Calculus of Variations (Canonical equations)

[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?

Thanks in advance

 

[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.

 

[ElDavidas]

Differentiate wrt "t" the eqn involving the first derivative of "y"

Sorry, what do you mean by this?

 

[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.