jango_IN_DTOWN
  • jango_IN_DTOWN
Non linear dynamics problem , stability analysis.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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jango_IN_DTOWN
  • jango_IN_DTOWN
\[\frac{ 1 }{ N } \frac{ dN }{ dt}=r-sN\]
jango_IN_DTOWN
  • jango_IN_DTOWN
r,s>0
jango_IN_DTOWN
  • jango_IN_DTOWN
@ganeshie8

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jango_IN_DTOWN
  • jango_IN_DTOWN
@MatthewPFDS hi
jango_IN_DTOWN
  • jango_IN_DTOWN
@ParthKohli
jango_IN_DTOWN
  • jango_IN_DTOWN
@dan815
jango_IN_DTOWN
  • jango_IN_DTOWN
@nincompoop
jango_IN_DTOWN
  • jango_IN_DTOWN
@Compassionate
alekos
  • alekos
no takers?
jango_IN_DTOWN
  • jango_IN_DTOWN
@alekos
alekos
  • alekos
do you have any boundary conditions?
jango_IN_DTOWN
  • jango_IN_DTOWN
no here r and s>0 given
jango_IN_DTOWN
  • jango_IN_DTOWN
We need to show its stability
jango_IN_DTOWN
  • jango_IN_DTOWN
I did the problem . will you see its correct or not?
alekos
  • alekos
yeah sure!
jango_IN_DTOWN
  • jango_IN_DTOWN
see clearly N is not zero . For equilibrium dN/dt=0 whence N=r/s
jango_IN_DTOWN
  • jango_IN_DTOWN
Name this N as N* and put N=N*+n in the given equation, where n is very small
alekos
  • alekos
ok, post your work and i'll be back in around 1 hour
jango_IN_DTOWN
  • jango_IN_DTOWN
or let me post the picture? @alekos
jango_IN_DTOWN
  • jango_IN_DTOWN
1 Attachment
jango_IN_DTOWN
  • jango_IN_DTOWN
@alekos check the solution
jango_IN_DTOWN
  • jango_IN_DTOWN
@zepdrix
jango_IN_DTOWN
  • jango_IN_DTOWN
@ganeshie8 is the solution correct?
alekos
  • alekos
ok, i'm back! give me ten minutes
jango_IN_DTOWN
  • jango_IN_DTOWN
ok
alekos
  • 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
  • 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
  • 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
  • jango_IN_DTOWN
yeah like that only.
alekos
  • alekos
OK, give me ten minutes
alekos
  • 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
  • jango_IN_DTOWN
ok... then you draw in paint and then upload a jpeg
alekos
  • alekos
this might not be altogether right, but it's a start
alekos
  • alekos
I'm not quite sure where to put the x(t) which is the forcing function
alekos
  • alekos
because it's not part of the original diff eqn
Michele_Laino
  • 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
  • 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.
jango_IN_DTOWN
  • jango_IN_DTOWN
@SithsAndGiggles can you help me in drawing the phase portrait? I am comfused
anonymous
  • 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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