Mathematics in Applied Sciences and Engineering https://ojs.lib.uwo.ca/mase Online First, pp.1-19 https://doi.org/10.5206/mase/xxxxx A STUDY ON THE MILD SOLUTION OF SPECIAL RANDOM IMPULSIVE FRACTIONAL DIFFERENTIAL EQUATIONS SAYOOJ ABY JOSE1,2, VARUN BOSE C. S.3, BIJESH P. BIJU4, AND ABIN THOMAS5 Abstract. In this article, we deal with mild solution of special random impulsive fractional differential equations. Initially, we present the existence of the mild solution via Leray-Schauder fixed point method. After that, we establish the exponential stability of the system. Finally, we give examples to illustrate the effectiveness of the theoretical results. 1. Introduction Impulsive differential Equations are very adaptive Mathematical model that replicate the evolution of large classes of real process. Recently, in the fields of science and technology, we use fractional differential equations and impulsive fractional differential equations as a great mathematical tool in modelling. The stabilities like continuous dependence Mittag – Leffler Stability, Hyers Ulam stability and Hyers-Ulam- Rassins stability for fractional differential equations and impulsive fractional differential equations made curiosity in the mind of many researchers in the field of mathematics [10, 8, 14]. For impulsive differential systems, most researchers concentrate on the problems related to fixed time impulses [5, 21, 29]. But the actual jumps happen mostly at random points. The solutions of the random impulsive differential equations are a stochastic process. Now a day, the characteristics of solutions to some integer order differential equations with random impulses have been analysed [25, 2, 15, 28]. Anguraj et al. [2] established the existence and exponential stability of semilinear functional differential equations with random impulses under non-uniqueness. Yong and Wu [28] investigated the solutions of stochastic differential equations with Random impulse using Lipschitz condition. Wu et al. [26, 27] discussed the exponential stability and boundedness of differential equations with random impulses. Sayooj et al.[17] have studied some characteristics of random integro differential equations with non local conditions. In [16, 18, 19], the author found sufficient conditions for the existence as well as stability of special random impulsive differential system with non local conditions using contraction mapping principle and continuous dependence on initial conditions. Now a days, fractional calculus has a lot of advanced research work has been done. And also it have proved to be valuable tools in the modeling on many phenomina in various field of science and engineering [4, 12, 13, 20, 22, 30]. The study about impulsive fractional differential equations have a great attention, Akram Ben Alissa et al. [1] study the impulsive wave equation and analysis of this problem from different angles to prove some results about impulsive controlability and obervability without any geometrical condition on space Ω. In Many researchers studies about the existence, stability and uniqueness of fractional differential equations without random impulses [5, 9]. In this paper we make Received by the editors 25 May 2022; 26 November 2022; published online 15 December 2022. 2010 Mathematics Subject Classification. 35R12; 60H99; 34D20; 35B10. Key words and phrases. Existence; Leray-Schauder alternative fixed point; fractional differential equation; random impulses; exponential stability. 1 https://ojs.lib.uwo.ca/mase https://dx.doi.org/https://doi.org/10.5206/mase/xxxxx 2 X. ONE, X. TWO, X. THREE, AND X. FOUR a first attempt to study the existence and exponential stability results for special random impulsive fractional differential systems by make use of the Leray – Schauder alternative fixed point theorem. The main contributions of this work are given below: ↪→ We substantiate sufficient conditions for the existence of solutions for special random impulsive fractional differential equations entangling the Caputo fractional derivative. ↪→ We prove the results on existence of solutions of special random impulsive fractional differential equations by the use of the Leray – Schauder alternative fixed point theorem. This problem helps to solve many complicated random impulsive fractional systems. ↪→ We find exponential stability in the quadratic mean of special random impulsive fractional differential equations. ↪→ We provide examples of special random impulsive fractional differential systems as well as random impulsive fractional differential systems. It helps to interpret the effectiveness of the proposed results. And the remaining work is constructed as follows: this paper consist of 4 sections. In Section 1 we present few preliminaries, hypotheses results about fractional derivatives. Section 2 will be concerned with existence and followed by exponential stability in the quadratic mean of special random impulsive fractional differential equations in Section 3. The last section is allocated to examples illustrating the applicability of the imposed conditions. 2. Preliminaries Consider a real separable Hilbert space X and a non empty set Ω. Let %k be a random variable and %k maps Ω to Dk, where Dk = (0,dk) for every k ∈ N ( collection of natural numbers ) and 0 < dk < +∞. Also for i,j = 1, 2, . . . assume that if i 6= j then %i and %j are independent with each other. Also assume %k follow Erlang distribution. Let % be a real constant, denote <% = [%, +∞), <+ = [0, +∞). Consider the semilinear functional special random impulsive differential equations of the form cDat x(t) = Ax(t) + f(t,x(t),Ux(t),V x(t)) t 6= ξk, t ≥ t0, x(ξk) = bk(%k)x(ξ − k ),k = 1, 2, 3, . . . , (2.1) x(t0) = xt0 cDat is the Caputo fractional derivative of order 0 < a < 1. A is the infinitesimal generator of a strongly continuous semi group of bounded linear operators T(t), T ∈ X. The function f : <% ×X × X × X → X,bk : Dk → X for each k ∈ N; ξ0 = t0 and ξk = ξk−1 + %k for each k ∈ N. Obviously 0 < t0 = ξ0 < ξ1 < ξ2 < ξ3 · · · < ξk < ... ; x(ξk−) = limt↑ξk x(t) according to their path with the norm ‖x‖ = supt0≤u≤t | x(u) | for every t satisfying t ∈ [t0,T]. Ux(t) = ∫ t t0 K (t,r)x(r)dr, K ∈ C[D,<+], V x(t) = ∫ T t0 H (t,r)x(r)dr, H ∈ C[D0,<+], where D = {(t,r) ∈ <2 : t0 ≤ r ≤ t ≤ T}, D0 = {(t,r) ∈ <2 : t0 ≤ t,r ≤ T}. Let {Bt, t ≥ 0} be the simple counting process generated by {ξn}, this implies {Bt ≥ t} = {ξn ≤ t}, also Ft is the notation for the σ− algebra generated by {Bt, t ≥ 0}. The probability space denoted as (Ω,P,{Ft}). And the Hilbert space of all {Ft}− measurable square integrable random variables with values in X is denoted as L2 = L2(Ω,{Ft},X). YOUR RUNNING TITLE 3 Let B represent Banach space B([t0,T], L2), the family of all {Ft}- measurable random variable ψ with the norm ‖ψ‖2 = sup t∈[t0,T] E‖ψ(t)‖2 Definition 2.1. The fractional integral of the order a with the lower limit 0 for a function f is defined as Iaf(t) = 1 Γ(a) ∫ t 0 f(r) (t−r)1−a dr, t > 0,a > 0, provided the right-hand side is pointwise defined on [0,∞), where Γ is a gamma function. Definition 2.2. The Riemann–Liouville derivative of order a with the lower limit 0 for a function f : [0,∞) → R can be written as LDaf(t) = 1 Γ(n−a) dn dtn ∫ t 0 f(r) (t−r)a+1−n dr, t > 0,n− 1 < a < n. Definition 2.3. The Caputo derivative of order a for a function f : [0,∞) → R can be written as cDaf(t) =L Da [ f(t) − n−1∑ k=0 tk k! f(k)(0) ] , t > 0,n− 1 < a < n. Definition 2.4. A semigroup {T(t), t ≥ t0} is said to be uniformly bounded if ‖T(t)‖≤W for all t ≥ t0, where W ≥ 1 is some constant. If W = 1, then the semigroup is said to be contraction semigroup. Definition 2.5. For a given T ∈ (t0, +∞), a stochastic process {x(t) ∈ B, 0 < t0 ≤ t ≤ T} is called a solution to the equation (2.1) in (Ω,P,{Ft}), if (i) x(t) ∈ B is Ft- adapted; (ii) x(t) = +∞∑ k=0 [ k∏ i=1 bi(%i)T(t− t0)xt0 + 1 Γ(a) k∑ i=1 k∏ j=i bj(%j) ∫ ξi ξi−1 (t−r)a−1T(t−r)f(r,x(r),Ux(r),V x(r))dr + 1 Γ(a) ∫ t ξk (t−r)a−1T(t−r)f(r,x(r),Ux(r),V x(r))dr ] I[ξk,ξk+1)(t), t ∈ [t0,T], where T ∈ (t0, +∞), n∏ j=m (·) = 1 as m > n, k∏ j=i bj(%j) = bk(%k)bk−1(%k−1) . . .bi(%i), and IA(·) is the index function. Remark: The proof of mild solution similar to [3, 23, 29], so we omit it. Hypotheses. Some hypotheses which are used for proving the main results are given below; (H1) There exist a continuous non-decreasing function H : <+ → (0,∞) and L1, L2, L3 ∈ L1([%,T],<+) so that E‖f(t,ζ1,ζ2,ζ3)‖2 ≤ L1(t)H ( E‖ζ1‖ )2 + L2(t)H ( E‖ζ2‖ )2 + L3(t)H ( E‖ζ3‖ )2 4 X. ONE, X. TWO, X. THREE, AND X. FOUR (H2) E { maxi,k {∏k j=i‖bj(%j)‖ }} is uniformly bounded if, E { max i,k {∏k j=i‖bj(%j)‖ }} ≤ Θ, for each %j ∈ Dj,j ∈ N, Θ > 0 a constant (H3) Define L ,K ∗ and H∗ such that, L = max{L1, L2, L3}, K∗ = sup t∈[t0,T] ∫ t t0 |K (t,r)|2dt < ∞, and H∗ = sup t∈[t0,T] ∫ T t0 |H (t,r)|2dt < ∞. Our existence and exponential stability theorems are based on the succeeding theorem, which is a version of the topological transversal theorem. Lemma 2.1. Let E be a convex subset of a Banach space X, and assume that 0 ∈ E. Let F : E → E be a completely continuous operator, and let U(F) = {x ∈ E : x = λFx for some 0 < λ < 1}, then either U(F) is unbounded or F has a fixed point. 3. Existence Here, we presents the results on existence of solutions of special random impulsive fractional differ- ential equations by make use of the Leray – Schauder alternative fixed point theorem. Theorem 3.1. Assume (H1), (H2), and (H3) hold, then the system (2.1) has mild solution x(t), defined on [t0,T], provided the following inequality is satisfied: Γ ∫ T t0 L (r)dr < ∫ ∞ γ1 dr H(r) , (3.1) where Γ = 2W2 max{1, Θ2}(T−t0) 2a−1(1+K∗+H∗) (2a−1)Γ(a) ,γ1 = 2W 2Θ2E‖ϕ‖2 and WΘ ≥ 1√ 2 . Proof. Let Ψ be an operator from B to B and the arbitrary positive number T ∈ (t0,∞): Ψx(t) = +∞∑ k=0 [ k∏ i=1 bi(%i)T(t− t0)xt0 + 1 Γ(a) k∑ i=1 k∏ j=i bj(%j) ∫ ξi ξi−1 (t−r)a−1T(t−r)f(r,x(r),Ux(r),V x(r))dr + 1 Γ(a) ∫ t ξk (t−r)a−1T(t−r)f(r,x(r),Ux(r),V x(r))dr ] I[ξk,ξk+1)(t), t ∈ [t0,T] YOUR RUNNING TITLE 5 First we deduce the solution of the integral equation and assume λ ∈ (0, 1): x(t) = λ +∞∑ k=0 [ k∏ i=1 bi(%i)T(t− t0)xt0 + 1 Γ(a) k∑ i=1 k∏ j=i bj(%j) ∫ ξi ξi−1 (t−r)a−1T(t−r)f(r,x(r),Ux(r),V x(r))dr + 1 Γ(a) ∫ t ξk (t−r)a−1T(t−r)f(r,x(r),Ux(r),V x(r))dr ] I[ξk,ξk+1)(t), t ∈ [t0,T] Hence by (H1), (H2) and (H3) ‖x(t)‖2 ≤ λ2 [ +∞∑ k=0 [∥∥∥∥ k∏ i=1 bi(%i) ∥∥∥∥‖T(t− t0)‖‖xt0‖ + 1 Γ(a) k∑ i=1 ∥∥∥∥ k∏ j=i bj(%j) ∥∥∥∥ ∫ ξi ξi−1 (t−r)a−1‖T(t−r)f(r,x(r),Ux(r),V x(r))‖dr + 1 Γ(a) ∫ t ξk (t−r)a−1‖T(t−r)f(r,x(r),Ux(r),V x(r))‖dr ] I[ξk,ξk+1)(t) ]2 ≤ 2 [ +∞∑ k=0 [∥∥∥∥ k∏ i=1 bi(%i) ∥∥∥∥2‖T(t− t0)‖2‖xt0‖2 ] + 2 [ ∞∑ k=0 1 Γ(a) k∑ i=1 ∥∥∥∥ k∏ j=i bj(%j) ∥∥∥∥ ∫ ξi ξi−1 (t−r)a−1‖T(t−r)f(r,x(r),Ux(r),V x(r))‖dr + 1 Γ(a) ∫ t ξk (t−r)a−1‖T(t−r)f(r,x(r),Ux(r),V x(r))‖dr ] I[ξk,ξk+1)(t) ]2 ≤ 2W2Θ2 ∥∥∥∥xt0 ∥∥∥∥2 + 2W2 max{1, Θ2}(T − t0)2a−1Γ(a)(2a− 1) ∫ t t0 ∥∥∥∥f(r,x(r),Ux(r),V x(r)) ∥∥∥∥2dr, ‖x(t)‖2 ≤ 2W2Θ2 ∥∥∥∥ϕ ∥∥∥∥2 + 2W2 max{1, Θ2}(T − t0)2a−1Γ(a)(2a− 1) ∫ t t0 ∥∥∥∥f(r,x(r),Ux(r),V x(r)) ∥∥∥∥2dr, and E‖x(t)‖2 ≤ 2W2Θ2E [ ‖ϕ‖2 ] + 2W2 max{1, Θ2} (T − t0)2a−1 Γ(a)(2a− 1) ∫ t t0 E [ ‖f(r,x(r),Ux(r),V x(r))‖2 ] dr ≤ 2W2Θ2E [ ‖ϕ‖2 ] + 2W2 max{1, Θ2} (T − t0)2a−1 Γ(a)(2a− 1) ∫ t t0 [ L1(r)H ( E‖x(r)‖2 ) + L2(r)H ( E‖Ux(r)‖2 ) + L3(r)H ( E‖V x(r)‖2 )] dr, sup t0≤υ≤t E‖x(υ)‖2 ≤ 2W2Θ2E‖ϕ‖2 + 2W2 max{1, Θ2} (T − t0)2a−1 Γ(a)(2a− 1) ∫ t t0 L (r)(1 + K∗ + H∗)H ( sup t0≤υ≤r E‖x(υ)‖2 ) dr, ≤ 2W2Θ2E‖ϕ‖2 + 2W2 max{1, Θ2} (T − t0)2a−1(1 + K∗ + H∗) Γ(a)(2a− 1) ∫ t t0 L (r)H(ω(r))dr 6 X. ONE, X. TWO, X. THREE, AND X. FOUR where ω(t) = sup t0≤υ≤t E [ ‖x(υ)‖2 ] , t ∈ [t0,T]. Moreover, for any t ∈ [t0,T], ω(t) ≤ 2W2Θ2E [ ‖ϕ‖2 ] + 2W2 max{1, Θ2} (T − t0)2a−1(1 + K∗ + H∗) Γ(a)(2a− 1) ∫ t t0 L (r)H(ω(r))dr. Represent by the right hand side of the above inequality as V (t), then ω(t) ≤ V (t) for t ∈ [t0,T], V (t0) = 2W2Θ2E‖ϕ‖2 = γ1 and V ′(t) = 2W2 max{1, Θ2} (T − t0)2a−1(1 + K∗ + H∗) Γ(a)(2a− 1) L (t)H(ω(t)) ≤ 2W2 max{1, Θ2} (T − t0)2a−1(1 + K∗ + H∗) Γ(a)(2a− 1) L (t)H(ω(t)), t ∈ [t0,T]. Then V ′(t) H(V (t)) ≤ 2W2 max{1, Θ2} (T − t0)2a−1(1 + K∗ + H∗) Γ(a)(2a− 1) L (t), t ∈ [t0,T] Apply the change of variable and integrate the previous inequality from t0 to t, we get ∫ V (t) V (t0) dr H(r) ≤ 2W2 max{1, Θ2} (T − t0)2a−1(1 + K∗ + H∗) Γ(a)(2a− 1) ∫ t t0 L (r)dr ≤ 2W2 max{1, Θ2} (T − t0)2a−1(1 + K∗ + H∗) Γ(a)(2a− 1) ∫ T t0 L (r)dr < ∫ ∞ γ1 dr H(r) = ∫ ∞ V (t0) dr H(r) . By the mean value theorem and the above inequality, there is a constant Υ such that V (t) ≤ Υ, and therefore ω(t) ≤ Υ. Where as supt0≤υ≤T E‖x(υ)‖ 2 = ω(t) hold for each t ∈ [t0,T], we have supt0≤υ≤T E‖x(υ)‖ 2 ≤ Υ, where Υ depends on the function L and H and on T, therefore E‖x(t)‖2 = sup t0≤υ≤T E‖x(υ)‖2 ≤ Υ In the following steps, we will show that Ψ is continuous and completely continuous. Step 1: We show that Ψ is continuous. For every t ∈ [t0,T] and consider {xn} be a convergent sequence of elements of x ∈ B, then Ψxn(t) = +∞∑ k=0 [ k∏ i=1 bi(%i)T(t− t0)ϕ(0) + 1 Γ(a) k∑ i=1 k∏ j=i bj(%j) ∫ ξi ξi−1 (t−r)a−1T(t−r)f(r,xn(r),Uxn(r),V xn(r))dr + 1 Γ(a) ∫ t ξk (t−r)a−1T(t−r)f(r,xn(r),Uxn(r),V xn(r))dr ] I[ξk,ξk+1)(t). YOUR RUNNING TITLE 7 So Ψxn(t) − Ψx(t) = +∞∑ k=0 [ 1 Γ(a) k∑ i=1 k∏ j=i bj(%j) ∫ ξi ξi−1 (t−r)a−1T(t−r) [ f(r,xn(r),Uxn(r),V xn(r)) −f(r,x(r),Ux(r),V x(r)) ] dr + 1 Γ(a) ∫ t ξk (t−r)a−1T(t−r) [ f(r,xn(r),Ux(r),V x(r)) −f(r,x(r),Ux(r),V x(r)) ] dr ] I[ξk,ξk+1)(t), and E‖Ψxn − Ψx‖2 ≤W2 max{1, Θ2} (T − t0)2a−1 Γ(a)(2a− 1) ∫ t t0 E‖f(r,xn(r),Uxn(r),V xn(r)) −f(r,x(r),Ux(r),V x(r))‖2dr. So Ψ is continuous. Step 2: We show that Ψ is completely continuous operator. Represent Θm = {x ∈ B | ‖x‖2 ≤ m} where m ≥ 0. Step 2.1: We prove that Ψ maps to Θm into an equicontinuous family. Let t1, t2 ∈ [t0,T] and x ∈ Θm. Whenever t0 < t1 < t2 < T, by (H1), (H2), (H3) and condition (3.1), we obtain Ψx(t2) − Ψx(t1) = +∞∑ k=0 [ k∏ i=1 bi(%i)T(t2 − t0)xt0 + 1 Γ(a) k∑ i=1 k∏ j=i bj(%j) ∫ ξ ξi−1 (t2 −r)a−1T(t2 −r)f(r,x(r),Ux(r),V x(r))dr + 1 Γ(a) ∫ t2 ξk (t2 −r)a−1T(t2 −r)f(r,x(r),Ux(r),V x(r))dr ] I[ξk,ξk+1)(t2) − +∞∑ k=0 [ k∏ i=1 bi(%i)T(t1 − t0)xt0 + 1 Γ(a) k∑ i=1 k∏ j=i bj(%j) ∫ ξi ξi−1 (t1 −r)a−1T(t1 −r)f(r,x(r),Ux(r),V x(r))dr + 1 Γ(a) ∫ t1 ξk (t1 −r)a−1T(t1 −r)f(r,x(r),Ux(r),V x(r))dr ] I[ξk,ξk+1)(t1) = +∞∑ k=0 [ k∏ i=1 bi(%i)T(t2 − t0)xt0 + 1 Γ(a) k∑ i=1 k∏ j=i bj(%j) ∫ ξi ξi−1 (t2 −r)a−1T(t2 −r)f(r,x(r),Ux(r),V x(r))dr 8 X. ONE, X. TWO, X. THREE, AND X. FOUR + 1 Γ(a) ∫ t2 ξk (t2 −r)a−1T(t2 −r)f(r,x(r),Ux(r),V x(r))dr ][ I[ξk,ξk+1)(t2) − I[ξk,ξk+1)(t1) ] + +∞∑ k=0 [ k∏ i=1 bi(%i) [ T(t2 − t0) −T(t1 − t0) ] xt0 + 1 Γ(a) k∑ i=1 k∏ j=i bj(%j) ∫ ξi ξi−1 [ (t2 −r)a−1T(t2 −r) − (t1 −r)a−1T(t1 −r) ] f(r,x(r),Ux(r),V x(r))dr + 1 Γ(a) ∫ t1 ξk [ (t2 −r)a−1T(t2 −r) − (t1 −r)a−1T(t1 −r) ] f(r,xn(r),Ux(r),V x(r))dr + 1 Γ(a) ∫ t2 t1 (t2 −r)a−1T(t2 −r)f(r,x(r),Ux(r),V x(r))dr ] I[ξk,ξk+1)(t1). Moreover E‖Ψx(t2) − Ψx(t1)‖2 ≤ 2E‖I1‖2 + 2E‖I2‖2, where I1 = +∞∑ k=0 [ k∏ i=1 bi(%i)T(t2 − t0)xt0 + 1 Γ(a) k∑ i=1 k∏ j=i bj(%j) ∫ ξi ξi−1 (t2 −r)a−1T(t2 −r)f(r,x(r),Ux(r),V x(r))dr + 1 Γ(a) ∫ t ξk (t2 −r)a−1T(t2 −r)f(r,x(r),Ux(r),V x(r))dr ][ I[ξk,ξk+1)(t2) − I[ξk,ξk+1)(t1) ] and I2 = +∞∑ k=0 [ k∏ i=1 bi(%i) [ T(t2 − t0) −T(t1 − t0) ] xt0 1 Γ(a) k∑ i=1 k∏ j=i bj(%j) ∫ ξi ξi−1 [ (t2 −r)a−1T(t2 −r) − (t1 −r)T(t1 −r) ] f(r,x(r),Ux(r),V x(r))dr + 1 Γ(a) ∫ t1 ξk (t2 −r)a−1T[(t2 −r) − (t1 −r)a−1T(t1 −r)]f(r,x(r),Ux(r),V x(r))dr + 1 Γ(a) ∫ t2 t1 (t2 −r)a−1T(t2 −r)f(r,x(r),Ux(r),V x(r))dr ] I[ξk,ξk+1)(t1) Besides, E‖I1‖2 ≤ E [ +∞∑ k=0 [ k∏ i=1 ‖bi(%i)‖‖T(t2 − t0)‖‖xt0‖ + 1 Γ(a) k∑ i=1 k∏ j=i ‖bj(%j)‖ ∫ ξi ξi−1 (t2 −r)a−1‖T(t2 −r)‖‖f(r,x(r),Ux(r),V x(r))‖dr + 1 Γ(a) ∫ t2 ξk (t2 −r)a−1‖T(t2 −r)‖‖f(r,x(r),Ux(r),V x(r))‖dr ][ I[ξk,ξk+1)(t2) − I[ξk,ξk+1)(t1) ]]2 ≤2W2Θ2E‖xt0| 2E(I[ξk,ξk+1)(t2) − I[ξk,ξk+1)(t1)) YOUR RUNNING TITLE 9 + 2 max{1, Θ2} (T − t0)2a−1 Γ(a)(2a− 1) E ∫ t2 t0 ‖T(t2 −r)‖‖f(r,x(r),Ux(r),V x(r))‖dr ×E(I[ξk,ξk+1)(t2) − I[ξk,ξk+1)(t1)) ≤2W2Θ2E‖xt0‖ 2E ( I[ξk,ξk+1)(t2) − I[ξk,ξk+1)(t1) ) + 2W2 max{1, Θ2} (T − t0)2a−1(1 + K∗ + H∗) Γ(a)(2a− 1) ∫ t2 t0 L (r)H(E‖x(r)‖2)dr ×E ( I[ξk,ξk+1)(t2) − I[ξk,ξk+1)(t1) ) ≤2W2Θ2E‖xt0‖ 2E ( I[ξk,ξk+1)(t2) − I[ξk,ξk+1)(t1) ) + 2W2 max{1, Θ2} (T − t0)2a−1(1 + K∗ + H∗) Γ(a)(2a− 1) ∫ t2 t0 L ∗H(E(m))drE ( I[ξk,ξk+1)(t2) − I[ξk,ξk+1)(t1) ) → 0, as t1 → t2. where L ∗ = sup { L (t) : t ∈ [t0,T] } . E‖I2‖2 ≤E [ +∞∑ k=0 [ k∏ i=1 ‖bi(%i)‖‖T(t2 − t0) −T(t1 − t0)‖‖xt0‖ + 1 Γ(a) k∑ i=1 k∏ j=i ‖bj(%j)‖ ∫ ξi ξi−1 ‖(t2 −r)a−1T(t2 −r) − (t1 −r)a−1T(t1 −r)‖‖f(r,x(r),Ux(r),V x(r))‖dr + 1 Γ(a) ∫ t1 ξk ‖(t2 −r)a−1T(t2 −r) − (t1 −r)a−1T(t1 −r)‖‖f(r,x(r),Ux(r),V x(r))‖dr + 1 Γ(a) ∫ t2 t1 (t2 −r)a−1‖T(t2 −r)‖‖f(r,x(r),Ux(r),V x(r))dr‖ ] I[ξk,ξk+1)(t1) ]2 ≤3Θ2‖T(t2 − t0) −T(t1 − t0)‖2E‖xt0‖ 2 + 3 max{1, Θ2}(t1 − t0) 1 Γ(a) E ∫ t1 t0 ‖(t2 −r)a−1T(t2 −r) − (t1 −r)a−1T(t1 −r)‖2‖f(r,x(r),Ux(r),V x(r))‖2dr + 3W2 (T − t0)2a−1 Γ(a)(2a− 1) E ∫ t2 t1 ‖f(r,x(r),Ux(r),V x(r))‖2dr ≤3Θ2‖T(t2 − t0) −T(t1 − t0)‖2E‖xt0‖ 2 + 3 max{1, Θ2}(t1 − t0) 1 Γ(a) ∫ t1 t0 ‖(t2 −r)a−1T(t2 −r) − (t1 −r)a−1T(t1 −r)‖2L ∗H(m)dr + 3W2 (T − t0)2a−1(1 + K∗ + H∗) Γ(a)(2a− 1) ∫ t2 t1 L ∗H(m)dr → 0 as t1 → t2. So, Ψ maps Θm into an equicontinuous family of functions. Step 2.2: We prove that ΨΘm is uniformly bounded. 10 X. ONE, X. TWO, X. THREE, AND X. FOUR By (3.1), ‖x‖2 ≤ m, (H1), (H2) and (H3), we get ‖Ψx(t)‖2 ≤ [ +∞∑ k=0 [ k∏ i=1 ‖bi(%i)‖‖T(t− t0)‖‖xt0‖ + 1 Γ(a) k∑ i=1 k∏ j=i ‖bj(%j)‖ ∫ ξi ξi−1 (t−r)a−1‖T(t−r)‖‖f(r,x(r),Ux(r),V x(r))‖dr + 1 Γ(a) ∫ t ξi−1 (t−r)a−1‖T(t−r)‖‖f(r,x(r),Ux(r),V x(r))‖dr ] I[ξk,ξk+1)(t) ]2 . ≤ 2W2Θ2‖ϕ(0)‖2 + 2W2 max{1, Θ2} (T − t0)2a−1 Γ(a)(2a− 1) ∫ t t0 ‖f(r,x(r),Ux(r),V x(r))‖2dr. Thus, E‖Ψx(t)‖2 ≤ 2W2Θ2E‖ϕ(0)‖2 + 2W2 max{1, Θ2} (T − t0)2a−1 Γ(a)(2a− 1) ∫ t t0 E‖f(r,x(r),Ux(r),V x(r))‖2dr. ≤ 2W2Θ2E‖ϕ(0)‖2 + 2W2 max{1, Θ2} (T − t0)2a(1 + K∗ + H∗) Γ(a)(2a− 1) ‖bm‖L. Therefore {(Ψx(t)),‖x‖2 ≤ m} is uniformly bounded, so does {ΨΘm}. Then by the Arzela – Ascoli theorem, Ψ maps Θm into a precompact set in X. Step 2.3: We prove that ΨΘm is compact. Let t ∈ (t0,T] be fixed, and let � be a real number such that � ∈ (0, t− t0) for x ∈ Θm, we establish (Ψ�x)(t) = +∞∑ k=0 [ k∏ i=1 bj(%j)T(t− t0)xt0 + 1 Γ(a) k∑ i=1 k∏ j=i bj(%j) ∫ ξi ξi−1 (t−r)a−1T(t−r)f(r,x(r),Ux(r),V x(r))dr + 1 Γ(a) ∫ t−� ξk (t−r)a−1T(t−r)f(r,x(r),Ux(r),V x(r))dr ] I[ξk,ξk+1)(t), t ∈ (t0, t− �). Being T(t) is a compact operator, the set H�(t) = {(Ψ�x)(t) : x ∈ Θm} is precompact in X for each � ∈ (0, t− t0). Furthermore, for each x ∈ Θm, we attain (Ψx)(t) − (Ψ�x)(t) = +∞∑ k=0 [ 1 Γ(a) ∫ t ξk (t−r)a−1T(t−r)f(r,x(r),Ux(r),V x(r))dr ] I[ξk,ξk+1)(t) − +∞∑ k=0 [ 1 Γ(a) ∫ t−� ξk (t−r)a−1T(t−r)f(r,x(r),Ux(r),V x(r))dr ] I[ξk,ξk+1)(t). By making use of (H1), (H2), (H3), condtion 4.1, and ‖x(B)‖2 ≤ m, we obtain E‖(Ψx)(t) − (Ψ�x)(t))‖2t ≤W 2 (T − t0) 2a−1(1 + K∗ + H∗) Γ(a)(2a− 1) ∫ t t−� L ∗H(m)dr. Hence, there exist precompact sets arbitrarily close to the set {(Ψx)(t) : x ∈ Θm} is precompact in X. So, Ψ is completely continuous operator. YOUR RUNNING TITLE 11 Furthermore, the set U(Ψ) = {x ∈ B : x = λΨx for some 0 < λ < 1} is bounded. Hence, by lemma 2.1, the operator Ψ has a fixed point in B. So, system (2.1) has a mild solution. � 4. Exponential stability in the quadratic mean This section, we establish the exponential stability of a second moment of mild solution of system. For an Ft - adapted process, Ψ(t) : [0,∞) → R is almost continuous in t. In order to attain the stability, we suppose that f(t, 0) ≡ 0 for any t ≤ t0 thus the system (2.1) accept a trivial solution. Furthermore, E‖Ψ‖2t → 0 as t →∞. Definition 4.1. System (2.1) is said to be exponentially stable in the quadratic mean if there exist positive constants K1 > 0 and ν > 0 such that E‖x(t)‖≤ K1E‖ϕ‖2e−ν(t−t0), t ≥ t0. Now we introduce the following hypothesis used in our discussion: (H4) µH(ψ) ≤ H(µψ) for all ψ ∈ R+, where µ > 1. (H5) ‖T(t)‖≤We−ξ(t−t0), t ≥ 1. Theorem 4.1. Assume that the hypothesis of Theorem 2.1 and (H4) − (H5) hold. If the following inequality is satisfied, then the system (2.1) is exponentially stable in the quadratic mean: Γ∗ ∫ T t0 L (r)dr < ∫ ∞ γ2 dr H(r) , (4.1) where Γ∗ = 2W2 max{1, Θ2}(T−t0) 2a−1(1+K∗+H∗) Γ(a)(2a−1) ,γ2 = 2W 2Θ2E‖ϕ‖2, and WΘ ≥ 1√ 2 . Proof. Let Ψ be defined in Theorem 2.1. Making use of hypotheses (H1) − (H5), we get ‖x(t)‖2 ≤ λ2 ( +∞∑ k=0 [∥∥∥∥ k∏ i=1 bi(%i) ∥∥∥∥‖T(t− t0)‖‖xt0‖ + 1 Γ(a) k∑ i=1 ∥∥∥∥ k∏ j=i bj(%j) ∥∥∥∥ ∫ ξi ξi−1 (t−r)a−1‖T(t−r)f(r,x(r),Ux(r),V x(r))‖dr + 1 Γ(a) ∫ t ξk (t−r)a−1‖T(t−r)f(r,x(r),Ux(r),V x(r))‖dr ] I[ξk,ξk+1)(t) )2 ≤ 2 +∞∑ k=0 [∥∥∥∥ k∏ i=1 bi(%i) ∥∥∥∥2‖W2e−2k(t−t0)‖‖xt0‖2I[ξk,ξk+1)(t) ] + 2   ∞∑ k=0   1 Γ(a) k∑ i=1 ∥∥∥∥ k∏ j=i bj(%j) ∥∥∥∥ ∫ ξi ξi−1 (t−r)a−1‖We−ξ(t−r)f(r,x(r),Ux(r),V x(r))‖dr + 1 Γ(a) ∫ t ξk (t−r)a−1W2e−ξ(t−r)‖f(r,x(r),Ux(r),V x(r))‖dr ] I[ξk,ξk+1)(t) )2 12 X. ONE, X. TWO, X. THREE, AND X. FOUR ≤ 2  max k { k∏ j=i ‖bj(%j)‖2 }2 W2e−2k(t−t0)‖xt0‖2 + 2  max i,k { 1, k∏ j=i ‖bj(%j)‖ }2 · · 1 Γ(a) +∞∑ k=0 ∫ t t0 (t−r)a−1We−ξ(t−r)‖f(r,x(r),Ux(r),V x(r))‖dr · I2[ξk,ξk+1)(t) ≤ 2W2Θ2e−2k(t−t0)‖xt0‖ 2 + 2W2 max{1, Θ2} (T − t0)2a−1 Γ(a)(2a− 1) ∫ t t0 e−2ξ(t−r)‖f(r,x(r),Ux(r),V x(r))‖2dr, ‖x(t)‖2 ≤ 2W2Θ2e−2ξ(t−t0)‖ϕ‖2 + 2W2 max{1, Θ2} (T − t0)2a−1 Γ(a)(2a− 1) ∫ t t0 e−2k(t−r)‖f(r,x(r),Ux(r),V x(r))‖2dr, zE‖x(t)‖2 ≤ 2W2Θ2e−2k(t−t0)E‖ϕ‖2 + 2W2 max{1, Θ2} (T − t0)2a−1(1 + K∗ + H∗) Γ(a)(2a− 1) ∫ t t0 e−2ξ(t−r)L (r)H ( E‖x(r)‖2 ) dr, = 2W2Θ2e−2ξ(t−t0)E‖ϕ‖2 + 2W2 max{1, Θ2} (T − t0)2a−1(1 + K∗ + H∗) Γ(a)(2a− 1) e−2k(t−t0) ∫ t t0 e2ξ(r−t0)L (r)H ( E‖x(r)‖2 )2 dr. Thus, e2ξ(t−t0)E‖x(t)‖2 ≤ 2W2Θ2E‖ϕ‖2 + 2W2 max{1, Θ2} (T − t0)2a−1(1 + K∗ + H∗) Γ(a)(2a− 1) ∫ t t0 e2ξ(t−r)L (r)H ( E‖x(r)‖2 )2 dr. Furhtermore, sup t0≤υ≤t e2ξ(υ−t0)E‖x(t)‖2 ≤ 2W2Θ2E‖ϕ‖2 + 2W2 max{1, Θ2} (T − t0)2a−1(1 + K∗ + H∗) Γ(a)(2a− 1) ∫ t t0 L (r)H ( sup t0≤υ≤t e2ξ(υ−t0)E‖x(r)‖2 ) dr. Take ω1(t) = sup t0≤υ≤t e2ξ(υ−t0)E‖x‖2, t ∈ [t0,T]. Also, for any t ∈ [t0,T], we have ω1(t) ≤ 2W2Θ2E‖ϕ‖2 + max{1, Θ2} (T − t0)2a−1(1 + K∗ + H∗) Γ(a)(2a− 1) ∫ t t0 L (r)H ( ω1(r) ) dr. YOUR RUNNING TITLE 13 Denote the right hand side of the above inequality V1(t), we obtain ω1(t) ≤ V1(t), t ∈ [t0,T], V1(t0) = 2W2Θ2E‖ϕ‖2 = γ2 and V ′1 (t) ≤ 2W 2 max{1, Θ2} (T − t0)2a−1(1 + K∗ + H∗) Γ(a)(2a− 1) L (t)H ( ω1(t) ) ≤ 2W2 max{1, Θ2} (T − t0)2a−1(1 + K∗ + H∗) Γ(a)(2a− 1) L (t)H ( V1(t) ) , t ∈ [t0,T]. That is, V ′1 (t) H ( V1(t) ) ≤ 2W2 max{1, Θ2}(T − t0)2a−1(1 + K∗ + H∗) Γ(a)(2a− 1) L (t), t ∈ [t0,T]. Apply the change of variable and integrate the previous inequality from t0 to t, we get∫ V1(t) V1(t0) dr H ( r ) ≤ 2W2 max{1, Θ2}(T − t0)2a−1(1 + K∗ + H∗) Γ(a)(2a− 1) ∫ t t0 L (r)dr ≤ 2W2 max{1, Θ2} (T − t0)2a−1(1 + K∗ + H∗) Γ(a)(2a− 1) ∫ T t0 L (r)dr < ∫ ∞ γ2 dr H ( r ) = ∫ ∞ V1(t0) dr H ( r ), t ∈ [t0,T]. By the mean value theorem and above inequality there exist a constant Υ1 such that V1(t) ≤ Υ1, and therefore ω1(t) ≤ Υ1. Whereas supt0≤υ≤t e 2ξ(υ−t0)E‖x‖2 = ω1(t) holds for each t ∈ [t0,T], we have supt0≤υ≤t e 2ξ(υ−t0)E‖x‖2 ≤ K1, where Υ1 depends on the function L and H. Therefore, e2ξ(t−t0)E‖x‖2 = sup t0≤υ≤T e2ξ(υ−t0)E‖x‖2 ≤ Υ1. As in the previous theorem, we will prove that Ψ is completely continuous operator through the follow- ing steps. Step 1: We show that Ψ is continuous. For every t ∈ [t0,T] and consider {xn} be a convegent sequence of element of x ∈ B, we obtain E‖Ψxn(t) − Ψx(t)‖2 ≤W2 max{1, Θ2} (T − t0)2a−1 Γ(a)(2a− 1) e−2k(t−t0) ∫ t t0 e2ξ(r−t0)E‖f(r,xn(r),Uxn(r),V xn(r)) −f(r,x(r),Ux(r),V x(r))‖2dr → 0, as n →∞. So Ψ is continuous. Step 2: We show that Ψ is completely continuous operator. Represent Θm1 = {x ∈ B | ‖x‖ 2 ≤ m1} 14 X. ONE, X. TWO, X. THREE, AND X. FOUR where m1 ≥ 0. Step 2.1: We prove that Ψ maps Θm1 into an equicontinuous family. Let x ∈ Θm1 and t1, t2 ∈ [t0,T]. If t0 < t1 < t2 < T , then by making use of (H1) − (H5) and condition (3) and pursuing the similar process of Step 2.1 of Theorem 3.1, we obtain E‖Ψx(t2) − Ψx(t1)‖2 → 0 as t2 → t1. So, Ψ maps to Θm1 into an equicontinuous family of functions. Step 2.2: We prove that ΨΘm1 is uniformly bounded. By the condition (3.1) and (H1) − (H5), we get ‖Ψx(t)‖2 ≤ 2 [ max k { k∏ j=i ‖bj(%j)‖2 }] W2e−2k(t−t0)‖xt0‖ 2 + 2 [ max i,k { 1, k∏ j=i ‖bj(%j)‖ }]2 × ( 1 Γ(a) +∞∑ k=0 ∫ t t0 (t−r)a−1We−ξ(t−r)‖f(r,x(r),Ux(r),V x(r))‖dr ) I[ξk,ξk+1)(t) 2. So E‖Ψx(t)‖2 ≤ 2W2Θ2e−2k(t−t0)E‖xt0‖ 2 + 2W2 max{1, Θ2} (T − t0)2a−1(1 + K∗ + H∗) Γ(a)(2a− 1) e−2k(t−t0) × ∫ t t0 e2ξ(r−t0)L ∗H(m)dr, where L ∗ = sup{L (t) : t ∈ [t0,T]}. Being e−2k(t−t0) → 0, the right hand side of the previous inequality tends to 0 as t →∞. ie, ‖(Ψx)‖2 → 0 t →∞. Therefore {(Ψx(t)),‖x‖2B ≤ m1} is uniformly bounded, thus {ΨΘm1} is uniformly bounded. Step 2.3: We prove that ΨΘm1 is compact. Let t ∈ (t0,T] be fixed and � be a real number such that � ∈ (0, t− t0), for x ∈ Θm1 , we establish (Ψ�x)(t) = +∞∑ k=0 [ k∏ i=0 bj(%j)T(t− t0)xt0 + 1 Γ(a) k∑ i=1 k∏ i=i bj(%j) ∫ ξi ξi−1 (t−r)a−1T(t−r)f(r,x(r),Ux(r),V x(r))dr YOUR RUNNING TITLE 15 + 1 Γ(a) ∫ t−� ξk (t−r)a−1T(t−r)f(r,x(r),Ux(r),V x(r))dr ] I[ξk,ξk+1)(t), t ∈ (t0, t− �). Being T(t) is a compact operator, the set H�(t) = {(Ψ�x)(t) : x ∈ Θm1} is precompact in X for each � ∈ (0, t− t0). Using (H1) − (H5), condition (3), and ‖x‖2 ≤ m1, we obtain E‖Ψxn(t) − Ψx(t)‖2 ≤W2 (T − t0)2a−1 Γ(a)(2a− 1) e−2k(t−t0) ∫ t t−� e2ξ(r−t0)L ∗H ( E‖x(r)‖2 ) dr. Hence, there exist precompact sets arbitrarily close to the set {(Ψx)(t) : x ∈ Θm1}. Thus the set {(Ψx)(t) : x ∈ Θm1} is precompact in X. So, Ψ is a completely continuous operator. Furthermore, the set U(Ψ) = {x ∈ B : x = λΨx for some 0 < λ < 1} is bounded. Hence, by Lemma 2.1, the operator Ψ has a fixed point in B. So the system 2.1 has a mild solution and E‖Ψ(t)‖2 → 0 as t →∞. Hence the proof. � 5. Applications Example 5.1. Consider random impulsive fractional differential equations,  cDat z(t,x) = zxx(x,t) + F1(t,z(t,x)) t 6= ξk, t ≥ % z(x,ξk) = q(k)%kz(x,ξ − k ) as x ∈4̂ z(t, 0) = z(t,π) = 0 z(t0,x) = z0(x), x ∈ ∂4̂ (5.1) Consider 4̂⊂ zn, z ∈ D(A) Here zn(ζ) = √ 2 π Sin(nζ),n = 1, 2, . . . , forms the orthonormal set of eigenvectors of A. Also for every z ∈ X,S(t)z = ∑∞ n=1 e (−n2t) < z,zn > zn, which holds ‖S(t)‖ ≤ e(−π 2(t−t0)), t ≥ t0. Therefore S(t) is 16 X. ONE, X. TWO, X. THREE, AND X. FOUR a semigroup. Consider that the following assumptions: (i) f : <% ×X → X, is a continuous function defined by f(t,z)(x) = F1(t,z(x)) t0 ≤ t ≤ T, 0 ≤ x ≤ π and also ∃ a continuous non-decreasing function H : <+ → (0,∞)X and L ∈ L1([%,T],<+) therefore E‖f(t,z)‖2 ≤ L (t)H ( E‖z‖2 ) (ii) E { maxi,k {∏k j=i‖bj(%j)‖ }} is uniformly bounded if, E { max i,k {∏k j=i‖bj(%j)‖ }} ≤ θ, for each %j ∈ Dj,j ∈ N,θ > 0 a constant (iii) Γ ∫ T t0 L (r)dr < ∫ ∞ γ1 dr H(r) , (5.2) where Γ = 2W2 max{1, Θ2}(T−t0) 2a−1 (2a−1)Γ(a) ,γ1 = 2W 2Θ2E‖ϕ‖2 and WΘ ≥ 1√ 2 . Assume that assumptions (i),(ii) and (iii) are satisfied, then the problem (5.1) becomes a random impulsive fractional differential equation. From all the above facts, in view of Theorem 3.1, we conclude that (5.1) has a mild solution. Remark 5.2. Let the conditions of Example 5.1 along with (H4) − (H5) be hold. If the following inequality is satisfied, Γ∗ ∫ T t0 L (r)dr < ∫ ∞ γ2 dr H(r) , (5.3) where Γ∗ = 2W2 max{1, Θ2}(T−t0) 2a−1 Γ(a)(2a−1) ,γ2 = 2W 2Θ2E‖ϕ‖2, and WΘ ≥ 1√ 2 . Then the mild solution z of the Example 5.1 is exponentially stable in the quadratic mean. Example 5.3. Consider special random impulsive fractional differential equations,  cDat zt(t,x) = zxx(x,t) + F1(t,z(t,x)) + ∫T 0 F2(θ,z(tsinθ,x))dθ t 6= ξk, t ≥ % z(x,ξk) = q(k)%kz(x,ξ − k ) as x ∈4̂ z(t, 0) = z(t,π) = 0 z(t0,x) = z0(x), x ∈ ∂4̂ (5.4) Consider 4̂⊂ zn, z ∈ D(A) Here zn(ζ) = √ 2 π Sin(nζ),n = 1, 2, . . . , forms the orthonormal set of eigenvectors of A. Also for every z ∈ X,S(t)z = ∑∞ n=1 e (−n2t) < z,zn > zn, which holds ‖S(t)‖ ≤ e(−π 2(t−t0)), t ≥ t0. Therefore S(t) is a semigroup. Consider that the following assumptions: (i) f : <% ×X → X,f1 : <% ×X → X is a continuous function defined by f(t,z)(x) = F1(t,z(x)) t0 ≤ t ≤ T, 0 ≤ x ≤ π f1(θ,x(t + θ))dθ = ∫ T 0 F2(θ,z(tsinθ,x))dθ and also function f and f1 satisfies the Lipschitz condition. (ii) E { maxi,k {∏k j=i‖bj(%j)‖ }} is uniformly bounded if, E { max i,k {∏k j=i‖bj(%j)‖ }} ≤ θ, for each %j ∈ Dj,j ∈ N,θ > 0 a constant (iii) Γ ∫ T t0 L (r)dr < ∫ ∞ γ1 dr H(r) , (5.5) where Γ = 2W2 max{1, Θ2}(T−t0) 2a−1(1+K∗+H∗) (2a−1)Γ(a) ,γ1 = 2W 2Θ2E‖ϕ‖2 and WΘ ≥ 1√ 2 . Assume that assumptions (i), (ii) and (iii) are satisfied, then the problem (5.1) becomes a random impulsive fractional differential equation. From all the above facts, in view of Theorem 3.1, we conclude that 5.4 has a mild solution. Remark 5.4. Let the conditions of Example 5.3 along with (H4) − (H5) be hold. If the following inequality is satisfied, Γ∗ ∫ T t0 L (r)dr < ∫ ∞ γ2 dr H(r) , (5.6) where Γ∗ = 2W2 max{1, Θ2}(T−t0) 2a−1(1+K∗+H∗) Γ(a)(2a−1) ,γ2 = 2W 2Θ2E[‖ϕ‖2], and WΘ ≥ 1√ 2 . Then the mild solution z of the Example 5.3 is exponentially stable in the quadratic mean. 6. Conclusion In this article we mainly focused on the existence and stability of the random impulsive fractional differential equations via Leray-Schauder fixed point method. Firstly, we established the existence of mild solution and continued to prove the exponential stability of the system. Finally, we provided an application to assist of our theory. In future, we will study controllability of random impulsive fractional differential system via fixed point approach. 18 X. ONE, X. TWO, X. THREE, AND X. FOUR References [1] A. B. Alissa and W. Zouhair, Qualitative properties for 1-D impulsive wave equation: controllability and observability, Quaestiones Mathematicae, (2021). [2] A. Anguraj, S. J. Wu and A. 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Zhou, Basic Theory of Fractional Differential Equations, World Scientific, Singapore, (2014). 1 Ramanujan Centre for Higher Mathematics, Alagappa University, Karaikudi-630 003, India., 2 Depart- ment of Mathematics, Alagappa University, Karaikudi-630 003, India. 3 Department of Mathematics, School of Advanced Sciences, Vellore Institute of Technology, Vellore 632 014, Tamil Nadu, India 4 Department of Mathematics, Muslim Association College of Engineering, Trivandrum, India. 5 Corresponding author, Department of Mathematics, Marian Engineering College, Trivandrum, India. Email address: thomas.abin49@gmail.com 1. Introduction 2. Preliminaries Hypotheses 3. Existence 4. Exponential stability in the quadratic mean 5. Applications 6. Conclusion References