<p>See Romer (2006, Sec 2.3) for details of notation.
Dynamics are given by
\begin{align}
\dot k &amp;= f(k) - (n+g)k - c\\
\dot c &amp;= \frac{c}{\theta}(f'(k) - \rho - \theta g)
\end{align}</p>

<p>Presume you are located somewhere on the $\dot k = 0$ locus. Now if we increase $c$ one gets
\begin{align}
\frac{\partial \dot k}{\partial c} = -1 &lt; 0
\end{align}</p>

<p>That is, if we are above $\dot k = 0$ capital decreaes over time. Hence at point $A$ we move to the left.</p>

<p>Now presume you are located somewhere on the $\dot c = 0$ locus. Now if we increase $k$ one gets
\begin{align}
\frac{\partial \dot c}{\partial k} = \frac{c}{\theta} f''(k) &lt; 0
\end{align}</p>

<p>That is, if we are right of $\dot c = 0$ consumption decreaes over time. Hence at point $A$ we move up in addition (left of $\dot c = 0$).</p>

CC BY-SA 3.0
2016-04-12T20:36:23.323Z
2016-04-12T20:36:23.323Z
clueless
<p>Consider the Ramsey-Cass-Koopmans model in discrete time. Specifically, consider the following phase diagram.</p>

<p><a href="https://i.sstatic.net/Y86my.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/Y86my.jpg" alt="enter image description here"></a></p>

<p>For simplicity, let's consider the trajectory marked A which is in the northwest corner of this diagram. </p>

<p>In this location, supposedly $K_{t+1}&lt;K_t$ and $C_{t+1}&gt;C_t$. But how does one show this? Up to this point, all I have seen is that "supposedly" this trajectory veers off, but I don't see how it necessarily does so. </p>

<p>Let $$C_t = K_t^\alpha + (1-\delta)K_t - K_{t+1}$$</p>

<h2>My Quesiton</h2>

<p>Can someone explain why A veers off towards the NW direction? Specifically, why must $K_{t+1}&lt;K_t$ and $C_{t+1}&gt;C_t$? </p>

CC BY-SA 3.0
2016-04-12T20:16:58.597Z
Actually, this is Romer (2006, p. 61), right? How come you just switched to discrete time? As far as I can tell phase diagramms are constructed from differential equation. Anyway, you may explictly state the differential equations for consumption $\dot c$ and capital $\dot k$. Differentiate them with respect to $c$ and $k$ and the dynamics will become clear.
clueless
CC BY-SA 3.0
2016-04-12T20:44:23.757Z
Regarding the title of your post, it seems you are rather interested in understanding the dynamics, but convergence properties of trajectories. Actually only the saddle path is converging towards the fixed point.
clueless
CC BY-SA 3.0
2016-04-12T23:29:23.140Z
Is this about something more about than the standard method to deduce the dynamic tendencies of a phase diagram? If not, this may be prove useful to you, https://alecospapadopoulos.wordpress.com/2015/11/13/phase-diagrams-tutorial/
Alecos Papadopoulos
<macroeconomics>
How does one prove the convergence of a trajectory in the Ramsey model?
Stan Shunpike
# How does one prove the convergence of a trajectory in the Ramsey model?

(Asked by: username_0 on 2016-04-12)

Consider the Ramsey-Cass-Koopmans model in discrete time. Specifically, consider the following phase diagram.

[![enter image description here](https://i.sstatic.net/Y86my.jpg)](https://i.sstatic.net/Y86my.jpg)

For simplicity, let's consider the trajectory marked A which is in the northwest corner of this diagram.

In this location, supposedly $K\_{t+1}C\_t$. But how does one show this? Up to this point, all I have seen is that "supposedly" this trajectory veers off, but I don't see how it necessarily does so.

Let $$C\_t = K\_t^\alpha + (1-\delta)K\_t - K\_{t+1}$$

## My Quesiton

Can someone explain why A veers off towards the NW direction? Specifically, why must $K\_{t+1}C\_t$?

**Tags:** macroeconomics
> **Comments:**
> **username_1**: Actually, this is Romer (2006, p. 61), right? How come you just switched to discrete time? As far as I can tell phase diagramms are constructed from differential equation. Anyway, you may explictly state the differential equations for consumption $\dot c$ and capital $\dot k$. Differentiate them with respect to $c$ and $k$ and the dynamics will become clear.  
> **username_1**: Regarding the title of your post, it seems you are rather interested in understanding the dynamics, but convergence properties of trajectories. Actually only the saddle path is converging towards the fixed point.  
> **username_2**: Is this about something more about than the standard method to deduce the dynamic tendencies of a phase diagram? If not, this may be prove useful to you, https://alecospapadopoulos.wordpress.com/2015/11/13/phase-diagrams-tutorial/  

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## 1 answers

### Answer by username_1 on 2016-04-12. Score: 2

See Romer (2006, Sec 2.3) for details of notation.
Dynamics are given by
\begin{align}
\dot k &= f(k) - (n+g)k - c\\
\dot c &= \frac{c}{\theta}(f'(k) - \rho - \theta g)
\end{align}

Presume you are located somewhere on the $\dot k = 0$ locus. Now if we increase $c$ one gets
\begin{align}
\frac{\partial \dot k}{\partial c} = -1 < 0
\end{align}

That is, if we are above $\dot k = 0$ capital decreaes over time. Hence at point $A$ we move to the left.

Now presume you are located somewhere on the $\dot c = 0$ locus. Now if we increase $k$ one gets
\begin{align}
\frac{\partial \dot c}{\partial k} = \frac{c}{\theta} f''(k) < 0
\end{align}

That is, if we are right of $\dot c = 0$ consumption decreaes over time. Hence at point $A$ we move up in addition (left of $\dot c = 0$).


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