# The relation between the Euler-Lagrange equation and the Beltrami identity

This question is specifically about deriving the Beltrami identity.

Just to give this question context I provide an example of a problem that is solved with Calculus of Variations: find the shape of a soap film that stretches between two coaxial rings.

For the surface area the expression to be integrated from start point to end point:

$$F = 2 \pi \int_{x_0}^{x_1} y \ \sqrt{1 + (y')^2} \ dx \tag{1}$$

For the purpose of finding the function that minimizes that surface area the Euler-Lagrange equation is applied.

As we know, since the value of F does not depend directly on the x-coordinate the Beltrami identity is applicable.

Comparison of the EL-equation and the Beltrami identity:

Euler-Lagrange:

$$\frac{\partial F}{\partial y} - \frac{d}{dx}\left(\frac{\partial F}{\partial y'}\right) = 0 \tag{2}$$

Beltrami:

$$F - y' \frac{\partial F}{\partial y'} = C \tag{3}$$

We see that the process of conversion from EL-eq. to Beltrami consists of integration with respect to the y-coordinate

For the first term:

$$\int \frac{\partial F}{\partial y} dy = F + C \tag{4}$$

with $$C$$ an arbitrary integration constant.

Question:
Is there a transparent way to evaluate the same integral for the second term?

$$\int \frac{d}{dx}\left(\frac{\partial F}{\partial y'}\right)dy + C \quad = ? = \quad \frac{dy}{dx} \frac{\partial F}{\partial y'} \tag{5}$$

The usual way to arrive at the Beltrami identity is quite a Rube Goldberg derivation.

The thing is: showing that (5) is indeed correct is worthwhile only if it can be done in a way that is more accessible than the usual way of obtaining the Beltrami identity.

• Comment to the post (v1): Eq. (4) does not seem correct if $F$ is supposed to depend on $y'$. Oct 30, 2023 at 11:02
• @Qmechanic You raise a point that I had previously failed to consider. The Euler-Lagrange equation and the Beltrami identity both have two terms, but I should not have assumed that the distribution of information is the same for each. I suppose I should try the other way round: take partial derivative with respect to $y$ of the Beltrami identity, and if that reproduces the Euler-Lagrange equation then that shows that the relation is one of integration/differentiation. Oct 30, 2023 at 17:10
• @Qmechanic I submitted a self-answer, answering the question to my own satisfaction. (Now that I have this understanding I am baffled at how this understanding eluded me before.) There is on my own website an exposition of (single variable) calculus of variations. I will replace the existing discussion of the Beltrami identity with the narrative presented here. Nov 2, 2023 at 2:26

The relation between the Euler-Lagrange equation and the Beltrami identity is: the Beltrami identity is the Euler-Lagrange equation with a differentiation with respect to $$x$$ backed out.

In order to back out a differentiation (here a differentiation with respect to $$x$$), one must arrive at an expression of the following form:

$$\frac{dF}{dx} = \frac{dG}{dx} \tag{1}$$

Rearrange:

$$\frac{dF}{dx} - \frac{dG}{dx} = 0 \tag{2}$$

Differentiation is distributive, so we can convert to:

$$\frac{d}{dx} (F - G) = 0 \tag{3}$$

At this point we can forego the differentiation $$\tfrac{d}{dx}$$ and state:

$$(F - G) = C \tag{4}$$

Where $$C$$ is a constant, to be determined.

In further preparation:
The product rule:

$$\frac{d(f(x)g(x))}{dx} = \frac{d\ f(x)}{dx} \ g(x) + f(x) \ \frac{d\ g(x)}{dx} \tag{5}$$

Here the product rule will be used in reverse: it will be used to collapse two terms into one.

$$\frac{d\ f(x)}{dx} \ g(x) + f(x) \ \frac{d\ g(x)}{dx} = \frac{d(f(x)g(x))}{dx} \tag{6}$$

That completes the preparations.

The general expression for derivative of an expression $$F$$ with respect to $$x$$

$$\frac{dF}{dx} = \frac{\partial F}{\partial x}\frac{dx}{dx} + \frac{\partial F}{\partial y}\frac{dy}{dx} + \frac{\partial F}{\partial y'}\frac{dy'}{dx} \tag{7}$$

Omit the term for partial derivative with respect to $$x$$:

$$\frac{dF}{dx} = \frac{\partial F}{\partial y}\frac{dy}{dx} + \frac{\partial F}{\partial y'}\frac{dy'}{dx} \tag{8}$$

In order to go from (8) to the Beltrami identity we need to accomplish the following three objectives:

• Combine with the Euler-Lagrange equation
• Collapse the two terms on the right hand side of of (8) into one term
• That one term must be one of differentiation with respect to $$x$$

$$\frac{\partial F}{\partial y} = \frac{d}{dx} \left( \frac{\partial F}{\partial y'} \right) \tag{9}$$

(9) is the Euler-Lagrange equation, we use it to substitute the factor $$\tfrac{\partial F}{\partial y}$$ on the right hand side of (8).

The substitution gives (10).

$$\frac{dF}{dx} = y' \frac{d}{dx} \left( \frac{\partial F}{\partial y'} \right) + y'' \frac{\partial F}{\partial y'} \tag{10}$$

The substitution has accomplished all three of the objectives:
The right hand side of (10) has the same pattern as the left hand side of (6), so those two terms fold into one.

$$\frac{dF}{dx} = \frac{d}{dx} \left( y' \frac{\partial F}{\partial y'} \right) \tag{11}$$

(11) has the same form as (1), so we can back out a differentiation with respect to $$x$$

$$\frac{d}{dx} \left(F - y' \frac{\partial F}{\partial y'} \right) = 0 \tag{12}$$

$$F - y' \frac{\partial F}{\partial y'} = C \tag{13}$$

The key point:
The relation between the Euler-Lagrange equation and the Beltrami identity is one of integration-differentiation. The difference is differentiation with respect to $$x$$.