Theorem: Noether's Theorem (Calculus of Variations)

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

Proof:
Apply Taylor's Theorem to the transformations $X$, $\mathbf Y$ at the point $\epsilon = 0$:
{{begin-eqn}}
{{eqn | l = X
      | r = \map \Phi {x, \mathbf y, \mathbf y'; \epsilon}
}}
{{eqn | r = \map \Phi {x, \mathbf y, \mathbf y'; 0} + \valueat {\epsilon \frac {\partial \map \Phi {x, \mathbf y, \mathbf y'; \epsilon} } {\partial \epsilon} } {\epsilon \mathop = 0} + \map \OO \epsilon
}}
{{eqn | r = x + \epsilon \phi + \map \OO \epsilon
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \mathbf Y
      | r = \map {\mathbf \Psi} {x, \mathbf y, \mathbf y'; \epsilon}
}}
{{eqn | r = \map {\mathbf \Psi} {x, \mathbf y, \mathbf y'; 0} + \valueat {\epsilon \frac {\partial \map {\mathbf \Psi} {x, \mathbf y, \mathbf y'; \epsilon} } {\partial \epsilon} } {\epsilon \mathop = 0} + \map \OO \epsilon
}}
{{eqn | r = y + \epsilon \boldsymbol \psi + \map \OO \epsilon
}}
{{end-eqn}}
Use the general variation formula, and suppose that the curve $\mathbf y = \map {\mathbf y} x$ is an extremal of $J \sqbrk {\mathbf y}$:
:$\delta x = \epsilon \phi$
:$\delta y = \epsilon \psi$
:$\delta J = \epsilon \bigintlimits {\nabla_{\mathbf y'} F \cdot \boldsymbol \psi + \paren {F - \mathbf y' \cdot \nabla_{\mathbf y'} F }\phi} {x \mathop = x_0} {x \mathop = x_1}$
Since $J \sqbrk {\mathbf y}$ is invariant under the transformation, the variation vanishes.
Then for any $\epsilon \ne 0$ we have:
:$\bigvalueat {\nabla_{\mathbf y'} F \cdot \boldsymbol \psi + \paren {F - \mathbf y'\cdot \nabla_{\mathbf y'} F} \phi} {x \mathop = x_0} = \bigvalueat {\nabla_{\mathbf y'} F \cdot \boldsymbol \psi + \paren {F - \mathbf y' \cdot \nabla_{\mathbf y'} F}\phi} {x \mathop = x_1}$
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.
{{qed}}
{{Namedfor|Emmy Noether|cat = Noether E}}