Let a 2 T be a fixed point. We consider the perturbation of the form f (t) := F(t, x(t)), where F is a function defined on Ta Rn such that F(t, 0) = 0 for all t 2 Ta. Then Equation (3.1) is rewritten as follows
Es(t)xD(t) = A(t)x(t) + F(t, x(t)), t a. (3.11) Since F(t, 0) = 0 for all t 2 Ta, Equation (3.11) has the trivial solution x(t) 0.
By using variable changes u := Px, v := Qx, and transforma-tion techniques in Section 3.1, we get the equations
uD = (P D + PsG 1 ¯ 1F(t, u + v), (3.12)
A)u + PsG
v = HQsG 1 ¯ 1F(t, u + v). (3.13) Au + HQsG
Assume that HQsG 1F(t, ) is Lipschitz continuous with Lipschitz constant gt< 1, i.e.,
kHQsG 1F(t,y) HQsG 1F(t, z)k gtky zk,
for all t a. Since HQsG 1 does not depend on the choice of H and Q, the Lipschitz property of HQsG 1F(t, ) does, too.
Fix u 2 Rn and choose t 2 Ta, we consider the mapping Gt : im Q(t) ! im Q(t) defined by
Gt(v) := H(t)Qs(t)G 1 ¯ 1(t)F(t, u + v). (t)A(t)u + H(t)Qs(t)G
It can be seen directly that
kGt(v) Gt(v0)k gtkv v0k
for all v, v0 2 im Q(t). Since gt < 1, Gt is a contractive mapping. Therefore, by the Fixed-point Theorem, there exists a mapping gt : im P(t) ! im Q(t), given by
1 ¯ (3.14)
gt(u) := H(t)Qs(t)G (t) A(t)u + F(t,u + gt(u)) .
Moreover, we have
kgt(u) gt(u0)k btku u0k+ gt ku u0k+ kgt(u) gt(u0)k ,
61
where bt= kH(t)Qs(t)G 1 ¯
(t)A(t)k. Hence, we get kgt(u) gt(u0)kgt+ bt ku u0k.
1 gt
This proves that the function gt is Lipschitz continuous with Lipschitz con- stant L := gt+bt . Substituting v = g (u) into (3.12) obtains
t 1 gt t
uD = (PD + PsG 1 A¯
)u + PsG 1F t, u + gt(u) . (3.15) Suppose that (3.15) is solvable. We get the solution u(t) from Equation (3.15). Therefore, the unique solution of Equation (3.11) is
x(t) = u(t) + gt(u(t)), t 2 Ta. (3.16) Definition 3.8. The IDE (3.11) is said to be exponentially stable if there exist numbers M > 0, a > 0 such that a 2 R+ and
kx(t, t0, x0)kMe a(t, t0)kP(t0)x0k, for all t t0 a, x0 2 Rn.
By the classical way, we see that the uniform boundedness and the exponen-tial stability of Equation (3.6) are characterized by Cauchy operator F(t, s) as follows:
Theorem 3.9. The implicit dynamic equation (3.6) is exponentially stable if and only if there exist numbers M > 0 and positively regressive a such that
kF(t, s)k Me a(t, s), for all t s a. (3.17) Proof. Similarly to the proof of [23, Theorem 3.14].
From the equality (3.8), and Assumption 3.1 we get
kF0(t, s)P(s)k = kP(t)F(t, s)k kP(t)kkF(t,s)k K0kF(t, s)k.
Thus, from the inequality (3.17), there exists a positive constant M such that kF0(t,s)P(s)k Me a(t,s), for all t s a. (3.18) We are now in position to consider the robust stability of IDEs under small perturbations. The following theorem will show that the exponential stabil- ity is also preserved under some integrable perturbations or small enough Lipschitz perturbation.
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Theorem 3.10. Assume that Equation (3.6) is of index-1, exponential stable and
i) L = sup
t2TaL
t <
¥, and
ii) the function Ps(t)G 1(t)F(t, x) is Lipschitz continuous with Lipschitz con- stant kt, such that one of the following conditions hold
Z ¥ k
a) N = a 1 am ( t )Dt < ¥.
b) lim supt!¥ kt(1 + Lt) = d < LMa , with a, M in Definition 3.8.t Then, there exist the constants K > 0 and positively regressivea1
such that kx(t)k Ke a1 (t, s)kP(s)x(s)k,
for all t s a, where x( ) is a solution of (3.11). That is, the perturbed equation (3.11) preserves the exponential stability.
Proof. We now prove this theorem with the condition a). By the variation of constants formula (3.9), the solution of Equation (3.15) is
Z t
u(t) = F0(t, s)u(s) + F0(t, s(t))PsG 1F(t, u(t) + gt(u(t)))Dt,
s
for all t > s a. Therefore, by estimate (3.18) we have
ku(t)k kF0(t,s)u(s)k
Z t
+ kF0(t,s(t))PskkPsG 1F(t, u(t) + gt(u(t)))kDt (3.19)
s Z t
e a(t, s(t))kt(1 + Lt)ku(t)kDt.
Me a(t, s)ku(s)k+ M s
Multiplying both sides of (3.19) by e a(t,s) yields
1
u(t)k M u(s) + M t kt(1 + Lt) u(t)k Dt. (3.20)
k k
am(t))ek
ek
a(t, s) Z
s (1 a(t, s)
Using Gronwall’s inequality, we obtain
ku(t)k Mku(s)k e M(1+L)k(t, s),
e a(t, s) 1 am( ).
Since M(1+L)k. is positive, by the definition of the exponential funtion it fol-
1 am( )
lows that
e M(1+L)k. eR st M(1+L)k t Dt R¥ M(1+L)k
e s t Dt=eMN(1+L).
1 am(t) 1 am(t)
1 am( )
Therefore, there exists a number M1> 0 such that ku(t)k M1e a(t, s)ku(s)k. By (3.16) we get
kx(t)k ku(t)k + kgt(u(t))k (1 + L)ku(t)k (1 + L)M1ea(t, s)ku(s)k, or,
kx(t)kK0e a(t, s)kP(s)x(s)k,
for all t > s a, where K0 = (1 + L)M1. We have the proof in the first case.
Next, in the case condition b) is satisfied, let #0 be a positive number such that d + #0 LMa . Then, follow from the second assumption, there exists an element T0 > a such that
kt(1 + Lt)< d + #0, for all t > T0. (3.21) By the solutions’ continuity of (3.15) with the initial condition, we can find a positive constant MT0 which depends only on T0 such that
ku(t)k MT0 ku(s)k, for alla s < t T0. (3.22) First, we consider the case t > T0 > s a. In the same way as (3.19) and (3.20), it follows that
ku(t)k M u(T ) e Mk (1+L) (t, T ) M u(T ) e M(d+# ) (t, T ).
e a (t,s) k 0 k 1 am( ). . 0 k 0k 1 am( )0 0
or equivalently,
ku(t)k Mku(T0)ke aM(d+#0) ( t, T
0) = M k u ( T
0)k e
a+M(d+#0)( t, T
0) .
1 am( )
It is clear that L regressive, so is
> 1. Set a1 := aM(d + #0)> 0. Since a is positively a1. Therefore, by (3.22), we have
ku(t) Mku(T0)ke (t, s) Me (T , t)e (t,s)k u(T )k,
k e a1(T0, s) a1 ( a1) 0 0 a1 0
64
or
ku(t)k MMT0 e ( a1)(T0, t0)e a1(t, s)ku(s)k. Thus,
ku(t)k K1ea1 (t,s)ku(s)k, where K
1= MM
T0 e
(a1)( T
0, t
0)
.
Next, we consider the case t > sT0. We have the estimate kPsG 1F(t, u + gt(u))k (d + #0)kuk
for all t s. Therefore, by the similar arguments as above, we obtain ku(t)k K2ku(s)ke a1(t, s).
Finally, we consider the remaining case as < t T0. With the number a1> 0 defined above, we have
ku(t)k MT0 ku(s)k MT0 ea1 (T0, t0)ea1 (t,s)ku(s)k. Put K3= maxfK1, K2, MT0 ea1(T0, t0)g, we get
ku(t)k K3e a1(t, s)ku(s)k. Pay attention to (3.16), we obtain
kx(t)kKe a1(t, s)kP(s)x(s)k,
for all t s a, where K = (1 + L)K3. The proof is complete. Remark 3.11. If Es is the identity matrix then from Theorem 3.10 we can obtain results about robust stability of the dynamic systems on time scales
xD(t) = A(t)x(t) + F(t, x) in [29].
Remark 3.12. Assume that the perturbation F(t, x) is linear, i.e. F(t, x) = S(t)x with S(t) 2 Rn n. Then the perturbed equation (3.11) has the form
EsxD= (A(t) + S(t))x(t).
In this case, it is easy to see that g(t) = kH(t)Qs(t)G 1(t)S(t)k < 1 if S(t) is small enough and kt = kPs(t)G 1(t)S(t)k. By Theorem 3.10, we can derive bounds for the perturbation S(t) such that the perturbed equation (3.11) is still exponentially stable. This can be used to evaluate the robust stability of DAEs, respectively T = R, and implicit difference equations, respectively T = Z, which arise in many applications, see [46, 48, 51, 55].
Next, we prove Bohl-Perron type theorem for linear time-varying IDEs, i.e., investigate the relation between the solutions’ boundedness of the nonho- mogenous Equation (3.1) and the exponential stability of the IDE (3.6).
We note that, in solving equation (3.1), the function f is split into two com- ponents PsG 1 f and HQsG 1 f . Therefore, for any t02 Ta we consider the function f as an element of the set
Lt0 ( 2 and supt t0 kPs(t)G 1(t) f (t)k < ¥) .
( ) = f C ([t0, ¥ ], Rn) : supt t0 kH(t)Qs(t)G 1(t) f (t)k < ¥
We can directly see that L(t0) is a Banach space eqiupped with the norm k f k = sup kPs(t)G 1(t) f (t)k + kH(t)Qs(t)G 1(t) f (t)k .
t t0
Denote by x(t, s, f ) the solution, associated with f , of Equation (3.1) with the initial condition P(s)x(s, s) = 0. For notational convenience, we will write x(t, s) or x(t) for x(t, s, f ) if that causes no confusion, and IVP-1 for
"Equation (3.1) with the initial condition P(s)x(s, s) = 0".
Lemma 3.13. If for any function f ( ) 2 L(t0), the solution x( , t0) of the IVP- 1 is bounded, then for all t1 t0, there is a constant k > 0, independent of t1, such that supt t1kx(t, t1)k kk f k.
Proof. Define a family of operators fVtgt t0 as follows Vt : L(t0)! Rn
f 7! Vt( f ) = x(t, t0).
From the assumption of this lemma, we have supt t0 kVtf k < ¥ for any f 2 L(t0). By the Uniform Boundedness Principle, there exists a constant k > 0 such that
sup kx(t, t0)k= kVt f k kk f k, (3.23)
t t0
for all t t0. Let f be an arbitrary funtion in L(t1). We define a function f¯
in L(t0) as follows: if t < t1 then f¯
(t) = 0, else f¯
(t) = f (t). Hence, by the variation of constants formula, for any t t1 we have
Z t
x(t, t0, f¯
) = F(t, s(t))Ps(t)G 1(t) f¯
(t)dt + H(t)Qs(t)G 1(t) f¯ (t)
t0
66
Z t
= F(t, s(t))Ps(t)G 1(t) f (t)dt + H(t)Qs(t)G 1(t) f (t).
t1
This means that
x(t, t0, f¯
) = x(t, t1, f ) for all t t1 t0.
Therefore, from (3.23) we have the relation sup kx(t, t1, f )k= sup kx(t, t0, f¯
)k kk f¯
k= kk f k.
t t1 t t0
The proof is complete.
We are now in the position to derive the Bohl-Perron type stability theorem for linear time-varying IDEs.
Theorem 3.14. All solutions of the IVP-1, associated with an arbitrary function f in L(t0), are bounded if and only if the index-1 IDE (3.6) is exponentially stable.
Proof. The proof is divided into two parts.
Necessity. We prove that if all solutions of the IVP-1 associated with f 2 L(t0), are bounded, then the IDE (3.6) is exponentially stable. Indeed, for any t1 t0, let
c(t) := kF(s(t), t1)k, t t1. For any y 2 Rn, we consider the function
f( t ) = Es( t )
F( s ( t ) , t
1) y
, t t1.c(t)
It is obvious that
1( t ) Es( t ) F c (
( s t ) t , t1 y
kPs(t)G1(t) f (t)k = Ps(t)G ( ) )
F(s(t), t1)
= Ps(t)
c(t)
y K0kyk. Moreover,
kH(t)Qs(t)G 1(t) f (t)k = H(t)Qs(t)G 1(t)Es(t)F c(
(st)t , t1 ) y = 0.
( )
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Thus, f 2 L(t1) and K0kyk. k f k = sup ( ) 1( ) ()k + kH( ) ( ) 1( ) ( ) k
t t1 kPs t G t f t t Qs t G t f t Moreover, Z
x(t, t t (t) f (t)Dt + H(t)Q (t)G 1(t) f (t) ) = F(t, s(t))P (t)G 1
1 t1 s s
t F(s(t), )y t F( , t )y
= Zt1 F(t, s(t))Ps(t)
t1
Dt = Z
t1
t 1
c(t) c(t) Dt.
Put Y(t) = t 1t c(t) 1 Dt > 0, we have x(t, t1) = F(t, t1)Y(t)y. From Lemma
3.13, we obta in
R
kx(t, t1)k = kF(t, t1)Y(t)yk = kF(t,t1)ykY(t) kk f k kK0kyk, which implies that
kF(t, t1)k h (3.24)
Y(t) , where h = kK0. On the other hand,
1 = c(t) = kF(s(t), t1)k h .
YD(t) Y(s(t))
Therefore,
YD(t) 1h Y(s(t)).
By Theorem 1.38, we get Y(t) Y(c)e (1h)(t, c),
(3.24) we have
kF(s(t), t1)k Yh
e 1(s(t), (c)
h
for all t c. This estimate leads to
for all t c. Hence, by c),
h h (t,c) = h
kF(t,t 1)k
Y(c)e Y(c)e 1 (c, t1)e
1
for all t c. Set a =h , N1= Y(c)e 1 (c,t1) and h
1 h
h kea(t, t1)k N max N1,t1 t c
F(t, t1)
= max ,
1 (t, t1)
h
68
we obtain the desired estimate kF(t, t1)k Ne a(t, t1) for all t t1.
Sufficiency. To complete the proof, we will show that, if (3.6) is exponen- tially stable, then all solutions of the IVP-1 associated with f in L(t0), are bounded. Indeed, let f 2 L(t0) and suppose that
sup kPs(t)G 1(t) f (t)k= C1, sup kH(t)Qs(t)G 1(t) f (t)k= C2.
t t0 t t0
By the variant of constants formula (3.10), we have
Z t
kx(t)k kF(t, s(t))PsG 1f (t)kDt + kHQsG 1f (t)k
t0 Z t
MC1 e a(t, s(t))Dt + C2
t0
Z t
= MC1e a(t, t0) e ( a)(s(t), t0)Dt + C2.
t0
By using the L’Hôspital rule, we get
t!¥ e a t t
0
Z t ( a) t
0
t!¥R t e
( a)(s(t), t0)Dt
t0e t0 e ( a)(t,t
0)
lim ( , ) (s(t), )Dt = = limlim e ( a)(s(t), t0) 1. ( a)e ( a)(t, t0) =
t!¥ a
Z t
Thus, supt t0 t0 e a(t, s(t))Dt < ¥, which implies that solutions of Equa-tion (3.1) associated with f are bounded. The proof is complete.
Remark 3.15. The above results extended the Bohl-Perron type stability the-orem with bounded input/output operators for differential and difference equations [3, 18, 64], for differential-algebraic equations [34], and implicit difference equations [26, 52], corresponding to the case T = R or T = Z for dynamic systems on time scales [29].
Example 3.16. Consider the simple circuit on time scales consists of a volt-age source vV = v(t), a resistor with conductance R and a capacitor with capacitance C > 0, see Figure 3.1. As in [71], this model can be written in the form EsxD= Ax + f , with
Es = 20 C 03, A = 2 R R 03, x = 2e23 , f = 203 .
6 0 0 0 7 6 R R 1 7 6e1 7 60 7
0 0 0 1 0 0 iv v
4 5 4 5 4 5 4 5
69
Figure 3.1: A simple circuit In this case, it is easy to see that 2
0103
Q = 20003 ,P=I Q= ,H=I.
6 1 0 0 7 60 0 0 7
0 0 1 0 0 0
Therefore, 4 5 4 5
AHQ¯s = 2 3 2 CR 3
G = Es R C 0 , G = 0 C .
6 R 0 1 7 1 6 0 0 1 7
1
1 0 0 1 0 R
4 5 4 5
This implies that k f k =q1 +R2 (C2+1) v . On the other hand, the spectral
set C2
R s(Es, A) =fl : det(A lEs) = 0g= n
o.
C
Therefore, if 1 m(t)RC > 0, or equivalently CR 2 R+then the homoge- nous equation EsxD = Ax is exponentially stable. By Theorem 3.14, if v is bounded then e1, e2, iv are bounded.