Minimal time crisis versus minimum time to reach a viability kernel : a case study in the prey-predator model

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Minimal time crisis versus minimum time to reach a viability kernel : a case study in the prey-predator model

Térence Bayen, Alain Rapaport

To cite this version:

Térence Bayen, Alain Rapaport. Minimal time crisis versus minimum time to reach a viability kernel : a case study in the prey-predator model. Optimal Control Applications and Methods, Wiley, 2019, 40 (2), pp.330-350. �10.1002/oca.2484�. �hal-01943636v2�

DOI: xxx/xxxx

RESEARCH ARTICLE

Minimal time crisis versus minimum time to reach a viability kernel: a case study in the prey-predator model

Terence Bayen¹ | Alain Rapaport²

1IMAG, Univ. Montpellier, CNRS, Montpellier, France

2MISTEA, Univ. Montpellier, INRA, Montpellier SupAgro, Montpellier, France

Correspondence

Terence Bayen, IMAG, Univ. Montpellier, Place Eugène Bataillon, 34090 Montpellier, France. Email:

[email protected]

Summary

In the present work, we provide a reformulation of the minimal time crisis problem associated with a constraint set as a free terminal time control problem. The proof requires the existence of a non-empty viability kernel that is reachable from the state space. In addition, we suppose a uniform lower bound between two consecu- tive crossing times of the constraint set. Thanks to this result, we compute an optimal synthesis for the minimal time crisis problem governed by the prey-predator dynam- ics, with a controlled mortality on the predators. Finally, we compare the time spent in the crisis set by optimal trajectories of the minimal time crisis problem with the minimum time problem to reach the viability kernel. This is made possible by means of an exact characterization of the viability kernel.

KEYWORDS:

Optimal Control, Viability Theory, Pontryagin Maximum Principle, Prey-predator model.

1 INTRODUCTION

This paper focuses on the notion of time crisis which represents the time spent by a solution of a controlled system outside a given domain𝐾. In many engineering problems, control systems are subject to state constraints (see¹), and the goal is to design admissible controls that bring the system in certain environments, and which are also optimal with respect to a certain criterion.

Since it is sometimes not possible to drive the system in such a way to satisfy state constraints, the study of minimal time crisis problem is an interesting alternative. The crisis set is by definition the complementary of the domain𝐾, and we then wish to find control strategies minimizing the total time spent in this set. In the context of viability theory (see^2,3), when initial conditions of the system are outside the viability kernel^†of𝐾, finding a control for which the associated trajectory spends the least possible time outside𝐾is a crucial issue.

The minimal time crisis problem was introduced in the context of viability theory in⁴(see also^5,6) where a characterization of the value function as a generalized solution of an Hamilton-Jacobi equation has been proposed. Necessary optimality conditions have been given in^7,8using a hybrid maximum principle (see^9,10,11) since the time crisis problem possesses a discontinuous Hamiltonian. For the well-posedness of the adjoint state, it is assumed in⁷ that optimal trajectories hit the boundary of𝐾 transversally (see also¹¹).

In the present paper, we consider a case study in the context of the prey-predator systems to compare two different approaches measuring the time spent by the system outside a constraint set, defined as a prey density above a given threshold (see^12,13).

Assuming that the control acts only on the mortality of predators (^14,15,16), we wish to compare optimal strategies for the two following criteria:

†The viability kernel is defined as the largest subset of𝐾composed of initial conditions for which there exists an admissible control such that the associated solution of the system remains in𝐾for all time𝑡≥0, see^2,3.

1. the minimal time crisis,

2. the minimum time to reach the viability kernel (this domain being here characterized analytically).

Prey-predator models have been widely studied in the literature (seee.g.,13,12,17,18,19,14,15,16 among others) with various objec- tives: optimal control, stabilization, impulse control and others. However, to our best knowledge, the question of determining a viability kernel together with the syntheses for the time crisis and the minimum time problem to reach this set, have not been previously addressed in the literature. In general, it is difficult to characterize a viability kernel. Note also that finding an optimal synthesis for the time crisis problem presents technicalities with respect to the classical application of the Pontryagin Maximum Principle due to the discontinuity of the characteristic function.

The paper is structured as follows. In section 2, we propose in a general framework an equivalent formulation of the minimal time crisis problem over a finite horizon when the viability kernel is non-empty and reachable from the state space. We also assume that the duration between two consecutive crossing times is uniformly bounded by below. This assumption appears to be strong, however we show in this work that it is satisfied in prey-predator systems. This reformulation allows us to obtain optimality conditions on the minimal time crisis problem in this general framework using the hybrid maximum principle. In section 3, we provide an analytical description of the viability kernel for the constraint set (defined as a prey density above a given threshold) under the prey-predator dynamics. In Section 4, we compute an optimal synthesis for the minimum time control problem to reach this set using the Pontryagin Maximum Principle. Finally, we compute in Section 5 an optimal synthesis for the minimal time crisis problem governed by the prey-predator dynamics, using the reformulation obtained in Section 2. In particular, we exhibit a subset of the state space which enjoys the following remarkable property: for any initial condition in this set, the time crisis is strictly less that the time spent in the crisis set by the minimal time strategy to reach the viability kernel.

2 OPTIMALITY CONDITIONS FOR THE TIME CRISIS FUNCTION

In this section, we provide optimality conditions on the time crisis function in a general setup. We first introduce some notations.

Given a non-empty subset𝐶ofℝ^𝑛, we denote byInt(𝐶)its interior and by𝜕𝐶its boundary. We also denote by|⋅|the euclidean norm inℝ^𝑛(𝑛≥1). If𝐶is a non-empty closed convex subset ofℝ^𝑛, the normal cone to𝐶at a point𝑥∈𝐶is defined as

𝑁_𝐶(𝑥) ∶= {𝑝∈ℝ^𝑛; 𝑝⋅(𝑦−𝑥)≤0, ∀𝑦∈𝐶},

where𝑎⋅𝑏denotes the standard scalar product of two vectors𝑎, 𝑏∈ℝ^𝑛. Throughout this section, we consider a control system:

̇𝑥=𝑓(𝑥, 𝑢), (2.1)

where𝑓 ∶ℝ^𝑛×ℝ^𝑚→ℝ^𝑛is the dynamics,𝑥is the state, and𝑢the control that takes values in a non-empty closed subset𝑈of ℝ^𝑚. The admissible control set is classically

 ∶= {𝑢∶ [0,+∞)→𝑈 ; 𝑢meas.}.

We assume the usual regularity assumptions on the dynamics (seee.g.,²⁰):

(H1) The function𝑓 is continuous w.r.t.(𝑥, 𝑢), of class𝐶¹w.r.t.𝑥and satisfies the linear growth condition: there exist𝑐₁>0 and𝑐₂>0such that for all𝑥∈ℝ^𝑛and all𝑢∈𝑈, one has:

|𝑓(𝑥, 𝑢)|≤𝑐₁|𝑥|+𝑐₂, (2.2)

(H2) For any𝑥∈ℝ^𝑛, the velocity set𝐹(𝑥) ∶= {𝑓(𝑥, 𝑢) ; 𝑢∈𝑈}is a non-empty compact convex set.

Under hypotheses (H1)-(H2), for any initial condition𝑥₀ ∈ ℝ^𝑛 and any𝑇 ≥ 0, there exists a unique absolutely continuous solution of (2.1) defined over [0, 𝑇] such that𝑥(0) = 𝑥₀, denoted by𝑥_𝑢(⋅, 𝑥₀)hereafter (this is a consequence of Cauchy- Lipschitz’s Theorem). Given a closed subset𝐾 ofℝ^𝑛 (that plays the role of constraint set), let us now introduce the viability kernel of𝐾under the dynamics𝑓as

Viab(𝐾) ∶= {𝑥₀∈𝐾 ; ∃𝑢∈, 𝑥_𝑢(𝑡, 𝑥₀) ∈𝐾, ∀𝑡≥0}.

If non-empty, the viability kernel is a closed subset of𝐾 and it enjoys several properties. In particular, it can be reached from outside only at its boundary in common with the boundary of the set𝐾. We refer to^2,3for more detailed properties of this set

and its fundamental role in several economic and biological models. Hereafter, we are interested in studying properties of the minimal time crisis function𝑇 ∶ℝ^𝑛→ℝ₊∪ {+∞}which states as

𝑇(𝑥₀) ∶= inf

𝑢∈

+∞

∫

0

𝟙_𝐾𝑐(𝑥_𝑢(𝑡, 𝑥₀)) d𝑡, (2.3)

where 𝟙_𝐾𝑐 is the indicator function of𝐾^𝑐:

𝟙_𝐾𝑐(𝑥) ∶=

{0 if 𝑥∈𝐾, 1 if 𝑥∉𝐾,

and𝐾^𝑐 ∶=ℝ^𝑛∖𝐾 denotes the complementary of the set𝐾. It has been introduced in⁴within the context of viability theory^2,3 typically when the initial state does not belong to the viability kernel. In particular, one can show that it is a lower semi-continuous function (see^4,7,8). Our main concern in this section is to investigate a reformulation of (2.3) over a finite horizon that involves the viability kernel. This will allow us to obtain necessary optimality conditions on the time crisis function, using the hybrid maximum principle⁹. To do so, let us introduce an additional hypothesis:

(H3) The viability kernel of𝐾 under the dynamics𝑓,Viab(𝐾), is non-empty and for any initial condition inℝ^𝑛, there exists a control𝑢∈ steering𝑥₀toViab(𝐾)in finite time.

Consider now the auxiliary optimal control problem:

𝑇̂(𝑥₀) ∶= inf

𝜏≥0,𝑢∈

𝜏

∫

0

𝟙_𝐾𝑐(𝑥_𝑢(𝑡, 𝑥₀)) d𝑡 s.t. 𝑥_𝑢(𝜏, 𝑥₀) ∈ Viab(𝐾). (2.4) Under the additional hypothesis (H3), the existence of an optimal control for both problems (2.3) and (2.4) follows from standard argumentation (see^4,7). Moreover, for any𝑥₀ ∈ℝ^𝑛, one has𝑇(𝑥₀)< +∞and𝑇̂(𝑥₀)<+∞. In order to relate (2.3) and (2.4), we introduce the notion of crossing time.

Definition 1. Given a solution𝑥_𝑢(⋅, 𝑥₀)of (2.1), we say that a time𝑡_𝑐 > 0is acrossing timefor𝑥_𝑢(⋅, 𝑥₀)from𝐾to𝐾^𝑐 if the control𝑢is left- and right-continuous at𝑡_𝑐,𝑥_𝑢(𝑡_𝑐, 𝑥₀) ∈𝜕𝐾, and there exists𝜂 >0such that for any time𝑡∈ [𝑡_𝑐−𝜂, 𝑡_𝑐], resp.

𝑡∈ (𝑡_𝑐, 𝑡_𝑐+𝜂],𝑥_𝑢(𝑡, 𝑥₀) ∈𝐾, resp.𝑥_𝑢(𝑡, 𝑥₀) ∈𝐾^𝑐. Similarly, we define the notion of crossing time from𝐾^𝑐to𝐾.

Next, we consider the following hypothesis:

(H4) There exists a number𝜂 >0such that for any pair of consecutive crossing times𝑡₁< 𝑡₂from𝐾 to𝐾^𝑐or from𝐾^𝑐 to𝐾, one has𝑡₂−𝑡₁≥𝜂,

which allows us to state the following equivalence result.

Proposition 1. Suppose that (H1)-(H2)-(H3)-(H4) are satisfied. Then, for any𝑥₀∈ℝ^𝑛one has 𝑇(𝑥₀) =𝑇̂(𝑥₀).

Furthermore, for any𝑥₀∈ℝ^𝑛, the infimum in (2.4) is reached for a finite𝜏.

Proof. As the value functions𝑇 and𝑇̂ are clearly identically equal to zero inViab(𝑥), we take𝑥₀∉ Viab(𝐾). It is known (see⁴) that one has𝑇(𝑥₀)≤𝑉̃(𝑥₀), where

𝑉̃(𝑥₀) ∶= inf

𝑢∈𝑇_𝑢 s.t. 𝑥_𝑢(𝑇_𝑢, 𝑥₀) ∈ Viab(𝐾).

Moreover, Hypothesis (H3) implies 𝑉̃(𝑥₀) < +∞. Let𝑢^∗(⋅)be an optimal control for 𝑇(𝑥₀) and𝑥^∗ the associated solution starting from𝑥₀. Define the time𝜏(𝑥₀) ∈ℝ₊∪ {+∞}by:

𝜏(𝑥₀) ∶= sup{𝑡≥0 ; 𝑥^∗(𝑡) ∈𝐾^𝑐},

and suppose by contradiction that𝜏(𝑥₀) = +∞. As𝑇(𝑥₀)<+∞, there exists𝑡₀ ≥0such that𝑥^∗(𝑡₀) ∈𝐾. Now, as𝜏(𝑥₀) = +∞, there exists𝑡₁ ≥𝑡₀such that𝑡₁is a crossing time from𝐾 to𝐾^𝑐. We now define𝑡₂as the first entry time𝑡 > 𝑡₁of𝑥^∗from𝐾^𝑐 into𝐾 (𝑡₂exists as𝑇(𝑥₀) < +∞). From (H4), we deduce that𝑡₂−𝑡₁ ≥ 𝜂 >0. By the Dynamic Programming Principle,𝑥^∗ is also optimal from𝑥^∗(𝑡₂)and one has𝑇(𝑥^∗(𝑡₂))≤𝑇(𝑦₀)< +∞together with𝜏(𝑥^∗(𝑡₂)) =𝜏(𝑦₀) = +∞. Therefore, the same argument can be applied from(𝑡₂, 𝑥^∗(𝑡₂))and we obtain two increasing sequences of times(𝑡_𝑖,𝑛)_𝑛,𝑖 = 1,2, such that one has

𝑡_2,𝑛−𝑡_1,𝑛 ≥𝜂 >0for any𝑛 ∈ℕ. It follows that one has𝑇(𝑦₀) = +∞and thus a contradiction. Therefore, we necessary have 𝜏(𝑥₀)<+∞which implies that𝑥^∗(𝑡) ∈𝐾for any time𝑡≥𝜏(𝑥₀)i.e.,𝑥^∗(𝜏(𝑥₀)) ∈ Viab(𝐾). It follows that

𝑇(𝑥₀) =

𝜏(𝑥₀)

∫

0

𝟙_𝐾𝑐(𝑥^∗(𝑡)) d𝑡≥𝑇̂(𝑥₀),

using that𝑥^∗(𝜏(𝑥₀)) ∈ Viab(𝐾). On the other hand, let(𝜏, ̂̂ 𝑢(⋅)) ∈ ℝ₊× be an optimal pair for𝑇̂(𝑥₀). If𝑥(⋅)̂ denotes the associated trajectory, we then have𝑥(̂ 𝜏) ∈ Viab(𝐾̂ )with𝑢̂defined over[0, ̂𝜏]. Hence, we can extend𝑢̂to a control function

̄

𝑢∈ defined on[0 + ∞)such that the associated trajectory𝑢(⋅)̄ satisfies𝑥(𝑡) ∈ Viab(𝐾)̄ for any time𝑡≥𝜏. We then havê 𝑇̂(𝑥₀) =

̂ 𝜏

∫

0

𝟙_𝐾𝑐(𝑥(𝑡)) d𝑡̂ =

+∞

∫

0

𝟙_𝐾𝑐(𝑥(𝑡)) d𝑡̄ ≥𝑇(𝑥₀), and the conclusion follows.

Finally, to prove that the infimum in (2.4) is reached for a finite𝜏, we suppose by contradiction that this is not the case. As previously, we can then find two increasing sequences of times(𝑡_𝑖,𝑛)_𝑛,𝑖= 1,2, such that one has𝑡_2,𝑛−𝑡_1,𝑛≥𝜂for any𝑛∈ℕ.

Similarly, we find a contradiction with the fact that𝑇̂(𝑥₀)is finite, which ends the proof.

Remark 1. Hypothesis (H4) is about trajectories of the system. It appears to be a strong hypothesis, nevertheless, we shall see that it is satisfied for the prey-predator model (Section 5).

In connection with Hypothesis (H4), it is relevant to recall thechattering phenomenon^‡.

Definition 2. Given𝑥₀∈ℝ^𝑛and a solution𝑥_𝑢(⋅, 𝑥₀)of (2.1) defined over[0,+∞), we say that a chattering phenomenon occurs if there exist two sequences of times(𝑡^𝑜𝑢𝑡_𝑛 )_𝑛≥0,(𝑡^𝑖𝑛_𝑛)_𝑛≥0satisfying:

(i) For any𝑛∈ℕ,𝑡^𝑜𝑢𝑡_𝑛 and𝑡^𝑖𝑛_𝑛 are two consecutive crossing times for𝑥_𝑢(⋅, 𝑥₀)from𝐾to𝐾^𝑐and from𝐾^𝑐 to𝐾respectively.

(ii) For any𝑛∈ℕ, one has𝑡^𝑜𝑢𝑡_𝑛 −𝑡^𝑖𝑛_𝑛 >0, and𝑡^𝑜𝑢𝑡_𝑛 −𝑡^𝑖𝑛_𝑛 →0when𝑛→+∞.

Let us finally recall the definition oftransversecrossing times⁷(for convex sets𝐾).

Definition 3. Given an admissible trajectory𝑥_𝑢(⋅, 𝑥₀), a crossing time𝑡_𝑐 (from𝐾to𝐾^𝑐 or from𝐾^𝑐to𝐾) is transverse when there exists𝜈, 𝜈^′∈𝑁_𝐾(𝑥_𝑢(𝑡_𝑐))such that ̇𝑥_𝑢(𝑡⁻_𝑐)⋅𝜈≠0and ̇𝑥_𝑢(𝑡⁺_𝑐)⋅𝜈^′≠0.

In other words, a transverse crossing time𝑡_𝑐 is such that the trajectory does not hit the boundary of𝐾 tangentially while crossing𝐾 (however, the control could switch at time𝑡_𝑐). We are now in a position to state optimality conditions for problem (2.3). Let𝐻 ∶ℝ^𝑛×ℝ^𝑛×ℝ×ℝ^𝑚→ℝbe the Hamiltonian associated with the system defined by:

𝐻 =𝐻(𝑥, 𝑝, 𝑝₀, 𝑢) ∶=𝑝⋅𝑓(𝑥, 𝑢) +𝑝₀𝟙_𝐾𝑐(𝑥).

Using the previous proposition, we can apply the hybrid principle on the time crisis problem which allows to state the following necessary optimality conditions (see^7,8,11).

Theorem 1. Suppose that𝐾is convex and that (H1)-(H2)-(H3)-(H4) hold true. Let𝑢be an optimal control for (2.3) and𝑥the associated trajectory.

(1) There exists𝜏_𝑓 >0,𝑛 ∈ ℕand𝑡₀ = 0< 𝑡₁ <⋯ < 𝑡_𝑛 < 𝑡_𝑛+1 =𝜏_𝑓 such that𝑥(𝜏_𝑓) ∈ Viab(𝐾)and𝑡_𝑖,1≤𝑖≤ 𝑛is a crossing time of𝑥.

(2) In addition, if we suppose that each crossing time is transverse, there exists a function𝑝∶ [0, 𝜏_𝑓]→ℝ^𝑛which is absolutely continuous in each interval(𝑡_𝑖, 𝑡_𝑖+1),0≤𝑖≤𝑛− 1and𝑝₀≤0satisfying the following conditions:

(i) The pair(𝑝(⋅), 𝑝₀)is non-zero.

(ii) The function𝑝satisfies the adjoint equation

̇𝑝(𝑡) = −∇_𝑥𝐻(𝑥(𝑡), 𝑝(𝑡), 𝑝₀, 𝑢(𝑡)) a.e. 𝑡∈ (𝑡_𝑖, 𝑡_𝑖+1), 0≤𝑖≤𝑛− 1. (2.5)

‡In general, this terminology is used when an optimal control has an infinite number of switching points over a finite horizon (in presence of a singular arc of order 2); see for instance the Fuller’s example^21,22.

(iii) The control𝑢satisfies the Hamiltonian condition 𝑢(𝑡) ∈ arg max

𝜔∈𝑈 𝐻(𝑥(𝑡), 𝑝(𝑡), 𝑝₀, 𝜔) a.e. 𝑡∈ [0, 𝜏_𝑓].

(iv) At every crossing time𝑡_𝑖,1≤𝑖≤𝑛, the co-vector satisfies:

𝑝(𝑡⁺_𝑖) =𝑝(𝑡⁻_𝑖) +𝑝(𝑡⁻_𝑖)⋅(𝑓(𝑥(𝑡_𝑖), 𝑢(𝑡⁻_𝑖)) −𝑓(𝑥(𝑡_𝑖), 𝑢(𝑡⁺_𝑖))) +𝜎𝑝₀

𝑛(𝑥(𝑡_𝑖))⋅𝑓(𝑥(𝑡_𝑖), 𝑢(𝑡⁺_𝑖)) 𝑛(𝑥(𝑡_𝑖)), (2.6) where 𝜎 = −1, resp. 𝜎 = +1 if 𝑡_𝑐 is a regular crossing time from 𝐾 to 𝐾_𝑐, resp. from 𝐾_𝑐 into 𝐾 and 𝑛(𝑥(𝑡_𝑖)) ∈𝑁_𝐾(𝑥(𝑡_𝑖))).

(v) The transversality condition holds true at the terminal time:

𝑝(𝜏_𝑓) ∈ −𝑁_Viab(𝐾)(𝑥(𝜏_𝑓)). (2.7)

Remark 2.

- In the previous Theorem, we can relax the convexity of𝐾 by assuming only that its boundary is smooth.

- The jump condition on the co-vector (2.6) follows using that the Hamiltonian is conserved along any extremal trajectory (since the system is autonomous).

- In (2.7),𝑁_Viab(𝐾)(𝑥(𝜏_𝑓))denotes the Mordukhovich (limiting) normal cone toViab(𝐾)(see¹). IfViab(𝐾)is convex (such as for the prey-predator problem, see Section 3), then it coincides with the normal cone in the sense of convex analysis.

Hereafter, we say that an extremal trajectory(𝑥, 𝑝, 𝑝₀, 𝑢)isnormalif𝑝₀ ≠0andabnormalif𝑝₀= 0. Whenever an extremal trajectory is normal, we can always assume that𝑝₀= −1. We shall next use this result in order to characterize optimal trajectories for the time crisis problem governed by the prey-predator dynamics (see Section 5).

3 VIABILITY KERNEL FOR THE PREY-PREDATOR SYSTEM In the rest of the paper, we shall focus on the prey-predator model:

{ ̇𝑥 = 𝑟𝑥−𝑥𝑦,

̇𝑦 = −𝑚𝑦+𝑥𝑦−𝑢𝑦, (3.1)

where𝑟 >0and𝑚 >0, and𝑢is a measurable control function taking values within the set defined as

 ∶= {𝑢∶ [0,+∞)→[0, ̄𝑢] ; 𝑢(⋅) meas.},

with𝑢 >̄ 0. We set∶=ℝ^∗₊×ℝ^∗₊which is invariant by (3.1). Our aim is to address the problem of preserving the preys from the predators, maintaining as much as possible their density above a given threshold𝑥 > 0, which amounts to have the state belonging to the set

𝐾(𝑥) ∶= {(𝑥, 𝑦) ∈; 𝑥≥𝑥}.

Although the mathematical analysis of (3.1) (for a constant control𝑢) predicts that the preys cannot be extinct in finite time, one may consider that practically having a small density of preys expose them to a danger of disappearance that should be avoided as much as possible. This is why, we wish to study in the rest of the paper the minimal time crisis problem governed by (3.1) associated with the set𝐾(𝑥). We first determine the viability kernel of𝐾(𝑥)under the dynamics (3.1) and its attainability from the set.

3.1 Computation of the viability hernel

For a fixed𝑢∈ [0, ̄𝑢], we define the function𝑊_𝑢∶→ℝby:

𝑊_𝑢(𝑥, 𝑦) ∶=𝑥− (𝑚+𝑢) ln𝑥+𝑦−𝑟ln𝑦, (𝑥, 𝑦) ∈, together with the number𝑐(𝑢) ∈ℝdefined by

𝑐(𝑢) ∶=𝑊_𝑢(𝑚+𝑢, 𝑟) = (𝑚+𝑢)(1 − ln(𝑚+𝑢)) +𝑟(1 − ln𝑟),

and the positive equilibrium point𝐸^⋆(𝑢)for (3.1)

𝐸^⋆(𝑢) ∶= (𝑥^⋆(𝑢), 𝑦^⋆) = (𝑚+𝑢, 𝑟).

For a given number𝑐 ≥𝑐(𝑢), we denote by𝐿_𝑢(𝑐), resp. by𝑆_𝑢(𝑐), the level set, resp. the sub-level set of𝑊_𝑢defined by 𝐿_𝑢(𝑐) ∶= {(𝑥, 𝑦) ∈, 𝑊_𝑢(𝑥, 𝑦) =𝑐}, resp.𝑆_𝑢(𝑐) ∶= {(𝑥, 𝑦) ∈, 𝑊_𝑢(𝑥, 𝑦)≤𝑐}.

We recall in the two following Lemmas classical results about the model (3.1) with constant control.

Lemma 1. For a constant control𝑢, a trajectory of (3.1) belongs to a level set𝐿_𝑢(𝑐)with𝑐 ≥𝑐(𝑢). The sets𝐿_𝑢(𝑐)are closed curves that surround the steady state𝐸^⋆(𝑢).

Proof. By differentiating𝑊_𝑢w.r.t.𝑥and𝑦, one finds𝜕_𝑥𝑊_𝑢(𝑥, 𝑦) = 1 −^𝑚+𝑢

𝑥 and𝜕_𝑥𝑊_𝑢(𝑥, 𝑦) = 1 −^𝑟

𝑦 for(𝑥, 𝑦) ∈. If(𝑥(⋅), 𝑦(⋅)) is a solution of (3.1) with the constant control𝑢, a direct computation gives

𝑑

𝑑𝑡𝑊_𝑢(𝑥(𝑡), 𝑦(𝑡)) =𝜕_𝑥𝑊_𝑢̇𝑥+𝜕_𝑦𝑊_𝑢̇𝑦= 0.

So, any solution of (3.1) with the constant control 𝑢belongs to a level set of the function 𝑊_𝑢. As 𝑊_𝑢(𝑥, 𝑦) → +∞when

|(𝑥, 𝑦)|→+∞, each level set𝐿_𝑢(𝑐)is bounded. For a constant control𝑢, one can check that the single equilibrium of the dynamics inis𝐸^⋆(𝑢), and that the level set𝐿_𝑢(𝑐(𝑢))is the singleton{𝐸^⋆(𝑢)}. Therefore, for any initial condition in⧵{𝐸^⋆(𝑢)}, the trajectory belongs to a level set𝐿_𝑢(𝑐)with𝑐 > 𝑐(𝑢)(recall that𝑊_𝑢(𝑥, 𝑦)→+∞when|(𝑥, 𝑦)|→+∞). As𝐿_𝑢(𝑐)is a compact set that does not contain any equilibrium point, Poincaré-Bendixon Theorem allows to state that the trajectory converges to a limit cycle that belongs to the same level set𝐿_𝑢(𝑐). Therefore, the trajectory is periodic and𝐿_𝑢(𝑐)is a closed curve which surrounds 𝐸^⋆(𝑢).

For𝑢∈ [0, ̄𝑢], we define two functions𝜙_𝑢∶ℝ₊→ℝand𝜓∶ℝ₊→ℝby 𝜙_𝑢(𝑥) ∶=𝑥− (𝑚+𝑢) ln𝑥, 𝑥∈ℝ₊ and 𝜓(𝑦) ∶=𝑦−𝑟ln𝑦, 𝑦∈ℝ₊. Lemma 2. Given𝑢∈ [0, ̄𝑢], one has the following properties

(i) For any𝑥 > 𝜙_𝑢(𝑚+𝑢), there exists unique𝑥⁺_𝑢(𝑥) ∈ (𝑚+𝑢,+∞)and𝑥⁻_𝑢(𝑥) ∈ (0, 𝑚+𝑢)such that𝜙_𝑢(𝑥⁺_𝑢(𝑥)) =𝜙_𝑢(𝑥⁻_𝑢(𝑥)) = 𝑥.

(ii) If𝑝 > 𝜓(𝑟), the equation𝜓(𝑦) =𝑝has exactly two roots𝑦⁻(𝑝),𝑦⁺(𝑝)that satisfy𝑦⁻(𝑝)< 𝑟 < 𝑦⁺(𝑝).

Proof. One can easily check thatlim_𝑥→+∞𝜙_𝑢(𝑥) = lim_𝑥→0𝜙_𝑢(𝑥) = +∞. Moreover, by differentiating𝜙_𝑢w.r.t.𝑥, one finds 𝜙^′_𝑢(𝑥) = 1 − ^𝑚+𝑢

𝑥 . So the function𝜙_𝑢is decreasing from+∞down to𝜙_𝑢(𝑚+𝑢)and increasing up to+∞. Therefore, for any 𝑥 > 𝜙_𝑢(𝑚+𝑢), the equation𝜙_𝑢(𝑧) =𝑥has exactly two solutions𝑥⁻_𝑢(𝑥),𝑥⁺_𝑢(𝑥), with𝑥⁻_𝑢(𝑥)< 𝑚+𝑢and𝑥⁺_𝑢(𝑥)> 𝑚+𝑢which proves (i). Similarly, the function𝜓is decreasing from+∞down to𝜓(𝑟)and increasing up to+∞, which gives (ii).

For𝑐∈ℝ, we consider the subsets o𝐿⁺_𝑢(𝑐),𝐿⁻_𝑢(𝑐),𝑆_𝑢⁺(𝑐)and𝑆_𝑢⁻(𝑐)of, defined by:

𝐿⁺_𝑢(𝑐) ∶=𝐿_𝑢(𝑐) ∩ {𝑦≥𝑟}, 𝐿⁻_𝑢(𝑐) ∶=𝐿_𝑢(𝑐) ∩ {𝑦≤𝑟}, 𝑆_𝑢⁺(𝑐) ∶=𝑆_𝑢(𝑐) ∩ {𝑦≥𝑟}, 𝑆_𝑢⁻(𝑐) ∶=𝑆_𝑢(𝑐) ∩ {𝑦≤𝑟}

and let𝑟₋∈ (0, 𝑟]be

𝑟₋∶=𝑦⁻(𝑊₀(𝑥⁺_𝑢̄(𝑥), 𝑟) −𝜙₀(𝑥)).

The next Proposition provides a description of the viability kernelViab(𝑥)of𝐾(𝑥)for (3.1).

Proposition 2. One has the following characterization of the viability kernel:

(i) If𝑚+𝑢 < 𝑥, the set̄ Viab(𝑥)is empty.

(ii) If𝑢̄≥𝑥−𝑚, the viability kernel is non-empty and is given by Viab(𝑥) =𝑆_𝑢⁺_̄(𝑊_𝑢̄(𝑥, 𝑟)) ∪ (

𝑆₀⁻(𝑊₀(𝑥⁺_𝑢̄(𝑥), 𝑟)) ∩𝐾(𝑥)) ,

where𝑥⁺_𝑢̄(𝑥)is given by Lemma 2. Its boundary is the union of the three curves 𝐵⁺(𝑥) ∶= 𝐿⁺_𝑢̄(𝑊_𝑢̄(𝑥, 𝑟)),

𝐵⁻(𝑥) ∶= 𝐿⁻₀(𝑊₀(𝑥⁺_𝑢̄(𝑥), 𝑟)) ∩ {𝑥≥𝑥}, 𝐵⁰(𝑥) ∶= {𝑥} × [𝑟₋, 𝑟].

Proof. Let us assume that𝑢 < 𝑥̄ −𝑚and let𝜀 >0be such that𝑚+𝑢 < 𝑥−𝜀. Consider a trajectory(𝑥(⋅), 𝑦(⋅))that stays in the set𝐾(𝑥)for any time𝑡≥0. As0≤𝑢(𝑡)≤𝑢, we deduce that̄

̇𝑦=𝑦(𝑥−𝑚−𝑢)≥𝑦(𝑥−𝑥+𝜀)≥𝜀𝑦,

using that𝑥(𝑡)≥𝑥for any time𝑡≥0. Therefore𝑦(⋅)is increasing, unbounded and thus there exists𝑡₁>0such that𝑦(𝑡)> 𝑟for any time𝑡≥𝑡₁. It follows that

̇𝑥(𝑡) =𝑥(𝑡)(𝑟−𝑦(𝑡))< 𝑥(𝑡)(𝑟−𝑦(𝑡₁)) ∀𝑡 > 𝑡₁,

implying that for𝑡 > 𝑡₁one has ̇𝑥(𝑡)<−𝐶𝑥(𝑡)with𝐶 ∶=𝑦(𝑡₁) −𝑟 >0. Thus, there exists𝑡₂ > 𝑡₁such that𝑥(𝑡₂)< 𝑥. So the trajectory(𝑥(⋅), 𝑦(⋅))must escape the set𝐾(𝑥)in finite time, and we have a contradiction. Thus, the viability kernelViab(𝑥)is empty which proves (i).

Assume now that one has𝑢̄≥𝑥−𝑚and let us prove (ii). Notice first that the three curves𝐵⁺(𝑥),𝐵⁻(𝑥)and𝐵⁰(𝑥)belong to the set𝐾(𝑥)and that their union𝑈(𝑥)defines the boundary of a compact subset𝑍(𝑥)of𝐾(𝑥), which is such that

𝑍(𝑥) =𝑆_𝑢⁺_̄(𝑊_𝑢̄(𝑥, 𝑟)) ∪ (

𝑆₀⁻(𝑊₀(𝑥⁺_𝑢̄(𝑥), 𝑟)) ∩𝐾(𝑥)) .

When𝑢̄=𝑥−𝑚, the set𝑍(𝑥)is reduced to the single point𝐸^⋆(̄𝑢)that is an equilibrium of (3.1) for the constant control𝑢.̄ Thus,𝑍(𝑥)is a viable set.

When𝑢 > 𝑥̄ −𝑚, we first show that for any initial condition in𝑈(𝑥), there exists a trajectory that stays in𝐾(𝑥)for any time𝑡≥0. Consider an initial condition in the set𝐵⁺(𝑥). With the control𝑢=𝑢, the corresponding solution of (3.1) remains̄ on the level set𝐿_𝑢̄(𝑊_𝑢̄(𝑥, 𝑟))which is contained in𝐾(𝑥)as its extreme left point is(𝑥, 𝑟). Take now an initial condition in 𝐵⁻(𝑥). With the control𝑢= 0, the corresponding solution of (3.1) remains on𝐵⁻(𝑥)until it reaches in finite time the boundary point(𝑥⁺_𝑢̄(𝑥), 𝑟)that belongs to𝐵⁺(𝑥). From this point, we come back to the previous case. Finally, take an initial condition in Int(𝐵⁰(𝑥))(if not empty). OnInt(𝐵⁰(𝑥)), one has ̇𝑥 >0for any control as one has𝑟−𝑦 >0. Thus the trajectory enters the subset 𝑍(𝑥)and cannot evade from𝐾(𝑥)onInt(𝐵⁰(𝑥)). If the trajectory touches𝐵⁺∪𝐵⁻, we face one of the two previous cases. So we conclude that𝑍(𝑥)is a viable domain.

We now show that𝑍(𝑥)is the largest viable domain included in𝐾(𝑥), that is, the viability kernelViab(𝑥), or equivalently that any trajectory with initial condition in𝐾(𝑥)⧵𝑍(𝑥)leaves the set𝐾(𝑥)in a finite horizon. For convenience we consider the two subsets of𝐾(𝑥)⧵𝑍(𝑥),𝐶⁺(𝑥)and𝐶⁻(𝑥)defined by:

𝐶⁺(𝑥) =(

𝐾(𝑥)⧵𝑍(𝑥))

∩ {𝑦≥𝑟} and 𝐶⁻(𝑥) =(

𝐾(𝑥)⧵𝑍(𝑥))

∩ {𝑦≤𝑟}.

Consider now an initial condition (𝑥₀, 𝑦₀) ∈ 𝐶⁺(𝑥), and let (𝑥(⋅), 𝑦(⋅)) a solution of (3.1) starting from(𝑥₀, 𝑦₀). One has 𝑊_𝑢̄(𝑥₀, 𝑦₀)> 𝑊_𝑢̄(𝑥, 𝑟), and by differentiating w.r.t̄ 𝑡one finds:

𝑑

𝑑𝑡𝑊_𝑢̄(𝑥(𝑡), 𝑦(𝑡)) = (𝑦(𝑡) −𝑟)(𝑢̄−𝑢(𝑡))≥0.

Therefore no trajectory can reach the level set𝐿_𝑢̄(𝑊_𝑢̄(𝑥, 𝑟))̄ from𝐾(𝑥)⧵𝑍(𝑥). When𝑦(𝑡) =𝑟, one has𝑥(𝑡)> 𝑚+𝑢̄and thus

̇𝑦(𝑡) =𝑟(𝑥(𝑡) −𝑚−𝑢(𝑡))>0as𝑢(𝑡)≤𝑢. We deduce that if there exists a trajectory with an initial condition̄ (𝑥₀, 𝑦₀)in𝐶⁺(𝑥) that stays in𝐾(𝑥), it has to stay in𝐶⁺(𝑥). By differentiating w.r.t𝑡, we obtain on𝐶⁺(𝑥)that

𝑑

𝑑𝑡𝑊₀(𝑥(𝑡), 𝑦(𝑡)) = −𝑢(𝑡)(𝑦(𝑡) −𝑟)≤0,

and thus𝑊₀(𝑥(𝑡), 𝑦(𝑡)) ≤ 𝑊₀(𝑥₀, 𝑦₀). It follows that the trajectory is bounded. Moreover, one has ̇𝑥 ≤ 0.i.e., the function 𝑡→𝑥(𝑡)is non-increasing and thus converges to a certain𝑥_∞>0. By Barbalat’s Lemma (see for instance²³), ̇𝑥(𝑡)converges to 0which implies that𝑦(𝑡)tends to𝑟. Then𝑊_𝑢̄(𝑥(𝑡), 𝑦(𝑡))converges to𝑊_𝑢̄(𝑥_∞, 𝑟)> 𝑊_𝑢̄(𝑥, 𝑟)̄ which implies that𝑥_∞< ̄𝑥. Thus, the trajectory necessarily leaves the set𝐾(𝑥)and we have a contradiction.

Consider now an initial condition (𝑥₀, 𝑦₀) ∈ 𝐶⁻(𝑥). Similarly, one can show that no trajectory can reach the level set 𝐿₀(𝑊₀(𝑥⁺(𝑥), 𝑟))from𝐾(𝑥)⧵𝑍(𝑥). It follows that a trajectory with an initial condition in𝐶⁻(𝑥)that stays in𝐾(𝑥)has to stay in𝐶⁻(𝑥)(otherwise, it reaches𝐶⁺(𝑥)and we have shown above that its has to escape𝐾(𝑥)). As previously, one can show that a trajectory that stays in𝐶⁻(𝑥)is bounded and as𝑡→𝑥(𝑡)is increasing, one obtains the convergence of𝑥(𝑡)to a certain𝑥_∞> 𝑥.

As before, by Barbalat’s Lemma, ̇𝑥(𝑡)converges to0when𝑡→+∞, which implies that𝑦(𝑡)→𝑟, and thus𝑥_∞≥𝑥⁺_𝑢̄(𝑥)> 𝑚+𝑢.̄ Therefore there exists𝜀 >0and𝑡₀>0such that ̇𝑦(𝑡) = (𝑥(𝑡) −𝑚−𝑢)𝑦(𝑡)> 𝜀𝑦(𝑡)for any𝑡 > 𝑡₀. This gives a contradiction with the convergence of𝑦(𝑡)to𝑟when𝑡→+∞. So, the trajectory has to enter𝐶⁺(𝑥)and then leaves𝐾(𝑥). This ends the proof of (ii).

The viability kernel is depicted on Fig. 1 in case (ii) of Proposition 2 together with the three curves𝐵⁺(𝑥),𝐵⁻(𝑥),𝐵⁰(𝑥)that define its boundary (see the Appendix for the numerical values of the parameters). It is worth noting that𝐵⁺(𝑥)is a semi-orbit of (3.1) with𝑢=𝑢̄passing trough the point(𝑥, 𝑟). Similarly,𝐵⁻(𝑥)is a semi-orbit of (3.1) with𝑢= 0passing though the point (𝑥, 𝑟₋). Note that whenever𝑚+𝑢̄=𝑥, then the viability kernel is reduced to one pointi.e.,

𝑚+𝑢̄=𝑥 ⇒ 𝑉 𝑖𝑎𝑏(𝑥) = {(𝑥, 𝑟)},

as the semi-orbit𝐵⁺(𝑥)is then reduced to the positive equilibrium point of (3.1) with𝑢=𝑢. The viability kernel of̄ 𝐾(𝑥)enjoys

r

r

c

−

− 0

K ( )x

B ( )x

x B ( )

x

Viab( )x

+x B ( )

xu+

FIGURE 1Viability kernel when𝑢 > 𝑥̄ −𝑚(the numerical values of the parameters are given in the appendix).

the following properties.

Proposition 3. Suppose that𝑢 > 𝑥̄ −𝑚.

(i) Consider the unique solution of (3.1) backward in time with𝑢= 0from(𝑥⁺_𝑢̄(𝑥), 𝑟), and let𝑡^′>0be the first time where this trajectory intersects the line{𝑦=𝑟}. Then, we have:

𝑥(𝑡^′)≤𝑥. (3.2)

(ii) The set𝑉 𝑖𝑎𝑏(𝑥)is a compact convex set with non-empty interior.

(iii) Suppose that𝑡₁ < 𝑡₂are two consecutive crossing times from𝐾(𝑥)to𝐾(𝑥)^𝑐 and from𝐾(𝑥)^𝑐 to𝐾(𝑥)respectively, that (𝑥(𝑡₁), 𝑦(𝑡₁)) ∉ Viab(𝑥), and that(𝑥(𝑡₂), 𝑦(𝑡₂)) ∉ Viab(𝑥). Then, one has:

𝑡₂−𝑡₁ ≥ ln(_𝑟

𝑟₋

)

𝑚+𝑢̄ . (3.3)

Proof. To prove (i), suppose by contradiction that𝑥(𝑡^′)> 𝑥. Denote by(𝑥₀(⋅), 𝑥₀(⋅))the unique solution of (3.1) with𝑢= 0and such that(𝑥₀(0), 𝑥₀(0)) = (𝑥(𝑡^′), 𝑟). By construction, the point(𝑥⁺_𝑢̄(𝑥), 𝑟)is on the graph of the parameterized curve(𝑥₀(⋅), 𝑦₀(⋅)).

Thus we denote by 𝑡₀ ∶= inf {𝑡 ≥ 0 ; (𝑥₀(𝑡), 𝑥₀(𝑡)) = (𝑥⁺_𝑢̄(𝑥), 𝑟)}, and let𝛾₀be the parametrized curve(𝑥₀(⋅), 𝑥₀(⋅))on the interval[0, 𝑡₀].

Now, consider the unique solution(𝑥₁(⋅), 𝑦₁(⋅))of (3.1) with𝑢 = 𝑢̄starting from the point(𝑥(𝑡^′), 𝑟)and let𝑡₁ ∶= inf {𝑡≥ 0 ; 𝑦₁(𝑡) =𝑟}. We define𝛾₁as the parametrized curve(𝑥₁(⋅), 𝑦₁(⋅))on the interval[0, 𝑡₁]. From (3.1), we know that the graph

of𝛾₁is below the graph of𝛾₀. To conclude, we consider the unique solution(𝑥̃₁(⋅), ̃𝑦₁(⋅))of (3.1) with𝑢=𝑢̄starting from(𝑥, 𝑟) until the first time𝑡^′₁ > 0where it reaches the segment line{𝑦 =𝑟}. It can be noticed that this curve passes through the point (𝑥⁺_𝑢̄(𝑥), 𝑟)and that its graph̃𝛾₁is also below the graph of𝛾₀. Hence, both graphs𝛾₁and̃𝛾₁should intersecti.e., there must exist a time𝜏 ∈ (0, 𝑡₀)such that(𝑥₁(𝜏), 𝑦₁(𝜏)) = (𝑥̃₁(𝜏), ̃𝑦₁(𝜏)). By Cauchy-Lipschitz’s Theorem, both solutions(𝑥₁(⋅), 𝑦₁(⋅))and (𝑥̃₁(⋅), ̃𝑦₁(⋅))should coincide everywhere which is a contradiction as(𝑥₁(0), 𝑦₁(0))≠(𝑥̃₁(0), ̃𝑦₁(0)).

Let us now show (ii). From Proposition 2 and the previous point, we know that𝑉 𝑖𝑎𝑏(𝑥)is a compact subset ofℝ² with non-empty interior. Now, one can easily verify that solutions of (3.1) in the plane(𝑥, 𝑦)satisfy:

𝑑²𝑦 𝑑𝑥²

||||

|^𝑢=0(𝑥) = 𝑦(𝑟(𝑥−𝑚)²+𝑚(𝑟−𝑦)²)

(𝑟−𝑦)³𝑥² and 𝑑²𝑦 𝑑𝑥²

||||

|^𝑢=̄𝑢(𝑥) = 𝑦(𝑟(𝑥−𝑚− 1)²+ (𝑚+ 1)(𝑟−𝑦)²) (𝑟−𝑦)³𝑥² . It follows that for𝑦 < 𝑟, resp.𝑦 > 𝑟, one has^𝑑2𝑦

𝑑𝑥²|𝑢=0

(𝑥)>0, resp._𝑑𝑥^𝑑2𝑦₂

|𝑢=̄𝑢

(𝑥)<0, which guarantees that the set𝑉 𝑖𝑎𝑏(𝑥)is convex.

Finally, to prove (iii), we integrate the equation ̇𝑦(𝑡) =𝑦(𝑡)(𝑥(𝑡) −𝑚−𝑢(𝑡))over[𝑡₁, 𝑡₂], which gives:

(𝑚+𝑢)(𝑡̄ ₂−𝑡₁)≥

𝑡₂

∫

𝑡₁

(𝑚+𝑢(𝑡)) d𝑡=

𝑡₂

∫

𝑡₁

𝑥(𝑡) d𝑡−

𝑦(𝑡₂)

∫

𝑦(𝑡₁)

d𝑦 𝑦 ≥−

𝑦(𝑡₂)

∫

𝑦(𝑡₁)

d𝑦 𝑦 ≥ln

( 𝑟 𝑟₋

) , using that𝑦(𝑡₁)> 𝑟and𝑦(𝑡₂)< 𝑟₋. This ends the proof.

Remark 3.

- Whereas when𝑚 < 𝑥, it is clear that (3.2) holds true (as the equilibrium point for (3.1) with𝑢= 0is(𝑚, 𝑟)), the previous Proposition shows that this property remains valid whenever𝑚 > 𝑥. Note also that in the latter case, the set𝑉 𝑖𝑎𝑏(𝑥)is always non-empty as𝑢 >̄ 0.

- We know that for a given initial state in𝑉 𝑖𝑎𝑏(𝑥), any control𝑢can be chosen until that the corresponding trajectory reaches the boundary of𝑉 𝑖𝑎𝑏(𝑥)(seee.g.,^2,3). If(𝑥₀, 𝑦₀) ∈ 𝐵⁻(𝑥), resp.(𝑥₀, 𝑦₀) ∈𝐵⁺(𝑥), then only the control𝑢= 0, resp.𝑢=𝑢̄is admissible in order to stay in𝑉 𝑖𝑎𝑏(𝑥).

- If𝑥 > 𝑚, then any point of the segment[𝑥, ̄𝑢+𝑚] × {𝑟}is a steady-state point for (3.1) with a prescribed constant control whereas if𝑥≤𝑚, then any point of the segment[𝑚, ̄𝑢+𝑚] × {𝑟}is a steady-state point for (3.1) with a prescribed constant control.

Remark 4. Inequality (3.3) is in line with Hypothesis (H4). A similar property between consecutive crossing times could also be obtained for more general prey-predator models taking into account oscillatory trajectories (such as in Rosenzweig-MacArthur’s model¹⁷).

3.2 Attainability of the viability kernel

In order to show the attainability of the target set, it is convenient to introduce the following feedback control:

𝐮[𝑥, 𝑦] ∶=

{𝑢̄ if 𝑦≥𝑟,

0 if 𝑦 < 𝑟. (3.4)

Given an initial condition(𝑥₀, 𝑦₀) ∈ (ℝ^∗₊×ℝ^∗₊)∖Viab(𝑥), we denote by(𝑥_𝑚(⋅), 𝑦_𝑚(⋅))the unique solution of (3.1) starting from (𝑥₀, 𝑦₀)at time0associated with the control𝑢_𝑚(⋅)defined by𝑢_𝑚(𝑡) ∶=𝐮[𝑥_𝑚(𝑡), 𝑦_𝑚(𝑡)].

Proposition 4. For any initial condition(𝑥₀, 𝑦₀) ∈ ∖Viab(𝑥), there exists a control𝑢 ∈  steering(𝑥₀, 𝑦₀)to the viability kernelViab(𝑥).

Proof. Suppose first that one has𝑚+𝑢 > 𝑥, hencē Viab(𝑥)has a non-empty interior.

First step. We show that it is enough to prove the result for any initial condition of type(𝑥₀, 𝑟)with𝑥₀ > 𝑥⁺_𝑢̄(𝑥). If the initial̄ condition(𝑥₀, 𝑦₀)is such that𝑥₀< 𝑟, then it is enough to replace𝑥₀by𝑥_𝑚(𝑡_𝑐)where𝑡_𝑐is the first time𝑡 >0such that𝑦_𝑚(𝑡_𝑐) =𝑟.

If(𝑥_𝑚(𝑡_𝑐), 𝑦_𝑚(𝑡_𝑐)) ∈ Viab(𝑥), then the result is proved. Otherwise, we have𝑥_𝑚(𝑡_𝑐)> 𝑥⁺_𝑢̄(𝑥). If now̄ 𝑥₀> 𝑟, we apply the control 𝑢_𝑚 until the first time𝑡^′_𝑐 > 0such that𝑦_𝑚(𝑡^′_𝑐) = 𝑟. Then, for𝑡 > 𝑡^′_𝑐 (close to𝑡^′_𝑐)one has𝑦_𝑚(𝑡^′_𝑐) < 𝑟, and we conclude by the previous case.