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## Main Question or Discussion Point

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

"Take the functional to be

[itex] J(Y) = \int_{a}^{b} \( \alpha Y'^2 + \beta Y^2) dx [/itex]

For this

[itex] F(x,y,y') = \alpha y'^2 + \beta y^2 [/itex]

and [itex] p = \frac{ \partial F}{\partial y'} = 2 \alpha y' [/itex]

[itex] \Rightarrow y' = \frac{p}{2 \alpha}[/itex]

The Hamiltonian H is

[itex] H = py' - F = \frac {p^2}{4 \alpha} - \beta y^2[/itex]

So the canonical equations are

[itex] \frac{dy}{dx} = \frac{ \partial H}{ \partial p} = \frac{p}{2 \alpha} [/itex]

and

[itex]- \frac{dp}{dx} = \frac{\partial H} {\partial y} = -2 \beta y [/itex]

I've also got the Euler Lagrange equation as

[itex] 2 \beta y - \frac{d}{dx} (2 \alpha y') = 0[/itex]

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

Thanks in advance

"Take the functional to be

[itex] J(Y) = \int_{a}^{b} \( \alpha Y'^2 + \beta Y^2) dx [/itex]

For this

[itex] F(x,y,y') = \alpha y'^2 + \beta y^2 [/itex]

and [itex] p = \frac{ \partial F}{\partial y'} = 2 \alpha y' [/itex]

[itex] \Rightarrow y' = \frac{p}{2 \alpha}[/itex]

The Hamiltonian H is

[itex] H = py' - F = \frac {p^2}{4 \alpha} - \beta y^2[/itex]

So the canonical equations are

[itex] \frac{dy}{dx} = \frac{ \partial H}{ \partial p} = \frac{p}{2 \alpha} [/itex]

and

[itex]- \frac{dp}{dx} = \frac{\partial H} {\partial y} = -2 \beta y [/itex]

I've also got the Euler Lagrange equation as

[itex] 2 \beta y - \frac{d}{dx} (2 \alpha y') = 0[/itex]

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

Thanks in advance