It would be good to test PSD and POD for Hamiltonians with quadratic coupling, i.e. $H(\mathbf{q},\mathbf{p}) = 0.5\mathbf{p}^T\mathbf{p} + 0.5\mathbf{q}^TU\mathbf{q}$, with $U$ being symmetric. PSD should then find a diagonalization of the potential.
The coupling can also be made cubic by taking $\sum_{i,j}0.5a_{ij}|q^i-q^j|^3$, where the coefficients $a_{ij}$ should again be symmetric.
For the quadratic coupling the resulting differential equations are: $\dot{q} = p$ and $\dot{p} = -Uq$.
For the cubic coupling the resulting equations are: $\dot{q^i} = p_i$ and $\dot{p_i} = -1.5a_{ij}|q^i - q^j|(q^i - q^j)$ (sum over $j=1,\ldots,n$).
It would be good to test PSD and POD for Hamiltonians with quadratic coupling, i.e.$H(\mathbf{q},\mathbf{p}) = 0.5\mathbf{p}^T\mathbf{p} + 0.5\mathbf{q}^TU\mathbf{q}$ , with $U$ being symmetric. PSD should then find a diagonalization of the potential.$\sum_{i,j}0.5a_{ij}|q^i-q^j|^3$ , where the coefficients $a_{ij}$ should again be symmetric.
The coupling can also be made cubic by taking
For the quadratic coupling the resulting differential equations are:$\dot{q} = p$ and $\dot{p} = -Uq$ .$\dot{q^i} = p_i$ and $\dot{p_i} = -1.5a_{ij}|q^i - q^j|(q^i - q^j)$ (sum over $j=1,\ldots,n$ ).
For the cubic coupling the resulting equations are: