The second theorem stated and proved

Plain-language counterpart: The second theorem, and why it is not the first.

Identities among the field equations from a local symmetry

The statement

Theorem II. If the integral $I$ is invariant under a ${𝔊}_{\mathrm{\infty }\varrho }$, in which the arbitrary functions occur up to the $\sigma$th derivative, then there exist $\varrho$ identical relations between the Lagrange expressions and their derivatives up to the $\sigma$th order; and here too the converse holds.

Three words in that sentence do the work. Identical — the relations hold in all arguments, not on solutions. Between the Lagrange expressions — not between conserved quantities. Up to the $\sigma$th order — the depth of the output is the depth of the group.

The trunk shared with the first theorem

Everything up to and including (N12) is common to both proofs and is not repeated here. It is on the first theorem's page and runs: the invariance condition (N1), the local form (N11) obtained through the barred variation (N9), and then the identity (N3) applied to $\overline{\delta }$, giving

$\sum {\psi }_i\overline{\delta }{u}_i=\mathrm{D}\mathrm{i}\mathrm{v}B,(B=A-f\cdot \Delta x)\mathrm{(}\mathrm{N}\mathrm{12}\mathrm{)}$

The two proofs diverge at exactly one point, and it is here. For a ${𝔊}_{\varrho }$, $\overline{\delta }u$ is linear in constant parameters, so (N12) splits into $\varrho$ separate relations by equating coefficients. For a ${𝔊}_{\mathrm{\infty }\varrho }$, $\overline{\delta }u$ is linear in arbitrary functions and their derivatives, which cannot be equated coefficient by coefficient because the derivatives of a function are not independent of it. Removing them is the whole work of this proof.

The proof

Step 1. Write out the linear dependence on the arbitrary functions. With $𝔊$ an infinite continuous group, $\overline{\delta }u$ and its derivatives, and hence $B$, are linear in the $p(x)$ and their derivatives. Substituting the values of $\overline{\delta }u$, independently of (N12),

$\sum {\psi }_i\overline{\delta }{u}_i=\sum _{\lambda ,i}{\psi }_i\{a_i^{(\lambda )}(x,u,\dots )p^{(\lambda )}(x)+b_i^{(\lambda )}(x,u,\dots )\frac{\partial p^{(\lambda )}}{\partial x}+\dots +c_i^{(\lambda )}(x,u,\dots )\frac{{\partial }^{\sigma }{p}^{(\lambda )}}{\partial x^{\sigma }}\}\mathrm{(}\mathrm{1}\mathrm{)}$

Step 2. Trade derivatives of $p$ for derivatives of the coefficients. Analogously to the formula for partial integration, the identity

$\phi (x,u,\dots )\frac{{\partial }^{\tau }p(x)}{\partial x^{\tau }}=(-1{)}^{\tau }\cdot \frac{{\partial }^{\tau }\phi }{\partial x^{\tau }}\cdot p(x)\mathrm{(}\mathrm{2}\mathrm{)}$

holds modulo divergences. Applying it term by term replaces the derivatives of $p$ by $p$ itself, at the cost of divergences that are themselves linear in $p$ and its derivatives. Collecting those into a single $\mathrm{D}\mathrm{i}\mathrm{v}\Gamma$,

$\sum {\psi }_i\overline{\delta }{u}_i=\sum _{\lambda }\{\left(a_i^{(\lambda )}{\psi }_i\right)-\frac{\partial }{\partial x}\left(b_i^{(\lambda )}{\psi }_i\right)+\dots +(-1{)}^{\sigma }\frac{{\partial }^{\sigma }}{\partial x^{\sigma }}\left(c_i^{(\lambda )}{\psi }_i\right)\}{p}^{(\lambda )}+\mathrm{D}\mathrm{i}\mathrm{v}\Gamma \mathrm{(}\mathrm{N}\mathrm{14}\mathrm{)}$

Step 3. Combine with the trunk. Setting (N14) against (N12),

$\sum \{\left(a_i^{(\lambda )}{\psi }_i\right)-\frac{\partial }{\partial x}\left(b_i^{(\lambda )}{\psi }_i\right)+\dots +(-1{)}^{\sigma }\frac{{\partial }^{\sigma }}{\partial x^{\sigma }}\left(c_i^{(\lambda )}{\psi }_i\right)\}{p}^{(\lambda )}=\mathrm{D}\mathrm{i}\mathrm{v}(B-\Gamma )\mathrm{(}\mathrm{N}\mathrm{15}\mathrm{)}$

Step 4. Kill the right-hand side by choosing the arbitrary functions. Form the $n$-fold integral of (N15) over any region, and choose the $p(x)$ so that they vanish on the boundary together with all derivatives occurring in $B-\Gamma$. The integral of a divergence reduces to a boundary integral, so the right-hand side integrates to zero. Therefore the integral of the left-hand side vanishes, for arbitrary $p(x)$ subject only to that boundary condition. By the standard argument, the integrand vanishes for every $p(x)$, and the $\varrho$ relations follow:

$\sum \{\left(a_i^{(\lambda )}{\psi }_i\right)-\frac{\partial }{\partial x}\left(b_i^{(\lambda )}{\psi }_i\right)+\dots +(-1{)}^{\sigma }\frac{{\partial }^{\sigma }}{\partial x^{\sigma }}\left(c_i^{(\lambda )}{\psi }_i\right)\}=0(\lambda =1,2\dots \varrho )\mathrm{(}\mathrm{N}\mathrm{16}\mathrm{)}$

These are the sought dependencies between the Lagrange expressions and their derivatives. Their linear independence follows as in the finite case, since the converse leads back to (N12).

Step 5. A corollary that comes free. From (N15) and (N16) together,

$\mathrm{D}\mathrm{i}\mathrm{v}(B-\Gamma )=0\mathrm{(}\mathrm{3}\mathrm{)}$

so a function system that does not vanish identically has an identically vanishing divergence. This is the non-uniqueness noted on the current page, arriving here as an output rather than a caveat.

What Step 4 required, and what it forbids

Step 4 is where the arbitrary functions earn their name. The choice of $p(x)$ vanishing on the boundary but arbitrary inside is available only because they are functions. It is unavailable for a parameter, and that is why no amount of increasing the parameter count of a ${𝔊}_{\varrho }$ produces this theorem.

Nothing in this proof used the field equations. (N16) is an identity, and it stays one. Setting ${\psi }_i=0$ does not convert it into a conservation law; it converts it into a statement about the field equations themselves, which is a different sentence with a different meaning.

Audit. The proof used: (N12) from the first theorem's page; (N3) from the Lagrange-expressions page, through (N12); the definition of ${𝔊}_{\mathrm{\infty }\varrho }$ and the linearity in $p$ from the previous page; and nothing else.

Provenance: the theorem statement is a paraphrase of Noether's Satz II. (N14), (N15) and (N16) are her equations (14), (15) and (16), and displays (1), (2), (3) are hers, set as running text in the original. Reproduced in her notation from the public-domain German original. Two deviations from the electronic text. Its (14) carries the unbarred variation where the surrounding argument and its own derivation require the barred one, and its sign factor in (14) is punctuated with the exponent inside the parenthesis, where the same factor in (16) is punctuated outside. Both are written here as the argument requires; whether the slips are the 1918 printing's or the transcription's needs the page images. See Sources.