How to draw phase diagram of this hamiltonian?
$begingroup$
I am learning how to plot gradient and Hamiltonian systems. The system is thus.
$$
left{begin{array}{l}
x' = −sin^2 xsin y\
y' = −2 sin x cos x cos y.
end{array}right.
$$
I know that because this is a Hamiltonian system, the phase diagram is just the level sets of the Hamiltonian, the problem is, I'm not sure how to find the actual Hamiltonian. Integrating the x and y prime expressions in terms of x and y respectively don't seem to give me something that makes much sense.
ordinary-differential-equations
edited Dec 5 '18 at 3:42
AVK
2,0961517
asked Dec 5 '18 at 2:39
MathGuyForLifeMathGuyForLife
1007
$endgroup$
add a comment |
$begingroup$
I am learning how to plot gradient and Hamiltonian systems. The system is thus.
$$
left{begin{array}{l}
x' = −sin^2 xsin y\
y' = −2 sin x cos x cos y.
end{array}right.
$$
I know that because this is a Hamiltonian system, the phase diagram is just the level sets of the Hamiltonian, the problem is, I'm not sure how to find the actual Hamiltonian. Integrating the x and y prime expressions in terms of x and y respectively don't seem to give me something that makes much sense.
ordinary-differential-equations
edited Dec 5 '18 at 3:42
AVK
2,0961517
asked Dec 5 '18 at 2:39
MathGuyForLifeMathGuyForLife
1007
$endgroup$
add a comment |
$begingroup$
I am learning how to plot gradient and Hamiltonian systems. The system is thus.
$$
left{begin{array}{l}
x' = −sin^2 xsin y\
y' = −2 sin x cos x cos y.
end{array}right.
$$
I know that because this is a Hamiltonian system, the phase diagram is just the level sets of the Hamiltonian, the problem is, I'm not sure how to find the actual Hamiltonian. Integrating the x and y prime expressions in terms of x and y respectively don't seem to give me something that makes much sense.
ordinary-differential-equations
edited Dec 5 '18 at 3:42
AVK
2,0961517
asked Dec 5 '18 at 2:39
MathGuyForLifeMathGuyForLife
1007
$endgroup$
I am learning how to plot gradient and Hamiltonian systems. The system is thus.
$$
left{begin{array}{l}
x' = −sin^2 xsin y\
y' = −2 sin x cos x cos y.
end{array}right.
$$
I know that because this is a Hamiltonian system, the phase diagram is just the level sets of the Hamiltonian, the problem is, I'm not sure how to find the actual Hamiltonian. Integrating the x and y prime expressions in terms of x and y respectively don't seem to give me something that makes much sense.
ordinary-differential-equations
ordinary-differential-equations
edited Dec 5 '18 at 3:42
AVK
2,0961517
asked Dec 5 '18 at 2:39
MathGuyForLifeMathGuyForLife
1007
edited Dec 5 '18 at 3:42
AVK
2,0961517
asked Dec 5 '18 at 2:39
MathGuyForLifeMathGuyForLife
1007
edited Dec 5 '18 at 3:42
AVK
2,0961517
edited Dec 5 '18 at 3:42
AVK
2,0961517
edited Dec 5 '18 at 3:42
AVK
2,0961517
2,0961517
asked Dec 5 '18 at 2:39
MathGuyForLifeMathGuyForLife
1007
asked Dec 5 '18 at 2:39
MathGuyForLifeMathGuyForLife
1007
asked Dec 5 '18 at 2:39
MathGuyForLifeMathGuyForLife
1007
1007
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
A Hamiltonian system is
$$
x'=-frac{partial H}{partial y},quad y'=frac{partial H}{partial x},
$$
thus, you need to integrate $x'$ with respect to $y$ or $y'$ with respect to $x$:
$$
H(x,y)=-int (-sin^2 xsin y),dy=-sin^2xsin y+c(x).$$
In order to obtain $c(x)$, we can differentiate $H(x,y)$:
$$
H_x= -2sin xcos xsin y+c'(x).
$$
Hence, $c'(x)=0$ and, if we take $c=0$, $H(x,y)=-sin^2xsin y$.
answered Dec 5 '18 at 3:15
AVKAVK
2,0961517
$endgroup$
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
A Hamiltonian system is
$$
x'=-frac{partial H}{partial y},quad y'=frac{partial H}{partial x},
$$
thus, you need to integrate $x'$ with respect to $y$ or $y'$ with respect to $x$:
$$
H(x,y)=-int (-sin^2 xsin y),dy=-sin^2xsin y+c(x).$$
In order to obtain $c(x)$, we can differentiate $H(x,y)$:
$$
H_x= -2sin xcos xsin y+c'(x).
$$
Hence, $c'(x)=0$ and, if we take $c=0$, $H(x,y)=-sin^2xsin y$.
answered Dec 5 '18 at 3:15
AVKAVK
2,0961517
$endgroup$
add a comment |
$begingroup$
A Hamiltonian system is
$$
x'=-frac{partial H}{partial y},quad y'=frac{partial H}{partial x},
$$
thus, you need to integrate $x'$ with respect to $y$ or $y'$ with respect to $x$:
$$
H(x,y)=-int (-sin^2 xsin y),dy=-sin^2xsin y+c(x).$$
In order to obtain $c(x)$, we can differentiate $H(x,y)$:
$$
H_x= -2sin xcos xsin y+c'(x).
$$
Hence, $c'(x)=0$ and, if we take $c=0$, $H(x,y)=-sin^2xsin y$.
answered Dec 5 '18 at 3:15
AVKAVK
2,0961517
$endgroup$
add a comment |
$begingroup$
A Hamiltonian system is
$$
x'=-frac{partial H}{partial y},quad y'=frac{partial H}{partial x},
$$
thus, you need to integrate $x'$ with respect to $y$ or $y'$ with respect to $x$:
$$
H(x,y)=-int (-sin^2 xsin y),dy=-sin^2xsin y+c(x).$$
In order to obtain $c(x)$, we can differentiate $H(x,y)$:
$$
H_x= -2sin xcos xsin y+c'(x).
$$
Hence, $c'(x)=0$ and, if we take $c=0$, $H(x,y)=-sin^2xsin y$.
answered Dec 5 '18 at 3:15
AVKAVK
2,0961517
$endgroup$
A Hamiltonian system is
$$
x'=-frac{partial H}{partial y},quad y'=frac{partial H}{partial x},
$$
thus, you need to integrate $x'$ with respect to $y$ or $y'$ with respect to $x$:
$$
H(x,y)=-int (-sin^2 xsin y),dy=-sin^2xsin y+c(x).$$
In order to obtain $c(x)$, we can differentiate $H(x,y)$:
$$
H_x= -2sin xcos xsin y+c'(x).
$$
Hence, $c'(x)=0$ and, if we take $c=0$, $H(x,y)=-sin^2xsin y$.
answered Dec 5 '18 at 3:15
AVKAVK
2,0961517
answered Dec 5 '18 at 3:15
AVKAVK
2,0961517
answered Dec 5 '18 at 3:15
AVKAVK
2,0961517
answered Dec 5 '18 at 3:15
AVKAVK
2,0961517
2,0961517
add a comment |
add a comment |
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