Proving that the semi-implicit Euler method is symplectic

Question

I'm having trouble understanding the following proof that the semi-implicit Euler method

$$\begin{cases} p^{k+1} = p^k-\Delta t \frac{\partial H}{\partial q}(p^{k+1},q^k) \\ q^{k+1} = q^k + \Delta t \frac{\partial H}{\partial p}(p^{k+1},q^k) \end{cases}$$

is symplectic.

Proof: Let $\Phi$ be the numerical flow and

$$\Phi' = \frac{\partial (p^{k+1}, q^{k+1})}{\partial (p^k, q^k)} $$

its jacobian. From

$$\begin{pmatrix} I+ \Delta t \frac{\partial^2H}{\partial p \partial q} & 0 \\ -\Delta t \frac{\partial^2H}{\partial p^2} & I \end{pmatrix} \Phi' = \begin{pmatrix} I & -\Delta t \frac{\partial^2H}{\partial q^2} \\ 0 & I + \Delta t \frac{\partial^2H}{\partial p \partial q}\end{pmatrix}, \ (I)$$

where the matrices are evaluated at $(p^{k+1},p^k)$ one can verify that

$$\Phi'^T J\Phi' = J$$

holds, where

$$J = \begin{pmatrix} 0 & I \\ -I & 0 \end{pmatrix}.$$

My problems with the proof are the following:

1. When I compute the jacobian I get

$$ \Phi' = \begin{pmatrix} I + \Delta t\frac{\partial^2 H}{\partial p^k \partial q} & -\Delta t \frac{\partial^2H}{\partial q^k \partial q} \\ \Delta t\frac{\partial^2 H}{\partial p^{k} \partial p} & I + \Delta t \frac{\partial^2H}{\partial q^k \partial q} \end{pmatrix}, $$

but it doesn't satisfy equation (I). Is my jacobian wrong?

2. Where does equation (I) come from and how does it imply that the method is symplectic?

Thanks in advance for any help!

Answer 1

You are applying the logic of explicit methods, two old components get transformed into two new components. However, with implicit equations you need to apply a less straightforward frame of mind. You have 4 variables/components that are coupled by 2 equations.

Then computing the derivatives of the equations proceeds by the same logic, equations first, time propagation second. So you get $$ dp^{k+1}=dp^k-Δt\,\frac{∂^2H}{∂p∂q}(p^{k+1},q^k)\,dp^{k+1}-Δt\,\frac{∂^2H}{∂q^2}(p^{k+1},q^k)\,dq^k\\ dq^{k+1}=dq^k+Δt\,\frac{∂^2H}{∂p^2}(p^{k+1},q^k)\,dp^{k+1}+Δt\,\frac{∂^2H}{∂p∂q}(p^{k+1},q^k)\,dq^k\\ $$ Now arranging to have $dp^{k+1}$ terms all on the left side gives the claimed linear system.

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In terms of symplectic geometry, the semi-implicit Euler step is a canonical transformation given by $S(p,q)=pq+Δt\,H(p,q)$ with the coordinate transformations $$ p^k=\frac{∂S}{∂q}(p^{k+1},q^k)\\ q^{k+1}=\frac{∂S}{∂p}(p^{k+1},q^k) $$