math - 如何实现自适应步长 Runge-Kutta Cash-Karp？

Question

尝试实现自适应步长 Runge-Kutta Cash-Karp 但失败并出现此错误:

home/anaconda/lib/python3.6/site-packages/ipykernel_launcher.py:15: RuntimeWarning: divide by zero encountered in double_scalars from ipykernel import kernelapp as app

我试图解决的 ODE(并在下面的示例中使用，从高阶转换为一阶 ODE 系统)如下:



这是我正在使用的实现:

def rkck(f, x, y, h, tol):
    #xn = x + h
    err = 2 * tol
    while (err > tol):
        xn = x + h
        k1 = h*f(x,y)
        k2 = h*f(x+(1/5)*h,y+((1/5)*k1)) 
        k3 = h*f(x+(3/10)*h,y+((3/40)*k1)+((9/40)*k2))
        k4 = h*f(x+(3/5)*h,y+((3/10)*k1)-((9/10)*k2)+((6/5)*k3))
        k5 = h*f(x+(1/1)*h,y-((11/54)*k1)+((5/2)*k2)-((70/27)*k3)+((35/27)*k4))
        k6 = h*f(x+(7/8)*h,y+((1631/55296)*k1)+((175/512)*k2)+((575/13824)*k3)+((44275/110592)*k4)+((253/4096)*k5))
        yn4 = y + ((37/378)*k1)+((250/621)*k3)+((125/594)*k4)+((512/1771)*k6)
        yn5 = y + ((2825/27648)*k1)+((18575/48384)*k3)+((13525/55296)*k4)+((277/14336)*k5)+((1/4)*k6)
        err = yn4[-1]-yn5[-1]
        if (err != 0):
            h = 0.8 * h * (tol/err)**(1/float(5))
        yn = yn4
    return xn, yn

def integrate_sStepControl(f, t0, y0, tend, h, tol):
    T = [t0]
    Y = [y0]
    t = t0
    y = y0 
    while (t < tend):
        h = min(h, tend-t)
        t, y = rkck(f, t, y, h, tol)
        T.append(t)
        Y.append(y)
    return np.array(T), np.array(Y)

def f_1(t,y): 
    return np.array([ y[1], -y[0]-(y[0])**3 ])

Y0_f1 = np.array([1.0,1.0])


# Execution
h = 0.05
tv, yv = integrate_sStepControl(f=f_1, t0=0.0, y0=Y0_f1, tend=100.0, h=h, tol=1.0E-05)
print("[ %20.15f, %20.15f]"%(yv[-1,0], yv[-1,1]) )
plt.plot(tv, yv)

得到上面的错误，但它被绘制出来。不知道我在这里做错了什么:-/


编辑:
添加了对 err == 0 的检查

Answer 1

您需要实际传回计算出的新步长 h以便它可以在第一步的主循环中使用。
误差应作为范数计算。添加一些小数以避免被零除。
主要误差项是 C*h^5对于四阶方法。这必须与所需的本地错误 tol*h 进行比较.总的来说，这会导致计算最佳 h 的第 4 个根。 .取 5 次方提供了某种抑制作用，但是对全局误差的影响并不是很直接。

def rkck(f, x, y, h, tol):
    #xn = x + h
    err = 2 * tol
    while (err > tol):
        xn = x + h
        k1 = h*f(x,y)
        k2 = h*f(x+(1/5)*h,y+((1/5)*k1)) 
        k3 = h*f(x+(3/10)*h,y+((3/40)*k1)+((9/40)*k2))
        k4 = h*f(x+(3/5)*h,y+((3/10)*k1)-((9/10)*k2)+((6/5)*k3))
        k5 = h*f(x+(1/1)*h,y-((11/54)*k1)+((5/2)*k2)-((70/27)*k3)+((35/27)*k4))
        k6 = h*f(x+(7/8)*h,y+((1631/55296)*k1)+((175/512)*k2)+((575/13824)*k3)+((44275/110592)*k4)+((253/4096)*k5))
        dy4 = ((37/378)*k1)+((250/621)*k3)+((125/594)*k4)+((512/1771)*k6)
        dy5 = ((2825/27648)*k1)+((18575/48384)*k3)+((13525/55296)*k4)+((277/14336)*k5)+((1/4)*k6)
        err = 1e-2*tol+max(abs(dy4-dy5))
        # h = 0.95 * h * (tol/err)**(1/5)
        h = 0.8 * h * (tol*h/err)**(1/4)
        yn = y+dy4
    return xn, yn, h

def integrate_sStepControl(f, t0, y0, tend, h, tol):
    T = [t0]
    Y = [y0]
    t = t0
    y = y0 
    while (t < tend):
        h = min(h, tend-t)
        t, y, h = rkck(f, t, y, h, tol)
        T.append(t)
        Y.append(y)
    return np.array(T), np.array(Y)

有了这个，你的例子给出了解决方案、时间步长和错误/tol的以下图