4
$\begingroup$

For quite some time I have been trying to solve the following system of differential equations for the two functions $G$ and $H$ defined on the interval $[0,1]$: $$ \begin{align}x G''(x)=&\mathscr{F}''(x)H(x)-\mathscr{F}(x)H''(x)\tag{1}\\ 4x G(x)+x^2G'(x)+\frac13\left(2+x^3\right)G''(x)=&3x\mathscr{F}(x)H'(x)-\frac13\left(2+x^3\right)\mathscr{F}'(x)H''(x)\tag{2}. \end{align} $$ Initial conditions are $$ G(1)=G'(1)=0,\quad H(1)=\frac15,\quad H'(1)=0. $$ Here $\mathscr{F}$ is a hypergeometric function satisfying the differential equation $$ x \mathscr{F}(x)+x^2 \mathscr{F}'(x)+\frac13\left(2+x^3\right)\mathscr{F}''(x)=0,\quad \mathscr{F}(1)=1,\quad \mathscr{F}'(1)=0.\tag{3} $$ The solution $\mathscr{F}$ to this equation can be given explicitly as a sum of two Gauss hypergeometric function, but writing out the explicit form seems to obscure the structure inherent in equations (1) and (2).

Using integration by parts, equation (1) can also be written as $$ G(x)-xG'(x)=\mathscr{F}(x)H'(x)-H(x)\mathscr{F}'(x),\tag{1'} $$ where the left-hand side is the Wronskian of $G$ and the identity function, and the right-hand side is the Wronskian of $\mathscr{F}$ and $H$. The second equation can then be rewritten as $$ x G(x)+4x^2 G'(x)+\frac13\left(2+x^3\right)G''(x) = \mathscr{F}'(x)\left[3x H(x)-\frac13\left(2+x^3\right)H''(x)\right],\tag{2'} $$ reducing the number of appearances of the hypergeometric factor $\mathscr{F}$ to one.

Differentiating (2) three times and substituting $G''$ and higher derivatives from (1) eliminates $G$ but leaves one with a fifth-order linear ODE for $H$ with quite horrendous hypergeometric coefficients.

I can also solve the equations numerically, but I'm really interested in a "closed form" solution if one exists. I also know series expansions of $G$ and $H$ around 1 up to order 75 or so and feeding these expansions into FriCAS' guessADE function suggests that neither $G$ nor $H$ satisfy a simple algebro-differential equation of low degree. Unfortunately, this rules out the possibility that $H$ is a polynomial and $G$ a hypergeometric function.

Question 1: What tricks can be used to solve (1)-(2) for $G$ and $H$? Or is there a reason to believe that a solution in terms of classical special functions does not exist?

There is a substantial literature about linear ODE's with polynomial coefficients such as (3).

Question 2: Are there any results about linear ODE's with hypergeometric coefficients, such as (1)-(2)?

It's frustrating that the equations seem quite simple, yet have eluded any attempt at solving them so far.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.