<p>Well, you can actually solve <span class="math-container">$k$</span> analytically. This is just standard first-order linear ODE technique.</p>
<p>The solution of the differential equation <span class="math-container">$ \dot{c}(t) / c(t) = z(t) - \rho $</span> is</p>
<p><span class="math-container">$$ c(t) = c_0 \exp\left( \int_0^t z(s) ds - \rho t \right) $$</span></p>
<p>Plugging this into <span class="math-container">$ \dot{k}(t) - z(t)k(t) = - c(t) $</span>, multiplying both sides by <span class="math-container">$ \exp\left( -\int_0^t z(s) ds \right) $</span> and using product rule, we get</p>
<p><span class="math-container">$$ \frac{d}{dt} \left[ k(t) \exp\left( -\int_0^t z(s) ds \right) \right] = -c_0 e^{-\rho t} $$</span></p>
<p>The solution is</p>
<p><span class="math-container">$$ k(t) = \left[ k_0 + \frac{c_0}{\rho} (e^{-\rho t} - 1) \right] \exp\left( \int_0^t z(s) ds \right) $$</span></p>
<p>What remains is to solve for <span class="math-container">$c_0$</span>. This is actully the trickiest part of the exercise. First, note that the steady state <span class="math-container">$(c^*, k^*)$</span> satisfies <span class="math-container">$ c^* = \rho k^* $</span>. Next, the transversality condition gives us <span class="math-container">$ (c(t), k(t)) \to (c^*, k^*) $</span>, so</p>
<p><span class="math-container">$$
\begin{align}
c^* &amp;= c_0 \zeta \\
k^* &amp;= \left( k_0 - \frac{c_0}{\rho} \right) \exp\left( \int_0^\infty z(t) dt \right) + \frac{c_0}{\rho} \zeta \\
\text{where} \quad \zeta &amp;= \lim_{t \to \infty} \exp\left( \int_0^t z(s) ds - \rho t \right)
\end{align}
$$</span></p>
<p>Finally, plugging the above into <span class="math-container">$ c^* = \rho k^* $</span> and subtracting <span class="math-container">$c_0 \zeta$</span> from both sides, we get <span class="math-container">$ c_0 = \rho k_0 $</span>. Thus, we actually have a nice closed form for <span class="math-container">$k$</span>:</p>
<p><span class="math-container">$$ k(t) = k_0 \exp\left( \int_0^t z(s) ds - \rho t \right) $$</span></p>

CC BY-SA 4.0
2021-12-04T12:10:55.187Z
Thank you for this. Much appreciated.
Denis Dinand
CC BY-SA 4.0
2021-12-04T12:05:49.383Z
2021-12-04T12:05:49.383Z
user141240
<p>Consider a representative agent model where the representative agent has standard preferences</p>
<p><span class="math-container">$\int^{\infty}_te^{-\rho t}ln(c(t))dt$</span></p>
<p>where <span class="math-container">$\rho&gt;0$</span> and <span class="math-container">$c(t)$</span> is consumption in period t as usual. And the production technology is something of the form</p>
<p><span class="math-container">$y(t) = z(t)f(k(t))$</span></p>
<p>For example, to simplify the matters I take <span class="math-container">$f(k(t)) =k(t)$</span>. By assumption take <span class="math-container">$z_t = z^*$</span> in even periods and <span class="math-container">$z_t= z^{**}$</span> for odd periods and further assume that <span class="math-container">$z^{*}-\rho &gt;0$</span> and <span class="math-container">$z^{**}-\rho&lt;0$</span> and <span class="math-container">$0\% (\delta= 0)$</span> depreciation for further simplication.</p>
<p>I want to solve for the planners optimal allocation and to show that the capital stock, output, and consumption increase in even periods and decrease in odd periods. Clearly the assumptions and simplifications are done with an example in mind, but I know the community guidelines such that I am not asking for the specific answer to this question. However, I am asking for a general strategy of how to proceed from the Euler's equation to the law of motion for capital. To put more clearly, let me demonstrate:</p>
<p>Clearly <span class="math-container">$\dot k(t) = z(t) k(t) - c(t)$</span>. And we somehow need to link the Euler's equation with this law of motion to derive a general relationship between <span class="math-container">$\frac{\dot k(t)}{k(t)}$</span> and <span class="math-container">$\frac{\dot c(t)}{c(t)}$</span>. Solving for <span class="math-container">$\frac{\dot c(t)}{c(t)}$</span> is easy since we can set up the current value Hamiltonian and get the result <span class="math-container">$\frac{\dot c(t)}{c(t)} = z(t) - \rho$</span>. This gives the relationship between <span class="math-container">$z(t)-\rho$</span> and <span class="math-container">$\frac{\dot c(t)}{c(t)}$</span>, so we can then plug in the assumption given by the question to derive the relationship that <span class="math-container">$\frac{\dot c(t)}{c(t)}&gt;0$</span> in even periods and <span class="math-container">$\frac{\dot c(t)}{c(t)}&lt;0$</span> in the odd periods. This much is clear.</p>
<p>However, I am consufed as to how to proceed from this relationship to the law of motion for capital. To see why we need link, we can use the production function <span class="math-container">$f(k(t))$</span> and differentiate with respect to time to see how it evolves. So we have</p>
<p><span class="math-container">$y(t) = z(t)k(t)$</span> and <span class="math-container">$\frac{\dot y(t)}{y(t)} = \frac{\dot z(t)}{z(t)} + \frac{\dot k(t)}{k(t)}$</span></p>
<p>In general <span class="math-container">$\frac{\dot z(t)}{z(t)}$</span> will be exogeneous in simple models like this. In the context of this model it is given by the assumptions, <span class="math-container">$ \frac{\dot z(t)}{z(t)}&gt;0$</span> in even periods and <span class="math-container">$ \frac{\dot z(t)}{z(t)}&lt;0$</span> for the odd periods. Then we need to find out how <span class="math-container">$\frac{\dot k(t)}{k(t)}$</span> evolves with <span class="math-container">$\frac{\dot c(t)}{c(t)}$</span>. Then I get stuck because we only know that</p>
<p><span class="math-container">$\dot k(t) = z(t)k(t) -c(t)$</span></p>
<p>which makes</p>
<p><span class="math-container">$\frac{\dot k(t)}{k(t)} = z(t) - \frac{c(t)}{k(t)}$</span>.</p>
<p>If I am told to look for the steady-state then I know where to look, I can use <span class="math-container">$\frac{\dot k(t)}{k(t)}=0$</span> and proceed. But the question is more general. How should one proceed in questions like this where we are not concerned about the steady-state?</p>

<macroeconomics><ramsey>
How to derive a general relationship between the Ramsey-Keynes rule and the law of motion for capital?
Denis Dinand
# How to derive a general relationship between the Ramsey-Keynes rule and the law of motion for capital?

(Asked by: username_0 on 2021-12-04)

Consider a representative agent model where the representative agent has standard preferences

$\int^{\infty}\_te^{-\rho t}ln(c(t))dt$

where $\rho>0$ and $c(t)$ is consumption in period t as usual. And the production technology is something of the form

$y(t) = z(t)f(k(t))$

For example, to simplify the matters I take $f(k(t)) =k(t)$. By assumption take $z\_t = z^\*$ in even periods and $z\_t= z^{\*\*}$ for odd periods and further assume that $z^{\*}-\rho >0$ and $z^{\*\*}-\rho<0$ and $0\% (\delta= 0)$ depreciation for further simplication.

I want to solve for the planners optimal allocation and to show that the capital stock, output, and consumption increase in even periods and decrease in odd periods. Clearly the assumptions and simplifications are done with an example in mind, but I know the community guidelines such that I am not asking for the specific answer to this question. However, I am asking for a general strategy of how to proceed from the Euler's equation to the law of motion for capital. To put more clearly, let me demonstrate:

Clearly $\dot k(t) = z(t) k(t) - c(t)$. And we somehow need to link the Euler's equation with this law of motion to derive a general relationship between $\frac{\dot k(t)}{k(t)}$ and $\frac{\dot c(t)}{c(t)}$. Solving for $\frac{\dot c(t)}{c(t)}$ is easy since we can set up the current value Hamiltonian and get the result $\frac{\dot c(t)}{c(t)} = z(t) - \rho$. This gives the relationship between $z(t)-\rho$ and $\frac{\dot c(t)}{c(t)}$, so we can then plug in the assumption given by the question to derive the relationship that $\frac{\dot c(t)}{c(t)}>0$ in even periods and $\frac{\dot c(t)}{c(t)}<0$ in the odd periods. This much is clear.

However, I am consufed as to how to proceed from this relationship to the law of motion for capital. To see why we need link, we can use the production function $f(k(t))$ and differentiate with respect to time to see how it evolves. So we have

$y(t) = z(t)k(t)$ and $\frac{\dot y(t)}{y(t)} = \frac{\dot z(t)}{z(t)} + \frac{\dot k(t)}{k(t)}$

In general $\frac{\dot z(t)}{z(t)}$ will be exogeneous in simple models like this. In the context of this model it is given by the assumptions, $ \frac{\dot z(t)}{z(t)}>0$ in even periods and $ \frac{\dot z(t)}{z(t)}<0$ for the odd periods. Then we need to find out how $\frac{\dot k(t)}{k(t)}$ evolves with $\frac{\dot c(t)}{c(t)}$. Then I get stuck because we only know that

$\dot k(t) = z(t)k(t) -c(t)$

which makes

$\frac{\dot k(t)}{k(t)} = z(t) - \frac{c(t)}{k(t)}$.

If I am told to look for the steady-state then I know where to look, I can use $\frac{\dot k(t)}{k(t)}=0$ and proceed. But the question is more general. How should one proceed in questions like this where we are not concerned about the steady-state?

**Tags:** macroeconomics, ramsey

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## 1 answers

### Answer by username_1 on 2021-12-04. Score: 3 | Accepted answer

Well, you can actually solve $k$ analytically. This is just standard first-order linear ODE technique.

The solution of the differential equation $ \dot{c}(t) / c(t) = z(t) - \rho $ is

$$ c(t) = c\_0 \exp\left( \int\_0^t z(s) ds - \rho t \right) $$

Plugging this into $ \dot{k}(t) - z(t)k(t) = - c(t) $, multiplying both sides by $ \exp\left( -\int\_0^t z(s) ds \right) $ and using product rule, we get

$$ \frac{d}{dt} \left[ k(t) \exp\left( -\int\_0^t z(s) ds \right) \right] = -c\_0 e^{-\rho t} $$

The solution is

$$ k(t) = \left[ k\_0 + \frac{c\_0}{\rho} (e^{-\rho t} - 1) \right] \exp\left( \int\_0^t z(s) ds \right) $$

What remains is to solve for $c\_0$. This is actully the trickiest part of the exercise. First, note that the steady state $(c^\*, k^\*)$ satisfies $ c^\* = \rho k^\* $. Next, the transversality condition gives us $ (c(t), k(t)) \to (c^\*, k^\*) $, so

$$
\begin{align}
c^\* &= c\_0 \zeta \\
k^\* &= \left( k\_0 - \frac{c\_0}{\rho} \right) \exp\left( \int\_0^\infty z(t) dt \right) + \frac{c\_0}{\rho} \zeta \\
\text{where} \quad \zeta &= \lim\_{t \to \infty} \exp\left( \int\_0^t z(s) ds - \rho t \right)
\end{align}
$$

Finally, plugging the above into $ c^\* = \rho k^\* $ and subtracting $c\_0 \zeta$ from both sides, we get $ c\_0 = \rho k\_0 $. Thus, we actually have a nice closed form for $k$:

$$ k(t) = k\_0 \exp\left( \int\_0^t z(s) ds - \rho t \right) $$

> **Comments:**
> **username_0**: Thank you for this. Much appreciated.  

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