# Noether's Theorem (Calculus of Variations)

## Theorem

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

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

Let $\mathbf y = \sequence {y_i}_{1 \mathop \le i \mathop \le n}$ and $\mathbf \Psi = \sequence{\Psi_i}_{1 \mathop \le i \mathop \le n}$ 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:

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

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 transformations $x \rightarrow X$ and $\mathbf y \rightarrow \mathbf Y$ 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:

$\map {\boldsymbol \psi} {x, \mathbf y, \mathbf y'} = \dfrac {\partial \map {\mathbf \Psi} {x, \mathbf y, \mathbf y'; \epsilon} } {\partial \epsilon} \Bigg\rvert_{\epsilon \mathop = 0}$
$\map \phi {x, \mathbf y, \mathbf y'} = \dfrac {\partial \map \Phi {x, \mathbf y, \mathbf y'; \epsilon} } {\partial \epsilon} \Bigg\rvert_{\epsilon \mathop = 0}$

## Proof

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

 $\ds X$ $=$ $\ds \map \Phi {x, \mathbf y, \mathbf y'; \epsilon}$ $\ds$ $=$ $\ds \map \Phi {x, \mathbf y, \mathbf y'; 0} + \epsilon \frac {\partial \map \Phi {x, \mathbf y, \mathbf y'; \epsilon} } {\partial \epsilon} \Bigg\rvert_{\epsilon \mathop = 0} + \map \OO \epsilon$ $\ds$ $=$ $\ds x + \epsilon \phi + \map \OO \epsilon$
 $\ds \mathbf Y$ $=$ $\ds \map {\mathbf \Psi} {x, \mathbf y, \mathbf y'; \epsilon}$ $\ds$ $=$ $\ds \map {\mathbf \Psi} {x, \mathbf y, \mathbf y'; 0} + \epsilon \frac {\partial \map {\mathbf \Psi} {x, \mathbf y, \mathbf y'; \epsilon} } {\partial \epsilon} \Bigg \rvert_{\epsilon \mathop = 0} + \map \OO \epsilon$ $\ds$ $=$ $\ds y + \epsilon \boldsymbol \psi + \map \OO \epsilon$

Use the general variation formula, and suppose that the curve $\mathbf y = \map {\mathbf y} x$ is an extremal of $J \sqbrk {\mathbf y}$:

$\delta x = \epsilon \phi, \quad \delta y = \epsilon\psi$
$\delta J = \epsilon \bigintlimits {\nabla_{\mathbf y'} F \cdot \boldsymbol \psi + \paren {F - \mathbf y' \cdot \nabla_{\mathbf y'} F }\phi} {x \mathop = x_0} {x \mathop = x_1}$

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

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

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

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.

$\blacksquare$

## Source of Name

This entry was named for Emmy Noether.