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DEtools[symmetric_product] - obtain the homomorphic image of the tensor product
Calling Sequence
symmetric_product(L1, L2, .., Ln, domain)
Parameters
L1, L2, .., Ln
-
differential operators
domain
list containing two names
Description
Let L1, L2, .., Ln be differential operators. The output of this procedure is a linear differential operator of minimal order such that for every solution of L1, of L2, ..., of Ln, the product is a solution of M.
Note that "symmetric product" is not a proper mathematical name for this construction on the solution space; it is a homomorphic image of the tensor product. The reason for choosing the name symmetric_product is the resemblance with the function symmetric_power.
The order of the returned operator is to some extent predictable. Given L1,L2 respectively admitting n1,n2 independent solutions, the returned admits at most solutions, when all products of solutions are different, and at least when some of these products are repeated. So M has differential order such that .
The argument domain describes the differential algebra. If this argument is the list , then the differential operators are notated with the symbols and . They are viewed as elements of the differential algebra where is the field of constants.
If the argument domain is omitted then the differential specified by the environment variable _Envdiffopdomain is 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 symmetric_product(..) only after executing the command with(DEtools). However, it can always be accessed through the long form of the command by using DEtools[symmetric_product](..).
Examples
A solution of is so the solutions of the following operator equal .
and since is of order 1, M has the same order as L. As an example where the order of M is smaller than n1 * n2 (the respective orders of L1 and L2) consider L1 and L2 the following 2nd and 3rd order differential operators:
The symmetric product of L1,L2 is not of order 6. It is of order 4, that is, equal to :
The solution of M is the product of the solutions of L1 and L2; to see that let's compute first the solutions to L1 and L2 - formally - using DESol:
See Also
DEtools/symmetric_power, diffop
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