The linearization at an equilibrium point of a planar Hamiltonian system has eigenvalues that are either +-a...
$begingroup$
This is a proposition in our textbook, but it is never explained as to why (we take it as given), and it doesn't seem very intuitive. Is there an elegant way to show this that I'm missing?
ordinary-differential-equations
asked Dec 2 '18 at 23:58
MathGuyForLifeMathGuyForLife
1007
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add a comment |
$begingroup$
This is a proposition in our textbook, but it is never explained as to why (we take it as given), and it doesn't seem very intuitive. Is there an elegant way to show this that I'm missing?
ordinary-differential-equations
asked Dec 2 '18 at 23:58
MathGuyForLifeMathGuyForLife
1007
$endgroup$
add a comment |
$begingroup$
This is a proposition in our textbook, but it is never explained as to why (we take it as given), and it doesn't seem very intuitive. Is there an elegant way to show this that I'm missing?
ordinary-differential-equations
asked Dec 2 '18 at 23:58
MathGuyForLifeMathGuyForLife
1007
$endgroup$
This is a proposition in our textbook, but it is never explained as to why (we take it as given), and it doesn't seem very intuitive. Is there an elegant way to show this that I'm missing?
ordinary-differential-equations
ordinary-differential-equations
asked Dec 2 '18 at 23:58
MathGuyForLifeMathGuyForLife
1007
asked Dec 2 '18 at 23:58
MathGuyForLifeMathGuyForLife
1007
asked Dec 2 '18 at 23:58
MathGuyForLifeMathGuyForLife
1007
asked Dec 2 '18 at 23:58
MathGuyForLifeMathGuyForLife
1007
asked Dec 2 '18 at 23:58
MathGuyForLifeMathGuyForLife
1007
1007
add a comment |
add a comment |
1 Answer
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$begingroup$
A planar Hamiltonian system looks like
$dot q = dfrac{partial H(p, q)}{partial p} = H_p, tag 1$
$dot p = -dfrac{partial H(p, q)}{partial q} = -H_q. tag 2$
The Jacobean matrix $J_H$ of (1)-(2) is
$J_H = begin{bmatrix} H_{pq} & H_{pp} \ -H_{qq} & -H_{qp} end{bmatrix}. tag 3$
If we allow that
$H in C^2(U, Bbb R), tag 4$
where $U subset Bbb R^2$ is an open set on which $H(p, q)$ is defined, then
$H_{pq} = H_{qp}, tag 5$
so it follows that the trace of $J_H$,
$text{Tr}(J_H) = H_{pq} - H_{qp} = 0; tag 6$
thus the characteristic polynomial $chi_H(lambda)$ of $J_H$ is easily seen to be
$chi_H(lambda) = lambda^2 - text{Tr}(J_H) lambda + det(J_H) = lambda^2 + det(J_H) = 0; tag 7$
it follows that
$det(J_H) le 0 Longrightarrow lambda =pm sqrt {-det(J_H)} in Bbb R,tag 8$
and
$det(J_H) > 0 Longrightarrow lambda =pm isqrt {det(J_H)} in iBbb R, tag 9$
as per request. $OEDelta$
answered Dec 3 '18 at 1:49
Robert LewisRobert Lewis
44.8k22964
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1 Answer
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1 Answer
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$begingroup$
A planar Hamiltonian system looks like
$dot q = dfrac{partial H(p, q)}{partial p} = H_p, tag 1$
$dot p = -dfrac{partial H(p, q)}{partial q} = -H_q. tag 2$
The Jacobean matrix $J_H$ of (1)-(2) is
$J_H = begin{bmatrix} H_{pq} & H_{pp} \ -H_{qq} & -H_{qp} end{bmatrix}. tag 3$
If we allow that
$H in C^2(U, Bbb R), tag 4$
where $U subset Bbb R^2$ is an open set on which $H(p, q)$ is defined, then
$H_{pq} = H_{qp}, tag 5$
so it follows that the trace of $J_H$,
$text{Tr}(J_H) = H_{pq} - H_{qp} = 0; tag 6$
thus the characteristic polynomial $chi_H(lambda)$ of $J_H$ is easily seen to be
$chi_H(lambda) = lambda^2 - text{Tr}(J_H) lambda + det(J_H) = lambda^2 + det(J_H) = 0; tag 7$
it follows that
$det(J_H) le 0 Longrightarrow lambda =pm sqrt {-det(J_H)} in Bbb R,tag 8$
and
$det(J_H) > 0 Longrightarrow lambda =pm isqrt {det(J_H)} in iBbb R, tag 9$
as per request. $OEDelta$
answered Dec 3 '18 at 1:49
Robert LewisRobert Lewis
44.8k22964
$endgroup$
add a comment |
$begingroup$
A planar Hamiltonian system looks like
$dot q = dfrac{partial H(p, q)}{partial p} = H_p, tag 1$
$dot p = -dfrac{partial H(p, q)}{partial q} = -H_q. tag 2$
The Jacobean matrix $J_H$ of (1)-(2) is
$J_H = begin{bmatrix} H_{pq} & H_{pp} \ -H_{qq} & -H_{qp} end{bmatrix}. tag 3$
If we allow that
$H in C^2(U, Bbb R), tag 4$
where $U subset Bbb R^2$ is an open set on which $H(p, q)$ is defined, then
$H_{pq} = H_{qp}, tag 5$
so it follows that the trace of $J_H$,
$text{Tr}(J_H) = H_{pq} - H_{qp} = 0; tag 6$
thus the characteristic polynomial $chi_H(lambda)$ of $J_H$ is easily seen to be
$chi_H(lambda) = lambda^2 - text{Tr}(J_H) lambda + det(J_H) = lambda^2 + det(J_H) = 0; tag 7$
it follows that
$det(J_H) le 0 Longrightarrow lambda =pm sqrt {-det(J_H)} in Bbb R,tag 8$
and
$det(J_H) > 0 Longrightarrow lambda =pm isqrt {det(J_H)} in iBbb R, tag 9$
as per request. $OEDelta$
answered Dec 3 '18 at 1:49
Robert LewisRobert Lewis
44.8k22964
$endgroup$
add a comment |
$begingroup$
A planar Hamiltonian system looks like
$dot q = dfrac{partial H(p, q)}{partial p} = H_p, tag 1$
$dot p = -dfrac{partial H(p, q)}{partial q} = -H_q. tag 2$
The Jacobean matrix $J_H$ of (1)-(2) is
$J_H = begin{bmatrix} H_{pq} & H_{pp} \ -H_{qq} & -H_{qp} end{bmatrix}. tag 3$
If we allow that
$H in C^2(U, Bbb R), tag 4$
where $U subset Bbb R^2$ is an open set on which $H(p, q)$ is defined, then
$H_{pq} = H_{qp}, tag 5$
so it follows that the trace of $J_H$,
$text{Tr}(J_H) = H_{pq} - H_{qp} = 0; tag 6$
thus the characteristic polynomial $chi_H(lambda)$ of $J_H$ is easily seen to be
$chi_H(lambda) = lambda^2 - text{Tr}(J_H) lambda + det(J_H) = lambda^2 + det(J_H) = 0; tag 7$
it follows that
$det(J_H) le 0 Longrightarrow lambda =pm sqrt {-det(J_H)} in Bbb R,tag 8$
and
$det(J_H) > 0 Longrightarrow lambda =pm isqrt {det(J_H)} in iBbb R, tag 9$
as per request. $OEDelta$
answered Dec 3 '18 at 1:49
Robert LewisRobert Lewis
44.8k22964
$endgroup$
A planar Hamiltonian system looks like
$dot q = dfrac{partial H(p, q)}{partial p} = H_p, tag 1$
$dot p = -dfrac{partial H(p, q)}{partial q} = -H_q. tag 2$
The Jacobean matrix $J_H$ of (1)-(2) is
$J_H = begin{bmatrix} H_{pq} & H_{pp} \ -H_{qq} & -H_{qp} end{bmatrix}. tag 3$
If we allow that
$H in C^2(U, Bbb R), tag 4$
where $U subset Bbb R^2$ is an open set on which $H(p, q)$ is defined, then
$H_{pq} = H_{qp}, tag 5$
so it follows that the trace of $J_H$,
$text{Tr}(J_H) = H_{pq} - H_{qp} = 0; tag 6$
thus the characteristic polynomial $chi_H(lambda)$ of $J_H$ is easily seen to be
$chi_H(lambda) = lambda^2 - text{Tr}(J_H) lambda + det(J_H) = lambda^2 + det(J_H) = 0; tag 7$
it follows that
$det(J_H) le 0 Longrightarrow lambda =pm sqrt {-det(J_H)} in Bbb R,tag 8$
and
$det(J_H) > 0 Longrightarrow lambda =pm isqrt {det(J_H)} in iBbb R, tag 9$
as per request. $OEDelta$
answered Dec 3 '18 at 1:49
Robert LewisRobert Lewis
44.8k22964
answered Dec 3 '18 at 1:49
Robert LewisRobert Lewis
44.8k22964
answered Dec 3 '18 at 1:49
Robert LewisRobert Lewis
44.8k22964
answered Dec 3 '18 at 1:49
Robert LewisRobert Lewis
44.8k22964
44.8k22964
add a comment |
add a comment |
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