Python 在一个图形上绘制具有相同驱动力的两个解决方案
我试图在同一个图上绘制两个解,同时只改变这些解的初始条件。我希望初始条件是y=0的一个值和y=-pi/2的另一个值。我不知道如何让它们位于同一个图形上,而不重新复制和粘贴所有内容并创建新数组Python 在一个图形上绘制具有相同驱动力的两个解决方案,python,numpy,plot,graph,Python,Numpy,Plot,Graph,我试图在同一个图上绘制两个解,同时只改变这些解的初始条件。我希望初始条件是y=0的一个值和y=-pi/2的另一个值。我不知道如何让它们位于同一个图形上,而不重新复制和粘贴所有内容并创建新数组 import numpy as np import matplotlib.pyplot as plt t = 0.0 y = 0.0 u = 0.0 F = 1.073 Wd = 2*np.pi w0 = 1.5*Wd b = w0/4 ts =[] ys =[] h= 0.001 while (t
import numpy as np
import matplotlib.pyplot as plt
t = 0.0
y = 0.0
u = 0.0
F = 1.073
Wd = 2*np.pi
w0 = 1.5*Wd
b = w0/4
ts =[]
ys =[]
h= 0.001
while (t <= 30):
m1 = u
k1 = (-w0**2)*np.sin(y) + u*(-2*b) + F*(w0**2)*np.cos(Wd*t)
m2 = u + (h / 2.) * k1
t_2 = t + (h / 2.)
y_2 = y +(h / 2.) * m1
u_2 = m2
k2 = (-w0**2)*np.sin(y_2) + u_2*(-2*b) + F*(w0**2)*np.cos(Wd*t_2)
m3 = u + (h / 2.) * k2
t_3 = t + (h / 2.)
y_3 = y + (h / 2.) * m2
u_3 = m3
k3 = (-w0**2)*np.sin(y_3) + u_3*(-2*b) + F*(w0**2)*np.cos(Wd*t_3)
m4 = u + h * k3
t_4 = t + h
y_4 = y + h * m3
u_4 = m4
k4 = (-w0**2)*np.sin(y_4) + u_4*(-2*b) + F*(w0**2)*np.cos(Wd*t_4)
t = t + h
y = y + (h / 6.) * (m1 + (2 * m2) + (2 * m3) + m4)
u = u + (h / 6.) * (k1 + (2 * k2) + (2 * k3) + k4)
ts.append(t)
ys.append(y)
plt.plot(ts,ys)
plt.xlabel('$t$',fontsize=18)
plt.ylabel('$\phi$',fontsize=18)
plt.axhline(y=0, color = 'black')
plt.show()
将numpy导入为np
将matplotlib.pyplot作为plt导入
t=0.0
y=0.0
u=0.0
F=1.073
Wd=2*np.pi
w0=1.5*Wd
b=w0/4
ts=[]
ys=[]
h=0.001
而(t将numpy作为np导入
将matplotlib.pyplot作为plt导入
def plotit(y):
t=0.0
#y=0.0
u=0.0
F=1.073
Wd=2*np.pi
w0=1.5*Wd
b=w0/4
ts=[]
ys=[]
h=0.001
while(t
import numpy as np
import matplotlib.pyplot as plt
def plotit(y):
t = 0.0
#y = 0.0
u = 0.0
F = 1.073
Wd = 2*np.pi
w0 = 1.5*Wd
b = w0/4
ts =[]
ys =[]
h= 0.001
while (t <= 30):
m1 = u
k1 = (-w0**2)*np.sin(y) + u*(-2*b) + F*(w0**2)*np.cos(Wd*t)
m2 = u + (h / 2.) * k1
t_2 = t + (h / 2.)
y_2 = y +(h / 2.) * m1
u_2 = m2
k2 = (-w0**2)*np.sin(y_2) + u_2*(-2*b) + F*(w0**2)*np.cos(Wd*t_2)
m3 = u + (h / 2.) * k2
t_3 = t + (h / 2.)
y_3 = y + (h / 2.) * m2
u_3 = m3
k3 = (-w0**2)*np.sin(y_3) + u_3*(-2*b) + F*(w0**2)*np.cos(Wd*t_3)
m4 = u + h * k3
t_4 = t + h
y_4 = y + h * m3
u_4 = m4
k4 = (-w0**2)*np.sin(y_4) + u_4*(-2*b) + F*(w0**2)*np.cos(Wd*t_4)
t = t + h
y = y + (h / 6.) * (m1 + (2 * m2) + (2 * m3) + m4)
u = u + (h / 6.) * (k1 + (2 * k2) + (2 * k3) + k4)
ts.append(t)
ys.append(y)
plt.plot(ts,ys)
plt.xlabel('$t$',fontsize=18)
plt.ylabel('$\phi$',fontsize=18)
plt.axhline(y=0, color = 'black')
plotit(y=0)
plotit(y=-np.pi/2.0)
plt.show()