Differential Equations, Vol. 40, No. 12, 2004, pp. 1694–1708. Translated from Differentsial’nye Uravneniya, Vol. 40, No. 12, 2004, pp. 1615–1628. c 2004 by Kiselev, Orlov. Original Russian Text Copyright

ORDINARY DIFFERENTIAL EQUATIONS

A Study of One-Dimensional Optimization Models with Infinite Horizon Yu. N. Kiselev and M. V. Orlov Moscow State University, Moscow, Russia Received June 7, 2004

1. INTRODUCTION We consider a number of models arising in economics and microbiology. In these optimal control problems, the phase variable is one-dimensional and the control occurs linearly in the differential equation of controlled motion and in the functional. We seek possible singular modes and justify the optimality of solutions suggested. The main technique used in our analysis is to represent the functional as the sum of a term not containing integrals (independent of the control and the phase variable) and an integral with integrand having a special structure and not containing the control. We also consider the general scheme covering a number of particular control problems interesting from the viewpoint of applications. 2. A MODEL OF A ONE-SECTOR ECONOMY 2.1. Statement of the Problem. Introductory Remarks Consider the problem k˙ = u(t)e%t f (k) − λk, 0 ≤ t < +∞, Z+∞ J= e−δt (1 − u(t))e%t f (k)dt → max,

k(0) = k0 , (1)

u(·)

0

where k is a one-dimensional phase variable, u is a one-dimensional control satisfying the constraints 0 ≤ u(t) ≤ 1, k0 , δ, λ, and % are known positive constants, and α ∈ (0, 1).

f (k) = kα , We assume that

(2)

δ − (1 − α)−1 % > 0.

(3)

A description of the economic meaning of problem (1) can be found in [1–3]. To analyze problem (1) completely, we perform the change η = e−%t k1−α /(1 − α)

(4)

of the phase variable, after which we obtain η˙ = −µη + u(t), 0 ≤ t < +∞, Z+∞ I= e−νt (1 − u(t))η β dt → max,

η(0) = η0 ,

0 ≤ u(t) ≤ 1, (5)

u(·)

0

where

β = α/(1 − α) > 0, µ = (1 − α)λ + % > 0,

ν = δ − (1 + β)% > 0, η0 = k01−α /(1 − α) > 0.

c 2004 MAIK “Nauka/Interperiodica” 0012-2661/04/4012-1694

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Then we give a complete solution of the optimal control problem (5), (6) and present a classification of optimal solution types. The introduction of the new phase variable η is expedient owing to the fact that the form of problem (5) is simpler than that of the original problem (1). Moreover, the possible singular segment η = ηsing of an optimal trajectory in problem (5) proves to be independent of time t, and the control u = using on the singular part is also independent of time. For problem (5), we find an optimal solution, justify its optimality, and list the types of optimal solutions depending on the problem parameters µ, ν, and η0 . The analysis is based on a special representation of the functional I in the form of a sum of two terms, one not containing integrals and independent of the control and the second being the integral of the product of the exponential e−νt by a function W (η) depending only on the phase variable η. This representation is possible owing to the fact that the control u occurs linearly in the differential equation of motion and in the integrand of the functional. 2.2. Introduction of a New Phase Variable. Reduction of Problem (1) to the Form (5) Let us rewrite the main differential equation of problem (1)–(3) in the form k−α k˙ = −λk1−α + e%t u(t).

(7)

ξ = k1−α/(1 − α),

(8)

Setting

˙ therefore, Eq. (7) acquires the linear form we find that ξ˙ = k−α k; ξ˙ = −(1 − α)λξ + e%t u(t).

(9)

η = e−%t ξ,

(10)

Further, by setting we obtain

  η˙ = −%e−%t ξ + e−%t ξ˙ = −%η + e−%t −(1 − α)λξ + e%t u(t) = −%η − (1 − α)λη + u(t). Finally, Eq. (9) acquires the form η˙ = −µη + u(t),

η(0) = η0 ,

(11)

where µ = % + (1 − α)λ and η0 = ξ(0) = (k(0))1−α /(1 − α) = k01−α /(1 − α) are positive constants. We have derived the differential equation of problem (5). Let us now transform the functional. The functional J of problem (1) in terms of the phase variable (8) acquires the form Z+∞ J= e−(δ−%)t (1 − u(t))[(1 − α)ξ]α/(1−α) dt.

(12)

0

We use the notation

β = α/(1 − α),

(13)

so that β > 0 for α ∈ (0, 1). Then α = β/(1 + β), 1 − α = 1/(1 + β), 1 + β = 1/(1 − α), and the functional (12) can be represented in the form −β

J = (1 + β)

Z+∞ e−(δ−%)t (1 − u(t))ξ β dt. 0

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After the change of variables (10), the functional (14) acquires the form −β

J = (1 + β)

Z+∞ e−(δ−(1+β)%)t (1 − u(t))η β dt.

(15)

0

Therefore, by (15) and (6), the functional to be maximized becomes Z+∞ I= e−νt (1 − u(t))η β dt.

(16)

0

The positiveness of the discount factor [see (6)] ν = δ − (1 + β)% in the integral (16) follows from assumption (3). We have passed from problem (1)–(3) to problem (5), (6). The latter problem contains four positive parameters µ, ν, β, and η0 . 2.3. Main Results for Problem (5) Lemma 1. Each admissible trajectory η(t) of problem (5) admits the two-sided estimate η− (t) ≤ η(t) ≤ η+ (t), where the lower bound

t ≥ 0,

(17)

η− (t) = η0 e−µt

(18)

corresponds to the control u(t) ≡ 0 and the upper bound   1 1 η+ (t) = η0 − e−µt + µ µ

(19)

corresponds to the control u(t) ≡ 1. The attainability set Y (T ) at time T ≥ 0 is the closed interval whose endpoints are the lower and upper bounds in (17) at time T : Y (T ) = [η− (T ), η+ (T )] .

(20)

Proof. Inequality (17) follows from the linearity of the differential equation (11) and the constraint 0 ≤ u(t) ≤ 1 for the control function. The inclusion Y (T ) ⊂ [η− (T ), η+ (T )] is a consequence of (17). On the other hand, for a constant control u(t) ≡ v ∈ [0, 1], we have η(T, v) = (η0 − v/µ) e−µT + v/µ ≡ h(v). The function h(v) is linear in v; moreover, h(0) = η− (T ) and h(1) = η+ (T ). The proof of relation (20) is complete. Figure 1 shows the location of the attainability set Y (T ) and the bounds (17) for η0 ∈ (0, 1/µ) and for η0 > 1/µ. Lemma 2 (the main lemma). The functional I of problem (5) admits the representation η 1+β I[η(·)] = 0 + 1+β

Z+∞ e−νt W (η(t))dt,

(21)

0

where

W (η) = η β (1 − ση),

η ≥ 0,

(22)

with the positive constant σ = µ + ν/(1 + β). DIFFERENTIAL EQUATIONS

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Fig. 1. The attainability set: (a) for η0 ∈ (0, 1/µ) and (b) η0 > 1/µ.

Proof. Since u = η˙ + µη in problem (5), it follows that the functional I can be represented in the form Z+∞ I[η] = e−νt (1 − η˙ − µη) η β dt. (24) 0

By taking account of the relation −νt

−e

  1+β d η 1+β −νt η ηη ˙ = −e − e−νt ν dt 1+β 1+β β

and by performing integration by parts, we obtain −νt

I[η] = −e

t=+∞ Z+∞   [η(t)]1+β ν −νt β 1+β 1+β η − µη dt, + e − η 1 + β t=0 1+β 0

which finally implies that η 1+β I[η] = 0 + 1+β

  Z+∞ e−νt 1 − µ +

ν 1+β

  η η β dt.

(25)

0

By taking account of notation (23), from (25), we obtain (21) and (22). The proof of Lemma 2 is complete. Lemma 3. The function W (0) = 0 has the following properties : W (1/σ) = 0, W (+∞) = −∞, W (η) > 0, η ∈ (0, 1/σ), W (η) < 0, η ∈ (1/σ, +∞), W 0 (η) > 0, η ∈ (0, η∗ ) , W 0 (η∗ ) = 0 and W 0 (η) < 0, η ∈ (η∗ , +∞) , where the unique maximizer η∗ = arg max W (η) = arg max W (η) 0≤η<+∞

(26)

0<η<1/σ

of W (·) is given by the formula η∗ = β/[(1 + β)σ] ∈ (0, 1/σ). Proof. All assertions of Lemma 3 about the properties of the function W (·) follow from (22), and W 0 (η) = η β−1 [β − (1 + β)ση]. Figure 2 represents the graphs of the function (22) for β ∈ (0, 1) and for β > 1. Theorem 1. The optimal trajectory ηopt (t) in problem (5) is given by the formula ηopt (t) = arg max W (η), η∈Y (t)

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t ≥ 0.

(27)

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Fig. 2. The graphs of the function W (·): (a) for β ∈ (0, 1) and (b) β > 1.

The optimal control uopt (t) in problem (5) has the form uopt (t) = η˙opt (t) + µηopt (t)

(28)

at the points of differentiability of the trajectory (27). Proof. The assertions of Theorem 1 follow from Lemmas 1–3. Indeed, by setting η¯(t) = arg max W (η)

(29)

η∈Y (t)

and by choosing an arbitrary admissible trajectory η(t) ∈ Y (t), on the basis of the representation (21), we write out the following expression for the increment of the functional I : Z+∞ ∆I ≡ I[η] − I [¯ η] = e−νt [W (η(t)) − W (¯ η (t))] dt.

(30)

0

It follows from the definition of the function (29) that W (η(t)) − W (¯ η (t)) ≤ 0 for all t ≥ 0, which, together with (30), implies that ∆I ≤ 0 for the increment of the functional and the trajectory (29) is optimal. Therefore, the trajectory (27) maximizes the functional I in problem (5). Relation (28) follows from the differential equation of problem (5). The proof of Theorem 1 is complete. Remark 1. The optimal trajectory in problem (5) is unique. In the following, we elaborate on the concise expression (27) for the optimal trajectory and further use it to obtain a complete classification of types of optimal solutions in problem (5). We show that the optimal trajectory has at most one breaking point. Remark 2. For the optimal trajectory, we have ( η+ (t) for η∗ > η+ (t) ηopt (t) = η∗ for η∗ ∈ [η− (t), η+ (t)] η− (t) for η∗ < η− (t), or

ηopt (t) = η± (t) + ∆η(t) sat ((η∗ − η± (t))/∆η(t)) , 

where sat(s) =

t > 0,

s for |s| ≤ 1 sgn(s) for |s| > 1

is the saturation function, η± (t) = (η+ (t) + η− (t))/2, ∆η(t) = (η+ (t) − η− (t))/2, ηopt (t) = K∗ (t, η∗ ), and   η+ (t) for η∗ > η+ (t) for η∗ ∈ [η− (t), η+ (t)] K∗ (t, η∗ ) = η∗  η (t) for η < η (t). − ∗ − DIFFERENTIAL EQUATIONS

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The character of dependence of the optimal trajectory ηopt (t) on time for various combinations of parameter values will be described below. 2.4. Analysis of the Singular Mode in Problem (5) We consider the Hamilton–Pontryagin function K = e−νt (1 − u)η β + ψ(−µη + u), the adjoint equation ψ˙ = −Kη0 = −β(1 − u)e−νt η β−1 + µψ, and the switching function π ≡ Ku0 = −e−νt η β + ψ. The identity π ≡ 0 is equivalent to ψ ≡ e−νt η β ,

(31)

and the identity π˙ ≡ 0 is equivalent to the identity ψ˙ ≡ −νe−νt η β + βe−νt η β−1 η, ˙ which, together with the main and adjoint equations, implies that µψ − β(1 − u)e−νt η β−1 ≡ −νe−νt η β + βe−νt η β−1 (−µη + u) or, by (31), µe−νt η β − βe−νt η β−1 + uβe−νt η β−1 ≡ −νe−νt η β − µβe−νt η β + uβe−νt η β−1 . The two terms containing the control cancel each other, and the multiplication by the positive factor exp(νt) term by term gives the relation [ν + (1 + β)µ]η β = βη β−1 .

(32)

Since η > 0, it follows from (32) that η ≡ β/[ν + (1 + β)µ]. Therefore, along a possible singular mode, the trajectory preserves the constant value η ≡ ηsing , where ηsing =

(33)

β β = = η∗ (1 + β)[µ + ν/(1 + β)] (1 + β)σ

is the maximizer (26) of the function W (η) (see Lemma 3). By differentiating relation (33) with respect to time, we obtain η˙ = 0, which, in view of the main equation, determines the singular control u = using ,

using = µη∗ ∈ (0, 1).

(34)

2.5. A Classification of Types of Optimal Solutions of Problem (5) Depending on the initial state η(0) = η0 > 0, we consider two cases, η0 ∈ (0, 1/µ) and η0 ∈ [1/µ, +∞). Throughout this section, we assume that ηsing ≡ η∗ and using ≡ u∗ . Note that η∗ < 1/µ by (34). We split the analysis of the case in which η0 ∈ (0, 1/µ) into three subcases: η∗ > η0 , η∗ = η0 , and η∗ ∈ (0, η0 ). For η∗ > η0 , we have one singular segment [τ, +∞), where τ is the breaking point of the optimal trajectory,  n 1 if 0 ≤ t < τ η (t) if 0 ≤ t < τ ηopt (t) = η+ (t) = u opt u if τ ≤ t < +∞, if τ ≤ t < +∞, sing

sing

and the switching point τ is determined by the formula τ = tion η∗ = η+ (τ ). DIFFERENTIAL EQUATIONS

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1 η0 − 1/µ ln > 0 from the equaµ η∗ − 1/µ

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Fig. 3. The graphs of the optimal phase trajectory ηopt (·): (a) for η∗ > η0 and (b) η0 ∈ [1/µ, +∞).

For η∗ = η0 , the singular segment fills the entire interval [0, +∞) : ηopt (t) ≡ ηsing ,

uopt (t) ≡ using .

For η∗ ∈ (0, η0 ), there is one singular segment [τ, +∞), where τ is the breaking point of the optimal trajectory, and  n 0 if 0 ≤ t < τ η (t) if 0 ≤ t < τ ηopt (t) = η− u opt (t) = using if τ ≤ t < +∞, if τ ≤ t < +∞, sing where the switching point τ is determined by the formula τ =

1 η0 > 0 from the equation ln µ η∗

η∗ = η− (τ ). The investigation of the case in which η0 ∈ [1/µ, +∞) is similar to that of the case in which η∗ ∈ (0, η0 ). Figure 3 represents the graphs of optimal trajectories depending on the position of the point η0 . We have thereby considered all cases and classified all types of optimal solutions of problem (5) : we have shown that the optimal solution has at most one switching (breaking) point and at most one singular part; moreover, the singular part is necessarily infinite. 3. A MODEL OF RESOURCE ALLOCATION IN A COLONY OF MICROORGANISMS 3.1. Statement of the Problem. Introductory Remark We consider the optimal control problem x˙ = a − (a + x)u(t), 0 ≤ t < +∞, +∞ Z x(t) J= e−νt u(t)dt → max . u(·) 1 + x(t)

x(0) = x0 > 0, (35)

0

Here x is a one-dimensional phase variable, and u is a one-dimensional control. A measurable function u(·) satisfying the condition u(t) ∈ [0, 1] for almost all t ≥ 0 is referred to as an admissible control if the corresponding integral J of problem (35) is convergent. The positive parameters a, x0 , and ν are assumed to be given. Problem (35) is of interest as a useful example [4] of application of mathematical optimal control theory [5] to resource allocation modeling for a trichome, a microbiological cell colony of special structure. A detailed discussion of the model can be found in [6], where this model was represented in the form of a two-dimensional control system with a terminal type functional and was considered on a finite time interval. Problem (35) is a control problem with bilinear dynamics and with an integral-type functional; the optimal solution of such DIFFERENTIAL EQUATIONS

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a problem can contain singular segments. The optimal solution is constructed on the basis of the Pontryagin maximum principle [5], which is a theorem on necessary optimality conditions. Some additional analysis is usually required for finding possible singular modes. In what follows, we give a complete solution of the optimal control problem (35) and classify the types of optimal solutions. 3.2. Main Results for Problem (35) Lemma 4. For every admissible control u(·), the solution of the Cauchy problem x˙ = a − (a + x)u(t),

x(0) = x0 > 0

is strictly positive for t ≥ 0. Proof. Since x(·) is continuous and x(0) = x0 > 0, it follows that there exists a number τ > 0 such that x(t) > 0 for all t ∈ [0, τ ). Let us show that x(τ ) > 0. Suppose the contrary: x(τ ) = 0. On the time interval [0, τ ], we have a − (a + x(t))u(t) ≥ a − (a + x(t)) · 1 = −x(t),

t ∈ [0, τ ],

which, together with the Chaplygin theorem on differential inequalities [7], implies that x(t) ≥ x− (t)

∀t ∈ [0, τ ],

where x− (t) is the solution of the Cauchy problem x˙ = −x,

x(0) = x0 > 0.

But then 0 = x(τ ) ≥ x− (τ ) = x0 e−τ > 0. The contradiction thus obtained completes the proof of Lemma 4. Lemma 5. For every admissible control u(·), the corresponding trajectory x(·) in problem (35) satisfies the two-sided inequality x− (t) ≤ x(t) ≤ x+ (t), where the lower bound

t ≥ 0,

x− (t) = x0 e−t

(36) (37)

corresponds to the control u(·) ≡ 1 and the upper bound x+ (t) = x0 + a · t

(38)

corresponds to the control u(·) ≡ 0 . The attainability set X(T ) at time T ≥ 0 is given by the formula X(T ) = [x− (T ), x+ (T )] . (39) Proof. By using Lemma 4, one can readily see that for an arbitrary admissible control the right-hand side of the dynamical system in problem (35) satisfies the two-sided estimate −x ≡ a − (a + x) · 1 ≤ a − (a + x)u ≤ a − (a + x) · 0 ≡ a. Inequality (36) is a corollary to the Chaplygin theorem [7] on differential inequalities. The inclusion X(T ) ⊂ [x− (T ), x+ (T )] follows from inequality (36). On the other hand, for a constant control u(t) ≡ v ∈ (0, 1], we have x(T, v) = x0 e−vT + a


1−v 1 − e−vT ≡ h(v). v

By performing straightforward computations, we obtain h(+0) = x+ (T ) and h(1) = x− (T ). Defining the function h(v) at zero to be equal to its right limit, we find that h(v) is continuous in the argument v; moreover, h(0) ≤ h(v) ≤ h(1). By using the property of a continuous function to pass through all intermediate values, we obtain (39). DIFFERENTIAL EQUATIONS

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Corollary 1. The functional of problem (35) is well defined for every admissible control u(·). Indeed, the integrand e−νt (x(t)/(1 + x(t)))u(t) admits the majorant e−νt , which implies that the integral J is convergent for each admissible control. Lemma 6 (the main lemma). The functional of problem (35) admits the representation Z+∞ J[x(·)] = W (x0 ) + e−νt G(x(t))dt,

(40)

0

where the functions W (·) and G(·) are given by the formulas Zx W (x) =

ξ dξ, (1 + ξ)(a + ξ)

x ≥ 0,

(41)

0

G(x) = aW 0 (x) − νW (x),

x ≥ 0.

(42)

Proof. Since u = (a − x) ˙ /(a + x) in problem (35), it follows that the functional J can be represented in the form Z+∞ J[x] = e−νt

x a − x˙ × dt = 1+x a+x

Z+∞ e−νt

0 +∞ Z

Z+∞

0

0

0

e−νt aW 0 (x(t))dt +

=

ax dt + (1 + x)(a + x)

Z+∞ e−νt

−xx˙ dt (1 + x)(a + x)

0

 −νt d −e W (x(t)) dt. dt

By taking account of the relation −e−νt

 d d  −νt W (x(t)) = −e W (x(t)) − νe−νt W (x(t)) dt dt

and by performing integration by parts, we obtain t=+∞ Z+∞ J[x] = −e−νt W (x(t)) + e−νt [aW 0 (x(t)) − νW (x(t))] dt, t=0

0

which, together with the inequality W (x) ≤ x, x ≥ 0, −νt W (x(t))e ≤ (x0 + at) e−νt , finally implies that

x(t) ≤ x+ (t) = x0 + at, t ≥ 0,

t ≥ 0,

Z+∞ J[x] = W (x(0)) + e−νt [aW 0 (x) − νW (x)] dt.

(43)

0

From relation (43) and notation (42), we obtain (40). The proof of Lemma 6 is complete. Lemma 7. The functions (41) and (42) have the following properties. First, W (0) = 0, W 0 (x) > 0

W (x) > 0

∀x > 0,

∀x > 0, W 00 (x) > 0,
√ W 00 (x) < 0, x ∈ a, +∞ ;

W 0 (0) = 0,  √
x ∈ 0, a ,

W 0 (+∞) = 0, √
W 00 a = 0,

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Fig. 4. The graphs of the functions: (a) W 0 (·) and (b) G(·).

i.e.,

√ a = arg max W 0 (x) is the unique maximizer of the function W 0 (·). Second, 0≤x<+∞

G(0) = 0, G0 (x) > 0,

G(+∞) < 0, G0 (0) = aW 00 (0) > 0, x ∈ [0, x∗ ) , G0 (x∗ ) = 0, G0 (x) < 0,

G0 (+∞) = 0, x ∈ (x∗ , +∞) ;

i.e., x∗ = arg max G(x) is the unique maximizer of the function G(·). 0≤x<+∞

Proof. All assertions of the lemma about the properties of the functions W (·) and G(·) follow from (41), (42), and the relations a − x2 , (1 + x)2 (a + x)2 a2 − ax2 − νx(1 + x)(a + x) G0 (x) = aW 00 (x) − νW 0 (x) = . (1 + x)2 (a + x)2

W 0 (x) =

x , (1 + x)(a + x)

W 00 (x) =

Figure 4 represents the graphs of the functions W 0 (·) and G(·). Theorem 2. The optimal trajectory xopt (t) of problem (35) is given by the formula xopt (t) = arg max G(x),

x ≥ 0.

(44)

x∈X(t)

The optimal control uopt (t) in problem (35) has the form uopt (t) =

a − x˙ opt (t) a + xopt (t)

(45)

at the points of differentiability of the trajectory (44). Proof. The assertions of Theorem 2 follow from Lemmas 5–7. Indeed, by setting x ¯(t) = arg max G(x)

(46)

x∈X(t)

and by choosing an arbitrary admissible trajectory x(t) ∈ X(t), on the basis of the representation (40) we write out the following expression for the increment of the functional J : Z+∞ ∆J ≡ J[x] − J [¯ x] = e−νt [G(x(t)) − G (¯ x(t))] dt. 0

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(47)

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It follows from the definition of the function (46) that G(x(t)) − G (¯ x(t)) ≤ 0 for all t ≥ 0, which, together with (47), implies that ∆J ≤ 0 for the increment of the functional, and hence we find that the trajectory (46) is optimal. Therefore, the trajectory (44) provides the maximum of the functional J in problem (35). Relation (45) follows from the differential equation (35). The proof of Theorem 2 is complete. Note that, in this case, we have Remark 1 with formulas (5) and (27) replaced by (35) and (44), respectively, and Remark 2 with η replaced by x. 3.3. Analysis of the Singular Mode in Problem (35) In problem (35), the Hamilton–Pontryagin function, the adjoint equation, and the switching function have the form x u + ψ[a − (a + x)u], 1+x  1 ψ˙ = −Kx0 ≡ − e−νt − ψ u, (1 + x)2 x π ≡ Ku0 = e−νt − ψ(a + x). 1+x

K(x, ψ, u) = e−νt

From the identities π ≡ 0 and π˙ ≡ 0, we obtain ψ ≡ e−νt

x = e−νt W 0 (x), (1 + x)(a + x)

ψ˙ ≡ −νe−νt W 0 (x) + e−νt W 00 (x)x. ˙

(48)

By virtue of the main and adjoint equations, the second identity can be rewritten in the form   1 −νt − e − ψ u ≡ −νe−νt W 0 (x) + e−νt W 00 (x)(a − (a + x)u) (1 + x)2 or, by (48),

 −νt

− e

 1 −νt 0 − e W (x) u (1 + x)2 ≡ −νe−νt W 0 (x) + ae−νt W 00 (x) − e−νt

a − x2 u, (1 + x)2 (a + x)

whence, after collecting similar terms and multiplying by the positive factor exp(νt) term by term, we obtain G0 (x) = 0. By using Lemma 7, we find that the trajectory preserves the constant value x ≡ xsing = x∗

(49)

along a possible singular mode; moreover, the parameter x∗ is the unique maximizer the function G(·). By differentiating relation (49) with respect to time, we obtain x˙ ≡ 0, which, together with the basic equation, implies that the singular control has the form u = using ,

using = a/(a + x∗ ) ∈ (0, 1).

(50)

3.4. Classification of Types of Optimal Solutions of Problem (35) Depending on the initial state x(0) = x0 > 0, we consider various cases, x0 ∈ (0, x∗ ), x0 = x∗ , and x0 > x∗ . Throughout this section, we assume that xsing ≡ x∗ and using ≡ u∗ . √ The assertion of Lemma 7 implies that x∗ < a. DIFFERENTIAL EQUATIONS

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Fig. 5. The graphs of the optimal phase trajectory xopt (·): (a) for x0 ∈ (0, x∗ ) and (b) x0 > x∗ .

If x0 ∈ (0, x∗ ), then we have one singular segment [τ, +∞), where τ is a breaking point of the optimal trajectory,   0 if 0 ≤ t < τ x+ (t) if 0 ≤ t < τ xopt (t) = uopt (t) = using if τ ≤ t < +∞; xsing if τ ≤ t < +∞, moreover, the switching point τ = (x∗ − x0 )/a > 0 is found from the equation x∗ = x+ (τ ). If x0 = x∗ , then the singular segment fills the entire interval [0, +∞) : xopt (t) ≡ xsing and uopt (t) ≡ using . If x0 > x∗ , then we have one singular segment [τ, +∞), where τ is the breaking point of the optimal trajectory, and  n 1 if 0 ≤ t < τ x (t) if 0 ≤ t < τ xopt (t) = x− u opt (t) = u if τ ≤ t < +∞, if τ ≤ t < +∞, sing

sing

x0 > 0 is found from the equation x∗ = x− (τ ). x∗ Figure 5 represents the graphs of optimal trajectories depending on the position of the point x0 .

moreover, the switching point τ = ln

4. THE GENERAL CASE 4.1. Statement of the Problem. Main Result We consider the optimal control problem x˙ = f0 (x) + f1 (x)u, 0 ≤ t < +∞, Z+∞ J= e−νt [h0 (x) + h1 (x)u] dt → max .

x(0) = x0 > 0, (51)

u(·)

0

Here x and u are the one-dimensional phase variable and control, respectively. A measurable function u(·) satisfying the condition u(t) ∈ [0, 1] for almost all t ≥ 0 is referred to as an admissible control if the corresponding integral J in problem (51) is convergent. The positive parameters x0 and ν are assumed to be given. The following assertion is valid for problem (51). Theorem 3. Let the following conditions be satisfied in problem (51) : 10 the functions fi (x) and hi (x), i = 0, 1, are defined on [0, +∞) and are smooth on (0, +∞); moreover, |f1 (x)| > 0 for all x > 0 [i.e., f1 (x) preserves its sign for x > 0]; 20 the functions x− (t) and x+ (t) are defined for all t ≥ 0, where either x− (·) is the trajectory corresponding to the control u ≡ 0 and x+ (·) is the trajectory corresponding to the control u ≡ 1 DIFFERENTIAL EQUATIONS

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[for f1 (·) > 0] or x− (·) is the trajectory corresponding to the control u ≡ 1 and x+ (·) is the trajectory corresponding to the control u ≡ 0 [for f1 (·) < 0]; Rx 30 the function w(x) = 0 (h1 (ξ)/f1 (ξ)) dξ is defined for all x > 0; 40 the limit relation limt→+∞ {e−νt w(x(t))} = 0 is valid for each admissible trajectory x(·); 50 the function g(x) = h0 (x)−w0 (x)f0 (x)+νw(x) has a global maximum on the half-line (0, +∞), which is attained at a unique point x ˆ ∈ (0, +∞); moreover, g0 (x) > 0 for x ∈ (0, xˆ) and g0 (x) < 0 for x ∈ (ˆ x, +∞) ; 60 the dual inequality 0 ≤ −f0 (x)/f1 (x)|x=ˆx ≤ 1 is valid. Then the optimal trajectory xopt (·) in problem (51) is given by the formula xopt (t) = arg max g(x), x∈X(t) 0 ≤ t < +∞, and the optimal control uopt (·) has the form uopt (t) = (x˙ opt (t) − f0 (xopt (t))) /f1 (xopt (t)) at the points of differentiability of the trajectory xopt (·). Remark 3. The relations obtained for the optimal solution can be further clarified in specific cases (see Sections 2 and 3). Proof. The use of the Chaplygin theorem under the assumptions of Theorem 3 permits one to prove the two-sided inequality x− (t) ≤ x(t) ≤ x+ (t)

∀t ≥ 0,

(52)

where x(·) is the trajectory corresponding to an arbitrary admissible control u(·). Moreover, the attainability set X(t) at time t ≥ 0 has the form X(t) = [x− (t), x+ (t)]; i.e., the attainability set is a segment whose endpoints satisfy the inequalities 0 < x− (t) < x+ (t) < +∞,

t > 0.

Then, by taking account of the relations u=

x˙ − f0 (x) , f1 (x)

w0 (x) =

h1 (x) , f1 (x)

d ˙ [w(x)] = w0 (x)x, dt

one can represent the functional J of problem (51) in the form     Z+∞ Z+∞ x˙ − f0 (x) h1 (x) h1 (x) −νt J= h0 (x) + h1 (x) e e−νt h0 (x) − dt = f0 (x) + x˙ dt f1 (x) f1 (x) f1 (x) 0 +∞ Z

0

e−νt [h0 (x) − w0 (x)f0 (x) + w0 (x)x] ˙ dt.

= 0

Since Z+∞ Z+∞ Z+∞
−νt 0 −νt −νt t=+∞ −νe−νt w(x(t))dt e [w (x)x] ˙ dt = e dw(x(t)) = e w(x(t))|t=0 − 0

0

0

Z+∞

νe−νt w(x(t))dt,

= −w (x0 ) + 0

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it follows that

1707

Z+∞ J = −w (x0 ) + e−νt g(x(t))dt, 0

which, together with assumptions 50 and 60 of the theorem, implies the desired assertion about the form of the optimal solution. Remark 4. By virtue of (52), it suffices to verify the limit relation in assumption 40 in Theorem 3 on the two trajectories x− (·) and x+ (·). 4.2. Examples In this section, we consider examples for which the general scheme of Theorem 3 can be used. Example 1 [6, 8]. Consider problem (35). Then x˙ = a − (a + x)u, 0 ≤ t < +∞, +∞ Z x J= e−νt u dt → max . u(·) 1+x

x(0) = x0 > 0,

0

Here f0 (x) = a, f1 (x) = −(a + x), h0 (x) = 0, h1 (x) = x/(1 + x), Zx

0

w (x) = h1 (x)/f1 (x) = −x/[(1 + x)(a + x)],

w(x) = −

ξ dξ, (1 + ξ)(a + ξ)

0

g(x) = −aw0 (x) + νw(x),

g0 (x) =

−νx − (a + aν + ν)x2 − aνx + a2 , (1 + x)2 (a + x)2 3

and the maximizer x ˆ > 0 exists and is uniquely determined by the equation g0 (x) = 0. Example 2 (the Ramsey model [2, 9]). In this case, we have x˙ = −µx + f (x)u, 0 ≤ t < +∞, R +∞ x(0) = x0 > 0, and J = 0 e−νt f (x)(1 − u)dt → maxu(·) . Here the function f (x) satisfies the neoclassical conditions [2], w(x) = −x, g(x) = f (x) − (µ + ν)x, and the maximizer x ˆ > 0 exists and is uniquely determined by the equation g0 (x) = 0. Example 3. Consider problem (5) for β = 1. Then x˙ = −µx + u, 0 ≤ t < +∞, +∞ Z J= e−νt x(1 − u)dt → max .

x(0) = x0 ,

u(·)

0

Here w(x) = −x2 /2, g(x) = x(1 − (µ + ν/2)x) = x(1 − σx), and x ˆ = 1/(2σ) = 1/(2µ + ν). Example 4. Let us separately consider problem (1) for α = 0.5, % = 0, and δ = ν. In this case, we have √ x˙ = −µx + u x, 0 ≤ t < +∞, x(0) = x0 , Z+∞ √ J= e−νt x(1 − u)dt → max . u(·)

0

Here w(x) = −x, g(x) =

√ √ x (1 − (µ + ν) x), and x ˆ = 1/(2(µ + ν))2 .

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ACKNOWLEDGMENTS The work was financially supported by the program “Universities of Russia” (project no. UR.03.03.008) and the program of support for leading scientific schools (project no. NSh-1846.2003.1). REFERENCES 1. Ashmanov, S.A., Vvedenie v matematicheskuyu ekonomiku (Introduction to Mathematical Economics), Moscow, 1984. 2. Ashmanov, S.A., Matematicheskie modeli i metody v ekonomike (Mathematical Models and Methods in Economics), Moscow, 1980. 3. Essays on the Theory of Optimal Economic Growth, Shell, K., Ed., The M.I.T., 1967. 4. Kiselev, Yu.N. and Orlov, M.V., Vestn. Mosk. Univ. Ser. 15. Vychislit. Matematika i Kibernetika, 1998, no. 3, pp. 23–26. 5. Pontryagin, L.S., Boltyanskii, V.G., Gamkrelidze, R.V., and Mishchenko, E.F., Matematicheskaya teoriya optimal’nykh protsessov (Mathematical Theory of Optimal Processes), Moscow, 1961. 6. Berg, H. van den, Kiselev, Yu.N., and Kooijman, S.A.L.M., Orlov, M.V., J. Math. Biol., 1998, vol. 37, pp. 28–48. 7. Berezin, I.S. and Zhidkov, N.P., Metody vychislenii (Numerical Methods), Moscow, 1961. 8. Kiselev, Yu.N. and Orlov, M.V., Materialy nauchnogo seminara “Matematicheskie modeli v ekonomike i biologii” (Proc. Sci. Sem. “Mathematical Models in Economics and Biology”), Moscow, 2003, pp. 72–74. 9. Kiselev, Yu.N. and Orlov, M.V., Materialy nauchnogo seminara “Matematicheskie modeli v ekonomike i ekologii” (Proc. Sci. Sem. “Mathematical Models in Economics and Ecology”), Moscow, 2004, pp. 49–52.

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