Control Strategies for Sterile Insect Techniques...(1) r Primary sex ratio ˆ Mean number of eggs...

Model Periodic releases Feedback control Mixed strategies Sparse measures

Control Strategies for Sterile InsectTechniques

Pierre-Alexandre Bliman* (Inria & LJLL, Paris)Daiver Cardona-Salgado (UAO, Cali, Colombia)

Yves Dumont (CIRAD, Montpellier & U. Pretoria, South Africa)Olga Vasilieva (Univalle, Cali, Colombia)

June 14, 2019

1 / 42


Dengue epidemicsI ' 400 millions infected per year

(96 millions symptomatic) – 30times more than 1960!

I ' 3.9 billion humans at risk in128 countries

I ' 500 000 people with severedengue, fatality ' 2.5% (≤1%with proper medical care)

I 4 different strains, complexcross-immunity: no treatment(except symptoms), no vaccine

I Aedes aegypti and Ae.albopictus main vectors (and forzika, chikungunya. . . )

[World Health Organization, 2016]

2 / 42


Vector repartition (1/2)

Ae. Aegypti

Presence probability(from 0 to 1)

Spatial resolution:5 km × 5 km

Ae. Albopictus

[MUG Kraemer et al., eLife, June 2015]

3 / 42


Vector repartition (2/2)

Year of classification ofAe. albopictus “implantedand active” by department

[Direction générale de la santé, May 2019]

4 / 42


NEW Control methods. . .

[NL Achee et al., PLoS Negl Trop Dis, March 2019]5 / 42


. . . and NEW Control theory problems

I Issues: Modeling, Observation, Control in PopulationDynamics

I Spatial aspects, but not only: age, sex and genotypicstructure may be important

I Methods: Control theory for (Sub-)Monotone Systems

Today’s talk:

Control Strategies for Sterile Insect Techniques

6 / 42


A controlled sex-structured entomological model

Impulse periodic releases

Feedback control approach

Mixed control strategies

Sparse measurements

7 / 42






Sparse measurements

8 / 42


A sex-structured model

M,F : number of males, females

{Ṁ = rρe−β(M+F )F − µMM,Ḟ = (1− r)ρe−β(M+F )F − µFF .

(1)

r Primary sex ratioρ Mean number of eggs deposited per female per day (day−1)

µM , µF Mean death rate for male, female per day (day−1)

β Characteristic of the competition effect per individual

9 / 42


Equilibria of the modelI All trajectories are ultimately uniformly bounded.I Denote E ∗0 = (0, 0) the mosquito-free equilibrium of (1).

Define the basic offspring numbers for males and females:

NF :=(1− r)ρµF

, NM :=rρ

µM(2)

Theorem 1 (Equilibria of the entomological model)

• If NF < 1, then (1) has E ∗0 as unique equilibrium.• If NF > 1, then (1) also has a unique positive equilibrium E ∗:

F ∗ =NF

NF +NM1

βlogNF , M∗ =

NMNF +NM

1

βlogNF

10 / 42


Stability of the equilibria

Theorem 2 (Stability properties of the entomological model)

I If NF < 1, then E ∗0 is Globally Asymptotically Stable(GAS).

I If NF > 1, then E ∗0 is unstable, and E ∗ is GAS inR2+ \ {(M, 0),M ∈ R+}

0 M*0

F* Phase portrait for NF > 1.

11 / 42


Entomological model controlled by sterile males

MS : number of sterile males.Ṁ = rρ MM+γMS (t)e

−β(M+F )F − µMM,Ḟ = (1− r)ρ MM+γMS (t)e

−β(M+F )F − µFFṀS = Λ− µSMS

(3)

Λ Number of sterile males released per day (day−1)µS Mean death rate for sterile male per day (day

−1)γ Relative reproductive efficiency

12 / 42


General assumptions

In all the sequel, we make the following hypotheses:

I NF > 1I µS ≥ µM ≥ µFI γ < 1

13 / 42






Sparse measurements

14 / 42


Impulse periodic releases of constant amplitudeFor Λ > 0 constant, consider the release policy

Ṁ = rρFM

M + γMSe−β(M+F ) − µMM, (4a)

Ḟ = (1− r)ρ FMM + γMS

e−β(M+F ) − µFF , (4b)

ṀS = τΛ∑n∈N

δnτ − µSMS (4c)

When t → +∞, MS converges towards the periodic function

MperS (t) =τΛe−µS(t−b

tτcτ)

1− e−µSτ. (5)

15 / 42


Behaviour of the asymptotic periodic system

We thus consider the periodic system:

Ṁ = rρFM

M + γMperS (t)e−β(M+F ) − µMM, (6a)

Ḟ = (1− r)ρ FMM + γMperS (t)

e−β(M+F ) − µFF . (6b)

Theorem 3 (Stability for constant periodic impulses)

Assume

Λ ≥ Λcritper :=cosh (µSτ)− 1

µSτ21

eβγ

×min{

2NM , 2NF ,max{r , 1− r}max{NMr,NF

1− r

}}. (7)

Then E ∗0 is GAS for system (6).16 / 42


Proof of Theorem 3 (extracts)

Ṁ =

(rρ

F

M + γMperSe−β(M+F ) − µM

)M (8a)

Ḟ =

((1− r)ρ M

M + γMperSe−β(M+F ) − µF

)F (8b)

• MM+γMperS

e−β(M+F ) ≤ MM+γMperS

e−βM ≤ αM+γMperS

≤ αγMperS

, where

α := max{xe−βx : x ≥ 0

}= 1eβ . Thus

F((n + 1)τ

)≤ e

((1−r)ραγ

〈1

MperS

〉−µF

)τF (nτ)

• Let V(M,F ) := 12 (M2 + F 2), then

V̇ ≤(

max{r , 1− r}α 1γMperS

− 2 min{µM , µF})V

17 / 42


Constant impulsive releases

0 100 200 300 400 500 600 700

time (days)

-3

-2

-1

0

1

2

3

4

5

Log

10(N

um

ber

of In

div

iduals

)

Male

Female

Sterile Male

0 100 200 300 400 500 600 700 800

time (days)

-3

-2

-1

0

1

2

3

4

5

Log

10(N

um

ber

of In

div

iduals

)

Male

Female

Sterile Male

τ = 7 days τ = 14 days

Period (days) Total # released males # weeks

τ = 7 924 627 84

τ = 14 942 869 84

18 / 42






Sparse measurements

19 / 42


Framework

Instead of (4), we consider, for Λn ≥ 0, n ∈ N,

Ṁ = rρFM

M + γMSe−β(M+F ) − µMM, (9a)

Ḟ = (1− r)ρ FMM + γMS

e−β(M+F ) − µFF , (9b)

ṀS = τ∑n∈N

Λnδnτ − µSMS (9c)

Assume available periodic measurements M(nτ),F (nτ), n ∈ N.

How to choose Λn, n ∈ N,to have M(t),F (t)→ 0 for t → +∞?

20 / 42


Step 1 - Acting directly on MSProposition 4 (Stabilization by direct control of MS)

Let k ∈ (0, µF(1−r)ρ). If M(t)M(t) + γMS(t)

< k , t ≥ 0 (10)

i.e. γMS(t) ≥(

1k − 1

)M(t) for any t ≥ 0, then E ∗0 is GAS for

system (4).

Proof of Proposition 4.When (10) holds, then(

Ṁ

Ḟ

)≤ A

(MF

), A :=

(−µM rρk

0 −µF + (1− r)ρk

)(11)

A is Metzler and Hurwitz: for

(Ṁ ′

Ḟ ′

)= A

(M ′

F ′

),

(M ′(0)F ′(0)

)=

(M(0)F (0)

),

0 ≤(M(t)F (t)

)≤(M ′(t)F ′(t)

)→ 0 for t → +∞

21 / 42


Step 2 - Shaping control compliant with Step 1

To obtain (10), that is: γMS(t) ≥(

1k − 1

)M(t) on (nτ, (n + 1)τ ],

choose Λn such that

∀t ∈ (nτ, (n + 1)τ ], γMS(t) ≥(

1

k− 1)M ′(t)

that is: ∀ s ∈ (0, τ ],

γ(Λnτ + MS(nτ))e−µS s ≥

(1

k− 1)(

1 0)eAs

(M(nτ)F (nτ)

)(12)

It turns out that (12) has to be verified only for s = τ .

22 / 42


Stabilization by impulsive feedback control

Theorem 5 (Stabilization by feedback control)For a given k ∈

(0, µF(1−r)ρ

), assume that for any n ∈ N:

τΛn ≥∣∣∣∣K (M(nτ)F (nτ)

)−MS(nτ)

∣∣∣∣+

(13a)

K :=1

γ

(1−kk e

(µS−µM )τ

rρ(1−k)µM−µF +(1−r)ρk

(e(µS−µF +(1−r)ρk)τ − e(µS−µM )τ

))T (13b)Then E∗0 is GAS for (9). If moreover

τΛn ≤ K(M(nτ)F (nτ)

)(13c)

then+∞∑n=0

Λn < +∞.

23 / 42


Proof of Theorem 5

I One first verifies that if Λn is chosen as in (13a), then∀t ∈ (nτ, (n + 1)τ ], γMS(t) ≥

(1k − 1

)M(t).

I Therefore, using the linear comparison system

0 ≤(M(t)F (t)

)≤(M ′(t)F ′(t)

)→ 0 for t → +∞

and F and M vanish exponentially.

I Due to this exponential decrease, (13c) implies+∞∑n=0

Λn < +∞.

24 / 42


Closed-loop periodic impulsive control, kNF = 0.2

0 100 200 300 400 500 600

time (days)

-2

-1

0

1

2

3

4

5

6

7

Lo

g10

(Nu

mb

er

of

Ind

ivid

ua

ls)

Male

Female

Sterile Male

0 100 200 300 400 500 600

time (days)

-2

-1

0

1

2

3

4

5

6

7

Lo

g10

(Nu

mb

er

of

Ind

ivid

ua

ls)

Male

Female

Sterile Male



τ = 7 2 251 052 64

τ = 14 2 390 676 64

25 / 42


Closed-loop periodic impulsive control, kNF = 0.99

0 200 400 600 800 1000 1200 1400 1600 1800

time (days)

-2

-1

0

1

2

3

4

5

6

Lo

g10

(Nu

mb

er

of

Ind

ivid

ua

ls)

Male

Female

Sterile Male

0 200 400 600 800 1000 1200

time (days)

-2

-1

0

1

2

3

4

5

6

Lo

g10

(Nu

mb

er

of

Ind

ivid

ua

ls)

Male

Female

Sterile Male



τ = 7 794 807 240

τ = 14 909 344 130

26 / 42






Sparse measurements

27 / 42



Theorem 6 (Stabilization by saturated control)Assume that, for any n ∈ N and Λ̄ positive constant,

Λn ≥ min{

Λ̄ ;

∣∣∣∣K (M(nτ)F (nτ))−MS(nτ)

∣∣∣∣+

}(14)

and that one of the following conditions is fulfilled:

• Case 1.Λ̄ = 2

(cosh (µSτ)− 1)µSτ 2

1

eβγNF , k ∈

(0,

µF(1− r)ρ

)(15)

• Case 2.

Λ̄ =(cosh (µSτ)− 1)

µSτ 21

eβγmax{r , 1− r}max

{NMr,NF

1− r

},

k ∈

0, 2 µMρ

1− rr2

√1 + µFµM

(r

1− r

)2− 1

(16)Then E∗0 is GAS for (9). 28 / 42


Proof of Theorem 6

Using the proof of Theorem 3, one shows that, along thetrajectories and whatever the mode,

I Case 1. F((n + 1)τ

)≤ e

((1−r)ρα

γ

〈1

MperS

〉−µF

)τF (nτ).

I Case 2. V̇ ≤ −εV, for V(M,F ) = 12 (M2 + F 2) and some

∃ε > 0.

29 / 42


Mixed periodic impulsive control, kNF = 0.2

0 100 200 300 400 500 600 700

time (days)

-2

-1

0

1

2

3

4

5

Lo

g10

(Nu

mb

er

of

Ind

ivid

ua

ls)

Male

Female

Sterile Male

0 100 200 300 400 500 600 700

time (days)

-2

-1

0

1

2

3

4

5

Lo

g10

(Nu

mb

er

of

Ind

ivid

ua

ls)

Male

Female

Sterile Male



τ = 7 450 668 72

τ = 14 465 187 72

30 / 42


Mixed periodic impulsive control, kNF = 0.99

0 200 400 600 800 1000 1200 1400 1600 1800 2000

time (days)

-2

-1

0

1

2

3

4

5

Lo

g10

(Nu

mb

er

of

Ind

ivid

ua

ls)

Male

Female

Sterile Male

0 200 400 600 800 1000 1200

time (days)

-2

-1

0

1

2

3

4

5

Lo

g10

(Nu

mb

er

of

Ind

ivid

ua

ls)

Male

Female

Sterile Male



τ = 7 457, 489 69

τ = 14 427, 701 74

31 / 42






Sparse measurements

32 / 42


Feedback control with sparse measurementsMeasurements are long and costly. We now adapt Theorem 5 tomeasurements achieved with period pτ , for given p ∈ N∗.

Theorem 7 (Impulsive control with sparse measurements)For given k ∈

(0, µF(1−r)ρ

), assume for any n ∈ N, m = 0, 1, . . . , p − 1,

τΛnp+m ≥

∣∣∣∣∣Kp(M(nτ)F (nτ)

)−MS(npτ)e−mµSτ −

m−1∑i=0

Λnp+ie−µS (m−i)τ

∣∣∣∣∣+

(17a)

Kp :=eµSτ

γ

(1−kk e−(m+1)µMτ

rρ(1−k)µM−µF +(1−r)ρk

(e−(µF−(1−r)ρk)(m+1)τ − e−µM (m+1)τ

))T (17b)Then E∗0 is GAS for (4).

If moreover τΛnp+m ≤ Kp(M(nτ)F (nτ)

), then

+∞∑n=0

Λn < +∞.

33 / 42


Closed-loop control, sparse measures, kNF = 0.2, p = 4

0 50 100 150 200 250 300 350 400 450 500

time (days)

-3

-2

-1

0

1

2

3

4

5

6

7

Lo

g10

(Nu

mb

er

of

Ind

ivid

ua

ls)

Male

Female

Sterile Male

0 100 200 300 400 500 600

time (days)

-3

-2

-1

0

1

2

3

4

5

6

7

Lo

g10

(Nu

mb

er

of

Ind

ivid

ua

ls)

Male

Female

Sterile Male


Period (days) # released males # weeks # releases

τ = 7 4 363 430 54 34

τ = 14 2 896 835 56 17

34 / 42


Closed-loop control, sparse measures, kNF = 0.99, p = 4

0 100 200 300 400 500 600

time (days)

-2

-1

0

1

2

3

4

5

6

Lo

g10

(Nu

mb

er

of

Ind

ivid

ua

ls)

Male

Female

Sterile Male

0 100 200 300 400 500 600

time (days)

-2

-1

0

1

2

3

4

5

6

Lo

g10

(Nu

mb

er

of

Ind

ivid

ua

ls)

Male

Female

Sterile Male



τ = 7 1 221 593 58 37

τ = 14 1 043 107 62 20

35 / 42


Mixed control, sparse measures, kNF = 0.2, p = 4

0 100 200 300 400 500 600

time (days)

-3

-2

-1

0

1

2

3

4

5

Lo

g10

(Nu

mb

er

of

Ind

ivid

ua

ls)

Male

Female

Sterile Male

0 100 200 300 400 500 600

time (days)

-2

-1

0

1

2

3

4

5

Lo

g10

(Nu

mb

er

of

Ind

ivid

ua

ls)

Male

Female

Sterile Male



τ = 7 534 849 65 53

τ = 14 499 497 66 25

36 / 42


Mixed control, sparse measures, kNF = 0.99, p = 4

0 100 200 300 400 500 600

time (days)

-2

-1

0

1

2

3

4

5

Lo

g10

(Nu

mb

er

of

Ind

ivid

ua

ls)

Male

Female

Sterile Male

0 100 200 300 400 500 600 700

time (days)

-2

-1

0

1

2

3

4

5

Lo

g10

(Nu

mb

er

of

Ind

ivid

ua

ls)

Male

Female

Sterile Male



τ = 7 450, 077 69 53

τ = 14 449, 059 74 28

37 / 42


Thank you for your attention

For more details:

I “Implementation of Control Strategies for Sterile InsectTechniques”, to appear in Mathematical Biosciences

I [email protected]

38 / 42

mailto:[email protected]


Kamke’s theoremDefinition 1 (Function of type K )f : Rm → Rm is of type K in a set S if: ∀x , y ∈ S , ∀i ∈ {1, . . . ,m},

xi = yi ∧ x ≥ y ⇒ fi (x) ≥ fi (y)Proposition 8 (Differential characterisation of type K )When S convex, f C1 is of type K if ∂fi∂xj (x) ≥ 0, x ∈ S , i 6= j.

Theorem 9 (e.g. [1965, Coppel])Let f (t, ·) be of type K , ∀t ∈ [a, b], let x fulfil ẋ = f (t, x) on [a, b].

I If y C0 satisfies D−(y)(t) := lim suph→0+

y(t)−y(t−h)h > f (t, y(t)) on

(a, b] and y(a) > x(a), then y(t) > x(t), t ∈ [a, b]

I If y C0 satisfies D−(y)(t) := lim infh→0+

y(t+h)−y(t)h < f (t, y(t)) on

(a, b] and y(a) < x(a), then y(t) < x(t), t ∈ [a, b]

In other words: a differential inequality induces comparison results.39 / 42


Metzler matrices and linear positive systems

Definition 2 (Metzler matrix)A ∈ Rm×m is called Metzler (or essentially nonnegative) matrix if

i 6= j ⇒ aij ≥ 0

Theorem 10 (e.g. [2010, Haddad et al.])A is Metzler iff componentwise: ∀t ≥ 0, eAt ≥ 0m

Definition 3 (Positive system)Let A ∈ Rm×m,B ∈ Rm×p,C ∈ Rq×m, the linear input/output system

ẋ = Ax + Bu, y = Cx , x(0) = x0 (18)

is positive if: x ≥ 0m, y ≥ 0q whenever x0 ≥ 0m and u ≥ 0p.

Corollary 11 (e.g. [2000, Farina, Rinaldi])System (18) is positive iff A Metzler and B,C ≥ 0.

40 / 42


Switched systems and stability

Definition 4 (e.g. [2003, Liberzon])

Given systems ẋ = fi (x(t)), fi : Rm → Rm locally Lipschitz,i ∈ I, and σ : [0,+∞)→ I piecewise constant, one defines aswitched system as:

ẋ = fσ(t)(x(t)), x(0) = x0 (19)

Stability of all ẋ = fi (x(t)) does not imply stability for (19)!

41 / 42


Common Lyapunov functions

Let a common equilibrium exist, e.g. fi (0m) = 0m, i ∈ I.

Definition 5 (e.g. [2003, Liberzon])

A positive definite C1 map V : Rm → R+ is called commonLyapunov function for the family of systems ẋ = fi (x(t)),i ∈ I, if exists a positive definite C0 map W : Rm → R+ with

∀i ∈ I, ∀x , ∂V∂x

fi (x) ≤ −W (x)

Theorem 12 (e.g. [2003, Liberzon])

Let the family of systems ẋ = fi (x(t)), i ∈ I, possess aradially unbounded common Lyapunov function. Then theswitched system ẋ = fσ(t)(x(t)) is GAS, uniformly wrt σ.

42 / 42

bbleu A controlled sex-structured entomological modelbbleu Impulse periodic releasesbbleu Feedback control approachbbleu Mixed control strategiesbbleu Sparse measurements

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