The $\delta$ notation in calculus of variations

Question

On page 224 of the 5th edition of Classical Dynamics of Particles and Systems by Stephen T. Thornton and Jerry B. Marion, the authors introduced the $\delta$ notation (in section 6.7). This notation is given by Equations (6.88) which are as follows:

$$\delta J = \frac{\partial J}{\partial \alpha}d\alpha$$ $$\delta y = \frac{\partial y}{\partial \alpha}d\alpha$$

I know that the $\delta$ notation stands for the variation from the actual path, but I cannot relate the geometrical interpretation to the above equation. Can anyone please explain the above terms and provide an explanation on why do the right-hand sides of these relations represent the variation (varied path) from the actual path?

Any help is much appreciated. Thank you so much.

Answer 1

I strongly recommand Calculus of Variations, from Gelfand & Fomin

Given a functional, by example: $$ J[y]=\int_a^b F(x,y(x),y'(x))dx $$ then $\delta J$ is nothing more than the differential $dJ$.

In the cited book you can read, page 12:

is called the variation (or differential) of $J[y]$ and is denoted by $\delta J[h]$

More details:

$$\delta J[y].h \equiv dJ[y].h$$ is the action of differential of $J$ evaluated at $y$ on the vector $h$ (here $y$ and $h$ are functions from a functional space).

In short this is only differential calculus but instead of working with finite vector spaces like $\mathbb{R}^n$ you are working with infinite vector spaces of functions.

An example of such space (always from the cited book) is $\mathcal{D}_1$ the space of all functions which are continuously differentiable in the interval $[a,b]$.

If you are not familiar with differential calculus a wonderful book is: Differential Calculus on Normed Spaces: A Course in Analysis, from H. Cartan

Answer 2

Typically $ \delta x $ refer to the change in position after $ \delta t $ have been taken into account. This makes life much easier at the Boundaries when e.g. deducing Transversality Conditions and similar. The displacement only looking in position direction are typically refered to as $ h $ or $ dx $. The two displacements $ dx $ and $ \delta x $ therefore have different meaning and are related by the formula:

$ \delta x = dx + \dot{x} \delta t $

Let a variational problem be given by:

$ J = \int_{t_0}^{t_1} \mathcal{L} (t,x(t)) dt $

If the endpoints $ t_0 $ and $t_1$ are fixed then $ \delta t $ are set to zero everywhere and $ dx = \delta x $. If the endpoints are free then $ \delta t $ have a non-zero value at the boundaries. This imply that $ dJ $ and $ \delta J $ are only equal if the end-points are fixed...

See attached figure to see the geometry

figure

The full formula for a variation in the path are given below. Note that $ dx $ are used in the integral while $ \delta x $ and $ \delta t $ are used at the boundaries. $ \delta J = \sum_{\alpha} \int_{t_0}^{t_1} \left( \frac{\partial L}{\partial x_{\alpha}} -\frac{d}{dt} \left( \frac{\partial L}{\partial \dot{x}_{\alpha}} \right) \right) dx_{\alpha} dt + \left[\sum_{\alpha} \frac{\partial L}{\partial \dot{x}_{\alpha}} \delta x_{\alpha}\right]_{t_0}^{t_1} + \left[ \left( L-\sum_{\alpha} \frac{\partial L}{\partial \dot{x}_{\alpha} }\dot{x}_{\alpha} \right) \delta t \right]_{t_0}^{t_1}$