Mixing
Reduction of coupled ODEs based on symmetry
Clash Royale CLAN TAG#URR8PPP
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I am considering the following question:
Suppose I have a coupled ODEs, described by the following scheme:
$$dotx_1 = f(x_1)+H(x_2); dotx_2 = f(x_2)+H(x_3)+H(x_1); dotx_3 = f(x_3)+H(x_2); $$
where $f$ and $H$ could be nonlinear function.
Now based on that scheme, obviously, node $1$ and $3$ are symmetric and both of them make no difference for the node $2$. So reduce the above three ODEs to the following by renaming a different variable $y$:
$$doty_1 = f(y_1)+H(y_2); doty_2 = f(y_2)+2H(y_1); $$
i.e., I combine nodes $1$ and $3$ to a new node $1$. More precisely, $x_1$ and $x_3$ are replaced by $y_1$, $x_2$ is replaced by $y_2$
My question is:
- Can I say both systems ($x$ and $y$) are equivalent?
- Solution behavior around equilibrium points are the same, such as region of attraction?
Can anyone please give me some directions or related article about questions?
Thanks!
differential-equations
asked Jul 28 at 20:59
sleeve chen
2,77531646
add a comment |Â
up vote
1
down vote
favorite
I am considering the following question:
Suppose I have a coupled ODEs, described by the following scheme:
$$dotx_1 = f(x_1)+H(x_2); dotx_2 = f(x_2)+H(x_3)+H(x_1); dotx_3 = f(x_3)+H(x_2); $$
where $f$ and $H$ could be nonlinear function.
Now based on that scheme, obviously, node $1$ and $3$ are symmetric and both of them make no difference for the node $2$. So reduce the above three ODEs to the following by renaming a different variable $y$:
$$doty_1 = f(y_1)+H(y_2); doty_2 = f(y_2)+2H(y_1); $$
i.e., I combine nodes $1$ and $3$ to a new node $1$. More precisely, $x_1$ and $x_3$ are replaced by $y_1$, $x_2$ is replaced by $y_2$
My question is:
- Can I say both systems ($x$ and $y$) are equivalent?
- Solution behavior around equilibrium points are the same, such as region of attraction?
Can anyone please give me some directions or related article about questions?
Thanks!
differential-equations
asked Jul 28 at 20:59
sleeve chen
2,77531646
2
Your replacement only makes sense if $x_1$ and $x_3$ have the same initial condition.
– Michael Lee
Jul 28 at 21:02
@MichaelLee Thanks! I will be careful on this.
– sleeve chen
Jul 28 at 21:03
@MichaelLee One more question, should the region of attraction be the same? for example $x_2$ and $y_2$. I think ROA depends on $f$ and $H$, looks like $x_2$ and $y_2$ should have the same ROA.
– sleeve chen
Jul 28 at 21:12
1
The system that you have written is absolutely correct on the invariant manifold $x_1 = x_3$. If you want to study what happens on that invariant manifold, you can use it without any hesitation. You also probably see that symmetry of equations implies the symmetry of solutions: for any solution $(X_1(t), X_2(t), X_3(t))$ follows that $(X_3(t), X_2(t), X_1(t))$ is also a solution. By the way, you probably will want to study system on invariant manifold later or sooner. The good property is that $x_1 = x_3$ divides the space: a trajectory can't go from $x_1 > x_3$ to $x_1 < x_3$ because of it.
– Evgeny
Jul 29 at 10:04
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am considering the following question:
Suppose I have a coupled ODEs, described by the following scheme:
$$dotx_1 = f(x_1)+H(x_2); dotx_2 = f(x_2)+H(x_3)+H(x_1); dotx_3 = f(x_3)+H(x_2); $$
where $f$ and $H$ could be nonlinear function.
Now based on that scheme, obviously, node $1$ and $3$ are symmetric and both of them make no difference for the node $2$. So reduce the above three ODEs to the following by renaming a different variable $y$:
$$doty_1 = f(y_1)+H(y_2); doty_2 = f(y_2)+2H(y_1); $$
i.e., I combine nodes $1$ and $3$ to a new node $1$. More precisely, $x_1$ and $x_3$ are replaced by $y_1$, $x_2$ is replaced by $y_2$
My question is:
- Can I say both systems ($x$ and $y$) are equivalent?
- Solution behavior around equilibrium points are the same, such as region of attraction?
Can anyone please give me some directions or related article about questions?
Thanks!
differential-equations
asked Jul 28 at 20:59
sleeve chen
2,77531646
I am considering the following question:
Suppose I have a coupled ODEs, described by the following scheme:
$$dotx_1 = f(x_1)+H(x_2); dotx_2 = f(x_2)+H(x_3)+H(x_1); dotx_3 = f(x_3)+H(x_2); $$
where $f$ and $H$ could be nonlinear function.
Now based on that scheme, obviously, node $1$ and $3$ are symmetric and both of them make no difference for the node $2$. So reduce the above three ODEs to the following by renaming a different variable $y$:
$$doty_1 = f(y_1)+H(y_2); doty_2 = f(y_2)+2H(y_1); $$
i.e., I combine nodes $1$ and $3$ to a new node $1$. More precisely, $x_1$ and $x_3$ are replaced by $y_1$, $x_2$ is replaced by $y_2$
My question is:
- Can I say both systems ($x$ and $y$) are equivalent?
- Solution behavior around equilibrium points are the same, such as region of attraction?
Can anyone please give me some directions or related article about questions?
Thanks!
differential-equations
asked Jul 28 at 20:59
sleeve chen
2,77531646
asked Jul 28 at 20:59
sleeve chen
2,77531646
asked Jul 28 at 20:59
sleeve chen
2,77531646
asked Jul 28 at 20:59
sleeve chen
2,77531646
2,77531646
2
Your replacement only makes sense if $x_1$ and $x_3$ have the same initial condition.
– Michael Lee
Jul 28 at 21:02
@MichaelLee Thanks! I will be careful on this.
– sleeve chen
Jul 28 at 21:03
@MichaelLee One more question, should the region of attraction be the same? for example $x_2$ and $y_2$. I think ROA depends on $f$ and $H$, looks like $x_2$ and $y_2$ should have the same ROA.
– sleeve chen
Jul 28 at 21:12
1
The system that you have written is absolutely correct on the invariant manifold $x_1 = x_3$. If you want to study what happens on that invariant manifold, you can use it without any hesitation. You also probably see that symmetry of equations implies the symmetry of solutions: for any solution $(X_1(t), X_2(t), X_3(t))$ follows that $(X_3(t), X_2(t), X_1(t))$ is also a solution. By the way, you probably will want to study system on invariant manifold later or sooner. The good property is that $x_1 = x_3$ divides the space: a trajectory can't go from $x_1 > x_3$ to $x_1 < x_3$ because of it.
– Evgeny
Jul 29 at 10:04
add a comment |Â
2
Your replacement only makes sense if $x_1$ and $x_3$ have the same initial condition.
– Michael Lee
Jul 28 at 21:02
@MichaelLee Thanks! I will be careful on this.
– sleeve chen
Jul 28 at 21:03
@MichaelLee One more question, should the region of attraction be the same? for example $x_2$ and $y_2$. I think ROA depends on $f$ and $H$, looks like $x_2$ and $y_2$ should have the same ROA.
– sleeve chen
Jul 28 at 21:12
1
The system that you have written is absolutely correct on the invariant manifold $x_1 = x_3$. If you want to study what happens on that invariant manifold, you can use it without any hesitation. You also probably see that symmetry of equations implies the symmetry of solutions: for any solution $(X_1(t), X_2(t), X_3(t))$ follows that $(X_3(t), X_2(t), X_1(t))$ is also a solution. By the way, you probably will want to study system on invariant manifold later or sooner. The good property is that $x_1 = x_3$ divides the space: a trajectory can't go from $x_1 > x_3$ to $x_1 < x_3$ because of it.
– Evgeny
Jul 29 at 10:04
2
2
Your replacement only makes sense if $x_1$ and $x_3$ have the same initial condition.
– Michael Lee
Jul 28 at 21:02
Your replacement only makes sense if $x_1$ and $x_3$ have the same initial condition.
– Michael Lee
Jul 28 at 21:02
@MichaelLee Thanks! I will be careful on this.
– sleeve chen
Jul 28 at 21:03
@MichaelLee Thanks! I will be careful on this.
– sleeve chen
Jul 28 at 21:03
@MichaelLee One more question, should the region of attraction be the same? for example $x_2$ and $y_2$. I think ROA depends on $f$ and $H$, looks like $x_2$ and $y_2$ should have the same ROA.
– sleeve chen
Jul 28 at 21:12
@MichaelLee One more question, should the region of attraction be the same? for example $x_2$ and $y_2$. I think ROA depends on $f$ and $H$, looks like $x_2$ and $y_2$ should have the same ROA.
– sleeve chen
Jul 28 at 21:12
1
1
The system that you have written is absolutely correct on the invariant manifold $x_1 = x_3$. If you want to study what happens on that invariant manifold, you can use it without any hesitation. You also probably see that symmetry of equations implies the symmetry of solutions: for any solution $(X_1(t), X_2(t), X_3(t))$ follows that $(X_3(t), X_2(t), X_1(t))$ is also a solution. By the way, you probably will want to study system on invariant manifold later or sooner. The good property is that $x_1 = x_3$ divides the space: a trajectory can't go from $x_1 > x_3$ to $x_1 < x_3$ because of it.
– Evgeny
Jul 29 at 10:04
The system that you have written is absolutely correct on the invariant manifold $x_1 = x_3$. If you want to study what happens on that invariant manifold, you can use it without any hesitation. You also probably see that symmetry of equations implies the symmetry of solutions: for any solution $(X_1(t), X_2(t), X_3(t))$ follows that $(X_3(t), X_2(t), X_1(t))$ is also a solution. By the way, you probably will want to study system on invariant manifold later or sooner. The good property is that $x_1 = x_3$ divides the space: a trajectory can't go from $x_1 > x_3$ to $x_1 < x_3$ because of it.
– Evgeny
Jul 29 at 10:04
add a comment |Â
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2
Your replacement only makes sense if $x_1$ and $x_3$ have the same initial condition.
– Michael Lee
Jul 28 at 21:02
@MichaelLee Thanks! I will be careful on this.
– sleeve chen
Jul 28 at 21:03
@MichaelLee One more question, should the region of attraction be the same? for example $x_2$ and $y_2$. I think ROA depends on $f$ and $H$, looks like $x_2$ and $y_2$ should have the same ROA.
– sleeve chen
Jul 28 at 21:12
1
The system that you have written is absolutely correct on the invariant manifold $x_1 = x_3$. If you want to study what happens on that invariant manifold, you can use it without any hesitation. You also probably see that symmetry of equations implies the symmetry of solutions: for any solution $(X_1(t), X_2(t), X_3(t))$ follows that $(X_3(t), X_2(t), X_1(t))$ is also a solution. By the way, you probably will want to study system on invariant manifold later or sooner. The good property is that $x_1 = x_3$ divides the space: a trajectory can't go from $x_1 > x_3$ to $x_1 < x_3$ because of it.
– Evgeny
Jul 29 at 10:04