Modeling Fast Diffusion Processes in Time Integration of Stiff Stochastic Differential Equations

Abstract

Numerical algorithms for stiff stochastic differential equations are developed using linear approximations of the fast diffusion processes, under the assumption of decoupling between fast and slow processes. Three numerical schemes are proposed, all of which are based on the linearized formulation albeit with different degrees of approximation. The schemes are of comparable complexity to the classical explicit Euler-Maruyama scheme but can achieve better accuracy at larger time steps in stiff systems. Convergence analysis is conducted for one of the schemes, that shows it to have a strong convergence order of 1/2 and a weak convergence order of 1. Approximations arriving at the other two schemes are discussed. Numerical experiments are carried out to examine the convergence of the schemes proposed on model problems.

Appendix A Details of Proofs in Convergence Analysis

Throughout the analysis in this section, the notations of $c_T$ and $C_T$ are used for generic constants dependent on T but not $\Delta t$, that may change from line to line.

1.1 A.1 Proof of Lemma 1

We first consider the case with $p=2$. Note that due to (5) and the Itô isometry we have

$$\begin{aligned} {\mathbb {E}}\left[ \left| {\hat{X}}(s) - x_{n_s}\right| ^2\right]\leqslant & {} M {\mathbb {E}}\left[ \sum ^M_{k=1} \Big (\int ^s_{t_{n_s}} g_k({\hat{X}}(\tau )) \mathrm {d}W_k(\tau )\Big )^2\right] \\\leqslant & {} M \sum ^M_{k=1} {\mathbb {E}}\left[ \int ^s_{t_{n_s}} g^2_k({\hat{X}}(\tau )) \mathrm {d}\tau \right] , \quad s \geqslant 0. \end{aligned}$$

Then by (41) we have

$$\begin{aligned} g^2_k({\hat{X}}(\tau )) \leqslant 2 \left( L^2_k \left| {\hat{X}}(\tau ) - x_{n_s}\right| ^2 + 2 g_k^2(0) + 2 L^2_k x_{n_s}^2\right) , \quad \tau \in [t_s, s), \end{aligned}$$

and thus due to Assumption (A2) we have

$$\begin{aligned}&\quad{\mathbb {E}}\left[ \left| {\hat{X}}(s) - x_{n_s}\right| ^2\right] \\&\leqslant 2 M \sum ^M_{k=1} \int ^s_{t_{n_s}} \left( L^2_k {\mathbb {E}}\left[ \left| {\hat{X}}(\tau ) - x_{n_s}\right| ^2\right] +2 g_k^2(0) + 2 L^2_k {\mathbb {E}}\left[ x_{n_s}^2\right] \right) \mathrm {d}\tau \\&\leqslant 2 M \sum ^M_{k=1}L^2_k \int ^s_{t_{n_s}} {\mathbb {E}}\left[ \left| {\hat{X}}(\tau ) - x_{n_s}\right| ^2\right] \mathrm {d}\tau + 4M \sum ^M_{k=1}\left( g^2_k(0) + L^2_k \Lambda _T \right) \Delta t \end{aligned}$$

for $s \geqslant 0.$ It then follows from Gronwall’s inequality that

$$\begin{aligned} {\mathbb {E}}\left[ \left| {\hat{X}}(s) - x_{n_s}\right| ^2\right] \leqslant 4M \sum ^M_{k=1}\left( g^2_k(0) + L^2_k \Lambda _T \right) \Delta t \text{e}^{ 2 M \sum ^M_{k=1}L^2_k \Delta t} \leqslant c_T \Delta t \end{aligned}$$

(A1)

for $s \geqslant 0$, where $c_T$ is a constant depends on M, $\Lambda _T$, T, $L_k$, and $g_k(0)$ for $k=1, \cdots , M$, but independent of $\Delta t$.

For $p> 2$, using again (5), the Itô isometry, Hölder’s inequality, and the Cauchy-Schwarz, we have

$$\begin{aligned} {\mathbb {E}}\left[ \left| {\hat{X}}(s) - x_{n_s}\right| ^p\right]\leqslant & {} c_{p, M} {\mathbb {E}}\left[ \sum ^M_{k=1} \Big (\int ^s_{t_{n_s}} g_k({\hat{X}}(\tau )) \mathrm {d}W_k(\tau )\Big )^p\right] \\\leqslant & {} c_{p, M} \sum ^M_{k=1} {\mathbb {E}}\left[ \left( \int ^s_{t_{n_s}} g^2_k({\hat{X}}(\tau )) \mathrm {d}\tau \right) ^{p/2}\right] \\\leqslant & {} \cdots \cdots \cdot \leqslant c_{p, M} (\Delta t)^{p/2-1} {\mathbb {E}}\left[ \int ^s_{t_{n_s}} g^p_k({\hat{X}}(\tau )) \mathrm {d}\tau \right] , \quad s \geqslant 0, \end{aligned}$$

where here and below $c_{p,M}$ is a generic constant depending on p and M that may change from line to line. On the other hand by (41) we have

$$\begin{aligned} g^p_k({\hat{X}}(\tau ))\leqslant & {} c_{p, M}\left( L^2_k \left| {\hat{X}}(\tau ) - x_{n_s}\right| ^2 + 2 g_k^2(0) + 2 L^2_k x_{n_s}^2\right) ^{p/2} \\\leqslant & {} \cdots \\\leqslant & {} c_{p, M} \left( L^p_k \left| {\hat{X}}(\tau ) - x_{n_s}\right| ^p + g_k^p(0) + L^p_k x_{n_s}^p\right) , \end{aligned}$$

and thus

$$\begin{aligned}&\quad{\mathbb {E}}\left[ \left| {\hat{X}}(s) - x_{n_s}\right| ^p\right] \\& \leqslant c_{p, M} (\Delta t)^{p/2-1} \sum ^M_{k=1} \int ^s_{t_{n_s}} \left( L^p_k {\mathbb {E}}\left[ \left| {\hat{X}}(\tau ) - x_{n_s}\right| ^p\right] + g_k^p(0) + L^p_k {\mathbb {E}}\left[ x_{n_s}^p\right] \right) \mathrm {d}\tau \\& \leqslant c_{p, M} (\Delta t)^{p/2-1} \sum ^M_{k=1}L^p_k \int ^s_{t_{n_s}} {\mathbb {E}}\left[ \left| {\hat{X}}(\tau ) - x_{n_s}\right| ^p\right] \mathrm {d}\tau \\ &\quad + c_{p, M} (\Delta t)^{p/2} \sum ^M_{k=1}\left( g^p_k(0) + L^p_k \Lambda _T\right) . \end{aligned}$$

Similar to (A1), applying Gronwall’s Lemma to the above inequality results in

$$\begin{aligned} {\mathbb {E}}\left[ \left| {\hat{X}}(s) - x_{n_s}\right| ^p\right] \leqslant c_{p, M} (\Delta t)^{p/2} \sum ^M_{k=1}\left( g^p_k(0) + L^p_k \Lambda _T\right) \text{e}^{(\Delta t)^{p/2} \sum ^M_{k=1}L^p_k }, \end{aligned}$$

which implies the desired assertion for all every number $p \geqslant 2$ with

$$\begin{aligned} C_{p,T} = c_{p, M} \sum ^M_{k=1}\left( g^p_k(0) + L^p_k \Lambda _T\right) \text{e}^{ \sum ^M_{k=1}L^p_k }, \end{aligned}$$

and the assumption without loss of generality that $\Delta t \leqslant 1$. The proof is complete.

1.2 A.2 Proof of Lemma 2

First, by Hölder’s inequality and the Lipschitz condition on f,

$$\begin{aligned} \sup _{0 \leqslant t \leqslant T}{\mathcal {E}}_1^2(t)\leqslant & {} T \sup _{0 \leqslant t \leqslant T}\int ^t_0 \big \vert f(X(s))-f({\tilde{x}}(s))\big \vert ^2 \mathrm {d}s \\\leqslant & {} T L_f^2 \sup _{0 \leqslant t \leqslant T} \int ^t_0 \big \vert X(s)-{\tilde{x}}(s)\big \vert ^2 \mathrm {d}s . \end{aligned}$$

Taking the expectation of the above inequality and using Doob’s maximal inequality gives

$$\begin{aligned} {\mathbb {E}}\left[ \sup _{0 \leqslant t \leqslant T}{\mathcal {E}}_1^2(t)\right]\leqslant & {} 4 T L^2_f {\mathbb {E}}\left[ \int ^T_0 \big \vert X(s)-{\tilde{x}}(s)\big \vert ^2 \mathrm {d}s\right] \nonumber \\\leqslant & {} 4 T L^2_f \left( \int ^T_0 {\mathbb {E}}\Bigg [\sup _{0 \leqslant t \leqslant s} {\mathcal {E}}_0^2(t)\Bigg ] \mathrm {d}s +\int ^T_0 {\mathbb {E}}\left[ \big \vert y(s)-{\tilde{x}}(s)\big \vert ^2 \right] \mathrm {d}s\right) . \end{aligned}$$

(A2)

Using (42) to obtain the term $y(s)-{\tilde{x}}(s)$ and squaring it, gives

$$\begin{aligned} \vert y(s)-{\tilde{x}}(s)\vert ^2\leqslant & {} (m+1) \left( f^2(x_{n_s}) (\Delta t)^2 + \sum ^m_{k=M+1} g^2_k(x_{n_s}) \left( W_k(s) - W_k(t_{n_s})\right) ^2 \right) \\&+ (m+1) \sum ^M_{k=1} \left( \int ^s_{t_{n_s}}\Big ( g_k(x_{n_s}) + g^\prime _k(x_{n_s})({\hat{X}}(\tau ) - x_{n_s})\Big)\mathrm {d}W_k(\tau )\right) ^2 . \end{aligned}$$

Then taking the expectation of the above inequality, and using (41) and the Itô isometry, we deduce

$$\begin{aligned}&\quad \, {\mathbb {E}}\left[ \vert y(s)-{\tilde{x}}(s)\vert ^2\right] \nonumber \\&\leqslant 2 (m+1) \left( (\Delta t)^2 \big (f^2(0) + L^2_f {\mathbb {E}}[x_{n_s}^2]\big ) + \sum ^m_{k=M+1}{\mathbb {E}}\Big [\big (g_k^2(0) + L^2_k x_{n_s}^2 \big ) \big (W_k(s) - W_k(t_{n_s})\big )^2 \Big ]\right) \\& \quad + 2 (m+1) \sum ^M_{k=1} {\mathbb {E}}\left[ \int ^s_{t_{n_s}}\Big ( g^2_k(x_{n_s}) + (g^\prime _k(x_{n_s}))^2({\hat{X}}(\tau ) - x_{n_s})^2\Big ) \mathrm {d}\tau \right] \nonumber \\&\leqslant 2 (m+1) \left( (\Delta t)^2 \big (f^2(0) + L^2_f {\mathbb {E}}[x_{n_s}^2]\big ) + \sum ^m_{k=1}\Big (\big (g_k^2(0) + L^2_k {\mathbb {E}}[x_{n_s}^2] \big ) \Delta t \Big )\right) \nonumber \\& \quad + 2 (m+1) {\mathbb {E}}\left[ \int ^s_{t_{n_s}} (g^\prime _k(x_{n_s}))^2({\hat{X}}(\tau ) - x_{n_s})^2 \mathrm {d}\tau \right] . \end{aligned}$$

(A3)

Using Hölder’s inequality, Lemma 1, Assumptions (A2) and (A3), the last term of (A3) satisfies

$$\begin{aligned}&\qquad \,{\mathbb {E}}\left[ \int ^s_{t_{n_s}} (g^\prime _k(x_{n_s}))^2({\hat{X}}(\tau ) - x_{n_s})^2 \mathrm {d}\tau \right] \nonumber \\&\quad \leqslant \left( {\mathbb {E}}\big [ \big (g^\prime _k(x_{n_s})\big )^4 \big ]\right) ^{1/2} \left( {\mathbb {E}}\Big [ \Big (\int ^s_{t_{n_s}} ({\hat{X}}(\tau ) - x_{n_s})^2 \mathrm {d}\tau \Big )^2\Big ] \right) ^{1/2} \nonumber \\&\quad \leqslant \lambda ^2 \left( {\mathbb {E}}\big [ \big (1 + \vert x_{n_s}\vert ^ \gamma \big )^4\big ]\right) ^{1/2} \left( \Delta t \int ^s_{t_{n_s}} {\mathbb {E}}\Big [ ({\hat{X}}(\tau ) - x_{n_s})^4 \Big ] \mathrm {d}\tau \right) ^{1/2} \nonumber \\&\quad \leqslant c_T (\Delta t)^2, \end{aligned}$$

(A4)

where $c_T$ depends on T, $\lambda$, $\gamma$, $\Lambda _T$, $L_k$, and $g_k(0)$ for $k=1, \cdots , m$, but independent of $\Delta t$.

Now inserting (A4) into (A3), using Assumption (A2), and integrating from 0 to T gives

$$\begin{aligned} \int ^T_0 {\mathbb {E}}\left[ \vert y(s)-{\tilde{x}}(s)\vert ^2\right] \mathrm {d}s\leqslant & {} 2(m+1) T \Bigg [ (\Delta t)^2 \big (f^2(0) + L^2_f \Lambda _T + c_T \big ) \nonumber \\&+ \Delta t \sum ^m_{k=1}\Big ( g_k^2(0) + L^2_k \Lambda _T \Big )\Bigg ]. \end{aligned}$$

(A5)

Consequently, (A2) can be further estimated to satisfy

$$\begin{aligned} {\mathbb {E}}\left[ \sup _{0 \leqslant t \leqslant T}{\mathcal {E}}_1^2(t)\right] \leqslant 4T L^2_f \int ^T_0 {\mathbb {E}}\Bigg [\sup _{0 \leqslant t \leqslant s} {\mathcal {E}}_0^2(t)\Bigg ] \mathrm {d}s + 2(m+1) T^2 L^2_f\left( (\Delta t)^2 c_1 + \Delta t c_2 \right) \end{aligned}$$

with $c_1 = f^2(0) + L^2_f \Lambda _T + c_T$ and $c_2 = \sum ^m_{k=1} \big (g_k^2(0) + L^2_k \Lambda _T\big )$. Setting

$$C_T = 2(m+1) T^2 L^2_f \max \{c_1, c_2\}$$

implies the desired assertion. The proof is complete.

1.3 A.3 Proof of Lemma 3

First by the Cauchy-Schwarz, the Doob’s martingale maximal inequality, the Itô’s isometry, and Assumption (A1) we have

$$\begin{aligned} {\mathbb {E}}\left[ \sup _{0 \leqslant t \leqslant T}{\mathcal {E}}_2^2(t)\right]&\leqslant (m-M) {\mathbb {E}}\left[ \sup _{0 \leqslant t \leqslant T} \sum ^m_{k=M+1} \left| \int ^t_0 \big (g_k(X(s) - g_k({\tilde{x}}(s))\big ) \mathrm {d}W_k(s)\right| ^2 \right] \nonumber \\&\leqslant 4(m-M) {\mathbb {E}}\left[ \sum ^m_{k=M+1} \left| \int ^T_0 \big (g_k(X(s) - g_k({\tilde{x}}(s))\big ) \mathrm {d}W_k(s)\right| ^2\right] \nonumber \\&= 4(m-M) {\mathbb {E}}\left[ \sum ^m_{k=M+1} \int ^T_0 \big (g_k(X(s) - g_k({\tilde{x}}(s))\big )^2 \mathrm {d}s \right] \nonumber \\&\leqslant 4(m-M) \sum ^m_{k=M+1} L_{k}^2 {\mathbb {E}}\left[ \int ^T_0 \big \vert X(s)-{\tilde{x}}(s)\big \vert ^2 \mathrm {d}s\right] . \end{aligned}$$

Similar to (A2) in Lemma 2,

$$\begin{aligned} {\mathbb {E}}\left[ \int ^T_0 \big \vert X(s)-{\tilde{x}}(s)\big \vert ^2 \mathrm {d}s\right] \leqslant \int ^T_0 {\mathbb {E}}\Bigg [\sup _{0 \leqslant t \leqslant s} {\mathcal {E}}_0^2(t)\Bigg ] \mathrm {d}s + {\mathbb {E}}\left[ \int ^T_0 \big \vert y(s)-{\tilde{x}}(s)\big \vert ^2 \mathrm {d}s \right] \end{aligned}$$

and then it follows from the estimate (A5) that

$$\begin{aligned} {\mathbb {E}}\left[ \int ^T_0 \big \vert X(s)-{\tilde{x}}(s)\big \vert ^2 \mathrm {d}s\right]\leqslant & {} \int ^T_0 {\mathbb {E}}\Bigg [\sup _{0 \leqslant t \leqslant s} {\mathcal {E}}_0^2(t)\Bigg ] \mathrm {d}s \\&+ 2(m+1) T \sum ^m_{k=M+1} L_{k}^2\left( (\Delta t)^2 c_1 + \Delta t c_2 \right) , \end{aligned}$$

where $c_1$ and $c_2$ are the same as in Lemma 2. Setting $C_T = 4(m-M)(2m+1)T \cdot\!\!\!\!{\sum ^m_{k=M+1} L_{k}^2 \max \{c_1, c_2\}}$ implies the desired assertion. The proof is complete.

1.4 A.4 Proof of Lemma 4

First, it follows from the Cauchy-Schwarz, Doob’s martingale maximal inequality, and Itô’s isometry that

$$\begin{aligned} {\mathbb {E}}\left[ \sup _{0 \leqslant t \leqslant T}{\mathcal {E}}_3^2(t)\right] \leqslant 4 M \sum ^M_{k=1} \int ^T_0 {\mathbb {E}}\left[ \Big ( g_k(X(s)) - g_k({\tilde{x}}(s)) - g^\prime _k({\tilde{x}}(s))({\hat{X}}(s) - {\tilde{x}}(s))\Big )^2 \right] \mathrm {d}s . \end{aligned}$$

(A6)

Note that due to Rolle’s theorem, and then by Assumption (A4), for every $s \in {\mathbb {R}}$ there exists $y_s$ between ${\hat{X}}(s)$ and $x_{n_s}$ such that

$$\begin{aligned}&\quad \left| g_k({\hat{X}}(s)) - g_k(x_{n_s} ) - g^\prime _k({\tilde{x}}(s))({\hat{X}}(s) - x_{n_s}) \right| \nonumber \\&= \left| \big (g^{\prime }(y_s) - g^\prime _k(x_{n_s}) \big ) ({\hat{X}}(s) - x_{n_s}) \right| \leqslant D_k ({\hat{X}}(s) - x_{n_s}) ^2. \end{aligned}$$

(A7)

Writing $g_k(X(s)) - g_k({\tilde{x}}(s))$ in (A6) as $g_k(X(s)) - g_k({\hat{X}}(s)) +g_k({\hat{X}}(s)) - g_k(x_{n_s})$, it then follows from (A7), Lemma 1, and Assumption (A1) that

$$\begin{aligned} {\mathbb {E}}\left[ \sup _{0 \leqslant t \leqslant T}{\mathcal {E}}_3^2(t)\right]&\leqslant 8 M \sum ^M_{k=1} \left( \int ^T_0 {\mathbb {E}}\Big [\big ( g_k(X(s)) - g_k({\hat{X}}(s)) \big )^2 \Big ] \mathrm {d}s \right. \nonumber \\&\quad \left. + D_k^2\int ^T_0{\mathbb {E}}\Big [({\hat{X}}(s) - x_{n_s}) ^4\Big ] \mathrm {d}s \right) \nonumber \\&\leqslant 8 M \sum ^M_{k=1} \Bigg ( L_k^2 \int ^T_0 {\mathbb {E}}\left[ \big (X(s) - {\hat{X}}(s)\big )^2 \right] \mathrm {d}s \nonumber \\&\quad + D_k^2 T C_{4,T} (\Delta t)^2 \Bigg ), \end{aligned}$$

(A8)

where $C_{4,T}$ is the constant in Lemma 1 which is independent of $\Delta t$.

It remains to estimate ${\mathbb {E}}\left[ \big \vert X(s) - {\hat{X}}(s)\big \vert ^2 \right]$. In fact, by (1) and (5), the Cauchy inequality, and the Hölder inequality we have

$$\begin{aligned}&\quad\, {\mathbb {E}}\left[ \vert X(s) - {\hat{X}}(s)\big \vert ^2\right] \nonumber \\&\leqslant (m+1) {\mathbb {E}}\left[ \Bigg \vert \int ^s_{t_{n_s}} f(X(\tau )) \mathrm {d}\tau \Bigg \vert ^2 \right] \nonumber \\& \quad \,+ (m+1) \sum ^m_{k=M+1}{\mathbb {E}}\left[ \Bigg \vert \int ^s_{t_{n_s}} g_k(X(\tau )) \mathrm {d}W_k(\tau ) \Bigg \vert ^2 \right] \nonumber \\& \quad + (m+1) \sum ^M_{k=1}{\mathbb {E}}\left[ \Bigg \vert \int ^s_{t_{n_s}} \big (g_k(X(\tau ))-g_k({\hat{X}}(\tau ))\big ) \mathrm {d}W_k(\tau ) \Bigg \vert ^2\right] \nonumber \\& \leqslant (m+1) \left( \Delta t \int ^s_{t_{n_s}} {\mathbb {E}}\big [ f^2(X(\tau )) \big ]\mathrm {d}\tau + \sum ^m_{k=M+1} \int ^s_{t_{n_s}} {\mathbb {E}}\big [ g^2_k(X(\tau )) \big ] \mathrm {d}\tau \right) \nonumber \\& \quad + (m+1) \sum ^M_{k=1}\int ^s_{t_{n_s}} {\mathbb {E}}\big [(g_k(X(\tau ))-g_k({\hat{X}}(\tau )))^2 \big ] \mathrm {d}\tau . \end{aligned}$$

(A9)

By Assumption (A1), $f^2(X(\tau )) \leqslant 2 f^2(x_{n_s}) + 2 L^2_f \vert X(\tau ) - x_{n_s}\vert ^2$ and

$$g^2_k(X(\tau )) \leqslant 2 g_k^2(x_{n_s}) + 2 L^2_k \vert X(\tau ) - x_{n_s}\vert ^2$$

and thus

$$\begin{aligned} \int ^s_{t_{n_s}} {\mathbb {E}}\big [ f^2(X(\tau )) \big ]\mathrm {d}\tau\leqslant & {} 2\Delta t \left( 2 f^2(0) + 2L^2_f \Lambda _T + L^2_f {\mathbb {E}}\left[ \sup _{0 \leqslant t \leqslant s} {\mathcal {E}}^2_0(t)\right] \right) , \quad \end{aligned}$$

(A10)

$$\begin{aligned} \int ^s_{t_{n_s}} {\mathbb {E}}\big [ g_k^2(X(\tau )) \big ]\mathrm {d}\tau\leqslant & {} 2\Delta t \left( 2 g_k^2(0) + 2L^2_k \Lambda _T + L^2_k {\mathbb {E}}\left[ \sup _{0 \leqslant t \leqslant s} {\mathcal {E}}^2_0(t)\right] \right) . \quad \end{aligned}$$

(A11)

Inserting (A10)–(A11) into (A9) and using Assumption (A1) again we obtain

$$\begin{aligned} {\mathbb {E}}\left[ \vert X(s) - {\hat{X}}(s)\big \vert ^2\right]\leqslant & {} c_3 \left[ \sup _{0 \leqslant t \leqslant s} {\mathcal {E}}^2_0(t)\right] + c_4 \\&+ (m+1) \sum ^M_{k=1} L_k^2 \int ^s_{t_{n_s}} {\mathbb {E}}\big [\vert X(\tau ) - {\hat{X}}(\tau )\vert ^2\big ] \mathrm {d}\tau , \end{aligned}$$

where due to Assumption (A2)

$$\begin{aligned} c_3= & {} 2(m+1) \Delta t \left( \Delta t L_f^2 + \sum ^m_{M+1} L_k^2 \right) , \\ c_4= & {} 2(m+1) \Delta t \left( \Delta t + 2 \Delta t( f^2(0) + L^2_f \Lambda _T) + 2 \sum ^m_{M+1}(g^2_k(0) + L^2_k \Lambda _T )\right) . \end{aligned}$$

It then follows from Gronwall’s inequality that

$$\begin{aligned} {\mathbb {E}}\left[ \vert X(s) - {\hat{X}}(s)\big \vert ^2\right]\leqslant & {} c_3 \left[ \sup _{0 \leqslant t \leqslant s} {\mathcal {E}}^2_0(t)\right] + c_4 \nonumber \\&+ (m+1) \sum ^M_{k=1} L_k^2 \int ^s_{t_{n_s}} \left( c_3 \Big [ \sup _{0 \leqslant t \leqslant \tau } {\mathcal {E}}^2_0(t)\Big ] + c_4\right) \text{e}^{(m+1) \sum ^M_{k=1} L_k^2 (s - \tau )} \mathrm {d}\tau \nonumber \\\leqslant & {} \left( c_3 \left[ \sup _{0 \leqslant t \leqslant s} {\mathcal {E}}^2_0(t)\right] + c_4 \right) \text{e}^{(m+1) \sum ^M_{k=1} L_k^2 \Delta t} \nonumber \\\leqslant & {} c \Delta t \left[ \sup _{0 \leqslant t \leqslant s} {\mathcal {E}}^2_0(t)\right] + c_T \Delta t, \end{aligned}$$

(A12)

where $c = 2(m+1) (\Delta t L^2_f + \sum _{k=M+1}^M L^2_k)\text{e}^{(m+1) \sum ^M_{k=1} L_k^2 }$ and $c_T$ is a generic constant dependent on $\Lambda _T$, m, $f^2(0)$, $g_k^2(0)$, $L_f^2$, $L_k^2$ but independent of $\Delta t$. Finally, inserting (A12) into (A8) results in the desired assertion by setting

$$\begin{aligned} C= 16 M (m+1) \left( \Delta t L^2_f + \sum _{k=M+1}^M L^2_k \right) \text{e}^{(m+1) \sum ^M_{k=1} L_k^2 } \sum _{k=1}^M L_k^2 . \end{aligned}$$

(A13)

The proof is complete.

1.5 A.5 Proof of Lemma 6

Consider integral representation of the piecewise continuous process $\eta (t)$

$$\begin{aligned} \eta (t) = \phi _n(t) \sum _{k=1}^M \left( b_{k,n} \left( - a_{k,n} \int ^t_{t_n} \mathrm {d}s + \int ^t_{t_n} \mathrm {d}W_k(s) \right) \right) , \quad t \in [t_n, t_{n+1}), \end{aligned}$$

(A14)

where $\phi _n(t)$ is defined in (24). Then

$$\begin{aligned} \vert \eta (t) - Y_n(t)\vert= & {} \phi _n(t) \sum _{k=1}^M \left| b_{k,n} \left( - a_{k,n} \int ^t_{t_n} (\phi ^{-1}_n(s) - 1) \mathrm {d}s \right. \right. \\&+ \left. \left. \int ^t_{t_n} (\phi ^{-1}_n(s) - 1) \mathrm {d}W_k(s) \right) \right| , \quad t \in [t_n, t_{n+1}), \end{aligned}$$

and it follows from the Cauchy-Bunyakovsky-Schwarz inequality and Hölder’s inequality that

$$\begin{aligned} \Big ({\mathbb {E}}\vert \eta (t) - Y_n(t)\vert \Big )^2\leqslant & {} 2 {\mathbb {E}}\left[ \phi ^2_n(t)\right] {\mathbb {E}}\left[ \sum ^M_{k=1} \left| b_{k,n} \left( - a_{k,n} \int ^t_{t_n} (\phi ^{-1}_n(s) - 1) \mathrm {d}s \right. \right. \right. \nonumber \\&\left. \left. \left. + \int ^t_{t_n} (\phi ^{-1}_n(s) - 1) \mathrm {d}W_k(s) \right) \right| ^2 \right] \nonumber \\\leqslant & {} 2 M {\mathbb {E}}\left[ \phi ^2_n(t)\right] \sum ^M_{k=1} {\mathbb {E}}\left[ b^2_{k,n} \left( a_{k,n}^2 \Delta t \int ^t_{t_n} (\phi ^{-1}_n(s) - 1)^2 \mathrm {d}s \right. \right. \nonumber \\&\left. \left. + \Bigg ( \int ^t_{t_n} (\phi ^{-1}_n(s) - 1) \mathrm {d}W_k(s) \Bigg )^2\right) \right] . \end{aligned}$$

(A15)

Noting that at each t, $\phi _n(t)$ follows a log-normal distribution, i.e.,

$$\begin{aligned} \ln \phi _n(t) \sim {\mathcal {N}}\left( -\frac{1}{2} \sum _{k=1}^M a_{k,n}^2(t-t_n), \sum _{k=1}^M a_{k,n}^2(t-t_n)\right) , \end{aligned}$$

(A16)

then by using Assumptions (A3) and (A2), the first term on the right-hand side of (A15) satisfies

$$\begin{aligned} {\mathbb {E}}\left[ \phi ^2_n(t)\right]= & {} \text{e}^{\sum _{k=1}^M a_{k,n}^2(t-t_n)} \leqslant 1 + \sum _{k=1}^M {\mathbb {E}}\left[ (g^\prime _k(x_n))^2\right] \Delta t + {\mathcal {O}}(\Delta t^2) \nonumber \\\leqslant & {} 1 + \lambda ^2 \sum _{k=1}^M \left( {\mathbb {E}}[(1 +\vert x_n\vert ^\gamma )^4]\right) ^{1/2} \Delta t + {\mathcal {O}}(\Delta t^2) \leqslant c. \end{aligned}$$

(A17)

Then using (A16) again, the two integrals on the right-hand side of (A15) satisfy respectively,

$$\begin{aligned} {\mathbb {E}}\left[ \int ^t_{t_n} (\phi ^{-1}_n(s) - 1)^2 \mathrm {d}s \right]\leqslant & {} 2 \int ^t_{t_n} {\mathbb {E}}\left[ \phi ^{-2}_n(s) + 1 \right] \mathrm {d}s \nonumber \\\leqslant & {} 2 (c + 1) \Delta t, \end{aligned}$$

(A18)

$$\begin{aligned} {\mathbb {E}}\left[ \Big ( \int ^t_{t_n} (\phi ^{-1}_n(s) - 1) \mathrm {d}W_k(s) \Big )^2 \right]= & {} {\mathbb {E}}\left[ \int ^t_{t_n} (\phi ^{-1}_n(s) - 1)^2 \mathrm {d}s\right] \nonumber \\\leqslant & {} 2(c+1) \Delta t. \end{aligned}$$

(A19)

Inserting (A17)–(A19) into (A15), and using again the boundedness of ${\mathbb {E}}[b_{k,n}^2]$ and ${\mathbb {E}}[a_{k,n}^2]$ implied by Assumptions (A1)–(A3) we obtain

$$\begin{aligned} \Big ({\mathbb {E}}[\vert \eta (t) - Y_n(t)\vert ]\Big )^2\leqslant & {} 2 M c \left\{ \Delta t {\mathbb {E}}[b_{k,n}^2] {\mathbb {E}}[a_{k,n}^2] {\mathbb {E}}\left[ \int ^t_{t_n} (\phi ^{-1}_n(s) - 1)^2 \mathrm {d}s \right] \right. \\&\left. + {\mathbb {E}}[b_{k,n}^2] {\mathbb {E}}\left[ \left( \int ^t_{t_n} (\phi ^{-1}_n(s) - 1) \mathrm {d}W_k(s) \right) ^2 \right] \right\} \nonumber \\\leqslant & {} 2 M c \left( (\Delta t)^2 + \Delta t \right) , \quad t \in [t_n, t_{n+1}), \,\, n = 0, \cdots , N, \end{aligned}$$

in which c is a generic constant independent of $\Delta t$ and may be different from line to line. It follows immediately that

$$\begin{aligned} \sup _{0 \leqslant t \leqslant T} {\mathbb {E}}[\vert \eta (t) - Y(t)\vert ] \leqslant C (\Delta t)^{1/2}, \end{aligned}$$

in which c is a constant depending on M, T, $\Lambda _T$, $L_k$, $D_k$, but is independent of $\Delta t$. The proof is complete.

1.6 A.6 Proof of Theorem 2

Note that since $x(t_n) = x_n$ for $n = 0, 1, \cdots , N$, the weak discretization error above satisfies

$$\begin{aligned} {{\mathfrak {E}}^w}&: =\big \vert {\mathbb {E}}[ \psi (x(T)) ] - {\mathbb {E}}[\psi (X(T))] \big \vert \\&\leqslant {} \big \vert {\mathbb {E}}[ \psi (y(T)) ] - {\mathbb {E}}[\psi (X(T))] \big \vert + \big \vert {\mathbb {E}}[ \psi (x(T)) ] - {\mathbb {E}}[\psi (y(T))] \big \vert , \end{aligned}$$

where x(t) and y(t) satisfy (33) and (39), respectively. Similar to the proof of strong convergence, we will first estimate

$$\begin{aligned} {{\mathfrak {E}}^w_1} := \big \vert {\mathbb {E}}[ \psi (y(T)) ] - {\mathbb {E}}[\psi (X(T))] \big \vert . \end{aligned}$$

To that end, let u(t, y) be a solution of the following Feynman-Kac partial differential equation:

$$\begin{aligned} u_t(t, y) + f(y)u_y(t, y) +\frac{1}{2} u_{yy}(t, y) \sum ^m_{k=1} g^2_k(y) = 0 \quad \text{ for } \, t\in [0, T],\,\, y \in {\mathbb {R}} \end{aligned}$$

with $u(T, y) = \psi (y).$ Applying Itô’s formula to u(t, y(t)) with y(t) satisfying (39) and using the above equation yields

$$\begin{aligned} \mathrm {d}u(t, y(t))= & {} \left( u_t(t, y(t)) + u_y(t, y(t)) f({\tilde{x}}(t)) \right) \mathrm {d}t \nonumber \\&+ \frac{1}{2} u_{yy}(t, y(t)) \left( \sum ^m_{k=M+1} g^2_k({\tilde{x}}(t)) \right. \\&\left. + \sum ^M_{k=1} \Big ( g_k({\tilde{x}}(t)) + g^\prime _k({\tilde{x}}(t))({\hat{X}}(t) - {\tilde{x}}(t))\Big )^2 \right) \mathrm {d}t \nonumber \\&+ u_y(t, y(t)) \left( \sum ^m_{k=M+1} g_k({\tilde{x}}(t)) \mathrm {d}W_k(t) \right. \\&\left. + \sum ^M_{k=1} \Big ( g_k({\tilde{x}}(t)) + g^\prime _k({\tilde{x}}(t))({\hat{X}}(t) - {\tilde{x}}(t))\Big ) \mathrm {d}W_k(t)\right), \\ \mathrm {d}u(t, y(t))= & {} \Bigg ( u_y(t, y(t)) \left( f({\tilde{x}}(t)) - f(y(t) \right) \\&+ \frac{1}{2} u_{yy}(t, y(t)) \left( \sum ^m_{k=M+1} \Big (g_k^2({\tilde{x}}(t)) - g_k^2(y(t)\Big ) \right) \Bigg ) \mathrm {d}t \\&+ \frac{1}{2} u_{yy}(t, y(t)) \sum _{k=1}^M \Big ( \Big ( g_k({\tilde{x}}(t)) + g^\prime _k({\tilde{x}}(t))({\hat{X}}(t) - {\tilde{x}}(t))\Big )^2 \\&- g^2_k(y(t)) \Big ) \mathrm {d}t \nonumber \\&+ u_y(t, y(t)) \Bigg (\sum ^m_{k=M+1} g_k({\tilde{x}}(t)) \mathrm {d}W_k(t) \\&+ \sum ^M_{k=1} \Big ( g_k({\tilde{x}}(t)) + g^\prime _k({\tilde{x}}(t))({\hat{X}}(t) - {\tilde{x}}(t))\Big ) \mathrm {d}W_k(t)\Bigg ). \end{aligned}$$

Notice that due to the Feynman-Kac formula, we have $u(0, x_0) = {\mathbb {E}}[\psi (X(T))]$. Then, integrating the above equation from 0 to T using $u(T, x_N) = \psi (x_N)$ and taking the expectation of the resulting equation gives

$$\begin{aligned}&\quad \, {\mathbb {E}}[ \psi (x_N) ] - {\mathbb {E}}[\psi (y(T))]\\& = {\mathbb {E}}\left[ \int ^T_0 \Bigg (u_y(t, y(t)) \left( f({\tilde{x}}(t)) - f(y(t)) \right) \right. \\& \quad \left. + \frac{1}{2} u_{yy}(t, y(t)) \sum ^m_{k=M+1} \Big (g_k^2({\tilde{x}}(t)) - g_k^2(y(t))\Big ) \Bigg ) \mathrm {d}t \right] \\& \quad + \frac{1}{2}{\mathbb {E}}\left[ \int ^T_0 u_{yy}(t, y(t)) \sum _{k=1}^M \left( \Big ( g_k({\tilde{x}}(t)) + g^\prime _k({\tilde{x}}(t))({\hat{X}}(t) - {\tilde{x}}(t))\Big )^2 - g^2_k(y(t)) \right) \mathrm {d}t \right] , \end{aligned}$$

which implies that

$$\begin{aligned} {{\mathfrak {E}}^w_1} \leqslant \int ^T_0 \left| {\mathbb {E}}[{\mathfrak {e}}_1(t, y(t))]\right| \mathrm {d}t + \int ^T_0 \left| {\mathbb {E}}[{\mathfrak {e}}_2(t, y(t))]\right| \mathrm {d}t +\int ^T_0 \left| {\mathbb {E}}[{\mathfrak {e}}_3(t, y(t))]\right| \mathrm {d}t , \end{aligned}$$

(A20)

where

$$\begin{aligned} {\mathfrak {e}}_1(t, y(t))= & {} u_y(t, y(t)) \left( f({\tilde{x}}(t)) - f(y(t)) \right) , \\ {\mathfrak {e}}_2(t, y(t))= & {} u_{yy}(t, y(t)) \sum ^m_{k=M+1} \Big (g_k^2({\tilde{x}}(t)) - g_k^2(y(t))\Big ), \\ {\mathfrak {e}}_3(t, y(t))= & {} u_{yy}(t, y(t)) \sum _{k=1}^M \left( \Big ( g_k({\tilde{x}}(t)) + g^\prime _k({\tilde{x}}(t))({\hat{X}}(t) - {\tilde{x}}(t))\Big )^2 - g^2_k(y(t)) \right) . \end{aligned}$$

Note that ${\mathfrak {e}}_1(t_n, y(t_n)) = {\mathfrak {e}}_2(t_n, y(t_n)) = {\mathfrak {e}}_3(t_n, y(t_n)) = 0.$ We next estimate each of ${\mathfrak {e}}_1$, ${\mathfrak {e}}_2$, and ${\mathfrak {e}}_3$.

First apply the Itô formula to ${\mathfrak {e}}_1(t, x(t))$ to obtain

$$\begin{aligned} \mathrm {d}{\mathfrak {e}}_1(t, y(t))= & {} \left( \frac{\partial {\mathfrak {e}}_1}{\partial t} + \frac{\partial {\mathfrak {e}}_1}{\partial y} f({\tilde{x}}(t)) \right. \nonumber \\&+ \frac{1}{2} \frac{\partial ^2 {\mathfrak {e}}_1}{\partial y^2} \Big ( \sum ^m_{k=M+1} g^2_k({\tilde{x}}(t)) \nonumber \\&\left. + \sum ^M_{k=1} \big ( g_k({\tilde{x}}(t)) + g^\prime _k({\tilde{x}}(t))({\hat{X}}(t) - {\tilde{x}}(t))\big )^2 \Big ) \right) \mathrm {d}t \nonumber \\&+ \frac{\partial {\mathfrak {e}}_1}{\partial y} \sum _{k=1}^m {\mathcal {G}}_k \mathrm {d}W_k(t), \end{aligned}$$

(A21)

where

$$\begin{aligned} {\mathcal {G}}_k = \left\{ \begin{array}{ll} g_k({\tilde{x}}(t)) + g^\prime _k({\tilde{x}}(t))({\hat{X}}(t) - {\tilde{x}}(t)) &{} \,\, \text{ for }\,\, k=1, \cdots , M, \\ g_k({\tilde{x}}(t)) &{} \,\, \text{ for }\,\, k=M+1, \cdots , m . \end{array} \right. \end{aligned}$$

Integrating (A21) from $t_n$ to $t \in [t_n, t_{n+1})$ using ${\mathfrak {e}}_1(t_n, y(t_n)) = 0$ then taking expectation of the resulting equation gives

$$\begin{aligned} {\mathbb {E}}[{\mathfrak {e}}_1(t, y(t)]= & {} \int ^t_{t_n} \Bigg \{{\mathbb {E}}\Bigg [ \frac{\partial {\mathfrak {e}}_1}{\partial t}(s, y(s)) \Bigg ] + {\mathbb {E}}\Bigg [ \frac{\partial {\mathfrak {e}}_1}{\partial y}(s, y(s)) f(x_n) \Bigg ] \\&+ \frac{1}{2} {\mathbb {E}}\left[ \frac{\partial ^2 {\mathfrak {e}}_1}{\partial y^2} \sum ^m_{k=M+1} g^2_k(x_n) \right] \Bigg \}\mathrm {d}s \nonumber \\&+ \frac{1}{2} \int ^t_{t_n} {\mathbb {E}}\left[ \frac{\partial ^2 {\mathfrak {e}}_1}{\partial y^2}\sum ^M_{k=1} \big ( g_k(x_n) + g^\prime _k(x_n)({\hat{X}}(s) - x_n)\big )^2\right] \mathrm {d}s. \end{aligned}$$

Then by the Cauchy inequality, the Itô isometry, and (5) we have

$$\begin{aligned} {\mathbb {E}}[{\mathfrak {e}}_1(t, y(t)]\leqslant & {} \int ^t_{t_n} \left\{ {\mathbb {E}}\Bigg [ \frac{\partial {\mathfrak {e}}_1}{\partial t}(s, y(s)) \Bigg ] + {\mathbb {E}}\Bigg [ \frac{\partial {\mathfrak {e}}_1}{\partial x}(s, y(s)) f(x_n) \Bigg ] \right. \\&+ \left. {\mathbb {E}}\left[ \frac{\partial ^2 {\mathfrak {e}}_1}{\partial x^2} \left( \sum ^m_{k=1} g^2_k(x_n) + \sum _{k=1}^M (g^\prime (x_n))^2 \int ^s_{t_n} \sum ^M_{k=1} g^2_k({\hat{X}}(\tau )) \mathrm {d}\tau \right) \right] \right\} \mathrm {d}s. \end{aligned}$$

For simplicity, assume that the functions f and $g_k$ satisfy conditions such that all expectations appearing in the above inequality are bounded. Then there exists $C_1 > 0$ such that

$$\begin{aligned} \vert {\mathbb {E}}[{\mathfrak {e}}_1(t, x(t)]\vert \leqslant C_1 \Delta t . \end{aligned}$$

(A22)

Following similar analysis we can obtain that there exist $C_2 > 0$ and $C_3 > 0$ such that

$$\begin{aligned} \vert {\mathbb {E}}[{\mathfrak {e}}_2(t, x(t)]\vert \leqslant C_2 \Delta t , \quad \vert {\mathbb {E}}[{\mathfrak {e}}_3(t, x(t)]\vert \leqslant C_3 \Delta t. \end{aligned}$$

(A23)

In the end, inserting (A22)–(A23) into (A20) we have that there exists $C_T>0$ such that ${\mathfrak {E}}_1 \leqslant C_T \Delta t$.

Note that in the FPM-LP scheme, x(t) is essentially an EM approximation of y(t), along with an exponential approximation of a strong convergence order 3/2 implied by (45). It then follows immediately that there exists $C_T>0$ such that the error

$$\begin{aligned} {{\mathfrak {E}}^w_2}: = \big \vert {\mathbb {E}}[ \psi (x(T)) ] - {\mathbb {E}}[\psi (y(T))] \big \vert \leqslant C_T \Delta t, \end{aligned}$$

which implies that the FPM-LP scheme has a weak order of convergence 1. The proof is complete.