Recap: minimization and the Euler-Lagrange equation

Last time we learned about path minimization, in which we want to find some path that is, generally, a function \(y(x)\) that minimizes some other quantity related to the path; examples include the path length, travel time along the path, area enclosed by the path, etc. To accomplish our goal, we defined some ideal path \(y(0,x)\) that accomplishes our minimization goal and then added some weighted deviation \(\alpha\eta(x)\), where \(\eta(x)\) is required to have a value of 0 at its end points and \(\alpha\) is a variable parameter value. The path to minimize is then \(y(\alpha,x) = y(0,x) + \alpha\eta(x)\) and we can minimize our quantity that is some integral of a function \(f(y,y',x)\) that implicitly depends on the parameter \(\alpha\). We were able to show that in order to have a minimum, we must meet the condition

\(\frac{\partial f}{\partial y} - \frac{d}{dx}\left( \frac{\partial f}{\partial y'}\right) = 0\),

which is called the Euler-Lagrange equation. Today we will apply this condition to a few examples and derive a useful identity from a special case of the optimization problem.

The shortest path between two points

We set up the following problem, but were not able to solve it without first knowing the Euler-Lagrange equation: find the path \(y(x)\) that has the shortest distance between any two points \((x_1, y_1)\) and \((x_2, y_2)\). This means we will minimize the path length integral \(S = \int_1^2 ds = \int_{x_1}^{x_2} \sqrt{1 + y'^2}dx\). We can immediately apply the Euler-Lagrange equation using the function \(f(y, y', x) = \sqrt{1 + y'^2}\).

Solving for the two terms in the E-L equation,

\(\frac{\partial f}{\partial y} = 0\) because there is no direct dependence on \(y\). This means that we must have \(\frac{d}{dx} \left( \frac{\partial f}{\partial y'} \right) = 0\), and we can say that \(\frac{\partial f}{\partial y'}=c\) where \(c\) is some constant.

\(\frac{d}{dx} \left( \frac{\partial f}{\partial y'} \right) = \frac{y'}{\sqrt{1+y'^2}} = c\). We could rewrite this to find \(y' = m\) where \(m\) is some other constant. Recall that \(y'\) is the slope of \(y\) with respect to \(x\), so we have found the equation of a line \(y = mx+b\). The shortest distance between two points is a line!

Minimal surface of rotation (soap film problem)

Suppose we want to use a piece of string to connect two points A and B, and then rotate the path the string makes around some axis coplanar to A and B. This will create some kind of surface of rotation. Find the minimal surface area of this problem.

We can solve the minimization problem using Euler-Lagrange. Let the path between A and B be defined as \(y(x)\). We will make our axis of rotation the \(y\)-axis.

The surface area can be built up of slices of our 3-d structure in \(y\): \(dA = ds\cdot 2\pi r = ds \cdot 2\pi x = \sqrt{1+y'^2} 2\pi x\). Then \(A = \int_{x_1}^{x_2} 2\pi x \sqrt{1 + y'^2} dx\) and \(f(y, y', x) = x\sqrt{1+y'^2}\). Using this function in the Euler-Lagrange equation, \(\frac{\partial f}{\partial y} = 0\) so the second term \(\frac{d}{dx}\frac{\partial f}{\partial y'} = 0\) and \(\frac{\partial f}{\partial y'}\) is constant.

\[\begin{align*} \frac{\partial f}{\partial y'} = \frac{xy'}{\sqrt{1+y'^2}} &= c \\ x^2y'^2 &= c^2(1+y'^2) \\ x^2y'^2 - c^2y'^2 &= c^2 \\ y' &= c\sqrt{\frac{1}{x^2/c^2 - 1}} \end{align*}\]

We would then integrate both sides in \(x\) to find a formula for \(y(x)\). This happens to be a known integral that we can look up: \(\frac{d}{dx}\cosh^{-1}(x) = \sqrt{\frac{1}{x^2-1}}\). So the result is \(y(x) = c\cosh^{-1}(x/c) + b\). We can invert this to find \(x = c \cosh\left(\frac{y-b}{c}\right)\), which is a caternary curve.

Beltrami’s Identity

Let’s consider an important case of path minimization and the Euler-Lagrange equation: what happens if \(f\) has no explicit dependence on \(x\)? We have already seen this several times, and it will keep coming up.

We know that even if \(f\) doesn’t depend on \(x\) explicitly (\(\frac{\partial f}{\partial x} = 0\)) there should still be some implicit dependence on \(x\) through \(y(x)\). Looking at the total derivative,

\[\begin{align*} \frac{df}{dx} &= \frac{\partial f}{\partial y}\frac{\partial y}{\partial x} + \frac{\partial f}{\partial y'}\frac{\partial y'}{\partial x} + \frac{\partial f}{\partial x} \\ &= \frac{\partial f}{\partial y} y' + \frac{\partial f}{\partial y'}\frac{\partial y'}{\partial x} + \frac{\partial f}{\partial x} \\ \frac{\partial f}{\partial y}y' &= -\frac{\partial f}{\partial y'}\frac{\partial y'}{\partial x} - \frac{\partial f}{\partial x} + \frac{df}{dx}\\ \end{align*}\]

If we multiply the E-L equation by \(y'\) we can substitute in the result above:

\[\begin{align*} \frac{\partial f}{\partial y}y' - \frac{x}{dx}\frac{\partial f}{\partial y'}y' &= 0 \\ -\frac{\partial f}{\partial y'}\frac{\partial y'}{\partial x} - \frac{\partial f}{\partial x} + \frac{df}{dx} - \frac{x}{dx}\frac{\partial f}{\partial y'}y' &= 0 \\ \frac{df}{dx} -\frac{\partial f}{\partial y'}\frac{\partial y'}{\partial x} - \frac{x}{dx}\frac{\partial f}{\partial y'}y' - \frac{\partial f}{\partial x} &= 0 \end{align*}\]

The second two terms are the distributed chain-rule form of \(\frac{d}{dx}\left[ y' \frac{\partial f}{\partial y'} \right]\) and the last term will be zero since \(f\) has no explicit dependence on \(x\). This leaves

\[\begin{align*} \frac{df}{dx} - \frac{d}{dx}\left[ y' \frac{\partial f}{\partial y'} \right] &= 0 \\ \frac{d}{dx}\left[ f - y' \frac{\partial f}{\partial y'} \right] &= 0 \\ f - y' \frac{\partial f}{\partial y'} &= c \\ \end{align*}\]

This result is called the Beltrami Identity, or the second form of Euler-Lagrange. Use it any time you meet the condition that \(f\) has no explicit dependence on \(x\).