# Calculus of Variations (Canonical equations)

• ElDavidas
Then, use the second equation to substitute for dy/dx. The resulting equation should be the same as the Euler-Lagrange equation for the given Lagrangian.In summary, the conversation is about a specific example involving a functional and its corresponding Lagrangian and Hamiltonian equations. The individual is asking how to determine if the Euler-Lagrange equation is equivalent to the canonical Euler equations in this example. The expert suggests differentiating the first equation with respect to "x" and substituting the second equation, resulting in an equation that should be equivalent to the Euler-Lagrange equation for the given Lagrangian.

#### 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?