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 $\mathrm{y1}$ of L1, $\mathrm{y2}$ of L2, ..., $\mathrm{yn}$ of Ln, the product $\mathrm{y1}\mathrm{y2}\dots \mathrm{yn}$ 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 $\mathrm{n1}\mathrm{n2}$ solutions, when all products of solutions are different, and at least $\mathrm{by}\mathrm{n1}+\mathrm{n2}-1$ when some of these products are repeated. So M has differential order $d$ such that $\mathrm{n1}+\mathrm{n2}-1\le d\le \mathrm{n1}\mathrm{n2}$.

The argument domain describes the differential algebra. If this argument is the list $\left[\mathrm{Dt}\,t\right]$, then the differential operators are notated with the symbols $\mathrm{Dt}$ and $t$. They are viewed as elements of the differential algebra $C\left(t\right)$ $\left[\mathrm{Dt}\right]$ 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](..).

$\mathrm{with}\left(\mathrm{DEtools}\right)\:$

$\mathrm{\_Envdiffopdomain}≔\left[\mathrm{Dx}\,x\right]\:$

$L≔{\mathrm{Dx}}^2+a\left(x\right)\mathrm{Dx}+b\left(x\right)$

$L≔{\mathrm{Dx}}^2+a\left(x\right)\mathrm{Dx}+b\left(x\right)$

$M≔\mathrm{symmetric\_product}\left(L\,\mathrm{Dx}-c\left(x\right)\right)$

$M≔{\mathrm{Dx}}^2+\left(a\left(x\right)-2c\left(x\right)\right)\mathrm{Dx}-a\left(x\right)c\left(x\right)+{c\left(x\right)}^2-\frac{\ⅆ}{\ⅆx}c\left(x\right)+b\left(x\right)$

$\mathrm{L1}≔{\mathrm{Dx}}^2+a\left(x\right)$

$\mathrm{L1}≔{\mathrm{Dx}}^2+a\left(x\right)$

$\mathrm{L2}≔{\mathrm{Dx}}^3+4a\left(x\right)\mathrm{Dx}+2\mathrm{diff}\left(a\left(x\right)\,x\right)$

$\mathrm{L2}≔{\mathrm{Dx}}^3+4a\left(x\right)\mathrm{Dx}+2\frac{\ⅆ}{\ⅆx}a\left(x\right)$

$M≔\mathrm{symmetric\_product}\left(\mathrm{L1}\,\mathrm{L2}\right)$

$M≔{\mathrm{Dx}}^4+10a\left(x\right){\mathrm{Dx}}^2+10\left(\frac{\ⅆ}{\ⅆx}a\left(x\right)\right)\mathrm{Dx}+9{a\left(x\right)}^2+3\frac{{\ⅆ}^2}{\ⅆx^2}a\left(x\right)$

$\mathrm{diffop2de}\left(\mathrm{L1}\,y\left(x\right)\right)$

$a\left(x\right)y\left(x\right)+\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)$

$\mathrm{sol\_L1}≔\mathrm{dsolve}\left(\right)$

$\mathrm{sol\_L1}≔y\left(x\right)=\mathrm{DESol}\left(\left\{a\left(x\right)\mathrm{\_Y}\left(x\right)+\frac{{\ⅆ}^2}{\ⅆx^2}\mathrm{\_Y}\left(x\right)\right\}\,\left\{\mathrm{\_Y}\left(x\right)\right\}\right)$

$\mathrm{sol\_L2}≔\mathrm{dsolve}\left(\mathrm{diffop2de}\left(\mathrm{L2}\,y\left(x\right)\right)\,y\left(x\right)\right)$

$\mathrm{sol\_L2}≔y\left(x\right)={\mathrm{DESol}\left(\left\{a\left(x\right)\mathrm{\_Y}\left(x\right)+\frac{{\ⅆ}^2}{\ⅆx^2}\mathrm{\_Y}\left(x\right)\right\}\,\left\{\mathrm{\_Y}\left(x\right)\right\}\right)}^2$

$\mathrm{dsolve}\left(\mathrm{diffop2de}\left(M\,y\left(x\right)\right)\,y\left(x\right)\right)$

$y\left(x\right)={\mathrm{DESol}\left(\left\{a\left(x\right)\mathrm{\_Y}\left(x\right)+\frac{{\ⅆ}^2}{\ⅆx^2}\mathrm{\_Y}\left(x\right)\right\}\,\left\{\mathrm{\_Y}\left(x\right)\right\}\right)}^3$