Calculus of Variations

It is a well-known fact, first enunciated by Archimedes, that the shortest distance between two points in a plane is a straight-line. However, suppose that we wish to demonstrate this result from first principles. Let us consider the length, $l$ , of various curves, $y(x)$ , that run between two fixed points, $A$ and $B$ , in a plane. Now, $l$ takes the form

$$l = \int_A^B [dx^{ 2} + dy^{ 2}]^{1/2} = \int_a^b [1 + y'^{ 2}(x)]^{1/2} dx,$$ (B.29)

where $y'\equiv dy/dx$ . Note that $l$ is a function of the function $y(x)$ . In mathematics, a function of a function is termed a functional.

In order to find the shortest path between points $A$ and $B$ , we need to minimize the functional $l$ with respect to small variations in the function $y(x)$ , subject to the constraint that the end points, $A$ and $B$ , remain fixed. In other words, we need to solve

$$\delta l = 0.$$ (B.30)

The meaning of the previous equation is that if $y(x)\rightarrow y(x)+\delta y(x)$ , where $\delta y(x)$ is small, then the first-order variation in $l$ , denoted $\delta l$ , vanishes. In other words, $l\rightarrow l + {\cal O}(\delta y^{ 2})$ . The particular function $y(x)$ for which $\delta l =0$ obviously yields an extremum of $l$ (i.e., either a maximum or a minimum). Hopefully, in the case under consideration, it yields a minimum of $l$ .

Consider a general functional of the form

$$I = \int_a^b F(y, y',x) dx,$$ (B.31)

where the end points of the integration are fixed. Suppose that $y(x)\rightarrow y(x)+\delta y(x)$ . The first-order variation in $I$ is written

$$\delta I = \int_a^b\left(\frac{\partial F}{\partial y} \delta y+ \frac{\partial F}{\partial y'} \delta y'\right)dx,$$ (B.32)

where $\delta y' = d(\delta y)/dx$ . Setting $\delta I$ to zero, we obtain

$$\int_a^b\left(\frac{\partial F}{\partial y} \delta y+ \frac{\partial F}{\partial y'} \delta y'\right) dx = 0.$$ (B.33)

This equation must be satisfied for all possible small perturbations, $\delta y(x)$ .

Integrating the second term in the integrand of the previous equation by parts, we get

$$\int_a^b\left[\frac{\partial F}{\partial y}- \frac{d}{dx}\!\left(... ...ight]\delta y dx +\left[\frac{\partial F}{\partial y'} \delta y\right]_a^b=0.$$ (B.34)

If the end points are fixed then $\delta y=0$ at $x=a$ and $x=b$ . Hence, the last term on the left-hand side of the previous equation is zero. Thus, we obtain

$$\int_a^b\left[\frac{\partial F}{\partial y}- \frac{d}{dx}\!\left(\frac{\partial F}{\partial y'}\right)\right]\delta y dx =0.$$ (B.35)

The previous equation must be satisfied for all small perturbations $\delta y(x)$ . The only way in which this is possible is for the expression enclosed in square brackets in the integral to be zero. Hence, the functional $I$ attains an extremum value whenever

$$\frac{d}{dx}\!\left(\frac{\partial F}{\partial y'}\right)-\frac{\partial F}{\partial y} = 0.$$ (B.36)

This condition is known as the Euler-Lagrange equation.

Let us consider some special cases. Suppose that $F$ does not explicitly depend on $y$ . It follows that $\partial F/\partial y = 0$ . Hence, the Euler-Lagrange equation, (B.36), simplifies to

$$\frac{\partial F}{\partial y'} = {\rm constant}.$$ (B.37)

Next, suppose that $F$ does not depend explicitly on $x$ . Multiplying Equation (B.36) by $y'$ , we obtain

$$y' \frac{d}{dx}\!\left(\frac{\partial F}{\partial y'}\right)-y' \frac{\partial F}{\partial y} = 0.$$ (B.38)

However,

$$\frac{d}{dx}\!\left(y' \frac{\partial F}{\partial y'}\right) = y... ...eft(\frac{\partial F}{\partial y'}\right)+ y'' \frac{\partial F}{\partial y'}.$$ (B.39)

Thus, we get

$$\frac{d}{dx}\!\left(y' \frac{\partial F}{\partial y'}\right) = y' \frac{\partial F}{\partial y} + y'' \frac{\partial F}{\partial y'}.$$ (B.40)

If $F$ is not an explicit function of $x$ then the right-hand side of the previous equation is the total derivative of $F$ , namely $dF/dx$ . Hence, we obtain

$$\frac{d}{dx}\!\left(y' \frac{\partial F}{\partial y'}\right) = \frac{dF}{dx},$$ (B.41)

which yields

$$y' \frac{\partial F}{\partial y'} - F = {\rm constant}.$$ (B.42)

Returning to the case under consideration, we have $F = (1+y'^{ 2})^{1/2}$ , according to Equations (B.29) and (B.31). Hence, $F$ is not an explicit function of $y$ , so Equation (B.37) yields

$$\frac{\partial F}{\partial y'} = \frac{y'}{\sqrt{1+y'^{ 2}}} = c,$$ (B.43)

where $c$ is a constant. So,

$$y' = \frac{c}{\sqrt{1-c^{ 2}}} = {\rm constant}.$$ (B.44)

Of course, $y' = {\rm constant}$ is the equation of a straight-line. Thus, the shortest distance between two fixed points in a plane is indeed a straight-line.