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 gen_exp
 generalized exponents of a linear homogeneous ODE

 Calling Sequence gen_exp(L, domain, T, opt) gen_exp(eqn, dvar, T, opt)

Parameters

 L - differential operator domain - list containing two names T - name opt - (optional) sequence of options eqn - homogeneous linear differential equation dvar - dependent variable

Description

 • The input is a differential operator L or a linear ODE (ordinary differential equation) eqn having rational function coefficients.
 • The output is a list of lists. Each of these lists contains one equivalence class of generalized exponents.
 • Let $x$ be the independent variable. If a differential operator is specified, then $x$ is the second element of the list domain. If an ODE is specified, then $x$ is implicitly given in $\mathrm{dvar}$ which is of the form $y\left(x\right)$.
 • An element $e$ in $C$ $\left[{x}^{-\frac{1}{r}}\right]$ is called a generalized exponent of L if there exists a formal solution $y$ of the form $y=s{ⅇ}^{\int \frac{e}{x}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}ⅆx}$ where $s$ is an element of $C$ $\left[\left[{x}^{\frac{1}{r}}\right]\right]$ $\left[\mathrm{log}\left(x\right)\right]$ with valuation 0, which means that the coefficient of $1$ in $s$ is not zero, for more details see the help page of formal_sol. If $r$ is the smallest positive integer for which $e$ is in $C$ $\left[{x}^{-\frac{1}{r}}\right]$ then $r$ is called the ramification index of $e$.
 • The name T, which must be specified in the input, is used to denote ${x}^{-\frac{1}{r}}$ times a constant. This procedure computes the generalized exponents and expresses them in terms of T. The relation between T and $x$ is given in the output as well, in each equivalence class of generalized exponents.
 • If the option restrict_to=S where $S$ is a subset of {minimal, integer, ramification1, rational}, then only a subset of the generalized exponents is given. If the option minimal is in $S$, then only the minimal generalized exponent in each equivalence class will be given. If the option integer or rational is given then only the generalized exponents in Z or Q, respectively, are given. If the option ramification1 is given, then only the generalized exponents with ramification index $1$ (i.e. the generalized exponents in $C$ $\left[\frac{1}{x}\right]$) are given.
 • If a generalized exponent $e$ is a constant (if $e$ is in $C$) then $e$ is an exponent. The exponents are the solutions of the indicial equation. If not all generalized exponents are constants, then the ODE is called irregular singular.
 • If the optional argument $x=p$ where $p$ in ${P}^{1}\left(=C\mathrm{union}\left\{\mathrm{infinity}\right\}\right)$ is given, then this procedure first applies a transformation DEtools[translate] to move the point $p$ to the point $0$, then computes the generalized exponents, and then substitutes $x=x-p$ in the result (or $x=\frac{1}{x}$, if $p=\mathrm{\infty }$). Note that this substitution only affects the part of the output that gives the relation between T and $x$.
 • The generalized exponents $e$ in $C$ $\left[{x}^{-\frac{1}{r}}\right]$ are computed only up to conjugation over the field $k\left(x\right)$, where $k$ is the minimal field of constants over which the input is defined. A larger field $k$ can be specified by the option groundfield = list of RootOfs.
 • The argument domain describes the differential algebra. If this argument is the list $\left[\mathrm{Dx},x\right]$, then the differential operators are notated with the symbols $\mathrm{Dx}$ and $x$. They are viewed as elements of the differential algebra $C\left(x\right)$ $\left[\mathrm{Dx}\right]$ where $C$ is the field of constants, and $\mathrm{Dx}$ denotes the differentiation operator.
 • If the argument domain is omitted then the differential algebra specified by the environment variable _Envdiffopdomain is used. If this environment variable is not set then the argument domain may not be omitted.
 • Instead of a differential operator, the input can also be a linear homogeneous ODE having rational function coefficients. In this case, the second argument dvar must be the dependent variable.
 • This function is part of the DEtools package, and so it can be used in the form gen_exp(..) only after executing the command with(DEtools). However, it can always be accessed through the long form of the command by using DEtools[gen_exp](..).

Examples

 > $\mathrm{with}\left(\mathrm{DEtools}\right):$
 > $L≔{\mathrm{Dx}}^{5}+2{x}^{3}{\mathrm{Dx}}^{3}+6{x}^{2}{\mathrm{Dx}}^{2}+6x\mathrm{Dx}+{x}^{6}\mathrm{Dx}+1$
 ${L}{≔}{{x}}^{{6}}{}{\mathrm{Dx}}{+}{2}{}{{x}}^{{3}}{}{{\mathrm{Dx}}}^{{3}}{+}{{\mathrm{Dx}}}^{{5}}{+}{6}{}{{x}}^{{2}}{}{{\mathrm{Dx}}}^{{2}}{+}{6}{}{x}{}{\mathrm{Dx}}{+}{1}$ (1)
 > $\mathrm{gen_exp}\left(L,\left[\mathrm{Dx},x\right],T,x=\mathrm{\infty }\right)$
 $\left[\left[{0}{,}{T}{=}\frac{{1}}{{x}}\right]{,}\left[\frac{{9}}{{4}}{+}\frac{{1}}{{{T}}^{{5}}}{,}\frac{{11}}{{4}}{+}\frac{{1}}{{{T}}^{{5}}}{,}{-}{{T}}^{{2}}{=}\frac{{1}}{{x}}\right]\right]$ (2)
 > $\mathrm{gen_exp}\left(L,\left[\mathrm{Dx},x\right],T,x=\mathrm{\infty },'\mathrm{restrict_to}'=\left\{'\mathrm{minimal}'\right\}\right)$
 $\left[\left[{0}{,}{T}{=}\frac{{1}}{{x}}\right]{,}\left[\frac{{9}}{{4}}{+}\frac{{1}}{{{T}}^{{5}}}{,}{-}{{T}}^{{2}}{=}\frac{{1}}{{x}}\right]\right]$ (3)
 > $\mathrm{gen_exp}\left(L,\left[\mathrm{Dx},x\right],T,x=\mathrm{\infty },'\mathrm{restrict_to}'=\left\{'\mathrm{ramification1}'\right\}\right)$
 $\left[\left[{0}{,}{T}{=}\frac{{1}}{{x}}\right]\right]$ (4)

Note: The quotes around the names in the options may be omitted unless a value has been assigned to those names.

 > $\mathrm{ode}≔\mathrm{diff}\left(y\left(x\right),x,x\right)+\frac{y\left(x\right)}{{x}^{4}}=0$
 ${\mathrm{ode}}{≔}\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){+}\frac{{y}{}\left({x}\right)}{{{x}}^{{4}}}{=}{0}$ (5)

In this example the field of definition is $Q$, so the generalized exponents at $x=0$ will be given up to conjugation over $Q$:

 > $\mathrm{gen_exp}\left(\mathrm{ode},y\left(x\right),T,x=0\right)$
 $\left[\left[\frac{{\mathrm{RootOf}}{}\left({{\mathrm{_Z}}}^{{2}}{+}{1}\right)}{{T}}{+}{1}{,}{T}{=}{x}\right]\right]$ (6)

Now specify the field $Q\left(i\right)$. Since both generalized exponents are defined over this field, both will appear in the output:

 > $\mathrm{gen_exp}\left(\mathrm{ode},y\left(x\right),T,x=0,'\mathrm{groundfield}'=\left[\mathrm{RootOf}\left({x}^{2}+1\right)\right]\right)$
 $\left[\left[{-}\frac{{\mathrm{RootOf}}{}\left({{\mathrm{_Z}}}^{{2}}{+}{1}\right)}{{T}}{+}{1}{,}{T}{=}{x}\right]{,}\left[\frac{{\mathrm{RootOf}}{}\left({{\mathrm{_Z}}}^{{2}}{+}{1}\right)}{{T}}{+}{1}{,}{T}{=}{x}\right]\right]$ (7)

Each generalized exponent gives the dominant term (ignoring logarithmic factors) in a formal solution. For example, consider the formal solutions of the following ode:

 > $\mathrm{ode}≔{x}^{5}\mathrm{diff}\left(y\left(x\right),\mathrm{}\left(x,3\right)\right)-{x}^{2}\mathrm{diff}\left(y\left(x\right),\mathrm{}\left(x,2\right)\right)+3x\mathrm{diff}\left(y\left(x\right),x\right)-\frac{15}{4}y\left(x\right)$
 ${\mathrm{ode}}{≔}{{x}}^{{5}}{}\left(\frac{{{ⅆ}}^{{3}}}{{ⅆ}{{x}}^{{3}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){-}{{x}}^{{2}}{}\left(\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){+}{3}{}{x}{}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){-}\frac{{15}{}{y}{}\left({x}\right)}{{4}}$ (8)
 > $\mathrm{formal_sol}\left(\mathrm{ode},y\left(x\right),T,x=0\right)$
 $\left[\left[{{T}}^{{3}}{{2}}}{}\left({-1}{+}\frac{{3}}{{16}}{}{{T}}^{{2}}{+}\frac{{105}}{{512}}{}{{T}}^{{4}}{+}{O}{}\left({{T}}^{{6}}\right)\right){,}{{T}}^{{3}}{{2}}}{}\left({T}{+}\frac{{5}}{{16}}{}{{T}}^{{3}}{+}\frac{{315}}{{512}}{}{{T}}^{{5}}{+}{O}{}\left({{T}}^{{6}}\right)\right){,}{T}{=}{x}\right]{,}\left[{{T}}^{{3}}{}{{ⅇ}}^{{-}\frac{{1}}{{2}{}{{T}}^{{2}}}}{}\left({1}{-}\frac{{9}}{{8}}{}{{T}}^{{2}}{+}\frac{{465}}{{128}}{}{{T}}^{{4}}{+}{O}{}\left({{T}}^{{6}}\right)\right){,}{T}{=}{x}\right]\right]$ (9)

The above solutions can be rewritten as ${T}^{3}{2}}\left(\mathrm{constant}+\mathrm{...}\right),{T}^{5}{2}}\left(\mathrm{constant}+\mathrm{...}\right)$ and ${T}^{3}{ⅇ}^{-\frac{1}{2{T}^{2}}}\left(\mathrm{constant}+\mathrm{...}\right)$ where the dots refer to higher order terms. Note that $T=x$ because we took the point $x=0$, and because the ramification index is $r=1$ in this example. Thus, the dominant terms are ${x}^{3}{2}},{x}^{5}{2}}$, and ${x}^{3}{ⅇ}^{-\frac{1}{2{x}^{2}}}$. Rewriting each of these in the form ${ⅇ}^{\int \frac{e}{x}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}ⅆx}$ for some $e$ in $C$ $\left[{x}^{-\frac{1}{r}}\right]$ one finds the following possible $e$'s: $\frac{3}{2},\frac{5}{2}$, and $3+\frac{1}{{x}^{2}}$. So those must be the generalized exponents of the ode at $x=0$. Indeed:

 > $\mathrm{gen_exp}\left(\mathrm{ode},y\left(x\right),T,x=0\right)$
 $\left[\left[\frac{{3}}{{2}}{,}\frac{{5}}{{2}}{,}{T}{=}{x}\right]{,}\left[\frac{{1}}{{{T}}^{{2}}}{+}{3}{,}{T}{=}{x}\right]\right]$ (10)

References

 Cluzeau, T., and van Hoeij, M. "A Modular Algorithm to Compute the Exponential Solutions of a Linear Differential Operator." J. Symb. Comput. Vol. 38, 2004: 1043-1076.
 Ince, E.L. Ordinary Differential Equations, Chap. XVI-XVII. New York: Dover Publications, 1956.
 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.
 * More information on generalized exponents is in the help page for DEtools[formal_sol].