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:


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*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?
 A: I must remark that it is not a good idea to mix the notations. Either choose $\delta$ to represent the variational operator, or choose $\delta$ to represent the Dirac-delta function. In an attempt o answer your question, I am going to use the latter.
I will try my best to make it clear that you cannot use the Dirac-delta function as a "shape function" with the purpose of a variation and expect to obtain a nice result.
I think the best way is to look at a concrete example:
Let $$\mathcal{J}[x(t)] = \int_a^b \frac{1}{2}x^2(t) dt \tag{1}\label{1} \,,$$
and suppose that $x^*(t)$ is a function that minimizes $\mathcal{J}[x(t)]$, such that
$$\mathcal{J}[x^*(t)] \leq \mathcal{J}[x(t)]$$
Note: In this case it is easy to see that the solution will be zero everywhere, but let's pretend that we don't know that at the moment.
Now let $x(t) = x^*(t) + \epsilon \eta(t)$, be a perturbation of the optimal solution.
We know that the minimum of $\mathcal{J}[x(t)]$ is found when
$$ \dfrac{ d \mathcal{J}[x^*(t) + \epsilon \eta(t)] }{d \epsilon} \Biggr\vert_{\epsilon =0}= 0 $$
We can derive this quantity step by step:
\begin{align}
 \dfrac{d}{d \epsilon} \Biggl( \int_a^b \frac{1}{2} \bigl( x^*(t) + \epsilon \eta(t) \bigr)^2 dt \Biggr) \Biggr\vert_{\epsilon =0} & = 0 \\
 \int_a^b \dfrac{d}{d \epsilon} \Biggl(  \frac{1}{2} \bigl( x^*(t) + \epsilon \eta(t) \bigr)^2 \Biggr) \Biggr\vert_{\epsilon =0} dt & = 0 \\
 \int_a^b \bigl( x^*(t) + \epsilon \eta(t) \bigr) \eta(t) \Biggr\vert_{\epsilon =0} dt & = 0 \\
 \int_a^b x^*(t) \eta(t) dt & = 0
\end{align}
Which implies
$$ x^*(t) \eta(t) = 0 \,, \tag{2} \label{2}$$
for arbitrary $\eta(t)$.
Note: Here "arbitrary" means that the equality must hold for any possible function $\eta(t)$. It does not mean "arbitrary" in the sense that one may decide what it is!
This is really important, because if $\eta(t)$ is arbitrary, then Eq. \eqref{2} allows us to infer that the solution must be identically equal to zero,
$$ x^*(t) = 0 \,,$$
as this is the only way for the equality to hold in that case.
Now the punchline...
Suppose we indeed select $\eta(t)$ to be the Dirac-delta function, $\eta(t) = \delta(t-c)$, where $c$ may be within the interval $[a,b]$ or not. Notice that this function is identically zero (almost) everywhere.
In this case, what information about the solution, $x^*(t)$, can we extract from \eqref{2} ?
Absolutely nothing!
The fact that $\eta(t)$ is zero everywhere will satisfy Eq. \eqref{2} without any implications on what $x^*(t)$ must be. And this does not help at all!
A: You can't proceed naively because $\epsilon \delta(x)$ is not continuos for any finite $\epsilon$.
