Chương 4. Stability radius for implicit dynamic equations 81 4.1. Stability of IDEs under causal perturbations
4.2. Stability radius under dynamic perturbations
Let Assumptions 4.1, 4.2 hold. The trivial solution of Equation (4.6) is said to be globally Lp-stable if there exist positive constants M3, M4 such that
kP(t)x(t; t0, x0)kKn M3 kP(t0)x0kKn, (4.12)
k x ( t; t
0, x
0)k
Lp(Tt0,Kn)M
4 kP(t
0)x
0k
Kn
for all t t0, x02 Kn.
Next, we extend the definition of stability radius introduced in [38, 45, 68]
for linear IDEs on time scales.
Definition 4.6. Let Assumptions 4.1, 4.2 hold. The complex (real) struc- tured stability radius of Equation (4.2) subject to linear, dynamic and causal perturbations in Equation (4.6) is defined by
rK(E , A; B, C; T) := inf (kSk, the trivial solution of (4.6) is not .
s globally Lp-stable or (4.6) is not of index-1 )
87
For every t0 2 Ta, we define the following operators
Lbt0 u := C( )Mct0 u, Let0 u := C( )Mft0 u, and Lt0 u := C( )Mt0 u.
The operator Lt0 is called a input-output operator associated with the per- turbed equation (4.6). It can be seen directly that Lt0 , Lt0 are the operators from Lp(Tt0;Km) to Lp(Tt0; Kq) , and kLt0k,kLt0 k are the decreasing func-
b b
tions in t0. Furthermore, kLt0 k .
kLt0 k = D- esssupt t0 CHQsG 1B e
Note that kLtk is decreasing in t. Therefore, there exists the limit kL
¥k := lim Ltk.
t !¥ k
Denote
1, g := kLeak 1,
b := kL¥k (4.13)
with the convention 0 1=+¥.
We say that, the causal operator Q 2 L(Lp(Ta; Km), Lp(Ta; Kq)) has finite memory, if there exists a function Y : [a, ¥) ![a, ¥) such that Y(t) t and (I pY(t))Qpt= 0, for all t a. The function Y is called a finite memory function associated with the operator Q.
Since Lt0 u defined by (Lt0 u)(t) := C(t)H(t)Qs(t)G 1 (t)B(t)u(t) is a causal
and e e
finite memory operator, we can adopt the arguments in [45], and get the following lemma.
Lemma 4.7. For any number # > 0, there exists a causal operator Q# 2 L Lp(Ta, Km), Lp(Ta, Kq)
with finite memory such that kLa Q#k < #.
To derive the main result in this section, we prove the following lemma.
Lemma 4.8. Suppose that b < ¥ and a > b, where b is defined in (4.13).
Then, there exist an operator
S 2 L Lp(Ta, Kq), Lp(Ta, Km) ,
the functions y˜, z˜ 2 Llocp(Ta, Kq) and a natural number N0 > 0 such that
88
i) kSk < a, S is causal and has finite memory;
ii) Sh(t) = 0 for every t 2 [0, N0] and all h 2 Lp(Ta, Kq); iii) y˜ 2 Llocp(Ta, Kq) n Lp(Ta, Kq) and supp z˜ [0, N0]; iv) (I LaS)y˜ = z˜.
Proof. Set # := a2abb . By Lemma 4.7, there is a causal operator Q# with finite memory, Q#2 L(Lp(Ta, Km), Lp(Ta, Kq)), such that
kLa Q#
Set Q#,tu := Q#[u]t. It is seen that kQ#,t
Therefore,
kQ#,tk >
for all t 2 Ta. Since
kQ#,ak >
k < #.
Ltk is a decreasing operator in t.
1
b #, 1
b #, there exists a function f˜
02 Lp(Ta, Kq), such that kQ#,af˜
0k > 1b #.
Therefore, we can choose an element t12 Ta, such that 1
kpt1Q#,af˜
0k > b #.
Let Y be a finite memory function associated with Q#. We define
f0 :=
pt1 f˜
0 and N0 := Y(t1), kpt1 f˜
0k and get
k f0k = 1, kQ#,a f0k > 1 #, supp f0 [a, N0], supp Q#,a f0 [a, N0]. b
Set K1 = N0+ 1. We also have
Q#,K1 > 1 #,
b
89
which implies that there are N1 > K1, the function f1 2 Lp(Ta, Km) such that
k f1k = 1, kQ#,K1f1k > 1 #, supp f1 [K1, N1], supp Q#,K1 f1 [K1, N1]. b
Continuing this way, we can set up sequences f fng Lp(Tt0 , Km), fKng, and fNng such that Kn < Nn < Kn+1 satisfying
k fnk = 1, kQ#,Kn fnk > 1 #, supp fn [Kn, Nn], supp Q#,Kn fn [Kn, Nn], b
where n = 0, 1, 2... and K0 := a. We define the function f := ồƠn=0 fn, and see that f , Q# f 2 Llocp(Tt0 , Kq)n Lp(Tt0 , Kq). Let Ln be a linear functional defined on Lp([Kn, Nn]; Km) such that
kLnk = 1,
Ln( Q
# f
nj[Kn,Nn]) =
kQ
# f
nk.
We define the operator
¥ f
S#(h) : = ồ k n+1 kLnhj[Kn,Nn]
n=0 Q
# f
n
for every h 2 Lp(Tt0 , Kq). We can directly see that the operator S#2 L(Lp(Tt0 , Kq), Lp(Tt0 , Km))
is causal and has finite memory. Moreover,
kS#k < b , S#(h)(t) = 0
1 #
b
for h 2 Lp(Tt0 , Kq), t 2[a, N0], and S#(Q# fn) = fn+1. These properties of the operator S# imply that (I S#Q#) f = f0. On the other hand, since
k(Q# La)S#k < #b < 1, 1 #b
the operator I (Q L )S is invertible in L (T , Kq), (T , Km).
We now define # a # Lp t0 Lp t0
S := S#[I (Q# La)S#] 12 L Lp(Tt0 , Kq), Lp(Tt0 , Km) .
Since the operator S# is causal and has finite memory, it is clear that S is causal and has finite memory as well. Moreover, we have
¥
S = ồ S#[(Q# La)S#]k,
k=0
it implies that S(h)(t) = 0 for all t 2[a, N0], h 2 Lp(Tt0 , Kq) and
¥ k b ¥ #b k b
S ồ S# (Q# La)S# < ồ = = a.
k k kk k 1 1 #b 1 2#b
k=0 k #b
k=0
Let us define
y˜ := (I (Q# La)S#)Q# f , z˜ := Q# f0. Since Q# f 2 Llocp(Tt0 , Kq)n Lp(Tt0 , Kq)
and I (Q# La)S# is invertible, we get y˜ 2 Llocp(Tt0 , Kq)n Lp(Tt0 , Kq). Moreover, supp z˜ supp f0 [a, N0] and
(I LaS)y˜ = (I Q#S#)Q# f = Q# f0= z˜.
The proof is complete.
We are now in position to derive the main result in this section.
Theorem 4.9. Let Assumptions 4.1, 4.2 hold. Then
rK(Es, A; B, C; T) = minfb, gg. (4.14) where b, g are defined in (4.13).
Proof. The proof is divided into three steps.
Step 1. We will prove that rK(Es, A; B, C; T) minfb, gg.
Consider the first case where b < ¥, g < ¥. Assume that S is a linear and causal perturbation with kSk< minfb, gg. Then, we have
kSk < g = kLak 1 = esssupt a kCHQsG1Bk 1 . e
Therefore,
kCHQsG 1BSk < 1, almost t 2 Ta,
which implies that the matrix I CHQsG 1BS is invertible, and so, by Lemma 4.2 and Definition 4.3, it is clear that Equation (4.6) is of index-1.
Consequently, it admits a unique mild solution x(t; t0, x0) for all t0 a, x0 2 Kn. On the other hand,
kSk < b = lim kLtk 1 ,
t!¥
91
which implies that there exits a number T > a such that kSkkLTk < 1.
From the formula (4.8), it follows that
C(t)x(t; t0, x0) = C(t)F(t, T)P(T)x(T; t0, x0) + LT(S(C( )[x( )]t0 ))(t) for all t T. Therefore, by Assumption 4.1, we have
kC( )x( ; t0, x0)kLp(TT,Kq) kC( )F( , T)P(T)x(T; t0, x0)kLp(TT,Kq) + kL
TkkS(C( )[x( ; t
0, x
0)]
t0 )
kLp(TT,Kq)
M5 kP(t
0)x
0k+
kL
TkkSkkC( )[x(;t
0,x
0)]
t0kLp(Tt0 ,Kq)= M5
kP(t0)x0k + kLTkkSkkC( )x( ; t0, x0)kLp(Tt0 ,Kq). Hence,
k C ( ) x ( ; t
0, x
0)k
Lp(Tt0 ,Kq)
k C ( ) x ( ; t
0, x
0)k
Lp([t0,T],Kq)+ k C ( ) x ( ; t
0, x
0)k
Lp(TT,Kq)
(M1 + M5) kP(t0)x0k + kLTkkSkkC( )x( ; t0, x0)kLp(Tt0 ,Kq). This proves that
k C( )x( ; t , x) q (M1 + M5) kP(t0)x0k .
0 0 k
Lp(Tt0 ,K ) 1 k LTkkSk
Note that by Assumption 4.1, the constants Ki, i = 0, 1, . . . , 6 in the proof of Theorem 4.4 do not depend on t. Thus, similar to the second part of this proof, (4.12) holds and the perturbed equation (4.6) is globally Lp-stable.
In the second case b = ¥ or g = ¥, the above arguments still hold true for kSk < minfb, gg.
Step 2. We will prove that rK(Es, A; B, C; T) g.
Without loss of generality, we assume that essential supremum, for any # > 0, there D-positive measure, such that
g < ¥. Due to the definition of exists a closed set J Ta with kC(t)H(t)Qs(t)G 1(t)B(t)k (g + #) 1, for all t 2 J.
We consider the linear mapping G : L¥(J, Kq) ! L¥(J, Kq) defined by (Gu)(t) := C(t)H(t)Qs(t)G 1(t)B(t)u(t), for all t 2 J.
92
It is clear that
kGk = D-esssupt2J kC(t)H(t)Qs(t)G 1(t)B(t)k > (g + #) 1.
By Kuratowski and Ryll-Nardzewski Theorem [70, Theorem 5.2.1], we can find a measurable function v defined in J with condition kv(t)k= 1, for all t 2 J, such that
kC(t)H(t)Qs(t)G 1(t)B(t)v(t)k = kC(t)H(t)Qs(t)G 1(t)B(t)k,
for all t 2 J. This means that kGvk= kGk. By using Hahn-Banach Theorem we can find a linear functional L on L¥(J, Kq), such that kLk= 1 and L(Gv)
= kGvk= kGk. We define
(S0u)(t) := v(t)L(u) , kGk for all t 2 J, u 2 L¥(J, Km). It is clear that
(I S0C( )H( )Qs( )G 1( )B( ))v (t) = v(t) ()L(G )()
v t v t
= 0.
kGk
Thus, I S0CHQsG 1B is not invertible in J. Let S be a causal perturbation operator defined by
(Su)(t) := 8S0u(t), if t 2 J
<0, if t / J.
It is clear that kSk= kS0k < g +:
#and I 2
S0CHQsG 1B is not invertible, which implies that Equation (4.6) is not of index-1 which is a contradiction.
Since # > 0 is arbitrary small, we get
rK(Es, A; B, C; T) g.
Step 3. We will prove that rK(Es, A; B, C; T) b.
Without loss of generality, we assume that b < ¥. Indeed, if b g, then this is evident by Step 2. Therefore, we can assume that b < g. On the contrary, suppose that
b < rK(Es, A; B, C; T) = a < g.
Then, we can find a number N0> a, a causal perturbation operator S 2 L(Lp(Ta, Kq), Lp(Ta, Km)),
93
and the functions y˜, z˜ 2 Llocp(Ta, Kq) which satisfy the conditions in Lemma 4.8. Define
f := y˜j[a,N0] and y := y˜j[N0,¥), we have that
y(t) = y˜(t) = (L0Sy˜)(t) + z˜(t) = (L0Sy˜)(t)
= (LN Sy˜)(t) + (L0pN Sy˜)(t)
0 0
= LN S(pN y˜ + [y˜]N ))(t) + L0pN SpN y˜ (t)
0 0 0 0 0
= (LN0 S)(t) + (LN0Sy)(t). for all t N0. Let
xy(t) := (MN0 S f )(t) + (MN0 Sy)(t), t N0. (4.15) It is clear that
C( )xy( ) = (LN0 S f )( ) + (LN0 Sy)( ) = y( )2 Llocp(Tt0 , Kq) n Lp(Tt0 , Kq). Thus, xy( ) is a solution of the equation
Es(t)xyD
(t) = A(t)xy(t) + B(t)S C( )[xy( )]N0 (t) + B(t)S( f )(t),
with the initial condition P(N0)xy(N0) = 0. Similar to the decomposition into the equations (4.9) and (4.10), we see that P(t)xy(t) is the unique solu-tion of the equation (Pxy)D = WPxy + Psh, where W is defined in (4.11) and h is defined by
h := G 1BSC(I HQsG 1BSC)1 HQsG 1BS f + G 1BS f . By Remark 4.5, we have
Z t
P(t)xy(t) = P(t)x(t; s(s), h(s))Ds.
N0
It is clear that the assumption C( ) 2 L¥(Tt0 , Kq n) implies xy( )2 Llocp(Tt0 , Kn)n Lp(Tt0 , Kn).
Since S as well as (I CHQsG 1BS) 1 = ồƠk=0(CHQsG 1BS)k are the finite memory operators, so is (I HQsG 1BSC) 1, by Lemma 4.2. Fur-thermore, since f has a compact support, so does h.
94
Now, we suppose that the trivial solution of Equation (4.6) is globally Lp- stable. This implies that Pxy( ) 2 Lp(Tt0 ; Kn). To this end, we use the esti- mate
Lp([N0,¥);Kn)= ¥ t p 1
Pxy(t) p
¥ t p 1
p
ZN0 Z
N0 kP(t)x(t; s(s), h(s))k Ds Dt
Z ¥ 1
N0 Zs ¥ kP(t)x(t;s(s), h(s))kp Dt pDs M3 Z ¥ kh(s)k Ds < +¥.
N0
Consequently, both CPxy( ) and CQxy( ) belong to Lp(Tt0 , Kq), which is contradicted to the fact that
Cxy( ) 2 Llocp(Tt0 , Kq) n Lp(Tt0 , Kq).
Thus, the trivial solution of Equation (4.6) is not globally Lp-stable. The proof is complete.
Remark 4.10. In case T = R, (4.14) gives the stability radius formula in [24, Theorem 2], and in case T = Z we obtain the stability radius formula in [69, Theorem 4.6]. However, the above proof has some modifications using different techniques and it is essentially simpler than the proofs in [24, 69].
Remark 4.11. In case T = R and E = I, (4.14) gives the stability radius formula in [45, Theorem 4.1]. However, since the operator of the left shift may not exist on an arbitrary time scale, we have derived Lemma 4.8 in order to illustrate that causal perturbations may destroy global Lp-stability.
This fact is different from [45].
Example 4.12. Consider Equation (4.2) with
03
E(t) = 20003 , A(t) = 2p1t ) p 1
61 1 07 6 ( (t) 0 7
0 0 0 0 0 1
¥ 4 ¥ 5 4 5
on time scale T = f3kgkS0 [3k + 1, 3k + 2], where
kS
0= =
(4.16) p(t) =8 1
< 14 if t if t [3k + 1, 3k + 2]. = 3k,
2
: 2
95
In this case, we can choose and compute that
P=P= 22 2 03 ,H=I,G = 22 2 03 .
1 1 0 1 1 0
62 2 7 1 62 1 2 7
e 1 10 0 0 10 0 1
4 5 4 5
Simple calculations yield that the transition matrices of the equation (E, A) are given by
F0(t, s) = 1 2ep(t, s) 1 ep(t, s) + 1 03, 6ep(t, s) + 1 ep(t, s) 1 0 7
2 0 0 2
4 5
F(t, s) = 1 2ep(t, s) ep(t, s) 03 . 6ep(t, s) ep(t, s) 0 7
2 0 0 0
4 5
Assume that B = C = I are the matrices defining the structure of perturba- tion in the perturbed equation (4.6). Then, we have
Z t Z t T
(Lt0 u)(t) = ep(t, s(s))u1(s)Ds, ep(t, s(s))u1(s)Ds, 0 ,
t0 t0
where u( ) = (u1( ), u2( ),
k Lt
0 u k
L1(Tt0;R3)
u3( ))T2 L1(Tt0 , R3). Therefore,
= 2Z
t0¥ Z t
0t
ep(t, s(s))u1(s)Ds Dt
¥ t
2 ep(t, s(s))u1(s) Ds Dt
Zt0 Zt0
¥ ¥
Z Z
= 2ep(t, s(s))Dt u1(s)Ds
t0 s(s)
¥ u1(s) ¥
= 2 Z
t0 Z
t0
Ds 8 u1(s)Ds p(s(s))
8 k u k
L1(Tt0 ,R3).
This implies that kLt0 k 8. Moreover, if we choose u( ) = (u1( ), 0, 0)T with
<18 if t = 3k,
u1(t) =
:0 if t 6= 3k, for some k
satisfying 3k > t
0, then we get kL
t0 u
kL1(Tt0 ,R3) = 8 k u
kL1(Tt0 ,R3). Therefore, kLt0 k = 8 for all t0, and it follows that b = 18 . On the other hand,
by (4.13),
2 1 3
kLt0 k = 00 2 00 = 1,for allt0.
0 0 2 1
e 6 1 7
4 5
This implies that g = 1. Thus, by Theorem 4.9, we obtain 1
rK(Es, A; B, C; T) = 8.
Let S( ) be a linear and causal operator from Lp(Tt0 , Kq) to Lp(Tt0 , K m), i.e., S( ) 2 L¥(Tt0 , Km q), defined by (Su)(t) = S(t)u(t). Moreover, we have
kSk = esssupt0 t ¥ kS(t)k.
Therefore, we get a necessary condition for global Lp-stability of the equa- tion (4.6) stated in the following corollary:
Corollary 4.13. Let Assumptions 4.1, 4.2 hold true. If rK(Es, A; B, C; T) > esssupt0 t ¥kS(t)k, then the perturbed equation (4.6) is globally Lp-stable.
Remark 4.14. In case T = R and E = I and S( ) 2 L¥(Rt0 , Km q), the above corollary implies a lower bound for the stability radius in [39].
Remark 4.15. By the Fourier-Plancherel transformation technique as in [38, 58], if E, A, B, C are constant matrices and p = 2, then we can prove the equality
kLt0 k = sup kC(A lE)1Bk,
l2ảS
where S is the domain of uniform exponential stability of the time scale T, S := fl 2 C : xD= lx is uniformly exponentially stableg.
Moreover,
L = lim ( A l E ) 1Bk. ket0 k l!¥ kC
Thus, in this case, we obtain the radius of stability formula in [28]
r(E, A; B, C; T) = 1 .
supl2ảS[Ơ kC(A lE) 1Bk
97