Novel phase-fitted symmetric splitting methods for chemical oscillators

Abstract

Splitting strategy is considered for the numerical solution of oscillatory chemical reaction systems. When applied to the harmonic oscillator, traditional splitting methods with constant coefficients are shown to be have some order of phase lag though they are zero-dissipative. Phase-fitted symmetric splitting methods of order two and order four are constructed. The result of the numerical experiment on the Lotka–Volterra system shows that the new phase-fitted symmetric splitting methods are more effective than their prototype splitting methods and can preserve the invariant of the system in long-term compared with the classical Runge–Kutta method.

Appendix

1.1 Proof of Theorem 2.1

Define $\alpha _i,\beta _i, i=1,\ldots ,s$ recursively by

$$\begin{aligned} \begin{array}{l} \beta _1=a_1, \alpha _1=b_1-\beta _1,\\ \beta _i =a_i-\alpha _{i-1}, \alpha _i=b_i-\beta _i, i=2,\ldots ,s. \end{array} \end{aligned}$$

(31)

With $a_{s+1}=\alpha _s$, the scheme (9) is equivalent to the following scheme

$$\begin{aligned} \Psi _h=\Phi _{\alpha _{s}h}\circ \Phi ^{*}_{\beta _{s}h}\circ \cdots \circ \Phi _{\alpha _{2}h}\circ \Phi ^{*}_{\beta _{2}h}\circ \Phi _{\alpha _{1}h}\circ \Phi ^{*}_{\beta _{1}h}, \end{aligned}$$

where $\Phi _{\alpha _{i}h}=\varphi ^{[1]}_{\alpha _{i}h}\circ \varphi ^{[2]}_{{\alpha _{i}h}}$ and $\Phi ^{*}_{\beta _{i}h}=\varphi ^{[2]}_{{\beta _{i}h}}\circ \varphi ^{[1]}_{{\beta _{i}h}}.$

The first to fourth order conditions are given as follow (see Example 3.15 in Chapter III of Hairer et al. [15])

$$\begin{aligned} \begin{array}{ll} \mathrm{Order\ one\ requires}&{}\sum \limits _{i=1}^{s}(\alpha _i+\beta _i)=1;\\ \mathrm{Order\ two\ requires\ in\ addition}&{}\sum \limits _{i=1}^{s}\left( \alpha _i^2-\beta _i^2\right) =0;\\ \mathrm{Order\ three\ requires\ in\ addition} &{} \sum \limits _{i=1}^{s}\left( \alpha _i^3+\beta _i^3\right) =0,\\ &{} \sum \limits _{i=1}^{s}\left( \alpha _i^2-\beta _i^2\right) \left( \sum \limits _{l=1}^{i-1}\alpha _l+ \sum \limits _{l=1}^{i}\beta _l\right) =0;\\ \end{array} \end{aligned}$$

$$\begin{aligned} \begin{array}{ll} \mathrm{Order\ four\ requires\ in\ addition} &{} \sum \limits _{i=1}^{s}\left( \alpha _i^4-\beta _i^4\right) =0,\\ &{} \sum \limits _{i=1}^{s}\left( \alpha _i^3+\beta _i^3\right) \left( \sum \limits _{l=1}^{i-1}\alpha _l+ \sum \limits _{l=1}^{i}\beta _l\right) =0,\\ &{} \sum \limits _{i=1}^{s}\left( \alpha _i^2-\beta _i^2\right) \left( \sum \limits _{l=1}^{i-1}\alpha _l+ \sum \limits _{l=1}^{i}\beta _l\right) ^2=0.\\ \end{array} \end{aligned}$$

(32)

From the relation (31), we have

$$\begin{aligned} \alpha _i+\beta _i=b_i, \end{aligned}$$

and

$$\begin{aligned}&\alpha _1=b_1-a_1,\\&\alpha _i-\alpha _{i-1}=b_i-a_i,\quad i\ge 2. \end{aligned}$$

which imply that

$$\begin{aligned} \alpha _i=\sum \limits _{l=1}^{i}(b_l-a_l):=d_i, \quad \beta _i=b_i-d_i\quad \mathrm{and}\quad \sum \limits _{l=1}^{i-1}\alpha _l+ \sum \limits _{l=1}^{i}\beta _l=\sum \limits _{l=1}^{i}a_l:=c_i. \end{aligned}$$

(33)

Substituting (33) into (32) gives the conditions (10) in Theorem 2.1. The proof is complete.

1.2 Proof of Theorem 3.1

Necessity. If the phase lag $PL(\nu )=c_{r+1}\nu ^{r+1}+\mathcal {O}(\nu ^{r+2})$, that is

$$\begin{aligned} \nu -\arccos \frac{\mathrm{tr}(R(\nu ))}{2\sqrt{\det (R(\nu ))}}=c_{r+1}\nu ^{r+1}+\mathcal {O}(\nu ^{r+2}),\ c_{r+1}\ne 0, \quad \nu \rightarrow 0, \end{aligned}$$

then

$$\begin{aligned} \frac{\mathrm{tr}(R(\nu ))}{2\sqrt{\det (R(\nu ))}}=\cos \big (\nu -c_{r+1}\nu ^{r+1}+\mathcal {O}(\nu ^{r+2})\big ), \end{aligned}$$

and the pseudo-phase lag

$$\begin{aligned} \begin{array}{rl} {\textit{PPL}}(\nu )&{}=\cos (\nu )-\frac{\mathrm{tr}(R(\nu ))}{2\sqrt{\det (R(\nu ))}} \\ &{}=\cos (\nu )-\cos \big (\nu -c_{r+1}\nu ^{r+1}+\mathcal {O}(\nu ^{r+2})\big )\\ &{}=\cos (\nu )-\big [\cos (\nu )+\sin (\nu )\big (c_{r+1}\nu ^{p+1}+\mathcal {O}(\nu ^{r+2})\big )+\mathcal {O}(\nu ^{2r+2})\big ]\\ &{}=-\sin (\nu )\big (c_{r+1}\nu ^{r+1}+\mathcal {O}(\nu ^{r+2})\big )+\mathcal {O}(\nu ^{2r+2})\\ &{}=-\big (\nu +\mathcal {O}(\nu ^3)\big )\big (c_{r+1}\nu ^{r+1}+\mathcal {O}(\nu ^{r+2})\big )+\mathcal {O}(\nu ^{2r+2})\\ &{}=-c_{r+1}\nu ^{r+2}+\mathcal {O}(\nu ^{r+3}). \end{array} \end{aligned}$$

Sufficiency. Conversely assume that ${\textit{PPL}}(\nu )=-c_{r+1}\nu ^{r+2}+\mathcal {O}(\nu ^{r+3})$, that is

$$\begin{aligned} \cos (\nu )-\frac{\mathrm{tr}(R(\nu ))}{2\sqrt{\det (R(\nu ))}}=-c_{r+1}\nu ^{r+2}+\mathcal {O}(\nu ^{r+3}), \end{aligned}$$

then

$$\begin{aligned} \frac{\mathrm{tr}(R(\nu ))}{2\sqrt{\det (R(\nu ))}}=\cos (\nu )+c_{r+1}\nu ^{r+2}+\mathcal {O}(\nu ^{r+3}), \end{aligned}$$

as $\nu \rightarrow 0$ ($\nu =\omega h>0$), and

$$\begin{aligned} \begin{array}{rl} PL(\nu )&{}= \nu -\arccos \left( \frac{\mathrm{tr}(R(\nu ))}{2\sqrt{\det (R(\nu ))}}\right) \\ &{}=\nu -\arccos \left( \cos (\nu )+c_{r+1}\nu ^{r+2}+\mathcal {O}(\nu ^{r+3})\right) \\ &{}=\nu -\arccos \big (\cos (\nu )\big )+\frac{1}{\sqrt{1-\cos (\nu )^2}}\big (c_{r+1}\nu ^{r+2}+\mathcal {O}(\nu ^{r+3})\big )+\mathcal {O}(\nu ^{r+2})\\ &{}=\frac{1}{\sin (\nu )}\big (c_{r+1}\nu ^{r+2}+\mathcal {O}(\nu ^{r+3})\big )+\mathcal {O}(\nu ^{r+2})\\ &{}=c_{r+1}\nu ^{r+1}+\mathcal {O}(\nu ^{r+2}), \end{array}\end{aligned}$$

The proof is complete.

1.3 Proof of Lemma 3.1

We only prove the lemma for the method SSP1h. The proof for the other methods is similar. The exact flows of the vector fields $F^{[1]}$ and $F^{[2]}$ in (19) are given respectively by

$$\begin{aligned} \varphi _h^{[1]}= \left( \begin{array}{cc} 1 &{}\quad \nu \\ 0 &{}\quad 1 \\ \end{array} \right) \quad {\text{ and }} \quad \varphi _h^{[2]}= \left( \begin{array}{cc} 1 &{}\quad 0 \\ -\nu &{}\quad 1 \\ \end{array} \right) ,\qquad \nu =h\omega . \end{aligned}$$

Applying the method SSP1 to the decomposed system (18)–(19) gives

$$\begin{aligned} \begin{array}{rl} \left( \begin{array}{c}q_{n+1}\\ p_{n+1}\end{array}\right) &{}=\Psi _h^\mathrm{SSP1h}(\nu )\left( \begin{array}{c}q_{n}\\ p_{n}\end{array}\right) \\ &{}= \varphi _{h/2}^{[1]}\circ \varphi _h^{[2]}\circ \varphi _{h/2}^{[1]}\left( \begin{array}{c}q_{n}\\ p_{n}\end{array}\right) \\ &{}=\left( \begin{array}{c} (1- \nu ^2/2)q_n+ ( \nu -\nu ^3/4 )p_n \\ -\nu q_n +(1- \nu ^2/2)p _n \end{array} \right) .\end{array} \end{aligned}$$

Then the stability matrices of SSP1h is

$$\begin{aligned} R^\mathrm{SSP1h}(\nu )= \left( \begin{array}{cc} 1- \nu ^2/2 &{}\quad \nu -\nu ^3/4 \\ -\nu &{}\quad 1- \nu ^2/2 \\ \end{array}\right) . \end{aligned}$$

Thus

$$\begin{aligned} {\textit{LTE}}_q^\mathrm{SSP1h}= & {} q_{n+1}-q(t_{n+1})\\= & {} \big ( (1- \nu ^2/2)q_n+ ( \nu -\nu ^3/4 )p_n\big )-\big (\cos (\nu )q_n+\sin (\nu )p_n\big )\\= & {} \big ( (1- \nu ^2/2)-\cos (\nu )\big )q_{n}+\big ((\nu -\nu ^3/4)-\sin (\nu )\big ) p_n\\= & {} -\frac{\omega ^3q_{n}}{2}h^3-\frac{\omega ^4q_{n}}{24}h^4+\mathcal {O}(h^5),\\ {\textit{LTE}}_p^\mathrm{SSP1h}= & {} p_{n+1}-p(t_{n+1})\\= & {} (-\nu q_{n}+(1-\nu ^2) p_n)-\big (-\sin (\nu )q_n+\cos (\nu )p_n\big )\\= & {} \big (\sin (\nu )-\nu ) q_{n}+\big (1-\nu ^2-\cos (\nu ) p_n)\\= & {} -\frac{\omega ^3q_{n}}{6}h^3-\frac{\omega ^4p_{n}}{24}h^4+\mathcal {O}(h^5). \end{aligned}$$

By Definition 3.1 and Taylor series expansion, we obtain

$$\begin{aligned} \begin{aligned} {\textit{PPL}}^\mathrm{SSP1h}(\nu )&=\cos (\nu )-\big (1-\frac{\nu ^2}{2}\big )=\frac{\nu ^4}{24}+\mathcal {O}(\nu ^{6}) \end{aligned} \end{aligned}$$

and

$$\begin{aligned} D^\mathrm{SSP1h}(\nu )=0. \end{aligned}$$

Therefore the method SSP1h has phase lag of order two and it is zero-dissipative. The proof is complete.