我正在尝试使用以下代码来解决ivp y'=-y-5 * exp(-t)* sin(5 t),y(0)= 1

%pylab inline
%matplotlib inline
from scipy.integrate import odeint

def mif(t, y):
    return -y-5*exp(-t)*sin(5*t)

tspan = np.arange(0, 3, 0.000001)
y0 = 1.0
y_result = odeint(mif, y0, tspan)
y_result = y_result[:, 0]  # convert the returned 2D array to a 1D array
plt.figure()
plt.plot(tspan, y_result)
plt.show()


但是,我得到的图是错误的,它与我从Matlab或Mathematica获得的图不匹配。它实际上不同于以下替代集成:

from scipy.integrate import ode

# initialize the 4th order Runge-Kutta solver
solver = ode(mif).set_integrator('dop853')

# initial value
y0 = 1.0
solver.set_initial_value(y0, 0)

values = 1000
t = np.linspace(0.0001, 3, values)
y = np.zeros(values)

for ii in range(values):
    y[ii] = solver.integrate(t[ii])[0] #z[0]=u


确实会产生正确的结果。我的odeint做错了什么?

最佳答案

函数参数在ode和odeint之间更改。对于odeint,您需要

def mif(y, t):


和颂歌

def mif(t, y):


例如

%pylab inline
%matplotlib inline
from scipy.integrate import odeint

def mif(t,y):
    return y

tspan = np.arange(0, 3, 0.000001)
y0 = 0.0
y_result = odeint(mif, y0, tspan)
plt.figure()
plt.plot(tspan, y_result)
plt.show()




from scipy.integrate import ode

def mif(y, t):
    return y

# initialize the 4th order Runge-Kutta solver
solver = ode(mif).set_integrator('dop853')

# initial value
y0 = 0.000000
solver.set_initial_value([y0], 0.0)

values = 1000
t = np.linspace(0.0000001, 3, values)
y = np.zeros(values)

for ii in range(values):
    y[ii] = solver.integrate(t[ii]) #z[0]=u
plt.figure()
plt.plot(t, y)
plt.show()

关于python - 来自scipy.in的odeint在Python中给出错误的结果?,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/28780770/

10-16 08:02