Solving a second-order recurrence relation / Series expansion of a confluent Heun equation

Question

I would like to know whether it is possible to solve (in "closed form") either one of the following two second-order recurrence relations, which are closely related to each other. The first is $$n(n-1)c_n-(n-1)^2c_{n-1}-\omega^2(c_{n-2}-c_{n-1})=0,$$ and the other is $$n^2b_n-(n-1)^2b_{n-1}+\omega^2b_{n-2}=0,$$ where $\omega^2$ is an arbitrary real parameter.

Background:

These equations arise from the second-order linear ODE $$F''(u)+\frac{F'(u)}{u-1}+\frac{\omega^2}{u}F(u)=0.\label{1}\tag{$*$}$$ This ODE is of confluent Heun type and admits the two linearly independent solutions $$F_1(u)=\sum_{n=0}^\infty c_nu^n=u\ Hc(\omega^2-1,\omega^2,2,1,0,u),$$ and $$F_2(u)=\sum_{n=0}^\infty b_n(1-u)^n=Hc(0,-\omega^2,1,0,0,1-u),$$ where $Hc(q,\alpha,\gamma,\delta,\epsilon,z)$ denotes the confluent Heun function as implemented in Mathematica. That is, it is the solution to $$ z(z-1)y''+[\gamma(z-1)+\delta z+z(z-1)\epsilon]y'+(\alpha z-q)y=,0 $$ which is normalized to one at $z=0$. These solutions may become linearly dependent for special discrete values of $\omega$.

"Closed form":

I am hoping to be able to find an "explicit" form of the series coefficients $b_n$ or $c_n$ in the expansion of the above Heun functions. I found an old paper by Harold Exton ("New Solutions of the Confluent Heun Equation", Le Matematiche (Catania) 53, No. 1, 11-20 (1998), MR1681613, Zbl 0929.34008) which is supposed to provide an explicit triple series representation of the confluent Heun function, but Exton's results appear to require $\epsilon\neq0$.

I realize the general Heun equation is complicated in general, but in this case many of the parameters are zero or take simple values, so I am hoping it may be possible to obtain a direct formula for $b_n$ or $c_n$ (possibly as a sum over many terms) that does not involve the preceding terms. Or if such a thing is not possible, that would also be great to know.

Thank you!

Edit: a new approach

By inspecting the equation, and in the great physics tradition of inventing a new expansion parameter when none is available, it seems fruitful to introduce a new coordinate $\rho$ and parameter $\Omega$ by letting $$u=\Omega^2\rho^2,\quad\omega=\frac{\lambda}{2\Omega}.$$ With these rescalings, the original ODE $(*)$ becomes $$F''(\rho)-\frac{1}{\rho}\left(\frac{1+\Omega^2\rho^2}{1-\Omega^2\rho^2}\right)F'(\rho)+\lambda^2F(\rho)=0,$$ and the first solution becomes $$F_1(\rho)=\Omega^2\rho^2\ Hc\left(\frac{\lambda^2}{4\Omega^2}-1,\frac{\lambda^2}{4\Omega^2},2,1,0,\Omega^2\rho^2\right).$$ The advantage is that the ODE and its solutions can now be expanded in (inverse) powers of $\omega=2\Omega/\lambda$. Indeed, one can show that $$F_1(\rho)=8\Omega\rho\sum_{k=1}^\infty\left(\frac{\Omega}{4\lambda}\right)^kf_k(\Omega\rho)J_k(\lambda\rho),$$ where the $J_k(x)$ are the Bessel functions of the first kind, and the functions $f_n(X)$ are recursively defined by $$f_1(X)=\frac{1}{\sqrt{1-X^2}},$$ and $$f_n(X)=-\frac{2}{\sqrt{1-X^2}}\times\int\left(\sqrt{1-X^2}f_{n-1}''(X)-\frac{2n\left(1-X^2\right)-3+5X^2}{X\sqrt{1-X^2}}f_{n-1}'(X)+\frac{(n-2)\left[n\left(1-X^2\right)+2X^2\right]}{X^2\sqrt{1-X^2}}f_{n-1}(X)\right)\mathrm{d}X.$$ In particular, $$f_2(X)=\frac{\left(1-X^2\right)\mathrm{arctanh}(X)-X}{\left(1-X^2\right)^{3/2}},$$ $$f_3(X)=\frac{\left(1-X^2\right) ^2\mathrm{arctanh}^2(X)+\frac{12-14X^2}{X}\left(1-X^2\right)\mathrm{arctanh}(X)-3\left(4-7X^2\right)}{2\left(1-X^2\right)^{5/2}},\ldots$$ Does anyone know how to find a direct formula for $f_n(X)$ or the full sum $F_1(\rho)$?

Answer 1

This is not an answer, only some observations:

The denominators $d_n$ of $b_n$ are given by ($n=0,1,\ldots$) \begin{align} d_n &= \{1,1, 4, 9, 64, 900, 20736,\ldots\} \\ &\stackrel{n\neq1}{=} \frac{n!^2}{(n-1)^2}. \end{align} They can be removed from the sequence $b_n$ by defining a modified sequence \begin{align} b_0'&=1 \\ b_1'&=1 \\ b_{n}' &= n^2b_{n-1}' - \omega^2 (n-1)^2 b_{n-2}'\,, \end{align} such that \begin{align} -\frac{n!^2}{\omega^2(n-1)^2} b_n=b_{n-2}'\, , \end{align} and \begin{align} b_2' &= 4-\omega^2 \\ b_3' &= 36-13\omega^2 \\ b_4' &= 576-244\omega^2+9\omega^4 \\ &\ldots\,. \end{align} Observe that the constant term in $b_n'$ is $n!^2$, while the coefficients of $\omega^2$ are given by A203156.

A similar transformation can be done with $c_n$: Define \begin{align} c_{-1}'&=1 \\ c_0'&=1 \\ c_{n}' &= (n^2+\omega^2)c_{n-1}' + \omega^2 n(n-1) c_{n-2}' \end{align} to get \begin{align} \frac{n!^2}{n} c_n=c_{n-1}' \end{align} and \begin{align} c_1' &= 1-\omega^2 \\ c_2' &= 4-3\omega^2+\omega^4 \\ c_3' &= 36-25\omega^2+6\omega^4-\omega^6 \\ c_4' &= 576-388\omega^2+85\omega^4-10\omega^6+\omega^8 \\ &\ldots\,. \end{align}

Answer 2

A methodology to find the corresponding formula for the coefficients of $\omega^{2n}$ from @FredHucht's response.

When we take $F_1(u)=\sum_{n\geq0}c_nu^n$ it follows that $c_0=0$, $c_1$ is our constant of integration, $c_2=\frac12 c_1$, and \begin{align} n(n-1)c_n+(\omega^2-(n-1)^2)c_{n-1}=\omega^2c_{n-2};\quad [n\geq3]. \end{align}

If we then take $c_n=nc_{n-1}'/n!^2$ it must follow that $c_{-1}'=0$, $c_0'=c_1$, and $c_1'=c_1$, but not $c_{-1}'=1$! So we are left with the relation \begin{align} c'_n=(n^2+\omega^2)c_{n-1}'+\omega^2n(n-1)c_{n-2}';\quad[n\geq2] \end{align} $c_1$ can be set to 1 without loss of generality, giving \begin{align} &c_0'=1\\ &c_1'=1\\ &c_2'=4+3\omega^2\\ &c_3'=36+37\omega^2+3\omega^4\\ &c_4'=576+676\omega^2+121\omega^4+3\omega^6\\ &\vdots \end{align}

Now for something neat! Let $c_n'=\sum_{m=0}^{n-1}\kappa_{n,m}\omega^{2m}$, so \begin{align} \sum_{m=0}^{n-1}\kappa_{n,m}\omega^{2m}=\sum_{m=0}^{n-2}n^2\kappa_{n-1,m}\omega^{2m}+\sum_{m=1}^{n-1}\kappa_{n-1,m-1}\omega^{2m}+\sum_{m=1}^{n-2}n(n-1)\kappa_{n-2,m-1}\omega^{2m}, \end{align} and so we have \begin{align} \kappa_{n,n}&=\kappa_{n-1,n-1},\qquad \kappa_{n,0}=n^2\kappa_{n-1,0},\\ \kappa_{n,m}&=n^2\kappa_{n-1,m}+\kappa_{n-1,m-1}+n(n-1)\kappa_{n-2,m-1};\quad n\geq4, m\in [1,n-2] \end{align} From the first two we recover $\kappa_{n,n-1}=3$ (for $n\geq2$) and $\kappa_{n,0}= n!^2$ (for $n\geq0$).

If we wish to find the coefficient of the second highest order term of $\omega^2$ we set $m=n-2$ to get \begin{align} \kappa_{n,n-2}=n^2\kappa_{n-1,n-2}+\kappa_{n-1,n-3}+n(n-1)\kappa_{n-2,n-3},\\ \kappa_{n,n-2}-\kappa_{n-1,n-3}=3n^2+3n(n-1)=3n(2n-1),\\ \sum_{k=4}^n(\kappa_{k,k-2}-\kappa_{k-1,k-3})=\sum_{k=4}^n 3k(2k-1),\\ \kappa_{n,n-2}=\frac12n(4n-1)(n+1)-29,\quad [n\geq3]. \end{align} Similarly, we can find the coefficients $\kappa_{n,1}$: let $m=1$ to get \begin{align} \kappa_{n,1}=n^2\kappa_{n-1,1}+\kappa_{n,0}+n(n-1)\kappa_{n-2,0},\\ \kappa_{n,1}-n^2\kappa_{n-1,1}=(n-1)!^2+n(n-1)(n-2)!^2,\\\\ \frac{\kappa_{n,1}}{n!^2}-\frac{\kappa_{n-1,1}}{(n-1)!^2}=\frac{1}{n^2}+\frac{1}{n(n-1)},\\\\ \sum_{k=3}^n\left(\frac{\kappa_{k,1}}{k!^2}-\frac{\kappa_{k-1,1}}{(k-1)!^2}\right)=\sum_{k=3}^n\left(\frac{1}{k^2}+\frac{1}{k(k-1)}\right),\\\\ \kappa_{n,1}=n!^2\left(\frac34+\sum_{k=3}^n\left(\frac{1}{k^2}+\frac{1}{k(k-1)}\right)\right),\quad [n\geq3]. \end{align} As far as a general formula for $\kappa_{n,m}$, I have been unable to find one, perhaps something involving induction might work? I imagine a similar method for $b_n'$ would work which I will leave as an exercise for the reader.