The set-up: I was reading 2.3 First-order necessary conditions for weak extrema and I got to the variable endpoint section and I got a bit confused and was hoping that someone could clarify. In 2.3.1 The author goes through minimizing the distance between two points (which I understand) by showing that the length functional (where $y=y(x)$) is

$$ I = \int_a^b ds = \int_a^b dx \sqrt{1+\dot{y}^2} $$ hence the Lagrangian is $$ L = \sqrt{1+\dot{y}^2} $$. We know that $$ \frac{d}{dx}\bigg(\frac{\partial L}{\partial \dot{y}}\bigg) - \frac{\partial L}{\partial y} = 0 $$ and it's plain to see that $\frac{\partial }{\partial y}L=0$, thus $$ \frac{d}{dx}\bigg(\frac{\partial L}{\partial \dot{y}}\bigg) = \frac{d}{dx}\bigg(\frac{\dot{y}}{\sqrt{1+\dot{y}^2}} \bigg) = 0 $$ or $$ \frac{\dot{y}}{\sqrt{1+\dot{y}^2}} = C $$. Thus $$ \dot{y}=a=\frac{c}{\sqrt{1-c^2}} $$. Integrating it's plain to see that $y=ax + b $, which is just the equation for a straight line.

This I all get, but here's where I get lost and where my question comes in. When the author gets to variable endpoints, he brings up in ex 2.4 that we should now consider the shortest path between a point and a vertical line:


I don't understand how this is different from what was done in the point-point distance situation... I'd just like it if someone could walk me through how the set-up is different and how we come to the conclusion a perpendicular line to the vertical line to the desired point is the shortest distance.

tl;dr: How does point-line distance differ from point-point distance, and how do we manipulate the Euler-Lagrange differential equation to express that? Walking me through this would be appreciated. Thank you for any help in advance.


1 Answer 1


I assume you are familiar with the proof of Euler-Lagrange (EL) eq. The main trick is to integrate by parts. In particular, this produces boundary terms (BTs).

  1. Now in the point-point distance minimization problem, these BTs vanish automatically because of essential/Dirichlet boundary conditions (BCs).

  2. Not so for the point-line distance minimization problem. Instead, this induces a natural BC.

  • $\begingroup$ thanks for the comment, I went through it slower and I think I figured it out for myself $\endgroup$
    – Illari
    Commented Oct 12, 2017 at 16:08
  • $\begingroup$ Great. I hope I didn't spoiled the fun. $\endgroup$
    – Qmechanic
    Commented Oct 12, 2017 at 21:35

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