# Study onH- Index of Stochastic Linear  Continuous-Time Systems

Authors: Yan Li; Tianliang Zhang; Xikui Liu; Xiushan Jiang
Journal: Mathematical Problems in Engineering
 (2015)

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## Abstract

This paper studies theH- index problem. We obtain a necessary and sufficient condition of H- index larger than γ>0. A generalized differential equation is introduced and it is proved that its solvability and the feasibility of the H- index are equivalent. We extend the deterministic cases to the stochastic models. Our results can be used to fault detection filter analysis. Finally, the effectiveness of the proposed results is illustrated by an example.


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## Body

## 1. Introduction

It is well known that many control and filtering problems have been discussed based on a certain performance index of a system, such asH2 norm, H∞ norm, and H- index; see [1–9]. H∞ norm is the measure of the worst-case disturbance inputs on the controlled outputs [1–4]. The H- index is a measure of the minimum sensitivity of system outputs to system inputs. H∞ norm and H- index with specific application to fault detection filter have been carried out in [10–17]. To ensure robustness, H- index should be maximized and H∞ norm should be minimized. Using H-/H∞ performance can make certain that the residual signal is maximally sensitive to faults and highly robust to disturbance inputs; see [16, 17].In [12], H- index was defined as the minimum nonzero singular value in zero frequency. In [10], the authors extended the results of [12] to all frequency range. By means of LMIs, a necessary and sufficient condition was given for the infinite frequency range. The case for finite frequency range was concluded through frequency weighting. In recent decades, a great deal of attention has been attracted to H- index in time domain. A fault residual generator was designed to maximize the fault sensitivity in the finite time domain [16–20]. Based on H- index, results on optimal fault detection can be found in [17, 18] and the references. The lower bound of H- index for linear time-varying systems was proposed in [19, 20]. A sliding mode observer was designed for sensor fault diagnosis of nonlinear time-delay systems; see [21]. In [22], a fault-tolerant controller was projected to compensate nonlinear faults by using a fuzzy adaptive fault observer.Although there is much work on theH- index problem, to the best of our knowledge, very little work was concerned with the H- index in stochastic systems. In this paper, the H- index for stochastic linear continuous-time systems is discussed. The definition of the H- index is extended to the stochastic case. We present a necessary and sufficient condition of the H- index. A generalized differential equation is introduced and it is proved that its solvability and the feasibility of the H- index are equivalent. Comparing our results with the bounded real lemma [2, 9], it shows that the H- index is not completely dual to H∞ norm. The H- index discussed in this paper is only fortall orsquare systems. The reason for this is that H- index is zero forwide systems. But bounded real lemma for H∞ is applicable to any systems. Finally, the effectiveness of the given methods is illustrated by numerical example.The outline of the paper is organized as follows. In Section2, some efficient criteria are given for the H- index of stochastic linear systems in finite horizon. Section 3 contains an example provided to show the efficiency of the proposed results. Finally, we conclude this paper in Section 4.Notations.
R is the field of real numbers. Rm×n is the vector space of all m×n matrices with entries in R. Sn(R) is the set of all real symmetric matrices Rn×n. A′ is the transpose of matrix A. A-1 is the inverse of A. Given positive semidefinite (positive definite) matrix A, we denote it by A≥0 (A>0). E is the mathematical expectation. I is identity matrix. 0n is n×n zero matrix. LF2([0,T],Rp) is the space of nonanticipative stochastic process y(t)∈Rp with respect to increasing σ-algebras Ft (t≥0) satisfying y(t)[0,T]2<∞, where y(t)[0,T]2=E∫0T‍yt′y(t)dt=E∫0T‍y(t)2dt. Asquare (wide ortall) system denotes a system when the number of inputs equals (is more than or less than) the outputs number.


## 2. Finite Horizon StochasticH- Index

In this section, we will discuss theH- index problem of stochastic linear continuous-time systems. We give a necessary and sufficient condition of the H- index larger than γ>0 in finite horizon.Consider the following stochastic linear time-varying systemG:(1)dxt=Atxt+Btvtdt+A0txt+B0tvtdωt,zt=Ctxt+Dtvt,x0=x0.In the above, ω(t) is the one-dimensional standard Wiener process defined on the complete probability space (Ω,F,P), with the natural filter Ft generated by ω(t) up to time t. Consider x(t)∈Rn, v(t)∈LF2([0,T],Rp), and z(t)∈Rl are the system state, control input, and regulated output, respectively. A(t), B(t), A0(t), B0(t), C(t), and D(t) are coefficients with appropriate dimensions. For any 0<T<∞ and (v(t),x0)∈LF2([0,T],Rp)×Rn, there exists unique solution x(t)=x(t;v,x0)∈LF2([0,T],Rn) of (1).The finite horizon stochasticH- index of system (1) can be stated as follows.Definition 1.
For stochastic system (1), given 0<T<∞, its H- index in [0,T] is defined as(2)G-0,T=infv≠0,x0=0⁡zt0,Tvt0,T=infv≠0,x0=0⁡E∫0Tzt′ztdt1/2E∫0Tvt′vtdt1/2,where v(t)∈LF2([0,T],Rp).Remark 2.
Ifv is fault signal and z is the residual, then the H- index describes the smallest fault sensitivity of system (1). In this paper, we suppose that system (1) istall orsquare because the H- index is zero forwide system.Givenγ>0 and 0<T<∞, let(3)JTγx0,v=zt0,T2-γ2vt0,T2=E∫0Tzt′zt-γ2vt′vtdt.We will study the following optimal problem:(4)minv∈LF20,T,Rp⁡JTγx0,v.Remark 3.
It can be shown thatG-[0,T]>γ is equivalent to the following inequality(5)JTγ0,v=zt0,T2-γ2vt0,T2=E∫0Tzt′zt-γ2vt′vtdt>0,∀v(t)∈LF2([0,T],Rp), v(t)≠0.Remark 4.
WhenT=∞, (2) corresponds to the infinite horizon case.Lemma 5.
SupposeP(t):[0,T]↦Sn(R) is continuously differentiable, T>0. Then, for every x0∈Rn, v(t)∈LF2([0,T],Rp),(6)JTγx0,v=x0′P0x0-ExT′PTxT+E∫0Txtvt′Mt,Ptxtvtdt,where M(t,P(t))=L(P(t))+P˙(t)K(P(t))KPt′Hγ(P(t))∈Sn+l(R),(7)LPt=PtAt+At′Pt+A0t′PtA0t+Ct′Ct,KPt=PtBt+A0t′PtB0t+Ct′Dt,HγPt=B0t′PtB0t+Dt′Dt-γ2I.Proof.
Letx0∈Rn, v(t)∈LF2([0,T],Rp), and x(t)=x(t;v,x0) denote the corresponding solution of (1). Applying Ito’s formula to xt′P(t)x(t) and taking expectations, we have that, for any T>0,(8)ExT′PTxT-x0′P0x0=E∫0Tdxt′Ptxt=E∫0Txtvt′Qt,Ptxtvtdt,where(9)Qt,Pt=PtAt+At′Pt+A0t′PtA0t+P˙tPtBt+A0t′PtB0tBt′Pt+B0t′PtA0tB0t′PtB0t.So(10)JTγx0,v=E∫0TCxt+Dvt2-γ2vt2dt+x0′P0x0-ExT′PTxT+E∫0Txtvt′Qt,Ptxtvtdt=x0′P0x0-ExT′PTxT+E∫0Txtvt′Mt,Ptxtvtdt,which ends the proof.Below, we prove the following theorem which is necessary in this paper.Theorem 6.
For (1) and some given γ>0, if the following differential Riccati equation(11)LPt+P˙t=KPtHγPt-1KPt′,HγPt>0,PT=0admits solution PT(t) on [0,T], then G-[0,T]>γ.Proof.
By Lemma5, for every v(t)∈LF2([0,T],Rp), v≠0, x0=0, we conclude that(12)JTγ0,v=E∫0Txtvt′Mt,PTtxtvtdt.By using completion of squares argument and the first equality in (11), we have(13)JTγ0,v=E∫0Txt′LPTt+P˙Tt-KPTt·HγPTt-1KPTt′xtdt+E∫0Tvt+HγPTt-1KPTt′xt′·HγPTt·vt+HγPTt-1KPTt′xtdt=E∫0Tvt-v∗t′HγPTt·vt-v∗tdt,where v∗(t)=-Hγ(PTt)-1KPTt′x(t).
FromHγ(PT(t))>0, JTγ(0,v)≥0, to show JTγ(0,v)>0, we define the operator L: LV(t)=v(t)-v∗(t) with its realization:(14)dxt=Atxt+Btvtdt+A0txt+B0tvtdωt,x0=0,vt-v∗t=vt+HγPTt-1KPTt′xt.
ThenL-1 exists, which is determined by(15)dxt=At-BtHγPTt-1Bt′PTt+B0t′PTtA0t+Dt′Ctxtdt+A0t-B0tHγPTt-1Bt′PTt+B0′tPTtA0t+Dt′Ctxtdωt+Btvt-v∗tdt+B0tvt-v∗tdωt,x0=0,where v(t)=-Hγ(PTt)-1KPTt′x(t)+(v(t)-v∗(t)).
We assume thatHγ(PT(t))≥ϵI, ϵ>0, so there exists constant C0>0, such that(16)JTγ0,v=E∫0Tvt-v∗t′HγPTtvt-v∗tdt≥ϵvt-v∗t0,T2=ϵLVt0,T2≥C0vt0,T2>0.That is, G-[0,T]>γ.
Now, we consider the following equation:(17)X˙t+LXt+KXtFt+Ft′KXt′+Ft′HγXtFt=0,t∈0,T,XT=0,where F(t)∈C[0,T] and this equation has unique solution X(t)=PFγ(t), t∈[0,T].
It is easy to see that (17) satisfies the following equation:(18)P˙Fγt+IFt′LPFγtKPFγtKPFγt′HγPFγtIFt=0,t∈0,T,PFγT=0.Lemma 7.
SupposeF(t)∈C[0,T] and PFγ(t) is the solution of (18). Then if v(t)∈LF2([0,T],Rp), one obtains(19)JTγx0,v+FxF=x0′PFγ0x0+E∫0Tvt′GtxFt+xFt′Gt′vt+vt′HγPFγtvtdt,where xF(t)=x(t,F(t)xF(t)+v(t),x0) is the solution of(20)dxFt=At+BtFtxFtdt+A0t+B0tFtxFtdwt+B0tvtdwt+Btvtdtwith xF(0)=x0 and(21)Gt=KPFγt′+HγPFγtFt.As v=0, then(22)JTγx0,FxF=x0′PFγ0x0.Proof.
In terms of Lemma5 with P(t)=PFγ(t) and F(t)xF(t)+v(t) for v(t),(23)JTγx0,v+FxF=x0′PFγ0x0+E∫0TxFtFtxFt+vt′Mt,PFγt·xFtFtxFt+vtdt=x0′PFγ0x0+E∫0TxFt′IFt′·LPFγt+P˙FγtKPFγtKPFγt′HγPFγtIFtxFtdt+E∫0Tvt′GtxFt+xFt′Gt′vt+vt′HγPFγtvtdt=x0′PFγ0x0+E∫0Tvt′GtxFt+xFt′Gt′vt+vt′HγPFγtvtdt.This means that (19) holds. Let v=0 in (19); we obtain (22).
Now we are in a position to prove thatHγ(PFγ(t)) is invertible for t∈[0,T].Lemma 8.
For system (1), if  G-[0,T]>γ for some given γ>0, Ft∈C0,T, T>0, and PFγ(t) satisfies (18). Then,(24)HγPFγt≥G-0,T2-γ2I>0,t∈0,T.Proof.
Let us first prove thatHγ(PFγ(t))≥0. Suppose this is false; then there exists t∗∈[0,T), u∈Rl, u=1 such that u′Hγ(PFγ(t∗))u≤-η for some η>0. Then, for sufficiently small δ>0,(25)u′HγPFγtu≤-η2,t∈t∗,t∗+δ⊂0,T.Define(26)vt=0,t∈0,t∗∪t∗+δ,T,u,t∈t∗,t∗+δ.Using Lemma 7 with this v(t) and x0=0, we can derive that xF(t)=0 for t∈[0,t∗] and(27)JTγ0,v=E∫0TCtxFt+Dtvt2-γ2vt2dt=E∫0Tvt′GtxFt+xFt′Gt′vt+vt′HγPFγtvtdt≤E∫t∗t∗+δ2Gt′uxFt-η2dt.Since xF(t) is continuous and xF(t∗)=0, (27) is negative. Moreover, the condition G-[0,T]>γ implies JTγ(0,v)≥0. As a result, this is a contradiction. If t∗=T, we can replace [t∗,t∗+δ] by [T-δ,T] and use a similar proof.
Next, letG-[0,T]>(γ2+ρ2)1/2 for any ρ>0 and λ=(γ2+ρ2)1/2. Replacing γ with λ in (18), we obtain the corresponding solution PFλ(t). Applying the previous step, we can deduce that Hλ(PFλ(t))≥0. For any t0∈[0,T), set Ft0=F(t+t0), t∈[0,T-t0]. Let PFt0λ(t) be the solution of (18) with γ replaced by λ and F replaced by Ft0 on [0,T-t0]. Then, PFt0λ(t)=PFλ(t+t0), t∈[0,T-t0]. By (22), for any t0∈[0,T), x0∈Rn,(28)x0′PFλt0x0=x0′PFt0λ0x0=JT-t0λx0,Ft0xFt0≤JT-t0γx0,Ft0xFt0=x0′PFγt0x0,and so Hλ(PFγ(t0))≥Hλ(PFλ(t0))≥0. By continuity, Hγ(PFγ(t))≥ρ2I for all t∈[0,T]. As this holds for arbitrary ρ2<(G-[0,T])2-γ2, it follows that Hγ(PFγ(t))≥[(G-[0,T])2-γ2]I>0. This completes the proof.Remark 9.
Whent=T, (24) becomes Hγ(PFγ(T))=DT′D(T)-γ2I>0. If system (1) is time-invariant, then(29)D′D-γ2I>0.Remark 10.
By the equalityA(I-BA)-1=(I-AB)-1A, we have that C′[I-D(D′D-γ2I)-1D′]C=C′(I-γ-2DD′)-1C. If system (1) is time-invariant and square, by (29),(30)C′I-DD′D-γ2I-1D′C=C′I-γ-2DD′-1C≤0.Now, we present the following theorem which is important in this paper.Theorem 11.
Suppose system (1) is time-invariant and square and satisfies G-[0,T]>γ for given γ≤0. Then (11) has a unique solution PT(t)≤0 on [0,T] for every T>0. Moreover, JTγ(x0,v) is minimized by the feedback control:(31)v∗t=FTtxFTt,FTt=-HγPTt-1KPTt′,where xFT(t) satisfies (32)dxFTt=A+BFTtxFTtdt+A0+B0FTtxFTtdwt,xFT0=x0and the optimal cost is(33)minv∈LF20,T,Rp⁡JTγx0,v=x0′PT0x0.Proof.
We prove thatG-[0,T]>γ implies the existence of solution PT(t) of (11) on [0,T]. Using a contradiction argument, we suppose that (11) does not admit a solution. By the standard theory of differential equations, there exists unique solution PT(t) backward in time on maximal interval [T0,T] (T0≥0), and as t→T0, PT(t) becomes unbounded.
Let0<δ<T-T0, x(T0+δ)=xT0,δ∈Rn, Q(t)=B′PT(t)+B0′PT(t)A0+D′C, R(t)=B0′PTtB0+D′D-γ2I, by completing the squares; then(34)Jγx,v,xT0,δ,T0+δ=E∫T0+δTzt′zt-γ2vt′vtdt=E∫T0+δTzt′zt-γ2vt′vtdt+E∫T0+δTdxt′PTtxtdt+xT0,δ′PTT0+δxT0,δ=E∫T0+δTxt′C′C+A′PTt+PTtA+A0′PTtA0+P˙Tt-Qt′Rt-1Qtxtdt+E∫T0+δTvt+Rt-1Qtxt′Rtvt+Rt-1Qtxtdt+xT0,δ′PTT0+δxT0,δ=E∫T0+δTvt+Rt-1Qtxt′Rtvt+Rt-1Qtxtdt+xT0,δ′PTT0+δxT0,δ.Obviously,(35)minv∈LF2T0+δ,T,Rp⁡Jγx,v,xT0,δ,T0+δ=Jγx,v∗,xT0,δ,T0+δ=xT0,δ′PTT0+δxT0,δ,where v∗(t)=-R(t)-1Q(t)x(t).
Furthermore, we can see that(36)Jγx,v,xT0,δ,T0+δ=E∫T0+δTzt′zt-γ2vt′vtdt=E∫T0+δTCxt+Dvt′Cxt+Dvt-γ2vt′vtdt=E∫T0+δTxt′C′I-DD′D-γ2I-1D′Cxtdt+E∫T0+δTvt+D′D-γ2I-1D′Cxt′D′D-γ2Ivt+D′D-γ2I-1D′Cxtdt.From Remark 10,(37)Jγx,v~,xT0,δ,T0+δ=E∫T0+δTxt′·C′I-DD′D-γ2I-1D′Cxtdt≤0,where v~(t)=-(D′D-γ2I)-1D′Cx(t). Considering (35), we get(38)xT0,δ′PTT0+δxT0,δ≤Jγx,v~,xT0,δ,T0+δ≤0,which implies that(39)PTT0+δ≤0.By linearity, the solution of (1) with initial state xT0,δ satisfies(40)xt,v,xT0,δ,T0+δ=xt,0,xT0,δ,T0+δ+xt,v,0,T0+δ.Suppose Φ(t) is the solution of(41)C′C+A′Φt+ΦtA+A0′ΦtA0+Φ˙t=0,ΦT=0,and we have(42)Jγx,v,xT0,δ,T0+δ-Jγx,v,0,T0+δ=xT0,δ′ΦT0+δxT0,δ+E∫T0+δTvt′·B0′Φt+D′Cxt,0,xT0,δ,T0+δdt+E∫T0+δTxt,0,xT0,δ,T0+δ′B0′Φt+D′C′·vtdt.
TakeG-[0,T]≥(γ2ϵ2)1/2,(43)v¯t=v~t=-D′D-γ2I-1D′Cxt,t∈0,T0+δvt,t∈T0+δ,T,and it is easy to show that(44)Jγx,v,0,T0+δ=E∫0Tzt2-γ2v¯t2dt-E∫0T0+δzt2-γ2v~t2dt=E∫0Tzt2-γ2v¯t2dt-E∫0T0+δxt′·C′I-DD′D-γ2I-1D′Cxtdt-E∫0T0+δv~t+D′D-γ2I-1D′Cxt′·D′D-γ2I·v~t+D′D-γ2I-1D′Cxtdt≥E∫0Tzt2-γ2v¯t2dt≥ϵ2v¯t0,T2≥ϵ2vtT0+δ,T2.It follows that(45)Jγx,v,xT0,δ,T0+δ≥E∫T0+δTϵ2vt2dt+xT0,δ′ΦT0+δxT0,δ+E∫T0+δTvt′B0′Φt+D′Cxt,0,xT0,δ,T0+δdt+E∫T0+δTxt,0,xT0,δ,T0+δ′B0′Φt+D′C′·vtdt=xT0,δ′ΦT0+δxT0,δ+E∫T0+δTϵv-ϵ-2B0′Φt+D′Cxt,0,xT0,δ,T0+δ2dt-E∫T0+δTϵ-1B0′Φt+D′Cxt,0,xT0,δ,T0+δ2dt≥xT0,δ′ΦT0+δxT0,δ-E∫T0+δTϵ-1B0′Φt+D′Cxt,0,xT0,δ,T0+δ2dt.It is obvious that there exists constant C0>0 such that(46)C2xT0,δ2≥E∫T0+δTxt,0,xT0,δ,T0+δ2dt.So, there is constant C1>0 such that(47)E∫T0+δTϵ-1B0′Φt+D′Cxt,0,xT0,δ,T0+δ2dt≤C1xT0,δ2.In addition,(48)xT0,δ′ΦT0+δxT0,δ=-E∫T0+δTdxt′Φtxt=E∫T0+δTxt,0,xT0,δ,T0+δC2dt≥0.From (45), we have(49)Jγx,v,xT0,δ,T0+δ≥-C1xT0,δ2.In view of (35) and (39), it yields(50)-C1≤PTT0+δ≤0.So, PT(T0+δ) can not become unbounded as δ→0, which means that (11) has unique solution PT(t) on [0,T].
SettingF(t)=FT(t), t∈[0,T], in (17), from (31), we obtain(51)P˙Tt+LPTt+KPTtFt+Ft′KPTt′+Ft′HγPTtFt=0.Hence PT(t) satisfies (17), or equivalently (18). So (52)PFTγt=PTt,t∈0,T.By (31), (53)Gt=KPTt′+HγPTtFTt=0,and, in terms of Lemma 7, (54)JTγx0,v+FTx=x0′PT0x0+E∫0Tvt′HγPTtvtdt.But by Lemma 8, (55)HγPTt=HγPFTγt⪰G-0,T2-γ2I≻0,t∈0,T.Hence, v∗(t)=FT(t)x(t) minimizes JTγ(x0,v) and minv∈LF20,T,RpJTγ(x0,v)=x0′PT0x0.According to Theorems6 and 11, we get the following theorem.Theorem 12.
If system (1) is time-invariant and square, for given γ>0, the following are equivalent:(i)
ConsiderG-[0,T]>γ.(ii)
The following equation(56)PtA+A′Pt+A0′PtA0+C′C+P˙t=PtB+A0′PtB0+C′D·B0′PtB0+D′D-γ2I-1·PtB+A0′PtB0+C′D′,B0′PtB0+D′D-γ2I>0,PT=0has unique solution PT(t)≤0 on [0,T]. Moreover, minv∈LF2([0,T],Rp)JTγ(x0,v)=x0′PT(0)x0.Remark 13.
For givenγ>0, if we replace B, C, D, and v(t) with Bδ=B0n×n, Cδ=CδIn′, Dδ=D0l×n0n×l0n×n, and vδ(t)=vt0n×n′, respectively, and z(t) with zδ(t) in (1), we deduce the corresponding H- index Gδ-[0,T] and(57)JT,δγx0,v=E∫0Tzδt2-γ2vδt2dt=E∫0Tzt2-γ2vt2+δ2Idt.When G-[0,T]>γ, then Gδ-[0,T]>γ. Using Theorem 12 to the modified data, it is easy to see that the following equation(58)PtA+A′Pt+A0′PtA0+C′C+δ2I+P˙t=PtB+A0′PtB0+C′D·B0′PtB0+D′D-γ2I-1·PtB+A0′PtB0+C′D′,B0′PtB0+D′D-γ2I>0,PT=0has unique solution Pδ,T(t)≤0 on [0,T]. Moreover, minv∈LF2([0,T],Rp)JT,δγ(x0,v)=x0′Pδ,T(0)x0.Now, we are to show what happens asT increases.Theorem 14.
If system (1) is time-invariant and square, G-[0,T]>γ for some γ>0. Then PT(t) in (56) decreases as T increases for every t∈[0,T].Proof.
SupposeT¯>T, t∈[0,T], and x0∈Rn. Let vT-t∗ be optimal for x0 on [0,T-t], and set(59)vτ=vT-t∗τ,τ∈0,T-t-D′D-γ2I-1D′Cxτ,τ∈T-t,T¯-t.By the time invariance of PT(t), PT-t(0)=PT(t). Then,(60)x0′PT¯tx0=x0′PT¯-t0x0≤JT¯-tγx0,v=JT-tγx0,vT-t∗+E∫T-tT¯-tzτ2-γ2vτ2dτ=JT-tγx0,vT-t∗+E∫T-tT¯-tx′τ·C′I-DD′D-γ2I-1D′Cxτdτ+E∫T-tT¯-tvτ+D′D-γ2I-1D′Cxτ′·D′D-γ2Ivτ+D′D-γ2I-1D′Cxτdτ=JT-tγx0,vT-t∗+E∫T-tT¯-tx′τ·C′I-DD′D-γ2I-1D′Cxτdτ≤JT-tγx0,vT-t∗=x0′PTtx0.This means that PT(t) decreases as T increases for every t∈[0,T].


## 3. A Numerical Example

Below, we give a numerical example to illustrate the rightness of Theorems12 and 14.Example 1.
In system (1), we consider a two-dimensional linear stochastic system with the following parameters:(61)A=0112,B=2113,C=1123,D=1001,A0=1221,B0=1000.Set γ=0.5, T=2,3; by solving (56), we can obtain the solutions of (62)P2t=p211tp212tp212tp222t,P3t=p311tp312tp312tp322t,for which their trajectories are shown in Figure 1. If we set t=1, then it yields(63)P21=-0.0892-0.0953-0.0953-0.1446,P31=-0.1044-0.0889-0.0889-0.1489.It is easy to see that P2(1)>P3(1), which verifies the rightness of Theorem 14.Figure 1
The trajectories ofP2(t) and P3(t).


## 4. Conclusion

In this paper, we have solved theH- index problem where both stochastic and deterministic perturbations are present. Necessary and sufficient condition for the lower bound of H- index is given by means of the solvability of a generalized differential equation. The proposed results are not completely dual to H∞ norm, and the effectiveness of the given methods is illustrated by numerical example.



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