Goodwin’s growth cycle model with the Bhaduri–Marglin accumulation function: a note on the C.E.S. case

Abstract

This paper investigates the stability properties of the Goodwin growth cycle model with the Bhaduri–Marglin accumulation function, when a constant elasticity of substitution production function is considered. It found that the substitution possibilities between the ‘factors’ of production simplify the local dynamic behaviour of the system.

Appendix: The mathematical proofs of Eqs. (11)–(13)

Proof of Eq. (11)

From Eq. (10) and the definition of $u$, we obtain

$$wy^{P} y^{ - 1} = (1 - \alpha )[(1 - \alpha )L^{ - \varepsilon } + \alpha K^{ - \varepsilon } ]^{ - 1 - 1/\varepsilon } L^{ - 1 - \varepsilon }$$

Hence, from Eq. (9) yields

$$w[(1 - \alpha )L^{ - \varepsilon } + \alpha K^{ - \varepsilon } ]^{ - 1/\varepsilon } y^{ - 1} = (1 - \alpha )[(1 - \alpha )L^{ - \varepsilon } + \alpha K^{ - \varepsilon } ]^{ - 1 - 1/\varepsilon } L^{ - 1 - \varepsilon }$$

Finally, recalling that $k \equiv KL^{ - 1}$, $h = 1 - wLy^{ - 1}$ and $e_{1} \equiv - (1 - h)h^{ - 1}$, we obtain

$$k = (\alpha e_{1} )^{1/\varepsilon } (\alpha - 1)^{ - 1/\varepsilon }$$

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Proof of Eq. (12)

Given the definition of $\pi_{K}$, from Eq. (9) it follows that $\pi_{K} = K^{ - 1} [(1 - \alpha )L^{ - \varepsilon } + \alpha K^{ - \varepsilon } ]^{ - 1/\varepsilon }$ or, recalling that $k \equiv KL^{ - 1}$,

$$\pi_{K} = [(1 - \alpha )k^{\varepsilon } + \alpha ]^{ - 1/\varepsilon }$$

or, recalling Eq. (11),

$$\pi_{K} = ( - \alpha e_{1} + \alpha )^{ - 1/\varepsilon }$$

or, recalling that $e_{1} \equiv - (1 - h)h^{ - 1}$,

$$\pi_{K} = (h\alpha^{ - 1} )^{1/\varepsilon }$$

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Proof of Eq. (13)

Given the definition of $\pi_{L}$, from Eq. (9) it follows that $\pi_{L} = L^{ - 1} [(1 - \alpha )L^{ - \varepsilon } + \alpha K^{ - \varepsilon } ]^{ - 1/\varepsilon } u$ or, recalling that $k \equiv KL^{ - 1}$,

$$\pi_{L} = [(1 - \alpha ) + \alpha k^{ - \varepsilon } ]^{ - 1/\varepsilon } u$$

or recalling Eq. (11),

$$\pi_{L} = (1 - h)^{1/\varepsilon } (1 - \alpha )^{ - 1/\varepsilon } u$$

or, recalling that $e_{1} \equiv - (1 - h)h^{ - 1}$,

$$\pi_{L} = (he_{1} )^{1/\varepsilon } (\alpha - 1)^{ - 1/\varepsilon } u$$

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