我是从Python开始的,所以我可能会问一个不那么微妙的问题,但是经过大量的研究,我无法解决这个错误。 实际上,我正试图用Gray-Scott模型来解决一个物理问题,但我一直停留在代码的最后:结果不被视为数字,在square、add、multiply和substract中会遇到溢出。
这里有人知道这是从哪里来的吗?
谢谢!
以下是我试图解决的问题的初始条件:
n = 192
Du, Dv, F, k = 0.00016, 0.00008, 0.035, 0.065
dh = 5/(n-1)
T = 8000
dt = .9 * dh**2 / (4*max(Du,Dv))
nt = int(T/dt)
uvinitial = numpy.load('./uvinitial.npz')
Uin = uvinitial['U']
Vin = uvinitial['V']
下面是我的功能:
def Nd1(U,V) :
return - U*(V)**2 + F*(1-U)
def Nd2(U,V) :
return U*(V)**2 -(F+k)*V
def gray_scott_solve(Du, Dv, dh, dt, nt, Uin, Vin, Nd1, Nd2):
Uplus = Uin.copy()
Vplus = Vin.copy()
for n in range(nt):
U = Uplus.copy()
V = Vplus.copy()
Uplus[1:-1,1:-1] = ( Nd1(U[1:-1,1:-1], V[1:-1,1:-1]) + Du/(dh**2) \
*(U[2:,1:-1] + U[:-2,1:-1] - 4*U[1:-1,1:-1]) \
+ U[1:-1,2:] + U[1:-1,:-2] )*dt \
+ U[1:-1,1:-1]
Uplus[:,-1] = Uplus[:,-2]
Uplus[-1,:] = Uplus[-2,:]
Uplus[:,0] = Uplus[:,1]
Uplus[0,:] = Uplus[1,:]
Vplus[1:-1,1:-1] = ( Nd2(U[1:-1,1:-1], V[1:-1,1:-1]) + Du/(dh**2) \
*(V[2:,1:-1] + V[:-2,1:-1] - 4*V[1:-1,1:-1]) \
+ V[1:-1,2:] + V[1:-1,:-2] )*dt \
+ V[1:-1,1:-1]
Vplus[:,-1] = Vplus[:,-2]
Vplus[-1,:] = Vplus[-2,:]
Vplus[:,0]= Vplus[:,1]
Vplus[0,:]= Vplus[1,:]
return U, V
我现在要打印我要查找的结果:
U, V = gray_scott_solve(Du, Dv, dh, dt, nt, Uin, Vin, Nd1, Nd2)
print(U[100,::40])
最后我得到了这个错误:
[ nan nan nan nan nan]
C:\Users\Anto\Anaconda3\lib\site-packages\ipykernel\__main__.py:2: RuntimeWarning: overflow encountered in square from ipykernel import kernelapp as app
C:\Users\Anto\Anaconda3\lib\site-packages\ipykernel\__main__.py:2: RuntimeWarning: overflow encountered in multiply from ipykernel import kernelapp as app
C:\Users\Anto\Anaconda3\lib\site-packages\ipykernel\__main__.py:5: RuntimeWarning: overflow encountered in square
C:\Users\Anto\Anaconda3\lib\site-packages\ipykernel\__main__.py:5: RuntimeWarning: overflow encountered in multiply
C:\Users\Anto\Anaconda3\lib\site-packages\ipykernel\__main__.py:11: RuntimeWarning: invalid value encountered in add
C:\Users\Anto\Anaconda3\lib\site-packages\ipykernel\__main__.py:11: RuntimeWarning: overflow encountered in multiply
C:\Users\Anto\Anaconda3\lib\site-packages\ipykernel\__main__.py:11: RuntimeWarning: invalid value encountered in subtract
C:\Users\Anto\Anaconda3\lib\site-packages\ipykernel\__main__.py:18: RuntimeWarning: invalid value encountered in add
C:\Users\Anto\Anaconda3\lib\site-packages\ipykernel\__main__.py:18: RuntimeWarning: overflow encountered in multiply
C:\Users\Anto\Anaconda3\lib\site-packages\ipykernel\__main__.py:18: RuntimeWarning: invalid value encountered in subtract
正如您所写的,在python 2中,空间步
dh
将等于零:如果您使用的是python 3,那么dh将被正确地视为一个浮点数。
否则,正如@WarrenWeckesser所说,您使用的是直线法,并与前向Euler方法进行了时间积分,该方法与您的约束集(您说您的讲师指定了您的时间步长和其他参数值)显然是不稳定的。但是,使用Runge-Kutta Two方法对您的
dt
有效(我已经验证了这一点),但是您的讲师可能提到了您应该使用的时间积分。不管怎样,如果Runge Kutta Two看起来让人望而生畏,那么使用二阶中心空间方法:
其中
f(t,u)
是右手边,u_{n-1}
是时间t_{n-1}
时u
的值,或者是向后euler方法。相关问题 更多 >
编程相关推荐