How should the Calculus of Variations deal with $\delta(t-t_0)$ variations? I'm familiar with using the Calculus of variations to find the condition for which first order variations of a functional wrt a function are zero:


We start with a functional $J[x]= \int_{t_f}^{t_i}L(x(t),\dot x(t),t)dt$, where $t_i, t_f$ are constants and $\dot x(t)= dx/dt$
If $J[f]$ is stationary at $f$, we add to $f$ another continuous function $\epsilon\eta(t)$ where $\eta(t)$ is arbitrary and $\epsilon$ is a positive number
For any small number $\epsilon$ close to 0, the first order variation of the functional is zero around $f$, giving the Euler-Lagrange equations as the necessary condition.


If $\delta(t)$ is what physicists/engineers call the Dirac-Delta function, even though mathematicians wouldn't define it as a function:

Can the above procedure be applied to $\delta(t-t_0)$ variations of f at $t = t_0$ to again yield the Euler-Lagrange equations?

I've had a go by adding $\epsilon\delta(t-t_0)$ to $f$ as the variation, getting the differential of $L$ in terms of the usual partial derivatives and differentials of the independent variables. But the differential of $x$, for example, then becomes $dx = \epsilon\delta(t-t_o)$ which appears to be infinite for any $\epsilon$, making my expression for $dL$ nonsense.

Again, according to an answer given on PSE that motivated me to ask my question here, the change in $L$ to an $\epsilon\delta(t-t_0)$ variation in $\dot x$ is:

$$dL= {\partial L \over \partial \dot{x_1}} \delta \dot{x_1} = {\partial L \over \partial \dot{x_1}}\epsilon\delta(t-t_0)$$

Is this correct?
You can't proceed naively because $\epsilon \delta(x)$ is not continuos for any finite $\epsilon$.
You can't proceed naively because $\epsilon \delta(x)$ is not continuos for any finite $\epsilon$.