Mixing
Prove that no periodic orbits exist
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Please help!
I need to prove for the system below that no periodic orbits exist when $V_0=0$:
$frac{d^2x}{dt^2}+zeta_1 frac{dx}{dt}+(k_{11}+k_{12}x^2)x-2gammafrac{dy}{dt}=V_0 cos(omega t),$
$frac{d^2y}{dt^2}+zeta_2 frac{dy}{dt}+(k_{21}+k_{22}y^2)y+2gammafrac{dx}{dt}=0$.
x and y are displacements, and for mechanical understanding, $zeta_1$ and $zeta_2$ are damping parameters, $k_{11}, k_{12}, k_{21}, k_{22}$ are stiffness parameters, $V_0$ is a prescribed input voltage and $gamma$ is the rotation rate. Can someone help me out here?
nonlinear-system
asked Jan 9 at 9:23
DennisDennis
12
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add a comment |
$begingroup$
Please help!
I need to prove for the system below that no periodic orbits exist when $V_0=0$:
$frac{d^2x}{dt^2}+zeta_1 frac{dx}{dt}+(k_{11}+k_{12}x^2)x-2gammafrac{dy}{dt}=V_0 cos(omega t),$
$frac{d^2y}{dt^2}+zeta_2 frac{dy}{dt}+(k_{21}+k_{22}y^2)y+2gammafrac{dx}{dt}=0$.
x and y are displacements, and for mechanical understanding, $zeta_1$ and $zeta_2$ are damping parameters, $k_{11}, k_{12}, k_{21}, k_{22}$ are stiffness parameters, $V_0$ is a prescribed input voltage and $gamma$ is the rotation rate. Can someone help me out here?
nonlinear-system
asked Jan 9 at 9:23
DennisDennis
12
$endgroup$
$begingroup$
Which techniques to prove such results do you know / have you tried?
$endgroup$
– Did
Jan 9 at 9:25
$begingroup$
I tried using the Poincaré-Bendixson criterion, but couldn't find how to apply this on a coupled system. I use the book Nonlinear systems of Hassan K. Khalil, the 3rd edition.
$endgroup$
– Dennis
Jan 9 at 9:30
add a comment |
$begingroup$
Please help!
I need to prove for the system below that no periodic orbits exist when $V_0=0$:
$frac{d^2x}{dt^2}+zeta_1 frac{dx}{dt}+(k_{11}+k_{12}x^2)x-2gammafrac{dy}{dt}=V_0 cos(omega t),$
$frac{d^2y}{dt^2}+zeta_2 frac{dy}{dt}+(k_{21}+k_{22}y^2)y+2gammafrac{dx}{dt}=0$.
x and y are displacements, and for mechanical understanding, $zeta_1$ and $zeta_2$ are damping parameters, $k_{11}, k_{12}, k_{21}, k_{22}$ are stiffness parameters, $V_0$ is a prescribed input voltage and $gamma$ is the rotation rate. Can someone help me out here?
nonlinear-system
asked Jan 9 at 9:23
DennisDennis
12
$endgroup$
Please help!
I need to prove for the system below that no periodic orbits exist when $V_0=0$:
$frac{d^2x}{dt^2}+zeta_1 frac{dx}{dt}+(k_{11}+k_{12}x^2)x-2gammafrac{dy}{dt}=V_0 cos(omega t),$
$frac{d^2y}{dt^2}+zeta_2 frac{dy}{dt}+(k_{21}+k_{22}y^2)y+2gammafrac{dx}{dt}=0$.
x and y are displacements, and for mechanical understanding, $zeta_1$ and $zeta_2$ are damping parameters, $k_{11}, k_{12}, k_{21}, k_{22}$ are stiffness parameters, $V_0$ is a prescribed input voltage and $gamma$ is the rotation rate. Can someone help me out here?
nonlinear-system
nonlinear-system
asked Jan 9 at 9:23
DennisDennis
12
asked Jan 9 at 9:23
DennisDennis
12
asked Jan 9 at 9:23
DennisDennis
12
asked Jan 9 at 9:23
DennisDennis
12
asked Jan 9 at 9:23
DennisDennis
12
12
$begingroup$
Which techniques to prove such results do you know / have you tried?
$endgroup$
– Did
Jan 9 at 9:25
$begingroup$
I tried using the Poincaré-Bendixson criterion, but couldn't find how to apply this on a coupled system. I use the book Nonlinear systems of Hassan K. Khalil, the 3rd edition.
$endgroup$
– Dennis
Jan 9 at 9:30
add a comment |
$begingroup$
Which techniques to prove such results do you know / have you tried?
$endgroup$
– Did
Jan 9 at 9:25
$begingroup$
I tried using the Poincaré-Bendixson criterion, but couldn't find how to apply this on a coupled system. I use the book Nonlinear systems of Hassan K. Khalil, the 3rd edition.
$endgroup$
– Dennis
Jan 9 at 9:30
$begingroup$
Which techniques to prove such results do you know / have you tried?
$endgroup$
– Did
Jan 9 at 9:25
$begingroup$
Which techniques to prove such results do you know / have you tried?
$endgroup$
– Did
Jan 9 at 9:25
$begingroup$
I tried using the Poincaré-Bendixson criterion, but couldn't find how to apply this on a coupled system. I use the book Nonlinear systems of Hassan K. Khalil, the 3rd edition.
$endgroup$
– Dennis
Jan 9 at 9:30
$begingroup$
I tried using the Poincaré-Bendixson criterion, but couldn't find how to apply this on a coupled system. I use the book Nonlinear systems of Hassan K. Khalil, the 3rd edition.
$endgroup$
– Dennis
Jan 9 at 9:30
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
You start with the standard trick to get from dynamic equations to energy equations, which is multiply first equation by $dot x$, second by $dot y$ and sum them:
$$
ddot xdot x+zeta_1dot x^2+(k_{11}+k_{12}x^2)xdot x -2gamma dot ydot x=
frac{d}{dt}left(frac12dot x^2+frac{k_{11}}2x^2+frac{k_{12}}4x^4right)+zeta_1dot x^2-2gammadot xdot y=0,\
ddot ydot y+zeta_2dot y^2+(k_{21}+k_{22}y^2)ydot y +2gamma dot xdot y=
frac{d}{dt}left(frac12dot y^2+frac{k_{21}}2y^2+frac{k_{22}}4y^4right)+zeta_2dot y^2+2gammadot xdot y=0,\
frac{dE}{dt}=-zeta_1dot x^2-zeta_2dot y^2, qquadtext{where}qquad E=frac14left(2dot x^2+2dot y^2+2k_{11}x^2+2k_{21}y^2+k_{12}x^4+k_{22}y^4right).
$$
So now we know for sure, that if both $zeta_1,zeta_2>0$ (or both $<0$), then energy is strictly decreasing (or increasing) and therefore there are no periodic orbits.
However, if $zeta_1=zeta_2=0$ or $zeta_1zeta_2<0$, then with some parameters there can be periodic orbits.
For example, if $k_{11}=k_{21}=1$ and everything else is zero, it's just two harmonic oscillator and every orbit is periodic. Or if $zeta_2=-zeta_1=1$, all $k=0$ and $gamma=1$, then for $z=(dot x, dot y)$ it is linear equation $dot z=Az$ with pure imaginary eigenvalues. Again, every orbit is periodic.
answered Jan 9 at 11:14
Vasily MitchVasily Mitch
2,1491311
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1 Answer
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$begingroup$
You start with the standard trick to get from dynamic equations to energy equations, which is multiply first equation by $dot x$, second by $dot y$ and sum them:
$$
ddot xdot x+zeta_1dot x^2+(k_{11}+k_{12}x^2)xdot x -2gamma dot ydot x=
frac{d}{dt}left(frac12dot x^2+frac{k_{11}}2x^2+frac{k_{12}}4x^4right)+zeta_1dot x^2-2gammadot xdot y=0,\
ddot ydot y+zeta_2dot y^2+(k_{21}+k_{22}y^2)ydot y +2gamma dot xdot y=
frac{d}{dt}left(frac12dot y^2+frac{k_{21}}2y^2+frac{k_{22}}4y^4right)+zeta_2dot y^2+2gammadot xdot y=0,\
frac{dE}{dt}=-zeta_1dot x^2-zeta_2dot y^2, qquadtext{where}qquad E=frac14left(2dot x^2+2dot y^2+2k_{11}x^2+2k_{21}y^2+k_{12}x^4+k_{22}y^4right).
$$
So now we know for sure, that if both $zeta_1,zeta_2>0$ (or both $<0$), then energy is strictly decreasing (or increasing) and therefore there are no periodic orbits.
However, if $zeta_1=zeta_2=0$ or $zeta_1zeta_2<0$, then with some parameters there can be periodic orbits.
For example, if $k_{11}=k_{21}=1$ and everything else is zero, it's just two harmonic oscillator and every orbit is periodic. Or if $zeta_2=-zeta_1=1$, all $k=0$ and $gamma=1$, then for $z=(dot x, dot y)$ it is linear equation $dot z=Az$ with pure imaginary eigenvalues. Again, every orbit is periodic.
answered Jan 9 at 11:14
Vasily MitchVasily Mitch
2,1491311
$endgroup$
add a comment |
$begingroup$
You start with the standard trick to get from dynamic equations to energy equations, which is multiply first equation by $dot x$, second by $dot y$ and sum them:
$$
ddot xdot x+zeta_1dot x^2+(k_{11}+k_{12}x^2)xdot x -2gamma dot ydot x=
frac{d}{dt}left(frac12dot x^2+frac{k_{11}}2x^2+frac{k_{12}}4x^4right)+zeta_1dot x^2-2gammadot xdot y=0,\
ddot ydot y+zeta_2dot y^2+(k_{21}+k_{22}y^2)ydot y +2gamma dot xdot y=
frac{d}{dt}left(frac12dot y^2+frac{k_{21}}2y^2+frac{k_{22}}4y^4right)+zeta_2dot y^2+2gammadot xdot y=0,\
frac{dE}{dt}=-zeta_1dot x^2-zeta_2dot y^2, qquadtext{where}qquad E=frac14left(2dot x^2+2dot y^2+2k_{11}x^2+2k_{21}y^2+k_{12}x^4+k_{22}y^4right).
$$
So now we know for sure, that if both $zeta_1,zeta_2>0$ (or both $<0$), then energy is strictly decreasing (or increasing) and therefore there are no periodic orbits.
However, if $zeta_1=zeta_2=0$ or $zeta_1zeta_2<0$, then with some parameters there can be periodic orbits.
For example, if $k_{11}=k_{21}=1$ and everything else is zero, it's just two harmonic oscillator and every orbit is periodic. Or if $zeta_2=-zeta_1=1$, all $k=0$ and $gamma=1$, then for $z=(dot x, dot y)$ it is linear equation $dot z=Az$ with pure imaginary eigenvalues. Again, every orbit is periodic.
answered Jan 9 at 11:14
Vasily MitchVasily Mitch
2,1491311
$endgroup$
add a comment |
$begingroup$
You start with the standard trick to get from dynamic equations to energy equations, which is multiply first equation by $dot x$, second by $dot y$ and sum them:
$$
ddot xdot x+zeta_1dot x^2+(k_{11}+k_{12}x^2)xdot x -2gamma dot ydot x=
frac{d}{dt}left(frac12dot x^2+frac{k_{11}}2x^2+frac{k_{12}}4x^4right)+zeta_1dot x^2-2gammadot xdot y=0,\
ddot ydot y+zeta_2dot y^2+(k_{21}+k_{22}y^2)ydot y +2gamma dot xdot y=
frac{d}{dt}left(frac12dot y^2+frac{k_{21}}2y^2+frac{k_{22}}4y^4right)+zeta_2dot y^2+2gammadot xdot y=0,\
frac{dE}{dt}=-zeta_1dot x^2-zeta_2dot y^2, qquadtext{where}qquad E=frac14left(2dot x^2+2dot y^2+2k_{11}x^2+2k_{21}y^2+k_{12}x^4+k_{22}y^4right).
$$
So now we know for sure, that if both $zeta_1,zeta_2>0$ (or both $<0$), then energy is strictly decreasing (or increasing) and therefore there are no periodic orbits.
However, if $zeta_1=zeta_2=0$ or $zeta_1zeta_2<0$, then with some parameters there can be periodic orbits.
For example, if $k_{11}=k_{21}=1$ and everything else is zero, it's just two harmonic oscillator and every orbit is periodic. Or if $zeta_2=-zeta_1=1$, all $k=0$ and $gamma=1$, then for $z=(dot x, dot y)$ it is linear equation $dot z=Az$ with pure imaginary eigenvalues. Again, every orbit is periodic.
answered Jan 9 at 11:14
Vasily MitchVasily Mitch
2,1491311
$endgroup$
You start with the standard trick to get from dynamic equations to energy equations, which is multiply first equation by $dot x$, second by $dot y$ and sum them:
$$
ddot xdot x+zeta_1dot x^2+(k_{11}+k_{12}x^2)xdot x -2gamma dot ydot x=
frac{d}{dt}left(frac12dot x^2+frac{k_{11}}2x^2+frac{k_{12}}4x^4right)+zeta_1dot x^2-2gammadot xdot y=0,\
ddot ydot y+zeta_2dot y^2+(k_{21}+k_{22}y^2)ydot y +2gamma dot xdot y=
frac{d}{dt}left(frac12dot y^2+frac{k_{21}}2y^2+frac{k_{22}}4y^4right)+zeta_2dot y^2+2gammadot xdot y=0,\
frac{dE}{dt}=-zeta_1dot x^2-zeta_2dot y^2, qquadtext{where}qquad E=frac14left(2dot x^2+2dot y^2+2k_{11}x^2+2k_{21}y^2+k_{12}x^4+k_{22}y^4right).
$$
So now we know for sure, that if both $zeta_1,zeta_2>0$ (or both $<0$), then energy is strictly decreasing (or increasing) and therefore there are no periodic orbits.
However, if $zeta_1=zeta_2=0$ or $zeta_1zeta_2<0$, then with some parameters there can be periodic orbits.
For example, if $k_{11}=k_{21}=1$ and everything else is zero, it's just two harmonic oscillator and every orbit is periodic. Or if $zeta_2=-zeta_1=1$, all $k=0$ and $gamma=1$, then for $z=(dot x, dot y)$ it is linear equation $dot z=Az$ with pure imaginary eigenvalues. Again, every orbit is periodic.
answered Jan 9 at 11:14
Vasily MitchVasily Mitch
2,1491311
answered Jan 9 at 11:14
Vasily MitchVasily Mitch
2,1491311
answered Jan 9 at 11:14
Vasily MitchVasily Mitch
2,1491311
answered Jan 9 at 11:14
Vasily MitchVasily Mitch
2,1491311
2,1491311
add a comment |
add a comment |
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$begingroup$
Which techniques to prove such results do you know / have you tried?
$endgroup$
– Did
Jan 9 at 9:25
$begingroup$
I tried using the Poincaré-Bendixson criterion, but couldn't find how to apply this on a coupled system. I use the book Nonlinear systems of Hassan K. Khalil, the 3rd edition.
$endgroup$
– Dennis
Jan 9 at 9:30