<p>If you have are trying to discretize the continuous time model
$$
\dot{\textbf{x}} = A\textbf{x},
$$
then in discrete time you will have
$$
\textbf{x}_{t+1} = B \textbf{x}_t
$$
but $A\neq B$, since $A$ describes the change in $\textbf{x}$ while $B$ describes the next value of $\textbf{x}$, not just the change. However
$$
\Delta\textbf{x}_t = \textbf{x}_{t+1} - \textbf{x}_t = B \textbf{x}_t - \textbf{x}_t = (B-I) \textbf{x}_t
$$
This new matrix $B-I$ would correspond to $A$.</p>

<p>(From this you can also see why the stability criterion is different.)</p>

<p>Given your equations
\begin{eqnarray*}
\dot{k} &amp; = &amp;  \frac{1}{1+\frac{1}{\rho}} - \frac{1}{1+\lambda} \\
\\
\dot{\lambda} &amp; = &amp;  \rho \lambda - \frac{1}{k}
\end{eqnarray*}
the equations for the discrete modell would be
\begin{eqnarray*}
k_{t+1} &amp; = &amp;  k_t + \frac{1}{1+\frac{1}{\rho}} - \frac{1}{1+\lambda_t} \\
\\
\lambda_{t+1} &amp; = &amp;  \lambda_t + \rho \lambda_t - \frac{1}{k_t}.
\end{eqnarray*}
You will have to calculate the Jacobian of this system to determine stability.</p>

CC BY-SA 3.0
2015-04-19T13:36:59.123Z
It looks like $E$ should be the Identity matrix, $I$.
Alecos Papadopoulos
CC BY-SA 3.0
2015-04-19T18:46:45.593Z
You are right. I learned linear algebra in another language, and our notation is sometimes different. Corrected.
Giskard
CC BY-SA 3.0
2015-04-21T09:43:12.497Z
The system equations are formulated in a related question [link](http://economics.stackexchange.com/questions/5120/splittet-value-function-and-hamilton-jacobi-bellman-equation?noredirect=1#comment5504_5120)
clueless
CC BY-SA 3.0
2015-04-19T12:48:08.083Z
2015-04-21T11:05:57.400Z
2015-04-21T11:05:57.400Z
Giskard
<p>I find that the system is saddle-path stable.  </p>

<p>Setting $z\equiv  1/k$ we get</p>

<p>\begin{eqnarray*}
\dot{z} &amp; = &amp;  -z^2\left(\frac{\rho}{1+\rho} - \frac{1}{1+\lambda}\right) \\
\\
\dot{\lambda} &amp; = &amp;  \rho \lambda - z
\end{eqnarray*}</p>

<p>The fixed point is $E=\{z^*, \lambda^*\} = \{1, 1/\rho\}$</p>

<p>The Jacobian of this system evaluated at the steady state is</p>

<p>\begin{align}
J_E = \begin{bmatrix}0&amp; -\frac{\rho^2}{(1+\rho)^2}\\
-1&amp; \rho\end{bmatrix}
\end{align}</p>

<p>The determinant is</p>

<p>$${\rm det}(J_E) = 0-\frac{\rho^2}{(1+\rho)^2} &lt; 0$$</p>

<p>and when the determinant of the Jaconbian in a two-by-two system is negative, the system is saddle-path stable (and so <em>mathematically speaking</em>, unstable), <em>irrespective</em> of whether the trace of the Jacobian (here equal to $\rho&gt;0$) is positive, negative, or zero.</p>

<p>Perhaps <a href="https://alecospapadopoulos.wordpress.com/2014/04/13/stability-of-systems-of-differential-equations/" rel="nofollow">this</a> could be helpful.</p>

CC BY-SA 3.0
2015-04-21T16:14:58.040Z
Thanks. Say, I don't know whether I have a real or complex eigenvalue. But I know $\det(J_E)<0$ and is real. Is that sufficient to inference on a saddle and/or real eigenvalues? From the link you posted I'd guess so, since  $\det(J_E)<0$ cannot hold for complex eigenvalues?!
clueless
CC BY-SA 3.0
2015-04-21T16:23:34.410Z
The point is, that I want to save space and do not want to write down those long eigenvalues and show that they are real and opposite in sign for all $\rho$. And the $\det(J_E)<0$ is way shorter and *smoother*. I actually do have an additional parameter which is implicity set to unity here, but would occur as well.
clueless
CC BY-SA 3.0
2015-04-21T16:31:31.190Z
@clueless That's right, if the determinant of the Jacobian is negative, then you don't get complex roots, that's very easy to show, if you look at how the _discriminant_ of the characteristic equation is calculated. Having a positive determinant is necessary although not sufficient, for complex roots.
Alecos Papadopoulos
CC BY-SA 3.0
2015-04-21T15:38:53.163Z
2015-04-21T15:38:53.163Z
Alecos Papadopoulos
<p><strong>UPDATE</strong><br>
<em>I was not thinking straight anymore and got totally confused after working hours on my equations. The point is, I have an unstable system, but I force it on the stable path. After realizing that crucial point everything made perfect sense.</em></p>

<h2>Problem</h2>

<p><em>(I fixed some major errors)</em></p>

<p>I'm having a two dimensional system with dynamics
\begin{align}
\dot{k} &amp;= \frac{1}{1+\frac{1}{\rho}} - \frac{1}{1+\lambda}\\
\dot{\lambda} &amp;= \rho\lambda - \frac{1}{k}.
\end{align}
Where $k\in[0,2]$ is the state, and $\lambda$ the costate. There is a (symmetric) fixed point $E(\tilde{k},\tilde{\lambda})$ at $\tilde{\lambda} = \frac{1}{k\rho}$ which yields $\tilde{k}=1$. The Jacobian at the fixed point is given by
\begin{align}
J_E = \begin{bmatrix}0,&amp; \frac{1}{\left(\frac{1}{\rho}+1\right)^2}\\
1,&amp; \rho\end{bmatrix}_E
\end{align}
I get two eigenvalues which are opposite in sign
\begin{align}
(\mu_1,\mu_2)=\left(\frac{p(p + \sqrt{p^2 + 2p + 5}+ 1}{2(p + 1)},\frac{p(p - \sqrt{p^2 + 2p + 5}+ 1}{2(p + 1)}\right)
\end{align}
where $\rho\in\mathbb{R}_{++}$ (time preference rate). The first eigenvalue is always positive ($\mu_1&gt;0$) and the second one is always negative ($\mu_2&lt;0$). So it is an unstable saddle,right? Nonetheless the system is stable. How is that possible, cause I used to think that it must be unstable?  The image shows the evolution of the state and the control which is a function of the costate. ($\rho = 0.05$)
<img src="https://i.sstatic.net/jy5OO.png" alt="enter image description here"></p>

<p>For reference I add figures with $\rho=2,5$. 
The first one seems to converge to a different fixed point $k&lt;1$ (I didn't solve for that one, cause I deal with a symmetric situation; I think there are three in total).</p>

<p><img src="https://i.sstatic.net/MLV6T.png" alt="enter image description here"> </p>

<p>And the second picture shows a strange attractor? Which I actually quite like cause of its chaotic stability. For $k\in[0,3]$ the policy function $\tau_1(k)$ is quite weird.</p>

<p><img src="https://i.sstatic.net/Mp4uf.png" alt="enter image description here"></p>

CC BY-SA 3.0
2015-04-19T09:58:15.413Z
Perhaps you could include the equations describing the dynamics?

A guess:
You are in a discrete time model, not a continuous one. The stability criterion for discrete time is that the absolute value of the eigenvalues are smaller than one.
See
http://en.wikipedia.org/wiki/Lyapunov_stability#Stability_for_linear_state_space_models
Giskard
CC BY-SA 3.0
2015-04-19T10:30:16.537Z
You've nailed it.
clueless
CC BY-SA 3.0
2015-04-21T15:22:05.823Z
Dont delete the question, there's valuable information here that might be useful for future visitors - and that's all StackExchange is about, generating a stock of information. Perhaps instead just accept the answer and move on :)
FooBar
<steady-state>
Real Positive Eigenvalue, but Stable Dynamics
clueless
# Real Positive Eigenvalue, but Stable Dynamics

(Asked by: username_0 on 2015-04-19)

**UPDATE**  
*I was not thinking straight anymore and got totally confused after working hours on my equations. The point is, I have an unstable system, but I force it on the stable path. After realizing that crucial point everything made perfect sense.*

## Problem

*(I fixed some major errors)*

I'm having a two dimensional system with dynamics
\begin{align}
\dot{k} &= \frac{1}{1+\frac{1}{\rho}} - \frac{1}{1+\lambda}\\
\dot{\lambda} &= \rho\lambda - \frac{1}{k}.
\end{align}
Where $k\in[0,2]$ is the state, and $\lambda$ the costate. There is a (symmetric) fixed point $E(\tilde{k},\tilde{\lambda})$ at $\tilde{\lambda} = \frac{1}{k\rho}$ which yields $\tilde{k}=1$. The Jacobian at the fixed point is given by
\begin{align}
J\_E = \begin{bmatrix}0,& \frac{1}{\left(\frac{1}{\rho}+1\right)^2}\\
1,& \rho\end{bmatrix}\_E
\end{align}
I get two eigenvalues which are opposite in sign
\begin{align}
(\mu\_1,\mu\_2)=\left(\frac{p(p + \sqrt{p^2 + 2p + 5}+ 1}{2(p + 1)},\frac{p(p - \sqrt{p^2 + 2p + 5}+ 1}{2(p + 1)}\right)
\end{align}
where $\rho\in\mathbb{R}\_{++}$ (time preference rate). The first eigenvalue is always positive ($\mu\_1>0$) and the second one is always negative ($\mu\_2<0$). So it is an unstable saddle,right? Nonetheless the system is stable. How is that possible, cause I used to think that it must be unstable? The image shows the evolution of the state and the control which is a function of the costate. ($\rho = 0.05$)
![enter image description here](https://i.sstatic.net/jy5OO.png)

For reference I add figures with $\rho=2,5$.
The first one seems to converge to a different fixed point $k<1$ (I didn't solve for that one, cause I deal with a symmetric situation; I think there are three in total).

![enter image description here](https://i.sstatic.net/MLV6T.png)

And the second picture shows a strange attractor? Which I actually quite like cause of its chaotic stability. For $k\in[0,3]$ the policy function $\tau\_1(k)$ is quite weird.

![enter image description here](https://i.sstatic.net/Mp4uf.png)

**Tags:** steady-state

> **Comments:**
> **username_1**: Perhaps you could include the equations describing the dynamics?  A guess: You are in a discrete time model, not a continuous one. The stability criterion for discrete time is that the absolute value of the eigenvalues are smaller than one. See http://en.wikipedia.org/wiki/Lyapunov_stability#Stability_for_linear_state_space_models  
> **username_0**: You've nailed it.  
> **username_2**: Dont delete the question, there's valuable information here that might be useful for future visitors - and that's all StackExchange is about, generating a stock of information. Perhaps instead just accept the answer and move on :)  

---

## 2 answers

### Answer by username_1 on 2015-04-19. Score: 2

If you have are trying to discretize the continuous time model
$$
\dot{\textbf{x}} = A\textbf{x},
$$
then in discrete time you will have
$$
\textbf{x}\_{t+1} = B \textbf{x}\_t
$$
but $A\neq B$, since $A$ describes the change in $\textbf{x}$ while $B$ describes the next value of $\textbf{x}$, not just the change. However
$$
\Delta\textbf{x}\_t = \textbf{x}\_{t+1} - \textbf{x}\_t = B \textbf{x}\_t - \textbf{x}\_t = (B-I) \textbf{x}\_t
$$
This new matrix $B-I$ would correspond to $A$.

(From this you can also see why the stability criterion is different.)

Given your equations
\begin{eqnarray\*}
\dot{k} & = & \frac{1}{1+\frac{1}{\rho}} - \frac{1}{1+\lambda} \\
\\
\dot{\lambda} & = & \rho \lambda - \frac{1}{k}
\end{eqnarray\*}
the equations for the discrete modell would be
\begin{eqnarray\*}
k\_{t+1} & = & k\_t + \frac{1}{1+\frac{1}{\rho}} - \frac{1}{1+\lambda\_t} \\
\\
\lambda\_{t+1} & = & \lambda\_t + \rho \lambda\_t - \frac{1}{k\_t}.
\end{eqnarray\*}
You will have to calculate the Jacobian of this system to determine stability.

> **Comments:**
> **username_3**: It looks like $E$ should be the Identity matrix, $I$.  
> **username_1**: You are right. I learned linear algebra in another language, and our notation is sometimes different. Corrected.  
> **username_0**: The system equations are formulated in a related question [link](http://economics.stackexchange.com/questions/5120/splittet-value-function-and-hamilton-jacobi-bellman-equation?noredirect=1#comment5504_5120)  

---

### Answer by username_3 on 2015-04-21. Score: 1

I find that the system is saddle-path stable.

Setting $z\equiv 1/k$ we get

\begin{eqnarray\*}
\dot{z} & = & -z^2\left(\frac{\rho}{1+\rho} - \frac{1}{1+\lambda}\right) \\
\\
\dot{\lambda} & = & \rho \lambda - z
\end{eqnarray\*}

The fixed point is $E=\{z^\*, \lambda^\*\} = \{1, 1/\rho\}$

The Jacobian of this system evaluated at the steady state is

\begin{align}
J\_E = \begin{bmatrix}0& -\frac{\rho^2}{(1+\rho)^2}\\
-1& \rho\end{bmatrix}
\end{align}

The determinant is

$${\rm det}(J\_E) = 0-\frac{\rho^2}{(1+\rho)^2} < 0$$

and when the determinant of the Jaconbian in a two-by-two system is negative, the system is saddle-path stable (and so *mathematically speaking*, unstable), *irrespective* of whether the trace of the Jacobian (here equal to $\rho>0$) is positive, negative, or zero.

Perhaps [this](https://username_3.wordpress.com/2014/04/13/stability-of-systems-of-differential-equations/) could be helpful.

> **Comments:**
> **username_0**: Thanks. Say, I don't know whether I have a real or complex eigenvalue. But I know $\det(J_E)<0$ and is real. Is that sufficient to inference on a saddle and/or real eigenvalues? From the link you posted I'd guess so, since  $\det(J_E)<0$ cannot hold for complex eigenvalues?!  
> **username_0**: The point is, that I want to save space and do not want to write down those long eigenvalues and show that they are real and opposite in sign for all $\rho$. And the $\det(J_E)<0$ is way shorter and *smoother*. I actually do have an additional parameter which is implicity set to unity here, but would occur as well.  
> **username_3**: @username_0 That's right, if the determinant of the Jacobian is negative, then you don't get complex roots, that's very easy to show, if you look at how the _discriminant_ of the characteristic equation is calculated. Having a positive determinant is necessary although not sufficient, for complex roots.  

---

