Variational calculus with the Lagrangian function not explicitly dependent on $y'$

Question

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.

Answer 1

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.

Answer 2

I've read this question a few times over the last couple of months and, while the existing answer gives the technical answer, I had trouble gaining an intuition of why it doesn't work. That is, why don't the standard Euler-Lagrange equations give an answer to this seemingly valid problem as given in the question?

Given the functional

$F[f]=\int_a^b \textrm{d}x f(x, y),$ where $y\equiv y(x)$,

find $y(x)$ that extremizes $F[f]$ under the boundary conditions: $y(a)=y_a$ and $y(b)=y_b$.

And, as the question initially asked, is there a way to modify the formalism to get a solution?

My understanding is that the issue is that the problem doesn't pose any constraint on $y(x)$'s continuity. So if the solution to the problem without the boundary conditions is $y(x)=Y(x)$, then the solution with the boundary conditions is

$$ y(x) = \begin{cases} y_a & \text{if } x=a \\ y_b & \text{if } x=b \\ Y(x) & \text{otherwise,} \end{cases} $$

which is not terribly helpful.

If you add a derivative term to $F$ that prevents the derivative from going too high, like $\frac{1}{2}my'^2$, then that effectively enforces differentiability and therefore continuity.