Find one domain of attraction for this system
$begingroup$
Assume the system:
begin{align}
begin{pmatrix}
x \
y \
end{pmatrix}' &= begin{pmatrix}
-(1-y)x \
-(1-x)y \
end{pmatrix}
end{align}
and the Lyapunov function
$$
V(x,y)=x^2+y^2$$
To begin with, the origin is a point of equilibrium and:
$bullet,$$V$ is positive definite since $$V(x,y) >0, quad forall x,y in mathbb{R^2}setminus(0,0)$$ and $V(0,0)=0$.
$bullet,$
$$
dot{V}(x,y)=-2left[x^2-(x^2y+xy^2)+y^2right]=-2left[x^2(1-y)+y^2(1-x)right]$$
For $dot{V}$ to be negative definite, we must have
$$
1-y>0 iff y<1 quadtext{and}quad 1-x<0 iff x<1
$$
Moreover, for the circular disks $x^2+y^2=c$ to lie inside the region:
$$
R=Big{ (x,y)in mathbb{R^2}: x<1,y<1Big}$$
we must have $c<1$. Therefore, one of the system's attraction domains is:
$$
S=Big{(x,y)in mathbb{R^2}: x^2+y^2 < 1Big}
$$
Question: It's very likely that this domain of attraction is not the maximum possible. Are there any ways to obtain the maximum one from the inequality $dot{V}(x,y)<0$? Is it necessary to find the maximum domain or finding a smaller one, which can be derived easily, does the job?
differential-equations dynamical-systems stability-in-odes lyapunov-functions
asked Nov 28 '18 at 19:05
JevautJevaut
1,015112
$endgroup$
add a comment |
$begingroup$
Assume the system:
begin{align}
begin{pmatrix}
x \
y \
end{pmatrix}' &= begin{pmatrix}
-(1-y)x \
-(1-x)y \
end{pmatrix}
end{align}
and the Lyapunov function
$$
V(x,y)=x^2+y^2$$
To begin with, the origin is a point of equilibrium and:
$bullet,$$V$ is positive definite since $$V(x,y) >0, quad forall x,y in mathbb{R^2}setminus(0,0)$$ and $V(0,0)=0$.
$bullet,$
$$
dot{V}(x,y)=-2left[x^2-(x^2y+xy^2)+y^2right]=-2left[x^2(1-y)+y^2(1-x)right]$$
For $dot{V}$ to be negative definite, we must have
$$
1-y>0 iff y<1 quadtext{and}quad 1-x<0 iff x<1
$$
Moreover, for the circular disks $x^2+y^2=c$ to lie inside the region:
$$
R=Big{ (x,y)in mathbb{R^2}: x<1,y<1Big}$$
we must have $c<1$. Therefore, one of the system's attraction domains is:
$$
S=Big{(x,y)in mathbb{R^2}: x^2+y^2 < 1Big}
$$
Question: It's very likely that this domain of attraction is not the maximum possible. Are there any ways to obtain the maximum one from the inequality $dot{V}(x,y)<0$? Is it necessary to find the maximum domain or finding a smaller one, which can be derived easily, does the job?
differential-equations dynamical-systems stability-in-odes lyapunov-functions
asked Nov 28 '18 at 19:05
JevautJevaut
1,015112
$endgroup$
1
$begingroup$
Whether it's necessary depends on what the "job" is.
$endgroup$
– Robert Israel
Nov 28 '18 at 19:37
add a comment |
$begingroup$
Assume the system:
begin{align}
begin{pmatrix}
x \
y \
end{pmatrix}' &= begin{pmatrix}
-(1-y)x \
-(1-x)y \
end{pmatrix}
end{align}
and the Lyapunov function
$$
V(x,y)=x^2+y^2$$
To begin with, the origin is a point of equilibrium and:
$bullet,$$V$ is positive definite since $$V(x,y) >0, quad forall x,y in mathbb{R^2}setminus(0,0)$$ and $V(0,0)=0$.
$bullet,$
$$
dot{V}(x,y)=-2left[x^2-(x^2y+xy^2)+y^2right]=-2left[x^2(1-y)+y^2(1-x)right]$$
For $dot{V}$ to be negative definite, we must have
$$
1-y>0 iff y<1 quadtext{and}quad 1-x<0 iff x<1
$$
Moreover, for the circular disks $x^2+y^2=c$ to lie inside the region:
$$
R=Big{ (x,y)in mathbb{R^2}: x<1,y<1Big}$$
we must have $c<1$. Therefore, one of the system's attraction domains is:
$$
S=Big{(x,y)in mathbb{R^2}: x^2+y^2 < 1Big}
$$
Question: It's very likely that this domain of attraction is not the maximum possible. Are there any ways to obtain the maximum one from the inequality $dot{V}(x,y)<0$? Is it necessary to find the maximum domain or finding a smaller one, which can be derived easily, does the job?
differential-equations dynamical-systems stability-in-odes lyapunov-functions
asked Nov 28 '18 at 19:05
JevautJevaut
1,015112
$endgroup$
Assume the system:
begin{align}
begin{pmatrix}
x \
y \
end{pmatrix}' &= begin{pmatrix}
-(1-y)x \
-(1-x)y \
end{pmatrix}
end{align}
and the Lyapunov function
$$
V(x,y)=x^2+y^2$$
To begin with, the origin is a point of equilibrium and:
$bullet,$$V$ is positive definite since $$V(x,y) >0, quad forall x,y in mathbb{R^2}setminus(0,0)$$ and $V(0,0)=0$.
$bullet,$
$$
dot{V}(x,y)=-2left[x^2-(x^2y+xy^2)+y^2right]=-2left[x^2(1-y)+y^2(1-x)right]$$
For $dot{V}$ to be negative definite, we must have
$$
1-y>0 iff y<1 quadtext{and}quad 1-x<0 iff x<1
$$
Moreover, for the circular disks $x^2+y^2=c$ to lie inside the region:
$$
R=Big{ (x,y)in mathbb{R^2}: x<1,y<1Big}$$
we must have $c<1$. Therefore, one of the system's attraction domains is:
$$
S=Big{(x,y)in mathbb{R^2}: x^2+y^2 < 1Big}
$$
Question: It's very likely that this domain of attraction is not the maximum possible. Are there any ways to obtain the maximum one from the inequality $dot{V}(x,y)<0$? Is it necessary to find the maximum domain or finding a smaller one, which can be derived easily, does the job?
differential-equations dynamical-systems stability-in-odes lyapunov-functions
differential-equations dynamical-systems stability-in-odes lyapunov-functions
asked Nov 28 '18 at 19:05
JevautJevaut
1,015112
asked Nov 28 '18 at 19:05
JevautJevaut
1,015112
asked Nov 28 '18 at 19:05
JevautJevaut
1,015112
asked Nov 28 '18 at 19:05
JevautJevaut
1,015112
asked Nov 28 '18 at 19:05
JevautJevaut
1,015112
1,015112
1
$begingroup$
Whether it's necessary depends on what the "job" is.
$endgroup$
– Robert Israel
Nov 28 '18 at 19:37
add a comment |
1
$begingroup$
Whether it's necessary depends on what the "job" is.
$endgroup$
– Robert Israel
Nov 28 '18 at 19:37
1
1
$begingroup$
Whether it's necessary depends on what the "job" is.
$endgroup$
– Robert Israel
Nov 28 '18 at 19:37
$begingroup$
Whether it's necessary depends on what the "job" is.
$endgroup$
– Robert Israel
Nov 28 '18 at 19:37
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
In most cases, exercises like this are properly created so you can find the true corresponding attraction domain by using the Lyapunov Function given, since if the Lyapunov Function does the trick then a strictly precise inequality (or conclusion) about its sign leads to the correct assumption. Note that in any general, complicated model, finding a Lyapunov Function is in the first place a very hard and tricky task, let alone finding the exact one that can give you the maximum domain of attraction.
In your exercise though, which is precisely created, the domain of attraction that you have yielded is indeed correct, as it can be seen by the streamplot below, since for $x>1$ and $y>1$ the phase portrait arrows "drag" away :
$qquad qquad qquad$
answered Nov 28 '18 at 19:19
RebellosRebellos
14.5k31246
$endgroup$
$begingroup$
Are you an ex-student of Staurakakis?!
$endgroup$
– Jevaut
Nov 28 '18 at 19:27
$begingroup$
Yes, that would be true. Why are you asking ?
$endgroup$
– Rebellos
Nov 28 '18 at 19:55
$begingroup$
Sorry for the excess excitement lol. I'm asking cause I'm also one of his students now. I checked your profile and it got me like: "what a small world".
$endgroup$
– Jevaut
Nov 28 '18 at 20:07
$begingroup$
@Jevaut Haha, that's true, indeed a coincidence. How come you remove the accept mark by the way ? Oh, a note for such problems : In this case, as Robert Israel perfectly elaborated below, since you can solve the system in closed form, you can make a safe conclusion. In cases where the system cannot be solved in closed form (which is the point of the dynamical systems course) then what happens is the elaboration of my answer.
$endgroup$
– Rebellos
Nov 28 '18 at 20:11
$begingroup$
@Jevaut Where are you coming from?
$endgroup$
– Magic K. Mamba
Nov 28 '18 at 20:14
|
show 1 more comment
$begingroup$
This particular system has closed form solutions. The integral curves are
$x-ln(|x|)-y+ln(|y|) =$constant. In particular, the separatrices forming the
boundary of the "maximal" basin of attraction of $(0,0)$ are the stable manifold of the saddle point $(1,1)$, namely the branches of $x - ln(x) - y + ln(y) = 0$ that are not on $x=y$.
answered Nov 28 '18 at 19:28
Robert IsraelRobert Israel
319k23209459
$endgroup$
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
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active
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votes
$begingroup$
In most cases, exercises like this are properly created so you can find the true corresponding attraction domain by using the Lyapunov Function given, since if the Lyapunov Function does the trick then a strictly precise inequality (or conclusion) about its sign leads to the correct assumption. Note that in any general, complicated model, finding a Lyapunov Function is in the first place a very hard and tricky task, let alone finding the exact one that can give you the maximum domain of attraction.
In your exercise though, which is precisely created, the domain of attraction that you have yielded is indeed correct, as it can be seen by the streamplot below, since for $x>1$ and $y>1$ the phase portrait arrows "drag" away :
$qquad qquad qquad$
answered Nov 28 '18 at 19:19
RebellosRebellos
14.5k31246
$endgroup$
$begingroup$
Are you an ex-student of Staurakakis?!
$endgroup$
– Jevaut
Nov 28 '18 at 19:27
$begingroup$
Yes, that would be true. Why are you asking ?
$endgroup$
– Rebellos
Nov 28 '18 at 19:55
$begingroup$
Sorry for the excess excitement lol. I'm asking cause I'm also one of his students now. I checked your profile and it got me like: "what a small world".
$endgroup$
– Jevaut
Nov 28 '18 at 20:07
$begingroup$
@Jevaut Haha, that's true, indeed a coincidence. How come you remove the accept mark by the way ? Oh, a note for such problems : In this case, as Robert Israel perfectly elaborated below, since you can solve the system in closed form, you can make a safe conclusion. In cases where the system cannot be solved in closed form (which is the point of the dynamical systems course) then what happens is the elaboration of my answer.
$endgroup$
– Rebellos
Nov 28 '18 at 20:11
$begingroup$
@Jevaut Where are you coming from?
$endgroup$
– Magic K. Mamba
Nov 28 '18 at 20:14
|
show 1 more comment
$begingroup$
In most cases, exercises like this are properly created so you can find the true corresponding attraction domain by using the Lyapunov Function given, since if the Lyapunov Function does the trick then a strictly precise inequality (or conclusion) about its sign leads to the correct assumption. Note that in any general, complicated model, finding a Lyapunov Function is in the first place a very hard and tricky task, let alone finding the exact one that can give you the maximum domain of attraction.
In your exercise though, which is precisely created, the domain of attraction that you have yielded is indeed correct, as it can be seen by the streamplot below, since for $x>1$ and $y>1$ the phase portrait arrows "drag" away :
$qquad qquad qquad$
answered Nov 28 '18 at 19:19
RebellosRebellos
14.5k31246
$endgroup$
$begingroup$
Are you an ex-student of Staurakakis?!
$endgroup$
– Jevaut
Nov 28 '18 at 19:27
$begingroup$
Yes, that would be true. Why are you asking ?
$endgroup$
– Rebellos
Nov 28 '18 at 19:55
$begingroup$
Sorry for the excess excitement lol. I'm asking cause I'm also one of his students now. I checked your profile and it got me like: "what a small world".
$endgroup$
– Jevaut
Nov 28 '18 at 20:07
$begingroup$
@Jevaut Haha, that's true, indeed a coincidence. How come you remove the accept mark by the way ? Oh, a note for such problems : In this case, as Robert Israel perfectly elaborated below, since you can solve the system in closed form, you can make a safe conclusion. In cases where the system cannot be solved in closed form (which is the point of the dynamical systems course) then what happens is the elaboration of my answer.
$endgroup$
– Rebellos
Nov 28 '18 at 20:11
$begingroup$
@Jevaut Where are you coming from?
$endgroup$
– Magic K. Mamba
Nov 28 '18 at 20:14
|
show 1 more comment
$begingroup$
In most cases, exercises like this are properly created so you can find the true corresponding attraction domain by using the Lyapunov Function given, since if the Lyapunov Function does the trick then a strictly precise inequality (or conclusion) about its sign leads to the correct assumption. Note that in any general, complicated model, finding a Lyapunov Function is in the first place a very hard and tricky task, let alone finding the exact one that can give you the maximum domain of attraction.
In your exercise though, which is precisely created, the domain of attraction that you have yielded is indeed correct, as it can be seen by the streamplot below, since for $x>1$ and $y>1$ the phase portrait arrows "drag" away :
$qquad qquad qquad$
answered Nov 28 '18 at 19:19
RebellosRebellos
14.5k31246
$endgroup$
In most cases, exercises like this are properly created so you can find the true corresponding attraction domain by using the Lyapunov Function given, since if the Lyapunov Function does the trick then a strictly precise inequality (or conclusion) about its sign leads to the correct assumption. Note that in any general, complicated model, finding a Lyapunov Function is in the first place a very hard and tricky task, let alone finding the exact one that can give you the maximum domain of attraction.
In your exercise though, which is precisely created, the domain of attraction that you have yielded is indeed correct, as it can be seen by the streamplot below, since for $x>1$ and $y>1$ the phase portrait arrows "drag" away :
$qquad qquad qquad$
answered Nov 28 '18 at 19:19
RebellosRebellos
14.5k31246
answered Nov 28 '18 at 19:19
RebellosRebellos
14.5k31246
answered Nov 28 '18 at 19:19
RebellosRebellos
14.5k31246
answered Nov 28 '18 at 19:19
RebellosRebellos
14.5k31246
14.5k31246
$begingroup$
Are you an ex-student of Staurakakis?!
$endgroup$
– Jevaut
Nov 28 '18 at 19:27
$begingroup$
Yes, that would be true. Why are you asking ?
$endgroup$
– Rebellos
Nov 28 '18 at 19:55
$begingroup$
Sorry for the excess excitement lol. I'm asking cause I'm also one of his students now. I checked your profile and it got me like: "what a small world".
$endgroup$
– Jevaut
Nov 28 '18 at 20:07
$begingroup$
@Jevaut Haha, that's true, indeed a coincidence. How come you remove the accept mark by the way ? Oh, a note for such problems : In this case, as Robert Israel perfectly elaborated below, since you can solve the system in closed form, you can make a safe conclusion. In cases where the system cannot be solved in closed form (which is the point of the dynamical systems course) then what happens is the elaboration of my answer.
$endgroup$
– Rebellos
Nov 28 '18 at 20:11
$begingroup$
@Jevaut Where are you coming from?
$endgroup$
– Magic K. Mamba
Nov 28 '18 at 20:14
|
show 1 more comment
$begingroup$
Are you an ex-student of Staurakakis?!
$endgroup$
– Jevaut
Nov 28 '18 at 19:27
$begingroup$
Yes, that would be true. Why are you asking ?
$endgroup$
– Rebellos
Nov 28 '18 at 19:55
$begingroup$
Sorry for the excess excitement lol. I'm asking cause I'm also one of his students now. I checked your profile and it got me like: "what a small world".
$endgroup$
– Jevaut
Nov 28 '18 at 20:07
$begingroup$
@Jevaut Haha, that's true, indeed a coincidence. How come you remove the accept mark by the way ? Oh, a note for such problems : In this case, as Robert Israel perfectly elaborated below, since you can solve the system in closed form, you can make a safe conclusion. In cases where the system cannot be solved in closed form (which is the point of the dynamical systems course) then what happens is the elaboration of my answer.
$endgroup$
– Rebellos
Nov 28 '18 at 20:11
$begingroup$
@Jevaut Where are you coming from?
$endgroup$
– Magic K. Mamba
Nov 28 '18 at 20:14
$begingroup$
Are you an ex-student of Staurakakis?!
$endgroup$
– Jevaut
Nov 28 '18 at 19:27
$begingroup$
Are you an ex-student of Staurakakis?!
$endgroup$
– Jevaut
Nov 28 '18 at 19:27
$begingroup$
Yes, that would be true. Why are you asking ?
$endgroup$
– Rebellos
Nov 28 '18 at 19:55
$begingroup$
Yes, that would be true. Why are you asking ?
$endgroup$
– Rebellos
Nov 28 '18 at 19:55
$begingroup$
Sorry for the excess excitement lol. I'm asking cause I'm also one of his students now. I checked your profile and it got me like: "what a small world".
$endgroup$
– Jevaut
Nov 28 '18 at 20:07
$begingroup$
Sorry for the excess excitement lol. I'm asking cause I'm also one of his students now. I checked your profile and it got me like: "what a small world".
$endgroup$
– Jevaut
Nov 28 '18 at 20:07
$begingroup$
@Jevaut Haha, that's true, indeed a coincidence. How come you remove the accept mark by the way ? Oh, a note for such problems : In this case, as Robert Israel perfectly elaborated below, since you can solve the system in closed form, you can make a safe conclusion. In cases where the system cannot be solved in closed form (which is the point of the dynamical systems course) then what happens is the elaboration of my answer.
$endgroup$
– Rebellos
Nov 28 '18 at 20:11
$begingroup$
@Jevaut Haha, that's true, indeed a coincidence. How come you remove the accept mark by the way ? Oh, a note for such problems : In this case, as Robert Israel perfectly elaborated below, since you can solve the system in closed form, you can make a safe conclusion. In cases where the system cannot be solved in closed form (which is the point of the dynamical systems course) then what happens is the elaboration of my answer.
$endgroup$
– Rebellos
Nov 28 '18 at 20:11
$begingroup$
@Jevaut Where are you coming from?
$endgroup$
– Magic K. Mamba
Nov 28 '18 at 20:14
$begingroup$
@Jevaut Where are you coming from?
$endgroup$
– Magic K. Mamba
Nov 28 '18 at 20:14
|
show 1 more comment
$begingroup$
This particular system has closed form solutions. The integral curves are
$x-ln(|x|)-y+ln(|y|) =$constant. In particular, the separatrices forming the
boundary of the "maximal" basin of attraction of $(0,0)$ are the stable manifold of the saddle point $(1,1)$, namely the branches of $x - ln(x) - y + ln(y) = 0$ that are not on $x=y$.
answered Nov 28 '18 at 19:28
Robert IsraelRobert Israel
319k23209459
$endgroup$
add a comment |
$begingroup$
This particular system has closed form solutions. The integral curves are
$x-ln(|x|)-y+ln(|y|) =$constant. In particular, the separatrices forming the
boundary of the "maximal" basin of attraction of $(0,0)$ are the stable manifold of the saddle point $(1,1)$, namely the branches of $x - ln(x) - y + ln(y) = 0$ that are not on $x=y$.
answered Nov 28 '18 at 19:28
Robert IsraelRobert Israel
319k23209459
$endgroup$
add a comment |
$begingroup$
This particular system has closed form solutions. The integral curves are
$x-ln(|x|)-y+ln(|y|) =$constant. In particular, the separatrices forming the
boundary of the "maximal" basin of attraction of $(0,0)$ are the stable manifold of the saddle point $(1,1)$, namely the branches of $x - ln(x) - y + ln(y) = 0$ that are not on $x=y$.
answered Nov 28 '18 at 19:28
Robert IsraelRobert Israel
319k23209459
$endgroup$
This particular system has closed form solutions. The integral curves are
$x-ln(|x|)-y+ln(|y|) =$constant. In particular, the separatrices forming the
boundary of the "maximal" basin of attraction of $(0,0)$ are the stable manifold of the saddle point $(1,1)$, namely the branches of $x - ln(x) - y + ln(y) = 0$ that are not on $x=y$.
answered Nov 28 '18 at 19:28
Robert IsraelRobert Israel
319k23209459
edited Nov 28 '18 at 19:35
answered Nov 28 '18 at 19:28
Robert IsraelRobert Israel
319k23209459
answered Nov 28 '18 at 19:28
Robert IsraelRobert Israel
319k23209459
answered Nov 28 '18 at 19:28
Robert IsraelRobert Israel
319k23209459
319k23209459
add a comment |
add a comment |
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$begingroup$
Whether it's necessary depends on what the "job" is.
$endgroup$
– Robert Israel
Nov 28 '18 at 19:37