Constant trajectory in Hamiltonian system
$begingroup$
Let the (non-canonical) Hamiltonian system be given in the form
$$dot{x}=J(x)DH(x)$$
where $H(x)$ is the Hamiltonian function, $J(x)$ is the skew-symmetric symplectic matrix, and $DH$ denotes the vector of partial derivatives of $H$ (the gradient of $H$). Let $J(x)$ have a nonempty null space, and let a function $C(x)$ satisfy $J(x)DC(x) = 0$ for all $x$. Show that value of $C(x)$ is constant on any trajectory of the system regardless of the choice of $H$.
Here is my proposed solution. I use the fact that $J(x)$ is the skew-symmetric in the fourth equality $big(J(x)^{-1} = J(x)^T = -J(x)big)$. The fact that $J(x)DC(x) = 0$ is used in the fifth equality.
begin{equation}
begin{split}
Cdot{(x)} & = nabla{C}boldsymbol{cdot}{x} \
& = nabla{C}boldsymbol{cdot}J(x)DH(x) \
& = Big((J(x))^T(nabla{C})^TBig)^Tboldsymbol{cdot}DH(x) \
& = Big(-J(x)DC(x)Big)^Tboldsymbol{cdot}DH(x) \
& = 0boldsymbol{cdot}DH(x) \
& = 0
end{split}
end{equation}
Therefore, as $Cdot{(x)}=0$, this implies that $C(x)$ is constant along any trajectory.
I'm having trouble justifying the first equality and fourth equality. Is this the right approach to this problem? Is there an alternative approach that would lead to the right answer?
proof-verification hamilton-equations
asked Dec 19 '18 at 16:01
Axion004Axion004
374313
$endgroup$
add a comment |
$begingroup$
Let the (non-canonical) Hamiltonian system be given in the form
$$dot{x}=J(x)DH(x)$$
where $H(x)$ is the Hamiltonian function, $J(x)$ is the skew-symmetric symplectic matrix, and $DH$ denotes the vector of partial derivatives of $H$ (the gradient of $H$). Let $J(x)$ have a nonempty null space, and let a function $C(x)$ satisfy $J(x)DC(x) = 0$ for all $x$. Show that value of $C(x)$ is constant on any trajectory of the system regardless of the choice of $H$.
Here is my proposed solution. I use the fact that $J(x)$ is the skew-symmetric in the fourth equality $big(J(x)^{-1} = J(x)^T = -J(x)big)$. The fact that $J(x)DC(x) = 0$ is used in the fifth equality.
begin{equation}
begin{split}
Cdot{(x)} & = nabla{C}boldsymbol{cdot}{x} \
& = nabla{C}boldsymbol{cdot}J(x)DH(x) \
& = Big((J(x))^T(nabla{C})^TBig)^Tboldsymbol{cdot}DH(x) \
& = Big(-J(x)DC(x)Big)^Tboldsymbol{cdot}DH(x) \
& = 0boldsymbol{cdot}DH(x) \
& = 0
end{split}
end{equation}
Therefore, as $Cdot{(x)}=0$, this implies that $C(x)$ is constant along any trajectory.
I'm having trouble justifying the first equality and fourth equality. Is this the right approach to this problem? Is there an alternative approach that would lead to the right answer?
proof-verification hamilton-equations
asked Dec 19 '18 at 16:01
Axion004Axion004
374313
$endgroup$
add a comment |
$begingroup$
Let the (non-canonical) Hamiltonian system be given in the form
$$dot{x}=J(x)DH(x)$$
where $H(x)$ is the Hamiltonian function, $J(x)$ is the skew-symmetric symplectic matrix, and $DH$ denotes the vector of partial derivatives of $H$ (the gradient of $H$). Let $J(x)$ have a nonempty null space, and let a function $C(x)$ satisfy $J(x)DC(x) = 0$ for all $x$. Show that value of $C(x)$ is constant on any trajectory of the system regardless of the choice of $H$.
Here is my proposed solution. I use the fact that $J(x)$ is the skew-symmetric in the fourth equality $big(J(x)^{-1} = J(x)^T = -J(x)big)$. The fact that $J(x)DC(x) = 0$ is used in the fifth equality.
begin{equation}
begin{split}
Cdot{(x)} & = nabla{C}boldsymbol{cdot}{x} \
& = nabla{C}boldsymbol{cdot}J(x)DH(x) \
& = Big((J(x))^T(nabla{C})^TBig)^Tboldsymbol{cdot}DH(x) \
& = Big(-J(x)DC(x)Big)^Tboldsymbol{cdot}DH(x) \
& = 0boldsymbol{cdot}DH(x) \
& = 0
end{split}
end{equation}
Therefore, as $Cdot{(x)}=0$, this implies that $C(x)$ is constant along any trajectory.
I'm having trouble justifying the first equality and fourth equality. Is this the right approach to this problem? Is there an alternative approach that would lead to the right answer?
proof-verification hamilton-equations
asked Dec 19 '18 at 16:01
Axion004Axion004
374313
$endgroup$
Let the (non-canonical) Hamiltonian system be given in the form
$$dot{x}=J(x)DH(x)$$
where $H(x)$ is the Hamiltonian function, $J(x)$ is the skew-symmetric symplectic matrix, and $DH$ denotes the vector of partial derivatives of $H$ (the gradient of $H$). Let $J(x)$ have a nonempty null space, and let a function $C(x)$ satisfy $J(x)DC(x) = 0$ for all $x$. Show that value of $C(x)$ is constant on any trajectory of the system regardless of the choice of $H$.
Here is my proposed solution. I use the fact that $J(x)$ is the skew-symmetric in the fourth equality $big(J(x)^{-1} = J(x)^T = -J(x)big)$. The fact that $J(x)DC(x) = 0$ is used in the fifth equality.
begin{equation}
begin{split}
Cdot{(x)} & = nabla{C}boldsymbol{cdot}{x} \
& = nabla{C}boldsymbol{cdot}J(x)DH(x) \
& = Big((J(x))^T(nabla{C})^TBig)^Tboldsymbol{cdot}DH(x) \
& = Big(-J(x)DC(x)Big)^Tboldsymbol{cdot}DH(x) \
& = 0boldsymbol{cdot}DH(x) \
& = 0
end{split}
end{equation}
Therefore, as $Cdot{(x)}=0$, this implies that $C(x)$ is constant along any trajectory.
I'm having trouble justifying the first equality and fourth equality. Is this the right approach to this problem? Is there an alternative approach that would lead to the right answer?
proof-verification hamilton-equations
proof-verification hamilton-equations
asked Dec 19 '18 at 16:01
Axion004Axion004
374313
asked Dec 19 '18 at 16:01
Axion004Axion004
374313
asked Dec 19 '18 at 16:01
Axion004Axion004
374313
asked Dec 19 '18 at 16:01
Axion004Axion004
374313
asked Dec 19 '18 at 16:01
Axion004Axion004
374313
374313
add a comment |
add a comment |
1 Answer
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$begingroup$
I think you are thinking on the right line. I have the following solution.
Write, along the solution trajectories of the system
$begin{align*}
frac{d}{dt} H(x(t)) &= frac{partial H}{partial x} dot{x} \
&= (DH(x))^T J(x) DH(x)\
&= DH(x)^T (-J(x))^T DH(x)\
text{Adding the two equations,}\
2 frac{d}{dt} H(x(t)) &= 0\
Rightarrow frac{d}{dt} H(x(t)) &= 0
end{align*}$
Hence, you can write $H(x(t)) = c$, for a constant $c$. So, $H(x)$ is constant along the trajectory of the system.
answered Feb 14 at 4:07
wootmanwootman
246
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
I think you are thinking on the right line. I have the following solution.
Write, along the solution trajectories of the system
$begin{align*}
frac{d}{dt} H(x(t)) &= frac{partial H}{partial x} dot{x} \
&= (DH(x))^T J(x) DH(x)\
&= DH(x)^T (-J(x))^T DH(x)\
text{Adding the two equations,}\
2 frac{d}{dt} H(x(t)) &= 0\
Rightarrow frac{d}{dt} H(x(t)) &= 0
end{align*}$
Hence, you can write $H(x(t)) = c$, for a constant $c$. So, $H(x)$ is constant along the trajectory of the system.
answered Feb 14 at 4:07
wootmanwootman
246
$endgroup$
add a comment |
$begingroup$
I think you are thinking on the right line. I have the following solution.
Write, along the solution trajectories of the system
$begin{align*}
frac{d}{dt} H(x(t)) &= frac{partial H}{partial x} dot{x} \
&= (DH(x))^T J(x) DH(x)\
&= DH(x)^T (-J(x))^T DH(x)\
text{Adding the two equations,}\
2 frac{d}{dt} H(x(t)) &= 0\
Rightarrow frac{d}{dt} H(x(t)) &= 0
end{align*}$
Hence, you can write $H(x(t)) = c$, for a constant $c$. So, $H(x)$ is constant along the trajectory of the system.
answered Feb 14 at 4:07
wootmanwootman
246
$endgroup$
add a comment |
$begingroup$
I think you are thinking on the right line. I have the following solution.
Write, along the solution trajectories of the system
$begin{align*}
frac{d}{dt} H(x(t)) &= frac{partial H}{partial x} dot{x} \
&= (DH(x))^T J(x) DH(x)\
&= DH(x)^T (-J(x))^T DH(x)\
text{Adding the two equations,}\
2 frac{d}{dt} H(x(t)) &= 0\
Rightarrow frac{d}{dt} H(x(t)) &= 0
end{align*}$
Hence, you can write $H(x(t)) = c$, for a constant $c$. So, $H(x)$ is constant along the trajectory of the system.
answered Feb 14 at 4:07
wootmanwootman
246
$endgroup$
I think you are thinking on the right line. I have the following solution.
Write, along the solution trajectories of the system
$begin{align*}
frac{d}{dt} H(x(t)) &= frac{partial H}{partial x} dot{x} \
&= (DH(x))^T J(x) DH(x)\
&= DH(x)^T (-J(x))^T DH(x)\
text{Adding the two equations,}\
2 frac{d}{dt} H(x(t)) &= 0\
Rightarrow frac{d}{dt} H(x(t)) &= 0
end{align*}$
Hence, you can write $H(x(t)) = c$, for a constant $c$. So, $H(x)$ is constant along the trajectory of the system.
answered Feb 14 at 4:07
wootmanwootman
246
answered Feb 14 at 4:07
wootmanwootman
246
answered Feb 14 at 4:07
wootmanwootman
246
answered Feb 14 at 4:07
wootmanwootman
246
246
add a comment |
add a comment |
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