Locational disadvantage of the hub

Abstract

We show how spatial evolution is different between the two representative models of economic geography: [Krugman 99:483–499, 1991] and [Ottaviano et al. 43:409–436, 2002]. We analyze the impacts of falling transport costs on the spatial distribution of economic activities and welfare for a network economy consisting of three regions located on a line. It is normally considered that a hub city, i.e., a central region, always has locational advantage and manufacturing workers gain from trade. This is true in the former model, but not in the latter when markets are opened up to trade. This is because the price competition is so keen in the central region that the manufacturing sector moves to the peripheral regions, which aggravates the social welfare. We then show that when goods are close substitutes and share of manufacturing is of an intermediate level, the manufacturing activities completely disappear from the central region leading to a full agglomeration in one peripheral region.

Appendices

Appendix A: Comparative statics

Comparative statics

• 1. Applying the implicit function theorem to Eq. 4 with $\lambda{\lambda}^{\ast }=(1/3,1/3,1/3)$ and $w_{r}=1$, $f_{4}\equiv V_{1}-w_{1}/G_{1}^{\mu }=0$ and $f_{5}\equiv V_{3}-w_{3}/G_{3}^{\mu }=0$, we have

  $${\left( {\begin{array}{*{20}l} {{\frac{{\partial w_{1} }}{{\partial \lambda _{1} }}} \hfill} & {{\frac{{\partial w_{1} }}{{\partial \lambda _{2} }}} \hfill} \\ {{\frac{{\partial w_{2} }}{{\partial \lambda _{1} }}} \hfill} & {{\frac{{\partial w_{2} }}{{\partial \lambda _{2} }}} \hfill} \\ {{\frac{{\partial w_{3} }}{{\partial \lambda _{1} }}} \hfill} & {{\frac{{\partial w_{3} }}{{\partial \lambda _{2} }}} \hfill} \\ {{\frac{{\partial V_{1} }}{{\partial \lambda _{1} }}} \hfill} & {{\frac{{\partial V_{1} }}{{\partial \lambda _{2} }}} \hfill} \\ {{\frac{{\partial w_{2} }}{{\partial \lambda _{1} }}} \hfill} & {{\frac{{\partial V_{2} }}{{\partial \lambda _{2} }}} \hfill} \\ \end{array} } \right)} = {\left( {\begin{array}{*{20}l} {{\frac{{\partial f_{1} }}{{\partial w_{1} }}} \hfill} & {{\frac{{\partial f_{1} }}{{\partial w_{2} }}} \hfill} & {{\frac{{\partial f_{1} }}{{\partial w_{3} }}} \hfill} & {{\frac{{\partial f_{1} }}{{\partial V_{1} }}} \hfill} & {{\frac{{\partial f_{1} }}{{\partial V_{3} }}} \hfill} \\ {{\frac{{\partial f_{2} }}{{\partial w_{1} }}} \hfill} & {{\frac{{\partial f_{2} }}{{\partial w_{2} }}} \hfill} & {{\frac{{\partial f_{2} }}{{\partial w_{3} }}} \hfill} & {{\frac{{\partial f_{2} }}{{\partial V_{1} }}} \hfill} & {{\frac{{\partial f_{2} }}{{\partial V_{3} }}} \hfill} \\ {{\frac{{\partial f_{3} }}{{\partial w_{1} }}} \hfill} & {{\frac{{\partial f_{3} }}{{\partial w_{2} }}} \hfill} & {{\frac{{\partial f_{3} }}{{\partial w_{3} }}} \hfill} & {{\frac{{\partial f_{3} }}{{\partial V_{1} }}} \hfill} & {{\frac{{\partial f_{3} }}{{\partial V_{3} }}} \hfill} \\ {{\frac{{\partial f_{4} }}{{\partial w_{1} }}} \hfill} & {{\frac{{\partial f_{4} }}{{\partial w_{2} }}} \hfill} & {{\frac{{\partial f_{4} }}{{\partial w_{3} }}} \hfill} & {{\frac{{\partial f_{4} }}{{\partial V_{1} }}} \hfill} & {{\frac{{\partial f_{4} }}{{\partial V_{3} }}} \hfill} \\ {{\frac{{\partial f_{5} }}{{\partial w_{1} }}} \hfill} & {{\frac{{\partial f_{5} }}{{\partial w_{2} }}} \hfill} & {{\frac{{\partial f_{5} }}{{\partial w_{3} }}} \hfill} & {{\frac{{\partial f_{5} }}{{\partial V_{1} }}} \hfill} & {{\frac{{\partial f_{5} }}{{\partial V_{3} }}} \hfill} \\ \end{array} } \right)}^{{ - 1}} {\left( {\begin{array}{*{20}l} {{\frac{{\partial f_{1} }}{{\partial \lambda _{1} }}} \hfill} & {{\frac{{\partial f_{1} }}{{\partial \lambda _{2} }}} \hfill} \\ {{\frac{{\partial f_{2} }}{{\partial \lambda _{1} }}} \hfill} & {{\frac{{\partial f_{2} }}{{\partial \lambda _{2} }}} \hfill} \\ {{\frac{{\partial f_{3} }}{{\partial \lambda _{1} }}} \hfill} & {{\frac{{\partial f_{3} }}{{\partial \lambda _{2} }}} \hfill} \\ {{\frac{{\partial f_{1} }}{{\partial \lambda _{1} }}} \hfill} & {{\frac{{\partial f_{4} }}{{\partial \lambda _{2} }}} \hfill} \\ {{\frac{{\partial f_{5} }}{{\partial \lambda _{1} }}} \hfill} & {{\frac{{\partial f_{5} }}{{\partial \lambda _{2} }}} \hfill} \\ \end{array} } \right)} = {\left( {0\begin{array}{*{20}c} {{\mu - 1}} & {0} & {0} & {0} & {0} \\ {0} & {{\mu - 1}} & {0} & {0} & {0} \\ {0} & {0} & {{\mu - 1}} & {0} & {0} \\ {{3^{{\frac{\mu }{{1 - \sigma }}}} {\left( {\mu - 1} \right)}}} & {0} & {0} & {1} & {0} \\ {0} & {{3^{{\frac{\mu }{{1 - \sigma }}}} {\left( {\mu - 1} \right)}}} & {0} & {0} & {1} \\ \end{array} } \right)}^{{ - 1}} {\left( {\begin{array}{*{20}c} {{3{\left( {\mu - 1} \right)}}} & {0} \\ {0} & {{3{\left( {\mu - 1} \right)}}} \\ {{3{\left( {1 - \mu } \right)}}} & {{3{\left( {1 - \mu } \right)}}} \\ {{_{{\frac{\mu }{{1 - \sigma }}}} 3^{{\frac{{\sigma - 1 - \mu }}{{\sigma - 1}}}} }} & {0} \\ {0} & {{_{{\frac{\mu }{{1 - \sigma }}}} 3^{{\frac{{\sigma - 1 - \mu }}{{\sigma - 1}}}} }} \\ \end{array} } \right)}$$

  By rearranging it, we get

  $$\frac{{\partial V_{1} }}{{\partial \lambda _{1} }} = \frac{{\partial V_{2} }}{{\partial \lambda _{2} }} = \frac{{3^{{\frac{{1 - \sigma + \mu }}{{1 - \sigma }}}} }}{{\sigma - 1}}{\left( {1 - \sigma + \mu \sigma } \right)}$$

  (13)
  $$\frac{{\partial V_{1} }}{{\partial \lambda _{2} }} = \frac{{\partial V_{2} }}{{\partial \lambda _{1} }} = 0$$

  (14)

• 2. From (1), let $f_{6}\equiv V_{1}-V_{2}$ and $f_{7}\equiv V_{1}-V_{3}$. Then, we similarly have

  $$\begin{aligned} & T^{\sigma } {\left( {\begin{array}{*{20}l} {{\frac{{\partial w_{1} }}{{\partial T}}} \hfill} \\ {{\frac{{\partial w_{2} }}{{\partial T}}} \hfill} \\ {{\frac{{\partial w_{3} }}{{\partial T}}} \hfill} \\ {{\frac{{\partial \lambda _{1} }}{{\partial T}}} \hfill} \\ {{\frac{{\partial \lambda _{2} }}{{\partial T}}} \hfill} \\ \end{array} } \right)} = T^{\sigma } {\left( {\begin{array}{*{20}l} {{\frac{{\partial f_{1} }}{{\partial w_{1} }}} \hfill} & {{\frac{{\partial f_{1} }}{{\partial w_{2} }}} \hfill} & {{\frac{{\partial f_{1} }}{{\partial w_{3} }}} \hfill} & {{\frac{{\partial f_{1} }}{{\partial \lambda _{1} }}} \hfill} & {{\frac{{\partial f_{1} }}{{\partial \lambda _{2} }}} \hfill} \\ {{\frac{{\partial f_{2} }}{{\partial w_{1} }}} \hfill} & {{\frac{{\partial f_{2} }}{{\partial w_{2} }}} \hfill} & {{\frac{{\partial f_{2} }}{{\partial w_{3} }}} \hfill} & {{\frac{{\partial f_{2} }}{{\partial \lambda _{1} }}} \hfill} & {{\frac{{\partial f_{2} }}{{\partial \lambda _{1} }}} \hfill} \\ {{\frac{{\partial f_{3} }}{{\partial w_{1} }}} \hfill} & {{\frac{{\partial f_{3} }}{{\partial w_{1} }}} \hfill} & {{\frac{{\partial f_{3} }}{{\partial w_{1} }}} \hfill} & {{\frac{{\partial f_{3} }}{{\partial \lambda _{1} }}} \hfill} & {{\frac{{\partial f_{3} }}{{\partial \lambda _{1} }}} \hfill} \\ {{\frac{{\partial f_{6} }}{{\partial w_{1} }}} \hfill} & {{\frac{{\partial f_{6} }}{{\partial w_{1} }}} \hfill} & {{\frac{{\partial f_{6} }}{{\partial w_{1} }}} \hfill} & {{\frac{{\partial f_{6} }}{{\partial \lambda _{1} }}} \hfill} & {{\frac{{\partial f_{6} }}{{\partial \lambda _{1} }}} \hfill} \\ {{\frac{{\partial f_{7} }}{{\partial w_{1} }}} \hfill} & {{\frac{{\partial f_{7} }}{{\partial w_{1} }}} \hfill} & {{\frac{{\partial f_{7} }}{{\partial w_{1} }}} \hfill} & {{\frac{{\partial f_{7} }}{{\partial \lambda _{1} }}} \hfill} & {{\frac{{\partial f_{7} }}{{\partial \lambda _{1} }}} \hfill} \\ \end{array} } \right)}^{{ - 1}} {\left( {\begin{array}{*{20}l} {{\frac{{\partial f_{1} }}{{\partial T}}} \hfill} \\ {{\frac{{\partial f_{2} }}{{\partial T}}} \hfill} \\ {{\frac{{\partial f_{3} }}{{\partial T}}} \hfill} \\ {{\frac{{\partial f_{6} }}{{\partial T}}} \hfill} \\ {{\frac{{\partial f_{7} }}{{\partial T}}} \hfill} \\ \end{array} } \right)} \\ & = {\left( {\begin{array}{*{20}c} {{\mu - 1}} & {0} & {0} & {{3{\left( {\mu - 1} \right)}}} & {0} \\ {0} & {{\mu - 1}} & {0} & {0} & {{3{\left( {\mu - 1} \right)}}} \\ {0} & {0} & {{\mu - 1}} & {{3{\left( {1 - \mu } \right)}}} & {{3{\left( {1 - \mu } \right)}}} \\ {{3^{{\frac{\mu }{{1 - \sigma }}}} {\left( {\mu - 1} \right)}}} & {{3^{{\frac{\mu }{{1 - \sigma }}}} {\left( {\mu - 1} \right)}}} & {0} & {{_{{\frac{\mu }{{1 - \sigma }}}} 3^{{\frac{{\sigma - 1 - \mu }}{{\sigma - 1}}}} }} & {{_{{\frac{\mu }{{1 - \sigma }}}} 3^{{\frac{{\sigma - 1 - \mu }}{{\sigma - 1}}}} }} \\ {{3^{{\frac{\mu }{{1 - \sigma }}}} {\left( {\mu - 1} \right)}}} & {0} & {{3^{{\frac{\mu }{{1 - \sigma }}}} {\left( {\mu - 1} \right)}}} & {{_{{\frac{{2\mu }}{{1 - \sigma }}}} 3^{{\frac{{\sigma - 1 - \mu }}{{\sigma - 1}}}} }} & {{_{{\frac{\mu }{{1 - \sigma }}}} 3^{{\frac{{\sigma - 1 - \mu }}{{\sigma - 1}}}} }} \\ \end{array} } \right)}^{{ - 1}} \\ & \cdot {\left( {\begin{array}{*{20}c} {0} \\ {0} \\ {0} \\ {{3^{{\frac{\mu }{{1 - \sigma }}}} \mu }} \\ {0} \\ \end{array} } \right)} \\ \end{aligned} $$

  where each element is evaluated at $T\rightarrow \infty$, $\lambda _{r}^{\ast }=1/3$ and $w_{r}^{\ast }=1$ for all $r$. By rearranging it, we obtain

  $$\begin{array}{*{20}c} {T^{\sigma } \frac{{\partial \lambda _{1} }}{{\partial T}}}{ = \frac{{\mu {\left( {\sigma - 1} \right)}}}{{9{\left( {\sigma - 1 - \mu \sigma } \right)}}} > 0} \\ {T^{\sigma } \frac{{\partial \lambda _{1} }}{{\partial T}}}{ = - \frac{{2\mu {\left( {\sigma - 1} \right)}}}{{9{\left( {\sigma - 1 - \mu \sigma } \right)}}} > 0} \\ \end{array} $$

Appendix B: Proof of Lemma 1

Let

$${\mathop \mu \limits^ \vee } \equiv \frac{{4\sigma }}{{11\sigma + 18}}\widehat{\mu }_{2} \equiv \frac{{4\sigma {\left( {5\sigma + 18} \right)}}}{{37\sigma ^{2} + 252\sigma + 324}}\overline{\mu } _{2} \equiv \frac{{2\sigma }}{{4\sigma + 9}}$$

where $\check{\mu}<\hat{\mu}_{2}.$ Domains of the four trade patterns (C1)–(C4) are illustrated in Fig. 4. We show that when for $\tau$ is just below $\tau _{\rm{trade}},$ the stable path enters the domain of neighboring trade between $1$ and $2$ and between $2$ and $3$ (C5) in Fig. 5 for $\mu<\check{\mu}$ (Lemma 2) and the domain of one-way trade from $1$ and $3$ to $2$ (C1) for $\check{\mu}<\mu<\hat{\mu}_{2}$ (Lemma 3), but that it does not enter the domains of (C2), (C3), and (C4) (Lemmas 4–6). The shaded triangular domain of (C5) is given by the three boundary conditions $\lambda _{r}<2\left( \alpha -\tau \right) /\sigma \tau$ ($r=1,2,3$), where $\lambda _{r}$ is no longer $1/3$ since $\tau$ is slightly less than $\tau _{\rm{trade}}.$ The vertexes of the triangle in Fig. 5 are computed as:

$${\left( {\frac{{2{\left( {a - \tau } \right)}}}{{\sigma \tau }},\frac{{2{\left( {a - \tau } \right)}}}{{\sigma \tau }}} \right)},{\left( {\frac{{2{\left( {a - \tau } \right)}}}{{\sigma \tau }},\frac{{{\left( {\sigma + 4} \right)}\tau - 4a}}{{\sigma \tau }}} \right)},{\left( {\frac{{{\left( {\sigma + 4} \right)}\tau - 4a}}{{\sigma \tau }},\frac{{2{\left( {a - \tau } \right)}}}{{\sigma \tau }}} \right)}$$

(15)

Also, define the RHS of Eq. 2 as

$$\lambda _{r} \equiv \lambda _{r} {\left[ {V_{r} {\left( {\lambda ^{*} } \right)} - {\sum\limits_{s = 1}^3 {\lambda _{s} V_{s} {\left( {\lambda ^{*} } \right)}} }} \right]}$$

for $r=1,2,3.$

Fig. 4

(C1) one-way trade from 1 and 3 to 2, (C2) one-way trade from 1 to 2 and neighboring trade between 2 and 3, (C3) autarky in 1 and one-way trade from 2 to 3, (C3’) autarky in 3 and one-way trade from 2 to 1, (C4) one-way trade from 2 to 1 and 3

Fig. 5

(C1)–(C4) same as Fig. 4, (C5) neighboring trade

Lemma 2

If $\mu<\check{\mu},$ the stable equilibrium path $\lambda{ \lambda }^{\ast }(\tau )$ passing through $\lambda{\lambda }^{\ast }=(1/3,1/3,1/3)$ at $\tau =\tau _{\rm{trade}}$ enters the domain of neighboring trade (C5) with

$$\frac{{\partial \lambda _{1} }}{{\partial \tau }} = \frac{{\partial \lambda _{3} }}{{\partial _{\tau } }} < 0$$

Proof

In the case of neighboring trade, we compute the Jacobian of Eq. 2 and evaluate it at $\lambda{\lambda }^{\ast }=(1/3,1/3,1/3)$ with $\tau =\tau _{\rm{trade}}$. Then, the stability condition is given by

$$\mu < \widetilde{\mu } \equiv \frac{{2\sigma {\left( {7\sigma + 18} \right)}}}{{34\sigma ^{2} + 153\sigma + 162}}$$

(16)

Applying the implicit function theorem to $y_{1}=y_{3}=0$ at $\lambda{ \lambda }^{\ast }=(1/3,1/3,1/3)$ with $\tau =\tau _{\rm{trade}}$, we get

$$\frac{{\partial \lambda _{1} }}{{\partial \tau }} = \frac{{\partial \lambda _{3} }}{{\partial \tau }} = \frac{{{\left( {2 - \mu } \right)}\sigma {\left( {\sigma + 6} \right)}}}{{18a{\left[ {\mu {\left( {34\sigma ^{2} + 162} \right)} - 2\sigma {\left( {7\sigma + 18} \right)}} \right]}}} < 0$$

(17)

where the inequality is due to Eq. 16. Therefore, the slope of the stable equilibrium path is $1.$

On the other hand, the marginal change in the first vertex of Eq. 15 at $\tau =\tau _{\rm{trade}}$ is

$$\frac{{\partial \lambda _{1} }}{{\partial \tau }} = \frac{{\partial \lambda _{3} }}{{\partial \tau }} = - \frac{{{\left( {\sigma + 6} \right)}^{2} }}{{18a\sigma }} < 0$$

(18)

Hence, the vertex moves toward the same direction as the stable equilibrium path for a marginal decrease in$\ \tau$. For the stable equilibrium path $\lambda{\lambda}^{\ast }(\tau )$ to enter the domain of neighboring trade (C5), the change in the stable path Eq. 17 is smaller then the latter Eq. 18 in absolute value, i.e.

$$ -\frac{(\sigma +6)^{2}}{18\alpha \sigma }<\frac{\left( 2-\mu \right) \sigma \left( \sigma +6\right) ^{2}}{18\alpha \left[ \mu \left( 34\sigma ^{2}+153\sigma +162\right) -2\sigma \left( 7\sigma +18\right) \right] }<0$$

The first inequality is equivalent to $\mu<\check{\mu}$. However, since $\check{\mu}<\tilde{\mu}$, this lemma holds for $\mu<\check{\mu}.$

Lemma 3

If $\check{\mu}<\mu<\hat{\mu}_{2},$ the stable equilibrium path $\lambda{\lambda }^{\ast }(\tau )$ passing through $\boldsymbol{\lambda } ^{\ast }=(1/3,1/3,1/3)$ at $\tau =\tau _{\rm{trade}}$ enters the domain (C1) with

$$ \frac{\partial \lambda _{1}}{\partial \tau }=\frac{\partial \lambda _{3}}{ \partial \tau }<0$$

Proof

Computing and evaluating the Jacobian of Eq. 2 at $\lambda{ \lambda }^{\ast }=(1/3,1/3,1/3)$ with $\tau =\tau _{\rm{trade}}$ for one-way trade to the center, we have the stability condition

$$ \mu<\hat{\mu}_{2}$$

(19)

The comparative statics at $\boldsymbol{\lambda }^{\ast }=(1/3,1/3,1/3)$ with $\tau =\tau _{\rm{trade}}$ is

$$ \frac{\partial \lambda _{1}}{\partial \tau }=\frac{\partial \lambda _{3}}{ \partial \tau }=\frac{2\sigma (\sigma +6)^{2}(2-\mu )}{9\alpha \left[ \mu \left( 37\sigma ^{2}+252\sigma +324\right) -4\sigma \left( 5\sigma +18\right) \right] }<0$$

where the inequality is due to Eq. 19.

For the equilibrium path to enter the domain of the one-way trade to the center (C1), it is necessary that

$$ \frac{2\sigma (\sigma +6)^{2}(2-\mu )}{9\alpha \left[ \mu \left( 37\sigma ^{2}+252\sigma +324\right) -4\sigma \left( 5\sigma +18\right) \right] }<- \frac{(\sigma +6)^{2}}{18\alpha \sigma }<0$$

which is equivalent to $\check{\mu}<\mu$. Hence, this lemma is true for $\check{\mu}<\mu<\hat{\mu}_{2}.$

Lemma 4

There is no stable equilibrium path $\lambda{\lambda }^{\ast }(\tau )$ passing through $\lambda{\lambda }^{\ast }=(1/3,1/3,1/3)$ at $\tau =\tau _{\rm{trade}}$ that enters domain (C2).

Proof

Computing and evaluating the Jacobian of Eq. 2 at $\lambda{ \lambda }^{\ast }=(1/3,1/3,1/3)$ with $\tau =\tau _{\rm{trade}}$ for one-way trade from 1 to 2 and neighboring trade between 2 and 3, we have the stability condition

$$\eqalign {z\equiv (53\sigma ^{3}+453\sigma ^{2}+1188\sigma +972)\mu ^{2}-12\sigma (5\sigma ^{2}+28\sigma +36)\mu \cr +16\sigma ^{2}\left( \sigma +3\right) >0}$$

The comparative statics at $\lambda{\lambda }^{\ast }=(1/3,1/3,1/3)$ with $\tau =\tau _{\rm{trade}}$ is

$$ \frac{\partial \lambda _{3}}{\partial \tau }=\frac{(\mu -2)\sigma (\sigma +6)^{2}}{9\alpha z}<0$$

where inequality holds from $z>0$. Thus, any slight decrease in $\tau$ leads to $\lambda _{3}>1/3\!$, which does not pass through domain (C2).

Lemma 5

If $\mu<\bar{\mu}_{2}$, then there is no stable equilibrium path $\lambda{\lambda }^{\ast }(\tau )$ passing through $\lambda{\lambda } ^{\ast }=(1/3,1/3,1/3)$ at $\tau =\tau _{\rm{trade}}$ that enters domain (C3).

Proof

Computing and evaluating the Jacobian of Eq. 2 at $\lambda{ \lambda }^{\ast }=(1/3,1/3,1/3)$ with $\tau =\tau _{\rm{trade}}$ autarky in $1$ and one-way trade from $2$ to $3$, we have the stability condition

$$ \mu<\frac{4\sigma (2\sigma +9)}{13\sigma ^{2}+99\sigma +162}$$

(20)

Since $\bar{\mu}_{2}<\frac{4\sigma (2\sigma +9)}{13\sigma ^{2}+99\sigma +162}$, stability condition Eq. 20 is met. The comparative statics at $\lambda{\lambda }^{\ast }=(1/3,1/3,1/3)$ with $\tau =\tau _{\rm{trade }}$ is

$$ \frac{\partial \lambda _{3}}{\partial \tau }=\frac{(\mu -2)\sigma (\sigma +6)^{2}}{9\alpha \lbrack (13\sigma ^{2}+99\sigma +162)\mu -4\sigma \left( 2\sigma +9\right) ]}>0$$

(21)

where the inequality holds from Eq. 20. We also have $d\lambda _{3}/d\lambda _{1}=-2$ along the equilibrium path $\lambda{\lambda } ^{\ast }(\tau )$ at $\tau =\tau _{\rm{trade}}.$

For the equilibrium path to enter domain (C3), $\lambda{\lambda} ^{\ast }(\tau )$ should not enter the neighboring trade triangle (C5). This is shown to be equivalent that Eq. 21 is smaller than the partial derivative of the vertical axis of the southeast vertex ${\left( {{{\left( {{\text{2}}\alpha - \tau } \right)}} \mathord{\left/ {\vphantom {{{\left( {{\text{2}}\alpha - \tau } \right)}} {\sigma \tau ,}}} \right. \kern-\nulldelimiterspace} {\sigma \tau ,}} \right)}{\left[ {{{\left( {\sigma + 4} \right)}\tau - 4\alpha } \mathord{\left/ {\vphantom {{{\left( {\sigma + 4} \right)}\tau - 4\alpha } {\sigma \tau }}} \right. \kern-\nulldelimiterspace} {\sigma \tau }} \right]}$ in absolute value. A straightforward computation yields $\mu >\bar{\mu}_{2}.$

Lemma 6

There is no stable equilibrium path $\lambda{\lambda }^{\ast }(\tau )$ passing through $\lambda{\lambda }^{\ast }=(1/3,1/3,1/3)$ at $\tau =\tau _{\rm{trade}}$ that enters domain (C4).

Proof

Computing and evaluating the Jacobian of Eq. 2 at $\lambda{ \lambda }^{\ast }=(1/3,1/3,1/3)$ with $\tau =\tau _{\rm{trade}}$ for one-way trade from the center, we have the stability condition

$$ \mu<\bar{\mu}_{2}$$

(22)

The comparative statics at $\lambda{\lambda }^{\ast }=(1/3,1/3,1/3)$ with $\tau =\tau _{\rm{trade}}$ is

$$ \frac{\partial \lambda _{1}}{\partial \tau }=\frac{\partial \lambda _{3}}{ \partial \tau }=\frac{2(\mu -2)\sigma (\sigma +6)^{2}}{18\alpha \lbrack (14\sigma ^{2}+99\sigma +162)\mu -2\sigma \left( 5\sigma +18\right) ]}>0$$

where the inequality holds from Eq. 22. For the equilibrium path to enter the domain of the one-way trade from the center (C4), it is necessary that

$$ \frac{(\mu -2)\sigma (\sigma +6)^{2}}{18\alpha \lbrack (14\sigma ^{2}+99\sigma +162)\mu -2\sigma \left( 5\sigma +18\right) ]}>\frac{(\sigma +6)^{2}}{36\alpha \sigma }$$

or equivalently $\mu >\bar{\mu}_{2}$, which contradicts Eq. 22.

Appendix C: Lemma 7

Lemma 7

In the OTT model with $\mu<\mu _{2}^{\ast }$, the stable equilibrium path $\lambda{\lambda }^{\ast }(\tau )$ passing through $\lambda{\lambda }=(1/3,1/3,1/3)$ at $\tau =\tau _{\rm{trade}}$ is always axisymmetric for all $\lambda _{1}^{\ast }=\lambda _{3}^{\ast }\in (0,1/2).$

Proof

Let $y_{rs}\equiv \left. \partial y_{r}/\partial \lambda _{s}\right\vert _{\lambda _{1}^{\ast }=\lambda _{3}^{\ast }}$. Since $y_{11}=y_{33}$ and $y_{13}=y_{31},$ the two eigenvalues are real and given by $y_{11}\pm y_{13}$ in an axisymmetric equilibrium $\lambda _{1}^{\ast }=\lambda _{3}^{\ast }.$ Let $X\equiv y_{13}(\lambda _{1}-\lambda _{3})$, then

$${\mathop X\limits^ \cdot } = y_{{13}} {\left( {{\mathop \lambda \limits^ \cdot }_{1} - {\mathop \lambda \limits^ \cdot }_{3} } \right)} = {\left( {y_{{11}} - y_{{13}} } \right)}X$$

in the vicinity of an axisymmetric equilibrium. We show $y_{11}-y_{13}<0$ below so that $X\$goes to zero, which necessarily results in the axisymmetric configuration $\lambda _{1}=\lambda _{3}$. Let

$$ \bar{\tau}_{\rm{trade}}\equiv \frac{2\sigma }{\sigma +4}<\tau _{\rm{ trade}}$$

at which full trade starts.

• 1. Neighboring trade $0<\mu \leq \check{\mu}$ and $\bar{ \tau}_{\rm{trade}}\leq \tau \leq \tau _{\rm{trade}}.$

  Noting that $y_{1r}=\partial \lambda _{1}\left( V_{1}-\sum_{s=1}^{3}\lambda _{s}V_{s}\right) /\partial \lambda _{r}$ about an axisymmetric equilibrium because $V_{1}-\sum_{s=1}^{3}\lambda _{s}V_{s}=0$ holds, we have

  $$ \rm{sgn}\left( {\it{y}}_{11}-{\it{y}}_{13}\right) =\rm{sgn}\left( \tau \left( \lambda _{1}\right) -\tau \right)$$

  (23)

  where

  $$ \tau \left( \lambda _{1}\right) =\frac{2\alpha \left( 4\sigma -18\mu -19\mu \sigma +27\mu \sigma \lambda _{1}\right) }{\sigma \left( 2\lambda _{1}-1\right) \left( 4\sigma -6\mu -13\mu \sigma +21\mu \sigma \lambda _{1}\right) }$$

  (24)

  It is readily shown that $\tau ^{\prime }\left( \lambda _{1}\right)<0$ for all $\mu<\hat{\mu}_{2}$. Four cases may arise.

  • 1a. For $0<\mu \leq \hat{\mu}_{1}$ and $\bar{\tau}_{ \rm{trade}}\leq \tau \leq \tau _{\rm{trade}}.$

    Solving $\tau \left( 0\right) =\bar{\tau}_{\rm{trade}}$ yields sgn$\left( \bar{\tau}_{\rm{trade}}-\tau \left( 0\right) \right) =$sgn$\left( 2\sigma /\left( 8\sigma +9\right) -\mu \right)$. If $\mu<2\sigma /\left( 8\sigma +9\right) ,$ then we have $\tau \geq \bar{\tau}_{\rm{trade} }>\tau \left( 0\right) >\tau \left( \lambda _{1}\right) ,$ where the last inequality is due to $\tau ^{\prime }\left( \lambda _{1}\right)<0$. That is, if $\mu<2\sigma /\left( 8\sigma +9\right) ,$ then $y_{11}<y_{13}$ holds from Eq. 23. Therefore, we only have to show $\hat{\mu}_{1}<2\sigma /\left( 8\sigma +9\right) .$

    Let

    $$f{\left( {\lambda _{1} } \right)} \equiv 36\mu ^{2} \sigma ^{2} {\left( {\lambda _{1} } \right)}^{4} - 12\mu \sigma {\left( {108\mu + 2\sigma + 73\mu \sigma } \right)}{\left( {\lambda _{1} } \right)}^{3} + {\left( {1981\mu ^{2} \sigma ^{2} + 5580\mu ^{2} \sigma + 3888\mu ^{2} + 202\mu \sigma ^{2} + 360 + 4\sigma ^{2} } \right)}{\left( {\lambda _{1} } \right)}^{2} - 6{\left( {164\mu ^{2} \sigma ^{2} + 552\mu ^{2} \sigma + 414\mu ^{2} + 9\mu \sigma ^{2} - 8\mu \sigma - 36\mu + 4\sigma ^{2} + 8\sigma } \right)}\lambda _{1} + 4{\left( {35\mu ^{2} \sigma ^{2} + 118\mu ^{2} \sigma + 99\mu ^{2} - \mu \sigma ^{2} - 14\mu \sigma - 18\mu + 2\sigma ^{2} + 4\sigma \lambda } \right)}$$

    (25)

    Let $J\left( \mu ,\sigma \right)$ be the discriminant of the fourth-order polynomial $f(\lambda _{1}).$ By definition, we have $J\left( \hat{\mu} _{1},\sigma \right) =0$ and $J\left( 2\sigma /\left( 8\sigma +9\right) ,\sigma \right)<0$, which implies $\hat{\mu}_{1}<2\sigma /\left( 8\sigma +9\right)$.

  • 1b. Neighboring trade $\hat{\mu}_{1}<\mu<\check{\mu}$ and $\bar{\tau}_{\rm{trade}}\leq \tau \leq \tau _{\rm{trade}}.$

  In this case, $\lambda _{1}$ lies in the interval of $[1/3,1/2]$. Since $\tau ^{\prime }\left( \lambda _{1}\right)<0$, $\tau \left( 1/3\right) >\tau \left( \lambda _{1}\right)$ for all $\lambda _{1}\in \lbrack 1/3,1/2]$. Furthermore, $\bar{\tau}_{\rm{trade}}>\tau \left( 1/3\right)$ for $\mu = \check{\mu},$ and hence $\tau >\tau \left( \lambda _{1}\right)$, which means $y_{11}<y_{13}$ at $\lambda _{1}=\lambda _{3}.$

• 2. One-way trade to the center $\check{\mu}<\mu<\hat{\mu}_{2}$ and $\bar{\tau}_{\rm{trade}}\leq \tau \leq \tau _{\rm{ trade}}.$

  A direct computation yields

  $$ \left. y_{11}-y_{13}\right\vert _{\lambda _{1}=\lambda _{3}}=\frac{\alpha ^{2}\sigma \,\,\left( 18\mu -4\sigma +4\mu \sigma +3\mu \sigma \lambda _{1}\right) }{\gamma \mu \left( \sigma \lambda _{1}+2\right) ^{3}}$$

  Since the RHS is increasing in $\mu$ and $\lambda _{1}$, substituting $\mu = \hat{\mu}_{2}$ and $\lambda _{1}=1/2$, we have

  $$ \left. y_{11}-y_{13}\right\vert _{\lambda _{1}=\lambda _{3}}<-\frac{2\alpha ^{2}\sigma ^{2}\left( 19\sigma +126\right) }{3\gamma \left( 5\sigma +18\right) \left( \sigma +4\right) ^{3}}<0$$

• 3. Full trade $\mu<\acute{\mu}\equiv \left( \sigma ^{2}-4\right) /\left( 10\sigma ^{2}+48\sigma +50\right)$ and $\tau<\bar{\tau}_{\rm{trade}}.$

The full trade condition $\mu<\acute{\mu}$ is equivalent to at $\lambda _{1}^{\ast }(\bar{\tau}_{\rm{trade}})>0$, which the interior equilibrium condition at $\tau =\bar{\tau}_{\rm{trade}}.$

Since the stable equilibrium path $\lambda{\lambda }^{\ast }(\tau )$ and its inverse function $\tau _{c}\left( \lambda _{1}\right)$are increasing functions, we get

$$\tau _{c}\left( \lambda _{1}\right) >\tau _{c}\left( 0\right) =\frac{4\alpha \left( 2+7\mu +\sigma +5\mu \sigma \right) }{4+4\sigma +\sigma ^{2}+2\mu +2\mu \sigma }>\bar{\tau}_{\rm{trade}} $$

(26)

On the other hand, the definition in Eq. 24 is revised by

$$ \tau \left( \lambda _{1}\right) =\frac{3\alpha \mu \left( 2\sigma +3\right) }{\mu +2\sigma +\sigma ^{2}+94\mu \sigma -\mu \sigma ^{2}+3\mu \sigma \left( \sigma +2\right) \lambda _{1}}$$

Since $\tau ^{\prime }\left( \lambda _{1}\right)<0$ holds, we have

$$\begin{array}{*{20}c} {\tau {\left( {\lambda _{1} } \right)}}{ < \tau {\left( 0 \right)} = \frac{{3a\mu {\left( {2\sigma + 3} \right)}}}{{\mu + 2\sigma + \sigma ^{2} + 94\mu \sigma - \mu \sigma ^{2} }}} \\ {}{ < \frac{{3a{\mathop \mu \limits^\prime }{\left( {2\sigma + 3} \right)}}}{{{\mathop \mu \limits^\prime } + 2\sigma + \sigma ^{2} + 94{\mathop \mu \limits^\prime }\sigma - {\mathop \mu \limits^\prime }\sigma ^{2} }}} \\ {}{ = \frac{{a{\left( {\sigma - 2} \right)}{\left( {2\sigma + 3} \right)}}}{{3\sigma ^{3} + 18\sigma ^{2} + 17\sigma - 6}}} \\ {}{ < \overline{\tau } _{{trade}} } \\ \end{array} $$

(27)

where the second inequality is due to $\partial \tau \left( 0\right) /\partial \mu >0$ and the last inequality is from $\sigma >2$ which is from $\mu<\acute{\mu}$. Putting Eqs. 26 and 27 together, we have $\tau \left( \lambda _{1}\right)<\bar{\tau}_{\rm{trade}}<\tau _{c}\left( \lambda _{1}\right)$, which is $y_{11}<y_{13}$ at $\lambda _{1}=\lambda _{3} .$

It is known in the two-dimensional dynamics that the equilibrium point is a node (resp. saddle point) if $\left\vert y_{11}\right\vert >\left\vert y_{13}\right\vert$ (resp. $\left\vert y_{11}\right\vert<\left\vert y_{13}\right\vert \!$).

Appendix D: Proposition 4

From Lemma 7, a stable interior equilibrium path should be on the curve

$$ \Delta V_{12}(\lambda _{1},\tau )\equiv \left. V_{1}-V_{2}\right\vert _{\lambda _{2}=1-2\lambda _{1},\lambda _{3}=\lambda _{1}}=0$$

which passes through $(\lambda _{1},\tau )=(1/3,\tau _{\rm{trade}})$. It intersects with $\lambda _{1}=1/2$ if $f(\lambda _{1})>0$ holds for all $\lambda _{1}\in \lbrack 1/3,1/2]$, which corresponds to disappearance of the center. This is equivalent to $f(\tilde{\lambda}_{1})>0$, where $\tilde{ \lambda}_{1}$ is the smallest solution of $f^{\prime }(\lambda _{1})=0$. In this way, we are able to classify the stable paths as follows.

Proposition 4

In the OTT model with $\mu<\mu _{2}^{\ast }$, the interior stable equilibrium path passing through $(\lambda _{1},\tau )=(1/3,\tau _{ \rm{trade}})$ is either one of the following:

If $\hat{\mu}_{1}<\mu<\mu _{2}^{\ast }$, $\lambda _{2}^{\ast }$ decreases monotonically from $1/3$ to $0$ at positive $\tau ;$

If $\mu<\hat{\mu}_{1},$ $\lambda _{2}^{\ast }$ decreases from $1/3$ and then increases to $1$.

(C6) stands for full trade between any pair of regions. Since the denominator of $\Delta V_{12}(\lambda _{1},\tau )$ is positive except $\lambda _{1}=\left( \sigma +2\right) /\sigma ,$ define the numerator of $\Delta V_{{12}} {\left( {\lambda _{1} ,\tau } \right)}$ by $\Delta V_{12}^{n}(\lambda _{1},\tau )$ for all $\lambda _{1}\neq \left( \sigma +2\right) /\sigma .$ Because the implicit function $\Delta V_{12}^{n}(\lambda _{1},\tau )=0$ is quadratic in $\tau$ in all cases of (C1), (C4), (C5) and (C6), it has two explicit solutions: $\tau _{a}\left( \lambda _{1}\right)$ and $\tau _{b}\left( \lambda _{1}\right) ,$ which are inverse functions of $\lambda _{1}^{\ast }(\tau ).$

Lemma 8

The function $\lambda _{1}^{\ast }(\tau )$ is continuous for all domain of $0\leq \lambda _{1}\leq 1/2$ and $0\leq \tau \leq \tau _{ \rm{trade}}$ irrespective of the trade patterns.

Proof

From the implicit functional theorem, $\lambda _{1}^{\ast }(\tau )$ is continuous inside the cases of (C1), (C4), (C5), and (C6). Therefore, we only have to check the continuity at the boundaries. However, say, at the boundaries of cases (C1) and (C4), if $\Delta V_{{12}} {\left( {\lambda _{1} ,\tau } \right)}$ holds in case (C1), then this also holds in case (C4). The same is true for all other boundaries.

Lemma 9

The functions $\tau _{a}\left( \lambda _{1}\right)$ and $\tau _{b}\left( \lambda _{1}\right)$ are real for all $\lambda _{1}\in \lbrack 1/2,\left( \sigma +2\right) /\sigma )$ except that there are at most $2$ discontinuous points in the interval.

Proof

First, we consider the number of discontinuous points. The denominator of $\tau _{a}\left( \lambda _{1}\right)$ and $\tau _{b}\left( \lambda _{1}\right)$ is common and quartic in $\lambda _{1}$ with the positive coefficient of $\left( \lambda _{1}\right) ^{4}$. Since, the denominator is shown to be negative at $\lambda _{1}=1/2$, $\left( \sigma +2\right) /\sigma$, there exist at most two discontinuous points.

Second, we show that $\tau _{a}\left( \lambda _{1}\right)$ and $\tau _{b}\left( \lambda _{1}\right)$ are real. The common terms in the square root of $\tau _{a}\left( \lambda _{1}\right)$ and $\tau _{b}\left( \lambda _{1}\right)$ are $f(\lambda _{1})$, which is defined by (25). $f(\lambda _{1})$ is quartic in $\lambda _{1}$ with a positive coefficient of $\left( \lambda _{1}\right) ^{4}$. It is readily shown that

$$ f^{\prime }(1/2)>0\qquad \lim_{\lambda _{1}\rightarrow \frac{\sigma +2}{ \sigma }-0}f^{\prime }\left( \frac{\sigma +2}{\sigma }\right) >0$$

and the larger inflexion point is outside the interval of $[1/2,\left( \sigma +2\right) /\sigma )\!\!\!$. These imply that both $\lambda _{1}=1/2$ and $\lambda _{1}=\left( \sigma +2\right) /\sigma -0$ are on the left-hand side of the quartic curve with a positive slope. Since $f(1/2)>0$, we have $f(\lambda _{1})>0$ for all $\lambda _{1}\in \lbrack 1/2,\left( \sigma +2\right) /\sigma )$, which means real $\tau _{a}\left( \lambda _{1}\right)$ and $\tau _{b}\left( \lambda _{1}\right) \!\!$.

Lemma 10

When a stable equilibrium lying on $\tau _{a}\left( \lambda _{1}\right)$ (resp. $\tau _{b}\left( \lambda _{1}\right)$) disappears, then it will jump to the same $\tau _{a}\left( \lambda _{1}\right)$ (resp. $\tau _{b}\left( \lambda _{1}\right)$) or the corners $\lambda _{1}=0,1/2.$

Proof

It is readily shown that $\Delta V_{12}(\lambda _{1},\tau )=0$ is quartic in $\lambda _{1}$ in cases (C1) and quadratic in $\lambda _{1}$ in case (C6). Since $\lim_{\lambda _{1}\rightarrow \pm \infty }\tau _{a}\left( \lambda _{1}\right) =$lim$_{\lambda _{1}\rightarrow \pm \infty }\tau _{b}\left( \lambda _{1}\right) =0$ holds for all cases of (C1), (C4), (C5) and (C6), $\tau _{a}\left( \lambda _{1}\right)$ and $\tau _{b}\left( \lambda _{1}\right)$ have at most one trough $(\check{\lambda}_{1},\check{\tau}),$ at which a jump may occur.

Case (C1)

Suppose $\tau _{a}\left( \lambda _{1}\right)$ has a peak $(\hat{\lambda}_{1},\hat{\tau}),$ such that $\check{\tau}<\hat{\tau}$ is in the interval of $[0,1/2]$, then a stable equilibrium path $\tau _{b}\left( \lambda _{1}\right)$ may jump at the trough $(\check{\lambda} _{1},\check{\tau}_{b})$ to the different curve $\tau _{a}\left( \lambda _{1}\right) .$ However, such a trough-peak combination violates the property in that there are more than $4$ solutions of $\lambda _{1}$ in $\Delta V_{12}(\lambda _{1},(\check{\tau}+\hat{\tau})/2)=0$ as shown below.

From Lemma 9, $\tau _{b}\left( \lambda _{1}\right)$ is discontinuous at most $2$ points.

• 1. If $\tau _{b}\left( \lambda _{1}\right)$ is continuous in $\lambda _{1}\in \lbrack 1/2,\left( \sigma +2\right) /\sigma )$, then $\tau _{b}\left( \lambda _{1}\right)$ necessarily crosses the horizontal line $\tau =(\check{\tau}+\hat{\tau})/2$ in $(\max \{\check{\lambda}_{1},\hat{ \lambda}_{1}\},\left( \sigma +2\right) /\sigma )$ in addition to $4$points of intersection in $[\min \{\check{\lambda}_{1},\hat{\lambda}_{1}\},\max \{ \check{\lambda}_{1},\hat{\lambda}_{1}\}],$ thus violating the property.

• 2. If $\tau _{b}\left( \lambda _{1}\right)$ is discontinuous only at $\bar{\lambda}_{1}\in (1/2,\left( \sigma +2\right) /\sigma )$, then $\tau _{b}\left( \lambda _{1}\right)$ goes to either $-\infty$ or $+\infty .$ However, the sign of $\tau _{b}\left( \lambda _{1}\right)$ does not change in the vicinity of $\bar{\lambda}_{1}$ $\tau _{b}\left( \lambda _{1}\right)$ since $\bar{\lambda}_{1}$ must be a repeated root of the quartic denominator of $\tau _{b}\left( \lambda _{1}\right) .$ Hence, the above case 1 applies.

• 3. If $\tau _{b}\left( \lambda _{1}\right)$ is discontinuous at two distinct $\bar{\lambda}_{1},\tilde{\lambda}_{1}$ with $1/2<\bar{\lambda} _{1}<\tilde{\lambda}_{1}<\left( \sigma +2\right) /\sigma$, then $\tau _{b}\left( \lambda _{1}\right)$ changes its sign in the vicinity of $\bar{ \lambda}_{1},\tilde{\lambda}_{1}.$ If $\lim_{\lambda _{1}\rightarrow \bar{ \lambda}_{1}-0}\tau _{b}\left( \lambda _{1}\right) =-\infty ,$ then $\tau _{b}\left( \lambda _{1}\right)$ crosses $\tau =(\check{\tau}+\hat{\tau})/2$ in $(\check{\lambda}_{1},\bar{\lambda}_{1})$, and hence the number of intersection points with $\tau =(\check{\tau}+\hat{\tau})/2$ exceeds $4.$

On the other hand, if $\lim_{\lambda _{1}\rightarrow \bar{\lambda} _{1}-0}\tau _{b}\left( \lambda _{1}\right) =+\infty ,$ then it must be that $\lim_{\lambda _{1}\rightarrow \bar{\lambda}_{1}+0}\tau _{b}\left( \lambda _{1}\right) =-\infty$ $\lim_{\lambda _{1}\rightarrow \tilde{\lambda} _{1}-0}\tau _{b}\left( \lambda _{1}\right) =-\infty$ and $\lim_{\lambda _{1}\rightarrow \tilde{\lambda}_{1}+0}\tau _{b}\left( \lambda _{1}\right) =+\infty .$ Thus, $\tau _{b}\left( \lambda _{1}\right)$ crosses the horizontal line $\tau =(\check{\tau}+\hat{\tau})/2$ in $(\tilde{\lambda} _{1},\left( \sigma +2\right) /\sigma ),$ and hence the number of intersection points with $\tau =(\check{\tau}+\hat{\tau})/2$ exceeds $4.$

Cases (C4) and (C5)

It is easily shown that the equilibrium path crosses at most once at the boundary between cases (C1) and (C4) and at the boundary between cases (C1) and (C5). Therefore, once a stable equilibrium path crosses one of the boundaries, it will never cross again, implying that the above trough-peak combination does not arise. Hence, it will be sure to hit the corner $\lambda _{1}=1/2$ (resp. $0$) in case (C1) [resp. (C4)].

Case (C6)

Since $\Delta V_{12}(\lambda _{1},\tau )=0$ is quadratic in $\lambda _{1},$ the above trough-peak combination cannot arise.

Lemma 10 together with Lemma 8 ensures that the stable equilibrium path starting from $\bf{\lambda }^{\ast }=(\rm 1/3,1/3,1/3)$ at $\tau =\tau _{\rm{trade}}$ will keep on the same curve or hit the corners. The only exception occurs when $d\tau _{a}\left( \lambda _{1}\right) /d\lambda _{1}=d\tau _{b}\left( \lambda _{1}\right) /d\lambda _{1}=\infty$, which corresponds to $d\lambda _{1}^{\ast }(\tau )/d\tau =0$. At this point, an equilibrium does not disappear, and the two stable curves $\tau _{a}\left( \lambda _{1}\right)$ and $\tau _{b}\left( \lambda _{1}\right)$ are connected. This is case 2 with $\mu<\hat{\mu}_{1},$ where the path first moves toward $\lambda _{1}=1/2$ and then moves back to $\lambda _{1}=0,$ resulting in $(0,1,0)$. Otherwise, the path moves according to case 1 with $\mu >\hat{\mu}_{1}$ and reaches $(1/2,0,1/2)$ monotonically. Finally, since there is no stable equilibrium path approaching an interior solution of $\lambda _{1}$ at $\tau =0,$ Proposition 4 is proven.

Appendix E: Proof of Proposition 3

• 1. In the case of neighboring trade, when $\lambda _{2}=0$, we have

  $$ \left. \frac{\partial \left( V_{1}-V_{3}\right) }{\partial \lambda _{1}} \right\vert _{\lambda _{1}=\lambda _{3}=1/2}=\frac{4\alpha ^{2}\sigma \left[ \mu \left( 11\sigma +36\right) -8\sigma \right] }{3\gamma \mu \left( \sigma +4\right) ^{3}}$$

  which is negative for all $\mu \leq \mu _{2}^{\ast }$, implying that ${\bf{\lambda} }^{\ast }=(1/2,0,1/2)$ is a stable during the neighboring trade of $\tau >\bar{\tau}_{\rm{trade}}.$ This equilibrium continues during $\tau >\bar{\tau}_{\rm{trade}}$ if $\left. V_{1}-V_{2}\right\vert _{\lambda _{1}=\lambda _{3}=1/2,\lambda _{2}=0,\tau = \bar{\tau}_{\rm{trade}}}\geq 0,$ which is equivalent to

  $$ \mu >\frac{2}{11}$$

  Otherwise, $(1/2,0,1/2)$ ceases to be an equilibrium, leading to the new equilibrium $(0,1,0)$ for a sufficiently large period of time.

• 2. In the case of full trade, when $\lambda _{2}=0$, we get

  $$ \left. \frac{\partial \left( V_{1}-V_{3}\right) }{\partial \lambda _{1}} \right\vert _{\lambda _{1}=\lambda _{3}=1/2}=\frac{2\sigma \tau \left[ \mu \left( \sigma ^{2}+14\sigma +18\right) +2\sigma (\sigma +2)\right] }{3\gamma \mu \left( \sigma +2\right) ^{2}}\left( \tau _{\rm{break}}-\tau \right)$$

  where

  $$\tau _{{break}} \equiv \frac{{6a\mu {\left( {2\sigma + 3} \right)}}}{{\mu {\left( {\sigma ^{2} + 14\sigma + 18} \right)} + 2\sigma {\left( {\sigma + 2} \right)}}}$$

  That is, $(1/2,0,1/2)$ is a stable equilibrium for $\tau >\tau _{\mathrm{ \rm break}}$. Since

  $$\frac{{\partial {{\mathop \lambda \limits^ \cdot }} \mathord{\left/ {\vphantom {{{\mathop \lambda \limits^ \cdot }} {\lambda _{2} \left| {_{{\lambda _{3} = 1 - \lambda _{1} }} } \right.}}} \right. \kern-\nulldelimiterspace} {\lambda _{2} \left| {_{{\lambda _{3} = 1 - \lambda _{1} }} } \right.}}}{{\partial \lambda _{1} }} = \frac{{18a^{2} \mu ^{2} \sigma ^{3} {\left( {2\sigma + 3} \right)}^{2} {\left( {2\lambda _{1} - 1} \right)}}}{{\gamma {\left( {\sigma + 2} \right)}^{2} {\left( {\mu \sigma ^{2} + 2\sigma ^{2} + 14\mu \sigma + 18\mu + 4\sigma } \right)}^{2} }}$$

  ${{\mathop \lambda \limits^ \cdot }_{2} } \mathord{\left/ {\vphantom {{{\mathop \lambda \limits^ \cdot }_{2} } {\left. {\lambda _{2} } \right|}}} \right. \kern-\nulldelimiterspace} {\left. {\lambda _{2} } \right|}_{{\lambda _{3} = 1\lambda _{1} }}$ is maximized at $\lambda _{1}=0,1.$ Solving ${{\mathop \lambda \limits^ \cdot }_{2} } \mathord{\left/ {\vphantom {{{\mathop \lambda \limits^ \cdot }_{2} } {\lambda _{2} }}} \right. \kern-\nulldelimiterspace} {\lambda _{2} } = 0$ at $\lambda _{1}=1$ yields $\mathrm{sgn }(\overset{\cdot }{\lambda _{2}})=\mathrm{sgn}\left( \check{\mu}_{1}-\mu \right) ,$ where

  $$ \check{\mu}_{1}\equiv \frac{\left( 5\sigma +2\right) \left( \sigma +2\right) }{8\sigma ^{2}+25\sigma +18+3\sqrt{32\sigma ^{4}+240\sigma ^{3}+489\sigma ^{2}+324\sigma +36}}$$

  The stability condition is therefore

  $$ \mu >\check{\mu}_{1}$$

  This equilibrium continues during $\tau >\bar{\tau}_{\rm{trade}}$ if $\left. V_{1}-V_{2}\right\vert _{\lambda _{1}=\lambda _{3}=1/2}\geq 0\!\!$, which is equivalent to

  $$ \bar{\tau}_{\rm{break}}\equiv \frac{8\alpha \left( 1-\mu \right) }{7\mu \sigma +6\mu +2\sigma +12}<\tau \leq \bar{\tau}_{\rm{trade}}$$

  where $\bar{\tau}_{\rm{break}}<\bar{\tau}_{\rm{trade}}$ holds for $\mu >2/11$. Solving ${{\mathop {\lambda _{2} }\limits^ \cdot }} \mathord{\left/ {\vphantom {{{\mathop {\lambda _{2} }\limits^ \cdot }} \lambda }} \right. \kern-\nulldelimiterspace} \lambda _{2} = 0$ at $\lambda _{1}=1$ and $\tau =\bar{\tau}_{\rm{break}}$ yields

  $$ \bar{\mu}_{1}\equiv \frac{3\sqrt{193\sigma ^{4}+1044\sigma ^{3}+1932\sigma ^{2}+1296\sigma +144}-13\sigma ^{2}-50\sigma -36}{98\sigma ^{2}+310\sigma +252}$$

  However, it can be shown that $\check{\mu}_{1}>\bar{\mu}_{1}$ for all $\mu >2/11$.

• 3. Finally, we consider the case that $(1/2,0,1/2)$ becomes unstable when full trade starts at $\tau =\bar{\tau}_{\rm{trade}}.$ Since

  $$\frac{{{\partial {\mathop \lambda \limits^ \cdot }_{2} } \mathord{\left/ {\vphantom {{\partial {\mathop \lambda \limits^ \cdot }_{2} } {\lambda _{2} \left| {_{{\lambda _{3} = 1 - \lambda _{1} }} } \right.}}} \right. \kern-\nulldelimiterspace} {\lambda _{2} \left| {_{{\lambda _{3} = 1 - \lambda _{1} }} } \right.}}}{{\partial \lambda _{1} }} = \frac{{2\sigma \alpha ^{2} {\left( {8\sigma ^{2} - 17\mu \sigma ^{2} - 19\mu \sigma - 72\mu + 16\sigma } \right)}{\left( {2\lambda _{1} - 1} \right)}}}{{3\gamma \mu {\left( {\sigma + 4} \right)}^{2} {\left( {\sigma + 2} \right)}^{2} }}$$

  at $\tau =\bar{\tau}_{\rm{trade}}$. ${{\mathop {\lambda _{2} }\limits^ \cdot }} \mathord{\left/ {\vphantom {{{\mathop {\lambda _{2} }\limits^ \cdot }} {\left. {\lambda _{2} } \right|_{{\lambda _{3} = 1 - \lambda _{1} ,\tau = \overline{\tau } _{{_{{trade}} }} }} }}} \right. \kern-\nulldelimiterspace} {\left. {\lambda _{2} } \right|_{{\lambda _{3} = 1 - \lambda _{1} ,\tau = \overline{\tau } _{{_{{trade}} }} }} }$ is maximized at $\lambda _{1}=0,1/2$ or $1.$ Solving ${\lambda _{2} } \mathord{\left/ {\vphantom {{\lambda _{2} } {{\mathop {\lambda _{2} }\limits^ \cdot }}}} \right. \kern-\nulldelimiterspace} {{\mathop {\lambda _{2} }\limits^ \cdot }} = 0$ at $\lambda _{1}=1/2$ and $1$ yields the stability conditions

  $$ \mu >\tilde{\mu}_{1}\qquad \rm{and}\qquad \mu >\frac{2}{11}$$

  where

  $$ \tilde{\mu}_{1}\equiv \frac{\left( 5\sigma +2\right) \left( \sigma +2\right) }{2\left( 7\sigma ^{2}+30\sigma +29\right) }$$

  Hence, the conditions for the stable path reaching $(1,0,0)$ or $(0,0,1)$ are summarized as

  $$ \mu _{1}^{\ast }\equiv \max \left\{ \frac{2}{11},\min \left\{ \tilde{\mu} _{1},\check{\mu}_{1}\right\} \right\}$$