When tackling a system of equations with RosenfeldGroebner, in each of the cases returned, there is no one equation that is the consequence of the other equations of the case and all the integrability conditions are taken into account. If during this simplification process the system proves to be inconsistent (say arriving at $1=0$) an empty list is returned meaning the system has no solution. To illustrate this consider the following equations

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

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

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

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

Indicate the dependent and independent variables and an ordering (ranking) for them. Because the purpose in this example is to only remove consequences and take into account integrability conditions, use an orderly ranking, that could be $\left[\left[u\,v\right]\right]$ or $\left[\left[v\,u\right]\right]$ (see DifferentialRing)

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

$R≔\mathrm{differential\_ring}$

Consider now the system constructed with $\mathrm{p1}$ and $\mathrm{p2}$ in this way:

$\mathrm{sys}≔\mathrm{expand}\left(\left[\mathrm{p1}=0\,\mathrm{p2}=0\,\mathrm{diff}\left(\mathrm{p1}\mathrm{p2}\,x\right)=0\,\mathrm{diff}\left(\mathrm{p2}\,y\right)=0\right]\right)$

$\mathrm{sys}≔\left[\frac{\partial }{\partial y}v\left(x\,y\right)+u\left(x\,y\right)=0\,\frac{\partial }{\partial x}u\left(x\,y\right)-\frac{\partial }{\partial y}u\left(x\,y\right)=0\,\left(\frac{{\partial }^2}{\partial x\partial y}v\left(x\,y\right)\right)\left(\frac{\partial }{\partial x}u\left(x\,y\right)\right)-\left(\frac{{\partial }^2}{\partial x\partial y}v\left(x\,y\right)\right)\left(\frac{\partial }{\partial y}u\left(x\,y\right)\right)+{\left(\frac{\partial }{\partial x}u\left(x\,y\right)\right)}^2-\left(\frac{\partial }{\partial x}u\left(x\,y\right)\right)\left(\frac{\partial }{\partial y}u\left(x\,y\right)\right)+\left(\frac{\partial }{\partial y}v\left(x\,y\right)\right)\left(\frac{{\partial }^2}{\partial x^2}u\left(x\,y\right)\right)-\left(\frac{\partial }{\partial y}v\left(x\,y\right)\right)\left(\frac{{\partial }^2}{\partial x\partial y}u\left(x\,y\right)\right)+u\left(x\,y\right)\left(\frac{{\partial }^2}{\partial x^2}u\left(x\,y\right)\right)-u\left(x\,y\right)\left(\frac{{\partial }^2}{\partial x\partial y}u\left(x\,y\right)\right)=0\,\frac{{\partial }^2}{\partial x\partial y}u\left(x\,y\right)-\frac{{\partial }^2}{\partial y^2}u\left(x\,y\right)=0\right]$

By construction, we know that the third and fourth equations, diff(p1*p2, x) = 0 and diff(p2, y) = 0, bring no additional information. They are consequences of $\mathrm{p1}=0,\mathrm{p2}=0$ and so they are simplified away in the system returned

$\mathrm{simplified\_sys}≔\mathrm{RosenfeldGroebner}\left(\mathrm{sys}\,R\right)$

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

$\mathrm{Equations}\left(\mathrm{simplified\_sys}\right)$

$\left[\left[\frac{\partial }{\partial x}u\left(x\,y\right)-\frac{\partial }{\partial y}u\left(x\,y\right)\,\frac{\partial }{\partial y}v\left(x\,y\right)+u\left(x\,y\right)\right]\right]$

Detection of inconsistencies in a differential system

Add now a third equation

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

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

What RosenfeldGroebner tells us about the system {p1, p2, p3}

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

This result indicates that the system bears a contradiction; $\mathrm{p1}$, $\mathrm{p2}$, and $\mathrm{p3}$ have no common solution. In this case it is possible to visually detect the contradiction:

$\mathrm{diff}\left(\mathrm{p1}=0\,x\right)-\mathrm{diff}\left(\mathrm{p3}=0\,y\right)-\left(\mathrm{p2}=0\right)$

$1=0$

The three equations cannot be satisfied simultaneously. The system is inconsistent; it has no solution.

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 obtain differential chains containing an 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 $x\left(t\right)$. Such an elimination ranking is represented by $\left[z\,y\,x\right]$

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

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

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

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 $\left[\left[T\,x\,y\right]\right]$

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

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

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

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

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

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

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

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 $T$ with regards to $x,y$, so this ranking is $\left[T\,\left[x\,y\right]\right]$

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

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

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

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

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

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

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

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

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.

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

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

$\mathrm{BurgersSymmetry}≔\mathrm{PDEtools}:-\mathrm{DeterminingPDE}\left(\mathrm{Burgers\_pde}\,u\left(x\,t\right)\,\left[\mathrm{\xi }\,\mathrm{\phi }\,\mathrm{\eta }\right]\left(x\,t\,u\right)\right)$

$\mathrm{BurgersSymmetry}≔\left\{\frac{{\partial }^3}{\partial x^3}\mathrm{\xi }\left(x\,t\,u\right)=0\,\frac{\partial }{\partial t}\mathrm{\eta }\left(x\,t\,u\right)=\frac{\frac{{\partial }^2}{\partial x^2}\mathrm{\xi }\left(x\,t\,u\right)}{2}\,\frac{\partial }{\partial u}\mathrm{\eta }\left(x\,t\,u\right)=-\frac{\frac{\partial }{\partial x}\mathrm{\xi }\left(x\,t\,u\right)}{2}\,\frac{\partial }{\partial x}\mathrm{\eta }\left(x\,t\,u\right)=-\frac{\left(\frac{{\partial }^2}{\partial x^2}\mathrm{\xi }\left(x\,t\,u\right)\right)u}{2}\,\frac{\partial }{\partial t}\mathrm{\phi }\left(x\,t\,u\right)=\frac{\frac{\partial }{\partial x}\mathrm{\xi }\left(x\,t\,u\right)}{2}\,\frac{\partial }{\partial u}\mathrm{\phi }\left(x\,t\,u\right)=0\,\frac{\partial }{\partial x}\mathrm{\phi }\left(x\,t\,u\right)=\frac{\left(\frac{\partial }{\partial x}\mathrm{\xi }\left(x\,t\,u\right)\right)u}{2}+\mathrm{\eta }\left(x\,t\,u\right)\,\frac{\partial }{\partial t}\mathrm{\xi }\left(x\,t\,u\right)=0\,\frac{\partial }{\partial u}\mathrm{\xi }\left(x\,t\,u\right)=0\right\}$

Seek the ordinary differential equations with respect to $t$ satisfied by the solutions of this system enclosing the dependent variables (in this example, all of them) using lex

$R≔\mathrm{DifferentialRing}\left(\mathrm{blocks}=\left[\mathrm{lex}\left[\mathrm{\eta },\mathrm{\phi },\mathrm{\xi }\right]\right]\,\mathrm{derivations}=\left[u\,x\,t\right]\right)$

$R≔\mathrm{differential\_ring}$

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

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

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

$\left[\frac{\partial }{\partial u}\mathrm{\eta }\left(x\,t\,u\right)=-\frac{\partial }{\partial t}\mathrm{\phi }\left(x\,t\,u\right)\,\frac{\partial }{\partial u}\mathrm{\phi }\left(x\,t\,u\right)=0\,\frac{\partial }{\partial u}\mathrm{\xi }\left(x\,t\,u\right)=0\,\frac{\partial }{\partial x}\mathrm{\eta }\left(x\,t\,u\right)=-\left(\frac{\partial }{\partial t}\mathrm{\eta }\left(x\,t\,u\right)\right)u\,\frac{\partial }{\partial x}\mathrm{\phi }\left(x\,t\,u\right)=\left(\frac{\partial }{\partial t}\mathrm{\phi }\left(x\,t\,u\right)\right)u+\mathrm{\eta }\left(x\,t\,u\right)\,\frac{\partial }{\partial x}\mathrm{\xi }\left(x\,t\,u\right)=2\frac{\partial }{\partial t}\mathrm{\phi }\left(x\,t\,u\right)\,\frac{{\partial }^2}{\partial t^2}\mathrm{\eta }\left(x\,t\,u\right)=0\,\frac{{\partial }^2}{\partial t^2}\mathrm{\phi }\left(x\,t\,u\right)=0\,\frac{\partial }{\partial t}\mathrm{\xi }\left(x\,t\,u\right)=0\right]$

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

Consider the following Clairaut partial differential equation.

$u\left(x\,y\right)=x\mathrm{diff}\left(u\left(x\,y\right)\,x\right)+y\mathrm{diff}\left(u\left(x\,y\right)\,y\right)+\mathrm{diff}\left(u\left(x\,y\right)\,x\right)\mathrm{diff}\left(u\left(x\,y\right)\,y\right)$

$u\left(x\,y\right)=x\left(\frac{\partial }{\partial x}u\left(x\,y\right)\right)+y\left(\frac{\partial }{\partial y}u\left(x\,y\right)\right)+\left(\frac{\partial }{\partial x}u\left(x\,y\right)\right)\left(\frac{\partial }{\partial y}u\left(x\,y\right)\right)$

Represent it by the equation $p$ with a more compact syntax, called jet notation.

$p≔-u+xu\left[x\right]+yu\left[y\right]+u\left[x\right]u\left[y\right]$

$p≔x{u}_x+y{u}_y+u_x{u}_y-u$

Before any manipulation of this equation, define a differential polynomial ring it belongs to:

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

$R≔\mathrm{differential\_ring}$

The Clairaut equation under consideration has a singular solution. This is shown in the output of RosenfeldGroebner by the existence of two cases: the general case (always exists unless the system is inconsistent) and an additional - therefore - singular case; each one represented by a regular_differential_chain

$\mathrm{Clairaut}≔\mathrm{RosenfeldGroebner}\left(p\,R\right)$

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

The equations and inequations of each egular_differential_chain can be accessed using the Equations and Inequations commands. The general case is represented by the differential chain that involve some additional inequations while the second regular differential chain represents the singular case

$\mathrm{Equations}\left(\mathrm{Clairaut}\left[1\right]\right),\mathrm{Inequations}\left(\mathrm{Clairaut}\left[1\right]\right)$

$\left[x{u}_x+y{u}_y+u_x{u}_y-u\right],\left[u_y+x\right]$

$\mathrm{Equations}\left(\mathrm{Clairaut}\left[2\right]\right),\mathrm{Inequations}\left(\mathrm{Clairaut}\left[2\right]\right)$

$\left[xy+u\right],\left[\right]$

Analysis of singular solutions

All singular cases are present in the output of RosenfeldGroebner while only the essential singular zeros are present when the option singsol = essential is used. For first order differential equations:

The essential singular solutions are envelopes of the nonsingular (general) solutions; that is: they are not particular cases of the general solution.

The other singular solutions are particular cases of the general solution.

Illustrate this analysis on the equation $\mathrm{Wp}$(cf. WeierstrassP):

$\mathrm{Wp}≔{\mathrm{diff}\left(w\left(z\right)\,z\right)}^2-4{w\left(z\right)}^3+\mathrm{g2}w\left(z\right)+\mathrm{g3}$

$\mathrm{Wp}≔{\left(\frac{\ⅆ}{\ⅆz}w\left(z\right)\right)}^2-4{w\left(z\right)}^3+\mathrm{g2}w\left(z\right)+\mathrm{g3}$

Case ${\mathrm{g2}}^3\ne 27{\mathrm{g3}}^2$ :

$R≔\mathrm{DifferentialRing}\left(\mathrm{blocks}=\left[w\right]\,\mathrm{derivations}=\left[z\right]\,\mathrm{arbitrary}=\left[\mathrm{g2}\,\mathrm{g3}\right]\right)$

$R≔\mathrm{differential\_ring}$

$\mathrm{Equations}\left(\mathrm{RosenfeldGroebner}\left(\left[\mathrm{Wp}\right]\,R\right)\right)$

$\left[\left[{\left(\frac{\ⅆ}{\ⅆz}w\left(z\right)\right)}^2-4{w\left(z\right)}^3+\mathrm{g2}w\left(z\right)+\mathrm{g3}\right]\,\left[4{w\left(z\right)}^3-\mathrm{g2}w\left(z\right)-\mathrm{g3}\right]\right]$

There are three singular solutions: $w\left(z\right)=\mathrm{ri}$, where $\mathrm{ri}$ is one of the roots of $4r^3-\mathrm{g2}r-\mathrm{g3}$. There are all three essential singular solutions as we can see using the option singsol = essential

$\mathrm{Equations}\left(\mathrm{RosenfeldGroebner}\left(\mathrm{Wp}\,R\,\mathrm{singsol}=\mathrm{essential}\right)\right)$

$\left[\left[{\left(\frac{\ⅆ}{\ⅆz}w\left(z\right)\right)}^2-4{w\left(z\right)}^3+\mathrm{g2}w\left(z\right)+\mathrm{g3}\right]\,\left[4{w\left(z\right)}^3-\mathrm{g2}w\left(z\right)-\mathrm{g3}\right]\right]$

As such they are envelopes of the nonsingular (general) solution of $\mathrm{Wp}$. This can also be observed by plotting the real zeros for $\mathrm{g2}=4,\mathrm{g3}=\mathrm{-1}$:

$\mathrm{singular\_solutions}≔\left\{\mathrm{plot}\left(\left\{\mathrm{fsolve}\left(-4r^3+4r-1\,r\right)\right\}\,z=-3..3\,\mathrm{color}=\mathrm{red}\,\mathrm{thickness}=2\right)\right\}\:$

$\mathrm{general\_solution\_sample\_1}≔\left\{\mathrm{seq}\left(\mathrm{plots}\left[\mathrm{odeplot}\right]\left(\mathrm{dsolve}\left(\left\{\mathrm{diff}\left(w\left(z\right)\,z\,z\right)-6{w\left(z\right)}^2+2\,w\left(\mathrm{ic}\right)=-1\,\mathrm{D}\left(w\right)\left(\mathrm{ic}\right)=-1.0\right\}\,w\left(z\right)\,\mathrm{type}=\mathrm{numeric}\right)\,-3..3\,\mathrm{color}=\mathrm{green}\right)\,\mathrm{ic}=0..4\right)\right\}\:$

$\mathrm{general\_solution\_sample\_2}≔\left\{\mathrm{seq}\left(\mathrm{plots}\left[\mathrm{odeplot}\right]\left(\mathrm{dsolve}\left(\left\{\mathrm{diff}\left(w\left(z\right)\,z\,z\right)-6{w\left(z\right)}^2+2\,w\left(\mathrm{ic}\right)=2\,\mathrm{D}\left(w\right)\left(\mathrm{ic}\right)=-5.0\right\}\,w\left(z\right)\,\mathrm{type}=\mathrm{numeric}\right)\,\mathrm{ic}-0.15..\mathrm{ic}+1.7\,\mathrm{color}=\mathrm{blue}\right)\,\mathrm{ic}=-3..1\right)\right\}\:$

$\mathrm{plots}\left[\mathrm{display}\right]\left(\mathrm{singular\_solutions}\mathbf{union}\mathrm{general\_solution\_sample\_1}\mathbf{union}\mathrm{general\_solution\_sample\_2}\,\mathrm{title}=''g2\; =\; 4,\; g3\; =\; -\; 1\backslash n\; red:\; singular\; solutions\backslash n\; green\; and\; blue:\; samples\; of\; general\; solutions''\,\mathrm{view}=\left[-3..3\,-\frac{3}{2}..3\right]\right)$

Case ${\mathrm{g2}}^3=27{\mathrm{g3}}^2$; parametrize the equation with $g$ as in

$\mathrm{Wps}≔\mathrm{subs}\left(\mathrm{g2}=3g^2\,\mathrm{g3}=g^3\,\mathrm{Wp}\right)$

$\mathrm{Wps}≔{\left(\frac{\ⅆ}{\ⅆz}w\left(z\right)\right)}^2-4{w\left(z\right)}^3+3g^{2}w\left(z\right)+g^3$

$\mathrm{Rs}≔\mathrm{DifferentialRing}\left(R\,\mathrm{arbitrary}=g\right)\:$

$\mathrm{Equations}\left(\mathrm{RosenfeldGroebner}\left(\mathrm{Wps}\,\mathrm{Rs}\right)\right)$

$\left[\left[{\left(\frac{\ⅆ}{\ⅆz}w\left(z\right)\right)}^2-4{w\left(z\right)}^3+3g^{2}w\left(z\right)+g^3\right]\,\left[2w\left(z\right)+g\right]\,\left[w\left(z\right)-g\right]\right]$

There are two singular solutions: $w\left(z\right)=g$ and $w\left(z\right)=-\frac{g}{2}$ from which only $w\left(z\right)=g$ is an essential singular solution of $\mathrm{Wps}$ as is shown using the option singsol = essential

$\mathrm{Equations}\left(\mathrm{RosenfeldGroebner}\left(\mathrm{Wps}\,\mathrm{Rs}\,\mathrm{singsol}=\mathrm{essential}\right)\right)$

$\left[\left[{\left(\frac{\ⅆ}{\ⅆz}w\left(z\right)\right)}^2-4{w\left(z\right)}^3+3g^{2}w\left(z\right)+g^3\right]\,\left[w\left(z\right)-g\right]\right]$

Therefore $w\left(z\right)=g$ is an envelope of the nonsingular, general solution of $\mathrm{Wps}$. On the contrary, the singular solution $w\left(z\right)=-\frac{g}{2}$ of $\mathrm{Wps}$ is not essential and therefore it is a limiting case of the general solution of $\mathrm{Wps}$. Observe the situation for $g=\mathrm{-1}$:

$\mathrm{singular\_solutions}≔\mathrm{plot}\left(\left\{-1\,\frac{1}{2}\right\}\,\mathrm{color}=\mathrm{red}\,\mathrm{thickness}=2\right)\:$

$\mathrm{general\_solution\_sample\_1}≔\mathrm{plot}\left(\left\{\mathrm{seq}\left(-1+\frac{3}{2}{\tanh \left(\frac{1}{2}\left(C-z\right)\mathrm{sqrt}\left(6\right)\right)}^2\,C=-1..1\right)\right\}\,z=-3..3\,\mathrm{color}=\mathrm{green}\right)\:$

$\mathrm{general\_solution\_sample\_2}≔\mathrm{plot}\left(\left\{\mathrm{seq}\left(-1+\frac{3}{2}{\coth \left(\frac{1}{2}\left(C-z\right)\mathrm{sqrt}\left(6\right)\right)}^2\,C=-1..1\right)\right\}\,z=-3..3\,\mathrm{color}=\mathrm{blue}\right)\:$

$\mathrm{plots}\left[\mathrm{display}\right]\left(\left\{\mathrm{singular\_solutions}\,\mathrm{general\_solution\_sample\_1}\,\mathrm{general\_solution\_sample\_2}\right\}\,\mathrm{view}=\left[-3..3\,-\frac{3}{2}..3\right]\,\mathrm{title}=''g2\; =\; 3,\; g3\; =\; -\; 1\backslash n\; red:\; singular\; solutions\backslash n\; green\; and\; blue:\; samples\; of\; general\; solutions''\right)$

Case $\mathrm{g2}=0,\mathrm{g3}=0$. The equation is:

$\mathrm{Wp0}≔\mathrm{subs}\left(\mathrm{g2}=0\,\mathrm{g3}=0\,\mathrm{Wp}\right)$

$\mathrm{Wp0}≔{\left(\frac{\ⅆ}{\ⅆz}w\left(z\right)\right)}^2-4{w\left(z\right)}^3$

$\mathrm{R0}≔\mathrm{DifferentialRing}\left(\mathrm{blocks}=\left[w\right]\,\mathrm{derivations}=\left[z\right]\right)\:$

$\mathrm{Equations}\left(\mathrm{RosenfeldGroebner}\left(\mathrm{Wp0}\,\mathrm{R0}\right)\right)$

$\left[\left[{\left(\frac{\ⅆ}{\ⅆz}w\left(z\right)\right)}^2-4{w\left(z\right)}^3\right]\,\left[w\left(z\right)\right]\right]$