Modeling epidemic spreading in star-like networks

22 July 2021

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Modeling epidemic spreading in star-like networks

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Modeling epidemic spreading in star-like networks

Luca Ferreri, Paolo Bajardi, Mario Giacobini GECO - Group of Computational Epidemiology

Department of Veterinary Sciences CSU - Complex Systems Unit Molecular Biotechnology Center

ARC2_{S - Applied Research on Computational Complex Systems Group} Department of Computer Science University of Torino

Tick-Borne Encephalitis

I _{endemic in Eurasia from Europe, through Russia To China}

and Japan

I _{the virus causes potentially fatal neurological infecion} I _{in last years emergenge of the virus in new area and increase}

of morbidity

I _{maintained in nature by complex cycle involving Ixodid ticks}

(I. ricinus and I. persulcatus) and wild vertebrate hosts

Eggs Larvae Nymphs Adults Females Hosts Hosts Hosts

Systemic Transmission

Our research question

how do thenon-systemic transmissiontogether with the different aggregation patternsinfluence the pathogen spreading?

Spreading Model

I _{at time t a fraction, π(t), of}

passengers (ticks) are infectious

I _{P(k ) probability that a bus}

(mouse) transports k passengers (ticks) of them

I β transmission probability for infectious path

I µ recovery probability

⇒ π(t + 1) = f (π(t))

Spreading Model

I _{at time t a fraction, π(t), of}

passengers (ticks) are infectious

I _{P(k ) probability that a bus}

(mouse) transports k passengers (ticks) of them

I β transmission probability for infectious path

I µ recovery probability

⇒ π(t + 1) = f (π(t))

Spreading Model

I _{at time t a fraction, π(t), of}

passengers (ticks) are infectious

I _{P(k ) probability that a bus}

(mouse) transports k passengers (ticks) of them

I β transmission probability for infectious path

I µ recovery probability

⇒ π(t + 1) = f (π(t))

Spreading Model

I _{at time t a fraction, π(t), of}

passengers (ticks) are infectious

I _{P(k ) probability that a bus}

(mouse) transports k passengers (ticks) of them

I β transmission probability for infectious path

I µ recovery probability

⇒ π(t + 1) = f (π(t))

Spreading Model

I _{at time t a fraction, π(t), of}

passengers (ticks) are infectious

I _{P(k ) probability that a bus}

(mouse) transports k passengers (ticks) of them

I β transmission probability for infectious path

I µ recovery probability ⇒ π(t + 1) = f (π(t))

Analytical Framework

the probability that asusceptible passenger, having h travel mates, gets theinfection is

1 − (1 − β)h

Let π(t) be theprevalence of infectionamong passengers at time t, the probabilityy for a susceptible passenger on a bus transporting k individuals including himself to beinfectiousat time t + 1 is

Analytical Framework

the probability that asusceptible passenger, having h travel mates, gets theinfection is

1 − (1 − β)h

Let π(t) be theprevalence of infectionamong passengers at time t, the probabilityy for a susceptible passenger on a bus transporting k individuals including himself to beinfectiousat time t + 1 is

Math

Recalling that P(k) is the probability for a bus to have k passengers, the probability for a passenger to be on a k-bus is

#passengers on a k-bus #passengers = k · #k-bus #passengers = = k ·#k-bus #bus · #bus #passengers = k · P(k) · 1 hki

Math

thus, the probability for asusceptiblepassenger at time t to be infectiousat time t + 1 is ∞ X k=1 h 1 − (1 − β)(k−1)·π(t)i· k hki· P(k)

and therefore the prevalence among passenger at time t + 1 is π(t + 1) = f (π(t)) = = (1 − µ)·π(t)+[1 − π(t)]· ( 1 − ∞ X k=1 (1 − β)(k−1)·π(t)· k hki · P(k) )

Equilibria

imposing the stationary condition π(t + 1) = π(t) = x we can derive the equilibria as solutions of the following equation x = f (x) = (1 − µ)·x+[1 − x]· ( 1 − ∞ X k=1 (1 − β)(k−1)·x· k hki· P(k) ) . Now: I _{x = 0 is a solution,} I _{f (1) = 1 − µ ≤ 1,} I f00(x) < 0.

assuming hki and hk2_{i finite}

0.5 1

0.5

1 f0(0) > 1

f0(0) < 1 therefore conditions to have one, and only one, solution ˆx ∈ (0, 1) is that f0(0) > 1 or

−ln (1 − β) µ >

hki hk2_{i − hki}.

Equilibria

imposing the stationary condition π(t + 1) = π(t) = x we can derive the equilibria as solutions of the following equation x = f (x) = (1 − µ)·x+[1 − x]· ( 1 − ∞ X k=1 (1 − β)(k−1)·x· k hki· P(k) ) . Now: I _{x = 0 is a solution,} I _{f (1) = 1 − µ ≤ 1,} I f00(x) < 0.

assuming hki and hk2_{i finite}

0.5 1

0.5

1 f0(0) > 1

f0(0) < 1 therefore conditions to have one, and only one, solution ˆx ∈ (0, 1) is that f0(0) > 1 or

−ln (1 − β) µ >

hki hk2_{i − hki}.

Equilibria

imposing the stationary condition π(t + 1) = π(t) = x we can derive the equilibria as solutions of the following equation x = f (x) = (1 − µ)·x+[1 − x]· ( 1 − ∞ X k=1 (1 − β)(k−1)·x· k hki· P(k) ) . Now: I _{x = 0 is a solution,} I _{f (1) = 1 − µ ≤ 1,} I f00(x) < 0.

assuming hki and hk2_{i finite}

0.5 1

0.5

1 f0(0) > 1

f0(0) < 1

therefore conditions to have one, and only one, solution ˆx ∈ (0, 1) is that f0(0) > 1 or

−ln (1 − β) µ >

hki hk2_{i − hki}.

Equilibria

imposing the stationary condition π(t + 1) = π(t) = x we can derive the equilibria as solutions of the following equation x = f (x) = (1 − µ)·x+[1 − x]· ( 1 − ∞ X k=1 (1 − β)(k−1)·x· k hki· P(k) ) . Now: I _{x = 0 is a solution,} I _{f (1) = 1 − µ ≤ 1,} I f00(x) < 0.

assuming hki and hk2_{i finite}

0.5 1

0.5

1 f0(0) > 1

f0(0) < 1 therefore conditions to have one, and only one, solution ˆx ∈ (0, 1) is that f0(0) > 1 or

−ln (1 − β) µ >

hki hk2_{i − hki}.

Stability

recalling that f00(x) < 0 and that

I _f0_{(1) = −µ +}P_{k(1 − β)}k−1 k

hkiP(k ) > −µ > −1

I f0(ˆx) < 1

hence ˆx isasymptotically stablewhen it exists. Therefore:

I _{disease-free equilibrium is asymptotically stable when ˆx does}

not exist.

I _{disease-free is unstable when ˆx exists. Furthermore, when ˆx}

Conclusion and Discussions

I _{co-feeding transmission}

I _{spreading on star-like networks}

I _{spreading dynamic on dynamic bipartite networks} I analytical result confirmed by simulations

Acknowledgements

Source of inspirations for this work come from a scientific project hold with the Fondazione Edmund Mach, San Michele all’Adige (TN). In particular, we are grateful to Anna Paola Rizzoli, Roberto Ros`a e Luca Bolzoni.