Integral relations and new solutions to the double-confluent Heun equation

Integral relations and transformation rules are used to obtain, out of an asymptotic solution, a new group of four pairs of solutions to the double-confluent Heun equation. Each pair presents the same series coefficients but has solutions convergent in different regions of the complex plane. Integral relations are also established between solutions given by series of Coulomb wave functions. The Whittaker-Hill equation and another equation are studied as particular cases of both the doubleconfluent and the single-confluent Heun equations. Finally, applications for the Schrödinger equation with certain potentials are discussed, mainly for quasi-exactly solvable potentials which lead to the above special equations. 1. Preliminary remarks Here we deal with two groups of series solutions to the double-confluent Heun equation (DCHE). The first group, constituted by four pairs of solutions, is generated from an asymptotic expansion by means of integral relations and transformation rules, and the second group is given by pairs of solutions in series of Coulomb wave functions, already derived in [1]. For the latter, we show that an integral relation also exists between the members of each pair, and we provide additional properties for the solutions. After this, we analyse two differential equations which are special cases of both the DCHE and the single-confluent Heun equation and, finally, we use some results to solve the Schrödinger equation for certain potentials. Before proceeding, we set down some conventions, present the procedures used to obtain the solutions and outline the structure of the paper. For the DCHE we adopt the form z dU dz2 + (B1 +B2z) dU dz + ( B3 − 2ηωz + ω z ) U = 0, (B1 6= 0, ω 6= 0) (1) where z = 0 and z = ∞ are irregular singularities. B1 = 0 and/or ω = 0 are excluded because, in these cases, the equation reduces to a confluent hypergeometric equation or to an equation with constant coefficients. Thus, if B1 = 0 and ω 6= 0, the substitutions y = −2iωz, U(z) = eyf(y), α − (1− B2)α +B3 = 0 (2a)