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 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 $M$ 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](..).

Examples

 > $\mathrm{with}\left(\mathrm{DEtools}\right):$
 > $\mathrm{_Envdiffopdomain}≔\left[\mathrm{Dx},x\right]:$
 > $L≔{\mathrm{Dx}}^{2}+a\left(x\right)\mathrm{Dx}+b\left(x\right)$
 ${L}{≔}{{\mathrm{Dx}}}^{{2}}{+}{a}{}\left({x}\right){}{\mathrm{Dx}}{+}{b}{}\left({x}\right)$ (1)

A solution of $\mathrm{Dx}-c\left(x\right)$ is ${ⅇ}^{\int c\left(x\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}ⅆx}$ so the solutions of the following operator $M$ equal ${ⅇ}^{\int c\left(x\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}ⅆx}\mathrm{\left(the solutions of L\right)}$.

 > $M≔\mathrm{symmetric_product}\left(L,\mathrm{Dx}-c\left(x\right)\right)$
 ${M}{≔}{{\mathrm{Dx}}}^{{2}}{+}\left({a}{}\left({x}\right){-}{2}{}{c}{}\left({x}\right)\right){}{\mathrm{Dx}}{-}{a}{}\left({x}\right){}{c}{}\left({x}\right){+}{{c}{}\left({x}\right)}^{{2}}{-}\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{c}{}\left({x}\right){+}{b}{}\left({x}\right)$ (2)

and since $\mathrm{Dx}-c\left(x\right)$ 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:

 > $\mathrm{L1}≔{\mathrm{Dx}}^{2}+a\left(x\right)$
 ${\mathrm{L1}}{≔}{{\mathrm{Dx}}}^{{2}}{+}{a}{}\left({x}\right)$ (3)
 > $\mathrm{L2}≔{\mathrm{Dx}}^{3}+4a\left(x\right)\mathrm{Dx}+2\mathrm{diff}\left(a\left(x\right),x\right)$
 ${\mathrm{L2}}{≔}{{\mathrm{Dx}}}^{{3}}{+}{4}{}{a}{}\left({x}\right){}{\mathrm{Dx}}{+}{2}{}\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{a}{}\left({x}\right)$ (4)

The symmetric product of L1,L2 is not of order 6. It is of order 4, that is, equal to $\mathrm{n2}+\mathrm{n1}-1$:

 > $M≔\mathrm{symmetric_product}\left(\mathrm{L1},\mathrm{L2}\right)$
 ${M}{≔}{{\mathrm{Dx}}}^{{4}}{+}{10}{}{a}{}\left({x}\right){}{{\mathrm{Dx}}}^{{2}}{+}{10}{}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{a}{}\left({x}\right)\right){}{\mathrm{Dx}}{+}{9}{}{{a}{}\left({x}\right)}^{{2}}{+}{3}{}\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{a}{}\left({x}\right)$ (5)

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:

 > $\mathrm{diffop2de}\left(\mathrm{L1},y\left(x\right)\right)$
 ${a}{}\left({x}\right){}{y}{}\left({x}\right){+}\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)$ (6)
 > $\mathrm{sol_L1}≔\mathrm{dsolve}\left(\right)$
 ${\mathrm{sol_L1}}{≔}{y}{}\left({x}\right){=}{\mathrm{DESol}}{}\left(\left\{{a}{}\left({x}\right){}{\mathrm{_Y}}{}\left({x}\right){+}\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{\mathrm{_Y}}{}\left({x}\right)\right\}{,}\left\{{\mathrm{_Y}}{}\left({x}\right)\right\}\right)$ (7)
 > $\mathrm{sol_L2}≔\mathrm{dsolve}\left(\mathrm{diffop2de}\left(\mathrm{L2},y\left(x\right)\right),y\left(x\right)\right)$
 ${\mathrm{sol_L2}}{≔}{y}{}\left({x}\right){=}{{\mathrm{DESol}}{}\left(\left\{{a}{}\left({x}\right){}{\mathrm{_Y}}{}\left({x}\right){+}\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{\mathrm{_Y}}{}\left({x}\right)\right\}{,}\left\{{\mathrm{_Y}}{}\left({x}\right)\right\}\right)}^{{2}}$ (8)
 > $\mathrm{dsolve}\left(\mathrm{diffop2de}\left(M,y\left(x\right)\right),y\left(x\right)\right)$
 ${y}{}\left({x}\right){=}{{\mathrm{DESol}}{}\left(\left\{{a}{}\left({x}\right){}{\mathrm{_Y}}{}\left({x}\right){+}\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{\mathrm{_Y}}{}\left({x}\right)\right\}{,}\left\{{\mathrm{_Y}}{}\left({x}\right)\right\}\right)}^{{3}}$ (9)