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Homomorphisms

  

compute the homomorphisms between two solution spaces

 

Calling Sequence

Parameters

Description

Examples

References

Calling Sequence

Homomorphisms(L1, L2, domain)

Parameters

L1

-

differential operator

L2

-

differential operator

domain

-

list containing two names

Description

• 

The input L1, L2 are differential operators. Denote V(L1) and V(L2) as the solution spaces of these two operators. If a map from V(L1) to V(L2) can be given in the form of a differential operator, then we call that map a homomorphism from V(L1) to V(L2). The command Homomorphisms computes a basis (as vector space) of all such homomorphisms, that is, it computes a basis of all operators r for which r(V(L1)) is a subset of V(L2).

• 

In the special case L1=L2 the same can also be calculated with the command DEtools[eigenring].

• 

Invertible homomorphisms (isomorphisms) from V(L1) to V(L2) are also called gauge transformations. If L1 and L2 belong to the same family of differential operators and differ only in their parameter values, then the name ladder operators is used as well.

• 

The argument domain describes the differential algebra. If this argument is the list Dt,t then the differential operators are notated with the symbols Dt and t. They are viewed as elements of the differential algebra Ct Dt where C is the field of constants, and Dt refers to the differentiation operator ddt.

• 

If the argument domain is omitted then the differential specified by the environment variable _Envdiffopdomain will be used. If this environment variable is not set, then the argument domain may not be omitted.

• 

This function is part of the DEtools package, and so it can be used in the form Homomorphisms(..) only after executing the command with(DEtools).  However, it can always be accessed through the long form of the command by using DEtools[Homomorphisms](..).

Examples

withDEtools:

Take the differential ring Cx Dx:

_EnvdiffopdomainDx,x

_EnvdiffopdomainDx,x

(1)

L1x2xDx2+ax+bx+xcDx+ab

L1x2xDx2+ax+bxc+xDx+ab

(2)

L2subsc=c+1,L1

L2x2xDx2+ax+bxc+x1Dx+ab

(3)

Compute a basis for the homomorphismsr:VL1VL2.

vHomomorphismsL1,L2

vx1Dxc+a+b

(4)

Since this basis has precisely one element, there is, up to multiplication by constants, precisely one map VL1VL2 that can be presented by an operator rCx Dx.

In the following example, every linear map VL1VL2 can be presented by an operator. Thus, the dimension of all such maps will be orderL1orderL2=32=6. Since the output is a basis of these maps, it must have 6 elements.

L1Dx3;L2Dx2

L1Dx3

L2Dx2

(5)

HomomorphismsL1,L2

Dx,xDx2,Dx2,xDx2Dx,x2Dx22xDx+2,x3Dx22x2Dx+2x

(6)

References

  

van der Put, M., and Singer, M. F. Galois Theory of Linear Differential Equations, Vol. 328. Springer: 2003. An electronic version of this book is available at http://www4.ncsu.edu/~singer/ms_papers.html.

  

van Hoeij, M. "Rational Solutions of the Mixed Differential Equation and its Application to Factorization of Differential Operators." ISSAC '96 Proceedings. (1996): 219-225.

See Also

DEtools[eigenring]

diffop