Explicit solution of a Hamiltonian system












3












$begingroup$


It is well-known that the following Hamiltonian system
begin{eqnarray}
left{begin{array}{rcl}
frac{dx}{dt}&=&y,\
frac{dy}{dt}&=&x(-1+x^2),
end{array}right.
end{eqnarray}
with
$$ H(x,y)=frac{x^2}2+frac{y^2}{2}-frac{x^4}{4}$$
has the solution
$$ x=tanh(frac{t}{sqrt2}),y=frac1{sqrt2}text{sech}^2(frac{t}{sqrt2}) $$
such that $H(x,y)=frac14$.



Now consider the following Hamiltonian system
begin{eqnarray}
left{begin{array}{rcl}
frac{dx}{dt}&=&y,\
frac{dy}{dt}&=&x(-1+x^4),
end{array}right.tag{1}
end{eqnarray}
with
$$ H(x,y)=frac{x^2}2+frac{y^2}{2}-frac{x^6}{6}$$
and I want to find the explicit solution such that $H(x,y)=frac{1}{3}$. I tried many ways to get the solution but failed. Any help is appreciated.






dg.differential-geometry ca.classical-analysis-and-odes differential-equations







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  • $begingroup$
    Something is wrong here: when I plug $t=0$ in your solution of the first system,
    $endgroup$
    – Alexandre Eremenko
    6 hours ago
















3












$begingroup$


It is well-known that the following Hamiltonian system
begin{eqnarray}
left{begin{array}{rcl}
frac{dx}{dt}&=&y,\
frac{dy}{dt}&=&x(-1+x^2),
end{array}right.
end{eqnarray}
with
$$ H(x,y)=frac{x^2}2+frac{y^2}{2}-frac{x^4}{4}$$
has the solution
$$ x=tanh(frac{t}{sqrt2}),y=frac1{sqrt2}text{sech}^2(frac{t}{sqrt2}) $$
such that $H(x,y)=frac14$.



Now consider the following Hamiltonian system
begin{eqnarray}
left{begin{array}{rcl}
frac{dx}{dt}&=&y,\
frac{dy}{dt}&=&x(-1+x^4),
end{array}right.tag{1}
end{eqnarray}
with
$$ H(x,y)=frac{x^2}2+frac{y^2}{2}-frac{x^6}{6}$$
and I want to find the explicit solution such that $H(x,y)=frac{1}{3}$. I tried many ways to get the solution but failed. Any help is appreciated.






dg.differential-geometry ca.classical-analysis-and-odes differential-equations







share|cite|improve this question









New contributor




xpaul is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    Something is wrong here: when I plug $t=0$ in your solution of the first system,
    $endgroup$
    – Alexandre Eremenko
    6 hours ago














3












3








3





$begingroup$


It is well-known that the following Hamiltonian system
begin{eqnarray}
left{begin{array}{rcl}
frac{dx}{dt}&=&y,\
frac{dy}{dt}&=&x(-1+x^2),
end{array}right.
end{eqnarray}
with
$$ H(x,y)=frac{x^2}2+frac{y^2}{2}-frac{x^4}{4}$$
has the solution
$$ x=tanh(frac{t}{sqrt2}),y=frac1{sqrt2}text{sech}^2(frac{t}{sqrt2}) $$
such that $H(x,y)=frac14$.



Now consider the following Hamiltonian system
begin{eqnarray}
left{begin{array}{rcl}
frac{dx}{dt}&=&y,\
frac{dy}{dt}&=&x(-1+x^4),
end{array}right.tag{1}
end{eqnarray}
with
$$ H(x,y)=frac{x^2}2+frac{y^2}{2}-frac{x^6}{6}$$
and I want to find the explicit solution such that $H(x,y)=frac{1}{3}$. I tried many ways to get the solution but failed. Any help is appreciated.






dg.differential-geometry ca.classical-analysis-and-odes differential-equations







share|cite|improve this question









New contributor




xpaul is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




It is well-known that the following Hamiltonian system
begin{eqnarray}
left{begin{array}{rcl}
frac{dx}{dt}&=&y,\
frac{dy}{dt}&=&x(-1+x^2),
end{array}right.
end{eqnarray}
with
$$ H(x,y)=frac{x^2}2+frac{y^2}{2}-frac{x^4}{4}$$
has the solution
$$ x=tanh(frac{t}{sqrt2}),y=frac1{sqrt2}text{sech}^2(frac{t}{sqrt2}) $$
such that $H(x,y)=frac14$.



Now consider the following Hamiltonian system
begin{eqnarray}
left{begin{array}{rcl}
frac{dx}{dt}&=&y,\
frac{dy}{dt}&=&x(-1+x^4),
end{array}right.tag{1}
end{eqnarray}
with
$$ H(x,y)=frac{x^2}2+frac{y^2}{2}-frac{x^6}{6}$$
and I want to find the explicit solution such that $H(x,y)=frac{1}{3}$. I tried many ways to get the solution but failed. Any help is appreciated.






dg.differential-geometry ca.classical-analysis-and-odes differential-equations




dg.differential-geometry ca.classical-analysis-and-odes differential-equations






share|cite|improve this question









New contributor




xpaul is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




xpaul is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 3 hours ago







xpaul













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asked 8 hours ago









xpaulxpaul

1183




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New contributor




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xpaul is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






xpaul is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • $begingroup$
    Something is wrong here: when I plug $t=0$ in your solution of the first system,
    $endgroup$
    – Alexandre Eremenko
    6 hours ago


















  • $begingroup$
    Something is wrong here: when I plug $t=0$ in your solution of the first system,
    $endgroup$
    – Alexandre Eremenko
    6 hours ago
















$begingroup$
Something is wrong here: when I plug $t=0$ in your solution of the first system,
$endgroup$
– Alexandre Eremenko
6 hours ago




$begingroup$
Something is wrong here: when I plug $t=0$ in your solution of the first system,
$endgroup$
– Alexandre Eremenko
6 hours ago










1 Answer
1






active

oldest

votes


















6












$begingroup$

Edited.



Surprisingly, there is an elementary solutio, if I made no mistake in
the following computation.



Your equation is equivalent to
$$(x')^2=frac{1}{3}(x^6-3x^2+2),$$
(I just plugged $y=x'$ to your Hamiltonian, and used its value $1/3$.)
This equation is separable,
$$t/sqrt{3}=intfrac{dx}{sqrt{x^6-3x^2+2}}=I$$
and requires inversion of the integral.



To reduce it
to a standard integral, change $x^2=1/(u+1), ; dx=-(1/2)(u+1)^{-3/2}du,$ and the integral becomes
$$I=-frac{1}{sqrt{2}}intfrac{du}{usqrt{2u+3}}.$$
The inverse function is elementary.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Thanks a lot. This is what I wanted.
    $endgroup$
    – xpaul
    3 hours ago












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1 Answer
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1 Answer
1






active

oldest

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active

oldest

votes






active

oldest

votes









6












$begingroup$

Edited.



Surprisingly, there is an elementary solutio, if I made no mistake in
the following computation.



Your equation is equivalent to
$$(x')^2=frac{1}{3}(x^6-3x^2+2),$$
(I just plugged $y=x'$ to your Hamiltonian, and used its value $1/3$.)
This equation is separable,
$$t/sqrt{3}=intfrac{dx}{sqrt{x^6-3x^2+2}}=I$$
and requires inversion of the integral.



To reduce it
to a standard integral, change $x^2=1/(u+1), ; dx=-(1/2)(u+1)^{-3/2}du,$ and the integral becomes
$$I=-frac{1}{sqrt{2}}intfrac{du}{usqrt{2u+3}}.$$
The inverse function is elementary.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Thanks a lot. This is what I wanted.
    $endgroup$
    – xpaul
    3 hours ago
















6












$begingroup$

Edited.



Surprisingly, there is an elementary solutio, if I made no mistake in
the following computation.



Your equation is equivalent to
$$(x')^2=frac{1}{3}(x^6-3x^2+2),$$
(I just plugged $y=x'$ to your Hamiltonian, and used its value $1/3$.)
This equation is separable,
$$t/sqrt{3}=intfrac{dx}{sqrt{x^6-3x^2+2}}=I$$
and requires inversion of the integral.



To reduce it
to a standard integral, change $x^2=1/(u+1), ; dx=-(1/2)(u+1)^{-3/2}du,$ and the integral becomes
$$I=-frac{1}{sqrt{2}}intfrac{du}{usqrt{2u+3}}.$$
The inverse function is elementary.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Thanks a lot. This is what I wanted.
    $endgroup$
    – xpaul
    3 hours ago














6












6








6





$begingroup$

Edited.



Surprisingly, there is an elementary solutio, if I made no mistake in
the following computation.



Your equation is equivalent to
$$(x')^2=frac{1}{3}(x^6-3x^2+2),$$
(I just plugged $y=x'$ to your Hamiltonian, and used its value $1/3$.)
This equation is separable,
$$t/sqrt{3}=intfrac{dx}{sqrt{x^6-3x^2+2}}=I$$
and requires inversion of the integral.



To reduce it
to a standard integral, change $x^2=1/(u+1), ; dx=-(1/2)(u+1)^{-3/2}du,$ and the integral becomes
$$I=-frac{1}{sqrt{2}}intfrac{du}{usqrt{2u+3}}.$$
The inverse function is elementary.






share|cite|improve this answer











$endgroup$



Edited.



Surprisingly, there is an elementary solutio, if I made no mistake in
the following computation.



Your equation is equivalent to
$$(x')^2=frac{1}{3}(x^6-3x^2+2),$$
(I just plugged $y=x'$ to your Hamiltonian, and used its value $1/3$.)
This equation is separable,
$$t/sqrt{3}=intfrac{dx}{sqrt{x^6-3x^2+2}}=I$$
and requires inversion of the integral.



To reduce it
to a standard integral, change $x^2=1/(u+1), ; dx=-(1/2)(u+1)^{-3/2}du,$ and the integral becomes
$$I=-frac{1}{sqrt{2}}intfrac{du}{usqrt{2u+3}}.$$
The inverse function is elementary.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 3 hours ago

























answered 6 hours ago









Alexandre EremenkoAlexandre Eremenko

50.9k6140259




50.9k6140259








  • 1




    $begingroup$
    Thanks a lot. This is what I wanted.
    $endgroup$
    – xpaul
    3 hours ago














  • 1




    $begingroup$
    Thanks a lot. This is what I wanted.
    $endgroup$
    – xpaul
    3 hours ago








1




1




$begingroup$
Thanks a lot. This is what I wanted.
$endgroup$
– xpaul
3 hours ago




$begingroup$
Thanks a lot. This is what I wanted.
$endgroup$
– xpaul
3 hours ago










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