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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.

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  • $\begingroup$ Comment to the post (v1): Eq. (4) does not seem correct if $F$ is supposed to depend on $y'$. $\endgroup$
    – Qmechanic
    Oct 30, 2023 at 11:02
  • $\begingroup$ @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. $\endgroup$
    – Cleonis
    Oct 30, 2023 at 17:10
  • $\begingroup$ @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. $\endgroup$
    – Cleonis
    Nov 2, 2023 at 2:26

1 Answer 1

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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 $x$.

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