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I have a curve built up by two straight lines in $\mathbb{R}^n$ connected at the middle specified by three distinct points $x_{-1}, x_0, x_1$:

$$ x(t) = \left\{ \begin{array}{cc} x_0 + t(x_0 - x_{-1}) & -1 \leq t \leq 0 \\ x_0 + t(x_1 - x_0) & 0 <t \leq 1 \end{array} \right. $$

This function however is only $C^0$. For a given $\epsilon > 0$, I would like to be able to create a $C^2$ function $y_{\epsilon}(t)$ such that the following properties are satisfied:

  • $y_{\epsilon}(i) = x_i$ for $i \in \{-1, 0, 1\}$
  • $y_{\epsilon}(t) = x(t)$ in all but some $t \in (-\epsilon, \epsilon)$
  • $ \int_{-1}^1 \|y_{\epsilon}^{(2)}(t)\|^2 dt \rightarrow \|\frac{1}{2} (x_{-1} - 2 x_0 + x_1)\|^2 $ as $\epsilon \rightarrow 0$

Is this possible and if so, are there any standard ways of doing this?

Thanks.

Edit: I have changed the integral to a norm squared. My goal here is to show that there is a $C^2$ curve which approximates a certain discrete approximation. I have tried the parameterisation method explained in the comments however I can't seem to get the integral in terms of the square of $\| (x_{-1} - 2 x_0 + x_1)\|^2$ but rather as a combination of $\|x_{-1} - x_0\|^2$ and $\|x_{1} - x_0\|^2$ and moreover, the integral blows up with order $O \left(\frac{1}{\epsilon}\right)$

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A kind of cheating way would be to reparametrize $x$ in such a way that in $]-\epsilon,\epsilon[$ the travelling speed has all derivatives $0$ at time $0$. The integral of the second derivatives of the new parametrization would be $0$ if you choose the rampdown symmetrically, but I think by breaking the symmetry you can tweak in such a way as to achieve every vector in the positive cone spanned by $x_{-1} - x_0$ and $x_1 - x_0$. –  Thomas Mar 15 '13 at 12:16
    
Oh I never thought about doing that, thanks for the suggestion! I'll have a look at different reparameterisations and see if something gets what I want –  muzzlator Mar 15 '13 at 16:50

1 Answer 1

up vote 1 down vote accepted

This looks like a physics problem. A particle has initial velocity $x_0-x_{-1}$ and terminal velocity $x_1-x_0$. Therefore, the integral of its acceleration $a=y''$ must be equal to the difference of velocities, $x_1-2x_0+x_{-1}$. In order to minimize the integral of the square of acceleration, the acceleration must be constant (during the time that it is allowed to be nonzero), namely $a=(x_1-2x_0+x_{-1})/(2\epsilon)$. Thus, the smallest possible value of the integral of $\|a\|^2$ is $\|x_1-2x_0+x_{-1}\|^2/(2\epsilon)$, which is exactly the blow up you observe.

A little more formally: use the Cauchy-Schwarz inequality $$\int_{-\epsilon}^\epsilon \|a(t)\|^2\,dt \;\int_{-\epsilon}^\epsilon 1\,dt \ge \left\|\int_{-\epsilon}^\epsilon a(t)\,dt\right\|^2 = \|x_1-2x_0+x_{-1}\|^2$$ hence $$\int_{-\epsilon}^\epsilon \|a(t)\|^2\,dt \ge \|x_1-2x_0+x_{-1}\|^2/(2\epsilon)$$

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Thanks and welcome to math.se! I thought this question had washed away in the sea of time. I actually came to the same conclusion a week or so ago but forgot to update my problem. Nice way of showing it rigorously, my argument was a lot more handwavey. –  muzzlator Mar 27 '13 at 6:30

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