c Birkh¨
auser Verlag, Basel, 1998 NoDEA Nonlinear differ. equ. appl. 5 (1998) 491 – 515 1021-9722/98/040491-25 $ 1.50+0.20/0
Nonlinear Differential Equations and Applications NoDEA
Asymptotic solution of a singularly perturbed set of functional-differential equations of Riccati type encountered in the optimal control theory∗ Valery Y. GLIZER Department of Chemical Engineering, Technion-Israel Institute of Technology, 32000 Haifa, Israel
Abstract The boundary-value problem for the set of functional-differential equations with partial derivatives of Riccati type, associated with a singularly perturbed linear-quadratic optimal control problem with delay in state, is considered. The expression for a solution of the problem, which transforms it to the explicit singular perturbation form, is proposed. An asymptotic solution of this problem is constructed.
1
Introduction
One of the fundamental results of control theory is the solution of the linearquadratic optimal control problem with fixed initial and free terminal states. This solution reduces the control problem to solving either the matrix Riccati equation [1] (finite dimensional case) or the operator one [2]–[6] (infinite dimensional case). If the controled system is governed by a differential equation with delay in the state, the operator Riccati equation is reduced to a set of functional-differential equations with partial derivatives and boundary conditions [4]–[7]. The Riccati equation, associated with the optimal control problem of singularly perturbed system, is also singularly perturbed. The singularly perturbed matrix Riccati equation was well examined in many publications in the open literature (see, for instance, [8]– [10]). In the present paper we investigate the set of functional-differential equations of Riccati type, associated with a linear-quadratic optimal control problem for singularly perturbed system with delay in the state. Note that as far as it is known to the author, there are only few publications in the open literature which deal with the singularly perturbed linear-quadratic optimal control problem with delay in the state [11],[12]. ∗ The
research has been supported by the Giladi Fund, State of Israel.
492
Valery Y. Glizer
NoDEA
Consider the linear system with delay in state variables dx(t)/dt = A1 (t)x(t)+A2 (t)y(t)+H1 (t)x(t−εh)+H2 (t)y(t−εh)+B1 (t)u(t) (1.1) εdy(t)/dt = A3 (t)x(t)+A4 (t)y(t)+H3 (t)x(t−εh)+H4 (t)y(t−εh)+B2 (t)u(t) (1.2) x(τ ) = x0 (τ ),
y(τ ) = y 0 (τ ),
−εh ≤ τ ≤ 0
(1.3)
where x ∈ E n and y ∈ E m are state variables, u ∈ E r is a control, ε > 0 is a small parameter (ε << 1), h > 0 is some constant independent of ε, Ai , Hi , Bj , (i = 1, . . . , 4; j = 1, 2), are time-depending matrices of corresponding dimensions. The cost functional (to be minimized by selection of u(t)) is Z J=
T
0
0
0
0
[x (t)D1 (t)x(t)+2x (t)D2 (t)y(t)+y (t)D3 (t)y(t)+u (t)N (t)u(t)]dt (1.4) 0
where the prime denotes the transposition, T > 0 is a fixed instant of the time, Di , (i = 1, 2, 3), and N are time-depending matrices of corresponding dimensions. The objective of the present paper is to construct an asymptotic solution of the boundary-value problem for the set of functional-differential equations with partial derivatives associated with the optimal control problem (1.1)–(1.4). The paper is organized as follows. In Section 2 main assumptions are formulated and the set of functional-differential equations of Riccati type is written out. In Section 3 this set is transformed to the explicit singular perturbation form. In Section 4 an asymptotic solution of the obtained set of equations is constructed and its properties are established. In Section 5 a justification of the asymptotic solution is carried out.
2
Main assumptions and the set of functionaldifferential equations of Riccati type
In the sequal we shall consider the problem (1.1)–(1.4) under the following assumptions: 1) the matrices Ai , Hi , Bj , Dl , (i = 1, . . . , 4; j = 1, 2; l = 1, 2, 3), and N are continuously differentiable functions of t∈ [0, T ]; D1 (t) D2 (t) 0 2) the matrix D(t) = is symmetric and positive semidefinite D2 (t) D3 (t) ∀t ∈ [0, T ]; 3) the matrix N (t) is symmetric and positive definite ∀t ∈ [0, T ]; 4) rank[A4 (t) + H4 (t) exp(−hλ) − λIm , B2 (t)] = m ∀t ∈ [0, T ] and any complex number λ with Reλ ≥ 0, where Im is theidentity matrix of the dimension m; A4 (t) + H4 (t) exp(−hλ) − λIm 5) rank = m ∀t ∈ [0, T ] and any complex C(t) 0 number λ with Reλ ≥ 0, where C (t)C(t) = D3 (t).
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Asymptotic solution of equations of Riccati type
493
Let us introduce in the consideration the following matrices A1 (t) A2 (t) H1 (t) H2 (t) A(t, ε) = , H(t, ε) = A3 (t)/ε A4 (t)/ε H3 (t)/ε H4 (t)/ε
(2.1)
From results of [7], which are applicable to the problem (1.1)–(1.4) for all ε > 0, we obtain the optimal control u∗ and optimal value of the cost functional J ∗ ∗
u (t) = −N
−1
(t) ∗
B1 (t) B2 (t)/ε
0
0
[P (t)z(t) +
Q(t, τ )z(t + τ )dτ ],
0
0
0
Z
Z
0
x y
0
Q(0, τ )z 0 (τ )dτ
J = (z (0)) P (0)z (0) + 2(z (0)) Z
z=
−εh
0
0
Z
−εh 0
0
(z 0 (τ )) R(0, τ, ρ)z 0 (ρ)dτ dρ
+ −εh
−εh
0
0
where P (t) = P (t), R(t, τ, ρ) = R (t, ρ, τ ), and the matrices P, Q, R are a unique solution of the following set of functional-differential equations of Riccati type 0
dP (t)/dt = −P (t)A(t, ε) − A (t, ε)P (t) + P (t)S(t, ε)P (t) 0
−Q(t, 0) − Q (t, 0) − D(t)
(2.2)
0
(∂/∂t − ∂/∂τ )Q(t, τ ) = −[A(t, ε) − S(t, ε)P (t)] Q(t, τ ) − R(t, 0, τ ) 0
(∂/∂t − ∂/∂τ − ∂/∂ρ)R(t, τ, ρ) = Q (t, τ )SQ(t, ρ)
(2.3) (2.4)
satisfying the boundary conditions P (T ) = 0,
Q(T, τ ) = 0,
P (t)H(t, ε) = Q(t, −εh), 0 ≤ t ≤ T, Here
S(t, ε) =
3
B1 (t) B2 (t)/ε
R(T, τ, ρ) = 0
(2.5)
H (t, ε)Q(t, τ ) = R(t, −εh, τ )
(2.6)
0
−εh ≤ τ, ρ ≤ 0 N
−1
(t)
B1 (t) B2 (t)/ε
0 (2.7)
Explicit singular perturbation form of problem (2.2)–(2.6)
In order to transform the problem (2.2)–(2.6) to the explicit singular perturbation form, we shall seek the solution of this problem as follows P1 (t) εP2 (t) Q1 (t, τ ) Q2 (t, τ ) 0 P (t) = (3.1) , Q(t, τ ) = Q3 (t, τ ) Q4 (t, τ ) εP2 (t) εP3 (t)
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Valery Y. Glizer
R1 (t, τ, ρ) R2 (t, τ, ρ) 0 R2 (t, ρ, τ ) R3 (t, τ, ρ)
R(t, τ, ρ) = (1/ε)
NoDEA
(3.2)
where Pi (t) and Ri (t, τ, ρ), (i = 1, 2, 3), are matrices of dimensions n×n, n×m, m× m respectively; Qj (t, τ ), (j = 1, . . . , 4), are matrices of dimensions n×n, n×m, m× n, m × m respectively. Applying (2.1),(2.7),(3.1),(3.2), one can rewrite the problem (2.2)–(2.6) in the form 0
0
0
dP1 (t)/dt = −P1 (t)A1 (t) − A1 (t)P1 (t) − P2 (t)A3 (t) − A3 (t)P2 (t) 0
0
+P1 (t)S1 (t)P1 (t) + P1 (t)S2 (t)P2 (t) + P2 (t)S2 (t)P1 (t) 0
0
+P2 (t)S3 (t)P2 (t) − Q1 (t, 0) − Q1 (t, 0) − D1 (t)
(3.3)
0
0
εdP2 (t)/dt = −P1 (t)A2 (t) − P2 (t)A4 (t) − εA1 (t)P2 (t) − A3 (t)P3 (t) 0
+εP1 (t)S1 (t)P2 (t) + P1 (t)S2 (t)P3 (t) + εP2 (t)S2 (t)P2 (t) 0
+P2 (t)S3 (t)P3 (t) − Q2 (t, 0) − Q3 (t, 0) − D2 (t) 0
(3.4)
0
0
εdP3 (t)/dt = −εP2 (t)A2 (t) − εA2 (t)P2 (t) − P3 (t)A4 (t) − A4 (t)P3 (t) 0
0
0
+ε2 P2 (t)S1 (t)P2 (t) + εP2 (t)S2 (t)P3 (t) + εP3 (t)S2 (t)P2 (t) 0
+P3 (t)S3 (t)P3 (t) − Q4 (t, 0) − Q4 (t, 0) − D3 (t) 0
(3.5)
0
ε(∂/∂t − ∂/∂τ )Q1 (t, τ ) = −ε[A1 (t) − P1 (t)S1 (t) − P2 (t)S2 (t)]Q1 (t, τ ) 0
−[A3 (t) − P1 (t)S2 (t) − P2 (t)S3 (t)]Q3 (t, τ ) − R1 (t, 0, τ ) 0
(3.6)
0
ε(∂/∂t − ∂/∂τ )Q2 (t, τ ) = −ε[A1 (t) − P1 (t)S1 (t) − P2 (t)S2 (t)]Q2 (t, τ ) 0
−[A3 (t) − P1 (t)S2 (t) − P2 (t)S3 (t)]Q4 (t, τ ) − R2 (t, 0, τ ) 0
0
(3.7)
0
ε(∂/∂t − ∂/∂τ )Q3 (t, τ ) = −ε[A2 (t) − εP2 (t)S1 (t) − P3 (t)S2 (t)]Q1 (t, τ ) 0
0
0
−[A4 (t) − εP2 (t)S2 (t) − P3 (t)S3 (t)]Q3 (t, τ ) − R2 (t, τ, 0) 0
0
(3.8)
0
ε(∂/∂t − ∂/∂τ )Q4 (t, τ ) = −ε[A2 (t) − εP2 (t)S1 (t) − P3 (t)S2 (t)]Q2 (t, τ ) 0
0
−[A4 (t) − εP2 (t)S2 (t) − P3 (t)S3 (t)]Q4 (t, τ ) − R3 (t, 0, τ )
(3.9)
0
ε(∂/∂t − ∂/∂τ − ∂/∂ρ)R1 (t, τ, ρ) = ε2 Q1 (t, τ )S1 (t)Q1 (t, ρ) 0
0
0
0
+εQ3 (t, τ )S2 (t)Q1 (t, ρ) + εQ1 (t, τ )S2 (t)Q3 (t, ρ) + Q3 (t, τ )S3 (t)Q3 (t, ρ)
(3.10)
0
ε(∂/∂t − ∂/∂τ − ∂/∂ρ)R2 (t, τ, ρ) = ε2 Q1 (t, τ )S1 (t)Q2 (t, ρ) 0
0
0
0
+εQ3 (t, τ )S2 (t)Q2 (t, ρ) + εQ1 (t, τ )S2 (t)Q4 (t, ρ) + Q3 (t, τ )S3 (t)Q4 (t, ρ) 0
ε(∂/∂t − ∂/∂τ − ∂/∂ρ)R3 (t, τ, ρ) = ε2 Q2 (t, τ )S1 (t)Q2 (t, ρ)
(3.11)
Vol. 5, 1998
Asymptotic solution of equations of Riccati type
0
0
0
495
0
+εQ4 (t, τ )S2 (t)Q2 (t, ρ) + εQ2 (t, τ )S2 (t)Q4 (t, ρ) + Q4 (t, τ )S3 (t)Q4 (t, ρ)
(3.12)
Pi (T ) = 0, Qj (T, τ ) = 0, Ri (T, τ, ρ) = 0, (i = 1, 2, 3; j = 1, . . . , 4)
(3.13)
P1 (t)Hk (t) + P2 (t)Hk+2 (t) = Qk (t, −εh), 0
εP2 (t)Hl−2 (t) + P3 (t)Hl (t) = Ql (t, −εh), 0
(k = 1, 2)
(3.14)
(l = 3, 4)
(3.15)
0
εH1 (t)Qk (t, τ ) + H3 (t)Qk+2 (t, τ ) = Rk (t, −εh, τ ), 0
0
0
0
(k = 1, 2)
(3.16)
εQ1 (t, τ )H2 (t) + Q3 (t, τ )H4 (t) = R2 (t, τ, −εh)
(3.17)
εH2 (t)Q2 (t, τ ) + H4 (t)Q4 (t, τ ) = R3 (t, −εh, τ ) 0 ≤ t ≤ T,
(3.18)
−εh ≤ τ, ρ ≤ 0
0
0
Here S1 (t) = B1 (t)N −1 (t)B1 (t), S2 (t) = B1 (t)N −1 (t)B2 (t), 0 S3 (t) = B2 (t)N −1 (t)B2 (t).
4
Zero-order asymptotic solution of the problem (3.3)–(3.18)
We shall seek the zero-order asymptotic solution Pi0 (t, ε), Qj0 (t, τ, ε), Ri0 (t, τ, ρ, ε), (i = 1, 2, 3; j = 1, . . . , 4), of (3.3)–(3.18) in the form t P10 (t, ε) = P¯10 (t), Pk0 (t, ε) = P¯k0 (t) + Pk0 (ξ), (k = 2, 3), ξ = (t − T )/ε (4.1)
Qj0 (t, τ, ε) = Qτj0 (t, η) + Qt,τ j0 (ξ, η), (j = 1, . . . , 4), η = τ /ε
(4.2)
τ,ρ t,τ,ρ (t, η, χ) + Ri0 (ξ, η, χ), (i = 1, 2, 3), χ = ρ/ε Ri0 (t, τ, ρ, ε) = Ri0
(4.3)
Equations and conditions for the asymptotic solution are obtained by substituting (4.1)–(4.3) into (3.3)–(3.18) and equating terms of zero power of ε on both sides of the equations, separately terms depending on (t, η, χ) and terms depending on (ξ, η, χ). 4.1 Terms depending on (t, η, χ). For the terms, depending on (t, η, χ), we have 0
0
0
dP¯10 (t)/dt = −P¯10 (t)A1 (t) − A1 (t)P¯10 (t) − P¯20 (t)A3 (t) − A3 (t)P¯20 (t) 0
0
+P¯10 (t)S1 (t)P¯10 (t) + P¯10 (t)S2 (t)P¯20 (t) + P¯20 (t)S2 (t)P¯10 (t) 0 0 +P¯20 (t)S3 (t)P¯20 (t) − Qτ10 (t, 0) − [Qτ10 (t, 0)] − D1 (t), P¯10 (T ) = 0
(4.4)
0
0 = −P¯10 (t)A2 (t) − P¯20 (t)A4 (t) − A3 (t)P¯30 (t) + P¯10 (t)S2 (t)P¯30 (t) 0 +P¯20 (t)S3 (t)P¯30 (t) − Qτ20 (t, 0) − [Qτ30 (t, 0)] − D2 (t) 0
0 = −P¯30 (t)A4 (t) − A4 (t)P¯30 (t) + P¯30 (t)S3 (t)P¯30 (t)
(4.5)
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Valery Y. Glizer
NoDEA
0
−Qτ40 (t, 0) − [Qτ40 (t, 0)] − D3 (t)
(4.6)
0 τ,ρ ∂Qτ10 (t, η)/∂η = [A3 (t) − P¯10 (t)S2 (t) − P¯20 (t)S3 (t)]Qτ30 (t, η) + R10 (t, 0, η) (4.7) 0 τ,ρ ∂Qτ20 (t, η)/∂η = [A3 (t) − P¯10 (t)S2 (t) − P¯20 (t)S3 (t)]Qτ40 (t, η) + R20 (t, 0, η) (4.8) 0 0 τ,ρ (t, η, 0)] ∂Qτ30 (t, η)/∂η = [A4 (t) − P¯30 (t)S3 (t)]Qτ30 (t, η) + [R20 0
τ,ρ ∂Qτ40 (t, η)/∂η = [A4 (t) − P¯30 (t)S3 (t)]Qτ40 (t, η) + R30 (t, 0, η) 0
τ,ρ (∂/∂η + ∂/∂χ)R10 (t, η, χ) = −[Qτ30 (t, η)] S3 (t)Qτ30 (t, χ) 0
τ,ρ (t, η, χ) = −[Qτ30 (t, η)] S3 (t)Qτ40 (t, χ) (∂/∂η + ∂/∂χ)R20 0
τ,ρ (∂/∂η + ∂/∂χ)R30 (t, η, χ) = −[Qτ40 (t, η)] S3 (t)Qτ40 (t, χ)
P¯10 (t)Hk (t) + P¯20 (t)Hk+2 (t) = Qτk0 (t, −h), P¯30 (t)Hl (t) = Qτl0 (t, −h), 0
(k = 1, 2)
(l = 3, 4)
τ,ρ H3 (t)Qτk+2,0 (t, η) = Rk0 (t, −h, η),
(k = 1, 2)
0
τ,ρ [Qτ30 (t, η)] H4 (t) = R20 (t, η, −h) 0
τ,ρ H4 (t)Qτ40 (t, η) = R30 (t, −h, η)
(4.9) (4.10) (4.11) (4.12) (4.13) (4.14) (4.15) (4.16) (4.17) (4.18)
Note 4.1 The problem ((4.4)–(4.18) can be partitioned into four simpler problems solved successively. The First Problem consists of (4.6),(4.10),(4.13),(4.15) with l = 4, (4.18); the Second Problem consists of (4.9),(4.12),(4.15) with l = 3, (4.16) with k = 2, (4.17); the Third Problem consists of (4.11),(4.16) with k = 1; the Fourth Problem consists of (4.4),(4.5),(4.7),(4.8)(4.14). Lemma 4.1 The First Problem (see Note 4.1) has a unique solution P¯30 (t), τ,ρ Qτ40 (t, η), R30 (t, η, χ), t ∈ [0, T ], η ∈ [−h, 0], χ ∈ [−h, 0] such that the matrix
P¯30 (t) Qτ40 (t, χ) 0 τ,ρ (t, η, χ) (Qτ40 (t, η)) R30
defines a linear bounded self-adjoint nonnegative operator mapping the space E m × L2 [−h, 0; E m ] into itself. Moreover, all roots λ of the equation det[λIm − A4 (t) + S3 (t)P¯30 (t) − H4 (t) exp(−λh) Z +S3 (t)
0 −h
Qτ40 (t, η) exp(λη)dη] = 0
lie inside the left-hand half-plane for all t ∈ [0, T ].
Vol. 5, 1998
Asymptotic solution of equations of Riccati type
497
Proof. Let us consider the following optimal control problem d˜ y (ζ)/dζ = A4 (t)˜ y (ζ) + H4 (t)˜ y (ζ − h) + B2 (t)˜ u(ζ), y˜(θ) = y˜0 (θ), −h ≤ θ ≤ 0 (4.19) Z +∞ 0 0 [˜ y (ζ)D3 (t)˜ y (ζ) + u ˜ (ζ)N (t)˜ u(ζ)]dζ → min (4.20) J˜ = u ˜
0
where t is a parameter. It is obvious that the First Problem is associated with the control problem (4.19),(4.20) by the optimality conditions for each t ∈ [0, T ] [4],[6]. Now, the statements of the lemma directly follow from the assumptions 2)–5), results of [4],[6] on properties of solutions of the operator Riccati equation and closed-loop system, and results of [6],[13],[14] on conditions for stabilizability and detectability in linear functional-differential equations. Lemma 4.2 The Second Problem and the Third Problem (see Note 4.1) have τ,ρ τ,ρ unique solutions Qτ30 (t, η), R20 (t, η, χ) and R10 (t, η, χ), respectively, for t ∈ [0, T ], η ∈ [−h, 0], χ ∈ [−h, 0]. Proof. Rewriting the problem (4.12),(4.16) with k = 2, (4.17) in an equivalent form, one has Z η 0 τ,ρ R20 (t, η, χ) = Φ(t, η − χ) − [Qτ30 (t, s)] S3 (t)Qτ40 (t, s − η + χ)ds max(η−χ−h,−h)
(4.21) where
Φ(t, σ) =
0
H3 (t)Qτ40 (t, −σ − h), −h ≤ σ ≤ 0 0 (Qτ30 (t, σ − h)) H4 (t), 0<σ≤h
(4.22)
From (4.15) we immediately obtain that Φ(t, σ) is continuous at σ = 0 ∀t ∈ [0, T ]. Substituting (4.21),(4.22) into (4.9) leads to the linear integral-differential equation for Qτ30 (t, η) 0 ∂Qτ30 (t, η)/∂η = [A4 (t) − P¯30 (t)S3 (t)]Qτ30 (t, η)
+[Qτ40 (t, −η
0
Z
− h)] H3 (t) −
η
−h
0
[Qτ40 (t, s − η)] S3 (t)Qτ30 (t, s)ds
(4.23)
Equation (4.23) with the initial condition (4.15) (l = 3) has a unique solution for t ∈ [0, T ], η ∈ [−h, 0], which completes the proof of the lemma’s statement concerning the Second Problem. The existence and the uniqueness of solution of the Third Problem is obvious. Moreover, this solution has the form Z η 0 τ,ρ R10 (t, η, χ) = Ψ(t, η − χ) − [Qτ30 (t, s)] S3 (t)Qτ30 (t, s − η + χ)ds max(η−χ−h,−h)
(4.24)
498
Valery Y. Glizer
where
NoDEA
0
H3 (t)Qτ30 (t, −σ − h), −h ≤ σ ≤ 0 0 0<σ≤h (Qτ30 (t, σ − h)) H3 (t),
Ψ(t, σ) =
(4.25)
So, the lemma is completely proved. Now let us proceed to the analysis of the Fourth Problem (see Note 4.1). Equations (4.7),(4.8),(4.14) can be rewritten in an equivalent form as follows 0 Qτk0 (t, η) = P¯10 (t)Hk (t) + P¯20 (t)Hk+2 (t) + [A3 (t) − P¯10 (t)S2 (t)
Z −P¯20 (t)S3 (t)]
Z
η −h
Qτk+2,0 (t, η)dη
η
+ −h
τ,ρ Rk0 (t, 0, η)dη, (k = 1, 2)
(4.26)
Substituting (4.26) (k = 2) into equation (4.5) and solving it relative P¯20 (t), one has P¯20 (t) = −[P¯10 (t)L1 (t) + L2 (t)] (4.27) where
L1 (t) = [A2 (t) + H2 (t) − S2 (t)G(t)]M −1 (t) Z 0 0 0 τ,ρ R20 (t, 0, η)dη + [Qτ30 (t, 0)] + D2 (t)}M −1 (t) L2 (t) = {A3 (t)G(t) + −h
Z
M (t) = A4 (t) + H4 (t) − S3 (t)G(t), G(t) = P¯30 (t) +
(4.28) (4.29)
0
−h
Qτ40 (t, η)dη
(4.30)
From Lemma 4.1 we directly obtain that the matrix M (t) is invertible for all t ∈ [0, T ]. Proposition 4.1 The following result holds: Z
0
−h
τ,ρ R20 (t, 0, η)dη
+
0 [Qτ30 (t, 0)]
Z
0
= H3 (t)G(t) +
0
−h
0
[Qτ30 (t, η)] dηM (t), t ∈ [0, T ]
Proof. Integrating (4.9) from η = −h to η = 0 and taking into account (4.15) (l = 3), one has Z
0
Qτ30 (t, 0) − P¯30 (t)H3 (t) = [A4 (t) − P¯30 (t)S3 (t)] Z
0
+ −h
0 −h
Qτ30 (t, η)dη
0
τ,ρ [R20 (t, η, 0)] dη
(4.31)
Using (4.12), (4.16) (k = 2), (4.17), we obtain Z
0
−h
0
τ,ρ [R20 (t, η, 0)] dη =
Z
0
−h
0
τ,ρ [R20 (t, η, −h)] dη +
Z
0
−h
Z
0
−h
0
τ,ρ [∂R20 (t, η, χ)/∂χ] dχdη
Vol. 5, 1998
Asymptotic solution of equations of Riccati type
Z = [H4 (t) − S3 (t)
0 −h
0
Q40 (t, χ)dχ] Z
0
Z
499
Z
0
−h
Qτ30 (t, η)dη
−
0
−h
0
τ,ρ [R20 (t, 0, χ)] dχ
0
+ −h
[Qτ40 (t, χ)] dχH3 (t)
(4.32)
Substituting (4.32) into (4.31) leads after some rearrangement to the statement of the proposition. Applying Proposition 4.1 to equation (4.29), one has Z L2 (t) = L3 (t) +
0
−h
0
0
[Qτ30 (t, η)] dη, L3 (t) = {[A3 (t) + H3 (t)] G(t) + D2 (t)}M −1 (t) (4.33)
Proposition 4.2 The following result holds: Z
−h
Z
0
+ −h
Z
0 τ,ρ R10 (t, 0, η)dη
0 [Qτ30 (t, η)] dηH3 (t)
0
+ −h
Z −
0
−h
τ,ρ R10 (t, η, 0)dη
Z
0
= H3 (t)
0 [Qτ30 (t, η)] dηS3 (t)
Z
0 −h
Qτ30 (t, η)dη
0 −h
Qτ30 (t, η)dη, t ∈ [0, T ]
Proof. The statement of the proposition is obtained similarly to equation (4.32) using (4.11),(4.16) (k = 1). Substituting (4.26) (k = 1), (4.27) into (4.4) and applying (4.33) and Proposition 4.2, we obtain after some rearrangement the following problem for P¯10 (t) ¯ − A¯0 (t)P¯10 (t) + P¯10 (t)S(t) ¯ P¯10 (t) − D(t), ¯ P¯10 (T ) = 0 dP¯10 (t)/dt = −P¯10 (t)A(t) (4.34) where ¯ = A1 (t) + H1 (t) − L1 (t)[A3 (t) + H3 (t)] + S2 (t)L03 (t) − L1 (t)S3 (t)L03 (t) A(t) −1 ¯ = B(t)N ¯ ¯ 0 (t), B(t) ¯ = B1 (t) − L1 (t)B2 (t) S(t) (t)B 0 0 0 ¯ D(t) = D1 (t) − L3 (t)[A3 (t) + H3 (t)] − [A3 (t) + H3 (t)] L3 (t) − L3 (t)S3 (t)L3 (t)
Let us assume that: ¯ 6) the matrix D(t) is positive semidefinite for all t ∈ [0, T ]. This assumption provides the existence and uniqueness of solution of the problem (4.34) [1]. So, we have proved the following lemma. Lemma 4.3 The Fourth Problem (see Note 4.1) has a unique solution P¯10 (t), P¯20 (t), Qτ10 (t, η), Qτ20 (t, η), t ∈ [0, T ], η ∈ [−h, 0].
500
Valery Y. Glizer
NoDEA
Note 4.2 Let us present two important cases when the assumption 6) is valid. a) D2 (t) = 0, D3 (t) = 0 t ∈ [0, T ]. τ,ρ In this case P¯30 (t) = 0, Qτ40 (t, η) = 0, R30 (t, η, χ) = 0 t ∈ [0, T ], η ∈ [−h, 0], ¯ χ ∈ [−h, 0] which leads to D(t) = D1 (t) ∀t ∈ [0, T ]. b) The matrix G(t) is nonsingular for all t ∈ [0, T ]. ¯ In this case the positive semidefiniteness of the matrix D(t) is shown similarly to the finite dimensional case [8] using that the matrix G(t) satisfies the following equation 0
0
0
−G (t)[A4 (t) + H4 (t)] − [A4 (t) + H4 (t)] G(t) + G (t)S3 (t)G(t) − D3 (t) = 0 Proposition 4.3 Let A(µ), H(µ), Q(η, µ) be matrices of dimension m × m depending on µ and η. Let A(µ), H(µ) be continuous in µ at µ = 0 and Q(η, µ) be continuous in η ∈ [−h, 0] for all sufficiently small |µ| and continuous in µ at µ = 0 uniformly in η ∈ [−h, 0]. Let for µ = 0 all roots λ of the equation Z det[λIm − A(µ) − H(µ) exp(−λh) −
0
−h
Q(η, µ) exp(λη)dη] = 0
(4.35)
lie inside the left-hand half-plane. Then for all sufficiently small |µ| > 0 all roots λ(µ) of this equation lie inside the left-hand half-plane. Proof (by contradiction). Let the statement of the proposition be wrong. Then there exists a sequence {µq }, (q = 1, 2, . . .), such that limq→+∞ µq = 0 and Reλ(µq ) ≥ 0. Taking into account the well-known property of solutions of the characteristic equation for a linear functional-differential equation [15], one can show that the sequence {λ(µq )} is bounded. Hence, there exists its subsequence ¯ and Reλ ¯ ≥ 0. The corresponding subsequence of {µq } tends to which has a limit λ zero. For the sake of simplicity we retain the same notation for the subsequences as for the sequences. Substituting µ = µq into (4.35) and computing its limit as ¯ is a root of (4.35) for µ = 0. This contradicts to q → +∞, one obtains that λ the assumption of the proposition that for µ = 0 all roots of (4.35) lie inside the left-hand half-plane. This contradiction proves the proposition. τ,ρ Lemma 4.4 The matrices P¯30 (t), Qτ40 (t, η), R30 (t, η, χ) are continuous functions of (t, η, χ) ∈ [0, T ] × [−h, 0] × [−h, 0]. τ,ρ Proof. First, let us note that the matrices P¯30 (t), Qτ40 (t, η), R30 (t, η, χ) are the solution of the First Problem and, consequently, they are continuous in (η, χ) ∈ [−h, 0] × [−h, 0] for each t ∈ [0, T ], [4]. Now we shall show that these matrices are continuous in t ∈ [0, T ] uniformly in (η, χ) ∈ [−h, 0] × [−h, 0]. Let t0 be an arbitrary but fixed point on the segment [0, T ] and ∆t 6= 0 is an arbitrary number such that t0 + ∆t ∈ [0, T ]. Let us denote
∆A4 = A4 (t0 + ∆t) − A4 (t0 ), ∆S3 = S3 (t0 + ∆t) − S3 (t0 ), ∆D3 = D3 (t0 + ∆t)
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Asymptotic solution of equations of Riccati type
501
−D3 (t0 ), ∆P¯30 = P¯30 (t0 + ∆t)− P¯30 (t0 ), ∆Qτ40 (η) = Qτ40 (t0 + ∆t, η)− Qτ40 (t0 , η), τ,ρ τ,ρ τ,ρ ∆R30 (η, χ) = R30 (t0 + ∆t, η, χ) − R30 (t0 , η, χ)
Applying the First Problem at t = t0 and t = t0 + ∆t, one has the following τ,ρ problem for ∆P¯30 , ∆Qτ40 (η), ∆R30 (η, χ) 0
0
−∆P¯30 α(t0 ) − α (t0 )∆P¯30 − ∆Qτ40 (η) − [∆Qτ40 (η)] − δP (∆P¯30 ) = 0
(4.36)
0 τ,ρ d∆Qτ40 (η)/dη = α (t0 )∆Qτ40 (η) + ∆P¯30 β(t0 , η) + ∆R30 (0, η) + δQ (∆P¯30 , ∆Qτ40 (η)) (4.37) 0 0 τ,ρ (η, χ) = [∆Qτ40 (η)] β(t0 , χ) + β (t0 , η)∆Qτ40 (χ) (∂/∂η + ∂/∂χ)∆R30
+δR (∆Qτ40 (η), ∆Qτ40 (χ)) ∆P¯30 H4 (t0 ) = ∆Qτ40 (−h),
(4.38) τ,ρ ∆R30 (−h, η)
(4.39)
β(t, η) = −S3 (t)Qτ40 (t, η)
(4.40)
H4 (t0 )∆Qτ40 (η)
=
where α(t) = A4 (t) − S3 (t)P¯30 (t),
0
δP (∆P¯30 ) = [P¯30 (t0 ) + ∆P¯30 ]∆A4 + ∆A4 [P¯30 (t0 ) + ∆P¯30 ] −[P¯30 (t0 ) + ∆P¯30 ]∆S3 [P¯30 (t0 ) + ∆P¯30 ] − ∆P¯30 S3 (t0 )∆P¯30 + ∆D3 0 δQ (∆P¯30 , ∆Qτ40 (η)) = {∆A4 − ∆S3 [P¯30 (t0 ) + ∆P¯30 ]} [Qτ40 (t0 , η) + ∆Qτ40 (η)] −∆P¯30 S3 (t0 )∆Qτ40 (η) 0
δR (∆Qτ40 (η), ∆Qτ40 (χ)) = −[Qτ40 (t0 , η) + ∆Qτ40 (η)] ∆S3 [Qτ40 (t0 , χ) + ∆Qτ40 (χ)] 0
−[∆Qτ40 (η)] S3 (t0 )∆Qτ40 (χ) Applying techniques of [4],[7] and taking into account Lemma 4.1 and (4.40), we can rewrite the problem (4.36)–(4.39) in the equivalent form Z +∞ 0 [K (t0 , s)δP (∆P¯30 )K(t0 , s) ∆P¯30 = 0
Z
0
+ −h
Z
0
+ Z
0
Z
−h 0
+ −h
−h
0 K (t0 , s)δQ (∆P¯30 , ∆Qτ40 (η))K(t0 , s + η)dη
0 0 K (t0 , s + η)δQ (∆P¯30 , ∆Qτ40 (η))K(t0 , s)dη
0
K (t0 , s + η)δR (∆Qτ40 (η), ∆Qτ40 (χ))K(t0 , s + χ)dηdχ]ds Z ∆Qτ40 (η) Z
=
+∞
0 ˜ 0 , s, η) [K (t0 , s)δP (∆P¯30 )K(t
0 0
+ −h
0
˜ 0 , s + χ, η)dχ K (t0 , s)δQ (∆P¯30 , ∆Qτ40 (χ))K(t
(4.41)
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Valery Y. Glizer
Z
0
+ Z
0
Z
−h 0
+ −h
−h
0 0 ˜ 0 , s, η)dχ K (t0 , s + χ)δQ (∆P¯30 , ∆Qτ40 (χ))K(t
0 ˜ 0 , s + χ1 , η)dχdχ1 ]ds K (t0 , s + χ)δR (∆Qτ40 (χ), ∆Qτ40 (χ1 ))K(t
Z
η+h
0
Z
0
+ −h
0
K (t0 , s + χ)δR (∆Qτ40 (χ), ∆Qτ40 (η − s))dχ]ds Z
τ,ρ ∆R30 = 0
+ −h
Z
0
+ Z
−h 0
+ −h
−h
+∞
(4.42)
˜ 0 (t0 , s, η)δP (∆P¯30 )K(t ˜ 0 , s, χ) [K
0
Z
0
0
[K (t0 , s)δQ (∆P¯30 , ∆Qτ40 (η − s))
+
Z
NoDEA
˜ 0 (t0 , s, η)δQ (∆P¯30 , ∆Q40 (χ1 ))K(t ˜ 0 , s + χ1 , χ)dχ1 K 0 ˜ 0 , s, χ)dχ1 ˜ 0 (t0 , s + χ1 , η)δQ (∆P¯30 , ∆Qτ40 (χ1 ))K(t K
˜ 0 (t0 , s + χ1 , η)δR (∆Qτ40 (χ1 ), ∆Qτ40 (χ2 ))K(t ˜ 0 , s + χ2 , χ)dχ1 dχ2 ]ds K Z
η+h
Z
0 0
+ −h
˜ 0 , s + χ1 , χ)dχ1 ]ds δR (∆Q40 (χ1 ), ∆Q40 (η − s))K(t Z
χ+h
+ Z
0
˜ 0 , s, χ) [δQ (∆P¯30 , ∆Qτ40 (η − s))K(t
+
˜ 0 (t0 , s, η)δQ (∆P¯30 , ∆Qτ40 (χ − s)) [K
0 0
+ −h
˜ 0 (t0 , s + χ1 , η)δR (∆Qτ40 (χ1 ), ∆Qτ40 (χ − s))dχ1 ]ds K
Z
min(η+h,χ+h)
δR (∆Qτ40 (η − s), ∆Qτ40 (χ − s))ds
+
(4.43)
0
where K(t, s) is a solution of the following problem Z 0 ∂K(t, s)/∂s = α(t)K(t, s) + H4 (t)K(t, s − h) + β(t, η)K(t, s + η)dη, s > 0 −h
K(t, 0) = Im ,
K(t, s) = 0 ∀s < 0
(4.44) (4.45)
˜ s, η) is defined as follows and K(t, Z ˜ s, η) = K(t, s − η − h)H4 (t) + K(t,
h
−η
K(t, s − η − χ)β(t, −χ)dχ
(4.46)
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503
Let us denote ν = max(k∆A4 k, k∆S3 k, k∆D3 k), where k · k denotes the Euclidean norm of matrix. Now, applying the procedure of successive approximations to the set (4.41)–(4.43) and taking into account the assumptions 1), 3), one can show that for all sufficiently small |∆t| there exists a unique solution of this set, such that τ,ρ max[k∆P¯30 k, k∆Qτ40 (η)k, k∆R30 (η, χ)k] ≤ cν ∀(η, χ) ∈ [−h, 0] × [−h, 0] (4.47)
where c > 0 is some constant independent of ν. Since (4.41)–(4.43) is a set of nonlinear equations and may have multiple solutions, we must show that its solution, satisfying (4.47), indeed is a difference between the solutions of the First Problem at the points t = t0 + ∆t and t = t0 satisfying Lemma 4.1. It is obvious that the set of the matrices τ,ρ τ,ρ {P¯30 (t0 ) + ∆P¯30 , Qτ40 (t0 , η) + ∆Qτ40 (η), R30 (t0 , η, χ) + ∆R30 (η, χ)} τ,ρ satisfies the First Problem for t = t0 + ∆t. Here {∆P¯30 , ∆Qτ40 (η), ∆R30 (η, χ)} is the solution of (4.41)–(4.43) satisfying (4.47). Applying the approach, proposed in [16], and taking into account results of [4],[17], Lemma 4.1, Proposition 4.3 and (4.47), we obtain that the matrix ¯ P30 (t0 ) + ∆P¯30 Qτ40 (t0 , χ) + ∆Qτ40 (χ) 0 τ,ρ τ,ρ (t0 , η, χ) + ∆R30 (η, χ) (Qτ40 (t0 , η) + ∆Qτ40 (η)) R30
defines a linear bounded self-adjoint nonnegative operator mapping the space E m × L2 [−h, 0; E m ] into itself. Hence (Lemma 4.1), P¯30 (t0 ) + ∆P¯30 = P¯30 (t0 + τ,ρ τ,ρ τ,ρ ∆t), Qτ40 (t0 , η)+∆Qτ40 (η) = Qτ40 (t0 +∆t, η), R30 (t0 , η, χ)+∆R30 (η, χ) = R30 (t0 + ∆t, η, χ). Since lim∆t→0 ν = 0, the solution of the First Problem is continuous in t at t = t0 and, consequently, on [0, T ] uniformly in (η, χ) ∈ [−h, 0] × [−h, 0]. This leads immediately to the statement of the lemma. τ,ρ Lemma 4.5 The matrices dP¯30 (t)/dt, ∂Qτ40 (t, η)/∂t, ∂R30 (t, η, χ)/∂t are continuous functions of (t, η, χ) ∈ [0, T ] × [−h, 0] × [−h, 0].
Proof. The lemma is proved similarly to [16] using the proof of Lemma 4.4. τ,ρ (t, η, χ)/∂t, where Corollary 4.1 The matrices dP¯k0 (t)/dt, ∂Qτi0 (t, η)/∂t, ∂Rk0 (k = 1, 2; i = 1, 2, 3), are continuous functions of (t, η, χ) ∈ [0, T ]×[−h, 0]×[−h, 0].
Proof. The statement of the corollary follows immediately from Lemma 4.5 and equations (4.21)–(4.30),(4.34). 4.2 Terms depending on (ξ, η, χ). For the terms, depending on (ξ, η, χ), we have 0 t t t t (ξ)/dξ = −P20 (ξ)A40 − A30 P30 (ξ) + P¯20 (T )S30 P30 (ξ) dP20 0 t,τ t t t +P20 (ξ)S30 P¯30 (T ) + P20 (ξ)S30 P30 (ξ) − Qt,τ 20 (ξ, 0) − [Q30 (ξ, 0)] 0
t t t t dP30 (ξ)/dξ = −P30 (ξ)A40 − A40 P30 (ξ) + P¯30 (T )S30 P30 (ξ)
(4.48)
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Valery Y. Glizer
0
t,τ t t t +P30 (ξ)S30 P¯30 (T ) + P30 (ξ)S30 P30 (ξ) − Qt,τ 40 (ξ, 0) − [Q40 (ξ, 0)]
NoDEA
(4.49)
0
t,τ t ¯ (∂/∂ξ − ∂/∂η)Qt,τ 10 (ξ, η) = −[A30 − (P20 (T ) + P20 (ξ))S30 ]Q30 (ξ, η) t,τ,ρ t (ξ)S30 Qτ30 (T, η) − R10 (ξ, 0, η) +P20
(4.50)
0
t,τ t ¯ (∂/∂ξ − ∂/∂η)Qt,τ 20 (ξ, η) = −[A30 − (P20 (T ) + P20 (ξ))S30 ]Q40 (ξ, η) t,τ,ρ t (ξ)S30 Qτ40 (T, η) − R20 (ξ, 0, η) +P20
(∂/∂ξ −
∂/∂η)Qt,τ 30 (ξ, η)
0
= −[A40 − (P¯30 (T ) +
(4.51)
t P30 (ξ))S30 ]Qt,τ 30 (ξ, η) 0
t,τ,ρ t (ξ)S30 Qτ30 (T, η) − [R20 (ξ, η, 0)] +P30
(4.52)
0
t,τ t ¯ (∂/∂ξ − ∂/∂η)Qt,τ 40 (ξ, η) = −[A40 − (P30 (T ) + P30 (ξ))S30 ]Q40 (ξ, η) t,τ,ρ t +P30 (ξ)S30 Qτ40 (T, η) − R30 (ξ, 0, η)
(∂/∂ξ − ∂/∂η −
t,τ,ρ ∂/∂χ)R10 (ξ, η, χ)
=
(4.53)
0 [Qτ30 (T, η)] S30 Qt,τ 30 (ξ, χ)
0
0
t,τ t,τ τ +[Qt,τ 30 (ξ, η)] S30 Q30 (T, χ) + [Q30 (ξ, η)] S30 Q30 (ξ, χ)
(4.54)
0
t,τ,ρ (∂/∂ξ − ∂/∂η − ∂/∂χ)R20 (ξ, η, χ) = [Qτ30 (T, η)] S30 Qt,τ 40 (ξ, χ) 0
0
t,τ t,τ τ +[Qt,τ 30 (ξ, η)] S30 Q40 (T, χ) + [Q30 (ξ, η)] S30 Q40 (ξ, χ)
(∂/∂ξ − ∂/∂η −
t,τ,ρ (ξ, η, χ) ∂/∂χ)R30
0
= [Qτ40 (T, η)]
0
(4.55)
S30 Qt,τ 40 (ξ, χ)
0
t,τ t,τ τ +[Qt,τ 40 (ξ, η)] S30 Q40 (T, χ) + [Q40 (ξ, η)] S30 Q40 (ξ, χ)
(4.56)
t,τ,ρ τ,ρ t τ Pk0 (0) = −P¯k0 (T ), Qt,τ j0 (0, η) = −Qj0 (T, η), Ri0 (0, η, χ) = −Ri0 (T, η, χ) (4.57) (k = 2, 3; j = 1, . . . , 4; i = 1, 2, 3) t P20 (ξ)Hk+2,0 = Qt,τ k0 (ξ, −h), t P30 (ξ)Hl0 = Qt,τ l0 (ξ, −h), 0
(k = 1, 2) (l = 3, 4)
t,τ,ρ H30 Qt,τ k+2,0 (ξ, η) = Rk0 (ξ, −h, η),
(k = 1, 2)
0
t,τ,ρ [Qt,τ 30 (ξ, η)] H40 = R20 (ξ, η, −h) 0
t,τ,ρ H40 Qt,τ 40 (ξ, η) = R30 (ξ, −h, η)
(4.58) (4.59) (4.60) (4.61) (4.62)
where Ak0 = Ak (T ), S30 = S3 (T ), Hl0 = Hl (T ), (k = 2, 3, 4; l = 3, 4). Note 4.3 The problem (4.48)–(4.62) can be partitioned into four simpler problems solved successively. The First Boundary Layer Problem (BLP) consists of (4.49),(4.53),(4.56),(4.57) with k = 3, j = 4, i = 3, (4.59) with l = 4, (4.62); the Second BLP consists of (4.52),(4.55), two last equations of (4.57) with j = 3, i = 2, (4.59) with l = 3, (4.60) with k = 2, (4.61); the Third BLP consists of (4.54), the last equation of (4.57) with i = 1, (4.60) with k = 1; the Fourth BLP consists of (4.48),(4.50),(4.51), two first equations of (4.57) with k = 2, j = 1, 2, (4.58).
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Asymptotic solution of equations of Riccati type
505
t Lemma 4.6 The First BLP (see Note 4.3) has a unique solution P30 (ξ), Qt,τ 40 (ξ, η) t,τ,ρ R30 (ξ, η, χ), ξ ∈ (−∞, 0], η ∈ [−h, 0], χ ∈ [−h, 0]. This solution satisfies the following inequality as ξ → −∞ t,τ,ρ t max[kP30 (ξ)k, kQt,τ 40 (ξ, η)k, kR30 (ξ, η, χ)k] ≤ c exp(γξ) ∀(η, χ) ∈ [−h, 0]×[−h, 0]
where c > 0 and γ > 0 are some constants. t Proof. Transforming the variables in the First BLP as follows P30 (ξ) = P˜30 (ξ) − t,τ t,τ,ρ τ ¯ ˜ ˜ P30 (T ), Q40 (ξ, η) = Q40 (ξ, η) − Q40 (T, η), R30 (ξ, η, χ) = R30 (ξ, η, χ) − τ,ρ R30 (T, η, χ) and taking into account the First Problem, one obtains 0 dP˜30 (ξ)/dξ = −P˜30 (ξ)A40 − A40 P˜30 (ξ) + P˜30 (ξ)S30 P˜30 (ξ)
˜ 40 (ξ, 0) − [Q ˜ 40 (ξ, 0)]0 − D3 (T ) −Q
(4.63)
0
˜ 40 (ξ, η) − R ˜ 30 (ξ, η, χ) ˜ 40 (ξ, η) = −[A40 − P˜30 (ξ)S30 ]Q (∂/∂ξ − ∂/∂η)Q 0
˜ 30 (ξ, η, χ) = [Q ˜ 40 (ξ, η)] S30 Q ˜ 40 (ξ, χ) (∂/∂ξ − ∂/∂η − ∂/∂χ)R ˜ 40 (0, η) = 0, R ˜ 30 (0, η, χ) = 0 P˜30 (0) = 0, Q ˜ 40 (ξ, −h), P˜30 (ξ)H40 = Q
0
˜ 40 (ξ, η) = R ˜ 30 (ξ, −h, η) H40 Q
(4.64) (4.65) (4.66) (4.67)
Let us consider the following optimal control problem ˜(ξ), y˜(θ) = y˜0 (θ), ξ0 −h ≤ θ ≤ ξ0 (4.68) d˜ y (ξ)/dξ = A40 y˜(ξ)+H40 y˜(ξ −h)+B20 u Z
0
0
0
[˜ y (ξ)D30 y˜(ξ) + u ˜ (ξ)N0 u ˜(ξ)]dξ → min
J˜ =
u ˜
ξ0
(4.69)
where B20 = B2 (T ), D30 = D3 (T ), N0 = N (T ), ξ0 < 0 is an arbitrary but fixed number. It is obvious that the problem (4.63)–(4.67) is associated with the control problem (4.68),(4.69) by the optimality conditions [7]. Hence, taking into account the assumptions 2), 3) and results of [7], we obtain the existence and uniqueness of solution of the problem (4.63)–(4.67) and, consequently, of the First BLP for (ξ, η, χ) ∈ [ξ0 , 0] × [−h, 0] × [−h, 0]. Since ξ0 < 0 is arbitrary, this leads to the first statement of the lemma. Now, let us proceed to the proof of the second statement. Applying results of [4],[7], Lemma 4.1 and the problem (4.63)–(4.67), we can rewrite the First BLP in the equivalent form Z
0
t (ξ) = P30
Z
0
[K (T, s)FP (s, ξ)K(T, s) + +∞
Z
0
+ −h
0
0
0
−h
0
K (T, s)FQ (s, ξ, η)K(T, s + η)dη
K (T, s + η)FQ (s, ξ, η)K(T, s)dη
506
Valery Y. Glizer
Z
Z
0
0
0
+ −h
−h
K (T, s + η)FR (s, ξ, η, χ)K(T, s + χ)dηdχ]ds Z
0
Qt,τ 40 (ξ, η) = 0
0
˜ K (T, s)FQ (s, ξ, χ)K(T, s+χ, η)dχ+
−h
Z
Z
0
0
−h
Z
0
Z
[K (T, s)FQ (s, ξ, η − s) + η+h
Z t,τ,ρ R30 (ξ, η, χ) =
Z
−h
Z
0
+ 0
Z
−h 0
+ −h 0
−h
˜ 0 (T, s, η)FP (s, ξ)K(T, ˜ [K s, χ)
˜ ˜ 0 (T, s + χ1 , η)FQ0 (s, ξ, χ1 )K(T, s, χ)dχ1 K
0
˜ (T, s + χ1 , η)FR (s, ξ, χ1 , χ2 )K(T, ˜ K s + χ2 , χ)dχ1 dχ2 ]ds Z
0
−h
Z
0
˜ (T, s, η)FQ (s, ξ, χ − s) + [K
+
0
K (T, s + χ)FR (s, ξ, χ, η − s)dχ]ds (4.71)
˜ ˜ 0 (T, s, η)FQ (s, ξ, χ1 )K(T, s + χ1 , χ)dχ1 K
η+h 0
0
0
0 ˜ [FQ (s, ξ, η − s)K(T, s, χ) +
+ Z
−h
0
+∞ 0
+
Z
−h
0
˜ K (T, s+χ)FQ (s, ξ, χ)K(T, s, η)dχ
˜ K (T, s + χ)FR (s, ξ, χ, χ1 )K(T, s + χ1 , η)dχdχ1 ]ds
−h
0
+
0
0
+
Z
Z
0
+
(4.70)
˜ [K (T, s)FP (s, ξ)K(T, s, η) +∞
Z
NoDEA
χ+h
Z
0
−h
˜ FR (s, ξ, χ1 , η − s)K(T, s + χ1 , χ)dχ1 ]ds ˜ 0 (T, s + χ1 , η)FR (s, ξ, χ1 , χ − s)dχ1 ]ds K
0
FR (s, ξ, η − s, χ − s)ds
+
(4.72)
min(η+h,χ+h)
˜ where K(T, s), K(T, s, η) are defined by (4.44)–(4.46) and t t (s + ξ), 0 ≤ s ≤ −ξ P30 (s + ξ)S30 P30 FP (s, ξ) = P¯30 (T )S30 P¯30 (T ) + D30 , −ξ < s < +∞ t P30 (s + ξ)S30 Qt,τ 0 ≤ s ≤ −ξ 40 (s + ξ, η), FQ (s, ξ, η) = P¯30 (T )S30 Qτ40 (T, η), −ξ < s < +∞ 0 t,τ (Qt,τ 0 ≤ s ≤ −ξ 40 (s + ξ, η)) S30 Q40 (s + ξ, χ), FR (s, ξ, η, χ) = 0 τ τ −ξ < s < +∞ (Q40 (T, η)) S30 Q40 (T, χ),
(4.73) (4.74) (4.75)
Now, applying the procedure of successive approximations to the set (4.70)–(4.75) and taking into account that kK(T, s)k is exponentially decaying as s → +∞ (see Lemma 4.1 and [4]), one directly obtains the second statement of the lemma.
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Lemma 4.7 The Second BLP and the Third BLP (see Note 4.3) have unique solut,τ,ρ t,τ,ρ tions Qt,τ 30 (ξ, η), R20 (ξ, η, χ) and R10 (ξ, η, χ), respectively, for ξ ∈ (−∞, 0], η ∈ [−h, 0], χ ∈ [−h, 0]. These solutions satisfy the following inequality as ξ → −∞ t,τ,ρ t,τ,ρ max[kQt,τ 30 (ξ, η)k, kR20 (ξ, η, χ)k, kR10 (ξ, η, χ)k] ≤ c1 exp(γ1 ξ)
where (η, χ) ∈ [−h, 0] × [−h, 0]; c1 > 0 and γ1 > 0 are some constants. Proof. Let us rewrite the problem, consisting of (4.55), the third equation of (4.57) with i = 2, (4.60) with k = 2, (4.61), in the equivalent form Z η t,τ,ρ R20 (ξ, η, χ) = Θ(ξ + η, ξ + χ) − V (ξ + η − s, s, χ − η + s)ds (4.76) g(ξ,η,χ)
where V (ξ, η, χ) is the right-hand part of (4.55), g(ξ, η, χ) = max(ξ + η, η − χ − h, −h), τ,ρ −h ≤ σ1 ≤ 0, −h ≤ σ2 ≤ 0 −R 20 (T, σ1 , σ2 ), 0 t,τ H Q (σ1 + h, σ2 − σ1 − h), σ1 ≤ −h, σ2 ≤ 0, σ1 ≤ σ2 Θ(σ1 , σ2 ) = 0 30t,τ 40 (Q30 (σ2 + h, σ1 − σ2 − h)) H40 , σ1 ≤ 0, σ2 ≤ −h, σ2 ≤ σ1 (4.77) From (4.16),(4.17),(4.57),(4.59) we immediately obtain that Θ(σ1 , σ2 ) is continuous for σ1 ≤ 0, σ2 ≤ 0. Let us substitute (4.76) into the problem, consisting of (4.52), the second equation of (4.57) with j = 3, (4.59) with l = 3, and transform the variable Qt,τ 30 (ξ, η) as ˆ t,τ Qt,τ (4.78) 30 (ξ, η) = Q30 (ξ, η) exp(−γξ/2) where γ is the constant from Lemma 4.6. As a result, we obtain the following ˆ t,τ (ξ, η) problem for Q 30 Z
0 ˆ t,τ (ξ, η) = −ˆ ˆ t,τ (ξ, η) + ϕ(ξ, η) (∂/∂ξ − ∂/∂η)Q α0 (ξ)Q 30 30
η
0 ˆ t,τ (ξ + η − s, s)]ds (4.79) [ψ(ξ + η − s, s, s − η) − βˆ0 (ξ + η − s, s − η)Q 30
+ g(ξ,η,0)
ˆ t,τ (0, η) = −Qτ30 (T, η), Q 30
t ˆ t,τ (ξ, −h) = Pˆ30 Q (ξ)H30 30
(4.80)
where βˆ0 (ξ, χ) = [β(T, χ)
t α ˆ 0 (ξ) = α(T ) − S30 P30 (ξ) + (γ/2)Im ,
−S30 Qt,τ 40 (ξ, χ)] exp(−γχ/2), 0
t ϕ(ξ, η) = [P30 (ξ)S30 Qτ30 (T, η) 0
τ −Θ (ξ +η, ξ)] exp(−γξ/2), ψ(ξ, η, χ) = [Qt,τ 40 (ξ, χ)] S30 Q30 (T, η) exp(−γ(ξ +χ)/2) t t Pˆ30 (ξ) = P30 (ξ) exp(−γξ/2)
Taking into account Lemma 4.6 and (4.40), (4.77), one has that α ˆ0 (ξ), βˆ0 (ξ, χ) t are bounded for (ξ, χ) ∈ (−∞, 0] × [−h, 0], and kϕ(ξ, η)k, kψ(ξ, η, χ)k, kPˆ30 (ξ)k
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NoDEA
are exponentially decaying, as ξ → −∞, uniformly in (η, χ) ∈ [−h, 0] × [−h, 0]. Rewriting the problem (4.79),(4.80) in an equivalent form, we obtain Z ˆ t,τ (ξ, η) = W (ξ, 0)Ω(ξ + η) + Q 30
ξ
W (ξ, ζ){ϕ(ζ, ξ + η − ζ) min(ξ+η+h,0)
Z
ξ+η−ζ
[ψ(ξ + η − s, s, s − ξ − η + ζ)
+ g(ζ,ξ+η−ζ,0)
0 ˆ t,τ (ξ + η − s, s)]ds}dζ −βˆ0 (ξ + η − s, s − ξ − η + ζ)Q 30
(4.81)
where W (ξ, ζ) is a solution of the following problem 0
∂W (ξ, ζ)/∂ξ = −ˆ α0 (ξ)W (ξ, ζ), ξ < ζ ≤ 0, W (ζ, ζ) = Im Ω(σ) =
−h ≤ σ ≤ 0 −Qτ30 (T, σ), −1 t ˆ W (σ + h, 0)P30 (σ + h)H30 , σ < −h
From (4.15), (4.57) we immediately obtain that Ω(σ) is continuous at σ = −h. It is also clear, taking into account the second statement of Lemma 4.6, that the matrix W (ξ, 0)W −1 (ξ + η + h, 0) is bounded for ξ + η ≤ −h, η ∈ [−h, 0], and the matrix W (ξ, ζ) is bounded for ξ ∈ (−∞, 0], η ∈ [−h, 0], ζ ∈ [ξ, min(ξ + η + h, 0)]. Now, applying the procedure of successive approximations to the equation ˆ t,τ (ξ, η) exists, is unique and is (4.81), one directly obtains that its solution Q 30 bounded for ξ ∈ (−∞, 0], η ∈ [−h, 0]. Hence (see (4.76)–(4.78)), the solution of t,τ,ρ the Second BLP Qt,τ 30 (ξ, η), R20 (ξ, η, χ) exists, is unique and is exponentially decaying, as ξ → −∞, uniformly in (η, χ) ∈ [−h, 0] × [−h, 0]. Now, the statements of the lemma, concerning the Third BLP, become obvious. The expression for t,τ,ρ R10 (ξ, η, χ) is obtained similarly to (4.76), (4.77). So, the lemma is completely proved. t (ξ), Lemma 4.8 The Fourth BLP (see Note 4.3) has a unique solution P20 t,τ t,τ Q10 (ξ, η), Q20 (ξ, η), ξ ∈ (−∞, 0], η ∈ [−h, 0]. This solution satisfies the following inequality as ξ → −∞ t,τ t (ξ)k, kQt,τ max[kP20 10 (ξ, η)k, kQ20 (ξ, η)k] ≤ c2 exp(γ2 ξ)
where η ∈ [−h, 0]; c2 > 0 and γ2 > 0 are some constants. Proof. Let us rewrite the problem, consisting of (4.50),(4.51), the second equation of (4.57) with j = 1, 2, (4.58), in the equivalent form Z Qt,τ k0 (ξ, η) = Υk (ξ + η) +
ξ
Zk (ζ, ξ + η − ζ)dζ, (k = 1, 2) min(ξ+η+h,0)
(4.82)
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where Zk (ξ, η), (k = 1, 2), are the right-hand parts of equations (4.50) and (4.51), respectively, and −h ≤ σ ≤ 0 −Qt,τ k0 (T, σ), Υk (σ) = t P20 (σ + h)Hk+2,0 , σ < −h From (4.14),(4.57) we immediately obtain that Υk (σ), (k = 1, 2), is continuous at σ = −h. Let us substitute (4.82) with k = 2 into (4.48). We obtain a functionalt differential equation for P20 (ξ). First, let us consider this equation on the segment [−h, 0]. It becomes a linear integral-differential equation. We don’t write out this t equation for the sake of simplicity. This equation with the initial condition P20 (0) = t ˘ ¯ −P20 (T ) has a unique solution denoted in the sequel as P20 (ξ), ξ ∈ [−h, 0]. t Now, let us consider the functional-differential equation for P20 (ξ) on the interval (−∞, −h). It has the form Z
ξ
t t t dP20 (ξ)/dξ = −P20 (ξ)α0 − P20 (ξ + h)H40 +
t P20 (ζ)β0 (ξ − ζ)dζ + f (ξ) (4.83) ξ+h
where α0 = α(T ), β0 (η) = β(T, η) and 0 0 t t f (ξ) = −[A30 − (P¯20 (T ) + P20 (ξ))S30 ]P30 (ξ) − [Qt,τ 30 (ξ, 0)]
Z
ξ
+
0 t,τ,ρ t {[A30 − (P¯20 (T ) + P20 (ζ))S30 ]Qt,τ 40 (ζ, ξ − ζ) + R20 (ζ, 0, ξ − ζ)}dζ (4.84)
ξ+h
Applying results of [18], we can rewrite the equation (4.83) with the initial condit t tion P20 (ζ) = P˘20 (ζ), ζ ∈ [−h, 0] in the equivalent form Z t t (ξ) = P˘20 (−h)K0 (−ξ − h) + P20
Z
0
−h
t ˘ 0 (ξ, ζ)dζ + P˘20 (−ζ − h)K
ξ
−h
f (s)K0 (s − ξ)ds (4.85)
where K0 (s) = K(T, s) (K(t, s) is defined by (4.44),(4.45)), Z ˘ 0 (ξ, ζ) = H40 K0 (−ξ − ζ − 2h) + K
h
−ζ
β0 (−χ)K0 (−ξ − ζ − χ − h)dχ
Now, applying the procedure of successive approximations to the set (4.84),(4.85) and taking into account that kK0 (s)k is exponentially decaying as s → +∞, one t (ξ) of this set exists, is unique and is exponentially directly has that the solution P20 decaying as ξ → −∞. This statement and (4.82) immediately prove the lemma.
5
Justification of the asymptotic solution
First, let us present auxiliary propositions on which the justification of the asymptotic solution is based.
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Lemma 5.1 Let Y (s, t, ε) be the solution of the following problem ε∂Y (s, t, ε)/∂s = α(s)Y (s, t, ε) + H4 (s)Y (s − εh, t, ε) Z
0
+(1/ε)
β(s, τ /ε)Y (s+τ, t, ε)dτ, s > t; −εh
Y (t, t, ε) = Im ;
Y (s, t, ε) = 0, s < t
where α(s) and β(s, η) are defined by (4.40). Then this matrix satisfies the following inequality for all sufficiently small ε > 0 kY (s, t, ε)k ≤ c exp(−γ(s − t)/ε), 0 ≤ t ≤ s ≤ T where c > 0 and γ > 0 are some constants independent of ε. Proof. The lemma is a generalization of [18], (Lemma 2, pp. 498–500), and is proved in the similar way, using the assumption 1 and Lemmas 4.1, 4.4. Let Λ(s, t, ε) be the solution of the following problem ˜ ε)Λ(s, t, ε) + H(s, ε)Λ(s − εh, t, ε) ∂Λ(s, t, ε)/∂s = A(s, Z
0
+ −εh
˜ τ, ε)Λ(s + τ, t, ε)dτ, s > t; Λ(t, t, ε) = In+m ; Λ(s, t, ε) = 0, s < t C(s,
where ˜ ε) = A(s, ε) − S(s, ε)P0 (s, ε), C(s, ˜ τ, ε) = −S(s, ε)Q0 (s, τ, ε) A(s, P0 (s, ε) =
P10 (s, ε) εP20 (s, ε) 0 εP20 (s, ε) εP30 (s, ε)
, Q0 (s, τ, ε) =
Q10 (s, τ, ε) Q20 (s, τ, ε) Q30 (s, τ, ε) Q40 (s, τ, ε)
and the blocks of the matrices P0 (s, ε) and Q0 (s, τ, ε) are defined by (4.1) and (4.2). Lemma 5.2 Let matrices Λ1 (s, t, ε), Λ2 (s, t, ε), Λ3 (s, t, ε) and Λ4 (s, t, ε) are upper left-hand, upper right-hand, lower left-hand and lower right-hand blocks of the matrix Λ(s, t, ε) of dimensions n × n, n × m, m × n and m × m respectively. These matrices satisfy the following inequalities for all sufficiently small ε > 0 kΛj (s, t, ε)k ≤ c, (j = 1, 3);
kΛ2 (s, t, ε)k ≤ cε
kΛ4 (s, t, ε)k ≤ c[ε + exp(−γ(s − t)/ε)];
0≤t≤s≤T
where c > 0 and γ > 0 are some constants independent of ε. Proof. The lemma is proved similarly to [12], (Lemma A.5), using Lemma 5.1.
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Theorem 5.1 The solution Pi (t, ε), Qj (t, τ, ε), Ri (t, τ, ρ, ε), (i = 1, 2, 3; j = 1, . . . , 4), of the problem (3.3)–(3.18) satisfies the following inequalities for (t, τ, ρ) ∈ [0, T ] × [−εh, 0] × [−εh, 0] and all sufficiently small ε > 0 kPi (t, ε) − Pi0 (t, ε)k ≤ cε,
kQj (t, τ, ε) − Qj0 (t, τ, ε)k ≤ cε
kRi (t, τ, ρ, ε) − Ri0 (t, τ, ρ, ε)k ≤ cε where Pi0 (t, ε), Qj0 (t, τ, ε), Ri0 (t, τ, ρ, ε) are defined by (4.1)–(4.3) and c > 0 is some constant independent of ε. Proof. Let us transform the variables in the problem (3.3)–(3.18) as follows Pi (t, ε) = Pi0 (t, ε) + θP i (t, ε), Qj (t, τ, ε) = Qj0 (t, τ, ε) + Ej (t, ε) + θQj (t, τ, ε) R1 (t, τ, ρ, ε) = R10 (t, τ, ρ, ε) + F1 (t, ρ, ε) + θR1 (t, τ, ρ, ε) 0
R2 (t, τ, ρ, ε) = R20 (t, τ, ρ, ε) + F2 (t, ρ, ε) + G1 (t, τ, ε) − F2 (t, −εh, ε) + θR2 (t, τ, ρ, ε) R3 (t, τ, ρ, ε) = R30 (t, τ, ρ, ε) + G2 (t, ρ, ε) + θR3 (t, τ, ρ, ε) t where (i = 1, 2, 3; j = 1, . . . , 4), and Ek (t, ε) = P20 ((t − T )/ε)(Hk+2 (t) − Hk+2,0 ) 0
t El (t, ε) = P30 ((t − T )/ε)(Hl (t) − Hl0 ) + εP20 (t, ε)Hl−2 (t) 0
0
t Fk (t, ρ, ε) = (H3 (t) − H30 ) Qt,τ k+2,0 ((t − T )/ε, ρ/ε) + H3 (t)P30 ((t − T )/ε)(Hk+2 (t) 0
0
t −Hk+2,0 ) + ε[H1 (t)Qk0 (t, ρ, ε) + H1 (t)P20 ((t − T )/ε)(Hk+2 (t) − Hk+2,0 ) 0
0
0
+H3 (t)P20 (t, ε)Hk (t)], Gk (t, ρ, ε) = (H4 (t) − H40 ) Qt,τ k+2,0 ((t − T )/ε, ρ/ε) 0
0
t ((t − T )/ε)(Hk+2 (t) − Hk+2,0 ) + ε[H2 (t)Qk0 (t, ρ, ε) +H4 (t)P30 0
0
0
t +H2 (t)P20 ((t−T )/ε)(Hk+2 (t)−Hk+2,0 )+H4 (t)P20 (t, ε)Hk (t)], (k = 1, 2; l = 3, 4)
It can be shown by application of the assumption 1) and Lemmas 4.6-4.8 that the following inequalities are satisfied for all sufficiently small ε > 0 kEj (t, ε)k ≤ cε, kFk (t, ρ, ε)k ≤ cε, kGk (t, ρ, ε)k ≤ cε, (t, ρ) ∈ [0, T ] × [−εh, 0] (5.1) where (j = 1, . . . , 4; k = 1, 2); c > 0 is some constant independent of ε. The transformation, introduced above, yields the following problem for the new variables θP i (t, ε), θQj (t, τ, ε), θRi (t, τ, ρ, ε), (i = 1, 2, 3; j = 1, . . . , 4) 0
0
˜ ε) − A˜ (t, ε)θP (t, ε) − θQ (t, 0, ε) − θQ (t, 0, ε) dθP (t, ε)/dt = −θP (t, ε)A(t, ˜ P (t, ε) + θP (t, ε)S(t, ε)θP (t, ε) −D 0
(5.2)
˜ τ, ε) − θR (t, 0, τ, ε) (∂/∂t − ∂/∂τ )θQ (t, τ, ε) = −A˜ (t, ε)θQ (t, τ, ε) − θP (t, ε)C(t,
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Valery Y. Glizer
˜ Q (t, τ, ε) + θP (t, ε)S(t, ε)θQ (t, τ, ε) −D 0
NoDEA
(5.3) 0
˜ ρ, ε) − C˜ (t, τ, ε)θQ (t, ρ, ε) (∂/∂t − ∂/∂τ − ∂/∂ρ)θR (t, τ, ρ, ε) = −θQ (t, τ, ε)C(t, 0
˜ R (t, τ, ρ, ε) + θQ (t, τ, ε)S(t, ε)θQ (t, ρ, ε) −D
(5.4)
θP (T, ε) = 0, θQ (T, τ, ε) = 0, θR (T, τ, ρ, ε) = 0
(5.5)
0
θP (t, ε)H(t, ε) = θQ (t, −εh), H (t, ε)θQ (t, τ, ε) = θR (t, −εh, τ, ε) 0 ≤ t ≤ T, Here
(5.6)
−εh ≤ τ, ρ ≤ 0
θQ1 (t, τ, ε) θQ2 (t, τ, ε) θQ3 (t, τ, ε) θQ4 (t, τ, ε) θR1 (t, τ, ρ, ε) θR2 (t, τ, ρ, ε) 0 θR (t, τ, ρ, ε) = (1/ε) θR2 (t, ρ, τ, ε) θR3 (t, τ, ρ, ε) ˜ P 2 (t, ε) ˜ P 1 (t, ε) D ˜ Q2 (t, τ, ε) ˜ Q1 (t, τ, ε) D D D ˜ Q (t, τ, ε) = ˜ P (t, ε) = , D D 0 ˜ Q3 (t, τ, ε) D ˜ Q4 (t, τ, ε) ˜ P 3 (t, ε) ˜ (t, ε) D D D P2 ˜ R1 (t, τ, ρ, ε) D ˜ R2 (t, τ, ρ, ε) D ˜ R (t, τ, ρ, ε) = D 0 ˜ R3 (t, τ, ρ, ε) ˜ (t, ρ, τ, ε) D D R2 θP (t, ε) =
θP 1 (t, ε) εθP 2 (t, ε) 0 εθP 2 (t, ε) εθP 3 (t, ε)
, θQ (t, τ, ε) =
˜ Qj (t, τ, ε), D ˜ Ri (t, τ, ρ, ε) are expressed in a known form ˜ P i (t, ε), D The matrices D by Pi0 (t, ε), Qj0 (t, τ, ε), Ri0 (t, τ, ρ, ε), (i = 1, 2, 3; j = 1, . . . , 4). These matrices are continuous in (t, τ, ρ) ∈ [0, T ] × [−εh, 0] × [−εh, 0]. They satisfy the following inequalities for all sufficiently small ε > 0 ˜ P 1 (t, ε)k ≤ c exp(γ(t − T )/ε), kD ˜ P k (t, ε)k ≤ cε, (k = 2, 3) kD
(5.7)
˜ Qj (t, τ, ε)k ≤ c, kD ˜ Ri (t, τ, ρ, ε)k ≤ c/ε, (i = 1, 2, 3; j = 1, . . . , 4) kD
(5.8)
where c > 0 and γ > 0 are some constants independent of ε. Let us denote ˜ P (t, ε) − θP (t, ε)S(t, ε)θP (t, ε) ΓP (t, ε, θP ) = D
(5.9)
˜ Q (t, τ, ε) − θP (t, ε)S(t, ε)θQ (t, τ, ε) ΓQ (t, τ, ε, θP , θQ ) = D
(5.10)
0
˜ R (t, τ, ρ, ε) − θQ (t, τ, ε)S(t, ε)θQ (t, ρ, ε) ΓR (t, τ, ρ, ε, θQ ) = D ˜ t, τ, ε) = Λ(s, t + τ + εh, ε)H(t + τ + εh, ε) Λ(s, Z εh ˜ + τ + ρ, −ρ, ε)dρ Λ(s, t + τ + ρ, ε)C(t +
(5.11)
(5.12)
−τ
˜ t, τ, ε) satisfies the Applying Lemma 5.2, it can be shown that the matrix Λ(s, following inequalities for all sufficiently small ε > 0 ˜ k (s, t, τ, ε)k ≤ c, kΛ
˜ l (s, t, τ, ε)k ≤ c[1 + exp(−γ(s − t)/ε)/ε] kΛ
(k = 1, 2; l = 3, 4),
0≤t≤s≤T
(5.13)
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˜ j (s, t, τ, ε), (j = 1, . . . , 4) are upper left-hand, upper right-hand, lower leftwhere Λ hand and lower right-hand blocks of this matrix of dimensions n × n, n × m, m × n and m × m respectively; c > 0 and γ > 0 are some constants independent of ε. Applying results of [7],[19] and equations (5.9)–(5.12), we can rewrite the problem (5.2)–(5.6) in the equivalent form Z
T
θP (t, ε) =
0
[Λ (s, t, ε)ΓP (s, ε, θP )Λ(s, t, ε) t
Z
0
0
+ −εh
Z
0
0
+ Z
−εh
Z
0
0
0
Λ (s + τ, t, ε)ΓQ (s, τ, ε, θP , θQ )Λ(s, t, ε)dτ
0
Λ (s + τ, t, ε)ΓR (s, τ, ρ, ε, θQ )Λ(s + ρ, t, ε)dτ dρ]ds
+ −εh
Λ (s, t, ε)ΓQ (s, τ, ε, θP , θQ )Λ(s + τ, t, ε)dτ
−εh
Z
T
t
Z
0
+ −εh
Z
0
+ −εh
Z
0
0
−εh
Z
0
˜ + ρ, t, τ, ε)dρ Λ (s, t, ε)ΓQ (s, ρ, ε, θP , θQ )Λ(s 0 0 ˜ t, τ, ε)dρ Λ (s + ρ, t, ε)ΓQ (s, ρ, ε, θP , θQ )Λ(s,
0 ˜ + ρ1 , t, τ, ε)dρdρ1 ]ds Λ (s + ρ, t, ε)ΓR (s, ρ, ρ1 , ε, θQ )Λ(s
+ −εh
0
˜ t, τ, ε) [Λ (s, t, ε)ΓP (s, ε, θP )Λ(s,
θQ (t, τ, ε) =
Z
(5.14)
min(t+τ +εh,T )
0
[Λ (s, t, ε)ΓQ (s, t − s + τ, ε, θP , θQ )
+ t
Z
0
+ −εh
0
Λ (s + ρ, t, ε)ΓR (s, ρ, t − s + τ, ε, θQ )dρ]ds Z
θR (t, τ, ρ, ε) = Z
−εh 0
+ Z
0
Z
−εh 0
+ −εh
−εh
˜ 0 (s, t, τ, ε)ΓP (s, ε, θP )Λ(s, ˜ t, ρ, ε) [Λ
t 0
+ Z
T
(5.15)
˜ 0 (s, t, τ, ε)ΓQ (s, ρ1 , ε, θP , θQ )Λ(s ˜ + ρ1 , t, ρ, ε)dρ1 Λ ˜ 0 (s + ρ1 , t, τ, ε)Γ0Q (s, ρ1 , ε, θP , θQ )Λ(s, ˜ t, ρ, ε)dρ1 Λ
˜ 0 (s + ρ1 , t, τ, ε)ΓR (s, ρ1 , ρ2 , ε, θQ )Λ(s ˜ + ρ2 , t, ρ, ε)dρ1 dρ2 ]ds Λ
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Valery Y. Glizer
Z
min(t+τ +εh,T )
+ Z
0 ˜ t, ρ, ε) [ΓQ (s, t − s + τ, ε, θP , θQ )Λ(s,
t 0
+ Z
NoDEA
−εh
˜ + ρ1 , t, ρ, ε)dρ1 ]ds ΓR (s, ρ1 , t − s + τ, ε, θQ )Λ(s
min(t+ρ+εh,T )
0
˜ (s, t, τ, ε)ΓQ (s, t − s + ρ, ε, θP , θQ ) [Λ
+ Z
t 0
+ Z
−εh
˜ 0 (s + ρ1 , t, τ, ε)ΓR (s, ρ1 , t − s + ρ, ε, θQ )dρ1 ]ds Λ
min(t+τ +εh,t+ρ+εh,T )
ΓR (s, t − s + τ, t − s + ρ, ε, θQ )ds
+
(5.16)
t
It is obvious that 0 ≤ min(t + τ + εh, T ) − t ≤ εh, 0 ≤ min(t + ρ + εh, T ) − t ≤ εh 0 ≤ min(t + τ + εh, t + ρ + εh, T ) − t ≤ εh
(5.17) (5.18)
Now, applying the procedure of successive approximations to the set (5.14)–(5.16) and taking into account Lemma 5.2, equations (5.9)–(5.11) and inequalities (5.7), (5,8),(5.13),(5.17),(5.18), one directly obtains for all sufficiently small ε > 0 kθP i (t, ε)k kθRi (t, τ, ρ, ε)k
≤ cε, ≤ cε,
kθQj (t, τ, ε)k ≤ cε, (i = 1, 2, 3; j = 1, . . . , 4)
which together with (5.1) immediately leads to the statement of the theorem.
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