Python 如何使用odeint实现微分ode解决方案

我使用odeint在python中求解微分方程，我想计算ode解的导数。在MATLAB中，ode解的导数可计算如下：

ode_fcn = @(t,y) ...;                                       % ODE Function
[t,y] = ode15i(ode_fcn, ...);                               % Integrate ODE
dy = ode_fcn(t,y);                                          % Calculate Derivatives

derivative = dmdT(ode_solution, T_span, *parameters)

def diff(x, y):
    x = np.array(x)
    y = np.array(y)
    # Forward difference
    dydx_start = (y[1]-y[0])/(x[1]-x[0])
    # Central difference quotient
    dydx = np.diff(y)/np.diff(x)
    dydx_mid = (dydx[1:] + dydx[:-1])/2
    # Backward difference
    dydx_end = (y[-1]-y[-2])/(x[-1]-x[-2])    
    dydx_mid = np.append(dydx_start, dydx_mid)
    return np.append(dydx_mid, dydx_end)

然而，在python中，我试图使用以这种方式定义的ode函数，这导致了错误的答案。例如，我将dmdT函数定义如下：

dmdT(m, T, *parameters):
    pass
return [dm1dT, dm2dT, ...]

然后我解了ode函数

ode_solution = odeint(dmdT, m0, T_span, args=parameters)

我预计ode_解的导数计算如下：

ode_fcn = @(t,y) ...;                                       % ODE Function
[t,y] = ode15i(ode_fcn, ...);                               % Integrate ODE
dy = ode_fcn(t,y);                                          % Calculate Derivatives

derivative = dmdT(ode_solution, T_span, *parameters)

def diff(x, y):
    x = np.array(x)
    y = np.array(y)
    # Forward difference
    dydx_start = (y[1]-y[0])/(x[1]-x[0])
    # Central difference quotient
    dydx = np.diff(y)/np.diff(x)
    dydx_mid = (dydx[1:] + dydx[:-1])/2
    # Backward difference
    dydx_end = (y[-1]-y[-2])/(x[-1]-x[-2])    
    dydx_mid = np.append(dydx_start, dydx_mid)
    return np.append(dydx_mid, dydx_end)

但这有时可能是错误的，我不知道为什么

所以我的问题是：我能用我定义的ode函数来计算ode解的导数吗

另一种方法是在Numpy中使用diff函数，但它会将数据长度缩短1。因此，我定义了一个新的差分函数来解决这个问题。但是这种方法并不像在MATLAB中那样优雅

我定义的diff函数如下：

ode_fcn = @(t,y) ...;                                       % ODE Function
[t,y] = ode15i(ode_fcn, ...);                               % Integrate ODE
dy = ode_fcn(t,y);                                          % Calculate Derivatives

derivative = dmdT(ode_solution, T_span, *parameters)

def diff(x, y):
    x = np.array(x)
    y = np.array(y)
    # Forward difference
    dydx_start = (y[1]-y[0])/(x[1]-x[0])
    # Central difference quotient
    dydx = np.diff(y)/np.diff(x)
    dydx_mid = (dydx[1:] + dydx[:-1])/2
    # Backward difference
    dydx_end = (y[-1]-y[-2])/(x[-1]-x[-2])    
    dydx_mid = np.append(dydx_start, dydx_mid)
    return np.append(dydx_mid, dydx_end)

任何建议都会有帮助，非常感谢

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根据沃伦·韦克瑟的建议，我试图说明我提到的问题。这是我使用的代码

def dmdT(m, T, beta, A1, E1, n1, k1, A2, E2, n2, k2):
    R = 8.314
    """
    FR -> char + volitale
    Polyester -> char + volitale
    """
    if len(m) == 4:
        m_FR= m[0]
        m_Polyester = m[1]
        m_residue = m[2]
        m_total = m[3]    

    
        rxn1 = -m[0]*A1 * np.exp(-E1/R/T) * (m_FR/m[0])**n1 / beta
        rxn2 = -m[1]*A2 * np.exp(-E2/R/T) * (m_Polyester/m[1])**n2 / beta
    else:
        m_FR= m[:, 0]
        m_Polyester = m[:, 1]
        m_residue = m[:, 2]  
        m_total = m[:, 3]
        
        rxn1 = -m[0,0] * A1 * np.exp(-E1/R/T) * (m_FR/m[0,0])**n1 / beta 
        rxn2 = -m[0,1] * A2 * np.exp(-E2/R/T) * (m_Polyester/m[0,1])**n2 / beta 
    
    r1 = rxn1
    r2 = rxn2
    r3 = -k1 * rxn1 - k2 * rxn2

    
    dm_FRdT = r1
    dm_PolyesterdT = r2
    dm_residuedT = r3
    dm_totalT = r1 + r2 + r3 
    
    return [dm_FRdT,  dm_PolyesterdT, dm_residuedT, dm_totalT]    

m0 = [0.0954577747, 0.9045422253, 0, 1]
T = np.arange(300, 1000, 0.1)
parameters = [5, 9.42351217e+04, 5.08266345e+04, 9.87252218e-01,
       6.90982451e-01, 8.28675597e+16, 2.29312948e+05, 6.94773512e-01,
       4.00782605e-01]
ode_solution = odeint(dmdT, m0, T, args = tuple(parameters))

MLR = dmdT(ode_solution, T, *parameters)[-1]
MLR2 = diff(T, ode_solution[:, -1])
plt.figure()
plt.plot(T, MLR)
plt.plot(T, MLR2)

这两种方法的解决方案是不同的。当我使用dmdT获得导数时，警告出现：

__main__:22: RuntimeWarning: invalid value encountered in power
__main__:23: RuntimeWarning: invalid value encountered in power

所以我想知道为什么会发生这种情况，以及如何解决这个问题？答案是错的吗

---

谢谢大家!

“但这有时可能是错误的，…”怎么会错呢？准确地展示你所做的，并解释为什么你认为这是错误的。谢谢你的建议，我已经添加了错误解决方案的细节