Relaxation Runge–Kutta Methods for Hamiltonian Problems

Abstract

The recently-introduced relaxation approach for Runge–Kutta methods can be used to enforce conservation of energy in the integration of Hamiltonian systems. We study the behavior of implicit and explicit relaxation Runge–Kutta methods in this context. We find that, in addition to their useful conservation property, the relaxation methods yield other improvements. Experiments show that their solutions bear stronger qualitative similarity to the true solution and that the error grows more slowly in time. We also prove that these methods are superconvergent for a certain class of Hamiltonian systems.

Superconvergence Theorem Proofs

To facilitate the proof of the general result 1, we first present and prove a version restricted to linear problems.

Theorem 2

Consider the linear Euclidean Hamiltonian system

$$\begin{aligned} {\left\{ \begin{array}{ll} u'(t) = L u(t), \\ u(0) = u_0, \end{array}\right. } \quad u(t) = \begin{pmatrix} q(t) \\ p(t) \end{pmatrix}, \quad L = \alpha \begin{pmatrix} 0 &{} \mathrm{I}\\ -\mathrm{I}&{} 0 \end{pmatrix}. \end{aligned}$$

(27)

and an explicit Runge–Kutta method of order $p \in {\mathbb {N}}$ with $s \ge p$ stages. The corresponding RRK scheme conserving the Hamiltonian has an order of accuracy $p+1$ if p is odd.

In the proof of Theorem 2, the following lemma will be used.

Lemma 1

For $s,m \in {\mathbb {N}}$, $2m \le s+1$,

$$\begin{aligned} \sum _{n=\max \left\{ 1-m, m-s\right\} }^{\min \left\{ m-1, s-m\right\} } (-1)^n \frac{1}{(m-n)! (m+n)!} = -2 (-1)^m \frac{1}{(2m)!}. \end{aligned}$$

(28)

Proof

Use some explicit calculations or the Mathematica [45] notebook Combinatorial_Lemma.nb in the repository [36]. $\square$

Proof of Theorem 2

It suffices to consider the first step. The baseline RK method starting from $u_0$ yields

$$\begin{aligned} u_+ = u_0 + \sum _{k=1}^s \alpha _k {\varDelta t}^k L^k u_0, \end{aligned}$$

(29)

where $\alpha _k$ are the monomial coefficients of the corresponding stability polynomial. The relaxation method results in the new value

$$\begin{aligned} u_+^\gamma = u_0 + \gamma \sum _{k=1}^s \alpha _k {\varDelta t}^k L^k u_0 \end{aligned}$$

(30)

with squared Euclidean norm

(31)

The second sum can also be written as

$$\begin{aligned}&{\mathop {\mathop {\sum }\limits _{j,k=1}}\limits _{j+k\text { even}}^{s}} \alpha _j \alpha _k (-1)^{(j-k)/2} h^{j+k} \left\| L^{(j+k)/2} u_0\right\| ^2 \nonumber \\&\quad = \sum _{m=1}^s \sum _{n=\max \left\{ 1-m, m-s\right\} }^{\min \left\{ m-1, s-m\right\} } (-1)^n \alpha _{m-n} \alpha _{m+n} \left\| L^m u_0\right\| ^2. \end{aligned}$$

(32)

The non-zero value of the relaxation parameter conserving the Euclidean norm is

$$\begin{aligned} \gamma = \frac{ - 2 \sum _{l=1}^{\lfloor s/2\rfloor } (-1)^l \alpha _{2l} {\varDelta t}^{2l} \left\| L^l u_0\right\| ^2 }{ \sum _{m=1}^s \sum _{n=\max \left\{ 1-m, m-s\right\} }^{\min \left\{ m-1, s-m\right\} } (-1)^n \alpha _{m-n} \alpha _{m+n} \left\| L^m u_0\right\| ^2 }. \end{aligned}$$

(33)

For a p-th order baseline scheme,

$$\begin{aligned} \alpha _k = \frac{1}{k!}, \qquad k \in \left\{ 1, \ldots , p\right\} . \end{aligned}$$

(34)

Hence, for odd p, the numerator of $\gamma$ (33) is

$$\begin{aligned}&- 2 \sum _{l=1}^{\lfloor s/2\rfloor } (-1)^l \alpha _{2l} {\varDelta t}^{2l} \left\| L^l u_0\right\| ^2 \nonumber \\&\quad = - 2 \sum _{l=1}^{\lfloor p/2\rfloor } (-1)^l \frac{1}{(2l)!} {\varDelta t}^{2l} \left\| L^l u_0\right\| ^2 \nonumber \\&\qquad - 2 (-1)^{(p+1)/2} \alpha _{p+1} {\varDelta t}^{p+1} \left\| L^{(p+1)/2} u_0\right\| ^2 + {\mathcal {O}}({\varDelta t}^{p+3}). \end{aligned}$$

(35)

Because of Lemma 1, the denominator of $\gamma$ (33) for odd p is

$$\begin{aligned}&\sum _{m=1}^s \sum _{n=\max \left\{ 1-m, m-s\right\} }^{\min \left\{ m-1, s-m\right\} } (-1)^n \alpha _{m-n} \alpha _{m+n} \left\| L^m u_0\right\| ^2 \nonumber \\&\quad = - 2 \sum _{m=1}^{\lfloor p/2\rfloor } (-1)^m \frac{1}{(2m)!} \left\| L^m u_0\right\| ^2 \nonumber \\&\qquad - 2 (-1)^{(p+1)/2} \frac{1}{(p+1)!} {\varDelta t}^{p+1} \left\| L^{(p+1)/2} u_0\right\| ^2 + {\mathcal {O}}({\varDelta t}^{p+3}). \end{aligned}$$

(36)

Thus,

$$\begin{aligned} \gamma = 1 - 2 (-1)^{(p+1)/2} \left( \alpha _{p+1} - \frac{1}{(p+1)!} \right) {\varDelta t}^{p-1} \frac{\left\| L^{(p+1)/2} u_0\right\| ^2}{\left\| L u_0\right\| ^2} + {\mathcal {O}}({\varDelta t}^{p+1}). \end{aligned}$$

(37)

Comparing the analytical solution

$$\begin{aligned} u(\gamma {\varDelta t}) = \left( \sum _{k=0}^\infty \frac{\gamma ^k {\varDelta t}^k}{k!} L^k \right) u_0 \end{aligned}$$

(38)

with the RRK solution,

$$\begin{aligned} u_+^\gamma - u(\gamma {\varDelta t})= & {} \left( \sum _{k=1}^p \frac{1}{k!} (\gamma - \gamma ^k) {\varDelta t}^k L^k \right) u_0 \nonumber \\&+ \left( \alpha _{p+1} \gamma - \frac{\gamma ^{p+1}}{(p+1)!} \right) {\varDelta t}^{p+1} L^{p+1} u_0 + {\mathcal {O}}({\varDelta t}^{p+2}). \end{aligned}$$

(39)

Inserting $\gamma$ (37) and expanding the term $(\gamma - \gamma ^2) {\varDelta t}^2 L^2 u_0$ results in

$$\begin{aligned}&u_+^\gamma - u(\gamma {\varDelta t}) = \left( \alpha _{p+1} - \frac{1}{(p+1)!} \right) \nonumber \\&\quad \left( (-1)^{(p+1)/2} \frac{\left\| L^{(p+1)/2} u_0\right\| ^2}{\left\| L u_0\right\| ^2} L^2 u_0 + L^{p+1} u_0 \right) {\varDelta t}^{p+1} + {\mathcal {O}}({\varDelta t}^{p+2}). \end{aligned}$$

(40)

Finally, the second term in brackets vanishes because of the special structure of L (27). $\square$

To prove Theorem 1, expansions using rooted trees will be applied, cf. [4, Chapter 3]. The following structural results will be used.

Lemma 2

For the Euclidean Hamiltonian system (23), $m \in {\mathbb {N}}$, and $n \in \left\{ 0, \ldots , m\right\}$, there exists a smooth function $h_{m,n}$ such that

$$\begin{aligned} f^{(m)}(\underbrace{f, \ldots , f}_{n \text { terms}}, \underbrace{u, \ldots , u}_{m-n \text { terms}}) = {\left\{ \begin{array}{ll} h_{m,n}(\left\| u\right\| ^2/2) f(u), &{} \text {if } n \text { is even}, \\ h_{m,n}(\left\| u\right\| ^2/2) u, &{} \text {if } n \text { is odd}. \end{array}\right. } \end{aligned}$$

(41)

Proof (Proof by induction)

The induction hypothesis is fulfilled for the basic cases $m = 1$ and $n \in \left\{ 0,1\right\}$, since

$$\begin{aligned} f'(q, p)= & {} \begin{pmatrix} g' p q^T &{} g' p p^T + g \mathrm{I}\\ - g' q q^T - g \mathrm{I}&{} - g' q p^T \end{pmatrix}, \nonumber \\ f' f= & {} - g^2 \begin{pmatrix} q \\ p \end{pmatrix} \propto u, \quad f' u = \bigl ( g' (|q|^2 + |p|^2) + g \bigr ) \begin{pmatrix} p \\ -q \end{pmatrix} \propto f, \end{aligned}$$

(42)

and $g, g'$ depend on $\left\| u\right\| ^2$.

The induction step from m to $m+1$ can be carried out by differentiating the identity (41). For even n, the derivative with respect to $u^j$ of the right hand side is

$$\begin{aligned} \partial _j h_{m,n} f^i = h_{m,n}' u^j f^i + h_{m,n} f^i_j. \end{aligned}$$

(43)

Multiplication by $u^j$ and $f^j$, respectively, yield

$$\begin{aligned} \begin{aligned} (\partial _j h_{m,n} f^i) u^j&= (h_{m,n}' \left\| u\right\| ^2 + g' \left\| u\right\| ^2 + g) f^i, \\ (\partial _j h_{m,n} f^i) f^j&= -g^2 h_{m,n} u^i. \end{aligned} \end{aligned}$$

(44)

Similarly, for odd n, the derivative of the right hand side is

$$\begin{aligned} \partial _j h_{m,n} u^i = h_{m,n}' u^j u^i + h_{m,n} \delta ^i_j \end{aligned}$$

(45)

and multiplication by $u^j$ and $f^j$, respectively, result in

$$\begin{aligned} \begin{aligned} (\partial _j h_{m,n} u^i) u^j&= (h_{m,n}' \left\| u\right\| ^2 + h_{m,n}) u^i, \\ (\partial _j h_{m,n} u^i) f^j&= h_{m,n} f^i. \end{aligned} \end{aligned}$$

(46)

For all n, multiplication of the derivative of the right hand side by $u^j$ doesn’t change the direction while multiplication by $f^j$ flips the direction from u to f and vice versa.

The derivative with respect to $u^j$ of the left hand side of (41) is

$$\begin{aligned} \begin{aligned}&\partial _j f^i_{j_1 \ldots j_m} f^{j_{1}} \ldots f^{j_{n}} u^{j_{n+1}} \ldots u^{j_{m}} \\&\quad = f^i_{j_1 \ldots j_m j} f^{j_{1}} \ldots f^{j_{n}} u^{j_{n+1}} \ldots u^{j_{m}} \\&\qquad + f^i_{j_1 \ldots j_m} f^{j_{1}}_j f^{j_{2}} \ldots f^{j_{n}} u^{j_{n+1}} \ldots u^{j_{m}} + \cdots + f^i_{j_1 \ldots j_m} f^{j_{1}} \ldots f^{j_{n-1}} f^{j_{n}}_j u^{j_{n+1}} \ldots u^{j_{m}} \\&\qquad + f^i_{j_1 \ldots j_m} f^{j_{1}} \ldots f^{j_{n}} \delta ^{j_{n+1}}_j u^{j_{n+2}} \ldots u^{j_{m}} + \cdots + f^i_{j_1 \ldots j_m} f^{j_{1}} \ldots f^{j_{n}} u^{j_{n+1}} \ldots u^{j_{m-1}} \delta ^{j_{m}}_j. \end{aligned}\nonumber \\ \end{aligned}$$

(47)

Multiplication of all but the first term on the right hand side by $u^j$ doesn’t change the direction while multiplication by $f^j$ flips the direction from u to f and vice versa, exactly as for the derivative of the right hand side of (41). Hence, the term $f^i_{j_1 \ldots j_m j} f^{j_{1}} \ldots f^{j_{n}} u^{j_{n+1}} \ldots u^{j_{m}}$ must show the same behavior, proving the induction hypothesis (41) for $m+1$ instead of m.

$\square$

Lemma 3

For the Euclidean Hamiltonian system (23) and a rooted tree t,

$$\begin{aligned} {\left\{ \begin{array}{ll} F(t)(u_0) \parallel f(u_0), &{} \text {if } |t| \text { is odd}, \\ F(t)(u_0) \parallel u_0, &{} \text {if } |t| \text { is even}, \end{array}\right. } \end{aligned}$$

(48)

where $\parallel$ indicates that two vectors are parallel.

Proof

The result is proved by induction using the hypothesis “Consider a rooted tree $t = [t_1, \ldots , t_m]$ with elementary differential $F(t)(u_0) = f^{(m)}(u_0)\bigl ( F(t_1)(u_0), \ldots , F(t_m)(u_0) \bigr )$. If a leaf is added to one of the $t_i$ or the direction of one of the arguments of $f^{(m)}(u_0)$ is changed from $u_0$ to $f(u_0)$ or vice versa, the direction of $F(t)(u_0)$ changes from $u_0$ to $f(u_0)$ and vice versa. Additionally, (48) holds.”

The induction hypothesis is fulfilled for the base case $|t| = 1$ because of (42).

Induction step: Appending a leaf or changing the direction of one of the arguments flips the direction because of the induction hypothesis. The bushy trees behave as desired because of Lemma 2. $\square$

Proof of Theorem 1

To generalize the approach for the linear case in the proof of Theorem 2, the leading order term of $\gamma - 1$ has to be computed and $u_+^\gamma$ has to be compared with $u(\gamma {\varDelta t})$.

The relaxation parameter $\gamma$ can be written as [25, eq. (11) and Remark 4]

$$\begin{aligned} \gamma= & {} \frac{ 2 \sum _{i,j=1}^s b_i a_{ij} \left\langle {f_i,\, f_j}\right\rangle }{ \sum _{i,j=1}^s b_i b_j \left\langle {f_i,\, f_j}\right\rangle } = \frac{ 2 \left\| u_0\right\| ^2 - 2 \left\langle {u_+,\, u_0}\right\rangle }{ \left\| u_+ - u_0\right\| ^2 } \nonumber \\= & {} \frac{ 2 \left\| u_0\right\| ^2 - 2 \left\langle {u_+,\, u_0}\right\rangle }{ 2 \left\| u_0\right\| ^2 - 2 \left\langle {u_+,\, u_0}\right\rangle + \left( \left\| u_+\right\| ^2 - \left\| u_0\right\| ^2\right) . }. \end{aligned}$$

(49)

Hence, the denominator of $\gamma$ is the numerator plus a high order correction $\left\| u_+\right\| ^2 - \left\| u_0\right\| ^2$, exactly as in the linear case.

For $n \in {\mathbb {N}}$, the approximate solution of the baseline RK scheme after one step can be expanded as [4, eq. (313c)]

$$\begin{aligned} u_+ = u_0 + \sum _{|t| \le n} \frac{1}{\sigma (t)} \varPhi (t) {\varDelta t}^{|t|} F(t)(u_0) + {\mathcal {O}}({\varDelta t}^{n+1}). \end{aligned}$$

(50)

For an at least second order baseline scheme, the numerator of $\gamma$ is

$$\begin{aligned}&2 \left\| u_0\right\| ^2 - 2 \left\langle {u_+,\, u_0}\right\rangle = - 2 \sum _{|t| = 2} \frac{1}{\sigma (t)} \varPhi (t) {\varDelta t}^{|t|} F(t)(u_0) + {\mathcal {O}}({\varDelta t}^{3}) \nonumber \\&\quad = - 2 \underbrace{\sum _{i=1}^s b_i c_i}_{= 1/2} {\varDelta t}^2 \left\langle { f' f(u_0) ,\, u_0 }\right\rangle + {\mathcal {O}}({\varDelta t}^{3}). \end{aligned}$$

(51)

Because of (42),

$$\begin{aligned} \left\langle { f' f(u_0) ,\, u_0 }\right\rangle = - g^2 |q|^2 - g^2 |p|^2. \end{aligned}$$

(52)

This is in perfect agreement with the corresponding term $\left\| L u_0\right\| ^2$ in the linear case.

Since the baseline method has an order of accuracy p,

$$\begin{aligned} \left\| u_+\right\| ^2 - \left\| u_0\right\| ^2 = \ell _{\mathrm {ot}}{\varDelta t}^{p+1} + {\mathcal {O}}({\varDelta t}^{p+2}), \end{aligned}$$

(53)

where $\ell _{\mathrm {ot}}$ denotes the leading order term. In conclusion, the relaxation parameter $\gamma$ can be expanded as

$$\begin{aligned} \gamma = 1 - \frac{\ell _{\mathrm {ot}}}{g^2 (|q_0|^2 + |p_0|^2)} {\varDelta t}^{p-1} + {\mathcal {O}}({\varDelta t}^{p}). \end{aligned}$$

(54)

To compute the order of accuracy, the expansions [4, eqs. (311d) and (313c)]

$$\begin{aligned} u({\varDelta t})&= u_0 + \sum _{|t| \le p+1} \frac{1}{\sigma (t) t!} {\varDelta t}^{|t|} F(t)(u_0) + {\mathcal {O}}({\varDelta t}^{p+2}), \end{aligned}$$

(55)

$$\begin{aligned} u(\gamma {\varDelta t})&= u_0 + \sum _{|t| \le p+1} \frac{1}{\sigma (t) t!} \gamma ^{|t|} {\varDelta t}^{|t|} F(t)(u_0) + {\mathcal {O}}({\varDelta t}^{p+2}), \end{aligned}$$

(56)

$$\begin{aligned} u_+^\gamma = u_0 + \gamma (u_+ - u_0)&= u_0 + \sum _{|t| \le p+1} \frac{1}{\sigma (t)} \varPhi (t) \gamma {\varDelta t}^{|t|} F(t)(u_0) + {\mathcal {O}}({\varDelta t}^{p+2}), \end{aligned}$$

(57)

will be used. The order conditions

$$\begin{aligned} \varPhi (t) = \frac{1}{t!}, \qquad |t| \le p, \end{aligned}$$

(58)

are satisfied for the baseline RK scheme. Hence,

$$\begin{aligned} u_+^\gamma - u(\gamma {\varDelta t})= & {} (\gamma - \gamma ^2) \sum _{|t| = 2} \frac{1}{\sigma (t)} \varPhi (t) {\varDelta t}^2 F(t)(u_0) \nonumber \\&+ \sum _{|t| = p+1} \frac{1}{\sigma (t)} \left( \varPhi (t) - \frac{1}{t!} \right) {\varDelta t}^{p+1} F(t)(u_0) + {\mathcal {O}}({\varDelta t}^{p+2}). \end{aligned}$$

(59)

Inserting the value of $\gamma$ and the only rooted tree of order two,

$$\begin{aligned} (\gamma - \gamma ^2) \sum _{|t| = 2} \frac{1}{\sigma (t)} \varPhi (t) {\varDelta t}^2 F(t)(u_0) = \frac{{\varDelta t}^{p+1}}{2} \frac{\ell _{\mathrm {ot}}}{g^2 (|q_0|^2 + |p_0|^2)} f'f(u_0). \end{aligned}$$

(60)

Using $\Vert u_0\Vert ^2 = \Vert u({\varDelta t})\Vert ^2$ and the expansions (50) and (55),

$$\begin{aligned} \ell _{\mathrm {ot}}= & {} 2 \sum _{|t| = p+1} \frac{1}{\sigma (t)} \left( \varPhi (t) - \frac{1}{t!} \right) \left\langle { F(t)(u_0) ,\, u_0 }\right\rangle \nonumber \\&+ \underbrace{\sum _{|t_1| + |t_2| = p+1} \frac{1}{\sigma (t_1) \sigma (t_2)} \left( \varPhi (t_1) \varPhi (t_2) - \frac{1}{t_1! \, t_2!} \right) \left\langle { F(t_1)(u_0) ,\, F(t_2)(u_0) }\right\rangle }_{= 0}.\quad \end{aligned}$$

(61)

The last sum on the right hand side vanishes, since $|t_1|, |t_2| \in \left\{ 1, \ldots , p\right\}$ and consequently $\varPhi (t_i) = \nicefrac {1}{t_i!}$ because of the order conditions.

Finally, using Lemma 3 and inserting $f' f(u_0)$, noticing that $\Vert u_0\Vert ^2 = |q_0|^2 + |p_0|^2$,

$$\begin{aligned} \begin{aligned}&\frac{1}{2} \frac{\ell _{\mathrm {ot}}}{g^2 (|q_0|^2 + |p_0|^2)} f'f(u_0) + \sum _{|t| = p+1} \frac{1}{\sigma (t)} \left( \varPhi (t) - \frac{1}{t!} \right) F(t)(u_0) \\&\quad = - \frac{1}{2} \ell _{\mathrm {ot}}\frac{1}{|q_0|^2 + |p_0|^2} u_0 + \sum _{|t| = p+1} \frac{1}{\sigma (t)} \left( \varPhi (t) - \frac{1}{t!} \right) \frac{1}{|q_0|^2 + |p_0|^2} \left\langle { F(t)(u_0) ,\, u_0 }\right\rangle u_0 \\&\quad = 0. \end{aligned}\nonumber \\ \end{aligned}$$

(62)

Hence, the RRK method has an order of accuracy $p+1$. $\square$