Euler-Lagrange Equation

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, $y(x)$ , which run between two fixed points, $A$ and $B$ , in a plane, as illustrated in Figure E.1. Now, $l$ takes the form

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

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.

Figure: Different paths between points $A$ and $B$ .

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

$$\delta l = 0.$$ (E.2)

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

$$I = \int_a^b F(y, y', x)\,dx,$$ (E.3)

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

$$\delta I = \int_a^b\left(\frac{\partial F}{\partial y}\,\delta y+ \frac{\partial F}{\partial y'}\,\delta y'\right)dx,$$ (E.4)

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

$$\int_a^b\left(\frac{\partial F}{\partial y}\,\delta y+ \frac{\partial F}{\partial y'}\,\delta y'\right)\,dx = 0.$$ (E.5)

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

$$\int_a^b\left[\frac{\partial F}{\partial y}- \frac{d}{dx}\!\left(... ...ight]\delta y\,dx +\left[\frac{\partial F}{\partial y'}\,\delta y\right]_a^b=0.$$ (E.6)

However, 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

$$\int_a^b\left[\frac{\partial F}{\partial y}- \frac{d}{dx}\!\left(\frac{\partial F}{\partial y'}\right)\right]\delta y\,dx =0.$$ (E.7)

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

$$\frac{d}{dx}\!\left(\frac{\partial F}{\partial y'}\right)-\frac{\partial F}{\partial y} = 0.$$ (E.8)

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 (E.8) simplifies to

$$\frac{\partial F}{\partial y'} = {\rm const}.$$ (E.9)

Next, suppose that $F$ does not depend explicitly on $x$ . Multiplying Equation (E.8) by $y'$ , we obtain

$$y'\,\frac{d}{dx}\!\left(\frac{\partial F}{\partial y'}\right)-y'\,\frac{\partial F}{\partial y} = 0.$$ (E.10)

However,

$$\frac{d}{dx}\!\left(y'\,\frac{\partial F}{\partial y'}\right) = y... ...eft(\frac{\partial F}{\partial y'}\right)+ y''\,\frac{\partial F}{\partial y'}.$$ (E.11)

Thus, we get

$$\frac{d}{dx}\!\left(y'\,\frac{\partial F}{\partial y'}\right) = y'\,\frac{\partial F}{\partial y} + y''\,\frac{\partial F}{\partial y'}.$$ (E.12)

Now, 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

$$\frac{d}{dx}\!\left(y'\,\frac{\partial F}{\partial y'}\right) = \frac{dF}{dx},$$ (E.13)

which yields

$$y'\,\frac{\partial F}{\partial y'} - F = {\rm const}.$$ (E.14)

Returning to the case under consideration, we have $F = \sqrt{1+y'^{\,2}}$ , according to Equation (E.1) and (E.3). Hence, $F$ is not an explicit function of $y$ , so Equation (E.9) yields

$$\frac{\partial F}{\partial y'} = \frac{y'}{\sqrt{1+y'^{\,2}}} = c,$$ (E.15)

where $c$ is a constant. So,

$$y' = \frac{c}{\sqrt{1-c^{\,2}}} = {\rm const}.$$ (E.16)

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.