with distinct amortization factors. The problem can be reduced to a canonical form by scaling the state variables and time. We prove the optimality of an extremal ...
c Pleiades Publishing, Ltd., 2012. ISSN 0012-2661, Differential Equations, 2012, Vol. 48, No. 12, pp. 1607–1622. c Yu.N. Kiselev, M.V. Orlov, 2012, published in Differentsial’nye Uravneniya, 2012, Vol. 48, No. 12, pp. 1642–1657. Original Russian Text
CONTROL THEORY
Optimal Resource Distribution Program in a Two-Sector Economic Model with a Cobb–Douglas Production Function with Distinct Amortization Factors Yu. N. Kiselev and M. V. Orlov Moscow State University, Moscow, Russia Received May 30, 2012
Abstract—We consider a resource distribution problem on a finite time interval with a terminal functional for a two-sector economic model with a two-factor Cobb–Douglas production function with distinct amortization factors. The problem can be reduced to a canonical form by scaling the state variables and time. We prove the optimality of an extremal solution constructed with the use of the maximum principle. For the case in which the initial state of the plant lies above the singular ray, the solution of the boundary value problem of the maximum principle is presented in closed form. DOI: 10.1134/S0012266112120075
1. STATEMENT OF THE PROBLEM 1.1. Canonical Form of the Problem In the present paper, we study the following two-dimensional optimal control problem on a fixed time interval: u F (x) − μ1 x1 , x1 (0) = x10 > 0, x˙ 1 = ε1 1−u (1) F (x) − μ2 x2 , x2 (0) = x20 > 0, x˙ 2 = ε2 0 ≤ u ≤ 1, J ≡ x2 (T ) → max, u(·)
where x1 and x2 are positive state variables, u is a scalar control, u ∈ [0, 1], and T > 0 is the given planning horizon. The differential equations of controlled motion in problem (1) contain the function x = (x1 , x2 )T ∈ R2+ , (2) F (x) = xε11 xε22 , that is, the Cobb–Douglas production function, where the output elasticities ε1 and ε2 (see [1–3]) satisfy the conditions ε2 > 0, ε1 + ε2 = 1 (3) ε1 > 0, and μ1 and μ2 are positive amortization factors. The set of input data of the control problem (1) consists of the positive numbers μ1 , μ2 , x10 , x20 , and T and the parameters (3). The planning horizon T is assumed to be “sufficiently large;” this assumption will be made more specific in what follows. The present paper is close to [4–12]. The case in which μ1 = μ2 was considered in [8]. The passage to the case of μ1 = μ2 proves to be nontrivial, and we consider this case in the present paper. 1.2. Reduction of a More General Control Problem to the Canonical Form (1) Problem (1) will be referred to as the problem in canonical form for distinct amortization factors μ1 and μ2 . A similar form (with the linear terms eliminated) proved to be convenient when 1607
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studying the problem in the case of equal amortization coefficients (μ1 = μ2 > 0). One can reduce the following more general optimal control problem to the form (1) : dX1 = ua1 F (X) − μ1 X1 , X1 |t =0 = X10 > 0, dt dX2 = (1 − u)a2 F (X) − μ2 X2 , X2 |t =0 = X20 > 0, dt 0 ≤ u ≤ 1, 0 ≤ t ≤ T , J ≡ X2 (T ) → max,
(4)
u(·)
where X1 and X2 are positive state variables, u is a one-dimensional bounded control, and a1 , a2 , μ1 , μ2 , and T are known positive parameters. The passage from problem (4) to problem (1) is performed by scaling the phase variables X1 and X2 and time t according to the formulas Xi = εi ai xi ,
μi = μi A,
t = At ∈ [0, T ].
i = 1, 2;
Here the positive parameters A and T have the form A = F (a)F (ε) ≡ aε11 aε22 εε11 εε22 , Indeed,
T = AT .
dxi dt 1 1 dXi dxi = = , i = 1, 2, dt dt dt A εi ai dt 1 u μ1 (ua1 F (X1 , X2 ) − μ1 X1 ) = X1ε1 X2ε2 − ε1 a1 x1 x˙ 1 = Aε1 a1 Aε1 Aε1 a1 u μ u = (ε1 a1 x1 )ε1 (ε2 a2 x2 )ε2 − 1 x1 = F (x) − μ1 x1 , Aε1 A ε1 1 1−u ((1 − u)a2 F (X1 , X2 ) − μ2 X2 ) = F (x) − μ2 x2 , x˙ 2 = Aε2 a2 ε2 J = a2 ε2 J. x˙ i =
In what follows, we study the canonical resource distribution problem (1). 2. ANALYSIS OF PROBLEM (1) WITH THE USE OF THE PONTRYAGIN MAXIMUM PRINCIPLE. COMPUTATION OF POSSIBLE SINGULAR MODES OF THE PROBLEM. BOUNDARY VALUE PROBLEM OF THE MAXIMUM PRINCIPLE 2.1. Analysis of the Condition of Maximum. Costate System The Hamilton–Pontryagin function u 1−u F (x) − μ1 x1 + ψ2 F (x) − μ2 x2 H(x, ψ, u) = ψ1 ε1 ε2 of problem (1) can be represented in the form H(x, ψ, u) = F (x)[uπ + ψ2 /ε2 ] − μ1 x1 ψ1 − μ2 x2 ψ2 , where the factor
π = π(ψ) ≡ ψ1 /ε1 − ψ2 /ε2
(5)
multiplying the control is the switch function. Since the Cobb–Douglas production function (2) is positive, it is obvious that the maximizer of the function H with respect to the argument u ∈ [0, 1] for π = 0 has the form 1 for π > 0 (6) u∗ (ψ) = arg max H(x, ψ, u) = h(π) = 0 for π < 0, u∈[0,1] DIFFERENTIAL EQUATIONS
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where h(·) is the Heaviside function. If π = 0, then every point u ∈ [0, 1] maximizes the function H. The function t ∈ [0, T ], Π(t) = π(ψ)|ψ=ψ(t) , of time, where ψ(t) = (ψ1 (t), ψ2 (t))T = 0 is the costate variable, will be called the preset switch function. The costate system of problem (1), −ψ˙ = Hx ,
− ψ˙ i = Hx i ,
is equivalent to the system
i = 1, 2;
in detailed notation, it acquires the form −ψ˙ 1 = Fx 1 (x)[uπ + ψ2 /ε2 ] − μ1 ψ1 , −ψ˙ 2 = Fx 2 (x)[uπ + ψ2 /ε2 ] − μ2 ψ2 ,
ψ1 (T ) = 0, ψ2 (T ) = 1.
(7)
The costate equations (7) of the Mayer problem are supplemented with the transversality conditions ψ(T ) = (0, 1)T . By taking into account the formulas Fx 1 (x) = ε1 F (x)/x1 ,
Fx 2 (x) = ε2 F (x)/x2
for the partial derivatives of the function (2), one can rewrite the costate system (7) in the form F (x) ψ1 ψ2 ψ˙ 1 = ψ1 (T ) = 0, uπ + − μ1 , − ε1 x1 ε2 ε1 (8) F (x) ψ2 ψ2 ψ˙ 2 = ψ2 (T ) = 1. uπ + − μ2 , − ε2 x2 ε2 ε2 By setting pi = ψi /εi ,
i = 1, 2,
(9)
we reduce the costate system (8) to the form F (x) [uπ + p2 ] − μ1 p1 , x1 F (x) [uπ + p2 ] − μ2 p2 , −p˙ 2 = x2 −p˙ 1 =
p1 (T ) = 0, 1 p2 (T ) = , ε2
(10)
where now, in the variables (9), the switch function is π = p1 − p2 . 2.2. Computation of Possible Singular Modes of Problem (1) Suppose that the following condition is satisfied: π ≡ 0,
t ∈ (α, β),
α < β,
(α, β) ⊂ [0, T ].
(11)
Recall that the maximum condition H ⇒ maxu∈[0,1] does not permit one to determine the maximizer (6) uniquely for π = 0. In more detailed notation, condition (11) has the form Π(t) ≡ π(ψ(t)) ≡ 0,
t ∈ (α, β).
(12)
An analysis of identity (12) and two of its differential corollaries permits one to determine the singular values usng of the control and xsng of the trajectory along the possible singular mode of problem (1). By virtue of (5), identity (12) is equivalent to the identity ψ2 ψ1 ≡ , ε1 ε2 DIFFERENTIAL EQUATIONS
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(13)
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By differentiating identity (13) with respect to time t, we obtain the relation ψ˙ 1 /ε1 = ψ˙ 2 /ε2 .
(14)
F (x) ψ2 ψ1 ψ2 F (x) ψ2 − μ1 ≡ − μ2 . x1 ε2 ε1 x2 ε2 ε2
(15)
From (14), (13), and (8), we have
Since the costate variable ψ = (ψ1 , ψ2 )T is nonzero, we have ψ1 = 0 and ψ2 = 0 on the singular part. Therefore, it follows from (15) and (13) that F (x) F (x) − μ1 ≡ − μ2 . x1 x2
(16)
From relations (2) and conditions (3), we obtain F (x) = x1
x2 x1
ε2 ,
F (x) = x2
x2 x1
−ε1 .
(17)
Then, by setting z = x2 /x1 ,
(18)
from relations (16) and (17), we obtain the equation z ε2 − μ1 = z −ε1 − μ2 for the unknown positive ratio (18); this equation can be represented in the convenient form g(z) ≡ z ε2 − z −ε1 = ν ≡ μ1 − μ2 .
(19)
Lemma 1. For arbitrary μ1 and μ2 , Eq. (19) has a unique positive root z = zsng ≡ zsng (ε1 , ν) > 0;
(20)
moreover , zsng |ν=0 = 1,
zsng > 1 if ν > 0, zsng |ε1 =ε2 =1/2 = (ν/2 + (ν/2)2 + 1 )2 , (zsng − 1)/ν → 1 as ν → 0.
zsng ∈ (0, 1)
if
ν < 0,
(21) (22) (23)
Proof. The function g(z) = z ε2 − z −ε1 on the left-hand side in Eq. (19) has the properties g(+0) = −∞,
g (z) = ε2 z g (z) = −
−ε1
g(+∞) = +∞, + ε1 z
ε1 z ε1 +2
−ε1 −1
> 0,
(ε2 z + ε1 + 1) < 0,
g(1) = 0, z > 0, z > 0.
This implies relations (21)–(23). √ Let us √ present brief explanations for relations (22) and (23). If ε1 = 1/2, then ε2 = 1/2, g(z) = z − 1/ z, and Eq. (19) acquires the form √ √ z − 1/ z = ν. After the substitution z = el , the last equation can be rewritten in the form sinh(l/2) = ν/2, DIFFERENTIAL EQUATIONS
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whence we obtain l = 2 arc sinh(l/2) = 2 ln(ν/2 +
1611
z = el = (ν/2 +
(ν/2)2 + 1 ),
(ν/2)2 + 1 )2 .
The proof of relation (22) is complete. Let us verify the limit relation (23), which is important in forthcoming considerations. The root z(ν) of Eq. (19) depends smoothly on the parameter ν; moreover, z(0) = 1, z (0) = 1, and z (0) = 2ε1 ; then, by the Taylor formula, we have z(ν) − 1 = z(ν) − z(0) = z (0)ν + z (ξ) Hence we obtain the limit relation lim
ν→0
ν2 , 2
ξ = θν,
θ ∈ [0, 1].
z(ν) − 1 = 1. ν
The proof of Lemma 1 is complete. Remark 1. The parameter zsng [see (20)] plays an important role in forthcoming constructions. The subscript sng indicates that this parameter belongs in the singular mode of problem (1). If ε1 = 1/2, then it is represented by the analytic expression (22). In the general case, for given ν and ε1 ∈ (0, 1), the root (20) of Eq. (19) can be found approximately by numerical methods. Let us introduce the notation ε2 −ε1 − μ1 = zsng − μ2 . Bsng ≡ zsng
Remark 2. The parameter Bsng can be positive, negative, or zero. For example, if ν = 0, which is equivalent to μ1 = μ2 , we have > 0 for μ1 ∈ (0, 1) Bsng = 1 − μ1 = 0 for μ1 = 1 < 0 for μ1 > 1. The root (20) of Eq. (19) is given by the formula ε1 ε1 ≡ 1 + νzsng . zsng = 1 + (μ1 − μ2 )zsng
(24)
Therefore, along a possible singular mode, we have the relation x2 = zsng ; x1
(25)
i.e., the singular part of the trajectory lies on the singular ray 2 : x2 /x1 = zsng }. Lsng = {x = (x1 , x2 )T ∈ R+
Consequently, x2 = x1 zsng ,
t ∈ (α, β),
2 x ∈ R+ .
By differentiating the last identity with respect to time t, we obtain the equation x˙ 2 = x˙ 1 zsng , whence, by virtue of the differential equations for the controlled motion of problem (1), we obtain u 1−u F (x) − μ2 x2 = F (x) − μ1 x1 zsng . ε2 ε1 By solving the resulting equation for u, we obtain the control along the singular mode, u= DIFFERENTIAL EQUATIONS
F (x)/ε2 − μ2 x2 + μ1 x1 zsng . F (x)(zsng /ε1 + 1/ε2 )
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By virtue of the relation x2 = x1 zsng , this expression admits a substantial simplification, u=
1/ε2 + (μ1 − μ2 )x2 /F (x) . zsng /ε1 + 1/ε2
Since, by virtue of relations (17) and (25), x2 = F (x)
x2 x1
ε1 ε1 = zsng ,
from relations (24) and (3), we have the following chain of relations for the singular value of the control: u=
ε1 1/ε2 + νzsng 1/ε2 + zsng /ε1 1/ε2 + zsng − 1 ε1 /ε2 + zsng = = = ε1 = ε1 , zsng /ε1 + 1/ε2 1/ε2 + zsng /ε1 1/ε2 + zsng /ε1 1/ε2 + zsng /ε1
(26)
where ε1 ∈ (0, 1). Therefore, under assumption (11), we have obtained relations (25) and (26); i.e., we have justified the following assertion. Lemma 2. Under assumption (11), the following relations hold along a possible singular mode : x2 = zsng , x1 The obtained result permits one 1 u∗ (ψ) = 0 usng = ε1
u = usng ≡ ε1 ∈ (0, 1),
t ∈ (α, β).
(27)
to define the function (6) for π = 0 by setting for π > 0 for π < 0 for π = 0,
or
u∗ (ψ) =
for π = 0 for π = 0,
h(π) ε1
(28)
and pass to the construction of the boundary value problem of the maximum principle. 2.3. Boundary Value Problem of the Maximum Principle By relations (1), (8), and (28), the boundary value problem of the maximum principle for the optimal control problem (1) can be represented in the form u F (x) − μ1 x1 , x1 (0) = x10 > 0, ε1 1−u F (x) − μ2 x2 , x2 (0) = x20 > 0, x˙ 2 = ε2 F (x) ψ1 ψ2 ψ˙ 1 = ψ1 (T ) = 0, uπ + − μ1 , − ε1 x1 ε2 ε1 F (x) ψ2 ψ2 ψ˙ 2 = ψ2 (T ) = 1, uπ + − μ2 , − ε2 x2 ε2 ε2 ψ1 ψ2 π= − . u = u∗ (ψ), ε1 ε2 x˙ 1 =
(29)
Here the maximizer u∗ (ψ) of the Hamilton–Pontryagin function has the form (28). Suppose that we have found a solution x(t) = (x1 (t), x2 (t))T ,
ψ(t) = (ψ1 (t), ψ2 (t))T ,
0 ≤ t ≤ T,
of the boundary value problem (29). By using the function (28) and the above-found costate variable ψ(t), we set 0 ≤ t ≤ T. (30) u(t) = u∗ (ψ(t))|ψ=ψ(t) , DIFFERENTIAL EQUATIONS
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The process (the trajectory–control pair) (x(t), u(t)),
0 ≤ t ≤ T,
(31)
is referred to as an extremal process; it satisfies necessary optimality conditions in the form of the Pontryagin maximum principle. The presence of singular parts is not excluded. The proof of the optimality of the extremal process (31) requires additional considerations and is carried out in the following section. Since Π(T ) = π(ψ(T )) = −1/ε2 < 0, it follows that the preset switch function Π(t) satisfies the condition Π(t) < 0,
t ∈ (θ, T ],
for some nonnegative θ < T . This implies that the extremal control (30) is zero on the terminal time interval (θ, T ]. 2.4. Proof of the Optimality of the Extremal Solution (31) Along with the extremal process (31), we consider an arbitrary admissible process (ˆ x(t), uˆ(t)),
0 ≤ t ≤ T,
(32)
such that d x ˆ(t) = Cuˆ(t)/ε1 (1 − u ˆ(t))/ε2 F (ˆ x(t)) − Cμμ12xx12 , dt
20 x ˆ(0) = Cxx10 ,
u ˆ ∈ [0, 1].
Let us introduce the increments ΔJ = J[ˆ u(·)] − J[u(·)] = x ˆ2 (T ) − x2 (T )
Δx(t) = x ˆ(t) − x(t),
for the trajectory and the functional. To justify the optimality of the extremal process (31), one should prove the inequality ΔJ ≤ 0 (33) for any admissible process (32). In the proof of inequality (33), we use the special integral representation derived below for the increment ΔJ of the functional. This representation contains the maximum function M (x, ψ) = H(x, ψ, u)|u=u∗ (ψ) = F (x)[h(π)π + ψ2 /ε2 ] − μ1 x1 ψ1 − μ2 x2 ψ2 = F (x)G(ψ) − μ1 x1 ψ1 − μ2 x2 ψ2 , where G(ψ) = h(π)π + ψ2 /ε2 and the switch function is defined in (5). Therefore, M (x, ψ) = F (x)G(ψ) − μ1 x1 ψ1 − μ2 x2 ψ2 .
(34)
We rewrite the costate equation (8) with u = u∗ (ψ) in the form ψ˙ = −F (x)G(ψ) + diag(μ1 , μ2 )ψ,
ψ(T ) = (0, 1)T ,
or, by virtue of (34), in the form ψ˙ = −Mx (x, ψ),
ψ(T ) = (0, 1)T .
(35)
The transversality condition ψ(T ) = (0, 1)T for the costate variable ψ(t) and the obvious relation Δx(0) = 0 permit one to represent the increment of the functional in the form ΔJ = (ψ(t), Δx(t))|t=T t=0 . DIFFERENTIAL EQUATIONS
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This, together with the Newton–Leibniz formula, implies that
T
d (ψ(t), Δx(t)) dt = dt
ΔJ = 0
T dˆ x(t) dx(t) ˙ − dt. (ψ(t), Δx(t)) + ψ(t), dt dt 0
By virtue of (35), the first term in the last integral can be rewritten in the form ˙ (ψ(t), Δx(t)) = −(Mx (x(t), ψ(t)), Δx(t)), and, by the differential equations of controlled motion of problem (1), the second term can be represented in the form dˆ x(t) dx(t) − = H(ˆ x(t), ψ(t), uˆ(t)) − H(x(t), ψ(t), u(t)). ψ(t), dt dt By taking into account the relation H(x(t), ψ(t), u(t)) = M (x(t), ψ(t)) for the extremal triple (x(t), u(t), ψ(t)), we obtain the following integral representation of the increment of the functional:
T ΔJ =
{H(ˆ x(t), ψ(t), uˆ(t)) − M (x(t), ψ(t)) − (Mx (x(t), ψ(t)), Δx(t))} dt.
(36)
0
By using the obvious inequality H(ˆ x(t), ψ(t), uˆ(t)) ≤ M (ˆ x(t), ψ(t)), from relation (36), we obtain an upper bound for the increment of the functional,
T ΔJ ≤
{M (ˆ x(t), ψ(t)) − M (x(t), ψ(t)) − (Mx (x(t), ψ(t)), Δx(t))} dt.
0
By using formula (34) for the maximum function M (x, ψ) and the formula for the gradient Mx (x, ψ) = F (x)G(ψ) − diag(μ1 , μ2 )ψ, which follows from it, we rewrite the resulting estimate in the form
T {F (ˆ x(t))G(ψ) − μ1 x ˆ1 ψ1 − μ2 x ˆ2 ψ2 − F (x(t))G(ψ) + μ1 x1 ψ1 + μ2 x2 ψ2
ΔJ ≤ 0
− (F (x(t)), Δx(t))G(ψ(t)) + μ1 ψ1 Δx1 + μ2 ψ2 Δx2 } dt. ˆi − xi , i = 1, 2, we obtain By taking into account the relations Δxi = x
T ΔJ ≤
{F (ˆ x(t)) − F (x(t)) − (F (x(t)), Δx(t))}G(ψ(t)) dt.
0
By virtue of the concavity of the Cobb–Douglas function (2), we have the inequality F (ˆ x) − F (x) − (F (x), xˆ − x) ≤ 0,
2 x, x ˆ ∈ R+ .
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By taking into account the last inequality and the positivity of the function G(ψ) (see [8]), we obtain the definitive upper bound ΔJ ≤ 0 for the increment of the functional, which implies the optimality of the extremal solution of problem (31). Thus, to construct an optimal solution of the control problem (1), it suffices to produce an extremal process obtained by solving the boundary value problem (29) of the maximum principle. It is in this way that we construct an optimal solution of problem (1) under the condition of a “sufficiently large” planning horizon T . 3. ANALYSIS OF THE TERMINAL TIME INTERVAL [θ, T ] WITH THE CONTROL u = 0 UNDER THE INITIAL CONDITION x(θ) ∈ Lsng . COMPUTATION OF THE COSTATE VARIABLES AND PROOF OF THEIR POSITIVITY. FINDING THE DURATION T − θ OF THE TERMINAL TIME INTERVAL 3.1. Solution of Differential Equations of Motion These equations with the control u = 0 and the initial condition x(θ) ∈ Lsng acquire the form x˙ 1 = −μ1 x1 ,
x1 (θ) > 0,
x˙ 2 =
1 F (x) − μ2 x2 , ε2
x2 (θ) = zsng x1 (θ) > 0.
(37)
Lemma 3. If ν ≡ μ1 − μ2 = 0, then the Cauchy problem (37) has the solution x1 (t) = x1 (θ)e−μ1 (t−θ) , −μ2 (t−θ) 1/ε1
x2 (t) = x1 (θ)e ε1 + q(t) = zsng
in addition,
x2 (t) x1 (t)
q
(38) (t),
(39)
−ε1 ν(t−θ)
1−e ε2 ν
;
(40)
ε1 = eε1 ν(t−θ) q(t).
(41)
Proof. The representation (38) is obvious. To obtain relation (39), we rewrite the second equation in (37) in the form 1 ε1 2 ˙2 = x − μ2 xε21 . x−ε 2 x ε2 1 By the change of variables ζ = xε21 ,
2 ζ˙ = ε1 x2ε1 −1 x˙ 2 = ε1 x−ε ˙ 2, 2 x
we reduce the last equation to the following linear inhomogeneous differential equation for the unknown function ζ : 1 ε 1 x (t) − μ2 ζ . (42) ζ˙ = ε1 ε2 1 It follows from the representation (38) that xε11 (t) = xε11 (θ)e−ε1 μ1 (t−θ) . By taking into account the last relation, we rewrite Eq. (42) in the form ε1 ζ˙ = −ε1 μ2 ζ + xε11 (θ)e−ε1 μ1 (t−θ) . ε2
(43)
Equation (43) with ν = 0 has the particular solution ¯ = De−ε1 μ1 (t−θ) , ζ(t) DIFFERENTIAL EQUATIONS
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xε11 (θ) . ε2 ν
(44)
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The general solution of the linear inhomogeneous differential equation (43) has the form ¯ ζ(t) = Ce−ε1 μ2 (t−θ) + ζ(t),
(45)
where C is an arbitrary constant. By setting t = θ in (45), we obtain C = ζ(θ) − D = xε21 (θ) +
xε11 (θ) . ε2 ν
Let us transform the representation (45) by using the above-obtained value of the arbitrary constant C, relation (44), and the formula x2 (θ) = zsng x1 (θ), ζ(t) = [ζ(θ) − D]e−ε1 μ2 (t−θ) + De−ε1 μ1 (t−θ) = e−ε1 μ2 (t−θ) [xε21 (θ) − D(1 − e−ε1 ν(t−θ) )] xε11 (θ) ε1 −ε1 μ2 (t−θ) ε1 −ε1 ν(t−θ) zsng x1 (θ) + (1 − e ) = e−ε1 μ2 (t−θ) xε11 (θ)q(t). =e ε2 ν Here the function q(t) has the form (40). Then ζ 1/ε1 (t) = e−μ2 (t−θ) x1 (θ)q 1/ε1 (t), which, together with the relation ζ = xε21 , implies the desired representation (39) for x2 (t). Relation (41) follows from the representations (38) and (39). The proof of Lemma 3 is complete. 3.2. Costate System In the variables (9), the costate system has the form (10). By virtue of the condition u(t) = 0 and relations (17), for t ∈ [θ, T ], system (10) acquires the form p˙ 1 = μ1 p1 − (x2 /x1 )ε2 p2 ,
p˙ 2 = μ2 p2 − (x2 /x1 )−ε1 p2 ,
p1 (T ) = 0,
p2 (T ) = 1/ε2 .
(46)
We seek the solution of the Cauchy problem (46) by using relation (41). We rewrite the second equation in system (46) in the form −ε1 x2 (t) 1 p2 (T ) = > 0. p2 , p˙ 2 = μ2 − x1 (t) ε2 We write out the solution p2 (t) of the resulting Cauchy problem in the form t
μ2 −
p2 (t) = p2 (T ) exp
x2 (s) x1 (s)
−ε1
ds ,
T
or 1 μ2 (t−T ) e exp p2 (t) = ε2
T
x2 (s) x1 (s)
−ε1
ds .
(47)
t
Let us transform the integral in the last formula by using relation (41),
T
x2 (s) x1 (s)
−ε1
T −ε1 ν (s−θ) e ds. ds = q(s)
t
(48)
t
Since, by virtue of (40), d[q(s)] = q(s) ˙ ds =
ε1 −ν(s−θ) e ds, ε2
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we have e−ε1 ν(s−θ) ds =
1617
ε2 d[q(s)]. ε1
Then the integral occurring in (48) acquires the form
T
x2 (s) x1 (s)
−ε1
T ε2 ε2 ε2 q(T ) d[q(s)] = . ds = ln[q(s)]|s=T ln s=t = ε1 q(s) ε1 ε1 q(t)
t
t
Consequently,
T exp
x2 (s) x1 (s)
−ε1 ds
q(T ) = q(t)
ε2 /ε1 .
t
By taking into account the last relation, we rewrite the representation (47) in the form ε /ε 1 μ2 (t−T ) q(T ) 2 1 e . p2 (t) = ε2 q(t)
(49)
ε /ε 1 μ2 (θ−T ) q(T ) 2 1 e . p2 (θ) = ε2 q(θ)
(50)
Now, by setting t = θ, we obtain
Now we seek a solution of the first equation in system (46). This equation is a linear inhomogeneous equation with known inhomogeneous term. The solution p1 (t) of this Cauchy problem can be represented in the form
T
p1 (t) = eμ1 (t−T ) p1 (T ) +
e−μ1 (s−T )
x2 (s) x1 (s)
ε2
p2 (s) ds ,
t
where p1 (T ) = 0. By using relations (41) and (49), we obtain
x2 (s) x1 (s)
ε2
p2 (s) = =
x2 (s) x1 (s)
ε1 ε2 /ε1 ε2 /ε1 ε1 ν(s−θ) ε2 /ε1 1 μ2 (s−T ) q(T ) p2 (s) = [e q(s)] e ε2 q(s)
1 [q(T )]ε2 /ε1 eμ2 (s−T ) eε2 ν(s−T ) . ε2
Then the solution p1 (t) acquires the form 1 p1 (t) = [q(T )]ε2 /ε1 eμ1 (t−T ) ε2
T
e−ν(s−T ) eε2 ν(s−θ) ds.
t
Let us compute the last integral,
T
−ν(s−T ) ε2 ν(s−T +T −θ)
e
e
ε2 ν(T −θ)
ds = e
t
e−ε1 ν(s−T ) ds
t ε2 ν(T −θ)
=e DIFFERENTIAL EQUATIONS
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Therefore, we finally obtain the representation p1 (t) =
[q(T )]ε2 /ε1 ε2 ν(T −θ) μ1 (t−T ) e−ε1 ν(t−T ) − 1 e e . ε2 ε1 ν
(51)
By setting t = θ in (51), we obtain p1 (θ) =
[q(T )]ε2 /ε1 −μ1 (T −θ) eν(T −θ) − eε2 ν(T −θ) e . ε2 ε1 ν
(52)
When carrying out these transformations, we have used the relation ε1 + ε2 = 1. We have thereby proved the following assertion. Lemma 4. The solution of the Cauchy problem (46) for t ∈ [θ, T ] is given by relations (51), (49). The values of the functions p1 (t) and p2 (t) at time t = θ have the form (52) and (50). 3.3. Computation of the Duration T − θ of the Terminal Time Interval We rewrite formula (50) in the form p2 (θ) =
[q(T )]ε2 /ε1 −μ2 (T −θ) e [q(θ)]−ε2 /ε1 . ε2
(53)
By setting t = θ in (40), we obtain the relation ε1 , q(θ) = zsng
(54)
ε2 . [q(θ)]ε2 /ε1 = zsng
(55)
which implies that Let us write out an equation for the computation of the duration T − θ of the terminal time interval. By taking into account the inclusion x(θ) ∈ Lsng and the formula π = p1 −p2 for the switch function, we find that the condition p1 (θ) = p2 (θ) should be satisfied at time t = θ. By virtue of relations (52), (53), and (55), after the division by the positive factor [q(θ)]ε2 /ε1 /ε2 , the last relation acquires the form eν(T −θ) − eε2 ν(T −θ) −ε2 = e−μ2 (T −θ) zsng , e−μ1 (T −θ) ε1 ν ε2 , whence we obtain the equation which is equivalent to the relation 1 − e−ε1 ν(T −θ) = ε1 ν/zsng
e−ε1 ν(T −θ) = 1 −
ε1 ν . ε2 zsng
(56)
ε2 −ε1 − zsng = ν We transform the right-hand side of Eq. (56) by using the known relation zsng [see Lemma 1 and Eq. (19)],
1−
ε2 −ε1 zsng − zsng ε1 ε1 + ε2 zsng ε1 ν = 1 − ε = 1 − ε1 + = . 1 ε2 ε2 zsng zsng zsng zsng
(57)
From relations (56) and (57), we obtain eε1 ν(T −θ) =
zsng , ε1 + ε2 zsng
whence we determine the desired duration T − θ of the terminal time interval, T −θ =
zsng 1 ln . ε1 ν ε1 + ε2 zsng DIFFERENTIAL EQUATIONS
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Remark 3. The right-hand side of (58) is positive for any sign of the difference μ1 − μ2 ≡ ν = 0. Indeed, if ν > 0, then zsng > 1, and therefore, ε1 zsng + ε2 zsng ε1 + ε2 zsng zsng = > = 1, ε1 + ε2 zsng ε1 + ε2 zsng ε1 + ε2 zsng which implies that the right-hand side of relation (58) is positive for ν > 0. If ν < 0, then zsng ∈ (0, 1), and hence we have ε1 zsng + ε2 zsng ε1 + ε2 zsng zsng = < = 1, ε1 + ε2 zsng ε1 + ε2 zsng ε1 + ε2 zsng which implies that the right-hand side of relation (59) is positive for ν < 0. Therefore, the expression (58) for T − θ is positive for any sign of ν = 0. Note that the relations T −θ =1
zsng = 1,
were obtained in the case of ν = 0, considered in [8]. The right-hand side of (58) satisfies the relation zsng 1 ln = 1. (59) lim ν→0 ε1 ν ε1 + ε2 zsng Indeed, by virtue of relations (3) and (24), the argument of the logarithm in (59) can be represented in the form ε1 zsng + ε2 zsng ε1 (zsng − 1) + ε1 + ε2 zsng zsng = = ε1 + ε2 zsng ε1 + ε2 zsng ε1 + ε2 zsng ε1 zsng zsng − 1 = 1 + ε1 = 1 + ε1 ν . ε1 + ε2 zsng ε1 + ε2 zsng Since zsng → 1 as ν → 0, we have ε1 zsng →1 ε1 + ε2 zsng
as
ν → 0.
The above-obtained formulas and limit relations imply relation (59). 4. CONSTRUCTION OF AN EXTREMAL SOLUTION. PROOF OF ITS OPTIMALITY When studying the optimal control problem (1) and the corresponding boundary value problem (29) of the maximum principle, we single out the following three cases of the position of the x2 2 : = zsng : initial state x0 with respect to the singular ray Lsng = x = Cxx12 ∈ R+ x1 x20 > zsng (the point x0 lies above the ray Lsng ); (C1 ) x10 x20 = zsng (the point x0 lies on the ray Lsng ); (Csng ) x10 x20 < zsng (the point x0 lies below the ray Lsng ). (C0 ) x10 The labeling of these three cases corresponds to the value of the corresponding optimal control on the initial time interval. In case (C1 ), the optimal control ⎧ for t ∈ [0, τ ] ⎨1 (60) u(t) = usng = ε1 for t ∈ (τ, θ] ⎩ 0 for t ∈ (θ, T ] DIFFERENTIAL EQUATIONS
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has two switch points
1 − ε1 νz0−ε2 x20 > zsng , > 0, z 0 ≡ −ε 1 − ε1 νzsng2 x10 zsng 1 ln < T. θ=T − ε1 ν ε1 + ε2 zsng
1 ln τ= ε2 ν
(61) (62)
The assumed “sufficiently large” duration of the control process implies that τ < θ; therefore, 0 < τ < θ < T. In case (Csng ), the optimal control u(t) =
for t ∈ [0, θ] for t ∈ (θ, T ]
usng = ε1 0
has two intervals of constant values, the singular interval [0, θ] and the terminal one (θ, T ]. In case (C0 ), the optimal control has the form ⎧ ⎨0 u(t) = usng = ε1 ⎩ 0
for t ∈ [0, τ0 ] for t ∈ (τ0 , θ] for t ∈ (θ, T ].
Therefore, in each of these cases, there is one singular time interval; in addition, the duration T − θ of the terminal time interval is the same [see (58)]. Below we restrict our considerations to case (C1 ). Let us introduce the functions x2 (t) , x1 (t)
z(t) =
z(0) ≡ z0 =
x20 > zsng , x10
(63)
eε2 νt − 1 , ε1 ν 1 − e−ε1 ν(t−θ) ε1 . + q(t) = zsng ε2 ν
p(t) = 1 + z0ε2
(64) (65)
Note that the functions p(t) and q(t) are the solutions of the Cauchy problems ε2 ε2 ε2 νt z e > 0, ε1 0 ε1 −ε1 ν(t−θ) e > 0, q(t) ˙ = ε2
p(t) ˙ =
p(0) = 1, ε1 q(θ) = zsng .
Theorem 1. In case (C1 ) with 1 ln T > ε2 ν
1 − ε1 νz0−ε2 −ε2 1 − ε1 νzsng
1 ln + ε1 ν
zsng ε1 + ε2 zsng
(a “sufficiently large” planning horizon 0 < τ < θ < T ), the solution x (t)
x(t) = Cx12(t) ,
ψ (t)
ψ(t) = Cψ12(t) ,
u(t),
0 ≤ t ≤ T,
of the boundary value problem (29) of the maximum principle has the following structure. 1. The control u(t) is given by relations (60)–(62). DIFFERENTIAL EQUATIONS
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2. The coordinates of the trajectory x(t) have the form ⎧ −μ1 t [p(t)]1/ε2 for t ∈ [0, τ ] ⎨ x10 e x1 (t) = x1 (τ )eBsng (t−τ ) for t ∈ (τ, θ] ⎩ −μ1 (t−θ) for t ∈ (θ, T ], x1 (θ)e ⎧ −μ2 t x e for t ∈ [0, τ ] ⎨ 20 Bsng (t−τ ) for t ∈ (τ, θ] x2 (t) = x1 (t)zsng ≡ x2 (τ )e ⎩ −μ2 (t−θ) 1/ε1 x1 (θ)e [q(t)] for t ∈ (θ, T ], where the functions p(t) and q(t) are given by relations (64) and (65) and the function (63) has the properties t ∈ [0, τ ) ∪ (θ, T ]; z(t) > zsng , in addition, the trajectory x(t) lies above the singular ray Lsng ; and if t ∈ [τ, θ], then it lies on the ray. On the singular time interval [τ, θ], the functions x1 (t) and x2 (t) are nondecreasing if Bsng > 0 and decreasing if Bsng < 0, and the state point x(t) remains fixed if Bsng = 0. 3. The coordinates of the costate variable ψ(t) have the form ψ1 (t) = ε1 p1 (t),
ψ2 (t) = ε2 p2 (t),
where the functions p1 (t) and p2 (t) are defined by the closed-form expressions ⎧ ε1 /ε2 ⎪ ⎪ μ1 (t−τ ) p(τ ) ⎪ p1 (τ )e for t ∈ [0, τ ] ⎪ ⎪ p(t) ⎨ p1 (t) = p1 (τ )e−Bsng (t−τ ) = p1 (θ)e−Bsng (t−θ) for t ∈ (τ, θ] ⎪ ⎪ ⎪ −ε1 ν(t−T ) ⎪ e −1 1 ⎪ ⎩ [q(T )]ε2 /ε1 eε2 ν(T −θ) eμ1 (t−τ ) for t ∈ (θ, T ], ε2 ε1 ν ⎧ ε ν(t−τ ) ⎪ p (τ ) 1 + z −ε1 [p(τ )]ε1 /ε2 e−ε1 ντ 1 − e 2 ⎪ for t ∈ [0, τ ] ⎪ 2 0 ⎪ ε2 ν ⎪ ⎨ for t ∈ (τ, θ] p2 (t) = p1 (t) ≡ p2 (τ )e−Bsng (t−τ ) = p2 (θ)e−Bsng (t−θ) ⎪ ε2 /ε1 ⎪ ⎪ ⎪ q(T ) 1 ⎪ ⎩ eμ2 (t−T ) for t ∈ (θ, T ]. ε2 q(t) Here the functions p(t) and q(t) are defined by relations (64) and (65). 4. The preset switch function Π(t) ≡ p1 (t) − p2 (t) ≡
ψ1 (t) ψ2 (t) − ε1 ε2
˙ and its derivative Π(t) have the following properties : > 0 for 0 ≤ t < τ
