Hamilton系统用MMA实现 [已关闭]

Question

D[Subscript[z, 1][Subscript[T, 1]], Subscript[T, 1]] = 
  Subscript[z, 2][Subscript[T, 
    1]] == \[PartialD]H/\[PartialD]Subscript[z, 2][Subscript[T, 1]];
D[Subscript[z, 2][Subscript[T, 1]], Subscript[T, 1]] = -Subscript[
\!\(\*OverscriptBox[\(\[Mu]\), \(~\)]\), 1] Subscript[z, 1][Subscript[
      T, 1]] + 
    Subscript[\[Eta], 1] Subscript[z, 1][Subscript[T, 1]]^3 + 
    2 Subscript[\[Beta], 12]
      Subscript[z, 1][Subscript[T, 1]] l[Subscript[T, 
      1]]^2 == -(\[PartialD]H/\[PartialD]Subscript[z, 1][Subscript[T, 
      1]]);
D[l[Subscript[T, 1]], Subscript[T, 1]] = 
  0 == \[PartialD]H/\[PartialD]\[Phi][Subscript[T, 1]];
l[Subscript[T, 1]] D[\[Phi][Subscript[T, 1]], Subscript[T, 
    1]] = -Subscript[\[Sigma], 2] l[Subscript[T, 1]] + 
    2 Subscript[\[Beta], 12]
      l[Subscript[T, 1]] Subscript[z, 1][Subscript[T, 1]]^2 + 
    3/2 Subscript[\[Beta], 24]
      l[Subscript[T, 1]]^3 == -(\[PartialD]H/\[PartialD]l[Subscript[T,
       1]]);

H = 1/2 Subscript[z, 2]^2 + 1/2 Subscript[
\!\(\*OverscriptBox[\(\[Mu]\), \(~\)]\), 1] Subscript[z, 1]^2 - 
  1/4 Subscript[\[Eta], 1] Subscript[z, 1]^4 + 
  1/2 Subscript[\[Sigma], 2] l^2 - 3/8 Subscript[\[Beta], 24] l^4 - 
  Subscript[\[Beta], 12] l^2 Subscript[z, 1]^2

 上面的方程组，怎么通过MMA求得最后的H的表达式