A LQ Vaccine Communication Game

Abstract

The vaccination issue is a crucial problem nowadays. We see the presence of an anti-vaccination movement, which takes actions to spread the idea that vaccines are ineffective and even dangerous. We propose a model for this public health problem using the differential game framework and aspire to help understanding the effectiveness of communication policies. One player of the game is the health-care system, which aims to minimize the number of unvaccinated people at minimum cost. The second player is a pharmaceutical firm, which produces and sells a given type of vaccine, and wants to maximize its profit. To pursue their objectives, the two players run suitable vaccination advertising campaigns. We study the open-loop Nash equilibrium advertising strategies of the two players and observe that the communication policy of the pharmaceutical firm helps the health-care system to decrease the number of unvaccinated people.

Appendix

Proof (Theorem 2)

Let u _s(t) ≥ 0, t ∈ [0, T], be an admissible control of player S, which determines the state function x(t) > 0 , t ∈ [0, T). We recall that the function u _f(t) is a known parameter in the present context.

Let us assume that

$$\displaystyle \begin{aligned} u_{s}(t)\ =\ 0\;,\qquad t \in [\tau, \tau + h)\;, \end{aligned}$$

for some τ ∈ [0, T − 2h], h > 0, and let us define the spike variation

$$\displaystyle \begin{aligned} u_{s}^\alpha(t)\ =\ \left\{ \begin{array}{ll} \alpha \;, & t \in [\tau, \tau + h)\;,\\ u_{s}(t)\;, \ & t \not\in [\tau, \tau + h)\;, \end{array} \right. \end{aligned}$$

of the control u _s(t), where α > 0. Let the admissible control $u_{s}^\alpha (t)$, jointly with u _f(t), determine the state function x ^α(t) , t ∈ [0, T], and assume that

$$\displaystyle \begin{aligned} x^\alpha(t)\ >\ 0\;,\qquad t \in [\tau, \tau + 2h)\;. \end{aligned}$$

Remark that

$$\displaystyle \begin{aligned} x^\alpha(\tau)\ =\ x(\tau)\;, \end{aligned}$$

$$\displaystyle \begin{aligned} x^\alpha(t)\ \le\ x(t)\;,\qquad t \in (\tau, T)\;, \end{aligned}$$

and

$$\displaystyle \begin{aligned} x^\alpha(t)\ <\ x(t) - \delta_s h\alpha \;,\qquad t\ \ge\ \tau + h\;. \end{aligned}$$

In order to compare the values of the objective functional J _s associated with the control pairs (u _s, u _f) and $(u_{s}^\alpha , u_{f})$, let us define

$$\displaystyle \begin{aligned} \varDelta J_s\ =\ J_s[u_{s}^\alpha, u_{f}] - J_s[u_{s}, u_{f}] \end{aligned}$$

and look for an upper bound of it. We observe that

$$\displaystyle \begin{aligned} \begin{array}{rl} \varDelta J_s &\le\ \varDelta J_s\mid_{[\tau, \tau+h)} \ +\ \varDelta J_s\mid_{[\tau+h, \tau+2h)} \\ {} &=\ \displaystyle \frac{\beta }{2}\int_{\tau}^{\tau + h} \left( (x^\alpha)^{2}\left( t\right) - x^{2}\left( t\right) \right) \mathrm{d} t\ \ +\ \frac{\kappa_s }{2}\int_{\tau}^{\tau + h} \alpha^2\, \mathrm{d} t\ \\ &\hskip 2 cm +\ \displaystyle \frac{\beta }{2}\int_{\tau+h}^{\tau + 2h} \left( (x^\alpha)^{2}\left( t\right) - x^{2}\left( t\right) \right) \mathrm{d} t\;, \end{array} \end{aligned}$$

because ΔJ _s∣_[0,τ) = 0, ΔJ _s∣_[τ+2h,T] ≤ 0, and because of the residual value function negative variation. The upper bound just obtained is strictly less than

$$\displaystyle \begin{aligned} \frac{\kappa_s }{2}\int_{\tau}^{\tau + h} \alpha^2\, \mathrm{d} t\ +\ \frac{\beta }{2}\int_{\tau + h}^{\tau + 2h} \left( (x^\alpha)^{2}\left( t\right) - x^{2}\left( t\right) \right) \mathrm{d} t\;, \end{aligned}$$

because

$$\displaystyle \begin{aligned} \frac{\beta }{2}\int_{\tau}^{\tau + h} \left( (x^\alpha)^{2}\left( t\right) - x^{2}\left( t\right) \right) \mathrm{d} t <\ 0\;. \end{aligned}$$

The last upper bound is strictly less than

$$\displaystyle \begin{aligned} \begin{array}{rl} &\displaystyle \frac{\kappa_s h}{2} \alpha^2\ +\ \frac{\beta }{2}\int_{\tau + h}^{\tau + 2h} \left( ( x(t) - \delta_s h\alpha )^2 - x^{2}\left( t\right) \right) \mathrm{d} t \\ &\displaystyle \hskip 1cm =\ \frac{h\alpha}{2} \left\{ \left( \kappa_s + \beta \delta_s^2 h^2 \right)\alpha\ -\ 2\beta \delta_s \int_{\tau + h}^{\tau + 2h}x(t)\,\mathrm{d} t \right\}\;. \end{array} \end{aligned}$$

Now, the last expression is negative if and only if

$$\displaystyle \begin{aligned} \alpha\ <\ \frac{2\beta \delta_s }{\kappa_s + \beta \delta^2_s h^2 }\, \int_{\tau + h}^{\tau + 2h}x(t)\,dt\;, \end{aligned}$$

which is true for suitably small α > 0, because the right-hand side of the above inequality is strictly positive. □