Symplecticness conditions of some low order partitioned methods for non-autonomous Hamiltonian systems

Abstract

We consider the application of partitioned Runge-Kutta (PRK) methods to non-autonomous Hamiltonian systems. Necessary and sufficient conditions for the symplecticness of PRK methods are given, more particularly for two low order PRK methods: the partitioned (explicit-implicit) Euler method and the 2-stage Lobatto IIIA-B PRK method. Both methods are often the basis of composition schemes of higher order. In particular for irreducible PRK methods we show the necessity for the nodes of the two underlying PRK methods to be equal.

Appendix: Main lines of a proof of Theorem 2.2

To prove the necessity of conditions (2.4b) we have assumed that the PRK method is irreducible in the sense given in [9, VI.7] and [9, Theorem VI.7.10, p. 222]. A reducible PRK method is defined as a method having equivalent stages (Q_i = Q_j and P_i = P_j for i≠j). We have also added the condition (2.8), i.e., that no index i exists where b_i = 0 and ${\widehat {b}_{i}}=0$. This eliminates methods having stages that have no influence on the numerical solution q₁, p₁ similar to the DJ-irreducibility of RK methods [10, Definition 12.15]. This is justified as follows. We already know that the conditions (2.4a), and (2.4c) are necessary for symplecticness. Assuming that there is an index i such that b_i = 0 or ${\widehat {b}_{i}}=0$, supposed to be unique for now, from the necessary conditions (2.4a) and (2.4c) we then obtain ${\widehat {b}_{i}}=0$, b_i = 0, a_ji = 0, and ${\widehat {a}_{ji}}=0$ for j ∈{1, … , s}∖{i}. Hence, clearly in this situation the internal stages Q_i, P_i will not influence the solution q₁, p₁ and the other internal stages Q_j, P_j for j≠i. If there is more than one index i with b_i = 0 and ${\widehat {b}_{i}}=0$, then one can easily show that all those internal stages can only influence each other, but they can influence neither the solution q₁, p₁, nor the other internal stages Q_j, P_j with coefficients $b_{j}={\widehat {b}}_{j}\not =0$. With that additional assumption (2.8) one is then in position to prove the necessity of (2.4b).

Proof

We can extend the sets of trees considered in [9, VI.7] and [9, Theorem VI.7.10, p. 222] by having an extra type of nodes, say grey nodes, standing for the value 1 of the scalar differential equation $\dot {t}=1$. We use the notation and definitions given in [9] though we exchange the role of q and p and f and g. No node is attached on top of a grey node since the partial derivatives of a constant vanish. A branch leading to a grey node stands for a partial differentiation with respect to t. For the order conditions of partitioned methods, grey nodes need not be indexed. When a grey node follows a black node with index j, then the sum in the order conditions over the index j = 1, … , s must contain the coefficients c_j. When a grey node follows a white node with index j, then the sum in the order conditions over the index j = 1, … , s must contain the coefficients ${\widehat {c}_{j}}$. Consider a P-series with coefficients a(u)

$$ \left( \begin{array}{c} {\sum}_{u\in TP_{q}}\frac{h^{|u|}}{\sigma(u)}a(u)F(u)(t_{0},q_{0},p_{0}) \\ {\sum}_{u\in TP_{p}}\frac{h^{|u|}}{\sigma(u)}a(u)F(u)(t_{0},q_{0},p_{0}) \end{array}\right) $$

For a P-series to be symplectic one of the necessary conditions for autonomous Hamiltonian systems is to have

$$ a(u) \text{ is independent of the color of the root of } u. $$

The same necessary condition holds for non-autonomous Hamiltonian systems for trees also containing grey nodes. This can be easily shown on a similar example given in the proof of [9, Theorem VI.7.4, p. 217] where the top black node of the tree u is replaced by a grey node. For that tree we take

$$ H(t,q,p)=q^{1}p^{2}p^{3}q^{4}+q^{3}t+p^{4} $$

and for that Hamiltonian we get

$$ F^{2}(u)(t,q,p)=(-1)^{\delta(u)}\sigma(u)q^{1},\qquad F^{1}({\bar u})(t,q,p)=(-1)^{\delta(u)}\sigma(u)p^{2} $$

where ${\bar u}$ is the tree obtained from u by replacing its black root with a white root. These elementary differentials are the only contribution to

$$ \left( \frac{\partial(q_{1},p_{1})}{\partial {q_{0}^{1}}}\right)^{T}J_{n} \left( \frac{\partial(q_{1},p_{1})}{\partial {p_{0}^{2}}}\right) $$

and we get

$$ 0=\left( \frac{\partial(q_{1},p_{1})}{\partial {q_{0}^{1}}}\right)^{T}J_{n} \left( \frac{\partial(q_{1},p_{1})}{\partial {p_{0}^{2}}}\right)= (-1)^{\delta(u)}h^{|u|}(a(u)-a({\bar u})). $$

Now to prove our statement we can consider the same PRK matrix Φ_PRK as given in [9, VI.7] and [9, Theorem VI.7.10, p. 222], we do not even need to consider trees with grey nodes in that matrix. We define the vector $d\in {\mathbb {R}}^{s}$ with elements $d_{i}:=b_{i}c_{i}-{\widehat {b}}_{i}{\widehat {c}}_{i}$ for i = 1, … , s. For irreducible PRK methods we already know that the condition ${\widehat {b}}_{i}=b_{i}$ for i = 1, … , s is necessary for symplecticness, hence we obtain $d_{i}=b_{i}(c_{i}-{\widehat {c}}_{i})$ for i = 1, … , s. The vector d satisfies

$$ d^{T}{\varPhi}_{PRK}=0 $$

since $d^{T}\phi (u)=a(v)-a({\bar v})$ where v is obtained from u by appending a grey node to its root and $a(v)=a({\bar v})$ for v ∈ TP_q as seen above. Since the matrix Φ_PRK is of maximal rank s we must have d = 0, hence its components satisfy $d_{i}=b_{i}(c_{i}-{\widehat {c}}_{i})=0$ for i = 1, … , s. Since b_i ≠ 0 for i = 1, … , s we obtain (2.4b). □