Euler-Lagrange equation for squared integral

Question

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?

Answer 1

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

2. 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.

3. Let us for simplicity additionally assume that

   1. $x\mapsto z(x)$ is continuous, and
   2. $h>0$ is positive (and leave the non-negative case $h\geq 0$ for the reader to work out).

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

5. 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 $$ \left. \frac{\delta^2 F[z]}{\delta z(x)\delta z(x^{\prime})}\right|_{z=c}~=~ 8c^2h(x)(\delta(x\!-\!x^{\prime})-h(x^{\prime})). \tag{E}$$

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