Find Hamiltonian from equation of motion
y^{..} + L(1-y^k)y^{.} + y =0
where y^{.}= dy/dt and y^{..) = second derivative with time
The Hamiltonian is just a name for the energy, which means that the equation should be consistent with energy being constant. Multiplying by dy/dt both sides shows, as expected, that the term proportional to dy/dt in the equation spoils this, since one finds that
(d/dt)( (1/2)(dy/dt)2 + (1/2)y2 ) + L (dy/dt)2(1-yk)=0
The first term is a total derivative with respect to time-the second term isn't.
Therefore the system isn't Hamiltonian.
Another way of seeing this is by defining dy/dt = z and writing the equation as the system
dy/dt = z = f(y,z)
dz/dt = -L(1-yk)z-y = g(y,z)
and showing that the divergence of the vector field (f,g) doesn't vanish, as it should, were the system Hamiltonian:
df/dy + dg/dz = -L(1-yk)≠0
the volume in phase space isn't conserved.
The Hamiltonian is just a name for the energy, which means that the equation should be consistent with energy being constant. Multiplying by dy/dt both sides shows, as expected, that the term proportional to dy/dt in the equation spoils this, since one finds that
(d/dt)( (1/2)(dy/dt)2 + (1/2)y2 ) + L (dy/dt)2(1-yk)=0
The first term is a total derivative with respect to time-the second term isn't.
Therefore the system isn't Hamiltonian.
Another way of seeing this is by defining dy/dt = z and writing the equation as the system
dy/dt = z = f(y,z)
dz/dt = -L(1-yk)z-y = g(y,z)
and showing that the divergence of the vector field (f,g) doesn't vanish, as it should, were the system Hamiltonian:
df/dy + dg/dz = -L(1-yk)≠0
the volume in phase space isn't conserved.
Once more: for the reasons mentioned a Hamiltonian doesn't exist, so no such expression can be given. Not to mention the obvious lack of any obligation for giving any expression, even did it exist. A mastery of English might be useful, along with a mastery of physics.
Dear Prof Stam Nicolis
For k=2 ; L=-1 ,the above eqn reduces to Van der Pol Oscillator. If you have interest,
give your views on Hamiltonian . RG is a forum for discussion .
Prof Rath
Just look at the expression provided in a previous message: for L non-zero, no matter what the value of k is, the first derivative term precludes the existence of a Hamiltonian-as the one-line calculation shows. it doesn't matter what it's called. So, for the third time: this system, for L non-zero is not a Hamiltonian system. Not all systems are Hamiltonian.
Incidentally, L is a dimensionful parameter and units can be chosen so that its numerical value isn't relevant-only its sign is. For L non-zero, the system isn't Hamiltonian, while for L=0 it is.
Dear Prof Damion
Thanks for reference .As I know Van der Pol oscillator gives stable solution.
Further this problem is very old in Physics . However I want to study it again .
Regard
Prof Rath
It is not possible to write a time independent Hamiltonian for the equation of motion you have mentioned because of the presence of attractor in the system. However, one can write a Hamiltonian for the system by extending the phase space of the system through the introduction of an auxiliary variable as shown in the attached publication. In the paper attached, Hamiltonian is written for van der Pol Oscillator. However, you can extend their formulation easily to write a general Hamiltonian for the equation you have mentioned.
In principle, it is possible to write a time dependent Hamiltonian for the equation but I don't think researchers have found it till now.
I hope that I have answered your question. If not, feel free to reply back .
Article Conservative perturbation theory for nonconservative systems
The term pq=qdq/dt is a total derivative. Therefore it contributes a surface term and does not affect the equations of motion. Using impressive words doesn't, necessarily, impart meaning, if one just states them and doesn't follow them through. indeed what the ``covariant derivative'' implies is that there doesn't exist a Hamiltonian, because the generator of time translations is the covariant derivative, not the ordinary derivative. So one reaches the same result in this way, too.
One must close the system consistently, which means provide a description of the degrees of freedom that contribute the non-Hamiltonian term, and one can obtain a Hamiltonian description for the full system. This description isn't unique, of course.
Dear Shah
thanks for paper in PRE.Give the explicit form of Hamiltonian in Eq(6) in !D.
Prof Rath
If the vibrabration is controlled or sustained ,why we are not able to write Hamiltonian
from the diff eqn of second order ?
Read the Shah paper in eqn(6) and write the Hamiltonian .
Because, in this case, the vibration is neither controlled, nor sustained, that's what the term proportional to the first derivative prevents from occurring.
The following papers discuss van der Pol Oscillators
(i) J.H.He " Determination of limit cycles for strongly Nonlinear Oscillators " Phys.Rev.Lett 90,174301(2003).
(ii) S. Rajendran et.al Comment on above paper in Phys.Rev.Lett ,93(6),069401(2004)
(iii) B.Rath and S.Agarwal "Controlled and uncontrolled vibrations " Proc.Nat.Aca.Sci(India) " 84A,83(2014) and references cited therein.
Any suggestion on above papers for modifications / rectifications /comments are highly accepted with thanks .
Prof Rath
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