Minimize $\frac{\int_0^1 f(x)\phi''(x)\phi''(x)\,\text{d}x} {\int_0^1 g(x) \phi'(x)\phi'(x)\,\text{d}x}$

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 $$ 0=[f(x)ϕ''(x)]''+λ[g(x)ϕ'(x)]'\implies C=[f(x)ϕ''(x)]'+λ[g(x)ϕ'(x)]. $$

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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 $k=1$.

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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.