我正试图使this example using the method of lines so solve a pde适应des系统。这不是我试图解决的系统,只是一个例子
如何合并第二个de?我已尝试将odefunc
修改为
def odefunc(u,t):
dudt = np.zeros(X.shape)
dvdt = np.zeros(X.shape)
dudt[0] = 0 # constant at boundary condition
dudt[-1] = 0
dvdt[0] = 0
dvdt[-1] = 0
# for the internal nodes
for i in range (1, N-1):
dudt[i] = k* (u[i+1] - 2*u[i] + u[i-1]) / h**2
dvdt[i] = u[i]
return [dudt,dvdt]
基于我发现的两个ODE系统的示例解决方案,但由于我已经有了dudt的数组,我怀疑这可能是我的问题。我也不知道我的初始条件现在应该是什么样子,所以它们是正确的尺寸等等
我得到的错误是
The array return by func must be one-dimensional, but got ndim=2.
在第sol = odeint(odefunc, init, tspan)
行
使用单个pde的示例
import numpy as np
from scipy.integrate import odeint
import matplotlib.pyplot as plt
plt.interactive(False)
N = 5 # number of points to discretise
L = 1.0 # length of the rod
X = np.linspace(0,L,N)
h = L/ (N - 1)
k = 0.02
def odefunc(u,t):
dudt = np.zeros(X.shape)
dudt[0] = 0 # constant at boundary condition
dudt[-1] = 0
# for the internal nodes
for i in range (1, N-1):
dudt[i] = k* (u[i+1] - 2*u[i] + u[i-1]) / h**2
return dudt
init = 150.0 * np.ones(X.shape) # initial temperature
init[0] = 100.0 # boundary condition
init[-1] = 200.0 # boundary condition
tspan = np.linspace(0.0, 5.0, 100)
sol = odeint(odefunc, init, tspan)
for i in range(0, len(tspan), 5):
plt.plot(X,sol[i], label = 't={0:1.2f}'.format(tspan[i]))
# legend outside the figure
plt.legend(loc='center left', bbox_to_anchor=(1,0.5))
plt.xlabel('X position')
plt.ylabel('Temperature')
# adjust figure edges so the legend is in the figure
plt.subplots_adjust(top=0.89, right = 0.77)
plt.show()
从ode函数的odeintdocumentation中,y定义为向量,但您将其定义为向量列表
您可以简单地使用
ravel
函数从向量列表创建一个向量。下面是代码中的一个示例:相关问题 更多 >
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