# Numerical Picard Iteration Methods for Simulation of Non-Lipschitz Stochastic Differential Equations

## Abstract

**:**

## 1. Introduction

**Assumption**

**1.**

## 2. Numerical Analysis of the Splitting Approaches

- Homogeneous equation:$$\begin{array}{c}dX(t)=X(t){a}_{1}(t,X(t))\phantom{\rule{0.277778em}{0ex}}dt+X(t)b(t,X(t))d{W}_{t},\phantom{\rule{0.277778em}{0ex}}t\in \times [0,T],\hfill \end{array}$$$$\begin{array}{c}X(0)={X}_{0},\hfill \end{array}$$
- Inhomogeneous equation:$$\begin{array}{c}dX(t)=X(t)({a}_{1}(t,X(t))+{a}_{2}(t,X(t)))\phantom{\rule{0.277778em}{0ex}}dt+X(t)b(t,X(t))d{W}_{t},\phantom{\rule{0.277778em}{0ex}}t\in \times [0,T],\hfill \end{array}$$$$\begin{array}{c}X(0)={X}_{0},\hfill \end{array}$$

- Approximate the diffusion process,
- Picard iterations with Doléans-Dade solutions of the SDE,
- Discretisation of the Picard iterations in time.

- Discretisation of the Picard iterations in time for the inhomogeneous part,
- Approximation of the integral-formulation of the inhomogeneous part.

#### 2.1. Homogeneous Equation

#### 2.1.1. Approximate the Diffusion Process

#### 2.1.2. Picard Iterations with Doléans-Dade Solutions of the SDE

**Lemma**

**1.**

**Proof.**

**Assumption**

**2.**

#### 2.1.3. Discretisation of the Picard Iterations in Time

#### 2.2. Inhomogeneous Equation

#### 2.2.1. Discretisation of the Picard Iterations in Time for the Inhomogeneous Part

#### 2.2.2. Approximation of the Integral-Formulation of the Inhomogeneous Part

**Lemma**

**2.**

**Proof.**

**Example**

**1.**

## 3. Numerical Examples

- AB-splitting approaches (AB), see [26],

- Iterative Picard-splitting with Doléans-Dade exponential approach (Picard-Splitt-Doleans), see Section 2.

#### 3.1. First Example: Nonlinear SDE With Root-Function or Irrational Function

- Euler-Maruyama-Scheme$$\begin{array}{c}X({t}_{i+1})=X({t}_{i})+\left(\frac{2}{5}{X}^{3/5}({t}_{i})+5{X}^{4/5}({t}_{i})\right)\Delta t+{X}^{4/5}({t}_{i})\Delta {W}_{t},\hfill \end{array}$$We have $i=0,\dots ,N-1$ with $X({t}_{0})=1.0$.
- Milstein-Scheme$$\begin{array}{c}X({t}_{i+1})=X({t}_{i})+\left(\frac{2}{5}{X}^{3/5}({t}_{i})+5{X}^{4/5}({t}_{i})\right)\Delta t+{X}^{4/5}({t}_{i})\Delta {W}_{t}\hfill \\ +\frac{2}{5}{X}^{3/5}({t}_{i})({(\Delta {W}_{t})}^{2}-\Delta t),\hfill \end{array}$$We have $i=0,\dots ,N-1$ with $X({t}_{0})=1.0$.
- AB-splitting approach:We initialize ${t}_{i}$ with $i=0,\dots ,N-1$, while ${t}_{N}=T$ and we have $X(0)$ is the initial condition.We deal with the 2 steps:
- A-step:$$\begin{array}{c}\tilde{X}({t}_{i+1})=X({t}_{i})+\left(\frac{2}{5}{X}^{3/5}({t}_{i})+5{X}^{4/5}({t}_{i})\right)\Delta t,\hfill \end{array}$$
- B-part:$$\begin{array}{c}X({t}_{i+1})=\tilde{X}({t}_{i+1})+{\tilde{X}}^{4/5}({t}_{i+1})\Delta {W}_{t}+\frac{2}{5}{\tilde{X}}^{3/5}({t}_{i+1})({(\Delta {W}_{t})}^{2}-\Delta t),\hfill \end{array}$$

- ABA-splitting approach:We initialize ${t}_{i}$ with $i=0,\dots ,N-1$, while ${t}_{N}=T$ and we have $X(0)$ is the initial condition.We deal with the 2 steps:
- A-step ($\Delta t/2$):$$\begin{array}{c}\tilde{X}({t}_{i+1})=X({t}_{i})+\left(\frac{2}{5}{X}^{3/5}({t}_{i})+5{X}^{4/5}({t}_{i})\right)\Delta t/2,\hfill \end{array}$$
- B-part ($\Delta t$):$$\begin{array}{c}\widehat{X}({t}_{i+1})=\tilde{X}({t}_{i+1})+{\tilde{X}}^{4/5}({t}_{i+1})\Delta {W}_{t}+\frac{2}{5}{\tilde{X}}^{3/5}({t}_{i+1})({(\Delta {W}_{t})}^{2}-\Delta t),\hfill \end{array}$$
- A-step ($\Delta t/2$):$$\begin{array}{c}X({t}_{i+1})=\widehat{X}({t}_{i})+\left(\frac{2}{5}{\widehat{X}}^{3/5}({t}_{i})+5{\widehat{X}}^{4/5}({t}_{i})\right)\Delta t/2,\hfill \end{array}$$

- Iterative Picard approach:$$\begin{array}{c}dX(t)=(\frac{2}{5}{X}^{3/5}+5{X}^{4/5})dt+{X}^{4/5}d{W}_{t},\phantom{\rule{0.277778em}{0ex}}t\in \times [0,1],\hfill \\ X(0)=1,\hfill \end{array}$$$$\begin{array}{c}d{X}_{i}(t)=(\frac{2}{5}{X}_{i-1}^{3/5}+5{X}_{i-1}^{4/5})dt+{X}_{i-1}^{4/5}d{W}_{t},\phantom{\rule{0.277778em}{0ex}}t\in \times [0,1],\hfill \\ X(0)=1,\hfill \end{array}$$The algorithm is given as: We initialize ${t}_{n}$ with $n=0,\dots ,N-1$, while ${t}_{N}=T$ and we have $X(0)$ is the initial condition.We deal with the 2 loops (loop 1 is the computation over the full time-domain and loop 2 is the computation with $i=1,2,\dots $):
- $n=0,\dots ,N-1$:
- $i=1,\dots ,I$:
- Computation$$\begin{array}{c}X{({t}_{n+1})}^{i}=X({t}_{n})+\left(\frac{2}{5}{X}^{3/5}{({t}_{n+1})}^{i-1}+5{X}^{4/5}{({t}_{n+1})}^{i-1}\right)\Delta t\hfill \\ +{X}^{4/5}({t}_{n})\Delta {W}_{t}+\frac{2}{5}{X}^{3/5}({t}_{n})({(\Delta {W}_{t})}^{2}-\Delta t),\hfill \end{array}$$
- $i=i+1$, if $i=I+1$ then we are done, else we go to Step (c)
- $n=n+1$, if $n+1=N+1$ then we are done, else we go to Step (b)

- Iterative Picard with Doléans-Dade exponential approach:$$\begin{array}{c}dX(t)=(\frac{2}{5}{X}^{3/5}+5{X}^{4/5})dt+{X}^{4/5}d{W}_{t},\phantom{\rule{0.277778em}{0ex}}t\in \times [0,1],\hfill \\ X(0)=1,\hfill \end{array}$$$$\begin{array}{c}d{X}_{i}(t)=(\frac{2}{5}{X}_{i-1}^{-2/5}+5{X}_{i-1}^{-1/5}){X}_{i}dt+{X}_{i-1}^{-1/5}{X}_{i}d{W}_{t},\phantom{\rule{0.277778em}{0ex}}t\in \times [0,1],\hfill \\ X(0)=1,\hfill \end{array}$$The algorithm is given as: We initialize ${t}_{n}$ with $n=0,\dots ,N-1$, while ${t}_{N}=T$ and we have $X(0)$ is the initial condition.We deal with the 2 loops (loop 1 is the computation over the full time-domain and loop 2 is the computation with $i=1,2,\dots $):
- $n=0,\dots ,N-1$:
- $i=1,\dots ,I$:
- Computation$$\begin{array}{c}X{({t}_{n+1})}^{i}=X({t}_{n})exp\left(\Delta t\frac{2}{5}\frac{\left({X}^{-2/5}{({t}_{n+1})}^{i-1}+{X}^{-2/5}({t}_{n})\right)}{2}\right.\hfill \\ +\Delta t\phantom{\rule{0.277778em}{0ex}}5\frac{\left({X}^{-1/5}{({t}_{n+1})}^{i-1}+{X}^{-1/5}({t}_{n})\right)}{2}-\Delta t\frac{1}{2}\frac{\left({X}^{-2/5}{({t}_{n+1})}^{i-1}+{X}^{-2/5}({t}_{n})\right)}{2}\hfill \\ \left.+{X}^{-1/5}({t}_{n})\Delta {W}_{t}+\frac{1}{2}{X}^{-2/5}({t}_{n})({(\Delta {W}_{t})}^{2}-\Delta t)\right),\hfill \end{array}$$
- $i=i+1$, if $i=I+1$ then we are done, else we go to Step (c)
- $n=n+1$, if $n+1=N+1$ then we are done, else we go to Step (b)

- Iterative Picard-Splitting with Doléans-Dade exponential approach:$$\begin{array}{c}dX(t)=(\frac{2}{5}{X}^{3/5}+5{X}^{4/5})dt+{X}^{4/5}d{W}_{t},\phantom{\rule{0.277778em}{0ex}}t\in \times [0,1],\hfill \\ X(0)=1,\hfill \end{array}$$$$\begin{array}{c}A(X(t))=\frac{2}{5}{X}^{3/5}+5{X}^{4/5},\hfill \\ {A}_{1}(X(t))=5{X}^{4/5},\phantom{\rule{0.277778em}{0ex}}{A}_{2}(X(t))=\frac{2}{5}{X}^{3/5},\hfill \end{array}$$We apply the Picard-iterations with Doléans-Dade exponential approach and the splitting approach:$$\begin{array}{c}d{X}_{i}(t)=(5{X}_{i-1}^{-1/5}\phantom{\rule{0.277778em}{0ex}}{X}_{i}dt+{X}_{i-1}^{-1/5}\phantom{\rule{0.277778em}{0ex}}{X}_{i}d{W}_{t})+\frac{2}{5}{X}_{i-1}^{3/5}+,\phantom{\rule{0.277778em}{0ex}}t\in \times [0,1],\hfill \\ X(0)=1,\hfill \end{array}$$The algorithm is given as: We initialize ${t}_{n}$ with $n=0,\dots ,N-1$, while ${t}_{N}=T$ and we have $X(0)$ is the initial condition.We deal with the 2 loops (loop 1 is the computation over the full time-domain and loop 2 is the computation with $i=1,2,\dots $):
- $n=0,\dots ,N-1$:
- $i=1,\dots ,I$:
- Computation (we apply the full exp):$$\begin{array}{c}\hfill \begin{array}{c}\phantom{\rule{-42.67912pt}{0ex}}X{({t}_{n+1})}^{i}=X({t}_{n})exp(\Delta t\phantom{\rule{0.277778em}{0ex}}5\frac{\left({X}^{-1/5}{({t}_{n+1})}^{i-1}+{X}^{-1/5}({t}_{n})\right)}{2}-\Delta t\frac{1}{2}\frac{\left({X}^{-2/5}{({t}_{n+1})}^{i-1}+{X}^{-2/5}({t}_{n})\right)}{2}\hfill \\ +{X}^{-1/5}({t}_{n})\Delta {W}_{t}+\frac{1}{2}{X}^{-2/5}({t}_{n})({(\Delta {W}_{t})}^{2}-\Delta t))\hfill \\ +\Delta texp(\Delta t\phantom{\rule{0.277778em}{0ex}}5\frac{\left({X}^{-1/5}{({t}_{n+1})}^{i-1}+{X}^{-1/5}({t}_{n})\right)}{2}-\Delta t\frac{1}{2}\frac{\left({X}^{-2/5}{({t}_{n+1})}^{i-1}+{X}^{-2/5}({t}_{n})\right)}{2}\hfill \\ +{X}^{-1/5}({t}_{n})\Delta {W}_{t}+\frac{1}{2}{X}^{-2/5}({t}_{n})({(\Delta {W}_{t})}^{2}-\Delta t))\hfill \\ \xb7\frac{2}{5}\frac{\left({X}^{-2/5}{({t}_{n+1})}^{i-1}+{X}^{-2/5}({t}_{n})\right)}{2}.\hfill \end{array}\end{array}$$
- $i=i+1$, if $i=I+1$ then we are done, else we go to Step (c)
- $n=n+1$, if $n+1=N+1$ then we are done, else we go to Step (b)

- Mean value at $t={t}^{n}$ and J-sample paths:$$\begin{array}{c}\hfill E({X}_{method}({t}_{n}))=\frac{1}{J}\sum _{j=1}^{J}{X}_{method}^{j}({t}_{n}),\end{array}$$
- Local mean square error value at $t={t}_{n}$ and J-sample paths:$$\begin{array}{c}Var{({X}_{method}({t}_{n}))}_{local}=\frac{1}{J}\sum _{j=1}^{J}{|E({X}_{method}({t}_{n}))-{X}_{method}^{j}({t}_{n})|}^{2},\hfill \end{array}$$
- Global means square error over the full time-scale with different time-steps $\Delta t$ and J-sample path:$$\begin{array}{c}Var{({X}_{method})}_{global}=\sum _{n=0}^{N}\Delta t\phantom{\rule{0.277778em}{0ex}}Var{({X}_{method}({t}_{n}))}_{local},\hfill \end{array}$$

**Remark**

**1.**

**Remark**

**2.**

#### 3.2. Second Example: Linear/Nonlinear SDE with Potential Function

- Euler-Maruyama-Scheme$$\begin{array}{c}X({t}_{i+1})=X({t}_{i})+\left(2{X}^{1/2}({t}_{i})+1\right)\Delta t+2{X}^{1/2}({t}_{i})\phantom{\rule{0.277778em}{0ex}}\Delta {W}_{t},\hfill \end{array}$$We have $i=0,\dots ,N-1$ with $X({t}_{0})=1.0$.
- Milstein-Scheme$$\begin{array}{c}X({t}_{i+1})=X({t}_{i})+\left(2{X}^{1/2}({t}_{i})+1\right)\Delta t+2{X}^{1/2}({t}_{i})\phantom{\rule{0.277778em}{0ex}}\Delta {W}_{t}+\frac{1}{2}({(\Delta {W}_{t})}^{2}-\Delta t),\hfill \end{array}$$We have $i=0,\dots ,N-1$ with $X({t}_{0})=1.0$.
- AB-splitting approach:We initialize ${t}_{i}$ with $i=0,\dots ,N-1$, while ${t}_{N}=T$ and we have $X(0)$ is the initial condition.We deal with the 2 steps:
- A-step:$$\begin{array}{c}\tilde{X}({t}_{i+1})=X({t}_{i})+\left(2{X}^{1/2}({t}_{i})+1\right)\Delta t,\hfill \end{array}$$
- B-part:$$\begin{array}{c}X({t}_{i+1})=\tilde{X}({t}_{i+1})+2{\tilde{X}}^{1/2}({t}_{i+1})\phantom{\rule{0.277778em}{0ex}}\Delta {W}_{t}+\frac{1}{2}({(\Delta {W}_{t})}^{2}-\Delta t),\hfill \end{array}$$

- ABA-splitting approach:We deal with the 2 steps:
- A-step ($\Delta t/2$):$$\begin{array}{c}\hfill \tilde{X}({t}_{i+1})=X({t}_{i})+\left(2{X}^{1/2}({t}_{i})+1\right)\Delta t/2,\end{array}$$
- B-part ($\Delta t$):$$\begin{array}{c}\widehat{X}({t}_{i+1})=\tilde{X}({t}_{i+1})+2{\tilde{X}}^{1/2}({t}_{i+1})\phantom{\rule{0.277778em}{0ex}}\Delta {W}_{t}+2({(\Delta {W}_{t})}^{2}-\Delta t),\hfill \end{array}$$
- A-step ($\Delta t/2$):$$\begin{array}{c}\hfill X({t}_{i+1})=\widehat{X}({t}_{i+1})+\left(2{\widehat{X}}^{1/2}({t}_{i+1})+1\right)\Delta t/2,\end{array}$$

- Iterative Picard approach:$$\begin{array}{c}dX(t)=(2{X}^{1/2}+1)dt+2{X}^{1/2}d{W}_{t},\phantom{\rule{0.277778em}{0ex}}t\in \times [0,1],\hfill \\ X(0)=1,\hfill \end{array}$$$$\begin{array}{c}d{X}_{i}(t)=(2{X}_{i-1}^{1/2}+1)dt+2{X}_{i-1}^{1/2}d{W}_{t},\phantom{\rule{0.277778em}{0ex}}t\in \times [0,1],\hfill \\ X(0)=1,\hfill \end{array}$$The algorithm is given as:We initialize ${t}_{n}$ with $n=0,\dots ,N-1$, while ${t}_{N}=T$ and we have $X(0)$ is the initial condition.
- $n=0,\dots ,N-1$:
- $i=1,\dots ,I$:
- Computation$$\begin{array}{c}{X}_{i}({t}^{n+1})=X({t}^{n})+(2{X}_{i-1}{({t}^{n+1})}^{1/2}+1)\Delta t\hfill \\ +2{X}^{1/2}({t}^{n})\phantom{\rule{0.277778em}{0ex}}\Delta {W}_{t}+\frac{1}{2}({(\Delta {W}_{t})}^{2}-\Delta t),\phantom{\rule{0.277778em}{0ex}}t\in \times [0,1],\hfill \\ X(0)=1,\hfill \end{array}$$
- $i=i+1$, if $i=I+1$ then we are done, else we go to Step (c)
- $n=n+1$, if $n+1=N+1$ then we are done, else we go to Step (b)

- Iterative Picard-splitting with Doléans-Dade exponential approach:$$\begin{array}{c}dX(t)=(2{X}^{1/2}+1)dt+2{X}^{1/2}d{W}_{t},\phantom{\rule{0.277778em}{0ex}}t\in \times [0,1],\hfill \\ X(0)=1,\hfill \end{array}$$$$\begin{array}{c}A(X(t))=2{X}^{1/2}+1,\hfill \\ {A}_{1}(X(t))=2{X}^{1/2},\phantom{\rule{0.277778em}{0ex}}{A}_{2}(X(t))=1,\hfill \end{array}$$We apply an Picard with Doléans-Dade exponential approach and right hand side$$\begin{array}{c}d{X}_{i}(t)=(2{X}_{i-1}^{-1/2}){X}_{i}dt+2{X}_{i-1}^{-1/2}{X}_{i}d{W}_{t}+1\phantom{\rule{0.277778em}{0ex}}dt,\phantom{\rule{0.277778em}{0ex}}t\in \times [0,1],\hfill \\ X(0)=1,\hfill \end{array}$$
- $n=0,\dots ,N-1$:
- $i=1,\dots ,I$:
- Computation (we apply Version 1 or Version 2)
- Computation (we apply the full exp and integration of the RHS)$$\begin{array}{c}X{({t}_{n+1})}^{i}=X({t}_{n})exp(\Delta t\phantom{\rule{0.277778em}{0ex}}\frac{\left(2{X}^{-1/2}{({t}_{n+1})}^{i-1}+2{X}^{-1/2}({t}_{n})\right)}{2}\hfill \\ -\Delta t\frac{\left({X}^{-1}{({t}_{n+1})}^{i-1}+{X}^{-1}({t}_{n})\right)}{2}\hfill \\ +2\phantom{\rule{0.277778em}{0ex}}{X}^{-1/2}({t}_{n})\Delta {W}_{t}+\phantom{\rule{0.277778em}{0ex}}\frac{1}{2}{X}^{-1}({t}_{n})({(\Delta {W}_{t})}^{2}-\Delta t))+\Delta t.\hfill \end{array}$$
- $i=i+1$, if $i=I+1$ then we are done, else we go to Step (c)
- $n=n+1$, if $n+1=N+1$ then we are done, else we go to Step (b)

**Remark**

**3.**

**Remark**

**4.**

## 4. Conclusions

## Funding

## Acknowledgments

## Conflicts of Interest

## Appendix A

#### Appendix A.1. Additional Proofs

**Lemma**

**A1.**

**Proof.**

#### Appendix A.2. Approximated exp-Functions

- Version with reduced $exp$ in Example 1: We replace the Equation (27) with the reduced exponential function, see:$$\begin{array}{c}\hfill \begin{array}{c}\phantom{\rule{-56.9055pt}{0ex}}X{({t}_{n+1})}^{i}=\left(X({t}_{n})+\Delta t\phantom{\rule{0.277778em}{0ex}}\frac{2}{5}\frac{\left({X}^{-2/5}{({t}_{n+1})}^{i-1}+{X}^{-2/5}({t}_{n})\right)}{2}\right)\hfill \\ \xb7(1+\Delta t\phantom{\rule{0.277778em}{0ex}}5\frac{\left({X}^{-1/5}{({t}_{n+1})}^{i-1}+{X}^{-1/5}({t}_{n})\right)}{2}-\Delta t\frac{1}{2}\frac{\left({X}^{-2/5}{({t}_{n+1})}^{i-1}+{X}^{-2/5}({t}_{n})\right)}{2}\hfill \\ +{X}^{-1/5}({t}_{n})\Delta {W}_{t}+\frac{1}{2}{X}^{-2/5}({t}_{n})({(\Delta {W}_{t})}^{2}-\Delta t)),\hfill \end{array}\end{array}$$
- Version with reduced $exp$ in Example 2: We replace the Equation (34) with the reduced exponential function and integration of the RHS, see:$$\begin{array}{c}X{({t}_{n+1})}^{i}=X({t}_{n})(1+\Delta t\phantom{\rule{0.277778em}{0ex}}\frac{\left(2{X}^{-1/2}{({t}_{n+1})}^{i-1}+2{X}^{-1/2}({t}_{n})\right)}{2}\hfill \\ -\Delta t\frac{\left(1{X}^{-1}{({t}_{n+1})}^{i-1}+1{X}^{-1}({t}_{n})\right)}{2}\hfill \\ +2\phantom{\rule{0.277778em}{0ex}}{X}^{-1/2}({t}_{n})\Delta {W}_{t}+\frac{1}{2}{X}^{-1}({t}_{n})({(\Delta {W}_{t})}^{2}-\Delta t))+\Delta t.\hfill \end{array}$$

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**Figure 2.**Results of the mean values are computed with Equation (29) and they are presented for the different methods with the following number of time steps $N=512,1024,2048,4096$.

**Figure 3.**Results of mean square errors are computed with Equation (31) and they are presented for the different methods with the following number of time steps $N=512,1024,2048,4096$.

**Figure 4.**Performance of the different schemes, where we compute the mean error values of the different methods at $t=1.0$ with $J=100$ samples and different number of time-steps with $N=500,1000,1500,2000,2500,3000,3500,4000$.

**Figure 5.**Results of the mean values, which are computed with Equation (29) and they are presented for the different methods with the number of time steps $N=512,1024,2048,4096$.

**Figure 6.**Results of mean square errors, which are computed with Equation (31) and they are presented for the different methods with the number of time steps $N=512,1024,2048,4096$.

**Figure 7.**Performance of the different schemes, where we compute the mean error values of the different methods at $t=1.0$ with $J=100$ samples and different number of time-steps with $N=500,1000,1500,2000,2500,3000,3500,4000$.

N | EM- | Milstein | AB | ABA | Picard | Picard- | Picard-Splitt- |
---|---|---|---|---|---|---|---|

Doleans | Doleans | ||||||

$\mathit{i}=\mathbf{2}$ | $\mathit{i}=\mathbf{2}$ | $\mathit{i}=\mathbf{2}$ | |||||

${2}^{9}$ | $33.9535$ | $35.5878$ | $35.4419$ | $34.6244$ | $40.6004$ | $37.8312$ | $31.3049$ |

${2}^{10}$ | $34.4942$ | $35.4737$ | $34.1016$ | $34.529$ | $40.0732$ | $37.9911$ | $32.2207$ |

${2}^{11}$ | $34.7694$ | $34.2016$ | $35.2721$ | $34.0984$ | $40.6338$ | $37.9833$ | $30.5993$ |

${2}^{12}$ | $34.2202$ | $36.2405$ | $35.1938$ | $35.8434$ | $40.1492$ | $37.9433$ | $31.4823$ |

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**MDPI and ACS Style**

Geiser, J.
Numerical Picard Iteration Methods for Simulation of Non-Lipschitz Stochastic Differential Equations. *Symmetry* **2020**, *12*, 383.
https://doi.org/10.3390/sym12030383

**AMA Style**

Geiser J.
Numerical Picard Iteration Methods for Simulation of Non-Lipschitz Stochastic Differential Equations. *Symmetry*. 2020; 12(3):383.
https://doi.org/10.3390/sym12030383

**Chicago/Turabian Style**

Geiser, Jürgen.
2020. "Numerical Picard Iteration Methods for Simulation of Non-Lipschitz Stochastic Differential Equations" *Symmetry* 12, no. 3: 383.
https://doi.org/10.3390/sym12030383