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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

$\displaystyle l = \int_A^B (dx^{\,2} + dy^{\,2})^{1/2} = \int_a^b [1 + y'^{\,2}(x)]^{1/2}\,dx,$ (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$ .
\begin{figure}
\epsfysize =2.5in
\centerline{\epsffile{AppendixE/figD.01.eps}}
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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

$\displaystyle \delta l = 0.$ (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

$\displaystyle I = \int_a^b F(y, y', x)\,dx,$ (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

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

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

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

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

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

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

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

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

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

However,

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

Thus, we get

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

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

which yields

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

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

where $ c$ is a constant. So,

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


next up previous
Next: Conditional Variation Up: Calculus of Variations Previous: Indroduction
Richard Fitzpatrick 2016-03-31