Nash-MFG equilibrium in a SIR model with time dependent newborn vaccination

Abstract

We study the newborn, non compulsory, vaccination in a SIR model with vital dynamics. The evolution of each individual is modeled as a Markov chain. His/Her vaccination decision optimizes a criterion depending on the time-dependent aggregate (societal) vaccination rate and the future epidemic dynamics. We prove the existence of a Nash-Mean Field Games equilibrium among all individuals in the population. Then we propose a novel numerical approach to find the equilibrium and test it numerically.

Appendices

Clarke generalized gradients

To recall the definition of the Clarke generalized gradients we follow the presentation in [16, section 10.1 page 194] and [17]. Let X be a Banach space, $X^*$ its dual and $x \in X$; also take $f: X \rightarrow {\mathbb R}$ to be a functional which is Lipschitz with constant $L>0$ in a neighborhood of x, that is, for some $\epsilon > 0$, we have $\Vert f(y)-f(z) \Vert \le L \Vert y-z \Vert$ for all y, z in the ball of center x and radius $\epsilon$. The generalized directional derivative of f at x in the direction v, denoted $f^o(x ; v)$, is defined as

$$\begin{aligned} f^o(x,v) = \limsup _{y \rightarrow x, t \downarrow 0 } \frac{f(y+tv)-f(y)}{t}. \end{aligned}$$

(46)

Note that $\Vert f^o(x,v) \Vert \le L \Vert v \Vert$ for any $v \in X$; moreover, as function of v, the directional derivative $f^o(x,v)$ is subadditive i.e. $f^o(x,v+w) \le f^o(x,v)+ f^o(x,w)$, $\forall v,w \in X$. In particular it can be lower bounded by a linear functional in $X^*$. The (Clarke) generalized gradient of f at x denoted $\overline{\partial }{f}(x)$ or $\dot{f}(x)$ is the set of all such linear functionals; the formal definition is the following:

$$\begin{aligned} \overline{\partial }{f}(x) = \{ \xi \in X^* | f^o(x,v) \ge \langle v , \xi \rangle , \forall v \in X \}. \end{aligned}$$

(47)

It can be shown that the Clarke generalized gradient is a non empty, convex, (weakly-$*$) compact subset of $X^*$. In particular when $X={\mathbb R}^k$ for some $k \in {\mathbb N}^*$, $\overline{\partial }{f}(x)$ is the convex hull of the set $\{ \lim _{\ell \rightarrow \infty } \nabla f(x_\ell ) \}$ for any sequence $x_\ell$ converging to x such that:

• $\nabla f(x_\ell )$ exists $\forall \ell$ (recall that since f is Lipschitz it is differentiable a.e.) and

• the limit $\lim _{\ell \rightarrow \infty } \nabla f(x_\ell )$ exists.

Technical details concerning the probability of infection

Recall that $\phi _I^{u} \left( . \right)$ is a function from $\mathbb {R}_{+}$ to $\left[ 0, 1 \right]$ such that, for any $t \in \mathbb {R}_{+}$, $\phi _I^{u} \left( t \right)$ is the probability of infection during the life of an individual, born in t and not vaccinated, when the population follows the vaccination strategy u. In mathematical terms, for any individual born in $t \ge 0$,

$$\begin{aligned} \phi _I^{u} \left( t \right) = \mathbb {P} \left( \exists \tau \ge t \text { such that } M_{ \tau }^t = I \; \vert \; M_{ t }^t = S \right) . \end{aligned}$$

(48)

In order to compute $\phi _I^{u} \left( t \right)$ we introduce the probability $\varphi _I^{u,t} \left( . \right)$ of infection before $\tau$:

$$\begin{aligned} \varphi _I^{u,t} \left( \tau \right) = \mathbb {P} \left( \exists s \in \left[ t, \tau \right] \text { such that } M^{t}_{ s } = I \; \vert \; M^{t}_{ t } = S \right) . \end{aligned}$$

(49)

Hence, we have:

$$\begin{aligned} \varphi _I^{u,t} \left( \tau + \varDelta \tau \right)&= \mathbb {P} \left( M^{t}_{ \tau + \varDelta \tau } \in I \cup R \cup D_2 \; \vert \; M^{t}_{ t } = S \right) \\&= \mathbb {P} \left( M^{t}_{ \tau + \varDelta \tau } \in I \cup R \cup D_2 \; \vert \; M^{t}_{ \tau } = S \right) \times \mathbb {P} \left( M^{t}_{ \tau } = S \; \vert \; M^{t}_{ t } = S \right) \\&\quad + \mathbb {P} \left( M^{t}_{ \tau + \varDelta \tau } \in I \cup R \cup D_2 \; \vert \; M^{t}_{ \tau } = I \right) \times \mathbb {P} \left( M^{t}_{ \tau } = I \; \vert \; M^{t}_{ t } = S \right) \\&\quad + \mathbb {P} \left( M^{t}_{ \tau + \varDelta \tau } \in I \cup R \cup D_2 \; \vert \; M^{t}_{ \tau } = R \right) \times \mathbb {P} \left( M^{t}_{ \tau } = R \; \vert \; M^{t}_{ t } = S \right) \\&\quad + \mathbb {P} \left( M^{t}_{ \tau + \varDelta \tau } \in I \cup R \cup D_2 \; \vert \; M^{t}_{ \tau } = D_1 \right) \times \mathbb {P} \left( M^{t}_{ \tau } = D_1 \; \vert \; M^{t}_{ t } = S \right) \\&\quad + \mathbb {P} \left( M^{t}_{ \tau + \varDelta \tau } \in I \cup R \cup D_2 \; \vert \; M^{t}_{ \tau } = D_2 \right) \times \mathbb {P} \left( M^{t}_{ \tau } = D_2 \; \vert \; M^{t}_{ t } = S \right) , \end{aligned}$$

with

$$\begin{aligned} \mathbb {P} \left( M^{t}_{ \tau + \varDelta \tau } \in I \cup R \cup D_2 \; \vert \; M^{t}_{ \tau } = D_1 \right)&= 0 \\ \mathbb {P} \left( M^{t}_{ \tau + \varDelta \tau } \in I \cup R \cup D_2 \; \vert \; M^{t}_{ \tau } = I \right)&= 1 \\ \mathbb {P} \left( M^{t}_{ \tau + \varDelta \tau } \in I \cup R \cup D_2 \; \vert \; M^{t}_{ \tau } = R \right)&= 1 \\ \mathbb {P} \left( M^{t}_{ \tau + \varDelta \tau } \in I \cup R \cup D_2 \; \vert \; M^{t}_{ \tau } = D_2 \right)&= 1 \\ \mathbb {P} \left( M^{t}_{ \tau + \varDelta \tau } \in I \cup R \cup D_2 \; \vert \; M^{t}_{ \tau } = S \right)&= \beta I^{u} \left( \tau \right) \varDelta \tau + o(\varDelta \tau ). \end{aligned}$$

Hence,

$$\begin{aligned} \varphi _I^{u,t} \left( \tau + \varDelta \tau \right)&= \beta I^{u} \left( \tau \right) \varDelta \tau \times \mathbb {P} \left( M^{t}_{ \tau } = S \; \vert \; M^{t}_{ t } = S \right) \\&\quad + \mathbb {P} \left( M^{t}_{ \tau } \in I \cup R \cup D_2 \; \vert \; M^{t}_{ t } = S \right) + o(\varDelta \tau ) \\&= \beta I^{u} \left( \tau \right) \varDelta \tau \times \mathbb {P} \left( M^{t}_{ \tau } = S \; \vert \; M^{t}_{ t } = S \right) + \varphi _I^{u,t} \left( \tau \right) + o(\varDelta \tau ). \end{aligned}$$

We denote $r^{u,t} \left( \tau \right) = \mathbb {P} \left( M^{t}_{ \tau } = S \; \vert \; M^{t}_{ t } = S \right)$ the probability of staying susceptible between t and $\tau$. We compute this probability:

$$\begin{aligned} r^{u,t} \left( \tau + \varDelta \tau \right)&= \mathbb {P} \left( M^{t}_{ \tau + \varDelta \tau } = S \; \vert \; M^{t}_{ \tau } = S \right) \times \mathbb {P} \left( M^{t}_{ \tau } = S \; \vert \; M^{t}_{ t } = S \right) \\&= \left( 1 - \beta I^{u} \left( \tau \right) \varDelta \tau - \mu \varDelta \tau \right) r^{u,t} \left( \tau \right) + o(\varDelta \tau ). \end{aligned}$$

Hence, the probability of staying susceptible between t and $\tau$ is:

$$\begin{aligned} r^{u,t} \left( \tau \right) = e^{ - \mu \left( \tau - t \right) } \exp \left( - \int _t^{ \tau } \beta I^{u} \left( s \right) ds \right) . \end{aligned}$$

Hence,

$$\begin{aligned} \varphi _I^{u,t} \left( \tau + \varDelta \tau \right) = \beta I^{u} \left( \tau \right) \varDelta \tau e^{ - \mu \left( \tau - t \right) } \exp \left( - \int _t^{ \tau } \beta I^{u} \left( s \right) ds \right) + \varphi _I^{u,t} \left( \tau \right) + o(\varDelta \tau ), \end{aligned}$$

which leads to

$$\begin{aligned} \frac{d \varphi _I^{u,t} \left( \tau \right) }{d \tau }&= \beta I^{u} \left( \tau \right) e^{ - \mu \left( \tau - t \right) } \exp \left( - \int _t^{ \tau } \beta I^{u} \left( s \right) ds \right) \\&= e^{ - \mu \left( \tau - t \right) } \exp \left( \int _0^t \beta I^{u} \left( s \right) ds \right) \times \beta I^{u} \left( \tau \right) \exp \left( - \int _0^{ \tau } \beta I^{u} \left( s \right) ds \right) \\&= - e^{- \mu \left( \tau - t \right) } \exp \left( \int _0^t \beta I^{u} \left( s \right) ds \right) \dfrac{d \left[ \exp \left( - \int _0^{ \tau } \beta I^{u} \left( s \right) ds \right) \right] }{d \tau } \\&= - e^{- \mu \left( \tau - t \right) } \exp \left( \int _0^t \beta I^{u} \left( s \right) ds \right) \dfrac{d \psi _I^{u} \left( \tau \right) }{d \tau }, \\ \end{aligned}$$

by setting $\psi _I^{u} \left( \tau \right) = \exp \left( - \int _0^{ \tau } \beta I^{u} \left( s \right) ds \right)$.

Finally, we just have to compute the probability of infection during the life of an individual born in t who is not vaccinated, which is:

$$\begin{aligned} \phi _I^{u} \left( t \right)&= \int _t^{ + \infty } d \varphi _I^{u,t} \left( \tau \right) \\&= - \int _t^{ + \infty } \exp \left( - \mu \left( \tau - t \right) + \int _0^t \beta I^{u} \left( s \right) ds \right) \left[ \psi _I^{u} \left( \tau \right) \right] ' \; d \tau \\&= - \exp \left( \mu t + \int _0^t \beta I^{u} \left( s \right) ds \right) \int _t^{ + \infty } e^{ - \mu \tau } \left[ \psi _I^{u} \left( \tau \right) \right] ' \; d \tau . \end{aligned}$$

By an integration by parts, we obtain:

$$\begin{aligned} \phi _I^{u} \left( t \right)&= 1 - \int _t^{ + \infty } \mu \exp \left( - \int _t^{ \tau } \left( \mu + \beta I^{u} \left( s \right) \right) ds \right) \; d \tau . \end{aligned}$$

In fact, this probability is solution of the differential equation:

$$\begin{aligned} \dfrac{d \phi _I^{u} \left( t \right) }{dt}&= \mu \exp \left( - \int _t^t \left( \mu + \beta I^{u} \left( s \right) \right) ds \right) \\&\quad - \int _t^{ + \infty } \mu \dfrac{\partial }{\partial t} \left[ \exp \left( - \int _t^{ \tau } \left( \mu + \beta I^{u} \left( s \right) \right) ds \right) \right] \; d \tau \\&= \mu - \mu \int _t^{ + \infty } \left( \mu + \beta I^{u} \left( t \right) \right) \exp \left( - \int _t^{ \tau } \left( \mu + \beta I^{u} \left( s \right) \right) ds \right) \; d \tau \\&= \mu - \left( \mu + \beta I^{u} \left( t \right) \right) \left( 1 - \phi _I^{u} \left( t \right) \right) = \left( \mu + \beta I^{u} \left( t \right) \right) \phi _I^{u} \left( t \right) - \beta I^{u} \left( t \right) . \end{aligned}$$

In order to get an explicit form for $\phi _I^{u} \left( 0 \right)$, we define, for all $t \ge 0$:

$$\begin{aligned} f_I^{u} \left( t \right)&= \exp \left[ - \int _0^t \left( \mu + \beta I^{u} \left( \tau \right) \right) d \tau \right] \\ F_I^{u} \left( t \right)&= \int _0^t \mu f_I^{u} \left( \tau \right) d\tau \\ \mathcal {F}_I^{u} \left( t \right)&= 1 - \phi _I^{u} \left( t \right) . \end{aligned}$$

The last function satisfies the following differential equation:

$$\begin{aligned} \left[ \mathcal {F}_I^{u} \left( t \right) \right] '&= - \mu + \left( \mu + \beta I^{u} \left( t \right) \right) \mathcal {F}_I^{u} \left( t \right) \\ \mathcal {F}_I^{u} \left( 0 \right)&= \int _0^{ + \infty } \mu \exp \left( - \int _0^{ \tau } \left( \mu + \beta I^{u} \left( s \right) \right) ds \right) \; d \tau \\&= \int _0^{ + \infty } \mu f_I^{u} \left( \tau \right) d\tau = F_I^{u} \left( \infty \right) . \end{aligned}$$

Note that:

$$\begin{aligned} \left[ \mathcal {F}_I^{u} \left( t \right) \times f_I^{u} \left( t \right) \right] '&= \left\{ \left[ \mathcal {F}_I^{u} \left( t \right) \right] ' - \mathcal {F}_I^{u} \left( t \right) \left( \mu + \beta I^{u} \left( t \right) \right) \right\} \times f_I^{u} \left( t \right) = - \mu f_I^{u} \left( t \right) \end{aligned}$$

Thus, $\mathcal {F}_I^{u} \left( t \right) \times f_I^{u} \left( t \right) = \mathcal {F}_I^{u} \left( 0 \right) - \int _0^t \mu f_I^{u} \left( \tau \right)$ and therefore

$$\begin{aligned} \mathcal {F}_I^{u} \left( t \right) = \dfrac{1}{ f_I^{u} \left( t \right) } \left[ F_I^{u} \left( \infty \right) - F_I^{u} \left( t \right) \right] . \end{aligned}$$

Hence, for all $t \ge 0$, we obtain:

$$\begin{aligned} \phi _I^{u} \left( t \right) = 1 - \dfrac{ F_I^{u} \left( \infty \right) - F_I^{u} \left( t \right) }{ f_I^{u} \left( t \right) }, \ \phi _I^{u} \left( 0 \right) = 1 - F_I^{u} \left( \infty \right) . \end{aligned}$$