4
$\begingroup$

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?

$\endgroup$

2 Answers 2

0
$\begingroup$

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!

$\endgroup$
-1
$\begingroup$

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

$\endgroup$
1
  • 2
    $\begingroup$ 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. $\endgroup$ Aug 18, 2015 at 8:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.