On a Dynamic Model of Cooperative and Noncooperative R and D in Oligopoly with Spillovers

Abstract

We developed a dynamic model of oligopoly in which firms’ R and D investments accumulate as R and D capital and have spillover effects. We showed that there exists a symmetric stable open-loop Nash equilibrium for each of the differential games under noncooperative R and D and cooperative R and D. We then showed that for small spillovers, each firm’s R and D investments are larger under R and D competition than under R and D cooperation. We further demonstrated that, in the limit, when the discount rate goes to zero, the stability condition for our dynamic game approaches the stability condition for the static two-stage game in d’Aspremont and Jacquemin (Am Econ Rev 78:1133–1137, 1988). However, we also showed that at the Markov perfect equilibrium, cooperative R and D investments are larger than noncooperative investments for all possible values of spillovers.

Appendices

Appendix 1: Proof of Proposition 5

Proof

Let $V^i(y_i ,y_j )$ be the value function for firm $i$. Feedback Nash equilibrium strategies must satisfy a system of Hamilton-Jacobi-Bellman equations, $\forall i$,

$$\begin{aligned} rV^i(y_i ,y_j )&= \mathop {\max }\limits _{x_i } \left\{ \pi _i (c_i ,\,c_j )-\frac{\gamma }{2}x_i^2 +\frac{\partial V^i(y_i ,y_j )}{\partial y_i }(x_i +\beta x_j -\delta y_i )\right. \nonumber \\&\qquad \quad \quad +\left. \frac{\partial V^i(y_i ,y_j )}{\partial y_j }(x_j +\beta x_i -\delta y_j ) \right\} . \end{aligned}$$

(16)

Assuming symmetric strategies of the firms, we can set $V^i(y_i ,y_j)=V(y_i ,y_j )$.

We may now suppose the value function takes the following form:

$$\begin{aligned} V(y_1 ,y_2 )=J+K_1 y_1 +\frac{1}{2}L_1 y_1 ^2+K_2 y_2 +\frac{1}{2}L_2 y_2 ^2+My_1 y_2 . \end{aligned}$$

Solving the maximization of the right-hand side of Eq. (16) yields

$$\begin{aligned} x_i^*=\left( {\frac{1}{\gamma }}\right) \left( {\frac{\partial V^i(y_i ,y_j )}{\partial y_i }+\frac{\partial V^i(y_i ,y_j )}{\partial y_j }\beta }\right) \end{aligned}$$

(17)

Imposing symmetry, $K_1 =K_2 =K,\;L_1 =L_2 =L$ and $y_1 =y_2 =y$, we have

$$\begin{aligned} V(y_1 ,y_2 )=J+Ky+(L+M)y^2. \end{aligned}$$

Thus, we obtain

$$\begin{aligned} x_i^*=\left( {\frac{(1+\beta )}{\gamma }}\right) ( {K+Gy}),\text{ where }\,\,G\equiv L+M. \end{aligned}$$

(18)

Substituting Eqs. (17) and (18) into Eq. (16) results in

$$\begin{aligned} r\left[ {J+2Ky+Gy^2} \right]&= \frac{(a-c)^2}{9}-\left( {\frac{\gamma }{2}}\right) \left( {\left( {\frac{(1+\beta )}{\gamma }}\right) ( {K+Gy})}\right) ^2\\&\quad +( {K+Gy})\left[ {2( {K+Gy})\left( {\frac{(1+\beta )^2}{\gamma }}\right) -2\delta y} \right] \end{aligned}$$

This equation must hold for any $y$, and it follows that we have

$$\begin{aligned} rG&= \frac{\alpha ^2}{9}+\frac{3(1+\beta )^2}{2\gamma }G^2-2\delta G,\\ 2rK&= \frac{2\alpha (a-\bar{c})}{9}+\frac{3(1+\beta )^2KG}{\gamma }-2\delta K, \end{aligned}$$

and

$$\begin{aligned} rJ=\frac{(a-\bar{c})^2}{9}-\frac{3(1+\beta )^2K^2}{2\gamma }. \end{aligned}$$

From these equations, we have

$$\begin{aligned} G^*&= \frac{( {r+2\delta })\pm \sqrt{( {r+2\delta })^2-\frac{2\alpha ^2(1+\beta )^2}{3\gamma ^2}} }{\frac{3(1+\beta )^2}{\gamma ^2}},\\ K^*&= \frac{2\alpha (a-\bar{c})}{9( {2(r+\delta )-\frac{3(1+\beta )^2}{\gamma ^2}G^*})}, \end{aligned}$$

and

$$\begin{aligned} J^{*} =\frac{1}{r}\left[ {\frac{(a-\bar{c})^2}{9}}-{\frac{3(1+\beta )^2 K^{*2}}{2\gamma }} \right] . \end{aligned}$$

To ensure that $x^*\ge 0$ for any $y$, we must have $K^*\ge 0$. Thus, we take a root

$$\begin{aligned} G^*=\frac{( {r+2\delta })-\sqrt{( {r+2\delta })^2-\frac{2\alpha ^2(1+\beta )^2}{3\gamma ^2}} }{\frac{3(1+\beta )^2}{\gamma ^2}}. \end{aligned}$$

Next, we seek a stability condition for the Markov perfect equilibrium. Substituting Eq. (18) into Eq. (2) yields

$$\begin{aligned} \dot{y}-\left( {\frac{(1+\beta )^2G}{\gamma }-\delta }\right) y-\frac{(1+\beta )^2K}{\gamma }=0. \end{aligned}$$

(19)

The complete solution to Eq. (19) is

$$\begin{aligned} y(t)&= \tilde{y}+(y^0-\tilde{y})e^{-\left( {\delta -\frac{(1+\beta )^2G^*}{\gamma }}\right) t},\\ \text{ where }\,\,\tilde{y}&= \frac{\left( {\frac{(1+\beta )^2K^*}{\gamma }}\right) }{\left( {\frac{(1+\beta )^2G^*}{\gamma }-\delta }\right) }. \end{aligned}$$

For this state trajectory to be asymptotically stable, we must have $\delta >\frac{(1+\beta )^2G^*}{\gamma }.$ $\square$

Appendix 2: Proof of Proposition 8

The current-value Hamiltonian in this case is given by

$$\begin{aligned} H_i =\pi _i (c_i ,\,c_j )-\frac{1}{2}x_i^2 +\lambda _i (x_i +\beta _j x_j -\delta y_i )+\mu _i (x_j +\beta _i x_i -\delta y_j ),\;\,i,\,j=1,\,2,\,\;i\ne j. \end{aligned}$$

The necessary conditions for an open-loop Nash equilibrium are

$$\begin{aligned} \frac{\partial H_i }{\partial x_i }&= -x_i +\lambda _i +\beta _i \mu _i =0, \\ \dot{\lambda }_i&= r\lambda _i -\frac{\partial H_i }{\partial y_i } \\&= (r+\delta )\lambda _i -\frac{4( {a-2(\bar{c}_i -y_i )+(\bar{c}_j -y_j )})}{9},\, \\ \dot{\mu }_i&= r\mu _i -\frac{\partial H_i }{\partial y_j } \\&= (r+\delta )\mu _i +\frac{2( {a-2(\bar{c}_i -y_i )+(\bar{c}_j -y_j )})}{9},\, \\&\quad \mathop {\lim }\limits _{t\rightarrow \infty } \hbox {e}^{-rt}\lambda _i = 0,\, \\ \end{aligned}$$

and

$$\begin{aligned} \mathop {\lim }\limits _{t\rightarrow \infty } \hbox {e}^{-rt}\mu _i = 0. \\ \end{aligned}$$

At the steady state, we have $\dot{\lambda }_i =0, \quad \dot{\mu }_i =0,$ $\dot{y}_i =0,$ and $\dot{y}_j =0.$

Thus, we have

$$\begin{aligned} (r+\delta )\lambda _i&= \frac{4( {a-2(\bar{c}_i -y_i )+(\bar{c}_j -y_j )})}{9},\\ (r+\delta )\mu _i&= -\frac{2( {a-2(\bar{c}_i -y_i )+(\bar{c}_j -y_j )})}{9},\\ x_i&= \frac{2(2-\beta _i )}{9(r+\delta )}( {a-2(\bar{c}_i -y_i )+(\bar{c}_j-y_j )}),\\ y_i&= \frac{x_i +\beta _j x_j }{\delta }, \end{aligned}$$

and

$$\begin{aligned} y_j =\frac{x_j +\beta _i x_i }{\delta }. \end{aligned}$$

It follows that we get

$$\begin{aligned} x_i^*=\frac{2(a-\bar{c})E_i \delta ( {9\delta (r+\delta )-2E_j ^2+2E_j F_j })}{( {9\delta (r+\delta )-2E_i ^2})\,( {9\delta (r+\delta )-2E_j ^2})-4E_i F_i E_j F_j }, \end{aligned}$$

where $E_i \equiv 2-\beta _i\, \mathrm{and}\, F_i \equiv 2-\beta _i ,\,\;i,\,j=1,\,2,\,\;i\ne j.$

Suppose that firm $i$ has a smaller R and D spillover rate than firm $j$, that is, $\beta _i >\beta _j.$

Since $E_i -E_j <0$ and $F_j -F_i <0$, we have $9\delta (r+\delta )(E_i -E_j )+2E_i E_j ( {(E_i -E_j )+(F_j -F_i )})<0.$

Thus, $x_i <x_j$. $\square$

1.1 The Case of R and D Cooperation with Asymmetric Spillovers

The current-value Hamiltonian in this case is given by $H_i^C =\pi _i (c_i ,\,c_j )+\pi _j (c_i ,\,c_j )-\frac{1}{2}x_i^2 -\frac{1}{2}x_j^2 +\lambda _i (x_i +\beta _j x_j -\delta y_i )+\mu _i (x_j +\beta _i x_i -\delta y_j ),\;\,i,\,j=1,\,2,\,\;i\ne j.$ Then, following the procedure in the proof of Proposition 2, we have the desired results.

Appendix 3: Proof of Proposition 9

1.1 The Case of R and D Competition

The current-value Hamiltonian in this case is given by

$$\begin{aligned} H_i =\pi _i -\frac{1}{2}x_i^2 +\mathop \sum \limits _{k=1}^n \lambda _i^k \left( x_k +\beta \mathop \sum \limits _{j\in N_{-k} } x_j -\delta y_k\right) . \end{aligned}$$

The necessary conditions for an open-loop Nash equilibrium are

$$\begin{aligned} \frac{\partial H_i }{\partial x_i }=&-x_i +\lambda _i^i +\beta \mathop \sum \limits _{j\in N_{-i} } \lambda _i^j =0, \\ \dot{\lambda }_i^i =&r\lambda _i^i -\frac{\partial H_i }{\partial y_i }\\ =&(r+\delta )\lambda _i^i -\frac{\partial \pi _i }{\partial y_i },\, \\ \dot{\lambda }_i^j =&r\lambda _i^j -\frac{\partial H_i }{\partial y_j }\\ =&(r+\delta )\lambda _i^j -\frac{\partial \pi _i }{\partial y_j }\quad \hbox {for every } \,j, \\ \mathop {\lim }\limits _{t\rightarrow \infty } \hbox {e}^{-rt}\lambda _{i}^{i} =&0,\, \\&and \\ \mathop {\lim }\limits _{t\rightarrow \infty } \hbox {e}^{-rt}\lambda _i^j =&0\quad \hbox {for every } \,j, \\ \end{aligned}$$

where

$$\begin{aligned} \frac{\partial \pi _i }{\partial y_i }=\frac{2n\left( {a-nc_i +\mathop \sum \nolimits _{j\in N_{-i} } c_j }\right) }{(n+1)^2} \end{aligned}$$

and

$$\begin{aligned} \frac{\partial \pi _i }{\partial y_j }=-\frac{2\left( {a-nc_i +\mathop \sum \nolimits _{j\in N_{-i} } c_j }\right) }{(n+1)^2}. \end{aligned}$$

At the steady state, we have $\dot{\lambda }_i^i =0, \quad \dot{\lambda }_i^j =0$ for every $j$, and $\dot{y}_i =0$.

For a symmetric equilibrium, we get

$$\begin{aligned} y&= \frac{( {1+(n-1)\beta })\,x}{\delta },\\ (r+\delta )\lambda&= \frac{2n\left( {a-nc_i +\mathop \sum \nolimits _{j\in N_{-i} } c_j }\right) }{(n+1)^2}, \end{aligned}$$

and

$$\begin{aligned} (r+\delta )\lambda ^j=-\frac{2\left( {a-nc_i +\mathop \sum \nolimits _{j\in N_{-i} } c_j }\right) }{(n+1)^2}\quad \hbox { for every j }. \end{aligned}$$

It follows that we have the following equilibrium values:

$$\begin{aligned} x^{N*}=\frac{2\delta ( {n-(n-1)\beta })(a-\bar{c})}{(n+1)^2\delta (r+\delta )-2( {n-(n-1)\beta })\,( {1+(n-1)\beta })} \end{aligned}$$

and

$$\begin{aligned} y^{N*}=\frac{2( {n-(n-1)\beta })\,( {1+(n-1)\beta })(a-\bar{c})}{(n+1)^2\delta (r+\delta )-2( {n-(n-1)\beta })\,( {1+(n-1)\beta })}. \end{aligned}$$

$\square$

1.2 The Case of R and D Cooperation

The current-value Hamiltonian in this case is

$$\begin{aligned} \mathop \sum \limits _{i=1}^2 \left[ {\pi _i (c_i ,\,c_j )-\frac{\gamma }{2}x_i^2 (t)} \right] +\mathop \sum \limits _{k=1}^n \lambda _i^k (x_k +\beta \mathop \sum \limits _{j\in N_{-k} } x_j -\delta y_k ). \end{aligned}$$

The rest of the analysis is similar to the proof of Proposition 3.

Thus, we can show $x^{C*}=\frac{2\delta ( {1+(n-1)\beta })(a-\bar{c})}{(n+1)^2\delta (r+\delta )-2\,( {1+(n-1)\beta })^2}$.

Appendix 4: Proof of Proposition 12

At the open-loop Nash equilibrium, the R and D level $x_\theta ^C$ under R and D cooperation is given by

$$\begin{aligned} x_\theta ^C =\frac{2\alpha \delta (1+\theta )(a-\bar{c})}{9\delta (r+\delta )\gamma -2\alpha ^2(1+\theta )^2}. \end{aligned}$$

Thus,

$$\begin{aligned} x^N-x_\theta ^C =\frac{2\alpha \delta (2-\beta )(a-\bar{c})}{9\delta (r+\delta )\gamma -2\alpha ^2(2-\beta )(1+\beta )}-\frac{2\alpha \delta (1+\theta )(a-\bar{c})}{9\delta (r+\delta )\gamma -2\alpha ^2(1+\theta )^2}. \end{aligned}$$

Hence,

$$\begin{aligned} x^N\mathop =\limits _<^> x_\theta ^C \Leftrightarrow 9\delta (r+\delta )\gamma [1-(\beta +\theta )]-2\alpha ^2(1+\theta )(2-\beta )(\theta -\beta )\mathop =\limits _<^> 0. \end{aligned}$$

Therefore, if $\frac{1}{2}<\beta \le \theta$, then $1-(\beta +\theta )<0$. Thus, $x^N<x_\theta ^C$.

If $\beta \le \theta <\frac{1}{2}$, then $1-(\beta +\theta )>0$. Thus,

$$\begin{aligned} x^N\mathop =\limits _<^> x_\theta ^C \Leftrightarrow 9\delta (r+\delta )\gamma \mathop =\limits _<^> \frac{2\alpha ^2(1+\theta )(2-\beta )(\theta -\beta )}{1-(\beta +\theta )} \end{aligned}$$

$\square$