Explicit solution of a Hamiltonian system












5












$begingroup$


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
$$ x=tanhleft(frac{t}{sqrt2}right),qquad y=frac1{sqrt2}text{sech}^2left(frac{t}{sqrt2}right) $$
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









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xpaul is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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    5












    $begingroup$


    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
    $$ x=tanhleft(frac{t}{sqrt2}right),qquad y=frac1{sqrt2}text{sech}^2left(frac{t}{sqrt2}right) $$
    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.







    $endgroup$















      5












      5








      5


      1



      $begingroup$


      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
      $$ x=tanhleft(frac{t}{sqrt2}right),qquad y=frac1{sqrt2}text{sech}^2left(frac{t}{sqrt2}right) $$
      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
      $$ x=tanhleft(frac{t}{sqrt2}right),qquad y=frac1{sqrt2}text{sech}^2left(frac{t}{sqrt2}right) $$
      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 2 days ago









      Ivo Terek

      442211




      442211






      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.









      asked Mar 28 at 17:00









      xpaulxpaul

      1283




      1283




      New contributor




      xpaul is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      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.






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






















          1 Answer
          1






          active

          oldest

          votes


















          10












          $begingroup$

          Edited.



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



          Your equation is equivalent to
          $$left(frac{dx}{dt}right)^2=frac{1}{3}(x^6-3x^2+2),$$
          (I just plugged $y=dx/dt$ to your Hamiltonian, and used its value $H=1/3$.)
          This equation is separable,
          $$frac{t}{sqrt{3}}=intfrac{dx}{sqrt{x^6-3x^2+2}}=:I$$
          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 to this integral is elementary. Integrating and returning to the original variables, I obtained the general solution
          $$x(t)=sqrt{frac{1+cosh(2t+c)}{2+cosh(2t+c)}}.$$
          Can you check this computation?



          Of course, your value $1/3$ for the Hamiltonian is crucial here. With
          some other value you obtain a much more complicated integral which cannot be expressed in elementary functions. Where did this $1/3$ come from?






          share|cite|improve this answer











          $endgroup$









          • 2




            $begingroup$
            Thanks a lot. This is what I wanted.
            $endgroup$
            – xpaul
            Mar 28 at 21:29










          • $begingroup$
            Note it is much clearer than before. I really appreciate your help.
            $endgroup$
            – xpaul
            2 days ago






          • 1




            $begingroup$
            @xpaul: tell me where your $1/3$ comes from.
            $endgroup$
            – Alexandre Eremenko
            2 days ago






          • 1




            $begingroup$
            1/3 is from fact that when $H(x,y)=alpha$ passes through the separatrices $(pm1,0)$, $alpha=1/3$.
            $endgroup$
            – xpaul
            2 days ago












          • $begingroup$
            If I plug in your $x(t)$, I get $y=x'(t)=frac{sinh (c+2 t)}{sqrt{frac{cosh (c+2 t)+1}{cosh (c+2 t)+2}} (cosh (c+2 t)+2)^2}$ but I find $frac{x^2}2+frac{y^2}{2}-frac{x^6}6neqfrac13$.
            $endgroup$
            – xpaul
            2 days ago












          Your Answer





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          1 Answer
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          active

          oldest

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          10












          $begingroup$

          Edited.



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



          Your equation is equivalent to
          $$left(frac{dx}{dt}right)^2=frac{1}{3}(x^6-3x^2+2),$$
          (I just plugged $y=dx/dt$ to your Hamiltonian, and used its value $H=1/3$.)
          This equation is separable,
          $$frac{t}{sqrt{3}}=intfrac{dx}{sqrt{x^6-3x^2+2}}=:I$$
          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 to this integral is elementary. Integrating and returning to the original variables, I obtained the general solution
          $$x(t)=sqrt{frac{1+cosh(2t+c)}{2+cosh(2t+c)}}.$$
          Can you check this computation?



          Of course, your value $1/3$ for the Hamiltonian is crucial here. With
          some other value you obtain a much more complicated integral which cannot be expressed in elementary functions. Where did this $1/3$ come from?






          share|cite|improve this answer











          $endgroup$









          • 2




            $begingroup$
            Thanks a lot. This is what I wanted.
            $endgroup$
            – xpaul
            Mar 28 at 21:29










          • $begingroup$
            Note it is much clearer than before. I really appreciate your help.
            $endgroup$
            – xpaul
            2 days ago






          • 1




            $begingroup$
            @xpaul: tell me where your $1/3$ comes from.
            $endgroup$
            – Alexandre Eremenko
            2 days ago






          • 1




            $begingroup$
            1/3 is from fact that when $H(x,y)=alpha$ passes through the separatrices $(pm1,0)$, $alpha=1/3$.
            $endgroup$
            – xpaul
            2 days ago












          • $begingroup$
            If I plug in your $x(t)$, I get $y=x'(t)=frac{sinh (c+2 t)}{sqrt{frac{cosh (c+2 t)+1}{cosh (c+2 t)+2}} (cosh (c+2 t)+2)^2}$ but I find $frac{x^2}2+frac{y^2}{2}-frac{x^6}6neqfrac13$.
            $endgroup$
            – xpaul
            2 days ago
















          10












          $begingroup$

          Edited.



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



          Your equation is equivalent to
          $$left(frac{dx}{dt}right)^2=frac{1}{3}(x^6-3x^2+2),$$
          (I just plugged $y=dx/dt$ to your Hamiltonian, and used its value $H=1/3$.)
          This equation is separable,
          $$frac{t}{sqrt{3}}=intfrac{dx}{sqrt{x^6-3x^2+2}}=:I$$
          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 to this integral is elementary. Integrating and returning to the original variables, I obtained the general solution
          $$x(t)=sqrt{frac{1+cosh(2t+c)}{2+cosh(2t+c)}}.$$
          Can you check this computation?



          Of course, your value $1/3$ for the Hamiltonian is crucial here. With
          some other value you obtain a much more complicated integral which cannot be expressed in elementary functions. Where did this $1/3$ come from?






          share|cite|improve this answer











          $endgroup$









          • 2




            $begingroup$
            Thanks a lot. This is what I wanted.
            $endgroup$
            – xpaul
            Mar 28 at 21:29










          • $begingroup$
            Note it is much clearer than before. I really appreciate your help.
            $endgroup$
            – xpaul
            2 days ago






          • 1




            $begingroup$
            @xpaul: tell me where your $1/3$ comes from.
            $endgroup$
            – Alexandre Eremenko
            2 days ago






          • 1




            $begingroup$
            1/3 is from fact that when $H(x,y)=alpha$ passes through the separatrices $(pm1,0)$, $alpha=1/3$.
            $endgroup$
            – xpaul
            2 days ago












          • $begingroup$
            If I plug in your $x(t)$, I get $y=x'(t)=frac{sinh (c+2 t)}{sqrt{frac{cosh (c+2 t)+1}{cosh (c+2 t)+2}} (cosh (c+2 t)+2)^2}$ but I find $frac{x^2}2+frac{y^2}{2}-frac{x^6}6neqfrac13$.
            $endgroup$
            – xpaul
            2 days ago














          10












          10








          10





          $begingroup$

          Edited.



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



          Your equation is equivalent to
          $$left(frac{dx}{dt}right)^2=frac{1}{3}(x^6-3x^2+2),$$
          (I just plugged $y=dx/dt$ to your Hamiltonian, and used its value $H=1/3$.)
          This equation is separable,
          $$frac{t}{sqrt{3}}=intfrac{dx}{sqrt{x^6-3x^2+2}}=:I$$
          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 to this integral is elementary. Integrating and returning to the original variables, I obtained the general solution
          $$x(t)=sqrt{frac{1+cosh(2t+c)}{2+cosh(2t+c)}}.$$
          Can you check this computation?



          Of course, your value $1/3$ for the Hamiltonian is crucial here. With
          some other value you obtain a much more complicated integral which cannot be expressed in elementary functions. Where did this $1/3$ come from?






          share|cite|improve this answer











          $endgroup$



          Edited.



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



          Your equation is equivalent to
          $$left(frac{dx}{dt}right)^2=frac{1}{3}(x^6-3x^2+2),$$
          (I just plugged $y=dx/dt$ to your Hamiltonian, and used its value $H=1/3$.)
          This equation is separable,
          $$frac{t}{sqrt{3}}=intfrac{dx}{sqrt{x^6-3x^2+2}}=:I$$
          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 to this integral is elementary. Integrating and returning to the original variables, I obtained the general solution
          $$x(t)=sqrt{frac{1+cosh(2t+c)}{2+cosh(2t+c)}}.$$
          Can you check this computation?



          Of course, your value $1/3$ for the Hamiltonian is crucial here. With
          some other value you obtain a much more complicated integral which cannot be expressed in elementary functions. Where did this $1/3$ come from?







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Mar 29 at 1:33

























          answered Mar 28 at 18:42









          Alexandre EremenkoAlexandre Eremenko

          51k6141261




          51k6141261








          • 2




            $begingroup$
            Thanks a lot. This is what I wanted.
            $endgroup$
            – xpaul
            Mar 28 at 21:29










          • $begingroup$
            Note it is much clearer than before. I really appreciate your help.
            $endgroup$
            – xpaul
            2 days ago






          • 1




            $begingroup$
            @xpaul: tell me where your $1/3$ comes from.
            $endgroup$
            – Alexandre Eremenko
            2 days ago






          • 1




            $begingroup$
            1/3 is from fact that when $H(x,y)=alpha$ passes through the separatrices $(pm1,0)$, $alpha=1/3$.
            $endgroup$
            – xpaul
            2 days ago












          • $begingroup$
            If I plug in your $x(t)$, I get $y=x'(t)=frac{sinh (c+2 t)}{sqrt{frac{cosh (c+2 t)+1}{cosh (c+2 t)+2}} (cosh (c+2 t)+2)^2}$ but I find $frac{x^2}2+frac{y^2}{2}-frac{x^6}6neqfrac13$.
            $endgroup$
            – xpaul
            2 days ago














          • 2




            $begingroup$
            Thanks a lot. This is what I wanted.
            $endgroup$
            – xpaul
            Mar 28 at 21:29










          • $begingroup$
            Note it is much clearer than before. I really appreciate your help.
            $endgroup$
            – xpaul
            2 days ago






          • 1




            $begingroup$
            @xpaul: tell me where your $1/3$ comes from.
            $endgroup$
            – Alexandre Eremenko
            2 days ago






          • 1




            $begingroup$
            1/3 is from fact that when $H(x,y)=alpha$ passes through the separatrices $(pm1,0)$, $alpha=1/3$.
            $endgroup$
            – xpaul
            2 days ago












          • $begingroup$
            If I plug in your $x(t)$, I get $y=x'(t)=frac{sinh (c+2 t)}{sqrt{frac{cosh (c+2 t)+1}{cosh (c+2 t)+2}} (cosh (c+2 t)+2)^2}$ but I find $frac{x^2}2+frac{y^2}{2}-frac{x^6}6neqfrac13$.
            $endgroup$
            – xpaul
            2 days ago








          2




          2




          $begingroup$
          Thanks a lot. This is what I wanted.
          $endgroup$
          – xpaul
          Mar 28 at 21:29




          $begingroup$
          Thanks a lot. This is what I wanted.
          $endgroup$
          – xpaul
          Mar 28 at 21:29












          $begingroup$
          Note it is much clearer than before. I really appreciate your help.
          $endgroup$
          – xpaul
          2 days ago




          $begingroup$
          Note it is much clearer than before. I really appreciate your help.
          $endgroup$
          – xpaul
          2 days ago




          1




          1




          $begingroup$
          @xpaul: tell me where your $1/3$ comes from.
          $endgroup$
          – Alexandre Eremenko
          2 days ago




          $begingroup$
          @xpaul: tell me where your $1/3$ comes from.
          $endgroup$
          – Alexandre Eremenko
          2 days ago




          1




          1




          $begingroup$
          1/3 is from fact that when $H(x,y)=alpha$ passes through the separatrices $(pm1,0)$, $alpha=1/3$.
          $endgroup$
          – xpaul
          2 days ago






          $begingroup$
          1/3 is from fact that when $H(x,y)=alpha$ passes through the separatrices $(pm1,0)$, $alpha=1/3$.
          $endgroup$
          – xpaul
          2 days ago














          $begingroup$
          If I plug in your $x(t)$, I get $y=x'(t)=frac{sinh (c+2 t)}{sqrt{frac{cosh (c+2 t)+1}{cosh (c+2 t)+2}} (cosh (c+2 t)+2)^2}$ but I find $frac{x^2}2+frac{y^2}{2}-frac{x^6}6neqfrac13$.
          $endgroup$
          – xpaul
          2 days ago




          $begingroup$
          If I plug in your $x(t)$, I get $y=x'(t)=frac{sinh (c+2 t)}{sqrt{frac{cosh (c+2 t)+1}{cosh (c+2 t)+2}} (cosh (c+2 t)+2)^2}$ but I find $frac{x^2}2+frac{y^2}{2}-frac{x^6}6neqfrac13$.
          $endgroup$
          – xpaul
          2 days ago










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          xpaul is a new contributor. Be nice, and check out our Code of Conduct.
















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          迪纳利峰 位于迪纳利国家公园和保留区的迪纳利峰（旧名麦金利峰） 最高点 海拔 20,237英尺（6,168米）   NGVD 29 . [1] [2] 相對高度 20,073英尺（6,118米） 列表 各洲最高峰列表 特独立山峰 各国最高点列表 美国最高点  排名第一 [3] 坐标 63°04′10″N 151°00′27″W  /  63.06944°N 151.00750°W  / 63.06944; -151.00750  ( Mount McKinley ) 坐标： 63°04′10″N 151°00′27″W  /  63.06944°N 151.00750°W  / 63.06944; -151.00750  ( Mount McKinley )   地理 位置   美國阿拉斯加州迪纳利自治市镇 迪纳利国家公园和保留区 山脈 阿拉斯加山脉 攀山 首次登頂 （ 英语 ： First ascent ） 1913年6月7日 哈德逊·史塔克 哈里·卡斯坦斯 沃尔特·哈珀 罗伯特·塔特姆 最簡路線 西坡路线（攀登冰川/雪） 迪纳利 （北阿萨巴斯卡语支、 英语： Denali ， / d ᵻ ˈ n ɑː l i / ，直译“高山”），1917年至2015年官方名称为 麦金利山 （另译作 麦金利峰 ， 英语： Mount McKinley ），位于阿拉斯加州东南部、阿拉斯加山脉中段，海拔6194米，是北美洲最高峰，也是美国的最高峰。它的俄语名称是 Большая Гора ， 罗马化：Bolshaya Gora ，意思是“大山”。 迪纳利山地区拥有变幻莫测的高山风、典型的北极植被以及野生动植物。这里大部分地区终年积雪，山间经常浓雾不断，雾气在皑皑白雪中缭绕弥漫时，几百米之外的景物便不可见。夏季，迪纳利山的青青山坡上鲜花盛开，紫色的杜鹃和精巧的铃状石南花随处可见。迪纳利山还是世界登山爱好者的汇集之地。每年5月到7月有数百人登山，只有一半不到的人能登顶成功。一次登山约花3个星期。 2015年8月31日，美國政府宣佈將山名由“麦金利山”改回原名“迪纳利山”。美國內政部發表聲
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          南乌拉尔铁路局

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          南乌拉尔铁路局 Южно-Уральская железная дорога 成立 1934年1月4日 国家   蘇聯（1934年—1991年）   俄羅斯（1991年—） 总部 车里雅宾斯克 隶属 俄罗斯铁路股份公司 路局代码 80 铁路里程 4,934.9公里 奖项 网址 http://yuzd.rzd.ru/ 查 论 编 南乌拉尔铁路局 （俄语： Южно-Уральская железная дорога ， 羅馬化： Yuzhno-Uralskaya zheleznaya doroga ，简称 ЮУЖД ）是俄罗斯铁路股份公司属下的铁路管理局之一，总部位于车里雅宾斯克，负责管辖俄罗斯联邦乌拉尔南部地区的铁路系统，范围包括奥伦堡州、车里雅宾斯克州、库尔干州的大部分铁路，以及萨马拉州、斯维尔德洛夫斯克州、巴什科尔托斯坦共和国的少部分铁路，并有部分铁路延伸至或穿过哈萨克斯坦境内。南乌拉尔铁路局的营业里程为4,934.9公里，与伏尔加铁路局、古比雪夫铁路局、斯维尔德洛夫斯克铁路局、西西伯利亚铁路局相邻，并与南方的哈萨克斯坦铁路连接。 外部链接 （俄文） Южно-Уральская железная дорога — филиал ОАО РЖД （俄文） 南乌拉尔铁路局管辖路线图（一） （俄文） 南乌拉尔铁路局管辖路线图（二） 查 论 编 俄罗斯铁路（集团）股份公司 十月铁路局 · 莫斯科铁路局 · 北高加索铁路局 · 高尔基铁路局 · 古比雪夫铁路局 · 北方铁路局 · 伏尔加铁路局 · 东南铁路局 · 斯维尔德洛夫斯克铁路局 · 南乌拉尔铁路局 · 西西伯利亚铁路局 · 克拉斯诺亚尔斯克铁路局 · 东西伯利亚铁路局 · 后贝加尔铁路局 · 远东铁路局 · 加里宁格勒铁路局 This page is only for reference, If you need detailed information, please check here
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