# Existence of Solutions of a Riccati Differential System from a General Cumulant Control Problem

Authors: Stanley R. Liberty; Libin Mou
Journal: International Journal of Differential Equations
 (2011)

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## Abstract

We study a system of infinitely many Riccati equations that arise from a cumulant control problem, which is a generalization of regulator problems, risk-sensitive controls, minimal cost variance controls, andk-cumulant controls. We obtain estimates for the existence intervals of solutions of the system. In particular, new existence conditions are derived for solutions on the horizon of the cumulant control problem.


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## Body

## 1. Introduction

Consider a linear control system and a quadratic cost function:(1.1)dx=(Ax+Bu)dt+Gdw,t∈[t0,tf];x(t0)=x0,J(u)=xT(tf)Qfx(tf)+∫t0tf(xTQx+uTRu)dt,
where A(t)∈ℝn×n,B(t)∈ℝn×m,Q(t)∈𝕊n, and R(t)∈𝕊m are continuous matrix functions for t∈[t0,tf], Qf∈𝕊n (𝕊n is the set of n×n symmetric matrices.), x(t)∈ℝn is the state with known initial state x0, u(t)∈ℝm the control, and w(t)∈ℝp a standard Wiener process. Because x is completely determined by the first equation in (1.1) in terms of u, the cost function J is only a function of u.Forθ∈ℝ, denote by E(eθJ) the (general) expectation of the random variable eθJ. For i≥1, let
(1.2)κi=didθiln⁡E(eθJ)|θ=0
be the ith cumulant of the cost J(u). Let {μi}i=1∞ be a sequence of nonnegative real numbers. Consider the following combination of κi:
(1.3)κ=∑i=1∞μiκii!.The cumulant control problem, considered in [1], is to find a control u that minimizes the combined cumulant κ defined in (1.3). This problem leads to the following system of (infinitely many) equations of Riccati type:
(1.4)H1′+(A+BK)TH1+H1(A+BK)+Q+KTRK=0,H1(tf)=Qf,Hi′+(A+BK)THi+Hi(A+BK)+2∑j=1i-1HjWHi-j=0,Hi(tf)=0,i≥2,
where ′=d/dt denotes the derivative to t, A(t)∈ℝn×n,B(t)∈ℝn×m,Q(t)∈𝕊n, R(t)∈𝕊m are as in (1.1), and W(t)=G(t)GT(t)∈𝕊n. K(t)∈ℝm×n and Hi(t)∈𝕊n are the unknown matrix functions, which are required to be continuously differentiable for t∈[t0,tf]. For convenience, the time variable t is often suppressed. System (1.4) will be combined with the following equation
(1.5)K=-R-1BT∑i=1∞μiHi.Ifk≥1 is a given integer and μi=0 for i>k, then κ=∑i=1kμiκi/i! is the k-cost cumulant investigated in [2, 3]. In particular, if k=1, then κ=μ1E{J(u)} and the cumulant problem is the classical regulator problem that minimizes E(J). If k=2, then the cumulant problem is the minimal cost variance control considered in [4, 5]. Interested readers are referred to [1–6] for the investigations, generalizations, and applications of cumulant controls.Another important cumulant control occurs whenμi=θi-1 for i≥1. In this case, κ=(1/θ)lnE(eθJ) is precisely the cumulant generating function, and the cumulant problem is the risk-sensitive control; see, for example, [7]. In this case, (1.4) and (1.5) lead an equation for the matrix functionP(θ;t)=∑i=1∞θi-1Hi(t):
(1.6)P′+ATP+PA+P(2θW-BR-1B′)P+Q=0;P(θ;tf)=Qf.As shown in [1], the solution Hi(t) of (1.4) is related to P(θ;t) by the equation:
(1.7)Hi(t)=1i!∂i∂θi[θP(θ;t)]|θ=0
and the equations in (1.4) for Hi(t) can be obtained by differentiating (1.6) to θ at θ=0.For a feedback controlu=Kx with a given matrix function K, it was shown in [8] and [1, Theorem  2] that the ith cumulant κi of J(u) has the following representation:
(1.8)κi=i!(x0THi(t0)x0+∫t0tftr⁡(Hi(s)W(s))ds),
where {Hi}i=1∞ is the solution of (1.4). Consequently (1.3) can be written as
(1.9)κ=∑i=1∞μi(x0THi(t0)x0+∫t0tftr⁡(Hi(s)W(s))ds).In [1] the cumulant control problem was restated as minimizing κ in (1.9) with K as a control, {Hi}i=1∞ as a state, and (1.4) a state equation. Furthermore, the following result is proved in [1, Theorem  3].Theorem 1.1.
If the controlu=Kx is the optimal feedback control of (1.9), then the solution {Hi}i=1∞ of (1.4) and K must satisfy (1.5).By Theorem1.1, it is necessary to solve (1.4) and (1.5) in order to find a solution of a cumulant control problem. Because of the nonlinearity of (1.4) and (1.5) in {Hi}i=1∞, a global solution may not exist on the whole horizon [t0,tf] of the cumulant problem. This can be illustrated by a scalar case of (1.6). Suppose A=0,B=G=θ=R=Q=Qf=1, then (1.6) becomes P′+P2+1=0 and P(tf)=1. The solution is P(t)=tan (π/4+tf-t), which is defined on (s,tf] with s=tf-π/4. So (1.6) has no solution unless t0>tf-π/4.By the local existence theory of differential equations, the solutionsK and {Hi}i=1∞of (1.4) and (1.5) exist on a maximal subinterval (s,tf]⊂[t0,tf]. Our interest is to give an estimate for this interval. In particular, we will obtain conditions that guarantee s=t0. The idea of our approach is to show that the trace tr(H) of
(1.10)H=∑i=1∞μiHi
satisfies a scalar differential inequality:(1.11)z′+az2+bz+c≥0;z(tf)=tr⁡(Qf)
with some functions a,b,c on [t0,tf]. A key of the proof is Proposition 2.1 below. It follows that tr(H) is bounded by the solution of the Riccati equation:
(1.12)z′+az2+bz+c=0;z(tf)=zf,
where zf=tr(Qf). Consequently, the existence interval of (1.12) gives an estimate for that of system (1.4) and (1.5); see Theorems 2.4 and 3.5 below. By a similar argument, we prove that the cumulant problem is well posed under appropriate conditions; see Theorems 2.3 and 3.4 below.In [9] the norm of a solution of a coupled matrix Riccati equation was shown to satisfy a differential inequality similar to (1.11). Consequently, specific sufficient conditions were derived for the existence of solutions of the Riccati equation in [9]. Estimates for maximal existence interval of a classical Riccati equation had been obtained in [10] in terms of upper and lower solutions. For the coupled Riccati equation associated with the minimal cost variance control, some implicit sufficient conditions had been given in [11] for the existence of a solution. In this paper, we use the trace tr(H) to bound the solution of system (1.4) and (1.5), which generally leads to a better estimate for the existence interval.


## 2. Comparison Results for Traces

We start with an assumption and some preparations. In this paper we assume that(2.1)Qf,Q(t)≥0(positivesemidefinite)fort∈[t0,tf];R(t)>0(positivedefinite)fort∈[t0,tf].For the sequenceμ={μi}i=1∞ in (1.3), we will assume that
(2.2)μ1=1;μi≥0fori≥1;ρ(μ)<∞,
where
(2.3)ρ(μ)=sup⁡{μiμjμi-j,i≥2,j=1,…,i-1}.Note that the assumptionμ1=1 is not essential. The assumption that μi≥0 for i≥1 and Proposition 2.1 below imply that the matrix H in (1.4) is a positive semidefinite series. The requirement that ρ(μ)<∞ imposes some growth condition for the sequence μ; see the proofs of Theorems 2.3 and 2.4.Also note that ifθ=ρ(μ)<∞, then μi≤θμi-1 and μi≤θi-1 for all i≥1. Theorem 3.4 below shows that the cumulant control problem is well posed for any sequence μ with a small ρ(μ). Some examples of ρ(μ) are as follows:(i)
ρ({1,θ,θ2,θ3,0,…})=θ;(ii)
ρ({1,μ2,0,0,…})=μ2;(iii)
ρ({1,μ2,μ3,0,…})=max{μ2,μ3/μ2}.We need the following properties of{Hi}i=1∞.Proposition 2.1.
Suppose{Hi}i=1∞ is a solution of (1.4) on some interval [s,tf] with a given K, then each Hi(t)≥0 for t∈[s,tf].Proof.
The formula of theith cumulant κi in [8] implies that κi is nonnegative for all x0. It follows from the representation (1.9) that Hi(t0) must be positive semidefinite. This argument continues to hold with t0 replaced by any s∈[t0,tf].Next we verify some properties related to matrix trace that are needed for our analysis. ForA∈ℝn×n, denote by tr(A)=∑i=1naii the trace of A, and λ1(A) and λn(A) the smallest and largest eigenvalue of A, respectively.Proposition 2.2.
(a) For allA,B∈ℝn×n,  tr(A)=tr(AT),  tr(AB)=tr(BA).
(b)  For allA∈ℝn×n,  B∈𝕊n,B≥0,
(2.4)λ1(A)tr⁡(B)≤tr⁡(AB)≤λn(A)tr⁡(B).
(c)  IfA,B,C∈𝕊n, then
(2.5)tr⁡(ABC)=tr⁡(ACB).
(d)  IfA,B,C∈𝕊n are all  ≥0, then
(2.6)tr⁡(AB)≤tr⁡(A)tr⁡(B),(2.7)tr⁡(ABC)≤tr⁡(A)tr⁡(B)tr⁡(C),(2.8)1nλ1(A)tr⁡(B)2≤tr⁡(AB2).Proof.
The properties in (a) are obvious by the definitions of trace and matrix multiplication. Some of the inequalities in (b)–(d) might be known, but the authors were not able to find proofs in existing literature. So we include our proofs of (b)–(d) below for readers’ convenience.
To prove (b), letT be a unitary matrix such that B=T*DT where D is diagonal with eigenvalues λi(B) of B,i=1,…,n. Then
(2.9)tr⁡(AB)=tr⁡(AT*DT)=∑i=1n(TAT*)iiλi(B),
where (TAT*)ii is the entry of TAT* at (i,i). Let vi be the ith row of T, which is a unit vector. Then (TAT*)ii=viAvi*∈[λ1(A),λn(A)]. Since λi(B)≥0,we have
(2.10)λ1(A)λi(B)≤(TAT*)iiλi(B)≤λn(A)λi(B).
This implies (2.4).
To show (c), use the symmetry ofA,B, and C; we get that tr(ABC)=∑ijkaijbjkcki=∑ijkajicikbkj=tr(ACB).
Inequality (2.6) follows from part (b) and the fact that λn(A)≤tr(A) since A≥0.
To show (2.7), first note that tr(ABC)≤λn(AB)tr(C) by (2.4); then it remains to show that λn(AB)≤λn(A)λn(B). Choose a v∈ℝn withvTv=1 such that λn(AB)=vTABv. By Schwartz inequality,
(2.11)vTABv≤vTA2vvTB2v≤λn(A2)λn(B2).
Since A,B≥0, we have λn(A2)=λn(A)2 and λn(B2)=λn(B)2. Therefore, λn(AB)≤λn(A)λn(B).
For (2.8), using notations in the proof of (b), we first get
(2.12)tr⁡(AB2)=tr⁡(AT*D2T)=∑i=1n(TAT*)iiλi2(B)≥λ1(A)∑i=1nλi2(B).
Then (2.8) follows from the inequalities
(2.13)∑i=1nλi2(B)≥1n(∑i=1nλi(B))2≥1ntr⁡(B)2.Now we estimate the existence intervals of solutions of (1.4) and (1.5). First, let K be given and {Hi}i=1∞ be the solution of system (1.4). We have the following result, which will be used in the proof of Theorem 3.5Theorem 2.3.
Supposeρ(μ)<∞ and K in (1.4) is given. Let a1, b1, and c1 be functions on [t0,tf] satisfying
(2.14)a1≥2ρ(μ)tr⁡(W),b1≥2λn(A+BK),c1≥tr⁡(Q+KTRK).
(a)  If{Hi}i=1∞ is a solution of system (1.4), then z=tr(H) satisfies the differential inequality
(2.15)-z′≤a1z2+b1z+c1,z(tf)=tr⁡(Qf).
(b)  If the equation(2.16)-z′=a1z2+b1z+c1,z(tf)=tr⁡(Qf)
has a solution on [t0,tf], then system (1.4) has a solution {Hi}i=1∞ on [t0,tf] such that H=∑i=1∞μiHi is convergent.Proof.
(a)  DenoteF=A+BK. Multiplying the equation in (1.4) for Hi by μi and sum over i, we obtain
(2.17)H′+FTH+HF+Q+KTRK+2∑i=2∞μi∑j=1i-1HjWHi-j=0.
Taking traces of both sides of (2.17) gives
(2.18)-z′=tr⁡(FTH+HF)+tr⁡(Q+KTRK)+2∑i=2∞μi∑j=1i-1tr⁡(HjWHi-j).
Note that W=GGT≥0 and by Proposition 2.1, Hi, H≥0 for t∈[tf,tf]. Proposition 2.2 (a), (b), and (c) imply that
(2.19)tr⁡(FTH+HF)=2tr⁡(FH)≤2λn(F)tr⁡(H)=b1z,tr⁡(HjWHi-j)≤tr⁡(W)tr⁡(Hj)tr⁡(Hi-j).
By (1.4) and definition (2.3) of ρ(μ), we have
(2.20)2∑i=2∞μi∑j=1i-1tr⁡(HjWHi-j)≤2∑i=2∞μi∑j=1i-1tr⁡(Hj)tr⁡(W)tr⁡(Hi-j)≤2ρ(μ)tr⁡(W)∑i=2∞∑j=1i-1μjμi-jtr⁡(Hj)tr⁡(Hi-j)=2ρ(μ)tr⁡(W)(∑i=1∞μitr⁡(Hi))2=2ρ(μ)tr⁡(W)tr⁡(H)2≤a1z2.
Substituting (2.19) and (2.20) into (2.18) and using the definition of c1 in (2.14) we get
(2.21)-z′≤b1z+c1+a1z2.(b)  Suppose that (2.16) has a solution z*(t) on [t0,tf]. By local existence theory, system (1.4) has a solution {Hi}i=1∞ on a maximal interval (s,tf]⊂[t0,tf].By (a), tr(H) satisfies inequality (2.15); that is tr(H) is a lower solution of (2.16). By a comparison theorem of lower-upper solutions, tr(H)≤z*(t) on (s,tf]. Since series (1.10) is positive semidefinite, it follows that H and Hi are all bounded and (1.10) is convergent on (s,tf]. Since {Hi}i=1∞ satisfies system (1.4), each Hi is in fact continuously differentiable on [s,tf]. If s>t0, then the local existence theory implies that {Hi}i=1∞ can be extended further to the left of s, a contradiction to the maximality of (s,tf]. Therefore s=t0 and (1.4) has a solution on (t0,tf], which can be extended to [t0,tf].Now consider system (1.4) and (1.5). We haveTheorem 2.4.
DenoteR̂=BR-1BT. Let  a2,b2, and  c2 be functions on  [t0,tf] satisfying
(2.22)a2≥-1nλ1(R̂)+2ρtr⁡(W),b2≥2λn(A),c2≥tr⁡(Q).(a) Suppose K and {Hi}i=1∞ are solutions of (1.4) and (1.5) on some (s,tf]. Then on (s,tf],z=tr(H) satisfies
(2.23)-z′≤a2z2+b2z+c2,z(tf)=tr⁡(Qf).(b) Suppose the equation
(2.24)-z′=a2z2+b2z+c2,z(tf)=tr⁡(Qf)
has a solution on [t0,tf], then system (1.4) and (1.5) have solutions K and {Hi}i=1∞ on [t0,tf].Proof.
(a)  SubstitutingK=-R-1BTH into system (2.17) we get
(2.25)-H′=ATH+HA-HR̂H+Q+2∑i=2∞μi∑j=1i-1HjWHi-j,
where R̂=BR-1BT. Taking traces of both sides of (2.25) gives
(2.26)-tr⁡(H′)=tr⁡(ATH+HA)-tr⁡(HR̂H)+tr⁡(Q)+2∑i=2∞μi∑j=1i-1tr⁡(HjWHi-j).
As in the proof of previous theorem, we have(2.27)tr⁡(ATH+HA)=2tr⁡(AH)≤2λn(A)tr⁡(H)≤b2z,2∑i=2∞μi∑j=1i-1tr⁡(HjWHi-j)≤2ρ(μ)tr⁡(W)z2.
Using the fact that H,R̂≥0and (2.8), we have
(2.28)tr⁡(HR̂H)=tr⁡(R̂H2)≥λ1(R̂)tr⁡(H)2n.
By combining (2.28), (2.27), and (2.26) we obtain (2.23).
(b) Suppose that (2.24) has solution z*(t) on [t0,tf]. By local existence theory, system (1.5) and (1.4) have solutions K and {Hi}i=1∞ on a maximal interval (s,tf]⊂[t0,tf]. By part (a), tr(H) satisfies inequality (2.23) on (s,tf]. It follows that tr(H)≤z* on (s,tf], which implies that K and {Hi}i=1∞ are continuously differentiable on [s,tf].If s>t0, then local existence theory implies that K and {Hi}i=1∞ can be extended further to the left of s, a contradiction to the maximality of (s,tf]. Therefore s=t0, and system (1.4) and (1.5) have solutions on [t0,tf].


## 3. Well-Posedness and Sufficient Existence Conditions

In this section we will derive specific conditions that ensure that the scalar equations (2.16) and (2.24) have solutions on [t0,tf]. Consequently we will obtain sufficient conditions for the well-posedness of the cumulant control and the existence of solutions of (1.4) and (1.5).First we consider an autonomous scalar equation(3.1)z′+h(z)=0,t≤tf;z(tf)=zf,
where z is a polynomial with degree ≥2. Assume for some k≥0 that h(z) has k distinct zeros z1<⋯<zk. Let z0=-∞ and zk+1=∞. Since h(z) is locally Lipschitz, the solution z(t) of (3.1) exists and is unique for every zf for t in a maximal interval, say (σ,tf]. If zf=zi for some i=1,…,k, then z(t)=zi for all t. If zf∈(zi,zi+1) for some i=0,2,…,k, then z(t)∈(zi,zi+1) for t∈(σ,tf]. This implies that for t∈(σ,tf], -z′(t)=h(z(t)) has the same sign as h(zf).In particular, as t∈(σ,tf] decreases, z(t) is strictly increasing if h(zf)>0 and decreasing if h(zf)<0. Denote ẑf=limt→σ+z(t). Then
(3.2)ẑf={zi+1ifh(zf)>0,ziifh(zf)<0.The following is a well-known fact in stability theory of differential equations.(3.3)Ifh′(zi)<0,thenẑf=ziforeveryzf∈(zi-1,zi+1),
where ′=d/dz. Indeed, h′(zi)<0 implies that h(zf)<0 for zf∈(zi,zi+1) and h(zf)>0 for zf∈(zi-1,zi). In either cases, ẑf=zi by (3.2).Consider (1.12) as an example. We have the following.Proposition 3.1.
DenoteΔ=b2-4ac and  z1,2=(-b±Δ)/2a if  Δ≥0. Then
(a)ẑf=-c/b for all  zf if  a=0 and  b<0;
(b)ẑf=z2if  Δ≥0 and  a(zf-z1)<0;
(c)ẑf=sgn(a)∞ if  Δ<0 or  Δ≥0 and  a(zf-z1)>0.Proof.
Ifa=0, then h(z)=bz+c, which has root z1=-c/b. Since h′(z1)=b<0, ẑf=z1 by (3.3). This shows (a).
Next we prove (b) and (c). First assumea=1. If Δ<0, then h(z)>0 for all z. So ẑf=∞ by (3.2). Next consider the case Δ≥0, in which -∞<z2≤z1<∞. If zf>z1, then h(zf)>0, which implies that  ẑf=∞. If zf<z1, then we have either h(zf)>0 when zf<z2 or h(zf)<0 when z2<zf<z1. In either cases, we have ẑf=∞ by (3.2). This finishes the proof of (b) and (c) when a=1. If a≠1, then consider y=az, which satisfies y′+y2+by+ac=0 and y(tf)=azf, and the conclusions follow from the special case just proved.Write (3.1) as dz/h(z)+dt=0 and integrate it against tfrom tf to σ, then we get
(3.4)tf-σ=∫zfẑfdzh(z).Note that ifẑf is finite, then ẑf must be a zero of h(z). It follows that ∫zfẑf[1/h(z)]dz=∞ and (σ,tf]=(-∞,tf]. If ẑf is infinite, then ∫zfẑf(dz/(h(z))) must converge because h(z) has a degree ≥2. In summary, we have the following.Proposition 3.2.
Supposeh(z) is a polynomial of degree  ≥2. Then
(a)ẑf is finite if and only if the solution  z(t) of (3.1) exists on  (-∞,tf].
(b)ẑf=±∞ if and only the solution  z(t) of (3.1) exists on a finite maximal interval  (σ,tf] with length  tf-σ=∫zf±∞[1/h(z)]dz.Applying Proposition3.2 to (1.12) we obtain the following.Proposition 3.3.
(a)  If eithera=0 and  b<0, or  a(zf-z1)<0 and  Δ≥0, then the solution of (1.12) exists on  (-∞,tf].
(b)  If eitherΔ<0 or Δ≥0 and a(zf-z1)<0, then the solution of (1.12) exists on a finite interval (σ,tf] with
(3.5)tf-σ={2|Δ|(π2-arctan2azf+b|Δ|)ifΔ<0,22azf+bifΔ=0,a(zf-z1)>0,1Δln⁡|zf-z2zf-z1|ifΔ>0,a(zf-z1)>0.Proof.
Part (a) directly follows from Proposition3.1(a) and Proposition 3.2 (a) (b).
For part (b), first assume thatΔ<0. Then ẑf=sgn(a)∞ and az2+bz+c=a[(z-z1)2+d2], where d=|Δ|/2|a|,z1=-b/2a. So
(3.6)tf-σ=∫zfẑfdzaz2+bz+c=1adarctanz-z1d|zf±∞=1ad(±π2-arctanzf-z1d)=2|Δ|(π2-arctan2azf+b|Δ|).
Next assumeΔ=0 and a(zf-z1)>0. Then ẑf=sgn(a)∞ and az2+bz+c=a(z-z1)2, where z1=-b/(2a). We have
(3.7)tf-σ=∫zf±∞1a(z-z1)2dz=-1a(z-z1)|zf±∞=1a(zf-z1)=22azf+b.
Finally whenΔ>0 and ẑf=sgn(a)∞, we have az2+bz+c=a(z-z1)(z-z2) and
(3.8)tf-σ=∫zf±∞1a(z-z1)(z-z2)dz=1a(z1-z2)ln⁡|z-z1z-z2|zf±∞=1Δln⁡|zf-z2zf-z1|.Now we show that the cumulant control problem is well posed by Theorem2.3 and Proposition 3.3.Theorem 3.4.
For any numberL>0 there is  ρ0>0 such that the series  ∑i=1∞μiκi/i! in (1.3) converges for each matrix  K and sequence  μ with  ||K||∞<L and  ρ(μ)<ρ0.Proof.
Suppose thatK is a matrix function with ||K||∞<L. Choose a1, b1, and c1 as follows:
(3.9)a1=2ρ(μ)∥tr⁡(W)∥∞,b1=2‖λn(A+BK)‖∞+1,c1=‖tr⁡(Q+KTRK)‖∞.
Note that c1≤∥tr(Q)∥∞+L2∥R∥∞, which depends only on L. In addition, a1→0 as ρ(μ)→0. It follows that when ρ(μ) is sufficiently small, h(z)=a1z2+b1z+c1 has two real roots
(3.10)z1,2=-b1±b12-4a1c12a1<0
with z2→-c1/b1 and z1→-∞ as ρ(μ)→0. In particular, since zf≥0, we have h(zf)>0 and so ẑf=∞. Proposition 3.3 implies that
(3.11)tf-σ=1b12-4a1c1ln⁡|zf-z2zf-z1|.
Sotf-σ→∞ as ρ(μ)→0. In particular, (2.16) has a solution z(t) on [t0,tf] when ρ(μ) is sufficiently small. By Theorem 2.3 (b), system (1.4) has a solution {Hi}i=1∞ such that H converges.
Finally we apply Proposition3.3 to (2.24) to give a sufficient existence condition for (1.4) and (1.5) and the cumulant control problem. Choose
(3.12)a2=max⁡t0≤t≤tf(-1nλ1(R̂)+2ρtr⁡(W)),b2=2max⁡t0≤t≤tf⁡λn(A),c2=∥tr⁡(Q)∥∞.Theorem 3.5.
System (1.4) and (1.5) have solutions  K and  {Hi}i=1∞ on  [t0,tf] if
(3.13)tf-t0≤∫zfẑf1a2z2+b2z+c2dz,
where ẑf is defined as in (3.2) with h(z)=a2z2+b2z+c2. In particular, system (1.4) and (1.5) have solutions K and {Hi}i=1∞ on [t0,tf] if one of the following holds.
(a)a2=-λ1(R̂)/n+2ρtr(W)<0.
(b)a2>0,  b2<0,  Δ=b22-4a2c2≥0 and  zf<z1, where  z1=(-b2+b22-4a2c2)/2a2.Proof.
The general conclusion follows directly from Proposition3.3 and Theorem 2.4. In the case (a), h(z)=0 has two roots z1<0<z2. Since zf=tr(Qf)≥0, a2(zf-z1)<0. In the case (b), h(z)=0 has two solutions z2≤z1. So a2(zf-z1)<0 also holds. The conclusion follows from Propositions 3.1 and 3.2 and Theorem 2.4.Note that in Theorem3.5 condition (a) holds if B has full rank (i.e., λ(R̂)>0) and ρ(μ) is sufficiently small, while condition (b) holds if the system in (1.1) is stable (i.e., b2<0) and the product ρ(μ)∥tr(Q)∥∞ is relatively small. The cumulant control problem has an optimal control under each of these conditions.As an existence theorem, Theorem3.5 gives one of the very few existence results for a Riccati differential system of infinitely many equations. In terms of the cumulant controls that lead to the system (3) and (4), Theorem 3.5 generalizes the corresponding results in [1, 5] for risk-sensitive controls (where μ={1,θ,θ2,…}) and in [2, 4] for finite cumulant controls (where μ has only finite nonzero components). Numerical examples for risk-sensitive and finite cumulant controls satisfying the conditions in Theorem 3.5 may be found in [3–6].


## 4. Conclusions

In general it is very difficult to determine the existence interval of a differential Riccati equation (or system). By the approach in this paper, we can at least give an estimate for the existence interval of the Riccati system. Such an estimate leads to sufficient conditions for the existence of solutions to the Riccati system and the cumulant control problem.



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