ON THE STRONG STABILITY OF A NONLINEAR VOLTERRA INTEGRO-DIFFERENTIAL SYSTEM
A. DIAMANDESCU
Abstract. In this paper we provide sufficient conditions for strong stability of the trivial solution of the systems (1) and (2).
1. Introduction
In [3], T. Hara, T. Yoneyama and T. Itoh proved sufficient conditions for uniform stability, asymptotic stability, uniform asymptotic stability and exponential asymptotic stability of trivial solution of a nonlinear Volterra integro- differential system of the form
x0=A(t)x+ Z t
0
F(t, s, x(s))ds (1)
The purpose of our paper is to provide sufficient conditions for strong stability of trivial solution of (1), as a perturbed system of
x0=A(t)x.
(2)
We investigate conditions on the fundamental matrixY(t) for linear system (2) and on the functionF(t, s, x) under which the trivial solution of (1) or (2) is strongly stable onR+.
Received January 15, 2005.
2000Mathematics Subject Classification. Primary 45M10; Secondary 45J05.
Key words and phrases. Strong-stability, Volterra integro-differential systems.
2. Definitions, notations and hypotheses
LetRn denote the Euclideann-space. Forx∈Rn, letkxk be the norm ofx. For ann×nmatrixA, we define the norm|A|ofAby
|A|= sup
kxk≤1
kAxk.
In equation (1) we consider thatAis a continuousn×nmatrix onR+andF :D×Rn−→Rn,D={(t, s)∈R2; 0≤s≤t <∞}, is a continuousn-vector such thatF(t, s,0) = 0 for (t, s)∈D.
Definition 2.1. The solution x(t) of (1) is said to be strongly stable (Ascoli, [1]) on R+ if for every ε > 0, there existsδ=δ(ε)>0 such that any solutionx(t) of (1) which satisfies the inequalitye kx(te 0)−x(t0)k< δfor somet0≥0, exists and satisfies the inequalitykex(t)−x(t)k< εfor allt≥0.
Remark 2.1. For definitions of other types of stability, see [2, page 51].
Remark 2.2. It is easy to see that strong stability is not equivalent with none of these types of stability.
3. The Main Results The following result [2] is well-known.
Theorem 3.1. Let Y(t)be a fundamental matrix for (2). Then, the trivial solution of (2)is strongly stable onR+ if and only if there exists a positive constantK such that
|Y(t)Y−1(s)| ≤K for all 0≤s, t <∞ or, equivalently,
|Y(t)| ≤K and |Y−1(t)| ≤K for all t≥0.
LetY(t) be a fundamental matrix for (2). Consider the following hypotheses:
H1: There exist a continuous functionϕ:R+−→(0,∞) and the constantsp1≥1,K1>0 for Z t
0
ϕ(s)|Y(t)Y−1(s)|p1
ds≤K1, for allt≥0.
H2: There exist a continuous functionϕ:R+−→(0,∞) and the constantsp2≥1,K2>0 for Z t
0
ϕ(s)|Y−1(t)Y(s)|p2
ds≤K2, for allt≥0.
H3: There exist a continuous functionϕ:R+−→(0,∞) and the constantsp3≥1,K3>0 for Z t
0
ϕ(s)|Y−1(s)Y(t)|p3
ds≤K3, for allt≥0.
H4: There exist a continuous functionϕ:R+−→(0,∞) and the constantsp4≥1,K4>0 for Z t
0
ϕ(s)|Y(s)Y−1(t)|p4
ds≤K4, for allt≥0.
Theorem 3.2. Suppose that the fundamental matrix Y(t)for (2)satisfies one of the following conditions:
C1: H1 andH2 are true.
C2: H1 andH4 are true.
C3: H2 andH3 are true.
C4: H3 andH4 are true.
Then, the trivial solution of (2)is strongly stable onR+.
Proof. We will prove thatY(t) andY−1(t) are bounded onR+.
First of all, we consider the caseC2. For the beginning we prove thatY(t) is bounded onR+.
Letq(t) =ϕp1(t)|Y(t)|−p1 fort≥0. From the identity Z t
0
q(s)ds
Y(t) = Z t
0
(ϕ(s)Y(t)Y−1(s))(q(s)(ϕ(s))−1Y(s))ds, t≥0, it follows that
Z t 0
q(s)ds
|Y(t)| ≤ Z t
0
ϕ(s)|Y(t)Y−1(s)|
q(s)(ϕ(s))−1|Y(s)|
ds, t≥0.
(3)
In casep1= 1, we have that q(s)(ϕ(s))−1|Y(s)|= 1. From (3) and the hypothesisH1it follows that Z t
0
q(s)ds
|Y(t)| ≤ Z t
0
ϕ(s)|Y(t)Y−1(s)|ds≤K1, t≥0.
In casep1>1, we have that q(s)(ϕ(s))−1|Y(s)|= (q(s))q11, p1
1 +q1
1 = 1. From (3), it follows that Z t
0
q(s)ds
ϕ(t)(q(t))−p11 ≤ Z t
0
(ϕ(s)|Y(t)Y−1(s)|)(q(s))q11ds, for all t≥0.
Using the H¨older inequality, we obtain Z t
0
q(s)ds
ϕ(t)(q(t))−p11
≤ Z t
0
(ϕ(s)|Y(t)Y−1(s)|)p1ds
p11 Z t 0
q(s)ds q11
, t≥0.
Using the hypothesisH1, we obtain that Z t
0
q(s)ds p11
ϕ(t)(q(t))−p11 ≤K
1 p1
1 , t≥0
or
Z t 0
q(s)ds
|Y(t)|p1≤K1, t≥0.
Thus, forp1≥1, the function|Y(t)|satisfies the inequality
|Y(t)| ≤K
1 p1
1
Z t 0
q(s)ds −p11
, t≥0.
DenoteQ(t) = Z t
0
q(s)dsfort≥0. Thus, we have
|Y(t)|≤K
1 p1
1 (Q(t))−p11, fort≥0.
Because
Q0(t) =q(t)≥K1−1(ϕ(t))p1Q(t) fort≥0,
we have that
Q(t)≥Q(1) e
K1−1
Z t 1
ϕp1(s)ds
, fort≥1.
It follows that
|Y(t)| ≤K
1 p1
1 (Q(1))−p11e
−(p1K1)−1
Z t 1
ϕp1(s)ds
, fort≥1.
Because|Y(t)| is a continuous function on [0,1], it follows that there exists a positive constantM1 such that
|Y(t)| ≤M1fort≥0.
In what follows we prove thatY−1(t) is bounded onR+. Letq(t) =ϕp4(t)|Y−1(t)|−p4 fort≥0. From the identity
Z t 0
q(s)ds
Y−1(t)
= Z t
0
(q(s)(ϕ(s))−1Y−1(s))(ϕ(s)Y(s)Y−1(t))ds, t≥0 it follows that
Z t 0
q(s)ds
|Y−1(t)|
≤ Z t
0
q(s)(ϕ(s))−1|Y−1(s)|
ϕ(s)|Y(s)Y−1(t)|
ds, t≥0.
(4)
In casep4= 1, we have that q(s)(ϕ(s))−1|Y−1(s)|= 1.
From (4) and the hypothesis H4, it follows that Z t
0
q(s)ds
|Y−1(t)| ≤ Z t
0
ϕ(s)||Y(s)Y−1(t)|ds≤K4, t≥0.
In casep4>1, we have that
q(s)(ϕ(s))−1|Y−1(s)|= (q(s))q14, s≥0.
where 1 p4
+ 1 q4
= 1.
From (4) it follows that Z t
0
q(s)ds
|Y−1(t)| ≤ Z t
0
qq14(s) ϕ(s)|Y(s)Y−1(t)|
ds for allt≥0.
Using the H¨older inequality, we obtain that Z t
0
q(s)ds
|Y−1(t)|
≤ Z t
0
ϕ(s)|Y(s)Y−1(t)|p4
ds
p14 Z t 0
q(s)ds q14
, t≥0.
Using the hypothesis H4,we have Z t
0
q(s)ds
|Y−1(t)| ≤ Z t
0
q(s)ds q14
K
1 p4
4 , t≥0
or
Z t 0
q(s)ds p14
|Y−1(t)| ≤K
1 p4
4 , t≥0.
Thus, forp4≥1, the function|Y−1(t)|satisfies the inequality
|Y−1(t)| ≤K
1 p4
4
Z t 0
q(s)ds −p14
, t≥0.
DenoteQ(t) = Z t
0
q(s)dsfort≥0. Thus, we have
|Y−1(t)| ≤K
1 p4
4 (Q(t))−p14 , t≥0.
Because
Q0(t) =q(t)≥ϕp4(t)K4−1Q(t), t≥0,
we have
Q(t)≥Q(1)eK4−1R1tϕp4(s)ds, t≥1.
It follows that
|Y−1(t)| ≤K
1 p4
4 (Q(1))−p14e−(p4K4)−1R1tϕp4(s)ds, t≥ 1.
Because|Y−1(t)|is a continuous function on [0,1], it follows that there exists a positive constantM2 such that
|Y−1(t)| ≤M2fort≥0.
Hence, the conclusion follows immediately from Theorem3.1.
Finally, in the casesC1,C3orC4, the proof is similarly.
The proof is now complete.
Remark 3.1. The function ϕcan serve to weaken the required hypotheses on the fundamental matrixY. Theorem 3.3. If
1. the fundamental matrixY(t) of the equation (2)satisfies
|Y(t)Y−1(s)| ≤K for all0≤s, t <+∞, whereK is constant,
2. the function F satisfies the condition
kF(t, s, x)−F(t, s, y)k ≤f(t, s)kx−yk
for0≤s≤t <+∞and for all x, y∈Rn, wheref is a continuous nonnegative function on D such that M =
Z ∞ 0
Z t 0
f(t, s)ds dt < K−1,
then, for all t0 ≥0,x0 ∈Rn andρ >0, there exists a unique solution of (1) on R+ such that x(t0) =x0 and kx(t)k ≤ρfor all t∈[0, t0], ifk x0kis sufficiently small.
Proof. It is well-known that the problem x0=A(t)x+
Z t 0
F(t, s, x(s))ds, x(t0) =x0
can be reduced by means of variation of constants to the nonlinear integral system x(t) =Y(t)Y−1(t0)x0+
Z t t0
Y(t)Y−1(s) Z s
0
F(s, u, x(u))du ds, t≥0.
(5)
We introduce the Fr´echet space Cc of all continuous maps from R+ into Rn with the seminorms kx|τ = sup
0≤t≤τ
kx(t)k,τ ≥0. Thus, convergence inCc is equivalent to the usual convergence over all compact intervals of R+.
Fort0≥0 andρ >0, letx0∈Rn be such that kx0k< ρ(1−KM)K−1. LetSρ be the set Sρ ={x∈Cc ; kxkt0 ≤ρ, kxkτ ≤ρeKM forτ > t0}.
We consider the following operatorT fromSρ intoCc:
(T x)(t) =Y(t)Y−1(t0)x0+ Z t
t0
Y(t)Y−1(s) Z s
0
F(s, u, x(u))du ds, t≥0.
Forx∈Sρand t∈[0, t0], we have
k(T x)(t)k ≤ Kkx0k+K Z t0
t
Z s 0
f(s, u)kx(u)kdu ds
≤ Kkx0k+K sup
0≤t≤t0
kx(t)k Z t0
0
Z s 0
f(s, u)du ds
≤ Kρ(1−KM)K−1+KρM =ρ.
Forx∈Sρand t > t0, using the same kind of arguments as above, we obtain
k(T x)(t)k ≤ρeKM.
Thus,T Sρ⊂Sρ.
Letx, y∈Sρ. Fort∈[0, t0], we have k(T x)(t)−(T y)(t)k
=
Z t t0
Y(t)Y−1(s) Z s
0
(F(s, u, x(u))−F(s, u, y(u)))du ds
≤ Z t0
t
Y(t)Y−1(s)
Z s 0
kF(s, u, x(u))−F(s, u, y(u))kdu ds
≤ K Z t0
t
Z s 0
f(s, u)kx(u)−y(u)k du ds
≤ K sup
0≤u≤t0
kx(u)−y(u)k Z t0
t
Z s 0
f(s, u)du ds
≤ KMkx−ykt0.
Then,
kT x−T ykt0 ≤KMkx−ykt0.
Similarly, forτ >t0, we have
kT x−T ykτ ≤KMkx−ykτ.
Hence,T is a contraction. By the Banach’s Theorem for Fr´echet spaces [4], Sρ contains a unique fixed point ex=Tx, i. e., the equation (1) has a unique solutione ex(t) on R+ such that ex(t0) =x0 and kx(t)k ≤e ρ for all t
∈[0, t0] andkex(t)k ≤ρeKM for allt≥0, if kx0k is sufficiently small.
Now, we suppose that x(t) is a solution in Cc of (5) such that kx(t)k ≤ ρ for t ∈ [0, t0] and kx0k ≤ ρ(1−KM)K−1. Fort≥t0 we have
kx(t)k= kY(t)Y−1(t0)x0+ Z t
t0
Y(t)Y−1(s) Z s
0
F(s, u, x(u))du dsk
≤ Kkx0k+K Z t
t0
Z s 0
f(s, u)kx(u)kdu ds
= Kkx0k+K Z t
t0
Z t0 0
f(s, u)kx(u)kdu ds+K Z t
t0
Z s t0
f(s, u)kx(u)kdu ds
≤ Kkx0k+Kρ Z t
t0
Z t0 0
f(s, u)du ds+K Z t
t0
Z s t0
f(s, u)kx(u)kdu ds
≤ Kρ(1−KM)K−1+KρM+K Z t
t0
Z s t0
f(s, u)kx(u)kdu ds
= ρ+K Z t
t0
Z s t0
f(s,u)kx(u)kdu ds.
It is easy to see that the functionQ(t) = Z t
t0
Z s t0
f(s, u)kx(u)kdu dsis continuously differentiable and increasing on [t0,∞).
Fort≥t0, we have
Q0(t) = Z t
t0
f(t, u)kx(u)kdu
≤ Z t
t0
f(t, u)(ρ+KQ(u))du=ρ Z t
t0
f(t, u)du+K Z t
t0
f(t, u)Q(u)du.
Then,
Q(t)e
−K
Z t t0
Z s t0
f(s, u)duds
0
= e
−K
Z t t0
Z s t0
f(s, u)du ds
Q0(t)−KQ(t) Z t
t0
f(t, u)du
≤ e
−K
Z t t0
Z s t0
f(s, u)du ds ρ
Z t t0
f(t, u)du+K Z t
t0
f(t, u)(Q(u)−Q(t))du
≤ e
−K
Z t t0
Z s t0
f(s, u)du ds ρ
Z t t0
f(t, u)du
=
−ρK−1e
−K
Z t t0
Z s t0
f(s, u)du ds
0
.
By integrating fromt0 tot≥t0, we have
Q(t)e
−K
Z t t0
Z s t0
f(s, u)du ds
−Q(t0)≤ −ρK−1e
−K
Z t t0
Z s t0
f(s, u)duds
+ρK−1. We deduce that
kx(t)k ≤ρ+KQ(t) fort≥t0, and then
kx(t)k ≤ρeKM fort≥t0.
This shows thatx∈Sρand thenx=ex. Thus, for allt0≥0,x0∈Rnandρ >0, there exists a unique solution of (1) on R+ such that x(t0) = x0 and kx(t)k ≤ ρ for allt ∈ [0, t0], if kx0k is sufficiently small. The proof is
complete.
Theorem 3.4. If the hypotheses of Theorem3.3are satisfied, then the trivial solution of (1)is strongly stable onR+.
Proof. Letε >0 be arbitrary and let δ(ε) =ε(1−KM)K−1e−KM, t0≥0 and letx0∈Rnsatisfykx0k< δ(ε).
Applying Theorem3.3, we deduce that there exists a unique solution x(t) onR+ of (1) withx(t0) =x0such thatx∈Sεe−KM, i. e.,kx(t)k ≤εfort≥0.
This proves that the trivial solution of (1) is strongly stable onR+. The proof is complete.
Example 3.1. Leta, b:R+→Rbe continuous and let the system (2) with A(t) =
a(t) −b(t) b(t) a(t)
.
It is easy to see that
Y(t) =r(t)
−cosθ(t) −sinθ(t)
−sinθ(t) cosθ(t)
,
where
r(t) = eR0ta(u)du and θ(t) = Z t
0
b(u)du, is a fundamental matrix of (2).
We have
|Y(t)Y−1(s)| ≤√
2eRsta(u)du for allt, s≥0.
In [3], it is proved that if there exists λ >0 such that
a(t)≤ −λ for allt≥0.
then the system (2) is uniformly asymptotically stable onR+. We remark that if there existC≥0 andλ >0 such that
Z t s
a(u)du≤C−λ(t−s) for allt≥s≥0,
then we have the same conclusion.
In addition, if there existsL >0 such that
| Z t
s
a(u)du| ≤L for allt, s≥0,
then the system (2) is strongly stable on R+.
Now, we consider
F(t, s, x) = e−αt+s
sinx1+tarctanx2 ssinx1−arctanx2
, whereα∈R.
It is easy to see that the function F satisfies the conditions of Theorem3.3 for αsufficiently large positive number.
In these conditions forA(t) andF, for allt0≥0,x0∈Rn andρ >0, there exists a unique solutionx(t) of (1) onR+ such thatx(t0) =x0 andkx(t)k ≤ρfor allt∈[0, t0], ifkx0k is sufficiently small.
In addition, the trivial solution of (1) is strongly stable onR+.
Acknowledgment. The author would like to thank very much the referee of this paper for valuable comments and suggestions.
1. Ascoli G.,Osservazioni sopra alcune questioni di stabilit`a, Atti. Accad. Naz. Lincei Rend. Cl. Sci. Fis. Mat. Natur.8(9) (1950), 129–134.
2. Coppel W. A.,Stability and Asymptotic Behaviour of Differential Equations, Heath, Boston, 1965.
3. Hara T., Yoneyama T. and Itoh T.,Asymptotic stability criteria for nonlinear Volterra integro-differential equations, Funkcialaj Ekvacioj33(1990), 39–57.
4. Marinescu G.Spatii vectoriale topologice ¸si pseudotopologice,Edit. Acad. R. P. R., Bucuresti, 1959.
A. Diamandescu, Department of Applied Mathematics, University of Craiova, 13, “Al. I. Cuza” st., 200585 Craiova, Romania,e-mail: