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I have two particles simultaneously traveling a $\mathcal{C}^1$ curve $\gamma \colon [0, 1] \rightarrow \Bbb R$.

The first particle's position at time $t$ is $p_1(t) = \gamma\left( \phi(t) \right)$, where $\phi \in \mathcal{C}^1\left( \left[0, 1\right] \right)$ is strictly increasing onto $[0, 1]$.

The second particle's position at time $t$ is simply $p_2(t) = \gamma\left( t \right)$.

I would like to integrate the squared distance between the particles over the period of travel.

$$ I = \int_0^1 \left( p_1(t) - p_2(t) \right)^2 \, dt $$

Can I bound $I$ in terms of $\phi^\prime$ and say, the arclength of $\gamma$? Or the $L^2$-norm of $\gamma$?

I think I should be able to get something of the form

$$I = \int_0^1 ( \phi^\prime (x) - 1 )^2 \gamma'(x)^2 \, d x $$ $$ \text{or } \ \ \ I = \int_0^1 ( \phi^\prime (x) - 1 )^2 \gamma(x)^2 \, dx $$ modulo a change of variables somewhere.

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  • $\begingroup$ integrate the square of the distance... $\endgroup$
    – Jean Marie
    Sep 5, 2017 at 19:07
  • $\begingroup$ Another correction, on the first line $\gamma \colon [0, 1] \to \mathbb{R^2}$ (or $\mathbb{R^3}$ ?). $\endgroup$
    – Jean Marie
    Sep 5, 2017 at 19:27
  • $\begingroup$ @JeanMarie Nope, just $\Bbb R^1$. $\endgroup$ Sep 5, 2017 at 20:47
  • $\begingroup$ Surprising, it is not what is usually called a (parameterized) curve ! $\endgroup$
    – Jean Marie
    Sep 5, 2017 at 21:29
  • $\begingroup$ So your curve is "flattened" onto the $x$ axis ? $\endgroup$
    – Jean Marie
    Sep 5, 2017 at 21:49

1 Answer 1

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By fundamental theorem of calculus you have $$p_1(t)-p_2(t)=\gamma(\phi(t))-\gamma(t)=\int_t^{\phi(t)}\gamma'(z)\operatorname dz.$$ From here you have many options, for example using Cauchy--Bunyakovskii--Schwarz you may write $$|p_1(t)-p_2(t)|^2\leq|\phi(t)-t|\int_t^{\phi(t)}|\gamma'(z)|^2\operatorname dz$$ and note that for $t\geq0$ (assuming $\phi(0)=0$, but you can drop this with a bit of work) $$ |\phi(t)-t|^2=\left|\int_0^t\phi'(\tau)-1\operatorname d\tau\right|^2\leq t\int_0^t|\phi'(\tau)-1|^2\operatorname d\tau. $$ In both inequalities you can "get rid" of many $t$s at the cost of more generous gaps, by noting that $0\leq t,\phi(t)\leq1$. You can then arrange things to your liking.

Sorry for not being more precise, but if you clarify what you want I may be able to help you more.

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  • $\begingroup$ I got it, thanks for your help! $\endgroup$ Sep 20, 2017 at 20:39

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