Find one domain of attraction for this system












1












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Whether it's necessary depends on what the "job" is.
    $endgroup$
    – Robert Israel
    Nov 28 '18 at 19:37
















1












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Whether it's necessary depends on what the "job" is.
    $endgroup$
    – Robert Israel
    Nov 28 '18 at 19:37














1












1








1





$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











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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














  • 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










2 Answers
2






active

oldest

votes


















1












$begingroup$

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 :



$qquad qquad qquad$enter image description here






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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



















2












$begingroup$

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 $x=y$.



enter image description here






share|cite|improve this answer











$endgroup$













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    2 Answers
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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

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    active

    oldest

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    1












    $begingroup$

    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 :



    $qquad qquad qquad$enter image description here






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      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
















    1












    $begingroup$

    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 :



    $qquad qquad qquad$enter image description here






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      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














    1












    1








    1





    $begingroup$

    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 :



    $qquad qquad qquad$enter image description here






    share|cite|improve this answer









    $endgroup$



    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 :



    $qquad qquad qquad$enter image description here







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    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


















    • $begingroup$
      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











    2












    $begingroup$

    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 $x=y$.



    enter image description here






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      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 $x=y$.



      enter image description here






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        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 $x=y$.



        enter image description here






        share|cite|improve this answer











        $endgroup$



        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 $x=y$.



        enter image description here







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 28 '18 at 19:35

























        answered Nov 28 '18 at 19:28









        Robert IsraelRobert Israel

        319k23209459




        319k23209459






























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            Plaza Victoria

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            Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
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            Puebla de Zaragoza

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            Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
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            Musa

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            Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
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