Mixing
What does “Completely integrable” mean in the context of Hamiltonian systems?
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I am reading Ordinary Differential Equations and Dynamical Systems by Gerald Teschl am confused on what is exactly meant by "completely integrable" in the following picture. This is Chapter 8 section 4. Also, they refer to the "Hamilton structure" in the text below. Does this mean the Hamilton's equations listed here: 
?
dynamical-systems
edited Jan 23 at 18:53
Bernard
123k741116
asked Jan 23 at 18:44
NaltNalt
776
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add a comment |
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I am reading Ordinary Differential Equations and Dynamical Systems by Gerald Teschl am confused on what is exactly meant by "completely integrable" in the following picture. This is Chapter 8 section 4. Also, they refer to the "Hamilton structure" in the text below. Does this mean the Hamilton's equations listed here: ?
dynamical-systems
edited Jan 23 at 18:53
Bernard
123k741116
asked Jan 23 at 18:44
NaltNalt
776
$endgroup$
$begingroup$
It means that you can write the flow of the system as a system of independent rotations resp. independent oscillations. In other words, there exists a system or $n$ first integrals.
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– LutzL
Jan 23 at 19:10
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@LutzL Thank you for your answer. Is this what the author is meaning by talking about a "canonical form" for the Hamilton system?
$endgroup$
– Nalt
Jan 23 at 20:24
add a comment |
$begingroup$
I am reading Ordinary Differential Equations and Dynamical Systems by Gerald Teschl am confused on what is exactly meant by "completely integrable" in the following picture. This is Chapter 8 section 4. Also, they refer to the "Hamilton structure" in the text below. Does this mean the Hamilton's equations listed here: ?
dynamical-systems
edited Jan 23 at 18:53
Bernard
123k741116
asked Jan 23 at 18:44
NaltNalt
776
$endgroup$
I am reading Ordinary Differential Equations and Dynamical Systems by Gerald Teschl am confused on what is exactly meant by "completely integrable" in the following picture. This is Chapter 8 section 4. Also, they refer to the "Hamilton structure" in the text below. Does this mean the Hamilton's equations listed here: ?
dynamical-systems
dynamical-systems
edited Jan 23 at 18:53
Bernard
123k741116
asked Jan 23 at 18:44
NaltNalt
776
edited Jan 23 at 18:53
Bernard
123k741116
asked Jan 23 at 18:44
NaltNalt
776
edited Jan 23 at 18:53
Bernard
123k741116
edited Jan 23 at 18:53
Bernard
123k741116
edited Jan 23 at 18:53
Bernard
123k741116
123k741116
asked Jan 23 at 18:44
NaltNalt
776
asked Jan 23 at 18:44
NaltNalt
776
asked Jan 23 at 18:44
NaltNalt
776
776
$begingroup$
It means that you can write the flow of the system as a system of independent rotations resp. independent oscillations. In other words, there exists a system or $n$ first integrals.
$endgroup$
– LutzL
Jan 23 at 19:10
$begingroup$
@LutzL Thank you for your answer. Is this what the author is meaning by talking about a "canonical form" for the Hamilton system?
$endgroup$
– Nalt
Jan 23 at 20:24
add a comment |
$begingroup$
It means that you can write the flow of the system as a system of independent rotations resp. independent oscillations. In other words, there exists a system or $n$ first integrals.
$endgroup$
– LutzL
Jan 23 at 19:10
$begingroup$
@LutzL Thank you for your answer. Is this what the author is meaning by talking about a "canonical form" for the Hamilton system?
$endgroup$
– Nalt
Jan 23 at 20:24
$begingroup$
It means that you can write the flow of the system as a system of independent rotations resp. independent oscillations. In other words, there exists a system or $n$ first integrals.
$endgroup$
– LutzL
Jan 23 at 19:10
$begingroup$
It means that you can write the flow of the system as a system of independent rotations resp. independent oscillations. In other words, there exists a system or $n$ first integrals.
$endgroup$
– LutzL
Jan 23 at 19:10
$begingroup$
@LutzL Thank you for your answer. Is this what the author is meaning by talking about a "canonical form" for the Hamilton system?
$endgroup$
– Nalt
Jan 23 at 20:24
$begingroup$
@LutzL Thank you for your answer. Is this what the author is meaning by talking about a "canonical form" for the Hamilton system?
$endgroup$
– Nalt
Jan 23 at 20:24
add a comment |
1 Answer
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A "completely integrable" Hamiltonian system is one that satisfies the Liouville-Arnold theorem. Building off what LutzL said, there exist $n$ independent first integrals.
If any of the integrals are dependent (nonvanishing under the Poisson bracket), then you need $>n$ first integrals. How many depends on the relationships given by the Poisson bracket.
"Canonical form" refers to the fact that the position and velocities are "on the same level" as each other, as opposed to the Lagrangian-style of dynamics. You'll have $2n$ first-order differential equations as opposed to the $n$ second-order differential equations from Lagrange. On a deeper level, this relates to the symplectic manifold that the Hamiltonian system lives in.
answered Jan 25 at 17:45
Michael SparapanyMichael Sparapany
1666
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$begingroup$
A "completely integrable" Hamiltonian system is one that satisfies the Liouville-Arnold theorem. Building off what LutzL said, there exist $n$ independent first integrals.
If any of the integrals are dependent (nonvanishing under the Poisson bracket), then you need $>n$ first integrals. How many depends on the relationships given by the Poisson bracket.
"Canonical form" refers to the fact that the position and velocities are "on the same level" as each other, as opposed to the Lagrangian-style of dynamics. You'll have $2n$ first-order differential equations as opposed to the $n$ second-order differential equations from Lagrange. On a deeper level, this relates to the symplectic manifold that the Hamiltonian system lives in.
answered Jan 25 at 17:45
Michael SparapanyMichael Sparapany
1666
$endgroup$
add a comment |
$begingroup$
A "completely integrable" Hamiltonian system is one that satisfies the Liouville-Arnold theorem. Building off what LutzL said, there exist $n$ independent first integrals.
If any of the integrals are dependent (nonvanishing under the Poisson bracket), then you need $>n$ first integrals. How many depends on the relationships given by the Poisson bracket.
"Canonical form" refers to the fact that the position and velocities are "on the same level" as each other, as opposed to the Lagrangian-style of dynamics. You'll have $2n$ first-order differential equations as opposed to the $n$ second-order differential equations from Lagrange. On a deeper level, this relates to the symplectic manifold that the Hamiltonian system lives in.
answered Jan 25 at 17:45
Michael SparapanyMichael Sparapany
1666
$endgroup$
add a comment |
$begingroup$
A "completely integrable" Hamiltonian system is one that satisfies the Liouville-Arnold theorem. Building off what LutzL said, there exist $n$ independent first integrals.
If any of the integrals are dependent (nonvanishing under the Poisson bracket), then you need $>n$ first integrals. How many depends on the relationships given by the Poisson bracket.
"Canonical form" refers to the fact that the position and velocities are "on the same level" as each other, as opposed to the Lagrangian-style of dynamics. You'll have $2n$ first-order differential equations as opposed to the $n$ second-order differential equations from Lagrange. On a deeper level, this relates to the symplectic manifold that the Hamiltonian system lives in.
answered Jan 25 at 17:45
Michael SparapanyMichael Sparapany
1666
$endgroup$
A "completely integrable" Hamiltonian system is one that satisfies the Liouville-Arnold theorem. Building off what LutzL said, there exist $n$ independent first integrals.
If any of the integrals are dependent (nonvanishing under the Poisson bracket), then you need $>n$ first integrals. How many depends on the relationships given by the Poisson bracket.
"Canonical form" refers to the fact that the position and velocities are "on the same level" as each other, as opposed to the Lagrangian-style of dynamics. You'll have $2n$ first-order differential equations as opposed to the $n$ second-order differential equations from Lagrange. On a deeper level, this relates to the symplectic manifold that the Hamiltonian system lives in.
answered Jan 25 at 17:45
Michael SparapanyMichael Sparapany
1666
answered Jan 25 at 17:45
Michael SparapanyMichael Sparapany
1666
answered Jan 25 at 17:45
Michael SparapanyMichael Sparapany
1666
answered Jan 25 at 17:45
Michael SparapanyMichael Sparapany
1666
1666
add a comment |
add a comment |
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$begingroup$
It means that you can write the flow of the system as a system of independent rotations resp. independent oscillations. In other words, there exists a system or $n$ first integrals.
$endgroup$
– LutzL
Jan 23 at 19:10
$begingroup$
@LutzL Thank you for your answer. Is this what the author is meaning by talking about a "canonical form" for the Hamilton system?
$endgroup$
– Nalt
Jan 23 at 20:24