Euler-Lagrange equation for squared integral Can an Euler-Lagrange equation be derived for the following functional?
$$F[y'] = \int h(x) (y'(x))^4 dx - \big( \int h(x) (y'(x))^2 dx\big)^2.$$
Here $h(x)\geq 0$ and $\int h(x) dx = 1$.
Note, that the functional is not written as a single integral in this form. So my question is, how to solve this? Can it even be solved?
 A: *

*If we replace $z=y^{\prime}$ then OP's functional becomes
$$\begin{align} F[z]~:=~&H_4[z]-H_2[z]^2, \cr 
H_4[z]~:=~&\int_{\mathbb{R}}\! dx~h(x) z(x)^4, \cr 
H_2[z]~:=~&\int_{\mathbb{R}}\! dx~h(x) z(x)^2, \cr
\int_{\mathbb{R}}\! dx~h(x)~=~&1.
\end{align}\tag{A}$$


*The functional derivative is
$$ \frac{\delta F[z]}{\delta z(x)}~=~ 4h(x)z(x)(z(x)^2-H_2[z]). \tag{B}$$
OP's sought-for equation is asking when the functional derivative (B) is zero.


*Let us for simplicity additionally assume that

*

*$x\mapsto z(x)$ is continuous, and

*$h>0$ is positive (and leave the non-negative case $h\geq 0$ for the reader to work out).



*Then one may show that the stationary configurations are the constant configurations
$$z(x)=c~\in~\mathbb{R}.\tag{C}$$


*The 2nd functional derivative is
$$ \frac{\delta^2 F[z]}{\delta z(x)\delta z(x^{\prime})}~=~ 4h(x)\delta(x\!-\!x^{\prime})(3z(x)^2-H_2[z])-8h(x)z(x)h(x^{\prime})z(x^{\prime}). \tag{D}$$
Then
$$ \left. \frac{\delta^2 F[z]}{\delta z(x)\delta z(x^{\prime})}\right|_{z=c}~=~ 8c^2h(x)(\delta(x\!-\!x^{\prime})-h(x^{\prime})). \tag{E}$$


*One may show that the constant configurations are saddle points. (The case $c=0$ requires a bit more work.)
