Python解决ivp vs R lsod

Question

我正在尝试将R代码转换为Python。R代码使用lsoda函数，它是FORTRAN DOE解算器的包装器。Python对应的似乎是solve_ivp，它是FORTRAN ODEPACK的包装器。我在Python中使用method='LSODA'，这应该是R使用的等效方法。然而，我的结果是不同的高达1%的误差。在我的代码中没有什么是随机的，所以我相信我应该能够完全复制结果。你知道吗

知道吗？！你知道吗

这是R代码的一部分（之前的代码只是计算参数值：

val = c("A1" = 1, "A2" = 1, "A3" = 1, "A4" = 1, "A5" = 1, "A6" = 1, "A7" = 1)

hamberg_ode <- function(t,val,p) { 
    dA1 = p["ktr1"]*(1 - ((p["E_MAX"] * p["C_s_gamma"])/(p["EC_50_gamma"] + p["C_s_gamma"]))) - p["ktr1"]*val["A1"]
    dA2 = p["ktr1"]*val["A1"] - p["ktr1"]*val["A2"]
    dA3 = p["ktr1"]*val["A2"] - p["ktr1"]*val["A3"]
    dA4 = p["ktr1"]*val["A3"] - p["ktr1"]*val["A4"]
    dA5 = p["ktr1"]*val["A4"] - p["ktr1"]*val["A5"]
    dA6 = p["ktr1"]*val["A5"] - p["ktr1"]*val["A6"]
    dA7 = p["ktr2"]*(1 - ((p["E_MAX"] * p["C_s_gamma"])/(p["EC_50_gamma"] + p["C_s_gamma"]))) - p["ktr2"]*val["A7"]
    cat(val["A1"], dA1, '\n')
    list(c(dA1, dA2, dA3, dA4, dA5, dA6, dA7))
}

out = lsoda(val, times, hamberg_ode, p)

Python代码：

val = [1]*7

class hamberg_ode:
    def __init__(self, p):
        self.p = p

    def f(self, t, val, p=None):
        if p is None:
            p = self.p
        dA1=p["ktr1"]*(1 - ((p["E_MAX"] * p["C_s_gamma"]) /
                        (p["EC_50_gamma"] + p["C_s_gamma"]))) - p["ktr1"]*val[0]
        dA2=p["ktr1"]*val[0] - p["ktr1"]*val[1]
        dA3=p["ktr1"]*val[1] - p["ktr1"]*val[2]
        dA4=p["ktr1"]*val[2] - p["ktr1"]*val[3]
        dA5=p["ktr1"]*val[3] - p["ktr1"]*val[4]
        dA6=p["ktr1"]*val[4] - p["ktr1"]*val[5]
        dA7=p["ktr2"]*(1 - ((p["E_MAX"] * p["C_s_gamma"]) /
                        (p["EC_50_gamma"] + p["C_s_gamma"]))) - p["ktr2"]*val[6]
        print(val[0], dA1)

        return (dA1, dA2, dA3, dA4, dA5, dA6, dA7)


h_function = hamberg_ode(p).f
out = solve_ivp(h_function, (0, maxTime), val, t_eval=times, method='LSODA')

作为数字如何发散的示例，下面是两个代码的A1和dA1的几个第一个值： R

1至0.2289151

1至0.2289151

0.9997726至0.2287975

0.9997727至0.2287976

0.9995454至0.22868

0.9995455-0.2286801

0.9901534至0.2238221

0.9901523至0.2238215

0.9809609至0.2190673

0.9809587至0.2190662

0.9719626至0.214413

0.9719604至0.2144119

0.9493722至0.2027284

0.9493668至0.2027255

0.927996至0.1916717

0.9280039至0.1916758

0.9078033-0.1812272

0.9078049至0.181228

0.8887056-0.1713491

0.8887071-0.1713499

Python

1.0至0.22891514470392998

0.9868338969217406-0.2221050913875888

0.9872255785819792-0.22230768534978135

0.9744278526945864-0.2156881719597506

0.9748085754105683-0.2158850975024998

0.9069550726140441-0.18078845812498728

0.906362742770375-0.18048208061964116

0.8502494750308627-0.15145797661644517

0.8491489787959607至0.1508887542597866

0.8022897024620746-0.1266511977015548

0.8013657199203642-0.1261732756972218

0.7405758555885625-0.0947302422152

0.7400188154524862-0.09444211821383663

0.7145516960742005-0.08126947025955095

0.714884659705225-0.08144169282733643

Answer 1

正如@astoriko所指出的，scipy的默认相对公差（rtol）是1e-3，而R的默认相对公差（lsoda）是1e-6。你知道吗