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I am trying to understand the way Smith arrived from (1) and (2) to (6) and (8).

Could anyone give me a hint? Thank you!

It is well known that the dynamics per capita consumption $c$ and capital $k$ are determined by the following pair of differential equations: \begin{eqnarray} \dot{k} &=& k^\alpha - \delta k - c \tag{1} \\ \frac{{\dot c}}{c} &=& \sigma(\alpha k^{\alpha - 1} - \delta - \rho) \tag{2} \end{eqnarray} A Solution

The system of equations (1)-(2) is dauntingly nonlinear. To approach a solution, define the capital-output ratio (a Bernoulli transformation) by $z = k^{1-\alpha}$ and the consumption-capital ratio by $x = c/k$. Using these transformations the system can be rewritten as \begin{eqnarray} \dot{z} &=& (1-\alpha)[1 - (\delta + x)z] \tag{6} \\ \frac{{\dot x}}{x} &=& \frac{\sigma \alpha - 1}{z} + \delta (1 - \sigma) - \rho\sigma + x \tag{7} \end{eqnarray} The Bernoulli transformation converts the capital accumulation equation [Equation (1)] into the linear, albeit non-autonomous, differential equation in Equation (6). Despite this simplification, the system still does not allow an analytical solution. Equation (7) is still non-linear.

Suppose, however, that $\sigma = 1/\alpha$. In that case $z$ disappears from Equation (7), and the system becomes recursive: Equation (7) reduces to $$ \frac{\dot{x}}{x} = - \frac{\delta(1- \alpha) \rho}{\alpha} + x \tag{8} $$ while $z$ evolves according to Equation (6), given the forcing process $x$ determined in Equation (8). This is a simple, autonomous logistic equation.

Citation: Smith, William. (2006). A Closed Form Solution to the Ramsey Model. Contributions to Macroeconomics. 6.

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If $z = k^{1-\alpha}$ then

\begin{eqnarray} \dot{z} &=& (1-\alpha)k^{1-\alpha-1}\dot{k} = (1-\alpha)k^{-\alpha} \dot{k} \\ &=& (1-\alpha) k^{-\alpha} [k^{\alpha} - \delta k - c] \\ &=& (1-\alpha) [1 - \delta k^{1-\alpha} - c k^{-\alpha}\color{blue}{(k/k)}] \\ &=& (1-\alpha) [1 - \delta z - c(k^{1-\alpha})/k] \\ &=& (1-\alpha) [1 - \delta z - x z] \end{eqnarray}

I will leave $\dot{c}/c$ for you to work out

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  • $\begingroup$ Thank you very much. My problem was that I missed the chain rule in the first row. $\endgroup$ – OST_EE Jan 24 '18 at 19:15

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