the no-normie zone - 2017-11-27

Anonymous

6:27 PM

@BalarkaSen Let's continue here

Anonymous

So basically I know the form of the equations:

Anonymous

$\dot{x}=ax-bxy$

Anonymous

and $\dot{y}=dxy-cy$

Anonymous

I'm trying to plot the phase space diagram , y vs x

Anonymous

I can approximately plot it...but

Anonymous

6:30 PM

I'm not sure why they should be closed

Anonymous

We can write it in form of $\dot{v}=L\mathbf{x}$ I guess

Balarka Sen

Right, so I have stumbled upon this before, but never put a lot of thought to it. Let's think about it.

Anonymous

Let's see if the eigenvalues say anything (?)

Balarka Sen

You can't do that, because this is not a linear system

Anonymous

Prof Balakrishnan was saying we can locally linearize it and deduce things from there

Anonymous

6:33 PM

By transformation

Balarka Sen

Yes, I was thinking that.

So one thing

is that we can easily spot the equillibrium

Namely, that's when $x' = y' = 0$

Anonymous

Yup, the equilibrium point

Anonymous

Right

Balarka Sen

so $ax_0 - bx_0 y_0 = 0$ and $dx_0 y_0 - c y_0 = 0$

That is to say, $x_0 = c/d$ and $y_0 = a/b$

that's the coordinate of the equillibrium point

Anonymous

Yes. Now shift origin to that point

Anonymous

6:34 PM

And use local linearization

Balarka Sen

slow down

Anonymous

Let's do it for a simple case first

Anonymous

$\dot{x}=x-xy$

Anonymous

and $\dot{y}=xy-y$

Anonymous

a,b,c,d=1

Balarka Sen

6:36 PM

no, there's no more effort required in doing the general case

Anonymous

I realize. We don't have to keep track of the coefficients if we take all as 1. But okay, as you say!

Anonymous

Let's do it for the general case then

Anonymous

Hmm...equilibrium point is (c/d,a/b)

Anonymous

Say, $x=c/d+u$

Anonymous

and $y=a/b+v$

Balarka Sen

6:39 PM

So

Anonymous

Hmm, analyzing eigenvalues in general case is going to be difficult

Balarka Sen

$x'/x = a - by$, $y'/y = -c + dx$

Anonymous

mhm

Balarka Sen

Hmm

Let me fiddle with it for a while on pen and paper

Anonymous

Okay, but I think shifting the origin would be the first logical step

Anonymous

6:45 PM

Sure, take your time

Anonymous

This is an interesting problem

Anonymous

Even wiki doesn't have all the details

Balarka Sen

Think about it while I ponder on it in my way.

Let me know if you get something

Anonymous

Okaies

Anonymous

First of all of we are to linearlize things we need to neglect the higher order terms when we are sufficiently close to the equilibrium point

Balarka Sen

6:48 PM

No, this is subtle. Not everything can be linearized.

Only hyperbolic ODE's can be linearized (which is a hard theorem with a long proof)

You can linearize to get something with vastly different phase diagram

Anonymous

I see, didn't know that

Anonymous

Interesting

Balarka Sen

Mhm, that's why I was careful about it

I'll tell you the story later at some point. But, give me some time to ponder

Anonymous

Sure :)

Balarka Sen

You can think about other stuff if you have exams to prepare for; I'll ping if you I get something

Anonymous

6:52 PM

Alright. I'm just revising some classical mechanics lectures (where I came across this problem)

Balarka Sen

@Blue I might have something

You can actually do separation of variables on this system

Namely, $dx/dt = ax - bxy$ and $dy/dt = dxy - cy$

Then $dx/(ax - bxy) = dy/(dxy - cy)$

Ugh, this notation is bad because we have $d$ as a coefficient and $dx, dy$ etc both

Anonymous

Let's use A,B,C,D

Anonymous

But okay, go on

Balarka Sen

OK. We have $dx/(Ax - Bxy) = dy/(Dxy - Cy)$

Anonymous

I guess separation of variables (like that) can be done only under certain conditions (?)

Balarka Sen

7:03 PM

Cross-multiply to get $(Dx - C)/x \cdot dx = (A - By)/y \cdot dy$

@Blue Away from the equilibrium point we can do this

Anonymous

Okay so far

Balarka Sen

Because all the multiplications make sense

Anonymous

Oh, right. Cool

Anonymous

Then?

Balarka Sen

So $(D - C/x) dx = (A/y - B) dy$

Anonymous

7:04 PM

mhm

Balarka Sen

Integrate both sides to get $Dx - C\log(x) = A\log(y) - By + k$

So the integral curves are given by that equation

Anonymous

+ some constant

Anonymous

But okay

Anonymous

yes

Anonymous

Right

Balarka Sen

7:05 PM

Yeah plus some constant.

Maybe call it k

Anonymous

So we need to plot this now: $Dx - C\log(x) = A\log(y) - By + k$

Anonymous

(away from equilibrium point)

Balarka Sen

Right.

These should be closed curves

Anonymous

@BalarkaSen I'm not sure why

Anonymous

Could you explain?

Anonymous

7:07 PM

Oh wait

Balarka Sen

If I simplify it I get $x^C y^A = \exp(D)^x \exp(B)^y \cdot k$

Anonymous

Okay, then?

Anonymous

(Probably I'm missing something obvious)

Balarka Sen

No, you aren't. I am not sure why these are closed either. I'm thinking about it

Anonymous

Oh ok :)

Balarka Sen

7:17 PM

I'm confused. Desmos says if I take $A = 2, B = 2, C = 1, D = 3, k = 2$ it's not closed

Anonymous

Yeah, I was plotting on Desmos too!

Anonymous

For most values it doesn't seem to be closed

Balarka Sen

I am really confused

Anonymous

room mode changed to Public: anyone may enter and talk

0celo7

Spam

Balarka Sen

7:26 PM

room mode changed to Gallery: anyone may enter, but only approved users can talk

@0celo7 u ok there

0celo7

No

Balarka Sen

i take that as a yes

tell me what im doing wrong bro

Anonymous

@0celo7 So, give us your expert advice now :P

0celo7

I never said I remembered how to do this

I took ODEs years ago

Balarka Sen

u baited us

to give you access

unghgh

0celo7

7:28 PM

I did?

What did I say that constitutes baiting?

Balarka Sen

whatever, this room needs a wider audience and contributers anyway

0celo7

I don’t think I asked for or agreed to be here.

Balarka Sen

room topic changed to the no-normie zone : no normies allowed (no tags)

Anonymous

@0celo7 Calm down XD

Anonymous

We needed your help

Balarka Sen

7:30 PM

@0celo7 i was kidding. help us pls

how in the fricking frick are $Dx - C\log x = A \log y - By + K$ are closed curves?

Anonymous

$C\log(x)+A\log(y)=Dx+By\implies \log(x^{C}y^{A})=Dx+By\implies x^{C}y^{A}=e^{Dx+By}$ hmm

0celo7

I’m assuming you plotted this for many choices of the coefficients

Also I’m trying to pay attention in class and do research

A third thing might not work

Balarka Sen

@Blue Oh wait

I got a closed curve

after randomly fiddling with sliders

Anonymous

Yes, some of them are surely closed curves

Anonymous

Depends on the choice of coefficients

Anonymous

7:38 PM

There's surely something more to this

Balarka Sen

I'm getting lots of closed curves now

Anonymous

Let's think a bit. What do the coefficients actually stand for?

Balarka Sen

It's obvious that you'd get closed curves if you think about the physical interpretation

It's a predator-prey cycle

That's not a problem

Anonymous

@BalarkaSen Is that really "obvious"?

Balarka Sen

Yep, think about it.

Let me figure out why I got non-closed plots before

This is weird

@Blue I misreported you

Anonymous

7:43 PM

@BalarkaSen ?

Balarka Sen

For $A = B = 2, C = 1, D = 3, k = 2$ you do not get a non-closed plot. You get nothing at all.

Anonymous

I'm getting a non-closed plot for it though

Anonymous

$x^{C}y^{A}=e^{Dx+By}$

Anonymous

desmos.com/calculator/x5xoo1ssxj

Balarka Sen

Where is $k$???

In any case you'll get a non-closed plot even if you put the factor of $e^2$

The resolution is simple

You want the plot to be in the the first quarter of the coordinate plane

it makes 0 sense to talk about a predator prey model for $x < 0$ and $y < 0$

those represent number of predators/preys

at each time

You have no plot in the +ve xy-quarter, which is the important point.

Anonymous

7:48 PM

@BalarkaSen Yeah, even with $k$, it's a non-closed plot

Anonymous

@BalarkaSen That's true

Anonymous

Right, so we need to look at curves in +ve quarter

Balarka Sen

Non closed but does not belong to x > 0, y > 0

Anonymous

Yup, right

Balarka Sen

The right claim should be, whenever it belongs to x > 0, y > 0, the integral curve is closed

how do you share links to desmos plots?

Anonymous

7:50 PM

So we basically need to prove that: For x>0, y>0, the curves are closed (unless they pass through equilibrium point)

Balarka Sen

i have a plot you might want to take a look

Anonymous

Also they should not be intersecting

Anonymous

@BalarkaSen There is a "share symbol" on the top right

Balarka Sen

Thanks!

desmos.com/calculator/fwe7pergbh

Fiddle around

Make sure not to put the coefficients negative

Because that would also be physically nonsensical

they are growth coefficients; population growth cannot be negative

Anonymous

Right

Anonymous

7:52 PM

However, we still have to do the proof :P (for the first quadrant)

Balarka Sen

Can I be the physicist and nope out of the proof? :P

Or I can be a mathematician and say: left as an exercise to the reader

Anonymous

Lol :D

Balarka Sen

I'll think about it later. It should be high school algebra now that we have the curves

@Blue Do you see why it's obvious that these are closed?

Physically, I mean

Anonymous

Suppose we start initially with a large number of predators

Anonymous

And there is less number of preys

Anonymous

7:55 PM

The predators start dying

Anonymous

While the preys increase in number

Balarka Sen

That's the idea

Anonymous

As the preys increase, the predators can start eating them again

Anonymous

And then the prey population decreases again

Balarka Sen

mhm

Anonymous

7:58 PM

I'm confused. Suppose we start with 1 predator and 10 preys and the predator eats up all the preys?

Anonymous

@BalarkaSen

Anonymous

There's surely some condition for which this cycle holds :/

Balarka Sen

@Blue Hm? The predators are also populating

What do you mean, "the predator eats up all the preys"?

The rate at which they kill the preys is fixed

That's what the coefficients of the Lotka-Volterra model mean

Anonymous

If mean if the rate of predators populating is very very low and the rate at which they kill they preys is very very high

Anonymous

Someone could claim that all the preys would be killed off

Anonymous

8:02 PM

Unless we actually solve the equations and show they are closed curves

Anonymous

I mean it's not really "obvious"

Balarka Sen

Nah; all the preys cannot be killed off. You're think discretely like in real life. The preys are populating at a specific rate

And the predators are also dying at a specific rate

You can convince yourself that you can never fall into the equillibrium

Anonymous

@BalarkaSen Even if there is a specific rate, I'm not sure how you can claim that $x\neq 0$ or $y\neq 0$ at any instant

Balarka Sen

But you're right that mathematical justification is needed to be fully sure, I'll give you that

Anonymous

Without actually solving the equations (or plotting them)

Balarka Sen

8:06 PM

@Blue I am guessing you can show that the equillibrium is a unstable

Anonymous

@BalarkaSen Yes, that has to be done by linearizing (we need to prove why it can be linearized)

Anonymous

hyperbolic ode thing you said

Balarka Sen

You don't need to linearize to figure out stability, nope

Anonymous

Uh, ok?

Balarka Sen

Or, well, let me parse that differently

You don't need to compare the phase portrait of the linearization to figure out stability

In particular stability implies nothing about closedness of the integral curves, like you were trying to do before

Anonymous

8:10 PM

@BalarkaSen That is true

Balarka Sen

@Blue All it does is help you show this, i.e., that you never fall into the equillibrium. Which is correct

So, let's try that maybe?

Anonymous

Okay!

Balarka Sen

I have $(x', y') = f(x, y)$

Where $f(x, y) = (ax - bxy, -cy + dxy)$

Anonymous

Yup

Balarka Sen

Then $Df(x, y)$ is the matrix $[a - by, -bx ; dy, -c + dx]$

I hope

The equillibrium was $(x_0, y_0) = (c/d, a/b)$, right?

Anonymous

8:14 PM

Yes

Balarka Sen

So $Df(x_0, y_0) = [0, -bc/d; da/b, 0]$

Anonymous

yes

Balarka Sen

Trace is $0$, determinant is uh

$bc/d \cdot da/b$

which is $ac$, right?

Anonymous

Right!

Balarka Sen

For the characteristic polynomial $t^2 - \text{tr}(A) t + \det(A) = 0$, the discriminant is

$\Delta = \tr(A)^2 - 4\det(A)$

in this case thats $-4ac$

Anonymous

8:17 PM

I have a question. You used the Jacobian there. Right? Isn't that same as first order approximation (linearization) ?

Balarka Sen

I think $\Delta < 0$ means it's an elliptic fixed point.

Anonymous

Near the equilibrium point you're effectively assuming the transformation to be a linear transformation

Balarka Sen

No! I am taking the first order approximation, but in no way am I assuming the phase portrait look the same

Let me tell you the story

Say you have $\mathbf{x}' = f(\mathbf{x})$

With fixed point $f(\mathbf{0}) = 0$ at the origin, who cares

Then $\mathbf{x}' = A\mathbf{x}$ is the first order approximation/linearization when $A = Df(0)$, like you said

Anonymous

@BalarkaSen Yeah, that's what I was saying

Balarka Sen

In general there is not a homeomorphism (continuous bijective map with continuous inverse) from the phase space of $x' = f(x)$ to the phase space of the linearization $x' = Ax$ which sends the integral curves of the first portrait to the other

i.e., in general the phase portraits and integral curves of these two curves look very different, even near the equillibrium point

This is, however, true if $A$ is a special kind of matrix known as "hyperbolic"

Anonymous

8:23 PM

@BalarkaSen Right, so the question is : IS A the special kind of matrix, here? I guess so, because you used the same method

Balarka Sen

Nope, it isn't! $A$ is elliptic, not hyperbolic

That's why I did not use $x' = Ax$ to determine the nature of the integral curves of $x' = f(x)$

there is no guarantee that they look the same

@Blue For example, not all elliptic ODE's have integral curves as ellipses

The integral curves can be very non-closed, like archimedean spirals

Anonymous

Oh. You're just analysing $A$ (when $A$ is elliptic, and $\Delta<0$)

Anonymous

I don't know this stuff

Anonymous

That sort of gives some information about the type of equilibrium point (?)

Balarka Sen

Yeah I am just using $A$ to determine stability. That's just saying something infinitisimal (i.e, whether the integral curves come close or get repelled off by the equillibrium point) instead of speaking about the nature of the integral curves

@Blue Correct

I wish I had an example to show you. I am not sure what the phase diagram of $x' = Ax$ looks for the $A$ in this case

Anonymous

8:27 PM

Gotcha. I don't think I'll understand this elliptic, hyperbolic stuff unless I learn ODE properly first

Anonymous

But yeah, got the idea now

Balarka Sen

I can tell you about them later. It's quite subtle actually

Anonymous

@BalarkaSen Which book are you using for differential equations? (ODE and PDE)

Balarka Sen

If you perturb a hyperbolic ODE to second order terms, the phase portrait will remain and look the same

If you perturb an elliptic ODE to second order terms, the phase portrait will be a mess

Closed curves will become non-closed spirals

That's why hyperbolic ODE's are easier to analyze

@Blue I am officially using Arnol'd, but I really like Hirsch-Smale.

It's online if you want

math.upatras.gr/~bountis/files/def-eq.pdf

Anonymous

I really find this dynamical systems and chaos (physics) stuff quite interesting. Downloading the books you mentioned for now. After these exams get over I'll read them properly. Thanks :)

Balarka Sen

8:32 PM

Chaos is mathematics, not physics :)

Beautiful things

Anonymous

@BalarkaSen Well, yes :)

Balarka Sen

There are some beautiful physical examples, however, indeed

Anonymous

Are you still attending those reading sessions?

Anonymous

For DE

Balarka Sen

Yeah we meet next month

prof's away this week

Anonymous

8:34 PM

Oh. Accha. I see. What's his name btw?

Balarka Sen

Kingshook Biswas

Anonymous

I see. Cool!

Anonymous

Alright, see you then. I got to complete some homework for tomorrow :P

Anonymous

Thanks a ton

Balarka Sen

Byes

No problem

Transcript for

♦ $Mathematics$ the no-normie zone

Transcript for

♦ the no-normie zone

♦ $Mathematics$ the no-normie zone