A Hicksian Two-Sector Model of Cycles and Growth

Abstract

This chapter analyzes cycles and growth using a dynamic version of the Hicksian two-sector model. Two types of friction, nominal wage stickiness and non-shiftability of capital, are present. It is found that sectoral imbalances caused by the non-shiftablilty of capital are corrected in finite time through investment allocation and that the possibility of cycles depends on the speed of nominal wage adjustment. The arriving order of the turning points of some important economic variables is established.

This chapter was adapted from Hajime Hori, “A Hicksian Two-sector Model of Cycles and Growth,” Journal of Economic Dynamics and Control, vol. 22, 1998 with permission from Elsevier.

Appendices

Appendix

A.1 Proof of Lemma  11

The proof of Lemma 11 requires two additional lemmas. Note that, if $z = \left (k_{1},k_{2},w,v_{1},v_{2}\right )$ can be reached by a backward solution of System I which starts from $\mathcal{S}^{0}$, then a forward solution of System I which starts from such a z reaches $\mathcal{S}^{0}$ in finite time. The following two lemmas examine the totality of such z’s.

At the end of this appendix is attached a list collecting the symbols of the sets and functions used in the appendix.

Letting $y^{{\ast}} = \left (k^{{\ast}},w^{{\ast}},v^{{\ast}}\right )$ be a saddle-point stable long-run equilibrium of System II, let

$$\displaystyle{ v = g(k,w) }$$

(2.48)

denote its stable manifold, let $\mathcal{D}$ be the domain of definition of g, and let $\mathcal{S}^{0}$ be the stable manifold embedded in the five-dimensional z-space as in Lemma 11. Then, using (2.31), a $\zeta \in \mathcal{S}^{0}$ can be written

$$\displaystyle\begin{array}{rcl} \zeta & =& \zeta \left (k,w\right ) \\ & \equiv & \left (k_{1}\left (k,w,g\left (k,w\right )\right ),k_{2}\left (k,w,g\left (k,w\right )\right ),w,g\left (k,w\right ),g\left (k,w\right )\right ).{}\end{array}$$

(2.49)

Note that if $z = \left (k_{1},k_{2},w,v,v\right ) \in \mathcal{S}^{0},$ z satisfies π ₁ = π ₂ and v = g(k, w). Thus, to each $\left (k,w\right ) \in \mathcal{D}$, $\zeta \left (k,w\right )$ defines k ₁, k ₂, and v in such a way that π ₁ = π ₂ and v = g(k, w). Next, noting that a solution of System I that starts outside $\mathcal{S}^{0}$ and reaches it at some t (t may be positive or negative) is a solution of either $\dot{z} =\eta ^{+}\left (z\right )$ or $\dot{z} =\eta ^{-}(z)$, let $z^{j}\left (t;\bar{z}\right ),$ j = +, −, be a solution of $\dot{z} =\eta ^{j}(z)$ such that $z^{j}(0;\bar{z}) =\bar{ z}$, and let

$$\displaystyle{ \tilde{\mathcal{S}}^{j} = \left \{z: z = z^{j}\left (t;\zeta \right )\text{ for some }t\text{ and some }\zeta \in \mathcal{S}^{0}\right \},j = +,-, }$$

(2.50)

where time t may be positive or negative. Note that $z^{j}\left (t;\zeta \left (k,w\right )\right )$ defines a point $\left (k_{1},k_{2},w,,v_{1},v_{2}\right )$ which the differential equations $\dot{z} =\eta ^{j}(z)$ transfer to $\zeta \left (k,w\right )$ in time t. Finally, define $\xi _{1}^{j}: R \times \mathcal{D}\rightarrow R^{3}$ and $\xi _{2}^{j}: R \times \mathcal{D}\rightarrow R^{2},$ j = +, −, by

$$\displaystyle{ \left (\xi _{1}^{j}\left (t,k,w\right ),\xi _{ 2}^{j}\left (t,k,w\right )\right ) = z^{j}\left (t;\zeta \left (k,w\right )\right ). }$$

(2.51)

Thus $\xi _{1}^{j}\left (t,k,w\right )$ is the predetermined component and $\xi _{2}^{j}\left (t,k,w\right )$ is the forward-looking component of $z^{j}\left (t;\zeta \left (k,w\right )\right )$.

Lemma 16

If y ^∗ is saddle-point stable, there is a neighborhood O ₁ of $\left (k_{1}^{{\ast}},k_{2}^{{\ast}},w^{{\ast}}\right )$ satisfying the following:

• (i) $\left (\xi _{1}^{j}\right )^{-1}$ is well defined on O ₁ for both j = +,−.

• (ii) If $\left (k_{1},k_{2},w\right ) \in O_{1}$ , then for both j = +,−, the price vector defined by

  $$\displaystyle{ \left (v_{1}^{j},v_{ 2}^{j}\right ) = \left (\xi _{ 2}^{j} \circ \left (\xi _{ 1}^{j}\right )^{-1}\right )\left (k_{ 1},k_{2},w\right ) }$$

  (2.52)

  is the locally unique price vector satisfying

  $$\displaystyle{ \left (k_{1},k_{2},w,v_{1}^{j},v_{ 2}^{j}\right ) \in \tilde{\mathcal{S}}^{j}. }$$

  (2.53)

Proof

Let $J^{j}\left (t,k,w\right )$ be the Jacobian of $\xi _{1}^{j}\left (t,k,w\right )$. Then, since

$$\displaystyle{ \xi _{1}^{j}\left (t,k,w\right ) = \left (k_{ 1}^{j}\left (t;\zeta \left (k,w\right )\right ),k_{ 2}^{j}\left (t;\zeta \left (k,w\right )\right ),w\left (t;\zeta \left (k,w\right )\right )\right ), }$$

we have

$$\displaystyle{ J^{j} = \left \vert \begin{array}{ccc} \partial k_{1}^{j}/\partial t&\partial k_{1}^{j}/\partial k&\partial k_{1}^{j}/\partial w \\ \partial k_{2}^{j}/\partial t&\partial k_{2}^{j}/\partial k&\partial k_{2}^{j}/\partial w \\ \partial w/\partial t & \partial w/\partial k & \partial w/\partial w \end{array} \right \vert. }$$

Now evaluate J ^j(t, k, w) at $\left (0,k^{{\ast}},w^{{\ast}}\right )$. We have (i) ∂ w∕∂ t = 0 because $\dot{w} = 0$ at a long-run equilibrium whatever the allocation of investment may be, (ii) ∂ w∕∂ k = 0 because w and k are independent, (iii) ∂ w∕∂ w = 1, (iv) ∂ k ₁ ^j∕∂ k + ∂ k ₂ ^j∕∂ k = 1 because k ₁ ^j + k ₂ ^j = k, (v) ∂ k ₁ ⁺∕∂ t = k ₁ f ₁ − nk ₁ and ∂ k ₂ ^j∕∂ t = −nk ₂, (vi) ∂ k ₁ ⁻∕∂ t = −nk ₁ and ∂ k ₂ ⁻∕∂ t = k ₁ f ₁ − nk ₂, and (vii) $k_{1}f_{1} - n\left (k_{1} + k_{2}\right ) = 0$ because the evaluation is made at a long-run equilibrium. Combining these and expanding J ^j by Laplace’s expansion theorem, we obtain

$$\displaystyle{ J^{+} = \left \vert \begin{array}{cc} \partial k_{1}^{+}/\partial t&\partial k_{ 1}^{+}/\partial k \\ \partial k_{2}^{+}/\partial t&\partial k_{2}^{+}/\partial k \end{array} \right \vert = k_{1}f_{1}-nk_{1}> 0 }$$

and

$$\displaystyle{ J^{-} = \left \vert \begin{array}{cc} \partial k_{1}^{-}/\partial t&\partial k_{ 1}^{-}/\partial k \\ \partial k_{2}^{-}/\partial t&\partial k_{2}^{-}/\partial k \end{array} \right \vert = -nk_{1} <0. }$$

Thus, for both j = +, −, $\left (\xi _{1}^{j}\right )^{-1}$ is well defined in some neighborhood O ₁ of $\left (k_{1}^{{\ast}},k_{2}^{{\ast}},w^{{\ast}}\right ) =\xi _{ 1}^{j}\left (0,k^{{\ast}},w^{{\ast}}\right )$. Since (2.53) is equivalent to

$$\displaystyle{ \left (k_{1},k_{2},w,v_{1}^{j},v_{ 2}^{j}\right ) = \left (\xi _{ 1}^{j}\left (t^{j},k^{j},w^{j}\right ),\xi _{ 2}^{j}\left (t^{j},k^{j},w^{j}\right )\right ) }$$

for some t ^j and some $\left (k^{j},w^{j}\right ) \in \mathcal{D},$ the lemma follows. ■

This lemma shows that, to each $\left (k_{1},k_{2},w\right )$ which is close to $\left (k_{1}^{{\ast}},k_{2}^{{\ast}},w^{{\ast}}\right )$ but is not in the balanced stage, one can assign at most two price vectors $\left (v_{1}^{j},v_{2}^{j}\right ),$ j = +, −, using solutions z ^j(t) of $\dot{z} =\eta ^{j}(z)$ which start in $\mathcal{S}^{0}$. Thus it remains to show that one and only one of these price vectors is consistent with a convergent solution. For this purpose, note that, in the definition of $\tilde{\mathcal{S}}^{j}$, t is not required to be negative but that, if it takes a positive time for a solution z ^j(t) to reach $\left (k_{1},k_{2},w,v_{1}^{j},v_{2}^{j}\right )$, then a forward solution of System I which starts from $\left (k_{1},k_{2},w,v_{1}^{j},v_{2}^{j}\right )$ cannot reach $\mathcal{S}^{0}$.

Now let

$$\displaystyle{ \mathcal{S}^{j} = \left \{z: z = z^{j}\left (t;\zeta \right )\text{ for some }t <0\text{ and some }\zeta \in \mathcal{S}^{0}\right \},j = +,-, }$$

(2.54)

which is a subset of $\tilde{\mathcal{S}}^{j}$. If $\bar{z} \in \mathcal{S}^{j}$ for j = + or −, then a solution $z^{j}\left (t;\bar{z}\right )$ of $\dot{z} =\eta ^{j}\left (z\right )$ reaches $\mathcal{S}^{0}$ in finite positive time. Let $\mathcal{P}^{j},$ j = +, −, 0, be the projection of $\mathcal{S}^{j}$ into the $\left (k_{1},k_{2},w\right )$-space. It is clear that, for j = +, −, $\mathcal{P}^{j}$ is the set of $\left (k_{1},k_{2},w\right )$ such that $\left (k_{1},k_{2},w\right ) =\xi _{ 1}^{j}\left (t^{j},k^{j},w^{j}\right )$ for some t ^j < 0 and some $\left (k^{j},w^{j}\right ) \in \mathcal{D}$. Then one obtains

Lemma 17

If y ^∗ is saddle-point stable, there is a neighborhood O ₂ of $\left (k_{1}^{{\ast}},k_{2}^{{\ast}},w^{{\ast}}\right )$ with the following properties.

• (i) $\mathcal{P}^{0} \cap O_{2},\mathcal{P}^{+} \cap O_{2},$ and $\mathcal{P}^{-}\cap O_{2}$ are mutually disjoint.

• (ii) $\left (\mathcal{P}^{0} \cup \mathcal{P}^{+} \cup \mathcal{P}^{-}\right ) \cap O_{2} = O_{2}$ .

Proof

Define $\Pi \left (k_{1},k_{2},w\right )$ by

$$\displaystyle{ \Pi \left (k_{1},k_{2},w\right ) =\pi \left (k_{1},k_{2},w,g\left (k_{1} + k_{2},w\right ),g\left (k_{1} + k_{2},w\right )\right ), }$$

where π = π ₁ −π ₂. Then $d\Pi \left (\xi _{1}^{+}\left (0,k^{{\ast}},w^{{\ast}}\right )\right )/dt <0$ and $d\Pi \left (\xi _{1}^{-}\left (0,k^{{\ast}},w^{{\ast}}\right )\right )/dt> 0$ by (2.27), which imply

$$\displaystyle{ \Pi \left (\xi _{1}^{+}\left (t,k^{{\ast}},w^{{\ast}}\right )\right ) \gtrless 0\text{ and }\Pi \left (\xi _{ 1}^{-}\left (t,k^{{\ast}},w^{{\ast}}\right )\right ) \lessgtr 0\text{ if }t \lessgtr 0, }$$

which in turn imply that

$$\displaystyle{ \Pi \left (\xi _{1}^{+}\left (t,k^{{\prime}},w^{{\prime}}\right )\right ) \gtrless 0\text{ and }\Pi \left (\xi _{ 1}^{-}\left (t,k^{{\prime}},w^{{\prime}}\right )\right ) \lessgtr 0\text{ if }t \lessgtr 0, }$$

(2.55)

provided that $\left (t,k^{{\prime}},w^{{\prime}}\right )$ is close to $\left (0,k^{{\ast}},w^{{\ast}}\right )$. Let O ₁ be as in Lemma 16 and let O ₂ ⊂ O ₁ be such that if $\left (k_{1},k_{2},w\right ) \in O_{2}$ and if $\left (t,k^{{\prime}},w^{{\prime}}\right ) =$ $\left (\xi _{1}^{j}\right )^{-1}\left (k_{1},k_{2},w\right )$ for j = + or −, then (2.55) holds. Such an O ₂ exists by the continuity of $\left (\xi _{1}^{j}\right )^{-1}$. Now suppose that $\left (k_{1},k_{2},w\right ) \in \left (\mathcal{P}^{+} \cup \mathcal{P}^{-}\cup \mathcal{P}^{0}\right )\cap$ O ₂. Since t < 0 by the definition of $\mathcal{P}^{j}$ if $\left (k_{1},k_{2},w\right ) =\xi _{ 1}^{j}\left (t,k^{{\prime}},w^{{\prime}}\right ) \in \mathcal{P}^{j} \cap O_{2}$ for j = + or −, it follows from (2.55) that

$$\displaystyle{ \Pi \left (k_{1},k_{2},w\right )\left \{\begin{array}{c}> 0\text{ if }\left (k_{1},k_{2},w\right ) \in \mathcal{P}^{+} \cap O_{2}, \\ = 0\text{ if }\left (k_{1},k_{2},w\right ) \in \mathcal{P}^{0} \cap O_{2}, \\ <0\text{ if }\left (k_{1},k_{2},w\right ) \in \mathcal{P}^{-}\cap O_{2}. \end{array} \right. }$$

(2.56)

Statement (i) of the lemma follows from (2.56).

Next suppose that $\left (k_{1},k_{2},w\right ) \in \left (\mathcal{P}^{+} \cup \mathcal{P}^{-}\right )^{c} \cap O_{2}$. Since $\left (k_{1},k_{2},w\right ) \in O_{1},$ statement (i) of Lemma 16 implies that there exist $\left (t^{j},k^{j},w^{j}\right ),$ j = +, −, such that

$$\displaystyle{ \left (t^{j},k^{j},w^{j}\right ) = \left (\xi _{ 1}^{j}\right )^{-1}\left (k_{ 1},k_{2},w\right ),\text{ and }\left (k^{j},w^{j}\right ) \in \mathcal{D}, }$$

(2.57)

where t ^j ≥ 0 because $\left (k_{1},k_{2},w\right ) \in \left (\mathcal{P}^{+} \cup \mathcal{P}^{-}\right )^{c} \cap O_{2}$. If t ^j > 0 for both j = +, −, then

$$\displaystyle{ \Pi \left (k_{1},k_{2},w\right ) = \Pi \left (\xi _{1}^{+}\left (t^{+},k^{+},w^{+}\right )\right ) <0 }$$

and

$$\displaystyle{ \Pi \left (k_{1},k_{2},w\right ) = \Pi \left (\xi _{1}^{-}\left (t^{-},k^{-},w^{-}\right )\right ) <0 }$$

by (2.55), which is a contradiction. Thus t ^j = 0 for either j = + or −, which implies $\left (k_{1},k_{2},w\right ) \in \mathcal{P}^{0} \cap O_{2}$. Therefore statement (ii) of Lemma  5 holds. ■

One can now prove Lemma 11. Let O ₂ be as in Lemma 17 and define g ^j by

$$\displaystyle{ g^{j}\left (k_{ 1},k_{2},w\right ) = g\left (k_{1} + k_{2},w\right )\text{ for both }j = 1,2 }$$

if $\left (k_{1},k_{2},w\right ) \in \mathcal{P}^{0} \cap O_{2},$ and by

$$\displaystyle{ \left (g^{1}\left (k_{ 1},k_{2},w\right ),g^{2}\left (k_{ 1},k_{2}w\right )\right ) = \left (\xi _{2}^{j} \circ \left (\xi _{ 1}^{j}\right )^{-1}\right )(k_{ 1},k_{2},w) }$$

if $(k_{1},k_{2},w) \in \mathcal{P}^{j} \cap O_{2}$ for either j = + or −. Since $\xi _{2}^{j} \circ \left (\xi _{1}^{j}\right )^{-1}$ is well defined in O ₂ for both j = +, − by Lemma 16 and since each $\left (k_{1},k_{2},w\right ) \in O_{2}$ determines j = 0, +, or − uniquely by Lemma 17, g ¹ and g ² are well defined. If one lets $v_{j} = g^{j}\left (k_{1},k_{2},w\right ),$ j = 1, 2, then $\left (v_{1},v_{2}\right )$ clearly satisfies the requirements of Lemma 11. Thus Lemma 11 holds.

List of symbols for the sets and mappings used in Appendix  A.1

• $v = g\left (k,w\right ) \cdot \cdot \cdot \cdot \cdot$ Stable manifold of a long-run equilibrium $y^{{\ast}} = \left (k^{{\ast}},w^{{\ast}},v^{{\ast}}\right )$ of System II.

• $\mathcal{D}\cdot \cdot \cdot \cdot \cdot$ Domain of definition of the function g.

• $\mathcal{S}^{0} \cdot \cdot \cdot \cdot \cdot$ The stable manifold $v = g\left (k,w\right )$ of System II embedded in the five-dimensional z-space. An element ζ of $\mathcal{S}^{0}$ is written as

  $$\displaystyle{ \zeta =\zeta \left (k,w\right ) \equiv \left (k_{1}\left (k,w,g\left (k,w\right )\right ),k_{2}\left (k,w,g\left (k,w\right )\right ),w,g\left (k,w\right ),g\left (k,w\right )\right ), }$$

  where the $k_{j}\left (k,w,v\right )$ is defined by (2.31).

• $z^{j}\left (t;\bar{z}\right ) \cdot \cdot \cdot \cdot \cdot$ Solution of $\dot{z} =\eta ^{j}(z)$ with $z^{j}\left (0;\bar{z}\right ) =\bar{ z},\ j = +,-$.

• $\xi _{1}^{j}\left (t,k,w\right ) \cdot \cdot \cdot \cdot \cdot$ Predetermined component of $z^{j}\left (t;\zeta \left (k,w\right )\right )$.

• $\xi _{2}^{j}\left (t,k,w\right ) \cdot \cdot \cdot \cdot \cdot$ Forward-looking component of $z^{j}\left (t;\zeta \left (k,w\right )\right ).$

• $\tilde{\mathcal{S}}^{j} = \left \{z: z = z^{j}\left (t;\zeta \right )\exists t\text{ and }\exists \zeta \in \mathcal{S}^{0}\right \},$ j = 1, 2.

• $\mathcal{S}^{j} = \left \{z: z = z^{j}\left (t;\zeta \right )\exists t <0\text{ and }\exists \zeta \in \mathcal{S}^{0}\right \},$ j = 1, 2. $\mathcal{S}^{j}$ is a subset of $\tilde{\mathcal{S}}^{j}$.

• $\mathcal{P}^{j} \cdot \cdot \cdot \cdot \cdot$ Projection of $\mathcal{S}^{j}$ into the $\left (k_{1},k_{2},w\right )$-space, j = +, −, 0.

• O _j ⋅ ⋅ ⋅ ⋅ ⋅ Neighborhood of $\left (k_{1}^{{\ast}},k_{2}^{{\ast}},w^{{\ast}}\right )$ in the three-dimensional space of predetermined variables, j = 1, 2. O ₂ ⊂ O ₁.

A.2 Some Mathematical Expressions Relating to the Balanced Stage

2.3.1 A.2.1 Comparative Statics

Let

$$\displaystyle{ D = \left (x_{1}+\rho \right )k_{2}x_{2}^{{\prime}} + \left (x_{ 2}+\rho \right )k_{1}x_{1}^{{\prime}} + \left (x_{ 1} - x_{2}\right )k_{2}, }$$

and

$$\displaystyle{ b = \left (1 - s\right )\left (x_{1}+\rho \right ) + s\left (x_{2}+\rho \right )(> 0). }$$

Then the partial derivatives of the relevant endogenous variables in the balanced stage, namely $\rho \left (w,v\right ),$ $k_{1}\left (k,w,v\right ),$ and $i\left (k,w,v\right )$ defined by (2.31), are as follows.

$$\displaystyle\begin{array}{rcl} \frac{\partial \rho } {\partial w}& =& -\frac{x_{1}+\rho } {w}, {}\\ \frac{\partial \rho } {\partial v}& =& \frac{x_{1}+\rho } {v}, {}\\ \frac{\partial k_{1}} {\partial k} & =& \frac{k_{1}} {k} + \frac{a + m} {bkw}, {}\\ \frac{\partial k_{1}} {\partial w} & =& \frac{1} {bw}\left [\left (x_{1}+\rho \right )\left (\left (1 - s\right )k_{1}x_{1}^{{\prime}}- sk_{ 2}x_{2}^{{\prime}}\right ) - sk_{ 2}\left (x_{1} - x_{2}\right )\right ], {}\\ \frac{\partial k_{1}} {\partial v} & =& -\frac{1} {bv}\left [\left (x_{1}+\rho \right )\left (\left (1 - s\right )k_{1}x_{1}^{{\prime}}- sk_{ 2}x_{2}^{{\prime}}\right ) - sk_{ 2}\left (x_{1} - x_{2}\right ) -\frac{a + m} {w} \right ], {}\\ \frac{\partial i} {\partial k}& =& - \frac{1} {bkL_{1}}\left [bw\left (1 +\rho k\right ) + \left (m + a\right )\left (x_{1} - x_{2}\right )\right ]L_{2} -\frac{vL_{3}} {L_{1}}, {}\\ \frac{\partial i} {\partial w}& =& \frac{\left (x_{1}+\rho \right )DL_{2}} {bL_{1}}, {}\\ \frac{\partial i} {\partial v}& =& - \frac{1} {bvL_{1}}\left [bw\left (1 +\rho k\right ) + w\left (x_{1}+\rho \right )D + \left (x_{1} - x_{2}\right )\left (m + a\right )\right ]L_{2} -\frac{kL_{3}} {L_{1}}. {}\\ \end{array}$$

2.3.2 A.2.2 Coefficients of the Linear System

Let

$$\displaystyle\begin{array}{rcl} \dot{k}& =& h_{1}^{1}\left (k - k^{{\ast}}\right ) + h_{ 2}^{1}(w - w^{{\ast}}) + h_{ 3}^{1}(v - v^{{\ast}}), {}\\ \dot{w}& =& h_{1}^{2}\left (k - k^{{\ast}}\right ) + h_{ 2}^{2}(w - w^{{\ast}}) + h_{ 3}^{2}(v - v^{{\ast}}), {}\\ \dot{v}& =& h_{1}^{3}\left (k - k^{{\ast}}\right ) + h_{ 2}^{3}(w - w^{{\ast}}) + h_{ 3}^{3}(v - v^{{\ast}}) {}\\ \end{array}$$

be the linearization of System II around the long-run equilibrium $y^{{\ast}} = \left (k^{{\ast}},w^{{\ast}},v^{{\ast}}\right )$ and let

$$\displaystyle\begin{array}{rcl} E& =& \left (x_{1}+\rho \right )\left \{\left (x_{1}+\rho \right )k_{2}x_{2}^{{\prime}} + \left (x_{ 2}+\rho \right )k_{1}x_{1}^{{\prime}}\right \}\left (> 0\right ), {}\\ Q& =& sE + s\left (x_{1}+\rho \right )\left (x_{1} - x_{2}\right )k_{2} + \dfrac{\left (x_{1}+\rho \right )\left (m + a\right )} {w}, {}\\ R& =& E + sk_{2}\left (x_{1} - x_{2}\right )^{2} + \dfrac{\left (x_{1} - x_{2}\right )\left (m + a\right )} {w}. {}\\ \end{array}$$

Then the h _j ⁱ can be shown to be as follows:

$$\displaystyle{ \begin{array}{c} h_{1}^{1} = \dfrac{\left (x_{1}+\rho \right )\left (m + a\right )} {kvb}, \\ h_{2}^{1} = -\dfrac{s\left (x_{1}+\rho \right )D} {vb}, \\ h_{3}^{1} = \dfrac{wQ} {bv^{2}}, \\ h_{1}^{2} = \dfrac{\sigma w} {bk}\left \{b + \dfrac{\left (x_{1} - x_{2}\right )\left (m + a\right )} {w} \right \}, \\ h_{2}^{2} = -\dfrac{\sigma } {b}\left \{E + sk_{2}\left (x_{1} - x_{2}\right )^{2}\right \}, \\ h_{3}^{2} = \dfrac{\sigma wR} {bv}, \\ h_{1}^{3} = - \dfrac{vw} {bkL_{1}}\left [\left \{b\left (1 +\rho k\right ) + \dfrac{\left (x_{1} - x_{2}\right )\left (m + a\right )} {w} \right \}L_{2} + \dfrac{vkbL_{3}} {w} \right ], \\ h_{2}^{3} = x_{1} + \dfrac{v\left (x_{1}+\rho \right )DL_{2}} {bL_{1}}, \\ h_{3}^{3} = -\dfrac{x_{1}w} {v} - \dfrac{w} {bL_{1}}\left \{\left (x_{1}+\rho \right )D + b\left (1 +\rho k\right ) + \dfrac{\left (x_{1} - x_{2}\right )\left (m + a\right )} {w} \right \}L_{2} -\dfrac{vkL_{3}} {L_{1}}. \end{array} }$$

The functions are all evaluated at the long-run equilibrium so that, in particular, the relations k ₁ f ₁ = n and x ₁ k ₁ + x ₂ k ₂ = 1 are used to derive the expressions.

2.3.3 A.2.3 Coefficients of the Characteristic Polynomial

Write the characteristic polynomial of the coefficient matrix $H = \left [h_{j}^{i}\right ]$ in A.2.2 above as

$$\displaystyle{ \varphi \left (\lambda \right ) = -\lambda ^{3} +\alpha _{ 1}\lambda ^{2} -\alpha _{ 2}\lambda +\alpha _{3}. }$$

Since H and therefore $\varphi \left (\lambda \right )$ depend on σ, namely the speed of wage adjustment, we will also write $\varphi \left (\lambda;\sigma \right )$ and $\alpha _{j}\left (\sigma \right )$ when it is necessary to make the dependence explicit. By the relation between the h _j ⁱ and α _j , the $\alpha _{j}\left (\sigma \right )$ are as follows.

$$\displaystyle\begin{array}{rcl} \alpha _{1}\left (\sigma \right )& =& \alpha _{11}\sigma +\alpha _{12}, {}\\ \alpha _{11}& =& -\frac{E + sk_{2}\left (x_{1} - x_{2}\right )^{2}} {b}, {}\\ \alpha _{12}& =& \frac{\left (x_{1}+\rho \right )\left (m + a\right ) - bwx_{1}k} {bkv} {}\\ & & - \frac{1} {bL_{1}}\left [w\left \{\left (x_{1}+\rho \right )D + b\rho k + b + \frac{\left (x_{1} - x_{2}\right )\left (m + a\right )} {w} \right \}L_{2} + bkvL_{3}\right ], {}\\ \end{array}$$

$$\displaystyle\begin{array}{rcl} \alpha _{2}\left (\sigma \right )& =& \alpha _{21}\sigma +\alpha _{22}, {}\\ \alpha _{21}& =& \frac{w} {bkv}\left \{sw\left (x_{1}+\rho \right )D -\left (m + a\right )E - x_{1}k\left (x_{1} - x_{2}\right )\left (m + a\right )\right \} {}\\ & & + \frac{1} {bL_{1}}\left [\left \{w\left (1 +\rho k\right )E + vkk_{2}\left (n + a\right )\left (x_{1} - x_{2}\right )^{2}\right \}L_{ 2}\right. {}\\ & & +\left.kv\left \{E + sk_{2}\left (x_{1} - x_{2}\right )^{2}\right \}L_{ 3}\right ], {}\\ \alpha _{22}& =& -\frac{wx_{1}\left (x_{1}+\rho \right )\left (m + a\right )} {bkv^{2}} + \frac{w\left (x_{1}+\rho \right )D} {bL_{1}} \left \{\left (n + a\right )L_{2} + sL_{3}\right \}, {}\\ \end{array}$$

and

$$\displaystyle\begin{array}{rcl} \alpha _{3}\left (\sigma \right )& =& \alpha _{31}\sigma, {}\\ \alpha _{31}& =& -\alpha _{22}. {}\\ \end{array}$$

A.3 Proof of Theorem  12

Theorem 12 is proved following three lemmas which deal with the root condition, some useful identities involving characteristic roots, and the vector condition in turn.

Lemma 18

The root condition  (2.35) holds for some σ > 0 if and only if α ₂₂ < 0. If α ₂₂ < 0, then the root condition holds either (i) for all σ > 0, or (ii) for all small σ > 0 and all large σ > 0.

Proof

Since

$$\displaystyle{ \lambda _{1}\left (\sigma \right )\lambda _{2}\left (\sigma \right )\lambda _{3}\left (\sigma \right ) =\alpha _{3}\left (\sigma \right ) = -\alpha _{22}\sigma }$$

by Appendix A.2.3 above and the familiar relation between the α _j and the characteristic roots

$$\displaystyle\begin{array}{rcl} \lambda _{1} +\lambda _{2} +\lambda _{3}& =& \alpha _{1}, \\ \lambda _{2}\lambda _{3} +\lambda _{1}\lambda _{3} +\lambda _{1}\lambda _{2}& =& \alpha _{2}, \\ \lambda _{1}\lambda _{2}\lambda _{3}& =& \alpha _{3},{}\end{array}$$

(2.58)

(2.35) does not hold for any σ > 0 if α ₂₂ ≥ 0.

Suppose that α ₂₂ < 0. Then λ ₁ λ ₂ λ ₃ > 0 and therefore one of the characteristic roots and the product of the other two characteristic roots are positive for all σ > 0. Thus, without loss of generality, one can let λ ₃(σ) > 0 for all σ > 0. It follows that if α ₂₂ < 0, then

$$\displaystyle{ \lambda _{1}(\sigma )\lambda _{2}(\sigma )> 0\text{ and }\lambda _{3}(\sigma )> 0\text{ for all }\sigma> 0. }$$

Therefore, if α ₂₂ < 0, the root condition (2.35) holds if and only if

$$\displaystyle{ \lambda _{1}\left (\sigma \right ) +\lambda _{2}\left (\sigma \right ) <0. }$$

(2.59)

Now, by (2.58), we have

$$\displaystyle{ \lambda _{1} +\lambda _{2} +\lambda _{3} =\alpha _{1} =\alpha _{11}\sigma +\alpha _{12} }$$

and

$$\displaystyle{ \lambda _{3}(\lambda _{1} +\lambda _{2}) +\lambda _{1}\lambda _{2} =\alpha _{2} =\alpha _{21}\sigma +\alpha _{22}. }$$

Since λ ₁ λ ₂ > 0 and λ ₃ > 0,  (2.59) holds if

$$\displaystyle{ \alpha _{1} \leq 0\text{ or }\alpha _{2} \leq 0. }$$

(2.60)

Now, if α ₁₂ ≤ 0 or α ₂₁ ≤ 0, then, by Appendix A.2.3 and the assumption that α ₂₂ < 0, one of the above two inequalities holds for all σ > 0. Suppose that α ₁₂ > 0 and α ₂₁ > 0. Then we obtain

$$\displaystyle{ \alpha _{1}\left (\sigma \right ) \leq 0\text{ if }\sigma \geq -\frac{\alpha _{12}} {\alpha _{11}} }$$

and

$$\displaystyle{ \alpha _{2}\left (\sigma \right ) \leq 0\text{ if }\sigma \leq -\frac{\alpha _{22}} {\alpha _{21}}. }$$

Thus (2.60) holds for all $\sigma \in \left (0,-\dfrac{\alpha _{22}} {\alpha _{21}}\right ) \cup \left (-\dfrac{\alpha _{12}} {\alpha _{11}},\infty \right )$, which completes the proof of the lemma. ■

In order to state and prove the next lemma, let

$$\displaystyle{ \varphi ^{1}\left (\lambda \right ) =\lambda ^{2} -\alpha _{ 12}\lambda +\alpha _{22} }$$

and

$$\displaystyle{ \varphi ^{2}\left (\lambda \right ) =\lambda ^{2} -\frac{\alpha _{21}} {\alpha _{11}}\lambda + \frac{\alpha _{31}} {\alpha _{11}}. }$$

These functions are independent of σ. Using $\varphi ^{1}\left (\lambda \right )$ and $\varphi ^{2}\left (\lambda \right ),$ the characteristic polynomial of the linear system $\dot{y} = H\left (y - y^{{\ast}}\right )$ can be written as

$$\displaystyle{ \varphi \left (\lambda;\sigma \right ) = -\lambda \varphi ^{1}\left (\lambda \right ) +\alpha _{ 11}\sigma \varphi ^{2}\left (\lambda \right ). }$$

(2.61)

Also, with E, Q, and R as defined in Appendix A.2.2, let

$$\displaystyle{ U \equiv \left (\frac{wQ -\left (m + a\right )E} {vk} \right )^{2} +\alpha _{ 12} \times \left (-\frac{R\left (wQ -\left (m + a\right )E\right )} {vk} \right ) + R^{2}\alpha _{ 22}, }$$

(2.62)

which amounts to

$$\displaystyle{ U \equiv R^{2}\varphi ^{1}\left (-\frac{wQ -\left (m + a\right )E} {vkR} \right ) }$$

(2.63)

if R ≠ 0. U can also be written as

$$\displaystyle{ U = Q\tilde{U} }$$

(2.64)

for some $\tilde{U}$. The proof of (2.64) is just a tedious calculation and is omitted. The sign of U turns out to be crucial for the existence of cycles, as will be shown later. As was noted in Appendix A.2.2, E > 0.

Lemma 19

Let

$$\displaystyle{ T(\sigma ) = s\left (m + a\right )\left (x_{1}+\rho \right )ED\sigma - bkvQ\alpha _{22} }$$

and

$$\displaystyle{ \Phi \left (\sigma \right ) = wQ\left (wQ -\left (m + a\right )E\right ) - vk\left (m + a\right )ER\sigma, }$$

where b and D are defined in Appendix  A.2.1.Then

$$\displaystyle{ \mathop{\prod }\limits _{j=1}^{3}\left (wQ\lambda _{ j}\left (\sigma \right ) + \left (m + a\right )E\sigma \right ) = \frac{wT\left (\sigma \right )\Phi \left (\sigma \right )\sigma } {bvk} }$$

(2.65)

and

$$\displaystyle{ Q\mathop{\prod }\limits _{j=1}^{3}\left (vkR\lambda _{ j}\left (\sigma \right ) + wQ -\left (m + a\right )E\right ) = \frac{v^{2}k^{2}U\Phi \left (\sigma \right )} {w}. }$$

(2.66)

Proof

We will first prove (2.65). If Q = 0, then it can be shown that

$$\displaystyle{ D = -\frac{m + a} {sw} \text{ and }R = \frac{bE} {x_{1}+\rho }, }$$

and that, therefore,

$$\displaystyle{ T\left (\sigma \right ) = -\frac{\left (m + a\right )^{2}\left (x_{1}+\rho \right )E\sigma } {w} \text{ and }\Phi \left (\sigma \right ) = -\frac{vkb\left (m + a\right )E^{2}\sigma } {x_{1}+\rho }, }$$

by which (2.65) holds. If Q ≠ 0,  (2.65) can be derived by letting $\lambda = -\left (m + a\right )E\sigma /wQ$ in the identity

$$\displaystyle{ \mathop{\prod }\limits _{j=1}^{3}\left (\lambda _{ j}\left (\sigma \right )-\lambda \right ) =\varphi \left (\lambda;\sigma \right ). }$$

(2.67)

We next turn to the proof of (2.66). If Q = 0, then U = 0 by (2.64), and therefore (2.66) holds trivially. Therefore we assume that Q ≠ 0 in the rest of the proof.

If R = 0, then

$$\displaystyle{ U = \left (\frac{wQ -\left (m + a\right )E} {vk} \right )^{2}\text{ and }\Phi \left (\sigma \right ) = wQ\left (wQ -\left (m + a\right )E\right ), }$$

by which (2.66) holds.

Suppose R ≠ 0 and let

$$\displaystyle{ \bar{\lambda }= -\frac{wQ -\left (m + a\right )E} {vkR} \text{ and }\bar{\sigma } = - \frac{wQ\bar{\lambda }} {\left (m + a\right )E}. }$$

Then it holds that

$$\displaystyle{ \varphi \left (\bar{\lambda };\bar{\sigma }\right ) = 0. }$$

(2.68)

To see this, note that $\Phi \left (\bar{\sigma }\right ) = 0$ by the definition of $\Phi$. Therefore by (2.65) which was just proved,

$$\displaystyle\begin{array}{rcl} 0& =& \Pi _{j=1}^{3}\left (wQ\lambda _{ j}\left (\bar{\sigma }\right ) + \left (m + a\right )E\bar{\sigma }\right ) {}\\ & =& \Pi _{j=1}^{3}\left (wQ\lambda _{ j}\left (\bar{\sigma }\right ) - wQ\bar{\lambda }\right ) {}\\ & =& \left (wQ\right )^{3}\varphi \left (\bar{\lambda };\bar{\sigma }\right ), {}\\ \end{array}$$

where the last equality is due to (2.67). Since Q ≠ 0 by assumption, (2.68) holds.

Now,

$$\displaystyle{ \varphi ^{2}\left (\bar{\lambda }\right ) = -\frac{\left (m + a\right )E} {\alpha _{11}wQ} \varphi ^{1}\left (\bar{\lambda }\right ) }$$

(2.69)

by (2.61) and (2.68), and

$$\displaystyle{ \varphi ^{1}\left (\bar{\lambda }\right ) = \frac{U} {R^{2}} }$$

by the definition of U. Therefore by (2.61) for an arbitrary σ and by the definition of $\Phi,$

$$\displaystyle{ \varphi \left (\bar{\lambda };\sigma \right ) = -\bar{\lambda }\varphi ^{1}\left (\bar{\lambda }\right ) +\alpha _{ 11}\sigma \varphi ^{2}\left (\bar{\lambda }\right ) = \frac{U\Phi \left (\sigma \right )} {vkR^{3}wQ}. }$$

Since $\varphi \left (\bar{\lambda };\sigma \right )$ is equal to

$$\displaystyle{ \Pi _{j=1}^{3}\left (\lambda _{ j}\left (\sigma \right ) + \frac{wQ -\left (m + a\right )E} {vkR} \right ) = \frac{1} {\left (vkR\right )^{3}}\Pi _{j=1}^{3}\left (vkR\lambda _{ j}\left (\sigma \right ) + wQ -\left (m + a\right )E\right ), }$$

we obtain (2.66). ■

Lemma 20

If T(σ) ≠ 0 and $wQ\lambda _{3}\left (\sigma \right ) +\sigma \left (m + a\right )E\neq 0,$ then the vector condition is satisfied and

$$\displaystyle{ \begin{array}{c} g_{1} = \dfrac{v} {kR}\left [ \dfrac{bv^{2}k^{2}U} {wQ\left (vkR\lambda _{3} + wQ -\left (m + a\right )E\right )} -\left \{b + \dfrac{\left (x_{1} - x_{2}\right )\left (m + a\right )} {w} \right \}\right ], \\ g_{2} = \dfrac{v\left (sw\left (x_{1}+\rho \right )D\lambda _{3} + kvb\alpha _{22}\right )} {w\left (wQ\lambda _{3} +\sigma \left (m + a\right )E\right )}. \end{array} }$$

Moreover, the number of values of σ at which either $T\left (\sigma \right ) = 0$ or $wQ\lambda _{3}(\sigma ) +\sigma \left (m + a\right )E = 0$ is finite.

Proof

We will first show that, if $T\left (\sigma \right )\neq 0,$ then the characteristic vectors corresponding to $\lambda _{j}\left (\sigma \right )$ are given by

$$\displaystyle\begin{array}{rcl} \mu _{1}^{j}\left (\sigma \right )& =& 1, \\ \mu _{2}^{j}\left (\sigma \right )& =& \frac{\sigma w\left (vkR\lambda _{j}\left (\sigma \right ) + wQ -\left (m + a\right )E\right )} {k\left (wQ\lambda _{j}\left (\sigma \right ) +\sigma \left (m + a\right )E\right )},{}\end{array}$$

(2.70)

and

$$\displaystyle{ \mu _{3}^{j}\left (\sigma \right ) = \left \{\begin{array}{c} -\dfrac{\left (h_{1}^{1} -\lambda _{j}\left (\sigma \right )\right )\mu _{1}^{j}\left (\sigma \right ) + h_{2}^{1}\mu _{2}^{j}\left (\sigma \right )} {h_{3}^{1}} \text{ if }h_{3}^{1}\neq 0, \\ -\dfrac{h_{1}^{2}\mu _{1}^{j}\left (\sigma \right ) + \left (h_{2}^{2} -\lambda _{j}\left (\sigma \right )\right )\mu _{2}^{j}\left (\sigma \right )} {h_{3}^{2}} \text{if }h_{3}^{1} = 0. \end{array} \right. }$$

(It can be shown that h ₃ ² ≠ 0 if h ₃ ¹ = 0.) In the above, if $wQ\lambda _{j}\left (\sigma \right ) +\sigma \left (m + a\right )E = 0$ at $\sigma =\bar{\sigma },$ $\mu _{2}^{j}\left (\bar{\sigma }\right )$ should be understood to mean $\lim _{\sigma \rightarrow \bar{\sigma }}\mu _{2}^{j}\left (\sigma \right )$. If $\mu _{2}^{j}\left (\sigma \right )$ is well defined, it is trivial to check that the $\mu ^{j}\left (\sigma \right ) = \left (\mu _{1}^{j}\left (\sigma \right ),\mu _{2}^{j}\left (\sigma \right ),\mu _{3}^{j}\left (\sigma \right )\right )$ thus defined is a characteristic vector. Thus our first task is to show that the above limit exists.

For this purpose suppose that $wQ\lambda _{j}\left (\bar{\sigma }\right ) +\bar{\sigma } \left (m + a\right )E = 0$ and that $T\left (\bar{\sigma }\right )\neq 0$. Then, noting that $T\left (\sigma \right )\neq 0$ if σ is close to $\bar{\sigma }$ and that Q ≠ 0 if $wQ\lambda _{j}\left (\bar{\sigma }\right ) +\bar{\sigma } \left (m + a\right )E = 0,$ use (2.65) to obtain

$$\displaystyle\begin{array}{rcl} & & \frac{vkR\lambda _{j}\left (\sigma \right ) + wQ -\left (m + a\right )E} {wQ\lambda _{j}\left (\sigma \right ) +\sigma \left (m + a\right )E} {}\\ & =& \frac{vk} {wQ}\left [R + \frac{b} {\sigma wT\left (\sigma \right )}\mathop{\prod }\limits _{i\neq j}\left (wQ\lambda _{i}\left (\sigma \right ) +\sigma \left (m + a\right )E\right )\right ]. {}\\ \end{array}$$

Since the right-hand side of this equation is well defined and continuous in σ at $\sigma =\bar{\sigma },$ $\lim _{\sigma \rightarrow \bar{\sigma }}\mu _{2}^{j}\left (\sigma \right )$ exists.

Substituting the $\mu ^{j}\left (\sigma \right )$ thus defined in (2.37) and using (2.65) and (2.66), one arrives at the formulae for g ₁ and g ₂ provided in the statement of the lemma. We have to show that the g ₁ and g ₂ thus arrived at are well defined.

g ₂ is well defined because its denominator is non-zero by assumption. Thus it remains to show that g ₁ is also well defined, namely that

$$\displaystyle{ R\neq 0\text{ and }wQ\left (vkR\lambda _{3} + wQ -\left (m + a\right )E\right )\neq 0. }$$

(2.71)

This will be achieved for the two cases of (i) $\Phi \left (\sigma \right ) = 0$ and (ii) $\Phi \left (\sigma \right )\neq 0$ separately.

• (i) Suppose $\Phi \left (\sigma \right ) = 0$. Then

  $$\displaystyle{ vkR = \frac{wQ\left (wQ -\left (m + a\right )E\right )} {\left (m + a\right )E\sigma }, }$$

  (2.72)

  which in turn implies that

  $$\displaystyle{ \begin{array}{c} vkR\lambda _{3}(\sigma ) + wQ -\left (m + a\right )E = \dfrac{\left (wQ -\left (m + a\right )E\right )\left (wQ\lambda _{3}\left (\sigma \right ) + \left (m + a\right )E\sigma \right )} {\left (m + a\right )E\sigma }. \end{array} }$$

  (2.73)

  Since $wQ\lambda _{3}\left (\sigma \right ) + \left (m + a\right )E\sigma \neq 0$ by assumption, (2.71) will hold if

  $$\displaystyle{ Q\neq 0\text{ and }wQ -\left (m + a\right )E\neq 0. }$$

  (2.74)

  The proof will consist in deriving a contradiction from the assumption that Q = 0 or $wQ -\left (m + a\right )E = 0$.

  First suppose Q = 0 so that

  $$\displaystyle{ E = -\frac{x_{1}+\rho } {s} \left (sk_{2}\left (x_{1} - x_{2}\right ) + \frac{m + a} {w} \right ). }$$

  (2.75)

  Since Q = 0 implies R = 0 by (2.72), we also have

  $$\displaystyle{ E = -\left (x_{1} - x_{2}\right )\left (sk_{2}\left (x_{1} - x_{2}\right ) + \frac{m + a} {w} \right ). }$$

  (2.76)

  These two equations imply

  $$\displaystyle{ \left (\left (x_{1}+\rho \right ) + s\left (x_{2} - x_{1}\right )\right )\left (sk_{2}\left (x_{1} - x_{2}\right ) + \frac{m + a} {w} \right ) = 0, }$$

  which is a contradiction because $\left (x_{1}+\rho \right ) + s\left (x_{2} - x_{1}\right ) = b> 0$ and $sk_{2}\left (x_{1} - x_{2}\right ) + \dfrac{m + a} {w}$ $= - \dfrac{sE} {x_{1}+\rho }\neq 0$ by (2.75) and E > 0. Thus Q ≠ 0.

  Next suppose that $wQ -\left (m + a\right )E = 0$ so that

  $$\displaystyle{ \left (s -\frac{m + a} {w} \right )E + \left (x_{1}+\rho \right )\left (sk_{2}\left (x_{1} - x_{2}\right ) + \frac{m + a} {w} \right ) = 0. }$$

  (2.77)

  The assumption that $wQ -\left (m + a\right )E = 0$ again implies R = 0 because of (2.72) so that (2.76) holds as well. Substitution of (2.76) in (2.77) yields

  $$\displaystyle{ \left (sk_{2}\left (x_{1} - x_{2}\right ) + \frac{m + a} {w} \right )\left (b + \frac{\left (x_{1} - x_{2}\right )\left (m + a\right )} {w} \right ) = 0. }$$

  (2.78)

  Now, the first factor in the left-hand side of this equation is non-zero as was just shown. We also have

  $$\displaystyle{ b + \frac{\left (x_{1} - x_{2}\right )\left (m + a\right )} {w}> 0 }$$

  (2.79)

  because, by virtue of (2.31),

  $$\displaystyle\begin{array}{rcl} b + \frac{\left (x_{1} - x_{2}\right )\left (m + a\right )} {w} & =& \frac{1} {wk_{2}}\left [wk_{2}b + k_{2}\left (x_{1} - x_{2}\right )\left (m + a\right )\right ] {}\\ & =& \frac{k} {wk_{2}}\left [\left (1 - s\right )w\left (x_{1}+\rho \right ) + \left (a + m\right )x_{1} + av\right ]> 0, {}\\ \end{array}$$

  where one of the long-run equilibrium conditions, namely x ₁ k ₁ + x ₂ k ₂ = 1, was used to derive the second equality. Thus (2.78) cannot hold, which implies wQ − (m + a)E ≠ 0. Therefore (2.74) holds if $\Phi \left (\sigma \right ) = 0$.

• (ii) Suppose next that $\Phi \left (\sigma \right )\neq 0$. In this case one can use (2.66) and rewrite g ₁ to obtain

  $$\displaystyle\begin{array}{rcl} g_{1}& =& \frac{v} {k\Phi }\left [\left (vk\right )^{2}bR\lambda _{ 1}\lambda _{2} + bvk\left (wQ -\left (m + a\right )E\right )\left (\lambda _{1} +\lambda _{2}\right )\right. {}\\ & & \left.-\left (m + a\right )\left \{\left (x_{1}+\rho \right )\left (wQ -\left (m + a\right )E\right ) + vkE\sigma \right \}\right. {}\\ & & \left.\times \left (b + \frac{\left (m + a\right )\left (x_{1} - x_{2}\right )} {w} \right )\right ]. {}\\ \end{array}$$

  Thus g ₁ is well defined in this case as well.

  Finally, one can easily see from the definition of D, Q, R, and α ₂₂ that each of $T\left (\sigma \right )$ and $\Phi \left (\sigma \right )$ can vanish at most at one value of σ. Thus by (2.65), $wQ\lambda _{3}\left (\sigma \right ) + \left (m + a\right )E\sigma$ can vanish at most at two values of σ > 0. ■

One can now prove Theorem 12. By Lemma 20, the vector condition is satisfied except possibly at a finite number of values of σ. Thus Lemma 18 implies that y ^∗ is saddle-point stable for some σ > 0 if and only if α ₂₂ < 0 and that, if α ₂₂ < 0, then it is saddle-point stable for all small σ > 0 and all large σ > 0.

Now let $\Psi \left (\rho \right )$ and $\Gamma \left (\rho \right )$ be as in the proof of Theorem 7. Then one can show that

$$\displaystyle{ \Psi ^{{\prime}}(\rho ) = \frac{L_{1}b\left (m + a\right )\alpha _{22}} {vk\left (\Gamma \left (\rho \right )\right )^{2}}. }$$

Thus α ₂₂ < 0 if and only if $\Psi ^{{\prime}}> 0$. The assertion concerning the number of saddle-point stable equilibria follows from this and (2.24).

A.4 Proof of Lemma  13

Lemma 13 is proved following another lemma. Let D, Q, R, E, and T(σ) be as in Appendix A.2 and Lemma 19. Then we have

Lemma 21

If y ^∗ satisfies (2.39),then the following holds.

• (i) D ≥ 0,Q > 0,α ₂₂ < 0, and R > 0.

• (ii) $T(\sigma )> 0,wQ\lambda _{3}\left (\sigma \right ) +\sigma \left (m + a\right )E> 0,$ and $vkR\lambda _{3}\left (\sigma \right ) + wQ -\left (m + a\right )E> 0$ for all σ > 0.

(The last inequality in (ii) is used for the proofs of Lemma 22 and (2.47).)

Proof

The first three inequalities in (i) are trivial. The fourth inequality is also trivial if x ₁ − x ₂ ≥ 0. If x ₁ − x ₂ < 0, use (2.31) to rewrite R to obtain

$$\displaystyle{ R = \left (x_{1}+\rho \right )D -\left (x_{1} - x_{2}\right )k\left \{\left (1 - s\right )\left (x_{1}+\rho \right ) + \frac{av} {w} \right \}> 0. }$$

The first two inequalities in (ii) follow from (i) and $\lambda _{3}\left (\sigma \right )> 0$. The third inequality follows from R > 0 if $wQ -\left (m + a\right )E \geq 0$. Suppose $wQ -\left (m + a\right )E <0$. Then $\Phi \left (\sigma \right ) <0$ for all σ > 0, where $\Phi \left (\sigma \right )$ is defined in Lemma 19, and therefore the sign of the right-hand side of (2.66) is independent of σ. Now, by Lemma 18, $\mathop{\mathrm{Re}}\nolimits \left (\lambda _{j}\left (\sigma \right )\right ) <0$ for j = 1, 2 for some σ > 0 and therefore

$$\displaystyle{ \mathop{\prod }\limits _{j=1}^{2}\left (vkR\lambda _{ j}\left (\sigma \right ) + wQ -\left (m + a\right )E\right )> 0 }$$

for some σ > 0. Therefore, by (2.66), $vkR\lambda _{3}\left (\sigma \right ) + wQ -\left (m + a\right )E$ and U are of the opposite sign for all σ > 0, where U is defined by (2.62). But

$$\displaystyle{ U <0\text{ if }wQ -\left (m + a\right )E \leq 0 }$$

(2.80)

because U can be rewritten as

$$\displaystyle\begin{array}{rcl} U& =& \frac{w\left (wQ -\left (m + a\right )E\right )} {v^{2}k^{2}} \left [\frac{Q} {b} \left (b + \frac{\left (m + a\right )\left (x_{1} - x_{2}\right )} {w} \right )\right. {}\\ & & \left.-\frac{vkR} {w} \left \{ \frac{w} {bL_{1}}\left (\left (x_{1}+\rho \right ) + b\rho k + b + \frac{\left (m + a\right )\left (x_{1} - x_{2}\right )} {w} \right )L_{2} + \frac{vkL_{3}} {L_{1}} \right \}\right ] {}\\ & & -\frac{w^{2}Rx_{1}Q} {bv^{2}k} \left (b + \frac{\left (m + a\right )\left (x_{1} - x_{2}\right )} {w} \right ) {}\\ & & +\frac{w\left (x_{1}+\rho \right )DR^{2}} {bL_{1}} \left \{\left (a + n\right )L_{2} + sL_{3}\right \}, {}\\ \end{array}$$

which is negative if $wQ -\left (m + a\right )E \leq 0$ due to (2.79). The third inequality in (ii) now follows. ■

Lemma 13 is now proved. Suppose that (2.39) holds. Then the assertion of Lemma 13 concerning the root condition is true by Lemma 18 because α ₂₂ < 0 by Lemmas 18 and 21, and the assertion concerning the vector condition is true by Lemmas 20 and 21. Thus Lemma 13 holds.

A.5 Proof of Theorem  14

Theorem 14 is proved following two lemmas. Lemma 22 deals with a property of the stable manifold while Lemma 23 derives some technical relations that are useful for characterizing characteristic roots.

Lemma 22

Suppose that y ^∗ satisfies  (2.39) . Then

$$\displaystyle{ \frac{\partial e} {\partial k} = \frac{kv^{2}} {w\left (vkR\lambda _{3} + wQ -\left (m + a\right )E\right )}\left \{\begin{array}{c}>\\ =\\ <\end{array} \right \}0\text{ as }U\left \{\begin{array}{c}>\\ =\\ <\end{array} \right \}0 }$$

and

$$\displaystyle{ \frac{\partial e} {\partial w} = -\frac{b(m + a)E\left (wQ\lambda _{3} +\sigma \left (m + a\right )E\right ) + wRT} {bw^{2}Q\left (wQ\lambda _{3} +\sigma \left (m + a\right )E\right )} <0. }$$

Proof

The derivation of the expressions is straightforward from (2.43) and Lemma 20. The signs of ∂ e∕∂ k and ∂ e∕∂ w are due to Lemma 21. ■

Lemma 23

If y ^∗ satisfies  (2.39) , then for j = 1,2, there is a unique ν _j ⁻ < 0 such that

$$\displaystyle{ \varphi ^{j}\left (\lambda \right )\left \{\begin{array}{c}>\\ = \\ <\end{array} \right \}0\text{ as }\lambda \left \{\begin{array}{c} <\\ = \\> \end{array} \right \}\nu _{j}^{-}\text{ if }\lambda <0. }$$

(2.81)

Moreover,

$$\displaystyle{ \varphi ^{1}(\nu _{ 2}^{-})\left \{\begin{array}{c}>\\ = \\ <\end{array} \right \}0\text{ as }U\left \{\begin{array}{c}>\\ = \\ <\end{array} \right \}0. }$$

(2.82)

Proof

The unique existence of a ν _j ⁻ < 0 satisfying (2.81) follows from φ ¹(0) = α ₂₂ < 0 and φ ²(0) = −α ₂₂∕α ₁₁ < 0.

To see (2.82), note from $\varphi \left (\lambda;\sigma \right ) = \Pi _{j=1}^{3}\left (\lambda _{j}\left (\sigma \right )-\lambda \right ),$ (2.65), and Lemma 21 that $\varphi \left (-\sigma \left (m + a\right )E/\left (wQ\right );\sigma \right )$ has the same sign as $\Phi \left (\sigma \right )$. Therefore, since

$$\displaystyle{ \Phi \left (\sigma \right )\left \{\begin{array}{c}>\\ =\\ <\end{array} \right \}0\text{ as }-\frac{\sigma \left (m + a\right )E} {wQ} \left \{\begin{array}{c}>\\ =\\ <\end{array} \right \}-\frac{wQ -\left (m + a\right )E} {vkR}, }$$

we obtain

$$\displaystyle{ \varphi \left (-\frac{\sigma \left (m + a\right )E} {wQ};\sigma \right )\left \{\begin{array}{c}>\\ =\\ <\end{array} \right \}0\text{ as }-\frac{\sigma \left (m + a\right )E} {wQ} \left \{\begin{array}{c}>\\ =\\ <\end{array} \right \}-\frac{wQ -\left (m + a\right )E} {vkR}. }$$

Noting that this relation holds for any σ, let

$$\displaystyle{ \sigma = - \frac{wQ\lambda } {\left (m + a\right )E}. }$$

Then

$$\displaystyle{ \varphi \left (\lambda;- \frac{wQ\lambda } {\left (m + a\right )E}\right )\left \{\begin{array}{c}>\\ =\\ <\end{array} \right \}0\text{ as }\lambda \left \{\begin{array}{c}>\\ =\\ <\end{array} \right \}-\frac{wQ -\left (m + a\right )E} {vkR}. }$$

(2.83)

Now suppose U > 0 so that

$$\displaystyle{ \varphi ^{2}\left (-\frac{wQ -\left (m + a\right )E} {vkR} \right )> 0\text{ and }\frac{wQ -\left (m + a\right )E} {vkR}> 0, }$$

where the first inequalty is due to (2.69) and the second is due to (2.80). Then by (2.83),

$$\displaystyle{ -\frac{wQ -\left (m + a\right )E} {vkR} <\nu _{ 2}^{-}. }$$

Therefore by (2.83) and $\varphi ^{2}\left (\nu _{2}^{-}\right ) = 0,$

$$\displaystyle{ 0 <\varphi \left (\nu _{2}^{-};- \frac{wQ\nu _{2}^{-}} {\left (m + a\right )E}\right ) = -\nu _{2}^{-}\varphi ^{1}\left (\nu _{ 2}^{-}\right ), }$$

which implies $\varphi ^{1}\left (\nu _{2}^{-}\right )> 0$. The other cases can be proved similarly. ■

One can now proceed to the proof of Theorem 14. Since the cyclicity of System III is determined by whether $\lambda _{1}\left (\sigma \right )$ and $\lambda _{2}\left (\sigma \right )$ are real or complex, the following proof concerns how $\lambda _{1}\left (\sigma \right )$ and $\lambda _{2}\left (\sigma \right )$ vary as σ varies.

First suppose that U > 0. Then, by Lemma 23, we have (1) ν ₂ ⁻ < ν ₁ ⁻, (2) $\varphi ^{j}\left (\lambda \right ) <0$ for j = 1, 2 if ν ₁ ⁻ < λ < 0, (3) φ ¹(λ) > 0 and φ ²(λ) < 0 if ν ₂ ⁻ < λ < ν ₁ ⁻, and (4) φ ^j(λ) > 0 for j = 1, 2 if λ < ν ₂ ⁻. Therefore, in terms of $\varphi \left (\lambda;\sigma \right ) = -\lambda \varphi ^{1}\left (\lambda \right ) +\alpha _{11}\sigma \varphi ^{2}\left (\lambda \right )$, we have

$$\displaystyle{ \begin{array}{c} \varphi \left (\nu _{2}^{-};\sigma \right ) =\nu _{ 2}^{-}\varphi ^{1}\left (\nu _{2}^{-}\right )> 0\text{ for all }\sigma> 0, \\ \lim _{\sigma \rightarrow 0}\varphi \left (\lambda;\sigma \right ) = -\lambda \varphi ^{1}\left (\lambda \right ) <0\text{ if }\nu _{1}^{-} <\lambda <0, \end{array} }$$

(2.84)

and

$$\displaystyle{ \lim _{\sigma \rightarrow \infty }\varphi \left (\lambda;\sigma \right ) =\alpha _{11}\lim _{\sigma \rightarrow \infty }\sigma \varphi ^{2}\left (\lambda \right ) = \left \{\begin{array}{c} \infty \text{ if }\nu _{2}^{-} <\lambda, \\ -\infty \text{ if }\lambda <\nu _{ 2}^{-} \end{array}.\right. }$$

We also have

$$\displaystyle{ \lim _{\lambda \rightarrow -\infty }\varphi \left (\lambda;\sigma \right ) = \infty \text{ and }\varphi \left (0;\sigma \right ) = -\sigma \alpha _{22}> 0\text{ for all }\sigma> 0. }$$

(2.85)

Since $\varphi \left (\lambda;\sigma \right )$ has at most two negative roots, (2.84) and (2.85) imply the existence of a σ ₁ and σ ₂, 0 < σ ₁ < σ ₂, such that the equation $\varphi \left (\lambda;\sigma \right ) = 0$ has two distinct negative roots in $\left (\nu _{1}^{-},0\right )$ if $\sigma \in \left (0,\sigma _{1}\right ) \cup (\sigma _{2},\infty ),$ complex roots if $\sigma \in \left (\sigma _{1},\sigma _{2}\right ),$ and a double negative root if σ = σ ₁ or σ ₂.

Next suppose that U < 0. Then $\varphi ^{1}\left (\nu _{2}^{-}\right ) <0$ by Lemma 23 and therefore $\varphi \left (\nu _{2}^{-};\sigma \right ) = -\nu _{2}^{-}\varphi ^{1}\left (\nu _{2}^{-}\right ) <0$ for all σ > 0. It follows that, because of (2.85), $\varphi \left (\lambda;\sigma \right ) = 0$ has two distinct negative roots for all σ > 0.

Combining this result with Lemma 18 and noting that the sign of ∂ e∕∂ k coincides with that of U by Lemma 22, one obtains the theorem. Statement (ii) of the theorem corresponds to the possible saddle-point instability of y ^∗ indicated in Lemma 18.

Figure 2.2 illustrates the above proof for the case where U > 0. In the figure, three curves are drawn, representing a quadratic function $\sigma \alpha _{11}\varphi ^{2}\left (\lambda \right )$ and two cubic functions −λ φ ¹(λ) and $\varphi \left (\lambda;\sigma \right )$. The curve representing $\varphi \left (\lambda;\sigma \right )$ is obtained by vertically adding the other two curves. For σ = 0, $\varphi \left (\lambda;\sigma \right ) = 0$ has two distinct (non-positive) roots λ = 0 and λ = ν ₁ ⁻. As σ increases from zero, the positive influence of $\sigma \alpha _{11}\varphi ^{2}\left (\lambda \right )$ for each $\lambda \in \left (\nu _{2}^{-},0\right )$ increases and the roots of $\varphi \left (\lambda;\sigma \right )$ move to the interior of $\left (\nu _{1}^{-},0\right ),$ disappearing eventually. At the same time, to the left of ν ₂ ⁻, the negative influence of $\sigma \alpha _{11}\varphi ^{2}\left (\lambda \right )$ increases and eventually $\varphi \left (\lambda;\sigma \right )$ acquires roots in the interval $\left (-\infty,\nu _{2}^{-}\right )$.

Fig. 2.2

Location of characteristic roots

A.6 Proof of (2.47)

It is straightforward to show that

$$\displaystyle\begin{array}{rcl} \!\!& & \frac{\partial e} {\partial w} \frac{\partial \omega } {\partial k} - \frac{\partial e} {\partial k} \frac{\partial \omega } {\partial w} {}\\ \!\!& =& -bwvkQR\left [\left (m + a\right )\left (x_{1} - x_{2}\right )\left (vkR\lambda _{3} + wQ -\left (m + a\right )E\right )^{-1}kU\right. {}\\ \!\!& & \!\left.+\!\left (b\left (m + a\right )E + \left (wQ\lambda _{3} +\sigma \left (m + a\right )E\right )^{-1}wQT\right ) \times \left (b +\! \frac{\left (x_{1} - x_{2}\right )\left (m + a\right )} {w} \right )\!\right ]\!, {}\\ \end{array}$$

which is negative if x ₁ − x ₂ ≥ 0 by (2.79) and Lemma 21. To see the sign of this expression for the case where x ₁ − x ₂ < 0, note from (2.44) that it is negative if ∂ ω∕∂ k = −wg ₁∕v ² > 0. Therefore, it suffices to show that if x ₁ − x ₂ < 0, then g ₁ < 0.

g ₁ can be rewritten as

$$\displaystyle{ g_{1} = - \frac{v^{2}} {vkR\lambda _{3} + wQ -\left (m + a\right )E}\left (b + \frac{\left (x_{1} - x_{2}\right )\left (m + a\right )} {w} \right )\left (\lambda _{3} - \Lambda \right ), }$$

where

$$\displaystyle{ \Lambda = \frac{bvkU} {wQR}\left (b + \frac{\left (x_{1} - x_{2}\right )\left (m + a\right )} {w} \right )^{-1} -\frac{wQ -\left (m + a\right )E} {vkR}. }$$

Therefore by (2.79), g ₁ < 0 if $\lambda _{3} - \Lambda> 0$. Since this inequality holds trivially if $\Lambda \leq 0,$ assume that $\Lambda> 0$. Then, since λ ₃ is the unique positive root of $\varphi \left (\lambda \right ) = 0,$ $\lambda _{3} - \Lambda$ > 0 is equivalent to $\varphi \left (\Lambda \right )> 0$.

Now a tedious calculation shows that

$$\displaystyle{ \varphi \left (\Lambda \right ) = - \frac{kU} {wQ^{2}}\left (b + \frac{\left (x_{1} - x_{2}\right )\left (m + a\right )} {w} \right )^{-2}\left (\Lambda A +\sigma QB\right ), }$$

where A > 0 and

$$\displaystyle\begin{array}{rcl} B& =& x_{1}w\left (b + \frac{\left (x_{1} - x_{2}\right )\left (m + a\right )} {w} \right ) {}\\ & & -\frac{vkwL_{2}} {L_{1}} \left (\rho E -\left (x_{1} - x_{2}\right )\left (x_{1}+\rho \right )x_{2}k_{2} + \frac{avk_{2}\left (x_{1} - x_{2}\right )^{2}} {w} \right ) {}\\ & & +\frac{\alpha _{11}bv^{2}kL_{3}} {L_{1}}. {}\\ \end{array}$$

Since B > 0 if x ₁ − x ₂ < 0 due to (2.79), the desired result follows.