Which shape does an elastic rod take as its ends are getting closer?

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Which shape does an idealized elastic rod take whose ends are moved towards each other? (The rod is supposed to be straight in relaxed state and to be deformable perpendicular to its direction, i.e. bendable. It's supposed to be not deformable in its own direction, i.e. not stretchable.)

Mathematically, this means:

How do I find in a systematic way (without guessing) the function $f:[0,a] \rightarrow \mathbb{R}^+$ that fulfills these conditions ($a > 0$):

  1. $f(0) = f(a) = 0$

  2. $\int_0^a \sqrt{1+ f'(x)^2}dx = L$ with $L > a$, i.e. the length of the function graph seen as a curve is $L$

  3. $\int_0^a |f''(x)|dx $, i.e. the overall "curvature" (= "energy") of the curve is minimal

Not obviously false guesses (but of course strongly depending on the right definition of "curvature" resp."energy"):

  • $f(x)$ a circle arc
    enter image description here

  • $f(x)$ a parabola
    enter image description here

  • $f(x)$ a half period of the sine function
    enter image description here

Note that in the case of the circle arc you cannot bring $a$ to $0$ closer than $a = L/\pi$ without leaving the realm of functions $y=f(x)$ - for parabolas and sine functions you can.


Assuming that @Rahul's answer is correct and that - with the correct curvature and energy - the rod is shaped like a circle arc, this is what an oscillating rod would look like (roughly) after you release its ends (but constrain them to the horizontal line):

enter image description here

Note that this doesn't look at all like a sinoidal wave.

Note further that you cannot simply apply Fourier transformation to the semicircle because the endpoints are not fixed:

enter image description here

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The energy as defined in the question, $\int_0^1|f''(x)|\,\mathrm dx$, is the total variation of the slope $f'(x)$. If $f$ is concave*, as the figure suggests is desired, then $f''$ is negative and the energy is simply $f'(0)-f'(1)$. The optimal curve must be (two edges of) a triangle, otherwise one can reduce $f'(0)$ without changing $f'(1)$ by replacing the curve with the triangle $ABC$ where $A=(0,0)$, $C=(1,0)$, $BC$ has slope $f'(1)$, and $B$ is chosen so $|AB|+|BC|=L$. Among all triangles with $|AB|+|BC|=L$, the energy $f'(0)-f'(1)=\tan\angle A+\tan\angle C$ is minimized by an isosceles triangle with sides $L/2$, $L/2$, and $1$.

Since this is not anything close to what an elastic rod does, there must be some problem with the formulation. I advise reconsidering the choice of the energy function.

*I expect that for any curve with non-monotonic slope, one can find another curve of the same length but with lower energy, but I don't have a proof yet.