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I want to minimize $I = \int |\dot{x}|^2 dt$ subject to the constraint $|x|^2=1$ (sphere) which gives an Euler equation of $\lambda x - \ddot{x} = 0$.

I have to show that the Euler equation is actually $|\dot{x}|^2 x - \ddot{x} = 0$. Is it right to assume that $\lambda=|\dot{x}|^2$ simply by the fact that it minimizes $I^* = \int |\dot{x}|^2- \lambda (|x|^2-1) dt$ which is $\geq 0$, so the $\lambda$ that minimizes $I^*$ is $|\dot{x}|^2$?

If I then try to integrate the Euler equation, then I get a SHM equation:

$x1= A \cos(|\dot{x}| t - C)$ where A, C are constants and similarly for x2, x3

But how do I combine them to give the equation of a great circle, since I don't know the $C$'s?

Thank you for any enlightenment!

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The dot is the derivative with respect to time. I tried to improve the formatting. – Raskolnikov Jul 26 '11 at 9:45
Derp, I should have figured. – anon Jul 26 '11 at 9:53
If you write $I^*$ the way you did, your Euler Lagrange equation is wrong. When you integrate by parts you should pick up an additional minus sign, which should give $\lambda x + \ddot{x} = 0$ to be you ELE. – Willie Wong Jul 26 '11 at 10:08
up vote 5 down vote accepted

Firstly, your expression for $I^*$ does not agree with your Euler equation: if you write $I^*$ as you did, the Euler equation should be $\ddot{x} + \lambda x = 0$. (In fact, you have a general sign error. The correct Euler equation for the geodesic is $\ddot{x} + |\dot{x}|^2 x = 0$, if the sign were as you wrote, the solution would not be a trigonometric function, but rather an exponential function.)

Now, to compute $\lambda$, you need to use the constraint $|x|^2 = 1$ twice. First, take the dot product of the Euler equation with $x$, you get that

$$ x\cdot \ddot{x} + \lambda |x|^2 = x \cdot\ddot{x} + \lambda = 0$$

Second, take the second time derivative of the constraint

$$ 0 = \frac{d^2}{dt^2}(|x|^2 - 1) = \frac{d}{dt} (x\cdot \dot{x}) = \dot{x}\cdot\dot{x} + x \cdot \ddot{x} $$

Comparing the two equations and solving for $\lambda$, you have that $\lambda = |\dot{x}|^2$. Hence the correct Euler equation is in fact

$$ \ddot{x} + |\dot{x}|^2 x = 0 $$

For your second question, the $A$s and $C$s (a total of 6 free variables) are fixed by the initial data: the initial position and initial velocity of the geodesic. In other words, you have that

$$ x_i(t) = A_i \cos\left( |\dot{x}|^2 t - C_i\right) $$

and you want to solve for $A_i, C_i$ for prescribed values $x_i(0)$ and $\dot{x}_i(0)$. (Of course, the initial data must satisfy the constraints that the velocity vector is orthogonal to the position vector, and that the position vector has norm 1.)

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Thanks! This is very helpful :) – Brooke Knight Jul 26 '11 at 21:25

Let the initial data be given by:

$x(t=0) = x_0,\quad\dot{ x}(t=0) = v_0 $

The equations of motion of the constrained Lagrangian define a simple harmonic motion. After the substitution of the initial data, the solution has the form

$x = x_0 \cos(\sqrt{\lambda} t) + \frac{ v_0}{\sqrt{\lambda}} \sin(\sqrt{\lambda} t) $

The requirement that at all times $t$

$|x|^2 = 1$

Implies (by the substitution of the solution into the constraint equation)

$|x_0|^2 = 1, v_0. x_0 = 0, | v_0|^2 = \lambda$

Therefore (again by substitution)

$|\dot{x}(t)|^2 = \lambda = |v_0|^2$


$\lambda = |\dot{ x}(t)|^2$

To see that the trajectory is a great circle, we notice that it lies on the plane (passing through the sphere's center):

$ w. x(t)= 0$


$\vec w = x_0 \times v_0$

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Thanks! this is very helpful :) – Brooke Knight Jul 26 '11 at 21:25

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