所以我试图用Python来解以下微分方程组
System of differential equations
如您所见,对于{0,1,2,3,…}中的每一个n,系统取决于前一个系统的解
我已经尝试过求解n=0的系统,并找到了一个可以插入R(1 | t)中的解决方案R(0 | t),Python可以毫无问题地解决该系统。我已将解决方案R(0 | t)定义为r0(t),并实现了n=1的解决方案,如下所示:
def model(z,t):
dxdt = -3.273*z[0] + 3.2*z[1] + r0(t)
dydt = 3.041*z[0] - 3.041*z[1]
dzdt = [dxdt, dydt]
return dzdt
z0 = [0,0]
t = np.linspace(0,90, 90)
z1 = odeint(model, z0, t)
但是,我想通过在求解n时调用n-1系统的解来推广这个解决方案。由于微分方程在矩阵的右上角只有与零不同的条目,我们只需要上一个解的z1解。我试过了
def model0(z,t):
dxdt = -3.273*z[0] + 3.2*z[1]
dydt = 3.041*z[0] - 3.041*z[1]
dzdt = [dxdt, dydt]
return dzdt
z0 = [1,1]
t = np.linspace(0,90)
def model1(z,t):
dxdt = -3.273*z[0] + 3.2*z[1] + 0.071*odeint(model0, z0, t)[t,1]
dydt = 3.041*z[0] - 3.041*z[1]
dzdt = [dxdt, dydt]
return dzdt
z1 = [0,0]
z = odeint(model1, z1, t)
没有任何运气。有没有人有用Python解决这些递归ODE系统的经验
提前谢谢
更新了6x6矩阵和6函数的代码:
A = np.array([[h1h1, h1h2, h1h3, h1a1, h1a2, h1a3],
[h2h1, h2h2, h2h3, h2a1, h2a2, h2a3],
[h3h1, h2h3, h3h3, h3a1, h3a2, h3a3],
[a1h1, a1h2, a1h3, a1a1, a1a2, a1a3],
[a2h1, a2h2, a2h3, a2a1, a2a2, a2a3],
[a3h1, a3h2, a3h3, a3a1, a3a2, a3a3]
])
B = np.array([[0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, h3a0, 0],
[0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0],
])
def model0n(u,t):
Ra = u.reshape([-1,6])
n = len(Ra) - 1
dRa = np.zeros(Ra.shape)
dRa[0] = A @ Ra[0]
for i in range(1,n+1):
dRa[i] = A @ Ra[i] + B @ Ra[i-1]
return dRa.flatten()
u0 = [1,1,1,1,1,1,0,0,0,0,0,0]
t = np.linspace(0,90,90+1)
u = odeint(model0n,u0,t)
以上结果为u[:,0]的以下曲线图: Plot for u[:,0] which is supposed to be probabilities
对于n=0,它提供矩阵乘积“手动”的结果:
def modeln0manually(z,t):
d1dt = h1h1*z[0] + h1h2 * z[1] + h1h3*z[2] + h1a1*z[3] + h1a2*z[4] + h1a3*z[5]
d2dt = h2h1*z[0] + h2h2 * z[1] + h2h3*z[2] + h2a1*z[3] + h2a2*z[4] + h2a3*z[5]
d3dt = h3h1*z[0] + h3h2 * z[1] + h3h3*z[2] + h3a1*z[3] + h3a2*z[4] + h3a3*z[5]
d4dt = a1h1*z[0] + a1h2 * z[1] + a1h3*z[2] + a1a1*z[3] + a1a2*z[4] + a1a3*z[5]
d5dt = a2h1*z[0] + a2h2 * z[1] + a2h3*z[2] + a2a1*z[3] + a2a2*z[4] + a2a3*z[5]
d6dt = a3h1*z[0] + a3h2 * z[1] + a3h3*z[2] + a3a1*z[3] + a3a2*z[4] + a3a3*z[5]
drdt = [d1dt, d2dt, d3dt, d4dt, d5dt, d6dt]
return drdt
u0 = [1,1,1,1,1,1]
t = np.linspace(0,90)
z = odeint(modeln0manually, u0, t)
生成u[:,0]的绘图: Plot of u[:,0] as it is supposed to be
你们的系统是耦合的,即使是以三角形的方式。因此,最简洁的方法是将其作为一个耦合系统来求解
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