Coupled non-linear differential equations

[leothorn]

Homework Statement

x'= E - sin x + K sin (y-x)
y'= E + sin y + K sin (x-y)

E and K >0

Find fixed points for this system of equations

Homework Equations

This system is the form of coupled oscillators described in Strogatz.

θ1'= ω1 + K sin (θ2-θ1)
θ2'= ω2 + K sin (θ1-θ2)

The Attempt at a Solution

I made three different approaches to the problem

1) I compared to the system given in strogatz. By equating ω1 = E - sin x and ω2 = E + sin y.
But then i didnt know how to proceed because i didnt know if the ωs can be a function of x and y
as they represent the natural frequency of the system.

The approach meant i go ψ' = θ1'-θ2' = 0

Ending with sin x +sin y = -2K sin ψ
I am stumped after this
Do i compare to damped osciallator equation ?

bΩ' + mgL sinΩ = λ

if so how do i get rid of θ1 and θ2 ?


I don't know how to find stable points from this point and am hopelessly stuck

 

[voko]

At fixed points, x' = y' = 0, so the system of differential equations is reduced to a system of ordinary equations, and you need to find all the pairs of (x, y) satisfying it.

 

[leothorn]

Yes that wht i did with phi'=0
but i didnt get points instead getting the question shown above it. Nvm ill put up my whole worked sheet as scans.
typing it wil take an era

 

[voko]

If typing will take an era, then it is not worth it. Consider the following change of variables $u = \frac {x + y} 2, \ v = \frac {x - y} 2$.

 

[leothorn]

I get u' = -cos u sin v

so either sinv =0 or cosu = 0 so v =arcsin 0 +- npi , u =arccos 0 +- npi

Ordinarily we take the jacobian of the systems to get eigen values and eigen vectors to state something about the stabiiy of the points.
What do i do in this case ?

 

[voko]

I get u' = -cos u sin v

I do not think this is correct. You may want to show your steps.

Ordinarily we take the jacobian of the systems to get eigen values and eigen vectors to state something about the stabiiy of the points.
What do i do in this case ?

Once you get the correct fixed points, you linearize the system in their neighborhoods. Which may be what you mean by "taking the Jacobian". Once you get the linear system, you examine it for stability.

 

[leothorn]

from the transformation equations we get

x = u +v
y = u -v

so u'=x'+y' and v'=x'-y'

Now the values are put into get the equations

u'= E-cos u sin v (1)
v'= -sin u cos v - K sin 2v (2)

Now setting the system to zero for fixed points

we get,

E = (-1/4K) sin 2u

sin v = (-1/2K) sin u

Now,

Taking the jacobian for (1) and (2)...

J11 = sinu sin v
J12 = -cos u cos
J21 = -cosu cos v
J22 = sinu sinv - 2K cos 2v

..And complicates ...

 

[voko]

I think you need to find what u and v are at the fixed points, and you also need to compute the Jacobian at those points. Your objective is to get a linear system with constant coefficients.

 

[leothorn]

I tried to solve the problem just using phi = x - y
this is the worked out solution.
Wanted to check if the method in is acceptable. Solution in the first 3 pages is actual manipulation the remaining are the stability interpretation of the result


Page 1 : http://postimg.org/image/orpkchn8f/
Page 2 : http://postimg.org/image/l9n69rf3n/
Page 3 : http://postimg.org/image/p610cntir/
Page 4 : http://postimg.org/image/627owbgoj/
Page 5 : http://postimg.org/image/517g76zoz/
Page 6 : http://postimg.org/image/tizjv329f/
Page 7 : http://postimg.org/image/qq6cb21wz/
Page 8 : http://postimg.org/image/91eljfq5v/

 

[voko]

In the step following (3) and (4) on page 1, you found only a subset of possible solutions. You have an equation

-a sin x - a sin y = sin x cos y - cos x sin y

which is equivalent to

(cos y + a) sin x = (cos x - a) sin y

The pair (a = -cos y, a = cos x) is indeed a set of solutions, but not the entire set.

First of all, if a = ±1, then cos y = -1 is another set of solutions independent of x; likewise, cos x = 1 is a set of solutions independent of y.

Second, sin x = sin y = 0 is yet another set of solutions.

If none of the conditions above apply, then sin x / (cos x - a) = sin y / (cos y + a) has another set of solutions.

Another remark is that all of the above is only obtained from the analysis of x' - y' = 0. Solutions obtained from that consideration only may not be solutions for the entire system. For example, if E is large enough (and K is not as large), there are no solutions at all.

 

[leothorn]

Sorry I am a bit confused by what you said . Let me reiterate what you said to confirm that I understood you.

I have the equation

sin (x-y) = -a sin x - a sin y where a = 1/2K

I compared this to (sin x cos y - cos x sin y) and concluded

1 )cos y = (-a) and cos x = (a) [ Point to note here is that K >0 therefore a>0 ]

[ Also K is not defined between [0 and 1/2] because then arcsin /arcos become out of bounds ..out of [-1 1] range ]

so if a = +1 that is K = 1/2 is one solution set (?) i.e cos y = -1 and cos x = 1 which makes y = pi rad / x = 0

I discussed this condition at : page 7 where I concluded that at that condition we have star node.

2) If both sin x and sin y are zero the whole system becomes zero doesn't it ? Then its has fixed points all over the plane . That would be a trivial solution (?)

3) Even this involves setting sin(x-y) = 0 that is setting the system zero (?) . Didnt understand this argument too well

4) I agree when E is large then we have all degenerate points. Since this system is better explained on a torus , i didnt know how to proceed with just linearisation

E disappears on linearisation by this method

 

[voko]

1 )cos y = (-a) and cos x = (a) [ Point to note here is that K >0 therefore a>0 ]

[ Also K is not defined between [0 and 1/2] because then arcsin /arcos become out of bounds ..out of [-1 1] range ]

This is not correct; you should say that solution 1) is only possible if K > 1/2.

so if a = +1 that is K = 1/2 is one solution set (?) i.e cos y = -1 and cos x = 1 which makes y = pi rad / x = 0

I discussed this condition at : page 7 where I concluded that at that condition we have star node.

If a = 1, and cos y = -1, then sin y = 0, so you have a zero solution no matter what x is. cos x = 1 results in a similar situation. So instead of isolated fixed points, you have infinite lines.

2) If both sin x and sin y are zero the whole system becomes zero doesn't it ? Then its has fixed points all over the plane . That would be a trivial solution (?)

No, and this is where you should consider the other equations, or just plug that into the original system, where you would get x' = E, y ' = E - does that define a fixed point?

3) Even this involves setting sin(x-y) = 0 that is setting the system zero (?) . Didnt understand this argument too well

Are you referring to the general solution sin x / (cos x - a) = sin y / (cos y + a)?

Observe that sin y = sin (π - y), and cos y = - cos (π - y), so if you let f(x) = sin x / (cos x - a), then the equation is f(x) = -f(π - y). Note further that f(-x) = -f(x), so the equation simplifies to f(x) = f(y - π). Now, taking into account that f(x) is naturally 2π-periodic, what are the solutions?

Note the solutions exist for any value of a.

 

[leothorn]

before correcting my solution , are u saying that the transformation you suggested would yield broader generalized results ?

 

[voko]

before correcting my solution , are u saying that the transformation you suggested would yield broader generalized results ?

I believe so, but you don't have to listen to me, see for yourself - you will have to, anyway.

 

[leothorn]

I used the transformation suggested. I am stuck at the page 2.
The un-elegance of the jacobian prompts me to believe there could be a better substitution or I am wrong somewhere in the beginning.

Can u suggest a way out ?Page 1 : http://postimg.org/image/81ljfhhdf/
Page 2 : http://postimg.org/image/gxwbjf7zn/

Basically once gets stuck unable to find u and v without going into multuiple arcsin transformations
and the Jacobian becomes too complicated to handle analytically.

 

[voko]

Quite frankly, I am surprised that you are switching to u and v at this stage.

Anyway, let $z = \sin u$, then $\cos u = \sqrt {1 - z^2}$, and you obtain $\sin u$ and $\cos u$, from which you can obtain $\sin v$ and $\cos v$.

 

[leothorn]

Trying it out

 

[leothorn]

Quite frankly, I am surprised that you are switching to u and v at this stage.

Anyway, let $z = \sin u$, then $\cos u = \sqrt {1 - z^2}$, and you obtain $\sin u$ and $\cos u$, from which you can obtain $\sin v$ and $\cos v$.

I don't know what to do .

Where should i have started substitution ?
I am beyond hope right now and i am not able to cross the mental block.

Hints and suggestions are wasted on me right now. You will have to explain to me explicitly if I am to solve this :)

Thanks

 

[voko]

Hmm. This is pre-calculus stuff, you should be able to do that cold! You have: $$

\sin v = - \frac 1 {2K} \sin u

\\

\sin u \cos u = - 2EK

$$ Let $z = \sin u$, then $$

z \sqrt {1 - z^2} = - 2EK

\\

z^2(1 - z^2) = 4(EK)^2

\\

(z^2)^2 - z^2 + 4(EK)^2 = 0

\\

z^2 = \frac {1 \pm \sqrt {1 - 16(EK)^2} } 2

$$ Thus $$

\sin u = \pm \sqrt {\frac {1 \pm \sqrt {1 - 16(EK)^2} } 2}

\\

\cos u = \mp \sqrt {\frac {1 \mp \sqrt {1 - 16(EK)^2} } 2}

\\

\sin v = \mp \frac 1 {2K} \sqrt {\frac {1 \pm \sqrt {1 - 16(EK)^2} } 2}

$$

Continue from here. Note also that $\cos v = 0$ that you dismissed on page 2 may also yield a solution set.

 

[leothorn]

I don't know about doing this cold. This occurred to me . But I will be knocked out cold by the time I solve this matrix analytically and then calculate its eigen values :P
There seems to be no respite to the complexity . There has to be a better substitution . Thanks

 

[voko]

I do not see impenetrable difficulties here. You can continue with u and v or go back to x and y.