Noether's Theorem (Calculus of Variations)

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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 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

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

Proof

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

\(\displaystyle X\) \(=\) \(\displaystyle \map {\Phi} {x,\mathbf y,\mathbf y';\epsilon}\)
\(\displaystyle \) \(=\) \(\displaystyle \map {\Phi} {x,\mathbf y,\mathbf y';0}+\epsilon \frac{\partial \map{\Phi} {x,\mathbf y,\mathbf y';\epsilon} }{\partial\epsilon}\Bigg\rvert_{\epsilon=0}+\map {\mathcal O} {\epsilon}\)
\(\displaystyle \) \(=\) \(\displaystyle x+\epsilon\phi+\map {\mathcal O} {\epsilon}\)
\(\displaystyle \mathbf Y\) \(=\) \(\displaystyle \map {\mathbf\Psi} {x,\mathbf y,\mathbf y';\epsilon}\)
\(\displaystyle \) \(=\) \(\displaystyle \map {\mathbf\Psi} {x,\mathbf y,\mathbf y';0}+\epsilon\frac{\partial \map {\mathbf\Psi} {x,\mathbf y,\mathbf y';\epsilon} }{\partial\epsilon}\Bigg\rvert_{\epsilon=0}+\map {\mathcal O} {\epsilon}\)
\(\displaystyle \) \(=\) \(\displaystyle y+\epsilon\boldsymbol\psi+\map {\mathcal O} {\epsilon}\)

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

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

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.


$\blacksquare$

Source of Name

This entry was named for Emmy Noether.


Sources