- #1
Cleonis
Gold Member
- 732
- 39
- TL;DR Summary
- 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.