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Consider the differential equation $dx/dt= \left(1 - x\right)\left(2 - x\right)\left(3 - x\right)$. Which of its equilibria is unstable?

  1. $x=0$
  2. $x=1$
  3. $x=2$
  4. $x=3$
  5. None of the above.
asked in Calculus by Boss (41k points) | 308 views

2 Answers

+5 votes
Best answer

Here, dx/dt is neither constant nor function of variable 't'. So on solving for 'x', we get exponential decay function something similar to radioactive decay dN/dt = -KN ;K>0.

For any  equilibrium point, we have dx/dt = 0 => x = 1,2,3.

Now if we consider sign scheme of dx/dt for different ranges of x, we have :

dx/dt > 0 for x < 1 ;

dx/dt  < 0 for 1 < x < 2;

dx/dt  > 0 for 2 < x < 3;

dx/dt  < 0 for  x > 3;

For x -> 2 , dx/dt becomes -ve to +ve. So x = 2 is point of unstable equilibrium.

Since here dx/dt is function of 'x' not 't' , so condition for checking stable / unstable points (aka local minima / local maxima) is opposite to what we apply in case of dy/dx = f(x). Here we have dy/dx = f(y) which is actually exponential function in y and x.

We can understand equilibrium in terms of radioactive decay.

Let dN/dt = -KN ;K>0 its significance is that an element is loosing energy so it is getting stability because we know more energy an element gets,more de-stability it gains and vice versa.

answered by Active (3.4k points)
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0
Hello shashank

from where you studied this ?

because as per my knowledge both your answer and concept behind answer is wrong.

Tell me if i'm missing something.
0
@shashank mind blowing explanation
+1 vote

Equilibrium points for any curve y=f(x)

points where $\frac{\mathrm{d}y}{\mathrm{d} x}=0$ (which is a function of y) are called Equilibrium points.

// y'=f(y)=0

Now Equilibrium points can be of two types

1) Stable :- from stable points if we move little bit right or left , in both cases we will come closer to equilibrium. in more simple words you can think like some ⋃ kind bowl , in which you places some glass pebbles , then you can think they will always reach to equilibrium so thins is stable point of equilibrium.

Mathematically stable points are those where $\frac{\mathrm{d} y}{\mathrm{d} x}$ would be -ve if you move bit right from point of equilibrium and $\frac{\mathrm{d} y}{\mathrm{d} x}$ would be +ve if you move bith left from the point of equilibrium.

2) Unstable :-From unstable points if we move lil bit right or left , we're going away from equilibrium in both cases.Now think any bowl of shape ⋂ , place any glass pebble , it will always move away from equilibrium so this is unstable point fo equilibrium.Mathematically unstable points are those where $\frac{\mathrm{d} y}{\mathrm{d} x}$ would be +ve if you move bit right from point of equilibrium and $\frac{\mathrm{d} y}{\mathrm{d} x}$ would be -ve if you move bith left from the point of equilibrium.

Now from this theory we can easily see x=2 is an unstable point of equilibrium while x=1 and x=3 are both stable points of equilibrium.

answered by Boss (12.8k points)
edited by
0

Hi Rupendra,

You have explained just opposite to the actual concept of finding maxima(unstable) & minima(stable) equilibrium points.

For stable equilibrium, dy/dx is -ve on left and +ve on right of the equilibrium point. This actually tells us that curve is decreasing on left & increasing on right of the equilibrium point So we get U shaped curve. Hence we can say it is stable equilibrium point.

For unstable equilibrium, dy/dx is +ve on left and -ve on right of the equilibrium point. This actually tells us that curve is increasing on left & decreasing on right of the equilibrium point. So we get  ⋂ shaped curve. Hence we can say it is unstable equilibrium point(local maximum).

But here in our question, above concepts are not directly applicable. So we have to think differently because dependency between x & t is not related as per the above concepts.

Just try to find put stable/unstable points using another method e.g. using 2nd order derivative.

So finding d^2x/dt^2, how would you find using above concepts, it will get complicated. Here variable is 't' not 'x' in our question.

For more reference, read this article where I have copied the image for explanation purpose:

https://www.ma.utexas.edu/users/davis/375/popecol/lec9/equilib.html

0

Hello shashank

pardon me that i forget to write that $\frac{\mathrm{d} y}{\mathrm{d} x}$ is here a function of y (means no independent variable should appear).

and by ⋂ and ⋃ shape , i was not taking about maxima and minima , i was here just trying to get familiar with , what does equilibrium means in physical sense.

As now here to find equilibria so y' when it's function of x is of no use here. i don't think we can relate it in any manner with maxima/minima.

and now after edit , i find nothing wrong in your answer.

i studied this concept from here

http://www.math.psu.edu/tseng/class/Math251/Notes-1st%20order%20ODE%20pt2.pdf

0
Hi Rupendra,

Please have a look at the page-5 of the pdf you shared:

"All the steps are really the same, only the interpretation of the result differs. A result that would indicate a local minimum now means that the equilibrium solution/critical point is unstable; while that of a local maximum result now means an asymptotically stable equilibrium solution."

In fact we can explain any concepts like Potential Energy, Entropy, Force acting on a particle, Equilibrium, Stablity etc in terms of maxima & minima concepts. e.g when we say if a system is in a stable state, it means system has minimum energy within it and we can represent it on energy curve. So it depends how you interpret the things.

Whether y' is function of 'x' or not , we can always relate concept of local minima/ maxima to system stability. It's just that our calculation for finding these values becomes complex if y' is function of y.

Moreover if we directly draw curve from given diff. equation ( which is somewhat tricky as we need to solve diff. equation first) and then find local min/max, we can predict the stability of the system which should be consistent with the above solution.

If you rotate the graph drawn in page-3 of your pdf, you can clearly see that local minima is nothing but stable equilibrium configuration.
0
Hello shashank

Thanks for paying attention.Yes! maxima/minima playing role. and here we're doing it wrt y.

But as i'm able to see if in our original curve b/w x and t.when dx/dt becomes 0 at x axis ,its equilibrium point and if it's minima here then it will lead to unstable point of equilibrium.

but in your last line you wrote that minima will correspond to stable point of equilibrium?
0
Hi Rupendra,

In nutshell, If we consider sign of dy/dx w.r.t to 'x', dy/dx is -ve on left & +ve on right for the equilibrium point i.e. we get minima at the point which becomes stable point because graph of y = f(x) becomes decreasing on LHS  & increasing on RHS of equilibrium point.

Now if we consider say dy/dx w.r.t. to 'y', things get reversed because entire curves get reversed because of change in axis. So we can say what was earlier minima is now actually maxima and hence becomes unstable point.
0

Hello Shashank

I respect your time Shashank.If you think it's trivial question which i'm taking too long to understand , you can ignore this. But i really didn't understand your maxima/minima point.

Let suppose $y=x^2$ then $\frac{\mathrm{d} y}{\mathrm{d} x}$=$2x$

Now as we're interested in such a $\frac{\mathrm{d} y}{\mathrm{d} x}$ which is a function of $y$ so lets make it.

$\frac{\mathrm{d} y}{\mathrm{d} x}$= $2 x$ =$2 \sqrt{y}$

It's graph of $y=x^2$ which have global minima at $x=0$

Now below curve is of $\frac{\mathrm{d} y}{\mathrm{d} x}$ vs $y$

Still i don't know how minima became maxima.

I played lil bit with our original differential equation and got relation b/w x and y

$\int \frac{dy}{(1-y)(2-y)(3-y)} =\int dx$

$\int ( \frac{1}{2(1-y)}-\frac{1}{2-y}+\frac{1}{2(3-y)})dy $

so x=ln$((2-y)*\sqrt{\frac{3-y}{1-y}})$

now even this baffled me more. this says x is not define at y>=1

if i study $\frac{dy}{dx}$ curve then it says that at 2+ dy/dx=+ve and at 2- it's -ve and this is exactly what we use for minima.if we go bit right of minima point then dy/dx slop will increase and if we go bit left , slope will fo on decreasing...

0
@Rupendra Choudhary, i think you messed up with logic of maxima and minima.. if in U shape from bottom most point if you move little further right then dy/dx will be +ive  i.e means local minima which you explained opposite in you answer.

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