<p>If $z = k^{1-\alpha}$ then</p>

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

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

CC BY-SA 3.0
2018-01-24T19:15:22.200Z
Thank you very much. My problem was that I missed the chain rule in the first row.
OST_EE
CC BY-SA 3.0
2018-01-22T20:15:47.177Z
2018-01-22T20:15:47.177Z
caverac
<p>I am trying to understand the way Smith arrived from (1) and (2) to (6) and (8). </p>

<p>Could anyone give me a hint? Thank you!</p>

<blockquote>
  <p>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} &amp;=&amp; k^\alpha - \delta k - c \tag{1} \\
\frac{{\dot c}}{c} &amp;=&amp; \sigma(\alpha k^{\alpha - 1} - \delta - \rho) \tag{2}
\end{eqnarray}
  <strong>A Solution</strong></p>
  
  <p>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} &amp;=&amp; (1-\alpha)[1 - (\delta + x)z] \tag{6} \\
\frac{{\dot x}}{x} &amp;=&amp;  \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 <em>linear</em>, 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.</p>
  
  <p>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.</p>
</blockquote>

<p>Citation: <a href="https://www.researchgate.net/publication/4744308_A_Closed_Form_Solution_to_the_Ramsey_Model" rel="nofollow noreferrer">Smith, William. (2006). A Closed Form Solution to the Ramsey Model. Contributions to Macroeconomics. 6.</a></p>

<economic-growth>
Bernoulli transformation in " A Close Form Solution to the Ramsey model" by Smith (2006)
OST_EE
# Bernoulli transformation in " A Close Form Solution to the Ramsey model" by Smith (2006)

(Asked by: username_0 on 2018-01-22)

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.](https://www.researchgate.net/publication/4744308_A_Closed_Form_Solution_to_the_Ramsey_Model)

**Tags:** economic-growth

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## 1 answers

### Answer by username_1 on 2018-01-22. Score: 1 | Accepted answer

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

> **Comments:**
> **username_0**: Thank you very much. My problem was that I missed the chain rule in the first row.  

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