1. ## Calculus of variations

I have $S = \int\limits_{x_1}^{x_2} \sqrt{F[x,x',s]} ds$. I want to find $x(s)$ which "extremises" S. I must take the variational derivative, etc. ... ( $\sqrt{F[x,x',s]}$ is in fact the norm of a speed)
Does someone know if I can alternatively "extremise" $S' = \int\limits_{x_1}^{x_2} F[x,x',s] ds$ and get the same $x(s)$ ?
I tried to prove that but I got an awful restriction.

2. Originally Posted by vincisonfire
I have $S = \int\limits_{x_1}^{x_2} \sqrt{F[x,x',s]} ds$. I want to find $x(s)$ which "extremises" S. I must take the variational derivative, etc. ... ( $\sqrt{F[x,x',s]}$ is in fact the norm of a speed)
Does someone know if I can alternatively "extremise" $S' = \int\limits_{x_1}^{x_2} F[x,x',s] ds$ and get the same $x(s)$ ?
I tried to prove that but I got an awful restriction.
Before you do anything, what space are you working with? (there are certain spaces where $\int f$ is defined but $\int f^2$ isn't, for example $L^2$, so you first need to ask if your new functional works in the same space).

3. This is just a little bit of math we are doing on the side in Classical Mechanics, so I am not able to answer your question. I not really aware of any subtleties in this context. Maybe my question doesn't even make sense. However, I someone would like to fill in the gaps, it would be greatly appreciated.
Here is the question :
Consider the surface in three dimensional (x,y,z)-space defined by the equation z = f(x,y) where f(x,y) is some function of x and y. Consider a curve (x(s),y(s),z(s)) on this surface, which connects two points (x0, y0, z0) and (x1, y1, z1). Here s is the parameter which labels points on this curve and runs from 0 to 1. Write down an expression for the length of this curve as an integral over s. Using the equation for the surface z = f(x,y) write this length in terms of x, y, f and their derivatives.
Write down the Euler-Lagrange equations which determine x(s) and y(s) for a curve of minimum length.

4. Originally Posted by vincisonfire
This is just a little bit of math we are doing on the side in Classical Mechanics, so I am not able to answer your question. I not really aware of any subtleties in this context. Maybe my question doesn't even make sense. However, I someone would like to fill in the gaps, it would be greatly appreciated.
Here is the question :
Consider the surface in three dimensional (x,y,z)-space defined by the equation z = f(x,y) where f(x,y) is some function of x and y. Consider a curve (x(s),y(s),z(s)) on this surface, which connects two points (x0, y0, z0) and (x1, y1, z1). Here s is the parameter which labels points on this curve and runs from 0 to 1. Write down an expression for the length of this curve as an integral over s. Using the equation for the surface z = f(x,y) write this length in terms of x, y, f and their derivatives.
Write down the Euler-Lagrange equations which determine x(s) and y(s) for a curve of minimum length.
Okay, so without worrying about the more subtle aspects what do you have so far?

5. I thought I should start by assuming solutions $

x(s)
$
and $

y(s)
$

for
$

S = \int\limits_{x_1}^{x_2} F[x,x',s] ds
$
.
Then
$

\frac{d}{ds}\frac{\partial F}{\partial x'} - \frac{\partial F}{\partial x}=0
$

Similarly for $

y(s)
$
.
Then we want
$

u(s)
$
and $

v(s)
$

such that
$

\frac{d}{ds}\frac{\partial \sqrt{F}}{\partial u'} - \frac{\partial \sqrt{F}}{\partial u}=0
$
.
Similarly for $

v(s)
$
.
Then I got
$-\frac{1}{4(F)^{\frac{3}{2}}}\left(
\frac{\partial F}{\partial u}u'+\frac{\partial F}{\partial u'}u''+\frac{\partial F}{\partial v}v'+\frac{\partial F}{\partial v'}v''+\frac{\partial F}{\partial s}
\right)+
\frac{1}{2\sqrt{F}}
\left(\frac{d}{ds}\frac{\partial \sqrt{F}}{\partial u'} - \frac{\partial \sqrt{F}}{\partial u}\right)=0
$

I thought that it would come out nicely and that putting $
u(x)=
x(s)
$
everything should work.

6. Originally Posted by vincisonfire
I thought I should start by assuming solutions $

x(s)
$
and $

y(s)
$

for
$

S = \int\limits_{x_1}^{x_2} F[x,x',s] ds
$
.
Then
$

\frac{d}{ds}\frac{\partial F}{\partial x'} - \frac{\partial F}{\partial x}=0
$

Similarly for $

y(s)
$
.
Then we want
$

u(s)
$
and $

v(s)
$

such that
$

\frac{d}{ds}\frac{\partial \sqrt{F}}{\partial u'} - \frac{\partial \sqrt{F}}{\partial u}=0
$
.
Similarly for $

v(s)
$
.
Then I got
$-\frac{1}{4(F)^{\frac{3}{2}}}\left(
\frac{\partial F}{\partial u}u'+\frac{\partial F}{\partial u'}u''+\frac{\partial F}{\partial v}v'+\frac{\partial F}{\partial v'}v''+\frac{\partial F}{\partial s}
\right)+
\frac{1}{2\sqrt{F}}
\left(\frac{d}{ds}\frac{\partial \sqrt{F}}{\partial u'} - \frac{\partial \sqrt{F}}{\partial u}\right)=0
$

I thought that it would come out nicely and that putting $
u(x)=
x(s)
$
everything should work.
It should be kind of obvious in the Euler-Lagrange equations for x,y that it wasn't going to be easy (because of the chain rule). If you were trying to solve the questions you posed on your second post here's how I would do it:

First note that if $z=f(x,y)$ is you parametrization of the surface (which we'll call M) with $(x,y)\in \Omega \subset \mathbb{R} ^2$ then curves in $M$ are given by curves $(u(t),v(t)) \in \Omega$ with $t\in [0,1]$ (reparametrize if necessary) and so the curve in $M$ is given by $(u(t),v(t),f(u(t),v(t)))=:M(t) \in \Omega \times \mathbb{R}$ so the lenght of the curve becomes $\int_{0}^{1} \| M'(t)\|dt =\int_{0}^{1} \|M_uu'+M_vv' \|dt$ $= \int_{0}^{1} (Au'^2+2Bu'v'+Cv'^2)^{\frac{1}{2} } dt$ where $A=\| M_u \| ^2 = (1+f_u^2),$ $B= \langle M_u,M_v \rangle$ and $C= \| M_v \| ^2 =(1+f_v^2)$.

Now, without worrying about differentiability, regularity of the surface or other stuff we can easily arrive at the Euler-Lagrange equations:

$\frac{d}{dt} \left( \frac{Au'+Bv'}{ \| M'\| } \right) - \frac{A_uu'^2+2B_uu'v'+C_uv'^2}{2\| M'\| }$

and we obtain an analogous one for $v$, now just replace $M=(u,v,f(u,v))$ and you have all your answers (although this won't be a pretty task either, maybe someone else can provide a more efficient way to do this).

7. I can do the question I put on my second post (or think I can, I got the answer you gave me), but I was wondering if it was equivalent to mininize $
\int_{0}^{1} \| M'(t)\|^2 dt
$
instead of $
\int_{0}^{1} \| M'(t)\| dt
$
.

8. Not really. You will get a different Euler-Lagrange equation.

Now, kudos to Jose for all that manual labor.
If we wished to make things a notch simpler, we could let the plane curve be expressed locally as a graph, $x(u)=(u,v(u)), a and proceed from there. Anyhow, you possibly won't get anything better than what wikipedia and wolfram have to offer under the lemma "Geodesic". That is term assigned to curves that satisfy the Euler-Lagrange equations concerning the variation of length on a surface (and, more generally, on a manifold).

Here's food for thought. Is every geodesic a length minimizing curve between any two of its points?

9. Originally Posted by Rebesques
Not really. You will get a different Euler-Lagrange equation.

Now, kudos to Jose for all that manual labor.
If we wished to make things a notch simpler, we could let the plane curve be expressed locally as a graph, $x(u)=(u,v(u)), a and proceed from there. Anyhow, you possibly won't get anything better than what wikipedia and wolfram have to offer under the lemma "Geodesic". That is term assigned to curves that satisfy the Euler-Lagrange equations concerning the variation of length on a surface (and, more generally, on a manifold).

Here's food for thought. Is every geodesic a length minimizing curve between any two of its points?
But for example, if $F: \Omega \rightarrow \mathbb{R}$ is such that $\Omega$ is bounded and $F$ positive and $\sqrt{F} \in L^2(\Omega )$ then Hölder's inequality give $\| \sqrt{F} \| _1 \leq \| F \| _1 |\Omega |^{\frac{1}{2} }$ so maybe we could obtain another such inequality that could give what is asked, specially since we're restricting $F$ to $[0,1]$ (we are integrating the curve...more or less) and possibly continous curves.

Edit: Well, this doesn't relate the minimums of the two functionals, but my point is if we restrict to some spaces couldn't we have that although the Euler-Lagrange eq are distinct they have at least one solution in common (and that this could be indeed a minimum for both).

vincisonfire: You're really better off not trying this to simplify things, since as you can see it doesn't. However it would be interesting to give necessary and sufficient conditions for these two functionals to have at least a critical point in common.

10. so maybe we could obtain another such inequality that could give what is asked

Feel free to try man. If you succeed, you 'll get your name coined to this new theory.

11. Originally Posted by Rebesques
Feel free to try man. If you succeed, you 'll get your name coined to this new theory.
Not going to, since as I said this says nothing about what I wanted () but it would be interesting to know how to relate the critical points (at least) of two 'alike' functionals.

12. Originally Posted by Rebesques
Here's food for thought. Is every geodesic a length minimizing curve between any two of its points?
My guess is that not necessarily, since say we have a point z in a geodesic joining x,y such that there is a strictly 'shorter' curve joining x,z then it could happen that this curve is in your function space but the joint x,z and z,y isn't (having a singular point at z or something)

13. Originally Posted by Jose27
My guess is that not necessarily, since say we have a point z in a geodesic joining x,y such that there is a strictly 'shorter' curve joining x,z then it could happen that this curve is in your function space but the joint x,z and z,y isn't (having a singular point at z or something)

Yes, no one guarantees that could not happen (and by example, you can show it does).

The easiest example of non-minimization I can think of, involves the sphere and a major circle (which is ofcourse a geodesic).