# Distance between two particles traveling the same curve

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.

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

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.