# Stability Analysis of Stochastic Differential Equation Driven by G-Brownian Motion under Non-Lipschitz Condition

Authors: Li Ma; Yujing Li
Journal: Mathematical Problems in Engineering
 (2022)

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## Abstract

This paper is devoted to studying the p-th moment exponential stability for a class of stochastic differential equation (SDE) driven by G-Brownian motion under non-Lipschitz condition. The delays considered in this paper are time-varying delaysτit1≤i≤3. Since the coefficients are non-Lipschitz, the normal enlargement on the coefficients is not available and the Gronwall inequality is not suitable in this case. By Bihari inequality and Itô integral formula, it is pointed out that there exists a constant τ∗ such that the p-th moment exponential stability holds if the time-varying delays are smaller than τ∗.


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## Body

## 1. Introduction

In recent years, much effort has been made to develop the theory of sublinear expectations connected with the volatility uncertainty and the so-called G-Brownian motion, which is an efficient way to incorporate the unknown volatility into financial model. G-Brownian motion has a very rich and interesting new structure which non-trivially generalizes the classical one. Stochastic dynamic equations based on G-Brownian motion have been studied by several authors [1–5]. Among them, the existence-uniqueness, stability, moment estimations, continuity, and differentiability of solution with respect to the initial value were studied in detail. In [6], Zhu et al. got the p-th moment exponential stability for a class of stochastic delay systems driven by G-Brownian motion under Lipschitz condition and linear growth condition. In [7], Gao et al. got the almost sure exponential stability of stochastic differential delay equations driven by Brownian motion under Lipschitz condition and linear growth condition.Letp>0. In this paper, we will prove the p-th moment exponential stability problem of solutions for a class of stochastic delay non-linear systems driven by G-Brownian motion under non-Lipschitz condition. Throughout this paper, the following notations will be used. Let ℝd be the d-dimensional Euclidean space and x be the Euclidean norm of a vector x. Let R=−∞,+∞ and R+=0,∞. Let Ftt⩾0 be a filtration generated by G-Brownian motion Bt,t⩾0. Let τ>0 and C−τ,0;ℝd denote the family of continuous functions ϕ from −τ,0 to ℝd with the uniform norm ϕτ=sup−τ⩽θ⩽0ϕθ. Let E∧ be the G-expectation. For t⩾0 and p>0, denote by LFtpΩ;ℝd the family of all Ft measurable ℝd-valued stochastic variables ξ such that E∧ξp<∞ and denote by LFtpΩ;C−τ,0;ℝd the family of all Ft measurable, C−τ,0;ℝd-valued stochastic variables ξ=ξθ:−τ⩽θ⩽0 such that sup−τ⩽θ⩽0E∧ξθp<∞. Define MGp,0−τ,T,ℝd:=ηtw:=∑j=1N−1ξtjwItj,tj+1;ξtj∈LFtpΩ;ℝd,−τ=t0<t1<⋯<tN=T,∀T⩾0. Let MGp−τ,T,ℝd be the completion of MGp,0−τ,T,ℝd under the norm ηMGp−τ,T,ℝd=1/T∫−τTE∧ηtpdt1/p. Let Cb,Lipℝd be the real-valued space of bounded and Lipschitz continuous functions. Let ℝ0d=ℝd−0 and V be a set of all Borel measures on ℝ0d,Bℝ0d. Let C12ℝd×ℝ+;ℝ+ denote the family of all non-negative functions Vy,t on ℝd×ℝ+ which are continuously twice differentiable in y and once differentiable in t.In this paper, we will discuss the following stochastic delay non-linear system driven by G-Brownian motion:(1)dxt=bt,xt−τ1tdt+ht,xt−τ2tdBt+σt,xt−τ3tdBt,for t⩾t0⩾0 with the initial data xt0+θ=ξθ for −τ⩽θ⩽0, where ξ∈LFt0pΩ;C−τ,0;ℝd, b,h,σ∈MGp−τ,T,ℝd. B⋅ is the quadratic variation process of the G-Brownian motion B⋅ with Ga:=1/2E∧aB12=1/2σ¯2a++σ¯2a−, for a∈R, where a+=maxa,0,a−=max−a,0,σ¯2=E∧B12,σ¯2=−E∧−B12, τ1t,τ2t, τ3t are time delays satisfying 0⩽τ1t⩽τ1,0⩽τ2t⩽τ2,0⩽τ3t⩽τ3,τ=maxτ1,τ2,τ3.We will study the stability of system 1 under the following conditions.Assumption 1.
Letp>0. There exists a concave function ρx:R+⟶R+ such that the following conditions hold.H1ρx is a continuous non-decreasing function satisfying ρ0+=0 and ∫011/r+ρrdr=+∞.H2 There is a constant C>0 such that for all t≥0,x,y∈ℝd,(2)bt,x−bt,yp+ht,x−ht,yp+σt,x−σt,yp⩽Cρx−yp.H3bt,0=0,ht,0=0,σt,0=0.Remark 1.
In fact, under conditionsH1–H3, we havelimr⟶0+r+ρr=0,+∞=∫011/r+ρrdr⩽∫011/ρrdr; therefore,∫011/ρrdr=+∞.Remark 2.
H1–H3 are valid if p=1, ρx=x, and σt,x=ht,x=bt,x≡x on 0,∞. By 8, Example 1.12], for all x,y⩾0, we have x−y/x−y⩽C. Hence, σ,b,h are not Lipschitz continuous but 1/2-Hölder continuous.Remark 3.
Whenρx≡x, then H2 and H3 just mean that σ,b,h satisfy Lipschitz condition and linear growth condition, so the existence and uniqueness of solutions to equation (1) are obvious. Moreover, under Lipschitz condition and other similar conditions, Zhu and Huang [6] got the exponential stability of the solution. When Btis a Brownian motion, ρx≡x [7] got the exponential stability of the solution.
As we know, G-Brownian motion is a generalization of Brownian motion and concave functions are more general than the identity function, so the innovation of this paper is that under more loosely condition, namely, Assumption1, there is a τ∗ such that the solution of (1) is p-th moment exponential stable when time delays are smaller than τ∗, which generalizes [6, 7]. The difficulties in the proof come from the unknown expression of ρx, and the Gronwall inequality could not be used to get the necessary estimations. Those difficulties are overcome by using Bihari inequality doubly and L’H o∧ pital’s rule.


## 2. Main Results and Proofs

It is known from [9] that (1) has a unique solution under the conditions of H1–H3. In what follows, we present an important lemma to give some moment estimations on the solution xt.Lemma 1.
Let Assumption1 hold. For any initial data ξ∈LFt0pΩ;C−τ,0;ℝn, let xt;t0,ξ=xt be the solution of 1. Then, for all t⩾t0, the following holds.(1)
Forp⩾2,(3)E∧xtp≤G1−1G12E∧ξτp+Ap,t−t0,(4)E∧sup0⩽s⩽τxt+s−xtp⩽M1p,τρG1−1G12E∧ξτp+Ap,t+τ−t0,(5)E∧supt0⩽s⩽txsp⩽G2−1G24p−1E∧ξτp+A2p,t−t0.(2)
For0<p<2,(6)E∧xtp⩽G1−1G12E∧ξτ2+A2,t−t0p/2,(7)E∧sup0⩽s⩽τxt+s−xtp⩽M12,τ⋅ρG1−1G12E∧ξτ2+A2,t+τ−t0p/2,(8)E∧supt0⩽s⩽txsp⩽G2−1G24E∧ξτ2+A22,t−t0p/2,whereG1r=∫1r1/s+ρsds, G2r=∫1r1/ρsds, for r>0, and Gi−1 is the inverse function of Gi,i=1,2, C1p=p−1+1/2pp−1σ¯2,C2p=C1+pσ¯2,Ap,t−t0=maxC1p,C2p⋅t−t0, M1p,τ=3p−1Cτp1+σ¯2pτ+σ¯2pτp3/2p−1p/2,A2p,t−t0=M2p,t−t0⋅t−t0, M2p,t=4p−1Ctp−11+σ¯2pt+σ¯2ptp3/2p−1p/2.Proof.
(1) By G-Itô formula, for anyu>t0,(9)xup−xt0p=p∫t0uxsp−2xTsbs,xs−τ1sds+∫t0upxsp−2xTsσs,xs−τ3sdBs+∫t0upxsp−2xTshs,xs−τ2sdBs+p2∫t0uxsp−2σs,xs−τ3s2dBs+12pp−2∫t0uxsp−4xTsσs,xs−τ3s2dBs=I1+I2+I3+I4+I5.
By Young inequality,H1–H3, Jensen inequality, and 6, Lemma 2.4(iii)],(10)E∧I1=pE∧∫t0uxsp−2xTsbs,xs−τ1sds⩽pE∧∫t0up−1pxsp−1⋅pp−1+1pbs,xs−τ1spds⩽p−1∫t0uE∧xspds+∫t0uE∧bs,xs−τ1spds⩽p−1∫t0uE∧xspds+C∫t0uE∧ρxs−τ1spds⩽p−1∫t0uE∧xspds+C∫t0uρE∧xs−τ1spds.
Similarly, by Young inequality,H1–H3, and 6, Lemma 2.4(i)],(11)E∧I3⩽pσ¯2E∧∫t0uxsp−1hs,xs−τ2sds⩽pσ¯2E∧∫t0up−1pxsp+1phs,s−τ2spds⩽p−1σ¯2∫t0uE∧xspds+σ¯2C∫t0uρE∧xs−τ2spds,(12)E∧I4=12pp−2E∧∫t0uxsp−2σs,xs−τ3s2dBs⩽12pp−2σ¯2E∧∫t0up−2pxsp+2pσs,xs−τ3spds⩽12p−22σ¯2∫t0uE∧xspds+Cp−2σ¯2∫t0uρE∧xs−τ3spds,(13)E∧I5=p2E∧∫t0uxsp−2σs,xs−τ3s2dBs⩽p2σ¯2E∧∫t0uxsp−2σs,xs−τ3s2ds⩽p−22σ¯2∫t0uE∧xspds+Cσ¯2∫t0uρE∧xs−τ3spds.
By10, Lemma 3.5, E∧I2=0, and hence(14)E∧xup−E∧xt0p⩽p−1+12pp−1σ¯2∫t0uE∧xspds+C∫t0tρE∧xs−τ1spds+Cσ¯2∫t0uρE∧xs−τ2spds+Cp−1σ¯2∫t0uρE∧xs−τ3spds,and(15)supt0⩽u⩽tE∧xup⩽E∧xt0p+maxC1p,C2p∫t0tsupt0−τ⩽u⩽sE∧xup+ρsupt0−τ⩽u⩽sE∧xupds,where C1p=p−1+1/2pp−1σ¯2, C2p=C1+pσ¯2. Let Kr:=r+ρr; then,(16)supt0−τ⩽u⩽tE∧xup⩽supt0⩽u⩽tE∧xup+E∧ξτp⩽2E∧ξτp+maxC1p,C2p∫t0tKsupt0−τ⩽u⩽sE∧xupds.
ByH1, K:R+⟶R+ is a continuous non-decreasing function such that Kt>0 for all t>0. Define G1r=∫1r1/Ksds on r>0 and denote its inverse function by G1−1. By H1, one sees that limε↓0G1ε=−∞ and G1+∞=+∞, that is, DomG1−1=−∞,+∞. So, by 11, Theorem 2.3 (Bihari’s inequality)] or 12, Theorem D],(17)supt0−τ⩽u⩽tE∧xup⩽G1−1G12E∧ξτp+Ap,t−t0,where Ap,t−t0=maxC1p,C2p⋅t−t0. Inequality 3 is proved.
Next, we will prove (4). By the elementary inequality a+b+cp⩽3p−1ap+bp+cp, for any u>0,(18)xt+u−xtp⩽3p−1∫tt+ubs,xs−τ1sdsp+3p−1∫tt+uhs,xs−τ2sdBsp+3p−1∫tt+uσs,xt−τ3sdBsp=I1+I2+I3.
By Hölder inequality,(19)E∧sup0⩽u⩽τI1⩽3p−1τp−1E∧∫tt+τbs,xs−τ1spds⩽C3p−1τp−1∫tt+τρE∧xs−τ1spds⩽C3p−1τpsupt0−τ⩽u⩽t+τρE∧xup.
By6, Lemma 2.5],(20)E∧sup0⩽u⩽τI2=3p−1E∧sup0⩽u⩽τ∫tt+uhs,xs−τ2sdBsp⩽3p−1σ¯2pτpE∧∫tt+τhs,xs−τ2spds⩽3p−1Cσ¯2pτp∫tt+τρE∧xs−τ2spds⩽3p−1Cσ¯2pτp+1supt0−τ⩽u⩽t+τρE∧xup.
By6, Lemma 2.6],(21)E∧sup0⩽u⩽τI3=3p−1E∧sup0⩽u⩽τ∫tt+uσs,xt−τ3sdBsp⩽3p−1σ¯2pτpp32p−1p/2E∧∫tt+τσs,xt−τ3spds⩽C3p−1σ¯2pτpp32p−1p/2∫tt+τρE∧xs−τ3spds⩽C3p−1σ¯2pτp+1p32p−1p/2supt0−τ⩽u⩽t+τρE∧xup.
Hence, by (3) and (17),(22)E∧sup0⩽u⩽τxt+u−xtp⩽E∧sup0⩽u⩽τI1+E∧sup0⩽u⩽τI2+E∧sup0⩽u⩽τI3⩽M1p,τsupt0−τ⩽u⩽t+τρE∧xup⩽M1p,τρsupt0−τ⩽u⩽t+τE∧xup⩽M1p,τρG1−1G12E∧ξτp+Ap,t+τ−t0,where M1p,τ=3p−1Cτp1+σ¯2pτ+σ¯2pτp3/2p−1p/2, so inequality (4) is proved.
Finally, we will prove (5). Similar to the proof of (3), it is easy to get that(23)E∧supt0⩽u⩽txup⩽4p−1E∧xt0p+4p−1Ct−t0p−1∫t0tρE∧xs−τ1spds+4p−1Cσ¯2pt−t0p∫t0tρE∧xs−τ1spds+4p−1Cσ¯2pt−t0pp32p−1p/2∫t0tρE∧xs−τ3spds⩽4p−1E∧xt0p+M2p,t−t0∫t0tρE∧supt0−τ⩽r⩽sxrpds,where M2p,t=4p−1Ctp−11+σ¯2pt+σ¯2ptp3/2p−1p/2. Define G2r=∫1r1/ρsds, on r>0, and let G2−1 be the inverse function of G2. By H1, one sees that limε↓0G2ε=−∞ and G2+∞=+∞, and hence DomG2−1=−∞,+∞. So, by 11, Theorem 2.3 (Bihari’s inequality)] or 12, Theorem D],(24)supt0⩽u⩽tE∧xup⩽G2−1G24p−1E∧ξτp+M2p,t−t0⋅t−t0=G2−1G24p−1E∧ξτp+A2p,t−t0,where A2p,t−t0=M2p,t−t0⋅t−t0.
(2) Firstly assume thatξ∈C−τ,0;ℝd. Notice 0<p<2 and by Hölder inequality,(25)E∧|xt|p⩽E∧|xt|2p/2⩽G1−1G12E∧ξτ2+A2,t−t0p/2.
Next, for anyξ∈LFt0pΩ;C−τ,0;ℝd, it follows from the property of conditional expectation that for t⩾t0,(26)E∧xtp⩽E∧E∧xtp|Ft0,which verifies (6). Similarly, (7) and (8) hold by Hölder inequality. The proof is complete.
Consider the following equation:(27)dyt=bt,ytdt+ht,ytdBt+σt,ytdBt,where the initial data yt0=y0∈LFt0pΩ;ℝd and t⩾t0. Denote the solution of (27) by yt;t0,yt0. In particular, yt;t0+τ,xt0+τ means the solution of (27) with initial data xt0+τ and initial time t0+τ.Definition 1.
(see [13]). For each V∈C12ℝd×ℝ+;ℝ+, define an operator L from ℝd×ℝ+ to ℝ by(28)LVt,y=Vtt,y+Vyt,ybt,y+G2Vyt,yht,y+σTt,yVyyt,yσt,y,where Vyt,y=∂Vt,y/∂y1⋯∂Vy,t/∂yd,Vtt,y=∂Vt,y/∂t,Vyyt,y=∂2Vt,y/∂yi∂yjd×d.Lemma 2.
(see9, Theorem 4.1]). Assume that there exist V∈C12ℝd×ℝ+;ℝ+ and constants C4>C3>0 and λ1>0 such that(1)C3yp⩽Vt,y⩽C4yp for all t⩾t0,y∈ℝd.(2)LVt,yt⩽−λ1Vt,yt.Then, the solution of system (28) is p-th moment exponentially stable in the following sense: E∧ytp⩽C4/C3e−λ1t−t0E∧yt0p,limsupt⟶∞logE∧ytp/t⩽−λ1.Next, we give a lemma, which plays an important role in the proof of our main results.Lemma 3.
Let Assumption1 hold and xt;t0,ξ=xt be the solution of 1 with initial data ξ∈LFt0pΩ;C−τ,0;ℝn. Then, for all t⩾t0+τ,(1)
p⩾2,(29)E∧yt;t0+τ,xt0+τ−xtp⩽G1−1G1Sp,t−t0,τ+M4p⋅t−t0−τ.(2)
0<p<2,(30)E∧yt;t0+τ,xt0+τ−xtp⩽G1−1G1S2,t−t0,τ+M42⋅t−t0−τp/2,where Sp,t−t0,τ=C6p⋅t−t0−τ⋅ρM3p,τ⋅τ⋅ρG1−1G12E∧ξτp+Ap,t−t0,M3p,τ=C3p−1τp−1+σ¯2pτp+σ¯2pτpp3/2p−1p/2,C5p=p−1+p2−p/2σ¯2,C6p=C2p−1pσ¯2+1,M4p=maxC5p,C6p.Proof.
(1) For simplicity, letyt;t0+τ,xt0+τ=yt. Then, by G-Itô formula, for t⩾t0+τ,(31)xt−ytp=p∫t0+τtxs−ysp−2xs−ysTbs,xs−τ1s−bs,ysds+p∫t0+τtxs−ysp−2xs−ysTσs,xs−τ3s−σs,ysdBs+p∫t0+τtxs−ysp−2xs−ysThs,xs−τ2s−hs,ysdBs+pp−22p∫t0+τtxs−ysp−4xs−ysTσs,xs−τ3s−σs,ys2dBs+p2E∧p∫t0+τtxs−ysp−2σs,xs−τ3s−σs,ys2dBs=I1+I2+I3+I4+I5.
By Young inequality, Assumption1, and 6, Lemma 2.4(iii)],(32)E∧I1=pE∧∫t0+τtxs−ysp−2xs−ysTbs,xs−τ1s−bs,ysds⩽pE∧∫t0+τtp−1pxs−ysp−1⋅p/p−1+1pbs,xs−τ1s−bs,yspds,(33)⩽p−1∫t0+τtE∧xs−yspds+2p−1∫t0+τtE∧bs,xs−τ1s−bs,xspds+2p−1∫t0+τtE∧bs,xs−bs,yspds⩽p−1∫t0+τtE∧xs−yspds+C2p−1∫t0+τtE∧ρxs−xs−τ1spds+C2p−1∫t0+τtE∧ρ|xs−ys|pds⩽p−1∫t0+τtE∧xs−yspds+C2p−1∫t0+τtρE∧xs−xs−τ1spds+C2p−1∫t0+τtρE∧xs−yspds.
On the other hand, by (3), for t0+τ⩽s⩽t,(34)E∧xs−xs−τ1sp⩽3p−1E∧∫s−τ1ssbu,xu−τ1udup+3p−1E∧∫s−τ1sshu,xu−τ2udBup+3p−1E∧∫s−τ1ssσu,xu−τ3udBup⩽3p−1τp−1C∫s−τ1ssρE∧xu−τ1updu+3p−1Cσ¯2pτp∫s−τ1ssρE∧xu−τ2updu+C3p−1σ¯2pτpp32p−1p/2∫s−τ1ssρE∧xu−τ3updu⩽M3p,τ∫s−τ1ssρsupt0−τ⩽v⩽uE∧xvpdu⩽M3p,τ⋅τ⋅ρG1−1G12E∧ξτp+Ap,s−t0,where M3p,τ=C3p−1τp−1+σ¯2pτp+σ¯2pτpp3/2p−1p/2. Similarly, for i=2,3 and t0+τ⩽s⩽t,(35)E∧xs−xs−τisp⩽M3p,τ⋅τ⋅ρG1−1G12E∧ξτp+Ap,s−t0.
Notice thatAp,s−t0 is increasing with respect to s, so(36)E∧I1⩽p−1∫t0+τtE∧xs−yspds+C2p−1∫t0+τtρE∧xs−yspds+C2p−1t−t0−τρM3p,τ⋅τ⋅ρG1−1G12E∧ξτp+Ap,t−t0.
Notice thatE∧I2=0. By 6, Lemma 2.4(i)],(37)E∧I3=pE∧∫t0+τtxs−ysp−2xs−ysThs,xs−τ2s−hs,ysdBs⩽pσ¯2E∧∫t0+τtxs−ysp−1hs,xs−τ2s−hs,ysds⩽pσ¯2E∧∫t0+τtp−1pxs−ysp+1phs,xs−τ2s−hs,ysp0ds⩽p−1σ¯2∫t0+τtE∧xs−yspds+σ¯22p−1∫t0+τtE∧hs,xs−τ2s−hs,yspds+2p−1σ¯2∫t0+τtE∧hs,xs−hs,yspds⩽p−1σ¯2∫t0+τtE∧xs−yspds+C2p−1σ¯2∫t0+τtE∧ρxs−xs−τ2spds+C2p−1σ¯2∫t0+τtE∧ρxs−yspds,(38)⩽p−1σ¯2∫t0+τtE∧xs−yspds+C2p−1σ¯2∫t0+τtρE∧xs−xs−τ2spds+C2p−1σ¯2∫t0+τtρE∧xs−yspds⩽p−1σ¯2∫t0+τtE∧xs−yspds+C2p−1σ¯2∫t0+τtρE∧xs−yspds+C2p−1σ¯2t−t0−τρM3p,τ⋅τ⋅ρG1−1G12E∧ξτp+Ap,t−t0.
Similarly,(39)E∧I4=12pp−2E∧∫t0+τtxs−ysp−4xs−ysTσs,xs−τ3s−σs,ys2dBs≤12pp−2σ¯2E∧∫t0+τtp−2pxs−ysp+2pσs,xs−τ3s−σs,yspds≤12pp−2σ¯2∫t0+τtE∧xs−yspds+Cp−2σ¯22p−1∫t0+τtE∧xs−xs−τ3spds+C2p−1p−2σ¯2∫t0+τtρE∧xs−yspds≤12pp−22σ¯2∫t0+τtE∧xs−yspds+C2p−1p−2σ¯2∫t0+τtρE∧xs−yspds+C2p−1σ¯2p−2t−t0−τM3P,τ⋅τ⋅ρG1−1G12E∧ξτP+Ap,t−t0,E∧I5=p2E∧∫t0+τtxs−ysp−2σs,xs−τ3s−σs,ys2dBs≤p2σ¯2E∧∫t0+τtxs−ysp−2σs,xs−τ3s−σs,yspds≤p−22σ¯2∫t0+τtE∧xs−yspds+Cσ¯22p−1∫t0+τtρE∧xs−xs−τ3spds+C2p−1σ¯2∫t0+τtρE∧xs−yspds≤12p−22σ¯2∫t0+τtE∧xs−yspds+C2p−1σ¯2∫t0+τtρE∧xs−yspds+C2p−1σ¯2t−t0−τρM3P,τ⋅τ⋅ρG1−1G12E∧ξτP+Ap,t−t0.
Then,(40)E∧xt−ytp⩽C5p∫t0+τtE∧xs−yspds+C6p∫t0+τtρE∧xs−yspds+C6p⋅t−t0−τρM3p,τ⋅τ⋅ρG1−1G12E∧ξτp+Ap,t−t0⩽M4p∫t0+τtE∧xs−ysp+ρE∧xs−yspds+C6p⋅t−t0−τρM3p,τ⋅τ⋅ρG1−1G12E∧ξτp+Ap,t−t0,where M4p=maxC5p,C6p, C6p=C2p−1pσ¯2+1, C5p=p−1+p2−p/2σ¯2. By 12, Theorem D],(41)E∧xt−ytp⩽G1−1G1Sp,t−t0,τ+M4p⋅t−t0−τ,where Sp,t−t0,τ=C6⋅t−t0−τ⋅ρM3p,τ⋅τ⋅ρG1−1G12E∧ξτp+Ap,t−t0.
(2) Next, the proof of case0<p<2 is similar to that of Lemma 1 (2); hence, we omit the details.Theorem 1.
Let Assumption1 and the conditions of Lemma 2 hold. Then, there is a positive number τ∗ such that for any initial data ξ∈LFt0pΩ;C−τ,0;ℝn with E∧ξτp⩾1, the solution of system 1 is p-th moment exponentially stable provided τ<τ∗. In practice, we can find τ∗ according to the following three cases.
(1)p≥2, 0<δ<1 and take T satisfying the following equation:(42)4p−1C4C3e−λ1TE∧ξ0p=δ.
Then,τ∗ is the unique root to the equation (in τ).(43)1=4p−1C4C3e−λ1TG1−1G12E∧ξτp+Ap,τ+4p−1G1−1G1Sp,τ+T,τ+M4pT+2p−1⋅M1p,τ⋅ρG1−1G12E∧ξτp+Ap,T+2τ.
(2)1<p<2, 0<δ<1 and take T satisfying the following equation:(44)22.5p−2C4C3e−λ1TE∧ξ02=δ.
Then,τ∗ is the unique root to the equation (in τ)(45)1=22p−2C4C3e−λ1TG1−1G12E∧ξτ2+A2,τp/2+4p−1G1−1G1S2,τ+T,τ+M42⋅Tp/2+2p−1⋅M1p/22,τ⋅ρG1−1G12E∧ξτ2+A2,T+2τp/2.
(3)0<p≤1, 0<δ<1 and take T satisfying the following equation:(46)20.5pC4C3e−λ1TE∧ξ02=δ.
Then,τ∗ is the unique root to the equation (in τ)(47)1=C4C3e−λ1TG1−1G12E∧ξτ2+A2,τp/2+G1−1G1S2,τ+T,τ+M42⋅Tp/2+M1p/22,τ⋅ρG1−1G12E∧ξτ2+A2,T+2τp/2.Proof.
(1) For any initial dataξ∈LFt0pΩ;−τ,0;ℝn, write xt;t0,ξ=xt for t⩾t0. Write yt0+τ+T;t0+τ,xt0+τ=yt0+τ+T. Then, it follows from (3) and Lemma 2 that(48)E∧yt0+τ+Tp⩽C4C3e−λ1TE∧xt0+τp⩽C4C3e−λ1TG1−1G12E∧ξτp+Ap,τ.
By (28) and (48) and the elementary inequality a+bp⩽2p−1ap+bp,(49)E∧xt0+τ+Tp⩽2p−1E∧yt0+τ+Tp+2p−1E∧xt0+τ+T−yt0+τ+Tp⩽2p−1C4C3e−λ1TG1−1G12E∧ξτp+Ap,τ+2p−1G1−1G1Sp,τ+T,τ+M4pT.
DefineE∧xt0+2τ+Tτp:=supt0+τ+T⩽t⩽t0+2τ+TE∧xtp. Then, combining (4) and (48),(50)E∧xt0+2τ+Tτp⩽2p−1E∧xt0+τ+Tp+2p−1E∧sup0⩽s⩽τxt0+τ+T−xt0+τ+T+sp⩽4p−1C4C3e−λ1TG1−1G12E∧ξτp+Ap,τ+4p−1G1−1G1Sp,τ+T,τ+M4pT+2p−1⋅M1τ,p⋅ρG1−1G12E∧ξτp+Ap,T+2τ.
Let(51)Fτ=4p−1C4C3e−λ1TG1−1G12E∧ξτp+Ap,τ+4p−1G1−1G1Sp,τ+T,τ+M4pT+2p−1⋅M1τ,p=⋅ρG1−1G12E∧ξτp+Ap,T+2τ.
It is clear thatFτ is a continuously increasing function of τ>0, F+∞=+∞, and F0=4p−1C4/C3e−λ1TE∧ξ0p=δ<1, so the equation Fτ=1 (in τ) has a unique root τ∗>0.
Fixτ∈0,τ∗ and ξ∈LFt0pΩ;−τ,0;ℝn arbitrarily. Then, Fτ<1, and so logFτ<0. Take γ=−logFτ/2τ+T+logE∧ξτp/2τ+T, and then it is clear that γ>0 and Fτ=e−γ2τ+TE∧ξτp. From (49), we have(52)E∧xt0+2τ+Tτp⩽e−γ2τ+TE∧ξτp=e−γ2τ+Tsupt0−τ⩽t⩽t0E∧xtp,which means that the p-th moment of the solution at time t0+2τ+T can be dominant by the product of e−γ2τ+T and the upper bound of the p-th moment of the initial value.
Now, we consider the solutionxt of (1) when t⩾t0+2τ+T. By the uniqueness of the solution, we could look at xt0+22τ+T as the solution of (1) at time t0+2τ+T with initial value xt0+2τ+T and initial time t0+2τ+T, which has continuous paths. So, by (52),(53)E∧xt0+22τ+Tτp⩽e−γ2τ+TE∧xt0+2τ+Tτp⩽e−2γ2τ+TE∧ξτp,and repeating the iteration,(54)E∧xt0+i2τ+Tτp⩽e−iγ2τ+TE∧ξτp,for all i=1,2,⋯. Moreover, it is clear that (54) holds for i=0. Therefore, by (5) and (54), for all i⩾0,(55)E∧supt0+i⋅2τ+T⩽t⩽t0+i+1⋅2τ+Txtp⩽G2−1G24p−1E∧xt0+i2τ+Tτp+A2p,2τ+T⩽G2−1G24p−1e−iγ2τ+TE∧ξτp+A2p,2τ+T.
Wheni⟶∞, G24p−1e−iγ2τ+TE∧ξτp goes to G20+=−∞, and note that G2 is continuous and(56)limi⟶∞G2−1G24p−1e−iγ2τ+TE∧ξτp+A2p,2τ+TG2−1G24p−1e−iγ2τ+TE∧ξτp=1.
Hence,(57)limi⟶∞logG2−1G24p−1e−iγ2τ+TE∧ξτp+A2p,2τ+T−logG2−1G24p−1e−iγ2τ+TE∧ξτpt0+i2τ+T=limi⟶∞logG2−1G24p−1e−iγ2τ+TE∧ξτp+A2p,2τ+T/G2−1G24p−1e−iγ2τ+TE∧ξτpt0+i2τ+T=0.
So, we can easily prove that(58)limsupt⟶∞logE∧xtpt⩽limi⟶∞logE∧supt0+i⋅2τ+T⩽t⩽t0+i+1⋅2τ+Txtpt0+i2τ+T⩽limi⟶∞logG2−1G24p−1e−iγ2τ+TE∧ξτp+A2p,2τ+Tt0+i2τ+T=limi⟶∞logG2−1G24p−1e−iγ2τ+TE∧ξτpt0+i2τ+T⩽limi⟶∞−iγ2τ+Tt0+i2τ+T=−γ.
(2) By (6) and Lemma 2,(59)E∧yt0+τ+Tp⩽C4C3e−λ1TE∧xt0+τp⩽C4C3e−λ1TG1−1G12E∧ξτ2+A2,τp2.
By (29), for all t⩾t0+τ,(60)E∧yt;t0+τ,xt0+τ−xtp⩽G1−1G1S2,t−t0,τ+M42t−t0−τp2.
By (59) and (60) and the elementary inequality a+bp⩽2p−1ap+bp,(61)E∧xt0+τ+Tp⩽2p−1E∧yt0+τ+Tp+2p−1E∧xt0+τ+T−yt0+τ+Tp⩽2p−1C4C3e−λ1TG1−1G12E∧ξτ2+A2,τp/2+2p−1G1−1G1S2,τ+T,τ+M42⋅Tp/2.
By (61) and (7),(62)E∧xt0+2τ+Tτp⩽2p−1E∧xt0+τ+Tp+2p−1E∧sup0⩽s⩽τxt0+τ+T−xt0+τ+T+sp⩽22p−2C4C3e−λ1TG1−1G12E∧ξτ2+A2,τp/2+4p−1G1−1G1S2,τ+T,τ+M42⋅Tp/2+2p−1⋅M1p/22,τ⋅ρG1−1G12E∧ξτ2+A2,T+2τp/2.
Let(63)Fτ=22p−2C4C3e−λ1TG1−1G12E∧ξτ2+A2,τp/2+4p−1G1−1G1S2,τ+T,τ+M42⋅Tp/2+2p−1⋅M1p/22,τ⋅ρG1−1G12E∧ξτ2+A2,T+2τp/2.
Then,Fτ is a continuously increasing function of τ>0, F+∞=+∞, and F0=22.5p−2C4/C3e−λ1TE∧ξ02=δ<1. So, the equation Fτ=1 (in τ) has a unique root τ∗>0. The rest of the proof is the same as that in part (1) and is omitted.
(3) Clearly, (59) and (61) also hold for 0<p⩽1. Then, by (59) and (61) and a+bp⩽ap+bp,(64)E∧xt0+τ+Tp⩽E∧yt0+τ+Tp+E∧xt0+τ+T−yt0+τ+Tp⩽C4C3e−λ1TG1−1G12E∧ξτ2+A2,τp/2+G1−1G1S2,τ+T,τ+M42Tp/2,(65)E∧xt0+2τ+Tτp⩽E∧xt0+τ+Tp+E∧sup0⩽s⩽τxt0+τ+T−xt0+τ+T+sp⩽C4C3e−λ1TG1−1G12E∧ξτ2+A2,τp/2+G1−1G1S2,τ+T,τ+M42Tp/2+M1p/22,τ⋅ρG1−1G12E∧ξτ2+A2,T+2τp/2.
Let(66)Fτ=C4/C3e−λ1TG1−1G12E∧ξτ2+A2,τp/2+G1−1G1S2,τ+T,τ+M42Tp/2+M1p/22,τ⋅ρG1−1G12E∧ξτ2+A2,T+2τp/2.
It is clear thatFτ is a continuously increasing function of τ>0 and F0=20.5pC4/C3e−λ1TE∧ξ02=δ<1. So, the equation Fτ=1 (in τ) has a unique root τ∗>0. Similar to the proof of part (2), from the definition of Fτ, we know that (58) holds for 0<p≤1.
From the above discussion, system 1 isp-th moment exponentially stable and the proof is finished.Remark 4.
In [9], the existence and uniqueness of solution to stochastic delay non-linear systems driven by G-Lévy processes are proved under non-Lipschitz condition, which is similar to Assumption 1. So, we wonder whether the corresponding results similar to Theorem 1 hold if there is another Lévy term in equation (1). By careful calculation, we can get Lemmas 1 and 3 with a little revision of some parameters. However, we cannot prove the p-th moment exponential stability of the solution by the method presented here. The main reason is that in equation (53), we take xt0+2τ+T as an initial point of the new solution xt, which is continuous, so we could use (52). However, the solution of stochastic delay non-linear systems driven by G-Lévy processes is not continuous, so the formula cannot hold. In the future, we will look for other methods to study the p-th order moment exponential stability of stochastic delay non-linear system driven by G-Lévy process under the conditions of H1–H3.


## 3. Example

We consider the following one-dimensional stochastic differential equation driven by G-Brownian motion:(67)dyt=−2ytdt−12sint1+t2+sin2t21+t2ytdBt+121+sint1+t2ytdBt,y0=1,where B is a G-Brownian motion with σ¯=0.1,σ¯=0.01. Let bt,yt=−2yt, ht,yt=−1/2sint/1+t2+sin2t/21+t2yt, σt,yt=1+sint/1+t2yt, Vt,y=y2. Then,(68)Vyt,ytbt,yt=−4y2t,Vyt,ytht,yt=−sint1+t2−sin2t21+t2y2t,Vyyt,ytσ2t,yt=1+sint1+t22y2t,and by [9], we have(69)LVt,y=−4y2+G−2sint1+t2−sin2t1+t2y2+1+sint1+t22y2⩽−3y2.So, by takingC3=1/2,C4=2,p=2,λ1=3, the solution of system (67) is p-th moment exponentially stable by Lemma 2.Next, consider the following stochastic delay non-linear system driven by G-Brownian motion:(70)dxt=−2xt−τ1tdt−12sint1+t2+sin2t21+t2xt−τ2tdBt+121+sint1+t2xt−τ3tdBt,x0=1.By Remark2, we have ρx=x and C=1, so G1x=lnx/2,G1−1x=e2x. Let δ=100−1; then, T=ln1/400⋅−1/3=1.997155,G12=∫121/2sds=0.3465736, A2,τ=τ⋅max1+σ¯2,C1+2σ¯2=1+2σ¯2τ=1.02τ,C62=C4σ¯2+2=2.04,M32,τ=3τ+0.0005τ2, A2,τ+T=τ+T⋅max1+σ¯2,C1+2σ¯2=1.02τ+T,M42=max1+σ¯2,2.04=2.04, S2,τ+T,τ=2.04⋅T⋅3τ+0.0015τ2⋅τ⋅G1−1G12+2.04τ+T,M12,τ=3τ21+0.0005τ, A2,2τ+T=2τ+T⋅max1+σ¯2,C1+2σ¯2=1.022τ+T. By Theorem 1, the solution of system (67) is p-th moment exponentially stable provided τ:=maxτ1t,τ2t,τ3t<τ∗ and τ∗ is the unique root to the equation (in τ):(71)1=100−1expln2+2.04τ+4ln2.04⋅T⋅3τ+0.0015τ2⋅τ⋅expln2+36τ+Texp4.08⋅T+6⋅τ2⋅1+0.0005τ⋅expln2+2.042τ+T,so τ∗≈1.7×10−3.



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*Source: 7592535-2022-09-01.xml*