Analizing the stability of the equilibrium points of the system $ddot{x}=(x-a)(x^2-a)$












2














$require{amsmath}$
$DeclareMathOperator{Tr}{Tr}$
$DeclareMathOperator{Det}{Det}$




Investigate the stability of the equilibrium points of the system $ddot{x}=(x-a)(x^2-a)$ for all real values of the parameter $a$. (Hints: It might help to
graph the right-hand side. An alternative is to rewrite the equation as $ddot{x}=−V′(x)$ for a suitable potential energy function $V$ and then use your intuition about particles
moving in potentials.)




I am not really sure on how to approach the problem with the given hints, since it would require plotting the graph for different critical values for $a$, which I am not really sure how to find. Thus, I am wondering if the following is correct.



The system $ddot{x}=(x-a)(x^2-a)$ can be re-written as
$$begin{cases}
dot{x}=y\
dot{y}=(x-a)(x^2-a)
end{cases}$$

with fixed points $P_1(a,0),,P_2(sqrt{a},0)$ and $P_3(sqrt{a},0)$.



The Jacobian is
$$J(x,y)=begin{bmatrix}
0 &1\
3x^2-2ax-a &0
end{bmatrix}$$

and thus
$$J(a,0)=begin{bmatrix}
0 &1\
a^2-a &0
end{bmatrix};quad J(sqrt{a},0)=begin{bmatrix}
0 &1\
2a-2a^{3/2} &0
end{bmatrix}; quad J(-sqrt{a},0)=begin{bmatrix}
0 &1\
2a+2a^{3/2} &0
end{bmatrix}.$$

It can be noticed that the $Trleft[J(x,y)right]=0$ and that
$$begin{aligned}
&1.,Detleft[J(a,0)right]=a(1-a)implies text{Saddle for }a<0 wedge a>1, text{Center for }0<a<1.\
&2.,Detleft[J(sqrt{a},0)right]=2a(sqrt{a}-1)implies text{Saddle for }0<a<1, text{Center for }a>1.\
&3.,Detleft[J(-sqrt{a},0)right]=-2a(sqrt{a}+1)implies text{Saddle for }a>0.
end{aligned}$$



If $a=0$ the system reduces to $ddot{x}=x^3$ where the only fixed point is at $(0,0)$ and thus it is unstable. On the other hand, if $a=1$ then the system reduces to $ddot{x}=x^3-x^2-x+1$ with fixed points at $(1,0),(-1,0)$, both being unstable.



Is my work correct?










share|cite|improve this question
























  • Isn't this question better suited for Physics SE?
    – Bertrand Wittgenstein's Ghost
    Nov 29 '18 at 2:11






  • 4




    @BertrandWittgenstein'sGhost this is part of dynamical systems, so no. I could certainly ask there too, but there is no point really
    – DMH16
    Nov 29 '18 at 2:13












  • That's funny what you did, It was a genuine question. No need to get defensive. Cheers!
    – Bertrand Wittgenstein's Ghost
    Nov 29 '18 at 2:23










  • I haven't checked all your algebra carefully but it looks like you've go the right idea!
    – Robert Lewis
    Nov 29 '18 at 2:32






  • 1




    @AlexanderJ93 yes you're right the determinant is always negative. Thanks
    – DMH16
    Nov 29 '18 at 2:52
















2














$require{amsmath}$
$DeclareMathOperator{Tr}{Tr}$
$DeclareMathOperator{Det}{Det}$




Investigate the stability of the equilibrium points of the system $ddot{x}=(x-a)(x^2-a)$ for all real values of the parameter $a$. (Hints: It might help to
graph the right-hand side. An alternative is to rewrite the equation as $ddot{x}=−V′(x)$ for a suitable potential energy function $V$ and then use your intuition about particles
moving in potentials.)




I am not really sure on how to approach the problem with the given hints, since it would require plotting the graph for different critical values for $a$, which I am not really sure how to find. Thus, I am wondering if the following is correct.



The system $ddot{x}=(x-a)(x^2-a)$ can be re-written as
$$begin{cases}
dot{x}=y\
dot{y}=(x-a)(x^2-a)
end{cases}$$

with fixed points $P_1(a,0),,P_2(sqrt{a},0)$ and $P_3(sqrt{a},0)$.



The Jacobian is
$$J(x,y)=begin{bmatrix}
0 &1\
3x^2-2ax-a &0
end{bmatrix}$$

and thus
$$J(a,0)=begin{bmatrix}
0 &1\
a^2-a &0
end{bmatrix};quad J(sqrt{a},0)=begin{bmatrix}
0 &1\
2a-2a^{3/2} &0
end{bmatrix}; quad J(-sqrt{a},0)=begin{bmatrix}
0 &1\
2a+2a^{3/2} &0
end{bmatrix}.$$

It can be noticed that the $Trleft[J(x,y)right]=0$ and that
$$begin{aligned}
&1.,Detleft[J(a,0)right]=a(1-a)implies text{Saddle for }a<0 wedge a>1, text{Center for }0<a<1.\
&2.,Detleft[J(sqrt{a},0)right]=2a(sqrt{a}-1)implies text{Saddle for }0<a<1, text{Center for }a>1.\
&3.,Detleft[J(-sqrt{a},0)right]=-2a(sqrt{a}+1)implies text{Saddle for }a>0.
end{aligned}$$



If $a=0$ the system reduces to $ddot{x}=x^3$ where the only fixed point is at $(0,0)$ and thus it is unstable. On the other hand, if $a=1$ then the system reduces to $ddot{x}=x^3-x^2-x+1$ with fixed points at $(1,0),(-1,0)$, both being unstable.



Is my work correct?










share|cite|improve this question
























  • Isn't this question better suited for Physics SE?
    – Bertrand Wittgenstein's Ghost
    Nov 29 '18 at 2:11






  • 4




    @BertrandWittgenstein'sGhost this is part of dynamical systems, so no. I could certainly ask there too, but there is no point really
    – DMH16
    Nov 29 '18 at 2:13












  • That's funny what you did, It was a genuine question. No need to get defensive. Cheers!
    – Bertrand Wittgenstein's Ghost
    Nov 29 '18 at 2:23










  • I haven't checked all your algebra carefully but it looks like you've go the right idea!
    – Robert Lewis
    Nov 29 '18 at 2:32






  • 1




    @AlexanderJ93 yes you're right the determinant is always negative. Thanks
    – DMH16
    Nov 29 '18 at 2:52














2












2








2


1





$require{amsmath}$
$DeclareMathOperator{Tr}{Tr}$
$DeclareMathOperator{Det}{Det}$




Investigate the stability of the equilibrium points of the system $ddot{x}=(x-a)(x^2-a)$ for all real values of the parameter $a$. (Hints: It might help to
graph the right-hand side. An alternative is to rewrite the equation as $ddot{x}=−V′(x)$ for a suitable potential energy function $V$ and then use your intuition about particles
moving in potentials.)




I am not really sure on how to approach the problem with the given hints, since it would require plotting the graph for different critical values for $a$, which I am not really sure how to find. Thus, I am wondering if the following is correct.



The system $ddot{x}=(x-a)(x^2-a)$ can be re-written as
$$begin{cases}
dot{x}=y\
dot{y}=(x-a)(x^2-a)
end{cases}$$

with fixed points $P_1(a,0),,P_2(sqrt{a},0)$ and $P_3(sqrt{a},0)$.



The Jacobian is
$$J(x,y)=begin{bmatrix}
0 &1\
3x^2-2ax-a &0
end{bmatrix}$$

and thus
$$J(a,0)=begin{bmatrix}
0 &1\
a^2-a &0
end{bmatrix};quad J(sqrt{a},0)=begin{bmatrix}
0 &1\
2a-2a^{3/2} &0
end{bmatrix}; quad J(-sqrt{a},0)=begin{bmatrix}
0 &1\
2a+2a^{3/2} &0
end{bmatrix}.$$

It can be noticed that the $Trleft[J(x,y)right]=0$ and that
$$begin{aligned}
&1.,Detleft[J(a,0)right]=a(1-a)implies text{Saddle for }a<0 wedge a>1, text{Center for }0<a<1.\
&2.,Detleft[J(sqrt{a},0)right]=2a(sqrt{a}-1)implies text{Saddle for }0<a<1, text{Center for }a>1.\
&3.,Detleft[J(-sqrt{a},0)right]=-2a(sqrt{a}+1)implies text{Saddle for }a>0.
end{aligned}$$



If $a=0$ the system reduces to $ddot{x}=x^3$ where the only fixed point is at $(0,0)$ and thus it is unstable. On the other hand, if $a=1$ then the system reduces to $ddot{x}=x^3-x^2-x+1$ with fixed points at $(1,0),(-1,0)$, both being unstable.



Is my work correct?










share|cite|improve this question















$require{amsmath}$
$DeclareMathOperator{Tr}{Tr}$
$DeclareMathOperator{Det}{Det}$




Investigate the stability of the equilibrium points of the system $ddot{x}=(x-a)(x^2-a)$ for all real values of the parameter $a$. (Hints: It might help to
graph the right-hand side. An alternative is to rewrite the equation as $ddot{x}=−V′(x)$ for a suitable potential energy function $V$ and then use your intuition about particles
moving in potentials.)




I am not really sure on how to approach the problem with the given hints, since it would require plotting the graph for different critical values for $a$, which I am not really sure how to find. Thus, I am wondering if the following is correct.



The system $ddot{x}=(x-a)(x^2-a)$ can be re-written as
$$begin{cases}
dot{x}=y\
dot{y}=(x-a)(x^2-a)
end{cases}$$

with fixed points $P_1(a,0),,P_2(sqrt{a},0)$ and $P_3(sqrt{a},0)$.



The Jacobian is
$$J(x,y)=begin{bmatrix}
0 &1\
3x^2-2ax-a &0
end{bmatrix}$$

and thus
$$J(a,0)=begin{bmatrix}
0 &1\
a^2-a &0
end{bmatrix};quad J(sqrt{a},0)=begin{bmatrix}
0 &1\
2a-2a^{3/2} &0
end{bmatrix}; quad J(-sqrt{a},0)=begin{bmatrix}
0 &1\
2a+2a^{3/2} &0
end{bmatrix}.$$

It can be noticed that the $Trleft[J(x,y)right]=0$ and that
$$begin{aligned}
&1.,Detleft[J(a,0)right]=a(1-a)implies text{Saddle for }a<0 wedge a>1, text{Center for }0<a<1.\
&2.,Detleft[J(sqrt{a},0)right]=2a(sqrt{a}-1)implies text{Saddle for }0<a<1, text{Center for }a>1.\
&3.,Detleft[J(-sqrt{a},0)right]=-2a(sqrt{a}+1)implies text{Saddle for }a>0.
end{aligned}$$



If $a=0$ the system reduces to $ddot{x}=x^3$ where the only fixed point is at $(0,0)$ and thus it is unstable. On the other hand, if $a=1$ then the system reduces to $ddot{x}=x^3-x^2-x+1$ with fixed points at $(1,0),(-1,0)$, both being unstable.



Is my work correct?







differential-equations analysis dynamical-systems






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 29 '18 at 2:51

























asked Nov 29 '18 at 2:06









DMH16

578217




578217












  • Isn't this question better suited for Physics SE?
    – Bertrand Wittgenstein's Ghost
    Nov 29 '18 at 2:11






  • 4




    @BertrandWittgenstein'sGhost this is part of dynamical systems, so no. I could certainly ask there too, but there is no point really
    – DMH16
    Nov 29 '18 at 2:13












  • That's funny what you did, It was a genuine question. No need to get defensive. Cheers!
    – Bertrand Wittgenstein's Ghost
    Nov 29 '18 at 2:23










  • I haven't checked all your algebra carefully but it looks like you've go the right idea!
    – Robert Lewis
    Nov 29 '18 at 2:32






  • 1




    @AlexanderJ93 yes you're right the determinant is always negative. Thanks
    – DMH16
    Nov 29 '18 at 2:52


















  • Isn't this question better suited for Physics SE?
    – Bertrand Wittgenstein's Ghost
    Nov 29 '18 at 2:11






  • 4




    @BertrandWittgenstein'sGhost this is part of dynamical systems, so no. I could certainly ask there too, but there is no point really
    – DMH16
    Nov 29 '18 at 2:13












  • That's funny what you did, It was a genuine question. No need to get defensive. Cheers!
    – Bertrand Wittgenstein's Ghost
    Nov 29 '18 at 2:23










  • I haven't checked all your algebra carefully but it looks like you've go the right idea!
    – Robert Lewis
    Nov 29 '18 at 2:32






  • 1




    @AlexanderJ93 yes you're right the determinant is always negative. Thanks
    – DMH16
    Nov 29 '18 at 2:52
















Isn't this question better suited for Physics SE?
– Bertrand Wittgenstein's Ghost
Nov 29 '18 at 2:11




Isn't this question better suited for Physics SE?
– Bertrand Wittgenstein's Ghost
Nov 29 '18 at 2:11




4




4




@BertrandWittgenstein'sGhost this is part of dynamical systems, so no. I could certainly ask there too, but there is no point really
– DMH16
Nov 29 '18 at 2:13






@BertrandWittgenstein'sGhost this is part of dynamical systems, so no. I could certainly ask there too, but there is no point really
– DMH16
Nov 29 '18 at 2:13














That's funny what you did, It was a genuine question. No need to get defensive. Cheers!
– Bertrand Wittgenstein's Ghost
Nov 29 '18 at 2:23




That's funny what you did, It was a genuine question. No need to get defensive. Cheers!
– Bertrand Wittgenstein's Ghost
Nov 29 '18 at 2:23












I haven't checked all your algebra carefully but it looks like you've go the right idea!
– Robert Lewis
Nov 29 '18 at 2:32




I haven't checked all your algebra carefully but it looks like you've go the right idea!
– Robert Lewis
Nov 29 '18 at 2:32




1




1




@AlexanderJ93 yes you're right the determinant is always negative. Thanks
– DMH16
Nov 29 '18 at 2:52




@AlexanderJ93 yes you're right the determinant is always negative. Thanks
– DMH16
Nov 29 '18 at 2:52










1 Answer
1






active

oldest

votes


















2














Almost everything looks right.



For $a > 0$, $-2a < 0$ and $sqrt{a}+1 > 0$, so $Det[J(-sqrt{a},0)] < 0$. This equilibrium should be a saddle then for positive $a$.



Also, when analyzing the stability of a system with a parameter, it helps to draw the bifurcation diagram, i.e. the graph of the fixed points with respect to the parameter. The fixed point equations $x = a, x = sqrt{a}, x = -sqrt{a}$ can be graphed on the $ax$-plane. Intersections of these curves are the bifurcations, which are the only places where the behavior can change.



bifurcation diagram



This is the bifurcation diagram for your system. The lower black curve is the $x=-sqrt{a}$ equilibrium, which doesn't have any intersections for $a>0$, so there will be no change in behavior.





Edit:



This isn't specifically asked about in the question but it is an important feature of this equation. For $a>0, aneq 1$, there are centers at one fixed point and saddles at another nearby fixed point. When this happens, a periodic orbit will collide with the saddle point, creating a closed path from a fixed point back around to itself. This is called a homoclinic orbit, and usually has a "teardrop" shape, as seen below for $a = frac{1}{2}$.



homoclinic orbit






share|cite|improve this answer























  • Just out of curiosity, what software did you use to sketch the phase plane?
    – DMH16
    Nov 29 '18 at 4:49






  • 1




    pplane, which can be found here as a java application: math.rice.edu/~dfield/dfpp.html I strongly recommend it to any student in dynamics. It's limited to plane dynamics (obviously) but can be extremely helpful in learning and verifying results.
    – AlexanderJ93
    Nov 29 '18 at 5:03










  • Thanks you for the suggestion!
    – DMH16
    Nov 29 '18 at 5:47











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1 Answer
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1 Answer
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active

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2














Almost everything looks right.



For $a > 0$, $-2a < 0$ and $sqrt{a}+1 > 0$, so $Det[J(-sqrt{a},0)] < 0$. This equilibrium should be a saddle then for positive $a$.



Also, when analyzing the stability of a system with a parameter, it helps to draw the bifurcation diagram, i.e. the graph of the fixed points with respect to the parameter. The fixed point equations $x = a, x = sqrt{a}, x = -sqrt{a}$ can be graphed on the $ax$-plane. Intersections of these curves are the bifurcations, which are the only places where the behavior can change.



bifurcation diagram



This is the bifurcation diagram for your system. The lower black curve is the $x=-sqrt{a}$ equilibrium, which doesn't have any intersections for $a>0$, so there will be no change in behavior.





Edit:



This isn't specifically asked about in the question but it is an important feature of this equation. For $a>0, aneq 1$, there are centers at one fixed point and saddles at another nearby fixed point. When this happens, a periodic orbit will collide with the saddle point, creating a closed path from a fixed point back around to itself. This is called a homoclinic orbit, and usually has a "teardrop" shape, as seen below for $a = frac{1}{2}$.



homoclinic orbit






share|cite|improve this answer























  • Just out of curiosity, what software did you use to sketch the phase plane?
    – DMH16
    Nov 29 '18 at 4:49






  • 1




    pplane, which can be found here as a java application: math.rice.edu/~dfield/dfpp.html I strongly recommend it to any student in dynamics. It's limited to plane dynamics (obviously) but can be extremely helpful in learning and verifying results.
    – AlexanderJ93
    Nov 29 '18 at 5:03










  • Thanks you for the suggestion!
    – DMH16
    Nov 29 '18 at 5:47
















2














Almost everything looks right.



For $a > 0$, $-2a < 0$ and $sqrt{a}+1 > 0$, so $Det[J(-sqrt{a},0)] < 0$. This equilibrium should be a saddle then for positive $a$.



Also, when analyzing the stability of a system with a parameter, it helps to draw the bifurcation diagram, i.e. the graph of the fixed points with respect to the parameter. The fixed point equations $x = a, x = sqrt{a}, x = -sqrt{a}$ can be graphed on the $ax$-plane. Intersections of these curves are the bifurcations, which are the only places where the behavior can change.



bifurcation diagram



This is the bifurcation diagram for your system. The lower black curve is the $x=-sqrt{a}$ equilibrium, which doesn't have any intersections for $a>0$, so there will be no change in behavior.





Edit:



This isn't specifically asked about in the question but it is an important feature of this equation. For $a>0, aneq 1$, there are centers at one fixed point and saddles at another nearby fixed point. When this happens, a periodic orbit will collide with the saddle point, creating a closed path from a fixed point back around to itself. This is called a homoclinic orbit, and usually has a "teardrop" shape, as seen below for $a = frac{1}{2}$.



homoclinic orbit






share|cite|improve this answer























  • Just out of curiosity, what software did you use to sketch the phase plane?
    – DMH16
    Nov 29 '18 at 4:49






  • 1




    pplane, which can be found here as a java application: math.rice.edu/~dfield/dfpp.html I strongly recommend it to any student in dynamics. It's limited to plane dynamics (obviously) but can be extremely helpful in learning and verifying results.
    – AlexanderJ93
    Nov 29 '18 at 5:03










  • Thanks you for the suggestion!
    – DMH16
    Nov 29 '18 at 5:47














2












2








2






Almost everything looks right.



For $a > 0$, $-2a < 0$ and $sqrt{a}+1 > 0$, so $Det[J(-sqrt{a},0)] < 0$. This equilibrium should be a saddle then for positive $a$.



Also, when analyzing the stability of a system with a parameter, it helps to draw the bifurcation diagram, i.e. the graph of the fixed points with respect to the parameter. The fixed point equations $x = a, x = sqrt{a}, x = -sqrt{a}$ can be graphed on the $ax$-plane. Intersections of these curves are the bifurcations, which are the only places where the behavior can change.



bifurcation diagram



This is the bifurcation diagram for your system. The lower black curve is the $x=-sqrt{a}$ equilibrium, which doesn't have any intersections for $a>0$, so there will be no change in behavior.





Edit:



This isn't specifically asked about in the question but it is an important feature of this equation. For $a>0, aneq 1$, there are centers at one fixed point and saddles at another nearby fixed point. When this happens, a periodic orbit will collide with the saddle point, creating a closed path from a fixed point back around to itself. This is called a homoclinic orbit, and usually has a "teardrop" shape, as seen below for $a = frac{1}{2}$.



homoclinic orbit






share|cite|improve this answer














Almost everything looks right.



For $a > 0$, $-2a < 0$ and $sqrt{a}+1 > 0$, so $Det[J(-sqrt{a},0)] < 0$. This equilibrium should be a saddle then for positive $a$.



Also, when analyzing the stability of a system with a parameter, it helps to draw the bifurcation diagram, i.e. the graph of the fixed points with respect to the parameter. The fixed point equations $x = a, x = sqrt{a}, x = -sqrt{a}$ can be graphed on the $ax$-plane. Intersections of these curves are the bifurcations, which are the only places where the behavior can change.



bifurcation diagram



This is the bifurcation diagram for your system. The lower black curve is the $x=-sqrt{a}$ equilibrium, which doesn't have any intersections for $a>0$, so there will be no change in behavior.





Edit:



This isn't specifically asked about in the question but it is an important feature of this equation. For $a>0, aneq 1$, there are centers at one fixed point and saddles at another nearby fixed point. When this happens, a periodic orbit will collide with the saddle point, creating a closed path from a fixed point back around to itself. This is called a homoclinic orbit, and usually has a "teardrop" shape, as seen below for $a = frac{1}{2}$.



homoclinic orbit







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 29 '18 at 3:19

























answered Nov 29 '18 at 2:57









AlexanderJ93

6,093823




6,093823












  • Just out of curiosity, what software did you use to sketch the phase plane?
    – DMH16
    Nov 29 '18 at 4:49






  • 1




    pplane, which can be found here as a java application: math.rice.edu/~dfield/dfpp.html I strongly recommend it to any student in dynamics. It's limited to plane dynamics (obviously) but can be extremely helpful in learning and verifying results.
    – AlexanderJ93
    Nov 29 '18 at 5:03










  • Thanks you for the suggestion!
    – DMH16
    Nov 29 '18 at 5:47


















  • Just out of curiosity, what software did you use to sketch the phase plane?
    – DMH16
    Nov 29 '18 at 4:49






  • 1




    pplane, which can be found here as a java application: math.rice.edu/~dfield/dfpp.html I strongly recommend it to any student in dynamics. It's limited to plane dynamics (obviously) but can be extremely helpful in learning and verifying results.
    – AlexanderJ93
    Nov 29 '18 at 5:03










  • Thanks you for the suggestion!
    – DMH16
    Nov 29 '18 at 5:47
















Just out of curiosity, what software did you use to sketch the phase plane?
– DMH16
Nov 29 '18 at 4:49




Just out of curiosity, what software did you use to sketch the phase plane?
– DMH16
Nov 29 '18 at 4:49




1




1




pplane, which can be found here as a java application: math.rice.edu/~dfield/dfpp.html I strongly recommend it to any student in dynamics. It's limited to plane dynamics (obviously) but can be extremely helpful in learning and verifying results.
– AlexanderJ93
Nov 29 '18 at 5:03




pplane, which can be found here as a java application: math.rice.edu/~dfield/dfpp.html I strongly recommend it to any student in dynamics. It's limited to plane dynamics (obviously) but can be extremely helpful in learning and verifying results.
– AlexanderJ93
Nov 29 '18 at 5:03












Thanks you for the suggestion!
– DMH16
Nov 29 '18 at 5:47




Thanks you for the suggestion!
– DMH16
Nov 29 '18 at 5:47


















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