# Expanding to only first order in the functional minimisation derivation

Consider the functional minimisation derivation for $$F[y, y'] = \int_a^b f(y, y')$$.

This derivation begins by letting $$y$$ be our function which minimises $$F$$. We consider the variation in $$F$$, $$\delta F$$, due to a small deformation of our path $$\delta y$$. In particular, we have $$\delta F = F[y + \delta y, y' + \delta y'] - F[y, y']$$.

We expand $$f$$ to first order:

$$f(y + \delta y, y' + \delta y') = f(y, y') + \frac{\partial f}{\partial y} \delta y + \frac{\partial f}{\partial y'} \delta y'$$,

and then continue the derivation from there to find the Euler-Lagrange equation for $$F$$.

My question is about the expansion of $$f$$ to first order. It is surely the case that the $$\delta y$$ term in the expansion is much bigger than $$\delta y^2$$, since we have assumed that $$\delta y$$ is sufficiently small to allow us to do this. However, is it necessarily the case that the second order expansion term $$\frac{1}{2} \frac{\partial^2 f}{\partial y^2} \delta y^2$$ is necessarily much less than the first order term $$\frac{\partial f}{\partial y'} \delta y'$$?

For instance, what if $$\delta y$$ had a particularly small derivative? How can we dump the first of these terms, but keep the second?

I don't think the motivation can be that we need a nonzero $$\delta y$$ and $$\delta y'$$ term somewhere in there. Given that $$\int_a^b \delta y' = 0$$, then if $$\frac{\partial f}{\partial y'}$$ is constant, the term $$\frac{\partial f}{\partial y'} \delta y'$$ integrates to zero, and so makes no change to the value of $$F$$. The term $$\frac{1}{2} \frac{\partial^2 f}{\partial y'^2} \delta y'^2$$ could make a nonzero contribution to a change in $$F$$, but we insist it is 'small'. I'd say it's contribution is bigger than the $$\frac{\partial f}{\partial y'}$$ here, so why don't we keep it?

In my research, I saw this, but it was taking a more 'doesn't this propagate a lot of error?' approach, rather than specifically a 'sometimes the second order terms are bigger than the first?' approach.

Can anyone provide some insight? Thanks!

The approach based on the Taylor expansion is quite non-rigorous, but it can be replaced with a natural approach based on standard derivatives which yields exactly the same final result.

Consider the functional $$F[y] := \int_a^bf(y(t),y'(t))dt$$ where the $$C^2$$ curves $$y$$ are constrained to satisfy the boundary conditions $$y(a) =y_a$$, $$y(b)=y_b$$ for some constants $$y_a,y_b$$.

We say that a function $$y_0$$ is an extremal curve of $$F$$ if for every deformation $$y_0(t) + \alpha \delta y(t)$$ with $$\delta y$$ any curve with $$\delta y(a)=\delta y(b)=0$$ (in order to satisfy the boundary conditions) and $$\alpha \in \mathbb{R}$$, it holds $$\frac{d}{d\alpha}|_{\alpha=0} F[y_0+\alpha \delta y]=0\:.\tag{1}$$

The reason for this maybe apparently cumbersome definition relies upon the analogous fact in standard calculus: if $$f=f(x) \in \mathbb{R}$$ is $$C^1$$ with $$x \in \mathbb{R}^n$$, then $$x_0$$ is an extremal point if and only if $$\frac{d}{d\alpha}|_{\alpha =0} f(x_0 + \alpha \delta x)=0 \quad \mbox{for every \delta x \in \mathbb{R}^n}\tag{2}$$ In fact, (2) is equivalent to $$\delta x \cdot \nabla f(x_0)=0 \quad \mbox{for every \eta \in \mathbb{R}^n}$$ which means, because $$\delta x$$ is arbitrary, $$\nabla f(x_0)=0$$ which is nothing but the standard definition of extremal point $$x_0$$ for the function $$f: \mathbb{R}^n \to \mathbb{R}$$.

Coming back to (1), if $$F(y) = \int_a^b f(y(t),y'(t))dt$$, the equation can be rephrased to $$\frac{d}{d\alpha}|_{\alpha=0}\int_a^b f(y_0+ \alpha \delta y, y_0' + \alpha \delta y') dt=0,$$ that is $$\int_a^b \frac{\partial f}{\partial y}|_{y_0, y_0'} \delta y + \frac{\partial f}{\partial y'}|_{y_0,y'_0} \delta y' dt=0\:.\tag{3}$$ Using $$\delta y(a)=\delta y(b)=0$$ and integration by parts, (3) can be equivalently written $$\int_a^b \left(\frac{\partial f}{\partial y}|_{y_0,y'_0}- \frac{d}{dt}\frac{\partial f}{\partial y'}|_{y_0,y'_0}\right)\:\delta y(t) \: dt =0$$ The arbitrariness of $$\delta y$$ (and the fact that its factor is continuous) implies the usual E-L equations for $$y_0$$.

I stress that

(1) nowhere $$\delta y$$ is required to be small,

(2) all mathematical passages are justified just assuming the integrand function $$f$$ to be sufficiently regular ($$C^2$$ is sufficient with $$-\infty < a < b < +\infty$$),

(3) everyting arises from (is equivalent to) the definition (1) of extremal curve,

(4) this approach, with some technical hypotheses leads to the definition of the Gateaux derivative of functionals in functional analysis.