h-stability in p-th moment of neutral pantograph stochastic di erential equations with Markovian switching driven by L evy noise

h-stability in p-th moment of neutral pantograph stochastic differential equations with Markovian switching driven by L´evy noise Tom´as Caraballo Universidad de Sevilla, Dpto. Ecuaciones Diferenciales y An´alisis Num´erico, Facultad de Matem´aticas, c/ Tarfia s/n, 41012-Sevilla (Spain). [email protected] Mohsen Belfeki Department of Mathematics, Faculty of sciences of Sfax, Tunisia. [email protected] Lassaad Mchiri Department of Mathematics, Faculty of sciences of Sfax, Tunisia. lmc[email protected] Mohamed Rhaima Department of Mathematics, Faculty of sciences of Tunis El Manar, Tunisia. [email protected] Abstract In this paper we investigate the h-stability in p-th moment of neutral pantograph stochastic differential equations with Markovian switching driven by L´evy noise. The main tool used to prove the results is the Lyapunov method. We analyze two illustrative examples to show the interest and usefulness of the main results. Keywords: Neutral pantograph stochastic differential equations with Markovian switching, L´evy noise, h-stability, p-th moment. 2000 MSC: 60H10, 34K20, 34K50 1 1 Introduction Stochastic delay differential equations (SDDEs) have an important role in many branches of science, industry and economics. Such models have been used in epidemiology, biology, mechanics and finance. The stability of these stochastic delay differential equations has much interest especially for control and stabilization problems and it is still necessary to develop new methods to analyze the asymptotic properties, despite of the fact that there are many published results for the stability of ordinary stochastic differential equations and stochastic delay differential equations (see, e.g., [3], [8], [9], [14] and [15]). Neutral stochastic differential equations with Markovian switching (NSDEMS) is a particular case of stochastic delay differential equations. In the literature, many authors have studied (NSDEMS) (see [7] and [12]). Recently, as a special case of SDDEs, a class of nonlinear stochastic pantograph delay equations (NSPDEs) has received a great deal of attention and various studies have been carried out on the polynomial and exponential stability of NSPDEs (see [1], [2], [10], [11], [13] and [16]). One important class of stochastic delay systems, the neutral pantograph stochastic differential equations with Markovian switching (NPSDEwMS) has also been studied (see [11] and [13]). The almost sure stability with general decay rate of (NPSDEwMS) was first investigated by X. Mao et al. (see [11]). However, all equations of the above-mentioned works are driven by white noise perturbations with continuous initial data, and white noise perturbations are not always appropriate to interpret real data in a reasonable way. In real phenomena, the state of stochastic pantograph delay system may be perturbed by extreme events or abrupt impulses. A more natural mathematical framework for these phenomena takes into account other processes rather than Brownian motions. In particular, we use the L´evy noise with jumps into neutral pantograph stochastic differential equations with Markovian switching to model abrupt changes and, in particular, we generalize the results in [6] which were obtained for exponential stability and did not include Markovian switching. Nevertheless, as far as we know, there is no research on the h-stability on the stochastic case, although there are some papers on the deterministic case (see [4] and [5]). This type of stability provides new insights about the asymptotic behavior of solutions and it is well worth analyzing it. In this paper, we will study the h-stability in p-th moment of neutral pantograph stochastic differential equations with Markovian switching driven by L´evy noise. To do this, we have structured the paper as follows. In Section 2, we introduce some basic notations and assumptions. In Section 3, we establish some sufficient conditions ensuring h-stability in p-th moment of neutral pantograph stochastic differential equations with Markovian switching driven by L´evy noise by using the Lyapunov techniques and Itˆo’s formula. In Section 4, we analyze two illustrative examples to show our theoretical results. 2 2 Preliminaries and definitions Let {Ω,F,(Ft)t≥0,P}be a complete probability space with a filtration satisfying the usual conditions, i.e., the filtration is continuous on the right and F0contains all P-zero sets. W(t) is an m-dimensional Brownian motion defined on the probability space. Let t0>0, q∈(0,1) and C([qt0, t0]; Rn) denote the family of the continuous functions ϕfrom [qt0, t0] to Rnwith the norm kϕk= supqt0≤s≤t0|ϕ(s)|and |x|=√xTxfor any x∈Rn. If Ais a matrix, its trace norm is denoted by |A|=ptrace(ATA), while its operator norm is denoted by kAk= sup{|Ax|:|x|= 1}. Let p > 0, Lp Ft([qt, t]; Rn) denote the family of all Ft-measurable, C([qt, t]; Rn)-valued random variables ϕ={ϕ(θ) : qt ≤θ≤t}such that Ekϕkp<∞. For a∈R,F(a−) denotes the left-hand limit of function F(.) at a, i.e., F(a−) = liml→0−F(a+l). Let {r(t), t ∈R+= [0,+∞)}be a right-continuous Markov chain on the probability space {Ω,F,(Ft)t≥0,P}taking values in a finite state space S={1,2, . . . , N}with a generator Γ = (γij)N×Ngiven by P(r(t+ ∆) = j|r(t) = i) = γij∆ + o(∆),if i6=j 1 + γii∆ + o(∆),if i=j where ∆ >0. Here γij ≥0 is the transition rate from ito j, if i6=j, while γii =−X i6=j γij. We assume that the Markov chain r(.) is independent of the Brownian motion W(.). Consider the following neutral pantograph stochastic differential equation with Markovian switching driven by L´evy noise: d[x(t)−G(t, x(qt))] = f(t, x(t), x(qt), r(t))dt +g(t, x(t), x(qt), r(t))dW(t), +Z|z|<c H1(t, x(t−), x(qt−), r(t), z)e N(dt, dz) + Z|z|≥c H2(t, x(t−), x(qt−), r(t), z)N(dt, dz)t≥t0, (2.1) with the initial condition ξ∈Lp Ft([qt0, t0]; Rn), i.e. x(t) := x(t;t0, ξ) = ξ(t) for qt0≤t≤t0. Let u(t) = x(t)−G(t, x(qt)). Here, we furthermore assume that f: [t0,+∞)×Rn×Rn×S−→ Rn, g : [t0,+∞)×Rn×Rn×S−→ Rn×m, G: [t0,+∞)×Rn−→ Rn. Hi: [t0,+∞)×Rn×Rn×S×Rd−→ Rn(i=1, 2), and the constant c∈(0,+∞) is the maximum allowable jump size. N(., .) is a Poisson random measure defined on R+×(Rd\{0}) with compensator e Nand intensity measure ν(.). 3 It is always assumed that N(., .) is independent of W(.). Let ν(.) be a L´evy measure such that e N(dt, dz) = N(dt, dz)−ν(dz) and RRd\{0}(|z|2∧1)ν(dz) = λ < +∞. Usually, the pair (W(.), N(., .)) is called a L´evy noise, R|z|<c H1(t, x(t−), x(qt−), r(t), z)e N(dt, dz) is called a ‘small jump’ and R|z|≥cH2(t, x(t−), x(qt−), r(t), z)N(dt, dz) is called ‘large jump’. We denote by x(t;t0, ξ) the solution of Eq. (2.1). Let C1,2([qt0,+∞)×Rn×S;R+) be the family of all non-negative functions V(t, x, i) on [qt0,+∞)×Rn×S, which are twice continuously differentiable with respect to xand once continuously differentiable with respect to t. For any (t, x, y, i)∈[qt0,+∞)×Rn×Rn×S,u(t) = x(t)−G(t, y(t)), with y(t) = x(qt), by the generalized Itˆo’s formula we have, V(t, u(t), r(t)) = V(t0, u(t0), r(t0)) + Zt t0 LV (s, x(s), x(qs), r(s))ds +Mt,(2.2) where the operator LV (t, x, y, i):[qt0,+∞)×Rn×Rn×S→Rand the process Mtare defined respectively by LV (t, x, y, i) = Vt(t, u, i) + Vx(t, u, i)f(t, x, y, i) +1 2trace gT(t, x, y, i)Vxx(t, u, i)g(t, x, y, i) +Z|z|<c V(t, u +H1(t, x, y, i, z)) −V(t, u, i)−Vx(t, u, i)H1(t, x, y, i, z)ν(dz) +Z|z|≥cV(t, u +H2(t, x, y, i, z)) −V(t, u, i)ν(dz) + N X j=1 γijV(t, u, j) and Mt=Zt t0 Vx(s, u(s), r(s))g(s, x(s), x(qs), r(s))dW (s) +Zt t0Z|z|<c V(s, u(s) + H1(s, x(s−), x(qs−), r(s), z)) −V(s, u(s), r(s))e N(ds, dz), +Zt t0Z|z|≥cV(s, u(s) + H2(s, x(s−), x(qs−), r(s), z)) −V(s, u(s), r(s))N(ds, dz) where Vt=∂V (t, x, i) ∂t , Vx=∂V (t, x, i) ∂x1 ,...,∂V (t, x, i) ∂xn, Vxx =∂2V(t, x, i) ∂xi∂xjn×n . 4 For our purpose, we will state some assumptions which can ensure the existence and uniqueness of a solution x(t) = x(t;t0, ξ) on t≥t0, for equation (2.1). H1: (A local Lipschitz condition): For each integer d≥1 there is an ld>0 such that |f(t, x, y, i)−f(t, x, y, i)|∨|g(t, x, y, i)−g(t, x, y, i)|∨|Hb(t, x, y, i, z)−Hb(t, x, y, i, z)| ≤ ld|x−x|+|y−y|, for all t≥t0,i∈Sand x,x,y,y∈Rnwith |x|∨|x|∨|y|∨|y| ≤ dand b= 1,2. Besides, f(t, 0,0, i) = 0 , g(t, 0,0, i) = 0 and Hb(t, 0,0, z) = 0, G(t, 0) = 0 , for any t≥t0,z∈Rd and b= 1,2. H2: There exists a constant β∈(0,1 2) such that |G(t, y)−G(t, y)| ≤ β|y−y|,(2.3) for all t≥t0and y,y∈Rn. H3: There exist positive numbers c1,c2,c3,c4, and psuch that c3≥c4and for all (t, x, y, i)∈ [qt0,+∞)×Rn×Rn×S, we have c1|x|p≤V(t, x, i)≤c2|x|p.(2.4) LV (t, x, y, i)≤ −c3|x|p+c4q|y|p.(2.5) Recall now some useful inequalities which will be used in our analysis. Lemma 2.1. (i) Let 0< p ≤1and a,b∈R+. Then (a+b)p≤ap+bp. (ii) Let p > 1,ε > 0and a,b∈R+. Then (a+b)p≤1 + ε 1 p−1p−1ap+bp ε. Proof. See [9]. Remark 2.2. For p > 1, we can take ε= 1, then, for any a,b∈R+, we have (a+b)p≤2p−1(ap+bp).(2.6) Theorem 2.3. Let assumptions H1-H3hold. Then for any given initial data ξ, there is a unique global solution x(t)of equation (2.1) on t∈[t0,+∞). 5 Proof. Using Assumption H1, for any initial value ξ∈Lp Ft([qt0, t0]; Rn), there exists a unique maximal solution x(·) = x(·;t0, ξ) on [t0, σe) (see Theorem 1 in [12]), where σeis the explosion time. Let k0>0 be sufficiently large such that kξk< k0. For each integer k≥k0, define the stopping time τk= inf {t∈[t0, σe); |x(t)| ≥ k}. Clearly, τkis increasing as k→ ∞ and τk→τ∞≤σea.s.. If we can show τ∞=∞a.s., then we have σe=∞a.s. Therefore, we just need to show τ∞=∞a.s. Case 1: Assume 0 < p ≤1. By the generalized Itˆo formula (see, [14]) and Lemma 2.1 we have E(V(t∧τk, u(t∧τk), r(t∧τk))) =E(V(t0, u(t0), r(t0))) + EZt∧τk t0 LV (s, x(s), x(qs), r(s))ds ≤E(V(t0, u(t0), r(t0))) −c3EZt∧τk t0|x(s)|pds+c4qE Zt∧τk t0|x(qs)|pds ≤E(V(t0, u(t0), r(t0))) −c3EZt∧τk t0|x(s)|pds+c4E Zq(t∧τk) qt0|x(s)|pds! ≤E(V(t0, u(t0), r(t0))) −c3EZt∧τk t0|x(s)|pds+c4EZt∧τk qt0|x(s)|pds =E(V(t0, u(t0), r(t0))) −c3EZt∧τk t0|x(s)|pds+c4EZt0 qt0|x(s)|pds +c4EZt∧τk t0|x(s)|pds ≤E(V(t0, u(t0), r(t0))) −(c3−c4)EZt∧τk t0|x(s)|pds+c4t0(1 −q)Ekξkp ≤E(V(t0, u(t0), r(t0))) −(c3−c4)EZt∧τk t0|x(s)|pds+c4Ekξkp ≤E(V(t0, u(t0), r(t0))) + c4Ekξkp ≤c2E(|u(t0)|p)) + c4Ekξkp ≤c2E(|x(t0)|p) + c2βpE(|x(t0)|p) + c4Ekξkp ≤c2(1 + βp) + c4Ekξkp. Then, by assumption H3, E|u(t∧τk)|p≤1 c1c2(1 + βp) + c4Ekξkp. 6 By the definition of τk, we have |x(τk)|=kand |x(t∧τk)| ≤ k,|x(t∧τk)|p≤kp, including |x(q(t∧τk))|p≤kp. For τk≤t, we have |x(t∧τk)|=|x(τk)|=k, and we therefore obtain E|x(t∧τk)|p1τk≤t=kpP(τk≤t). By the inequality |x(t)|p≤ |u(t)|p+βp|x(qt)|p, we obtain |u(t∧τk)|p≥ |u(t∧τk)|p1{τk≤t} ≥ |x(t∧τk)|p1τk≤t−βp|x(q(t∧τk))|p1{τk≤t} ≥kp(1 −βp) 1{τk≤t}. Then, P(τk≤t)≤1 kp(1 −βp)E|u(t∧τk)|p ≤1 c1kp(1 −βp)c2(1 + βp) + c4Ekξkp. Letting k→ ∞ implies P(τ∞≤t) = 0 and, consequently, τ∞> t a.s. Letting t→ ∞, we can obtain τ∞=∞a.s, which implies that there exists a global solution x(t) to the system (2.1). Case 2: Assume now p > 1. Proceeding as in Case 1 and using (2.6), E(V(t∧τk, u(t∧τk), r(t∧τk))) ≤E(V(t0, u(t0), r(t0))) + c4Ekξkp ≤c2E(|u(t0)|p)) + c4Ekξkp ≤2p−1c2E(|x(t0)|p)+2p−1c2βpE(|x(qt0)|p) + c4Ekξkp ≤2p−1c2(1 + βp) + c4Ekξkp. From the inequality |x(t)|p≤2p−1|u(t)|p+ 2p−1βp|x(qt)|p, we obtain, by using the same techniques as in Case 1, |u(t∧τk)|p≥ |u(t∧τk)|p1{τk≤t} ≥kp1 2p−1−βp1{τk≤t}. Then, P(τk≤t)≤1 kp1 2p−1−βpE|u(t∧τk)|p ≤1 c1kp1 2p−1−βp2p−1c2(1 + βp) + c4Ekξkp. 7 Letting k→ ∞ implies that P(τ∞≤t) = 0, and, therefore, τ∞> t a.s. Letting t→ ∞, we deduce τ∞=∞a.s, which implies that there exists a global solution x(t) to the system (2.1). Definition 2.1. The function h:R+→(0,∞) is said to be an h-type function if the following conditions are satisfied: (i) It is continuous and nondecreasing in R+and continuously differentiable in R+. (ii) h(0) = 1, limt→∞ h(t) = ∞and φ= supt>0|h0(t) h(t)|<∞. (iii) For any t≥s≥0, h(t)≤h(s)h(t−s). Definition 2.2. Equation (2.1) is said to be h-stable in p-th moment, if there exist positive constants δand csuch that, for any ξ∈Lp Ft([qt0, t0]; Rn), the corresponding solution x(·;t0, ξ) of (2.1) with initial value ξsatisfies E(|x(t;t0, ξ)|p)≤cE(||ξ||p)h−δ(t)hδ(t0), t ≥t0.(2.7) We will impose the following condition on Gto obtain the h-stability of equation (2.1). H4: There exist constants β∈(0,1) and ≥0 such that for all y,y∈Rnand t≥t0, it holds |G(t, y)−G(t, y)|p≤βh−((1 −q)t)|y−y|p,(2.8) and G(t, 0) = 0. Theorem 2.4. Let assumptions H1,H3and H4hold except (2.5) which is replaced by LV (t, x, y, i)≤ −c3|x|p+c4qh−((1 −q)t)|y|p,(2.9) for all (t, x, y, i)∈[qt0,+∞)×Rn×Rn×S, > 0and 2p−1β < 1, for p > 1. Let 0< δ ≤c3−c4 2p−1c2φ1 + β q, for p > 1, and φis the positive constant which verifies condition (ii)in Definition 2.1. Then, Equation (2.1) is h-stable in p-th moment. Proof. Let ξ∈Lp Ft([qt0, t0]; Rn) be an initial function for Eq. (2.1) and denote by x(·) its corresponding solution. We split our analysis into two cases. 8 Case 1: Let δ∈(0, ) and 0 < p ≤1. For each k∈N∗, define the stopping time τk= inf{t≥ t0;|x(t)| ≥ k}. For any t≥t0, by the generalized Itˆo formula applied to hδ(t)V(t, u(t), r(t)), Lemma 2.1, H3and H4, we have Ehδ(t∧τk)V(t∧τk, u(t∧τk), r(t∧τk)) =hδ(t0)E(V(t0, u(t0), r(t0))) , +EZt∧τk t0 hδ(s)δh0(s) h(s)V(s, u(s), r(s)) + LV (s, x(s), x(qs), r(s))ds, ≤hδ(t0)E(V(t0, u(t0), r(t0))) + c2φδEZt∧τk t0 hδ(s)|u(s)|pds −c3EZt∧τk t0 hδ(s)|x(s)|pds+c4qE Zt∧τk t0 hδ(s)h−((1 −q)s)|x(qs)|pds ≤hδ(t0)E(V(t0, u(t0), r(t0))) + c2φδEZt∧τk t0 hδ(s)|x(s)|pds +c2φδβEZt∧τk t0 hδ(s)h−((1 −q)s)|x(qs)|pds−c3EZt∧τk t0 hδ(s)|x(s)|pds +c4qE Zt∧τk t0 hδ(s)h−((1 −q)s)|x(qs)|pds, ≤hδ(t0)E(V(t0, u(t0), r(t0))) + c2φδEZt∧τk t0 hδ(s)|x(s)|pds +c2φδβEZt∧τk t0 hδ(s)h−δ((1 −q)s)|x(qs)|pds−c3EZt∧τk t0 hδ(s)|x(s)|pds +c4qE Zt∧τk t0 hδ(s)h−δ((1 −q)s)|x(qs)|pds ≤hδ(t0)E(V(t0, u(t0), r(t0))) + c2φδEZt∧τk t0 hδ(s)|x(s)|pds +c2φδβEZt∧τk t0 hδ(qs)|x(qs)|pds−c3EZt∧τk t0 hδ(s)|x(s)|pds +c4qE Zt∧τk t0 hδ(qs)|x(qs)|pds. 9 LV (t, x(t), x(0.5t),2) ≤ −0.68x2(t)+0.682x2(t)+0.09 (ln(e+ 0.5t))−1x2(0.5t) +(−0.68)0.09 (ln(e+ 0.5t))−1x2(0.5t) ≤ −0.21x2(t)+0.05 (ln(e+ 0.5t))−1x2(0.5t) =−0.21x2(t) + (0.1)(0.5) (ln(e+ 0.5t))−1x2(0.5t). Then, for i∈S, we obtain LV (t, x(t), x(0.5t), i)≤ −0.21x2(t) + (0.1)(0.5)(ln(e+ 0.5t))−1x2(0.5t).(3.3) Thus, the assumptions of theorem 2.4 are satisfied with c1=1 2,c2= 1,c3= 0.21,c4= 0.1, p= 2,β= 0.3,q= 0.5,= 1,h(t) = ln(e+t),φ=1 eand δ= 0.09. Therefore system (3.1) is h-stable in mean square. Example 3.2. We consider the following neutral pantograph stochastic differential equation with Markovian switching driven by L´evy noise: dx(t)−0.2 (1 + 0.5t)−0.5e−0.5tx(0.5t)=f(t, x(t), x(0.5t), i)dt +g(t, x(t), x(0.5t), i)dW(t), +Z|z|<1 H1(t, x(t−), x(0.5t−), i, z)e N(dt, dz) + Z|z|≥1 H2(t, x(t−), x(0.5t−), i, z)N(dt, dz)t≥t0, (3.4) Where the initial data x0=ξ∈Lp Ft([0.5,1]; R),r(1) = 1,z∈R,W(t)is a one dimensional Brownian motions, N(., .)is a Poisson random measure defined on R+×(R−{0})with compensator e Nand intensity measure ν(.). It is always assumed that N(., .)is independent of W(.). Let ν(.) be a L´evy measure such that e N(dt, dz) = N(dt, dz)−ν(dz)and RR−{0}(|z|2∧1)ν(dz) = λ < +∞. Let f(t, x(t), x(0.5t),1) = −7 2x(t)−0.2 (1 + 0.5t)−0.5e−0.5tx(0.5t), f(t, x(t), x(0.5t),2) = −1 8x(t)−0.2 (1 + 0.5t)−0.5e−0.5tx(0.5t), g(t, x(t), x(0.5t),1) = 1 4x(t)−0.2 (1 + 0.5t)−0.5e−0.5tx(0.5t), g(t, x(t), x(0.5t),2) = 1 2x(t)−0.2 (1 + 0.5t)−0.5e−0.5tx(0.5t), H1(t, x(t), x(0.5t),1, z) = 1 2zx(t)−0.2 (1 + 0.5t)−0.5e−0.5tx(0.5t). H1(t, x(t), x(0.5t),2, z) = 1 3zx(t)−0.2 (1 + 0.5t)−0.5e−0.5tx(0.5t). H2(t, x(t), x(0.5t),1, z) = x(t)−0.2 (1 + 0.5t)−0.5e−0.5tx(0.5t). 16 H2(t, x(t), x(0.5t),2, z)=(√31 6−1) x(t)−0.2 (1 + 0.5t)−0.5e−0.5tx(0.5t). Let u(t) = x(t)−0.2 (1 + 0.5t)−0.5e−0.5tx(0.5t)and ν(dz) = dz 1 + z2. Let S={1,2}and the matrix Γ = (γij)1≤i,j≤2define by −1 1 1−1 We will prove that system (3.4) is exponentially stable in mean square. Indeed, as in the preceding example, let V(t, x, i) = ψix2, for i∈S, where ψ1=1 2and ψ2= 1. Then 1 2|x|2≤V(t, x, i)≤ |x|2(3.5) By the definition of LV, we have for i= 1 LV (t, x(t), x(0.5t),1) = −7 2u2(t) + 1 32u2(t) + 1 8u2(t)Z|z|<1 z2 1 + z2dz +3 2u2(t)Z|z|≥1 dz 1 + z2+1 2u2(t) =11π 16 −87 32u2(t) =−0.56 x(t)−0.2 (1 + 0.5t)−0.5e−0.5tx(0.5t)2. For i= 2, LV (t, x(t), x(0.5t),2) = −1 4u2(t) + 1 4u2(t) + 1 9u2(t)Z|z|<1 z2 1 + z2dz −5 36u2(t)Z|z|≥1 dz 1 + z2−1 2u2(t) =−9π 72 +5 18u2(t) =−0.67 x(t)−0.2 (1 + 0.5t)−0.5e−0.5tx(0.5t)2 Then, using 2ab ≤a2+b2, for all a,b∈R, we have LV (t, x(t), x(0.5t),1) ≤ −0.56x2(t)+0.562x2(t)+0.04 (1 + 0.5t)−1e−tx2(0.5t) + (−0.56)(0.04) (1 + 0.5t)−1e−tx2(0.5t) ≤ −0.24x2(t)+0.02 (1 + 0.5t)−1e−tx2(0.5t) =−0.24x2(t) + (0.04)(0.5) (1 + 0.5t)−1e−tx2(0.5t). 17 LV (t, x(t), x(0.5t),2) ≤ −0.67x2(t)+0.672x2(t)+0.04 (1 + 0.5t)−1e−tx2(0.5t) +(−0.67)0.04 (1 + 0.5t)−1e−tx2(0.5t) ≤ −0.09x2(t)+0.02 (1 + 0.5t)−1e−tx2(0.5t) =−0.09x2(t) + (0.04)(0.5) (1 + 0.5t)−1e−tx2(0.5t). 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