thanks for taking a look at my question. This is a homework problem from a section covering Euler-Lagrange equations. I'm asked to consider the arc length formula:
$S = \int\limits_{{t_1}}^{{t_2}} {\sqrt {{{\left( {{\textstyle{{du} \over {dt}}}} \right)}^2} + {{\cos }^2}u{{\left( {{\textstyle{{dv} \over {dt}}}} \right)}^2}} dt} $.
Then I'm asked to find the geodesic between two fixed points with respect to the arc length formula, shown above, by choosing $t=u$. The question also gives a hint that says:
$\int {{\textstyle{c \over {\cos u\sqrt {{{\cos }^2}u - {c^2}} }}}du = - \arccos } \left( {{\textstyle{c \over {\sqrt {1 - {c^2}} }}} \cdot {\textstyle{{\sin u} \over {\cos u}}}} \right)$, where $c$ is a constant.
So I've substituted $u$ for $t$ to arrive at:
$S = \int\limits_{{u_1}}^{{u_2}} {\sqrt {{{\left( {{\textstyle{{du} \over {du}}}} \right)}^2} + {{\cos }^2}u{{\left( {{\textstyle{{dv} \over {du}}}} \right)}^2}} du \to } \int\limits_{{u_1}}^{{u_2}} {\sqrt {1 + {{\cos }^2}u{{\left( {{\textstyle{{dv} \over {du}}}} \right)}^2}} du} $
Based on the integral obtained after the substitution, would I be correct if I used:
$\frac{d}{{du}}\left( {\frac{{\partial f}}{{\partial v'}}} \right) = \frac{{\partial f}}{{\partial v}}$, where $f = \sqrt {1 + {{\cos }^2}u{{\left( {{\textstyle{{dv} \over {du}}}} \right)}^2}} du$ and $v' = \frac{{dv}}{{du}}$
Thank you for your help!
So the functional you are optimizing here is
$$ \int_{u_1}^{u_2} \sqrt{1 + \cos(u)^2 \left( \frac{dv}{du} \right)^2} du $$
This is expressed as
$$ \int_{u_1}^{u_2} F(u,v, \frac{dv}{du}) du $$
ELE states that this is optimized by finding $v$ that satisfies
$$ \frac{\partial F}{\partial v} - \frac{d}{du} \frac{\partial f}{\partial \frac{dv}{du}} = 0 \rightarrow \frac{\partial F}{\partial v} = \frac{d}{du} \frac{\partial f}{\partial \frac{dv}{du}}$$
As you correctly noted.
SPOILER ALERT SOLUTION BELOW:
We can go ahead and solve this equation then, noting that $\frac{\partial F}{\partial v} = 0$, and
$$\frac{d}{du} \frac{\partial f}{\partial \frac{dv}{du}} = \frac{d}{du} \left[ \frac{ \cos(u)^2 \frac{dv}{du}}{\sqrt{1 + \cos(u)^2 \left( \frac{dv}{du} \right)^2}} \right] = -\frac{ 2\sin(u) \frac{dv}{du}}{\sqrt{1 + \cos(u)^2 \left( \frac{dv}{du} \right)^2}} + \frac{ \cos(u)^2 \frac{d^2v}{du^2}}{\sqrt{1 + \cos(u)^2 \left( \frac{dv}{du} \right)^2}} - \frac{ \cos(u)^2 \frac{dv}{du}\left( \cos(u)^2 \frac{dv}{du} \frac{d^2v}{du^2}- \sin(u) (\frac{dv}{du})^2 \right)}{\left(1 + \cos(u)^2 \left( \frac{dv}{du} \right)^2\right)^{\frac{3}{2}}}$$
So we are now finding $v(u)$ such that:
$$0 = \frac{ \cos(u)^2 \frac{d^2v}{du^2}}{\sqrt{1 + \cos(u)^2 \left( \frac{dv}{du} \right)^2}}-\frac{ 2\sin(u) \frac{dv}{du}}{\sqrt{1 + \cos(u)^2 \left( \frac{dv}{du} \right)^2}} - \frac{ \cos(u)^2 \frac{dv}{du}\left( \cos(u)^2 \frac{dv}{du} \frac{d^2v}{du^2}- \sin(u) (\frac{dv}{du})^2 \right)}{\left(1 + \cos(u)^2 \left( \frac{dv}{du} \right)^2\right)^{\frac{3}{2}}} $$
Multiplying out the bottom square root yields:
$$0 = \cos(u)^2 \frac{d^2v}{du^2}-2\sin(u) \frac{dv}{du} - \frac{ \cos(u)^2 \frac{dv}{du}\left( \cos(u)^2 \frac{dv}{du} \frac{d^2v}{du^2}- \sin(u) (\frac{dv}{du})^2 \right)}{1 + \cos(u)^2 \left( \frac{dv}{du} \right)^2} $$
Again forming common denominators
$$0 = (1 + \cos(u)^2 \left( \frac{dv}{du} \right)^2)(\cos(u)^2 \frac{d^2v}{du^2}-2\sin(u) \frac{dv}{du} ) - \cos(u)^2 \frac{dv}{du}\left( \cos(u)^2 \frac{dv}{du} \frac{d^2v}{du^2}- \sin(u) (\frac{dv}{du})^2 \right) $$
We reduce the first term:
$$0 = (1 + \cos(u)^2 \left( \frac{dv}{du} \right)^2)(-2\sin(u) \frac{dv}{du} ) - \cos(u)^2 \frac{dv}{du}\left(- \sin(u) (\frac{dv}{du})^2 \right) + \cos(u)^2 \frac{d^2v}{du^2} $$
And Again:
$$0 = \cos(u)^2 \frac{d^2v}{du^2} - 2 \sin (u) \frac{dv}{du} + \cos(u)^2 \sin(u) \left(\frac{dv}{du} \right)^3$$
But proceeding further seems... stuck.
Perhaps I made an algebraic mistake