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I have never solved numerically differential equations and in an optimal control problem I got this one that I cannot solve: $$1-f’(x)^2+f’(x)(x +1)+f’’(x)-f’’(x)f’(x)-f’’(x)f’(x)^2=f(x)$$ The initial conditions I have are the following: $$f(0)=0, f′(z)=−1$$ where $z$ is the first point at which $f′′(z)=0$ I hope what I wrote makes some sense. Thank you in advance for your help. This is a follow-up to sloppier version of this question I already posed: Nonlinear differential equation numerical solution+plot

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    $\begingroup$ You can't really solve this. How would you find the "first" point $z$ where $f''(x)$ is zero in order to use it for initial conditions on $f'(z)$? You'd need to know $f(z)$ first, but that needs to know $z$ first. But that is what you are solving for? $\endgroup$
    – Nasser
    Commented Apr 2, 2022 at 1:30
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    $\begingroup$ @Nasser I think that I am the reason that caused this confusion, because I wrote too quickly in the other post. Apologies for the mess and inconvenience to all. Just some comments: if you do asmpt=AsymptoticDSolveValue[{eqn}, f[x], {x, 0, 3}] // FullSimplify you get an answer. You can impose on that solution that for x=0 the answer goes to 0 and this is setting one of the constants of integration to zero. Then you can examine something like Solve[(D[asmpt, {x, 2}] /. C[1] -> 0) == 0] // Factor // FullSimplify // Apart to get three solutions. And then you can start experimenting. $\endgroup$
    – bmf
    Commented Apr 2, 2022 at 2:03
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    $\begingroup$ It's not very elegant and it is only at the level of the asymptotic expansion, but it might give some intuition for the full problem. $\endgroup$
    – bmf
    Commented Apr 2, 2022 at 2:04
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    $\begingroup$ It can't be solved by just brute-force trying a range of $f'(0)=k$ and checking the first time $f'(z)=-1$? $\endgroup$
    – josh
    Commented Apr 2, 2022 at 10:10
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    $\begingroup$ @Giuliosky: Ok, I'll try to set it up sometime today or tomorrow morning. $\endgroup$
    – josh
    Commented Apr 2, 2022 at 16:41

1 Answer 1

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There appears to be no solution for the ODE system in the question. To begin, obtain an explicit expression for f''[x].

First@SolveValues[eq, f''[x]]
(* (1 - f[x] + f'[x] + x*f'[x] - f'[x]^2)/(-1 + f'[x] + f'[x]^2) *)

where eq is the ODE in the question. The ODE therefore becomes singular for values f'[x]

SolveValues[Denominator[%] == 0, f'[x]]
(* {1/2 (-1 - Sqrt[5]), 1/2 (-1 + Sqrt[5])} *)

Next, plot f'[x] for a range of f'[0] values, and superimpose on the plot the singular values just determined, as well as the desired f'[x] = -1.

funp = ParametricNDSolveValue[{eq, f[0] == 0, f'[0] == y}, f'[x], {x, 0, 10}, {y}]
Plot[Evaluate@Table[funp[y], {y, -3, 2, .2}], {x, 0, 6}, PlotRange -> {-4, 4}]
Plot[{-1, 1/2 (-1 - Sqrt[5]), 1/2 (-1 + Sqrt[5])}, {x, 0, 6}, 
    PlotStyle -> Directive[Black, Dashed]]
Show[%, %%]

enter image description here

Visibly, where f'[x] = -1, f''[x] is nowhere near zero.

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  • $\begingroup$ Thank you for your answer! $\endgroup$
    – Giuliosky
    Commented Apr 3, 2022 at 3:10
  • $\begingroup$ @bbgodfrey: Supurb! Thank you. $\endgroup$
    – josh
    Commented Apr 3, 2022 at 12:21

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