鉴于我的老师创建的Matlab代码:
function [] = explicitWave(T,L,N,J)
% Explicit method for the wave eq.
% T: Length time-interval
% L: Length x-interval
% N: Number of time-intervals
% J: Number of x-intervals
k=T/N;
h=L/J;
r=(k*k)/(h*h);
k/h
x=linspace(0,L,J+1); % number of points = number of intervals + 1
uOldOld=f(x); % solution two time-steps backwards. Initial condition
disp(uOldOld)
uOld=zeros(1,length(x)); % solution at previuos time-step
uNext=zeros(1,length(x));
% First time-step
for j=2:J
uOld(j)=(1-r)*f(x(j))+r/2*(f(x(j+1))+f(x(j-1)))+k*g(x(j));
end
% Remaining time-steps
for n=0:N-1
for j=2:J
uNext(j)=2*(1-r)*uOld(j)+r*(uOld(j+1)+uOld(j-1))-uOldOld(j);
end
uOldOld=uOld;
uOld=uNext;
end
plot(x,uNext,'r')
end
我试图通过使用以下代码在Python中实现这一点:
import numpy as np
import matplotlib.pyplot as plt
def explicit_wave(f, g, T, L, N, J):
"""
:param T: Length of Time Interval
:param L: Length of X-interval
:param N: Number of time intervals
:param J: Number of X-intervals
:return:
"""
k = T/N
h = L/J
r = (k**2) / (h**2)
x = np.linspace(0, L, J+1)
Uoldold = f(x)
Uold = np.zeros(len(x))
Unext = np.zeros(len(x))
for j in range(1, J):
Uold[j] = (1-r)*f(x[j]) + (r/2)*(f(x[j+1]) + f(x[j-1])) + k*g(x[j])
for n in range(N-1):
for j in range(1, J):
Unext[j] = 2*(1-r) * Uold[j]+r*(Uold[j+1]+Uold[j-1]) - Uoldold[j]
Uoldold = Uold
Uold = Unext
plt.plot(x, Unext)
plt.show()
return Unext, x
但是,当我使用相同的输入运行代码时,在打印它们时会得到不同的结果。我的意见:
g = lambda x: -np.sin(2*np.pi*x)
f = lambda x: 2*np.sin(np.pi*x)
T = 8.0
L = 1.0
J = 60
N = 480
Python绘图结果与精确结果进行比较。x-es表示实际解决方案,红线表示函数:
你能看到我在翻译这段代码时可能犯的任何明显错误吗
如果有人需要确切的解决方案:
exact = lambda x,t: 2*np.sin(np.pi*x)*np.cos(np.pi*t) - (1/(2*np.pi))*np.sin(2*np.pi*x)*np.sin(2*np.pi*t)
我通过调试发现了错误。这里的主要问题是代码:
因此,在Python中,当您将一个新变量定义为与旧变量相等时,它们将成为彼此的引用(即相互依赖)。让我以列表为例说明这一点:
所以这里的解决方案是使用
.copy()
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