Using ParametricNDSolve with n subscripted functions

I want to calculate trajectories of an n-pendulum. For the number n=1 it's just a pendulum, for n=2 a double pendulum , for n=3 a triple and so on. The Solver has to solve n equations for the angle. The phi's are the angles and ParametricNDSolve has to solve these. The parameters of ParametricNDSolve are the initial conditions - position and velocity at t=0 of each pendulum. So for n=3 pendulums I have 6 Parameters (position and velocity three times).

The Problem is that I cannot figure out how to define these parameters later and plot the function. I tried the solution from the documentation, but I cannot adapt it to my subscripted functions (I guess - maybe the problem lies somewhere else).

(The whole thing works if I use NDSolve and define the starting conditions beforehand.)

my code:

n = 1; (*number of pendulums chained to each other*)

g = 10;
α =  0;
tend = 50;
(* g is gravity, alpha  is friction, tend is end-time of solve *)

m = Table[1, {i, n}]; (* masses of pendulums *)
l = Table[1 + 0 i, {i, n}]; (* lengths of pendulume *)

r[i_, t_] := Sum[l[[j]]*{Sin[ Subscript[ϕ, j][t]], -Cos[Subscript[ϕ, j][t]]}, {j, 1, i}]; (* positions of pendulums *)
T = Sum[m[[i]]*D[r[i, t], t].D[r[i, t], t]/2, {i, 1, n}];(* kinetic energy *)
U = Sum[m[[i]]*r[i, t].{0, g}, {i, 1, n}];(* potential energy *)
F = Sum[1/2*α*m[[i]]*D[r[i, t], t].D[r[i, t], t], {i, 1, n}]; (* friction term *)
L = T - U // Simplify // Evaluate (* lagrangian *)

sol = ParametricNDSolve[
Table[{D[D[L, Subscript[ϕ, i]'[t]], t] -
D[L, Subscript[ϕ, i][t]] == -D[F,
Subscript[ϕ, i]'[t]],
Subscript[ϕ, i] == Subscript[ϕ0s, i],
Subscript[ϕ, i]' == Subscript[ϕ0v, i]}, {i, n}],
Table[Subscript[ϕ, i][t], {i, 1, n}], {t, 0, tend + 1},
Flatten[Table[{Subscript[ϕ0s, i], Subscript[ϕ0v, i]}, {i,
n}], 1]]

(* ϕ0s is start value, ϕ0v is start velocity *)

f10 = Subscript[ϕ, 1][1, 0][t] /. sol
Plot[f10[t], {t, 0, 1}]

Help would be much appreciated. For anyone interested, I'm trying to find cyclic solutions of the pendulum.

Example for n=3

n = 3;(*number of pendulums chained to each other*)g = 10;
\[Alpha] = 0;
tend = 50;
(*g is gravity,alpha is friction,tend is end-time of solve*)

m = Table[1, {i, n}];(*masses of pendulums*)l =
Table[1 + 0 i, {i, n}];(*lengths of pendulume*)
r[i_, t_] :=
Sum[l[[j]]*{Sin[
Subscript[\[Phi], j][t]], -Cos[Subscript[\[Phi], j][t]]}, {j, 1,
i}];(*positions of pendulums*)T =
Sum[m[[i]]*D[r[i, t], t].D[r[i, t], t]/2, {i, 1,
n}];(*kinetic energy*)U =
Sum[m[[i]]*r[i, t].{0, g}, {i, 1, n}];(*potential energy*)F =
Sum[1/2*\[Alpha]*m[[i]]*D[r[i, t], t].D[r[i, t], t], {i, 1,
n}];(*friction term*)L =
T - U // Simplify // Evaluate (*lagrangian*);

sol = ParametricNDSolve[
Table[{D[D[L, Subscript[\[Phi], i]'[t]], t] -
D[L, Subscript[\[Phi], i][t]] == -D[F,
Subscript[\[Phi], i]'[t]],
Subscript[\[Phi], i] == Subscript[\[Phi]0s, i],
Subscript[\[Phi], i]' == Subscript[\[Phi]0v, i]}, {i, n}],
Table[Subscript[\[Phi], i][t], {i, 1, n}], {t, 0, tend + 1},
Flatten[
Table[{Subscript[\[Phi]0s, i], Subscript[\[Phi]0v, i]}, {i, n}], 1]]

(*\[Phi]0s is start value,\[Phi]0v is start velocity*)

f10 = Subscript[\[Phi], 1][t] /. sol
Plot[f10[1, .1, .1, 1, 0, .1], {t, 0, 1},
PlotLabel -> Row[{"n = ", n}]] To show several functions we use

Table[f[i] = Subscript[\[Phi], i][t] /. sol, {i, 1, n}]

Plot[Evaluate[Table[f[i][1, .1, .1, 1, 0, .1], {i, 1, n}]], {t, 0, 1},
PlotLabel -> Row[{"n = ", n}]] • wow thank you very much! It seems the exact notation of mathematica was a little too complicated for me. I have one more question (since you seem to know mathematica quite well): Is there a way to make the number of parameters that I give to the parametrized solution arbitrary? What I mean is the [1, .1, .1, 1,0,.1]-part in your solution. I'd like to use a vector of length n*2. Right now I have to give each individual parameter to the function. I have to change the expression when I change n. – exocortex Feb 28 at 15:35
• Parameters can be set separately, for example par = {1, .1, .1, 1, 0, .1}; Plot[Evaluate[Table[f[i] @@ par, {i, 1, n}]], {t, 0, 1}, PlotLabel -> Row[{"n = ", n}]] – Alex Trounev Feb 28 at 16:15
• Thanks a lot! This is TRULY helpful! I cannot upvote your solution unfortunatly! I am quite new here. Do you think I should change my original question a little bit so it might help others more? – exocortex Feb 28 at 21:51