (4) can be written as
z(t) = T2 (t)z 0
(9)
Substracting (9) from (8) yields
Z
z(t) − z(t) = T2 (t)(z0 − z 0 ) +
t
T2 (t − s)P ż(s)ds
(10)
0
Since P A = AP , we see that P T1 (t) = P T1 (t). It should be emphasized that
(I − P )P = 0 and T2 (t) = (I − P )T1 (t), and consequently,
Z t
Z t
T2 (t − s)P ż(s)ds =
(I − P )T1 (t − s)P ż(s)ds
0
0
Z t
T1 (t − s)(I − P )P ż(s)ds
=
0
=0
It can be obtained from (10) and (5) that
kz(t) − z(t)k ≤ kT2 (t)kkz0 − z 0 k ≤ M2 eω2 T kz0 − zk,
Since kz0 − z0 k ≤ δ, we have
kz(t) − z(t)k ≤ M2 eω2 T δ.
Thus,
limkz(t) − z 0 k = 0.
δ→0
The proof of the theorem is complete.
We see from the Theorem 3 that the actual sliding mode can be approximated by ideal sliding mode in any accuracy.
12
Xuezhang Hou
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