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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:

  1. 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 $\dot x(t)= dx/dt$

  2. 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

  3. 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?

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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!

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

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    $\begingroup$ I think it would improve this post a lot, if you elaborated on the pitfalls of the naïve approach. I guess you want to use bits and pieces of elementary theory of distributions, but not all your readers may not be aware of that let alone conversant with it. I am not asking you to reproduce the relevant definitions and such. Something like pretending to answer to a person who does know those bits and pieces, or whatever you feel like assuming from your reader. IMHO the important thing is to have a reader in mind. $\endgroup$ Commented Aug 18, 2015 at 8:30

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