How can steady states be found for discrete models?

In summary: I was saying I think.CBYes, so at the steady state we have \( N^*=f(N^*) \), so if we have \(N_t=N^*+v_t\) then we have:\[N_{t+1}=N^*+v_{t+1}=f(N^*+v_t)=f(N^*)+v_tf'(N^*)+\mathit{O}(v_t)^2\]But since \(N^*=f(N^*)\) we have:\[N_{t+1}=N^*+v_{t+1}=N^*+v_tf'(N^*)+\mathit{O}(v_t)^2\]So
  • #1
Dustinsfl
2,281
5
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:
Mathematics news on Phys.org
  • #2
dwsmith said:
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
 
  • #3
CaptainBlack said:
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?
 
  • #4
dwsmith said:
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:
  • #5
CaptainBlack said:
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:
  • #6
dwsmith said:
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
 
  • #7
CaptainBlack said:
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:
  • #8
dwsmith said:
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
 
  • #9
CaptainBlack said:
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.
 
  • #10
dwsmith said:
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

$\chi$ $\sigma$
 
  • #11
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...

View attachment 41

Kind regards

$\chi$ $\sigma$
 

Attachments

  • i50247726._szt5_.jpg
    i50247726._szt5_.jpg
    2 KB · Views: 58
  • MHB01.JPG
    MHB01.JPG
    5.5 KB · Views: 52
  • #12
chisigma said:
$\displaystyle \Delta_{n}=u_{n+1}-u_{n}= (r-1)\ u_{n} -r\ u^{2}_{n}= f(u_{n})\ ,\ 0<r<1$ (1)

I have no idea how you came up with this equality.
 
  • #13
dwsmith said:
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

$\chi$ $\sigma$
 
  • #14
dwsmith said:
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?
 

FAQ: How can steady states be found for discrete models?

1. What is a discrete model steady state?

A discrete model steady state is a state in which the variables of a discrete mathematical model remain constant over time. It is a stable equilibrium point, where the system is at rest and no longer changes.

2. How is a discrete model steady state different from a continuous model steady state?

A discrete model steady state is different from a continuous model steady state in that it only considers changes at specific time intervals, while a continuous model considers changes over an infinite number of time points. In a discrete model, the variables can only change in discrete steps, whereas in a continuous model, the variables can change at any point in time.

3. What factors can affect the stability of a discrete model steady state?

There are several factors that can affect the stability of a discrete model steady state, including the initial conditions, the values of the model parameters, and the structure of the model itself. Small changes in any of these factors can lead to significant changes in the steady state behavior of the system.

4. How can one determine the stability of a discrete model steady state?

The stability of a discrete model steady state can be determined by analyzing the eigenvalues of the model's Jacobian matrix. If all of the eigenvalues have a magnitude less than one, the steady state is stable. If any of the eigenvalues have a magnitude greater than one, the steady state is unstable.

5. Can a discrete model have multiple steady states?

Yes, a discrete model can have multiple steady states. This can occur when the model has multiple stable equilibrium points. In this case, the initial conditions will determine which steady state the system will reach. Additionally, a system can have a bifurcation point, where a small change in a parameter can cause the system to jump from one steady state to another.

Back
Top