Necessary Conditions for Liouvillian Solutions of (Third Order) Linear Differential Equations^{¶}

Michael F. Singer^{∗} and Felix Ulmer^{∗∗}^{††}

North Carolina State University Department of Mathematics Box 8205 Raleigh, N.C. 27695-8205

October 16, 1993

Abstract

In this paper we show how group theoretic information can be used to derive a set of necessary conditions on the coefficients of L(y) for L(y) = 0 to have a liouvillian solution. The method is used to derive (and improve in one case) the necessary conditions of the Kovacic algorithm and to derive an explicit set of necessary conditions for third order differential equations.

1 Introduction

In our previous work [20], [21], we have shown how group theoretic techniques can be used to develop effective algorithms to calculate Galois groups of second and third order homogeneous linear differential equations and to decide questions about the algebraic nature of the solutions of such equations (e.g., solvability in terms of liouvillian functions or in terms of linear differential equations of lower order).

¶A weaker version of these results were announcened in Liouvillian Solutions of Third Order Linear Differential Equations: New Bounds and Necessary Conditions, Proceedings of the 1992 International Symposium on Symbolic and Algebraic Computation, ACM Press

∗Partially supported by NSF Grant 90-24624 ∗∗Partially supported by Deutsche Forschungsgemeinschaft , while on leave from Universität

Karlsruhe. This paper was written during the two year visit of the second author at North Carolina State University. The second author would like to thank North Carolina State University for its hospitality and support during the preparation of this paper.

††Present Address of second author: Cornell University; Dept. of Math.; White Hall; Ithaca, NY 14853-7901

In [12], Kovacic gave an algorithm to decide if a second order homogeneous linear differential equation has liouvillian solutions. In the process of doing this, Kovacic derived a very strong set of necessary conditions for the existence of such solutions. In this paper, we show how one can derive these conditions from general group theoretic properties. We also derive similar conditions for the third order case.

We shall assume that the reader is familiar with the elementary concepts of differential Galois theory and elementary properties of linear operators. The necessary facts are reviewed in [20]. We shall need results from [21], but these will be stated in full for the convenience of the reader.

The rest of the paper is organized as follows. In section 2, we give some elementary results concerning exponents and monodromy groups. Section 3 contains the necessary conditions. The last section deals with linear differential equations having 3 singular points and examples.

We wish to thank the referees for pointing out a mistake in the original proof of Theorem 3.5 and for suggestions that allowed us to sharpen our original version of the necessary conditions for case 2.

2 Exponents and Monodromy Groups

In this section we review some facts about formal and analytic solutions of linear differential equations. There is a well developed general theory of formal solutions at any point (of both systems and single equations) as well as an asymptotic theory (c.f. [1, 5, 14]). We have restricted ourselves to the most basic facts in order to give an elementary and direct exposition of the results that we need.

In what follows we will frequently wish to consider solutions of a linear differential equation that can be expressed as a series in (not necessarily integral) powers of x. This motivates the following discussion.

Let

L(y) = a_{n}(x)y^{(n)} + a_{n−1}(x)y^{(n−1)} + … + a_{0}(x)y

be a linear differential equation with coefficients in the field C((x)) of formal Laurent series and let a_{i}(x) = ∑_{j≥i_{0}} a_{i,j}x^{j} where a_{i,i_{0}} ≠ 0. If L(y) = 0 has a solution of the form y = x^{ρ} ∑_{j≥0} c_{j}x^{j}, c_{0} ≠ 0, then formally substituting this expression into L(y) = 0 and examining the coefficient of the smallest power of x, one sees that ρ will satisfy the equation

P (ρ) = ∑ a_{k,k_{0}}(ρ)_{k} = 0

where (ρ)_{j} = ρ(ρ − 1) … (ρ − (j − 1)) and the sum is over all k with k_{0} − k = min_{0≤i≤n}{i_{0} − i}. P (ρ) is called the indicial equation of L(y) (at 0). We shall refer to the roots of the indicial polynomial as exponents of L(y) (at 0).

If L(y) has coefficients in C(x), then for any c ∈ C we can expand these coefficients in powers of (x − c). Analogously, we can define the indicial equation at c and exponents at c. Via the transformation x = 1/t, \frac{d}{dx} = −t^{2} \frac{d}{dt} we can also define the indicial equation and exponents at infinity by considering the point t = 0 of the transformed equation. We say that x = c is an ordinary point if, for 0 ≤ i ≤ n, a_{i}(x)/a_{n}(x) is analytic at x = c (for c = ∞ this property is related to the point t = 0 of the transformed equation). The following lemma contains facts that are easily verified by computation (cf. [5], Ch. 4,5; [14], Ch. IV,V).

Lemma 2.1 Let L(y) = 0 have coefficients in C(x) and let c ∈ C.

1. If L(y) = 0 has a formal solution of the form

y = (x − c)^{ρ} ∑_{i≥0} c_{i}(x − c)^{r_{i}}, r_{i} ∈ R, 0 = r_{0} < r_{1} < …, c_{0} ≠ 0,

then ρ is an exponent of L(y) at c.

2. The degree of P (ρ) is at most n.

3. If c is an ordinary point then the exponents at c are {0, …, n − 1}.

The converse of part 1 of the above lemma can be false in the case when log terms occur in formal solutions. For example, L(y) = x^{2}y′′ − (3 − 2x)xy′ + (3 − 6x)y has exponents {1, 3} at 0 and at this point it has a fundamental set of solutions of the form {x^{3}, −4x^{3} ln x + xϕ(x)}, where ϕ(x) is a power series in x with ϕ(0) ≠ 0. No linear combination of these two solutions can be of the form xψ(x) where ψ(x) is a power series with ψ(0) ≠ 0.

If a point c is not an ordinary point, it is called a singular point. If, in addition, the indicial polynomial at this point has degree n, we say that the point is a regular singular point. An obviously equivalent way of saying this is that the order of the pole of a_{n−i}(x)/a_{n}(x) at c is at most i. A linear differential equation with only regular singular points (including ∞) is called a fuchsian linear differential equation. We gather some facts about regular singular points in the following lemma, but first give the following definition:

Definition 2.1 Let {y_{1}, …, y_{n}} be a fundamental set of solutions of the linear differential equation L(y) = 0. The m^{th} symmetric power L^{◯ s m}(y) of L(y) is the differential equation of smallest order, such that the solution space of L^{◯ s m}(y) = 0 is spanned by {y_{1}^{m}, y_{1}^{m−1}y_{2}, …, y_{n}^{m}}.

Note that this definition is independent of the choice of fundamental solutions and that L^{◯ s m}(y) is the linear differential equation of smallest order satisified by all homogeneous forms F (y_{1}, …, y_{n}) of degree m with constant coefficients in solutions y_{1}, …, y_{n} of L(y) = 0. See [17] for a further discussion of this concept. An algorithm to construct the equation L^{◯ s m}(y) is given in [17] and [20], Section 3.2.2.

Lemma 2.2 Let L(y) = 0 be of order n and have coefficients in C(x) and let c ∈ C.

1. If c is a regular singular point and ρ is a root of multiplicity r of the indicial equation, then L(y) = 0 has r independent solutions of the form

y_{i} = (x − c)^{ρ} (∑_{j=0}^{i} ϕ_{i,j}(x) (log(x − c))^{j})  (1)

where i = 0, …, r−1, the ϕ_{i,j}(x) are analytic at c and for each i there is some j such that ϕ_{i,j}(c) ≠ 0. Furthermore, if ρ + N is not an exponent of L(y) = 0 for any positive N ∈ Z, then L(y) = 0 has a solution of the form (x−c)^{ρ}φ(x), where φ(x) is analytic at c and φ(c) ≠ 0.

2. c is a regular singular point if and only if in any open angular sector Ω at c, all solutions of L(y) = 0 analytic in Ω, satisfy lim_{x→c}(x − c)^{N}y = 0 for some N ≥ 0.

3. If all solutions of L(y) = 0 are algebraic over C(x), then L(y) = 0 is a fuchsian equation and at any singular point c there are n distinct rational exponents. Furthermore, at any point there is a fundamental set of solutions of the form y_{i} = (x − c)^{a_{i}}ϕ_{i}(x) where ϕ_{i}(x) is analytic at c, ϕ_{i}(c) ≠ 0, and a_{1}, …, a_{n} are the exponents.

4. Let L(y) = 0 be a fuchsian equation and a_{1}, …, a_{n} exponents at some point c. Then the m^{th} symmetric power L^{◯ s m}(y) is fuchsian and the exponents at c are in the set {(∑_{j=1}^{m} a_{i_{j}} ) + t | i_{j} ∈ {1, …, n}; t ∈ Z; t ≥ 0}.

Proof: 1. and 2. are proved in [5], Ch. 5, §1-5 and [14] Ch. V, §16,17. Lemma 2.2.2 is usually refered to as Fuchs’ Criterion (cf. [5], p. 124 and [14], pp. 65-68). Lemma 2.2.2 implies that if all solutions are algebraic, the equation must be fuchsian. To prove 3., let c = 0 and let y_{1}, …, y_{n} be solutions of L(y) = 0 as described in 1. Since these are algebraic over C(x), they are also algebraic over C((x)). Let k be the differential field C((x))(x^{ρ_{1}}, …, x^{ρ_{n}}), where ρ_{1}, …, ρ_{n} are the exponents at 0. We claim that for each solution y_{i} of the above form, we must have ϕ_{i,j} = 0 for j = 1, …, i. If not, then log x would be algebraic over k. The Kolchin-Ostrowski Theorem ([11]) implies that log x would be in C((x)). Since there is no element y in C((x)) such that y′/y = 1/x, we have a contradiction. Therefore each y_{i} is of the form x^{ρ_{i}}ϕ_{i}(x), for some ϕ_{i}(x), ϕ_{i}(0) ≠ 0 in C((x)). Since each y_{i} is algebraic over C((x)), we have that each ρ_{i} is rational. We now claim that all the exponents of L(y) are distinct. Consider the set of bases B = (y_{1}, …, y_{n}) where each y_{i} = x^{ρ_{i}}ϕ_{i}(x), ϕ_{i}(0) ≠ 0 where ρ_{1} ≤ … ≤ ρ_{n} and define an order on this set by letting B < B̃ if (ρ_{1}, …, ρ_{n}) < (ρ̃_{1}, …, ρ̃_{n}) in the lexicographical order. Let B̂ be a maximal element of this set, then for this set the ρ̂_{i} are distinct. This will show that there are n distinct exponents. Assume not and let j be the smallest index so that all ρ̂_{i} with i > j are distinct. We then have that ρ̂_{j} = ρ̂_{j+1}. Therefore, there is a constant d such that z = y_{j} − dy_{j+1} = x^{ρ}ψ(x) for some ρ > ρ̂_{j}. Replacing y_{j+1} by z gives a new basis which (after we perhaps rearrange the elements) is larger in the ordering. This contradicts the maximality of B̂.

To prove 4., consider a basis of the solution space of L(y) = 0 of the form y_{i} = (x − c)^{a_{i}} (∑_{j=0}^{k_{i}} ϕ_{i,j}(x) (log(x − c))^{j}) where the ϕ_{i,j}(x) are analytic at c, some ϕ_{i,j}(c) ≠ 0, and a_{1}, …, a_{n} are the exponents of L(y) = 0 at c. Note that L^{◯ s m}(y) has a basis of solutions of the form y_{1}^{i_{1}}y_{2}^{i_{2}} ⋯ y_{n}^{i_{n}}, where i_{1} + … + i_{n} = m. The second assertion of the Lemma implies that c is at worst a regular singular point of L^{◯ s m}(y). Let ρ be an exponent of L^{◯ s m}(y) at c. From 1. we get that L^{◯ s m}(y) = 0 has a solution of the form y = (x − c)^{ρ} (∑_{r=0}^{h} ψ_{r}(x) (log(x − c))^{r}), where ψ_{r}(x) is analytic at c, and some ψ_{r}(c) ≠ 0. Therefore there exist constants c_{I} such that

y = ∑_{I=(i_{1},... ,i_{n})} c_{I}y_{1}^{i_{1}} · . . . · y_{n}^{i_{n}}

= ∑_{I=(i_{1},... ,i_{n})} c_{I}(x − c)^{a_{1}i_{1}+...+a_{n}i_{n}} (∑_{j=0}^{k} φ_{I,j}(x) (log(x − c))^{j})

where the sum is over all I = (i_{1}, …, i_{n}) with i_{1} + … + i_{n} = m, the φ_{I,j}(x) are analytic at c and for some (I, j), φ_{I,j}(c) ≠ 0. Since the φ_{I,j}(x) contain only nonnegative powers of x, comparing the lowest power of log(x) whose coefficient is not zero and taking into account possible cancelation, yields the result.

For non fuchsian equations very little can be said regarding the relationship between the exponents of a linear differential equation and those of its symmetric power. The following example shows that one can have a linear differential equation with no exponents at 0 while the symmetric square has an exponent at 0. Let

L(y) = y′′ − \frac{2}{x(x − 2)}y′ + \frac{2x^{2} − 3x + 2}{x^{4}(x − 2)}y

This equation has a basis of solutions {x exp(1/x), exp(−1/x)} and no exponents at 0. The second symmetric power is

y′′′ − \frac{6}{x(x − 2)}y′′ + \frac{4(3x^{3} − 6x^{2} + 8x − 4)}{x^{4}(x − 2)^{2}}y′ − \frac{4(3x^{3} − 6x^{2} + 8x − 4)}{x^{5}(x − 2)^{2}}y

and this has 1 as an exponent since x is a solution.

We do note that the theory of formal solutions at a point does allow one to give apriori bounds on the exponents of the symmetric power even in the non-fuchsian case (c.f. [5], Ch. 5, Sec. 2 and [1, 19]). This involves calculating formal exponents, a task that is more difficult than calculating exponents defined above. Since this will not be needed, we have stated Lemma 2.2.4 just for the fuchsian case.

Consider the class of linear differential equation with coefficients in C(x). For these equations we can use analytic considerations to define a group called the monodromy group that is a subgroup of the differential Galois group G(L) of L(y) = 0 over C(x) (see e.g. [10, 20]). Let c_{1}, …, c_{m} be the singular points of L(y) = 0 (including infinity if it is a singular point) and let c_{0} be an ordinary point of the equation. We consider these points as lying on the Riemann Sphere S^{2}. Let {y_{1}, …, y_{n}} be a fundamental set of solutions of L(y) = 0 analytic at c_{0} and let γ be a closed path in S^{2} − {c_{1}, …, c_{m}} that begins and ends at c_{0}. One can analytically continue {y_{1}, …, y_{n}} along γ and get new fundamental solutions {ỹ_{1}, …, ỹ_{n}} analytic at c_{0}. These two sets must be related via (ỹ_{1}, …, ỹ_{n})^{T} = M_{γ}(y_{1}, …, y_{n})^{T} where M_{γ} ∈ GL(n, C). One can show that M_{γ} depends only on the homotopy class of γ and that the map γ ↦ M_{γ} defines a group homomorphism from π_{1}(S^{2} −{c_{1}, …, c_{m}}) to GL(n, C). The image of this map depends on the choice of c_{0} and {y_{1}, …, y_{n}} but is unique up to conjugacy and is called the monodromy group of L(y). In general the image of this group will be a proper subgroup of G(L), but when L(y) is fuchsian, the Zariski closure of this group will be the full differential Galois group G(L) (c.f. [23]). In particular if G(L) is finite (i.e., all solutions of L(y) = 0 are algebraic) then the map is surjective and the monodromy and Galois groups coincide. The fact that the monodromy group is a subgroup of the differential Galois group allows us to prove the following:

Lemma 2.3 Let L(y) = 0 be a linear differential equation with coefficients in C(x), assume all solutions of L(y) = 0 are algebraic over C(x). If a is an exponent at any point c, then Na ∈ Z for some integer N that is the order of an element of G(L).

Proof: Lemma 2.2 implies that there is a fundamental set of solutions of the form y_{i} = (x − c)^{a_{i}}ϕ_{i}(x) where ϕ_{i}(x) is analytic at c, ϕ_{i}(c) ≠ 0, and a_{1}, …, a_{n} are the exponents. If we continue analytically along a small loop around c, we get the solutions ỹ_{i} = e^{(2π\sqrt{−1})a_{i}}(x − c)^{a_{i}}ϕ_{i}(x). Therefore the diagonal matrix g = diag(e^{(2π\sqrt{−1})a_{1}}, …, e^{(2π\sqrt{−1})a_{n}}) is conjugate to an element of the differential Galois group. Since g^{N} = 1, we have Na_{i} ∈ Z for i = 1, …, n.

The relationship between the exponents of L(y) at c and the eigenvalues of the monodromy matrix M_{γ} corresponding to a simple loop γ around c, containing no other singular point, is given by the following:

Lemma 2.4 Let L(y) = 0 be a linear differential equation with coefficients in C(x), c a regular singular point of L(y) = 0 and a_{1}, …, a_{n} the exponents at c. The eigenvalues of the monodromy matrix M_{γ}, corresponding to a loop γ around c are e^{2π\sqrt{−1}a_{1}}, …, e^{2π\sqrt{−1}a_{n}}.

Proof: For convenience let c = 0. Separate the exponents into different sets S_{1}, …, S_{r}, such that the elements of each set differ by integers, and elements of different sets differ by non-integers. Let m_{j} = |S_{j}| and n = m_{1} + … + m_{r}. For each set S_{j}, we denote by a_{j} an element with the property that for any integer m > 1, a_{j} + m ∉ S_{j}. From Lemma 2.2 we get that there exists a solution of the form x^{a_{j}}ϕ_{j}(x), where ϕ_{j}(x) is analytic at 0 and ϕ_{j}(0) ≠ 0. Therefore λ_{j} = e^{2π\sqrt{−1}a_{j}} is an eigenvalue of the monodromy matrix. We claim that each λ_{j} is an eigenvalue of multiplicity at least m_{j}. Since n = m_{1}+…+m_{r}, this will imply that λ_{j} is of multiplicity m_{j}. Since for any a ∈ S_{j}, λ_{j} = e^{2π\sqrt{−1}a}, we will have established the conclusion of the Lemma. To prove the claim, let a_{1}, …, a_{t} be the distinct elements of S_{j} and assume that a_{i} ∈ S_{j} is a root of multiplicity s_{i} of the indicial polynomial. For each a_{i} there are s_{i} independent solutions of the form (1). If y = x^{a_{i}} (∑_{j=0}^{h} ϕ_{j}(x)(log(x))^{j})

with 0 ≤ h < s_{i} is one of these solutions, then (M_{γ} − e^{2π\sqrt{−1}a_{i}}Id)(y)

= e^{2π\sqrt{−1}a_{i}}x^{a_{i}} (∑_{j=0}^{h} ϕ_{j}(x)(log(x) + 2π\sqrt{−1})^{j}) − e^{2π\sqrt{−1}a_{i}}x^{a_{i}} (∑_{j=0}^{h} ϕ_{j}(x)(log(x))^{j})

= x^{a_{i}} (∑_{j=0}^{h−1} ψ_{j}(x)(log(x))^{j}) ,

where the ψ_{j}(x) are analytic at 0. This shows that (M_{γ} − e^{2π\sqrt{−1}a_{i}}Id)^{s_{i}−1}(y) = x^{a_{i}}ψ(x), where ψ(x) is analytic at 0, and thus we get (M_{γ} − e^{2π\sqrt{−1}a_{i}}Id)^{s_{i}}(y) = 0. Note that for distinct a_{i} we get independent solutions of the form (1). Therefore the generalized eigenspace corresponding to λ_{j} = e^{2π\sqrt{−1}a_{j}} has dimension at least s_{1} + … + s_{t} = m_{j}.

3 Necessary Conditions

In this section we show how the results of [21] can be used to derive strong necessary conditions for a third order linear differential equation with coefficients in C(x) to have a solution liouvillian over C(x) (see e.g. [10, 11, 12, 18, 20, 21] for definitions). We will also derive the necessary conditions of the Kovacic algorithm using our approach, and improve them.

In practice, one is given a linear differential equation L(y) = 0 whose coefficients lie in F (x), where F is a finitely generated extension of Q. In order to do our calcutations, we need to assume that one has algorithms to perform the field operations and to factor polynomials over F. When this is the case, it is known that the same is true for the algebraic closure \overline{F} of F ([7]). Although we give necessary conditions for certain behavior over C(x), it is clear from our procedures, that all calculations can be done in \overline{F} (x). We have not addressed the issue of finding the most efficient way to calculate in this field. In particular, we assume that all polynomials over the field F can be factored into linear factors whenever needed (for example, to consider individually each singular point of the differential equation). Reducing the need to factor (and work in algebraic extensions) would increase the efficiency of our procedures. Techniques for attacking this problem are discussed in [2, 3].

We shall continue using the philosophy of [20] and [21], that is, to distinguish 3 different cases (for second order differential equations these correspond to the first 3 cases of the Kovacic algorithm). If a third order equation L(y) = 0 has a liouvillian solution, then one of the following holds (cf. [20]):

1. The differential Galois group is a reducible subgroup of SL(3, C) (cf. case 1 in the Kovacic algorithm [12]). In this case the computation of the liouvillian solutions of L(y) = 0 can be reduced by factorisation to the computation of the liouvillian solutions of some second order linear differential equation over C(x) (cf. [21], Section 2).

2. The differential Galois group is an imprimitive subgroup of SL(3, C) (cf. case 2 in the Kovacic algorithm). In this case all solutions of the third order equation L(y) = 0 are Liouvillian, and L(y) = 0 has a fundamental system of solutions {y_{1}, y_{2}, y_{3}} whose logarithmic derivatives y′_{i}/y_{i} are algebraic of degree 3 over C(x) (cf. [21], Section 3).

3. The differential Galois group is a finite primitive subgroup of SL(3, C) (cf. case 3 in the Kovacic algorithm). In this case all solutions of L(y) = 0 are algebraic over C(x) (cf. [21], Section 4).

If none of the above cases holds, then the differential Galois group is an infinite primitive subgroup of SL(3, C) and L(y) = 0 has no liouvillian solutions. Case 1 must be tested first, since this case is assumed to not hold in the other cases. (Note that the above classification can be used for differential equations of arbitrary order.)

In the following, we will assume that the differential Galois group of L(y) = 0 is unimodular (i.e. G(L) ⊆ SL(3, C)), or equivalently (cf. [10], Ch. VI, Sec. 24) that L(y) = 0 is of the form

L(y) = y′′′ + Ay′′ + By′ + Cy = 0, A, B, C ∈ C(x),

where A is of the form ∑_{i} \frac{n_{i}}{x − α_{i}}, with n_{i} ∈ Z and α_{i} ∈ C. If G(L) is not unimod-

ular, we can always transform the equation by a change of variable to an equation where A = 0 (cf. [10], Ch. VI, Sec. 24). The new differential equation has liouvillian solutions if and only if the original differential equation does.

3.1 Reducible Case

In this section we give conditions on the coefficients of L(y) = 0 for this case to occur.

3.1.1 First step

We start by showing that for a given differential equation L(y) = 0 of arbitrary order n and coefficients in C(x), all solutions of the form P (x) ∏_{i}(x − c_{i})^{a_{i}}, where P (x) ∈ C[x], c_{i} ≠ ∞ are singular points of L(y) = 0 and a_{i} are exponents of L(y) = 0 at c_{i}, can be computed by only finding polynomial solutions of some n-th order linear differential equations with coefficients in C(x).

We first note the following necessary condition:

Lemma 3.1 (cf. [16], §178) If a linear differential equation L(y) = 0 of degree n has a solution of the form P (x) ∏_{i}(x−c_{i})^{a_{i}}, where P (x) ∈ C[x], c_{i} ≠ ∞ are the singular points of L(y) = 0, a_{i} are exponents at c_{i} and P (x) is a polynomial, then there is an exponent e_{∞} at ∞ such that the sum (∑_{i} a_{i}) + e_{∞} is a non-positive integer. In particular, if L(y) = 0 has no singularities other than ∞ and has such a solution, then it must have a non-positive integer exponent at ∞.

Proof: Each a_{i} is an exponent at the singular point c_{i}. Expanding P (x) ∏_{i}(x−c_{i})^{a_{i}} in increasing powers of x^{−1}, we see that −deg(P) − ∑_{i} a_{i} is the exponent of the leading term. Therefore this must be an exponent e_{∞} at infinity.

Lemma 3.2 Let c_{i}, a_{i} ∈ C and y an m-times differentiable function. then

(y ∏_{i}(x − c_{i})^{a_{i}})^{(m)} = (∑_{j=0}^{m} (q_{j}(x) y^{(j)})) ∏_{i}(x − c_{i})^{a_{i}},

where q_{j}(x) ∈ C(x).

Proof: The result is true for m = 0. For 1 ≤ m we get:

(y ∏_{i}(x − c_{i})^{a_{i}})^{(m)}

= (y′ ∏_{i}(x − c_{i})^{a_{i}})^{(m−1)} + (y ∑_{i} (a_{i}(x − c_{i})^{a_{i}−1} ∏_{j≠i}(x − c_{i})^{a_{i}} ))^{(m−1)}

= (y′ ∏_{i}(x − c_{i})^{a_{i}})^{(m−1)} + ((y ∑_{i} \frac{a_{i}}{x − c_{i}} ) ∏_{i}(x − c_{i})^{a_{i}})^{(m−1)}

Now the result follows by induction by using y′ and y ∑_{i} \frac{a_{i}}{x−c_{i}} instead of y.

For a given equation L(y) = 0 of degree n, there are only a finite number of singularities and at each singularity c_{i}, there are at most n possible exponents a_{i}. If L(y) = 0 has a solution of the form P (x) ∏_{i}(x − c_{i})^{a_{i}}, where c_{i} ≠ ∞ are singular points of L(y) = 0 and a_{i} are exponents of L(y) = 0 at c_{i}, then there are at most a finite number of possibilities for ∏_{i}(x − c_{i})^{a_{i}}. For each possible term ∏_{i}(x − c_{i})^{a_{i}}, we consider the differential equation:

L̃(y) := \frac{L (y ∏_{i}(x − c_{i})^{a_{i}})}{∏_{i}(x − c_{i})^{a_{i}}} = 0.

From Lemma 3.2 we get that the coefficients of L̃(y) belong to C(x). If L(y) = 0 has a solution of the form P (x) ∏_{i}(x − c_{i})^{a_{i}}, then L̃(y) = 0 has a solution P (x) ∈ C[x]. We thus have to compute a basis {p_{1}(x), ⋯, p_{k}(x)} (k ≤ n) of the polynomial solutions of L̃(y) = 0 to get a basis of the solutions of the form P (x) ∏_{i}(x − c_{i})^{a_{i}}.

3.1.2 Necessary conditions for case 1

If the differential Galois group of a differential equation L(y) = 0 of order n is a reducible subgroup of GL(n, C), then L(y) factors as a differential operator and algorithms performing such a factorisation are known (cf. [20], Section 3.2.1). For a third order equation, this leads either to a right factor of order one or to a left factor of order one. Since a left factor of order one leads to a right factor of order one of the adjoint differential operator L^{∗}(y) (cf. [14], p. 38), testing reducibility leads to the computation of a right factor of order one of L(y) or L^{∗}(y). If an equation has a right factor of order one, then it has a solution y whose logarithmic derivative u = y′/y is rational. If L(y) = 0 is of fuchsian type, then any solution whose logarithmic derivative is rational must be of the form P (x) ∏_{i}(x−c_{i})^{a_{i}}, where c_{i} ≠ ∞ are singular points of L(y) = 0 and a_{i} are exponents of L(y) = 0 at c_{i} (cf. [16], §178). Using Lemma 3.1 and the fact that the adjoint of a differential equation of fuchsian type is also of fuchsian type (This follows from the facts that the solutions of L^{∗}(y) = 0 are contained in a Picard-Vessiot extension K associated with L(y) = 0 (cf. [5], p. 101, Ex. 19; [14], p. 43, Ex. 12 and p. 38) and any element of K must have the growth properties described in Lemma 2.2.2) we get:

Corollary 3.3 Let L(y) = 0 be a third order differential equation which is of fuchsian type. If L(y) = 0 is reducible, then for either L(y) = 0 or L^{∗}(y) = 0 at each finite singular point c_{i} there are exponents a_{i} such that for some exponent e_{∞} at ∞, the sum (∑_{i} a_{i}) + e_{∞} is a non-positive integer.

We now consider the non fuchsian case. The riccati equation associated with L(y) = y′′′ + Ay′′ + By′ + Cy = 0 is

R(u) = u′′ + 3uu′ + Au′ + u^{3} + Au^{2} + Bu + C = 0.

and the adjoint L^{∗}(y) = y′′′ − Ay′′ + (B − 2A′)y′ + (−C + B′ − A′′)y of L(y) has riccati equation:

R^{∗}(u) = u′′ + 3uu′ − Au′ + u^{3} − Au^{2} + (B − 2A′)u + (−C + B′ − A′′) = 0.

In our computations we will usually assume that the finite singularity is at 0. For the Laurent series at 0 or ∞ we introduce the following notation:

A = αx^{a} + ⋯ (higher order terms)
B = βx^{b} + ⋯ (higher order terms)
C = γx^{c} + ⋯ (higher order terms)
A = α_{∞}x^{a_{∞}} + ⋯ (lower order terms)
B = β_{∞}x^{b_{∞}} + ⋯ (lower order terms)
C = γ_{∞}x^{c_{∞}} + ⋯ (lower order terms)

For A = A_{1}/A_{2} where A_{i} ∈ C[x], a_{∞} denotes deg_{x}(A_{1}) − deg_{x}(A_{2}).

If a Puiseux or Laurent series of a solution y of L(y) = 0 or a solution u of R(u) = 0 exists at the point 0 or ∞, we denote them by:

y = ρx^{r} + ⋯ (higher order terms)
u = ηx^{h} + ⋯ (higher order terms)
y = ρ_{∞}x^{r_{∞}} + ⋯ (lower order terms)
u = η_{∞}x^{h_{∞}} + ⋯ (lower order terms)

Necessary conditions for case 1 Let L(y) = y′′′ + Ay′′ + By′ + Cy be a third order linear differential equation with coefficients in C(x) such that L(y) = 0 and the adjoint L^{∗}(y) = 0 have no solutions of the form P (x) ∏_{i}(x − c_{i})^{a_{i}}, where P (x) ∈ C[x], c_{i} ≠ ∞ are singular points of L(y) = 0 and a_{i} are exponents of L(y) = 0 at c_{i}. If L(y) is reducible over C(x), then one of the following holds:

1. C = 0 and

L(y) = (\frac{d^{2}}{dx^{2}} + A\frac{d}{dx} + B) ((\frac{d}{dx}) (y))

2. C ≠ 0 and at some finite singular point of L(y) = 0 the coefficients A, B and C have exponents a, b and c such that one of the following holds:

(a) A ≠ 0, a ≤ −2, 3a ≤ c and (if B ≠ 0 then 2a ≤ b).

(b) B ≠ 0, b ≤ −4, b ∈ 2Z, b ≤ \frac{2}{3}c and (if A ≠ 0 then b ≤ 2a).

(c) c ≤ −6, c ∈ 3Z, (if A ≠ 0 then c ≤ 3a) and (if B ≠ 0 then c ≤ \frac{3}{2}b).

(d) AB ≠ 0, a < −2, b < −4, b − a < −1, 2a < b and 2b − a ≤ c.

(e) A ≠ 0, a < −2, c < −6, c − a < −4, (c − a) ∈ 2Z, 3a < c and (if B ≠ 0 then a + c ≤ 2b).

(f) B ≠ 0, c < −6, b < −4, c ≤ b−2, 3b < 2c and (if A ≠ 0 then 2b < a+c).

3. C ≠ 0 and for A, B and C, one of the following holds:

(a) A ≠ 0, a_{∞} ≥ 0, 3a_{∞} ≥ c_{∞} and (if B ≠ 0 then 2a_{∞} ≥ b_{∞}).

(b) B ≠ 0, 0 ≤ b_{∞}, b_{∞} ∈ 2Z, b_{∞} ≥ \frac{2}{3}c_{∞} and (if A ≠ 0 then b_{∞} ≥ 2a_{∞}).

(c) 0 ≤ c_{∞} ∈ 3Z, (if A ≠ 0 then c_{∞} ≥ 3a_{∞}) and (if B ≠ 0 then c_{∞} ≥ \frac{3}{2}b_{∞}).

(d) AB ≠ 0, 0 < a_{∞}, a_{∞} ≤ b_{∞} and 2b_{∞} − a_{∞} ≥ c_{∞}.

(e) A ≠ 0, 0 ≤ c_{∞} − a_{∞}, c_{∞} − a_{∞} ∈ 2Z, 0 < c_{∞} < 3a_{∞} and (if B ≠ 0 then a_{∞} + c_{∞} ≥ 2b_{∞}).

(f) B ≠ 0, 0 < b_{∞} ≤ c_{∞}, 3b_{∞} > 2c_{∞} and (if A ≠ 0 then 2b_{∞} > a_{∞} + c_{∞}).

Example: We use the above necessary conditions to show that for the following equation (which is the second symmetric power of the Airy equation)

L(y) = \frac{d^{3}y}{dx^{3}} − 4x\frac{dy}{dx} − 2y,

case 1 (of a reducible group G(L) ⊆ SL(3, C)) cannot occur. The point ∞ is the only singular point of L(y), and the only exponent at ∞ is 1/2. From Lemma 3.1 we get that L(y) = 0 has no solution of the form P (x) ∏_{i}(x − c_{i})^{a_{i}}. The first condition above cannot hold, since the coefficient of y is not 0. Since L(y) = 0 has no finite singular point, the above second condition does not hold. Finally, for A = 0, b_{∞} = 1 and c_{∞} = 0 the third condition above does not hold. Thus, since L(y) is selfadjoint, L(y) = 0 is an irreducible equation.

Proof of the Necessary conditions for case 1: We first note that if C = 0, then L(y) = 0 has a right factor of order one of the form stated. We assume from now on that C ≠ 0. If L(y) = 0 is a reducible third order differential equation, then either L(y) or L^{∗}(y) has a right factor of order 1 ([20]).

1. First assume L(y) = L_{1}(L_{2}(y)), where L_{2}(y) is of order 1, then R(u) has a solution u ∈ C(x).

(a) If u has a pole of order bigger than 1 then u must have a pole of order > 1 at some singularity of L(y) = 0 which we assume to be 0. We have u = ηx^{h} + … (higher order terms) and h < −1. Plugging u into R(u) = 0 we get:

(ηh(h − 1)x^{h−2} + …) + (3η^{2}hx^{2h−1} + …) + (ηαhx^{a+h−1} + …)

+ (η^{3}x^{3h} + …) + (η^{2}αx^{2h+a} + …) + (βηx^{b+h} + …) + (γx^{c} + …) = 0 Since h < −1 we get h − 2 > 3h, 2h − 1 > 3h and a + h − 1 > a + 2h. For the lowest term to cancel one of the following must hold:

i. If 3h is the lowest exponent, then:

A. if 3h = 2h + a, in which case A ≠ 0, then a ≤ −2, 3a ≤ c and (if

B ≠ 0 then 2a ≤ b). B. if 3h = h + b, in which case B ≠ 0, then b ≤ −4, b ∈ 2Z, b ≤ \frac{2}{3}c

and (if A ≠ 0 then b ≤ 2a). C. if 3h = c, then c ≤ −6, c ∈ 3Z, (if A ≠ 0 then c ≤ 3a) and (if

B ≠ 0 then c ≤ \frac{3}{2}b). ii. If a + 2h is the lowest exponent but not 3h, then:

A. if a + 2h = h + b, in which case AB ≠ 0, then a < −2, b < −4,

b − a < −1, 2a < b and 2b − a ≤ c. B. if a + 2h = c, in which case A ≠ 0, then a < −2, c < −6,

c − a < −4, (c − a) ∈ 2Z, 3a < c and (if B ≠ 0 then a + c ≤ 2b). iii. If a + 2h and 3h are not the lowest exponents, then B ≠ 0, c < −6,

b < −4, c ≤ b − 2, 3b < 2c and (if A ≠ 0 then 2b < a + c).

(b) If u has no pole of order bigger than 1, then u is of the form

u = ∑ \frac{γ_{i}}{x − c_{i}} + Q(x),

where Q(x) ∈ C[x]. If Q(x) = 0, then L(y) = 0 has a solution of the form P (x) ∏_{i}(x − c_{i})^{γ_{i}}, where P (x) ∈ C[x], c_{i} ≠ ∞ are singular points of L(y) = 0 and γ_{i} are exponents of L(y) = 0 at c_{i}. Since we assume that this is not the case, we get h_{∞} ≥ 0. Plugging u into R(u) = 0 we get an equation similar to the one in 1(a) above (the only difference is that we have added the subscript ∞). Since h_{∞} ≥ 0 we get h_{∞} − 2 < 3h_{∞}, 2h_{∞} − 1 < 3h_{∞}, a_{∞} + h_{∞} − 1 < a_{∞} +2h_{∞}. For the highest term to cancel one of the following must hold:

i. If 3h_{∞} is the highest exponent, then:

A. if 3h_{∞} = a_{∞}+2h_{∞}, in which case A ≠ 0, then a_{∞} ≥ 0, 3a_{∞} ≥ c_{∞}

and (if B ≠ 0 then 2a_{∞} ≥ b_{∞}). B. if 3h_{∞} = b_{∞} + h_{∞}, in which case B ≠ 0, then 0 ≤ b_{∞}, b_{∞} ∈ 2Z,

b_{∞} ≥ \frac{2}{3}c_{∞} and (if A ≠ 0 then b_{∞} ≥ 2a_{∞}). C. if 3h_{∞} = c_{∞}, then 0 ≤ c_{∞} ∈ 3Z, (if A ≠ 0 then c_{∞} ≥ 3a_{∞}) and

(if B ≠ 0 then c_{∞} ≥ \frac{3}{2}b_{∞}). ii. If a_{∞} + 2h_{∞} is the highest exponent but not 3h_{∞}, then:

A. if a_{∞} + 2h_{∞} = b_{∞} + h_{∞}, in which case AB ≠ 0, then 0 < a_{∞},

a_{∞} ≤ b_{∞} and 2b_{∞} − a_{∞} ≥ c_{∞}. B. if a_{∞} + 2h_{∞} = c_{∞}, in which case A ≠ 0, then 0 ≤ c_{∞} − a_{∞},

c_{∞} − a_{∞} ∈ 2Z, 0 < c_{∞} < 3a_{∞} and (if B ≠ 0 then a_{∞} + c_{∞} ≥ 2b_{∞}). iii. If b_{∞} + h_{∞} is the highest exponents, but not 3h_{∞} or a_{∞} + 2h_{∞}, then B ≠ 0, 0 < b_{∞} ≤ c_{∞}, 3b_{∞} > 2c_{∞} and (if A ≠ 0 then 2b_{∞} > a_{∞}+c_{∞}).

2. Now assume L^{∗}(y) has a right factor of order one. Note that L^{∗}(y) has the same singular points as L(y) = 0. When one expands the coefficients appearing in R^{∗}(u) and applies the above arguments, one gets the same conditions on a, b and c (resp. a_{∞}, b_{∞} and c_{∞}).

3.2 Case of an imprimitive unimodular differential Galois group

If a third order linear differential equation L(y) = 0 with coefficients in C(x) has a differential Galois group which is an imprimitive subgroup of SL(3, C), then all the solutions of L(y) = 0 are liouvillian. In fact in this case L(y) = 0 has a solution z such that u = z′/z is algebraic over C(x) of degree at most 3 (Theorem 3.3, [21]). The minimal polynomial of u can be computed by the method described in Section 3.2 of [21]. We now derive a necessary condition for this case to hold.

Since an imprimitive subgroup of prime degree (e.g. 3) is a monomial group (cf. [24], Definition 3.2), we will now derive necessary conditions for the differential Galois group to be monomial. For third (or just prime) order equations, these are necessary conditions for the imprimitive case.

Necessary conditions for case 2 Let L(y) = 0 be an irreducible linear differential equation of order n over C(x) with monomial differential Galois group G(L) ⊆ SL(n, C). The n-th symmetric power L^{◯ s n}(y) = 0 of L(y) = 0 must have a non trivial solution of the form P (x) ∏_{i}(x − c_{i})^{α_{i}}, where P (x) ∈ C[x], c_{i} ≠ ∞ are singular points of L^{◯ s n}(y) = 0 with exponents α_{i} ∈ \frac{1}{2}Z. In particular, any singularity of L^{◯ s n}(y) = 0 must have an exponent of the form b/2, where b ∈ Z.

If, in addition, L(y) = 0 is of fuchsian type, then at each of the m singular points a_{i} of L(y) = 0 on the Riemann Sphere there must be exponents α_{i,1}, α_{i,2}, …, α_{i,n} of L(y) = 0 at a_{i} such that

• (∑_{j=1}^{n} α_{i,j}) ∈ \frac{1}{2}Z, and

• (∑_{i=1}^{m} ∑_{j=1}^{n} α_{i,j}) ∈ Z and is non positive.

Note that the elements α_{i,1}, α_{i,2}, …, α_{i,n} are not necessarily distinct. Furthermore, note that the above conditions imply that L(y) = 0 has a solution whose square is a rational function. For third order equations, the condition that L^{◯ s 3}(y) = 0 has a solution whose square is rational is in fact necessary and sufficient (cf. [20], Theorem 4.6).

Example: We now use the above necessary conditions to show that, for the second symmetric power of the Airy equation

L(y) = \frac{d^{3}y}{dx^{3}} − 4x\frac{dy}{dx} − 2y,

the case 2 (of an imprimitive group G(L) ⊆ SL(3, C)) cannot hold. Since L(y) = 0 is not of fuchsian type, we have to compute L^{◯ s 3}(y) which gives:

L^{◯ s 3}(y) = \frac{d^{7}y}{dx^{7}} − 56x\frac{d^{5}y}{dx^{5}} − 140\frac{d^{4}y}{dx^{4}} + 784x^{2} \frac{d^{3}y}{dx^{3}} + 2352x\frac{d^{2}y}{dx^{2}}

−4(576x^{3} − 295)\frac{dy}{dx} − 3456x^{2}y.

The only possible exponent of L^{◯ s 3}(y) at the unique singular point ∞ is 3/2, which does not rule out the possibility of an imprimitive group G(L) ⊆ SL(3, C). We thus have to test the stronger condition that if G(L) ⊆ SL(3, C) is an imprimitive group, then L^{◯ s 3}(y) = 0 must have a solution of the form P (x) ∏_{i}(x − c_{i})^{a_{i}}, where P (x) ∈ C[x], c_{i} ≠ ∞ are singular points of L(y) = 0 and a_{i} ∈ \frac{1}{2}Z. Since ∞ is the only singular point and 3/2 is the only exponent, Lemma 3.1 implies that L^{◯ s 3}(y) = 0 has no non-zero solution of this form. Thus case 2 (in which case we always have a liouvillian solution) does not hold.

Proof of the Necessary conditions for case 2: The “non fuchsian” part of our necessary conditions follows from:

Theorem 3.4 (cf. Proposition 3.6, [20]) If an irreducible linear differential equation L(y) = 0 of order n with coefficients in C(x) has a monomial differential Galois group G(L) ⊆ SL(n, C), then the n-th symmetric power L^{◯ s n}(y) = 0 of L(y) = 0 has a solution which is the square root of an element of C(x). At any singularity, L^{◯ s n}(y) = 0 must have an exponent of the form a/2, where a ∈ Z.

When L(y) is fuchsian we can do better. If for some solutions y_{i} of L(y) = 0 we have (y_{1}y_{2} ⋯ y_{n})^{2} is a rational function, then (y_{1}y_{2} ⋯ y_{n}) must be of the form

P (x) ∏(x − a_{i})^{\frac{b_{i}}{2}} ,

where P (x) ∈ C[x], a_{i} are singular points of L(y) = 0 and \frac{b_{i}}{2} = e_{i} with b_{i} ∈ Z is an exponent of L^{◯ s n}(y) = 0 at a_{i} (the apparent singularities may also contribute to P (x)). Thus, for some exponent e_{∞} of L^{◯ s n}(y) = 0 at ∞ we have that −deg(P) − ∑_{i} e_{i} = e_{∞}. Thus e_{∞} + ∑_{i} e_{i} is a non-positive integer. Using Lemma 2.2 we can express the exponents of L^{◯ s n}(y) = 0 at a_{i} in terms of the exponents α_{k} of L(y) = 0 at a_{i}. We get that there exists non negative integers m_{1}, …, m_{n} with ∑_{j=1}^{n} m_{j} = n such that

(∑_{j=1}^{n} m_{j}α_{∞,j}) + t_{∞} + ∑_{i} ((∑_{j=1}^{n} m_{j}α_{i,j}) + t_{i})

(where t_{j} are positive integer) is non positive and in Z. In particular

(∑_{j=1}^{n} m_{j}α_{∞,j}) + ∑_{i} (∑_{j=1}^{n} m_{j}α_{i,j}) ,

is non positive and in Z. Since e_{i} = ∑_{j=1}^{n} m_{j}α_{i,j} + t_{i} and e_{i} = \frac{b_{i}}{2} with b_{i} ∈ Z, we get that ∑_{j=1}^{n} m_{j}α_{i,j} ∈ \frac{1}{2}Z. This proves the fuchsian part of the Theorem.

For a second order differential equation L(y) = y′′ − ry (r ∈ C(x)), one gets the condition of the Kovacic algorithm by looking at the riccati equation of the third order linear differential equation L^{◯ s 2}(y) = 0 which is (cf. [12], p. 10 and [6]):

θ′′ + 3θ′θ + θ^{3} = 4rθ + 2r′

For a third order differential equation L(y) = 0 the order of L^{◯ s 3}(y), which is less than or equal to 10, is not known in advance (from [20], Lemma 3.5 we get that the order can be 7, 9 or 10). This makes a general discussion as in [12] difficult.

3.3 Case of a finite primitive unimodular differential Galois group

There are, up to isomorphism, only 8 primitive finite subgroups of SL(3, C). Following [20] section 2.2, we denote them A_{6}^{SL_{3}}, A_{5}, A_{5} × C_{3}, G_{168}, G_{168} × C_{3}, H_{216}^{SL_{3}}, H_{72}^{SL_{3}} and F_{36}^{SL_{3}}. We note that the last 3 groups are solvable, G_{168} is the simple group of 168 elements and A_{6}^{SL_{3}} is a central extension of A_{6}. By Jordan’s Theorem, such a finite list exists for any degree.

The order of a one dimensional character ζ is the smallest integer i, such that (ζ)^{i} is the trivial character. If G is a group and V a G-module, then we denote the m^{th} symmetric power of V, which is also a G-module, by S^{m}(V) (c.f. [13], p. 586).

Our necessary conditions in this case are based on the following Theorem:

Theorem 3.5 Let L(y) = 0 be a differential equation of degree n whose differential Galois group is a finite subgroup G of SL(n, C). We denote by V the solution space of L(y) = 0. If S^{m}(V) has a G−summand of dimension 1 whose character has order i, then, at each point c including ∞, there there exists positive integers m_{1}, …, m_{n}, with ∑_{j=1}^{n} m_{j} = m, such that i(∑_{j=0}^{n} m_{j}e_{j}) ∈ Z, where e_{1}, …, e_{n} are the exponents of L(y) = 0 at c.

Proof: There is a natural map Φ_{m} of S^{m}(V) into the Picard-Vessiot extension for L(y) = 0 given by sending z_{1}◯ s … ◯ s z_{m} to z_{1} · . . . · z_{m}. Since a finite group is completely reducible, the solution space of L^{◯ s m}(y) = 0 is G-isomorphic to a direct summand of S^{m}(V) (cf. [20] Lemma 3.5). Let {y_{1}, …, y_{n}} be a fundamental set of solutions of L(y) = 0. Since G is a finite group and S^{m}(V) has a one dimensional summand, there is a homogeneous polynomial P (Y_{1}, …, Y_{n}) of degree m whose (possibly trivial) image P (y_{1}, …, y_{n}) under Φ_{m} is a semi-invariant of G and such that P (y_{1}, …, y_{n})^{i} is rational (cf. [21], Proposition 1.6).

If P (y_{1}, …, y_{n}) ≠ 0, then the exponents of P (y_{1}, …, y_{n}) at any point c are of the form (∑_{j=0}^{n} m_{j}e_{j}) + h, where h and m_{j} are positive integers and ∑_{j=1}^{n} m_{j} = m (cf. Lemma 2.2.4). Since (P (y_{1}, …, y_{n}))^{i} is rational, we have i ((∑_{j=0}^{n} m_{j}e_{j}) + h) ∈ Z, thus i(∑_{j=0}^{n} m_{j}e_{j}) ∈ Z.

Now assume P (y_{1}, …, y_{n}) = 0 and let ζ be the character of the one dimensional summand of S^{m}(V). In this case we will show that there is a linear differential equation L̃(y) = 0 with the following properties:

i. L(y) and L̃(y) have the same differential Galois group G, ii. The solution space Ṽ of L̃(y) = 0 is G-isomorphic to V, iii. For some fundamental set of solutions {ỹ_{1}, …, ỹ_{n}} of L̃(y) = 0, P (ỹ_{1}, …, ỹ_{n}) ≠ 0 and P (ỹ_{1}, …, ỹ_{n}) generates a one dimensional invariant subspace of the solution space of L̃^{◯ s m}(y) = 0 corresponding to the character ζ,

iv. For any point p on the Riemann Sphere, the sets of exponents E_{p} of L(y) and Ẽ_{p} of L̃(y) are the same mod Z

Assuming we have constructed L̃(y), we can finish the proof. We apply the argument of the preceeding paragraph to the equation L̃(y) and conclude that, at each point c there exist positive integers m_{1}, …, m_{n}, with ∑_{j=1}^{n} m_{j} = m, such that i(∑_{j=0}^{n} m_{j}ẽ_{j}) ∈ Z, where ẽ_{1}, …, ẽ_{n} are the exponents of L̃(y) = 0 at c. Since the exponents of L̃(y) and L(y) coincide mod Z, we achieve the conclusion of the theorem.

To construct L̃(y) we proceed as follows. Let R_{1}, …, R_{n} be new variables and consider the substitution

( Y_{1} ; ⋮ ; Y_{n} ) = ( \begin{matrix}y_{1} & y′_{1} & ⋯ & y_{1}^{(n−1)} \\ ⋮ & ⋮ & ⋮ & ⋮ \\ y_{n} & y′_{n} & ⋯ & y_{n}^{(n−1)} \end{matrix} ) ( R_{1} ; ⋮ ; R_{n} )

Under this substitution, the polynomial P (Y_{1}, …, Y_{n}) becomes P̂(R_{1}, …, R_{n}), a polynomial with coefficients in K, the Picard-Vessiot extension associated with L(y). Since the matrix (y_{i}^{(j)}) is nonsingular, we see that P̂(R_{1}, …, R_{n}) is a

nonzero polynomial. Let Q̂(R_{1}, …, R_{n}) be the image of the differential polynomial det(W r(Y_{1}, …, Y_{n})) (where W r is the wronskian matrix) under this substitution. One can select the r_{i} ∈ C(x) so that P̂(r_{1}, …, r_{n}) · Q̂(r_{1}, …, r_{n}) ≠ 0 (see [15], p.

35). Let

( ỹ_{1} ; ⋮ ; ỹ_{n} ) = ( \begin{matrix}y_{1} & y′_{1} & ⋯ & y_{1}^{(n−1)} \\ ⋮ & ⋮ & ⋮ & ⋮ \\ y_{n} & y′_{n} & ⋯ & y_{n}^{(n−1)} \end{matrix} ) ( r_{1} ; ⋮ ; r_{n} )

We have that ỹ_{1}, …, ỹ_{n} belong to K and, by construction, these are linearly independent over C. Applying elements of the Galois group G to ỹ_{1}, …, ỹ_{n}, one sees that the C-span Ṽ of these elements is left invariant by G. Furthermore, these elements form a fundamental set of solutions of the linear differential equation L̃(y) = det(W r(y, ỹ_{1}, …, ỹ_{n}))/det(W r(ỹ_{1}, …, ỹ_{n})) whose coefficients are left fixed by G and so lie in C(x).

Note that the matrix S = (W r(y_{1}, …, y_{n})) · (W r(ỹ_{1}, …, ỹ_{n}))^{−1} is left invariant by the differential Galois group and so has entries in C(x). Comparing first rows of (W r(y_{1}, …, y_{n})) and S · W r(ỹ_{1}, …, ỹ_{n}) we have that y_{i} = ∑_{j=0}^{n−1} s_{j}ỹ_{i}^{(j)} with s_{j} ∈ C(x). The map y ↦ ∑_{j=0}^{n−1} r_{j}y_{i}^{(j)} (resp. ỹ ↦ ∑_{j=0}^{n−1} s_{j}ỹ_{i}^{(j)}) will take solutions of L(y) = 0 to solutions of L̃(y) = 0 (resp. solutions of L̃(y) = 0 to solutions of L(y) = 0). Therefore, the extension C(x) < ỹ_{1}, …, ỹ_{n} > coincides with K. In particular, L(y) and L̃(y) have the same differential Galois group, so property i. holds. To verify that property ii. holds, one observes that the matrix of any σ ∈ G is the same with respect to the bases {y_{1}, …, y_{n}} and {ỹ_{1}, …, ỹ_{n}}. Property iii. follows by construction. To verify property iv., let p = 0 and let y = x^{ρ} ∑_{n} a_{n}x^{n} be a solution of L(y) = 0. We then have that ỹ = ∑_{j=0}^{n−1} r_{j}y^{(j)} will be a solution of L̃(y) and furthermore it will be of the form ỹ = x^{ρ} ∑_{n} b_{n}x^{n} (the r_{i} are at worst meromorphic at 0). Therefore, if ρ is an exponent of L(y) = 0 at 0, then ρ+N will be an exponent of L̃(y) at 0. Therefore elements of E_{0} are congruent to elements of Ẽ_{0} mod Z. Reversing the roles of L(y) and L̃(y) and arguing simillarly, one establishes the last property.

We now state the necessary conditions for third order equations in this case:

Necessary conditions for case 3 Let L(y) be an irreducible third order linear differential equation with Galois group a finite primitive group G(L) ⊂ SL(3, C). Then L(y) = 0 must be a differential equation of fuchsian type having 3 distinct rational exponents at any singularity. Furthermore, if c is any singularity of L(y) = 0 and e_{1}, e_{2}, e_{3} are the exponents at c, then one of the following holds:

1. If G(L) ≅ A_{6}^{SL_{3}}, then

(a) each e_{i} is of the form a_{i}/m_{i}, where (a_{i}, m_{i}) = 1, a_{i}, m_{i} ∈ Z and lcm(m_{1}, m_{2}, m_{3}) ∈ {1, 2, 3, 4, 5, 6, 12, 15} and

(b) there exists non negative integers n_{1}, n_{2}, n_{3}, such that ∑_{i=1}^{3} n_{i} = 6 and ∑_{i=1}^{3} n_{i}e_{i} ∈ Z.

2. If G(L) ≅ A_{5}, then

(a) each e_{i} is of the form a_{i}/m_{i}, where (a_{i}, m_{i}) = 1, a_{i}, m_{i} ∈ Z and lcm(m_{1}, m_{2}, m_{3}) ∈ {1, 2, 3, 5}, and

(b) there exists non negative integers n_{1}, n_{2}, n_{3}, such that ∑_{i=1}^{3} n_{i} = 2 and ∑_{i=1}^{3} n_{i}e_{i} ∈ Z.

3. If G(L) ≅ A_{5} × C_{3}, then

(a) each e_{i} is of the form a_{i}/m_{i}, where (a_{i}, m_{i}) = 1, a_{i}, m_{i} ∈ Z and lcm(m_{1}, m_{2}, m_{3}) ∈ {1, 2, 3, 5, 6, 15}, and

(b) there exists non negative integers n_{1}, n_{2}, n_{3}, such that ∑_{i=1}^{3} n_{i} = 2 and 3(∑_{i=1}^{3} n_{i}e_{i}) ∈ Z.

4. If G(L) ≅ G_{168}, then

(a) each e_{i} is of the form a_{i}/m_{i}, where (a_{i}, m_{i}) = 1, a_{i}, m_{i} ∈ Z and lcm(m_{1}, m_{2}, m_{3}) ∈ {1, 2, 3, 4, 7}, and

(b) there exists non negative integers n_{1}, n_{2}, n_{3}, such that ∑_{i=1}^{3} n_{i} = 4 and ∑_{i=1}^{3} n_{i}e_{i} ∈ Z.

5. G(L) ≅ G_{168} × C_{3}, then

(a) each e_{i} is of the form a_{i}/m_{i}, where (a_{i}, m_{i}) = 1, a_{i}, m_{i} ∈ Z and lcm(m_{1}, m_{2}, m_{3}) ∈ {1, 2, 3, 4, 6, 7, 12, 21}, and

(b) there exists non negative integers n_{1}, n_{2}, n_{3}, such that ∑_{i=1}^{3} n_{i} = 4 and 3(∑_{i=1}^{3} n_{i}e_{i}) ∈ Z.

6. G(L) ≅ H_{216}^{SL_{3}} , then

(a) each e_{i} is of the form a_{i}/m_{i}, where (a_{i}, m_{i}) = 1, a_{i}, m_{i} ∈ Z and lcm(m_{1}, m_{2}, m_{3}) ∈ {1, 2, 3, 4, 6, 9, 12, 18}, and

(b) there exists non negative integers n_{1}, n_{2}, n_{3}, such that ∑_{i=1}^{3} n_{i} = 6 and 3(∑_{i=1}^{3} n_{i}e_{i}) ∈ Z.

7. G(L) ≅ H_{72}^{SL_{3}}, then

(a) each e_{i} is of the form a_{i}/m_{i}, where (a_{i}, m_{i}) = 1, a_{i}, m_{i} ∈ Z and lcm(m_{1}, m_{2}, m_{3}) ∈ {1, 2, 3, 4, 6, 12}, and

(b) there exists non negative integers n_{1}, n_{2}, n_{3}, such that ∑_{i=1}^{3} n_{i} = 6 and ∑_{i=1}^{3} n_{i}e_{i} ∈ Z.

8. G(L) ≅ F_{36}^{SL_{3}}, then

(a) each e_{i} is of the form a_{i}/m_{i}, where (a_{i}, m_{i}) = 1, a_{i}, m_{i} ∈ Z and lcm(m_{1}, m_{2}, m_{3}) ∈ {1, 2, 3, 4, 6, 12}, and (b) there exists non negative integers n_{1}, n_{2}, n_{3}, such that ∑_{i=1}^{3} n_{i} = 3 and

4(∑_{i=1}^{3} n_{i}e_{i}) ∈ Z.

In order to exclude this case it is enough to show that none of the conditions on the exponents is satisfied.

Example: The second symmetric power of the Airy equation

L(y) = \frac{d^{3}y}{dx^{3}} − 4x\frac{dy}{dx} − 2y

has an unimodular differential Galois group and is not of fuchsian type. Thus none of the necessary conditions 1, 2 and 3 hold and case 1, 2 and 3 can not occur for L(y) = 0. This proves that L(y) = 0 has no liouvillian solutions.

Proof of Necessary conditions for case 3: The conditions that L(y) = 0 is of fuchsian type and that all exponents are rational are a consequence of the result that any solution of L(y) = 0 has to be algebraic in this case (cf. Lemma 2.2.3).

From Lemma 2.3 we get that all exponents are of the form a/m, where m is the order of an element of G(L). The possible set given in the necessary conditions is just the set of orders of elements of G(L).

For each irreducible three dimensional character χ of a finite primitive group G, using the recurence relation given in [20] Section 2.3 we can compute the character χ_{m} of S^{m}(V). The following are the decompositions of the χ_{m} into irreducible characters of G in which a one dimensional summand appears for the first time (For example, in 7., χ_{6} is the sum of the trivial character 1, 3 non trivial different one dimensional characters ζ_{1,1}, ζ_{1,2} and ζ_{1,3}, and 3 times the same irreducible eight dimensional character ζ_{8}):

1. For G(L) ≅ A_{6}^{SL_{3}} we get χ_{6} = 1 + ζ_{5,1} + ζ_{5,2} + ζ_{8} + ζ_{9}.

2. For G(L) ≅ A_{5} we get χ_{2} = 1 + ζ_{5}

3. For G(L) ≅ A_{5} × C_{3} we get χ_{2} = ζ_{1} + ζ_{5}, and ζ_{1} is of order 3.

4. For G(L) ≅ G_{168} we get χ_{4} = 1 + ζ_{6} + ζ_{8}

5. For G(L) ≅ G_{168} × C_{3} we get χ_{4} = ζ_{1} + ζ_{6} + ζ_{8} and ζ_{1} is of order 3.

6. For G(L) ≅ H_{216}^{SL_{3}} we get χ_{6} = ζ_{1} + ζ_{3} + ζ_{8,1} + 2ζ_{8,2}, and ζ_{1} is of order 3.

7. For G(L) ≅ H_{72}^{SL_{3}} we get χ_{6} = 1 + ζ_{1,1} + ζ_{1,2} + ζ_{1,3} + 3ζ_{8}, and ζ_{1,i} is of order 2.

8. For G(L) ≅ F_{36}^{SL_{3}} we get χ_{3} = ζ_{1,1} + ζ_{1,2} + ζ_{4,1} + ζ_{4,2}, and ζ_{1,i} is of order 4.

Theorem 3.5 now gives the result.

We now show how our approach can be used to get and to improve the necessary conditions of the Kovacic algorithm in this case (case 3 in [12]). The finite primitive subgroups of SL(2, C) are the tetrahedral, octahedral and icosahedral group, which are denoted resp. A_{4}^{SL_{2}}, S_{4}^{SL_{2}} and A_{5}^{SL_{2}} (see e.g. [20] section 2.2)

Theorem 3.6 Let L(y) be an irreducible second order linear differential equation whose differential Galois group is a finite primitive group G(L) ⊆ SL(2, C). Then L(y) = 0 must be a differential equation of fuchsian type having 2 distinct rational exponents at any singularity. Furthermore, if c is any singularity of L(y) = 0 and e_{1}, e_{2} are the exponents at c, then one of the following holds:

1. G(L) ≅ A_{5}^{SL_{2}}, then

(a) each e_{i} is of the form a_{i}/m_{i}, where (a_{i}, m_{i}) = 1, a_{i}, m_{i} ∈ Z and lcm(m_{1}, m_{2}) ∈ {1, 2, 3, 4, 5, 6, 10}, and

(b) there exists non negative integers n_{1}, n_{2}, such that n_{1} + n_{2} = 12 and n_{1}e_{1} + n_{2}e_{2} ∈ Z.

2. G(L) ≅ S_{4}^{SL_{2}}, then

(a) each e_{i} is of the form a_{i}/m_{i}, where (a_{i}, m_{i}) = 1, a_{i}, m_{i} ∈ Z and lcm(m_{1}, m_{2}) ∈ {1, 2, 3, 4, 6, 8}.

(b) there exists non negative integers n_{1}, n_{2}, such that n_{1} + n_{2} = 6 and 2(n_{1}e_{1} + n_{2}e_{2}) ∈ Z.

3. G(L) ≅ A_{4}^{SL_{2}}, then

(a) each e_{i} is of the form a_{i}/m_{i}, where (a_{i}, m_{i}) = 1, a_{i}, m_{i} ∈ Z and lcm(m_{1}, m_{2}) ∈ {1, 2, 3, 4, 6}.

(b) there exists non negative integers n_{1}, n_{2}, such that n_{1} + n_{2} = 4 and 3(n_{1}e_{1} + n_{2}e_{2}) ∈ Z.

Proof: The proof is similar to the proof of the Necessary Conditions 3. For any irreducible character χ of degree 2 of A_{4}^{SL_{2}}, S_{4}^{SL_{2}} and A_{5}^{SL_{2}}, we have to find the decomposition of the character χ_{m} of S^{m}(V) in which a one dimensional summand appears for the first time. Using the same notation as in proof of the Necessary Conditions 3 we get:

1. For A_{4}^{SL_{2}} we get χ_{4} = ζ_{1,1} + ζ_{1,2} + ζ_{3}, where ζ_{1,i} is of degree 1 and order 3.

2. For S_{4}^{SL_{2}} we get χ_{6} = ζ_{1} + 2ζ_{3}, where ζ_{1} is of degree 1 and order 2.

3. For A_{5}^{SL_{2}} we get χ_{12} = 1 + ζ_{3} + ζ_{4} + ζ_{5}.

We conclude as in Necessary Conditions 3.

The above improvement of the necessary conditions given in [12] (see also [6]), which impose divisibility conditions on the denominator of the exponents, can also be found in a paper of L. Fuchs (cf. [8]). We note that for the above result one does not need to compute the semi invariants, but only the order of a one dimensional character.

3.4 Solvability in terms of lower order equations

If a second order equation with unimodular differential Galois group has no nonzero liouvillian solutions, then its differential Galois group is SL(2, C) (cf. [12, 20]). If a third order equation L(y) = 0 with unimodular differential Galois group has no non-zero liouvillian solutions, then its differential Galois group G(L) is either SL(3, C) or is congugate to ρ_{2}(SL(2, C)) or ρ_{2}(SL(2, C)) × C_{3}, where

ρ_{2} ( \begin{matrix}a & b \\ c & d \end{matrix} ) = ( \begin{matrix}a^{2} & 2ab & b^{2} \\ ac & ad + bc & bd \\ c^{2} & 2cd & d^{2} \end{matrix} )

and C_{3} is the center of SL(3, C) which is cyclic of order 3 (cf. [20]; Note that ρ_{2}(SL(2, C)) ≅ P GL(2, C)). If G(L) is not SL(3, C), then L(y) = 0 is solvable in terms of second order equations (cf. [18]).

If λ_{1} and λ_{2} are the eigenvalues of ( \begin{matrix}a & b \\ c & d \end{matrix} ), then the eigenvalues of ρ_{2} ( \begin{matrix}a & b \\ c & d \end{matrix} )

are {λ_{1}^{2}, 1, λ_{2}^{2}} (since λ_{1}λ_{2} = 1). This observation yields:

Lemma 3.7 Let L(y) = 0 be a fuchsian third order linear differential equation whose differential Galois group is conjugate to a subgroup of ρ_{2}(SL(2, C)) × C_{3}. Then at each singular point, some exponent is of the form \frac{n}{3} for some n ∈ Z.

Proof: The monodromy matrix at any point is an element of the differential Galois group. Since the elements of C_{3} commute with the elements of ρ_{2}(SL(2, C)) one sees that the eigenvalues of a monodromy matrix are of the form {ωλ_{1}^{2}, ω, ωλ_{2}^{2}}, where λ_{1}λ_{2} = 1 and ω^{3} = 1. Since L(y) = 0 is fuchsian, Lemma 2.4 implies that these must be of the form \{e^{2π\sqrt{−1}a_{1}}, e^{2π\sqrt{−1}a_{2}}, e^{2π\sqrt{−1}a_{3}}\}. So a_{2} = \frac{n}{3} with n ∈ Z.

In the fuchsian case, the above yields necessary conditions useful in testing solvability in terms of second order equation and in computing differential Galois groups. This will be used in the next section.

4 Differential equations with three singular points

In the previous section we derived necessary conditions on the exponents of a differential equation L(y) = 0 with coefficients in C(x) to have a liouvillian solution. By assuming that we have only three singular points, we will be able to sharpen the necessary conditions for the finite primitive groups.

Theorem 4.1 Let L(y) = 0 be a third order linear differential equation over C(x) having three singular points s_{1}, s_{2} and s_{3} and whose differential Galois group G(L) is a finite primitive subgroup of SL(3, C). Then for each possible group, all lists [lcm_{1}, lcm_{2}, lcm_{3}] of possible least common multiples lcm_{i} of the denominators d_{i,1}, d_{i,2}, d_{i,3} of the rational exponents \frac{n_{i,1}}{d_{i,1}} , \frac{n_{i,2}}{d_{i,2}} , \frac{n_{i,3}}{d_{i,3}} , where (n_{i,j}, d_{i,j}) = 1, at s_{i} can be derived.

1. If G(L) ≅ A_{6}^{SL_{3}}, we get: [2, 4, 15], [2, 5, 12], [2, 5, 15], [2, 12, 15], [2, 15, 15], [3, 3, 4], [3, 3, 5], [3, 3, 12], [3, 3, 15], [3, 4, 5], [3, 4, 15] , [3, 5, 5], [3, 5, 12], [3, 5, 15], [3, 12, 15], [3, 15, 15], [4, 4, 4], [4, 4, 5], [4, 4, 15], [4, 5, 5], [4, 5, 6], [4, 5, 12], [4, 5, 15], [4, 6, 15], [4, 12, 12], [4, 12, 15], [4, 15, 15], [5, 5, 5], [5, 5, 6], [5, 5, 12], [5, 5, 15], [5, 6, 12], [5, 6, 15], [5, 12, 12], [5, 12, 15], [5, 15, 15], [6, 12, 15], [6, 15, 15], [12, 12, 12], [12, 12, 15], [12, 15, 15] or [15, 15, 15].

2. If G(L) ≅ A_{5}, we get: [2, 3, 5], [2, 5, 3], [2, 5, 5], [3, 3, 5], [3, 5, 5] or [5, 5, 5].

3. If G(L) ≅ A_{5} × C_{3}, we get: [2, 3, 15], [2, 15, 15], [3, 3, 5], [3, 3, 15], [3, 5, 6], [3, 5, 15], [3, 6, 15], [3, 15, 15], [5, 6, 15], [5, 15, 15], [6, 15, 15] or [15, 15, 15].

4. If G(L) ≅ G_{168}, we get: [2, 3, 7], [2, 4, 7], [2, 7, 7], [3, 3, 4], [3, 3, 7], [3, 4, 4], [3, 4, 7], [3, 7, 7], [4, 4, 4], [4, 4, 7], [4, 7, 7] or [7, 7, 7].

5. If G(L) ≅ G_{168} × C_{3}, we get: [2, 3, 21], [2, 12, 21], [2, 21, 21], [3, 3, 4], [3, 3, 7], [3, 3, 12], [3, 3, 21], [3, 4, 12], [3, 4, 21], [3, 6, 7], [3, 6, 21], [3, 7, 12], [3, 7, 21], [3, 12, 12], [3, 12, 21], [3, 21, 21], [4, 6, 21], [4, 12, 12], [4, 12, 21], [4, 21, 21], [6, 7, 12], [6, 7, 21], [6, 12, 21], [6, 21, 21], [7, 12, 12], [7, 12, 21], [7, 21, 21], [12, 12, 12], [12, 12, 21], [12, 21, 21] or [21, 21, 21].

6. If G(L) ≅ H_{216}^{SL_{3}}, we get: [3, 3, 4], [3, 3, 12], [3, 3, 18], [3, 4, 9], [3, 4, 18], [3, 9, 12], [3, 9, 18], [3, 12, 18], [4, 9, 18], [4, 18, 18], [9, 12, 18], [12, 18, 18] or [18, 18, 18].

7. If G(L) ≅ H_{72}^{SL_{3}}, we get: [4, 4, 4], [4, 4, 12], [4, 12, 12] or [12, 12, 12].

8. If G(L) ≅ F_{36}^{SL_{3}}, we get: [2, 4, 12], [2, 12, 12], [3, 4, 4], [3, 4, 12], [3, 12, 12], [4, 4, 6], [4, 6, 12] or [6, 12, 12].

Proof: We use the monodromy group of L(y) = 0, which is introduced in the first section. To the singular points at s_{1}, s_{2} and s_{3} correspond the matrices M_{s_{1}}, M_{s_{2}}, and M_{s_{3}} of the monodromy group. The product M_{s_{1}}M_{s_{2}}M_{s_{3}} corresponds to the zero path on the punctured Riemann Sphere and thus must be the identity. Since M_{s_{1}}, M_{s_{2}} and M_{s_{3}} generate the monodromy group and M_{s_{1}}M_{s_{2}} = M_{s_{3}}^{−1}, we get that the group G(L) is generated by two elements. Using the group theory system CAYLEY (see [4]) we can compute all possible sets of two generators for each finite primitive subgroup of SL(3, C) and compute their order and the order of their product. From the proof of Lemma 2.3 we get that the order of M_{s_{i}} is the least common multiple of the denominator of the exponents at s_{i}. This leads to the possibilities listed above.

We note that the above possibilities often allow one to distinguish the groups A_{5} from A_{5} × C_{3} and G_{168} from G_{168} × C_{3}.

Example: The following equation due to Hurwitz (see [9]):

H(y) = x^{2}(x − 1)^{2}y′′′ + (7x − 4)x(x − 1)y′′ +

(\frac{72}{7}(x^{2} − x) − \frac{20}{9}(x − 1) + \frac{3}{4}x)y′ + (\frac{792}{7^{3}}(x − 1) + \frac{5}{8} + \frac{2}{63})y

is of fuchsian type and has a unimodular differential Galois group. The exponents of H(y) = 0 at the regular singular points 0, 1 and ∞ are respectively \{0, −\frac{1}{3}, −\frac{2}{3}\}, \{\frac{1}{2}, 0, −\frac{1}{2}\} and \{\frac{11}{7}, \frac{9}{7}, \frac{8}{7}\}.

We now use the necessary conditions to test the possible structure of the differential Galois group G(H) ⊆ SL(3, C) of H(y) = 0:

1. Reducibility: Since H(y) is fuchsian, we can use Corollary 3.3. The sum of three exponents corresponding to three different singular points is never a non positive integer. Thus H(y) = 0 has no right factor of order one. The adjoint H^{∗}(y) = 0 of H(y) = 0 is:

y′′′ + (\frac{−7x + 4}{x(x − 1)}) y′′ + ( \frac{\frac{170}{7}x^{2} − \frac{6995}{252}x + \frac{92}{9}}{x^{2}(x − 1)^{2}} ) y′

+ ( \frac{\frac{12650}{343}x^{3} − \frac{1561655}{24696}x^{2} + \frac{1143403}{24696}x + \frac{112}{9}}{x^{3}(x − 1)^{3}} ) y

The exponents of H^{∗}(y) = 0 at 0, 1 and ∞ are respectively \{\frac{8}{3}, 2, \frac{7}{3}\}, \{\frac{5}{2}, 2, \frac{3}{2}\} and \{−\frac{22}{7}, −\frac{23}{7}, −\frac{25}{7}\}. The sum of three exponents corresponding to three different singular points is never a non positive integer. Thus H^{∗}(y) = 0 has no right factor of order one. This shows, that the differential equation H(y) = 0 is irreducible.

2. Imprimitivity: We note that at ∞, the only sum of three (possibly repeated) exponents which is in \frac{1}{2}Z is the sum of the three exponents \frac{11}{7}, \frac{9}{7} and \frac{8}{7} whose sum is 4. At 0 the possible sums of three exponents which are in \frac{1}{2}Z are 0, −1 and −2. Since the sum of three exponents at 1 is ≥ −\frac{3}{2}, we get that no sum of the form prescribed in the Necessary Conditions 2 will be a non positive element of Z. Thus the differential Galois group G(H) of this equation can not be an imprimitive group.

3. Finite primitive groups: Since the list of least common multiples of the denominators of the exponents at the singularities is [2, 3, 7], we get from the above theorem, that if G(H) is a finite primitive group, then G(H) is isomorphic to G_{168} (note that the Necessary Conditions 3 would lead to the two possibilities G(H) ≅ G_{168} and G(H) ≅ G_{168} × C_{3}).

4. Infinite primitive groups: From Lemma 3.7 it follows that the differential Galois group of H(y) = 0 cannot be a subgroup of ρ_{2}(SL(2, C)) × C_{3}, since at ∞ no exponent is one third of an integer.

Therefore our necessary conditions show that the group G(H) is isomorphic to either G_{168} or SL(3, C). In [22] we use the results of [20] and the fact that H^{◯ s 4}(y) = 0 is only of order 14 to deduce that G(H) is not isomorphic to SL(3, C) and thus that G(H) ≅ G_{168}.

Example: The general third order linear differential equation L_{p}(y) = 0 of fuchsian type having respectively exponents \{0, \frac{1}{6}, \frac{5}{6}\}, \{\frac{1}{12}, \frac{1}{3}, \frac{7}{12}\} and \{\frac{1}{12}, \frac{1}{6}, \frac{3}{4}\} at 0, 1 and ∞ is of the form:

L_{p}(y) = y′′′ + (\frac{2}{x − 1} + \frac{2}{x}) y′′ + (\frac{\frac{13}{48}}{(x − 1)^{2}} + \frac{\frac{43}{24}}{x − 1} + \frac{\frac{5}{36}}{x^{2}} + \frac{\frac{−43}{24}}{x} ) y′

+ (\frac{\frac{−7}{432}}{(x − 1)^{3}} + \frac{p + \frac{23}{864}}{(x − 1)^{2}} + \frac{−2p − \frac{23}{864}}{x − 1} + \frac{p}{x^{2}} + \frac{2p + \frac{23}{864}}{x} ) y

where p is an arbitrary parameter.

We now use the necessary conditions to test the possible structure of the differential Galois group G(L_{p}) ⊆ SL(3, C) of L_{p}(y) = 0:

1. Reducibility: Since L_{p}(y) is fuchsian, we can use Corollary 3.3. Since all exponents of L_{p}(y) = 0 are non-negative and at 1 and ∞ in fact positive, no sum of the prescribed form can be a negative integer. Thus L_{p}(y) = 0 has no right factor of order one. The adjoint L_{p}^{∗}(y) = 0 of L_{p}(y) = 0 is:

y′′′ + (\frac{−2}{x − 1} + \frac{−2}{x}) y′′ + (\frac{\frac{205}{48}}{(x − 1)^{2}} + \frac{\frac{43}{24}}{x − 1} + \frac{\frac{149}{36}}{x^{2}} + \frac{\frac{−43}{24}}{x} ) y′

+ (\frac{\frac{−1955}{432}}{(x − 1)^{3}} + \frac{−p − \frac{1571}{864}}{(x − 1)^{2}} + \frac{2p + \frac{23}{864}}{x − 1} + \frac{\frac{−67}{18}}{x^{3}} + \frac{−p + \frac{43}{24}}{x^{2}} + \frac{−2p − \frac{23}{864}}{x} ) y

The exponents of L_{p}^{∗}(y) = 0 at 0, 1 and ∞ are respectively \{\frac{11}{6}, 2, \frac{7}{6}\}, \{\frac{5}{3}, \frac{17}{12}, \frac{23}{12}\} and \{\frac{−25}{12}, \frac{−11}{4}, \frac{−13}{6}\}. The sum of three exponents corresponding to three different singular points is never a non positive integer. Thus L_{p}^{∗}(y) = 0 has no right factor of order one. This shows, that for any value of p the differential equation L_{p}(y) = 0 is irreducible.

2. Imprimitivity: According to the the Necessary Conditions 2 the differential Galois group G(L_{p}) of this equation can not be an imprimitive group, since, because all exponents are non-negative and at 1 positive, the sum of the (possibly) repeated sum of three exponents at each singular point will never be non positive.

3. Finite primitive groups: Since the list of least common multiples of the denominators of the exponents at the singularities is [6, 12, 12], we get from the above theorem, that if G(L_{p}) is a finite primitive group, then G(L_{p}) is isomorphic to F_{36}^{SL_{3}} (note that the Necessary Conditions 3 would only exclude the cases G(L_{p}) ≅ A_{5}, G(L_{p}) ≅ A_{5} × C_{3} and G(L_{p}) ≅ G_{168}).

4. Infinite primitive groups: From Lemma 3.7 it follows that the the differential Galois group of L_{p}(y) = 0 cannot be a subgroup of ρ_{2}(SL(2, C)) × C_{3}, since at ∞ no exponent is one third of an integer.

Therefore our necessary conditions show that for any p, the group G(L_{p}) is isomorphic to either F_{36}^{SL_{3}} or SL(3, C).

For second order differential equations with three singular points, the method of the above Theorem always leads to at most one possible finite primite group, as the conditions found are exclusive:

Theorem 4.2 Let L(y) = 0 be a second order linear differential equation over C(x) having three singular points s_{1}, s_{2} and s_{3} and whose differential Galois group G(L) is a finite primitive subgroup of SL(3, C). Then for each possible group, all lists [lcm_{1}, lcm_{2}, lcm_{3}] of possible least common multiples lcm_{i} of the denominators d_{i,1}, d_{i,2} of the rational exponents \frac{n_{i,1}}{d_{i,1}} , \frac{n_{i,2}}{d_{i,2}} , where (n_{i,j}, d_{i,j}) = 1, at s_{i} can be derived.

1. If G(L) ≅ A_{5}^{SL_{2}}, we get: [3, 3, 10], [3, 4, 5], [3, 4, 10], [3, 5, 5], [3, 5, 6], [3, 5, 10], [3, 10, 10], [4, 5, 5], [4, 5, 6], [4, 5, 10], [4, 6, 10], [4, 10, 10], [5, 5, 6], [5, 5, 10], [5, 6, 10], [6, 6, 10], [6, 10, 10] or [10, 10, 10].

2. If G(L) ≅ S_{4}^{SL_{2}}, we get: [3, 4, 8], [3, 8, 8], [4, 6, 8] or [6, 8, 8].

3. If G(L) ≅ A_{4}^{SL_{2}}, we get: [3, 3, 4],[3, 3, 6], [3, 4, 6], [4, 6, 6] or [6, 6, 6].

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