Arbitrary descent trajectory

The role of the function​​INPUT: Expression of orbit----y[x] , e.g.
1-
2
(x-1)
​​CALCULATE：the Lagrangian, and the left hand side of the Lagrange Equation
lhs=
d
dt
∂L
∂
•
q
-
∂L
∂q
，其中
q=x
​​OUTPUT：A second order ordinary differential equations with respect to the Generalized Coordinates​THE CORE PROBLEM：Solve the Lagrange equation, acquire the expression of x[t].
lagrangianEquation[expr_]:=Module{rule,lagrangian,left,eqn,solx,soly},​​rule={y[x]->expr,y'[x]->D[expr,x],y''[x]->D[expr,{x,2}]}/.{x->x[t]};​​(*Transformationrulestobeusedinthelaststep*)​​lagrangian=
1
2
m(
2
x'[t]
+
2
(y'[x[t]]x'[t])
)+mgy[x[t]];​​(*TheLagrangianiscompoundedusingx[t],andtheindependentvariablesinthisequationare:generalizedcoordinatesandgeneralizedvelocity.*)​​(left=(D[D[lagrangian,x'[t]],t]-D[lagrangian,x[t]])//Simplify);​​Simplify[(left=left/.rule)==0,Assumptions->{x[t]!=0,x'[t]!=0,x''[t]!=0}]​​​​lagrangianEquation[y[x]]//Simplify//TraditionalForm​​lagrangianEquation[x]​​lagrangianEquation[2x]​​lagrangianEquation[
x
]
Out[]//TraditionalForm=
m
′
y
(x(t))
2
′
x
(t)
′′
y
(x(t))-g+
′′
x
(t)
2
′
y
(x(t))
+
′′
x
(t)0
Out[]=
gm2m
′′
x
[t]
Out[]=
2gm5m
′′
x
[t]
Out[]=
m4g
3/2
x[t]
+
2
′
x
[t]
-2x[t]
′′
x
[t]-8
2
x[t]
′′
x
[t]0
In[]:=
ε=0.0001;​​tLast=3;​​Clear[x,y,m,g];​​tracks=Sqrt[x],x,
2
x
,Sin
π
2
x,
1-
2
(x-1)
,LegendreP[5,x],x+
1
10
Sin[10πx],BesselJ[1/2,x];​​solutionsOfLagrangeEquation=(NDSolve[{lagrangianEquation[#]/.{m->1,g->9.8},x[ε]==ε,x'[ε]==ε},x[t],{t,ε,tLast}][[1]])&/@tracks;​​Tracks=-tracks;​​Manipulate[Show[Plot[Tracks,{x,0,1.5},PlotTheme->"Business",PlotLabel->"time="<>ToString[tt],PlotLegends->Evaluate[ToString[#,TraditionalForm]&/@tracks],AspectRatio->1],​​Graphics[{Black,PointSize[0.03],​​Point[Evaluate[({x[t],-tracks[[#]]/.x->x[t]}/.solutionsOfLagrangeEquation[[#]])&/@Range[Length[tracks]]]/.t->tt]}]]​​,{tt,ε,tLast},SaveDefinitions->True]​​
Out[]=
​
tt
x
x
2
x
sin
πx
2
1-
2
(x-1)
1
8
(63
5
x
-70
3
x
+15x)
x+
1
10
sin(10πx)
2
π
sin(x)
x
In[]:=
Export["fall.gif",Table[​​Show[Plot[Tracks,{x,0,1},PlotTheme->"Business",PlotLabel->"time="<>ToString[ttt],PlotLegends->Evaluate[ToString[#,TraditionalForm]&/@tracks],AspectRatio->1],​​Graphics[{Black,PointSize[0.03],​​Point[Evaluate[({x[t],-tracks[[#]]/.x->x[t]}/.solutionsOfLagrangeEquation[[#]])&/@Range[Length[tracks]]]/.t->ttt]}]]​​,{ttt,ε,1.5,0.03}],"DisplayDurations"->0.1]
Out[]=
fall.gif

The shortcut line drops

Orbit

p
2
∫
p
1
2
′
y
(x(t))
+1
2gy(x(t))
x
In[]:=
F=
1+
2
y'[x]
2gy[x]
;​​(*Theformoftheequationiscalculated*)​​Simplify[(D[D[F,y'[x]],x]-D[F,y[x]])==0,Assumptions->{y[x]!=0,y'[x]!=0,g>0}]
Out[]=
1+
2
′
y
[x]
+2y[x]
′′
y
[x]
3/2
g1+
2
′
y
[x]

y[x]
0
In[]:=
ε=0.001;​​curve[X_,Y_]:=Module{sol,point,T,g=9.8},​​sol=NDSolve[{1+
2
′
y
[x]
+2y[x]
′′
y
[x]==0,y[1]==1,y[X+1]==Y+1},y[x],{x,1,X+1}];​​point=Table[{x-1,-y[x]+1}/.sol[[1,1]],{x,1,X+1,0.01}];​​T=NIntegrateEvaluate
1+
2
y'[x]
2gy[x]
/.sol[[1,1]]/.{y'[x]->D[y[x]/.sol[[1,1]],x]},{x,1,X+1};​​ListLinePlot[point,PlotRange->{{-0.5,5.5},{0.5,-1.5}},Epilog->{PointSize[Large],Red,Point[{{0,0},{X,-Y}}]},PlotLabel->"T="<>ToString[T]]​​​​Manipulate[curve[x,y],{x,1,5},{y,1,5}]
Out[]=
In[]:=
Export["2.jpg",curve[2,1]]
Out[]=
2.jpg

The time of fall

In[]:=
Time[X_,Y_]:=Module[{sol,T,g=9.8,y},sol=NDSolve[{1+Derivative[1][y][x]^2+2y[x](y''[x])==0,y[1]==1,y[X+1]==Y+1},y[x],{x,1,X+1}];​​T=NIntegrate[Sqrt[(1+y'[x]^2)/(2gy[x])]/.sol[[1,1]]/.{y'[x]->D[y[x]/.sol[[1,1]],x]},{x,1,X+1}]]​​redPoints=FindMinimum[Time[#,y],{y,1}]&/@Range[0.01,5,0.01]//Quiet;
points={#,MinimalBy[Table[{yy,Time[#,yy]},{yy,0.1,3,0.04}],Last][[1,1]]}&/@Range[0.2,5,0.1];​​Show[pic,Plot[Interpolation[{#[[1]],-#[[2]]}&/@points,x],{x,0.2,5},PlotStyle->Red]]
Out[]=

CITE THIS NOTEBOOK

Brachistochrone problem: shortest time to slide on a curve​
by Yaosheng Zhang​
Wolfram Community, STAFF PICKS, March 25, 2023
​https://community.wolfram.com/groups/-/m/t/2859015