[英]Convert code from matlab to python, problems occurred
我發現了用matlab編寫的向后Euler的實現, 這里計算步長的方程為:
function [t,y]=beul(f,df,y0,t0,tf,n)
h=(tf-t0)/n;
t=linspace(t0,tf,n+1);
y=zeros(n+1,length(y0));
y(1,:)=y0;
for i=1:n
x0=y(i,:)’;
x1=x0-inv(eye(length(y0))-h*feval(df,t(i),x0))*(x0-h*feval(f,t(i),x0)’-y(i,:)’);
while (norm(x1-x0)>0.0001)
x0=x1;
x1=x0-inv(eye(length(y0))-h*feval(df,t(i),x0))*(x0-h*feval(f,t(i),x0)’-y(i,:)’);
end
y(i+1,:)=x1’;
end
end
然后將此函數稱為函數和jacobian的定義系統:
function yp=volt(t,y)
a=4;
c=1;
yp(1)=a*(y(1)-y(1)*y(2));
yp(2)=-c*(y(2)-y(1)*y(2));
end
function y=dvol(t,x)
a=4;
c=1;
y(1,1)=a*(1-x(2));
y(1,2)=-a*x(1);
y(2,1)=c*x(2);
y(2,2)=-c*(1-x(1));
后向歐拉稱為:
[t,y]=beul(’volt’,’dvol’,[2,1],0,10,1000);
我已經翻譯了python中的代碼:
class Backward(Euler):
def solve(self):
for i in range(len(self.time)-1):
u0 = self.u[i].T
u1 = u0 - np.linalg.inv(np.eye(len(self.dydt.u0)) - self.dt * self.dydt.df(self.time[i+1],u0)) * (u0 - self.dt * self.dydt.f(self.time[i+1], u0).T - self.u[i].T )
error = np.array([1.0])
iters = 0
while True:
try:
u0 = u1.T
u1 = u0 - np.linalg.inv(np.eye(len(self.dydt.u0)) - self.dt * self.dydt.df(self.time[i+1],u0)) * (u0 - self.dt * self.dydt.f(self.time[i+1], u0).T - self.u[i].T )
iters += 1
error = np.abs(u1-u0)
if np.sum(np.abs(error)) <= toll:
break
except ValueError as ex:
print('Error occurred in Implicit-NR Euler Solvers: %s' %ex.args)
return None , None
self.u[i+1] = u1.T
然后我將jacobian矩陣定義如下:
def f(self,ti,ui):
return np.array([function(ti,ui) for function in self.func])
def df(self, t, u, **params):
eps = 1e-12
J = np.zeros([len(u), len(u)], dtype = np.float)
for i in range(len(u)):
u1 = u.copy()
u2 = u.copy()
u1[i] += eps
u2[i] -= eps
f1 = self.f(t, u1, **params)
f2 = self.f(t, u2, **params)
J[ : , i] = (f1 - f2) / (2 * eps)
return J
如果我嘗試運行單個方程式問題,則方法效果很好(我已與其他求解器進行了比較)
但是問題是matlab產品的行為有所不同! 所以我不知道如何在python中將產品固定為相同的,因為當我為系統運行代碼時(例如,由matlab解決的相同)
eq1 = lambda t,u : a*(u[0]-u[0]*u[1]);
eq2 = lambda t,u : -c*(u[1]-u[0]*u[1]);
func1 = np.array([eq1,eq2])
y0 = np.array([2.,1.])
我收到此錯誤:
Running Newton-Rapson Backward Euler ....
Error occurred in Implicit-NR Euler Solvers: could not broadcast input array from shape (2,2) into shape (2)
所以我該如何定義與matlab計算相同的產品(它的df也是2x2矩陣)以修復python的方法?
我解決了使用numpy.dot產品:
u1 = u0 - np.linalg.inv(np.eye(len(self.dydt.u0)) - self.dt * self.dydt.df(self.time[i+1],u0)).dot(u0 - self.dt * self.dydt.f(self.time[i+1], u0).T - self.u[i].T )
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