Agglomeration patterns in a multi-regional economy without income effects

Abstract

We study the long-run spatial distribution of industry using a multi-region core–periphery model with quasi-linear log utility Pflüger (Reg Sci Urban Econ 34:565–573, 2004). We show that a distribution in which industry is evenly dispersed among some of the regions, while the other regions have no industry, cannot be stable. A spatial distribution where industry is evenly distributed among all regions except one can be stable, but only if that region is significantly more industrialized than the other regions. When trade costs decrease, the type of transition from dispersion to agglomeration depends on the fraction of workers that are mobile. If this fraction is low, the transition from dispersion to agglomeration is catastrophic once dispersion becomes unstable. If it is high, there is a discontinuous jump to partial agglomeration in one region and then a smooth transition until full agglomeration. Finally, we find that mobile workers benefit from more agglomerated spatial distributions, whereas immobile workers prefer more dispersed distributions. The economy as a whole shows a tendency towards overagglomeration for intermediate levels of trade costs.

Appendices

Appendix A

This appendix contains the formal proofs pertaining to Sect. 2 of the paper.

Proof of Proposition 1

The average nominal wage is the weighted sum of nominal wages in each region, given by (12):

$$\begin{aligned} \bar{w}&=\sum _{i=1}^{n}h_{i}w_{i}=\dfrac{\mu }{\sigma } \sum _{i=1}^{n}h_{i}\sum _{j=1}^{n}\dfrac{\phi _{ij} \left( \lambda /n+h_{j}\right) }{\phi +(1-\phi )h_{j}}= \dfrac{\mu }{\sigma }\sum _{j=1}^{n}\sum _{i=1}^{n}h_{i}\dfrac{\phi _{ij} \left( \lambda /n+h_{j}\right) }{\phi +(1-\phi )h_{j}}\\&=\dfrac{\mu }{\sigma }\sum _{j=1}^{n}\dfrac{\lambda /n +h_{j}}{\phi +(1-\phi )h_{j}}\sum _{i=1}^{n}h_{i} \phi _{ij}=\dfrac{\mu }{\sigma }\sum _{j=1}^{n} \dfrac{\lambda /n+h_{j}}{\phi +(1-\phi )h_{j}} \left[ \phi +(1-\phi )h_{j}\right] , \end{aligned}$$

and the result $\bar{w}=\frac{\mu }{\sigma }(1+\lambda )$ follows. $\square$

Appendix B

This appendix contains all the proofs concerning both existence and local stability of equilibria (Sect. 3).

Proof of Proposition 2

Let i be the core region. We show that $V_{j}<V_{i}$ $\forall j\ne i$. We have, from (16), that:

$$\begin{aligned} V_{i}=\dfrac{\mu }{\sigma }(1+\lambda )+\eta \ \text { and }\ V_{j}=\frac{\mu }{\sigma n\phi }\left[ \lambda +\phi ^{2}(\lambda +n) +\lambda (n-2)\phi \right] +\frac{\mu }{\sigma -1}\ln \phi +\eta . \end{aligned}$$

A straightforward simplification of the inequality $V_{j}<V_{i}$ yields the desired result. $\square$

Proof of Proposition 3

Local stability of interior equilibria in $\triangle$ is given by the sign of the real part of the eigenvalues of the Jacobian matrix of the system in (17) at $\left( h_{1},h_{2},\ldots ,h_{n-1}\right) =\left( \tfrac{1}{n},\tfrac{1}{n},\ldots ,\tfrac{1}{n}\right)$. At symmetric dispersion, the average utility $\bar{V}$ is invariant in the permutation of any two coordinates, due to symmetry. If we interchange the distributions between region 1 and region n we then have $\bar{V}\left( \tfrac{1}{n}+\varepsilon ,\tfrac{1}{n},\ldots ,\tfrac{1}{n}\right) =\bar{V}\left( \tfrac{1}{n}-\varepsilon ,\tfrac{1}{n},\ldots ,\tfrac{1}{n}\right)$. But this implies that $\partial _{h_{i}}\bar{V}\left( \tfrac{1}{n},\tfrac{1}{n},\ldots ,\tfrac{1}{n}\right) =0$.²⁵ The argument of invariance extends to the indirect utility $V_{i}$ in the permutation of any two coordinates $j\ne i$, which implies that $\partial _{h_{j}}V_{i}\left( \tfrac{1}{n}, \tfrac{1}{n},\ldots ,\tfrac{1}{n}\right) =0,\forall j\ne i$. Finally, symmetry among regions establishes that we must have $\partial _{h_{i}}V_{i}\left( \tfrac{1}{n}, \tfrac{1}{n},\ldots ,\tfrac{1}{n}\right) = \partial _{h_{j}}V_{j}\left( \tfrac{1}{n},\tfrac{1}{n} ,\ldots ,\tfrac{1}{n}\right)$. The Jacobian matrix of (17) at the symmetric equilibrium is thus given by:

$$\begin{aligned} J=\begin{bmatrix} \dfrac{\partial V_{i}}{\partial h_{i}}&\quad 0&\quad \ldots&\quad 0\\ 0&\quad \dfrac{\partial V_{i}}{\partial h_{i}}&\quad \ldots&\quad 0\\ \vdots&\quad \vdots&\quad \ddots&\quad \vdots \\ 0&\quad 0&\quad \ldots&\quad \dfrac{\partial V_{i}}{\partial h_{i}} \end{bmatrix}, \end{aligned}$$

which has a repeated real eigenvalue with multiplicity $n-1$ given by $\partial _{h_{i}}V_{i}\left( \tfrac{1}{n}, \tfrac{1}{n},\ldots ,\tfrac{1}{n}\right) ,$ and total dispersion is stable if $\partial _{h_{i}}V_{i}\left( \tfrac{1}{n} ,\ldots ,\tfrac{1}{n}\right) <0$.

Replacing $h_{n}=1-\sum _{j=1}^{n-1}h_{j}$ in expression (16) and computing the partial derivative with respect to $h_{i}$, $\partial _{h_{i}}V_{i}\left( \tfrac{1}{n},\ldots ,\tfrac{1}{n}\right)$, we find that total dispersion is stable if:

$$\begin{aligned} \partial _{h_{i}}V_{i}\left( \tfrac{1}{n},\ldots , \tfrac{1}{n}\right) \equiv \frac{\mu n(1-\phi ) \left[ (2n-1)\sigma \phi -\lambda (\sigma -1)(1-\phi ) -n\phi +\sigma \right] }{(\sigma -1)\sigma \left[ (n-1) \phi +1\right] {}^{2}}<0\\ \Leftrightarrow \ (2n-1)\sigma \phi -n\phi +\sigma -\lambda (\sigma -1)(1-\phi )<0. \end{aligned}$$

Solving for $\phi$ finishes the proof. $\square$

Proof of Theorem 1

Two conditions are necessary for the study of stability of boundary dispersion. The first condition ensures that empty regions remain empty and, analogously to the proof of Proposition 2, demands that $\left. V_{i}\right| _{BD}-\left. V_{j}\right| _{BD}<0$, where $h_{i}=0$ and $h_{j}=1/k$. The second condition guarantees that along the boundary $(h_{i}=0;i=1,\ldots ,n-k)$ the configuration is stable. Similarly to the proof of Proposition 3, this condition requires that $\left. \partial _{h_{j}}V_{j}\right| _{BD}<0.$

These two conditions yield, respectively:

$$\begin{aligned}&n\sigma \phi \left[ (k-1)\phi +1\right] \ln \left[ \dfrac{k\phi }{(k-1)\phi +1}\right] +(\sigma -1)(1-\phi )\left[ \lambda (1-\phi ) -n\phi \right]&<0,\\&\quad k\left[ n(2\sigma -1)\phi -\lambda (\sigma -1)(1-\phi )\right] +n\sigma (1-\phi )&<0. \end{aligned}$$

Solving for $\lambda$, we obtain:

$$\begin{aligned}&\frac{n\left[ k(2\sigma -1)\phi +\sigma (1-\phi )\right] }{k(\sigma -1)(1-\phi )}\\&\quad<\lambda <\frac{n\phi \left\{ \sigma \left[ (k-1)\phi +1\right] \ln \left[ \dfrac{(k-1)\phi +1}{k\phi }\right] +(\sigma -1)(1-\phi )\right\} }{(\sigma -1)(1-\phi )^{2}}. \end{aligned}$$

The strict inequalities are incompatible, and thus boundary dispersion is unstable, if:

$$\begin{aligned}&\frac{n\phi \left\{ \sigma \left[ (k-1)\phi +1\right] \ln \left[ \dfrac{(k-1)\phi +1}{k\phi }\right] +(\sigma -1)(1-\phi )\right\} }{(\sigma -1)(1-\phi )^{2}}\\&\quad -\frac{n\left[ k(2\sigma -1)\phi +\sigma (1-\phi )\right] }{k(\sigma -1)(1-\phi )} <0. \end{aligned}$$

Eliminating common factors of positive sign and rearranging terms to our convenience, the above inequality becomes:

$$\begin{aligned}&\left[ (k-1)\phi +1\right] \left\{ k\phi \ln \left[ \dfrac{(k-1)\phi +1}{k\phi }\right] -(1-\phi )\right\} <0\\&\quad \Leftrightarrow \ \dfrac{1-\phi }{k\phi }-\ln \left( 1+\dfrac{1-\phi }{k\phi }\right)&>0. \end{aligned}$$

Define $g(x)=x-\ln (1+x)$. It is easy to see that $g(0)=0$ and that g is strictly increasing. Hence, $g(x)>0,\forall x>0$. Noting that $\tfrac{1-\phi }{k\phi }>0$ finishes the proof. $\square$

Proof of Proposition 4

Configurations in $\Delta _{\mathrm{inv}}$ satisfy $h_{1}\in [0,1],\ h_{j}=\tfrac{1-h_{1}}{n-1}$ for $j\ne 1.$ By replacing the values in (17) and solving for an equilibrium we obtain $\lambda =\lambda ^{*}(h_{1})$, given by (24) and reproduced here for convenience:

$$\begin{aligned} \lambda ^{*}(h_{1})\equiv \frac{n(\sigma -1)(1-\phi )\phi (h_{1}n-1)-n\sigma \left[ h_{1}(1-\phi )+\phi \right] \left[ \phi (h_{1}+n-2)-h_{1}+1\right] \nu }{(\sigma -1)(1-\phi )^{2}(h_{1}n-1)}, \end{aligned}$$

where $\nu =\ln \left\{ \frac{\phi (h_{1}+n-2)-h_{1}+1}{\left( n-1\right) \left[ h_{1}(1-\phi )+\phi \right] }\right\}$. Let $h_{1}\in (0,1/n)\cup (1/n,1)$ to exclude equilibria other than partial agglomeration. Notice that $\nu >0$ for $h_{1}\in (0,1/n)$, while $\nu >0$ for $h_{1}\in (1/n,1)$.

• (i) Write $\lambda ^{*}(h_{1})=\left( \alpha +\beta \nu \right) /\gamma$, where:

  $$\begin{aligned} \alpha&=n(\sigma -1)(1-\phi )\phi (h_{1}n-1)\\ \beta&=-n\sigma \left[ h_{1}(1-\phi )+\phi \right] \left[ \phi (h_{1}+n-2)+1-h_{1}\right] \\ \gamma&=(\sigma -1)(1-\phi )^{2}(h_{1}n-1), \end{aligned}$$

  and observe that $\alpha /\gamma >0,\beta <0$ and $\nu /\gamma <0$ to see that $\lambda ^{*}(h_{1})>0$.

• (ii) Calculating the first and second derivatives of $\lambda ^{*}(h_{1})$ we obtain:

  $$\begin{aligned} \dfrac{\partial \lambda ^{*}(h_{1})}{\partial h_{1}}&=\frac{\alpha _{1}+\beta _{1}\nu }{(\sigma -1)(1-\phi )^{2}(h_{1}n-1)^{2}},\\ \dfrac{\partial ^{2}\lambda ^{*}(h_{1})}{\partial h_{1}^{2}}&=\frac{n\sigma \left[ (n-1)\phi +1\right] ^{2}\left( \alpha _{2}+\beta _{2}\nu \right) }{(\sigma -1)(1-\phi )^{2}(h_{1}n-1)^{3}\left[ h_{1}(1-\phi )+\phi \right] \left[ \phi (h_{1}+n-2)-h_{1}+1\right] }, \end{aligned}$$

  where

  $$\begin{aligned} \alpha _{1}&=-(1-\phi )(1-h_{1}n)\left[ (n-1)\phi +1\right] \\ \beta _{1}&=h_{1}^{2}n(1-\phi )^{2}-2h_{1}(1-\phi )^{2}+\phi \left\{ n\left[ (n-3)\phi +2\right] +3\phi -4\right\} +1\\ \alpha _{2}&=(1-\phi )(1-h_{1}n)\left[ (3-2n)\phi -h_{1}(n-2)(1-\phi )-1\right] \\ \beta _{2}&=-2(n-1)\left[ h_{1}(1-\phi )+\phi \right] \left[ \phi (h_{1}+n-2)-h_{1}+1\right] . \end{aligned}$$

The sign of the numerator of $\tfrac{\partial ^{2}\lambda ^{*}(h_{1})}{\partial h_{1}^{2}}$ is the sign of $F(h_{1})\equiv \alpha _{2}+\beta _{2}\nu$. Computing $F''(h_{1})$, we find that $F(1/n)=0$, $F'(1/n)=0$, and $F''(1/n)=0$, and that $F(h_{1})$ is concave for $h_{1}\in (0,1/n)$ and convex for $h_{1}\in (1/n,1)$. This implies that the numerator of $\tfrac{\partial ^{2}\lambda ^{*}(h_{1})}{\partial h_{1}^{2}}$ is negative for $h_{1}\in (0,1/n)$ and positive for $h_{1}\in (1/n,1)$.

Since the denominator of $\tfrac{\partial ^{2}\lambda ^{*}(h_{1})}{\partial h_{1}^{2}}$ is positive for $h_{1}\in (0,1/n)$ and negative for $h_{1}\in (1/n,1)$, we conclude that $\partial _{h}^{2}\lambda ^{*}(h_{1})<0$ which means that $\lambda ^{*}(h_{1})$ is strictly concave. Hence, at most two partial agglomeration equilibria exist.

Calculate the limits of $\lambda ^{*}(h_{1})$ and its first derivative as $h_{1}$ approaches 1 / n:

$$\begin{aligned} \lim _{h_{1}\rightarrow \tfrac{1}{n}\pm }\lambda ^{*}(h_{1})&=\frac{\sigma (2n\phi +1)-n\phi }{(\sigma -1)(1-\phi )}>0.\\ \lim _{h_{1}\rightarrow \tfrac{1}{n}\pm } \dfrac{\partial \lambda ^{*}(h_{1})}{\partial h_{1}}&=\frac{(n-2)n\sigma }{2(n-1)(\sigma -1)}>0. \end{aligned}$$

From the first limit we establish the continuity of $\lambda ^{*}(h_{1})$. The fact that the second limit is positive, together with the concavity of $\lambda ^{*}(h_{1})$, guarantees that $\lambda ^{*}(h_{1})$ is increasing for $h_{1}\in (0,1/n)$. Therefore, at most one equilibrium exists for $h_{1}\in (0,1/n)$. If there are two equilibria, both may belong to (1 / n, 1). $\square$

Proof of Theorem 2

Analogously to the proof of Proposition 3, the symmetry of the problem ensures that the Jacobian matrix at the partial agglomeration equilibrium with coordinates $\left( h_{1},\tfrac{1-h_{1}}{1-n},\ldots ,\tfrac{1-h_{1}}{1-n}\right)$ is of the form:

$$\begin{aligned} J=\begin{bmatrix}\dfrac{\partial f_{1}}{\partial h_{1}}&\quad 0&\quad \ldots&\quad 0\\ 0&\quad \dfrac{\partial f_{j}}{\partial h_{j}}&\quad \ldots&\quad 0\\ \vdots&\quad \vdots&\quad \ddots&\quad \vdots \\ 0&\quad 0&\quad \ldots&\quad \dfrac{\partial f_{j}}{\partial h_{j}} \end{bmatrix}. \end{aligned}$$

The eigenvalues of J are given by:

$$\begin{aligned} \dfrac{\partial f_{1}}{\partial h_{1}}&=h\left( \dfrac{\partial V_{1}}{\partial h_{1}}-\dfrac{\partial \bar{V}}{\partial h_{1}}\right) ,\ \text { and }\ \dfrac{\partial f_{j}}{\partial h_{j}}=\dfrac{1-h}{1-n}\,\dfrac{\partial V_{j}}{\partial h_{j}}, \end{aligned}$$

which must both be negative at partial agglomeration for this configuration to be stable. Using (24) to calculate these derivatives, the stability conditions become, respectively:

$$\begin{aligned} {\left\{ \begin{array}{ll} \gamma (h_{1})\equiv \frac{(1-\phi )(1-h_{1}n)-(n-1)\left[ h_{1}(1-\phi )+\phi \right] \nu }{(1-h_{1}n)} &{}<0\\ \delta (h_{1})\equiv \frac{(1-\phi )(h_{1}n-1)\left[ (n-1)\phi +1\right] +\varPhi \nu }{(h_{1}n-1)} &{} <0. \end{array}\right. } \end{aligned}$$

We finish the proof by showing that:

• (i) When $h_{1}\in (0,1/n)$ we have $\gamma (h_{1})>0$, thus partial agglomeration is unstable regardless of the sign of $\delta (h_{1})$. To verify this, notice that, for $h_{1}\in (0,1/n)$, the denominator of $\gamma (h_{1})$ is positive so the sign of $\gamma (h_{1})$ is that of the numerator. Call $N(h_{1})$ this numerator. Direct calculation shows that:

  $$\begin{aligned} \dfrac{\partial N(h_{1})}{\partial h_{1}}<0\ \Leftrightarrow \ -\left( 1-\phi \right) \left[ \frac{(1-\phi )(1-h_{1}n)}{\phi (h_{1}+n-2)+1-h_{1}}+(n-1)\nu \right] <0, \end{aligned}$$

  which is always true since $\nu >0$, that is, the numerator of $\gamma (h_{1})$ decreases in $h_{1}.$ We also have, noticing that $\nu (1/n)=0,$ $\lim _{h\rightarrow \tfrac{1}{n}}N(h_{1})=0.$ Then, $N(h_{1})>0$ and $\gamma (h_{1})>0$.

• (ii) When $h_{1}\in \left( 1/n,1\right)$ we have $\gamma (h_{1})<0$ so that only $\delta (h_{1})<0$ needs to be verified for stability. To verify this, notice that, for $h_{1}\in \left( 1/n,1\right)$, the denominator of $\gamma (h_{1})$ is negative. From proof of Theorem 2, we have $\lim _{h_{1}\rightarrow \tfrac{1}{n}}N(h_{1})=0$. Also, since $\nu <0$ for $h_{1}\in (1/n,1)$, we find that $\tfrac{\hbox {d}N(h_{1})}{\hbox {d}h_{1}}>0$. Therefore, the numerator of $\gamma (h_{1})$ is positive for $h_{1}\in (1/n,1)$ and we conclude that $\gamma (h_{1})<0$ if $h_{1}\in (1/n,1)$. $\square$

Appendix C

This appendix contains the formal proofs concerning Sect. 4. We denote by $f_{i}(h)$ the right-hand side of Eq. (17).

Proof of Proposition 5

The conditions required for a transcritical bifurcation (Guckenheimer and Holmes 2002; pp. 149–150) are as follows:

• (T1.) For all values of the bifurcation parameter $\phi$, we must have $f_{i}\left( \tfrac{1}{n},\ldots ,\tfrac{1}{n};\phi \right) =0$.

  This condition is satisfied since total dispersion is always an equilibrium.

• (T2.) The Jacobian of $f_{i}(h)$ has a zero eigenvalue at total dispersion. This occurs at the break point $\phi _{b}$ given in (21).

• (T3.) At total dispersion and at the break point we must have $\tfrac{\partial ^{2}f_{i}}{\partial h_{i}^{2}}\left( \tfrac{1}{n},\ldots ,\tfrac{1}{n};\phi _{b}\right) \ne 0.$

  The second derivative of $f_{i}$ with respect to $h_{i}$ at the symmetric equilibrium is given by:

  $$\begin{aligned}&\dfrac{\partial ^{2}f_{i}}{\partial h_{i}^{2}}\left( \tfrac{1}{n},\ldots ,\tfrac{1}{n}\right) \\&\quad =\frac{\mu (n-2)(1-\phi )\left\{ \phi ^{2}\left[ n\sigma (2\lambda +4n-3)-2n(\lambda +n)+\sigma \right] +\sigma \phi \left[ (3-2\lambda )n-2\right] +2\lambda n\phi +\sigma \right\} }{(\sigma -1)\sigma \left[ (n-1)\phi +1\right] ^{3}}. \end{aligned}$$

  At $\phi =\phi _{b}$, we have:

  $$\begin{aligned} \dfrac{\partial ^{2}f_{i}}{\partial h_{i}^{2}}\left( \tfrac{1}{n},\ldots ,\tfrac{1}{n};\phi _{b}\right) =\frac{\mu (n-2)(1-2\sigma )^{2}}{(\lambda +1)^{2}(\sigma -1)^{3}}, \end{aligned}$$

  which is positive for $n\ge 3.$

• (T4.) At total dispersion and at the break point we must have $\tfrac{\partial ^{2}f_{i}}{\partial h_{i}\partial \phi }\left( \tfrac{1}{n},\ldots ,\tfrac{1}{n};\phi _{b}\right) \ne 0.$

Again, direct computation yields:

$$\begin{aligned} \dfrac{\partial ^{2}f_{i}}{\partial h_{i}\partial \phi }\left( \tfrac{1}{n},\ldots , \tfrac{1}{n};\phi _{b}\right) =\frac{\mu (2\sigma -1) \left[ \lambda -\sigma (\lambda +2n)+n+\sigma \right] ^{2}}{(\lambda +1)^{2}n(\sigma -1)^{3}\sigma }>0. \end{aligned}$$

Since all conditions are verified, we conclude that the model undergoes a transcritical bifurcation at the break point $\phi _{b}$. $\square$

Proof of Proposition 6

A primary branch satisfies $\lambda ^{*}(h_{1})$ in Eq. (24). We use the conditions for a saddle-node bifurcation given by Guckenheimer and Holmes (2002, Theorem 3.4.1). Applied to the QL model, they are as follows:

(SN1.) At partial agglomeration we must have $\dfrac{\hbox {d}f}{\hbox {d}h}\left( h_{1};\lambda ^{*}(h_{1});\phi _{f}\right) =0$.

In this instance, $f(h_{i})$ is the RHS of (17) and the proof of Theorem 10 gives:

$$\begin{aligned} \dfrac{\hbox {d}f}{\hbox {d}h_{1}}\left( h_{1};\lambda ^{*}(h_{1}); \phi _{f}\right)&=0\ \Leftrightarrow \\ \delta&=0, \end{aligned}$$

where $\delta$ is as in (25). We rewrite $\delta =0$ as:

$$\begin{aligned} \frac{(1-\phi _{f})(1-h_{1}n)\left[ (n-1)\phi _{f}+1\right] }{\varPhi (h_{1};\phi _{f})}=\nu (h_{1};\phi _{f}), \end{aligned}$$

(32)

where

$$\begin{aligned} \varPhi (h_{1},\phi _{f})&=h_{1}^{2}n(1-\phi _{f})^{2}-2h_{1}(1-\phi _{f})^{2}+\phi _{f}\left\{ n\left[ (n-3)\phi _{f}+2\right] \right. \\&\quad \left. +\,3\phi _{f}-4\right\} +1,\\ \nu (h_{1},\phi _{f})&=\ln \left\{ \frac{\phi _{f}(h_{1}+n-2)-h_{1}+1}{\left( n-1\right) \left[ h_{1}(1-\phi _{f})+\phi _{f}\right] }\right\} , \end{aligned}$$

and $\phi _{f}$ is the level of freeness of trade at which the interior equilibrium changes stability.

(SN2.) At partial agglomeration, $\dfrac{d^{2}f}{\hbox {d}h^{2}}\left( h_{1};\lambda ^{*}(h_{1}); \phi _{f}\right) \ne 0$.

From (24) and (32), we have:

$$\begin{aligned} \dfrac{d^{2}f}{\hbox {d}h^{2}}\left( h_{1};\lambda ^{*}(h_{1});\phi _{f}\right) =&\frac{(h_{1}-1)h_{1}\mu (1-\phi )^{2} \left[ (n-1)\phi +1\right] ^{2}\Gamma }{(\sigma -1)\left[ h_{1}(1-\phi )+\phi \right] ^{2}\left[ \phi (h_{1}+n-2)-h_{1}+1\right] ^{2}\varPhi }, \end{aligned}$$

where $\Gamma (h,\phi )=h_{1}^{2}(n-2)(1-\phi )^{2} +2h_{1}(1-\phi )\left[ (2n-3)\phi +1\right] -\phi \left\{ n\left[ (n-5)\phi +2\right] +5\phi -4\right\} -1$. The term $\varPhi$ is positive. The term $\Gamma (h_{1},\phi )$ has only one (meaningful) zero given by:

$$\begin{aligned} h_{1}=h_{1}^{*}\equiv -\frac{\phi (2n(1-\phi ) +3\phi -4)-\sqrt{n-1}(1-\phi )\left[ (n-1)\phi +1\right] +1}{(n-2)(1-\phi )^{2}}, \end{aligned}$$

which is not compatible with (SN1). By replacing $h_{1}=h_{1}^{*}$ in (25) we obtain:

$$\begin{aligned} \dfrac{\hbox {d}f}{\hbox {d}h_{1}}\left( h_{1}^{*};\lambda ^{*}(h_{1}); \phi _{f}\right) =\delta (h_{1}^{*})\equiv -&\dfrac{\left[ (n-1)\phi +1\right] {}^{2}}{(n-2)^{2}}\Xi , \end{aligned}$$

where

$$\begin{aligned} \Xi =&(n-2)\left[ \left( \sqrt{n-1}+2\right) n-2\right] +(n-1)\left( n+2\sqrt{n-1}\right) \ln (n-1). \end{aligned}$$

Since $\Xi >0,$ it follows that $\delta (h_{1}^{*})<0$. Thus $\tfrac{d^{2}f}{\hbox {d}h_{1}^{2}}\left( h_{1};\lambda ^{*} (h_{1});\phi _{f}\right) \ne 0$.

(SN3.) At partial agglomeration, $\dfrac{\hbox {d}f}{d\phi }\left( h_{1};\lambda ^{*}(h_{1});\phi _{f}\right) \ne 0$.

From (24) and (32), we have:

$$\begin{aligned} \dfrac{\hbox {d}f}{d\phi }\left( h_{1};\lambda ^{*}(h_{1});\phi _{f}\right)&=\frac{(1-h_{1})h_{1}\mu (h_{1}n-1)\times \Theta }{(\sigma -1)\sigma \left[ h_{1}(\phi -1)-\phi \right] \left[ \phi (h_{1}+n-2)-h_{1}+1\right] \times \varPhi }, \end{aligned}$$

where $\Theta =h_{1}^{2}n(1-\phi )^{2}-h_{1}(1-\phi )\left\{ (n-2)\sigma \left[ (n-1)\phi +1\right] -2\phi +2\right\} -\sigma \left[ (n-1)\phi +1\right] (2n\phi -3\phi +1)+\phi \left\{ n\left[ (n-3)\phi +2\right] +3\phi -4\right\} +1<0.$ Since the term $h_{1}n-1>0$ for $h_{1}\in (1/n,1)$, we can conclude that $\hbox {d}f/\hbox {d}\phi >0$ when evaluated at partial agglomeration and at $\phi _{f}$, ensuring that (SN3) is satisfied. This concludes the proof. $\square$

Proof of Proposition 7

We know that total dispersion is stable if $\lambda >\lambda _{b}$, whereas agglomeration is stable if $\lambda <\lambda _{s}$. As a result, both equilibria are simultaneously stable if $\left( \lambda _{b},\lambda _{s}\right)$ is non-empty. Using (20) and (23), simultaneity of stability then requires $\lambda _{s}-\lambda _{b}>0:$

$$\begin{aligned} \frac{n\phi \left[ (\sigma -1)(1-\phi )-\sigma \ln \phi \right] }{(\sigma -1)(\phi -1)^{2}}-\frac{\sigma (2n\phi +1-\phi )-n\phi }{(\sigma -1)(1-\phi )}&>0\ \Leftrightarrow \\ -\frac{(1-\phi )\left[ (n-1)\phi +1\right] +n\phi \log (\phi )}{(\sigma -1)(1-\phi )^{2}}&>0\ \Leftrightarrow \\ -\frac{(1-\phi )^{2}}{\phi (1-\phi +\ln \phi )}&<n, \end{aligned}$$

which concludes the proof. $\square$

Appendix D

This appendix contains the formal proofs concerning Sect. 5.

Proof of Theorem 3

The weighted average utility of entrepreneurs, obtained from (15), is given by:

$$\begin{aligned} \bar{V}=\dfrac{\mu }{\sigma }\left( 1+\lambda \right) +\dfrac{\mu }{\sigma -1}\sum _{i=1}^{n}h_{i}\ln \left[ \phi +(1-\phi )h_{i}\right] +\eta . \end{aligned}$$

Define $f(h_{i})\equiv h_{i}\ln \left[ \phi +(1-\phi )h_{i}\right]$. Observe that $f:[0,1]\rightarrow \mathbb {R}$ is continuous and twice differentiable, and that:

$$\begin{aligned} f^{\prime \prime }(h_{i})=\dfrac{2\left( 1 -\phi \right) \left[ \phi +\left( 1-\phi \right) h_{i} \right] -h_{i}\left( 1-\phi \right) ^{2}}{\left[ \phi +\left( 1-\phi \right) h_{i}\right] ^{2}}=\dfrac{h_{i} \left( 1-\phi \right) ^{2}+2\phi \left( 1-\phi \right) }{\left[ \phi +\left( 1-\phi \right) h_{i}\right] ^{2}}\,{>}\,0, \end{aligned}$$

which means that f is strictly convex.

Now define $g(h_{1},\ldots ,h_{n-1})\equiv f(1-\sum _{i=1}^{n-1}h_{i})$. The function $g:\mathbb {R}^{n-1}\rightarrow \mathbb {R}$ is convex because it is a composition of a convex function with an affine function.

Therefore, $\sum _{i=1}^{n-1}f(h_{i})+g(h)$ is convex as it is a sum of convex functions. We conclude that $\bar{V}:\mathbb {R}^{n-1}\rightarrow \mathbb {R}$ is a convex function of $(h_{1},\ldots ,h_{n-1})$. Since $\partial _{h_{i}}\bar{V}\left( \tfrac{1}{n},\ldots ,\tfrac{1}{n}\right) =0$ (see Proof of Proposition 3) and $\bar{V}$ is convex, it attains a global minimum at dispersion. $\square$

Proof of Theorem 4

Let $f(h_{i})\equiv \ln [\phi +(1-\phi )h_{i}]$. Then:

$$\begin{aligned} f^{\prime \prime }(h_{i})=-\frac{(1-\phi )^{2}}{\left[ h(1-\phi ) +\phi \right] {}^{2}}<0, \end{aligned}$$

which means that $f(h_{i})$ is strictly concave. Therefore, $\sum _{i=1}^{n-1}\ln [\phi +(1-\phi )h_{i}]$ is a strictly concave function of $(h_{1},\ldots ,h_{n-1})$.

Now define $g(h_{1},\ldots ,h_{n-1})\equiv f(1-\sum _{i=1}^{n-1}h_{i})$. The function $g:\mathbb {R}^{n-1}\rightarrow \mathbb {R}$ is concave because f is a concave monotonic transformation of a concave function of $(h_{1},\ldots ,h_{n-1})$. This implies that $\bar{V}^{L}$, given in (28), is strictly concave.

Each price index $P_{i}$ in (13) is invariant to the permutation of any two region’s coordinates. Therefore, we can assert that $\partial _{h_{i}}\bar{V}^{L}\left( \tfrac{1}{n},\ldots , \tfrac{1}{n}\right) =0$, $\forall i\in N$.²⁶ Given strict concavity, $\bar{V}^{L}$ attains a global maximum at $h=(1/n,\ldots ,1/n)$, which concludes the proof. $\square$

Proof of Proposition 8

Rewrite the social welfare function $\varOmega (h)$ in (30) as:

$$\begin{aligned} \varOmega (h)=\tfrac{\epsilon }{\lambda +1}+\tfrac{\mu }{(\lambda +1)(\sigma -1)}\left[ g(h_{1})+\cdots +g(h_{n})\right] , \end{aligned}$$

(33)

where $g(h_{i})\equiv \left( \tfrac{\lambda }{n} +h_{i}\right) \ln \left[ \phi +(1-\phi )h_{i}\right]$.

The optimization plan for $\varOmega (h)$ consists on maximizing $\sum _{i}g(h_{1})$ subject to $\sum _{j=1}^{n}h_{i}=1$. Write the Lagrangian as $\mathcal {L}=g(h_{1})+g(h_{2})+\cdots +g(h_{n})+\gamma (1-h_{1}-\cdots -h_{n})$, where $\gamma$ is the Lagrange multiplier. From the first-order conditions:

$$\begin{aligned} g^{\prime }(h_{1})=g^{\prime }(h_{2})=\cdots =\gamma . \end{aligned}$$

We must have $g^{\prime }(h_{1})=g^{\prime }(h_{2})=\cdots =g^{\prime }(h_{n})$. Each $g^{\prime }(h_{i})$ is given by:

$$\begin{aligned} g^{\prime }(h_{i})=\frac{(1-\phi )(\tfrac{\lambda }{n} +h_{i})}{h_{i}(1-\phi )+\phi }+\ln \left[ h_{i}(1-\phi )+\phi \right] . \end{aligned}$$

The second derivative $g^{\prime \prime }(h_{i})$ is given by:

$$\begin{aligned} g^{\prime \prime }(h_{i})=-\frac{(1-\phi ) \left[ \lambda -h_{i}n(1-\phi )-\phi (\lambda +2n)\right] }{n\left[ h_{i}(1-\phi )+\phi \right] ^{2}}, \end{aligned}$$

which has either one zero for $h\in [0,1]$ or none. Therefore, $g^{\prime }(h_{i})$ has at most one local extreme. This implies that at most two different values of $h_{i}\in [0,1]$ may satisfy $g^{\prime }(h_{i})=\gamma$. The consequence of this is that all potential interior maximizers of $\varOmega (h)$ are characterized by a vector $h=(h_{1},h_{2},\ldots ,h_{n})$ such that k of its elements correspond to a share of entrepreneurs equal to h / k and the remaining $n-k$ elements have a share equal to $(1-h)/(n-k)$. This concludes the proof. $\square$

Proof of Theorem 5

• (i) Define the summation term of $\varOmega (h)$ in (30) as $F(h)=f(h_{1})+f(h_{2})+\cdots +f(h_{n-1})+g(h_{n}),$ where $h_{n}:(h_{1},h_{2},\ldots ,h_{n-1})\mapsto 1-h_{1}-\cdots -h_{n-1}$. The second derivative of each $f(h_{i})$ evaluated at $h_{i}=1/n$ is given by:

  $$\begin{aligned} \left. \dfrac{\partial ^{2}f(h_{i})}{\partial ^{2}h_{i}} \right| _{h_{i}=\tfrac{1}{n}}=\frac{n(1-\phi ) \left[ \phi (\lambda +2n-1)-\lambda +1\right] }{\left[ (n-1) \phi +1\right] {}^{2}}, \end{aligned}$$

  which is negative if and only if:

  $$\begin{aligned} \phi <\phi _{w}\equiv \frac{\lambda -1}{\lambda +2n-1}. \end{aligned}$$

  Using (21), it is easily verified that $\phi _{b}<\phi _{w}$. If $\phi<\phi _{b}<\phi _{w},$ symmetric dispersion is stable and $f(h_{i})$ is concave. Given that $f:\mathbb {R\mapsto R}$ is strictly concave, replicating the reasoning from the proof of Theorem 3 allows us to conclude that $g(h_{n})=f\circ h_{n}:\mathbb {R}^{n-1}\mapsto \mathbb {R}$ is also strictly concave. Therefore, F(h) is strictly concave for $h=(1/n,\ldots ,1/n)$ and $\varOmega (h),$ a constant term plus F(h),  is also strictly concave at symmetric dispersion when the latter is stable. Since $\varOmega (h)$ attains a critical value at symmetric dispersion, we conclude that the latter always attains a local maximum when it is stable.

  Evaluating welfare at symmetric dispersion gives us:

  $$\begin{aligned} \varOmega \left( \dfrac{1}{n},\ldots ,\dfrac{1}{n}\right) =\dfrac{1}{\lambda +1}\left[ \varepsilon +\dfrac{\mu (\lambda +1)}{(\sigma -1)} \ln \left( \phi +\dfrac{1-\phi }{n}\right) \right] . \end{aligned}$$

  At agglomeration, welfare is given by:

  $$\begin{aligned} \varOmega (h_{i}=1)=\dfrac{1}{\lambda +1}\left[ \varepsilon +\dfrac{\mu \lambda (n-1)}{(\sigma -1)n}\ln \phi \right] . \end{aligned}$$

  This implies that agglomeration yields a higher welfare than dispersion if and only if:

  $$\begin{aligned} \Delta \varOmega \equiv \dfrac{\lambda (n-1)}{n}\ln \phi -(\lambda +1)\ln \left( \phi +\dfrac{1-\phi }{n}\right) >0. \end{aligned}$$

  The difference $\Delta \varOmega$ is concave in $\phi ,$ has a zero for $\phi \in (0,1)$ and another at $\phi =1$, and is negative at $\phi =\phi _{b}$. Symmetric dispersion is thus strictly better than agglomeration from a social point of view when the former is stable.

• (ii) It can be shown that $\varOmega ^{\prime }(1)$ is concave in $\phi$ with only one root $\phi _{z}\in (0,1)$ and another at $\phi =1$. Moreover, it is negative when evaluated at the sustain point $\phi _{s}$, which implies that $\phi _{s}<\phi _{z}$. Therefore, there exists a $\phi \in (\phi _{s},\phi _{z})$ where agglomeration is stable and $\varOmega ^{\prime }(1)<0$, meaning that welfare is higher at another less asymmetric distribution. This concludes the proof. $\square$