Welcome to our community

Be a part of something great, join today!

[SOLVED] Discrete Model steady states

dwsmith

Well-known member
Feb 1, 2012
1,673
So the book is showing an example about discrete steady states but neglected to show how the steady states were found. Here is what it has

$u_{t+1}=ru_{t}(1-u_t), \quad r>0$

where we assume $0<r<1$ and we are interested in solutions $u_t>0$

Then it list the steady states

$u^*=0, \quad \lambda=f'(0)=r$
$u^*=\dfrac{r-1}{r}, \quad \lambda=f'(u^*)=2-r$

How did they find those?

I don't understand how to find thee steady states for the discrete models.
 
Last edited:

CaptainBlack

Well-known member
Jan 26, 2012
890
So the book is showing an example about discrete steady states but neglected to show how the steady states were found. Here is what it has

$u_{t+1}=ru_{t}(1-u+t), \quad r>0$

where we assume $0<r<1$ and we are interested in solutions $u_t>0$

Then it list the steady states

$u^*=0, \quad \lambda=f'(0)=r$
$u^*=\dfrac{r-1}{r}, \quad \lambda=f'(u^*)=2-r$

How did they find those?

I don't understand how to find thee steady states for the discrete models.
First use the correct recurrence (probable error in LaTeX), also explain the extraneous notation.

In the steady state you have \(u_{t+1}=u_t=u^*\) substitute into the recurrence relation and solve the resulting equation.

CB
 

dwsmith

Well-known member
Feb 1, 2012
1,673
First use the correct recurrence (probable error in LaTeX), also explain the extraneous notation.

In the steady state you have \(u_{t+1}=u_t=u^*\) substitute into the recurrence relation and solve the resulting equation.

CB
It should be $u_{t+1}=ru_t(1-u_t)$

---------- Post added at 01:08 AM ---------- Previous post was at 12:44 AM ----------

I know we want to perturb around the steady state so $u_t=u^*+v_t$, where $|v_t| \ll 1$ and I need to do a Taylor Series expansion and only focus on the linear terms. But what do I do with $u_t$ now?
 

CaptainBlack

Well-known member
Jan 26, 2012
890
It should be $u_{t+1}=ru_t(1-u_t)$

---------- Post added at 01:08 AM ---------- Previous post was at 12:44 AM ----------

I know we want to perturb around the steady state so $u_t=u^*+v_t$, where $|v_t| \ll 1$ and I need to do a Taylor Series expansion and only focus on the linear terms. But what do I do with $u_t$ now?
You do nothing, the steady state/s satisfies/satisfy:

\[ u^*=ru^*(1-u^*)\]

Maybe you are interested in the stability analysis of the steady states, but if so you have not said so.

Stability analysis:

\[u_{t+1}=f(u_t)\]

with a steady state solution \(u^*\), now for \(u_t\) close to \(u^*\) write \(u_t=u^*+\varepsilon\), then to first order we have:

\[u_{t+1}=f(u^*)+\varepsilon f'(u^*)=u^*+\varepsilon f'(u^*)\]

That is \(u_{t+1}\) is \(\varepsilon f'(u^*)\) from \( u^* \) if \(u_t\) was \(\varepsilon\) from \(u^*\).

So we see that the steady state is stable if \( |f'(u^*)|<1 \), that is any perturbation on the steady state dies away, and unstable if \( |f'(u^*)|>1 \). If \( |f'(u^*)|=1 \) stability depends on higher order terms.

CB
 
Last edited:

dwsmith

Well-known member
Feb 1, 2012
1,673
You do nothing, the steady state/s satisfies/satisfy:

\[ u^*=ru^*(1-u^*)\]

Maybe you are interested in the stability analysis of the steady states, but if so you have not said so.

Stability analysis:

\[u_{t+1}=f(u_t)\]

with a steady state solution \(u^*\), now for \(u_t\) close to \(u^*\) write \(u_t=u^*+\varepsilon\), then to first order we have:

\[u_{t+1}=f(u^*)+\varepsilon f'(u^*)=u^*+\varepsilon f'(u^*)\]

That is \(u_{t+1}\) is \(\varepsilon f'(u^*)\) from \( u^* \) if \(u_t\) was \(\varepsilon\) from \(u^*\).

So we see that the steady state is stable if \( |f'(u^*)|<1 \), that is any perturbation on the steady state dies away, and unstable if \( |f'(u^*)|>1 \). If \( |f'(u^*)|=1 \) stability depends on the sign of \(f''(u^*)\)

CB
Ok so I was able to obtain the steady states. How were the lambdas determined for each state?

Also, how would I use the Taylor series to discuss their linear stability. I know I would truncate at the non-linear terms. But I am trying to understand the example well so I can do problem 1.
 
Last edited:

CaptainBlack

Well-known member
Jan 26, 2012
890
Ok so I was able to obtain the steady states. How were the lambdas determined for each state?

Also, how would I use the Taylor series to discuss their linear stability. I know I would truncate at the non-linear terms. But I am trying to understand the example well so I can do problem 1.
I have no idea without looking at the book what your notation means, but everything you need is in my previous post.

CB
 

dwsmith

Well-known member
Feb 1, 2012
1,673
I have no idea without looking at the book what your notation means, but everything you need is in my previous post.

CB
So in order to see if I understand, I am going to work on $N_{t+1} = N_t\left[1+r\left(1-\frac{N_t}{K}\right)\right]$.

We are looking for $N_{t+1} = N_t = N_*$. Then the steady states are $N_* = 0$ and $N_*=K$.
Now perturbing around the steady states. Let $|v_t|\ll 1$ and $N_{t+1}=f(N_t)$.

$N_t = N_*+v_t$ Then $N_{t+1}=N_*+v_{t+1}=f(N_*+v_t)=f(N_*)+v_tf'(N_*)+\mathit{O}(v_t)^2$

At the steady state, $N_*=f(N_*)$ so $v_{t+1}=v_tf'(N_*)$. (Here is where the lambda's come from) Then $v_{t+1}=v_tf'(N_*)=\lambda v_t$. So $v_t=\lambda v_t$.

$\lambda = f'(N_*)$ is the eigenvalue of the first iteration. Is this all correct? I feel like I need to use $N_{t+1} = N_t\left[1+r\left(1-\frac{N_t}{K}\right)\right]$ somewhere as maybe the $f(N_t)$.


Is it this $N_{t+1}=N_*+v_{t+1}=f(N_*+v_t)$ or $N_{t+1}=N_*+v_{t+1}=N_t\left[1+r\left(1-\frac{N_t}{K}\right)\right]$ and I make the substitution $N_t = N_* + v_t$?
 
Last edited:

CaptainBlack

Well-known member
Jan 26, 2012
890
So in order to see if I understand, I am going to work on $N_{t+1} = N_t\left[1+r\left(1-\frac{N_t}{K}\right)\right]$.

We are looking for $N_{t+1} = N_t = N_*$. Then the steady states are $N_* = 0$ and $N_*=K$.
Now perturbing around the steady states. Let $|v_t|\ll 1$ and $N_{t+1}=f(N_t)$.

$N_t = N_*+v_t$ Then $N_{t+1}=N_*+v_{t+1}=f(N_*+v_t)=f(N_*)+v_tf'(N_*)+\mathit{O}(v_t)^2$

At the steady state, $N_*=f(N_*)$ so $v_{t+1}=v_tf'(N_*)$. (Here is where the lambda's come from) Then $v_{t+1}=v_tf'(N_*)=\lambda v_t$. So $v_t=\lambda v_t$.

$\lambda = f'(N_*)$ is the eigenvalue of the first iteration. Is this all correct? I feel like I need to use $N_{t+1} = N_t\left[1+r\left(1-\frac{N_t}{K}\right)\right]$ somewhere as maybe the $f(N_t)$.


Is it this $N_{t+1}=N_*+v_{t+1}=f(N_*+v_t)$ or $N_{t+1}=N_*+v_{t+1}=N_t\left[1+r\left(1-\frac{N_t}{K}\right)\right]$ and I make the substitution $N_t = N_* + v_t$?


The general itteration is written:

\[ N_{t+1}=f(N_t)\]

so in this case: \[f(N) = N\left[1+r\left(1-\frac{N}{K}\right)\right] \]

CB
 

dwsmith

Well-known member
Feb 1, 2012
1,673
The general itteration is written:

\[ N_{t+1}=f(N_t)\]

so in this case: \[f(N) = N\left[1+r\left(1-\frac{N}{K}\right)\right] \]

CB
How do I continue from where I left off at then? I am not sure about what to do now.
 

chisigma

Well-known member
Feb 13, 2012
1,704
So the book is showing an example about discrete steady states but neglected to show how the steady states were found. Here is what it has

$u_{t+1}=ru_{t}(1-u_t), \quad r>0$

where we assume $0<r<1$ and we are interested in solutions $u_t>0$

Then it list the steady states

$u^*=0, \quad \lambda=f'(0)=r$
$u^*=\dfrac{r-1}{r}, \quad \lambda=f'(u^*)=2-r$

How did they find those?

I don't understand how to find thee steady states for the discrete models.
Let's write the recursive relation in terms [at least for me...] more 'familiar'...

$u_{n+1}=r\ u_{n}\ (1-u_{n})\ ,\ 0<r<1$ (1)

An important detail: the 'initial value' $u_{0}$ isn't specified. Now we can write the (1) as...

$\Delta_{n}= u_{n-1}-u_{n}= (r-1)\ u_{n} -r\ u^{2}_{n}= f(u_{n})$ (2)

As explained in my 'tutorial post' about difference equations in MHF, condition for an $x_{0}$ to be a 'steady point' is to be an 'attractive fixed point', i.e. it must be $f(x_{0})=0$ and $f^{'}(x_{0})<0$. From (2) we derive that the equation $f(x)= (r-1)\ x -r\ x^{2}$ has two solution in $x=0$ and $x=1-\frac{1}{r}$ and . because is $0<r<1$, is $f^{'}(0)= -1-r<0$, only $x_{0}=0$ is an 'attractive fixed point' and can be 'steady state'. Now, in order to proceed with the analysis, it is necessary to specify a little better what does it mean 'we are interested to solutions $u_{n}>0$'...

Kind regards

$\chi$ $\sigma$
 

chisigma

Well-known member
Feb 13, 2012
1,704
The reason of my question about the fact that solution $u_{n}>0$ are requested is now illustrated. Let's write again the recursive relation...

$\displaystyle \Delta_{n}=u_{n+1}-u_{n}= (r-1)\ u_{n} -r\ u^{2}_{n}= f(u_{n})\ ,\ 0<r<1$ (1)

In the previous post it has been demonstrated that there is only one 'attractive fixed point' in $x_{0}=0$, so that the only 'steady state' can be in $x_{0}=0$. Now we have to find the set of 'initial values' $u_{0}$ for which the sequence converges to $x_{0}$. Analysing we find the following...

a) if $\displaystyle u_{0}< x_{-}= 1-\frac{1}{r}$ the sequence diverges to $- \infty$...

b) if $\displaystyle u_{0}= x_{-}= 1-\frac{1}{r}$ the sequence converges to $x_{-}$...

c) if $\displaystyle x_{-}<u_{0}<0$ the sequence converges to the 'steady state' $x_{0}=0$ and $\forall n$ is $u_{n}<0$...

d) if $\displaystyle u_{0}=0$ the sequence converges [trivially...] to the 'steady state' $x_{0}=0$ and $\forall n$ is $u_{n}=0$...

e) if $\displaystyle 0<u_{0}< x_{1}=1$ the sequence converges to the 'steady state' $x_{0}=0$ and $\forall n$ is $u_{n}>0$...

f) if $\displaystyle u_{0}=x_{1}=1$ the sequence converges [trivially...] to the 'steady state' $x_{0}=0$ and $\forall n>0 $ is $u_{n}=0$...

g) if $\displaystyle 1<u_{0}< x_{2}=\frac{1}{2}+ \frac{\sqrt{r^{2}-4 r +4}}{2 r}$ the sequence converges to the 'steady state' $x_{0}=0$ and $\forall n>0 $ is $u_{n}<0$...

h) if $\displaystyle u_{0}= x_{2}=\frac{1}{2}+ \frac{\sqrt{r^{2}-4 r +4}}{2 r}$ the sequence converges to ' $x_{-}= 1-\frac{1}{r}$ and $\forall n>0 $ is $u_{n}= x_{-}$...

i) if $\displaystyle u_{0}> x_{2}=\frac{1}{2}+ \frac{\sqrt{r^{2}-4 r +4}}{2 r}$ the sequence to $- \infty$ and $\forall n>0 $ is $u_{n}<0$...

These result are more easily understandable looking at the figure, where f(x) for $r=\frac{1}{2}$ is represented...

MHB01.JPG

Kind regards

$\chi$ $\sigma$
 

Attachments

dwsmith

Well-known member
Feb 1, 2012
1,673
$\displaystyle \Delta_{n}=u_{n+1}-u_{n}= (r-1)\ u_{n} -r\ u^{2}_{n}= f(u_{n})\ ,\ 0<r<1$ (1)
I have no idea how you came up with this equality.
 

chisigma

Well-known member
Feb 13, 2012
1,704
I have no idea how you came up with this equality.
$u_{n+1}= r\ u_{n}\ (1-u_{n}) \implies \Delta_{n}=u_{n+1}-u_{n}= (r-1)\ u_{n}-r\ u_{n}^{2}$

Kind regards

$\chi$ $\sigma$
 

dwsmith

Well-known member
Feb 1, 2012
1,673
So in order to see if I understand, I am going to work on $N_{t+1} = N_t\left[1+r\left(1-\frac{N_t}{K}\right)\right]$.

We are looking for $N_{t+1} = N_t = N_*$. Then the steady states are $N_* = 0$ and $N_*=K$.
Now perturbing around the steady states. Let $|v_t|\ll 1$ and $N_{t+1}=f(N_t)$.

$N_t = N_*+v_t$ Then $N_{t+1}=N_*+v_{t+1}=f(N_*+v_t)=f(N_*)+v_tf'(N_*)+\mathit{O}(v_t)^2$

At the steady state, $N_*=f(N_*)$ so $v_{t+1}=v_tf'(N_*)$. (Here is where the lambda's come from) Then $v_{t+1}=v_tf'(N_*)=\lambda v_t$. So $v_t=\lambda v_t$.

$\lambda = f'(N_*)$ is the eigenvalue of the first iteration. Is this all correct? I feel like I need to use $N_{t+1} = N_t\left[1+r\left(1-\frac{N_t}{K}\right)\right]$ somewhere as maybe the $f(N_t)$.


Is it this $N_{t+1}=N_*+v_{t+1}=f(N_*+v_t)$ or $N_{t+1}=N_*+v_{t+1}=N_t\left[1+r\left(1-\frac{N_t}{K}\right)\right]$ and I make the substitution $N_t = N_* + v_t$?
So I set $f(x) = x\left[1+r\left(1-\frac{x}{K}\right)\right]$

Then I evaluated $|f'(0)| = |1+r|$ which is our first steady state correct?

Then $|f'(K)|=|1-r|$ which is our nontrivial steady state and our extreme values occur r=0,2. At r=0, we have a tangent bifurcation, and at r=2, we have a pitch fork bifurcation.

How do I discuss linearity stability? I have found all the bifurcations value right?