Resource Mobility and Market Performance

Abstract

We derive a feedback equilibrium of an infinite-horizon dynamic Cournot game where production requires exploitation of a renewable mobile resource, such as migratory fish, wildlife, and groundwater. We study how a small increase in the resource mobility parameter (starting from a position of no resource mobility) impacts on the equilibrium and the associated consumer’s surplus, firms’ profits and social welfare. We show that consumer’s surplus and social welfare increase in the short run but decrease in the long run, while firms’ profits may either increase or decrease in the short run, depending on initial conditions, and increase in the long run. Over the entire planning horizon, both the discounted consumer’s surplus and the discounted social welfare decrease, whereas the discounted profits increase. This result remains valid also in the presence of a per unit tax on extraction.

Appendices

Appendix A: Proof of Proposition 1

Feedback equilibrium strategies must satisfy the following Hamilton–Jacobi–Bellman equations for the value functions $V_{i}({\textbf{s}})$:

$$\begin{aligned}{} & {} rV_{i}({\textbf{s}})=\max _{q_{i}}\left\{ p\left( Q\right) q_{i}+\dfrac{dV_{i}( {\textbf{s}})}{\textrm{d}s_{i}}\left[ \alpha s_{i}-q_{i}-\gamma \left( s_{i}-s_{j}\right) \right] \right. \nonumber \\{} & {} \qquad \qquad \qquad \left. +\dfrac{dV_{i}({\textbf{s}})}{\textrm{d}s_{j}}\left[ \alpha s_{j}-\phi _{j}^{*}-\gamma \left( s_{j}-s_{i}\right) \right] \right\} \text {,} \end{aligned}$$

(A.1)

with $i,j=1,2$, $j\ne i$. The necessary (and sufficient) condition for an interior solution of the maximization of the right-hand side of (A.1) implies¹¹:

$$\begin{aligned} q_{i}=\dfrac{1}{2}\left[ 1-\phi _{j}^{*}-\dfrac{dV_{i}({\textbf{s}})}{ \textrm{d}s_{i}}\right] \text {.} \end{aligned}$$

(A.2)

Substitution of (A.2) into (A.1) gives a system of partial differential equations for $V_{1}({\textbf{s}})$ and $V_{2}({\textbf{s}})$. Given the linear-quadratic structure of the game, we guess value functions of the form:

$$\begin{aligned} V_{i}({\textbf{s}})=\kappa _{0}+\kappa _{1}s_{i}+\kappa _{2}s_{j}+\tfrac{1}{2} \left( \kappa _{11}s_{i}^{2}+2\kappa _{12}s_{i}s_{j}+\kappa _{22}s_{j}^{2}\right) \text {,} \end{aligned}$$

(A.3)

where $\kappa _{0},\kappa _{1},\kappa _{2},\kappa _{11},\kappa _{12},\kappa _{22}$ are coefficients to be identified. It follows that:

$$\begin{aligned} \dfrac{dV_{i}({\textbf{s}})}{\textrm{d}s_{i}}=\kappa _{1}+\kappa _{11}s_{i}+\kappa _{12}s_{j}\text {,} \end{aligned}$$

(A.4)

which implies the following linear strategy:

$$\begin{aligned} q_{i}=\frac{1}{3}\left[ 1-\kappa _{1}+\left( \kappa _{12}-2\kappa _{11}\right) s_{i}+\left( \kappa _{11}-2\kappa _{12}\right) s_{j}\right] \text {.} \end{aligned}$$

(A.5)

Consider the following solution:

$$\begin{aligned} \left. \begin{array}{l} \kappa _{0}=\dfrac{1}{32\alpha ^{2}\Delta ^{2}r}\left[ 288\alpha ^{6}+63\alpha ^{2}r^{4}+28r^{4}\gamma ^{2}-108\alpha \gamma r^{4}-252\alpha ^{3}r^{3}\right. \\ ~~~~~~~+96r^{3}\gamma ^{3}-176\alpha r^{3}\gamma ^{2}-8\alpha ^{2}r^{3}\gamma +603\alpha ^{4}r^{2}+48r^{2}\gamma ^{4}+160\alpha r^{2}\gamma ^{3} \\ ~~~~~~~-152\alpha ^{2}r^{2}\gamma ^{2}+392\alpha ^{3}r^{2}\gamma -702\alpha ^{5}r-64r\gamma ^{5}+160\alpha r\gamma ^{4}-416\alpha ^{2}r\gamma ^{3} \\ ~~~~~~~+208\alpha ^{3}r\gamma ^{2}-3r\sqrt{\Gamma }\left( 5\alpha ^{2}-5\alpha r+2r\gamma +4\gamma ^{2}-4\alpha \gamma \right) ^{2} \\ \ \ \ \ \ \ \ -148\alpha ^{4}r\gamma +128\alpha ^{4}\gamma ^{2}-128\alpha ^{5}\gamma ]\text {,} \end{array} \right. \end{aligned}$$

(A.6)

$$\begin{aligned}{} & {} \qquad \qquad \kappa _{1}=\frac{11}{8}+\frac{3}{16}\left[ \frac{\sqrt{\Gamma }-5r-4\gamma }{\alpha }-\frac{4\left( r-\alpha \right) \left( 3r-\sqrt{\Gamma }\right) }{ \Delta }\right] \text {,} \end{aligned}$$

(A.7)

$$\begin{aligned}{} & {} \kappa _{2} =\frac{1}{16\alpha \Delta }\left\{ -90\alpha ^{3}+220\alpha ^{2}\gamma -120\alpha \gamma ^{2}+80\gamma ^{3}-45\alpha r^{2}+2\gamma r^{2}+3\sqrt{\Gamma }\right. \nonumber \\{} & {} \qquad \qquad (5\alpha ^{2}-4\alpha \gamma +4\gamma ^{2}-5\alpha r+2\gamma r)+135\alpha ^{2}r-204\alpha \gamma r+44\gamma ^{2}r\}\text {,} \end{aligned}$$

(A.8)

$$\begin{aligned}{} & {} \qquad \qquad \qquad \quad \kappa _{22}=\frac{-3\left( 3r+12\gamma -6\alpha +\sqrt{\Gamma }\right) }{16} \text {,} \end{aligned}$$

(A.9)

$$\begin{aligned}{} & {} \qquad \qquad \qquad \quad \kappa _{11}=\frac{23r+28\gamma -46\alpha -3\sqrt{\Gamma }}{16}\text {,} \end{aligned}$$

(A.10)

and

$$\begin{aligned} \kappa _{12}=\frac{7r-4\gamma -14\alpha -3\sqrt{\Gamma }}{16}\text {.} \end{aligned}$$

(A.11)

By inserting (A.7), (A.10) and (A.11) into (A.5), we obtain $q_{i}=\phi _{i}^{*}$, with $\phi _{i}^{*}$ given in Proposition 1. It can be easily checked that for $q_{i}=\phi _{i}^{*}$, the value functions $V_{i}({\textbf{s}})$ with coefficients (A.6)–(A.11) satisfy the Hamilton–Jacobi–Bellman equations (A.1).

Appendix B: Proof of Corollary 1

Altogether, there exist six pairs of candidates for a feedback equilibrium, resulting from the standard application of the “undetermined coefficient technique,” with equilibrium strategies of the form $\phi _{i}^{*}=\lambda _{0}s_{i}+\lambda _{1}s_{j}+\lambda _{2}$, $i,j=1,2$, $j\ne i$. Within the class of linear, symmetric, stationary strategies, the candidates for a feedback equilibrium are:

$$\begin{aligned}{} & {} \qquad \quad q_{i}=\dfrac{1}{3}\text {,} \end{aligned}$$

(B.1)

$$\begin{aligned}{} & {} \qquad \quad q_{i}=-\frac{3\left( -2\alpha +r+2\gamma \right) }{4}s_{i}+\frac{2\left( \alpha +\gamma \right) -r}{4}s_{j}\nonumber \\ {}{} & {} \qquad +\frac{-2\alpha ^{2}+r^{2}+\alpha r-2r\gamma -4\gamma ^{2}-6\alpha \gamma }{4\alpha \left( \alpha +r-2\gamma \right) }\text {,} \nonumber \\\end{aligned}$$

(B.2)

$$\begin{aligned}{} & {} q_{i}=\frac{1}{24\alpha }\left[ 5r-2\alpha +9\alpha \left( 2\alpha -r\right) \left( s_{i}+s_{j}\right) \right] \text {,} \end{aligned}$$

(B.3)

$$\begin{aligned}{} & {} q_{i}=\frac{1}{4}\left[ 2+\frac{r}{\alpha -2\left( r+\gamma \right) }+\left( 2\alpha -r-4\gamma \right) \left( s_{i}-s_{j}\right) \right] \text {,} \end{aligned}$$

(B.4)

$$\begin{aligned}{} & {} \left. \begin{array}{l} q_{i}=\left[ \dfrac{\sqrt{\Gamma }+13\left( 2\alpha -r\right) -20\gamma }{16} \right] s_{i}+\left[ \dfrac{\sqrt{\Gamma }-3\left( 2\alpha -r\right) +12\gamma }{16}\right] s_{j} \\ ~~~~~~+\dfrac{2\Delta \left( 2\gamma -\alpha \right) +12\alpha r^{2}+r\left[ \sqrt{\Gamma }\left( 5\alpha -2\gamma \right) +5\Delta -12\alpha ^{2}\right] }{16\alpha \Delta } \\ ~~~~~~+\dfrac{\sqrt{\Gamma }\left( 4\gamma \alpha -4\gamma ^{2}-5\alpha ^{2}\right) }{16\alpha \Delta } \end{array} \right. \end{aligned}$$

(B.5)

and

$$\begin{aligned} \left. \begin{array}{l} q_{i}=\left[ \dfrac{-\sqrt{\Gamma }+13\left( 2\alpha -r\right) -20\gamma }{16 }\right] s_{i}+\left[ \dfrac{-\sqrt{\Gamma }-3\left( 2\alpha -r\right) +12\gamma }{16}\right] s_{j} \\ ~~~~~~+\dfrac{1}{16}\left[ \dfrac{4\left( r-\alpha \right) \left( \sqrt{ \Gamma }+3r\right) }{\Delta }+\dfrac{\sqrt{\Gamma }+5r+4\gamma }{\alpha }-2 \right] \text {.} \end{array} \right. \end{aligned}$$

(B.6)

As usual in the literature, out of these six pairs, we select the one(s) inducing trajectories of the asset stocks that converge to globally asymptotically stable steady states. It can be easily checked that the only pair of strategies stabilizing the states for every possible initial conditions is (B.5), corresponding to $\phi _{i}^{*}$ given in Proposition 1. Candidates (B.1) and (B.4) induce unstable trajectories, while candidates (B.2), (B.3) and (B.6) induce trajectories that converge to stationary points only for specific initial conditions. For candidates (B.2), (B.3) and (B.6), saddle path stability requires $\alpha >r+\gamma$, $\alpha >3r/2$, and $\alpha >r$, respectively.¹² Note that candidate (B.1) corresponds to the open-loop (state-independent) solution. All of the other candidates are clearly state-dependent.

The steady-state levels of firm i’s asset stock associated with $(\phi _{1}^{*},\phi _{2}^{*})$ are given by:

$$\begin{aligned} \left\{ \begin{array}{c} \overset{\cdot }{s}_{i}=\alpha s_{i}-\phi _{i}^{*}-\gamma \left( s_{i}-s_{j}\right) =\alpha s_{i}-\left( \lambda _{0}^{*}s_{i}+\lambda _{1}^{*}s_{j}+\lambda _{2}^{*}\right) -\gamma \left( s_{i}-s_{j}\right) =0 \\ \overset{\cdot }{s}_{j}=\alpha s_{j}-\phi _{j}^{*}-\gamma \left( s_{j}-s_{i}\right) =\alpha s_{j}-\left( \lambda _{0}^{*}s_{j}+\lambda _{1}^{*}s_{i}+\lambda _{2}^{*}\right) -\gamma \left( s_{j}-s_{i}\right) =0\text {,} \end{array} \right. \end{aligned}$$

(B.7)

with $i=1,2$. From (B.7), we obtain:

$$\begin{aligned} s_{i\infty }=s_{j\infty }=s_{\infty }=\frac{\lambda _{2}^{*}}{\alpha -\lambda _{0}^{*}-\lambda _{1}^{*}}\text {,} \end{aligned}$$

(B.8)

where considering (B.5), $\lambda _{0}^{*}$, $\lambda _{1}^{*}$, and $\lambda _{2}^{*}$ are the coefficient of $s_{i}$, the coefficient of $s_{j}$, and the constant, respectively. It is immediate to check that $s_{\infty }$ given in (B.8) corresponds to $s_{\infty }$ given in Corollary 1, and that $s_{\infty }>0$ for $\gamma$ sufficiently small. The (per firm) level of production associated with $s_{\infty }$ is:

$$\begin{aligned} \left. \phi _{i}^{*}\right| _{s_{\infty }}=\alpha s_{\infty }\text {.} \end{aligned}$$

By computing the eigenvalues of the Hessian matrix, it can be verified that the steady state $s_{\infty }$ is globally asymptotically stable if $\alpha >r$ and $\gamma <{\widehat{\gamma }}$, with ${\widehat{\gamma }}$

$$\begin{aligned} {\widehat{\gamma }}=\frac{\left( \alpha -r\right) \left( 2\alpha +r\right) }{2r }\text {,} \end{aligned}$$

or, equivalently, $\alpha >{\widehat{\alpha }}$, with ${\widehat{\alpha }}$ given in Corollary 1. Global asymptotic stability implies that for any initial asset stocks $s_{1,0}$, $s_{2,0}$ (such that an interior solution exists) the couple of equilibrium strategies $(\phi _{1}^{*},\phi _{2}^{*})$ induces trajectories of the asset stocks that converge asymptotically to $s_{\infty }$.

Appendix C: Proof of Proposition 2

As to (i), we have

$$\begin{aligned} \left. \frac{\partial \phi _{i}^{*}}{\partial \gamma }\right| _{\gamma =0,s_{i0}=s_{j0}=s_{0}}=-\frac{r+2\alpha \left( 9\alpha s_{0}-5\right) }{27\alpha ^{2}}\text {,} \end{aligned}$$

which is decreasing in $s_{0}$ and nil at $s_{0}={\widetilde{s}}_{0}$, with

$$\begin{aligned} {\widetilde{s}}_{0}=\frac{10\alpha -r}{18\alpha ^{2}}\text {.} \end{aligned}$$

Hence,

$$\begin{aligned} \left. \frac{\partial \phi _{i}^{*}}{\partial \gamma }\right| _{\gamma =0,s_{i0}=s_{j0}=s_{0}}\left\{ \begin{array}{c}<0\text { if }s_{0}>{\widetilde{s}}_{0} \\ >0\text { if }s_{0}<{\widetilde{s}}_{0}\text {.} \end{array} \right. \end{aligned}$$

However, since ${\widetilde{s}}_{0}>{\overline{s}}_{0}=(5\alpha -2r)/[6\alpha \left( 2\alpha -r\right) ]$, then

$$\begin{aligned} \left. \frac{\partial \phi _{i}^{*}}{\partial \gamma }\right| _{\gamma =0,s_{i0}=s_{j0}=s_{0}}>0\text {.} \end{aligned}$$

As to (ii), we have

$$\begin{aligned} \left. \frac{\partial cs^{*}}{\partial \gamma }\right| _{\gamma =0,s_{i0}=s_{j0}=s_{0}}=\frac{4\left[ \alpha \left( 1-6s_{0}\alpha \right) +r\left( 3\alpha s_{0}-1\right) \right] \left[ r+2\alpha \left( 9\alpha s_{0}-5\right) \right] }{81\alpha ^{3}}\text {,} \end{aligned}$$

which is concave in $s_{0}$ and nil at $s_{0}={\underline{s}}_{0,cs}$ and $s_{0}={\overline{s}}_{0,cs}$, with

$$\begin{aligned} {\underline{s}}_{0,cs}={\underline{s}}_{0}=\frac{\alpha -r}{3\alpha \left( 2\alpha -r\right) } \end{aligned}$$

and

$$\begin{aligned} {\overline{s}}_{0,cs}={\widetilde{s}}_{0}>{\overline{s}}_{0}\text {.} \end{aligned}$$

It follows that

$$\begin{aligned} \left. \frac{\partial cs^{*}}{\partial \gamma }\right| _{\gamma =0,s_{i0}=s_{j0}=s_{0}}>0\text {.} \end{aligned}$$

As to (iii), we have

$$\begin{aligned} \left. \frac{\partial \pi ^{*}}{\partial \gamma }\right| _{\gamma =0,s_{i0}=s_{j0}=s_{0}}=\frac{\left[ r+2\alpha \left( 9\alpha s_{0}-5\right) \right] \left[ r\left( 4-12\alpha s_{0}\right) +\alpha \left( 24\alpha s_{0}-7\right) \right] }{81\alpha ^{3}}\text {,} \end{aligned}$$

which is convex in $s_{0}$ and nil at $s_{0}={\underline{s}}_{0,\pi }$ and $s_{0}={\overline{s}}_{0,\pi }$, with

$$\begin{aligned} {\underline{s}}_{0,\pi }={\widehat{s}}_{0}=\frac{7\alpha -4r}{12\alpha \left( 2\alpha -r\right) }\in \left( {\underline{s}}_{0},{\overline{s}}_{0}\right) \text {,} \end{aligned}$$

and

$$\begin{aligned} {\overline{s}}_{0,\pi }={\widetilde{s}}_{0}>{\overline{s}}_{0}\text {.} \end{aligned}$$

It follows that

$$\begin{aligned} \left. \frac{\partial \pi ^{*}}{\partial \gamma }\right| _{\gamma =0,s_{i0}=s_{j0}=s_{0}}\left\{ \begin{array}{c}>0\text { if }s_{0}<{\widehat{s}}_{0} \\ <0\text { if }s_{0}>{\widehat{s}}_{0}\text {.} \end{array} \right. \end{aligned}$$

Finally, we have

$$\begin{aligned} \left. \frac{\partial w^{*}}{\partial \gamma }\right| _{\gamma =0,s_{i0}=s_{j0}=s_{0}}=\frac{2\left[ \alpha \left( 12\alpha s_{0}-5\right) +2r\left( 1-3\alpha s_{0}\right) \right] \left[ r+2\alpha \left( 9\alpha s_{0}-5\right) \right] }{81\alpha ^{3}}\text {,} \end{aligned}$$

which is convex in $s_{0}$ and nil at $s_{0}={\underline{s}}_{0,w}$ and $s_{0}= {\overline{s}}_{0,w}$, with

$$\begin{aligned} {\underline{s}}_{0,w}={\overline{s}}_{0}\text {,} \end{aligned}$$

and

$$\begin{aligned} {\overline{s}}_{0,w}={\widetilde{s}}_{0}>{\overline{s}}_{0}\text {.} \end{aligned}$$

Hence,

$$\begin{aligned} \left. \frac{\partial w^{*}}{\partial \gamma }\right| _{\gamma =0,s_{i0}=s_{j0}=s_{0}}>0\text {.} \end{aligned}$$

Appendix D: Proof of Proposition 3

As to (i), we have

$$\begin{aligned} \left. \frac{\partial s_{\infty }}{\partial \gamma }\right| _{\gamma =0}= \frac{r-4\alpha }{27\alpha ^{2}\left( \alpha -r\right) }<0\text {,} \end{aligned}$$

since $\alpha >r$ is required for the stability of the steady state.

As to (ii), we have

$$\begin{aligned} \left. \frac{\partial q_{\infty }}{\partial \gamma }\right| _{\gamma =0}= \frac{r-4\alpha }{27\alpha \left( \alpha -r\right) }<0\text {.} \end{aligned}$$

The steady-state price increases since p(Q) is decreasing in its argument.

As to (iii), we have

$$\begin{aligned} \left. \frac{\partial cs_{\infty }}{\partial \gamma }\right| _{\gamma =0}=\frac{4\left( r-4\alpha \right) }{81\alpha \left( \alpha -r\right) }<0 \text {.} \end{aligned}$$

As to (iv), we have

$$\begin{aligned} \left. \frac{\partial \pi _{\infty }}{\partial \gamma }\right| _{\gamma =0}=\frac{r-4\alpha }{81\alpha \left( r-\alpha \right) }>0\text {.} \end{aligned}$$

Finally, we have

$$\begin{aligned} \left. \frac{\partial w_{\infty }}{\partial \gamma }\right| _{\gamma =0}= \frac{2\left( r-4\alpha \right) }{81\alpha \left( \alpha -r\right) }<0\text {. } \end{aligned}$$

Appendix E: Proof of Proposition 4

As to (i), from (5), we have

$$\begin{aligned} \left. \frac{\partial CS}{\partial \gamma }\right| _{\gamma =0,s_{i0}=s_{j0}=s_{0}}=\frac{4\left( \alpha -r\right) \left[ -4\alpha +r\left( -5+18\alpha s_{0}\right) \right] }{81r\alpha ^{3}}<0\text {,} \end{aligned}$$

since $\alpha >r$ is required for the stability of the steady state, and the expression in square brackets is increasing in $s_{0}$ and nil at

$$\begin{aligned} s_{0,CS}=\frac{5r+4\alpha }{18r\alpha }>{\overline{s}}_{0}\text {.} \end{aligned}$$

As to (ii), from (2), we have

$$\begin{aligned} \left. \frac{\partial J_{i}}{\partial \gamma }\right| _{\gamma =0,s_{i0}=s_{j0}=s_{0}}=-\frac{1}{4}\left. \frac{\partial CS}{\partial \gamma }\right| _{\gamma =0,s_{i0}=s_{j0}=s_{0}}>0\text {.} \end{aligned}$$

As to (iii), from (6), we have

$$\begin{aligned} \left. \frac{\partial W}{\partial \gamma }\right| _{\gamma =0,s_{i0}=s_{j0}=s_{0}}= & {} \left. \frac{\partial CS}{\partial \gamma } \right| _{\gamma =0,s_{i0}=s_{j0}=s_{0}}+2\left. \frac{\partial J_{i}}{ \partial \gamma }\right| _{\gamma =0,s_{i0}=s_{j0}=s_{0}} \\= & {} \frac{1}{2}\left. \frac{\partial CS}{\partial \gamma }\right| _{\gamma =0,s_{i0}=s_{j0}=s_{0}}<0\text {.} \end{aligned}$$

Appendix F: Proof of Proposition 5

As to (i), from (5), we have

$$\begin{aligned} \left. \frac{\partial CS}{\partial \gamma }\right| _{\gamma =0,s_{i0}=s_{j0}=s_{0}}=\frac{4\left( 1-\tau \right) \left( \alpha -r\right) \left\{ r\left[ 18\alpha s_{0}-5\left( 1-\tau \right) \right] -4\alpha \left( 1-\tau \right) \right\} }{81r\alpha ^{3}}<0\text {,} \end{aligned}$$

since $\alpha >r$ is required for the stability of the steady state, and the expression in square brackets is increasing in $s_{0}$ and nil at

$$\begin{aligned} s_{0,CS}=\frac{\left( 1-\tau \right) \left( 5r+4\alpha \right) }{18r\alpha }> \overline{{\overline{s}}}_{0,\tau }\text {.} \end{aligned}$$

As to (ii), from (2), we have

$$\begin{aligned} \left. \frac{\partial J_{i}}{\partial \gamma }\right| _{\gamma =0,s_{i0}=s_{j0}=s_{0}}=-\frac{1}{4}\left. \frac{\partial CS}{\partial \gamma }\right| _{\gamma =0,s_{i0}=s_{j0}=s_{0}}>0\text {.} \end{aligned}$$

As to (iii) and (iv), from (6) and (8), we have

$$\begin{aligned} \left. \frac{\partial W}{\partial \gamma }\right| _{\gamma =0,s_{i0}=s_{j0}=s_{0}}= & {} \left. \frac{\partial CS}{\partial \gamma } \right| _{\gamma =0,s_{i0}=s_{j0}=s_{0}}+2\left. \frac{\partial J_{i}}{ \partial \gamma }\right| _{\gamma =0,s_{i0}=s_{j0}=s_{0}}+\left. \frac{ \partial G}{\partial \gamma }\right| _{\gamma =0,s_{i0}=s_{j0}=s_{0}} \\= & {} \frac{1}{2}\left. \frac{\partial CS}{\partial \gamma }\right| _{\gamma =0,s_{i0}=s_{j0}=s_{0}}+\left. \frac{\partial G}{\partial \gamma } \right| _{\gamma =0,s_{i0}=s_{j0}=s_{0}}\text {,} \end{aligned}$$

with

$$\begin{aligned} \left. \frac{\partial G}{\partial \gamma }\right| _{\gamma =0,s_{i0}=s_{j0}=s_{0}}=\frac{3\tau }{2\left( 1-\tau \right) }\left. \frac{ \partial CS}{\partial \gamma }\right| _{\gamma =0,s_{i0}=s_{j0}=s_{0}}<0 \text {.} \end{aligned}$$

Hence,

$$\begin{aligned} \left. \frac{\partial W}{\partial \gamma }\right| _{\gamma =0,s_{i0}=s_{j0}=s_{0}}<0\text {.} \end{aligned}$$