Variational Calculus : Geodesics w/ Constraints

[lstellyl]

Homework Statement

Consider the cylinder S in R³ defined by the equation x^2+y^2=a^2

(a). The points A=(a,0,0) \: and \: B = (a \cos{\theta}, a \sin{\theta}, b) both lie on S. Find the geodesics joining them.

(b). Find 2 different extremals of the length functional joining A=(a,0,0) and C = (a, 0 2 \pi, b). How many extremals join A and C?

Homework Equations

Euler-Lagrange generalized equations:
\frac{\partial}{\partial x_i } \{F+\sum_k \lambda_j (t) G_j \} - \frac{d}{dt} \{ \frac{\partial}{\partial x'_i } \{ F+\sum_k \lambda_j (t) G_j \} \}

The Attempt at a Solution

using F= \sqrt{x'^2+y'^2+z'^2} \: and \: G= x^2+y^2 and assuming that the geodesic path \gamma was parameterized by arc length, meaning that |u'(t)|=1, i was able to get the following equations:

\lambda (t) x(t) = x''(t)

\lambda (t) y(t) = y''(t)

z'(t) = C where C is a constant.

although for some reason mathematica wouldn't solve this for me... these are all simple harmonic oscillator equations... (hopefully, i say that and actually solve them correctly...)

forming the equations with constants and solving for the boundary conditions, i found the following using Mathematica's Solve function:

<br /> x(t) = A e^{k t} + D
y(t) = B e^{k t} + E
z(t) = \frac{b}{L} \:\: (trivial)

where

A = - \frac{2 a \sin{\theta/2}^2}{e^{ k L} - 1}

B = \frac{ a \sin{\theta}}{e^{ k L} - 1}

D = \frac{ a ( e^{k L} - \cos{\theta})}{e^{ k L} - 1}

E = B

where k = Sqrt[lambda] and L is the length of the curve (since it is parameterized by arc-length)

...

This solution doesn't look right to me, and I don't want to move on to part 2 with this incorrect. I feel like i am doing something implicitly wrong and should maybe be canceling out lambdas or something... because I still have a lambda in those constant values...

any help or direction would be appreciated... it has not been a good day for me and i am having trouble focusing.

 

[Dick]

You probably actually want to use G=(x^2+y^2-a^2). And yes, they are simple harmonic motion equations, since lambda is a constant. Not a function of t. Your next step after that shouldn't be to resort to Mathematica's Solve function. Those solutions actually look pretty wacky. Your next step should be to try to find lambda in terms of C and a^2. Hint: differentiate x^2+y^2=a^2 twice and use your |u'(t)|=1 condition and the other equations. Do you really have to solve this as a constraint problem? There are easier ways.