The standard problem in variational calculus is that given the functional
$F[f]=\int_a^b \textrm{d}x f(x, y, y'),$ where $y\equiv y(x)$,
find $y(x)$ that extremizes $F[f]$ under the boundary conditions: $y(a)=y_a$ and $y(b)=y_b$.
Extremizing $F[f]$ yields the Euler-Lagrange equations that are essentially ODEs that $y(x)$ satisfies which are then solved under the given boundary conditions to obtain the desired $y(x)$.
My question is: Will this formalism work for functions $f$ that do not have any dependence on the derivative of y?
In the context of my problem simply applying the above yields an equation in y(x) — so not an ODE and thus there is no reason to expect that such an equation will satisfy the boundary conditions and indeed it does not.
I want to know what the conceptual difficulty is in such cases. From what I understand it has to do with the fact that only when we have an explicit dependence on the derivative of $y$ in $f$ that we would obtain an ODE in $y$ and then it becomes a standard problem of solving the ODE under the given boundary conditions. However, I am not entirely sure.
Moreover, is there any other way to solve such problems?
Any remark or reference would be of great help.
Best Answer
More generally, variational problems with an $N$'th-order Lagrangian $$L(x,y(x), y^{\prime}(x), y^{\prime\prime}(x),\ldots, y^{(N)}(x)),$$ with 1 independent variable $x$, generically has an EL equation that is an ODE of order $2N$, and therefore requires $2N$ boundary conditions (BCs) to be well-posed. More BCs would overdetermine the solution to the EL equation.
Returning to OP's question, a $0$'th-order Lagrangian $L(x,y(x))$ [which corresponds to a so-called static problem if we identify $x$ with time] generically needs no BCs.