Question

Non linear dynamics problem , stability analysis.

Answer 1

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

Answer 2

r,s>0

Answer 3

@ganeshie8

Answer 4

@MatthewPFDS hi

Answer 5

@ParthKohli

Answer 6

@dan815

Answer 7

@nincompoop

Answer 8

@Compassionate

Answer 9

no takers?

Answer 10

@alekos

Answer 11

do you have any boundary conditions?

Answer 12

no here r and s>0 given

Answer 13

We need to show its stability

Answer 14

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

Answer 15

yeah sure!

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

or let me post the picture? @alekos

Answer 20

@alekos check the solution

Answer 21

@zepdrix

Answer 22

@ganeshie8 is the solution correct?

Answer 23

ok, i'm back! give me ten minutes

Answer 24

ok

Answer 25

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!!

Answer 26

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.

Answer 27

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?

Answer 28

yeah like that only.

Answer 29

OK, give me ten minutes

Answer 30

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

Answer 31

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

Answer 32

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

Answer 33

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

Answer 34

because it's not part of the original diff eqn

Answer 35

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

Answer 36

@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.

Answer 37

Phase portrait!
http://openstudy.com/users/sithsandgiggles#/updates/5621bf97e4b009d6565bd1e2

Answer 38

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

Answer 39

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 \(0<N<\dfrac{r}{s}\), you have \(\dfrac{dN}{dt}>0\). 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
\[\frac{dn}{dt}=N(r-sN)=\frac{2r}{s}\left(r-s\frac{2r}{s}\right)=-\frac{2r^2}{s}<0\]