Noether's Theorem (Calculus of Variations)

Theorem
Let $y_i$, $F$, $\Psi_i$, $\Phi$ be real functions.

Let $x, \epsilon \in \R$.

Let $ \mathbf y= \langle y_i \rangle_{1 \le i \le n}$, $ \mathbf \Psi= \langle \Psi_i \rangle_{1 \le i \le n}$ be vectors.

Let
 * $ \Phi=\Phi \left({ x, \mathbf y, \mathbf y'; \epsilon } \right), \quad \Psi_i=\Psi_i  \left({ x, \mathbf y, \mathbf y'; \epsilon  } \right)$

such that


 * $ \Phi \left({ x, \mathbf y, \mathbf y'; 0 } \right)=x, \quad \Psi_i \left({ x, \mathbf y, \mathbf y'; 0 } \right)=y_i$

where ${ x, \mathbf y, \mathbf y', \epsilon }$ are variables.

Let


 * $ \displaystyle J[ \mathbf y] = \int_{x_0}^{x_1} F \left({ x, \mathbf y, \mathbf y' } \right) \mathrm d x$

be a functional.

Let


 * $ X=\Phi \left({ x, \mathbf y, \mathbf y'; \epsilon } \right), \quad \mathbf Y= \mathbf \Psi  \left({ x, \mathbf y, \mathbf y'; \epsilon  } \right)$

Suppose, $J[ \mathbf y]$ is invariant under these transformations for arbitrary $x_0$ and $x_1$.

Then


 * $ \nabla_{ \mathbf y'} F \cdot \boldsymbol{ \psi} + \left({ F- \mathbf y' \cdot \nabla_{ \mathbf y'} F } \right) \phi = C$

where $C$ is a constant and


 * $ \displaystyle \boldsymbol{ \psi} \left({ x, \mathbf y, \mathbf y' } \right)= \frac{ \partial \mathbf \Psi \left({ x, \mathbf y, \mathbf y'; \epsilon} \right) }{ \partial \epsilon} \Bigg \rvert_{\epsilon = 0}$


 * $ \displaystyle \phi \left({ x, \mathbf y, \mathbf y' } \right)= \frac{ \partial \Phi \left({ x, \mathbf y, \mathbf y'; \epsilon} \right) }{ \partial \epsilon} \Bigg \rvert_{\epsilon = 0}$

Proof
Apply Taylor's theorem to the transformations $X$, $ \mathbf Y$ at the point $\epsilon=0$:

Use the general variation formula, and suppose that the curve $ \mathbf y= \mathbf y \left({ x } \right)$ is an extremal of $J[ \mathbf y ]$:


 * $ \delta x = \epsilon \phi, \quad \delta y = \epsilon \psi$


 * $ \displaystyle \delta J= \epsilon \left[{ \nabla_{ \mathbf y} F \cdot \boldsymbol \psi + \left({ F - \mathbf y' \cdot \nabla_{ \mathbf y} F } \right) \phi } \right]_{x=x_0}^{x=x_1}$

Since $J[ \mathbf y]$ is invariant under the transformation, variation vanishes.

Then for any $ \epsilon \ne 0$ we have

$\left[{ \nabla_{ \mathbf y} F \cdot \boldsymbol \psi + \left({ F - \mathbf y' \cdot \nabla_{ \mathbf y} F } \right) \phi } \right]_{x=x_0}=\left[{ \nabla_{ \mathbf y} F \cdot \boldsymbol \psi + \left({ F - \mathbf y' \cdot \nabla_{ \mathbf y} F } \right) \phi } \right]_{x=x_1}$

This has to hold for arbitrary $x_0$ and $x_1$.

Since only a constant mapping produces the same result for any input, the term in brackets has to be a constant.