NDSolve::ndnum: Encountered non-numerical value for a derivative at t == 1.`.

Problem 1. This will not work. See Help, Scope, example 1 (v. 12.1) S1=InterpolatingFunction[...] S1[0]=scalar InverseFunction[S1]

Fix 1: RR = InverseFunction[(S1[#]) &]

Problem 2: InverseFunction does not find an inverse. Therefore you cannot use it in NDSolve. You might alter S1 strategically so that InverseFunction can find an inverse. For this you need to expand S1 into a series.

Problem 3: RR does not contain h, so h on the left cannot be replaced with the right. RRnew = RR /. h -> Function[t, g[t]]

Problem 4: NDSolve won't solve unless function a[t] is less nebulous.

In[88]:= eqn = 0 == g'[t] - a[t] g[t]^2
Out[88]= 0 == -a[t] g[t]^2 + Derivative[1][g][t]
In[89]:= NDSolve[{eqn, g[1] == 0.1}, {g[t]}, {t, 1, 10}]
During evaluation of In[89]:= NDSolve::underdet: There are more dependent variables, {a[t],g[t]}, than equations, so the system is underdetermined.

A proposed solution (noting I am unfamiliar with the meaning behind the equations and do not state solutions are not lost):

RR = InverseFunction[(Normal[Series[S1[x], {x, 0, 5}]] /. x -> #) &]
(* note RR[domain] has infinity at R[1], is Complex valued *)
ComplexListPlot[Table[RR[x], {x, 1, 10}]]
(* (i assume) a diagonal line with +slope in complex plane *)
G[t_] := g'[t] - (RR[t])/6 + 2*g[t]^2
eqn = 0 == G[t]
sol=NDSolveValue[{G[t] == 0, g[.99] == 0.099}, g, {t, 1.1, 10}]
InterpolatingFunction[...]
ComplexListPlot[Table[sol[x], {x, 1.1, 10}]]
(* a short diagonal rise over [.03,.05] and a horizontal run *)
(* axis is roughly: Interval [.03,.1] + Interval [-10^(-17),-10^(-16)] i *)

I've found a problem however when checking my work: I cannot get

RR = InverseFunction[(Normal[Series[S1[x], {x, 0, 5}]] /. x -> #) &]

To "Inverse again back to S1" (not at the moment) - and I suspect that it may mean RR is "not usable". I found the below to be more believable...

RR = InverseSeries[Series[S1[x], {x, 0, 5}]];
(* the above re-inverses back to S1 in a "same plot" manner *)
G[t_] := g'[t] - ((Normal[RR] //. x -> #) &[t])/6 + 2*g[t]^2
NDSolve:  At t == 1.1`, step size is effectively zero; singularity or stiff ?
InterpolatingFunction[ . , Domain{{1.1, 1.1}}, scalar ]

I'm sorry I didn't check my steps more thoroughly the first time (i checked them but too briefly). But in the end - many more things must be checked still I feel, before the answer is usable (such as stiffness NDSolve mentioned).

Please disregard the previous file i send u. I noticed that i was doing something wrong, the ODE equation took the time as the series expansion of of the inverse and should be h. Also my new boundary condition was too large. But what i can see that there still an issue solving the secondry ODE with time

Attachments:

R power of 2new ...nb

understood and thank you for your time and affort Ezra