notes-archive/docs/physics/mechanics/6-calculus-of-variations.md

Chapter 6 - Calculus of Variations

Section 6.1 - Two Examples

Consider two points (x_1, y_1) and (x_2, y_2), and some path y = y(x) joining them. Then, we know that the length of a path segment is ds = \sqrt{dx^2 + dy^2}. Then, we know dy = \frac{dy}{dx}dx = y'(x) dx, so we can rewrite ds as

ds = \sqrt{1 + y'(x)^2} dx

This allows us to say that L = \int_1^2 ds = \int_{x_1}^{x_2} \sqrt{1 + y'(x)^2} dx. Now, to minimize L, we must find a value of y'(x) that makes the integral minimal.

Now, consider light. The time dt for light to travel ds is ds/v where n = c / v tells us that dt = n / c ds. Then, we see that

t = \frac{1}{c} \int_1^2 n ds

Now, we assume that n = n(x, y), so then

int_1^2 n(x, y)ds = \int_{x_1}^{x_2} n(x, y) \sqrt{1 + y'(x)^2} dx

Thus, to find the shortest path, we must minimize again.

We know that for some function f(x), a critical point is when df / dx = 0. A critical point is a minimum depending on the sign of df^2 / d^2x, or neither of it equals zero.

Definition. A stationary point x_0 is a critical point in which the minimum/maximum status is unknown.

Section 6.2 - The Euler-Lagrange Equation

We have an integral of the form

S = \int_{x_1}^{x_2} f[y(x), y'(x), x] dx

Here, f depends only on x and functions of x.

Now, let y(x) represent the path that minimizes S. Then, we can say that any other path can be represented as Y(x) = y(x) + \alpha\eta, with the boundary points forcing \eta(x_1) = \eta(x_2) = 0. Then, when \alpha = 0, Y(x) = y(x). We can then parameterize S as

S(\alpha) = \int_{x_1}^{x_2} f(Y, Y', x) dx = \int_{x_1}^{x_2} f(y + \alpha \eta, y' + \alpha \eta', x) dx

Then, we can find \partial S / \partial \alpha as \int \partial f / \partial \alpha dx. Applying the chain rule, \partial f / \partial \alpha = \eta \frac{\partial f}{\partial y} + \eta' \frac{\partial f}{\partial y'}. Then, we can minimize S by setting its derivative to zero so that

0 = \frac{dS}{d\alpha}\int_{x_1}^{x_2}\frac{\partial f}{\partial \alpha}dx = \int_{x_1}^{x_2}(\eta \frac{\partial f}{\partial y} + \eta' \frac{\partial f}{\partial y'})dx = 0

Then, we can rewrite the second term by integrating by parts so that

\int_{x_1}^{x_2} = \eta'(x) \frac{\partial f}{\partial y'} dx = [\eta(x) \frac{\partial f}{\partial y'}]_{x_1}^{x_2} - \int_{x_1}^{x_2} \eta(x) \frac{d}{dx}(\frac{\partial f}{\partial y'})dx

The first term is zero due to the conditions on \eta(x). Then, we see that

\int_{x_1}^{x_2} = \eta'(x) \frac{\partial f}{\partial y'} dx = - \int_{x_1}^{x_2} \eta(x) \frac{d}{dx}(\frac{\partial f}{\partial y'})dx

We can substitute this into the above equation for the minimal location of S to see that

\int_{x_1}^{x_2} \eta(x) (\frac{\partial f}{\partial y} - \frac{d}{dx} \frac{\partial f}{\partial y'}) dx = 0

This must be true for any \eta(x). If we let \eta(x) be an arbitrary function with the same sign as the differential terms over the integral, we see that the integral would be non-negative. This forces the differential terms to be zero.

Definition. The Euler-Lagrange Equation. The following criteria holds true for all f(y(x), y'(x), x):

\frac{\partial f}{\partial y} - \frac{d}{dx}\frac{\partial f}{\partial y'} = 0

Section 6.3 - Applications of the Euler-Lagrange Equation

Recall the distance between two points, in which f(y, y', x) = \sqrt{1 + y'^2}. Then, the Euler-Lagrange Equation becomes \frac{d}{dx} \frac{\partial f}{\partial y'} = 0, so \frac{\partial f}{\partial y'} = \frac{y'}{\sqrt{1 + y'^2}} is a constant with regards to x, which implies that y' is a constant and thus this is a straight line.

Now consider a case where a fixed, frictionless track is placed in a gravitational field. What is the most efficient track from point 1 to 2, assuming some horizontal displacement?

We know that due to conservation of energy, v = \sqrt{2gy}. From time equations, t = \int_1^2 \frac{ds}{v}, so that t = \int_0^{y_2} \frac{\sqrt{x'(y)^2 + 1}}{\sqrt{2gy}}dy. Then, f(x, x', y) = \frac{\sqrt{x'^2 + 1}}{\sqrt{y}}, and the Euler-Lagrange Equation tells us that y = a(1 - \cos \theta) and x = a(\theta - \sin \theta) + C, for some \theta.

Note that in no case did we check that the paths found were minimal and not maximal. We only know that these are statioary points.

Section 6.4 - More than T wo Variables

Consider x = x(u), y = y(u). Then, ds = \sqrt{x'(u)^2 + y'(u)^2}du, so that L = \int_{u_1}^{u_2} \sqrt{x'(u)^2 + y'(u)^2}du. Then, f = f(x, y, x', y', u) so that S = \int_{u_1}^{u_2} f du. Now, if x = x(u) + \alpha \xi(u) and y = y(u) + \beta \eta(u), we see that \partial S / \partial \alpha = 0 = \partial S / \partial \Beta, so that \frac{\partial f}{\partial x} = \frac{d}{dx} \frac{\partial f}{\partial x'}, and the same respectively with y.

Definition. Now, we can say that for a system with n coordinates, we have q_1, q_2, \ldots, q_n position vectors, otherwise known as generalized coordinates. Then, we can think of n generalized coordinates as defining a point in an $n$-dimensional configuration space.