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[Calculus] Beltrami Identity

2021-11-28 14:58 作者:AoiSTZ23  | 我要投稿

By: Tao Steven Zheng (鄭濤)

【Problem】

The Beltrami identity, named after the Italian mathematician Eugenio Beltrami (1835 - 1900), is a simplified and less general version of the Euler–Lagrange equation in the calculus of variations.

Show that the Euler-Lagrange equation

%5Cfrac%7Bd%7D%7Bdt%7D%20%5Cleft(%5Cfrac%7B%5Cpartial%20L%7D%7B%5Cpartial%20%5Cdot%7By%7D%7D%20%5Cright)%20-%20%5Cfrac%7B%5Cpartial%20L%7D%7B%5Cpartial%20y%7D%20%3D%200

can be written as

%20%5Cfrac%7Bd%7D%7Bdt%7D%20%5Cleft%5BL%20-%20%5Cdot%7By%7D%20%5Cfrac%7B%5Cpartial%20L%7D%7B%5Cpartial%20%5Cdot%7By%7D%7D%5Cright%5D%20-%20%5Cfrac%7B%5Cpartial%20L%7D%7B%5Cpartial%20t%7D%20%3D%200

Then show that if L does not explicitly depend on t, then

L%20-%20%5Cdot%7By%7D%20%5Cfrac%7B%5Cpartial%20L%7D%7B%5Cpartial%20%5Cdot%7By%7D%7D%20%3D%20C

where C is constant.

Hint:Use the shorthand %5Cfrac%7Bdy%7D%7Bdt%7D%20%3D%20%5Cdot%7By%7D and %5Cfrac%7Bd%5E2%20y%7D%7Bdt%5E2%7D%20%3D%20%5Cfrac%7Bd%5Cdot%7By%7D%7D%7Bdt%7D%20%3D%20%5Cddot%7By%7D%20.

【Solution】

Note that the total derivative

%20%5Cfrac%7BdL%7D%7Bdt%7D%20%3D%20%5Cfrac%7B%5Cpartial%20L%7D%7B%5Cpartial%20t%7D%20%2B%20%5Cfrac%7B%5Cpartial%20L%7D%7B%5Cpartial%20y%7D%20%5Cfrac%7Bdy%7D%7Bdt%7D%20%2B%20%5Cfrac%7B%5Cpartial%20L%7D%7B%5Cpartial%20%5Cdot%7By%7D%7D%20%5Cfrac%7Bd%20%5Cdot%7By%7D%7D%7Bdt%7D

can be expressed as

%5Cfrac%7BdL%7D%7Bdt%7D%20%3D%20%5Cfrac%7B%5Cpartial%20L%7D%7B%5Cpartial%20t%7D%20%2B%20%5Cfrac%7B%5Cpartial%20L%7D%7B%5Cpartial%20y%7D%20%5Cdot%7By%7D%20%2B%20%5Cfrac%7B%5Cpartial%20L%7D%7B%5Cpartial%20%5Cdot%7By%7D%7D%20%5Cddot%7By%7D%20

Also,

%5Cfrac%7Bd%7D%7Bdt%7D%20%5Cleft%5B%5Cdot%7By%7D%20%5Cfrac%7B%5Cpartial%20L%7D%7B%5Cpartial%20%5Cdot%7By%7D%7D%5Cright%5D%20%3D%20%5Cdot%7By%7D%20%5Cfrac%7Bd%7D%7Bdt%7D%20%5Cleft(%5Cfrac%7B%5Cpartial%20L%7D%7B%5Cpartial%20%5Cdot%7By%7D%7D%20%5Cright)%20%2B%20%5Cfrac%7B%5Cpartial%20L%7D%7B%5Cpartial%20%5Cdot%7By%7D%7D%20%5Cddot%7By%7D

Substituting the above two expressions into

%5Cfrac%7Bd%7D%7Bdt%7D%20%5Cleft%5BL%20-%20%5Cdot%7By%7D%20%5Cfrac%7B%5Cpartial%20L%7D%7B%5Cpartial%20%5Cdot%7By%7D%7D%5Cright%5D%20-%20%5Cfrac%7B%5Cpartial%20L%7D%7B%5Cpartial%20t%7D%20%3D%200

gives

%5Cfrac%7B%5Cpartial%20L%7D%7B%5Cpartial%20t%7D%20%2B%20%5Cfrac%7B%5Cpartial%20L%7D%7B%5Cpartial%20y%7D%20%5Cdot%7By%7D%20%2B%20%5Cfrac%7B%5Cpartial%20L%7D%7B%5Cpartial%20%5Cdot%7By%7D%7D%20%5Cddot%7By%7D%20-%20%5Cleft%5B%5Cdot%7By%7D%20%5Cfrac%7Bd%7D%7Bdt%7D%20%5Cleft(%5Cfrac%7B%5Cpartial%20L%7D%7B%5Cpartial%20%5Cdot%7By%7D%7D%20%5Cright)%20%2B%20%5Cfrac%7B%5Cpartial%20L%7D%7B%5Cpartial%20%5Cdot%7By%7D%7D%20%5Cddot%7By%7D%5Cright%5D%20-%20%5Cfrac%7B%5Cpartial%20L%7D%7B%5Cpartial%20t%7D%20%3D%200%20


Simplify this expression and factor out -%20%5Cdot%7By%7D:

%5Cfrac%7B%5Cpartial%20L%7D%7B%5Cpartial%20y%7D%20%5Cdot%7By%7D%20-%20%5Cdot%7By%7D%20%5Cfrac%7Bd%7D%7Bdt%7D%20%5Cleft(%5Cfrac%7B%5Cpartial%20L%7D%7B%5Cpartial%20%5Cdot%7By%7D%7D%20%5Cright)%20%3D%200%20

-%20%5Cdot%7By%7D%20%5Cleft%5B%5Cfrac%7Bd%7D%7Bdt%7D%20%5Cleft(%5Cfrac%7B%5Cpartial%20L%7D%7B%5Cpartial%20%5Cdot%7By%7D%7D%20%5Cright)-%5Cfrac%7B%5Cpartial%20L%7D%7B%5Cpartial%20y%7D%20%5Cright%5D%20%3D%200

Divide away -%20%5Cdot%7By%7D:

%5Cfrac%7Bd%7D%7Bdt%7D%20%5Cleft(%5Cfrac%7B%5Cpartial%20L%7D%7B%5Cpartial%20%5Cdot%7By%7D%7D%20%5Cright)-%5Cfrac%7B%5Cpartial%20L%7D%7B%5Cpartial%20y%7D%20%3D%200%20

Therefore,

%5Cfrac%7Bd%7D%7Bdt%7D%20%5Cleft%5BL%20-%20%5Cdot%7By%7D%20%5Cfrac%7B%5Cpartial%20L%7D%7B%5Cpartial%20%5Cdot%7By%7D%7D%5Cright%5D%20-%20%5Cfrac%7B%5Cpartial%20L%7D%7B%5Cpartial%20t%7D%20%3D%200

is equivalent to the Euler-Lagrange equation.

If L%20 does not explicitly depend on t, then %5Cfrac%7B%5Cpartial%20L%7D%7B%5Cpartial%20t%7D%20%3D%200 and%5Cfrac%7Bd%7D%7Bdt%7D%20%5Cleft%5BL%20-%20%5Cdot%7By%7D%20%5Cfrac%7B%5Cpartial%20L%7D%7B%5Cpartial%20%5Cdot%7By%7D%7D%5Cright%5D-%5Cfrac%7B%5Cpartial%20L%7D%7B%5Cpartial%20t%7D%20%3D%200%20

becomes

%5Cfrac%7Bd%7D%7Bdt%7D%20%5Cleft%5BL%20-%20%5Cdot%7By%7D%20%5Cfrac%7B%5Cpartial%20L%7D%7B%5Cpartial%20%5Cdot%7By%7D%7D%5Cright%5D%20%3D%200

So by integration

L%20-%20%5Cdot%7By%7D%20%5Cfrac%7B%5Cpartial%20L%7D%7B%5Cpartial%20%5Cdot%7By%7D%7D%20%3D%20C

where C%20 is constant.


[Calculus] Beltrami Identity的評(píng)論 (共 條)

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