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.


1 Answer 1


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

  • $\begingroup$ Thank you very much. My problem was that I missed the chain rule in the first row. $\endgroup$
    – OST_EE
    Jan 24, 2018 at 19:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.