Calculus of variation; Calculating First Variation

calculus-of-variationsclassical-mechanicseuler-lagrange-equation

so from my understanding of the subject there seems to be a whole deluge of differing definitions for things such as the First variation for a functional.

now i've been asked to calculate the first and second variation (i'll ask about just the first variation) of the arc-length given in parametric form. but first a bit of pre-reading i guess:

In my course the First variation has been defined as the function

$$\delta J(u;v) = \left.\frac{d}{dt} J(u+tv)\right|_{t=0} = \lim_{t\longrightarrow 0}\frac{J(u+tv)-J(u)}{t}$$ and is quite simply the partial derivative along some arbitrary function v (if i remember right it's a direction), it's then noted that if the above limit exists for every v then we call the functional $\delta(u;v)$ the first variation and denote it as $\delta(u;\cdot)$

its then shown later in the course that for a functional $J(u)$ defined as
$$J(u) = \int_{a}^{b} \Lambda(x,u,u') dx$$

$$\delta J(u;v) = \left.\frac{d}{dt} J(u+tv)\right|_{t=0} = \left.\frac{d}{dt}\int_{a}^{b}\Lambda(x,u+tv,u'+tv') dx\right|_{t=0} =\int_{a}^{b}\left.\frac{d}{dt}\Lambda(x,u+tv,u'+tv)\right|_{t=0} =\int_{a}^{b}\left[\frac{\partial \Lambda}{\partial u}(x,u,u')v+\frac{\partial \Lambda}{\partial u'}(x,u,u')v'\right]$$

so now as far as i'm aware all of these are equivalent definitions of the first variation along a given arbitrary function v of a functional J. But this obviously leads to the Euler-Lagrange equation. ie

$$\delta J(u;v) = \int_{a}^{b}\left[\frac{\partial \Lambda}{\partial u} – \frac{d}{dx}\left(\frac{\partial \Lambda}{\partial u'}\right)\right]v~dx$$ which we set to zero in order to find the extremal of a problem. I'm assuming all of these definitions remain equivalent and so finding the first variation of an functional is then simply finding the euler-legrange equation and multiplying it by some direction v

In which case for more complicated functions say
$$J(x,u,u_1, \cdots, u_n, u', \cdots , u'_n) = \int_{a}^{b} \Lambda(x,u,u_1, \cdots, u_n, u', \cdots , u'_n) dx$$ the first variation then becomes for directions $\mathbf{v} = (v_{1},\cdots,v_{n})$
$$\delta J( \mathbf{u},\mathbf{v}) = \int_{a}^{b}\sum_{i}^{b} v_{j} \left[ \frac{\partial \Lambda}{\partial u_j} – \frac{d}{dx} \frac{\partial \Lambda}{\partial u'_j}\right] dx$$
correct? but in practicality.. how do we choose $\mathbf{v}$ it's a direction so surely wouldn't picking the basis vectors themselves be smarter, in the same way that we get partial derivatives by using the basis vectors as the arbitary vector for directional derivatives? so for example in the above would we just pick

and particularly, how does this then effect parameterisations of functions?

for example finding the First variation of the arc length given by
$$\int_{a}^{b} \Phi(\mathbf{x},\mathbf{\dot{x}}) dt= \int_{a}^{b} \sqrt{\dot{x}^2(t) + \dot{y}^2(t)} dt$$ we know this is the reparameterisation of the functional $$\int_{a}^{b} \sqrt{1 – \frac{dy}{dx}} dx$$ and Gelfand and Fomin give the solution of the first variation as

$$\delta J(\mathbf{x},\mathbf{v}) = \int_{a}^{b}\left[\dot{x}\left( \frac{\partial\Phi}{\partial x} – \frac{d}{dt} \frac{\partial\Phi}{\partial \dot{x}} \right) + \dot{y}\left( \frac{\partial\Phi}{\partial y} – \frac{d}{dt} \frac{\partial\Phi}{\partial \dot{y}} \right)\right]$$

so essentially my question is, how do we choose these directions to be useful? why is it for the arc length theyve got $\dot(x),\dot(y)$ as the direction for the euler-legrange equation?..i dont know if this is coming through but this has confused me greatly.

Thanks for taking the time to read through, appreciate it.


Ok for clarification; i'm asking a few smaller questions in one large amount.
1st: How do we often choose v? is it simply just a small increment or do we often chose a "useful" direction?

is the first variation of a functional just the Euler-Legrange equation projected into the direction v?

Best Answer

One cannot choose $\vec v$, it is supposed to be arbitrary as long as it satisfies the boundary condition on the region of interest, say $D$, that $\forall \vec x \in \partial D$, $\vec v(\vec x)=\vec 0$.