Together, the lists blocks and derivations provide the ranking of R, that is, an ordering for the derivatives of the dependent variables with respect to the independent variables. In brief, the elements in blocks (dependent variables or sublists of them) are ranked between each other using an elimination raking and those within a sublist are ranked used an orderly ranking.

An elimination ranking is one where derivatives of leftmost dependent variables, or of all the dependent variables in a leftmost sublist within blocks, rank higher than derivatives of variables to their right. An orderly ranking is one where derivatives of higher differential order rank higher than derivatives of lower differential order, and if the order is the same, then derivatives of leftmost variables rank higher. Hence for example if blocks = [u, [v, w]] and derivations = [x, t],

ring := DifferentialAlgebra:-DifferentialRing(blocks = [u, [v, w]], derivations = [x, t]);

$\mathrm{ring}≔\mathrm{differential\_ring}$

the list of second and first order derivatives

[u[t, t], u[x, x], u[t, x], v[t, t], v[x, x], v[t, x], w[t, t], w[x, x], w[t, x], u[t], u[x], v[t], v[x], w[t], w[x]];

$\left[u_{t,t}\,u_{x,x}\,u_{t,x}\,v_{t,t}\,v_{x,x}\,v_{t,x}\,w_{t,t}\,w_{x,x}\,w_{t,x}\,u_t\,u_x\,v_t\,v_x\,w_t\,w_x\right]$

rank according to (see SortByRank)

DifferentialAlgebra:-Tools:-SortByRank((2), descending, ring);

$\left[u_{x,x}\,u_{x,t}\,u_{t,t}\,u_x\,u_t\,v_{x,x}\,v_{x,t}\,v_{t,t}\,w_{x,x}\,w_{x,t}\,w_{t,t}\,v_x\,v_t\,w_x\,w_t\right]$

When a system of differential polynomials is handled by DifferentialAlgebra commands, rankings are typically used to decide which functions and derivatives are expressed in terms of which other ones. When the system is nonlinear and handled by RosenfeldGroebner, taking into account the ranking will also frequently result in a splitting of the system into cases (differential chains), that then naturally depend on the ordering of variables within the ranking.

The ranking can be adjusted further, in different ways, using the keywords grlexA, grlexB, degrevlexA, and degrevlexB as explained in the Rankings section of the DifferentialAlgebra Glossary.

Consider the 3 following equations.

$\mathrm{p1}≔\mathrm{diff}\left(u\left(x\,y\right)\,x\right)+v\left(x\,y\right)-y$

$\mathrm{p1}≔\frac{\partial }{\partial x}u\left(x\,y\right)+v\left(x\,y\right)-y$

$\mathrm{p2}≔\mathrm{diff}\left(u\left(x\,y\right)\,y\right)+v\left(x\,y\right)$

$\mathrm{p2}≔\frac{\partial }{\partial y}u\left(x\,y\right)+v\left(x\,y\right)$

$\mathrm{p3}≔\mathrm{diff}\left(v\left(x\,y\right)\,x\right)-\mathrm{diff}\left(v\left(x\,y\right)\,y\right)$

$\mathrm{p3}≔\frac{\partial }{\partial x}v\left(x\,y\right)-\frac{\partial }{\partial y}v\left(x\,y\right)$

Before any manipulation of these equations, indicate the dependent and independent variables, and an ordering (ranking) for them variables all inside a list within a list, as in $\left[\left[u\,v\right]\right]$

$R≔\mathrm{DifferentialRing}\left(\mathrm{blocks}=\left[\left[u\,v\right]\right]\,\mathrm{derivations}=\left[x\,y\right]\right)$

$R≔\mathrm{differential\_ring}$

Call now the RosenfeldGroebner to simplify the system taking into account all the integrability conditions implied in it

$\mathrm{RosenfeldGroebner}\left(\left\{\mathrm{p1}\,\mathrm{p2}\,\mathrm{p3}\right\}\,R\right)$

This result indicates that the system bears a contradiction; $\mathrm{p1}$, $\mathrm{p2}$, and $\mathrm{p3}$ have no common solution, the system is inconsistent; see details in the Examples section of RosenfeldGroebner.

To perform a chain resolution of a system of ordinary differential equations, use an elimination ranking. Consider the differential system defined by the following set $S$ of differential polynomials in the unknown functions $x\left(t\right),y\left(t\right),z\left(t\right)$.

$S≔\left[\mathrm{diff}\left(x\left(t\right)\,t\right)-x\left(t\right)\left(x\left(t\right)+y\left(t\right)\right)\,\mathrm{diff}\left(y\left(t\right)\,t\right)+y\left(t\right)\left(x\left(t\right)+y\left(t\right)\right)\,{\mathrm{diff}\left(x\left(t\right)\,t\right)}^2+{\mathrm{diff}\left(y\left(t\right)\,t\right)}^2+{\mathrm{diff}\left(z\left(t\right)\,t\right)}^2-1\right]$

$S≔\left[\frac{\ⅆ}{\ⅆt}x\left(t\right)-x\left(t\right)\left(x\left(t\right)+y\left(t\right)\right)\,\frac{\ⅆ}{\ⅆt}y\left(t\right)+y\left(t\right)\left(x\left(t\right)+y\left(t\right)\right)\,{\left(\frac{\ⅆ}{\ⅆt}x\left(t\right)\right)}^2+{\left(\frac{\ⅆ}{\ⅆt}y\left(t\right)\right)}^2+{\left(\frac{\ⅆ}{\ⅆt}z\left(t\right)\right)}^2-1\right]$

Using an elimination ranking where $z\left(t\right)>y\left(t\right)>x\left(t\right)$, you rewrite the system in such a way that there is one equation in $x\left(t\right)$ alone, an equation determining $y\left(t\right)$ in terms of $x\left(t\right)$, and finally an equation determining $z\left(t\right)$ in terms of $y\left(t\right)$ and $x\left(t\right)$. To achieve this elimination use an elimination ranking, that is one where all the dependent variables are enclosed in a list in equal footing, and first the variables that you want to be expressed in terms of the next ones, so as in $\left[z\,y\,x\right]$.

$R≔\mathrm{DifferentialRing}\left(\mathrm{blocks}=\left[z\,y\,x\right]\,\mathrm{derivations}=\left[t\right]\right)\:$

To see these dependencies more clearly in the rewritten system use Equations command, which is an alternative to the Equations command.

$G≔\mathrm{RosenfeldGroebner}\left(S\,R\right)$

$G≔\left[\mathrm{regular\_differential\_chain}\,\mathrm{regular\_differential\_chain}\right]$

In the result above we see two cases. The equations of the first case satisfying this elimination ranking mentioned are

$\mathrm{Equations}\left(G\left[1\right]\,\mathrm{solved}\right)$

$\left[{\left(\frac{\ⅆ}{\ⅆt}z\left(t\right)\right)}^2=-\frac{{\left(\frac{\ⅆ}{\ⅆt}x\left(t\right)\right)}^4-2{\left(\frac{\ⅆ}{\ⅆt}x\left(t\right)\right)}^3{x\left(t\right)}^2+2{\left(\frac{\ⅆ}{\ⅆt}x\left(t\right)\right)}^2{x\left(t\right)}^4-{x\left(t\right)}^4}{{x\left(t\right)}^4}\,y\left(t\right)=-\frac{-\frac{\ⅆ}{\ⅆt}x\left(t\right)+{x\left(t\right)}^2}{x\left(t\right)}\,\frac{{\ⅆ}^2}{\ⅆt^2}x\left(t\right)=2\left(\frac{\ⅆ}{\ⅆt}x\left(t\right)\right)x\left(t\right)\right]$

The equations of the second case also satisfying the elimination ranking proposed are

$\mathrm{Equations}\left(G\left[2\right]\,\mathrm{solved}\right)$

$\left[{\left(\frac{\ⅆ}{\ⅆt}z\left(t\right)\right)}^2=-{y\left(t\right)}^4+1\,\frac{\ⅆ}{\ⅆt}y\left(t\right)=-{y\left(t\right)}^2\,x\left(t\right)=0\right]$

Consider the set of equations describing the motion of a pendulum, that is, a point mass $m$ suspended from a massless rod of length $l$ under the influence of gravity $g$, in Cartesian coordinates $x,y$. The Lagrangian formulation leads to two second order differential equations with an algebraic constraint. $T$ is the Lagrangian multiplier.

$P≔\left[m\mathrm{diff}\left(y\left(t\right)\,t\,t\right)+T\left(t\right)y\left(t\right)+g\,m\mathrm{diff}\left(x\left(t\right)\,t\,t\right)+T\left(t\right)x\left(t\right)\,{x\left(t\right)}^2+{y\left(t\right)}^2-l^2\right]$

$P≔\left[m\left(\frac{{\ⅆ}^2}{\ⅆt^2}y\left(t\right)\right)+T\left(t\right)y\left(t\right)+g\,m\left(\frac{{\ⅆ}^2}{\ⅆt^2}x\left(t\right)\right)+T\left(t\right)x\left(t\right)\,{x\left(t\right)}^2+{y\left(t\right)}^2-l^2\right]$

Orderly Ranking

To find the lowest order equations vanishing on the zeros of the system modeling the pendulum and in particular all the algebraic constraints (that is, the equations of order $0$), use an orderly ranking, with all the dependent variables in a list within a list, as in $\left[\left[T\,x\,y\right]\right]$.

$R≔\mathrm{DifferentialRing}\left(\mathrm{blocks}=\left[\left[T\,x\,y\right]\right]\,\mathrm{derivations}=\left[t\right]\,\mathrm{arbitrary}=\left[m\,l\,g\right]\right)\:$

$G≔\mathrm{RosenfeldGroebner}\left(P\,R\right)$

$G≔\left[\mathrm{regular\_differential\_chain}\,\mathrm{regular\_differential\_chain}\right]$

$\mathrm{Equations}\left(G\left[1\right]\,\mathrm{solved}\right)$

$\left[\frac{\ⅆ}{\ⅆt}T\left(t\right)=-\frac{3\left(\frac{\ⅆ}{\ⅆt}y\left(t\right)\right)g}{l^2}\,{\left(\frac{\ⅆ}{\ⅆt}y\left(t\right)\right)}^2=-\frac{T\left(t\right){y\left(t\right)}^2{l}^2-T\left(t\right)l^4+{y\left(t\right)}^{3}g-y\left(t\right)l^{2}g}{m{l}^2}\,{x\left(t\right)}^2=-{y\left(t\right)}^2+l^2\right]$

$\mathrm{Equations}\left(G\left[2\right]\,\mathrm{solved}\right)$

$\left[T\left(t\right)=-\frac{y\left(t\right)g}{l^2}\,x\left(t\right)=0\,{y\left(t\right)}^2=l^2\right]$

The second component in the output of RosenfeldGroebner corresponds to the equilibria of the pendulum. From the first component of the output, it can be seen that the actual motion is described by two first order differential equations with a constraint: the solution depends on only two arbitrary constants.

Mixed Ranking

To obtain the differential system satisfied by $x$ and $y$ alone, eliminate $T$, so enter a ranking where $T$ is in equal footing to the list of $\left[x\,y\right]$ and to its left, as in $\left[T\,\left[x\,y\right]\right]$

$\mathrm{RM}≔\mathrm{DifferentialRing}\left(R\,\mathrm{blocks}=\left[T\,\left[x\,y\right]\right]\right)$

$\mathrm{RM}≔\mathrm{differential\_ring}$

$\mathrm{GM}≔\mathrm{RosenfeldGroebner}\left(P\,\mathrm{RM}\right)$

$\mathrm{GM}≔\left[\mathrm{regular\_differential\_chain}\,\mathrm{regular\_differential\_chain}\right]$

$\mathrm{Equations}\left(\mathrm{GM}\left[1\right]\,\mathrm{solved}\right)$

$\left[T\left(t\right)=-\frac{{\left(\frac{\ⅆ}{\ⅆt}y\left(t\right)\right)}^{2}m{l}^2+{y\left(t\right)}^{3}g-y\left(t\right)l^{2}g}{{y\left(t\right)}^2{l}^2-l^4}\,\frac{{\ⅆ}^2}{\ⅆt^2}y\left(t\right)=-\frac{-{\left(\frac{\ⅆ}{\ⅆt}y\left(t\right)\right)}^{2}y\left(t\right)m{l}^2-{y\left(t\right)}^{4}g+2{y\left(t\right)}^2{l}^{2}g-l^{4}g}{{y\left(t\right)}^{2}m{l}^2-m{l}^4}\,{x\left(t\right)}^2=-{y\left(t\right)}^2+l^2\right]$

$\mathrm{Equations}\left(\mathrm{GM}\left[2\right]\,\mathrm{solved}\right)$

$\left[T\left(t\right)=-\frac{y\left(t\right)g}{l^2}\,x\left(t\right)=0\,{y\left(t\right)}^2=l^2\right]$

Note here that the motion is given by a second order differential equation in $y\left(t\right)$. Then the Lagrange multiplier is given explicitly in terms of $y\left(t\right)$ and its first derivative.

To solve overdetermined systems of partial differential equations it is interesting to compute first the ordinary differential equations satisfied by the solutions of the system. For this purpose, a lexicographic ranking must be chosen. Consider the differential system $S$ defining the infinitesimal generators of the symmetry group of the Burgers equations.

$\mathrm{Burgers\_pde}≔\mathrm{diff}\left(u\left(t\,x\right)\,t\right)=\mathrm{diff}\left(u\left(t\,x\right)\,x\,x\right)-u\left(t\,x\right)\mathrm{diff}\left(u\left(t\,x\right)\,x\right)$

$\mathrm{Burgers\_pde}≔\frac{\partial }{\partial t}u\left(t\,x\right)=\frac{{\partial }^2}{\partial x^2}u\left(t\,x\right)-u\left(t\,x\right)\left(\frac{\partial }{\partial x}u\left(t\,x\right)\right)$

$\mathrm{BurgersSymm}≔\mathrm{PDEtools}:-\mathrm{DeterminingPDE}\left(\mathrm{Burgers\_pde}\,u\left(t\,x\right)\,\left[\mathrm{V1}\,\mathrm{V2}\,\mathrm{V3}\right]\left(t\,x\,u\right)\right)$

$\mathrm{BurgersSymm}≔\left\{\frac{\partial }{\partial u}\mathrm{V1}\left(t\,x\,u\right)=0\,\frac{\partial }{\partial x}\mathrm{V1}\left(t\,x\,u\right)=0\,\frac{\partial }{\partial t}\mathrm{V2}\left(t\,x\,u\right)=\frac{u\left(\frac{\partial }{\partial t}\mathrm{V1}\left(t\,x\,u\right)\right)}{2}+\mathrm{V3}\left(t\,x\,u\right)\,\frac{\partial }{\partial u}\mathrm{V2}\left(t\,x\,u\right)=0\,\frac{\partial }{\partial x}\mathrm{V2}\left(t\,x\,u\right)=\frac{\frac{\partial }{\partial t}\mathrm{V1}\left(t\,x\,u\right)}{2}\,\frac{\partial }{\partial t}\mathrm{V3}\left(t\,x\,u\right)=-\frac{\left(\frac{{\partial }^2}{\partial t^2}\mathrm{V1}\left(t\,x\,u\right)\right)u}{2}\,\frac{\partial }{\partial u}\mathrm{V3}\left(t\,x\,u\right)=-\frac{\frac{\partial }{\partial t}\mathrm{V1}\left(t\,x\,u\right)}{2}\,\frac{\partial }{\partial x}\mathrm{V3}\left(t\,x\,u\right)=\frac{\frac{{\partial }^2}{\partial t^2}\mathrm{V1}\left(t\,x\,u\right)}{2}\,\frac{{\partial }^3}{\partial t^3}\mathrm{V1}\left(t\,x\,u\right)=0\right\}$

Seek the ordinary differential equations with respect to $t$ satisfied by the solutions of this system using lex at the time of raking the variables, as in $\left[{\mathrm{lex}}_{\mathrm{V3},\mathrm{V2},\mathrm{V1}}\right]$

$R≔\mathrm{DifferentialRing}\left(\mathrm{blocks}=\left[\mathrm{lex}\left[\mathrm{V3},\mathrm{V2},\mathrm{V1}\right]\right]\,\mathrm{derivations}=\left[u\,x\,t\right]\right)$

$R≔\mathrm{differential\_ring}$

$G≔\mathrm{RosenfeldGroebner}\left(\mathrm{map}\left(\mathrm{lhs}-\mathrm{rhs}\,\mathrm{BurgersSymm}\right)\,R\right)$

$G≔\left[\mathrm{regular\_differential\_chain}\right]$

$\mathrm{Equations}\left(G\left[1\right]\,\mathrm{solved}\right)$

$\left[\frac{\partial }{\partial u}\mathrm{V3}\left(t\,x\,u\right)=-\frac{\frac{\partial }{\partial t}\mathrm{V1}\left(t\,x\,u\right)}{2}\,\frac{\partial }{\partial u}\mathrm{V2}\left(t\,x\,u\right)=0\,\frac{\partial }{\partial u}\mathrm{V1}\left(t\,x\,u\right)=0\,\frac{\partial }{\partial x}\mathrm{V3}\left(t\,x\,u\right)=-\frac{\frac{\partial }{\partial t}\mathrm{V3}\left(t\,x\,u\right)}{u}\,\frac{\partial }{\partial x}\mathrm{V2}\left(t\,x\,u\right)=\frac{\frac{\partial }{\partial t}\mathrm{V1}\left(t\,x\,u\right)}{2}\,\frac{\partial }{\partial x}\mathrm{V1}\left(t\,x\,u\right)=0\,\frac{{\partial }^2}{\partial t^2}\mathrm{V3}\left(t\,x\,u\right)=0\,\frac{{\partial }^2}{\partial t^2}\mathrm{V1}\left(t\,x\,u\right)=-\frac{2\left(\frac{\partial }{\partial t}\mathrm{V3}\left(t\,x\,u\right)\right)}{u}\,\frac{\partial }{\partial t}\mathrm{V2}\left(t\,x\,u\right)=\frac{u\left(\frac{\partial }{\partial t}\mathrm{V1}\left(t\,x\,u\right)\right)}{2}+\mathrm{V3}\left(t\,x\,u\right)\right]$

There are three independent ordinary differential equations with respect to $t$ (the last three of the differential characteristic set). All other ordinary (in $t$) equations vanishing on the solutions of $\mathrm{BurgersSymm}$ can be written as linear combination of these three with their derivatives.

You can change any of the pieces of information constituting a differential ring by passing the ring itself as the first argument, followed by any of the keywords blocks = ..., derivations = ..., parameters = ..., and so on, where the right-hand sides specify the new information, or you can also add to the previous information by passing the keyword additionally.

$\mathrm{Get}\left(\mathrm{data}\,R\right)$

$\mathrm{*\; Partial\; match\; of\; \text{'}data\text{'}\; against\; keyword\; \text{'}differentialringdata\text{'}}$

$\mathrm{derivations}=\left[u\,x\,t\right],\mathrm{blocks}=\left[{\mathrm{lex}}_{\mathrm{V3},\mathrm{V2},\mathrm{V1}}\right],\mathrm{parameters}=\left[\right],\mathrm{arbitrary}=\left[\right]$

$\mathrm{R\_with\_replaced\_blocks}≔\mathrm{DifferentialRing}\left(R\,\mathrm{blocks}=\left[F\,\mathrm{V1}\,H\right]\right)$

$\mathrm{R\_with\_replaced\_blocks}≔\mathrm{differential\_ring}$

$\mathrm{Get}\left(\mathrm{data}\,\mathrm{R\_with\_replaced\_blocks}\right)$

$\mathrm{*\; Partial\; match\; of\; \text{'}data\text{'}\; against\; keyword\; \text{'}differentialringdata\text{'}}$

$\mathrm{derivations}=\left[u\,x\,t\right],\mathrm{blocks}=\left[F\,\mathrm{V1}\,H\right],\mathrm{parameters}=\left[\right],\mathrm{arbitrary}=\left[\right]$

$\mathrm{R\_additionally\_blocks\_and\_arbitrary}≔\mathrm{DifferentialRing}\left(R\,\mathrm{blocks}=\left[F\,H\right]\,\mathrm{arbitrary}=\left[\mathrm{V1}\,\mathrm{V2}\right]\,\mathrm{additionally}\right)$

$\mathrm{R\_additionally\_blocks\_and\_arbitrary}≔\mathrm{differential\_ring}$

$\mathrm{Get}\left(\mathrm{data}\,\mathrm{R\_additionally\_blocks\_and\_arbitrary}\right)$

$\mathrm{*\; Partial\; match\; of\; \text{'}data\text{'}\; against\; keyword\; \text{'}differentialringdata\text{'}}$

$\mathrm{derivations}=\left[u\,x\,t\right],\mathrm{blocks}=\left[{\mathrm{lex}}_{\mathrm{V3}}\,F\,H\,\left[\mathrm{V1}\,\mathrm{V2}\right]\right],\mathrm{parameters}=\left[\right],\mathrm{arbitrary}=\left[\mathrm{V1}\,\mathrm{V2}\right]$

$\mathrm{R\_with\_replaced\_arbitrary}≔\mathrm{DifferentialRing}\left(R\,\mathrm{arbitrary}=\left[F\,H\right]\right)$

$\mathrm{R\_with\_replaced\_arbitrary}≔\mathrm{differential\_ring}$

$\mathrm{Get}\left(\mathrm{data}\,\mathrm{R\_with\_replaced\_arbitrary}\right)$

$\mathrm{*\; Partial\; match\; of\; \text{'}data\text{'}\; against\; keyword\; \text{'}differentialringdata\text{'}}$

$\mathrm{derivations}=\left[u\,x\,t\right],\mathrm{blocks}=\left[{\mathrm{lex}}_{\mathrm{V3},\mathrm{V2},\mathrm{V1}}\,\left[F\,H\right]\right],\mathrm{parameters}=\left[\right],\mathrm{arbitrary}=\left[F\,H\right]$