How do I prove that the given system is globally asymptotically stable, using Lyapunov analysis?
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How do I prove that the given system is globally asymptotically stable, using Lyapunov analysis?
begin{equation}
left.begin{aligned}
dot{x_1} &= x_2 \
dot{x_2} &= -frac{x_1}{1 + x_2^2} label{eq:q2}
end{aligned}qquadright}
end{equation}
dynamical-systems control-theory nonlinear-system lyapunov-functions
asked Nov 12 at 16:48
Abhinav Sinha
31
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Abhinav Sinha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
0
down vote
favorite
How do I prove that the given system is globally asymptotically stable, using Lyapunov analysis?
begin{equation}
left.begin{aligned}
dot{x_1} &= x_2 \
dot{x_2} &= -frac{x_1}{1 + x_2^2} label{eq:q2}
end{aligned}qquadright}
end{equation}
dynamical-systems control-theory nonlinear-system lyapunov-functions
asked Nov 12 at 16:48
Abhinav Sinha
31
New contributor
Abhinav Sinha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Are you familiar with Lyapunov functions? And what have you tried yourself?
– Kwin van der Veen
Nov 12 at 16:58
I have tried general quadratic functions, such as (x1^2 + x2^2), but a term xy remains and I can't conclude properly.
– Abhinav Sinha
Nov 12 at 17:25
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
How do I prove that the given system is globally asymptotically stable, using Lyapunov analysis?
begin{equation}
left.begin{aligned}
dot{x_1} &= x_2 \
dot{x_2} &= -frac{x_1}{1 + x_2^2} label{eq:q2}
end{aligned}qquadright}
end{equation}
dynamical-systems control-theory nonlinear-system lyapunov-functions
asked Nov 12 at 16:48
Abhinav Sinha
31
New contributor
Abhinav Sinha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
How do I prove that the given system is globally asymptotically stable, using Lyapunov analysis?
begin{equation}
left.begin{aligned}
dot{x_1} &= x_2 \
dot{x_2} &= -frac{x_1}{1 + x_2^2} label{eq:q2}
end{aligned}qquadright}
end{equation}
dynamical-systems control-theory nonlinear-system lyapunov-functions
dynamical-systems control-theory nonlinear-system lyapunov-functions
asked Nov 12 at 16:48
Abhinav Sinha
31
New contributor
Abhinav Sinha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 12 at 16:48
Abhinav Sinha
31
New contributor
Abhinav Sinha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 12 at 16:48
Abhinav Sinha
31
New contributor
Abhinav Sinha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 12 at 16:48
Abhinav Sinha
31
asked Nov 12 at 16:48
Abhinav Sinha
31
31
New contributor
Abhinav Sinha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Abhinav Sinha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Abhinav Sinha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Are you familiar with Lyapunov functions? And what have you tried yourself?
– Kwin van der Veen
Nov 12 at 16:58
I have tried general quadratic functions, such as (x1^2 + x2^2), but a term xy remains and I can't conclude properly.
– Abhinav Sinha
Nov 12 at 17:25
add a comment |
Are you familiar with Lyapunov functions? And what have you tried yourself?
– Kwin van der Veen
Nov 12 at 16:58
I have tried general quadratic functions, such as (x1^2 + x2^2), but a term xy remains and I can't conclude properly.
– Abhinav Sinha
Nov 12 at 17:25
Are you familiar with Lyapunov functions? And what have you tried yourself?
– Kwin van der Veen
Nov 12 at 16:58
Are you familiar with Lyapunov functions? And what have you tried yourself?
– Kwin van der Veen
Nov 12 at 16:58
I have tried general quadratic functions, such as (x1^2 + x2^2), but a term xy remains and I can't conclude properly.
– Abhinav Sinha
Nov 12 at 17:25
I have tried general quadratic functions, such as (x1^2 + x2^2), but a term xy remains and I can't conclude properly.
– Abhinav Sinha
Nov 12 at 17:25
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
You could try to get the Lyapunov function in the form
$$
V=frac12x_1^2+g(x_2)
$$
so that
$$
dot V=x_1x_2+g'(x_2)frac{-x_1}{1+x_2^2}
$$
so that one would get a usable result with $g'(x_2)=x_2(1+x_2^2)$, integrating to,for example, $g(x_2)=frac14(1+x_2^2)^2$.
answered Nov 12 at 17:43
LutzL
53.4k41953
Thank you so much. I was taking something similar to g(x2) but it was constant and I was stuck. Thanks for your help :)
– Abhinav Sinha
Nov 12 at 18:55
1
@SampleTime : Then consider $V-V(0)$ to get the normalized value. I prefer the completed square for its simplicity.
– LutzL
Nov 12 at 21:18
1
@SampleTime : Could you elaborate? How to get a negative value from $V-V(0)=frac12x_1^2+frac12x_2^2+frac14x_2^4$?
– LutzL
Nov 12 at 21:27
@LutzL You are correct, nice! I made an error when writing down the equation, sorry for the confusion.
– SampleTime
Nov 12 at 21:40
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You could try to get the Lyapunov function in the form
$$
V=frac12x_1^2+g(x_2)
$$
so that
$$
dot V=x_1x_2+g'(x_2)frac{-x_1}{1+x_2^2}
$$
so that one would get a usable result with $g'(x_2)=x_2(1+x_2^2)$, integrating to,for example, $g(x_2)=frac14(1+x_2^2)^2$.
answered Nov 12 at 17:43
LutzL
53.4k41953
Thank you so much. I was taking something similar to g(x2) but it was constant and I was stuck. Thanks for your help :)
– Abhinav Sinha
Nov 12 at 18:55
1
@SampleTime : Then consider $V-V(0)$ to get the normalized value. I prefer the completed square for its simplicity.
– LutzL
Nov 12 at 21:18
1
@SampleTime : Could you elaborate? How to get a negative value from $V-V(0)=frac12x_1^2+frac12x_2^2+frac14x_2^4$?
– LutzL
Nov 12 at 21:27
@LutzL You are correct, nice! I made an error when writing down the equation, sorry for the confusion.
– SampleTime
Nov 12 at 21:40
add a comment |
up vote
1
down vote
accepted
You could try to get the Lyapunov function in the form
$$
V=frac12x_1^2+g(x_2)
$$
so that
$$
dot V=x_1x_2+g'(x_2)frac{-x_1}{1+x_2^2}
$$
so that one would get a usable result with $g'(x_2)=x_2(1+x_2^2)$, integrating to,for example, $g(x_2)=frac14(1+x_2^2)^2$.
answered Nov 12 at 17:43
LutzL
53.4k41953
Thank you so much. I was taking something similar to g(x2) but it was constant and I was stuck. Thanks for your help :)
– Abhinav Sinha
Nov 12 at 18:55
1
@SampleTime : Then consider $V-V(0)$ to get the normalized value. I prefer the completed square for its simplicity.
– LutzL
Nov 12 at 21:18
1
@SampleTime : Could you elaborate? How to get a negative value from $V-V(0)=frac12x_1^2+frac12x_2^2+frac14x_2^4$?
– LutzL
Nov 12 at 21:27
@LutzL You are correct, nice! I made an error when writing down the equation, sorry for the confusion.
– SampleTime
Nov 12 at 21:40
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You could try to get the Lyapunov function in the form
$$
V=frac12x_1^2+g(x_2)
$$
so that
$$
dot V=x_1x_2+g'(x_2)frac{-x_1}{1+x_2^2}
$$
so that one would get a usable result with $g'(x_2)=x_2(1+x_2^2)$, integrating to,for example, $g(x_2)=frac14(1+x_2^2)^2$.
answered Nov 12 at 17:43
LutzL
53.4k41953
You could try to get the Lyapunov function in the form
$$
V=frac12x_1^2+g(x_2)
$$
so that
$$
dot V=x_1x_2+g'(x_2)frac{-x_1}{1+x_2^2}
$$
so that one would get a usable result with $g'(x_2)=x_2(1+x_2^2)$, integrating to,for example, $g(x_2)=frac14(1+x_2^2)^2$.
answered Nov 12 at 17:43
LutzL
53.4k41953
answered Nov 12 at 17:43
LutzL
53.4k41953
answered Nov 12 at 17:43
LutzL
53.4k41953
answered Nov 12 at 17:43
LutzL
53.4k41953
53.4k41953
Thank you so much. I was taking something similar to g(x2) but it was constant and I was stuck. Thanks for your help :)
– Abhinav Sinha
Nov 12 at 18:55
1
@SampleTime : Then consider $V-V(0)$ to get the normalized value. I prefer the completed square for its simplicity.
– LutzL
Nov 12 at 21:18
1
@SampleTime : Could you elaborate? How to get a negative value from $V-V(0)=frac12x_1^2+frac12x_2^2+frac14x_2^4$?
– LutzL
Nov 12 at 21:27
@LutzL You are correct, nice! I made an error when writing down the equation, sorry for the confusion.
– SampleTime
Nov 12 at 21:40
add a comment |
Thank you so much. I was taking something similar to g(x2) but it was constant and I was stuck. Thanks for your help :)
– Abhinav Sinha
Nov 12 at 18:55
1
@SampleTime : Then consider $V-V(0)$ to get the normalized value. I prefer the completed square for its simplicity.
– LutzL
Nov 12 at 21:18
1
@SampleTime : Could you elaborate? How to get a negative value from $V-V(0)=frac12x_1^2+frac12x_2^2+frac14x_2^4$?
– LutzL
Nov 12 at 21:27
@LutzL You are correct, nice! I made an error when writing down the equation, sorry for the confusion.
– SampleTime
Nov 12 at 21:40
Thank you so much. I was taking something similar to g(x2) but it was constant and I was stuck. Thanks for your help :)
– Abhinav Sinha
Nov 12 at 18:55
Thank you so much. I was taking something similar to g(x2) but it was constant and I was stuck. Thanks for your help :)
– Abhinav Sinha
Nov 12 at 18:55
1
1
@SampleTime : Then consider $V-V(0)$ to get the normalized value. I prefer the completed square for its simplicity.
– LutzL
Nov 12 at 21:18
@SampleTime : Then consider $V-V(0)$ to get the normalized value. I prefer the completed square for its simplicity.
– LutzL
Nov 12 at 21:18
1
1
@SampleTime : Could you elaborate? How to get a negative value from $V-V(0)=frac12x_1^2+frac12x_2^2+frac14x_2^4$?
– LutzL
Nov 12 at 21:27
@SampleTime : Could you elaborate? How to get a negative value from $V-V(0)=frac12x_1^2+frac12x_2^2+frac14x_2^4$?
– LutzL
Nov 12 at 21:27
@LutzL You are correct, nice! I made an error when writing down the equation, sorry for the confusion.
– SampleTime
Nov 12 at 21:40
@LutzL You are correct, nice! I made an error when writing down the equation, sorry for the confusion.
– SampleTime
Nov 12 at 21:40
add a comment |
Abhinav Sinha is a new contributor. Be nice, and check out our Code of Conduct.
Abhinav Sinha is a new contributor. Be nice, and check out our Code of Conduct.
Abhinav Sinha is a new contributor. Be nice, and check out our Code of Conduct.
Abhinav Sinha is a new contributor. Be nice, and check out our Code of Conduct.
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Are you familiar with Lyapunov functions? And what have you tried yourself?
– Kwin van der Veen
Nov 12 at 16:58
I have tried general quadratic functions, such as (x1^2 + x2^2), but a term xy remains and I can't conclude properly.
– Abhinav Sinha
Nov 12 at 17:25