Path variations in deriving Euler-Lagrange equation

Question

This is a basic question about Euler-Lagrange equations that I could not find being addressed anywhere else (some similar questions: Q1 and Q2).

In deriving the Euler-Lagrange equations from the functional $$ S = \int_{x_a}^{x_b} f(y,y',x) \, dx $$ one takes the first variation of $S$ through varying the optimal path $y_c$ according to $y(x)=y_c(x) + \epsilon \eta(x)$ where $\eta$ is assumed to be an arbitrary function (with $\eta(x_a) = \eta(x_b) = 0$). One obtains $$\delta S = \epsilon \int_{x_a}^{x_b} \eta \left(\frac{\partial f}{\partial y}\bigg\rvert_{y_c} - \frac{d}{dx}\frac{\partial f}{\partial y'}\bigg\rvert_{y_c}\right) \, dx$$ which should vanish for arbitrary $\eta$ and from which Euler-Lagrange equation results.

First Question is about the arbitrariness of $\eta$; intuitively, I'd think $\eta$ should be 'perpendicular' to $y_c$ at each point, or otherwise $y_c + \epsilon \eta$ would not define a new path. Is this intuitive picture wrong?

I guess this is probably more relevant for the multidimensional case. For example, consider $f = f( \lbrace y_i \rbrace_{i=1}^n,x)$ where $f$ does not depend on $y'_i$; then following along the same steps, one arrives at $$ \delta S = \epsilon \int_{x_a}^{x_b} dx \, \sum_{i=1}^n \, \eta_i \, \frac{\partial f}{\partial y_i} \bigg\rvert_{y_c} = \epsilon \int_{x_a}^{x_b} dx \, \mathbf{\eta}\cdot\nabla f =0 .$$ This can either mean that $\nabla f(\mathbf{y}_c)=0$ at each point; alternatively if we are only considering $\eta$'s that are perpendicular to the optimising path, this means that the optimal path should be along the gradient of $f$ at each point. Are either of these two the correct conclusion here?

Second Question Regardless of the question about the nature of $\eta$, such a path that connects $x_a$ to $x_b$ and satisfies either of these conditions (ie $\nabla f=0$ or being along $\nabla f$) may not exist after all -- this seems to me to be the case especially when $f = f(y,x)$ is independent of $y'$; how can one find the optimizing $y$ then?

Answer 1

I am not sure if the following helps you but anyways. In ordinary multidimensional calculus if one has a differentiable, scalar function $f: \mathbb{R}^n \to \mathbb{R}$, one can define the directional derivative as \begin{equation} \nabla_{\vec{n}}f(\vec{x}) := \frac{d}{d\varepsilon}f(\vec{x}+ \epsilon \vec{n})\Big |_{\varepsilon = 0} = \nabla f \cdot \vec{n} \end{equation} for some unit vector $\vec{n}\in \mathbb{R}^n$. Now the functional is again just a scalar function $S: V \to \mathbb{R}$, just this time, $V$ is usually a infinite dimensional vector space, for example $C^{\infty}$. The paths $y : [x_a,x_b] \to \mathbb{R}$ are then the "input vectors" from $C^\infty$ and the variation $\delta S$ is really with respect to a "direction" $\eta \in C^\infty$ within this function space $C^\infty$. With this, analogous to the equation above \begin{equation} \frac{d}{d\varepsilon}F(y+ \epsilon \eta)\Big|_{\varepsilon=0} = \int_{x_a}^{x_b} \frac{\delta F[y]}{\delta y} \eta(x) dx = \langle \frac{\delta F[y]}{\delta y}, \eta(x) \rangle \end{equation} One can recognise the integral, as suggested by the notation as a scalar product in the infinite dimensional vector space. And if this integral vanishes (i.e. $\delta S=0$ for all variations), the "functional-derivative" $\frac{\delta F[y]}{\delta y}$ must vanish, since it is $orthogonal$ to every vector $\eta$, in the sense of vanishing scalar product. So yes, $\eta$ is perpendicular to $\frac{\delta F[y]}{\delta y}$, but the only vector that is perpendicular to every other vector is then recognised to be the $0$ vector. Thus $\frac{\delta F[y]}{\delta y}= \frac{\partial f}{\partial x}- \frac{d}{dx}\frac{\partial f}{\partial y'}=0$

Answer 2

In my opinion the Euler-Lagrange equation itself provides the answer to your question.

The process of deriving the Euler-Lagrange equation is a process of boiling down to the essence. The Euler-Lagrange equation specifies a particular differentiation. That specific differentiation is the core.

Example:
Application of the Euler-Lagrange equation in Classical Mechanics:

In theory of motion:
The Euler-Lagrange equation is populated with expressions for energy. Those expressions for energy are obtained with the Work-Energy theorem

Recapitulating:
Potential energy is a function of position; the value of the potential energy is obtained by integrating force with respect to the position coordinate. Kinetic energy is the integral of $ma$ (mass times acceleration) with respect to the position coordinate.

In theory of motion:
The Euler-Lagrange equation (populated with potential energy and kinetic energy) takes the derivative of the energy with respect to the position coordinate.

The next step is to take this result and use it to inform you about what the variation is. Whatever arbitrariness is attributed in the early stage: in the end the aspect of the variation that makes it to the Euler-Lagrange equation is variation of position coordinate.

(Conversely, everything that does not make it to the Euler-Lagrange euqation is manifestly superfluous.)

Linear independence

You used the expression 'perpendicular', I take it that you are referring to the property of linear independence. The direction of integration of the functional, and the direction of the variation, are independent.

Example:
Plot motion as a function of time using cartesian coordinates. When you sweep out variation: that is variation of the position coordinate. The direction of the variation sweep is perpendicular to the time axis.

Further reading:
There is an oktober 2021 discussion by me of Hamilton's stationary action. That discussion is illustrated with diagrams.

The derivative of kinetic energy with respect to the position coordinate:

$$ \frac{d(\tfrac{1}{2}mv^2)}{ds} = \tfrac{1}{2}m\left( 2v\frac{dv}{ds} \right) = m\frac{ds}{dt}\frac{dv}{ds} = m\frac{dv}{dt} = ma \tag{1} $$

The Euler-Lagrange does a chained differentiation; first differentiation with respect to velocity 'v', and then differentiation with respect to time 't'.

$$ \frac{d}{dt} \frac{d(\tfrac{1}{2}mv^2)}{dv} = \frac{d}{dt}mv = ma \tag{2} $$

Both (1) and (2) evaluate to $ma$, so in effect (1) and (2) are the same differentiation, only the order of operations is different.