Advertising a product to face a competitor entry: a differential game approach

Abstract

We analyze a market in which advertising is the dominant marketing tool to create market share. We assume that an incumbent firm dominates the market during an initial stage, and that a new competitor is going to enter the market. In particular, we analyze the different advertising policies that the incumbent firm can adopt, before and after the entry of the rival. We explore three possible behaviours. In the first scenario the firm knows that the competitor will arrive at a given instant. In the second one we assume the original firm to be surprised, in the sense that it does not anticipate the entry of the opponent either because it does not expect the competitor to arrive, or it is not prepared to react before the entry takes place. Finally, in the third scenario, the original firm knows that the competitor will enter at a constant rate. We characterize a differential game model and compare the firms’ behaviours in a strategic perspective.

Appendices

Proof of Theorem 1

The non-surprised case in duopoly stage corresponds to the symmetric case presented and solved in Prasad and Sethi (2004). The optimal Markovian advertising decisions are then

$$\begin{aligned} u_1^*(x)&= V_1'(x) \rho \sqrt{1-x}/2, \end{aligned}$$

(31)

$$\begin{aligned} u_2^*(x)&= -V_2'(x) \rho \sqrt{x}/2. \end{aligned}$$

(32)

where the value functions have the following forms:

$$\begin{aligned} V_1 =\alpha _1+\beta _1 x, \quad V_2 =\alpha _2+\beta _2(1-x) \end{aligned}$$

(33)

and the optimal coefficients $\alpha = \alpha _1= \alpha _2 ,\beta = \beta _1= \beta _2$ are determined by solving the following equations

$$\begin{aligned}&r \alpha =\beta ^2\rho ^2/4+\beta \delta , \end{aligned}$$

(34)

$$\begin{aligned}&r\beta =m-3\beta ^2\rho ^2/4-2\beta \delta . \end{aligned}$$

(35)

There exist two solutions for $\beta$ with opposite sign, the negative one has to be rejected because it would make (31) negative, and this would not satisfy the non-negativity constraint for the control. So that we obtain

$$\begin{aligned}&\alpha = \frac{ (r-\delta ) (r+2\delta -\sqrt{(r+2\delta )^2+3 m \rho ^2} ) +6m\rho ^2/4}{18\rho ^2r/4}, \end{aligned}$$

(36)

$$\begin{aligned}&\beta = \frac{\sqrt{(r+2\delta )^2+3m \rho ^2}-(r+2\delta )}{3 \rho ^2/2}. \end{aligned}$$

(37)

It is easy to check that $\beta >0$ represents the weight of the final goodwill and, as similarly to $\alpha$, it depends on the parameters $\rho$, m, $\delta$ and r. The optimal Markovian advertising decisions give (11) and (12).

In order to find the steady state we put

$$\begin{aligned} \dot{x}(t)= \beta \rho ^{2} ( 1- x(t)) /2 - \beta \rho ^{2} x(t) /2 - \delta ( 2 x(t) -1))=0 \end{aligned}$$

(38)

that gives $\hat{x}_{\mathrm{NS}}^D=0.5,$ and symmetrically for player 2. Equation (38) is linear in x with a negative coefficient, and therefore the obtained steady state is stable. So that the market share converges to the steady state for any initial value. Moreover, since $\ddot{x}=-2( \beta \rho ^{2} /2 +\delta )\dot{x},$ the market share is either increasing and concave (if $x_0<x_\mathrm{SS}$) or decreasing and convex, (if $x_0>x_\mathrm{SS}$) and the trajectory cannot have multiple peaks.

Following Dockner et al. (2000, p. 100) the Markovian Nash equilibrium is subgame perfect, and therefore called Markov perfect.

Proof of Theorem 2

The current value Hamiltonian function associated to the non-surprised incumbent problem in the monopoly stage is

$$\begin{aligned} H(x, u_1, \lambda , t) = m x -u_1^2+\lambda (\rho u_1 \sqrt{1-x} -\delta x). \end{aligned}$$

(39)

Maximizing the Hamiltonian function w.r.t. control we obtain

$$\begin{aligned} u_{1\mathrm{NS}}^{M}(t)=\frac{\lambda (t) \rho \sqrt{1-x(t)}}{2}. \end{aligned}$$

(40)

Since the Hamiltonian function is strictly concave w.r.t. control $u_{1\mathrm{NS}}^{M}(t)$ turns out to be a unique Maximum. Thus the advertising decision satisfying the Maximum Principle conditions is

$$\begin{aligned} u_{1\mathrm{NS}}^{*M}(t)=\max \left\{ {0, \frac{\lambda (t) \rho \sqrt{1-x(t)}}{2}}\right\} . \end{aligned}$$

(41)

The costate equation is

$$\begin{aligned} \dot{\lambda }(t)= & {} - \frac{\partial H}{\partial x}(t)+r\lambda (t) \nonumber \\= & {} -m+ \lambda (t)\rho u_1(t) \frac{1}{2\sqrt{1-x(t)}}+\lambda (t)\delta +r\lambda (t) \nonumber \\= & {} -m+\frac{\rho ^2 \lambda ^2(t) }{4}+( \delta +r)\lambda (t) \nonumber \\= & {} \frac{\rho ^{2}}{2} \lambda ^2(t) +U\lambda (t)-m, \end{aligned}$$

(42)

where (41) has been substituted and $U=\delta + r$, $R= \sqrt{U^2+ m \rho ^{2}}$ has been used. Recalling the salvage value function we obtain the following transversality condition

$$\begin{aligned} \lambda (T) = \frac{\partial S(x(T))}{\partial x} = \beta >0 \end{aligned}$$

(43)

which can be used to solve the above costate equation, i.e.

$$\begin{aligned} \lambda (t)= \frac{2 \left( \beta \rho ^2 \left( (R+U) e^{R (t-\text {Te})}+R-U\right) +2 (R-U) (R+U) \left( 1-e^{R (t-\text {Te})}\right) \right) }{\rho ^2 \left( \beta \rho ^2+e^{R (t-\text {Te})} \left( 2 (R-U)-\beta \rho ^2\right) +2 (R+U)\right) }. \end{aligned}$$

(44)

Observe that because of the transversality condition the costate function is positive at the final time and whenever it is null its derivative is negative, in fact

$$\begin{aligned} \dot{\lambda }(t)|_{\lambda (t)=0}=-m <0, \end{aligned}$$

so that it follows that $\lambda (t)$ is positive for all $t \in [0,T].$ Therefore the optimal control $u_{1\mathrm{NS}}^{*M}(t)$ in (41) is strictly positive for all t and it becomes (17).

Observe that the maximized Hamiltonian $H^*$

$$\begin{aligned} H^*(x,\lambda , t)= m x(t) - \frac{\lambda ^2(t) \rho ^2 (1-x(t))}{4}+\frac{ \lambda ^2(t) \rho ^2 (1-x(t)) }{2} -\lambda (t)\delta x(t), \end{aligned}$$

(45)

turns out to be linear with respect to x,  as well as the salvage function, so the sufficiency Arrow Theorem holds and the Maximum Principle conditions are also sufficient for optimality. Moreover, note that (41) and (44) imply that the solution of the optimal control problem is unique.

Proof of Theorem 3

The current value Hamiltonian function associated to the surprised incumbent is

$$\begin{aligned} H(x, u_1, \lambda , t) = m x - u_1^2 +\lambda (\rho u_1 \sqrt{1-x} -\delta x). \end{aligned}$$

(46)

Analogously to the proof before we are able to derive the optimal control value, which is

$$\begin{aligned} u_{1S}^{*M}(t)=\max \left\{ {0, \frac{\lambda (t) \rho \sqrt{1-x(t)}}{2}}\right\} . \end{aligned}$$

(47)

Strict concavity of the Hamiltonian implies that $u_{1S}^{*M}(t)$ is a unique Maximum.

For the costate equation we obtain [substituting (47) in (48) and $U= \delta + r$]

$$\begin{aligned} \dot{\lambda }(t)= & {} - \frac{\partial H}{\partial x}(t)+r\lambda (t)= -m+ \lambda (t)\rho u_1 (t)\frac{1}{2\sqrt{1-x(t)}}+\lambda (t) \delta +r\lambda (t) \nonumber \\= & {} \rho ^{2} \lambda (t)^2 /4+U \lambda (t)-m. \end{aligned}$$

(48)

Since the model has infinite time horizon the limiting transversality condition reads

$$\begin{aligned} \lim _{t \rightarrow \infty } e^{-rt} \lambda (t)=0. \end{aligned}$$

(49)

The steady state of the canonical system $( \dot{x} , \dot{\lambda } )$ can uniquely be derived, i.e.

$$\begin{aligned} \hat{x}_S= & {} 1-\frac{\delta }{R-U +\delta } \nonumber \\ \hat{\lambda }_S= & {} \frac{-U + \sqrt{U^2+m \rho ^{2}}}{\rho ^{2} /2}=\frac{(R-U)2}{\rho ^2}, \end{aligned}$$

(50)

where $R=\sqrt{U^2+m \rho ^{2}}$ and $\hat{x}_S \in [0,1]$. Its corresponding Jacobian matrix (evaluated in the steady state) is

$$\begin{aligned} J(x, \lambda )= \begin{pmatrix} \lambda \rho ^2 / 2+\delta +r &{} 0 \\ \rho ^2 (1-x) / 2 &{}- \rho ^2 \lambda / 2 -\delta . \\ \end{pmatrix} \end{aligned}$$

(51)

In the steady state $\det (J(\hat{x}_S, \hat{\lambda }_S))<0$ holds, therefore $(\hat{x}_S, \hat{\lambda }_S)$ is a saddle point. According to standard optimal control theory the optimal solution follows the stable trajectory of the (unique) saddle point equilibrium (see Grass et al. 2008, section 3).

Analogously to the proof before the maximized Hamiltonian is linear in x, thus the Arrow sufficiency Theorem holds and implies that the conditions of the Maximum Principle are sufficient. Furthermore the solution of the optimal control problem is unique.

Proof of Theorem 4

Recalling that $V_1(x(t))=\alpha + \beta x(t),$ the Pontryagin’s Maximum Principle for the infinite time horizon stochastic problem (25) and (26) gives the current value Hamiltonian function

$$\begin{aligned} H(x, u_1, \lambda , t) = m x - u_1^2 + \gamma (\alpha + \beta x)+\lambda (\rho u_1 \sqrt{1-x} -\delta x). \end{aligned}$$

(52)

Analogously to the proof of Theorem 2 before we are able to derive the optimal control value, which is

$$\begin{aligned} u_{1{ St}}^{*M} (t)= \frac{\lambda (t) \rho \sqrt{1-x(t)}}{2}. \end{aligned}$$

(53)

Strict concavity of the Hamiltonian implies that $u_{1{ St}}^{*M} (t)$ is a unique Maximum.

The costate equation we obtain

$$\begin{aligned} \dot{\lambda }(t)&= - \frac{\partial H}{\partial x}(t)+(r+\gamma )\lambda (t) \nonumber \\&= -m-\gamma \beta + \lambda (t) \rho u_1 (t)\frac{1}{2\sqrt{1-x(t)}}+\lambda (t) (\delta +r+\gamma ). \end{aligned}$$

(54)

with the limiting transversality condition $\lim _{t \rightarrow \infty } e^{-rt} \lambda (t)$.

Analogously to the proof before the unique steady-state solution can be derived, which equals

$$\begin{aligned} \hat{x}_{\mathrm{St}}= & {} 1- \frac{\delta }{-Q+\sqrt{Q^2+ \rho ^{2} (m+\gamma \beta )}+\delta } \nonumber \\ \hat{\lambda }_{\mathrm{St}}= & {} \frac{-Q+\sqrt{Q^2+ \rho ^{2}(m+\gamma \beta )}}{\rho ^{2} /2} \end{aligned}$$

(55)

where $Q=r+\delta +\gamma$. The associated Jacobian matrix (evaluated in the steady state) is

$$\begin{aligned} J(x, \lambda )= \begin{pmatrix} \rho ^{2} \lambda /2 +Q &{} 0 \\ \rho ^{2} (1-x)/2 &{} - \rho ^{2} \lambda /2 -\delta \\ \end{pmatrix} \end{aligned}$$

and its determinant is negative, hence $(\hat{x}_{\mathrm{St}}, \hat{\lambda }_{\mathrm{St}})$ is a saddle point (see Grass et al. 2008, section 2). The solution can exactly be found as in the case of the surprised monopoly (“Appendix C”).

Analogously to the proof before the maximized Hamiltonian is linear in x, thus the Arrow sufficiency Theorem holds and implies that the conditions of the Maximum Principle are sufficient. Furthermore the solution of the optimal control problem is unique.

Proof of Proposition 1

Let be $x_{\mathrm{St}}(t)=x_S(t)=x_{\mathrm{NS}}(t)=x,$ It can be easily proved that $N(t)<R-U,$ for all t,  so that the first inequality in (30) trivially holds.

The second condition is equivalent to

$$\begin{aligned} -(U+\gamma )+\sqrt{(U+\gamma )^2+(m+\gamma \beta ) \rho ^{2}} < -U+\sqrt{U^2+m \rho ^{2}}, \end{aligned}$$

(56)

and it is verified if and only if

$$\begin{aligned} m > \beta ( \beta \rho ^{2} /4+\delta +r). \end{aligned}$$

(57)

Recalling the value of $\beta$

$$\begin{aligned} \beta = \frac{\sqrt{(r+2\delta )^2+3m \rho ^{2}}-(r+2\delta )}{3 \rho ^{2} /2} \end{aligned}$$

(57) becomes

$$\begin{aligned} 3m \rho ^{2} -(2 r+ \delta ) \left( \sqrt{3m \rho ^{2} +(r+2 \delta )^2}-(r+2 \delta ) \right) >0, \end{aligned}$$

(58)

which is always verified.

Proof of Proposition 2

From the initial conditions we have that

$$\begin{aligned} x_{\mathrm{NS}}(0)=x_S(0)=x_0 \end{aligned}$$

and

$$\begin{aligned} \dot{x}_{\mathrm{NS}}(0)= N(0)(1-x_0)-\delta x_0<(R-U) (1-x_0)- \delta x_0=\dot{x}_S(0). \end{aligned}$$

So that $x_{\mathrm{NS}}(t)<x_S(t)$ for any $t\in [0,\epsilon [$ with $\epsilon \ge 0.$ We can prove that as soon as the two functions coincide at a given instant $\tilde{t}>0$ than at that point their derivatives satisfy the inequality

$$\begin{aligned} \dot{x}_{\mathrm{NS}}(\tilde{t})= N(\tilde{t})(1-x_0)-\delta x_0<(R-U) (1-x_0)- \delta x_0=\dot{x}_S(\tilde{t}), \end{aligned}$$

are therefore we can conclude that $x_{\mathrm{NS}}(t)<x_S(t)$ for any $t\in [0,T^e].$