Variational symmetries and Lagrangian multiforms

Abstract

By considering the closure property of a Lagrangian multiform as a conservation law, we use Noether’s theorem to show that every variational symmetry of a Lagrangian leads to a Lagrangian multiform. In doing so, we provide a systematic method for constructing Lagrangian multiforms for which the closure property and the multiform Euler–Lagrange (EL) both hold. We present three examples, including the first known example of a continuous Lagrangian 3-form: a multiform for the Kadomtsev–Petviashvili equation. We also present a new proof of the multiform EL equations for a Lagrangian k-form for arbitrary k.

Lagrangian k-form EL equations

The multiform EL equations for a Lagrangian k-form were first published in [8]. Here we present a new proof of those equations. We let

$$\begin{aligned} \textsf {L}=\sum _{1\le l_1<\ldots <l_k\le N}\mathscr {L}_{(l_1\ldots l_k)}\ \textsf {d}x_{l_1}\wedge \ldots \wedge \textsf {d}x_{l_k}. \end{aligned}$$

(A.1)

be a k-form on a manifold of N independent coordinates $x_1,\ldots ,x_N$ and dependent variable u. Therefore,

$$\begin{aligned} \textsf {dL}=\sum _{1\le i_1<\ldots <i_{k+1}\le N}A^{i_1\ldots i_{k+1}}\textsf {d}x_{i_1}\wedge \ldots \wedge \textsf {d}x_{i_{k+1}} \end{aligned}$$

(A.2)

where the $A^{i_1\ldots i_{k+1}}$ depend on the $\mathscr {L}_{(l_1\ldots l_k)}$ in the usual way, i.e.,

$$\begin{aligned} A^{i_1\ldots i_{k+1}}=\sum _{\alpha =1}^{k+1}(-1)^{k(\alpha +1)}\mathrm{D}\,_{x_{i_\alpha }}\mathscr {L}_{(i_{\alpha +1}\ldots i_{k+1}i_1\ldots i_{\alpha -1})}. \end{aligned}$$

(A.3)

For a fixed $i_1,\ldots ,i_{k+1}$, we shall write $\mathscr {L}_{(\bar{\alpha })}$ to denote $\mathscr {L}_{(i_{\alpha +1}\ldots i_{k+1}i_1\ldots i_{\alpha -1})}$. We define the variational derivative with respect to $u_I$acting on$\mathscr {L}_{(\bar{\alpha })}$

$$\begin{aligned} \frac{\delta \mathscr {L}_{(\bar{\alpha })}}{\delta u_{I}}=\sum _{\begin{array}{c} J\\ j_{i_\alpha }=0 \end{array}}(-\mathrm{D}\,)_J\frac{\partial \mathscr {L}_{(\bar{\alpha })}}{\partial u_{IJ}}, \end{aligned}$$

(A.4)

where I is the usual N component multi-index representing derivatives with respect to $x_1,\ldots ,x_N$, and the multi-indices J are such that components $j_i=0$ whenever $i\ne i_1,\ldots , i_{k+1}$, i.e., J represents derivatives with respect to $x_{i_1}, \ldots ,x_{i_{k+1}}$. We define that $\dfrac{\delta \mathscr {L}_{(\bar{i})}}{\delta u_{I}}=0$ in the case where any component of the multi-index I is negative. Note that by this definition, the variational derivative of $\mathscr {L}_{(i_{\alpha +1}\ldots i_{k+1}i_1\ldots i_{\alpha -1})}$ with respect to $u_I$ only sees derivatives of $u_I$ with respect to the variables $x_{i_{\alpha +1}},\ldots x_{i_{k+1}},x_{i_1}\ldots ,x_{ i_{\alpha -1}}$, even though derivatives with respect to other variables may appear in $\mathscr {L}_{(i_{\alpha +1}\ldots i_{k+1}i_1\ldots i_{\alpha -1})}$.

Theorem A.1

The dependent variable u is a critical point of the k-form $\textsf {L}$ as defined in (A.1) if and only if for all $i_1,\ldots i_{k+1}$ such that $1\le i_1<\ldots <i_{k+1}\le N$, and for all I,

$$\begin{aligned} \sum _{\alpha =1}^{k+1}(-1)^{\alpha k}\frac{\delta \mathscr {L}_{(\bar{\alpha })}}{\delta u_{I\backslash i_\alpha }}=0 \end{aligned}$$

(A.5)

In order to prove that these are the multiform EL equations, we will require the following lemma:

Lemma A.2

Let $1\le i_1<\ldots <i_{k+1}\le N$ be fixed. For all multi-indices I,

$$\begin{aligned} \frac{\partial \mathscr {L}_{(\bar{\alpha })}}{\partial u_I}=\sum _{\begin{array}{c} J\\ j_i\le 1\\ j_{i_\alpha }=0 \end{array}}\mathrm{D}\,_J\frac{\delta \mathscr {L}_{(\bar{\alpha })}}{\delta u_{IJ}} \end{aligned}$$

(A.6)

where the summation is over all multi-indices J as defined for (A.4), such that the $i_\alpha$th component of J is zero and the nonzero $j_i$ are equal to 1.

Proof

We first notice that the partial derivative on the left-hand side of (A.6) appears only once in the sum on the right-hand side. We now need to show that all other terms that appear on the right-hand side of (A.6), which are all of the form $\mathrm{D}\,_A\dfrac{\partial \mathscr {L}_{(\bar{\alpha })}}{\partial u_{IA}}$ for some multi-index A, sum to zero. To show this, we consider the term $\mathrm{D}\,_A\dfrac{\partial \mathscr {L}_{(\bar{\alpha })}}{\partial u_{IA}}$, and let r be the number of nonzero entries in A. We notice that this term appears exactly once when $|J|=0$ with a factor of $(-1)^{|A|}$, exactly ${r\atopwithdelims ()1}$ times with a factor of $(-1)^{|A|+1}$ when $|J|=1$, exactly ${r\atopwithdelims ()2}$ times with a factor of $(-1)^{|A|+2}$ when $|J|=2$, etc. In total, this term appears with a factor of ${\pm }\sum _{i=0}^{r}(-1)^i{r\atopwithdelims ()i}$. It can easily be seen that this sum is zero by considering the binomial expansion of $(1-1)^r$. $\square$

Proof of Theorem A.1

For the first part of this proof, we will show that $\delta \textsf {dL}=0$ by following the argument given in [16]. We assume that L contains terms up to nth-order derivatives of u (i.e., L depends on $u_I$ with $|I|\le n$). Let B be an arbitrary $k+1$-dimensional ball with surface $\partial B$. We consider the action functional S over the closed surface $\partial B$ such that

$$\begin{aligned} S[u]=\oint _{\partial B}\mathsf {L} \end{aligned}$$

(A.7)

We then apply Stokes’ theorem to write S in terms of an integral over B:

$$\begin{aligned} S[u]=\int _ B\mathsf {dL} \end{aligned}$$

(A.8)

and we look for solutions of

$$\begin{aligned} \delta S=\int _{B}\delta \textsf {d}\textsf {L}=0 \end{aligned}$$

(A.9)

Since this must hold for arbitrary variations (i.e., with no boundary constraints) for every ball B, it follows that u is a critical point of L if and only if the integrand $\delta \mathsf {dL}=0$, where

$$\begin{aligned} \delta \textsf {dL}=\sum _{1\le i_1<\ldots <i_{k+1}\le N}\sum _I\frac{\partial A^{i_1\ldots i_{k+1}}}{\partial u_{I}} \delta u_I\wedge \textsf {d}x_{i_1}\wedge \ldots \wedge \textsf {d}x_{i_{k+1}}. \end{aligned}$$

(A.10)

This is equivalent to the statement that for all $1\le i_1<\ldots <i_{k+1}\le N$, for all I,

$$\begin{aligned} \frac{\partial A^{i_1\ldots i_{k+1}}}{\partial u_{I}}=0 \end{aligned}$$

(A.11)

We could stop here and use (A.11) as our multiform EL equations. Indeed, there are occasions where this is the most convenient formulation to use. However, it is more illuminating to express this in terms of variational derivatives; by doing so, we see more clearly the interplay between the constituent $\mathscr {L}_{(l_1\ldots l_k)}$ and see that a consequence of $\delta \mathsf {dL}=0$ is that $\mathrm{E}\,(\mathscr {L}_{(l_1\ldots l_k)})=0$ for each $\mathscr {L}_{(l_1\ldots l_k)}$.

For the second part of this proof, we show that, for any choice of $1\le i_1<\ldots <i_{k+1}\le N$, (A.11) holds if and only if $\forall I$,

$$\begin{aligned} \sum _{\alpha =1}^{k+1}(-1)^{\alpha k}\frac{\delta \mathscr {L}_{(\bar{\alpha })}}{\delta u_{I\backslash i_\alpha }}=0. \end{aligned}$$

(A.12)

To do this, we first show that (A.12) holds for $|I|> n$. We then use an inductive argument to show that if (A.12) holds for $|I|> m$, then it also holds for $|I|=m$. The converse (that (A.12) $\implies$ (A.11)) is then easily seen from the intermediary steps of the proof.

We begin by (arbitrarily) fixing $1\le i_1<\ldots <i_{k+1}\le N$ and noticing that for $|I|\ge n+2$, (A.12) holds. In fact, all terms are zero since, by definition, there are no $n+1$th-order derivatives in our multiform. We now consider the relation $\dfrac{\partial A^{i_1\ldots i_{k+1}}}{\partial u_{I}}=0$ in the case where $|I|=n+1$. In this case, we find that

$$\begin{aligned} \frac{\partial A^{i_1\ldots i_{k+1}}}{\partial u_{I}}=\sum _{\alpha =1}^{k+1}(-1)^{\alpha k+1}\frac{\partial \mathscr {L}_{(\bar{\alpha })}}{\partial u_{I\backslash i_\alpha }} \end{aligned}$$

(A.13)

since there are no $n+1$th-order derivatives in the $\mathscr {L}_{(\bar{\alpha })}$. By setting this equal to zero, we see that (A.12) holds in the case where $|I|=n+1$.

Our inductive hypothesis is that (A.12) holds for $|I| > m$. We now consider the relation $\dfrac{\partial A^{i_1\ldots i_{k+1}}}{\partial u_{I}}=0$ in the case where $|I| =m$.

We now notice that

$$\begin{aligned} \frac{\partial A^{i_1\ldots i_{k+1}}}{\partial u_{I}}&=\sum _{\alpha =1}^{k+1}(-1)^{\alpha k+1}\frac{\partial }{\partial u_{I}}\mathrm{D}\,_{x_{i_\alpha }}\mathscr {L}_{(\bar{\alpha })}\nonumber \\&=\sum _{\alpha =1}^{k+1}(-1)^{\alpha k+1}\bigg \{\frac{\partial \mathscr {L}_{(\bar{\alpha })}}{\partial u_{I\backslash i_\alpha }}+\mathrm{D}\,_{x_{i_\alpha }}\frac{\partial \mathscr {L}_{(\bar{\alpha })}}{\partial u_I}\bigg \}\nonumber \\&=\sum _{\alpha =1}^{k+1}(-1)^{\alpha k+1}\bigg \{\frac{\partial \mathscr {L}_{(\bar{\alpha })}}{\partial u_{I\backslash i_\alpha }}+\sum _{\begin{array}{c} J\\ j_i\le 1\\ j_{i_\alpha }=0 \end{array}}\mathrm{D}\,_{Ji_\alpha }\frac{\delta \mathscr {L}_{(\bar{\alpha })}}{\delta u_{IJ}}\bigg \}\nonumber \\&=\sum _{\alpha =1}^{k+1}(-1)^{\alpha k+1}\bigg \{\frac{\partial \mathscr {L}_{(\bar{\alpha })}}{\partial u_{I\backslash i_\alpha }}+\sum _{\begin{array}{c} J\\ j_i\le 1\nonumber \\ j_{i_\alpha }=1 \end{array}}\mathrm{D}\,_{J}\frac{\delta \mathscr {L}_{(\bar{\alpha })}}{\delta u_{IJ\backslash i_\alpha }}\bigg \}\nonumber \\&=\sum _{\alpha =1}^{k+1}(-1)^{\alpha k+1}\bigg \{\frac{\partial \mathscr {L}_{(\bar{\alpha })}}{\partial u_{I\backslash i_\alpha }}\bigg \}+\sum _{ \begin{array}{c} J\\ j_i\le 1\\ |J|>0 \end{array}}~\sum _{\begin{array}{c} \alpha \\ j_{i_\alpha }>0 \end{array}}(-1)^{\alpha k+1}D_{J}\frac{\delta \mathscr {L}_{(\bar{\alpha })}}{\delta u_{IJ\backslash i_\alpha }}\nonumber \\ \end{aligned}$$

(A.14)

where we have made use of (A.6) in the third line, re-labeled J in the fourth line and changed the order of the summation in the last. We now apply the inductive hypothesis to get

$$\begin{aligned} \frac{\partial A^{i_1\ldots i_{k+1}}}{\partial u_{I}}= & {} \sum _{\alpha =1}^{k+1}(-1)^{\alpha k+1}\bigg \{\frac{\partial \mathscr {L}_{(\bar{\alpha })}}{\partial u_{I\backslash i_\alpha }}\bigg \}+\sum _{\begin{array}{c} J\\ j_i\le 1\\ |J|>0 \end{array}}~\sum _{\begin{array}{c} \alpha \\ j_{i_\alpha }=0 \end{array}}(-1)^{\alpha k}\mathrm{D}\,_{J}\frac{\delta \mathscr {L}_{(\bar{\alpha })}}{\delta u_{IJ\backslash i_\alpha }} \nonumber \\= & {} \sum _{\alpha =1}^{k+1}(-1)^{\alpha k+1}\left\{ \frac{\partial \mathscr {L}_{(\bar{\alpha })}}{\partial u_{I\backslash i_\alpha }}-\sum _{\begin{array}{c} J\\ j_i\le 1\\ j_{i_\alpha }=0\\ |J| >0 \end{array}}\mathrm{D}\,_{J}\frac{\delta \mathscr {L}_{(\bar{\alpha })}}{\delta u_{IJ\backslash i_\alpha }}\right\} =0. \end{aligned}$$

(A.15)

Finally, we use (A.6) to express this as

$$\begin{aligned} \begin{aligned} \frac{\partial A^{i_1\ldots i_{k+1}}}{\partial u_{I}}&=\sum _{\alpha =1}^{k+1}(-1)^{\alpha k+1}\frac{\delta \mathscr {L}_{(\bar{\alpha })}}{\delta u_{I\backslash i_\alpha }}=0 \end{aligned} \end{aligned}$$

(A.16)

and we have shown that (A.12) holds for $|I|=m$. By induction, it follows that (A.12) holds for all I. The converse can easily be seen to hold by following the steps taken in (A.14), (A.15) and (A.16) in reverse order.

We have shown that the multiform EL equations (A.5) for a given $1\le i_1<\ldots <i_{k+1}\le N$ are equivalent to $\delta A^{i_1\ldots i_{k+1}}=0$ for the same $1\le i_1<\ldots <i_{k+1}\le N$. It follows that the multiform EL equations holding for all $1\le i_1<\ldots <i_{k+1}\le N$ is equivalent to $\delta \textsf {dL}=0$. $\square$