在 Scilab 上求解非线性二阶微分方程？

如何解决在 Scilab 上求解非线性二阶微分方程？

我需要解决以下问题

Public Sub PivotFieldsChange()
Dim ValType As String,param As String
Dim pf As PivotField
Dim pt As PivotTable

Application.ScreenUpdating = False
Sheet = "mysheet"
'select between % calculated column or normal column
If Range("Z1").Value = 1 Then
    ValType = "%"
Else: ValType = ""
End If
Application.EnableEvents = False
For Each pt In Sheets(Sheet).PivotTables
    Select Case pt.Name
        Case "case1": param = "param1"
        Case "case2": param = "param2"
        Case "case3": param = "param3"
        Case Else: GoTo line1
End Select
pt.PivotFields("Values").PivotItems("X").Visible = False
pt.PivotFields("Values").PivotItems("Y").Visible = False

pt.PivotFields (param & ValType & "_X")
pt.PivotFields(param & ValType & "_X").Orientation = xlDataField
pt.PivotFields (param & ValType & "_Y")
pt.PivotFields(param & ValType & "_Y").Orientation = xlDataField
For Each pf In pt.DataFields
    pf.Function = xlAverage
    pf.Caption = Right(pf.Caption,1)

Next
line1:
Next pt
Application.EnableEvents = True
End Sub

我发现很难求解，因为 -cos(y)y''+sin(y)y'^2+sin(y)=0,y'(0)=y'(1)=0,such that y=y(t) 项和边界条件。

解决方法

这是您的 bvp 的 Scilab 代码

-cos(y)y''+sin(y)y'^2+sin(y)=0,y'(0)=y'(1)=0,y(0)=0,y (1)=1.5

但不同的边界条件没有给出简单的解决方案。您必须首先将 y'' 写为 y 和 y' 的函数。函数 fsub 计算 y'' 作为 u=[y,y']

的函数

function ysecond=fsub(x,u)
    y=u(1);
    yprime=u(2);
    ysecond = sin(y)/cos(y)*(1+yprime^2);
end

function g=gsub(i,u),y=u(1);    
    select i
      case 1 then  // x=zeta(1)=0
        g = y // y(0)=0
      case 2 then // x=zeta(2)=1
        g = y-1.5 // y(1)=1.5
    end
end

N=1;
m=2;
x_low=0
x_up=1;
xpoints=linspace(0,1,100);
zeta=[0,1];


u = bvodeS(xpoints,m,N,x_low,x_up,fsub,gsub,zeta)
plot(xpoints,u(1,:))