Noether's Theorem (Calculus of Variations)

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

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

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

Let
 * $\Phi=\map {\Phi} {x,\mathbf y,\mathbf y';\epsilon}, \quad \Psi_i=\map {\Psi_i} {x,\mathbf y,\mathbf y';\epsilon}$

such that


 * $\map {\Phi} {x,\mathbf y,\mathbf y';0}=x,\quad\map {\Psi_i} {x,\mathbf y,\mathbf y';0}=y_i$

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

Let


 * $\displaystyle J\sqbrk{\mathbf y}=\int_{x_0}^{x_1} \map F {x,\mathbf y,\mathbf y'}\rd x$

be a functional.

Let


 * $X=\map {\Phi} {x,\mathbf y,\mathbf y';\epsilon},\quad\mathbf Y=\map {\mathbf\Psi} {x,\mathbf y,\mathbf y';\epsilon}$

Suppose, $J\sqbrk{\mathbf y}$ is invariant under these transformations for arbitrary $x_0$ and $x_1$.

Then


 * $ \nabla_{\mathbf y'} F\cdot\boldsymbol{\psi}+\paren{F-\mathbf y'\cdot\nabla_{\mathbf y'}F}\phi=C$

where $C$ is a constant and


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


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

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=\map {\mathbf y} x$ is an extremal of $J\sqbrk{\mathbf y}$:


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


 * $\displaystyle\delta J=\epsilon \sqbrk{\nabla_{\mathbf y'} F\cdot\boldsymbol\psi+\paren{F-\mathbf y'\cdot\nabla_{\mathbf y'}F}\phi}_{x=x_0}^{x=x_1}$

Since $J\sqbrk{\mathbf y}$ is invariant under the transformation, variation vanishes.

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

$\sqbrk{\nabla_{\mathbf y'}F\cdot\boldsymbol\psi+\paren{F-\mathbf y'\cdot\nabla_{\mathbf y'}F}\phi}_{x=x_0}=\sqbrk{\nabla_{\mathbf y'}F\cdot\boldsymbol\psi+\paren{F-\mathbf y'\cdot\nabla_{\mathbf y'}F}\phi}_{x=x_1}$

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.