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I'm trying to see how we use the E-L equation

\begin{equation} L_x(t,q(t),q'(t))-\dfrac{d}{dt}L_v(t,q(t),q'(t))=0 \end{equation}

to find the shortest distance between two points in the Euclidean Plane (i.e. equation of a straight line?)

This is the proof I have done but I'm unsure as to whether I have executed it correctly? It needs to be precise

Let $\ell(f)$ be the length of the graph, such that:

\begin{equation} \ell(f)= \int_b^a \sqrt{1+(f'(x))^2}\,dx \end{equation}

Let $L(x, f(x), f'(x)) = L(x, y, y')$, such that:

\begin{equation} \int_b^a \sqrt{1+(f'(x))^2}\,dx =\large \int_b^a \sqrt{1+(y')^2}\,dx \end{equation}

Evaluating: \begin{equation} \quad \dfrac{\partial L}{\partial y'}=\dfrac{y'}{\sqrt{1+(y')^2}} \quad and \quad \dfrac{\partial L}{\partial y}=0 \end{equation}

By substituting these into the E-L equation, we now have:

\begin{align} \dfrac{d }{dx} \dfrac{y'}{\sqrt{1+(y')^2}} &= 0\\ \Rightarrow \dfrac{d }{dx} \dfrac{f'(x)}{\sqrt{1+(f'(x))^2}} &=0\\ \Rightarrow \displaystyle\int_b^a \dfrac{d }{dx} \dfrac{f'(x)}{\sqrt{1+(f'(x))^2}} \,dx &= constant = C\\ \Rightarrow f'(x) = \dfrac{C}{\sqrt{1-C^2}} &= m\\ \Rightarrow \quad f(x) &= mx+ \text{constant} \end{align}

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  • $\begingroup$ edit looks good. now just solve for your constants in terms of your initial two arbitrary points and youre all set. $\endgroup$ – faith_in_facts Mar 6 '14 at 19:50
  • $\begingroup$ Okay what's confusing me is although I follow this, doesn't integrating the Lagrangian defeat the object of using it? If that makes sense, that might be a completely stupid question $\endgroup$ – Sarah Jayne Mar 6 '14 at 19:55
  • $\begingroup$ Sorry and is l(f) the length of the graph or the curve? Or does it mean the same thing? (This is for my dissertation so I have to be precise!) Thanks so much for the help, I really appreciate it! $\endgroup$ – Sarah Jayne Mar 6 '14 at 19:56
  • $\begingroup$ l(f) is exactly[1] the length of the curve. And defeat the object of using it? the object is to have an equation which is solveable for our function of interest. integrating is part of this solving process. [1] mathwords.com/a/arc_length_of_a_curve.htm $\endgroup$ – faith_in_facts Mar 6 '14 at 19:58
  • $\begingroup$ Oh of course! I'm being daft! Thank you for helping me clarify this :-) $\endgroup$ – Sarah Jayne Mar 6 '14 at 20:00
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Suppose we have the two points $(x,y)=(0,a)$ and $(x,y) = (1,b)$. Suppose that the solution can be expressed as a function of one variable so that $y= y(x)$. We seek $y$ such that

$$D = \int_{0}^1 \sqrt{1+ (\frac{dy}{dx})^2} dx$$ such that y(0)=a, y(1)=b, is minimized. Of course what this equation represents is the arclength formula in 1D.

Then for the Functional D, the EL equation is :

$$ \frac{d}{dx} ( \frac{y'}{\sqrt{1+y'^2}})=0$$

Integrating this equation se have that $y' = $ constant. and so this implies that

$$y(x) = a +(b-a)x$$

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  • $\begingroup$ So can I just substitute the t's for x's in the E-L equation? As in the definition the functional is a function of 't' not of 'x' as above? $\endgroup$ – Sarah Jayne Mar 6 '14 at 19:40
  • $\begingroup$ yep. exactly. our function is a function of x,y and y'. It just so happens that the explicit y dependence is missing hence the first half of EL being equal to 0. $\endgroup$ – faith_in_facts Mar 6 '14 at 19:44
  • $\begingroup$ I'm just going to edit my post, can you check if my proof is correct please? I've confused myself a little with my x's and t's! Thank you so much! :-) $\endgroup$ – Sarah Jayne Mar 6 '14 at 19:45

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