Symplectic integrator

Remarks

For $n = 1$, a map of the plane is symplectic exactly when its Jacobian determinant is $1$, that is, when it preserves area. Example: for $H = \frac12 (p^2 + q^2)$, the symplectic Euler method $p_{k+1} = p_k - h q_k$, $q_{k+1} = q_k + h p_{k+1}$ gives $q_{k+1} = (1 - h^2) q_k + h p_k$, with Jacobian determinant $(1 - h^2) \cdot 1 - h \cdot (-h) = 1$. In general the symplectic Euler method is $p_{k+1} = p_k - h \nabla_q H(q_k, p_{k+1})$, $q_{k+1} = q_k + h \nabla_p H(q_k, p_{k+1})$, and it is symplectic for every $H$. Non-example: the explicit Euler method for the same $H$ is $q_{k+1} = q_k + h p_k$, $p_{k+1} = p_k - h q_k$, with Jacobian determinant $1 + h^2 \ne 1$, so it is not symplectic. A Runge–Kutta method with coefficients $a_{ij}$, $b_i$ is symplectic when $b_i a_{ij} + b_j a_{ji} = b_i b_j$ for all $i, j$; the implicit midpoint rule is an example.