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







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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.






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      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












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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










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










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
















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          數位音樂下載 為非以「實體」方式來販賣音樂的相關產品，而使用數位格式如mp3、AAC等文件格式來進行銷售販賣的一种音乐产业形式。著名的數位音樂下載服務商有iTunes Store、RecoChoku、Google Play音乐和微软Zune等。数字音乐下载在二十一世纪初一度风靡全球， [1] [2] 使得传统唱片产业走向衰落，但其自身也在2010年代流媒体音乐服务如Apple Music、Spotify等的挤压下迅速失去市场份额。 [3] [4] 数字音乐下载有便于携带、传播迅速、制作简单成本低等优点，与流媒体音乐相比也能更好保障音乐人权益； [5] 但它同时很可能带来侵犯版权和滥用的情形， [6] 也给音乐产业带来了诸多不利影响。数字音乐下载为音乐人和乐迷提供了更加便捷直接的交流渠道，也一定程度上推动了网络音乐人的出现和独立音乐的迅速发展。 [7] 目录 1 历史 1.1 技术积淀（1987-2000） 1.2 出现与兴起（2000-2004） 1.3 唱片市场的颠覆者（2004-2007） 1.4 巅峰和衰落（2005-2013） 1.5 摇摇欲坠的现状与不可预知的明天（2014至今） 2 主要文件格式 2.1 PCM编码格式 2.1.1 WAV 2.1.2 APE 2.1.3 FLAC 2.2 苹果推出的无损文件格式 2.2.1 AIFF 2.2.2 ALAC 2.3 有损压缩格式 2.3.1 MP3 2.3.2 AAC 3 主要市场 3.1 iTunes Store 3.2 Google Play音乐商店 3.3 亚马逊音乐商店 3.4 部分其他音乐商店 4 影响与评价 4.1 对音乐产业的影响 4.2 对科技行业的影响 4.3 正面评价 4.4 负面评价 4.4.1 侵权问题 4.4.2 经营模式不明晰 4.4.3 不适应移动时代 4.5 文化现象 5 延伸阅读 6 参考资料 历史 苹果iTunes+iPod的软硬件搭配让数字音
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          格利澤436b

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          格利泽436b 太陽系外行星 太陽系外行星列表 艺术家笔下的格利泽436b 母恆星 母恆星 Gliese 436 星座 獅子座 赤經 ( α ) 11 h  42 m  11.0941 s [1] 赤緯 ( δ ) +26° 42′ 23.652″ [1] 距離 33.4 ly (10.2 pc) 光譜類型 M2.5 V [1] 軌道參數 半长轴 ( a ) 0.0291±0.0004 [2] AU 軌道離心率 ( e ) 0.150±0.012 [2] 公轉週期 ( P ) 2.643904±0.000005 [3] d (0.00723849 y) 軌道傾角 ( i ) 85.8 +0.21 −0.25 [3] ° 角距 ( θ ) 2.794 mas 近星點時間 ( T 0 ) 2,451,551.716 ±0.01 JD 半振幅 ( K ) 18.68±0.8 m/s 物理性质 质量 ( m ) 22.2±1.0 [2] M ⊕ 半徑 ( r ) 4.327±0.183 [2] [4] R ⊕ 密度 ( ρ ) 1.51 g cm -3 表面重力 ( g ) 1.18 g 溫度 ( T ) 712±36 [2] K 發現 發現時間 2004 發現者 巴特勒、沃格特、 馬西 et al. 發現方法 徑向速度、凌日法 發現地點 美國加利福尼亞州 發表論文 出版 其他名稱 Ross 905 b, GJ 436 b, [5] LTT 13213 b, GCTP 2704.10 b, LHS 310, AC+27:28217 b, Vyssotsky 616 b, HIP 57087 b, GEN# +9.80120068 b, LP 319-75 b, G 121-7 b, LSPM J1142+2642 b, 1RXS J114211.9+264328 b, ASCC 683818 b, G 147-68 b, UCAC2 41198281 b, BP
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          When can things happen in Etherscan, such as the picture below?

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          2 I don't know why balance of the address is 0. when can this situation? etherscan balances share | improve this question asked yesterday Ru Hi Ru Hi 11 1 New contributor Ru Hi is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct. add a comment  | 
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