How to model a simple energy minimization using Euler-Lagrange equations

Suppose I have a randomly-generated set of numbers, where each number is indexed by $x$. This is equivalent to a 1-D scalar field, let's call that $u(x)$.

Let's say my objective is to minimize the distance between each subsequent pair of scalars, which may be expressed as minimizing the following energy functional:

$$E[u(x)] = \sum_{x}\left(\frac{du(x)}{dx}\right)^2$$

Thanks to comment/clarification by @BrianBorchers, setting $u(x)=0$ would ultimately achieve an unconstrained minimum. Thus, I introduce the following constraints: $u(x)$ is defined on $[a,b]$, so that $u(a)$ and $u(b)$ are fixed scalar values.

The first (or second) derivatives for the scalar field $u(x)$ can be approximated easily by taking finite differences. I would like to simply use gradient descent to get the scalar field u to converge to the desired state. Hence I need an update step for $u(x)$, something like:

$$u^{k+1} = u^{k} - \alpha\left(\frac{dE[u(x)]}{du(x)}\right)$$, where $\alpha$ is a constant update step/rate.

I'm doing this to gain thorough understanding of a research article which uses similar, albeit more complex energy functions expressed in terms of a vector field. The authors state that the update step, i.e. the equivalent of the term $\frac{dE[u(x)]}{du(x)}$ above is derived using the Euler-Lagrange equations. Links to the article and the supplementary material are given below.

I looked at numerous tutorials and some books, and if I understand correctly, the Euler-Lagrange equation for the said energy $E[u(x)]$ would be:

$$-\frac{d}{dx}E_{u'}[u(x)] + E_{u}[u(x)] = 0$$ or, using Lagrange's notation for derivatives, $$-\frac{d}{dx}\left(\frac{dE[u(x)]}{d\frac{du(x)}{dx}}\right) + \frac{dE[u(x)]}{du(x)} = 0$$

1. Is the above formulation of the Euler-Lagrange equation correct?
2. The Euler-Lagrange equation looks like it would directly yield a stationary point in the energy functional. I suppose, I cannot do that directly for some reason. How exactly does the Euler-Lagrange equation relate to the update step I should take in my gradient descent above?
3. How exactly do I formulate the update step as a function of first and second derivatives of $u(x)$ w.r.t. $x$?

The original research article (see sec 4.2) and supplementary material (see sec 2.3) :

• Your question is unclear- Simply setting $u(x)=0$ produces an unconstrained minimum of $E$. What additional constraints do you want to impose? – Brian Borchers Nov 14 '17 at 16:19
• @BrianBorchers, I'm sorry for my relative illiteracy in this, what is an unconstrained minimum and how would $u(x) = 0$ yield this minimum? - Thanks – Algomorph Nov 14 '17 at 16:23
• @BrianBorchers, on second thought, I think I understand what you're asking. Artificially setting $u(x) = 0$ would achieve a global minimum without additional constraints. I guess, I should introduce constraint that $u(x)$ is defined on an interval [a,b], and $u(a)$ and $u(b)$ cannot be modified. Should I modify the question like this? – Algomorph Nov 14 '17 at 16:28
• If your only constraints are that $u(a)$ and $u(b)$ are given, then the solution (with an objective value of 0) is $u(x)=u(a)+((x-a)/(b-a))(u(b)-u(a))$. – Brian Borchers Nov 14 '17 at 17:06
• Thanks, @BrianBorchers. I was looking for a way one would solve this with gradient descent, not directly, as I described in question #2. One of my colleagues, Nitin Sanket already provided the answer (big thanks to Nitin!). The update step $\frac{dE[u(x)]}{du(x)}$ may be difficult to compute. This gradient is exactly the second part of the Euler-Lagrange equation I wrote (after the plus sign). If we shift it to the right-hand side and solve for the left-hand side, we can compute this gradient locally, it becomes $-2\sum_{x} u_{xx}u_{x}$. – Algomorph Nov 14 '17 at 17:24

1) The Euler-Lagrange equation is essentially correct. Normally, with the exception of $-\frac{d}{dx}$ in the very beginning, we would have partial derivatives, but here the regular derivative works since $E$ is a function only in $u(x)$. There is a slight problem with the notation here, same as in the paper: the total energy needs to be expressed separately from the function inside the integral (in this case, sum). This way, it is clear that the update step at each point x of the scalar field does not involve the sum.
2) and 3) The update step $\frac{dE[u(x)]}{du(x)}$ can be obtained by moving the first term of the given Euler-Lagrange equation to the right-hand side, thereby solving for the step. Since first- and second-order derivatives are easy to compute on the scalar field, we can easily obtain the update at each step.