Pointwise agglomeration in continuous racetrack model

Abstract

Using analytical methods they devised, the authors intend to study economic aspects of the racetrack economy described as a continuous-space version of the new economic geography model. The main mathematical conclusions are summarized as follows: 1) Workers and firms agglomerate only to a finite number of cities, and the maximum number of possible emerging cities reduces as the preference for variety increases or the transport cost decreases. 2) The further apart cities are located from each other, the more stable the city configuration is against the increasing preference for variety or the decreasing of the transport cost.

Appendices

Appendices

It might be meaningful to supplement analytical comments for readers interested in knowing more about the arguments of our analysis.

1.1 Mathematical formulation

The measure space $\mathcal M(S)$ is isomorphic to the space of all continuous linear functionals on the continuous function space $\mathcal C(S)$ on S. The duality product of $m\in \mathcal M(S)$ and $f\in \mathcal C(S)$ is denoted by

$$\begin{aligned} \langle m,f \rangle =\int _S f(x)dm(x). \end{aligned}$$

Equipped with the norm

$$\vert\vert m\vert\vert_{\mathcal M}=\sup_{\vert\vert f\vert\vert_{\mathcal C}=1}\vert\langle m,f\rangle$$

(10)

$\mathcal M(S)$ becomes a Banach space. In the definition (3) of $\mathcal M_P(S)$, $m\ge 0$ means that $\langle m,f\rangle \ge 0$ holds true for any $f \in \mathcal C_+(S)$. For a positive measure $m>0$, its norm $\Vert m\Vert _{\mathcal M}$ is calculated by

$$\begin{aligned} \Vert m\Vert _{\mathcal M} = \langle m,1\rangle = \int _S dm(x), \end{aligned}$$

(11)

for $\left| \langle m, f \rangle \right| \le \langle m,1 \rangle$ holds for any $f\in \mathcal C(S)$ satisfying $\Vert f\Vert _{\mathcal C} = 1$ due to $m \ge 0$.

As for the decomposition (12), we refer to Brezis (2010, pp. 115-116) or Dieudonné (1976, Chapter XIII).

1.2 $\omega$-limit set

Weak${}^*$ limit

Let X be a Banach space, and let $X^*$ be its dual space. It is said that a sequence $\left\{ f_n\right\}$ of $X^*$ is convergent to a weak${}^*$ limit $f\in X^*$ if $\left\{ f_n\right\}$ satisfies that

$$\begin{aligned} \text {as} n \rightarrow \infty ,\langle f_n,x\rangle \rightarrow \langle f,x\rangle ,~\forall x \in X. \end{aligned}$$

Additionally, it is written that

$$\begin{aligned} \text {as}n \rightarrow \infty , f_n \overset{*}{\rightharpoonup } f. \end{aligned}$$

Confer (Brezis 2010, Subsection 3.4).

Outline of proof of Theorem 1

Let X be a separable Banach space, and let $X^*$ be its dual space. For any bounded sequence $\left\{ x_n\right\}$ of $\mathcal X^*$, one can choose a subsequence $\left\{ x_{n_k}\right\}$ that is convergent in the weak${}^*$ topology (see Brezis (2010, Corollary 3.30)). At the same time, it is well known that $\mathcal C(S)$ is a separable Banach space (see (Dieudonné 1976, Subsection 7.3.1)); therefore, this general result is applicable to the present case. As a result, one can choose a sequence $\left\{ \Lambda (t_n;\Lambda _0)\right\}$ for any initial distribusion $\Lambda _0\in \mathcal M_P(S)$ that converges to an element $\overline{\Lambda } \in \mathcal M(S)$ in the weak${}^*$ topology.

It is easy to see that $\overline{\Lambda }\ge 0$, because $\Lambda (t_n;\Lambda _0)\ge 0$ holds for any n. $\Vert \overline{\Lambda }\Vert _{\mathcal M}=1$ is also verified using $\Vert \overline{\Lambda }\Vert _{\mathcal M}=\langle \overline{\Lambda },1\rangle$ (see Equation (11)). Therefore, the limit $\overline{\Lambda }$ lies in $\mathcal M_P(S)$. Hence, the theorem is verified.

1.3 Heuristic arguments

For simplicity, we assume throughout this subsection that $\mu =1$.

Let $(\overline{w}(x),\overline{G}(x),\overline{\omega }(x),\overline{\Lambda }(x))$ be any stationary solution. According to the Lebesgue decomposition theorem (Dieudonné 1976, Theorem 13.18.4), $\overline{\Lambda }(x) \in \mathcal M_P(S)$ can be decomposed into a sum of an absolutely continuous part and a singular part, say, into the form

$$\begin{aligned} \overline{\Lambda }(x) = \overline{\lambda }(x)dx + \sum _{i=1}^M \overline{\Lambda }_i\delta _i(x). \end{aligned}$$

(12)

Here, $\overline{\lambda }(x)$ denotes an $L^1$-function on S, and $\delta _i(x)$ is a Dirac delta function with center $x_i \in S$ for $i=1,2,3,\ldots ,M$. We will here show by somewhat heuristic arguments that either $\overline{\Lambda }(x)=\overline{\lambda }(x)=(2\pi )^{-1}dx$ or $\overline{\Lambda }(x)=\sum _{i=1}^M \overline{\Lambda }_i\delta _i(x)$.

In view of (12), put

$$\begin{aligned} \overline{\omega } = \int _S \overline{\omega }(y)d\overline{\Lambda }(y) = \int _S \overline{\omega }(y)\overline{\lambda }(y)dy + \sum _{i=1}^M \overline{\Lambda }_i\overline{\omega }(x_i), \end{aligned}$$

and consider the subset $S_{\overline{\omega }} = \{x \in S;\; \overline{\omega }(x) = \overline{\omega }\}$ of S.

If $x \not \in S_{\overline{\omega }}$, then $\overline{\omega }(x)-\overline{\omega }\not =0$; consequently, it must hold that $\overline{\Lambda (x)}=0$. Hence, $\overline{\Lambda }(x)=0$ for any $x \not \in S_{\overline{\omega }}$. Meanwhile, if $x \in S_{\overline{\omega }}$, then it must hold that $\overline{w}(x)=\overline{\omega }\overline{G}(x)$ (due to $\mu =1$). Based on these facts, the two integral equations in (1) are rewritten on $S_{\overline{\omega }}$ into

$$\begin{aligned} \overline{G}(x)^\sigma=& \ \overline{\omega }^{1-\sigma } \int _{S_{\overline{\omega }}} \overline{G}(y)^\sigma e^{-\alpha |x-y|}d\overline{\Lambda }(y) \nonumber \\=& \ \overline{\omega }^{1-\sigma } \Big [\int _{S_{\overline{\omega }}} \overline{\lambda }(y)\overline{G}(y)^\sigma e^{-\alpha |x-y|}dy \nonumber \\ &+ \sum _{i=1}^M \overline{\Lambda }_i\overline{G}(x_i)^\sigma e^{-\alpha |x-x_i|}\Big ], \qquad x \in S_{\overline{\omega }}, \end{aligned}$$

(13)

and

$$\begin{aligned} \overline{G}(x)^{1-\sigma }=& \ \overline{\omega }^{1-\sigma } \int _{S_{\overline{\omega }}} \overline{G}(y)^{1-\sigma } e^{-\alpha |x-y|}d\overline{\Lambda }(y) \\=& \ \overline{\omega }^{1-\sigma } \Big [ \int _{S_{\overline{\omega }}}\overline{\lambda }(y)\overline{G}(y)^{1-\sigma } e^{-\alpha |x-y|}dy \\& + \sum _{i=1}^M \overline{\Lambda }_i\overline{G}(x_i)^{1-\sigma } e^{-\alpha |x-x_i|}\Big ], \qquad x \in S_{\overline{\omega }}, \end{aligned}$$

respectively. Therefore, $\overline{G}(x)^\sigma$ and $\overline{G}(x)^{1-\sigma }$ are seen to satisfy the same integral equation

$$\begin{aligned} u(x) =& \ \overline{\omega }^{1-\sigma } \Big [ \int _{S_{\overline{\omega }}} \overline{\lambda }(y)u(y) e^{-\alpha |x-y|}dy \\ &+ \sum _{i} \overline{\Lambda }_i u(x_i)e^{-\alpha |x-x_i|}\Big ], \qquad x \in S_{\overline{\omega }}. \end{aligned}$$

Assume that this integral equation has a unique solution¹. Then, we have $\overline{G}(x)^\sigma = \mathrm{const.}\overline{G}(x)^{1-\sigma }$ for $x \in S_{\overline{\omega }}$, which means that $\overline{G}(x)$ is constant on $S_{\overline{\omega }}$. It then follows from (13) that

$$\begin{aligned} \int _{S_{\overline{\omega }}} \overline{\lambda }(y) e^{-\alpha |x-y|}dy + \sum _{i=1}^M \overline{\Lambda }_i e^{-\alpha |x-x_i|} \equiv \overline{\omega }^{\sigma -1}, \qquad x \in S_{\overline{\omega }}. \end{aligned}$$

Moreover, it may be allowed to consider that both

$$\begin{aligned} \int _{S_{\overline{\omega }}} \overline{\lambda }(y) e^{-\alpha |x-y|}dy \quad \text {and}\quad \sum _{i=1}^M \overline{\Lambda }_i e^{-\alpha |x-x_i|} \end{aligned}$$

are constant on $S_{\overline{\omega }}$, i.e.,

$$\begin{aligned} \int _{S_{\overline{\omega }}} \overline{\lambda }(y) e^{-\alpha |x-y|}dy \equiv \overline{\omega }_1, \qquad x \in S_{\overline{\omega }} \end{aligned}$$

(14)

and

$$\begin{aligned} \sum _{i=1}^M \overline{\Lambda }_i e^{-\alpha |x-x_i|} \equiv \overline{\omega }_2, \qquad x \in S_{\overline{\omega }}. \end{aligned}$$

(15)

With the aid of (15), we can claim that either

$$\begin{aligned} S_{\overline{\omega }} = \{x_1,\, x_2,\, x_3,\ldots ,\, x_M\} \end{aligned}$$

(16)

is the case or $\overline{\Lambda }(x)$ has no singular points (i.e., $M=0$). In fact, consider the case where $\overline{\Lambda }(x)$ has at least one singular point. We already know that $\{x_1,x_2.x_3,\ldots ,x_M\} \subset S_{\overline{\omega }}$ because of $\overline{\Lambda }(x_i)=\infty$. Therefore it holds that

$$\begin{aligned} \sum _{i=1}^M \overline{\Lambda }_i e^{-\alpha |x_j-x_i|} = \overline{\omega }_2 \quad \text {for all }\; j=1,2,3,\ldots ,M. \end{aligned}$$

This means that the weights $\overline{\Lambda }_i$ are determined by the positions of $x_i$ and the constant $\overline{\omega }_2$ by means of M-linear equations concerning the $\overline{\Lambda }_i$. Suppose that there exists a point $x_{M+1} \in S_{\overline{\omega }}-\{x_1,x_2,x_3,\ldots ,x_M\}$. It then follows from (15) that

$$\begin{aligned} \sum _{i=1}^M \overline{\Lambda }_i e^{-\alpha |x_{M+1}-x_i|} = \overline{\omega }_2. \end{aligned}$$

This means that, as $\overline{\Lambda }_i$ must satisfy another linear equation, it is overdetemined for $\overline{\Lambda }_i$. Hence, (16) must be the case.

Furthermore, if $\overline{\Lambda }(x)$ has at least one singular point, then (16) yields that $\overline{\Lambda }(x)=0$ for $x \not = x_i\, (1 \le i \le M)$. Consequently, we have $\overline{\lambda }(x)=0$ for all $x \in S$. Hence, the stationary solution $(\overline{w}(x),\overline{G}(x),\overline{\omega }(x),\overline{\Lambda }(x))$ must be singular.

In the meantime, consider the case that $\overline{\Lambda }(x)$ has no sigular points — i.e., $\overline{\Lambda }(x) =\overline{\lambda }(x)dx$ with some $L^1$-function $\overline{\lambda }(x)$. Then, (14) implies that the function $\varphi (x) = \int _{S_{\overline{\omega }}} \overline{\lambda }(y) e^{-\alpha |x-y|}dy$ is constant for $x \in S_{\overline{\omega }}$. Writing as $y=(\cos \eta ,\sin \eta )$ with $-\pi \le \eta \le \pi$, express $\overline{\lambda }(y)$ by the Fourier expansion

$$\begin{aligned} \overline{\lambda }(y) = \sum _{-\infty< n < \infty } \overline{\lambda }_n e^{in\eta }, \qquad -\pi \le \eta \le \pi . \end{aligned}$$

Then, writing as $x=(\cos \xi ,\sin \xi )$ with $-\pi \le \xi \le \pi$, $\varphi (x)$ can be expressed as

$$\begin{aligned} \varphi (x)&= \int _S \overline{\lambda }(y) e^{-\alpha |x-y|}dy = \sum _{-\infty< n< \infty } \overline{\lambda }_n \int _{-\pi }^\pi e^{in\eta }e^{-\alpha |\xi - \eta |}d\eta \\&= \sum _{-\infty< n < \infty } \overline{\lambda }_n e^{in\xi } \int _{-\pi }^{\pi } e^{in\eta '}e^{-\alpha |\eta '|}d\eta ', \end{aligned}$$

where $\xi -\eta =-\eta '$. From this expression, we notice that $\varphi (x)$ can be constant on $S_{\overline{\omega }}$ having a positive measure only when $\overline{\lambda }_n = 0$ for all $n = \pm 1,\,\pm 2,\,\pm 3,\,\ldots$ except $n=0$. Hence, it must hold that $\overline{\lambda }(x) = \overline{\lambda }_0$ is constant for $x \in S$.

1.4 Calculation of the linearization matrix $J^\prime$

The calculation of $J'$ is carried out as follows (For the details, see Ohtake and Yagi (2019, Section 6). Let A, B, C, D, E, and F be $M \times M$ matrices given by

$$\begin{aligned}A_{ij} =& \ \frac{\mu }{\sigma }\overline{w}_i^{1-\sigma } \frac{\overline{\Lambda }_j e^{-\alpha |x_i-x_j|}}{\sum _k\overline{\Lambda }_k\overline{w}_k^{1-\sigma } e^{-\alpha |x_j-x_k|}}\nonumber \\&+\frac{\mu (\sigma -1)}{\sigma }\overline{w}_i^{1-\sigma }\overline{\Lambda }_j\overline{w}_j^{-\sigma }\sum _s \frac{\overline{\Lambda }_s\overline{w}_s e^{-\alpha |x_s-x_j|} e^{-\alpha |x_i-x_s|}}{\left[ \sum _k\overline{\Lambda }_k\overline{w}_k^{1-\sigma } e^{-\alpha |x_s-x_k|}\right] ^2}\end{aligned}$$

(17)

$$\begin{aligned}&+\frac{(1-\mu )(\sigma -1)}{2\pi \sigma }\overline{w}_i^{1-\sigma }\overline{\Lambda }_j\overline{w}_j^{-\sigma }\int _S \frac{e^{-\alpha |y-x_i|} e^{-\alpha |y-x_j|}}{\left[ \sum _k\overline{\Lambda }_k\overline{w}_k^{1-\sigma } e^{-\alpha |y-x_k|}\right] ^2}dy, \nonumber \\B_{ij} =&\ \frac{\mu }{\sigma }\overline{w}_i^{1-\sigma } \frac{\overline{w}_j e^{-\alpha |x_i-x_j|}}{\sum _k\overline{\Lambda }_k\overline{w}_k^{1-\sigma } e^{-\alpha |x_j-x_k|}}\nonumber \\&-\frac{\mu }{\sigma }\overline{w}_i^{1-\sigma }\sum _s \frac{\overline{\Lambda }_s\overline{w}_s\overline{w}_j^{1-\sigma } e^{-\alpha |x_s-x_j|} e^{-\alpha |x_i-x_s|}}{\left[ \sum _k\overline{\Lambda }_k\overline{w}_k^{1-\sigma } e^{-\alpha |x_s-x_k|}\right] ^2}\end{aligned}$$

(18)

$$\begin{aligned}&-\frac{1-\mu }{2\pi \sigma }\overline{w}_i^{1-\sigma }\overline{w}_j^{1-\sigma }\int _S \frac{e^{-\alpha |y-x_i|} e^{-\alpha |y-x_j|}}{\left[ \sum _k\overline{\Lambda }_k\overline{w}_k^{1-\sigma } e^{-\alpha |y-x_k|}\right] ^2}dy,\nonumber \\&C_{ij} =\overline{G}^{\sigma }_i\overline{\Lambda }_j\overline{w}_j^{-\sigma } e^{-\alpha |x_i-x_j|}, D_{ij} =-\frac{\overline{G}^{\sigma }_i}{\sigma -1}\overline{w}_j^{1-\sigma } e^{-\alpha |x_i-x_j|},\end{aligned}$$

(19)

$$\begin{aligned}&E=\text {diag}(\overline{G}^{-\mu }_1,\cdots ,\overline{G}^{-\mu }_M), F=-\mu \text {diag}(\overline{w}_1\overline{G}^{-\mu -1}_1,\cdots ,\overline{w}_M\overline{G}^{-\mu -1}_M), \end{aligned}$$

(20)

respectively. Furthermore, let K, L, and R be given by

$$\begin{aligned} K = \left( \begin{array}{ccc} \overline{\Lambda }_1 &{} \cdots &{} \overline{\Lambda }_M \\ \overline{\Lambda }_1 &{} \cdots &{} \overline{\Lambda }_M \\ \vdots &{} \vdots &{} \vdots \\ \overline{\Lambda }_1 &{} \cdots &{} \overline{\Lambda }_M \end{array}\right) ,~ L=\text {diag}(\overline{\Lambda }_1,\cdots ,\overline{\Lambda }_M),~ R=\overline{\omega }\left( \begin{array}{ccc} 1 &{} \cdots &{} 1 \\ 1 &{} \cdots &{} 1 \\ \vdots &{} \vdots &{} \vdots \\ 1 &{} \cdots &{} 1 \end{array}\right) , \end{aligned}$$

where $\overline{\omega }=\sum _{i=1}^M\overline{w}_i\overline{G}_i^{-\mu }\overline{\Lambda }_i$. Using these, we introduce the matrices

$$\begin{aligned} \Omega =E(I-A)^{-1}B+F\left\{ C(I-A)^{-1}B+D\right\} , \end{aligned}$$

and

$$\begin{aligned} J=L\left[ (I-K)\Omega -R\right] , \end{aligned}$$

I being the $M\times M$ identity matrix. Then, the matrix $J^\prime$ can be expressed by

$$\begin{aligned} J^\prime = P_1JP_2, \end{aligned}$$

where $P_1\in \mathbb {R}^{(M-1)\times M}$ and $P_2\in \mathbb {R}^{M\times (M-1)}$ are such that

$$\begin{aligned} P_1=\left( \begin{array}{ccccc} 1&{}0&{}\cdots &{}0&{}0\\ 0&{}1&{}0&{}0&{}0\\ \vdots &{}&{}\ddots &{}&{}\vdots \\ 0&{}0&{}\cdots &{}1&{}0 \end{array}\right) \text {~~and~~} P_2= \left( \begin{array}{ccccc} 1&{}0&{}\cdots &{}0&{}0\\ 0&{}1&{}0&{}\cdots &{}0\\ \vdots &{}&{}\ddots &{}&{}\\ 0&{}0&{}\cdots &{}0&{}1\\ -1&{}-1&{}\cdots &{}-1&{}-1 \end{array}\right) . \end{aligned}$$

Let us assume that $M=2$ or 3, $\overline{\Lambda }_i=1/M~(i=1,2,\cdots ,M)$, and the distance between any adjacent centers $x_i$ and $x_j$ on S is d — that is, $d\in (0,\pi ]$ when $M=2$ and $d=2\pi /3$ when $M=3$. Then, by the matrices X and Y defined by

$$\begin{aligned} X_{ij}= & {} e^{-\alpha |x_i-x_j|}=\left\{ \begin{array}{l} e^{-\alpha d} \text{ for } i\ne j,\\ 1 \text{ for } i=j \end{array} \right. ,\\ Y_{ij}= & {} \int _S\frac{e^{-\alpha |y-x_i|} e^{-\alpha |y-x_j|}}{\left[ \sum _{k=1}^M e^{-\alpha |y-x_k|}\right] ^2}dy, \end{aligned}$$

(21)

respectively, the matrices $A,\, B,\, C,\, D,\, E$, and F introduced above by (17), (18), (19), and (20) can simply be represented as

$$\begin{aligned}A&=\frac{\mu (\sigma -1)}{M^2\sigma } \overline{w}^{2-2\sigma }\overline{G}^{2\sigma -2}X^2 + \frac{\mu }{M\sigma } \overline{w}^{1-\sigma }\overline{G}^{\sigma -1}X + \frac{M(1-\mu )(\sigma -1)}{2\pi \sigma } \overline{w}^{-1}Y,\nonumber \\B&= -\frac{\mu }{M\sigma } \overline{w}^{3-2\sigma }\overline{G}^{2\sigma -2}X^2 + \frac{\mu }{\sigma }\overline{w}^{2-\sigma }\overline{G}^{\sigma -1}X - \frac{1-\mu }{2\pi \sigma }M^2Y,\nonumber \\C&= \frac{1}{M} \overline{G}^\sigma \overline{w}^{-\sigma }X,\quad D =-\frac{\overline{G}^\sigma \overline{w}^{1-\sigma }}{\sigma -1}X,\quad E = \overline{G}^{-\mu }I,\quad F=-\mu \overline{w}\overline{G}^{-\mu -1}I,\nonumber \end{aligned}$$

(22)

respectively. Then, $\Omega$ is observed to take the form

$$\begin{aligned} \begin{aligned}&\Omega = \begin{pmatrix} \Omega _1 &{} \Omega _2 \\ \Omega _2 &{} \Omega _1 \end{pmatrix} \text{ when } M=2,\\&\Omega = \begin{pmatrix} \Omega _1 &{} \Omega _2 &{} \Omega _2 \\ \Omega _2 &{} \Omega _1 &{} \Omega _2 \\ \Omega _2 &{} \Omega _2 &{} \Omega _1 \end{pmatrix}\text{ when } M=3, \end{aligned} \end{aligned}$$

respectively, where $\Omega _1$ and $\Omega _2$ can of course be calculated precisely. As a result, we have

$$\begin{aligned} \begin{aligned}&J^\prime =\frac{1}{2} (\Omega _1-\Omega _2)\text{ when } M=2,\\&J^\prime =\frac{1}{3} \left( \Omega _1-\Omega _2\right) \left( \begin{array}{cc}1&{}0\\ 0&{}1\end{array}\right) \text{ when } M=3. \end{aligned} \end{aligned}$$

(23)