Minimize $\frac{\int_0^1 f(x)\phi''(x)\phi''(x)\,\text{d}x} {\int_0^1 g(x) \phi'(x)\phi'(x)\,\text{d}x}$ Below is a situation I came across during solving problems using energy method in mechanics. I have formulated it in question form, omitting details I think are unnecessary. 

Question:
Given positive continuous diffenertiable functions $f(x)$ and $g(x)$, find non-trivial $\phi(x)$ that minimizes the functional $F$:
$$
F[\phi,\phi',\phi'';x]=
\frac{\displaystyle\int_0^1 f(x)\phi''(x)\phi''(x)\,\text{d}x}
{\displaystyle\int_0^1 g(x) \phi'(x)\phi'(x)\,\text{d}x}
$$
with boundary conditions $\phi(0)=\phi''(0)=\phi(1)=\phi''(1)=0$.


*

*For $f(x)=1$ and $g(x)=1$, prove that $\phi(x)=C\sin(\pi x)$, where $C$ is arbitrary constant, minimizes $F$.

*For $f(x)=1$ and $g(x) = x$, find $\phi(x)$. 

*For general $f(x)$ and $g(x)$, find $\phi(x)$.



Comments
One source claims that the functional $F$ is called the Rayleigh-Ritz quotient. Some special cases of the original problem can be solved exactly from differential equation, so there is some information regarding the expected solution:


*

*For $f(x)=1$ and $g(x)=1$, the function $\phi(x)=C\sin(\pi x)$ corresponds to the exact solution, but not proven in terms of minimizing the functional $F$.

*For $f(x)=1$ and $g(x)=x$, a similar case with different boundary condition has the exact solution $\phi(x)$ in the form of Bessel function of the first kind $J_{-1/3}$. Physically I believe Bessel function should also be the solution form in this case, but mathematically I am not sure. 

*For general $f(x)$ and $g(x)$, we normally resort to numerical methods such as finite element, because we believe that the closed-form solution is not possible. Here I am asking merely for curiosity. Would it be possible to express $\phi(x)$ in some form, such as using infinite series or special functions? May be imposing certain conditions for $f(x)$ and $g(x)$? Sorry for being vague, I can add more information if required.



Edited: 2019-07-07
I would like to add the numerical results for case 2 where $f(x)=1$ and $g(x)=x$.
The mode shape $\phi(x)$ is expressed in power series:
$$\phi(x)=a_0+a_1 x+a_2 x^2+\cdots + a_n x^n + \cdots$$
where
$$a_0=0,\quad a_1=1,\quad a_2=0,\quad a_3\approx -0.7272$$
$$a_{3n+1}=\frac{(-1)^n\xi^n}{(3n+1)!}\prod_{k=1}^n (3k-2),\quad n\ge 1$$
$$a_{3n+2}=0,\quad n\ge 1$$
$$a_{3n+3}=-6a_3\frac{(-1)^n\xi^n}{(3n+3)!}\prod_{k=1}^n (3k),\quad n\ge 1$$
$$\xi \approx 18.5687 $$
The functional $F$ is calculated as
$$F=\xi\approx 18.5687$$
It seems that my physical belief is wrong: the shape of $\phi(x)$ is not like a Bessel function, and not even symmetrical on domain $[0,1].$
For comparison, if $\phi(x)=\sin(\pi x)$, then $F=2\pi^2\approx1 9.7392.$
 A: You can unpack this to get a "normal" Lagrange functional in
$$
\text{minimize}\int_0^1f(x)ϕ''(x)^2dx \text{ such that } \int_0^1g(x)ϕ'(x)^2dx=1
$$
Then the Lagrange functional is
$$
L(ϕ,λ)=\frac12\int_0^1f(x)ϕ''(x)^2dx+\fracλ2\left(1-\int_0^1g(x)ϕ'(x)^2dx\right).
$$
The saddle point condition resp. Euler-Lagrange equations then lead to
\begin{align}
0=δL(ϕ,λ)&=\int_0^1f(x)ϕ''(x)δϕ''(x)dx-λ\int_0^1g(x)ϕ'(x)δϕ'(x)dx
\\
&=[f(x)ϕ''(x)δϕ'(x)]_0^1-\int_0^1[f(x)ϕ''(x)]'δϕ'(x)dx-λ[g(x)ϕ'(x)δϕ(x)]_0^1+λ\int_0^1[g(x)ϕ'(x)]'δϕ(x)dx
\\
&=-[f(x)ϕ''(x)]'δϕ(x)]_0^1+\int_0^1[f(x)ϕ''(x)]''δϕ(x)dx+λ\int_0^1[g(x)ϕ'(x)]'δϕ(x)dx
\end{align}
using the boundary conditions and that the variations have to leave the boundary conditions fixed, thus being zero at the same places in the same derivatives.
The resulting equation is
$$
0=[f(x)ϕ''(x)]''+λ[g(x)ϕ'(x)]'\implies C=[f(x)ϕ''(x)]'+λ[g(x)ϕ'(x)].
$$

For $f=g=1$ this gives $$0=ϕ^{(4)}+λϕ''$$ which has non-trivial solutions $A\cos(ωx)+B\sin(ωx)+Cx+D$, leading to $A=C=D=0$ and $ω=k\pi$, $λ=ω^2$, minimum with non-trivial solution for $k=1$.

For the second example $f=1$, $g(x)=x$, one gets 
$$
0=ϕ^{(4)}+λxϕ''+λϕ'\text{ or }C=ϕ'''+λxϕ'
$$
The homogeneous part of the second form has a solution in terms of Airy functions for $ϕ'$, then apply variation of constants and integrate to get $ϕ$, this looks like a long unintuitive formula.
A: I think I have a partial answer, hopefully someone can expand on it.
Let $\eta$ be a function satisfying the same boundary conditions. Then the first variation of the functional is 
$$ \delta F = \lim_{\epsilon \to 0} \frac{F[\phi + \epsilon \eta] - F[\phi]}{\epsilon} $$
If $\phi$ minimizes the functional, this first variation has to be zero for all $\eta$.
Working this out for the given functional we get
$$\frac{\delta F}{2} = \left( \int_0^1 f \phi'' \eta'' dx \right)\left( \int_0^1 g \phi'\phi' dx \right) - \left( \int_0^1 f \phi'' \phi'' dx \right) \left( \int_0^1 g \phi' \eta' dx \right)  $$


*

*Now let $\phi(x) = \sin(\pi x)$ and $f(x) = g(x) = 1$. Then the first variation turns into


$$ \frac{\delta F}{2} = -\left(\int_0^1 \pi^2 \sin(\pi x)\eta''\right)\frac{\pi^2}{2} - \frac{\pi^4}{2} \left( \int_0^1 \pi \cos(\pi x) \eta' dx \right)  $$
Partial integration then shows that this is zero, proving $\sin(\pi x)$ minimizes $F$.
