About Minimal Time Impulsive Control of Sequential Batch Reactors ...

Proceedings of the 2007 American Control Conference Marriott Marquis Hotel at Times Square New York City, USA, July 11-13, 2007

FrC19.3

About minimal time impulsive control of sequential batch reactors with several species A. Rapaport, P. Gajardo and H. Ramirez Abstract— We consider the optimal control problem of feeding in minimal time a tank where several species compete on a single resource. We allow controls to be bounded measurable functions of time as well as impulses. For the one species case, we extend former results given by Moreno [12] to the framework of impulse controls. For the two species case with increasing growth functions, we show that the solution of the minimal time problem consists in either one impulse control at initial time filling the whole tank, either reaching a singular arc. This last one is optimally reached with an impulse or waiting, and consists in keeping the system at a constant resource level until the tank is filled. We give conditions under which the first strategy is optimal, as well as an example where it is not.

I. I NTRODUCTION Sequential batch reactors (SBR) are often used in biotechnological industries, notably in waste-water treatments. Typically, a tank is filled with activated sludge or biological micro-organisms capable to degrade some undesirable substrate. The method consists then in a sequence of cycles composed of three phases: - Phase 1: filling the reactor with water to be treated, - Phase 2: waiting the concentration of the undesirable substrate to decrease until a given (low) concentration, - Phase 3: emptying the reactor from the ’clean’ water, leaving the sludge inside. The time necessary to achieve such cycles can be substantially long and have economical impact on the overall process. Manipulating the input flow during the filling phase has clearly an influence on the total duration of the cycle (more precisely the duration of phases 1 and 2, duration of phase 3 being fixed). But the nonlinear kinetics of the biological reactions does not make always obvious to determine which input flow strategy minimizes the total time. Very similar problems (optimizing the production of biomass at a fixed terminal time) have already been tackled with the help of optimal control theory [1], [7], [6] and has led to computational methods [20], [8]. For models with one biological species, a complete solution of the minimal time problem, for monotonic as well as for non-monotonic kinetics, has been proposed by Moreno in [12]. It has been proved that for monotonic growth functions, such as the wellknown Monod law, the optimal solution consists in a ’most This work was supported by the INRIA-CONICYT grant. A. Rapaport is with the UMR Analyse des Syst`emes et Biom´etrie, Montpellier, France [email protected] P. Gajardo is with the Departamento de Matem´atica, Universidad T´ecnica Federico Santa Mar´ıa, Valpara´ıso, Chile [email protected]. This author was partially supported by Fondecyt project N 3060068. H. Ramirez is with the CMM (CNRS UMI 2807), Universidad de Chile, Santiago, Chile. [email protected]

1-4244-0989-6/07/$25.00 ©2007 IEEE.

rapid approach’ filling as quick as possible the tank up to its maximum capacity and waiting. The optimality proof is based on a technique due to Miele [9], using the Green’s theorem. More precisely, the proof relies on a reformulation in a planar problem. In the present work, we consider minimal time problems where several species compete for the same substrate. For these cases, the problem cannot be reformulated into a planar one, and the technique mentioned above does not apply. Nevertheless, we are interested in characterizing biological systems for which the ’most rapid approach’ strategy is still optimal. In systems with two species, we show that the optimal strategy is the most rapid approach or to fill the tank keeping constant the concentration of the substrate, even when the species’s growth functions are both increasing. This type of strategy is related to the existence of a ’singular arc’. We allow also the control variable to be unbounded. A bounded measurable control can be assimilated to a device that tunes the speed of a pump over a certain range, while an unbounded control can be assimilated to an instantaneous dilution of a positive volume. II. F ORMULATION

OF THE PROBLEM

The dynamics of a SBR is given by the system (see [19])  F   x˙ i = µi (s)xi − xi , xi (t0 ) = yi (i = 1 · · · n)   v  n X F (1) s˙ = − µj (s)xj + (sin − s), s(t0 ) = z   v  j=1   v˙ = F, v(t0 ) = w

where xi , s and v stand respectively for the concentration of the ith species, the concentration of the substrate and the current volume of water present in the tank. The growth functions µi (·) are non-negative functions such that µi (0) = 0, and the input flow F is a non-negative control variable. Consider the domain D = IRn+ ×]0, sin [×]0, vmax [ and a target T = IRn+ ×]0, sout ] × {vmax }. From any initial condition ξ = (y, z, w) in D, the objective is to reach T in minimal time. Let us write V (·) the value function of the problem  V (ξ) = inf t − t0 | sξ,F (t) ≤ sout , v ξ,F (t) = vmax , F (·)

where s (·), v ξ,F (·) denote solutions of (1) with initial condition ξ ∈ D and control F (·). We allow here F (·) to be unbounded. The question of proper treatment of optimal control problems with unbounded or ”impulse” controls have already been studied in the literature (see [4], [5], [13], [14],

6116

ξ,F

FrC19.3 [15], [17] or [10], [11]). For our minimal time problem, with a scalar control entering linearly in the dynamics, this amounts to consider measures dF , that can be decomposed into a sum of a measure absolutely continuous with respect to the Lebesgue measure dt, and a singular or ”impulsive” part dF (t) = u(t)dt + dσ

A second strategy, analyzed in this paper, is defined as follows. Consider a time t0 , a state ξ = (y, z, w) ∈ D and a positive number s¯. The Singular Arc strategy SA(¯ s) consists in making 1. a. if z < s¯, an impulse of volume w(¯ s −z)/(sin − s¯) at t0 , i.e. r(τ ) = 0, u(τ ) = umax , for τ ∈ [t0 , t0 + w(sin − z)/umax(sin − s¯)] (then s and v jump to s¯ and v¯ = w(sin − z)/(sin − s¯) respectively). (see [22]). When an ’impulse’ dσ occurs a time t, the b. if z > s¯, no feeding until the concentration s(·) − + volume v jumps from v (t) to v (t), which implies that reaches the value s¯, i.e. r(τ ) = 1, u(τ ) = 0 for the concentrations xi and s also jump as follows y,z ¯ τ ∈ [t , (t) = s¯, and   0 t¯], where t¯ is such that s − − − v (t) v (t) v (t) y,z + − + − s (·) is the solution of the free dynamics xi (t) = xi (t) + , s (t) = s (t) + +sin 1 − + v (t) v (t) v (t)  xi (t0 ) = yi (i = 1 · · · n)  i,  x˙ i = µi (s)x This is equivalent to integrate the dynamics n X (3) µj (s)xj , s(t0 ) = z. u ds u dv dxi   s˙ = − = − xi , = (sin − s), =u j=1 dτ v dτ v dτ − + Let us denote by x ¯ the vector of species concentrations from τ to τ with any control u(·) such that at t¯, where t¯ is the first instant for which s(t¯+ ) = s¯. Z τ+ 2. a singular arc by taking r(τ ) = 1 and a suitable control u(τ )dτ = v + (t) − v − (t) . − τ u(·) ensuring s(τ ) = s¯ for any τ ∈ (t¯, T˜], where T˜ is The trajectories of system (1) can then be parameterized such that v(T˜) = vmax . by a fictitious time τ such that dt = rdτ , where r is a control 3. if s¯ > sout , no feeding is applied until the concentrathat takes values r = 1 when dF is regular at τ , and r = 0 tion s(·) reaches sout . when dF is impulsive :  dxi u IV. D ERIVATION FROM THE M AXIMUM P RINCIPLE   = rµi (s)xi − xi (i = 1 · · · n)   dτ v  Define the Hamiltonian H(x, s, v, u, r, p, k, q) :=  n  ds X ! u µj (s)xj + (sin − s) = −r n (2) P u  dτ v  xj µj (s) r + qu + k v (sin − s) − r j=1   j=1  dv  (4)  n = u. P dτ pj xj (rµj (s) − u/v) + j=1 Here, the controls r(·) and u(·) are sought among measurable functions w.r.t. τ , that take values in [0, 1] and [0, umax], The Pontryaguin Maximum Principle (PMP) says that if respectively. (x, s, v, u, r)(·) is a solution of the minimal time problem Remark 2.1: Since we can always take r = 0 and u = 0 associated to the system (2), then there exists a n dimensional on a arbitrarily large interval without modifying the optimal multiplier p(·) and scalar multipliers q(·) and k(·), such that time, this problem has no unique solution. Hence, in what  dpi follows, we will be only interested in solutions satisfying   = pi u/v − r(pi − k)µi (s)   r(τ ) 6= 0 or u(τ ) 6= 0 for all (fictitious) time τ . dτ    n  Notice also that dynamics (2) possesses an invariant : P   dk (pj − k)xj µ′j (s) + ku/v = −r      dτ n n j=1 X X   (5)  n yi + z − sin  := ρ(ξ). v xi + s − sin  = w   X  u dq  k(sin − s) −  pj xj  = j=1 j=1  2  dτ v   j=1  III. T HE ONE IMPULSE AND SINGULAR ARC STRATEGIES  pi (T ) = 0, k(T ) = 1 At a time t0 and a state ξ = (y, z, w) ∈ D, we consider the Immediate One Impulse strategy (IOI) which consists where T is the optimal terminal time. Moreover, the Hamiltonian (u, v) −→ H(x(τ ), s(τ ), v(τ ), ·, ·, p(τ ), k(τ ), q(τ )) is in making minimized in u(τ ) and r(τ ), at any τ ∈ [0, T ]. 1. an impulse of volume vmax − w at t0 , i.e. r(τ ) = 0, Define the variables p˜i := pi − k, whose dynamics are u(τ ) = umax , for τ ∈ [t0 , t0 + (vmax − w)/umax ], 2. no control until the concentration s(τ ) reaches sout , i.e. d˜ p = A(τ )˜ p, p˜i (T ) = −1, (6) r(τ ) = 1, u(τ ) = 0 for τ > t0 + (vmax − w)/umax . dτ Notice that the state of the system after an impulse of where A(τ ) is time dependent matrix. Consequently one has vmax − w from ξ = (y, z, w) ∈ D can be easily expressed as p˜(τ ) 6= 0 for any τ ∈ [0, T ]. One has also     w w w argmin H = argmin uφu (x, s, p, k, q)+rφr (x, s, p, k), (7) y , vmax . ,z + sin 1 − vmax vmax vmax u,r u,r

6117

FrC19.3

with

  k   φu (x, s, p, k, q) = q + v (sin − s) −    φr (x, s, p, k)

= 1+

n P

1 v

n P

pj xj

j=1

p˜j µj (s)xj

j=1

If we derive with respect to the fictitious time τ we obtain dφu (sin − s) dφr (sin − s) = −r h˜ p, mi, =u h˜ p, mi (8) dτ v dτ v  ′  µ1 (s(τ ))x1 (τ )   .. where m = m(τ ) =  (9) . . µ′n (s(τ ))xn (τ )

Finally, it is straightforward to check   dm d , h˜ p, mi = p˜, A⊤ m + dτ dτ

(10)

Assumption A0. The functions µi (·) are non decreasing. The proof of the following lemma is direct. Lemma 4.1: Under Assumption A0, the following assertions hold: i. the matrix A(τ ) has non-negative off-diagonal terms, i.e. the dynamical system (6) is cooperative (see [18]), ii. the vector m(τ ), defined in (9), lies in IRn+ . V. T HE COST

OF THE ONE IMPULSE STRATEGIES

We consider a family of functions ϕc (·) defined on IRn+ ×]0, sin [ and parameterized by c ∈ [sout , sin ] ϕc (y, z) = inf {t | sy,z (t) ≤ c} ,

j=1

(∂yj ϕc (y, z) − ∂z ϕc (y, z))µj (z)yj + 1 = 0

(11)

z ≤ c.

(12)

The time costs of the Immediate One Impulse strategy can be written in terms of functions ϕ(·) , as follows    w w w TIOI (ξ) = ϕsout y . ,z + sin 1 − vmax vmax vmax

∆r V (ξ) ≥ 0 ,

∆u V (ξ) ≥ 0 ,

For convenience, we shall denote by (˜ y , z˜) the concentrations obtained with an impulse of volume vmax − w from state ξ = (y, z, w) ∈ D    w w w . (17) ,z + sin 1 − (˜ y , z˜) = y vmax vmax vmax The reader can check that an easy computation leads to the following lemma. Lemma 6.1: Under Assumption A1, at any ξ ∈ D, one has ∆r TIOI (ξ) = n X
y , z˜) y˜j (µj (z) − µj (˜ z )) . y , z˜) − ∂z ϕsout (˜ ∂yj ϕsout (˜

(18) We then have the following result concerning the optimality of the IOI strategy for any initial condition. Proposition 6.1: Under Assumption A1, TIOI is solution of the HJB equation (13) with boundary condition (14) if and only if

∆u V (ξ) = −

j=1

(19)

Take ξ ∈ D. If ξ ∈ T , one has TIOI (ξ) = 0, thus boundary condition (14) is fulfilled. At ξ = (y, z, w) ∈ D\T , condition (19) is exactly the first variational inequality (15). It is easy to check the second variational inequality (16)

(∂yj V (ξ) − ∂z V (ξ))µj (z)yj + 1,

n X

∀ξ ∈ D \ T .

Proof: We proceed to show that the function TIOI is a C 1 solution of the two variational inequalities (15-16) with boundary condition (14).

Define

j=1

(15) (16)

Assumption A1. For any c ∈ [sout , sin ], the function ϕc (·) is C 1 , that is, continuously differentiable.

CHARACTERIZATION

n X

(14)

independently of the upper bound umax .

VI. T HE H AMILTON -JACOBI -B ELLMAN

∆r V (ξ) =

z ∈]0, sout ] .

It is straightforward to check that equation (13) is equivalent to the system of two variational inequalities

∆r TIOI (ξ) ≥ 0,

on the domain IRn+ ×]c, sin [ with boundary conditions ϕc (., z) = 0,

V (·, z, vmax ) = 0,

j=1

where sy,z (·) is solution of the free dynamics (3). Standard analysis of minimal time problems shows that ϕc (·) are Lipschitz-continuous functions and solutions, in the viscosity sense, of the partial differential equation (see for instance [2]) n X

The Hamilton-Jacobi-Bellman equation (HJB) is written as umax min (0, ∆r V (ξ)) + min (0, ∆u V (ξ)) = 0 (13) w for ξ ∈ D \ T , with the boundary condition

∂yj V (ξ)yj + ∂z V (ξ)(sin − z) + ∂w V (ξ)w.

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∆u TIOI (ξ) = X w (sin − z) − y , z˜) y , z˜)y˜j + ∂z ϕsout (˜ ∂yj ϕsout (˜ v max j X w y , z˜) y , z˜)y˜j − ∂z ϕsout (˜ ∂yj ϕsout (˜ + (sin − z) = 0 . vmax j

FrC19.3 Let us now consider the function ψ(ξ, c) = ϕc (y, z) + TIOI (x(tc ), c, w),

ξ ∈ D, c ∈]0, z],

y,z

y,z

where tc is such that s (tc ) = c, with s (·) the solution of the free dynamics (3). The reader can check easily the results of the next technical lemma. Lemma 6.2: Under Assumption A1, one has ∂c ψ(ξ, z) = −

∆r TIOI (ξ) , n X µj (z)yj

ξ∈D .

Proof: Consider p and k solutions of PMP-system (5), and m defined in (9). Define the auxiliary variable p˜ = p−k. Recall that since (6) it follows that p˜(τ ) 6= 0, for all time τ . If I is a singular arc, it follows that the first derivatives of φu and φr are null on I. Since controls u and r are not simultaneously null and via equations (8), it is equivalent to saying that h˜ p, mi = 0 on I. Deriving this last equation w.r.t. τ and using expression (10), it holds that   dm ⊤ = 0 on I. (21) h˜ p, mi = 0 and p˜, A m + dτ

Since p˜ is always non null and has dimension 2, equalities (21) are satisfied if and only if m and A⊤ m+ dm dτ are linearly From this Lemma, one can easily deduce the result. dependent on I. We easily verify that, under A2, this is equivalent to s˙ = 0 on I. Indeed, a simple computation Proposition 6.2: At states ξ ∈ D \ T such that leads to

∆r TIOI (ξ) < 0, the IOI strategy cannot be optimal.  

m × A⊤ m + dm = |s| ˙ x1 x2 (µ′1 (s)µ′′2 (s) − µ′′1 (s)µ′2 (s))

dτ VII. T HE ONE SPECIES CASE (n = 1) from which it is easy to conclude. We first consider increasing growth functions. Assumption A3. For any (s1 , s2 ) ∈ [sout , sin ], one has Proposition 7.1: Under Assumption A0, the IOI strategy {s2 ≥ s1 ⇒ µ2 (s2 )µ1 (s1 ) ≥ µ1 (s2 )µ2 (s1 )} . (22) is optimal for any initial condition in D. j=1

We consider now C1 function µ that attains a unique isolated maximum point s∗ ∈]0, sin [ and such that µ′ (s) > 0 for all s ∈ [0, s∗ [, µ′ (s) < 0 for all s > s∗ , and µ′ (s∗ ) = 0. One instance of such functions is typically the Haldane law: µmax s µ(s) = , K + s + s2 /R This kind of growth functions occurs in bioprocesses where the substrate is a toxic substance and, for big concentrations, inhibits the activity of the biomass x [19].

Remark 8.2: Assumption A3 is fulfilled for growth functions constant or linear on [sout , sin ]. For Monod laws (20), condition (22) is exactly fulfilled when K2 ≥ K1 . Theorem 8.3: Suppose that the initial volume v(0) = w is strictly lower than vmax . Under Assumptions A0, A1, A2 and A3, the solution of the minimal time problem associated to the system (2) consists in either the IOI strategy, or a SA(¯ s) strategy. The proof of Theorem 8.3 is based on two technical lemmas.

Proposition 7.2: For an initial condition in D, the SA(s∗ ) strategy is optimal. Because of lack of space, we do not provide proofs of these two propositions They lie on the verification of the HJB equation (13) and use of the PMP (see Section IV). These two results have been previously obtained in [12], but only for measurable and bounded control laws. VIII. T HE TWO

SPECIES CASE

(n = 2)

2

We first consider C functions µi (·) such that [µ′1 /µ′2 ](·) is strictly monotonic (without loss of generality strictly decreasing) : Assumption A2. µ′1 (s)µ′′2 (s) > µ′′1 (s)µ′2 (s), ∀s ∈ (0, sin ]. Remark 8.1: Assumption A2 is fulfilled for Monod laws with K2 > K1 µmax,i s . (20) µi (s) = Ki + s Lemma 8.1: Under Assumption A2, a singular arc1 I is characterized by s˙ = 0 on I. 1 See

Lemma 8.2: Under Assumptions A0 and A3, one has p˜1 ≥ 0 and φr = 0 imply h˜ p, mi ≤ 0.

(23)

Lemma 8.3: Under Assumption A0 and A2, for p˜(τ ) ∈ E(τ ) = {˜ p = (˜ p1 , p˜2 ) | h˜ p, m(τ )i = 0}, we have that sign

ds d h˜ p, mi = − sign p˜1 . dτ dτ

(24)

Proof: Recall first that one has ∆r V (ξ) = φr (x, s, p, k), ∆u V (ξ) = vφu (x, s, p, k, q) (see for instance Theorem 12.5.1 in [21]). Then, from (15), (16), we deduce the positivity of φu and φr . Moreover, from Remark 2.1, φu and φr cannot be both strictly positive. Recall that, under Assumption A0, we have that the matrix A in system (6) is cooperative. Then, since p˜ = (p1 − k, p2 − k) is equal to (−1, −1) at the final time T , we deduce that p˜ ∈ / IR2+ for every instant τ , and moreover, once p˜ reaches the negative octant IR2− it remains therein until T . Our proof splits into this two cases.

[3, Part III Ch. 2] for a definition

6119

FrC19.3 Case 1: p˜1 (t0 ) > 0. Let us discard the following case: a. v < vmax , p˜1 > 0, h˜ p, mi < 0, and φr = 0. In fact, if we have this situation, necessarily u = 0 and r = 1 in order to keep φr nonnegative. Thus, s decreases, φu increases and necessarily, in order to reach the target, there is an instant τ such that h˜ p, mi = 0 and just afterward d h˜ p, mi > 0 (cf. (24)). Since one has h˜ p, mi > 0, due to dτ φr = 0, we obtain a contradiction with Lemma 8.2. Thus, if v < vmax , p˜1 > 0, and h˜ p, mi < 0 then φr > 0 and φu = 0. In such case one has r = 0 and u = umax , which implies that s increases and φr decreases until a instant at which, necessarily in order to reach the target, φr = 0. Since d case (a) above is not possible and dτ h˜ p, mi < 0 (cf. (24)), we have at that instant the equality v = vmax . Afterward, since φr must be nonnegative, one has u = 0 and r = 1. The described situation is the IOI strategy. If v < vmax , p˜1 > 0, and h˜ p, mi > 0, from Lemma 8.2, the unique possibility is φr > 0, and consequently φu = 0. In such case one has r = 0 and u = umax . Hence s and φr increase. Necessarily, in order to reach the target, there exists an instant such that h˜ p, mi = 0. Note that afterward it holds d h˜ p, mi < 0. In this that h˜ p, mi < 0, due once again to dτ situation we have that φr decreases until φr = 0, and from Lemma 8.2, it coincides with the instant at which v = vmax . Afterward, since φr must be nonnegative, it follows u = 0 and r = 1. The described situation is also the IOI strategy. Hence, we have proven that if p˜1 (t0 ) > 0 is positive, then the optimal strategy is IOI. Case 2: p˜1 (t0 ) < 0. Clearly, p˜1 remains always negative. We now proceed to discard the following cases: b. v < vmax , (˜ p1 < 0,) φu > 0, and h˜ p, mi < 0. In fact, this situation implies u = 0 and r = 1. Thus, s decreases, and by (8) we have that φu increases. This d h˜ p, mi is together with (24) implies that the sign of dτ negative, obtaining that h˜ p, mi remains always negative and consequently φu always increases. This does not allow us to reach the target, we thus conclude that this situation cannot happen. c. v < vmax , (˜ p1 < 0,) and h˜ p, mi > 0 Indeed, if φu = 0 then, in order to keep φu nonnegative, one necessarily obtain r = 0 and u = umax . Hence, s and (by (8)) φr increase. This together with (24) implies that the d sign of dτ h˜ p, mi is positive, obtaining that h˜ p, mi remains always positive and consequently φr always increases. This does not permit us to reach the target, we thus conclude that this situation is not possible. Else if φu > 0, we have that u = 0 and r = 1. Then s and φu decrease until h˜ p, mi = 0. So, equation (24) allows us to say that the sign d of dτ h˜ p, mi is negative, obtaining just afterward h˜ p, mi < 0. This implies (in order to keep φr nonnegative) that u = 0, and consequently φu remains always positive. Which does not allow to reach the target. Let us finally analyze the remaining case v < vmax , (˜ p1 < 0,) φu = 0, and h˜ p, mi < 0. Here, in order to keep φu nonnegative, one necessarily has r = 0 and u = umax .

Therefore, s increases and (by (8)) φr decreases until an instant τ ∗ when one of the following three cases occur: - Case φr > 0 and h˜ p, mi = 0: This implies u = umax and r = 0, and consequently s increases. This together with (24) d h˜ p, mi is positive. Obtaining then implies that the sign of dτ that φr will always remain positive, which cannot allows us to reach the target. This case is thus discarded. - Case φr = 0 and h˜ p, mi < 0: This implies u = 0 and r = 1, and therefore (by (8)) φu becomes positive. Since the case (b) above has been discarded, it necessarily follows that v reaches vmax at the same instant τ ∗ , obtaining the IOI strategy. - Case φr = 0 and h˜ p, mi = 0: Due to equality φu = 0 holds at the same instant τ ∗ , this configuration represents a singular arc. In order to conclude the proof, we demonstrate that it corresponds to a singular arc strategy SA(s∗ ) for a given s∗ ∈]0, sin [. Let us prove that when one reaches the instant τ ∗ , then this configuration of values remains constant until v = vmax . Indeed, let us denote by τ¯ the instant so that τ¯ > τ ∗ and φu = φr = 0 on I = [τ ∗ , τ¯]. We claim that the only admissible configuration at this instant τ¯ is (˜ p1 < 0 and) h˜ p, mi < 0. By contradiction, if h˜ p, mi > 0 at τ¯ then, in order to keep φu nonnegative, one necessarily has r = 0 and u = umax . Hence s increases and consequently, by (24), the d h˜ p, mi is negative at τ¯. This contradicts the fact sign of dτ that h˜ p, mi > 0 at such instant. Therefore, at the instant τ¯ (which is the end of the singular arc I) we have h˜ p, mi < 0. In this case we have u = 0 and r = 1 in order to keep φr nonnegative (cf. (8)). Obtaining that s decreases and consequently, by (24), the sign of d p, mi is negative onwards. It yields to say that h˜ p, mi dτ h˜ will always remain negative and then the configuration u = 0 and r = 1 will be hold onwards. It necessarily implies that v = vmax at the instant τ¯. This concludes the proof. We focus now on sufficient conditions for which the IOI strategy is optimal. Assumption A4. µ2 (s) ≥ µ1 (s),

∀s ∈ [0, sin ].

Lemma 8.4: Under Assumptions A1 and A4, for any c ∈ [sout , sin ], (y, z) ∈ IR2+ ×]0, sin [, one has 1. ∂z ϕc (y, z) ≥ ∂y2 ϕc (y, z), 2. ∂y1 ϕc (y, z) ≥ ∂y2 ϕc (y, z), 3. ∂y2 ϕc (y, z) ≤ 0.

Proof: The proof is based on the cooperativity property of the reduced dynamics when t tends to +∞. Proposition 8.1: Under Assumptions A1, A3 and A4, the IOI strategy is optimal for any initial condition in D \ T . Proof: Let us write ∆1 TIOI (ξ) given in (18) as follows ∆1 TIOI (ξ) = 2 X
µj (z) − µj (˜ z) y , z˜) µj (˜ z )y˜j y , z˜) − ∂z ϕsout (˜ ∂yj ϕsout (˜ . µj (˜ z) j=1

6120

FrC19.3 Recall that z˜ ≥ z (17) and µi (·) are non-decreasing (Assumption A0). Then, by Assumption A4, one has

believe that this result should have an important impact for biotechnological applications.

µ1 (z) − µ1 (˜ z) µ2 (z) − µ2 (˜ z) ≤ ≤0 . µ2 (˜ z) µ1 (˜ z)

R EFERENCES

Lemma 8.4 gives the inequality y, z˜) ≤ 0 , y , z˜) − ∂z ϕsout (˜ ∂y2 ϕsout (˜ and consequently ∆1 TB (ξ) ≥ 2 X
µ1 (z) − µ1 (˜ z) y , z˜) µj (˜ z )y˜j y , z˜) − ∂z ϕsout (˜ ∂yj ϕsout (˜ µ1 (˜ z) j=1 µ1 (z) − µ1 (˜ z) =− ≥0. µ1 (˜ z) IX. A N E XAMPLE We consider functions µ1 (·), µ2 (·) that fulfill Assumptions A0, A2, and A3 but not A4 √ µ1 (s) = 5 s, µ2 (s) = s2 , for the values sout = 0.1 and sin = 5 (see Figure 1). 25

21

µ1 17

13

µ2

9

5

1 1

2

3

S out

Fig. 1.

4

5

S in

Graphs of the two growth functions.

We compare the strategies IOI and SA(s∗ ), where s∗ minimizes the cost of SA(¯ s) for s¯ ∈ (Sout , Sin ). For vmax = 10 and initial conditions with y1 = 1, z = 3 and w = 1, we have computed numerically s∗ for different values of y2 . Results are reported on Table I. TABLE I y2 0 10−4 10−3 10−2 0.05 0.1 0.5

T (IOI) 2.320765 2.126756 1.700494 1.186231 0.867009 0.746446 0.522361

s∗ not reached 1.678 1.97 2.46 3.05 3.34 not reached

T (SA(s∗ )) – 1.973976 1.515415 1.101530 0.842255 0.739046 –

gain – 7% 11% 7% 3% 1% –

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This example shows that in presence of a small population of a species more efficient for small substrate concentrations, the Singular Arc strategy may be better that the Immediate One Impulse one, differently from the one species case. We

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