解决未给出正确解决方案的bvp

2024-04-26 12:22:46 发布

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我正试图解决一个边界值问题,我知道它会有贝塞尔形状。我的条件是u[0]=1,u[40]=v[0]=v[40]=0。然而,对于U,它给了我一条x=1的平线。谁能告诉我我做错了什么?提前谢谢。这是我的密码:

#Parameters
mu=0
eta = 1 #0.3
Dlt= 0.1 #.5
lbd= -1
alp= 1
Vz=1 

def dU_dx(x, U):
    # Let's try to make U as vector such that u=U[0],y=U[1],v=U[2] and z=U[3].
    #This function should return [u',y',v', z']
    
    return [U[1], 
            lbd*Dlt*U[2]+alp*(U[3]+U[2]/x)+(Vz-mu)*U[0] - U[1]*(1/x),
            U[3],
            (-lbd*Dlt*U[0]-alp*(U[1])+(-Vz-mu)*U[2] -U[3]*(1/x)+1/x**2*U[2])/eta]

def bc1(ya1,yb1):

    return np.array([ya1[0]-1,yb1[0],ya1[1],yb1[1]-0.5])

#Define the initial mesh with 5 nodes:

x = np.linspace(0, 40, 5) #(0, 1, 10)

#This problem is known to have two solutions. To obtain both of them, we use two different initial guesses for y. We denote them by subscripts a and b.

U = np.zeros((4, x.size))

U[0] = 1
U[1] = 0
U[2] = 0
U[3] = 0.5

#Now we are ready to run the solver.

from scipy.integrate import solve_bvp

res_a = solve_bvp(dU_dx, bc1, x, U)

#Let’s plot the two found solutions. We take an advantage of having the solution in a spline form to produce a smooth plot.

x_plot = np.linspace(10**-6, 40, 1000)

y_plot_a = res_a.sol(x_plot)[0]
y_plot_b = res_a.sol(x_plot)[2]

import matplotlib.pyplot as plt
plt.plot(x_plot, y_plot_a, label='y_a')
plt.plot(x_plot, y_plot_b, label='y_b')
plt.legend()
plt.xlabel("x")
plt.ylabel("y")
plt.show()

我的代码基于scipy.integrate.solve_bvp站点的一个示例

我有以下警告:

RuntimeWarning: divide by zero encountered in true_divide

RuntimeWarning: invalid value encountered in multiply

RuntimeWarning: invalid value encountered in add


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1条回答
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1楼 · 发布于 2024-04-26 12:22:46

完整回溯的这一部分(您应该包括在内)

RuntimeWarning: invalid value encountered in true_divide
  lbd * Dlt * U[2]  + alp * (U[3] + U[2] / x) + (Vz - mu) * U[0] - U[1] * (1 / x),
RuntimeWarning: divide by zero encountered in true_divide
  lbd * Dlt * U[2]  + alp * (U[3] + U[2] / x) + (Vz - mu) * U[0] - U[1] * (1 / x),
...

为您提供完美的指针:在true_divide中遇到被零除的情况(提示:... / x

替换

x = np.linspace(0, 40, 5)

x = np.linspace(1, 40, 5)

看看当你不尝试除以0时会发生什么

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