Stability Systems - Duffing oscillator [closed]
$begingroup$
In the case a=1,b=-1 this is the system:
$$ dx=y $$
$$ dy=-x + x^3$$
I have to draw the phase space with the trajectories of the orbits. And I don´t know who to demonstrate the direction in the orbits. I only know is a circle for the $(0,0)$ and hyperbola for$(-1,0),(1,0)$.
ordinary-differential-equations
asked Dec 22 '18 at 10:34
MeliodasMeliodas
215
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closed as off-topic by José Carlos Santos, Brahadeesh, mrtaurho, Saad, Tianlalu Dec 22 '18 at 12:56
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Brahadeesh, mrtaurho, Saad, Tianlalu
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
In the case a=1,b=-1 this is the system:
$$ dx=y $$
$$ dy=-x + x^3$$
I have to draw the phase space with the trajectories of the orbits. And I don´t know who to demonstrate the direction in the orbits. I only know is a circle for the $(0,0)$ and hyperbola for$(-1,0),(1,0)$.
ordinary-differential-equations
asked Dec 22 '18 at 10:34
MeliodasMeliodas
215
$endgroup$
closed as off-topic by José Carlos Santos, Brahadeesh, mrtaurho, Saad, Tianlalu Dec 22 '18 at 12:56
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Brahadeesh, mrtaurho, Saad, Tianlalu
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
In the case a=1,b=-1 this is the system:
$$ dx=y $$
$$ dy=-x + x^3$$
I have to draw the phase space with the trajectories of the orbits. And I don´t know who to demonstrate the direction in the orbits. I only know is a circle for the $(0,0)$ and hyperbola for$(-1,0),(1,0)$.
ordinary-differential-equations
asked Dec 22 '18 at 10:34
MeliodasMeliodas
215
$endgroup$
In the case a=1,b=-1 this is the system:
$$ dx=y $$
$$ dy=-x + x^3$$
I have to draw the phase space with the trajectories of the orbits. And I don´t know who to demonstrate the direction in the orbits. I only know is a circle for the $(0,0)$ and hyperbola for$(-1,0),(1,0)$.
ordinary-differential-equations
ordinary-differential-equations
asked Dec 22 '18 at 10:34
MeliodasMeliodas
215
asked Dec 22 '18 at 10:34
MeliodasMeliodas
215
asked Dec 22 '18 at 10:34
MeliodasMeliodas
215
asked Dec 22 '18 at 10:34
MeliodasMeliodas
215
asked Dec 22 '18 at 10:34
MeliodasMeliodas
215
215
closed as off-topic by José Carlos Santos, Brahadeesh, mrtaurho, Saad, Tianlalu Dec 22 '18 at 12:56
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Brahadeesh, mrtaurho, Saad, Tianlalu
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by José Carlos Santos, Brahadeesh, mrtaurho, Saad, Tianlalu Dec 22 '18 at 12:56
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Brahadeesh, mrtaurho, Saad, Tianlalu
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
For the critical point at $(0, 0)$ just evaluate the dynamical system close to the origin. For example, take $x = 0.1$, and $y = 0$, you see that at that location ${rm d y}/{rm d}t < 0$, that means that a that location $y$ will be decreasing. In other words, the orbit going through $(0.1, 0)$ will rotate clock-wise. Same argument can be applied to the other critical points.
Here's a sketch to confirm it
answered Dec 22 '18 at 10:56
caveraccaverac
14.6k31130
$endgroup$
$begingroup$
Thank you very much! So clear explanaition.
$endgroup$
– Meliodas
Dec 22 '18 at 11:06
$begingroup$
Not a bad way but it's not such a rigorous method in my opinion.
$endgroup$
– Rebellos
Dec 22 '18 at 11:08
add a comment |
$begingroup$
Form the Jacobian of the system :
$$J(x,y) = begin{bmatrix} 0 & 1 \ -1 + 3x^2 & 0end{bmatrix}$$
For the origin $O(0,0)$ which is a critical point for the given system, it is :
$$J(0,0) = begin{bmatrix} 0 & 1 \ -1 & 0 end{bmatrix}$$
Then, the eigenvalues of the given Jacobian for the origin :
$$det(J(0,0) -lambda I) = 0 Rightarrow begin{vmatrix} - lambda & 1 \ -1 & -lambdaend{vmatrix} = 0 Leftrightarrow lambda^2 + 1 = 0 Leftrightarrow lambda = pm i$$
This truly ensures that the origin $O(0,0)$ is a center for the given system and clockwise.
Now, you also have the critical points $A(-1,0)$ and $B(1,0)$. For $A$, the Jacobian is
$$J(-1,0) = begin{bmatrix} 0 & 1 \ 2 & 0 end{bmatrix}$$
with eigenvalues :
$$det(J(-1,0) -lambda I) = 0 Rightarrow begin{vmatrix} - lambda & 1 \ 2 & -lambdaend{vmatrix} = 0 Leftrightarrow lambda^2 -2 = 0 Leftrightarrow lambda = pm sqrt{2}$$
Since $lambda_1 cdot lambda _2 < 0$ and the eigenvalues are purely real, the critical point $A$ will then be a saddle for the system, which by theory is unstable, thus the arrows are pointing away.
I will leave the case of $B$ up to you to work around.
answered Dec 22 '18 at 11:09
RebellosRebellos
15.1k31250
$endgroup$
$begingroup$
thank you!. Is also very clear.
$endgroup$
– Meliodas
Dec 22 '18 at 11:12
add a comment |
$begingroup$
Multiply
$$
ddot x+x-x^2=0
$$
with $2dot x$ and integrate to get
$$
dot x^2+x^2-frac12x^4=R^2
$$
and parametrize this as a circle equation to get polar coordinates
$$(Rcos phi(t),Rsin(phi))=left(dot x,xsqrt{1-frac12x^2}right).$$
Close to $x=0$ the equation $u=f(x)=xsqrt{1-frac12x^2}$ in the second component is monotone and can be inverted as $x=g(u)=g(Rsin(phi))$ with time derivative
$$
dot x = g'(Rsin(phi))Rcos phi(t),dotphi(t)
$$
so that $dotphi(t)=1+O(R^2)$. This means counter-clock wise rotation in a $(dot x,x)$ diagram or, diagonally reflected, clockwise rotation in a $(x,dot x)$ diagram.
answered Dec 22 '18 at 11:12
LutzLLutzL
59.2k42057
$endgroup$
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For the critical point at $(0, 0)$ just evaluate the dynamical system close to the origin. For example, take $x = 0.1$, and $y = 0$, you see that at that location ${rm d y}/{rm d}t < 0$, that means that a that location $y$ will be decreasing. In other words, the orbit going through $(0.1, 0)$ will rotate clock-wise. Same argument can be applied to the other critical points.
Here's a sketch to confirm it
answered Dec 22 '18 at 10:56
caveraccaverac
14.6k31130
$endgroup$
$begingroup$
Thank you very much! So clear explanaition.
$endgroup$
– Meliodas
Dec 22 '18 at 11:06
$begingroup$
Not a bad way but it's not such a rigorous method in my opinion.
$endgroup$
– Rebellos
Dec 22 '18 at 11:08
add a comment |
$begingroup$
For the critical point at $(0, 0)$ just evaluate the dynamical system close to the origin. For example, take $x = 0.1$, and $y = 0$, you see that at that location ${rm d y}/{rm d}t < 0$, that means that a that location $y$ will be decreasing. In other words, the orbit going through $(0.1, 0)$ will rotate clock-wise. Same argument can be applied to the other critical points.
Here's a sketch to confirm it
answered Dec 22 '18 at 10:56
caveraccaverac
14.6k31130
$endgroup$
$begingroup$
Thank you very much! So clear explanaition.
$endgroup$
– Meliodas
Dec 22 '18 at 11:06
$begingroup$
Not a bad way but it's not such a rigorous method in my opinion.
$endgroup$
– Rebellos
Dec 22 '18 at 11:08
add a comment |
$begingroup$
For the critical point at $(0, 0)$ just evaluate the dynamical system close to the origin. For example, take $x = 0.1$, and $y = 0$, you see that at that location ${rm d y}/{rm d}t < 0$, that means that a that location $y$ will be decreasing. In other words, the orbit going through $(0.1, 0)$ will rotate clock-wise. Same argument can be applied to the other critical points.
Here's a sketch to confirm it
answered Dec 22 '18 at 10:56
caveraccaverac
14.6k31130
$endgroup$
For the critical point at $(0, 0)$ just evaluate the dynamical system close to the origin. For example, take $x = 0.1$, and $y = 0$, you see that at that location ${rm d y}/{rm d}t < 0$, that means that a that location $y$ will be decreasing. In other words, the orbit going through $(0.1, 0)$ will rotate clock-wise. Same argument can be applied to the other critical points.
Here's a sketch to confirm it
answered Dec 22 '18 at 10:56
caveraccaverac
14.6k31130
answered Dec 22 '18 at 10:56
caveraccaverac
14.6k31130
answered Dec 22 '18 at 10:56
caveraccaverac
14.6k31130
answered Dec 22 '18 at 10:56
caveraccaverac
14.6k31130
14.6k31130
$begingroup$
Thank you very much! So clear explanaition.
$endgroup$
– Meliodas
Dec 22 '18 at 11:06
$begingroup$
Not a bad way but it's not such a rigorous method in my opinion.
$endgroup$
– Rebellos
Dec 22 '18 at 11:08
add a comment |
$begingroup$
Thank you very much! So clear explanaition.
$endgroup$
– Meliodas
Dec 22 '18 at 11:06
$begingroup$
Not a bad way but it's not such a rigorous method in my opinion.
$endgroup$
– Rebellos
Dec 22 '18 at 11:08
$begingroup$
Thank you very much! So clear explanaition.
$endgroup$
– Meliodas
Dec 22 '18 at 11:06
$begingroup$
Thank you very much! So clear explanaition.
$endgroup$
– Meliodas
Dec 22 '18 at 11:06
$begingroup$
Not a bad way but it's not such a rigorous method in my opinion.
$endgroup$
– Rebellos
Dec 22 '18 at 11:08
$begingroup$
Not a bad way but it's not such a rigorous method in my opinion.
$endgroup$
– Rebellos
Dec 22 '18 at 11:08
add a comment |
$begingroup$
Form the Jacobian of the system :
$$J(x,y) = begin{bmatrix} 0 & 1 \ -1 + 3x^2 & 0end{bmatrix}$$
For the origin $O(0,0)$ which is a critical point for the given system, it is :
$$J(0,0) = begin{bmatrix} 0 & 1 \ -1 & 0 end{bmatrix}$$
Then, the eigenvalues of the given Jacobian for the origin :
$$det(J(0,0) -lambda I) = 0 Rightarrow begin{vmatrix} - lambda & 1 \ -1 & -lambdaend{vmatrix} = 0 Leftrightarrow lambda^2 + 1 = 0 Leftrightarrow lambda = pm i$$
This truly ensures that the origin $O(0,0)$ is a center for the given system and clockwise.
Now, you also have the critical points $A(-1,0)$ and $B(1,0)$. For $A$, the Jacobian is
$$J(-1,0) = begin{bmatrix} 0 & 1 \ 2 & 0 end{bmatrix}$$
with eigenvalues :
$$det(J(-1,0) -lambda I) = 0 Rightarrow begin{vmatrix} - lambda & 1 \ 2 & -lambdaend{vmatrix} = 0 Leftrightarrow lambda^2 -2 = 0 Leftrightarrow lambda = pm sqrt{2}$$
Since $lambda_1 cdot lambda _2 < 0$ and the eigenvalues are purely real, the critical point $A$ will then be a saddle for the system, which by theory is unstable, thus the arrows are pointing away.
I will leave the case of $B$ up to you to work around.
answered Dec 22 '18 at 11:09
RebellosRebellos
15.1k31250
$endgroup$
$begingroup$
thank you!. Is also very clear.
$endgroup$
– Meliodas
Dec 22 '18 at 11:12
add a comment |
$begingroup$
Form the Jacobian of the system :
$$J(x,y) = begin{bmatrix} 0 & 1 \ -1 + 3x^2 & 0end{bmatrix}$$
For the origin $O(0,0)$ which is a critical point for the given system, it is :
$$J(0,0) = begin{bmatrix} 0 & 1 \ -1 & 0 end{bmatrix}$$
Then, the eigenvalues of the given Jacobian for the origin :
$$det(J(0,0) -lambda I) = 0 Rightarrow begin{vmatrix} - lambda & 1 \ -1 & -lambdaend{vmatrix} = 0 Leftrightarrow lambda^2 + 1 = 0 Leftrightarrow lambda = pm i$$
This truly ensures that the origin $O(0,0)$ is a center for the given system and clockwise.
Now, you also have the critical points $A(-1,0)$ and $B(1,0)$. For $A$, the Jacobian is
$$J(-1,0) = begin{bmatrix} 0 & 1 \ 2 & 0 end{bmatrix}$$
with eigenvalues :
$$det(J(-1,0) -lambda I) = 0 Rightarrow begin{vmatrix} - lambda & 1 \ 2 & -lambdaend{vmatrix} = 0 Leftrightarrow lambda^2 -2 = 0 Leftrightarrow lambda = pm sqrt{2}$$
Since $lambda_1 cdot lambda _2 < 0$ and the eigenvalues are purely real, the critical point $A$ will then be a saddle for the system, which by theory is unstable, thus the arrows are pointing away.
I will leave the case of $B$ up to you to work around.
answered Dec 22 '18 at 11:09
RebellosRebellos
15.1k31250
$endgroup$
$begingroup$
thank you!. Is also very clear.
$endgroup$
– Meliodas
Dec 22 '18 at 11:12
add a comment |
$begingroup$
Form the Jacobian of the system :
$$J(x,y) = begin{bmatrix} 0 & 1 \ -1 + 3x^2 & 0end{bmatrix}$$
For the origin $O(0,0)$ which is a critical point for the given system, it is :
$$J(0,0) = begin{bmatrix} 0 & 1 \ -1 & 0 end{bmatrix}$$
Then, the eigenvalues of the given Jacobian for the origin :
$$det(J(0,0) -lambda I) = 0 Rightarrow begin{vmatrix} - lambda & 1 \ -1 & -lambdaend{vmatrix} = 0 Leftrightarrow lambda^2 + 1 = 0 Leftrightarrow lambda = pm i$$
This truly ensures that the origin $O(0,0)$ is a center for the given system and clockwise.
Now, you also have the critical points $A(-1,0)$ and $B(1,0)$. For $A$, the Jacobian is
$$J(-1,0) = begin{bmatrix} 0 & 1 \ 2 & 0 end{bmatrix}$$
with eigenvalues :
$$det(J(-1,0) -lambda I) = 0 Rightarrow begin{vmatrix} - lambda & 1 \ 2 & -lambdaend{vmatrix} = 0 Leftrightarrow lambda^2 -2 = 0 Leftrightarrow lambda = pm sqrt{2}$$
Since $lambda_1 cdot lambda _2 < 0$ and the eigenvalues are purely real, the critical point $A$ will then be a saddle for the system, which by theory is unstable, thus the arrows are pointing away.
I will leave the case of $B$ up to you to work around.
answered Dec 22 '18 at 11:09
RebellosRebellos
15.1k31250
$endgroup$
Form the Jacobian of the system :
$$J(x,y) = begin{bmatrix} 0 & 1 \ -1 + 3x^2 & 0end{bmatrix}$$
For the origin $O(0,0)$ which is a critical point for the given system, it is :
$$J(0,0) = begin{bmatrix} 0 & 1 \ -1 & 0 end{bmatrix}$$
Then, the eigenvalues of the given Jacobian for the origin :
$$det(J(0,0) -lambda I) = 0 Rightarrow begin{vmatrix} - lambda & 1 \ -1 & -lambdaend{vmatrix} = 0 Leftrightarrow lambda^2 + 1 = 0 Leftrightarrow lambda = pm i$$
This truly ensures that the origin $O(0,0)$ is a center for the given system and clockwise.
Now, you also have the critical points $A(-1,0)$ and $B(1,0)$. For $A$, the Jacobian is
$$J(-1,0) = begin{bmatrix} 0 & 1 \ 2 & 0 end{bmatrix}$$
with eigenvalues :
$$det(J(-1,0) -lambda I) = 0 Rightarrow begin{vmatrix} - lambda & 1 \ 2 & -lambdaend{vmatrix} = 0 Leftrightarrow lambda^2 -2 = 0 Leftrightarrow lambda = pm sqrt{2}$$
Since $lambda_1 cdot lambda _2 < 0$ and the eigenvalues are purely real, the critical point $A$ will then be a saddle for the system, which by theory is unstable, thus the arrows are pointing away.
I will leave the case of $B$ up to you to work around.
answered Dec 22 '18 at 11:09
RebellosRebellos
15.1k31250
answered Dec 22 '18 at 11:09
RebellosRebellos
15.1k31250
answered Dec 22 '18 at 11:09
RebellosRebellos
15.1k31250
answered Dec 22 '18 at 11:09
RebellosRebellos
15.1k31250
15.1k31250
$begingroup$
thank you!. Is also very clear.
$endgroup$
– Meliodas
Dec 22 '18 at 11:12
add a comment |
$begingroup$
thank you!. Is also very clear.
$endgroup$
– Meliodas
Dec 22 '18 at 11:12
$begingroup$
thank you!. Is also very clear.
$endgroup$
– Meliodas
Dec 22 '18 at 11:12
$begingroup$
thank you!. Is also very clear.
$endgroup$
– Meliodas
Dec 22 '18 at 11:12
add a comment |
$begingroup$
Multiply
$$
ddot x+x-x^2=0
$$
with $2dot x$ and integrate to get
$$
dot x^2+x^2-frac12x^4=R^2
$$
and parametrize this as a circle equation to get polar coordinates
$$(Rcos phi(t),Rsin(phi))=left(dot x,xsqrt{1-frac12x^2}right).$$
Close to $x=0$ the equation $u=f(x)=xsqrt{1-frac12x^2}$ in the second component is monotone and can be inverted as $x=g(u)=g(Rsin(phi))$ with time derivative
$$
dot x = g'(Rsin(phi))Rcos phi(t),dotphi(t)
$$
so that $dotphi(t)=1+O(R^2)$. This means counter-clock wise rotation in a $(dot x,x)$ diagram or, diagonally reflected, clockwise rotation in a $(x,dot x)$ diagram.
answered Dec 22 '18 at 11:12
LutzLLutzL
59.2k42057
$endgroup$
add a comment |
$begingroup$
Multiply
$$
ddot x+x-x^2=0
$$
with $2dot x$ and integrate to get
$$
dot x^2+x^2-frac12x^4=R^2
$$
and parametrize this as a circle equation to get polar coordinates
$$(Rcos phi(t),Rsin(phi))=left(dot x,xsqrt{1-frac12x^2}right).$$
Close to $x=0$ the equation $u=f(x)=xsqrt{1-frac12x^2}$ in the second component is monotone and can be inverted as $x=g(u)=g(Rsin(phi))$ with time derivative
$$
dot x = g'(Rsin(phi))Rcos phi(t),dotphi(t)
$$
so that $dotphi(t)=1+O(R^2)$. This means counter-clock wise rotation in a $(dot x,x)$ diagram or, diagonally reflected, clockwise rotation in a $(x,dot x)$ diagram.
answered Dec 22 '18 at 11:12
LutzLLutzL
59.2k42057
$endgroup$
add a comment |
$begingroup$
Multiply
$$
ddot x+x-x^2=0
$$
with $2dot x$ and integrate to get
$$
dot x^2+x^2-frac12x^4=R^2
$$
and parametrize this as a circle equation to get polar coordinates
$$(Rcos phi(t),Rsin(phi))=left(dot x,xsqrt{1-frac12x^2}right).$$
Close to $x=0$ the equation $u=f(x)=xsqrt{1-frac12x^2}$ in the second component is monotone and can be inverted as $x=g(u)=g(Rsin(phi))$ with time derivative
$$
dot x = g'(Rsin(phi))Rcos phi(t),dotphi(t)
$$
so that $dotphi(t)=1+O(R^2)$. This means counter-clock wise rotation in a $(dot x,x)$ diagram or, diagonally reflected, clockwise rotation in a $(x,dot x)$ diagram.
answered Dec 22 '18 at 11:12
LutzLLutzL
59.2k42057
$endgroup$
Multiply
$$
ddot x+x-x^2=0
$$
with $2dot x$ and integrate to get
$$
dot x^2+x^2-frac12x^4=R^2
$$
and parametrize this as a circle equation to get polar coordinates
$$(Rcos phi(t),Rsin(phi))=left(dot x,xsqrt{1-frac12x^2}right).$$
Close to $x=0$ the equation $u=f(x)=xsqrt{1-frac12x^2}$ in the second component is monotone and can be inverted as $x=g(u)=g(Rsin(phi))$ with time derivative
$$
dot x = g'(Rsin(phi))Rcos phi(t),dotphi(t)
$$
so that $dotphi(t)=1+O(R^2)$. This means counter-clock wise rotation in a $(dot x,x)$ diagram or, diagonally reflected, clockwise rotation in a $(x,dot x)$ diagram.
answered Dec 22 '18 at 11:12
LutzLLutzL
59.2k42057
answered Dec 22 '18 at 11:12
LutzLLutzL
59.2k42057
answered Dec 22 '18 at 11:12
LutzLLutzL
59.2k42057
answered Dec 22 '18 at 11:12
LutzLLutzL
59.2k42057
59.2k42057
add a comment |
add a comment |
