3.1 Partial Euler operators

The partial Euler operator with respect to $x$ and $u^\alpha _{\textbf{I}}$ is obtained by varying each $u^\alpha _{\textbf{I}}$ independently, treating $x$ as the sole independent variable:

(3.1) \begin{equation}{\textbf{E}}_{u^\alpha _{\textbf{I}}}^{x}=(\!-\!D_x)^j\frac{\partial }{\partial u^\alpha _{\textbf{I},j}}\,. \end{equation}

Consequently, the standard Euler operator with respect to $u^\alpha$ amounts to

(3.2) \begin{equation} {\textbf{E}}_{u^\alpha }=(\!-\!D)_{\textbf{I}}{\textbf{E}}_{u^\alpha _{\textbf{I}}}^{x}\,. \end{equation}

Similarly, the partial Euler operator with respect to $x$ and $u^\alpha _{\textbf{I},k}$ is

(3.3) \begin{equation}{\textbf{E}}_{u^\alpha _{\textbf{I},k}}^{x}=(\!-\!D_x)^j\frac{\partial }{\partial u^\alpha _{\textbf{I},j+k}}\,. \end{equation}

Note that

(3.4) \begin{equation} {\textbf{E}}_{u^\alpha _{\textbf{I},k}}^{x}=\frac{\partial }{\partial u^\alpha _{\textbf{I},k}}\!-\!D_x{\textbf{E}}^x_{u^\alpha _{{\textbf{I}},k+1}}. \end{equation}

The following identities are easily verified; here, $f({\textbf{x}},[{\textbf{u}}])$ is an arbitrary function.

(3.5) \begin{align} {\textbf{E}}_{u^\alpha _{\textbf{I}}}^{x}(D_x f)=0, \end{align}

(3.6) \begin{align} {\textbf{E}}_{u^\alpha _{\textbf{I},k}}^{x}(D_x f)=\frac{\partial f}{\partial u^\alpha _{{\textbf{I}},k-1}}\,,\qquad k\geq 1, \end{align}

(3.7) \begin{align} {\textbf{E}}_{u^\alpha _{\textbf{I},k}}^{x}(D_i f)=D_i\left ({\textbf{E}}_{u^\alpha _{\textbf{I},k}}^{x}(f)\right )+{\textbf{E}}^x_{u^\alpha _{{\textbf{I}}-{\boldsymbol{1}}_i,k}}(f),\qquad x^i\neq x, \end{align}

where the last term in (3.7) is zero if $j^i=0$ .

3.2 Inversion of $\boldsymbol{D}_\boldsymbol{x}$

The identities (3.5), (3.6) and (3.7) are central to the inversion of total divergences, including the following inversion of $P=D_xF$ as an indefinite line integral.

Lemma 3.1. If $P(\boldsymbol{x},[\boldsymbol{u}])=D_xF$ , then, up to an (irrelevant) arbitrary function of all independent variables other than $x$ ,

(3.8) \begin{equation} F(\boldsymbol{x},[\boldsymbol{u}])=\int \left (P-\sum _{k\geq 1}u^\alpha _{\boldsymbol{I},k}\,\boldsymbol{E}_{u^\alpha _{\boldsymbol{I},k}}^{x}(P)\right )\!\,\mathrm{d} x+\sum _{k\geq 0}\boldsymbol{E}^x_{u^\alpha _{\boldsymbol{I},k+1}}(P)\,\mathrm{d}u^\alpha _{\boldsymbol{I},k}\,. \end{equation}

Proof. By the identity (3.6),

\begin{equation*} F({\textbf{x}},[{\textbf{u}}])=\int \frac {\partial F}{\partial x}\,\,\mathrm {d} x+\sum _{k\geq 0}{\textbf{E}}^x_{u^\alpha _{{\textbf{I}},k+1}}(D_xF)\,\mathrm {d}u^\alpha _{\textbf{I},k}\,. \end{equation*}

Moreover,

\begin{equation*} P=\frac {\partial F}{\partial x}+\sum _{k\geq 1}u^\alpha _{\textbf{I},k}\,\frac {\partial F}{\partial u^\alpha _{{\textbf{I}},k-1}}=\frac {\partial F}{\partial x}+\sum _{k\geq 1}u^\alpha _{\textbf{I},k}\,{\textbf{E}}_{u^\alpha _{\textbf{I},k}}^{x}(D_x F)\,. \end{equation*}

Substituting $P$ for $D_xF$ completes the proof.

Example 3.1. Locally, away from its singularities, the function

(3.9) \begin{equation} P=\frac{u_{xx}v_y-u_xv_{xy}}{v_y^2}+\frac{uv_x-u_xv}{v(u+v)}+\frac{1}{x} \end{equation}

belongs to $\mathrm{im}(D_x)$ , but cannot be inverted using the standard homotopy formula. Substituting

\begin{align*} &\boldsymbol{E}^x_{u_{,1}}(P)=\frac{\partial P}{\partial u_x}-D_x\frac{\partial P}{\partial u_{xx}}=-\frac{1}{u+v}\,, &&\boldsymbol{E}^x_{u_{,2}}(P)=\frac{\partial P}{\partial u_{xx}}=\frac{1}{v_y}\,,\\[5pt] &\boldsymbol{E}^x_{v_{,1}}(P)=\frac{\partial P}{\partial v_{x}}=\frac{u}{v(u+v)}\,, &&\boldsymbol{E}^x_{v_{y,1}}(P)=\frac{\partial P}{\partial v_{xy}}=-\frac{u_x}{v_y^2}\,, \end{align*}

into (3.8) yields the inversion:

(3.10) \begin{equation} F=\int \frac{\,\mathrm{d} x}{x} -\frac{\,\mathrm{d} u}{u+v}\,+\frac{\,\mathrm{d} u_x}{v_y} +\frac{u\,\mathrm{d} v}{v(u+v)}-\frac{u_x\,\mathrm{d} v_y}{v_y^2}\, =\, \ln \left |\frac{xv}{u+v}\right |+\frac{u_x}{v_y}\,. \end{equation}

3.3 Integration by parts

From here on, we will restrict attention to total divergences $\mathcal{C}$ whose fully expanded form has no terms that depend on $\textbf{x}$ only. Such terms can be inverted easily by evaluating an indefinite integral, as explained in the Introduction. Henceforth, all indefinite integrals denote antiderivatives with the minimal number of terms in their fully expanded form. Any arbitrary constants and functions that would increase the number of terms are set to zero. This restriction facilitates the search for minimal inversions.

The indefinite line integral formula (3.8) is closely related to integration by parts. To see this, we introduce a positive ranking on the variables $u^\alpha _{{\textbf{J}}}\,$ ; this is a total order $\preceq$ that is subject to two conditions:

\begin{equation*} (i)\,\ u^\alpha \prec u^\alpha _{{\textbf{J}}}\,,\quad {\textbf{J}}\neq \textbf{0},\qquad (ii)\,\ u^\beta _{{\textbf{I}}}\prec u^\alpha _{{\textbf{J}}}\Longrightarrow D_{{\textbf{K}} }u^\beta _{{\textbf{I}}}\prec D_{{\textbf{K}} }u^\alpha _{{\textbf{J}}}\,. \end{equation*}

The leading part of a differential function is the sum of terms in the function that depend on the highest-ranked $u^\alpha _{\textbf{J}}$ , and the rank of the function is the rank of its leading part (see Rust et al. [Reference Rust, Reid and Wittkopf12] for details and references). Let $f({\textbf{x}},[{\textbf{u}}])$ denote the leading part of the fully expanded form of $F$ and let $\mathrm{U}_{,k}$ denote the highest-ranked $u^\alpha _{\textbf{I},k}$ ; then the highest-ranked part of $P=D_xF$ is $\mathrm{U}_{,k+1}\partial f/\partial \mathrm{U}_{,k}$ . Then (3.8) includes the contribution

\begin{equation*} \int {\textbf{E}}^x_{\mathrm {U}_{,k+1}}(P)\,\mathrm {d} \mathrm {U}_{,k}=\int \frac {\partial f}{\partial \mathrm {U}_{,k}}\,\,\mathrm {d} \mathrm {U}_{,k}= f + \text{lower-ranked terms}. \end{equation*}

Integration by parts gives the same result. Subtracting $f$ from $F$ and iterating shows that evaluating the line integral (3.8) is equivalent to integrating by parts from the highest-ranked terms downwards.

Integration by parts is useful for splitting a differential expression $P({\textbf{x}},[{\textbf{u}}])$ , with $P({\textbf{x}},[0])=0$ , into $D_xF$ and a remainder, $R$ , whose $x$ -derivatives are of the lowest-possible order. The splitting is achieved by the following procedure.

1. Step 0. Choose a positive ranking in which $u^\alpha _{{\textbf{I}},0}\prec D_xu^\beta$ for all $\alpha,{\textbf{I}}$ and $\beta$ . (We call such rankings $x$ -dominant.) Initialize by setting $F\;:\!=\;0$ and $R\;:\!=\;0$ .

2. Step 1. Identify the highest-ranked $u^\alpha _{\textbf{I},k}$ in $P$ ; denote this by $\mathrm{U}_{,k}$ . If $k=0$ , add $P$ to $R$ and stop. Otherwise, determine the leading part, $g$ , of $P$ .

3. Step 2. Determine the sum $h\mathrm{U}_{,k}$ of all terms in the fully expanded form of $g$ that are of the form $\gamma \mathrm{U}_{,k}$ , where $\gamma$ is ranked no higher than $\mathrm{U}_{,k-1}$ , and let

   \begin{equation*} H=\int h\,\mathrm {d}\mathrm {U}_{,k-1}\,. \end{equation*}

4. Step 3. Update $F, R$ and $P$ , as follows:

   \begin{equation*} F\;:\!=\;F+H,\qquad R\;:\!=\;R+g-h\mathrm {U}_{,k},\qquad P\;:\!=\;P-g+h\mathrm {U}_{,k}\!-\!D_xH. \end{equation*}
   If $P\neq 0$ , return to Step 1. Otherwise output $F$ and $R$ , then stop.

The reason for choosing an $x$ -dominant ranking is to ensure that the derivative order with respect to $x$ outweighs all other ranking criteria. Consequently, the minimally ranked remainder cannot contain $x$ -derivatives of unnecessarily high order.

Example 3.2. To produce a concise inversion of a conservation law of the Harry Dym equation (see Example 4.1 below), it is necessary to split

\begin{equation*} P=-\frac {8}{3}u^2u_{,4}-\frac {16}{3}uu_{,1}u_{,3}-4uu_{,2}^2+4u_{,1}^2u_{,2}-u^{-1}u_{,1}^4\,. \end{equation*}

Procedure A gives the splitting

\begin{equation*} P=D_x\!\left \{-\frac {8}{3}u^2u_{,3}+\frac {4}{3}u_{,1}^3\right \}-4uu_{,2}^2-u^{-1}u_{,1}^4\,. \end{equation*}

Example 3.3. The ranking criterion in Step 2 of Procedure A ensures that there are no infinite loops. It is not enough that terms are linear in the highest-ranked $x$ -derivative, as shown by the following splitting of

\begin{equation*} P=\frac {v_{,3}}{u_{y}}+\frac {u_{,2}}{v_{y}} \end{equation*}

For the positive $x$ -dominant ranking defined by $v\prec u\prec v_y$ , Procedure A yields

\begin{equation*} P=D_x\!\left \{\frac {v_{,2}}{u_{y}}+\frac {u_{,1}}{v_{y}}\right \}+\frac {v_{,2}u_{y,1}}{u_{y}^2}+\frac {u_{,1}v_{y,1}}{v_{y}^2}\,. \end{equation*}

Both terms in the remainder are linear in their highest-ranked $x$ -derivatives, which are $v_{,2}$ and $v_{y,1}$ , respectively. However, further integration by parts would return $P$ to a form with a higher-ranked remainder.

3.4 Partial scalings

To investigate partial Euler operators further, it is helpful to use a variant of the homotopy approach. The partial scaling (by a positive real parameter, $\lambda$ ) of a function $f({\textbf{x}},[{\textbf{u}}])$ with respect to $x$ and $u^\alpha _{\textbf{I}}$ is the mapping

\begin{equation*} \sigma _{u^\alpha _{\textbf{I}}}^{x}\;:\;(f;\;\lambda )\mapsto f\big |_{\{u^\alpha _{\textbf{I},j}\mapsto \lambda u^\alpha _{\textbf{I},j},\ j\geq 0\}}. \end{equation*}

Again, each $u^\alpha _{\textbf{I}}$ is treated as a distinct dependent variable. Note the identity

(3.11) \begin{equation} \sigma _{u^\alpha _{\textbf{I}}}^{x} D_x= D_x \sigma _{u^\alpha _{\textbf{I}}}^{x}\,. \end{equation}

Definition 3.1. The partial scaling $\sigma _{u^\alpha _\boldsymbol{I}}^{x}$ is a good scaling for a given differential function $f(\boldsymbol{x},[\boldsymbol{u}])$ if

(3.12) \begin{equation} \sigma _{u^\alpha _\boldsymbol{I}}^{x}(f;\;\lambda )=\int \frac{\,\mathrm{d}}{\,\mathrm{d} \lambda }\!\left (\sigma _{u^\alpha _\boldsymbol{I}}^{x}(f;\;\lambda )\right )\,\mathrm{d} \lambda, \end{equation}

for all $\lambda$ in some neighborhood of $1$ .

By definition, the partial scaling $\sigma _{u^\alpha _{\textbf{I}}}^{x}$ fails to be a good scaling for $f$ if and only if there are terms that are independent of $\lambda$ in the fully expanded form of $\sigma _{u^\alpha _{\textbf{I}}}^{x}(f;\;\lambda )$ . The simplest cause of this is that the fully expanded form of $f$ has terms that are independent of $u^\alpha _{\textbf{I}}$ and its $x$ -derivatives. However, this is not the only cause, as the following example illustrates.

Example 3.4. The scalings $\sigma ^y_{u}$ and $\sigma ^y_v$ are not good scalings for

(3.13) \begin{equation} {\mathcal{C}}=u_{x}(2u+v_{y})-v_{x}(u_y+2v_{yy})+\frac{u_{x}}{u^2}+\frac{v_{yy}}{v_{y}}+\frac{2u_{y}}{u}\ln |u|, \end{equation}

because (in fully expanded form),

(3.14) \begin{align} \sigma ^y_{u}({\mathcal{C}};\;\lambda )=\lambda (2uu_{x}-v_{x}u_y)+\frac{u_{x}}{\lambda ^2 u^2}+\frac{2u_{y}}{u}\ln (\lambda )+\left \{u_{x}v_{y}-2v_{x}v_{yy}+\frac{v_{yy}}{v_{y}}+\frac{2u_{y}}{u}\ln |u|\right \}, \end{align}

(3.15) \begin{align} \sigma ^y_{v}({\mathcal{C}};\;\lambda )&=\lambda (u_{x}v_{y}-2v_{x}v_{yy}) +\left \{2uu_{x}-v_{x}u_y+\frac{u_{x}}{u^2}+\frac{v_{yy}}{v_{y}}+\frac{2u_{y}}{u}\ln |u|\right \}. \end{align}

The terms in braces are independent of $\lambda$ ; in (3.14) (resp. (3.15)), some of these depend on $u$ (resp. $v$ ) and/or its $y$ -derivatives. Part of the scaled logarithmic term is independent of $\lambda$ , though part survives differentiation. Note that $\sigma ^y_{u}$ is a good scaling for the term $u_{x}/u^2$ ; the singularity at $u=0$ is not an obstacle.

Lemma 3.2. The partial scaling $\sigma _{u^\alpha _\boldsymbol{I}}^{x}$ is a good scaling for $f(\boldsymbol{x},[\boldsymbol{u}])$ if and only if

(3.16) \begin{equation} f=\lim _{\lambda \rightarrow 1}\int \frac{\,\mathrm{d}}{\,\mathrm{d} \lambda }\!\left (\sigma _{u^\alpha _\boldsymbol{I}}^{x}(f;\;\lambda )\right )\,\mathrm{d} \lambda. \end{equation}

Proof. If $\sigma _{u^\alpha _\boldsymbol{I}}^{x}$ is a good scaling, (3.16) is a consequence of $f=\sigma _{u^\alpha _{\textbf{I}}}^{x}(f;\;1)$ and local smoothness. Conversely, suppose that (3.16) holds and let $\mu$ be a positive real parameter that is independent of $\lambda$ . Then for $\mu$ sufficiently close to $1$ ,

\begin{equation*} \sigma _{u^\alpha _{\textbf{I}}}^{x}(f;\;\mu )=\lim _{\lambda \rightarrow 1}\int \frac {\,\mathrm {d}}{\,\mathrm {d} \lambda }\!\left (\sigma _{u^\alpha _{\textbf{I}}}^{x}(f;\;\lambda \mu )\right )\,\mathrm {d} \lambda =\lim _{\lambda \rightarrow \mu }\int \frac {\,\mathrm {d}}{\,\mathrm {d} \lambda }\!\left (\sigma _{u^\alpha _{\textbf{I}}}^{x}(f;\;\lambda )\right )\,\mathrm {d} \lambda =\int \frac {\,\mathrm {d}}{\,\mathrm {d} \mu }\!\left (\sigma _{u^\alpha _{\textbf{I}}}^{x}(f;\;\mu )\right )\,\mathrm {d} \mu. \end{equation*}

Therefore, $\sigma _{u^\alpha _{\textbf{I}}}^{x}$ is a good scaling.

The use of the limit in (3.16) is needed to deal with any values of $u^\alpha _{\textbf{I},k}$ for which the integral is an indeterminate form. For other values, simple substitution of $\lambda =1$ gives the limit.

Given a partial scaling $\sigma _{u^\alpha _{\textbf{I}}}^{x}$ and a differential function $f({\textbf{x}},[{\textbf{u}}])$ , let

(3.17) \begin{equation} \pi _{u^\alpha _{\textbf{I}}}^{x}(f)=\lim _{\lambda \rightarrow 1}\int \frac{\,\mathrm{d}}{\,\mathrm{d} \lambda }\!\left (\sigma _{u^\alpha _{\textbf{I}}}^{x}(f;\;\lambda )\right )\,\mathrm{d} \lambda. \end{equation}

For $\mu$ sufficiently close to $1$ (using $\widehat{f}$ as shorthand for $\pi _{u^\alpha _{\textbf{I}}}^{x}(f)$ ),

\begin{equation*} \sigma _{u^\alpha _{\textbf{I}}}^{x}(\widehat {f};\;\mu )=\!\int \frac {\,\mathrm {d}}{\,\mathrm {d} \mu }\!\left (\sigma _{u^\alpha _{\textbf{I}}}^{x}(f;\;\mu )\right )\!\,\mathrm {d} \mu =\!\int \frac {\,\mathrm {d}}{\,\mathrm {d} \mu }\!\left \{ \int \frac {\,\mathrm {d}}{\,\mathrm {d} \mu }\!\left (\sigma _{u^\alpha _{\textbf{I}}}^{x}(f;\;\mu )\right )\,\mathrm {d} \mu \right \}\!\,\mathrm {d} \mu =\!\int \frac {\,\mathrm {d}}{\,\mathrm {d} \mu }\!\left (\sigma _{u^\alpha _{\textbf{I}}}^{x}(\widehat {f};\;\mu )\right )\!\,\mathrm {d} \mu ; \end{equation*}

the first equality comes from the proof of Lemma 3.2. Therefore, $\sigma _{u^\alpha _{\textbf{I}}}^{x}$ is a good scaling for $\pi _{u^\alpha _{\textbf{I}}}^{x}(f)$ . Moreover, there are no terms in the fully expanded form of the remainder, $f-\pi _{u^\alpha _{\textbf{I}}}^{x}(f)$ , for which $\sigma _{u^\alpha _{\textbf{I}}}^{x}$ is a good scaling, because

\begin{equation*} \frac {\,\mathrm {d}}{\,\mathrm {d} \mu }\!\left (\sigma _{u^\alpha _{\textbf{I}}}^{x}(f;\;\mu )\right )-\frac {\,\mathrm {d}}{\,\mathrm {d} \mu }\!\left (\sigma _{u^\alpha _{\textbf{I}}}^{x}(\widehat {f};\;\mu )\right )=0. \end{equation*}

So $\pi _{u^\alpha _{\textbf{I}}}^{x}$ is the projection that maps a given function onto the component which has $\sigma _{u^\alpha _{\textbf{I}}}^{x}$ as a good scaling.

For instance, both $\sigma ^y_{u}$ and $\sigma ^y_v$ are poor scalings for (3.13), as explained in Example 3.4. Section 4.3 addresses the problem of inverting divergences such as (3.13) that have poor scalings. First, we develop the inversion process for general divergences. The following results are fundamental.

Proof. All three statements are proved by expanding $\pi _{u^\alpha _{\textbf{I}}}^{x}(f)$ :

(3.20) \begin{align} \pi _{u^\alpha _{\textbf{I}}}^{x}(f) =\lim _{\lambda \rightarrow 1}\int \sum _{j\geq 0}u^\alpha _{\textbf{I},j}\,\sigma _{u^\alpha _{\textbf{I}}}^{x}\!\left (\frac{\partial f}{\partial u^\alpha _{\textbf{I},j}};\;\lambda \right )\,\mathrm{d} \lambda \end{align}

(3.21) \begin{align} =\lim _{\lambda \rightarrow 1}\int u^\alpha _{\textbf{I}}\,\sigma _{u^\alpha _{\textbf{I}}}^{x}\!\left ({\textbf{E}}_{u^\alpha _{\textbf{I}}}^{x}(f);\;\lambda \right )\,\mathrm{d} \lambda +D_xh. \end{align}

Here, $h({\textbf{x}},[{\textbf{u}}])$ is obtained by integrating by parts, using the identity (3.11).

If $f=D_x F$ , the identity (3.5) amounts to $f\in \mathrm{ker}({\textbf{E}}_{u^\alpha _{\textbf{I}}}^{x})$ . Replace $f$ by $F$ in (3.20) and use the identity (3.6) to obtain (3.18). Statements $2$ and $3$ come directly from (3.21).

Note that (3.18) is a homotopy formula for (at least partially) inverting $D_xF$ , giving a third way to do this. The line integral formula (3.8) carries out the full inversion in one step, but may take longer to compute.

4.1 A single iteration

The basic method for inverting a given total divergence one independent variable at a time works similarly to the example above. Suppose that after $n$ iterations the inversion process has yielded components $F^i_n$ and that an expression of the form ${\mathcal{C}}=D_if_n^i$ remains to be inverted. For the next iteration, let ${\textbf{E}}_{u}^{x}$ be the partial Euler operator that is applied to $\mathcal{C}$ . Here, $u$ is one of the variables $u^\alpha _{\textbf{I}}$ , which is chosen to ensure that for each $i$ such that $x^i\neq x$ ,

(4.1) \begin{equation} {\textbf{E}}_{u}^{x} D_if_n^i =D_i{\textbf{E}}_{u}^{x} f_n^i. \end{equation}

This requires care, in view of the identity (3.7). However, it is achievable by using the variables $u^\alpha _{\textbf{I}}$ in the order given by a ranking that is discussed in Section 4.2. This ranking is entirely determined by user-defined rankings of the variables $x^j$ and $u^\alpha$ .

Taking (3.5) into account leads to the identity

(4.2) \begin{equation} {\textbf{E}}_{u}^{x}({\mathcal{C}})=\sum _{x^i\neq x}D_i({\textbf{E}}_{u}^{x}(f_n^i)), \end{equation}

which, with together with Theorem 3.1, is the basis of the inversion method. The method works without modification provided that:

• • there are no poor scalings for any terms in $\mathcal{C}$ ;

• • the fully expanded form of $\mathcal{C}$ has no terms that are linear in $[{\textbf{u}}]$ .

We begin by restricting attention to divergences for which these conditions hold, so that

(4.3) \begin{equation} {\textbf{E}}_{u}^{x}({\mathcal{C}})=\sum _{x^i\neq x}D_i({\textbf{E}}_{u}^{x}(\pi _u^x f_n^i)), \end{equation}

where every term in ${\textbf{E}}_{u}^{x}(\pi _u^x f_n^i)$ depends on $[{\textbf{u}}]$ . The modifications needed if either condition does not hold are given in Sections 4.3 and 4.4.

The iteration of the inversion process runs as follows. Calculate ${\textbf{E}}_{u}^{x}({\mathcal{C}})$ , which is a divergence $D_iP^i$ with no $D_x$ term, by (4.3); it involves at most $p-1$ (but commonly, very few) nonzero functions $P^i$ . Invert this divergence, treating $x$ as a parameter. If it is possible to invert in more than one way, always invert into the $P^i$ for which $x^i$ is ranked as low as possible; the reason for this is given in the next paragraph. If ${\textbf{E}}_{u}^{x}({\mathcal{C}})$ has nonlinear terms that involve derivatives with respect to more than one $D_i$ (excluding $D_x$ ), this is accomplished by iterating the inversion process with as few independent variables as are needed. Otherwise, $P^i$ can be determined more quickly by using integration by parts (Procedure A, with $x$ replaced by the appropriate $x^i$ ), and/or the method for linear terms (see Procedure B in Section 4.4). Note that this shortcut can be used whenever there are only two independent variables.

At this stage, check that the fully expanded form of each $P^i$ has no terms that are ranked lower than $u$ . If any term is ranked lower than $u$ , stop the calculation and try a different ranking of the variables $x^j$ and/or $u^\alpha$ . This is essential because, to satisfy (4.1) and avoid infinite loops, the variables $u^\alpha _{\textbf{I}}$ that are chosen to be $u$ in successive iterations must progress upwards through the ranking. Where there is a choice of inversion, the rank of each term in $P^i$ is maximized by using the $x^i$ of minimum order; this avoids unnecessary re-ranking.

Having found and checked $P^i$ , use (3.21) to obtain

(4.4) \begin{equation} \pi _u^x(f_n^i)=\left \{\lim _{\lambda \rightarrow 1}\int u\,\sigma _u^x\!\left (P^i;\;\lambda \right )\,\mathrm{d} \lambda \right \} +D_xh^i, \end{equation}

for arbitrary functions $h^i({\textbf{x}},[{\textbf{u}}])$ . Apply Procedure A to the function in braces and choose $h^i$ to make the right-hand side of (4.4) equal the remainder from this procedure. This yields the representation of $\pi _u^x(f_n^i)$ that has the lowest-order derivatives (with respect to $x$ ) consistent with the inversion of $D_iP^i$ ; call this representation $f^i$ . Commonly, such a lowest-order representation is needed to obtain a minimal inversion.

By Theorem 3.1, there exists $\phi$ such that

(4.5) \begin{equation} \pi _u^x\left ({\mathcal{C}}-\sum _{x^i\neq x}D_if^i\right )=D_x\phi, \end{equation}

because (by construction) the expression in parentheses belongs to $\mathrm{ker}({\textbf{E}}_{u}^{x})$ . Use the line integral formula (3.8) or Procedure A to obtain $\phi$ , then set $f^i\;:\!=\;\phi$ for $x^i=x$ . Now update: set

\begin{equation*} {\mathcal {C}}\;:\!=\;{\mathcal {C}}\!-\!D_if^i,\qquad F^i_{n+1}\;:\!=\;F^i_n+f^i. \end{equation*}

4.2 Ranking and using the variables

Having described a single iteration, we now turn to the question of how to choose $x$ and $u$ effectively. The starting point is to construct a derivative-dominant ranking of the variables $u^\alpha _{\textbf{J}}$ . This is a positive ranking that is determined by:

• • a ranking of the independent variables, $x^1\prec x^2\prec \cdots \prec x^p$ ;

• • a ranking of the dependent variables, $u^1\prec u^2\prec \cdots \prec u^q$ .

(Later in this section, we give a heuristic for ranking the dependent and independent variables effectively.) The derivative-dominant ranking (denoted ${\textbf{u}}_p$ ) is constructed iteratively, as follows.

\begin{align*}{\textbf{u}}_0&=u^1\prec \cdots \prec u^q,\\[5pt] {\textbf{u}}_1&={\textbf{u}}_0\prec D_1{\textbf{u}}_0 \prec D_1^2{\textbf{u}}_0 \prec \cdots,\\[5pt] {\textbf{u}}_2&={\textbf{u}}_1\prec D_2{\textbf{u}}_1 \prec D_2^2{\textbf{u}}_1 \prec \cdots,\\[5pt] \vdots &\qquad \vdots \\[5pt] {\textbf{u}}_p&={\textbf{u}}_{p-1}\prec D_p{\textbf{u}}_{p-1} \prec D_p^2{\textbf{u}}_{p-1} \prec \cdots. \end{align*}

In practice, very few $u^\alpha _{\textbf{J}}$ are needed to carry out many inversions of interest, but it is essential that these are used in the order given by their ranking, subject to a constraint on $|{\textbf{I}}|$ that is explained below.

Given an independent variable, $x$ , we call $u^\alpha _{\textbf{I}}$ relevant if the updated $\mathcal{C}$ depends on $u^\alpha _{\textbf{I},k}$ for some $k\geq 0$ . The first set of iterations uses $x=x^1$ . For the initial iteration, $u$ is the lowest-ranked relevant $u^\alpha$ . In the following iteration, $u$ is the next-lowest-ranked relevant $u^\alpha$ and so on, up to and including $u^q$ . (From (3.7), the condition (4.1) holds whenever $u=u^\alpha,\ \alpha = 1,\ldots, q$ .) After these iterations, the updated $\mathcal{C}$ is independent of $\textbf{u}$ and its unmixed $x$ -derivatives.

If the updated $\mathcal{C}$ has any remaining $x$ -derivatives, these are mixed. Thus, as $\mathcal{C}$ has no linear terms, a necessary condition for the inversion to be minimal is that every $f_n^i$ is independent of $\textbf{u}$ and its unmixed $x$ -derivatives. Consequently, (4.1) holds for $u=u^\alpha _{\textbf{I}}$ whenever $|{\textbf{I}}|=1$ , because

\begin{equation*} {\textbf{E}}^x_{u^\alpha _{{\textbf{I}}-{\boldsymbol {1}}_i}}(f_n^i)=0,\qquad x^i\neq x. \end{equation*}

Therefore, the process can be continued using each relevant $u=u^\alpha _{\textbf{I}}$ with $|{\textbf{I}}|=1$ in the ranked order. Iterating, the same argument is used with $|{\textbf{I}}|=2,3,\ldots$ , until $\mathcal{C}$ is independent of $x$ -derivatives. Now set $x=x^2$ and iterate, treating $x^1$ as a parameter. In principle, this can be continued up to $x=x^p$ ; in practice, only a very few iterations are usually needed to complete the inversion. The best rankings invert many terms at each iteration. On the basis of some experience with conservation laws, the following heuristic for ranking the variables $x^j$ and $u^\alpha$ is recommended.

Ranking heuristic. Apply criteria for ranking independent variables, using the following order of precedence.

1. 1. Any independent variables that occur in the arguments of arbitrary functions of $\textbf{x}$ should be ranked as high as possible, if they multiply terms that are nonlinear in $[{\textbf{u}}]$ . For instance, if nonlinear terms in a divergence depend on an arbitrary function, $f(t)$ , set $x^p=t$ .

2. 2. If independent variables occur explicitly in non-arbitrary functions, they should be ranked as high as possible (subject to 1 above), with priority going to variables with the most complicated functional dependence. For instance, if $\mathcal{C}$ is linear in $x^i$ and quadratic in $x^j$ , then $x^i\prec x^j$ (so $i\lt j$ in our ordering).

3. 3. If an unmixed derivative of any $u^\alpha$ with respect to $x^i$ is the argument of a function other than a rational polynomial, rank $x^i$ as low as possible.

4. 4. Set $x^i\prec x^j$ if the highest-order unmixed derivative (of any $u^\alpha$ ) with respect to $x^i$ is of higher order than the highest-order unmixed derivative with respect to $x^j$ .

5. 5. Set $x^i\prec x^j$ if there are more occurrences of unmixed $x^i$ -derivatives (in the fully expanded divergence) than there are of unmixed $x^j$ -derivatives.

6. 6. Apply criteria $3$ , $4$ , and $5$ in order of precedence, replacing unmixed by ‘minimally mixed’ derivatives. Minimally mixed means that there are as few derivatives as possible with respect to any other variable(s).

The derivative indices $\textbf{J}$ in a derivative-dominant ranking are ordered according to the ranking of the independent variables. This can be used to rank the dependent variables; if there is more than one dependent variable in $\mathcal{C}$ , use the following criteria in order.

1. 1. Let $u^\alpha \prec u^\beta$ if $\mathcal{C}$ is linear in $[u^\alpha ]$ and nonlinear in $[u^\beta ]$ .

2. 2. Let $u^\alpha \prec u^\beta$ if the lowest-ranked derivative of $u^\alpha$ that occurs in $\mathcal{C}$ is ranked lower than the lowest-ranked derivative of $u^\beta$ in $\mathcal{C}$ . (In conservation laws, the lowest-ranked derivative of $u^\alpha$ is commonly the undifferentiated $u^\alpha$ , which corresponds to ${\textbf{J}}=0$ .)

3. 3. Let $u^\alpha \prec u^\beta$ if the lowest-ranked derivative of $u^\alpha$ in the fully expanded form of $\mathcal{C}$ occurs in more terms than the corresponding derivative of $u^\beta$ does.

These two sets of criteria are not exhaustive (allowing ties, which must be broken), but the aim that underlies them is to carry as few terms as possible into successive iterations of the procedure. Partial Euler operators with respect to unmixed derivatives are used in the earliest iterations; commonly, these are sufficient to complete the inversion.

4.3 How to deal with poor scalings

To remove an earlier restriction on the inversion process, we now address the problem of poor scalings, namely, that $u^\alpha _{\textbf{I}}$ and its $x$ -derivatives (denoted $[u^\alpha _{\textbf{I}}]_x$ ) may occur in terms that belong to ${\mathcal{C}}-\pi _{u^\alpha _{\textbf{I}}}^{x}({\mathcal{C}})$ . Such terms (when fully expanded) are products of homogeneous rational polynomials in $[u^\alpha _{\textbf{I}}]_x$ of degree zero and logarithms of a single element of $[u^\alpha _{\textbf{I}}]_x$ . We refer to these collectively as zero-degree terms.

To overcome this difficulty, we modify the approach used by Anco & Bluman [Reference Anco and Bluman3] to treat singularities. In our context, $[u^\alpha _{\textbf{I}}]_x$ is replaced by $[u^\alpha _{\textbf{I}}+U^\alpha _{\textbf{I}}]_x$ , where $U^\alpha _{\textbf{I}}$ is regarded as a new dependent variable that is ranked higher than $u^\alpha _{\textbf{I}}$ . This approach works equally well for logarithms, ensuring that $\pi _{u^\alpha _{\textbf{I}}}^{x}$ is a good scaling for all terms that depend on $[u^\alpha _{\textbf{I}}]_x$ , so that its kernel consists only of terms that are independent of these variables. At the end of the calculation, all members of $[U^\alpha _{\textbf{I}}]_x$ are set to zero. Note that there is no need to replace $u^\alpha _{\textbf{I}}$ in terms that are not zero-degree in $[u^\alpha _{\textbf{I}}]_x$ , as total differentiation preserves the degree of homogeneity.Footnote ²

Example 4.2. To illustrate the inversion process for divergences that have zero-degree terms, we complete Example 3.4 by inverting

\begin{equation*} {\mathcal {C}}=D_xF+D_yG=u_{x}(2u+v_{y})-v_{x}(u_y+2v_{yy})+\frac {u_{x}}{u^2}+\frac {v_{yy}}{v_{y}}+\frac {2u_{y}}{u}\ln |u|; \end{equation*}

this has no linear terms. The ranking heuristic gives $y\prec x$ and $u\prec v$ . The rest of the calculation goes as follows.

1. 1. ${\mathcal{C}}-\pi _u^y({\mathcal{C}})$ has just one zero-degree term (in $[u]_y$ ), namely $(2u_y/u)\ln |u|$ . Replace this term by $(2(u_y+U_y)/(u+U))\ln |u+U|$ .

2. 2. Calculate $\boldsymbol{E}_u^y({\mathcal{C}})=2u_x+v_{xy}-2u_xu^{-3}=D_x\{2u+v_y+u^{-2}\}=D_x\{\boldsymbol{E}_u^y(u^2+uv_y-u^{-1})\}$ . No term in $2u+v_y+u^{-2}$ is ranked lower than $u$ , as $u$ is the lowest-ranked variable.

3. 3. Then $\pi _u^y({\mathcal{C}}\!-\!D_x\{u^2+uv_y-u^{-1}\})=D_y\{-uv_x+(\ln |u+U|)^2\}$ .

4. 4. Now set $U=0$ to yield the remainder ${\mathcal{C}}_1={\mathcal{C}}\!-\!D_x\{u^2+uv_y-u^{-1}\}\!-\!D_y\{-uv_x+(\ln |u|)^2\}$ at the close of the first iteration. This amounts to ${\mathcal{C}}_1=-2v_xv_{yy}+v_{yy}/v_y$ .

5. 5. The second iteration starts with ${\mathcal{C}}_1-\pi _v^y({\mathcal{C}}_1)=v_{yy}/v_y$ ; as this term is zero-degree in $[v]_y$ , replace it by $(v_{yy}+V_{yy})/(v_y+V_y)$ .

6. 6. Calculate $\boldsymbol{E}_v^y({\mathcal{C}}_1)=-2v_{xyy}=D_x\{-2v_{yy}\}=D_x\{E_v^y(-vv_{yy})\}=D_x\{E_v^y(v_y^2)\}$ . Note that $-2v_{yy}$ is not ranked lower than $v$ .

7. 7. Then $\pi _v^y({\mathcal{C}}_1\!-\!D_x\{v_y^2\})=D_y\{-2v_xv_y+\ln |v_y+V_y|\}$ .

8. 8. Now set $V=0$ to yield the remainder ${\mathcal{C}}_2={\mathcal{C}}_1\!-\!D_x\{v_y^2\}\!-\!D_y\{-2v_xv_y+\ln |v_y|\}=0$ at the close of the second iteration. The inversion process stops, having yielded the output

   \begin{equation*} F=u^2+uv_y-u^{-1}+v_y^2,\qquad G=-uv_x+(\ln |u|)^2-2v_xv_y+\ln |v_y|. \end{equation*}

Note that the homotopy formula (3.18) for inverting $D_xF$ can be adjusted in the same way, whenever $\sigma _{u^\alpha _\boldsymbol{I}}^{x}$ is a poor scaling for $\boldsymbol{E}^x_{u^\alpha _{\boldsymbol{I},j+1}}(D_xF)$ .

4.4 Linear divergences

The inversion process runs into a difficulty when a given divergence has terms that are linear in $[{\textbf{u}}]$ , with mixed derivatives. Then it is possible to invert in more than one way, some of which may not produce a minimal result. To address this, it is helpful to invert using a different process. Suppose that $\mathcal{C}$ is a linear divergence (in fully expanded form). Instead of using the derivative-dominant ranking, integrate by parts, working down the total order $|{\textbf{J}}|$ of the derivatives $u^\alpha _{\textbf{J}}$ . For a given total order, we will invert the mixed derivatives first, though this is not essential.

Starting with the highest-order derivatives, one could integrate each term $f({\textbf{x}})u^\alpha _{\textbf{J}}$ in $\mathcal{C}$ by parts with respect to any $x^i$ such that $j^i\geq 1$ , yielding the remainder $\!-\!D_i(f)u^\alpha _{{\textbf{J}}-{\boldsymbol{1}}_i}$ . If $D_{\textbf{J}}$ is a mixed derivative, we seek to choose $D_i$ in a way that keeps the result concise. Here are some simple criteria that are commonly effective, listed in order of precedence.

1. 1. $f$ is independent of $x^i$ .

2. 2. $\mathcal{C}$ includes the term $D_i(f)u^\alpha _{{\textbf{J}}-{\boldsymbol{1}}_i}$ .

3. 3. $f$ is linear in $x^i$ .

These criteria can be used as an initial pass to invert $\mathcal{C}$ at least partially, leaving a remainder to be inverted that may have far fewer terms than $\mathcal{C}$ does.

Integrating the remainder by parts is straightforward if each $u^\alpha _{\textbf{J}}$ is an unmixed derivative. If $\textbf{J}$ denotes a mixed derivative, integrate with respect to each $x^i$ such that $j^i\geq 1$ in turn, multiplying each result by a parameter (with the parameters summing to $1$ ). Iterate until either the remainder has a factor that is zero for some choice of parameters or there is no remainder. The final stage is to choose the parameters so as to minimize the number of terms in the final expression. Although this produces a minimal inversion, it comes at the cost of extra computational time spent doing all possible inversions followed by the parameter optimization.

Example 4.3. Consider the linear divergence

\begin{equation*} {\mathcal {C}}=\left (\frac {1}{6}f'(t)y^3+f(t)xy\right )\!(u_{xt}-u_{yy}), \end{equation*}

where $f$ is an arbitrary function. The first of the simple criteria above yields

\begin{equation*} {\mathcal {C}}=D_x\!\left \{\frac {1}{6}f'y^3u_t\right \}+fxyu_{xt}-\left (\frac {1}{6}f'y^3+fxy\right )\!u_{yy}; \end{equation*}

The second criterion is not helpful at this stage, but the third criterion gives

\begin{equation*} {\mathcal {C}}=D_x\!\left \{\left (\frac {1}{6}f'y^3+fxy\right )\!u_t\right \}-fyu_{t}-\left (\frac {1}{6}f'y^3+fxy\right )\!u_{yy}. \end{equation*}

The remainder has no mixed derivatives; integrating it by parts produces the minimal inversion

(4.6) \begin{equation} {\mathcal{C}}=D_x\!\left \{\left (\frac{1}{6}f'y^3+fxy\right )\!u_t\right \}+D_y\!\left \{\left (\frac{1}{2}f'y^2+fx\right )\!u-\left (\frac{1}{6}f'y^3+fxy\right )\!u_y\right \}+D_t\!\left \{-fyu\right \}\!. \end{equation}

Example 4.4. To illustrate the parametric approach, consider

\begin{equation*} {\mathcal {C}}=\exp (t-x^2)(tu_{xtt}+2x(t+1)u_t). \end{equation*}

The simple criteria are irrelevant at present, so instead introduce parameters $\lambda _l$ and consider all possible inversions of the mixed derivative terms. Step-by-step, one obtains the following.

\begin{align*}{\mathcal{C}}&=D_x\!\left \{\lambda _1\exp (t-x^2)tu_{tt}\right \}+D_t\!\left \{(1-\lambda _1)\exp (t-x^2)tu_{xt}\right \}\\[5pt] &\quad +\exp (t-x^2)\{2\lambda _1xtu_{tt}-(1-\lambda _1)(t+1)u_{xt}+2x(t+1)u_t\}\\[5pt] &=D_x\!\left \{\exp (t-x^2)\{\lambda _1tu_{tt}-\lambda _2(1-\lambda _1)(t+1)u_{t}\}\right \}\\[5pt] &\quad +D_t\!\left \{\exp (t-x^2)\{(1-\lambda _1)tu_{xt}+2\lambda _1xtu_{t}-(1-\lambda _2)(1-\lambda _1)(t+1)u_{x}\}\right \}\\[5pt] &\quad +(1-\lambda _2)(1-\lambda _1)\exp (t-x^2)\{2x(t+1)u_{t}+(t+2)u_{x}\}. \end{align*}

As the remainder has a factor $(1-\lambda _2)(1-\lambda _1)$ , the inversion is complete if either parameter is set to $1$ . The minimal (two-term) inversion has $\lambda _1=1$ , which gives ${\mathcal{C}}=D_xF+D_tG$ , where

\begin{equation*} F=t\exp (t-x^2)u_{tt}\,,\qquad G=2xt\exp (t-x^2)u_{t}\,. \end{equation*}

For $\lambda _1\neq 1, \lambda _2=1$ , the inversion has three terms if $\lambda _1=0$ , or five terms otherwise.

Note the importance of factorizing the remainder to stop the calculation once a possible parameter choice occurs. If we had continued the calculation without setting either $\lambda _i$ to $1$ , it would have stopped at order zero, not one, giving an eleven-term inversion for general $\lambda _i$ .

To summarize, one can invert a divergence that is linear in $[{\textbf{u}}]$ by applying the following procedure.

1. Step 0. Identify the maximum derivative order, $N=|{\textbf{J}}|$ , of the variables $u^\alpha _{\textbf{J}}$ that occur in $\mathcal{C}$ . Set $F^i\;:\!=\;0,\ i=1\ldots,p$ .

2. Step 1. While $\mathcal{C}$ has at least one term of order $N$ , do the following. Select any such term, $f({\textbf{x}})u^\alpha _{\textbf{J}}$ , and determine a variable $x^i$ over which to integrate. If desired, parametrize for mixed derivatives, as described above. Set $F^i\;:\!=\;F^i+fu^\alpha _{{\textbf{J}}-{\boldsymbol{1}}_i}$ for the chosen $i$ (with the appropriate modification if parameters are used), and update the remainder by setting ${\mathcal{C}}\;:\!=\;{\mathcal{C}}-f({\textbf{x}})u^\alpha _{\textbf{J}}\!-\!D_i(f)u^\alpha _{{\textbf{J}}-{\boldsymbol{1}}_i}$ . Once $\mathcal{C}$ has no terms of order $N$ , continue to Step 2.

3. Step 2. If $\mathcal{C}$ is nonzero and cannot be set to zero by any choice of parameters, set $N\;:\!=\;N-1$ and return to Step 1; otherwise, set $\mathcal{C}$ to zero and carry out parameter optimization (if needed), give the output $F^i,\ i=1,\ldots,p$ , then stop.

4.5 Summary: a procedure for inverting Div

Having addressed potential modifications, we are now in a position to summarize the inversion process for any total divergence $\mathcal{C}$ whose fully expanded form has no terms depending on $\textbf{x}$ only.

1. Step 0. Let ${\mathcal{C}}_\ell$ be the linear part of $\mathcal{C}$ . If ${\mathcal{C}}_\ell =0$ , set ${\mathcal{C}}_0\;:\!=\;{\mathcal{C}}$ and $F_0^i\;:\!=\;0,\ i=1,\ldots,p$ . Otherwise, invert ${\mathcal{C}}_\ell$ using the technique described in Procedure B above, to obtain functions $F_0^i$ that satisfy ${\mathcal{C}}_\ell =D_iF_0^i$ . Set ${\mathcal{C}}_0\;:\!=\;{\mathcal{C}}-{\mathcal{C}}_\ell$ . Choose a derivative-dominant ranking for ${\mathcal{C}}_0$ (either using the ranking heuristic or otherwise). In the notation used earlier, set $x\;:\!=\;x^1$ and $u$ to be the lowest-ranked relevant $u^\alpha$ ; typically, $u\;:\!=\;u^1$ . Set $n\;:\!=\;0$ ; here $n+1$ is the iteration number.

2. Step 1. Calculate ${\mathcal{C}}_n-\pi _u^x({\mathcal{C}}_n)$ ; if this includes terms depending on $[u]_x$ , replace $[u]_x$ in these terms only by $[u+U]_x$ .

3. Step 2. Apply the process detailed in Section 4.1 (with ${\mathcal{C}}_n$ replacing $\mathcal{C}$ ). Provided that the ranking check is passed, this yields components $f^i$ , which may depend on $[U]_x$ . If the ranking check is failed, choose a different ranking of $x^j$ and $u^\alpha$ for the remainder of the inversion process and return to Step 1, starting with the lowest-ranked $x$ and (relevant) $u$ and working upwards at each iteration.

4. Step 3. Replace all elements of $[U]_x$ in $f^i$ by zero.

5. Step 4. Update: set $F^i_{n+1}\;:\!=\;F^i_n+f^i,\ {\mathcal{C}}_{n+1}\;:\!=\;{\mathcal{C}}_n\!-\!D_if^i$ and $n\;:\!=\;n+1$ . If ${\mathcal{C}}_{n+1}= 0$ , output $F^i=F^i_{n+1}$ and stop. Otherwise, update $u$ and $x$ as detailed in Section 4.2 and return to Step 1.

Example 4.5. As an example with linear terms, the Khokhlov–Zabolotskaya equation,

\begin{equation*} u_{xt}-uu_{xx}-u_x^2-u_{yy}=0, \end{equation*}

has a conservation law (see Poole & Hereman [11]) that involves an arbitrary function, $f(t)$ :

\begin{equation*} {\mathcal {C}}=D_xF+D_yG+D_tH=\left (\frac {1}{6}f'y^3+fxy\right )\!\left (u_{xt}-uu_{xx}-u_x^2-u_{yy}\right ). \end{equation*}

The linear part ${\mathcal{C}}_\ell$ of $\mathcal{C}$ is inverted by Procedure B, as shown in Example 4.3. From (4.6), Step 0 in Procedure C gives

\begin{equation*} F_0=\left (\frac {1}{6}f'y^3+fxy\right )\!u_t\,,\qquad G_0=\left (\frac {1}{2}f'y^2+fx\right )\!u-\left (\frac {1}{6}f'y^3+fxy\right )\!u_y\,, \qquad H_0=-fyu. \end{equation*}

The remainder ${\mathcal{C}}-{\mathcal{C}}_\ell$ is

\begin{equation*} {\mathcal {C}}_0=-\left (\frac {1}{6}f'y^3+fxy\right )\!\left (uu_{xx}+u_x^2\right ). \end{equation*}

With the derivative-dominant ranking $x\prec y\prec t$ , only one iteration is needed to complete the inversion, because $\boldsymbol{E}_{u}^{x}({\mathcal{C}}_0)=0$ . Then

\begin{equation*} \pi _u^x({\mathcal {C}}_0)=D_x\!\left \{\frac {1}{2}fyu^2-\left (\frac {1}{6}f'y^3+fxy\right )\!uu_x\right \}, \end{equation*}

and as ${\mathcal{C}}_0=\pi _u^x({\mathcal{C}}_0)$ , the calculation stops after updating, giving the output

\begin{equation*} F=\left (\frac {1}{6}f'y^3+fxy\right )\!\left (u_t-uu_x\right )+\frac {1}{2}fyu^2,\qquad G=\left (\frac {1}{2}f'y^2+fx\right )\!u-\left (\frac {1}{6}f'y^3+fxy\right )\!u_y,\qquad H=-fyu. \end{equation*}

This corresponds to the minimal result in Poole & Hereman [11].

Example 4.6. In every example so far, all iterations have used $u=u^\alpha$ , but not $u^\alpha _\boldsymbol{I}$ for $|I|\neq 0$ . The BBM equation from Example 2.1 illustrates the way that the process continues unhindered once the current $\mathcal{C}$ is independent of all elements of $[u^\alpha ]_x$ . The conservation law (2.6) is

\begin{equation*} {\mathcal {C}}=D_xF+D_tG=(u^2+2u_{xt})(u_{t}-uu_x-u_{xxt}); \end{equation*}

it has no linear part or zero-degree terms. The ranking heuristic gives $x\prec t$ . The first iteration using $\boldsymbol{E}_{u}^{x}$ is similar to what we have seen so far, giving the following updates:

\begin{equation*} {\mathcal {C}}_1=2u_{xt}(u_t-u_{xxt}),\qquad F_1=-\frac {1}{4}u^4-u^2u_{xt},\qquad G_1=\frac {1}{3}u^3. \end{equation*}

At this stage, ${\mathcal{C}}_1$ is independent of $[u]_x$ and equals $\pi ^x_{u_t}({\mathcal{C}}_1)$ . In the second iteration, the ranking requires us to apply $\boldsymbol{E}^x_{u_t}$ . This annihilates ${\mathcal{C}}_1$ ; consequently, ${\mathcal{C}}_1$ is a total derivative with respect to $x$ . Inverting this gives the final (minimal) result,

\begin{equation*} F=-\frac {1}{4}u^4-u^2u_{xt}-u_{xt}^2+u_t^2,\qquad G=\frac {1}{3}u^3, \end{equation*}

which was obtained by inspection in Example 2.1.

The Appendix lists inversions of various other divergences; in each case, the inversion produced by Procedure C and the ranking heuristic is minimal and takes very few iterations to complete.

4.6 Splitting a divergence using discrete symmetries

A given divergence may have discrete symmetries between various terms in its fully expanded form. If the divergence has very many terms that are connected by a particular discrete symmetry group, it can be worth splitting these into disjoint divergences that are mapped to one another by the group elements. Then it is only necessary to invert one of these divergences, using the symmetries to create inversions of the others without the need for much extra computation. However, to use Procedure C, it is necessary to check that all split terms are grouped into divergences; this check is done by using the criterion (2.3).

Polynomial divergences can first be split by degree, yielding divergences that are homogeneous in $[{\textbf{u}}]$ . Such splitting does not add significantly to the computation time, nor does it need to be checked using (2.3), which holds automatically. Splitting by degree can make it easy to identify terms that are linked by discrete symmetries, as illustrated by the following example.

Example 4.7. In Cartesian coordinates $(x,y)$ , the steady non-dimensionalized von Kármán equations for a plate subject to a prescribed axisymmetric load function, $p(x^2+y^2)$ , are ${\mathcal{A}}_\ell =0$ , $\ell =1,2$ , where

\begin{align*}{\mathcal{A}}_1&=u_{xxxx}+2u_{xxyy}+u_{yyyy}-u_{xx}v_{yy}+2u_{xy}v_{xy}-u_{yy}v_{xx}-p,\\[5pt] {\mathcal{A}}_2&=v_{xxxx}+2v_{xxyy}+v_{yyyy}+u_{xx}u_{yy}-u_{xy}^2\,. \end{align*}

Here $u$ is the displacement and $v$ is the Airy stress. This is a system of Euler–Lagrange equations. By Noether’s Theorem, the one-parameter Lie group of rotational symmetries yields the following conservation law (see Djondjorov and Vassilev [Reference Djondjorov, Vassilev, Bertram, Constanda and Struthers6]):

\begin{equation*} {\mathcal {C}}=(yv_x-xv_y){\mathcal {A}}_2-(yu_x-xu_y){\mathcal {A}}_1\,. \end{equation*}

This conservation law has linear, quadratic and cubic terms, so $\mathcal{C}$ can be inverted by summing the inversions of each of the following divergences:

\begin{align*}{\mathcal{C}}_\ell &=(yu_x-xu_y)p,\\[5pt] {\mathcal{C}}_q&=(yv_x-xv_y)( v_{xxxx}+2v_{xxyy}+v_{yyyy})-(yu_x-xu_y)(u_{xxxx}+2u_{xxyy}+u_{yyyy}),\\[5pt] {\mathcal{C}}_c&=(yv_x-xv_y)(u_{xx}u_{yy}-u_{xy}^2)+(yu_x-xu_y)(u_{xx}v_{yy}-2u_{xy}v_{xy}+u_{yy}v_{xx}). \end{align*}

The quadratic terms have an obvious discrete symmetry, $\Gamma _1\;:\;(x,y,u,v)\mapsto (\!-\!x,y,v,u)$ , which gives a splitting into two parts, each of which is a divergence:

\begin{equation*} {\mathcal {C}}_q=\overline {{\mathcal {C}}}_q+\Gamma _1(\overline {{\mathcal {C}}}_q), \end{equation*}

where

\begin{equation*} \overline {{\mathcal {C}}}_q=(yv_x-xv_y)(v_{xxxx}+2v_{xxyy}+v_{yyyy}). \end{equation*}

Consequently, we can invert ${\mathcal{C}}_q$ by inverting $\overline{{\mathcal{C}}}_q$ and applying the symmetry $\Gamma _1$ .

Note that $\overline{{\mathcal{C}}}_q$ has a discrete symmetry, $\Gamma _2\;:\;(x,y,v)\mapsto (y,-x,v)$ , which gives

\begin{equation*} \overline {{\mathcal {C}}}_q=g+\Gamma _2(g),\quad \text {where}\quad g=(yv_x-xv_y)(v_{xxxx}+v_{xxyy}). \end{equation*}

Checking (2.3) shows that this is not a valid splitting into divergences, because

\begin{equation*} \boldsymbol{E}_v(g)=-2v_{xxxy}-2v_{xyyy}\neq 0. \end{equation*}

If necessary, split divergences can be inverted using different rankings. However, in this simple example, a single ranking works for all (nonlinear) parts. One tie-break is needed: let $x\prec y$ . The ranking heuristic gives $v\prec u$ for the cubic terms; the variables $[u]$ are not relevant in the inversion of $\overline{{\mathcal{C}}}_q$ . Procedure C gives the following minimal inversions,

\begin{align*}{\mathcal{C}}_\ell &=D_x(yup)+D_y(\!-\!xup),\\[5pt] \overline{{\mathcal{C}}}_q&=D_x\{F_q(x,y,[v])\}+D_y\{G_q(x,y,[v])\},\\[5pt] {\mathcal{C}}_c&=D_xF_c+D_yG_c\,, \end{align*}

where

\begin{align*} F_q(x,y,[v])=&\ y\left \{v_xv_{xxx}-\frac{1}{2}v_{xx}^2+2v_xv_{xyy}+v_{xy}^2+vv_{yyyy}-\frac{1}{2}v_{yy}^2\right \}\\[5pt] &+x\left \{-v_yv_{xxx}+v_{xx}v_{xy}-2v_yv_{xyy}\right \}+v_yv_{xx}\,,\\[5pt] G_q(x,y,[v])=&\ y\left \{-2v_{xx}v_{xy}-vv_{xyyy}+v_yv_{xyy}\right \}+x\left \{-\frac{1}{2}v_{xx}^2+v_{xy}^2-v_yv_{yyy}+\frac{1}{2}v_{yy}^2\right \}+vv_{xyy}\,,\\[5pt] F_c=&\ y\left (u_xu_{yy}v_x+\frac{1}{2}u_x^2v_{yy}\right )+x\left (u_yu_{xy}v_y+\frac{1}{2}u_y^2v_{xy}\right )-\frac{1}{2}u_y^2v_y\,,\\[5pt] G_c=&\ -y\left (u_xu_{xy}v_x+\frac{1}{2}u_x^2v_{xy}\right )-x\left (u_yu_{xx}v_y+\frac{1}{2}u_y^2v_{xx}\right )+\frac{1}{2}u_x^2v_x\,. \end{align*}

Applying $\Gamma _1$ to $\overline{{\mathcal{C}}}_q$ gives the following minimal inversion for $\mathcal{C}$ :

\begin{equation*} {\mathcal {C}}=D_x\!\left \{yup +F_q(x,y,[v])-F_q(x,y,[u]) +F_c\right \}+D_y\!\left \{-xup +G_q(x,y,[v])-G_q(x,y,[u]) +G_c\right \}. \end{equation*}

Note that the tie-break $x\prec y$ causes the inversion to break the symmetry $\Gamma _2$ that is apparent in $\overline{{\mathcal{C}}}_q$ . As it turns out, this symmetry can be restored without increasing the overall number of terms, by adding components of a trivial conservation law (which are of the form (2.1)). It is an open question whether such preservation of symmetry and minimality is achievable in general.

In Djondjorov & Vassilev [Reference Djondjorov, Vassilev, Bertram, Constanda and Struthers6], the inversion of $\mathcal{C}$ for the special case $p=0$ has 62 terms, which are grouped according to their physical meaning. By contrast, the minimal inversion above has just 46 (resp. 48) terms when $p$ is zero (resp. nonzero), a considerable saving. Moreover, by exploiting the symmetry $\Gamma _1$ , only 28 (resp. 30) of these terms are determined using Procedure C. However, in seeking an efficient inversion, we have ignored the physics. It would be interesting to understand the extent to which the use of a minimal inversion obscures the underlying physics.