multivariable calculus - Deriving a high ordered Euler-Lagrange equation.

I've been able to derive the Euler-Lagrange equation for $$\int_a^b F(x,y,y')dx$$ relatively easily by using the total derivative and integration by parts. However, I was unable to apply the same methods whilst trying to derive the Euler-Lagrange for $$\int_a^b F(x,y,y',y'')dx$$ A proof would be great, but a proper name for this type of equation would also be appreciated (without a term I've found scouring the internet for a Euler-Lagrange of this form to be quite difficult).

Answer

Start by setting up a variation $y_\epsilon (x) = y(x) + \epsilon h(x)$ where $\epsilon$ is some small real number and $h(x)$ is zero on the boundaries. You now have a functional of the form

$$\int_a^b f(x,y_\epsilon,y'_\epsilon,y''_\epsilon)dx.$$

Differentiate with respect to $\epsilon$ and use the Leibniz rule.

$$\frac{d}{d \epsilon} \int_a^b f(x,y_\epsilon,y'_\epsilon,y''_\epsilon)dx = \int_a^b \frac{\partial}{\partial \epsilon} f(x,y_\epsilon,y'_\epsilon,y''_\epsilon)dx .$$

Note that

$$ \frac{\partial}{\partial \epsilon} f(x,y_\epsilon,y'_\epsilon,y''_\epsilon) = \frac{\partial f}{\partial x}\frac{\partial x}{\partial \epsilon} + \frac{\partial f}{\partial y_\epsilon}\frac{\partial y_\epsilon}{\partial \epsilon} + \frac{\partial f}{\partial y'_\epsilon}\frac{\partial y'_\epsilon}{\partial \epsilon} + \frac{\partial f}{\partial y''_\epsilon}\frac{\partial y''_\epsilon}{\partial \epsilon}$$

It follows that

$$ \frac{\partial}{\partial \epsilon} f(x,y_\epsilon,y'_\epsilon,y''_\epsilon) = \frac{\partial f}{\partial y_\epsilon}h(x) + \frac{\partial f}{\partial y'_\epsilon} h'(x) + \frac{\partial f}{\partial y''_\epsilon} h''(x)$$

Your functional now looks like

$$\int_a^b \left( \frac{\partial f}{\partial y_\epsilon}h(x) + \frac{\partial f}{\partial y'_\epsilon} h'(x) + \frac{\partial f}{\partial y''_\epsilon} h''(x) \right) dx.$$

Since you are looking for a stationary point, set it equal to zero and now evaluate it at $\epsilon = 0$.

$$0 = \int_a^b \left( \frac{\partial f}{\partial y}h(x) + \frac{\partial f}{\partial y'} h'(x) + \frac{\partial f}{\partial y''} h''(x) \right) dx.$$

Use integration by parts on the second and third terms. I will only perform it on the third term since you state you've derived the simpler case before. This term must be integrated by parts twice.

$$ \int_a^b \frac{\partial f}{\partial y''} h''(x) dx.$$

Choose $u= \frac{\partial f}{\partial y''}$ and $v'= h''(x) $ so $u' = \frac{d}{dx} \left( \frac{\partial f}{\partial y''} \right)$ and $v = h'(x)$.

The resulting integration by parts is

$$ \frac{\partial f}{\partial y''} h'(x) \bigg|_a^b - \int_a^b \frac{d}{dx} \left( \frac{\partial f}{\partial y''}\right) h'(x) dx$$

The first term is zero since $h(x)$ is zero on the boundaries. The second term must be integrated by parts again but now with

$$ u = \frac{d}{dx}\left( \frac{\partial f}{\partial y''} \right)$$

$$u' = \frac{d^2}{dx^2}\left( \frac{\partial f}{\partial y''} \right)$$

$$v = h(x)$$

$$v' = h'(x)$$

I will leave the substitutions to you. The functional now reads:

$$ 0= \int^a_b \left[ \frac{\partial f}{\partial y} - \frac{d}{dx} \left( \frac{\partial f}{\partial y'} \right) + \frac{d^2}{dx^2} \left( \frac{\partial f}{\partial y''} \right) \right] h(x) dx $$

This should work for any function $h(x)$ so your Euler-Lagrange equation is

$$ 0= \frac{\partial f}{\partial y} - \frac{d}{dx} \left( \frac{\partial f}{\partial y'} \right) + \frac{d^2}{dx^2} \left( \frac{\partial f}{\partial y''} \right) $$