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Prove that the system with $dotx = x+y-4x^3$ and $doty = y-x-4y^3$ has a limit cycle
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Prove that the system with $dotx = x+y-4x^3$ and $doty = y-x-4y^3$ has a limit cycle.
How do I begin diving in on this problem? Should it be solved through the use of a Lyapunov function, or is this a Poincare-Bendixon problem? Also some guidance on where to go after that would be helpful as well.
dynamical-systems mathematical-modeling nonlinear-system
edited Jul 18 at 13:48
Did
242k23208443
asked Jul 18 at 3:18
obewanjacobi
312110
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up vote
2
down vote
favorite
Prove that the system with $dotx = x+y-4x^3$ and $doty = y-x-4y^3$ has a limit cycle.
How do I begin diving in on this problem? Should it be solved through the use of a Lyapunov function, or is this a Poincare-Bendixon problem? Also some guidance on where to go after that would be helpful as well.
dynamical-systems mathematical-modeling nonlinear-system
edited Jul 18 at 13:48
Did
242k23208443
asked Jul 18 at 3:18
obewanjacobi
312110
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Prove that the system with $dotx = x+y-4x^3$ and $doty = y-x-4y^3$ has a limit cycle.
How do I begin diving in on this problem? Should it be solved through the use of a Lyapunov function, or is this a Poincare-Bendixon problem? Also some guidance on where to go after that would be helpful as well.
dynamical-systems mathematical-modeling nonlinear-system
edited Jul 18 at 13:48
Did
242k23208443
asked Jul 18 at 3:18
obewanjacobi
312110
Prove that the system with $dotx = x+y-4x^3$ and $doty = y-x-4y^3$ has a limit cycle.
How do I begin diving in on this problem? Should it be solved through the use of a Lyapunov function, or is this a Poincare-Bendixon problem? Also some guidance on where to go after that would be helpful as well.
dynamical-systems mathematical-modeling nonlinear-system
edited Jul 18 at 13:48
Did
242k23208443
asked Jul 18 at 3:18
obewanjacobi
312110
edited Jul 18 at 13:48
Did
242k23208443
edited Jul 18 at 13:48
Did
242k23208443
edited Jul 18 at 13:48
Did
242k23208443
242k23208443
asked Jul 18 at 3:18
obewanjacobi
312110
asked Jul 18 at 3:18
obewanjacobi
312110
asked Jul 18 at 3:18
obewanjacobi
312110
312110
add a comment |Â
add a comment |Â
1 Answer
1
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oldest
votes
up vote
5
down vote
accepted
Using the squared radius $V=x^2+y^2$ as Lyapunov function, one finds, similar to what was computed for the radius,
$$
dot V = 2V-8(x^4+y^4)
$$
and using
$$
frac12(x^2+y^2)^2+frac12(x^2-y^2)^2=x^4+y^4=(x^2+y^2)^2-2(xy)^2
$$
gives
$$
2V-8V^2le dot Vle 2V-4V^2
$$
This already tells us that the vector field points outwards, away from zero, for $Vlefrac14$ and inwards towards zero for $V>frac12$. This is already sufficient to apply Poincaré-Bendixon.
The differential inequality can be solved to give more information on the dynamic of the radius, as
$$
4le fracddt(V^-1)+2(V^-1)le 8,\~\
2(e^2t-1)le e^2tV(t)^-1-V_0^-1le 4(e^2t-1),\~\
fracV_0e^-2t+4V_0(1-e^-2t)le V(t) lefracV_0e^-2t+2V_0(1-e^-2t),
$$
which tells us that the dynamic moves every solution towards the annulus $frac14le V(t)lefrac12$,
$$
frac12le rle sqrtfrac12.
$$
To conclude that there is a limiting circle one now needs to check that there are no stationary points inside this annulus.
answered Jul 18 at 13:38
LutzL
49.8k31849
How is $x^4 + y^4 = x^2 + y^2 -2(xy)^2$? Or did you mean $x^4 + y^4 = (x^2 + y^2)^2 - 2(xy)^2$?
– Robert Lewis
Jul 18 at 15:32
1
@RobertLewis Of course the latter. Thanks for the hint.
– LutzL
Jul 18 at 16:03
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
Using the squared radius $V=x^2+y^2$ as Lyapunov function, one finds, similar to what was computed for the radius,
$$
dot V = 2V-8(x^4+y^4)
$$
and using
$$
frac12(x^2+y^2)^2+frac12(x^2-y^2)^2=x^4+y^4=(x^2+y^2)^2-2(xy)^2
$$
gives
$$
2V-8V^2le dot Vle 2V-4V^2
$$
This already tells us that the vector field points outwards, away from zero, for $Vlefrac14$ and inwards towards zero for $V>frac12$. This is already sufficient to apply Poincaré-Bendixon.
The differential inequality can be solved to give more information on the dynamic of the radius, as
$$
4le fracddt(V^-1)+2(V^-1)le 8,\~\
2(e^2t-1)le e^2tV(t)^-1-V_0^-1le 4(e^2t-1),\~\
fracV_0e^-2t+4V_0(1-e^-2t)le V(t) lefracV_0e^-2t+2V_0(1-e^-2t),
$$
which tells us that the dynamic moves every solution towards the annulus $frac14le V(t)lefrac12$,
$$
frac12le rle sqrtfrac12.
$$
To conclude that there is a limiting circle one now needs to check that there are no stationary points inside this annulus.
answered Jul 18 at 13:38
LutzL
49.8k31849
How is $x^4 + y^4 = x^2 + y^2 -2(xy)^2$? Or did you mean $x^4 + y^4 = (x^2 + y^2)^2 - 2(xy)^2$?
– Robert Lewis
Jul 18 at 15:32
1
@RobertLewis Of course the latter. Thanks for the hint.
– LutzL
Jul 18 at 16:03
add a comment |Â
up vote
5
down vote
accepted
Using the squared radius $V=x^2+y^2$ as Lyapunov function, one finds, similar to what was computed for the radius,
$$
dot V = 2V-8(x^4+y^4)
$$
and using
$$
frac12(x^2+y^2)^2+frac12(x^2-y^2)^2=x^4+y^4=(x^2+y^2)^2-2(xy)^2
$$
gives
$$
2V-8V^2le dot Vle 2V-4V^2
$$
This already tells us that the vector field points outwards, away from zero, for $Vlefrac14$ and inwards towards zero for $V>frac12$. This is already sufficient to apply Poincaré-Bendixon.
The differential inequality can be solved to give more information on the dynamic of the radius, as
$$
4le fracddt(V^-1)+2(V^-1)le 8,\~\
2(e^2t-1)le e^2tV(t)^-1-V_0^-1le 4(e^2t-1),\~\
fracV_0e^-2t+4V_0(1-e^-2t)le V(t) lefracV_0e^-2t+2V_0(1-e^-2t),
$$
which tells us that the dynamic moves every solution towards the annulus $frac14le V(t)lefrac12$,
$$
frac12le rle sqrtfrac12.
$$
To conclude that there is a limiting circle one now needs to check that there are no stationary points inside this annulus.
answered Jul 18 at 13:38
LutzL
49.8k31849
How is $x^4 + y^4 = x^2 + y^2 -2(xy)^2$? Or did you mean $x^4 + y^4 = (x^2 + y^2)^2 - 2(xy)^2$?
– Robert Lewis
Jul 18 at 15:32
1
@RobertLewis Of course the latter. Thanks for the hint.
– LutzL
Jul 18 at 16:03
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
Using the squared radius $V=x^2+y^2$ as Lyapunov function, one finds, similar to what was computed for the radius,
$$
dot V = 2V-8(x^4+y^4)
$$
and using
$$
frac12(x^2+y^2)^2+frac12(x^2-y^2)^2=x^4+y^4=(x^2+y^2)^2-2(xy)^2
$$
gives
$$
2V-8V^2le dot Vle 2V-4V^2
$$
This already tells us that the vector field points outwards, away from zero, for $Vlefrac14$ and inwards towards zero for $V>frac12$. This is already sufficient to apply Poincaré-Bendixon.
The differential inequality can be solved to give more information on the dynamic of the radius, as
$$
4le fracddt(V^-1)+2(V^-1)le 8,\~\
2(e^2t-1)le e^2tV(t)^-1-V_0^-1le 4(e^2t-1),\~\
fracV_0e^-2t+4V_0(1-e^-2t)le V(t) lefracV_0e^-2t+2V_0(1-e^-2t),
$$
which tells us that the dynamic moves every solution towards the annulus $frac14le V(t)lefrac12$,
$$
frac12le rle sqrtfrac12.
$$
To conclude that there is a limiting circle one now needs to check that there are no stationary points inside this annulus.
answered Jul 18 at 13:38
LutzL
49.8k31849
Using the squared radius $V=x^2+y^2$ as Lyapunov function, one finds, similar to what was computed for the radius,
$$
dot V = 2V-8(x^4+y^4)
$$
and using
$$
frac12(x^2+y^2)^2+frac12(x^2-y^2)^2=x^4+y^4=(x^2+y^2)^2-2(xy)^2
$$
gives
$$
2V-8V^2le dot Vle 2V-4V^2
$$
This already tells us that the vector field points outwards, away from zero, for $Vlefrac14$ and inwards towards zero for $V>frac12$. This is already sufficient to apply Poincaré-Bendixon.
The differential inequality can be solved to give more information on the dynamic of the radius, as
$$
4le fracddt(V^-1)+2(V^-1)le 8,\~\
2(e^2t-1)le e^2tV(t)^-1-V_0^-1le 4(e^2t-1),\~\
fracV_0e^-2t+4V_0(1-e^-2t)le V(t) lefracV_0e^-2t+2V_0(1-e^-2t),
$$
which tells us that the dynamic moves every solution towards the annulus $frac14le V(t)lefrac12$,
$$
frac12le rle sqrtfrac12.
$$
To conclude that there is a limiting circle one now needs to check that there are no stationary points inside this annulus.
answered Jul 18 at 13:38
LutzL
49.8k31849
edited Jul 18 at 17:11
answered Jul 18 at 13:38
LutzL
49.8k31849
answered Jul 18 at 13:38
LutzL
49.8k31849
answered Jul 18 at 13:38
LutzL
49.8k31849
49.8k31849
How is $x^4 + y^4 = x^2 + y^2 -2(xy)^2$? Or did you mean $x^4 + y^4 = (x^2 + y^2)^2 - 2(xy)^2$?
– Robert Lewis
Jul 18 at 15:32
1
@RobertLewis Of course the latter. Thanks for the hint.
– LutzL
Jul 18 at 16:03
add a comment |Â
How is $x^4 + y^4 = x^2 + y^2 -2(xy)^2$? Or did you mean $x^4 + y^4 = (x^2 + y^2)^2 - 2(xy)^2$?
– Robert Lewis
Jul 18 at 15:32
1
@RobertLewis Of course the latter. Thanks for the hint.
– LutzL
Jul 18 at 16:03
How is $x^4 + y^4 = x^2 + y^2 -2(xy)^2$? Or did you mean $x^4 + y^4 = (x^2 + y^2)^2 - 2(xy)^2$?
– Robert Lewis
Jul 18 at 15:32
How is $x^4 + y^4 = x^2 + y^2 -2(xy)^2$? Or did you mean $x^4 + y^4 = (x^2 + y^2)^2 - 2(xy)^2$?
– Robert Lewis
Jul 18 at 15:32
1
1
@RobertLewis Of course the latter. Thanks for the hint.
– LutzL
Jul 18 at 16:03
@RobertLewis Of course the latter. Thanks for the hint.
– LutzL
Jul 18 at 16:03
add a comment |Â
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