Liouvillian solutions for second order linear differential equations with Laurent polynomial coefficient

This paper is devoted to a complete parametric study of Liouvillian solutions of the general trace-free second order differential equation with a Laurent polynomial coefficient. This family of equations, for fixed orders at 0 and ∞\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\infty$$\end{document} of the Laurent polynomial, is seen as an affine algebraic variety. We prove that the set of Picard-Vessiot integrable differential equations in the family is an enumerable union of algebraic subvarieties. We compute explicitly the algebraic equations of its components. We give some applications to well known subfamilies, such as the doubly confluent and biconfluent Heun equations, and to the theory of algebraically solvable potentials of Shrödinger equations. Also, as an auxiliary tool, we improve a previously known criterium for a second order linear differential equations to admit a polynomial solution.


Introduction
There is no general agreement about the meaning of the term "explicit solution" for ordinary differential equations.Some authors allow certain special functions, some others consider only elementary functions.This paper is written inside the framework of the differential Galois theory of linear differential equations, also known as Picard-Vessiot theory.Therefore, our notion of explicit solution is that of Liouvillian function.In the particular case of linear differential equations with rational coefficients, solutions that can be expressed in terms of elementary functions and indefinite integrals (integration by quadratures) are always Liouvillian functions.It is also well known that a second order linear differential equation with rational coefficients admits a Liovillian solution if and only it is Picard-Vessiot integrable.Previous theoretical results on Picard-Vessiot theory by M. F. Singer [1] ensure that given a finite dimensional family of linear differential equations, the subfamily of Picard-Vessiot integrable equations admits a canonical description as union of algebraic subvarities.Our objective is to give an explicit description of the subfamily of integrable Picard-Vessiot equations inside the family of second order trace-free linear differential equations with Laurent polynomial coefficient, and the Liouvillian solutions attached to those integrable equations.Picard-Vessiot integrability of equations of some subfamilies of the families studied here, corresponding to confluences of hypergeometric and Heun equations, where already studied in detail by A.
We consider the family of second order trace-free linear differential equations, where L(x) ∈ C[x, x −1 ] is a monic Laurent polynomial with ℓ −r = 0. We say that L(x) has type (r, m).The space M (r,m) of monic Laurent polynomials of type (r, m) is an affine algebraic variety M (r,m) ≃ C * × C r+m−1 .The purpose of this paper is to classify Picard-Vessiot integrable equations in the family (1) and to study how their Liouvillian solutions depend algebraically on the coefficients of L(x) when it moves in the space M (r,m) .We thus introduce the spectral set.
Definition 1.The spectral set S (r,m) ⊂ M (r,m) is the set of monic Laurent polynomials of type (r, m) such that its corresponding Eq. ( 1) is Picard-Vessiot integrable (or equivalently, has a Liouvillian solution).
For the formal definition of the Galois group, Picard-Vessiot integrability, and Kovacic algorithm we refer the readers the original article [3] of J. Kovacic.
Nevertheless, we included an appendix with the parts of the algorithm that are relevant for the our results.
The Picard-Vessiot integrability analysis of Eq. ( 1) is done in two steps.
Theoretically, Kovacic algorithm admits three different cases of integrability.
However, by means of the D'Alembert transform, we show that, after a twosheet covering of the Riemann sphere, we can reduce the analysis to the first case.This reduces the problem to the existence of a polynomial solution of an attached auxiliary equation.To deal with this existence problem, we give an algebraic generalization (Theorem 4) of the asymptotic iteration method (AIM) due to Cifti, Hall and Saad [4].Here we discover a universal family of differential polynomials in two variables (Table 1) that controls the existence of polynomial solutions for second order differential equations with coefficients in arbitrary differential fields.
We arrive to a decomposition of the spectral sets as enumerable union of spectral varieties (Theorems 1, 2, 3 and Propositions 2, 3, 4) whose equations can be given explicitly by means of universal polynomials ∆ d and auxiliary equations (A 1 , A 2 , A 3 ).These results are summarized in the following: we have a decomposition of the spectral set as a enumerable union of spectral varieties.Moreover: (r,m) with r = 2 then the differential equation (1) has a solution of the form: where P (x) is a monic polynomial of degree d, and ω(x) is a Laurent polynomial.
In section 4 present some applications that involve the analysis of biconfluent Heun equation and doubly confluent Heun equation, as well Schrödinger Equations with Mie potentials (Laurent polynomial potentials, exponential potentials) and Inverse Square Root potentials, see [5,6,7].Finally, our results allow us to state that there are no new algebraically solvable Laurent polynomial potentials for the Shcrödinger equation beyond those previously known (Corolary 1) corresponding to m = 2 and r = 0, 1, 2.

A note on the general case
The assumptions of having a trace-free differential equation with a monic Laurent polynomial is done with the purpose of presenting clearer computations and formulae.In fact, our analysis does applies to the more general case of differential equations of the form, • First, we can always reduce Eq. ( 2) to trace-free from by means of the so called D'alembert transform.Namely, we look for a suitable scaling y = f (x)ỹ of the dependent variable, so that the second derivative of the new unknown function is: Here, the term in dỹ dx vanish if we take f to be a solution of the differential equation and a trace-free differential equation, depending of a Laurent polynomial whose coefficients are polynomials of degree 1 and 2 in the coefficients of L 0 (x) and L 1 (x).
• Second, the trace-free differential equation above with non necessarily monic Laurent polynomial, can be transformed into Eq.( 6) by scaling the independent variable x = m+2 √ a m x.We fall in the monic case, and the coefficients of the new Laurent polynomial are polynomials in a −r , . .., a m−1 , m+2 √ a m .
• Third, the assumption of r • m > 0 is not necessary.If r = 0 then we fall in the simpler polynomial case that has been studied exhaustively by the authors in [8] as a previous step for the analysis of the Laurent polynomial case.If m = 0 then we may just apply consider z = 1 x as a new independent variable and fall into the m • r > 0 case.

Kovacic algorithm analysis
The most relevant fact about Kovacic algorithm is that the Galois group of Eq. ( 1) with fixed L(x) is an algebraic subgroup of SL 2 (C), and the equation is integrable if and only if such group is conjugated to a subgroup of the Borel subgroup or it is finite and not contained in the former.These correspond to the so-called first three cases of Kovacic algorithm [3], the fourth being the non-integrable case.In what follows, for practical purposes related with the aplication of the algorithm, we split the set Z 2 >0 in four different disjoint subsets, corresponding to different orders of Laurent polynomials at 0 and ∞ that allow different possibilities for Liouvillian solutions of Eq. (1).Definition 2. We say that a pair (r, m) ∈ Z 2 >0 is of class: (1) r ∈ {1} ∪ {2k + 4 : k ∈ Z, k ≥ 0} and m is even; (2) r = 2 and m is odd; (3) r = 2 and m is even; (4) not in classes (1), (2) or (3).
So that, we have a disjoint partition Lemma 1.Let us assume that Eq. (1) with some fixed L(x) ∈ M (r,m) is Picard-Vessiot integrable.The following statements hold.
Proof.This lemma is a consequence of the necessary conditions in Kovacic algorithm (Theorem 6 in Appendix A).The order of L(x) at x = ∞ is −m, a negative integer.Thus, case 3 of Kovacic algorithm is discarded.Thus, the remaining possibilities are: the Galois group is either conjugated to a subgroup of B 2 or to a subgroup of D ∞ .Let us assume that the Galois group is conjugated to a subgroup of B 2 then, necessary conditions for case 1 must be satisfied.
Let us introduce the following notation: (a) B (r,m) ⊂ S (r,m) is the set of Laurent polynomials L(x) ∈ M (r,m) such that the Galois group of Eq. ( 1) is conjugated to a subgroup of B 2 .
(b) D (r,m) ⊂ S (r,m) is the set of Laurent polynomials L(x) ∈ M (r,m) such that the Galois group of Eq. ( 1) is conjugated to a subgroup of D ∞ .Remark 1. From Lemma 1 we have: Let us consider the case with r = 1 and m = 2p even.In such case we have a unique decomposition: where A(x) is a monic polynomial of degree p and quadratic residue B(x) is a polynomial of degree p − 1.We also have in mind that the map is an invertible polynomial map.
is Picard-Vessiot integrable, that is, L(x) ∈ S (1,2p) , if and only for a choice of sign s ∞ = ±1 the following conditions hold: 1.The quantity 2. There exist a polynomial P (x) of degree d such that: In such case, the liouvillian solution is an eigenvector of the Galois group.
Proof.red We are in case C 1 and, by Lemma 1, if the equation is Picard-Vessiot integrable then it corresponds to case 1 of Kovacic algorithm.
Step 1 of case 1 gives us conditions {c 1 , ∞ 3 }, in addition, regarding to x 1 , we got that For each choice of the sign s ∞ ∈ {+1, −1} we consider the complex number If none of them is a non-negative integer then we can discard case 1 and the Galois group SL 2 (C).SL 2 (C).Otherwise we proceed to step 2 of case 1 for each suitable value of d = d(s ∞ ).We set the rational x and search for a monic polynomial P of degree d, which satisfy the auxiliary differential equation, This last equation can be written in terms of the decomposition showed in Eq.
(3), obtaning Eq. (A 1 ) of the statement.If a pair (ω, P ) described as above can be found, Eq. ( 1) is Picard-Vessiot integrable.Moreover, Kovacic algorithm provides us the solution given in the statement.

Characterization of B (2,2p) .
Let us consider the case with r = 2 and m even.We have a unique decomposition: where A(x) is a monic polynomial of degree p and quadratic residue B(x) is a polynomial of degree p − 1.As in the above case, the map is an invertible polynomial map.
Theorem 2. The Galois group of Eq. ( 1) ) , if and only if for a combination of signs s ∞ = ±1 and s 0 = ±1 the following conditions hold: 1.The quatity is a non negative integer.
2. There exist a polynomial P (x) of degree d such that: In such case, the liouvillian solution is an eigenvector of the Galois group.
Proof.This proof is similar to that of Theorem 1.By Lemma 1 we are in case (r, m) ∈ C 2 .We proceed to step 1 of case 1 of Kovacic algorithm with conditions For each choice of s 0 = ±1 and s ∞ = ±1 we consider the complex number: If none of them is a non-negative integer then we discard case 1.Otherwise, for each non-negative integer value of d = d(s 0 , s ∞ ) we set the rational function and proceed to the step 2 of case 1 of Kovacic algorithm.We search for a monic polynomial P of degree d, which satisfy the auxiliary differential equation, And, using Eq. ( 4) we arrive to the expression of Eq. (A 2 ).If a pair (ω, P ) described as above can be found, Eq. ( 1) is Picard-Vessiot integrable.In addition, Kovacic algorithm provides us the solution given in the statement.
For Q(x) ∈ M (2q,2p) with q > 1 we look for a decomposition that takes into account the cuadratic residues at zero and infinity simultaneously.
In this case R(x) is not uniquely determined, as both R(x) and its reciprocal −R(x) can be used in the decomposition.The coefficients of R(x) are polynomials in some of the the coefficients of L(x) and r −q = ℓ −2q .Coefficients of B(x) and A(x) are polynomials in some of the coefficients of L(x).What we have is that the map, is a two-sheet cover, with the advantage that elements (R(x), B(x), A(x)) ∈ M(2q,2p) correspond to specific decomposition.Let us denote by S(2q,2p) = ) ) the pullback of the spectral set.
From now, let us fix a decomposition as in Eq. ( 5) and work with the differential equation, For each choice of s 0 = ±1 and s ∞ = ±1 let us define the following quantities and functions: Note that changing the sign of s 0 is equivalent to changing the choice of R(x) in the decomposition.
) , if and only if for a combination of signs s 0 = ±1 and s ∞ = ±1 the following conditions hold: 1. d is a non negative integer.
2. There exist a polynomial P (x) of degree d such that: In such case, the liouvillian solution is an eigenvector of the Galois group.
Proof.By Lemma 1, if the Eq. ( 6) is Picard-Vessiot integrable then it corresponds to case 1 of Kovacic algorithm.We proceed to step 1 of case 1 obtaining a case of type {c 3 , ∞ 3 }.
A necessary integrability condition, given in step 2 of case 1, is that some of the quantities d (as defined in Eq. ( 7)), corresponding to some choice of signs s 0 = ±1 and s ∞ = ±1, is a non-negative integer.If not, the Galois group of Eq. ( 6) is SL 2 (C).Otherwise, for each suitable choice of signs we set the rational x and search for a monic polynomial P of degree d, which satisfy the auxiliary differential equation, This equation assumes the form of Eq. (A 3 ) when we substitute the decomposition given in (5).If a pair (ω, P ) described as above can be found then Eq. ( 6) is Picard-Vessiot integrable.In addition, Kovacic algorithm provides us the solution given in the statement.Therefore, if we apply the change of variable x = w 2 in Eq. ( 1) we arrive to differential equation: whose associated Riccati equation has a Laurent polynomial solution.We may then transform the equation into a reduced form with the same associated Riccati equation by taking, This trace free equation has a Laurent polynomial coefficient whose leading term w 2m+2 has a coefficient 4. We scale the independent variable by taking w = m+2 √ 2w obtaining: that can be seen as a differential equation in the family B (2,2m+2) .
Proof.Let part from any L(x) ∈ M (2,m) and let us follow the change of variables and reduction from Eq. (1) to Eq. (10).It is clear that if the second equation has a Liouvillian solution, then we obtain a liouvillian solution of (1).
But, because of the above discussion, if the second equation is Picard-Vessiot integrable, it must be in B (2,2m+2) .
Finally let us note that the D'alembert transform and reductions process is a polynomial map in the coefficients of the involved Laurent polynomials.We define: , 0 .
It is clearly an affine embedding.
Proof.It is a direct consequence of Lemma 2.

Asymptotic iteration method for auxiliary equations
From Theorems 1, 2 and 3 we have that in order to characterize the spectral sets S (1,2p) , B 2,2p and S (2q,2p) we need to determine in which cases the auxiliary differential equations (A 1 ), (A 2 ) and (A 3 ) admit a polynomial solution.With this purpose we use an adapted version of the asymptotic iteration method or AIM, that was introduced by H. Ciftci et al in [4] and previously applied to a related problem in [8].

General considerations about the asymptotic iteration method
Let us consider Q{η, ζ} the ring of differential polynomials in two differential indeterminates η, ζ.Let us consider a differential equation, We also set: By derivation of Eq. ( 11) we obtain a sequence of differential equations, where {λ j (η, ζ)} j∈N and {s j (η, ζ)} j∈N are sequences of differential polynomials in η and ζ defined by the recurrence (cf.(21) in [8]), and a sequence of obstructions, Another way of looking at this sequence {∆ n } n∈N is to consider the recurrence law (13) as the iteration of a Q-linear operator in Q{η, ζ} 2 .Note that: where f and g are elements of K. Proof.For the first part of the proof we follow the same argument that [4, Theorem 2].Let us assume that Eq. ( 15) admits a polynomial solution P (x) = 0 of degree at most n.Note that from derivation of of Eq. ( 15) we obtain equations By taking y = P (x) we obtain: Now, let us assume ∆ n (f, g) = 0.The key point of the proof is that any homogeneous linear differential equation with coefficients in K of order r has a vector space of solutions of dimension r in a suitable extension of K. We have two different possibilities: (a) If λ n−1 (f, g) = 0 and s n−1 (f, g) = 0 then Eq. ( 16) yield y n+1 = 0.This implies that solutions of Eq. ( 1) are polynomials of degree at most n (b) In any other case then there is a unique h ∈ K such that hλ n−1 (f, g) = λ n (f, y) and hs n−1 (f, g) = s n (f, g).By substitution in Eq. ( 16) we obtain The general solution of this equation is (n + 2)-vector space over C containing the space of all polynomials of degree at most n as an hyperplane.
The space of solutions of Eq. ( 15) is a 2-subspace of such (n + 2)-vector space.It must intersect the hyperplane of polynomials at least along a line.Thus, it contains a polynomial of degree at most n.

Asymptotic iteration method for auxiliary equations (A 1 ), (A 2 ) and (A 3 )
Here we complete the integrability analysis of equation ( 1) by applying the asymptotic iteration method to auxiliary equations.We need to introduce the spectral varieties, some subsets of the spectral set that turn out to be algebraic varieties.
Definition 6.Let us call S(d±±) (2q,2p) to the set of (R(x), B(x), A(x)) ∈ M(2q,2p) such that the corresponding auxiliary equation (A 3 ) with s ∞ = ±1, s 0 = ±1 has a polynomial solution of degree d.Let us also set, By the above definitions we have that, for any (r, m) ∈ Z 2 + the spectral set decomposes as enumerable union of strata, By application of the asymptotic iteration method we will see that these strata S (d) (r,m) are algebraic varieties, that we call spectral varieties, and compute their equations.
Proof.It follows automatically from Theorem 1 by application of Theorem ( 4) to (A 1 ).Note that condition (a) implies that the degree of the polynomial solution of (A 1 ) is not strictly smaller than d.
Remark 2. Note that given a choice of the sign s ∞ then Therefore, the coefficients of are regular functions in M (1,2p) and, together with equation ±b p−1 = 2d + p + 2, determine the spectral variety S (d±) (1,2p) .Now, we apply the asymptotic iteration method to Eq. (A 2 ) in order to give necessary and sufficient conditions for the existence of a polynomial solution of degree d.Given L(x) ∈ M (2,2p) we consider its decomposition as in Eq. ( 4).
this determines the choice of the square root, that is, the sign s 0 , and therefore λ).
Proof.It follows automatically from Theorem 2 by application of Theorem 4 to Eq. (A 2 ).Note that condition (a) implies that the degree of the polynomial solution of Eq. (A 1 ) is not strictly smaller than d.
Remark 3. Note that given a choice of the sign s ∞ then Therefore, the coefficients of x are regular functions in M (1,2p) and, together with equation determine the spectral variety S (d±) (1,2p) .
Finally, we apply the asymptotic iteration method to Eq. (A 3 ).Given (R(x), B(x), A(x)) ∈ M(2q,2p) and a choice of the signs s ∞ = ±1 and s 0 = ±1 in Eq. (A 3 ) let us recall: Proof.It follows automatically from Theorem 3 by application of Theorem 4 to Eq. (A 2 ).Note that condition (a) implies that the degree of the polynomial solution of Eq. (A 1 ) is not strictly smaller than d.
Remark 4. Note that given a choice of the signs s 0 and s ∞ then The coefficients of ∆ d (ω(x), B(x), λ) as a Laurent polynomial in x are regular functions in M(2q,2p) and its zero locus is the spectral variety S(d±±) (2p,2q) .
Heun differential equations.Our approach recovers and extends the results of Duval et.al. in [2] about these equations, which in their normal form are particular cases of the following differential equation:

Biconfluent Heun equation
Consider the reduced form of biconfluent Heun equation, which is given by We see that Eq. ( 17) corresponds to Eq. ( 18) A galoisian analysis of Eq. ( 18) was developed by Duval et al. in [2], where the authors obtained some description of the spectral variety by treating the auxiliary equation by the method of undetermined coefficients.Using our approach we apply Theorem 4, after a coefficient decomposition as presented in Eq. ( 5).
Through our approach for Eq. ( 18), the AIM method help us to obtain in an explicit way the vanishing condition of the determinant provided in [2], which is equivalent to the vanishing of the universal polynomial ∆ d .We provide a criteria for the existence of the solution and we use d = 1, 2 to illustrate the method.Solutions are given in pairs and the Galois group is diagonalizable.
That is, we obtain the same universal differential polynomial for cases s ∞ = 1, s 0 = 1 and s ∞ = 1, s 0 = −1 as follows.
By the use of Asymptotic Iteration Method to Eq. ( 22) we get the conditions for the existence of such solutions.We compute these conditions for d = 1 and conditions for others small d are easily computable but extremely large.
We see that Eq. ( 17) corresponds to Eq. ( 25) when k = 1, a 2 = 1, a 1 = J, a 0 = a −1 = 0 and a −2 = 3 4 .We recall that Eq. (25) corresponds to the reduced form of biconfluent Heun equation (18) with α = ±2, β = J, γ = J 2 /4 and δ = 0. Applying theorem 2 we find out that a necessary condition to obtain On the other hand, sufficient condition comes from the polynomial solutions of the auxiliary equation where ω(z) = s ∞ (z + J 2 ).The existence of such polynomials solutions are guaranteed by the vanishing of the obstruction x .
Now, we can use our approach to apply Theorem 4 to Eq. ( 28), which corresponds to Eq. ( 17) being k = 3, By theorem 4 we obtain a necessary condition λ := −s 0 2 γ α 2 + 3 2 = −s∞4β− α method we search for polynomial solutions of the auxiliary equation where ω(z) = s ∞ z + s 0 In order to illustrate this process we compute the polynomial system which solutions are the suitable selection for α, β, γ and δ that vanishes ∆ z for each combination of signs s ∞ and s 0 : for s ∞ = 1 and s 0 = 1 we obtain the equation of the intersection of S (1++) (6,2) with the family (28) the solution of this system is the union of four algebraic curves: for s ∞ = 1 and s 0 = −1 we obtain the equation of the intersection of S with family (28) the solution of this system is the union of four algebraic curves: Further computation shows that for family Eq.( 18) the intersections with  (6,2) .This implies that two of the auxiliary equations have polynomial solutions simultaneously, there are two linearly independent solutions that are eigenvectors of the Galois group which is then diagonalizable.

Perturbed canonical equation
Another interesting example is the differential equation which was deeply analyzed in [8] for m = 0. Now, by application of theorem 2 we can determine an arithmetic condition on the parameters that needs to be fulfilled, that is This condition implies that we have four different cases to consider related to the selection of signs s 0 and s ∞ .Let us take for example case s 0 = s ∞ = 1, so we have the following auxiliary equation Through the change of variables L(z) = P d (x) and z = − 2x n+1 n+1 , the equation (36) turns into: (38) In general, the change of independent variable z = −s ∞ 2x n+1 n+1 transform the corresponding auxiliary equation into a generalized Laguerre equation.In the table 6 we summarize some relations of the parameters in equation (34) that must be satisfied in order to obtain integrability for each selection of signs.

Algebraically solvable potentials for Schrödinger equation
There is some interest in understanding Liouvillian solutions for Schrödinger equations with Mie type potentials that come from Supersymmetric Quantum Mechanics, Atomic and Molecular Physics, among others, see [5,13,6,14,7,15] among others.Algebraically solvable potentials correspond to those that have an infinite discrete spectrum and Liouvillian eigenfunctions.The following is an extension of Corollary 5.3 in [16], see also Corollary 2.2.3 in [17,18]. .Thus, we obtain that q = 0, otherwise we will have quasi-solvability in the potential (finite number of points in the algebraic curve in the spectral variety).
Thus, we conclude the proof.
Remark 6.We can notice that Eq. ( 17) not only include the reduced forms of biconfluent and doubly confluent Heun equations, also include some Schrödinger Equations with 3D algebraically solvable potentials and energy levels denoted by E such as follows: • Harmonic Oscillator, see , is relevant to complex oscillation theory.
)dx where P (x) is a monic polynomial of degree d, and ω(x) is a Laurent polynomial.(b) If L(x) ∈ S (d) (r,m) with r = 2 then the differential equation (1) has a solution of the form: (a) If (r, m) ∈ C 1 then the Galois group is conjugated to a subgroup of B 2 .(b) If (r, m) ∈ C 2 then the Galois group is conjugated to a subgroup of D ∞ .(c) If (r, m) ∈ C 3 then the Galois group is conjugated to a subgroup of B 2 or

2. 4 .
Characterization of S (2,m) in terms of B (2,2m+2) Let us take differential equation (1) with L(x) ∈ S (2,m) .It implies that the associated Riccati equation, u = y ′ y , u ′ = u 2 + L(x) has an algebraic solution u(x) of degree 1 (case B (2,m) ) or 2 (case D (2,m) ).Let us assume that L(x) ∈ D (2,m) .Now we have a look on the geometric properties of the solutions of the Riccati equation.It is well known (see [9] Proposition VIII.1.1)that given an initial condition (x 0 , u 0 ) ∈ C * × C and any simply connected open subset U of C * there is a solution u with u(x 0 ) = u 0 and meromorphic in U.This implies that the solutions of the Riccati equation may have poles but not ramification points outside of the singular locus {0, ∞}.If we compose the algebraic solution with the ramified cover, C → C, w → x = w 2 we obtain that the two determinations of the algebraic function u split in two rational functions of w with poles only in 0 and ∞.In other words, the Riccati equation has a solution which is a rational function on √ x.

Theorem 4 .
Differential equation (15) has a polynomial solution in C[x] of degree at most n if and only if ∆ n (f, g) = 0.

( 1 ,Definition 4 .
2p) to the set of L(x) ∈ M (1,2p) such that the corresponding auxiliary equation (A 1 ) with s ∞ = ±1 has a polynomial solution of degree d.Let us also set, Let us call B (d±) (2,2p) to the set of L(x) ∈ M (2,2p) such that the corresponding auxiliary equation (A 2 ) with s ∞ = ±1 and some choice of s 0 has a polynomial solution of degree d.Let us also set,

S 2 )
coincide.The same happens for the intersections with S

Corollary 1 .
Assume that V (x) ∈ M (r,m) is an algebraically solvable potential with m > 0.Then, m = 2 and r ∈ {0, 1, 2}.Proof.It follows directly that:• (r, m) = (2, 1) by applying of part (a) in Theorem 2, or • (r, m) = (2, 2) by applying of Theorem 3, part (a) due to the relation with E or part (b) due to it is valid ∀ℓ ∈ Z, or • (r, m) = (2, 0) by the applying of part (a) due to the relation with E and part (b) in Theorem 4

Table 1 :
First universal differential polynomials ∆ d

Table 4 :
Obstructions ∆ d for the case s∞ = 1

Table 6 :
Relations between parameters in equation (34), solutions y n,d .