求Julia库中ESRK1与RSwM1实现代码，排查ESRK1收敛异常

我正在用Python实现Rackauckas & Nie（2017）提出的SDE求解器ESRK1及自适应步长算法RSwM1，以此验证对算法的理解。但在ESRK1实现阶段出现问题：用几何布朗运动的简单SDE测试时，时间步长dt不断减小的情况下解并未收敛，说明代码存在错误。

我了解到这些算法已在Julia的DifferentialEquations.jl库中实现，希望通过查看Julia代码获得帮助，但无法定位对应实现。若有人能指出Julia库中ESRK1和RSwM1的实现（或其他可读性强的正确实现），我将非常感激。我曾在StochasticDiffEq.jl仓库中搜索ESRK和RSwM，但未找到与论文描述相符的实现。

更新：
我已找到ESRK1的代码，但仍未找到RSwM1的代码。

以下是我尚未修正的ESRK1 Python实现代码：

def ESRK1(U, t, dt, f, g, dW, dZ):
    # Implementation of ESRK1, following Rackauckas & Nie (2017)
    # Eq. (2), (3) and (4) and Table 1
    
    # Stochastic integrals, taken from Eqs. (25) - (30) in Rackauckas & Nie (2017)
    I1   = dW
    I11  = (I1**2 - dt) / 2
    I111 = (I1**3 - 3*dt*I1) / 6
    I10  = (I1 + dZ/np.sqrt(3))*dt / 2
    
    # Coefficients, taken from Table 1 in Rackauckas & Nie (2017)
    # All coefficients not included below are zero
    c0_2 = 3/4
    c1_2, c1_3, c1_4 = 1/4, 1, 1/4
    A0_21 = 3/4
    B0_21 = 3/2
    A1_21 = 1/4
    A1_31 = 1
    A1_43 = 1/4
    B1_21 = 1/2
    B1_31 = -1
    B1_41, B1_42, B1_43 = -5, 3, 1/2
    alpha1, alpha2 = 1/2, 2/3
    alpha_tilde1, alpha_tilde2 = 1/2, 1/2
    beta1_1, beta1_2, beta1_3 = -1, 4/3, 2/3
    beta2_1, beta2_2, beta2_3 = -1, 4/3, -1/3
    beta3_1, beta3_2, beta3_3 =  2, -4/3, -2/3
    beta4_1, beta4_2, beta4_3, beta4_4 = -2, 5/3, -2/3, 1
    
    # Stages in the Runge-Kutta approximation
    # Eqs. (3) and (4) and Table 1 in Rackauckas & Nie (2017)
    # First stages
    H0_1 = U # H^(0)_1
    H1_1 = U
    # Second stages
    H0_2 = U + A0_21 * f(t, H0_1)*dt + B0_21 * g(t, H1_1)*I10/dt
    H1_2 = U + A1_21 * f(t, H0_1)*dt + B1_21 * g(t, H1_1)*np.sqrt(dt)
    # Third stages
    H0_3 = U
    H1_3 = U + A1_31 * f(t, H0_1) * dt + B1_31 * g(t, H1_1) * np.sqrt(dt)
    # Fourth stages
    H0_4 = U
    H1_4 = U + A1_43 * f(t, H0_3) * dt + (B1_41 * g(t, H1_1) + B1_42 * g(t+c1_2*dt, H1_2) + B1_43 * g(t+c1_3*dt, H1_3)) * np.sqrt(dt)
    
    # Construct next position
    # Eq. (2) and Table 1 in Rackauckas & Nie (2017)
    U_ = U  + (alpha1*f(t, H0_1) + alpha2*f(t+c0_2*dt, H0_2))*dt \
            + (beta1_1*I1 + beta2_1*I11/np.sqrt(dt) + beta3_1*I10/dt ) * g(t, H1_1) \
            + (beta1_2*I1 + beta2_2*I11/np.sqrt(dt) + beta3_2*I10/dt ) * g(t + c1_2*dt, H1_2) \
            + (beta1_3*I1 + beta2_3*I11/np.sqrt(dt) + beta3_3*I10/dt ) * g(t + c1_3*dt, H1_3) \
            + (beta4_4*I111/dt ) * g(t + c1_4*dt, H1_4)
    
    # Calculate error estimate
    # Eq. (9) and Table 1 in Rackauckas & Nie (2017)
    E = -dt*(f(t, H0_1) + f(t + c0_2*dt, H0_2))/6  \
        + (beta3_1*I10/dt + beta4_1*I111/dt)*g(t, H1_1) \
        + (beta3_2*I10/dt + beta4_2*I111/dt)*g(t + c1_2*dt, H1_2) \
        + (beta3_3*I10/dt + beta4_3*I111/dt)*g(t + c1_3*dt, H1_3) \
        + (beta4_4*I111/dt)*g(t + c1_4*dt, H1_4)

    # Return next position and error
    return U_, E

参考论文：Rackauckas & Nie（2017）

内容的提问来源于stack exchange，提问作者Tor