Runge–Kutta Lawson schemes for stochastic differential equations

Abstract

In this paper, we present a framework to construct general stochastic Runge–Kutta Lawson schemes. We prove that the schemes inherit the consistency and convergence properties of the underlying Runge–Kutta scheme, and confirm this in some numerical experiments. We also investigate the stability properties of the methods and show for some examples, that the new schemes have improved stability properties compared to the underlying schemes.

Appendix

1.1 Platen strong order 1.5 SL scheme

Writing the explicit order 1.5 strong scheme by Platen for $M=1$ [18, Eq. 11.2.1] in the form (2.9), we see that the coefficients $z_i^{m,n}$ of the scheme are given by

	$m=0$	$m=1$
$i=1$	$z^{0,n}_1 = \frac{1}{2}h$	$z^{1,n}_1 = \Delta W^n - \frac{1}{h}[\Delta W^n h-\Delta Z^n]$
$i=2$	$z^{0,n}_2 = \frac{1}{2\sqrt{h}}\Delta Z^n + \frac{1}{4}h$	$z^{1,n}_2 = \frac{1}{h}\left[ \frac{1}{2}\Delta W^n h-\Delta Z^n - \frac{1}{4} \{\frac{1}{3}(\Delta W^n)^2 - h\}\Delta W^n\right]$
$i=3$	$z^{0,n}_3 =-\frac{1}{2\sqrt{h}}\Delta Z^n + \frac{1}{4}h$	$z^{1,n}_3 = \frac{1}{h}\left[ \frac{1}{2}\Delta W^n h-\Delta Z^n + \frac{1}{4} \{\frac{1}{3}(\Delta W^n)^2 - h\}\Delta W^n\right]$
$i=4$	$z^{0,n}_4 = 0$	$z^{1,n}_4 = \frac{1}{4h} \left[ \frac{1}{3}(\Delta W^n)^2 - h\right] \Delta W^n$
$i=5$	$z^{0,n}_5 = 0$	$z^{1,n}_4 = -\frac{1}{4h} \left[ \frac{1}{3}(\Delta W^n)^2 - h\right] \Delta W^n$

where

$$\begin{aligned} \Delta Z^n=\frac{1}{2}h\Delta W^n + \frac{\sqrt{3}}{6}h^{3/2}U^n \end{aligned}$$

with $U ^n\sim \mathcal {N}(0,1)$. Similarly, the coefficients of $Z_{ij}^{m,n}$ are given by

	$j=1$	$j=2$	$j=3$	$j=4$	$j=5$
$i=1$
$m=0$	$Z^{0,n}_{1,1} = 0$,	$Z^{0,n}_{1,2} = 0$,	$Z^{0,n}_{1,3} = 0$,	$Z^{0,n}_{1,4} = 0$,	$Z^{0,n}_{1,5} = 0$
$m=1$	$Z^{1,n}_{1,1} = 0$,	$Z^{1,n}_{1,2} = 0$,	$Z^{1,n}_{1,3} = 0$,	$Z^{1,n}_{1,4} = 0$,	$Z^{1,n}_{1,5} = 0$
$i=2$
$m=0$	$Z^{0,n}_{2,1} = h$,	$Z^{0,n}_{2,2} = 0$,	$Z^{0,n}_{2,3} = 0$,	$Z^{0,n}_{2,4} = 0$,	$Z^{0,n}_{2,5} = 0$
$m=1$	$Z^{1,n}_{2,1} = \sqrt{h}$,	$Z^{1,n}_{2,2} = 0$,	$Z^{1,n}_{2,3} = 0$,	$Z^{1,n}_{2,4} = 0$,	$Z^{1,n}_{2,5} = 0$
$i=3$
$m=0$	$Z^{0,n}_{1,1} = h$,	$Z^{0,n}_{1,2} = 0$,	$Z^{0,n}_{1,3} = 0$,	$Z^{0,n}_{1,4} = 0$,	$Z^{0,n}_{1,5} = 0$
$m=1$	$Z^{1,n}_{1,1} = -\sqrt{h}$,	$Z^{1,n}_{1,2} = 0$,	$Z^{1,n}_{1,3} = 0$,	$Z^{1,n}_{1,4} = 0$,	$Z^{1,n}_{1,5} = 0$
$i=4$
$m=0$	$Z^{0,n}_{2,1} = h$,	$Z^{0,n}_{2,2} = 0$,	$Z^{0,n}_{2,3} = 0$,	$Z^{0,n}_{2,4} = 0$,	$Z^{0,n}_{2,5} = 0$
$m=1$	$Z^{1,n}_{2,1} = \sqrt{h}$,	$Z^{1,n}_{2,2} = \sqrt{h}$,	$Z^{1,n}_{2,3} = 0$,	$Z^{1,n}_{2,4} = 0$,	$Z^{1,n}_{2,5} = 0$
$i=5$
$m=0$	$Z^{0,n}_{2,1} = h$,	$Z^{0,n}_{2,2} = 0$,	$Z^{0,n}_{2,3} = 0$,	$Z^{0,n}_{2,4} = 0$,	$Z^{0,n}_{2,5} = 0$
$m=1$	$Z^{1,n}_{2,1} = \sqrt{h}$,	$Z^{1,n}_{2,2} = -\sqrt{h}$,	$Z^{1,n}_{2,3} = 0$,	$Z^{1,n}_{2,4} = 0$,	$Z^{1,n}_{2,5} = 0$

Using the definitions of the coefficients $c_{m}^{n,i}$ and $\Delta L_{i}^n$ we calculate

	$m=0$	$m=1$		$\Delta L_i^n$
$i=1$	$c^{n,1}_0=0$	$c^{n,1}_1=0$	$\implies$	$\Delta L_1^n = 0$
$i=2$	$c^{n,2}_0=h$	$c^{n,2}_1=\sqrt{h}$	$\implies$	$\Delta L_2^n = (A_0 -\gamma ^\star A_1^2)h + A_1 \sqrt{h}$
$i=3$	$c^{n,3}_0=h$	$c^{n,3}_1=-\sqrt{h}$	$\implies$	$\Delta L_3^n = (A_0 -\gamma ^\star A_1^2)h - A_1 \sqrt{h}$
$i=4$	$c^{n,4}_0=h$	$c^{n,4}_1=2\sqrt{h}$	$\implies$	$\Delta L_4^n = (A_0 -\gamma ^\star A_1^2)h + 2A_1 \sqrt{h}$
$i=5$	$c^{n,5}_0=h$	$c^{n,5}_1=0$	$\implies$	$\Delta L_5^n = (A_0 -\gamma ^\star A_1^2)h$

Finally, using the definitions of $c_{m}^{n}$ and $\Delta L^n$ we calculate $c^{n}_0=h$, $c^{n}_1=\Delta W^n$ and thus $\Delta L^n = (A_0 -\gamma ^\star A_1^2)h + A_1\Delta W^n$.

With all the parameters of the scheme (2.12) in place, the resulting Platen strong order 1.5 SL scheme is given by

$$\begin{aligned} H_1= & {} {\bar{V}}^n_n, \\ H_2= & {} {\bar{V}}^n_n + \tilde{g}^n_0 (t_n,H_1) h + \tilde{g}^n_1 (t_n,H_1) \sqrt{h}, \\ H_3= & {} {\bar{V}}^n_n + \tilde{g}^n_0 (t_n,H_1) h - \tilde{g}^n_1 (t_n,H_1) \sqrt{h}, \\ H_4= & {} {\bar{V}}^n_n + \tilde{g}^n_0 (t_n,H_1) h + \tilde{g}^n_1 (t_n,H_1) \sqrt{h} + e^{-\Delta L_2^n}\tilde{g}^n_1 (t_n+h,e^{\Delta L_2^n}H_2) \sqrt{h}, \\ H_5= & {} {\bar{V}}^n_n + \tilde{g}^n_0 (t_n,H_1) h + \tilde{g}^n_1 (t_n,H_1) \sqrt{h} - e^{-\Delta L_2^n}\tilde{g}^n_1 (t_n+h,e^{\Delta L_2^n}H_2) \sqrt{h},\\ {\bar{V}}^n_{n+1}= & {} {\bar{V}}^n_n + \tilde{g}^n_1 (t_n,H_1) \Delta W^n\\&+ \Big [e^{-\Delta L_2^n}\tilde{g}^n_0 (t_n+h,e^{\Delta L_2^n}H_2) - e^{-\Delta L_3^n}\tilde{g}^n_0 (t_n+h,e^{\Delta L_3^n}H_3)\Big ]\frac{\Delta Z^n}{2\sqrt{h}} \\&+ \Big [e^{-\Delta L_2^n}\tilde{g}^n_0 (t_n+h,e^{\Delta L_2^n}H_2) + 2\tilde{g}^n_0 (t_n,H_1) + e^{-\Delta L_3^n}\tilde{g}^n_0 (t_n+h,e^{\Delta L_3^n}H_3)\Big ]\frac{h}{4} \\&+ \Big [e^{-\Delta L_2^n}\tilde{g}^n_1 (t_n+h,e^{\Delta L_2^n}H_2) - e^{-\Delta L_3^n}\tilde{g}^n_1 (t_n+h,e^{\Delta L_3^n}H_3)\Big ]\frac{(\Delta W^n)^2 - h}{4\sqrt{h}} \\&+ \Big [e^{-\Delta L_2^n}\tilde{g}^n_1 (t_n+h,e^{\Delta L_2^n}H_2) - 2\tilde{g}^n_1 (t_n,H_1) + e^{-\Delta L_3^n}\tilde{g}^n_1 (t_n+h,e^{\Delta L_3^n}H_3)\Big ]\frac{\Delta W^n h - \Delta Z^n}{2h}\\&+\Big [e^{-\Delta L_4^n}\tilde{g}^n_1 (t_n+h,e^{\Delta L_4^n}H_4) - e^{-\Delta L_5^n}\tilde{g}^n_1 (t_n+h,e^{\Delta L_5^n}H_5) - e^{-\Delta L_2^n}\tilde{g}^n_1 (t_n+h,e^{\Delta L_2^n}H_2) \\&+ e^{-\Delta L_3^n}\tilde{g}^n_1 (t_n+h,e^{\Delta L_3^n}H_3)\Big ] \frac{1}{4h}\left[ \frac{1}{3}(\Delta W^n)^2 - h \right] \Delta W^n \end{aligned}$$

and

$$\begin{aligned} Y_{n+1} = e^{\Delta L^n} {\bar{V}}^n_{n+1}. \end{aligned}$$

1.2 Platen weak order 2.0 SL scheme

Following the same steps as above, for the explicit order 2 weak scheme by Platen for $M=1$ [18, Eq. 15.1.1] as underlying scheme we obtain

$$\begin{aligned} \Delta L_1^n= & {} 0 \\ \Delta L_2^n= & {} (A_0 - \frac{1}{2} A_1^2) h + A_1 \Delta W^n \\ \Delta L_3^n= & {} (A_0 - \frac{1}{2} A_1^2) h + A_1 \sqrt{h} \\ \Delta L_4^n= & {} (A_0 - \frac{1}{2} A_1^2) h - A_1 \sqrt{h} \\ \Delta L^n= & {} (A_0 - \frac{1}{2} A_1^2) h + A_1 \Delta W^n \end{aligned}$$

and thus the corresponding Platen weak order 2.0 SL scheme is given by

$$\begin{aligned} H_1&= {\bar{V}}^n_n, \\ H_2&= {\bar{V}}^n_n + \tilde{g}^n_0 (t_n,H_1) h + \tilde{g}^n_1 (t_n,H_1) \Delta W^n, \\ H_3&= {\bar{V}}^n_n + \tilde{g}^n_0 (t_n,H_1) h + \tilde{g}^n_1 (t_n,H_1) \sqrt{h}, \\ H_4&= {\bar{V}}^n_n + \tilde{g}^n_0 (t_n,H_1) h - \tilde{g}^n_1 (t_n,H_1) \sqrt{h}, \\ \\ {\bar{V}}^n_{n+1}&= {\bar{V}}^n_n + \Big [e^{-\Delta L_2^n}\tilde{g}^n_0 (t_n+h,e^{\Delta L_2^n}H_2) + \tilde{g}^n_0 (t_n,H_1)\Big ] \frac{h}{2}\\&+ \Big [e^{-\Delta L_3^n}\tilde{g}^n_1 (t_n+h,e^{\Delta L_3^n}H_3) + 2\tilde{g}^n_1 (t_n,H_1) + e^{-\Delta L_4^n}\tilde{g}^n_1 (t_n+h,e^{\Delta L_4^n}H_4)\Big ] \frac{\Delta W^n}{4} \\&+ \Big [e^{-\Delta L_3^n}\tilde{g}^n_1 (t_n+h,e^{\Delta L_3^n}H_3) - e^{-\Delta L_4^n}\tilde{g}^n_1 (t_n+h,e^{\Delta L_4^n}H_4)\Big ]\frac{(\Delta W^n)^2 - h}{4\sqrt{h}} \\&Y_{n+1} = e^{\Delta L^n} {\bar{V}}^n_{n+1}. \end{aligned}$$