Non linear dynamics problem , stability analysis.

- jango_IN_DTOWN

Non linear dynamics problem , stability analysis.

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- jango_IN_DTOWN

\[\frac{ 1 }{ N } \frac{ dN }{ dt}=r-sN\]

- jango_IN_DTOWN

r,s>0

- jango_IN_DTOWN

@ganeshie8

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## More answers

- jango_IN_DTOWN

@MatthewPFDS hi

- jango_IN_DTOWN

@ParthKohli

- jango_IN_DTOWN

@dan815

- jango_IN_DTOWN

@nincompoop

- jango_IN_DTOWN

@Compassionate

- alekos

no takers?

- jango_IN_DTOWN

@alekos

- alekos

do you have any boundary conditions?

- jango_IN_DTOWN

no here r and s>0 given

- jango_IN_DTOWN

We need to show its stability

- jango_IN_DTOWN

I did the problem . will you see its correct or not?

- alekos

yeah sure!

- jango_IN_DTOWN

see clearly N is not zero . For equilibrium dN/dt=0 whence N=r/s

- jango_IN_DTOWN

Name this N as N* and put N=N*+n in the given equation, where n is very small

- alekos

ok, post your work and i'll be back in around 1 hour

- jango_IN_DTOWN

or let me post the picture? @alekos

- jango_IN_DTOWN

##### 1 Attachment

- jango_IN_DTOWN

@alekos check the solution

- jango_IN_DTOWN

@zepdrix

- jango_IN_DTOWN

@ganeshie8 is the solution correct?

- alekos

ok, i'm back! give me ten minutes

- jango_IN_DTOWN

ok

- alekos

OK. Looks good. Every step is correct mathematically.
I've never done non-linear dynamics, but I'm a degree qualified electronics engineer so I'm familiar with stability related to circuits. All the logic seems valid and the dN/dt = 0 equilibrium point looks right.
I reckon you've nailed it!!

- jango_IN_DTOWN

But I have one question, in non-linear dynamics we generally have to draw diagrams, of the system.. In this case I dont have any idea, how to draw it. By the way, non-linear dynamics has many applications.

- alekos

Wow, draw a diagram as well! OK when I think in terms of an analogy to this system I think about analog circuits with a signal input say x(t) and an output N(t).
Is this what you're after?

- jango_IN_DTOWN

yeah like that only.

- alekos

OK, give me ten minutes

- alekos

well i had the drawing done on the openstudy drawing feature but the website did a refresh and lost it!!! I seem to be having a lot of these issues with open study lately. I'll try attaching a file

- jango_IN_DTOWN

ok... then you draw in paint and then upload a jpeg

- alekos

this might not be altogether right, but it's a start

##### 1 Attachment

- alekos

I'm not quite sure where to put the x(t) which is the forcing function

- alekos

because it's not part of the original diff eqn

- Michele_Laino

please try this variable change:
\[\Large N = \frac{1}{z}\]
so, we have:
\[\Large \frac{{dN}}{{dt}} = \frac{{ - 1}}{{{z^2}}}\frac{{dz}}{{dt}}\]

- anonymous

@alekos I don't think that's the kind of diagram OP has in mind. I remember answering a very similar question about the kind of diagram I *think* is expected here, but I can't remember its name for the life of me... I'll see if I can find the old question.

- anonymous

- jango_IN_DTOWN

@SithsAndGiggles can you help me in drawing the phase portrait? I am comfused

- anonymous

|dw:1449352323909:dw|
Thie directions of the arrow indicate that \(N\) approaches the asymptote given by the dashed lines. The direction represents the sign of \(\dfrac{dN}{dt}\).
When \(n<0\), you have \(\dfrac{dN}{dt}=N(r-sN)<0\), since \(r-sN>0\) if \(r,s>0\).
When \(00\). To see why, take \(N=\dfrac{r}{2S}\) (the midpoint between \(0\) and \(\dfrac{r}{s}\)). Then
\[N(r-sN)=\frac{r}{2s}\left(r-s\frac{r}{2s}\right)=\frac{r^2}{2s}-\frac{r^2}{4s}=\frac{r^2}{4s}>0\]
When \(N>\dfrac{r}{s}\), say \(N=\dfrac{2r}{s}\), you have \(\dfrac{dN}{dt}<0\), since
\[\frac{dn}{dt}=N(r-sN)=\frac{2r}{s}\left(r-s\frac{2r}{s}\right)=-\frac{2r^2}{s}<0\]

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