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Let $s\in \left(0, 1\right)$, and consider the task of finding the 'nicest-possible' function $f$ such that

  1. $f(0) = 0, f^\prime (0) = s$, and
  2. $f(1) = 1, f^\prime (1) = 0$,

where 'niceness' is framed as having minimal mean-square 'acceleration', as measured by

$$\mathcal{E} \left( f \right) := \int_0^1 f^{\prime\prime} \left( t \right)^2 \, \mathrm{d}t .$$

In principle, I'd also like to constrain that $f^\prime (t) > 0$ for all $t \in (0, 1)$, but for more-or-less physical reasons, I suspect that the optimal solution will satisfy this constraint automatically. As such, I'm just looking for the solution to the unconstrained problem.

I have derived a partial solution, starting from a bang-bang ansatz (i.e. that for some $\tau \in (0, 1)$, the acceleration $f^{\prime\prime}$ takes one value on $[0, \tau]$ and another value on $[\tau, 1]$) and then optimising the free parameters of that ansatz. I'm not sure whether the solution which I obtain is globally optimal, or how I might justify that.

As such, I'm interested in identifying the optimal $f$ (as a function of $s$), and justifying its optimality convincingly.

(I added the 'spline' tag because I note a superficial similarity to the task of fitting smoothing splines to data, but I'm not sure that it's substantively useful for solving the problem)

(I'm also nominally interested in the solution for $s > 1$, on the off chance that one can solve both cases in one fell swoop)

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    $\begingroup$ What have you tried? Euler-Lagrange seems straightforward here $\endgroup$ Commented May 12 at 17:39
  • $\begingroup$ @whpowell96 As mentioned in the question, I tried an ansatz and focused on that (for my application, the optimality wasn't crucial; it's only after the fact that I'm interested). I had it in my head that the mixture of discrete constraints with the continuous objective would be a bit annoying, though with the benefit of hindsight, maybe this isn't the case - Euler-Lagrange basically says that $f^{\prime\prime\prime\prime} = 0$, so I guess that I just have to fit the cubic myself! Maybe the similarity with splines was sufficient after all. $\endgroup$ Commented May 12 at 17:43
  • $\begingroup$ The mixture of constraints just adds boundary conditions to the EL equation. Not a big deal for something this simple $\endgroup$ Commented May 12 at 18:27
  • $\begingroup$ @whpowell96 Yeah, I see this now; thanks for getting me back on track. $\endgroup$ Commented May 12 at 18:44

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As pointed out by @whpowell96, a standard Calculus of Variations approach suffices here.

The Euler-Lagrange equation works out to $f^{\prime\prime\prime\prime} = 0$, and so the solution can be derived by fitting a cubic function which satisfies the requisite constraints, namely

$$f_s (x) = s\cdot x \cdot \left( 1 - x \right)^2 + \left[ 3 \cdot x^2 - 2\cdot x^3 \right]. $$

Moreover, one checks that

$$f_s^\prime (x) = \left( 1 - x \right) \cdot \left( s + 3 \cdot \left( 2 - s \right)\cdot x\right)$$

which, for the claimed range of $s$, is plainly non-negative on $\left[ 0, 1 \right]$, as anticipated.

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