我定义了一个python类来计算微分方程组的解。为此,我定义了一个名为Rhs (right and side)的类,它应该表示dy/dt(i-th)的右边和右边。这个类包含一个浮点值(初始时间,初始值,最终时间)和一个函数(函数数组)为了定义这个数组,我简单地定义了3个λ函数,表示方程(I),并创建了这个函数的np.array
func1 = lambda t,u : 10 * (u[1] - u[0])
func2 = lambda t,u : 28 * u[0] - u[1] - u[0] * u[2]
func3 = lambda t,u : -8/3 * u[2] + u[0]*u[1]
然后以这种方式传递给rhs类:
func = np.array([func1,func2,func3])
y0 = np.array([1.,0.,0.])
problem3 = rhs.Rhs(func,0.0,100.0,y0,1000)
Rhs类如下所示:
class Rhs:
def __init__(self, fnum : np.ndarray , t0: np.float, tf: np.float, y0 : np.array, n: int , fanal = None ):
self.func = fnum
Rhs.solution = fanal
self.t0 = t0
self.tf = tf
self.n = n
self.u0 = y0
def createArray(self):
'''
Create the Array time and f(time)
- the time array can be create now
- the solution array can be just initialize with the IV
'''
self.t = np.linspace(self.t0, self.tf, self.n )
self.u = np.array([self.u0 for i in range(self.n) ])
return self.t,self.u
def f(self,ti,ui):
return self.func(ti,ui)
def Df(self,ti,ui):
eps = 10e-6
return ((self.func(ti,ui)+eps) - self.f(ti,ui))/eps
这里的问题是当euler类调用函数f
时
class Explicit:
def __init__(self, dydt: rhs.Rhs, save : bool=True, _print: bool=False, filename : str=None):
self.dydt = dydt
self.dt = (dydt.tf-dydt.t0)/dydt.n
self._print = _print
def solve(self):
self.time, self.u = self.dydt.createArray()
for i in range(len(self.time)-1):
self.u[i+1] = self.u[i] + self.dt*self.dydt.f(self.time[i],self.u[i])
Explicit.solved = True
print('here')
if self._print:
with open(filename) as f:
print('here')
for i in range(len(self.u)):
f.write('%.4f %4f %4f %4f' %(self.time ,self.u[0,i], self.u[1], self.u[2]))
if self.save:
return self.time,self.u
这里的问题是:将形状= 1000,3的向量u传递给函数的正确方法是什么(以便在lambda函数系统中使用应用于3向量索引的3函数。)我不明白的是为什么在C++中我没有得到这个问题,看看这里:,all the class hierarchy,我不知道用哪种方法计算这个东西
这是错误:
Traceback (most recent call last):
File "drive.py", line 94, in <module>
main()
File "drive.py", line 63, in main
fet,feu = fwdeuler_p1.solve()
File "/home/marco/Programming/Python/Numeric/OdeSystem/euler.py", line 77, in solve
self.u[i+1] = self.u[i] + self.dt*self.dydt.f(self.time[i],self.u[i])
File "/home/marco/Programming/Python/Numeric/OdeSystem/rhs.py", line 44, in f
return self.func(ti,ui)
TypeError: 'numpy.ndarray' object is not callable
编辑感谢回复..但不幸的是没有:(这是错误消息:
drive.py:31: RuntimeWarning: overflow encountered in double_scalars
func2 = lambda t,u : 28 * u[0] - u[1] - u[0] * u[2]
drive.py:32: RuntimeWarning: overflow encountered in double_scalars
func3 = lambda t,u : -8/3 * u[2] + u[0]*u[1]
drive.py:31: RuntimeWarning: invalid value encountered in double_scalars
func2 = lambda t,u : 28 * u[0] - u[1] - u[0] * u[2]
/home/marco/Programming/Python/Numeric/OdeSystem/euler.py:77: RuntimeWarning: invalid value encountered in add
self.u[i+1] = self.u[i] + self.dt*self.dydt.f(self.time[i],self.u[i])
非常感谢你的帮助!你有其他的解决方案吗?
@LutzL我不需要复制u0,self.u0是shape=3,self.u应该是1000,3 (1000行)这3列分别表示等式(函数数组)中的u[0],u[1],u[2]
,顺便说一下,如果我增加步数(减少增量)
发布于 2018-07-09 02:16:57
这应该是可行的
def f(self,ti,ui):
return np.array([function(ti,ui) for function in self.func])
发布于 2018-07-10 19:52:06
self.u = np.array([self.u0 for i in range(self.n) ])
这也会给你带来问题,因为这会产生一个以u0
为行的矩阵或2d数组,而不是预期的副本。你可能想要
self.u = np.array([self.u0[i] for i in range(self.n) ])
或者更简单
self.u = self.u0.copy()
https://stackoverflow.com/questions/51234469
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