IsIdeal - Maple Help

IsIdeal

check if the solutions of a LAVF are an ideal in the Lie algebra of another LAVF

 Calling Sequence IsIdeal(L1, L2)

Parameters

 L1, L2 - LAVF objects that are Lie algebras i.e. IsLieAlgebra(obj) returns true, see IsLieAlgebra.

Description

 • Let L1, L2 be LAVF objects that are Lie algebras. Then IsIdeal(L1,L2) checks if solutions of L1 are an ideal in the Lie algebra of solutions of L2.
 • Internally the method returns true if $\left[\mathrm{L1},\mathrm{L2}\right]\phantom{\rule[-0.0ex]{0.5ex}{0.0ex}}\subseteq \mathrm{L1}$ (i.e. IsInvariant(L1,L2) returns true), $\mathrm{L2}$ is Lie algebra (i.e. IsLieAlgebra(L2) returns true), and $\mathrm{L1}\subseteq \mathrm{L2}$ (i.e. IsSubspace(L1,L2) returns true). False otherwise.
 • This method is associated with the LAVF object. For more detail, see Overview of the LAVF object.

Examples

 > $\mathrm{with}\left(\mathrm{LieAlgebrasOfVectorFields}\right):$
 > $\mathrm{Typesetting}:-\mathrm{Settings}\left(\mathrm{userep}=\mathrm{true}\right):$
 > $\mathrm{Typesetting}:-\mathrm{Suppress}\left(\left[\mathrm{\xi }\left(x,y\right),\mathrm{\eta }\left(x,y\right)\right]\right):$
 > $V≔\mathrm{VectorField}\left(\mathrm{\xi }\left(x,y\right)\mathrm{D}\left[x\right]+\mathrm{\eta }\left(x,y\right)\mathrm{D}\left[y\right],\mathrm{space}=\left[x,y\right]\right)$
 ${V}{≔}{\mathrm{\xi }}{}\frac{{\partial }}{{\partial }{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{+}{\mathrm{\eta }}{}\frac{{\partial }}{{\partial }{y}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}$ (1)
 > $S≔\mathrm{LHPDE}\left(\left[\mathrm{diff}\left(\mathrm{\xi }\left(x,y\right),x\right)=0,\mathrm{diff}\left(\mathrm{\xi }\left(x,y\right),y\right)=0,\mathrm{diff}\left(\mathrm{\eta }\left(x,y\right),x\right)=0,\mathrm{diff}\left(\mathrm{\eta }\left(x,y\right),y\right)=0\right],\mathrm{indep}=\left[x,y\right],\mathrm{dep}=\left[\mathrm{\xi },\mathrm{\eta }\right]\right)$
 ${S}{≔}\left[{{\mathrm{\xi }}}_{{x}}{=}{0}{,}{{\mathrm{\xi }}}_{{y}}{=}{0}{,}{{\mathrm{\eta }}}_{{x}}{=}{0}{,}{{\mathrm{\eta }}}_{{y}}{=}{0}\right]{,}{\mathrm{indep}}{=}\left[{x}{,}{y}\right]{,}{\mathrm{dep}}{=}\left[{\mathrm{\xi }}{,}{\mathrm{\eta }}\right]$ (2)
 > $\mathrm{E2}≔\mathrm{LHPDE}\left(\left[\mathrm{diff}\left(\mathrm{\xi }\left(x,y\right),y,y\right)=0,\mathrm{diff}\left(\mathrm{\eta }\left(x,y\right),x\right)=-\mathrm{diff}\left(\mathrm{\xi }\left(x,y\right),y\right),\mathrm{diff}\left(\mathrm{\eta }\left(x,y\right),y\right)=0,\mathrm{diff}\left(\mathrm{\xi }\left(x,y\right),x\right)=0\right],\mathrm{indep}=\left[x,y\right],\mathrm{dep}=\left[\mathrm{\xi },\mathrm{\eta }\right]\right)$
 ${\mathrm{E2}}{≔}\left[{{\mathrm{\xi }}}_{{y}{,}{y}}{=}{0}{,}{{\mathrm{\eta }}}_{{x}}{=}{-}{{\mathrm{\xi }}}_{{y}}{,}{{\mathrm{\eta }}}_{{y}}{=}{0}{,}{{\mathrm{\xi }}}_{{x}}{=}{0}\right]{,}{\mathrm{indep}}{=}\left[{x}{,}{y}\right]{,}{\mathrm{dep}}{=}\left[{\mathrm{\xi }}{,}{\mathrm{\eta }}\right]$ (3)

We first construct these two LAVFs,

 > $L≔\mathrm{LAVF}\left(V,S\right)$
 ${L}{≔}\left[{\mathrm{\xi }}{}\frac{{\partial }}{{\partial }{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{+}{\mathrm{\eta }}{}\frac{{\partial }}{{\partial }{y}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\right]\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{&where}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\left\{\left[{{\mathrm{\xi }}}_{{x}}{=}{0}{,}{{\mathrm{\eta }}}_{{x}}{=}{0}{,}{{\mathrm{\xi }}}_{{y}}{=}{0}{,}{{\mathrm{\eta }}}_{{y}}{=}{0}\right]\right\}$ (4)
 > $\mathrm{LE2}≔\mathrm{LAVF}\left(V,\mathrm{E2}\right)$
 ${\mathrm{LE2}}{≔}\left[{\mathrm{\xi }}{}\frac{{\partial }}{{\partial }{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{+}{\mathrm{\eta }}{}\frac{{\partial }}{{\partial }{y}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\right]\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{&where}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\left\{\left[{{\mathrm{\xi }}}_{{y}{,}{y}}{=}{0}{,}{{\mathrm{\xi }}}_{{x}}{=}{0}{,}{{\mathrm{\eta }}}_{{x}}{=}{-}{{\mathrm{\xi }}}_{{y}}{,}{{\mathrm{\eta }}}_{{y}}{=}{0}\right]\right\}$ (5)

For the solutions of L be an ideal in the Lie algebra of LE2, the following conditions must be true..

 > $\mathrm{IsInvariant}\left(L,\mathrm{LE2}\right)$
 ${\mathrm{true}}$ (6)
 > $\mathrm{IsLieAlgebra}\left(\mathrm{LE2}\right)$
 ${\mathrm{true}}$ (7)
 > $\mathrm{IsSubspace}\left(L,\mathrm{LE2}\right)$
 ${\mathrm{true}}$ (8)

or by using a more direct method.

 > $\mathrm{IsIdeal}\left(L,\mathrm{LE2}\right)$
 ${\mathrm{true}}$ (9)

conversely will be false since LE2 is not subalgebra of L.

 > $\mathrm{IsIdeal}\left(\mathrm{LE2},L\right)$
 ${\mathrm{false}}$ (10)

Compatibility

 • The IsIdeal command was introduced in Maple 2020.