# If a curve $\gamma$ through two points $P,Q$ satisfy $\|Q-P\| = \int^{t_1}_{t_0} \| \gamma^{'} \| \, \text {d}t$, then $\gamma$ is a straight line?

In a theorem called "A straight line is the shortest curve through two given points", I prove that for any two points $P,Q \in \mathbb R^2$ and any curve $\gamma : (a,b) \rightarrow \mathbb R^2$ with $\gamma(t_0) = P$ and $\gamma(t_1) = Q$ we have $$\| Q-P \| \le \int^{t_1}_{t_0} \| \gamma^{'} \| \, \text {d}t.$$ Then by taking the curve $\xi: (-\infty, \infty) \rightarrow \mathbb R^2$ with $\xi(t) = (Q-P)t + P$, I see that a straight line curve does indeed satisfy the inequality with inequality (with $t_0 = 0$ and $t_1 = 1$) and is thus optimal.

However, my question is: how do I show that if $\xi$ satisfy the inequality above with equality, then $\xi$ must be of the form $\xi(t) = (Q-P)t + P$ (which the definition of a straight line in $\mathbb R^n$ ?), that is $\xi$ is a straight line ?

• There are several ways to show this, a more popular one being a typical calculus of variations problem: $\frac{d}{dr}\bigg|_{r = 0}\int_{t_0}^{t_1} \| \gamma'(s) + r\; \eta'(s) \| ds$ for a suitable class of functions $\eta$. Differentiate under the integral, $O(r^2)$ and other terms disappear. Integrate by parts and solve the corresponding ODE -- you'll get a straight curve. Conclude by an argument as in the "fundamental lemma of calculus of variations." en.wikipedia.org/wiki/… . For convenience use $\| \gamma'(s) + r\; \eta'(s) \|^2$. – snar Jan 27 '15 at 17:54
• Ohh, I see. This is pretty hard. I thought the statement itself implied it. – Shuzheng Jan 27 '15 at 18:03

I shall assume that your $$\gamma'(t)=\bigl(x'(t),y'(t)\bigr)\qquad(a\leq t\leq b)$$ is continuous. Furthermore it is no restriction of generality to assume $$P=(0,0),\qquad Q=(q,0),\quad q>0\ .$$ Then \eqalign{\int_a^b\|\gamma'(t)\|\>dt&=\int_a^b\sqrt{x'^2(t)+y'^2(t)}\>dt\cr &\geq\>\int_a^b\bigl|x'(t)\bigr|\>dt\geq\int_a^b x'(t)\>dt\cr &=x(b)-x(a)=q=\|Q-P\|\ . \cr}\ Claim: We have equality sign here iff $$y'(t)\equiv0\qquad(a\leq t\leq b)\tag{1}$$ and $$x'(t)\geq0\qquad(a\leq t\leq b)\ .\tag{2}$$ Proof. That $(1)\wedge(2)$ imply equality is obvious. For the converse assume that $y'(t_*)=p\ne0$ for some $t_*$. Then $y'^2(t)\geq {p^2\over2}$ on an interval $I$ of length $\delta>0$. This implies $$\sqrt{x'^2(t)+y'^2(t)}-\bigl|x'(t)\bigr|={y'^2(t)\over \sqrt{x'^2(t)+y'^2(t)}+\bigl|x'(t)\bigr|}\geq{p^2\over2M}\qquad(t\in I)\ ,$$ where $M:=\max_{t\in I}\sqrt{x'^2(t)+y'^2(t)}$. In this way we obtain strict inequality of the integrals, and similarly when $x'(t_*)<0$ for some $t_*$.

Now $(1)$ says that $\gamma$ has to lie on the $x$-axis, and $(2)$ says that the point $\gamma(t)$ must not move leftwards anytime for $t\in[a,b]$. That's all we can say, and confirms that $\gamma\bigl([a,b]\bigr)$ has to be the segment $[P,Q]$. But it is not required that the exact "timetable" of $\gamma$ is linear in $t$.

• Why $x(Q)$ and $x(P)$ ? :) It is not a function of points ? – Shuzheng Jan 27 '15 at 19:56
• And $y^{'}(t)$ could be nonzero for some $t$ without integral change ? – Shuzheng Jan 27 '15 at 19:58
• Since $y′$ is supposed continuous we have strict inequality as soon as $y′(t) \neq 0$ for some $t \in [a,b]$. How do I see this ? – Shuzheng Jan 27 '15 at 20:10

Actually, you can't. Maybe some background information is of interest. What you found is the special case of a parametrization parametrized 'proportional to arclength', that is the length of a segment of the curve is a constant times the difference of the parameter values. (That means that you can show the minimizing curve is a straight line, but you cannot show it's parametrization is linear)

Any reparametrization of the line will do (including nonlinear ones), that is, minimize length, since the length functional is independent of the parametrization (it's just the change of variables formula). This fact makes it rather difficult to minimize the length functional, because there are simply too many minimizers (minimizing sequences will not converge in a reasonable topology).

There is a trick to get around this, which is to consider the energy functional $$\frac{1}{2}\int|\gamma^\prime|^2 dt = E(\gamma)$$ which is minimized precisely by straight lines parametrized proportionally to arclength (which can be shown using methods from the calculus of variations). One can, in addition, show, that the length $\ell( \gamma)^2\le c E(\gamma)$ with equality at a minimizer of $E$, so minimizing $E$ also minimizes $\ell$ .

Edit: the inequality between $\ell$ and $E$ needed correction. If you want to check the details see http://en.wikipedia.org/wiki/Geodesic