Acta Math Vietnam DOI 10.1007/s40306-015-0122-3

Asymptotically Stable Solutions for a Nonlinear Functional Integral Equation L. T. P. Ngoc · N. T. Long

Received: 17 May 2013 / Revised: 22 January 2014 / Accepted: 10 February 2014 © Institute of Mathematics, Vietnam Academy of Science and Technology (VAST) and Springer Science+Business Media Singapore 2015

Abstract In this paper, the solvability and the existence of asymptotically stable solutions for a nonlinear functional integral equation are studied. The main tools are a fixed point theorem of Krasnosel’skii type, the dominated convergence theorem, and other results from functional analysis. In order to illustrate the result obtained here, an example is given. Keywords Fixed point theorem of Krasnosel’skii type · Contraction mapping · Completely continuous · Asymptotically stable solution Mathematics Subject Classification (2010) 47H10 · 45G10 · 47N20 · 65J15

1 Introduction In this paper, we consider the following nonlinear functional integral equation  x(t) = V



μ1 (t)

t, x(t), 

+

∞







V1 t, s, x(σ1 (s)), 0



V2 (t, s, r, x(σ2 (r))) dr ds 0





μ3 (s)

F1 (t, s, r, x(σ3 (r)))dr ds, (1)

F t, s, x(χ1 (s)), . . . , x(χq (s)), 0



μ2 (s)

0

t ∈ R+ , where E is a Banach space, the functions V : R+ × E 2 → E; V1 : 1 × E 2 → E; V2 : 2 × E → E; F : R2+ × E q+1 → E, F1 : R+ × 3 × E → E; μ1 , μ2 , μ3 , σ1 , σ2 , L. T .P. Ngoc Nha Trang Educational College, 01 Nguyen Chanh Street, Nha Trang, Vietnam e-mail: [email protected] N. T. Long () Department of Mathematics and Computer Science, University of Natural Science, Vietnam National University Ho Chi Minh City, 227 Nguyen Van Cu Street, District 5, Ho Chi Minh, Vietnam e-mail: [email protected]

L. T. P. Ngoc, N. T. Long

σ3 , χ1 , . . . , χq ∈ C(R+ ; R+ ) are continuous and 1 = {(t, s) ∈ R2+ : s ≤ μ1 (t)}, 2 = {(t, s, r) ∈ R3+ : r ≤ μ2 (s), s ≤ μ1 (t)}, 3 = {(s, r) ∈ R2+ : r ≤ μ3 (s)}. Since the pioneering work of Volterra up to our days, it is well known that integral equations have attracted the interest of scientists not only because of their mathematical context, but also because of their miscellaneous applications in various fields of science and technology, see [11]. The special cases of (1) occur in mechanics, population dynamics, engineering systems, the theory of ‘adiabatic tubular chemical reactors’, etc. Details on such problems can be found, for example, in Corduneanu [3, pp. 273 - 275], Deimling [4, pp. 240, 243, 244]. In the case E = Rd , some types of (1) have been studied by C. Avramescu and C. Vladimirescu [1, 2]. The authors have proved the existence of asymptotically stable solutions to the integral equation  t  ∞ x(t) = Q(t) + K(t, s, x(s))ds + G(t, s, x(s))ds, t ∈ R+ , 0

0

under suitable hypotheses. In the proofs, a fixed point theorem of Krasnosel’skii type is used, (see [2]). Also, applying a fixed point theorem of Krasnosel’skii type and giving the suitable assumptions, Dhage and Ntouyas [5], Purnaras [11] obtained results on the existence of solutions to the nonlinear functional integral equation  μ(t)  σ (t) x(t) = Q(t) + k(t, s)f (s, x(θ (s)))ds + v(t, s)g(s, x(η(s)))ds, t ∈ [0, 1], 0

0

(2) where E = R, 0 ≤ μ(t) ≤ t; 0 ≤ σ (t) ≤ t; 0 ≤ θ (t) ≤ t; 0 ≤ η(t) ≤ t for all t ∈ [0, 1]. Some more general equations than (2) were also studied in [11], such as  μ(t) x(t) = q(t) + k(t, s)f (s, x(θ (s)))ds α(t)    λ(t)  σ (s)  k0 (s, v, x(η(v)))dv ds, t ∈ [0, 1]. + k(t, s)F s, x(ν(s)), 0

β(t)

Recently, using the technique of the measure of noncompactness and the Darbo fixed point theorem, Z. Liu et al. [7] have proved the existence and asymptotic stability of solutions for the equation    t x(t) = f t, x(t), u(t, s, x(a(s)), x(b(s))) ds , t ∈ R+ . 0

In the case when the Banach space E is arbitrary, the existence of asymptotically stable solutions of the equation   s  t  V t, s, x(s), V1 (t, s, r, x(r)) dr ds x(t) = q(t) + f (t, x(t)) + 0 0   ∞   s + G t, s, x(s), G1 (t, s, r, x(r)) dr ds, t ∈ R+ , 0

0

have been proved in [9], by using the following fixed point theorem of Krasnosel’skii type. Theorem 1 [8] Let (X, |·|n ) be a Fr´echet space and U, C : X → X be two operators. Assume that U is a k−contraction operator, k ∈ [0, 1) (depending on n) with respect to a family of seminorms ·n equivalent with the family |·|n and C is completely continuous |Cx| such that lim|x|n →∞ |x| n = 0 ∀n ∈ N. Then, U + C has a fixed point. n

Asymptotically Stable Solutions for an Integral Equation

Because of mathematical context, motivated by the above mentioned works, we study the solvability and the existence of asymptotically stable solutions of (1). The main tools are Theorem 1, the dominated convergence theorem, and other results from functional analysis. This paper consists of four sections. The existence of solutions, the existence of asymptotically stable solutions for (1) will be presented in Sections 2, 3. Finally, we give an illustrative example.

2 Existence of Solutions Let X = C(R+ ; E) be the space of all continuous functions on R+ to E equipped with the countable family of seminorms |x|n = sup |x(t)| , n ≥ 1. t∈[0,n]

Then, (X, |·|n ) is complete in the metric d(x, y) =

∞  n=1

2−n

|x − y|n , 1 + |x − y|n

and X is a Fr´echet space, see [10]. In X, we also consider the family of seminorms defined by xn = |x|γn + |x|hn , n ≥ 1, where |x|γn = sup |x(t)| , |x|hn = sup e−hn (t−γn ) |x(t)| , 0≤t≤γn

γn ≤t≤n

γn ∈ (0, n) and hn > 0 are arbitrary numbers, which are equivalent to |·|n , since e−hn (n−γn ) |x|n ≤ xn ≤ 2 |x|n We make the following assumptions.

∀x ∈ X, ∀n ≥ 1.

(A1 ) The functions μ1 , μ2 , μ3 , σ1 , σ2 , σ3 , χ1 , . . . , χq ∈ C(R+ ; R+ ) are continuous such that μi (t), σi (t), χj (t) ∈ [0, t] for all t ∈ R+ , i = 1, 2, 3, j = 1, . . . q. (A2 ) There exist a constant L ∈ [0, 1) and a continuous function ω0 : R+ → R+ such that |V (t; x, y) − V (t; x, ¯ ¯ y)| ¯ ≤ L |x − x| ¯ + ω0 (t) |y − y|

∀ (t; x, y) , (t; x, ¯ y) ¯ ∈ R+ × E 2 ;

(A3 ) There exists a continuous function ω1 : 1 → R+ such that |V1 (t, s, x, y) − V1 (t, s, x, ¯ + |y − y|) ¯ ¯ y)| ¯ ≤ ω1 (t, s) (|x − x| for all (t, s, x, y), (t, s, x, ¯ y) ¯ ∈ 1 × E 2 , 1 = {(t, s) ∈ R2+ : s ≤ μ1 (t)}; (A4 ) There exists a continuous function ω2 : 2 → R+ such that |V2 (t, s, r, x) − V2 (t, s, r, x)| ¯ ≤ ω2 (t, s, r) |x − x| ¯ for all (t, s, r, x), (t, s, r, x) ¯ ∈ 2 × E, 2 = {(t, s, r) ∈ R3+ : r ≤ μ2 (s), s ≤ μ1 (t)}; (A5 ) F : R2+ × E q+1 → E is completely continuous such that for all I1 ⊂ R+ , I2 ⊂ R+ , J ⊂ E q , S ⊂ E, bounded subsets for all ε > 0, there exists δ > 0, such that for any (t1 , v1 ), (t2 , v2 ) ∈ I1 × S, |t1 − t2 | < δ, |v1 − v2 | < δ =⇒ F (t1 , s, u1 , . . . , uq , v1 ) − F (t2 , s, u1 , . . . , uq , v2 ) < ε

for all (s, u1 , . . . , uq ) ∈ I2 × J ;

L. T. P. Ngoc, N. T. Long

(A6 )

There exists a continuous function ω3 : R2+ → R+ such that for each bounded subset I of R+ ,  ∞ sup ω3 (t, s)ds < ∞, 0

t∈I

and F (t, s, u1 , . . . , uq , v) ≤ ω3 (t, s) (A7 )

∀(t, s, u1 , . . . , uq , v) ∈ I × R+ × E q+1 ;

˜ ⊂ 3 , S1 ⊂ E, F1 : R+ × 3 × E → E is continuous such that for all I1 ⊂ R+ ,  bounded subsets for all ε > 0, there exists δ1 > 0, such that ∀t1 , t2 ∈ I1 , |t1 − t2 | < δ1 =⇒ |F1 (t1 , s, r, v) − F1 (t2 , s, r, v)| < ε ˜ × S1 , 3 = {(s, r) ∈ R2+ : r ≤ μ3 (s)}. for all (s, r, v) ∈ 

Then, we have the next theorem Theorem 2 Let (A1 ) − (A7 ) hold. Then, (1) has a solution on R+ . Proof The proof consists of four steps. Step 1 In X, we consider the equation    μ1 (t)  V1 t, s, x(σ1 (s)), x(t) = V t, x(t), 0



μ2 (s)

V2 (t, s, r, x(σ2 (r))) dr 0

 ds , t ∈ R+ . (3)

Lemma 1 Let (A1 ) − (A4 ) hold. Then, the (3) has a unique solution x = ξ. This lemma is proved as follows. First, we rewrite (3) in the form x(t) = Φx(t), where

  Φx(t) = V t, x(t),

μ1 (t)

t ∈ R+ ,

  V1 t, s, x(σ1 (s)),

0

μ2 (s)

(4)

  V2 (t, s, r, x(σ2 (r))) dr ds , (5)

0

(t, x) ∈ R+ × X. By the assumptions (A1 ) − (A4 ), for all x, x¯ ∈ X for all t ∈ R+ , we have |Φx(t) − Φ x(t)| ¯

(6)



μ1 (t)

≤ L |x(t) − x(t)| ¯ + ω0 (t) 0



μ1 (t)

+ ω0 (t)



μ2 (s)

ω1 (t, s)ds 0

ω1 (t, s) |x(σ1 (s)) − x(σ ¯ 1 (s))| ds ω2 (t, s, r) |x(σ2 (r)) − x(σ ¯ 2 (r))| dr.

0

Now, let n ∈ N be fixed. For all t ∈ [0, γn ], where γn ∈ (0, n) is chosen later, we have  t |Φx(t) − Φ x(t)| |x(σ1 (s)) − x(σ ¯ ≤ L |x(t) − x(t)| ¯ + ω˜ 0n ω˜ 1n ¯ 1 (s))| ds 0  t  s |x(σ2 (r)) − x(σ + ω˜ 0n ω˜ 1n ω˜ 2n ds ¯ 2 (r))| dr 0 0 

¯ γn , (7) ≤ L + γn ω˜ 0n ω˜ 1n + γn2 ω˜ 0n ω˜ 1n ω˜ 2n |x − x|

Asymptotically Stable Solutions for an Integral Equation

with ⎧ ω˜ = sup{ω0 (t) : 0 ≤ t ≤ n}, ω˜ 1n = sup{ω1 (t, s) : (t, s) ∈ 1n }, ⎪ ⎪ 0n ⎪ ⎨ ω˜ = sup{ω (t, s, r) : (t, s, r) ∈  }, 2n 2 2n ⎪ 1n = {(t, s) : 0 ≤ s ≤ μ1 (t), 0 ≤ t ≤ n}, ⎪ ⎪ ⎩ 2n = {(t, s, r) : 0 ≤ r ≤ μ2 (s), 0 ≤ s ≤ μ1 (t), 0 ≤ t ≤ n}. This implies that

 |Φx − Φ x| ¯ γn ≤ L + γn ω˜ 0n ω˜ 1n + γn2 ω˜ 0n ω˜ 1n ω˜ 2n |x − x| ¯ γn .

(8)

For all t ∈ [γn , n], observe that |Φx(t) − Φ x(t)| ¯

(9)



γn

≤ L |x(t) − x(t)| ¯ + ω˜ 0n ω˜ 1n  + ω˜ 0n ω˜ 1n

|x(σ1 (s)) − x(σ ¯ 1 (s))| ds

0 t γn

|x(σ1 (s)) − x(σ ¯ 1 (s))| ds 



γn

+ ω˜ 0n ω˜ 1n ω˜ 2n

s

ds 

+ ω˜ 0n ω˜ 1n ω˜ 2n

|x(σ2 (r)) − x(σ ¯ 2 (r))| dr

0

0



t

γn

ds

 |x(σ2 (r)) − x(σ ¯ 2 (r))| dr +

0

γn

+ ω˜ 0n ω˜ 1n ω˜ 2n γn2 |x − x| ¯ γn   t  ¯ γn + + ω˜ 0n ω˜ 1n ω˜ 2n ds γn |x − x| γn

s

 |x(σ2 (r)) − x(σ ¯ 2 (r))| dr

γn



¯ γn + ω˜ 0n ω˜ 1n ≤ L |x(t) − x(t)| ¯ + ω˜ 0n ω˜ 1n γn |x − x|

s

t

|x(σ1 (s)) − x(σ ¯ 1 (s))| ds

γn

 |x(σ2 (r)) − x(σ ¯ 2 (r))| dr

γn



¯ γn + ω˜ 0n ω˜ 1n ≤ L |x(t) − x(t)| ¯ + ω˜ 0n ω˜ 1n γn |x − x|

t

|x(σ1 (s)) − x(σ ¯ 1 (s))| ds

γn

+ ω˜ 0n ω˜ 1n ω˜ 2n γn2 |x − x| ¯ γn + ω˜ 0n ω˜ 1n ω˜ 2n (n − γn )γn |x − x| ¯ γn  t  s |x(σ2 (r)) − x(σ + ω˜ 0n ω˜ 1n ω˜ 2n ds ¯ 2 (r))| dr γn

γn

= L |x(t) − x(t)| ¯ + γn ω˜ 0n ω˜ 1n (1 + nω˜ 2n ) |x − x| ¯ γn  t |x(σ1 (s)) − x(σ + ω˜ 0n ω˜ 1n ¯ 1 (s))| ds γn

+ ω˜ 0n ω˜ 1n ω˜ 2n





t

s

ds γn

|x(σ2 (r)) − x(σ ¯ 2 (r))| dr.

γn

By the inequalities 0 < e−hn (t−γn ) ≤ e−hn (σi (t)−γn )

∀t ∈ [γn , n], i = 1, 2,

(10)

L. T. P. Ngoc, N. T. Long

in which hn > 0 is also chosen later, we get |Φx(t) − Φ x(t)| ¯ e−hn (t−γn ) ≤ L |x(t) − x(t)| ¯ e−hn (t−γn ) + γn ω˜ 0n ω˜ 1n (1 + nω˜ 2n ) |x − x| ¯ γn e−hn (t−γn )  t |x(σ1 (s)) − x(σ + ω˜ 0n ω˜ 1n e−hn (t−γn ) ¯ 1 (s))| ds γn

+ ω˜ 0n ω˜ 1n ω˜ 2n e−hn (t−γn )





t

s

|x(σ2 (r)) − x(σ ¯ 2 (r))| dr

ds γn

γn

¯ γn + ω˜ 0n ω˜ 1n J1 + ω˜ 0n ω˜ 1n ω˜ 2n J2 , (11) ≤ L |x − x| ¯ hn + γn ω˜ 0n ω˜ 1n (1 + nω˜ 2n ) |x − x| where J1 = e−hn (t−γn ) J2 = e−hn (t−γn )



t

|x(σ1 (s)) − x(σ ¯ 1 (s))| ds,

γn  t



s

ds γn

|x(σ2 (r)) − x(σ ¯ 2 (r))| dr.

γn

We shall estimate the integrals J1 and J2 as follows. Estimating J1 :  t |x(σ1 (s)) − x(σ J1 = e−hn (t−γn ) ¯ 1 (s))| ds γn  |x(σ1 (s)) − x(σ ¯ 1 (s))| ds = e−hn (t−γn ) {s∈[γn ,t]:σ1 (s)<γn }  |x(σ1 (s)) − x(σ + e−hn (t−γn ) ¯ 1 (s))| ds {s∈[γn ,t]:σ1 (s)≥γn }

≡

(1) J1

(2) + J1 .

Estimating J1(1) = e−hn (t−γn ) J1(1) ≤ e−hn (t−γn )  ≤

t γn



¯ 1 (s))| ds: {s∈[γn ,t]:σ1 (s)<γn } |x(σ1 (s)) − x(σ

{s∈[γn ,t]:σ1 (s)<γn }

|x − x| ¯ γn ds ≤ e−hn (t−γn )

e−hn (s−γn ) ds |x − x| ¯ γn =

(2)

Estimating J1



= e−hn (t−γn )





t γn

|x − x| ¯ γn ds

 1

1 |x − x| 1 − e−hn (t−γn ) |x − x| ¯ γn ≤ ¯ γn . hn hn

{s∈[γn ,t]:σ1 (s)≥γn } |x(σ1 (s)) −

x(σ ¯ 1 (s))| ds:

(2)

J1

= e−hn (t−γn ) ≤ e−hn (t−γn ) ≤ e

 

{s∈[γn ,t]:σ1 (s)≥γn }

{s∈[γn ,t]:σ1 (s)≥γn }  t −hn (t−γn ) hn (σ1 (s)−γn )

e

γn

=

ehn (σ1 (s)−γn ) e−hn (σ1 (s)−γn ) |x(σ1 (s)) − x(σ ¯ 1 (s))| ds ¯ hn ds ehn (σ1 (s)−γn ) |x − x|

|x − x| ¯ hn ds ≤ e−hn (t−γn )

 1

1 |x − x| ¯ hn ≤ ¯ hn . 1 − e−hn (t−γn ) |x − x| hn hn



t

γn

¯ hn ds ehn (s−γn ) |x − x|

Asymptotically Stable Solutions for an Integral Equation

Hence (1)

(2)

J1 = J1 + J1 Estimating J2 : J2 = e

−hn (t−γn )

≤ e−hn (t−γn )



t



s



γn t

≤

|x(σ2 (r)) − x(σ ¯ 2 (r))| dr

ds γn t



|x(σ2 (r)) − x(σ ¯ 2 (r))| dr

ds γn

1 x − x ¯ n. hn

γn

≤ (n − γn )e−hn (t−γn )



t

|x(σ2 (r)) − x(σ ¯ 2 (r))| dr.

γn

Similarly, e

−hn (t−γn )



t

|x(σ2 (r)) − x(σ ¯ 2 (r))| dr ≤

γn

1 x − x ¯ n. hn

Thus, J2 ≤ (n − γn )

1 x − x ¯ n. hn

Consequently, |Φx(t) − Φ x(t)| ¯ e−hn (t−γn ) ≤ L |x − x| ¯ hn + γn ω˜ 0n ω˜ 1n (1 + nω˜ 2n ) |x − x| ¯ γn 1 1 ¯ n + ω˜ 0n ω˜ 1n ω˜ 2n (n − γn ) x − x ¯ n + ω˜ 0n ω˜ 1n x − x hn hn = L |x − x| ¯ hn + γn ω˜ 0n ω˜ 1n (1 + nω˜ 2n ) |x − x| ¯ γn   1 ¯ n. + ω˜ 0n ω˜ 1n 1 + (n − γn )ω˜ 2n x − x hn This implies that |Φx − Φ x| ¯ hn ≤ L |x − x| ¯ hn + γn ω˜ 0n ω˜ 1n (1 + nω˜ 2n ) |x − x| ¯ γn   1 + ω˜ 0n ω˜ 1n 1 + (n − γn )ω˜ 2n x − x ¯ n. hn Combining (8), (12), we deduce that Φx − Φ x ¯ n = |Φx − Φ x| ¯ γn + |Φx − Φ x| ¯ hn     1 ¯ n ≤ L+ ω˜ 0n ω˜ 1n 1 + (n − γn )ω˜ 2n x − x hn   + γn ω˜ 0n ω˜ 1n 2 + (n + γn )ω˜ 2n |x − x| ¯ γn    1 ≤ L+ ω˜ 0n ω˜ 1n 1 + ω˜ 2n (n − γn ) hn   ¯ n + γn ω˜ 0n ω˜ 1n 2 + (n + γn )ω˜ 2n x − x ˜ n, ≤ Ln y − y     where Ln = L + 1 + ω˜ 2n (n − γn ) + γn ω˜ 0n ω˜ 1n 2 + (n + γn )ω˜ 2n . Choose γn , hn such that 0 < γn < n, and     1 ω˜ 0n ω˜ 1n 1 + ω˜ 2n (n − γn ) + γn ω˜ 0n ω˜ 1n 2 + (n + γn )ω˜ 2n < 1 − L, hn 1 ˜ 0n ω˜ 1n hn ω

(12)

L. T. P. Ngoc, N. T. Long

therefore Ln < 1. So, Φ is a Ln −contraction operator on the Fr´echet space (X, ·n ). Then, Lemma 1 follows from the Banach contraction principle, see [10]. Step 2 By the transformation x = y + ξ, where ξ is the unique solution of the (4), we can write the (1) in the form y(t) = Uy(t) + Cy(t),

t ∈ R+ ,

(13)

where ⎧  ⎪ ⎪ Uy(t) = −ξ(t) + V t, y(t) ⎪ ⎪ ⎪ ⎪ ⎪   ⎪  μ2 (s)  μ1 (t)

⎪ ⎪ t, s, + ξ (σ V (s)), V (r))) dr ds , s, r, + ξ (σ + ξ(t), (y ) 1 (y ) 2 ⎪ 1 2 (t, 0 0 ⎪ ⎨  ∞ Cy(t) = 0 F t, s, (y + ξ ) (χ1 (s)), . . . , (y + ξ ) (χq (s)), ⎪ ⎪ ⎪ ⎪  ⎪ ⎪  μ3 (s) ⎪ ⎪ F (r))) dr ds, s, r, + ξ (σ (t, (y ) ⎪ 1 3 ⎪ 0 ⎪ ⎪ ⎩ t ∈ R+ . (14) Then, U is a Ln −contraction operator, Ln ∈ [0, 1) (depending on n), with respect to a family of seminorms ·n . Indeed, fixing an arbitrary positive integer n ∈ N, we have   μ1 (t)  V1 t, s, (y + ξ ) (σ1 (s)), Uy(t) − U y(t) ¯ = V t, y(t) + ξ(t), 0



μ2 (s)

  V2 (t, s, r, (y + ξ ) (σ2 (r))) dr ds

0

  − V t, y(t) ¯ + ξ(t),

μ1 (t)

 V1 t, s, (y¯ + ξ ) (σ1 (s)),

0



μ2 (s)

  V2 (t, s, r, (y¯ + ξ ) (σ2 (r))) dr ds ,

0

so using the similar estimates as in the proof of Lemma 1, the results are as follows. For all t ∈ [0, γn ], we have  μ1 (t) |Uy(t) − U y(t)| ¯ ≤ L |y(t) − y(t)| ¯ + ω0 (t) ω1 (t, s) |y(σ1 (s)) − y(σ ¯ 1 (s))| ds 

0 μ1 (t)



μ2 (s)

ω1 (t, s)ds ω2 (t, s, r) |y(σ2 (r)) − y(σ ¯ 2 (r))| dr +ω0 (t) 0 0 

¯ γn . ≤ L + γn ω˜ 0n ω˜ 1n + γn2 ω˜ 0n ω˜ 1n ω˜ 2n |y − y| This implies that 

|Uy − U y| ¯ γn . ¯ γn ≤ L + γn ω˜ 0n ω˜ 1n + γn2 ω˜ 0n ω˜ 1n ω˜ 2n |y − y| For all t ∈ [γn , n], we have |Uy(t) − U y(t)| ¯ γn ¯ hn + γn ω˜ 0n ω˜ 1n (1 + nω˜ 2n ) |y − y| ¯ e−hn (t−γn ) ≤ L |y − y|   1 ¯ n. + ω˜ 0n ω˜ 1n 1 + (n − γn )ω˜ 2n y − y hn

(15)

Asymptotically Stable Solutions for an Integral Equation

It follows that |Uy − U y| ¯ γn ¯ hn ≤ L |y − y| ¯ hn + γn ω˜ 0n ω˜ 1n (1 + nω˜ 2n ) |y − y|   1 + ω˜ 0n ω˜ 1n 1 + (n − γn )ω˜ 2n y − y ¯ n. hn Consequently, Uy − U y ¯ n, ¯ n ≤ Ln y − y

(16)

and then U is a Ln −contraction operator with respect to ·n . Step 3 We show that C : X → X is completely continuous. First, the following lemma for the relative compactness of a subset in X is useful to prove the main results. Lemma 2 Let X = C(R+ ; E) be the Fr´echet space defined as above and A be a subset of X. For each n ∈ N, let Xn = C([0, n]; E) be the Banach space of all continuous functions u : [0, n] → E with the norm |u|n = supt∈[0,n] |u(t)| and An = {x|[0,n] : x ∈ A}. The set A in X is relatively compact if and only if for each n ∈ N, An is equicontinuous in Xn and for every s ∈ [0, n], the set An (s) = {x(s) : x ∈ An } is relatively compact in E. Lemma 2 was proved in detail in appendix [8], by applying the Ascoli-Arzela’s Theorem, see [6, p. 211]. We now show that C is continuous. For any y0 ∈ X, let {ym }m be a sequence in X such that limm→∞ ym = y0 . ∞ Let n ∈ N be fixed. For any given ε > 0, by 0 supt∈[0,n] ω3 (t, s)ds < ∞, there exists Tn ∈ N (Tn is large enough), such that  ∞  ∞ ε ∀t ∈ [0, n]. (17) ω3 (t, s)ds ≤ sup ω3 (t, s)ds < 8 Tn Tn t∈[0,n] Put K1 = {(ym + ξ )(χ1(s)) : s ∈ [0, Tn ], m ∈ Z+ }, .. . Kq = {(ym + ξ )(χq (s)) : s ∈ [0, Tn ], m ∈ Z+ },   μ3 (s) F1 (t, s, r, (ym + ξ ) (σ3 (r))) dr : t ∈ [0, n], s ∈ [0, Tn ], m ∈ Z+ , Kq+1 = 0

K˜ q+1 = {(ym + ξ )(σ3 (r)) : r ∈ [0, n], m ∈ Z+ }. Then, K1 , . . . , Kq+1 , K˜ q+1 are compact in E, since limm→∞ |ym − y0 |Tn = 0.   (i) K1 is compact in E. Indeed, let (ymj + ξ )(χ1 (sj )) j be a sequence in K1 . We can assume that limj →∞ sj = s0 and that limj →∞ ymj + ξ = y0 + ξ. We have (ym + ξ )(χ1(sj )) − (y0 + ξ )(χ1 (s0 )) j ≤ (ymj + ξ )(χ1(sj )) − (y0 + ξ )(χ1 (sj )) + (y0 + ξ )(χ1 (sj )) − (y0 + ξ )(χ1(s0 )) ≤ ymj − y0 T + (y0 + ξ )(χ1 (sj )) − (y0 + ξ )(χ1(s0 )) , n

which shows that limj →∞ (ymj + ξ )(χ1 (sj )) = (y0 + ξ )(χ1(s0 )) in E. This means that K1 is compact in E and, in a similar manner, the same holds true for K2 , . . . , Kq , K˜ q+1 . (ii) Kq+1 is compact in E.

L. T. P. Ngoc, N. T. Long

 μ (s )   Let { 0 3 j F1 (tj , sj , r, ymj + ξ (σ3 (r)))dr}j be a sequence in Kq+1 . We can assume   that limj →∞ tj , sj = (t0 , s0 ) and that limj →∞ ymj + ξ = y0 + ξ. On the other hand, since F1 is continuous on the compact set [0, n] × Tn × K˜ q+1 , there exists Mn > 0 such that for every (t, s, r, v) ∈ [0, n] × Tn × K˜ q+1 , |F1 (t, s, r, v)| ≤ Mn = sup{|F1 (t, s, r, v)| : (t, s, r, v) ∈ [0, n] × Tn × K˜ q+1 }, where Tn = {(s, r) ∈ R2+ : r ≤ μ3 (s), 0 ≤ s ≤ Tn }. We have  μ3 (sj )  μ3 (s0 )     F1 tj , sj , r, ymj + ξ (σ3 (r)) dr − F1 (t0 , s0 , r, (y0 + ξ ) (σ3 (r))) dr 

0

0

μ3 (sj )

=



   F1 tj , sj , r, ymj + ξ (σ3 (r)) dr

μ3 (s0 )  μ3 (s0 ) 

+

0

     F1 tj , sj , r, ymj + ξ (σ3 (r)) − F1 (t0 , s0 , r, (y0 + ξ ) (σ3 (r))) dr.

Hence,   μ3 (s0 ) μ3 (sj )     F1 tj , sj , r, ymj + ξ (σ3 (r)) dr − F1 (t0 , s0 , r, (y0 + ξ ) (σ3 (r))) dr 0 0  μ3 (sj )     ≤ F1 tj , sj , r, ymj + ξ (σ3 (r)) dr μ3 (s0 )  μ3 (s0 )     F1 tj , sj , r, ym + ξ (σ3 (r)) − F1 (t0 , s0 , r, (y0 + ξ ) (σ3 (r))) dr + j 0 ≤ Mn μ3 (sj ) − μ3 (s0 )  μ3 (s0 )     F1 tj , sj , r, ym + ξ (σ3 (r)) − F1 (t0 , s0 , r, (y0 + ξ ) (σ3 (r))) dr → 0, + j 0

as j → +∞, which shows that  μ3 (sj )     F1 tj , sj , r, ymj + ξ (σ3 (r)) dr lim j →+∞ 0  μ3 (s0 )

=

F1 (t0 , s0 , r, (y0 + ξ ) (σ3 (r))) dr ∈ E.

0

This means that Kq+1 is compact in E. Since F is continuous on the compact set [0, n] × [0, Tn ] × K1 × · · · × Kq+1 , there exists δ > 0 such that for every (u1 , . . . , uq+1 ), (u¯ 1 , . . . , u¯ q+1 ) ∈ K1 ×· · ·×Kq+1 , |ui − u¯ i | < δ, i = 1, . . . , q + 1, F (t, s, u1 , . . . , uq+1 ) − F (t, s, u¯ 1 , . . . , u¯ q+1 ) < ε ∀(t, s) ∈ [0, n] × [0, Tn ]. 4Tn On the other hand, since F1 is continuous on the compact set [0, n] × Tn × K˜ q+1 , there ¯ < δ1 , exists δ1 > 0 such that for every v, v¯ ∈ K˜ q+1 , |v − v| |F1 (t, s, r, v) − F1 (t, s, r, v)| ¯ <

δ 2Tn

∀ (t, s, r) ∈ [0, n] × Tn .

Asymptotically Stable Solutions for an Integral Equation

With i = 1, . . . , q, since sup |(ym + ξ )(χi (s)) − (y0 + ξ )(χi (s))| ≤ 0≤s≤Tn

sup |(ym + ξ )(s) − (y0 + ξ )(s)| 0≤s≤Tn

= |ym − y0 |Tn → 0, and sup |(ym + ξ )(σ3 (r)) − (y0 + ξ )(σ3 (r))| ≤ 0≤r≤n

sup |(ym + ξ )(s) − (y0 + ξ )(s)| 0≤s≤Tn

= |ym − y0 |Tn → 0, as m → ∞, there exists m0 such that, for every m > m0 , |(ym + ξ )(σ3 (r)) − (y0 + ξ )(σ3 (r))| < min{δ, δ1 }, |(ym + ξ )(χi (s)) − (y0 + ξ )(χi (s))| < min{δ, δ1 } for all r ∈ [0, n], s ∈ [0, Tn ] ∀i = 1, . . . , q. This implies that for all t ∈ [0, n], s ∈ [0, Tn ], and for all m > m0 ,   μ3 (s) μ3 (s) F1 (t, s, r, (ym + ξ ) (σ3 (r))) dr − F1 (t, s, r, (y0 + ξ ) (σ3 (r))) dr 0 0  μ3 (s) |F1 (t, s, r, (ym + ξ ) (σ3 (r))) − F1 (t, s, r, (y0 + ξ ) (σ3 (r)))| dr ≤ 0

 ≤

0

μ3 (s)

δ δ dr = μ3 (s) ≤ δ. 2Tn 2Tn

On the other hand, we have for all t ∈ [0, n], and for all m > m0 ,  Tn   F t, s, (ym + ξ )(χ1 (s)), . . . , (ym + ξ )(χq (s)), Cym (t) − Cy0 (t) = 

0 μ3 (s) 0

 F1 (t, s, r, (ym + ξ ) (σ3 (r))) dr



−F t, s, (y0 + ξ )(χ1(s)), . . . , (y0 + ξ )(χq (s)),  

μ3 (s) 0

∞

F1 (t, s, r, (y0 + ξ ) (σ3 (r))) dr

 ds



F t, s, (ym + ξ )(χ1(s)), . . . , (ym + ξ )(χq (s)),

=

Tn  μ3 (s)

 F1 (t, s, r, (ym + ξ ) (σ3 (r))) dr

0

 −F t, s, (y0 + ξ )(χ1(s)), . . . , (y0 + ξ )(χq (s)), 

μ3 (s) 0

F1 (t, s, r, (y0 + ξ ) (σ3 (r))) dr

 ds.

L. T. P. Ngoc, N. T. Long

Hence,  F t, s, (ym + ξ )(χ1(s)), . . . , (ym + ξ )(χq (s)), 0   μ3 (s) F1 (t, s, r, (ym + ξ ) (σ3 (r))) dr 

|Cym (t) − Cy0 (t)| ≤

Tn

0

 − F t, s, (y0 + ξ )(χ1(s)), . . . , (y0 + ξ )(χq (s)), 

μ3 (s)

0



+2

 F1 (t, s, r, (y0 + ξ ) (σ3 (r))) dr ds

∞

ω3 (t, s)ds Tn

< Tn

ε ε ε + 2 = < ε, 4Tn 8 2

so |Cym − Cy0 |n < ε for all m > m0 , and the continuity of C is proved. It remains to show that C maps bounded sets into relatively compact sets. We will use Lemma 2. Now, let  be a bounded subset of X. We have to prove that for n ∈ N, (a) The set (C)n is equicontinuous in Xn . (b) For every t ∈ [0, n], the set (C)n(t) = {Cy|[0,n](t) : y ∈ } is relatively compact in E. Let n ∈ N be fixed. Consider any ε > 0 given. Then, there exists Tn ∈ N (Tn is large enough) such that (17) is valid. Proof of (a) Put S1 = {(y + ξ )(χ1(s)) : y ∈ , s ∈ [0, Tn ]}, .. . Sq = {(y + ξ )(χq (s)) : y ∈ , s ∈ [0, Tn ]},   μ3 (s) F1 (t, s, r, (y + ξ ) (σ3 (r))) dr : y ∈ , t ∈ [0, n], s ∈ [0, Tn ] , Sq+1 = 0

S˜q+1 = {(y + ξ )(σ3 (r)) : y ∈ , r ∈ [0, n]}. Then, S1 , . . . , Sq+1 , S˜q+1 are bounded subsets in E. By (A5 ), there exists δ > 0 such that, for every (t1 , v1 ), (t2 , v2 ) ∈ [0, n] × Sq+1 , ε |t1 − t2 | < δ, |v1 − v2 | < δ =⇒ F (t1 , s, u1 , . . . , uq , v1 )−F (t2 , s, u1 , . . . , uq , v2 ) < 4Tn for all (s, u1 , . . . , uq ) ∈ [0, Tn ] × K1 × · · · × Kq . Moreover, by (A7 ), there exists δ1 > 0 such that ∀t1 , t2 ∈ [0, n],

|t1 − t2 | < δ1 =⇒ |F1 (t1 , s, r, v) − F1 (t2 , s, r, v)| <

δ 2Tn

Asymptotically Stable Solutions for an Integral Equation

for all (t, s, v) ∈ Tn × S˜q+1 . Hence,   μ3 (s) μ3 (s) F1 (t1 , s, r, (y + ξ ) (σ3 (r))) dr − F1 (t2 , s, r, (y + ξ ) (σ3 (r))) dr 0 0  μ3 (s) |F1 (t1 , s, r, (y + ξ ) (σ3 (r))) − F1 (t2 , s, r, (y + ξ ) (σ3 (r)))| dr ≤ 0

 ≤

0

μ3 (s)

δ δ dr = μ3 (s) ≤ δ. 2Tn 2Tn

Therefore, for all t1 , t2 ∈ [0, n], |t1 − t2 | < min(δ, δ1 ), and for every y ∈ n ,  Tn  F t1 , s, (y + ξ )(χ1 (s)), . . . , (y + ξ )(χq (s)), |Cy(t1) − Cy(t2)| ≤ 

0

μ3 (s) 0



F1 (t1 , s, r, (y + ξ ) (σ3 (r))) dr



− F t2 , s, (y + ξ )(χ1 (s)), . . . , (y + ξ )(χq (s)), 

μ3 (s) 0

+



∞

 F1 (t2 , s, r, (y + ξ ) (σ3 (r))) dr ds (ω3 (t1 , s) + ω3 (t2 , s)) ds

Tn

< Tn

ε ε ε + 2 = < ε. 4Tn 8 2

Thus, (C)n is equicontinuous on Xn . Proof of (b) Let {Cyk |[0,n] (t)}k , yk ∈ , be a sequence in (C)n(t). We have to show that there exists a convergent subsequence of {Cyk |[0,n] (t)}k . Since F is completely continuous and S1 , . . . , Sq+1 are bounded in E, the set F ([0, n] × [0, Tn ] × S1 × · · · × Sq+1 ) is relatively compact in E. Put   Fk (t, s) = F t, s, (yk + ξ )(χ1 (s)), . . . , (yk + ξ )(χq (s)),

μ3 (s)

 F1 (t, s, r, (yk + ξ ) (σ3 (r))) dr .

0

It is clear that the sequence  {Fk (t, s)}  k belongs to F ([0, n] × [0, Tn ] × S1 × ... × Sq+1 ), so there exist a subsequence Fkj (t, s) j and θ (t, s) ∈ E such that Fk (t, s) − θ (t, s) → 0, as j → ∞. (18) j On the other hand, by (A6 ), Fkj (t, s) ≤ ω3 (t, s) ∀(t, s) ∈ [0, n] × [0, Tn ]. Hence, Fk (t, s) − θ (t, s) ≤ Fk (t, s) + |θ (t, s)| (19) j j ≤ 2ω3 (t, s)

∀(t, s) ∈ [0, n] × [0, Tn ], ω3 (t, ·) ∈ L1 (0, Tn ) .

Using the dominated convergence theorem, (18) and (19) yield  Tn Fk (t, s) − θ (t, s) ds → 0, as j → ∞. j 0

L. T. P. Ngoc, N. T. Long

This means that, for given ε > 0, there exists j0 such that for j > j0 ,  Tn Fk (t, s) − θ (t, s) ds < ε . j 2 0 Consequently, for j > j0 ,  Tn Cyk (t) − = θ (t, s)ds j

 ∞  Tn F (t, s)ds − θ (t, s)ds kj 0 0  ∞  Tn  Tn Fkj (t, s)ds − θ (t, s)ds + Fkj (t, s)ds ≤ Tn 0 0  ∞  Tn Fk (t, s) − θ (t, s) ds + Fk (t, s) dy ≤ j j

0

0

ε ≤ + 2 





∞

Tn

Tn

ε ε ω3 (t, s)ds < + < ε, 2 8

so Cykj (t) j is a convergent subsequence of {Cyk (t)}k ; hence, (C)n(t) is relatively compact in E. In view of Lemma 2, C() is relatively compact in X. Therefore, C is completely continuous. Step 4 Finally, we show that, for every n ∈ N, lim

|y|n →∞

|Cy|n = 0. |y|n

(20)

By the assumption (A6 ), for all t ∈ [0, n], we get  ∞  F t, s, (y + ξ )(χ1(s)), . . . , (y + ξ )(χq (s)), |Cy(t)| ≤ 0

 F1 (t, s, r, (y + ξ ) (σ3 (r))) dr ds 0  ∞  ∞ ω3 (t, s)ds ≤ sup ω3 (t, s)ds < ∞. ≤ 

μ3 (s)

0

0

0≤t≤n

It follows that lim

|y|n →∞

|Cy|n = 0. |y|n

By Theorem 1, the operator U + C has a fixed point y in X. Then, the (1) has a solution  x = y + ξ on R+ . Theorem 2 is proved. 3 Asymptotically Stable Solutions We now consider the asymptotically stable solutions for (1) defined as follows. Definition 1 A function x is said to be an asymptotically stable solution of (1) if for any solution  x of (1), limt→∞ |x(t) − x(t)| ˜ = 0.

Asymptotically Stable Solutions for an Integral Equation

In this section, we assume (A1 ) − (A7 ) hold, then the (1) has a solution on R+ by Theorem 2 On the other hand, if x is a solution of (1) then, as in step 1 in the proof of Theorem 2, y = x − ξ satisfies (13). This implies that, for all t ∈ R+ , |y(t)| ≤ |Uy(t)| + |Cy(t)| ,

(21)

where Uy(t), Cy(t) are as in (14). Using (A1 ) − (A7 ), for all t ∈ R+ , we have 

μ1 (t)

|y(t)| ≤ L |y(t)| + ω0 (t)

ω1 (t, s)|y(σ1 (s))|ds 0



μ1 (t)

+ω0 (t)



μ2 (s)

ω1 (t, s)ds 0

 ω2 (t, s, r)|y(σ2(r))|dr +

0

∞

ω3 (t, s)ds. (22) 0

It follows that 

μ1 (t)

z(t) ≤ ω˜ 0 (t)

(23)

ω1 (t, s)z(σ1 (s))ds 0





μ1 (t)

+ ω˜ 0 (t)

μ2 (s)

ω1 (t, s)ds 0

0

ω2 (t, s, r)z(σ2 (r))dr + a(t) ∀t ∈ R+ ,

with z(t) = |y(t)| ,

ω˜ 0 (t) =

1 ω0 (t), 1−L

a(t) =

1 1−L



∞

ω3 (t, s)ds 0

∀t ∈ R+ . (24)

Noting that μ1 (t), μ2 (t) ∈ [0, t] for all t ≥ 0, we deduce from (23) that 

t

z(t) ≤ ω˜ 0 (t)

 ω1 (t, s)z(σ1 (s))ds + ω˜ 0 (t)

0 t



t

ω1 (t, s)ds 0

s

ω2 (t, s, r)z(σ2 (r))dr + a(t)

0

  t  s ≤ ω˜ 0 (t) ω1 (t, s)¯z(s)ds + ω˜ 0 (t) ω1 (t, s)ds ω2 (t, s, r)¯z(r)dr + a(t) ∀t ∈ R+ , 0

0

0

where z¯ (t) = z(t) + z(σ1 (t)) + z(σ2 (t)).

(25)

This yields 

t

z(t) ≤ ω˜ 0 (t) 0

 ω¯ 1 (t, s)¯z(s)ds + ω˜ 0(t) 0

t

 ω¯ 1 (t, s)ds 0

s

ω¯ 2 (t, s, r)¯z(r)dr +a(t) ∀t ∈ R+ , (26)

in which 

ω¯ 1 (t, s) = ω1 (t, s) + ω1 (σ1 (t), s) + ω1 (σ2 (t), s), ω¯ 2 (t, s, r) = ω2 (t, s, r) + ω2 (σ1 (t), s, r) + ω2 (σ2 (t), s, r).

(27)

L. T. P. Ngoc, N. T. Long

By 0 ≤ σi (t) ≤ t, for all t ≥ 0, (23) gives  σi (t) ω1 (σi (t), s)¯z(s)ds z(σi (t)) ≤ ω˜ 0 (σi (t)) 0



(28)



σi (t)

s

ω1 (σi (t), s)ds ω2 (σi (t), s, r)¯z(r)dr + a(σi (t)) + ω˜ 0 (σi (t)) 0 0  t ≤ ω˜ 0 (σi (t)) ω1 (σi (t), s)¯z(s)ds 0  t  s + ω˜ 0 (σi (t)) ω1 (σi (t), s)ds ω2 (σi (t), s, r)¯z(r)dr + a(σi (t)) 0 0  t ≤ ω˜ 0 (σi (t)) ω¯ 1 (t, s)¯z(s)ds 0  t  s + ω˜ 0 (σi (t)) ω¯ 1 (t, s)ds ω¯ 2 (t, s, r)¯z(r)dr + a(σi (t)). 0

0



Put

ω¯ 0 (t) = ω˜ 0 (t) + ω˜ 0 (σ1 (t)) + ω˜ 0 (σ2 (t)), (29) a(t) ¯ = a(t) + a(σ1 (t)) + a(σ2 (t)). From (28), summing up with respect to i = 1, 2, afterwards, and adding to (26), we obtain for every t ∈ R+ that  t  t  s z¯ (t) ≤ ω¯ 0 (t) ω¯ 1 (t, s)¯z(s)ds + ω¯ 0 (t) ω¯ 1 (t, s)ds ω¯ 2 (t, s, r)¯z(r)dr + a(t). ¯ (30) 0

0

0

Note that  t  s  t  t ω¯ 1 (t, s)ds ω¯ 2 (t, s, r)¯z(r)dr = z¯ (r)dr ω¯ 1 (t, s)ω¯ 2 (t, s, r)ds 0 0 0 r   t  t = ω¯ 1 (t, s)ω¯ 2 (t, s, r)ds z¯ (r)dr. 0

Hence,

 z¯ (t) ≤ ω¯ 0 (t) 0

t

r

ω(t, r)¯z(r)dr + a(t) ¯ ∀t ∈ R+ ,

where

 ω(t, r) = ω¯ 1 (t, r) +

t

(31)

ω¯ 1 (t, s)ω¯ 2 (t, s, r)ds.

(32)

r 2(a 2 + b 2 )

∀a, b ∈ R, (31) leads to Due to the inequality (a + b)2 ≤  t  t z¯ 2 (t) ≤ 2ω¯ 02 (t) ω2 (t, r)dr z¯ 2 (r)dr + 2a¯ 2 (t). 0

Putting Z(t) = z¯ 2 (t), b(t) = 2ω¯ 02 (t)

(33)

0

t

ω2 (t, r)dr, we can rewrite (33) as follows

0

 Z(t) ≤ b(t)

t

Z(r)dr + 2a¯ 2 (t).

(34)

0

By (34), based on classical estimates, we obtain  t   t 2 2 2 b(r)dr ds z¯ (t) = Z(t) ≤ 2a¯ (t) + 2b(t) a¯ (s) exp 0

s

Therefore, we can conclude this section with the following result.

∀t ∈ R+ .

(35)

Asymptotically Stable Solutions for an Integral Equation

Theorem 3 Let (A1 ) − (A7 ) hold. If   t    t 2 2 lim a¯ (t) + b(t) a¯ (s) exp b(r)dr ds = 0, t→∞

where

0

⎧ a(t) ¯ ⎪ ⎪ ⎪ ⎪ ⎪ a(t) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ b(t) ω¯ 0 (t) ⎪ ⎪ ⎪ ⎪ ω(t, r) ⎪ ⎪ ⎪ ⎪ ⎪ ω¯ 1 (t, s) ⎪ ⎪ ⎩ ω¯ 2 (t, s, r)

(36)

s

= a(t) + a(σ1 (t)) + a(σ2 (t)), ∞ 1 = 1−L 0 ω3 (t, s)ds ∀t ∈ R+ , t 2 = 2ω¯ 0 (t) 0 ω2 (t, r)dr, =

[ω0 (t) + ω0 (σ1 (t)) + ω0 (σ2 (t))] , t = ω¯ 1 (t, r) + r ω¯ 1 (t, s)ω¯ 2 (t, s, r)ds, 1 1−L

(37)

= ω1 (t, s) + ω1 (σ1 (t), s) + ω1 (σ2 (t), s), = ω2 (t, s, r) + ω2 (σ1 (t), s, r) + ω2 (σ2 (t), s, r),

then every solution x to (1) is a asymptotically stable. Furthermore, lim |x(t) − ξ(t)| = 0.

(38)

t→∞

4 An Example Let us illustrate the results obtained by an example. Let E = C([0, 1]; R) be the Banach space of all continuous functions u : [0, 1] → R with the norm u = sup0≤η≤1 |u(η)| , u ∈ E. Then, for all x ∈ X = C(R+ ; E), for any t ∈ R+ , x(t) is an element of E and we denote x(t)(η) = x(t, η), 0 ≤ η ≤ 1. Consider (1) in the form    μ1 (t)  V1 t, s, x(σ1 (s)), x(t) = V t, x(t), 0

 +

∞



V2 (t, s, r, x(σ2 (r))) dr ds 0



 F t, s, x(χ1 (s)), . . . , x(χq (s)),

0



μ2 (s)

μ3 (s)

 F1 (t, s, r, x(σ3 (r))) dr ds, (39)

0

where σi (t) = σ¯ i t, 0 < σ¯ i ≤ 1, μi (t) = μ¯ i t, 0 < μ¯ i ≤ 1, i = 1, 2, 3; χj (t) = χ¯ j t, 0 < χ¯ j ≤ 1, j = 1, . . . , q. Define the continuous functions V , V1 , V2 , F, F1 as follows. V : R+ × E 2 → E, V (t, x, y)(η) = k0 x∗ (t, η) + k1 |x(η)| + e−t |y(η)| , 0 ≤ η ≤ 1 1, (t, x, y) ∈ R+ ×E 2 , with x∗ (t, η) = η+e t , k0 = 32(1−k1 ), and k1 is a given constant such that 0 < k1 < 1. (ii) V1 : 1 × E 2 → E, 1 = {(t, s) ∈ R2+ : s ≤ μ¯ 1 t},     π x(η) + e−t |y(η)| , V1 (t, s, x, y)(η) = e−2s x∗ (t, η) sin x∗ (σ1 (s), η)

(i)

0 ≤ η ≤ 1, (t, s, x, y) ∈ 1 × E 2 .

L. T. P. Ngoc, N. T. Long

(iii)

V2 : 2 × E → E, 2 = {(t, s, r) ∈ R3+ : s ≤ μ¯ 1 t, r ≤ μ¯ 2 s}, V2 (t, s, r, x)(η) = e

−t−2r

 2π x(η) , x∗ (s, η) sin x∗ (σ2 (r), η) 

0 ≤ η ≤ 1, (t, s, r, x) ∈ 2 × E. (iv) F : R2+ × E q+1 → E,  = k2

e−2s x

∗ (t, η)

F (t, s; u1 , . . . , uq , v)(η) q 

j =1

sin

 π 1 2 0

5

uj (ζ ) x∗ (χj (s),ζ ) dζ



+ sin



 3π 1 2s 4 0 (e

+ ζ )v(ζ )dζ

!  ,

0 ≤ η ≤ 1, (t, s; u1 , . . . , uq , v) ∈ R2+ × E q+1 , with k2 = 62 q (k1 − 1). 2 (v) F1 : R+ × 3 × E → E, 3 = {(s, r) ∈ R+ : r ≤ μ¯ 3 s},  π F1 (t, s, r, x)(η) = e x∗ (t, η) sin x(η) , x∗ (σ3 (r), η) −s



0 ≤ η ≤ 1, (t, s, r, x) ∈ R+ × 3 × E. We can prove that (A1 ) − (A7 ) hold. It is easy to see that (A1 ) holds. –

¯ y) ¯ ∈ R+ × E 2 , Assumption (A2 ) holds because for all (t; x, y) , (t; x, V (t; x, y) − V (t; x, ¯ y) ¯ ≤ k1 x − x ¯ + ω0 (t) y − y ¯ ,

–

with ω0 (t) = e−t . Assumption (A3 ) holds, since for all (t, s, x, y), (t, s, x, ¯ y) ¯ ∈ 1 × E 2 , 1 = {(t, s) ∈ R2+ : s ≤ μ¯ 1 t} ∀η ∈ [0, 1], |V1 (t, s, x, y)(η) − V1 (t, s, x, ¯ y)(η)| ¯ # 1 " −2s σ¯ 1 s ¯ ¯ + e−t |y(η) − y(η)| π(e + η) |x(η) − x(η)| ≤ e t η+e ¯ + y − y ¯ ], ≤ ω1 (t, s) [x − x

–

in which ω1 (t, s) = 2πe−t−s . Assumption (A4 ) holds, since for all (t, s, r, x), (t, s, r, x) ¯ ∈ 2 ×E, 2 = {(t, s, r) ∈ R3+ : s ≤ μ¯ 1 t, r ≤ μ¯ 2 s} ∀η ∈ [0, 1], 1 2π(η + eσ¯ 2 r ) |x(η) − x(η)| ¯ es + η ¯ = ω2 (s, r) x − x ¯ , ≤ 4πe−t−s−r x − x

|V2 (t, s, r, x)(η) − V2 (t, s, r, x)(η)| ¯ ≤ e−t−2r

–

with ω2 (s, r) = 4πe−t−s−r . Assumption (A5 ) is also fulfilled. The proof is as follows.

Asymptotically Stable Solutions for an Integral Equation

First, we show F : R2+ × E q+1 → E is continuous. For all (t, s; u1 , . . . , uq , v), ¯ ∈ R2+ × E q+1 , 0 ≤ η ≤ 1, (t¯, s¯ ; u¯ 1 , . . . , u¯ q , v) ¯ F (t, s; u1 , . . . , uq , v)(η) − F (t¯, s¯ ; u¯ 1 , . . . , u¯ q , v)(η)    1    1$  q uj (ζ ) 3π π 5 dζ + sin = k2 e−2s x∗ (t, η) sin (e2s + ζ )v(ζ )dζ 2 0 x∗ (χj (s), ζ ) 4 0 j =1 $       1  q u¯ j (ζ ) 3π 1 2¯s π −2¯s 5 dζ +sin −k2 e x∗ (t¯, η) sin (e + ζ )v(ζ ¯ )dζ 2 0 x∗ (χj (¯s ), ζ ) 4 0 j =1    $  q uj (ζ ) π 1 5 et¯ − et −2¯s e dζ sin = k2 t 2 0 x∗ (χj (s), ζ ) (e + η)(et¯ + η) j =1    3π 1 2s + sin (e + ζ )v(ζ )dζ 4 0     q    1$ uj (ζ ) 3π 1 2s e−2s − e−2¯s  π 5 +k2 dζ +sin sin (e + ζ )v(ζ )dζ et + η 2 0 x∗ (χj (s), ζ ) 4 0 j =1    $ !    $ q  uj (ζ ) u¯ j (ζ ) π 15 π 15 1 −2¯s e dζ −sin dζ +k2 t¯ sin 2 0 x∗ (χj (s), ζ ) 2 0 x∗ (χj (¯s ), ζ ) e +η j =1    !    3π 1 2¯s 1 3π 1 2s −2¯s +k2 t¯ e (e + ζ )v(ζ )dζ − sin (e + ζ )v(ζ ¯ )dζ . sin 4 0 4 0 e +η Then,

F (t, s; u1 , . . . , uq , v)(η) − F (t¯, s¯ ; u¯ 1 , . . . , u¯ q , v)(η) ¯ ≤ |k2 | (q + 1) t − t¯ + |k2 | (q + 1) |2s − 2¯s | $  $ q  uj (ζ ) u¯ j (ζ ) π 1 5 1 −2¯s 5 − + |k2 | t¯ e dζ 2 0 x∗ (χj (s), ζ ) x∗ (χj (¯s ), ζ ) e +η j =1 !   1 1 2s −2¯s 3π 2¯s + |k2 | t¯ e ¯ ) dζ (e + ζ )v(ζ ) − (e + ζ )v(ζ 4 0 e +η   ≤ |k2 | (q + 1) t − t¯ + 2 |s − s¯ | $ $ q  1  uj (ζ ) u¯ j (ζ ) π 5 5 + |k2 | − dζ 2 x∗ (χj (¯s ), ζ ) 0 x∗ (χj (s), ζ ) j =1

" # 3π |k2 | e2s − e2¯s v + (e2¯s + 1) v − v ¯ . + 4 By using the inequalities √ 5 5x−√ y ≤ 24/5 |x − y|1/5 ,

% % % 5 |x| + |y| ≤ 5 |x| + 5 |y| ∀x, y ∈ R,

L. T. P. Ngoc, N. T. Long

we have

$ $ uj (ζ ) u¯ j (ζ ) 1/5 uj (ζ ) u¯ j (ζ ) 5 4/5 5 − − ≤2 x∗ (χj (s), ζ ) x∗ (χj (¯s ), ζ ) x∗ (χj (s), ζ ) x∗ (χj (¯s ), ζ )

 1/5 

 = 24/5 eχj (s) + ζ uj (ζ ) − u¯ j (ζ ) + u¯ j (ζ ) eχj (s) − eχj (¯s )

1/5 & & & & ≤ 24/5 eχj (s) + 1 &uj − u¯ j & + &u¯ j & eχj (s) − eχj (¯s ) % 1/5  &1/5 & & & 5 χ (s) . ≤ 24/5 e j + 1 &uj − u¯ j & + 5 &u¯ j & eχj (s) − eχj (¯s )

This implies that F (t, s; u1 , . . . , uq , v)(η) − F (t¯, s¯ ; u¯ 1 , . . . , u¯ q , v)(η) ¯   ≤ 2 |k2 | (q + 1) t − t¯ + |s − s¯ | q %  &1/5 & π 5 χ (s) |k | +√ e j + 1 &uj − u¯ j & + 2 5 2 j =1

+

5

1/5  & & χ (s) &u¯ j & e j − eχj (¯s )

" # 3π |k2 | e2s − e2¯s v + (e2¯s + 1) v − v ¯ . 4

So & & &F (t, s; u1 , . . . , uq , v) − F (t¯, s¯ ; u¯ 1 , . . . , u¯ q , v) ¯ &   ≤ 2 |k2 | (q + 1) t − t¯ + |s − s¯ | q % 1/5   &1/5 & & χ (s) & π 5 χ (s) 5 & χj (¯s ) j & & & j | |k +√ u ¯ e + 1 u − u ¯ + − e 2 j j j e 5 2 j =1 +

" # 3π |k2 | e2s − e2¯s v + (e2¯s + 1) v − v ¯ 4

and the continuity of F is proved. Next, we show F : R2+ × E q+1 → E is compact. Let B be bounded in R2+ × E q+1 , we deduce from & & &F (t, s; u1 , . . . , uq , v)& ≤ ω3 (t, s) = (q + 1) |k2 | e−t−2s ≤ (q + 1) |k2 | ≡ M

∀(t, s; u1 , . . . , uq , v) ∈ B,

that F (B) is uniformly bounded in E. For any η1 , η2 ∈ [0, 1] and (t, s; u1 , . . . , uq , v) ∈ B, F (t, s; u1 , . . . , uq , v)(η1 ) − F (t, s; u1 , . . . , uq , v)(η2 )    $  q uj (ζ ) π 15 η2 − η1 −2s = k2 e dζ sin 2 0 x∗ (χj (s), ζ ) (et + η1 ) (et + η2 ) j =1    3π 1 2s + sin (e + ζ )v(ζ )dζ , 4 0

Asymptotically Stable Solutions for an Integral Equation

so

F (t, s; u1 , . . . , uq , v)(η1 ) − F (t, s; u1 , . . . , uq , v)(η2 ) |η2 − η1 | + η1 ) (et + η2 ) ≤ (q + 1) |k2 | |η2 − η1 |

≤ (q + 1) |k2 | e−2s

(et

this implies that F (B) is equicontinuous. Finally, for all bounded subsets I1 , I2 of R+ and for any bounded subsets J ⊂ E q , S ⊂ E, for all ε > 0, there exists δ > 0 such that ∀ (t1 , v1 ) , (t2 , v2 ) ∈ I1 × S, |t1 − t2 | + v1 − v2  < δ & & ⇒ &F (t1 , s; u1 , . . . , uq , v1 ) − F (t2 , s; u1 , . . . , uq , v2 )& < ε for all (s; u1 , . . . , uq ) ∈ I2 × J. We get the above property since & & &F (t1 , s; u1 , . . . , uq , v1 ) − F (t2 , s; u1 , . . . , uq , v2 )&   3π |k2 | [|t1 − t2 | + v1 − v2  ] ≤ q +1+ 2

(40)

for all (t1 , v1 ) , (t2 , v2 ) ∈ I1 × S, (s; u1 , . . . , uq ) ∈ I2 × J. Indeed, F (t1 , s; u1 , . . . , uq , v1 )(η) − F (t2 , s; u1 , . . . , uq , v2 )(η)     1$ q uj (ζ ) e t2 − e t1 π −2s 5 e dζ = k2 t sin (e 1 + η)(et2 + η) 2 0 x∗ (χj (s), ζ ) j =1    3π 1 2s + sin (e + ζ )v1 (ζ )dζ 4 0    !    3π 1 2s e−2s 3π 1 2s (e + ζ )v1 (ζ )dζ − sin (e + ζ )v2 (ζ )dζ , + k2 t sin e2 +η 4 0 4 0 so F (t1 , s; u1 , . . . , uq , v1 )(η) − F (t2 , s; u1 , . . . , uq , v2 )(η) t  e 2 − e t1 e−2s 3π 1 2s −2s | |k e + (e + ζ ) |v1 (ζ ) − v2 (ζ )| dζ ≤ |k2 | (q + 1) t 2 (e 1 + η)(et2 + η) e t2 + η 4 0 3π v1 − v2  ≤ |k2 | (q + 1) |t1 − t2 | + |k2 | 2 3π ) (|t1 − t2 | + v1 − v2  ) . ≤ |k2 | (q + 1 + 2 – Assumption (A6 ) is also clear. Indeed, for all bounded subsets I of R+ , η ∈ [0, 1], (t, s) ∈ I × R+ , (u1 , . . . , uq , v) ∈ E q+1 , −2s F (t, s; u1 , . . . , uq , v)(η) ≤ (q + 1) |k2 | e = (q + 1) |k2 | e−t−2s ≡ ω3 (t, s), t e +η  ∞  ∞ 1 sup ω3 (t, s)ds ≤ (q + 1) |k2 | e−2s ds ≤ (q + 1) |k2 | < ∞. 2 0 0 t∈I

L. T. P. Ngoc, N. T. Long

–

˜ ⊂ 3 , S1 ⊂ E, and ε > 0, Assumption (A7 ): For all bounded subsets I1 ⊂ R+ ,  there exists δ1 > 0, such that ∀t1 , t2 ∈ I1 , |t1 − t2 | < δ1 =⇒ F1 (t1 , s, r, v) − F1 (t2 , s, r, v) < ε ˜ × S1 . The property is fulfilled, since for all (s, r, v) ∈  |F1 (t1 , s, r, v)(η) − F1 (t2 , s, r, v)(η)| ≤

t   e 2 − e t1 π −s sin e v(η) t t 1 2 (e + η)(e + η) x∗ (σ3 (r), η)

≤ |t1 − t2 | .

On the other hand, the condition (36) is satisfied. Indeed, ω0 (t) = e−t ,

ω1 (t, s) = 2πe−t−s ,

ω2 (t, s, r) = 4πe−t−s−r ,

ω3 (t, s) = (q + 1) |k2 | e−t−2s .

We verify the assertions (i), (ii) as follows. (i) a(t) ¯ → 0 as t → +∞ : Indeed, a(t) = =

1 1−L



∞ 0

ω3 (t, s)ds =

1 (q + 1) |k2 | e−t 1 − k1

1 (q + 1) |k2 | −t e ≡ A¯ 1 e−t 2 1 − k1



∞

e−2s ds

0

∀t ∈ R+ ,

a(t) ¯ = a(t) + a(σ1 (t)) + a(σ2 (t))

 = A¯ 1 e−t + e−σ¯ 1 t + e−σ¯ 2 t ≤ 3A¯ 1 e−σ¯ ∗ t → 0

as t → +∞,

where σ¯ ∗ = min{1, σ¯ 1 , σ¯ 2 }. So, (i)is proved. t t (ii) b(t) 0 a¯ 2 (s) exp s b(r)dr ds → 0 as t → +∞ : For this, we verify the assertions (ii1), (ii2) below. (ii1) b(t) → 0 as t → +∞ : Calculating ω¯ 0 (t), ω¯ 1 (t, s), ω¯ 2 (t, s, r), ω(t, r) : 1 [ω0 (t) + ω0 (σ1 (t)) + ω0 (σ2 (t))] 1−L # 1 " −t 3 = e−σ¯ ∗ t , e + e−σ¯ 1 t + e−σ¯ 2 t ≤ 1 − k1 1 − k1

ω¯ 0 (t) =

ω¯ 1 (t, s) = ω1 (t, s) + ω1 (σ1 (t), s) + ω1 (σ2 (t), s) " # = 2πe−s e−t + e−σ¯ 1 t + e−σ¯ 2 t ≤ 6πe−s e−σ¯ ∗ t , ω¯ 2 (t, s, r) = ω2 (t, s, r) + ω2 (σ1 (t), s, r) + ω2 (σ2 (t), s, r) " # = 4πe−s−r e−t + e−σ¯ 1 t + e−σ¯ 2 t ≤ 12πe−s−r−σ¯ ∗ t ,

Asymptotically Stable Solutions for an Integral Equation

For 0 ≤ r ≤ t,



ω(t, r) = ω¯ 1 (t, r) +

t

ω¯ 1 (t, s)ω¯ 2 (t, s, r)ds

r

≤ 6πe−r e−σ¯ ∗ t + 72π 2 e−r−2σ¯ ∗ t

e−2r − e−2t 2

≤ 72π 2 e−r−σ¯ ∗ t ≡ A¯ 2 e−r−σ¯ ∗ t . 2   t  t 3 b(t) = 2ω¯ 02 (t) ω2 (t, r)dr ≤ 2 A¯ 22 e−4σ¯ ∗ t e−2r dr 1 − k1 0 0  2 3 ≤ A¯ 22 e−4σ¯ ∗ t ≡ b¯ 1 e−4σ¯ ∗ t → 0, as t → +∞. 1 − k1

  t t (ii2) 0 a¯ 2 (s) exp s b(r)dr ds is bounded:  t  b¯ 1 −4σ¯ ∗ s b¯ 1 b(r)dr ≤ − e−4σ¯ ∗ t ≤ , e 4 σ ¯ 4 σ¯ ∗ ∗ s  t   ¯   ¯   t  t 9A¯ 21 b1 b1 2 2 −2σ¯ ∗ s ¯ ds ≤ < ∞. a¯ (s) exp b(r)dr ds ≤ 9A1 e exp exp 4 σ ¯ 2 σ ¯ 4 σ¯ ∗ ∗ ∗ 0 s 0 So, (ii) is proved. Then Theorem 3 holds for (39). For more details, it is not difficult to show that the equation     μ1 (t)   μ2 (s) ξ(t) = V t, ξ(t), V1 t, s, ξ(σ1 (s)), V2 (t, s, r, ξ(σ2 (r))) dr ds , 0

0

t ≥ 0, has a unique solution ξ defined by ξ : R+ → E, ξ(t)(η) = ξ(t, η) =

et

32 +η

and x∗ : R+ → E, x∗ (t)(η) = x∗ (t, η) =

et

1 +η

∀η ∈ [0, 1], ∀η ∈ [0, 1],

(41)

(42)

is a solution of (39). Furthermore, lim x∗ (t) − ξ(t) = 31 lim e−t = 0.

t→∞

t→∞

Consequently, ξ and x∗ as in (41), (42) are asymptotically stable solutions of (39). Acknowledgements The authors wish to express their sincere thanks to the referees for the suggestions and valuable comments. The authors are also extremely grateful for the support given by Vietnam National University HoChiMinh City (VNU-HCM) under Grant no. B2013-18-05.

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L. T. P. Ngoc, N. T. Long 5. Dhage, B.C., Ntouyas, S.K.: Existence results for nonlinear functional integral equations via a fixed point theorem of Krasnoselskii-Schaefer type. Nonlinear Studies 9, 307–317 (2002) 6. Lang, S.: Analysis, vol. II. Addison - Wesley, Reading, Mass., California London (1969) 7. Liu, Z., Kang, S.M., Ume, J.S.: Solvability and asymptotic stability of a nonlinear functional-integral equation. Appl. Math. Lett. 24(6), 911–917 (2011) 8. Ngoc, L.T.P., Long, N.T.: On a fixed point theorem of Krasnosel’skii type and application to integral equations. Fixed Point Theory Appl. 2006, Article ID 30847, 24 pages (2006) 9. Ngoc, L.T.P., Long, N.T.: Existence of asymptotically stable solutions for a nonlinear functional integral equation with values in a general Banach space. Nonlinear Analysis, Theory, Methods & Applications. Series A: Theory and Methods 74(18), 7111–7125 (2011) 10. Ngoc, L.T.P., Long, N.T.: Solving a system of nonlinear integral equations and existence of asymptotically stable solutions. Differ. Equ. Appl. 4(2), 233–255 (2012) 11. Purnaras, I.K.: A note on the existence of solutions to some nonlinear functional integral equations. Electron. J. Qual. Theory Differ. Equ. 17, 1–24 (2006)