Calculus of Variations (Canonical equations)

1. Feb 25, 2007

ElDavidas

I've been looking at this example for a while now. Could someone help?

"Take the functional to be

$J(Y) = \int_{a}^{b} \( \alpha Y'^2 + \beta Y^2) dx$

For this

$F(x,y,y') = \alpha y'^2 + \beta y^2$

and $p = \frac{ \partial F}{\partial y'} = 2 \alpha y'$
$\Rightarrow y' = \frac{p}{2 \alpha}$

The Hamiltonian H is

$H = py' - F = \frac {p^2}{4 \alpha} - \beta y^2$

So the canonical equations are

$\frac{dy}{dx} = \frac{ \partial H}{ \partial p} = \frac{p}{2 \alpha}$
and

$- \frac{dp}{dx} = \frac{\partial H} {\partial y} = -2 \beta y$

I've also got the Euler Lagrange equation as

$2 \beta y - \frac{d}{dx} (2 \alpha y') = 0$

How can you tell that the Euler Lagrange equation is equivalent to the Canonical Euler equations in this set example?

2. Feb 25, 2007

dextercioby

Differentiate wrt "t" the eqn involving the first derivative of "y" and substitute the first derivative of p from the second and the resulting 2-nd order ODE in "y" will coincide with the Euler-Lagrange eqn for the lagrangian.

3. Feb 25, 2007

ElDavidas

Sorry, what do you mean by this?

4. Feb 25, 2007

dextercioby

Differentiate with respect to "x" (sorry, i thought it was "t", like in physics, where "t" stands for time) the first equation, the one involving dy/dx.