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MXB323 Dynamical Systems

tag 0 Download 0 Pages / 0 Words tag 18-06-2022

Answers:

Solution 1:

The phase portrait can be given as below. The four dimensional projection can only be observed in this case. The model’s simplicity authorizes the complete solution to be viewed in a single plot. This also makes it feasible to describe the geometrical description of significant biological singularities associated to neuronal excitability and mechanism of the spike-generation.

The above portrait depicts the below points:

x-nullicine, that is represented by a N-shaped curve found out of the condition x?=0 ,wherever we have the x value sign of passing over zero,
 
y-nullcline, represented by a straight line found out of the condition y?=0 ,wherever we have the y value sign of passing over zero,

The condition of the equilibrium is noticed if we have condition of x=0 and y=0 which is the intersection of the nullclines. The unstability is caused in case if we there is point onto the middle of the x nullcline.

Solution 2:

Again considering the equation we have

In case if the agitation out of the steady state is satisfactorily great there is a huge variables excursion in their phase space beforehand frequent coming back to the steady state.

There are three possible stability condition in this case. So we have below pairs

(x,y) => (0,0)
 
(x,y) => S1
 
(x,y) => S2

On solving this

Hence we have

Solution 3:

As system will be unstable if it does not satisfy the condition of

Now as if b<1 so this turns out to be the condition of

If we rely on same condition of b<1, so below inequality does not holds.

This proves that we have not equilibrium point for b<1 which satisfy above inequality.

So clearly this point in not under equilibrium because this will not fulfill the condition of the

Or we will have no value of  which could satisfy both.

Again considering same condition we can have

Which can be satisfied with b<1.

Solution 4:

Now using below condition in plot

Finding minimum and maximum of the below equations

We can have below conditions

Thus we have the desired roots and conditions

So this exemplify the result of

Now again considering b=1/2

And

So

Solution 5:

Hopf Bifurcation Theorem:

Foresees the advent of a limits cycle near some steady state which experiences a changeover from a stable to an unstable focus. This happens if we have some variations in the parameters. As given

 

Also to summarize

References:

Helena Unger Djamshid Tavangarian Steffen Silberbach " FitzHugh’s mode implementation" Fachbereich Informatik Universität Rostock
 
Hills D. B. Johnson "A basic modeling of FitzHugh’s mode " IEEE Pers. Commun. vol. 3 no. 1 pp. 56-63 Feb. 1996.
 
S. Chhaya S. Gupta "Performance of FitzHugh’s mode" IEEE Pers. Commun. vol. 3 no. 5 pp. 8-15 Oct. 1996
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