obtain the homomorphic image of the tensor product
symmetric_product(L1, L2, .., Ln, domain)
L1, L2, .., Ln
list containing two names
Let L1, L2, .., Ln be differential operators. The output of this procedure is a linear differential operator M of minimal order such that for every solution y1 of L1, y2 of L2, ..., yn of Ln, the product y1⁢y2⁢…⁢yn 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 M is to some extent predictable. Given L1,L2 respectively admitting n1,n2 independent solutions, the returned M admits at most n1⁢n2 solutions, when all products of solutions are different, and at least by⁢n1+n2−1 when some of these products are repeated. So M has differential order d such that n1+n2−1≤d≤n1⁢n2.
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 C⁡t Dt where C 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](..).
A solution of Dx−c⁡x is ⅇ∫c⁡xⅆx so the solutions of the following operator M equal ⅇ∫c⁡xⅆx⁢(the solutions of L).
and since Dx−c⁡x 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 n2+n1−1:
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:
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