IsIdeal
check if the solutions of a LAVF are an ideal in the Lie algebra of another LAVF
Calling Sequence
Parameters
Description
Examples
Compatibility
IsIdeal(L1, L2)
L1, L2
-
LAVF objects that are Lie algebras i.e. IsLieAlgebra(obj) returns true, see IsLieAlgebra.
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 L1,L2⊆L1 (i.e. IsInvariant(L1,L2) returns true), L2 is Lie algebra (i.e. IsLieAlgebra(L2) returns true), and L1⊆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.
with⁡LieAlgebrasOfVectorFields:
Typesetting:-Settings⁡userep=true:
Typesetting:-Suppress⁡ξ⁡x,y,η⁡x,y:
V≔VectorField⁡ξ⁡x,y⁢Dx+η⁡x,y⁢Dy,space=x,y
V≔ξ⁢ⅆⅆx+η⁢ⅆⅆy
S≔LHPDE⁡diff⁡ξ⁡x,y,x=0,diff⁡ξ⁡x,y,y=0,diff⁡η⁡x,y,x=0,diff⁡η⁡x,y,y=0,indep=x,y,dep=ξ,η
S≔ξx=0,ξy=0,ηx=0,ηy=0,indep=x,y,dep=ξ,η
E2≔LHPDE⁡diff⁡ξ⁡x,y,y,y=0,diff⁡η⁡x,y,x=−diff⁡ξ⁡x,y,y,diff⁡η⁡x,y,y=0,diff⁡ξ⁡x,y,x=0,indep=x,y,dep=ξ,η
E2≔ξy,y=0,ηx=−ξy,ηy=0,ξx=0,indep=x,y,dep=ξ,η
We first construct these two LAVFs,
L≔LAVF⁡V,S
L≔ξ⁢ⅆⅆx+η⁢ⅆⅆy&whereξx=0,ηx=0,ξy=0,ηy=0
LE2≔LAVF⁡V,E2
LE2≔ξ⁢ⅆⅆx+η⁢ⅆⅆy&whereξy,y=0,ξx=0,ηx=−ξy,ηy=0
For the solutions of L be an ideal in the Lie algebra of LE2, the following conditions must be true..
IsInvariant⁡L,LE2
true
IsLieAlgebra⁡LE2
IsSubspace⁡L,LE2
or by using a more direct method.
IsIdeal⁡L,LE2
conversely will be false since LE2 is not subalgebra of L.
IsIdeal⁡LE2,L
false
The IsIdeal command was introduced in Maple 2020.
For more information on Maple 2020 changes, see Updates in Maple 2020.
See Also
LieAlgebrasOfVectorFields (Package overview)
LAVF (Object overview)
LieAlgebrasOfVectorFields[VectorField]
LieAlgebrasOfVectorFields[LHPDE]
LieAlgebrasOfVectorFields[LAVF]
IsLieAlgebra
IsInvariant
IsSubspace
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