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Cleonis

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- The operation that converts the Euler-Lagrange equation to the Beltrami identity is integration with respect to the y-coordinate. I'm looking for a transparent way to perform that conversion.

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 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.

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 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.