Question

我试图在同一图表上绘制两个解决方案，同时仅改变这些解决方案的初始条件。我希望初始条件是y = 0的一个值和y = -pi / 2的另一个值。我不确定如何让它们在同一个图形上，而无需复制并粘贴所有内容并创建新数组。

<ListView x:Name="ClipboardList"
                  xmlns:m="using:QuickieEdit.Models"
                  ItemsSource="{x:Bind ViewModel.MemoryItems}">
            <ListView.ItemTemplate>
                <DataTemplate x:DataType="m:MemoryItem">
                        <StackPanel Orientation="Horizontal">
                            <Button x:Name="MemoryCopyBtn"
                                         Content="Copy"
                                         Click="How to Copy currently selected
                                         MemoryListItem.Text?"/> 
                            <TextBox x:Name="MemoryListItem"
                                           Text="{x:Bind Memory, Mode=TwoWay}">
                           </TextBox>                                  
                       </StackPanel>
                </DataTemplate>
            </ListView.ItemTemplate>

Answer 1

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()

在一个图上绘制具有相同驱动强度的两个解

1 个答案: