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Quasi-Periodic Solutions of Heun's Equation

By exploiting a recently developed connection between Heun's differential equation and the generalized associated Lam\'e equation, we not only recover the well known periodic solutions, but also obtain a large class of new, quasi-periodic solutions of Heun's equation. Each of the quasi-periodic solutions is doubly degenerate.

The so-called periodic (also termed as polynomial) solutions of Heun's equation have been well studied.
However, as emphasized in ref. [1], much less attention has been devoted to the quasi-periodic (also termed as non-polynomial) solutions. In this letter we obtain a large class of (mostly new) quasi-periodic solutions.
We shall show that each such solution is doubly degenerate.
The canonical form of Heun's equation is given by [1] where α, β, γ, δ, ǫ, q, c are parameters. The parameters are not all independent; the constraint relation is The four regular singular points of the differential equation are located at x = 0, 1, c ( = 0, 1), and the point at infinity.
Before describing the quasi-periodic solutions of eq. (3), it may be worthwhile explaining how we arrived at such solutions. A few years ago, we studied in some detail the Schrödinger equation for the associated Lamé (AL) potentials [11,12] V (y, m) = a(a + 1)msn 2 (y, m) + b(b + 1)msn 2 (y + K(m), m) = a(a + 1)msn 2 (y, m) which after a transformation can be shown to be a special case of eq. (3). In particular, the band edges and the mid-band states of several of these potentials were studied, when a and b were related in various specific ways. Further, when a, b are both integers, we showed that these potentials had the special feature of possessing only a finite number of band gaps. This study has been extended to the generalized associated Lamé (GAL) potentials [13,14] V (y, m) = a(a + 1)msn 2 (y, m) + b(b + 1)msn 2 (y + K(m), m) The point to note is that after a transformation, the Schrödinger equation for the potential given in eq.
(5) is identical in form to eq. (3). In particular, let us start from the Schrödinger equation with V (y, m) given by eq. (5). On substituting one can show that φ(y) satisfies Heun's equation (3). More precisely, φ(y) satisfies the differential equation φ ′′ (y)+2 mb sn(y, m)cn(y, m) dn(y, m) +f sn(y, m)dn(y, m) cn(y, m) −g cn(y, m)dn(y, m) where Thus, once we obtain solutions of the Schrödinger equation for the GAL potential (5), then we can immediately write the solutions of eq. (3) and hence that of the original Heun's eq. (1) with the identification A few comments are in order.
Our strategy is now clear. We shall first obtain solutions of the Schrödinger eq. (6) for the GAL potential (5) and then using the connections given in eqs. (10) and (11)  We now show that when either a + b + f + g and/or a − b − f − g is an arbitrary half-integer (but not an integer), then one can obtain doubly degenerate eigenstates of the GAL eq. (6) which correspond to the mid-band states (rather than band-edges) of this periodic problem. In the special case when a + b + f + g and/or a − b − f − g is an integer, then one obtains two distinct (non-degenerate) eigenstates which correspond to the band edge eigenstates. It turns out that depending on whether b or f or g is halfintegral (while the other two are integral), we need to use a different ansatz. Let us consider all three cases one by one.
We start from eq. (8) and substitute the ansatz where t is any real number. It follows that Z(y) satisfies the equation where R and Q are as given by eq. (9). Not surprisingly, Z(y) = constant is a solution with energy

One can build solutions for higher half-integer values of
while if b + f + g = 2M + 3/2, then we consider the ansatz (M = 0, 1, 2, ...) Substitution into eq. (13) followed by lengthy but straightforward algebra yields analytic expressions for the energy eigenvalues for arbitrary M and b = 1/2, 3/2. In particular, for b = 1/2, we find where both f, g are nonnegative integers satisfying it follows that for a given N , there are N + 1 different solutions. Similarly, when b = 3/2, we get where, f and g are again nonnegative integers. Note that, in this case too, for a given N , there are N + 1 different solutions. In all cases, the corresponding eigenfunctions have the form given in eqs. (14) and (15). For small values of N , the explicit coefficients A k , B k appearing in the eigenfunction expressions can be easily written down.
We can immediately write down the solutions of Heun's eq. (3) by making use of eq. (10). In particular, we find the following two classes of solutions: The corresponding eigenfunctions are generically as given by eqs. (12), (14) and (15) with F (y) ≡ φ(y).
As an illustration, for N = 1, the eigenfunction is Z(y) = Acn(y, m) + Bsn(y, m) with B A = it for Several remarks are in order at this stage.
1. Since the Schrödinger eq. (6) for the GAL potential (5)  4. We can generate new solutions of Heun's eq. (3) by considering the transformations y → y + K(m), y → y + iK ′ (m), y → y + K(m) + iK ′ (m) and the fact that the eigenvalues are invariant provided we interchange a, b, f, g appropriately as explained above. Obviously, each of these new solutions is also doubly degenerate. In particular, by starting from solution (18) and using eq. (11), the three sets of new solutions of Heun's eq. (3) are while the eigenfunctions corresponding to solutions (20) to (22) are given by 6. We can generate even more new solutions of Heun's eq. (3) by considering the transformations b → −b − 1, f → −f − 1 and g → −g − 1 either singly or in various combinations. As an illustration, by starting from solution (18) and using eq. (11), we get the following seven sets of new solutions. . .
7. Starting from these seven solutions, we can generate further new solutions by considering the transformations y → y + K(m), y → y + iK ′ (m), y → y + K(m) + iK ′ (m) and interchanging a, b, f, g appropriately. One can show that in this way, by starting from the solution (18) with b = 1/2 one has 20 independent solutions.
8. There is one more remarkable symmetry associated with eqs. (14) and (15). Note that eq. (15) is invariant under t → −t followed by i → −i. Under this transformation, ansatz (14) becomes and hence it follows that solutions with the ansatz (14) and (34)  Case 2: f half-integral We start from eq. (8) and substitute the ansatz where t is any real number and k = √ m. It follows that Z(y) satisfies the equation where R and Q are as given by eq. (9). Not surprisingly, Z(y) = constant is a solution with energy E = (4mt 2 + 1)/4 provided b = g = 0, f = 1/2, a = t − 1/2 (i.e. b + f + g = 1/2).
Similarly, when f = 3/2, a = t − 1/2, g = p, b + g = N we find that where, b and g are again nonnegative integers. In all these cases, the corresponding eigenfunctions have the form as given above in eqs. (37) and (38). For small values of N , the explicit coefficients A k , B k in the eigenfunction expressions can be easily written down.
Using eq. (9), we can write down the corresponding solutions of Heun's eq. (3). They are given by Corresponding to each of these two solutions, we can again write down 32 independent sets of solutions as shown above in Case 1 when b was half-integral.

Case 3: g half-integral
We start from eq. (8) and substitute the ansatz where t is any real number. It then follows that Z(y) satisfies the equation where R and Q are as given by eq. (9). Clearly, Z(y) = constant is a solution with energy E = (1 + m)/4 provided b = f = 0, g = 1/2, a = t − 1/2 (i.e. b + f + g = 1/2).
One can build solutions for higher values of b+f +g from here. In particular, when b+f +g = 2M +1/2, Substitution into eq. (44) leads to analytic expressions for the energy eigenvalues and eigenfunctions for arbitrary M when b = 1/2, 3/2. In particular, for b = 1/2, we find that In all these cases, the corresponding eigenfunctions have the form given above in eqs. We would like to thank K. Takemura for informing us about some related work and recent developments in the mathematics literature on GAL potentials and Heun's equation.