Defensive, Offensive, and Generic Advertising in a Lanchester Model with Market Growth

Abstract

The paper considers a duopolistic market in which firms compete over time through their respective advertising efforts. In contrast to earlier work in advertising competition, the paper supposes that each firm may choose among three types of advertising: offensive advertising which has the purpose of attracting customers from the rival firm, defensive advertising which aims at protecting a firm’s customer base from the competitors’ attacks, and generic advertising which aims at enhancing industry sales. We address questions like: How should an advertising strategy, for each of the three types of advertising effort, be designed? How would the corresponding time paths of sales look like? The paper uses differential game theory to answer these questions and provides closed-form analytical expressions for equilibrium advertising strategies and sales rate paths. It is found that advertising strategies can be expressed in terms of the shadow prices of the firms’ sales rates and the model parameters. Two combinations of theses advertising are optimal: all three advertising together and both offensive and defensive advertising. As to the latter, an essential assumption is that offensive advertising is more cost-effective than defensive advertising.

Appendix: Proofs

Proof of Proposition 1

It has already been proved in Lemma 1 that $\varphi (t)>\psi (t)$ for all $t$ and in both cases. In Case 1, use () to show that $\dot{\psi }(t)<0$ for all $t\in \left[ 0,T\right] .$ Since $\psi (T)=0$ must hold, it follows that $\psi (t)>0$. The result $\varphi (t)>0$ follows directly from (13). Positivity of both shadow prices leads to $\varphi (t)+\psi (t)>0$ which implies a positive generic advertising rate. In Case 2, we use (11) to show that $\dot{\psi }(t)>0$ for all $t.$ Since $\psi (T)=0,$ it follows that $\psi (t)<0$ for all $t.$ To prove the remaining results, we need the sign of

$$\begin{aligned} \varphi (t)+\psi (t)&=\frac{2mC_{2}\left( T-t\right) }{C_{2}-C_{1}} \\&\quad -\frac{\left( C_{1}+C_{2}\right) \sqrt{m\left( C_{2}-C_{1}\right) }\tanh \sqrt{m\left( C_{2}-C_{1}\right) }\left( T-t\right) }{\left( C_{1}-C_{2}\right) ^{2}} \end{aligned}$$

which, introducing three constants

$$\begin{aligned} B_{1}\triangleq \frac{2mC_{2}}{C_{2}-C_{1}}<0,\text { }\gamma \triangleq \sqrt{m\left( C_{2}-C_{1}\right) }>0,\text { }B_{2}\triangleq \frac{\left( C_{1}+C_{2}\right) \gamma }{\left( C_{1}-C_{2}\right) ^{2}}<0 \end{aligned}$$

and defining $h(t)=\varphi (t)+\psi (t),$ can be written more conveniently as

$$\begin{aligned} h(t)=B_{1}\left( T-t\right) -B_{2}\tanh \gamma \left( T-t\right) . \end{aligned}$$

Next we determine the essential properties of function $h.$ It holds that $h(T)=0$ and

$$\begin{aligned} h(0)\left( {\begin{array}{c}>\\ <\end{array}}\right) 0\Leftrightarrow \frac{2C_{2}}{C_{1}+C_{2}}\left( {\begin{array}{c}<\\ >\end{array}}\right) \frac{ \tanh \gamma T}{\gamma T} \end{aligned}$$

where $2C_{2}/\left( C_{1}+C_{2}\right) \in \left( 0,1\right) .$ The function $z\left( T\right) =\tanh \gamma T/(\gamma T)$ is decreasing and $z\left( T\right) \in \left( 0,1\right) .$ We conclude that there exists a value of $T,$ say $\tilde{T},$ defined by $2C_{2}/\left( C_{1}+C_{2}\right) =\tanh \gamma \tilde{T}/\left( \gamma \tilde{T}\right)$ such that $h\left( 0\right) >0$ if $T<\tilde{T}$ and $h\left( 0\right) \le 0$ if $T\ge \tilde{ T}$.⁹ Function $h$ is strictly concave since

$$\begin{aligned} \ddot{h}(t)=-2B_{2}\gamma ^{2}\left( \tanh \gamma \left( T-t\right) \right) \left( \tanh ^{2}\gamma \left( T-t\right) -1\right) <0\text { for }t\in [0,T). \end{aligned}$$

Using the properties derived for function $h(t)$ shows that if $T<\tilde{T},$ then $h(t)$ is positive for $t\in [0,T).$If $T>\tilde{T},$ then $h(t)$ is negative on an initial interval $[0,t^{*})$ and positive on a terminal interval $(t^{*},T).$ Q.E.D.

Proof of Proposition 3

We solve the inhomogeneous equations that apply during time intervals $\left[ 0,T\right]$ and $[t^{*},T],$ respectively$.$ The differential equations for $S_{1}\left( t\right)$ and $S_{2}\left( t\right)$ can be transformed into a second-order equation for $S_{1}\left( t\right)$:

$$\begin{aligned} \ddot{S}_{1}\left( t\right) +\left( 2p(t)-\frac{\dot{p}(t)}{p(t)}\right) \dot{S}_{1}\left( t\right) =\dot{q}(t)+2p(t)q(t)-\dot{p}(t)\frac{q(t)}{p(t)} \\ \Leftrightarrow \ddot{S}_{1}\left( t\right) +f(t)\dot{S}_{1}\left( t\right) =g(t) \end{aligned}$$

which admits the solution

$$\begin{aligned} S_{1}\left( t\right) =\kappa _{1}+\kappa _{2}\int \hbox {e}^{-F(t)}\hbox {d}t+\int \hbox {e}^{-F(t)}\left( \int \hbox {e}^{F(t)}g(t)\hbox {d}t\right) \hbox {d}t \end{aligned}$$

(25)

where $F(t)=\int f(t)\hbox {d}t$ and $\kappa _{1},\kappa _{2}$ are constants of integration. First we determine the integral $F:$

$$\begin{aligned} F&=\int \left( 2p(t)-\frac{\dot{p}(t)}{p(t)}\right) \hbox {d}t=\int \left( \frac{ 4\gamma }{3}\tanh \gamma \left( T-t\right) -\frac{\gamma \left( \tanh ^{2}\gamma \left( T-t\right) -1\right) }{\tanh \gamma \left( T-t\right) } \right) \hbox {d}t \\&=-\ln \left( \hbox {e}^{2\gamma \left( t-T\right) }-1\right) +\ln \left( \hbox {e}^{2\gamma \left( t-T\right) }+1\right) -\frac{4}{3}\ln \left( \hbox {e}^{2\gamma \left( t-T\right) }+1\right) -\frac{4\gamma }{3}\left( 2T-t\right) \end{aligned}$$

which leads to

$$\begin{aligned} \hbox {e}^{F}=\hbox {e}^{\frac{4\gamma }{3}\left( t-2T\right) }\frac{\left( \hbox {e}^{-2\gamma \left( T-t\right) }+1\right) ^{-\frac{1}{3}}}{\hbox {e}^{-2\gamma \left( T-t\right) }-1};\quad \hbox {e}^{-F}=\hbox {e}^{-\frac{4\gamma }{3}\left( t-2T\right) }\frac{ \hbox {e}^{-2\gamma \left( T-t\right) }-1}{\left( \hbox {e}^{-2\gamma \left( T-t\right) }+1\right) ^{-\frac{1}{3}}}. \end{aligned}$$

Hence

$$\begin{aligned} \kappa _{2}\int \hbox {e}^{-F(t)}\hbox {d}t=\kappa _{2}\frac{3\hbox {e}^{\frac{4\gamma }{3}\left( 2T-t\right) }\left( 1+\hbox {e}^{-2\gamma \left( T-t\right) }\right) ^{\frac{4}{3}}}{ 4\gamma }. \end{aligned}$$

(26)

The next task is to determine the integral $\int \hbox {e}^{-F(t)}\left( \int \hbox {e}^{F(t)}g(t)\hbox {d}t\right) \hbox {d}t.$ Here we have

$$\begin{aligned} \int \hbox {e}^{F(t)}g(t)\hbox {d}t= & {} \frac{\theta \left( T-t\right) \left( 1+\hbox {e}^{-2\gamma \left( T-t\right) }\right) +\gamma \eta \left( 1-\hbox {e}^{-2\gamma \left( T-t\right) }\right) }{\hbox {e}^{-\frac{4\gamma }{3}\left( t-2T\right) }\left( \hbox {e}^{4\gamma \left( t-T\right) }-1\right) \left( \hbox {e}^{-2\gamma \left( T-t\right) }+1\right) ^{\frac{1}{3}}},\\ \hbox {e}^{-F(t)}\int \hbox {e}^{F(t)}g(t)\hbox {d}t= & {} \frac{\theta \left( T-t\right) \left( 1+)\hbox {e}^{-2\gamma \left( T-t\right) }\right) +\gamma \eta \left( 1-\hbox {e}^{-2\gamma \left( T-t\right) }\right) }{\hbox {e}^{-2\gamma \left( T-t\right) }+1} \end{aligned}$$

and then

$$\begin{aligned} \int \hbox {e}^{-F(t)}\left( \int \hbox {e}^{F(t)}g(t)\hbox {d}t\right) \hbox {d}t=-2T\gamma \eta +\left( T\theta +\gamma \eta \right) t-\frac{\theta t^{2}}{2}-\eta \ln \left( 1+\hbox {e}^{2\gamma \left( t-T\right) }\right) . \end{aligned}$$

(27)

Using (25), (26), and (27) yields the equilibrium sales rate trajectories:

$$\begin{aligned} S_{1}\left( t\right)= & {} \kappa _{1}+\kappa _{2}\frac{3\hbox {e}^{\frac{4\gamma }{3} \left( 2T-t\right) }\left( 1+\hbox {e}^{-2\gamma \left( T-t\right) }\right) ^{\frac{4 }{3}}}{4\gamma } \\&\quad -2T\gamma \eta +\left( T\theta +\gamma \eta \right) t-\frac{\theta }{2} t^{2}-\eta \ln \left( 1+\hbox {e}^{2\gamma \left( t-T\right) }\right) \nonumber \\ S_{2}(t)= & {} \kappa _{1}-\kappa _{2}\frac{3\hbox {e}^{\frac{4\gamma }{3}\left( 2T-t\right) }\left( 1+\hbox {e}^{2\gamma \left( t-T\right) }\right) ^{\frac{4}{3}}}{ 4\gamma } \nonumber \\&\quad -2T\gamma \eta +\left( T\theta +\gamma \eta \right) t-\frac{\theta }{2} t^{2}-\eta \ln \left( 1+\hbox {e}^{2\gamma \left( t-T\right) }\right) . \nonumber \end{aligned}$$

(28)

Setting $\theta =\eta =0$ provides the solutions of the homogeneous equations on the interval $\left[ 0,t^{*}\right]$:

$$\begin{aligned} S_{1}\left( t\right)= & {} \kappa _{1}+\kappa _{2}\int \hbox {e}^{-F(t)}\hbox {d}t=\kappa _{1}+\kappa _{2}\frac{\hbox {e}^{2\mu \left( 2T-t\right) }\left( 1+\hbox {e}^{-2\gamma \left( T-t\right) }\right) ^{\frac{4}{3}}}{2\mu } \\ S_{2}(t)= & {} \kappa _{1}-\kappa _{2}\frac{\hbox {e}^{2\mu \left( 2T-t\right) }\left( 1+\hbox {e}^{2\gamma \left( t-T\right) }\right) ^{\frac{4}{3}}}{2\mu }. \nonumber \end{aligned}$$

(29)

It remains to determine the two constants of integration. First consider the sales rates on the time interval $\left[ 0,t^{*}\right] .$ Use the equations

$$\begin{aligned} s_{10}=\kappa _{1}+\kappa _{2}\frac{3\hbox {e}^{\frac{8\gamma }{3}T}\left( 1+\hbox {e}^{-2\gamma T}\right) ^{\frac{4}{3}}}{4\gamma };\quad \text { }s_{20}=\kappa _{1}-\kappa _{2}\frac{3\hbox {e}^{\frac{8\gamma }{3}T}\left( 1+\hbox {e}^{-2\gamma T}\right) ^{\frac{4}{3}}}{4\gamma } \end{aligned}$$

to get

$$\begin{aligned} \kappa _{1}=\frac{s_{10}+s_{20}}{2};\quad \kappa _{2}=\frac{2\gamma \left( s_{10}-s_{20}\right) }{3\hbox {e}^{\frac{8\gamma }{3}}\left( \hbox {e}^{-2T\gamma }+1\right) ^{\frac{4}{3}}} \end{aligned}$$

which provides the sales rate trajectories

$$\begin{aligned} S_{1}\left( t\right)= & {} \frac{s_{10}+s_{20}}{2}+\frac{(s_{10}-s_{20})\hbox {e}^{ \frac{4\gamma }{3}\left( 2T-t\right) }\left( 1+\hbox {e}^{-2\gamma \left( T-t\right) }\right) ^{\frac{4}{3}}}{2\hbox {e}^{\frac{8\gamma }{3}T}\left( \hbox {e}^{-2\gamma T}+1\right) ^{\frac{4}{3}}} \\ S_{2}(t)= & {} \frac{s_{10}+s_{20}}{2}-\frac{(s_{10}-s_{20})\hbox {e}^{\frac{4\gamma }{3 }\left( 2T-t\right) }\left( 1+\hbox {e}^{2\gamma \left( t-T\right) }\right) ^{\frac{4 }{3}}}{2\hbox {e}^{\frac{8\gamma }{3}T}\left( \hbox {e}^{-2\gamma T}+1\right) ^{\frac{4}{3}}} \end{aligned}$$

and their terminal values at time $t=t^{*}:$

$$\begin{aligned} S_{1}\left( t^{*}\right)= & {} \frac{s_{10}+s_{20}}{2}+\frac{ (s_{10}-s_{20})\hbox {e}^{\frac{4\gamma }{3}\left( 2T-t^{*}\right) }\left( 1+\hbox {e}^{-2\gamma \left( T-t^{*}\right) }\right) ^{\frac{4}{3}}}{2\hbox {e}^{\frac{ 8\gamma }{3}T}\left( \hbox {e}^{-2\gamma T}+1\right) ^{\frac{4}{3}}} \nonumber \\ S_{2}(t^{*})= & {} \frac{s_{10}+s_{20}}{2}-\frac{(s_{10}-s_{20})\hbox {e}^{\frac{ 4\gamma }{3}\left( 2T-t^{*}\right) }\left( 1+\hbox {e}^{2\gamma \left( t^{*}-T\right) }\right) ^{\frac{4}{3}}}{2\hbox {e}^{\frac{8\gamma }{3}T}\left( \hbox {e}^{-2\gamma T}+1\right) ^{\frac{4}{3}}}. \end{aligned}$$

(30)

The values in (30) serve as initial conditions for the two inhomogeneous equations on the time interval $\left[ t^{*},T\right] .$ Solutions of the inhomogeneous equations are

$$\begin{aligned} S_{1}\left( t\right)&=\frac{s_{10}+s_{20}}{2}+\frac{\left( s_{10}-s_{20}\right) \hbox {e}^{\frac{4\gamma }{3}\left( 2T-t\right) }\left( 1+\hbox {e}^{-2\gamma \left( T-t\right) }\right) ^{\frac{4}{3}}}{2\hbox {e}^{\frac{8\gamma }{ 3}T}\left( \hbox {e}^{-2T\gamma }+1\right) ^{\frac{4}{3}}} \\&\quad +\left( T\theta +\gamma \eta \right) t-\frac{\theta }{2}t^{2}+\eta \ln \left( \hbox {e}^{-2T\gamma }+1\right) -\eta \ln \left( 1+\hbox {e}^{2\gamma \left( t-T\right) }\right) \\ S_{2}(t)&=\frac{s_{10}+s_{20}}{2}-\frac{\left( s_{10}-s_{20}\right) \hbox {e}^{ \frac{4\gamma }{3}\left( 2T-t\right) }\left( 1+\hbox {e}^{2\gamma \left( t-T\right) }\right) ^{\frac{4}{3}}}{2\hbox {e}^{\frac{8\gamma }{3}T}\left( \hbox {e}^{-2T\gamma }+1\right) ^{\frac{4}{3}}} \\&\quad +\left( T\theta +\gamma \eta \right) t-\frac{\theta }{2}t^{2}+\eta \ln \left( \hbox {e}^{-2T\gamma }+1\right) -\eta \ln \left( 1+\hbox {e}^{2\gamma \left( t-T\right) }\right) \\ \text {for }t&\in \left[ 0,T\right] \end{aligned}$$

and

$$\begin{aligned} S_{1}\left( t\right)&=\frac{s_{10}+s_{20}}{2}+\eta \left( \ln \left( \hbox {e}^{2\gamma \left( t^{*}-T\right) }+1\right) -\ln \left( 1+\hbox {e}^{2\gamma \left( t-T\right) }\right) \right) +\frac{\theta }{2}\left( (t^{*})^{2}-t^{2}\right) \\&\quad +\left( T\theta +\gamma \eta \right) \left( t-t^{*}\right) +\frac{ (s_{10}-s_{20})\hbox {e}^{-\frac{8}{3}T\gamma }\left( \hbox {e}^{2\gamma \left( t-T\right) }+1\right) ^{\frac{4}{3}}\hbox {e}^{-\frac{4}{3}\gamma \left( t-2T\right) }}{2\left( \hbox {e}^{-2T\gamma }+1\right) ^{\frac{4}{3}}} \\ S_{2}(t)&=\frac{s_{10}+s_{20}}{2}+\eta \left( \ln \left( \hbox {e}^{2\gamma \left( t^{*}-T\right) }+1\right) -\ln \left( 1+\hbox {e}^{2\gamma \left( t-T\right) }\right) \right) +\frac{\theta }{2}\left( (t^{*})^{2}-t^{2}\right) \\&\quad +\left( T\theta +\gamma \eta \right) \left( t-t^{*}\right) -\frac{ (s_{10}-s_{20})\hbox {e}^{-\frac{8}{3}T\gamma }\left( \hbox {e}^{2\gamma \left( t-T\right) }+1\right) ^{\frac{4}{3}}\hbox {e}^{-\frac{4}{3}\gamma \left( t-2T\right) }}{2\left( \hbox {e}^{-2T\gamma }+1\right) ^{\frac{4}{3}}} \\ \text {for }t&\in \left[ t^{*},T\right] . \end{aligned}$$

Q.E.D.

Proof of Lemma 1

Introducing constants

$$\begin{aligned} C_{1}=\frac{\lambda ^{2}}{2c_{d}}-\frac{\beta ^{2}}{c_{a}},\, C_{2}= \frac{\beta ^{2}}{2c_{a}}-\frac{\lambda ^{2}}{c_{d}}, \end{aligned}$$

where $C_{1}<0,C_{1}+C_{2}<0,C_{1}-C_{2}<0,$ the system in (11) can be written as

$$\begin{aligned} \dot{\varphi }(t)&=-m-C_{1}\left( \varphi (t)-\psi (t)\right) ^{2} \end{aligned}$$

(31)

$$\begin{aligned} \dot{\psi }(t)&=-C_{2}\left( \varphi (t)-\psi (t)\right) ^{2}. \end{aligned}$$

(32)

Recall that Case 1 is the one where $C_{2}>0$ while $C_{2}<0$ in Case 2$.$ To solve (31) and (32), we employ a modification of a method used in Bass et al. [2]. Multiplying in (32) by $-C_{1}/C_{2}$ yields

$$\begin{aligned} -\frac{C_{1}}{C_{2}}\dot{\psi }(t)=C_{1}\left( \varphi (t)-\psi (t)\right) ^{2} \end{aligned}$$

and adding this to (31) provides

$$\begin{aligned} \dot{\varphi }(t)-\frac{C_{1}}{C_{2}}\dot{\psi }(t)=-m. \end{aligned}$$

(33)

Defining $y(t)=\varphi (t)-C_{1}\psi (t)/C_{2},$ (33) becomes

$$\begin{aligned} \dot{y}(t)=-m\Rightarrow y(t)=-mt+K \end{aligned}$$

and using the condition $\varphi (T)=\psi (T)=0$ implies $y(T)=0.$ Therefore

$$\begin{aligned} y(t)=m(T-t). \end{aligned}$$

Then we have

$$\begin{aligned} \varphi (t)=m(T-t)+\frac{C_{1}}{C_{2}}\psi (t) \end{aligned}$$

(34)

which is substituted into (32) to yield

$$\begin{aligned} \dot{\psi }(t)=-C_{2}\left( m(T-t)+\frac{C_{1}-C_{2}}{C_{2}}\psi (t)\right) ^{2}. \end{aligned}$$

This differential equation has the solution

$$\begin{aligned} \psi (t)=\frac{mC_{2}\left( T-t\right) }{C_{2}-C_{1}}-\frac{C_{2}\sqrt{ m\left( C_{2}-C_{1}\right) }}{\left( C_{1}-C_{2}\right) ^{2}}\tanh \sqrt{ m\left( C_{2}-C_{1}\right) }\left( T-t\right) \end{aligned}$$

and substituting $\psi (t)$ into (34) yields

$$\begin{aligned} \varphi (t)=\frac{mC_{2}\left( T-t\right) }{C_{2}-C_{1}}-\frac{C_{1}\sqrt{ m\left( C_{2}-C_{1}\right) }}{\left( C_{1}-C_{2}\right) ^{2}}\tanh \sqrt{ m\left( C_{2}-C_{1}\right) }\left( T-t\right) . \end{aligned}$$

Q.E.D.