Solving 1Odes with Elementary Functions

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Title of the software package: LeapS1ode – Liouvillian, elementary, algebraic and polynomial Solutions of 1ODEs (ﬁrst order ordinary diﬀerential equations).

Catalogue number: (supplied by Elsevier)

Software obtainable from: CPC Program Library, Queen’s University of Belfast, N. Ire- land.

Licensing provisions: none

Operating systems under which the program has been tested: Windows 10.

Programming languages used: Maple 17.

Memory required to execute with typical data: 200 Megabytes.

No. of lines in distributed program, including On-Line Help, etc.: 1037

Keywords: 1ODEs with functions, Polynomial vector ﬁelds, Liouvillian functions, S- function method.

Nature of mathematical problem Search for general solutions of ﬁrst order ordinary diﬀerential equations (1ODEs) with (elementary) functions.

Methods of solution The method of solution is based on an algorithm described in this paper.

Restrictions concerning the complexity of the problem If the 1ODE that is being analysed presents a very high degree in (x,y,θ), then the method may not work well.

Typical running time This depends strongly on the 1ODE that is being analysed.

Unusual features of the program Our implementation can ﬁnd solutions in many cases where the 1ODE under study can not be solved by other powerful solvers nor by other powerful methods. Besides that, the program can perform (in a very natural way) an analysis of the integrability region of the 1ODE’s parameters. Finally, the package presents some useful research commands.

2 LONG WRITE-UP Apart from their mathematical interest, dealing with first order differential equation (1ODEs) has its more direct application. Any physical (or, for that matter, any scientific question) that can be formulate as a relationship between the rate of change of some quantity of interest and the quantity itself can be formulated as solving an 1ODE (at some stage). There are many well known phenomena which fall into this category: Concentration and dilution problems are quickly remembered examples. Population models are another such example. More specifically, of non-linear 1ODEs and, as a final example, Astrophysics [4], etc. There are many others and there can be many more lurking around the corner of the many scientific endeavours being pursuit. In this sense, any novel mathematical approach, technique to deal with such 1ODEs are welcome and could prove vital to solve a funda- mental question. So, due to this great importance (theoretical and practical) of solving 1ODEs, many methods have been improved in the last decades. Regarding the search for elementary and Liouvillian solutions1, the methods that use the Darboux-Prelle-Singer approach stand out. In this ‘direction’ we can highlight: [5, 6, 7, 8, 9, 10, 11, 12, 14, 15, 16, 17, 18, 19, 20]. We have been studying the existence of (and methods to search for) elementary and Liouvillian first integrals of vector fields in R2 since 2002 [1, 2, 12, 35, 36, 37]. In particular, we tackled the problem of looking for general solutions for 1ODEs with functions in [1, 2]. However, the mathematical basis was not laid out in such a solid way and, in addition, the method did not deal well (it was computationally expensive) with the problem of determining Darboux polynomials in three or more variables. This time, we managed to combine the technique we used in [3]2 with a reasoning very close to the idea used in [1, 2]3. Our method basically starts with an assumption (the existence of a general solution of a specific type) to arrive at an algebraic result: the existence (or not) of a Liouvillian first integral (of a certain type) for a polynomial vector field in three variables and, in the case of a positive answer, the determination of that first integral (and, consequently,

1The methods that use the Lie symmetry approach have also been greatly improved. See, for example [21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34]. 2In that work we use the (so-called) S-function (introduced in [38] and used by us and other people [3, 39, 40, 37, 41, 42, 44, 45, 46, 47]) that improves greatly the eﬃciency on the searching for Liouvillian ﬁrst integrals of 2ODEs. 3The central idea is to ‘transform’ elementary functions into rational functions.

3 of the 1ODE’s general solution). Briefly, we associate the 1ODE (presenting an elementary function) with a polynomial vector field in three variables and then analyze the (possible) existence of a Liouvillian first integral. If the first integral is found we only have to perform the inverse transformation to obtain the general solution of the 1ODE in implicit form.

This paper is organized as follows: In the first section, we present some basic concepts involved in the Darboux- Prelle-Singer (DPS) approach and in the S-function method. In the second section, we show how we can associate a 1ODE with an elementary function to a polynomial vector field in three variables and develop a technique similar to the one we built in [3] (to search for Liouvillian first integrals of rational 2ODEs using the S-function method) in order to search for a Liouvillian first integral of the vector field. We use these two steps to propose a procedure that can succeed even in cases where the integrating factors have Darboux polynomials of very high degree. In the end of this section (after a few examples in order to clarify the steps of the procedure), we propose a semi algorithm to deal with solving 1ODEs presenting an elementary function and we describe it in a more formal way. In the third section, we present a computacional package (in Maple) to search for Liouvillian solutions of 1ODEs presenting an elementary function. We start by giving an overview of the package and presenting a summary of the commands. Then, we go deeper into the description of each command and, in the last subsection, we present an example of the usage of the commands to show how they work in practice. In the fourth section, we discuss the performance of the method. We begin by presenting a set of 1ODEs that the method solves without problems but, we believe, would not be easily solved by any other method. In the following, we present some special features of the package we use to solve / study more complicated cases. Finally, we show a case of a 2ODE (in principle, not treatable by the method) that the package can successfully handle. Finally, we present our conclusions.

1. Basic concepts and results In this section we will set the mathematical basis for presenting our method. In the ﬁrst subsection we will present the Darboux-Prelle-Singer ap-

4 proach in a nutshell and, in the second, the basics of the S-function method.

1.1. The Darboux-Prelle-Singer approach When we talk succinctly about the Darboux-Prelle-Singer approach to se- eking Liouvillian first integrals of autonomous systems of polynomial 1ODEs in the plane, we can summarize the central idea in one paragraph: Find the Darboux polynomials (DPs) associated with the 1ODE system and use them to determine an integrating factor. Without exaggeration, the determination of the DPs is the most important and (unfortunately) the most complex part (by far) of the whole process. Thus, with a couple of definitions and results, we can describe (briefly) the DPS approach when applied to vector fields in the plane. To begin, let’s define more formally the concepts we referred to. Consider the following system of polynomial 1ODEs in the plane:

x˙ = f(x, y) (1) ( y˙ = g(x, y), where f and g are coprime polynomials in (x, y) and the dot means derivative with respect to a parameter t (u ˙ ≡ du/dt).

Deﬁnition 1.1. A function I(x, y) is called a ﬁrst integral of the system (1) if I is constant over the solutions of (1).

Deﬁnition 1.2. The vector ﬁeld associated with the system (1),

X ≡ f(x, y) ∂x + g(x, y) ∂y, (2) is called Darboux operator associated with (1).

Remark 1.1. If I(x, y) is a ﬁrst integral of the system (1), then X(I)=0.

Deﬁnition 1.3. The polynomial p(x, y) ∈ C[x, y] is called a Darboux polynomial of the vector ﬁeld X if X(p)= qp, where q is a polynomial denomi- nated cofactor.

Theorem 1.1 (Prelle-Singer). If the system (1) presents an elementary ﬁrst integral I, then there exists an integrating factor R for the system (1) of the ni form R = i pi , where the pi are Darboux polynomials of system (1) and ni are rational numbers. Q 5 Proof. For a proof see [6].

X(R) ni Since R = − div(X)= −(fx + gy), substituting R = i pi we obtain (see [1]) X(p ) Q n i = n q = −(f + g ), (3) i p i i x y i i i X X where the qi are the cofactors of the pi. Therefore, a possible method to ﬁnd an elementary ﬁrst integral is:

Prelle-Singer method:(sketch)

• Determine the DPs pi associated with the system. • Find numbers ni that satisfy i ni qi = −(fx + gy).

• ni Construct the integrating factorP R = i pi and find a first integral I(x, y) by quadratures. Q Remark 1.2. If the system (1) presents a non elementary Liouvillian first integral the Prelle-Singer (PS) method needs some modifications (see the Christopher-Singer (CS) method in [12, 13]).However, the main premisse remains the same: determine the DPs of the system.

Remark 1.3. For vector fields in R3, things get a little more complicated: first, the integrating factor R is no longer a Jacobi multiplier (as it happens in R2). Thus, X(R)/R is no longer a known polynomial and depends on a rational function not determined a priori. Second, we need to determine Darboux polynomials in three variables (a very difficult task since it was no longer easy to determine them in two variables).

1.2. The S-function method The concept of S-function emerged when we were trying to adapt the DPS approach to rational second order ordinary diﬀerential equations (rational 2ODEs). The idea was to ’add’ a null 1-form to the 1-form that represented the 2ODE (see [38, 41]). However, the problem of calculating DPs was so costly (in computational terms) that we developed a mixed procedure we called ‘the S-function method’. Brieﬂy, the method did not use the computation of Darboux polynomials (in exchange for the resolution of

6 two 1ODEs). The good news is that the S-function method proved to be more eﬀective precisely in problems in which the Darboux polynomials hada relatively high degree, being a great alternative to be used in these cases. In this section, we will see what the method consists of and its basic operation.

1.2.1. Some basic definitions and results Consider the rational 2ODE given by dz M(x, y, z) z′ = = φ(x, y, z)= , (z ≡ y′), (4) dx N(x, y, z) where M and N are coprime polynomials in C[x, y, z]. Definition 1.4. A function I(x, y, z) is called first integral of the 2ODE (4) if I is constant over the solutions of (4). Remark 1.4. If I(x, y, z) is a first integral of the 2ODE (4) then, over the solution curves of (4), the exact 1-form ω := dI = Ix dx + Iy dy + Iz dz is null. Over the solution curves of (4), we have two independent null 1-forms: α := φ dx − dz, (5) β := z dx − dy. (6) So, the 1-form ω is in the vector space sppaned by the 1-forms α and β, i.e., we can write ω = r1(x, y, z) α + r2(x, y, z) β:

dI = Ix dx + Iy dy + Iz dy = r1 (φ dx − dz) + r2 (z dx − dy)

= (r1 φ + r2 z)dx +(−r2)dy +(−r1)dz. (7)

If we call r ≡ R and r2 ≡ S , we can write 1 r1 dI = R [(φ + zS)dx − S dy − dz] . (8)

Therefore, we have that Ix = R (φ + zS), Iy = −RS, Iz = −R. Definition 1.5. Let γ be a 1-form. We say that R is an integrating factor for the 1-form γ if Rγ is an exact 1-form. Definition 1.6. Let I be a first integral of the 2ODE (4). The function defined by S := Iy/Iz is called a S-function associated with the 2ODE through the first integral I.

7 Remark 1.5. From the above deﬁnitions and in view of (8) we can see that R is an integrating factor for the 1-form (φ + zS)dx − S dy − dz and S is a S-function associated with the 2ODE (4) through I .

Theorem 1.2. Let I(x, y, z) be a ﬁrst integral of the 2ODE (4). If S and R are as in (8), then we can write

Dx(R)+ R (S + φz)=0 , (9)

SDx(R)+ R (Dx(S)+ φy)=0 , (10)

−(Rz S + RSz)+ Ry =0 , (11) where Dx := ∂x + z ∂y + φ(x, y, z) ∂z .

Proof: See [41].

Corollary 1.1. Let S be a S-function associated with the 2ODE z′ = φ(x, y, z). Then S obeys the following equation:

2 Dx(S)= S + φz S − φy . (12)

Proof: See [3]

1.2.2. The Associated 1ODEs From (12) we can see that a S-function associated with the rational 2ODE (4) satisﬁes a quasilinear 1PDE in the variables (x, y, z). Over the solutions of the 2ODE (4) we have that y = y(x) and z = z(x) and, therefore, the d operator Dx is, formally, dx . So, formally, over the solutions of the 2ODE (4) 2 we can write the 1PDE (12) as a Riccati 1ODE: ds/dx = s + φz s − φy. It is r′(x) of common knowledge that the transformation y(x)= − f(x) r(x) changes the Riccati equation y′(x)= f(x) y(x)2 + g(x) y(x)+ h(x) into the linear 2ODE r′′ = ((f ′(x)+ g(x) f(x)) r′)/f(x) − f(x) h(x) r. So, with the transformation s(x) = −w′(x)/w(x), the Riccati 1ODE turns (over the solutions of the ′′ ′ 2ODE (4)) into the homogeneous linear 2ODE w = φz w +φy w. Therefore, d we can use the formal equivalence Dx ∼ dx to produce a connection between the S-functions and the symmetries (written in a particular form) of the 2ODE. Applying the transformation

D (ν) S = − x (13) ν

8 into equation (12). We obtain:

2 Dx(ν)= φz Dx(ν)+ φy ν. (14)

The equation (14) is the symmetry condition for ν to be an inﬁnitesimal that deﬁnes a symmetry generator in the evolutionary form. So, we have:

Theorem 1.3. Let ν be a function of (x, y, z) such that [0, ν] deﬁnes a symmetry of the 2ODE (4) in the evolutionary form, i.e., Xe := ν ∂y ge- nerates a symmetry transformation for (4). Then the function deﬁned by S = −Dx(ν)/ν is a S-function associated with the 2ODE (4).

Proof: See [3]

Corollary 1.2. Let S be a S-function associated with the 2ODE (4). Then, the function ν given by (−S) ν ≡ eRx , (15)

(where x is the inverse operator of Dx , i.e., x Dx = Dx x = 1) deﬁnes a symmetry of the 2ODE (4) in the evolutionary form. R R R Proof: See [3]

The theorem 1.3 and corollary 1.2 establish a connection between S- functions and Lie symmetries. It is this connection that allowed us to avoid the use of Darboux polynomials in the searching for first integrals for the 2ODE (4). The main concept is the associated 1ODE4, which is an 1ODE that has its general solution defined by one of the first integrals of the 2ODE.

Definition 1.7. Let I be a first integral of the 2ODE (4) and let S(x, y, z) be the S-function associated with (4) through I. The first order ordinary differential equation defined by dz = −S(x, y, z), (16) dy where x is taken as a parameter, is called 1ODE[1] associated with (4) through I.

4This concept was developed in [41], page 222.

9 Theorem 1.4. Let I be a ﬁrst integral of the 2ODE (4) and let S(x, y, z) be the S-function associated with (4) through I. Then I(x, y, z) = C is a general solution of the 1ODE[1] associated with (4) through I. Proof: See [3]

Remark 1.6. The variable x (the independent variable of the 2ODE (4)) is just a parameter in the 1ODE (16), so, since any function of x is an invariant for the operator Da ≡ ∂y − S ∂z, i.e., Da(x)=0, the relation between a general solution H(x, y, z) = K of the 1ODE (16) and the ﬁrst integral I(x, y, z) of the 2ODE (4) is I(x, y, z)= F (x, H), such that Dx(I)= ∂x(F)+(Hx + zHy + φHz) ∂H (F)=0.

in a short way, the S-function method consists in determining the S- function, solving the associated 1ODE and solving the 1ODE that represents the characteristic system of the partial diﬀerential equation for F.

Remark 1.7. Some comments:

• Since the S-function is deﬁned as the division of Iy by Iz, we can think if the divisions Ix/Iz and Ix/Iy would have any meaning or utility. The answer is yes: We can deﬁne the the S-functions {S1,S2,S3} with similar meanings (see [3]).

• The main advantage of the S-function method is that, in general, it is more eﬃcient to determine the S-function than computing Darboux polynomials precisely in the cases where they (the DPs) have high degrees.

• A great advantage of having three S-functions is that the diﬃculty associated with determining each one of them can be very diﬀerent, i.e., it can be much easier to calculate one of them than the others.

• for more details (and examples) of the application of the method we recommend that the reader see the reference [3].

10 2. A new method to solve 1ODEs with an elementary function In this section we will present a method to find the general solution of 1ODEs with an elementary function5. In the first subsection we will show an equivalence between a certain type of 1ODEs and polynomial vector fields in three variables. In the second, we will show how to use the technique developed in [3] to find a Liouvillian first integral of a polynomial vector field in three variables.

2.1. 1ODEs with functions × polynomial vector ﬁelds The method we build is applicable to 1ODEs that can be written in the form dy M(x, y, θ) y′ = = φ(x, y, θ)= (17) dx N(x, y, θ) where φ is a rational function of (x, y, θ), i.e., φ ∈ K = C(x, y, θ)6, θ is an elementary generator over C(x, y)7 and M and N are coprime polynomials in (x, y, θ), i.e., M, N ∈ C[x, y, θ]8. Furthermore, if I(x, y, θ) = c represents a general solution of 1ODE (17) and I is a Liouvillian function of (x, y, θ) then for the method to work it is suﬃcient that the partial derivatives of I in relation to x, y and θ can be expressed as

Ix = RP1 (18)

Iy = RP2 (19)

Iθ = RP3 (20) where R is an elementary function of (x, y, θ) and P1, P2, P3 ∈ C[x, y, θ] (i.e., P1, P2 and P3 are polynomials in (x, y, θ) ). The main idea is to perform the transformation z = θ(x, y) that allows the association of 1ODE (17) with a polynomial vector ﬁeld that has I(x, y, z) as a Liouvillian ﬁrst integral. Let’s see how this happens:

• In ﬁrst place, since Dx(I) = 0 where Dx ≡ ∂x + φ ∂y, we have that ∂I ∂I ∂θ ∂I ∂I ∂θ D (I)= ∂ (I)+ φ ∂ (I)= + + φ + . (21) x x y ∂x ∂θ ∂x ∂y ∂θ ∂y

5This version of the method will be limited to non algebraic elementary functions. 6K = C(x,y,θ) is the differential field of rational functions in the variables (x,y,θ). 7This is a formal way of saying that θ is an elementary function of (x, y) (see [48]). 8C[x,y,θ] is the differential ring of polynomial functions in the variables (x,y,θ).

11 • Second, since θ is an elementary generator over C(x, y), (i.e., θ is in C ∂θ ∂θ an elementary extension of (x, y)) then the derivatives ∂x , ∂y ∈ K = C(x, y, θ).

• In view of this, we can write Dx(I) as

∂I ∂I ∂θ ∂θ ∂I D (I) = + φ + + φ = x ∂x ∂y ∂x ∂y ∂θ = Ix + φIy +(θx + φ θy) Iθ, (22)

∂θ ∂θ where θx ≡ ∂x and θy ≡ ∂y are rational functions of (x, y, θ).

• From the fact that φ, θx and θy are rational functions of (x, y, θ), we can multiply Dx(I) by the lcm of the denominators of φ, θx and θy φ, obtaining

lcm(denominator(φ, θx, θy φ)) Dx(I)=(f ∂x+g ∂y+h ∂θ)(I(x, y, θ))=0, (23) where f, g and h are polynomials in (x, y, θ) and

f = lcm(denominator(φ, θx, θy φ)), (24)

g = lcm(denominator(φ, θx, θy φ)) φ, (25)

h = lcm(denominator(φ, θx, θy φ)) (θx + φ θy) . (26)

• If we make the substitution θ → z, we have that

(f(x, y, z) ∂x + g(x, y, z) ∂y + h(x, y, z) ∂z)(I(x, y, z))=0. (27)

Therefore, following these steps we can associate the polynomial vector ﬁeld

χ ≡ f(x, y, z) ∂x + g(x, y, z) ∂y + h(x, y, z) ∂z (28) with the 1ODE (17). Moreover, we can associate the first integral I(x, y, z) of χ (see equation (27)) with the general solution of the 1ODE (17) (i.e., I(x, y, θ)= c). So, if we can find a Liouvillian first integral I(x, y, z) for the vector field χ, we will have found the general solution of 1ODE (17): we only have to apply the substitution z → θ on I(x, y, z)= c.

12 2.2. S-function method adapted to polynomial vector fields in three variables In this section we will show how we can adapt the method described in [3] (S-function method) to polynomial vector fields in three variables. Consider that I(x, y, z) is a Liouvillian first integral of the vector field χ ≡ f ∂x+g ∂y +h ∂z, such that the derivatives Ix, Iy and Iz are given, respectively, by RP1, RP2 and RP3 (see the section 2.1). Now, supose that I(x, y, z) is a Liouvillian first integral of a (hypothetical) rational 2ODE such that x is the independent variable, y is the dependent variable and z is dy/dx. This hypothetical 2ODE would be given by I + zI P + zP M (x, y, z) z′ = Φ(x, y, z)= − x y = − 1 2 = 0 , (29) Iz P3 N0(x, y, z) where M0 and N0 are coprime polynomials in (x, y, z). As we saw in section 1.2, the S-functions S1, S2 and S3 associated with this 2ODE through the first integral I are given by the following rational functions Iy P2 Ix P1 Ix P1 S1 = = , S2 = = , S3 = = , (30) Iz P3 Iz P3 Iy P2 where S1, S2 and S3 obey the equations

2 Dx(S1) = S1 + Φz S1 − Φy, (31) 1 Φ D (S ) = − S 2 + Φ − S − Φ , (32) x 2 z 2 z z 2 x Φ Φ − z Φy D (S ) = − y S 2 + x S + z Φ , (33) x 3 Φ 3 Φ 3 x where Dx ≡ ∂x +z ∂y +Φ ∂z. Since we do not have the function Φ, we cannot use the equations (31), (32) and (33) to find the S-functions S1, S2 and S3 associated with the hypothetical 2ODE (29). So, in order to use the technique developed in [3], we will need some relationship between the (hypothetical) 2ODE and the vector field (28). We will establish this relationship using the fact that the 2ODE (29) and the vector field χ have in common the first integral I. This fact allows us to establish a result that will give us a way to apply the method used in [3] to our problem. Let’s begin with the following definition:

Definition 2.1. Let f ∂x + g ∂y + h ∂z be a polynomial vector field in three variables presenting a first integral I(x, y, z) and let z′ = Φ(x, y, z) be a

13 rational 2ODE such that it also has I(x, y, z) as a first integral. We say that the 2ODE and the vector field are associated through the first integral I. Remark 2.1. Note that, since the status of the variables (x, y, z) in the vector field f ∂x +g ∂y +h ∂z are the same, there are, in principle, six distinct 2ODEs associated with it:

Ix + zIy Φ1 = − , (34) Iz

Iy + zIx Φ2 = − , (35) Iz

Ix + yIz Φ3 = − , (36) Iy

Iz + yIx Φ4 = − , (37) Iy

Iy + xIz Φ5 = − , (38) Ix

Iz + xIy Φ6 = − . (39) Ix This will be useful in the future process of finding the first integral I through the S-function method (see section 2). Theorem 2.1. Let χ, defined by (28), be a polynomial vector field presenting a Liouvillian first integral I and let the rational 2ODE defined by (29) be associated with the vector field χ through I. If their derivatives are given by

Ix = RP1, (40)

Iy = RP2, (41)

Iz = RP3, (42) where R is a Darboux function of (x, y, z) and P1, P2, P3 ∈ C[x, y, z] then the S-function S1 associated with the rational 2ODE (29) is given by

M0 f − N0 h S1 = , (43) N0 (g − z f) where g − z f =06 .

14 Proof of Theorem 2.1: From the hypotheses we have that

D0(I) = N0 ∂x(I)+ z N0 ∂y(I) + M0 ∂z(I)=0, (44)

χ(I) = f ∂x(I)+ g ∂y(I) + h ∂z(I)=0. (45) where M0 and N0 are, respectively, the numerator and denominator of the 2ODE (29). Deﬁning D1 ≡ f D0 − N0 χ, and applying it to I we obtain:

D1(I)=(f D0−N0 χ)(I)=(z f−g) N0 ∂y(I)+(M0 f−N0 h) ∂z(I)=0. (46)

From eq.(46) we have (see [3]) that Iy/Iz = −(M0 f − N0 h)/(N0 (z f − g)) = S1.

Remark 2.2. We note that the function S1 as presented in the equation (43) ‘defines’, in a certain sense, the relationship between the vector field χ and the 2ODE (29). So, we can use it to produce an algorithm (semi) to find the 2ODE (M0 and N0) and, in view of (43), the S-function.

2.3. A possible method In this section, we will use the equation (43) in the process of building an algorithm to ﬁnd the general solution of 1ODE (17). Before describing the steps of a possible algorithm in more detail, let’s outline the main steps of the process and discuss its application to some examples in order to materialize (and clarify) the path we are trying to follow to achieve our goal. The Procedure: (Sketch)

• If the 1ODE can be placed in the form (17) then construct the operator D (22).

• Make the substitution θ → z on the D operator and ﬁnd the polynomial operator χ (28).

• Construct polynomial candidates Mc and Nc (with undetermined coef- ﬁcients {mi} and {nj}) of some chosen degree in the variables (x, y, z).

15 • Substitute Mc f − Nc h S1 = Nc (g − z f) in the equation for the S-function: ∂Φ ∂Φ D (S )= S 2 + c S − c , x 1 1 ∂z 1 ∂y

where Dx ≡ ∂x + z ∂y + Φc ∂z and Φc ≡ Mc/Nc.

• Solve the polynomial equation for the unknown coefficients {mi} and {nj}, obtaining S1. • Use the S-function method to find de Liouvillian first integral I of the 2ODE z′ = Φ(x, y, z). • Make the substitution z → θ on the first integral I and obtain the desired general solution of the 1ODE.

Before presenting a more formal algorithm let’s discuss some examples: Example 1: Consider the 1ODE given by dy exx3y2 + exx2y2 + 2 exx2y + exxy + exx + y2 + ex = (47) dx x2y2 + exx2 + xy +1 and let’s apply the procedure sketched above:

x x • Choosing θ = e , we have that θx = e = θ and θy = 0. So, the operator D will be: θ x3y2 + θ x2y2 +2 θ x2y + θ xy + θ x + y2 + θ D = ∂ + ∂ +θ ∂ . (48) x x2y2 + θ x2 + xy +1 y θ

• The polynomials f, g and h will be: f = x2y2 + zx2 + xy +1, (49) g = zx3y2 + zx2y2 +2 zx2y + zxy + zx + y2 + z, (50) h = z x2y2 + zx2 + xy +1 . (51)

The operator χ will be:

χ = f ∂x + g ∂y + h ∂z. (52)

16 • For the degrees degM = 4 and degN = 5 we obtain:

M0 = − (xz − y)(xz + y) , (53) 2 N0 = −x (xy + 1) , (54) x2y2 + x2z + xy +1 S1 = − (55) x (xy + 1)2

• Using S1 in the S-function method, we have the following ﬁrst integral:

1 I = e xy+1 (zx − y) (56)

• Making the substitution z = ex we ﬁnally arrive at the solution:

1 x e xy+1 (x e − y)= C (57)

Now let’s take a closer look at the derivatives of the ﬁrst integral I (of the vector ﬁeld χ):

1 e xy+1 I = zx2y2 + zxy + y2 + z , (58) x (xy + 1)2

1 e xy+1 I = − x2y2 + zx2 + xy +1 , (59) y (xy + 1)2

1 e xy+1 I = x(xy + 1)2. (60) z (xy + 1)2

From them we can determine (directly) the integrating factor R, the 2ODE associated (Φ), the polynomials P1, P2 and P3 and the S-functions S1, S2

17 and S3: x2z2 − y2 Φ = , (61) (xy + 1)2 x

1 e xy+1 R = , (62) (xy + 1)2

2 2 2 P1 = x y z + xyz + y + z, (63)

2 2 2 P2 = − x y + x z + xy +1 , (64)

2 P3 = x(xy + 1) , (65) x2y2 + zx2 + xy +1 S1 = − , (66) (xy + 1)2 x zx2y2 + zxy + y2 + z S2 = , (67) (xy + 1)2 x zx2y2 + zxy + y2 + z S = − . (68) 3 x2y2 + zx2 + xy +1 The components (f,g,h) of the vector ﬁeld χ are (see above) f = x2y2 + zx2 + xy +1, (69) g = zx3y2 + zx2y2 +2 zx2y + zxy + zx + y2 + z, (70) h = z x2y2 + zx2 + xy +1 . (71)

Remark 2.3. So, comparing (f,g,h) with the polynomials P1, P2 and P3 and with the numerators and denominators of the S-functions, we can see that: N0 = P3, f = −P2, h = −zP2. (72)

The ﬁrst of these equations may not surprise, since Φ = −(P1 + zP2)/P3. However, the following two equations are (at ﬁrst glance) intriguing.

Example 2: Consider now the 1ODE given by dy −ey (e2yx2y − 2 eyxy2 − xeyy + y3 − 1) = . dx e3yx3y+e3yx3 − 2e2yx2y2 − 3e2yx2y+ eyxy3 + eyxy2 + xeyy− xey +1 (73)

18 Applying the whole procedure again we obtain

f = x3yz3 +z3x3 −2x2y2z2 −3z2x2y+xy3z+zxy2 +xzy−zx+1, (74) g = −z z2x2y − 2 zxy2 − xzy + y3 − 1 , (75) h = −z2 z2x2y − 2 zxy2 − xzy + y3 − 1 . (76) The function S1 is z3x3 − 2 z2x2y + zxy2 + xzy +1 S = . (77) 1 x (z2x2y − 2 zxy2 − xzy + y3 − 1)

From S1 we can determine the ﬁrst integral (for the vector ﬁeld χ) and (making the substitution z = ey) the general solution of the 1ODE (73):

1 1 y I = e zx−y (xzy + 1) , e eyx−y (xe y +1) = C. (78)

From the derivatives of the ﬁrst integral I we can get Φ, the polynomials P1, P2 and P3 and the S-functions S1, S2 and S3:

(zx − y)2 (zx + y) z Φ = , (79) x (z2x2y − 2 zxy2 − xzy + y3 − 1)

2 2 2 3 P1 = z z x y − 2 zxy − xzy + y − 1 , (80)

3 3 2 2 2 P2 = z x − 2 z x y + zxy + xzy +1, (81)

2 2 2 3 P3 = x z x y − 2 zxy − xzy + y − 1 , (82) z3x3 − 2 z2x2y + zxy2 + xzy +1 S = , (83) 1 x (z2x2y − 2 zxy2 − xzy + y3 − 1) z S = , (84) 2 x z (z2x2y − 2 zxy2 − xzy + y3 − 1) S = . (85) 3 z3x3 − 2 z2x2y + zxy2 + xzy +1

19 Remark 2.4. Again, if we compare (f,g,h) with the polynomials P1, P2 and P3 and with the S-functions, we see that:

N0 = P3, g = −P1, h = −zP1. (86) Besides that, as the roles of x and z can be switched in the ﬁrst integral I (i.e., the transformation x → z, z → x is a symmetry transformation for I) the S-function S2 has a very simple format.

Example 3: Let’s see now an example where we have a function of x and y. Consider the 1ODE

x 5 2 3 3 4 2 2 2 dy xy ln y +(xy − 2 x y + x y + y + x y − 2 xy + x ) y = . dx 2 x 5 2 3 3 3 4 2 2 2 xy ln y + x (−xy +2 x y − x y +2 xy + y − 2 xy + x ) (87) So f = x xy5 − 2 x2y3 + x3y − 2 xy3 − y4 +2 xy2 − 2 zy2 − x2 , (88) g = − xy5 − 2 x2y3 + x3y + y4 + x2y − 2 xy2 + x2 + zx y, (89) h = 2 xy 5 − 4 x2y3 +2 x3y − 2 xy3 + x2y − 2 zy2 + zx. (90)

The function S1 is xy4 − 2 x2y2 + x3 − 2 xy2 − 2 yz S = . (91) 1 y4 − 2 xy2 + x2 The ﬁrst integral and the general solution of the 1ODE are

1 1 x I = e y2−x (xy + z) , e y2−x xy + ln = C. (92) y The polynomials P1, P2, P3 and Φ are:

5 3 2 P1 = y − 2 xy + x y + xy + z, (93)

4 2 2 3 2 P2 = xy − 2 x y + x − 2 xy − 2 yz, (94)

2 2 P3 = −y + x , (95) xy4z−2x2y2z+y5 +x3z−2xy3 −2xy2z+x2y−2yz2 +xy+z Φ = . (96) (−y2 + x)2

20 Remark 2.5. Some comments:

1. This time, looking at the polynomials P1, P2 and P3 in the example 3, we see that none of them is a member of {cf,cg,ch} where c is a constant. 2. A very noticeable difference between examples 1 and 2 and example 3 is that in examples 1 and 2 the elementary function present on the 1ODE was a function of only one variable (x or y) whereas in example 3 the function θ was a function of the two variables (x, y), i.e., θ = θ(x, y). 3. In general it is much simpler (computationally speaking) to calculate the S-function if we already know its numerator or denominator. In these cases the algorithm is much more efficient (i.e., faster and with less memory consumption). 4. What happened in examples 1 and 2 was not a fluke. We can show that, in almost all cases, if θ is an elementary function of only one variable, then one of the coefficients of the vector field χ will divide one of the polynomials (P1,P2,P3). 5. The good news is that, if θ is an elementary function of a rational function, in a lot of cases we can perform a variable transformation that takes θ(x, y) into θ(x) (or θ(y)).

2.4. A possible improvement What was stated in observation 4 (remark 2.5) can be demonstrated and (in conjunction with the fact presented in observation 5) used to greatly improving the efficiency of the algorithm in a wide variety of cases9. Before presenting the alternative method, let’s define (a little more) the type of 1ODE that these improvement can handle. Definition 2.2. Consider that an 1ODE as described by equation (17) has the following characteristics: 1. I(x, y, θ(x, y)) = c (c constant) represents a general solution of the 1ODE (17) and I is a Liouvillian function of (x, y, θ). 2. The function θ(x, y) is in an elementary extension E of the differential field (of characteristic zero) C(x, y) such that E = C(x, y, exp(r)) or E = C(x, y, ln(r)), where r ∈ C(x, y).

9However, although the improvement we are going to present is quite general, the ‘problem solution’ can, in certain cases, introduce complexity to the process, almost canceling the eﬃciency gain obtained.

21 3. The 1ODE is associated with a vector ﬁeld χ ≡ f ∂x + g ∂y + h ∂z such that f,g,h ∈ C[x, y, z] and

f = lcm(denominator(φ, ∂x(θ), ∂y(θ) φ))|θ→z,

g = lcm(denominator(φ, ∂x(θ), ∂y(θ) φ)) φ|θ→z,

h = lcm(denominator(φ, ∂x(θ), ∂y(θ) φ)) (∂x(θ)+ φ ∂x(θ)) |θ→z. 4. The derivatives of the ﬁrst integral I (of the vector ﬁeld χ) are of the ni form: ∂k(I) = exp(A/B) i pi Pk, k ∈{1, 2, 3}, where A,B,pi,Pk are polynomials in C[x, y, z] and ni are integers. 5. Let T be the transformationQ T = {u = r(x, y), v = y} (or T = {u = x, v = r(x, y)}), where r ∈ C(x, y) is the argument of θ. If T −1 = {x = −1 r1(u, v),y = v} (or T = {x = u,y = r1(u, v)}) denote its inverse, we are assuming that r1 is a rational function of (u, v), i.e., r1 ∈ C(u, v).

We will denote the set of 1ODEs that have these characteristics as LS .

Now, we are going to present a result that, if applied to an 1ODE ∈ LS , can be used to construct an improvement of the method sketched above.

Theorem 2.2. Let χ ≡ f ∂x + g ∂y + h ∂z be the polynomial vector field associated with the 1ODE (17) (φ = M/N, M,N ∈ C[x, y, θ]) and let the first integral I (of the vector field χ) be such that its derivatives are of the form: ni ∂k(I) = exp(A/B) i pi Pk, k ∈{1, 2, 3}, where A,B,pi,Pk are polynomials in C[x, y, z] and ni are rational numbers. We can state that: x Q i) If θ = e then f|P2, h = z f and g|(P1 + zP3). ii) If θ = ln(x) then h|P2, f = x h and g|(xP1 + P3). y iii) If θ = e then g|P1, h = zg and f|(P2 + zP3). iv) If θ = ln(y) then h|P1, g = y h and f|(yP2 + P3).

Proof. We have seen that we can write the operator D ≡ ∂x+φ ∂y – associated with the 1ODE (17) – in the form D = ∂x + φ ∂y +(θx + φ θy) ∂θ, where θx and θy are rational functions of (x, y, θ). In order to obtain the vector ﬁeld χ we have to multiply D by the lcm of the denominators of φ, θx and θy φ and make the substitution θ → z, obtaining χ = f ∂x + g ∂y + h ∂θ, where f, g and h are polynomials given by

f = lcm(denominator(φ, θx, θy φ))|θ→z,

g = lcm(denominator(φ, θx, θy φ)) φ|θ→z,

h = lcm(denominator(φ, θx, θy φ)) (θx + φ θy) |θ→z.

22 Now we can prove the statements: x i): If θ = e then θx = z and θy =0 and so

f = N(x, y, z), g = M(x, y, z), h = z N(x, y, z). (97)

We have that χ(I)=fIx+gIy+hIz = fRP1+gRP2+hRP3 (by hypothesis) = R (fP1 +gP2 +hP3) = 0. Substituting (97) and noting that R is not null we can write NP1 + MP2 + zNP3 = 0 implying that N (P1 + zP3)= −MP2. Therefore, we can write

MP N (P + zP ) P + zP = − 2 and P = − 1 3 . (98) 1 3 N 2 M

Since M and N are coprime, we have that N|P2 ⇒ f|P2 and M|(P1 + zP3) ⇒ g|(P1 + zP3). ii): If θ = ln(x) ⇒ θx =1/x and θy = 0. We have two cases:

f = x N(x, y, z), g = x M(x, y, z), h = N(x, y, z), (99) if x is not a factor of N or

n n−1 f = x PN (x, y, z), g = M(x, y, z), h = x PN (x, y, z), (100)

n where PN = N/x is a polynomial, n is a positive integer and x and PN are coprime. First case (x and N are coprime): N (xP1 + P3) = −xMP2. Therefore, N|P2 ⇒ h|P2 and x M|(xP1 + P3) ⇒ g|(xP1 + P3). n−1 Second case (x|N): x PN (xP1 + P3) = − MP2. Since M and N are n−1 coprime, we have that x PN (= h) and M are coprime. Therefore, h|P2 and g|(xP1 + P3). y iii): If θ = e then θx = 0 and θy = z and so

f = N(x, y, z), g = M(x, y, z), h = z M(x, y, z). (101)

We can write NP1+MP2+zMP3 = 0 implying that NP1 = −M (P2+zP3). Therefore, we can write

NP M (P + zP ) P + zP = − 1 and P = − 2 3 . (102) 2 3 M 1 N

23 Since M and N are coprime, we have that N|(P2 + zP3) ⇒ f|(P2 + zP3) and M|P1 ⇒ g|P1. iv): If θ = ln(y) ⇒ θx = 0 and θy =1/y. We have two cases:

f = y N(x, y, z), g = y M(x, y, z), h = M(x, y, z), (103) if y is not a factor of M or

n n−1 f = N(x, y, z), g = y PM (x, y, z), h = y PM (x, y, z), (104)

n where PM = M/y is a polynomial, n is a positive integer and y and PM are coprime. First case (y and M are coprime): NyP1 = −M (yP2 + P3). Therefore, M|P1 ⇒ h|P1 and y N|(yP2 + P3) ⇒ f|(yP2 + P3). n−1 Second case (y|M): y PM (yP2 + P3) = − NP1. Since M and N are n−1 coprime, we have that y PM (= h) and N are coprime. Therefore, h|P1 and f|(yP2 + P3).

Corollary 2.1. If the 1ODE (17) ∈ LS, then there is a rational transforma- −1 tion T (with rational inverse T ) such that the transformed 1ODE[tr] has an associated vector ﬁeld χ[tr] in which fi|Pj for some fi ∈{f,g,h} and for some Pj ∈{P1,P2}.

Proof. For any 1ODE ∈ LS we can perform a variable transformation that takes θ(x, y) into exp(x), ln(x), exp(y) or ln(y). Thus, any 1ODE ∈ LS can be transformed into an 1ODE that fulfills the premises of the theorem 2.2. Therefore, the conclusion is a direct consequence of the theorem 2.2. In this point we can use the compatibility conditions for the first integral I of the vector field χ to write the 1PDE for S in terms of f, g and h.

Theorem 2.3. Let χ ≡ f ∂x + g ∂y + h ∂z be the polynomial vector ﬁeld associated with the 1ODE (17) and let I be a ﬁrst integral it. Then the S-function associated with χ through I obeys the 1PDE given by

fg − gf gf − fg + fh − hf fh − hf χ(S)= S2 z z + S y y z z − y y . f f f (105)

24 Proof. See the proof of theorem 1.1 and corollary 1.1 of [3].

Since S = Iy/Iz = P2/P3 and the status of the variables (x, y, z) in the vector ﬁeld f ∂x + g ∂y + h ∂z is the same, we can perform the following variable transformations:

• In ﬁrst place we can use the transformation T1 such that its inverse is −1 given by T1 = {y1 = x, x1 = r(x, y)} (where r(x, y) is the argument of θ) to obtain an 1ODE[1] such that θ[1] = exp(x1) or θ[1] = ln(x1).

• If θ[1] = exp(x1) we can use the transformation T2 = {x1 = ln(x2),y1 = y2} and obtain an 1ODE[2] such that θ[2] = ln(x2). • For the 1ODE[2], the vector ﬁeld χ2 associated is given by f2 ∂x2 +

g2 ∂y2 + h2 ∂z2 , where (To make the notation less heavy we omitted index 2 in the following equations):

f = x N(x, y, z), g = x M(x, y, z), h = N(x, y, z), (106)

if x is not a factor of N or

n n−1 f = x PN (x, y, z), g = M(x, y, z), h = x PN (x, y, z), (107)

n where PN = N/x is a polynomial, n is a positive integer and x and PN are coprime (see case (ii) of theorem 2.2).

• Since the status of the variables (x2,y2, z2) in the vector ﬁeld χ2 is the same, we can make the transformation T3 = {x2 = y3,y2 = z3, z2 = x3} that will make P2 take the place of P3 (which is the denominator of the transformed S-function). The transformed vector ﬁeld χ3 will be given

by f3 ∂x3 + g3 ∂y3 + h3 ∂z3 , where

f3 = T3(h2),

g3 = T3(f2),

h3 = T3(g2).

At this point we can use the equation (105) to determine the numerator of the transformed S-function S1[3]. We can use S1[3] to find a first integral −1 I[3] for the vector field χ[3]. Applying the inverse trasformation T3 to I[3]

25 we obtain the ﬁrst integral I[2] for χ[2]. From I[2] we can write the general solution of 1ODE[2] (using z → ln(x)). If we apply the inverse trasformation −1 −1 −1 −1 T2 followed by T1 to I[2] = c, we will obtain I = T1 (T2 (I[2])) = c, which represents the general solution of the original 1ODE. Let’s make these steps clearer with an example:

Example 4: Consider the 1ODE given by

dy 2 (exy)2 x3y2 − exyx3y2 + exyxy3 − 5 exyx2y + exyy2 +2 x = . (108) dx (exy)2 x2y2 + exyx4y − exyx2y2 + exyx3 − 3 xyexy +1

• In ﬁrst place, let’s make the transformation T1 = {x = x/y, y = y} leading to:

dy y (2 e2 xx3 + exxy3 − exx3 + exy3 − 5 exx2 +2 x) = . (109) dx e2 xx2y3 + 2 e2 xx4 − 2 exxy3 − 4 exx3 + y3 +2 x2

• Next we’re going to apply the transformation T2 = {x = ln(x), y = y} obtaining:

dy y(2 x2 ln(x)3 + x ln(x)y3 − x ln(x)3 + xy3 − 5 x ln(x)2 + 2 ln(x)) = . dx (x2 ln(x)2y3 +2 x2 ln(x)4 −2 x ln(x)y3 −4 x ln(x)3 + y3 + 2 ln(x)2)x (110)

• The vector ﬁeld χ[2] associated with the 1ODE[2] is deﬁned by:

3 2 2 f2 = y +2 z (xz − 1) x, (111)

2 3 3 3 3 2 g2 = y 2 x z + zxy + xy − xz − 5 z x +2 z , (112)

3 2 2 h2 = y +2 z (xz − 1) . (113) • Let’s apply the transformation T3 = {x = y,y = z, z = x} obtaining:

3 2 2 f3 = z +2 x (xy − 1) , (114)

3 2 2 g3 = z +2 x (xy − 1) y, (115)

3 2 3 3 3 2 h3 = z 2 x y + xyz − x y + yz − 5 x y +2 x . (116)

26 • Constructing a polynomial P with arbitrary coeﬃcients in (x, y, z) and substituting in the equation (105), we obtain (for degree 5):

3 2 3 2 x z (−z + x ) P = x z −z + x → S1[3] = . (117) (z3 +2 x2)(xy − 1)2 • Using the S-function method we can find a first integral for χ[3]: 1 I = − ln z3 − x2 + 2 ln (z) − . (118) [3] xy − 1 −1 • Applying T3 to I[3] we obtain: 1 I = − ln y3 − z2 + 2 ln (y) − . (119) [2] xz − 1 • From I[2] we can write the general solution of 1ODE[2] (z → ln(x)): 1 − ln y3 − (ln (x))2 + 2 ln (y) − = C . (120) x ln (x) − 1 2 −1 x • Applying T2 = {x = e , y = y} to the general solution of 1ODE[2] we arrive at the general solution of 1ODE[1]: 1 − ln y3 − x2 + 2 ln (y) − = C . (121) xex − 1 1 −1 • Finally, applying T1 = {x = xy, y = y} to the general solution of 1ODE[1] we have the general solution of the original 1ODE: 1 − ln −x2y2 + y3 + 2 ln (y) − = C. (122) xyexy − 1 Remark 2.6. Although the application of a sequence of transformations followed by the sequence of inverse transformations (applied in reverse order) is a bit tiring to follow, in practice these steps have no algorithmic weight, i.e., they consume virtually no memory or CPU time. The important point is that these transformations lead to a vector field whose associated S-function (to be determined) depends on obtaining only one polynomial (instead of 2). This greatly simplifies the algorithm.

27 2.5. Two possible algorithms In this section we will present two possible algorithms to solve an 1ODE that presents an elementary function.The first is more generic, yet less efficient. The second can be more restricted, but much more efficient than the first. Algorithm 2.1 (Lsolv). This algorithm is based on the method described in the section 2.3.

Steps: 1. Construct the vector ﬁeld χ, i.e., determine [f,g,h] (by using deﬁnition 2.2).

2. Choose DegMNP (a list of three positive integers - [dM ,dN ,dP ]) for the degrees of the candidates Mc, Nc and Pc (candidates for the polynomials M0, N0 and P3). 3. Construct three polynomials Mc, Nc and Pc of degrees dM , dN and dP , respectively, with undetermined coeﬃcients.

4. Substitute them in the equation E1 : Mc f − Nc h +(z f − g) Pc =0. 5. Collect the equation E1 in the variables (x, y, z) obtaining a set of (li-

near) equations SE1 for the coeﬃcients of the polynomials Mc, Nc and Pc.

6. Solve SE1 to the undetermined coeﬃcients. 2 7. Substitute the solution of SE1in the equation E2 :Dx(Sc)−Sc −∂z(Φc) Sc+ ∂y(Φc)=0, where Dx ≡ ∂x + z ∂y + Φc ∂z, Φc = Mc/Nc and Sc = Pc/Nc.

8. Collect the numerator of E2 in the variables (x, y, z) obtaining a set of

equations SE2 for the remaining undetermined coeﬃcients of the polynomials Mc, Nc and Pc.

9. Solve SE2 for the remaining undetermined coeﬃcients. If no solution is found then FAIL.

10. Substitute the solution of SE2 in Mc and Nc to obtain M0, N0 and in Pc/Nc to obtain S1. 11. Use the S-function method to find the Liouvillian first integral I of the 2ODE z′ = Φ(x, y, z). 12. Make the substitution z → θ on the first integral I and obtain the desired general solution of the 1ODE.

28 Remark 2.7. Some comments:

• Since there is no bound for the degrees [dM ,dN ,dP ] the procedure may never get to a conclusion, i.e., more formally, Lsolv is a semi algorithm. • Some of the steps described above are obviously much more complicated than others. Some involve very complex algorithms in themselves, for example, the algorithm that solves nonlinear polynomial systems. Besi- des that, since we are adapting the method of the S function to solve the ﬁnal part, the penultimate item is itself a very complicated algorithm.

Algorithm 2.2 (F astLs). This algorithm is an improvement of the algorithm 2.1 and is based on the theorems 2.2 and 2.3, on the corollary 2.1 and on their consequences (see example 4). It is restricted to 1ODEs ∈ LS. Steps:

1. Use a variable transformation T1 = {x = r(x, y),y = y} where r(x, y) is the argument of θ to obtain an 1ODE[1] such that θ[1] = exp(x) or θ[1] = ln(x). 2. If θ[1] = exp(x) then make a transformation T2 = {x = ln(x),y = y} to obtain an 1ODE[2] such that θ[2] = ln(x). 3. Construct the vector ﬁeld χ2 = f2 ∂x + g2 ∂y + h2 ∂z associated with the 1ODE[2], where

f2 = x N(x, y, z), g2 = x M(x, y, z), h2 = N(x, y, z), (123) if x is not a factor of N or N(x, y, z) f = N(x, y, z), g = M(x, y, z), h = , (124) 2 2 2 x if x|N. 4. Apply the transformation T3 = {x = y,y = z, z = x} to the vector ﬁeld χ[2], obtaining χ3 = f3 ∂x + g3 ∂y + h3 ∂z, where

f3 = T3(h2),

g3 = T3(f2),

h3 = T3(g2).

29 5. Choose dP for the degree of the candidate P (candidate for the polynomial P2 – numerator of the S-function). 6. Construct a polynomial P in (x, y, z) with arbitrary coeﬃcients and substitute it in the equation E1 : (105).

7. Collect the equation E1 in the variables (x, y, z) obtaining a set of qua-

dratic equations SE1 for the coeﬃcients of the polynomial P .

8. Solve SE1 to the undetermined coeﬃcients. If no solution is found then FAIL.

9. Substitute the solution of SE1 in P/f to obtain S1[3]. 10. Use the S-function method to determine the first integral I[3] of the vector field χ[3]. −1 11. Apply the inverse trasformation T3 and obtain the first integral I[2] of the vector field χ[2]. From I[2] we can have (doing z → ln(x)) the general solution of the 1ODE[2]. −1 12. Aplly T2 to the general solution of the 1ODE[2] and obtain the general solution of the 1ODE[1]. −1 13. Aplly T1 to the general solution of the 1ODE[1] and obtain the general solution of the original 1ODE.

Remark 2.8. The algorithm F astLs uses the equation (105) to determine the polynomial P (the numerator of the S-function). Since (105) has a term of the form S2, we have that it can be written as (see the example 4):

P P 2 fg − gf P gf − fg + fh − hf fh − hf χ = z z + y y z z − y y . f f f f f f (125) We can isolate the equation (125) for P 2, obtaining P 2 = polynomial. Since the higher degree monomials of P squared cannot cancel, the maximum degree of the polynomial P is bounded. In this case, therefore, the part of F astLs that determines the S-function is a full algorithm.

Remark 2.9. The fact that h|P2, f = x h and g|(xP1 + P3) (see theorem 2.2, case (ii)) does not imply that f = cP2 and that g = −c (P1 + zP3) (c constant). If the polynomials P2 and (xP1 + P3) have a polynomial factor in common, the F astLs algorithm may lose this case.

30 3. The LeapS1ode Package In this section we will present a Maple implementation of the algorithms 2.1 and 2.2 (with some modiﬁcations in order to improve ﬂexibility and prac- ticality). Summary of the commands:

• Dx constructs the Dx operator (total derivative over the solutions) associated with the 1ODE.

• Xi constructs the χ operator associated with the 1ODE.

• Tr1ode applies a transformation to the 1ODE and returns the transformed 1ODE and the inverse transformation.

• Ode2 tries to determine the 2ODE associated with the vector ﬁeld χ.

• Sfunction tries to determine a S-function associated with the vector ﬁeld χ.

• Ode1a determines the rational 1ODE associated with the 2ODE (from the knowledge of a S-function).

• Hfunction tries to ﬁnd the general solution for an associated rational 1ODE.

• PDEassol constructs and tries to solve the 1PDEs that relate the ﬁrst integral I with the H-functions.

• Solde tries to determine a Liouvillian general solution of the 1ODE from solving the associated 1PDE.

3.1. Package commands Here we present a description of the main commands of the LeapS1ode package.

31 3.1.1. Command: Dx Feature: This command returns the operator Dx (total derivative over the solutions). Calling sequence: [> Dx(ode); Parameters: ode - The 1ODE.

Synopsis: The command Dx returns the total derivative operator over the solutions of ′ the 1ODE, i.e., the operator dx = ∂x + y ∂y restricted to bound I(x, y)= c (the solution-curves). Since, over the solutions y′ = φ(x, y) this implies that

′ ′ dy D =(∂ + y ∂ )| ′ = ∂ + φ ∂ , y = . (126) x x y y =φ x y dx

3.1.2. Command: Xi Feature: This command returns the vector ﬁeld χ. Calling sequence: [> Xi(ode); Parameters: ode - The 1ODE. Extra parameters: FGH - If this parameter is present in the command call then the command returns a list: [f,g,h]. Synopsis: The command Xi returns the vector ﬁeld χ associated with the 1ODE, i.e., the command returns an operator u → f ∂x(u)+ g ∂y(u)+ h ∂z(u), where f, g and h are polynomials given by (24,25,26). If the extra parameter FGH is present in the entries then the command returns a list with the polynomials f, g and h:[f,g,h].

32 3.1.3. Command: Tr1ode Feature: This command returns the transformed 1ODE and the inverse transformation.

Calling sequence:

[> Tr1ode(ode);

Parameters: ode - The 1ODE.

Extra parameters: TR=[x=F(x,y),y=G(x,y)] - Where [x=F(x,y),y=G(x,y)] is an in- vertible transformation. The default is the identity transformation.

Synopsis: The command Tr1ode returns a list with two entries: the transformed 1ODE and the inverse transformation of TR. If the inverse transformation cannot be obtained, the command returns only the 1ODE transformed. If no transformation is provided, the default is the identity transformation.

3.1.4. Command: Ode2 Feature: This command returns the 2ODE(s) associated with the 1ODE.

Calling sequence:

[> Ode2(ode);

Parameters: ode - The 1ODE.

Extra parameters: PS - If this parameter is present in the command call, then the command returns the polynomials P1, P2 and P3:[P1,P2,P3].

Synopsis:

33 The command Ode2 returns the function Φ(x, y, z)= M0(x, y, z)/N0(x, y, z) of the 2ODE associated with the 1ODE. This 2ODE is one of six possible 2ODEs, namely, the one that considers x as the independent variable, y as the dependent variable and z as the ﬁrst derivative. If the extra parameter PS is present in the entries then the command returns a list with the polynomials P1, P2 and P3:[P1,P2,P3].

3.1.5. Command: Sfunction

Feature: This command tries to ﬁnd the S-function S1.

Calling sequence:

[> Sfunction(ode);

Parameters: ode - The 1ODE.

Extra parameters: Deg = n - Where n is a positive integer denoting the degree of the polynomial P (the numerator or denominator of S1). The default is 3. DegMNP = [degM,degN,degP] - Where [degM,degN,degP] is a list with the degrees of the candidates for M0, N0 and P2, respectively. Den = deno - Where deno is the denominator (presumed) of the S- function.

Synopsis: The command Sfunction tries to find a S-function associated with the rational 2ODE (Φ = M0/N0) through a Liouvillian first integral I. The command computs (if possible) the rational function S1 or a polynomial P that is the numerator (or denominator) of S1 (this case only if we are using the F astLs algorithm). The S-function is the basis for finding the Liouvillian first integral of the vector field χ associated with the 1ODE (see section 2.1). In the case where the numerator or the denominator of S1 is (presumed) one of the polynomials {f,g,h}, we can use the parameter Deg to inform the program of the degree we will use for the candidate Pc. Associated with this case we have also the parameter Den that can acelerate the process by

34 indicating the numerator (or denominator – this case only if we are using the F astLs algorithm) of S1. In the case where none of the polynomials {f,g,h} is the numerator (or denominator) of S1, we use the parameter DegMNP which is, actually, a list with three integers [degM,degN,degP] that tell the command the degree of candidates Mc, Nc and Pc, respectively.

3.1.6. Command: Ode1a Feature: This command determines the associated 1ODE.

Calling sequence:

[> Ode1a(ode);

Parameters: The same as above. Extra parameters: The same as above and:

SF = S1 - Where S1 is a S-function S1.

Synopsis: The command Ode1a uses the S-function to construct the associated 1ODE. The ﬁrst extra parameters are the same of the Sfunction command and the last extra parameter (SF = S1) allows the user to pass the S-function S1 to the Ode1a command.

3.1.7. Command: Hfunction Feature: This command tries to ﬁnd the solution of the associated 1ODE.

Calling sequence, parameters and extra parameters: Same as above.

3.1.8. Command: PDEsol Feature: This command constructs and tries to solve the 1PDE that relates the H-function to the ﬁrst integral I.

Calling sequence, parameters and extra parameters: Same as above.

35 3.1.9. Command: Solde Feature: This command tries to find a Liouvillian general solution for the 1ODE (or, equivalently, first integral of the vector field χ ). It returns a I(x, y, θ)= c, where θ is an elementary function of (x, y).

Calling sequence: [> Solde(ode);

Parameters: The same as above. Extra parameters: The same as above and: II - If this parameter is present in the command call, then the command returns a list with the Liouvillian ﬁrst integral of the vector ﬁeld χ and the z → θ transformation: [I(x, y, z), z = θ(x, y)].

Synopsis: The command Solde is the final part of the process, i.e., it returns (if possible) the general solution of the 1ODE. With the parameter II the user can make the command to return the first integral of the vector field χ (together with z → θ transformation).

Remark 3.1. The extra parameters are not mandatory – some are even redundant. For example, if we are to provide the S-function we do not need to provide the degree of the polynomial candidate Pc. If these redundant parameters are supplied simultaneously, the package will take care of doing (hopefully) the best choice.

3.2. Example of the usage of the package commands In this section we will show the commands of the LeapS1ode package simulating its use on a Maple platform to exemplify in a practical way its operation. Consider the 1ODE (example presented in the section 2.3)

dy exx3y2 + exx2y2 + 2 exx2y + exxy + exx + y2 + ex = . (127) dx x2y2 + exx2 + xy +1 After opening a Maple session, we will load the following packages:

36 [> with(DEtools): read(‘LeapS1ode.txt‘):

The : sign after the command line avoids printing (on-screen) of the result. The DEtools package loads several commands to handle ODEs. The read (‘LeapS1ode.txt‘): command loads our package. We will load 1ODE (127) by typing

[> _1ode := diff(y(x),x) = (exp(x)*x^3*y(x)^2+exp(x)*x^2*y(x)^2 +2*exp(x)*x^2*y(x)+exp(x)*x*y(x)+exp(x)*x+y(x)^2+exp(x))/(x^2*y (x)^2+exp(x)*x^2+x*y(x)+1):

First, we can ﬁnd the associated vector ﬁeld:

[> X := Xi(_1ode): [> fgh := Xi(_1ode,FGH);

The command X := Xi( 1ode): assigns X as the vector ﬁeld and fgh := Xi( 1ode,FGH); assigns fgh as a list with its coeﬃcients [f,g,h]: fgh := [x2y2 +zx2 +xy+1, zx3y2 +zx2y2 +2 zx2y+zxy+zx+y2 +z, z x2y2+

zx2 + xy +1 ]

We can ﬁnd the S-function S1 by typing [> S1 := op(Sfunction(_1ode,Deg=5)); x2y2 + x2z + xy +1 S1 := − (128) x (xy + 1)2

From S1 we can determine the associated 2ODE: [> M0N0 := Ode2(_1ode,SF=S1);

M0N0 := [(zx − y)(zx + y) , x (xy + 1)2] (129)

[> _2odeas := M0N0 [1]/M0N0 [2]; (zx − y)(zx + y) 2odeas := (130) x (xy + 1)2

In order to obtain the polynomials P1, P2 and P3, we simply add the parameter PS to the entry of the Ode2 command:

37 [> PPP := Ode2(_1ode,SF=S1,PS); PPP := [zx2y2 + zxy + y2 + z, −x2y2 − zx2 − xy − 1, x (xy + 1)2] (131)

Using the function S1 we can (by applying the S-function method – see [3]) find the first integral of the vector field χ (by applying the command Solde with the parameter II): [> Inv := Solde(_1ode,SF=S1,II); −1 Inv := [e(xy+1) (zx − y) , z = ex] (132) Finally, we can (by doing the substitution z → θ or apllying directly the Solde command) obtain the general solution of the 1ODE (127): [> sol1ode := Solde(_1ode,SF=S1); −1 sol1ode := e(xy+1) (exx − y)= C1 (133) We can now test the solutions found: [> DX := Dx(_1ode): [> X(Inv[1]); [> DX(lhs(sol1ode)); 0 0

4. Performance The analysis presented in this section is divided in three parts10: • In the ﬁrst subsection, we solve (with our package) a set of 1ODEs that we could not solve using the traditional methods (and even non- traditional methods) implemented in the Maple computer algebra system (Maple CAS). • In the following, we present the special features of the LeapS1ode in action, to solve more complex 1ODEs. • Finally, we present an example of a 1ODE with parameters. In that case we use the integrability analysis provided by using the Par parameter in the Sfunction command.

10In this paper all the computational data (time of running etc) was obtained on the same computer with the following conﬁguration: Intel(R) Core(TM) i5-8265U @ 1.8 GHz.

38 4.1. A set of ‘hard’ 1ODEs In this section we present a set of 1ODEs which the powerful methods implemented in the Maple ODE solver (release 17) – dsolve command – cannot solve. Bellow, we show the CPU time and memory that the package LeapS1ode spends to ﬁnd the solutions (through the use of Solde command). Consider the following 1ODEs: 1. The following ﬁve 1ODEs have an elementary Liouvillian solution. We will refer to them as 1ODEs 1, 2, 3, 4, 5:

dy x 2 (ln (x))2 x4y2 + 2 ln (x) x2y + 2 ln (x) y2 − yx2 + y2 +2 = 2 4 2 4 2 dx (ln (x)) x y + ln (x) x + ln (x) x y +1 (134)

dy 3 (ex)2x4y4 +(y4 −y3x4 −x3y3 +6 x3y2 + xy4) ex +3 x2 = (135) dx (ex)2 x2y4 +(x4y2 − xy3 +2 xy2) ex − x3 + y +1

dy (2 x3y4 − 4 x2y2 − x2 + ln (y)+2 x) y = − (136) dx −2 x4y2 − x2y4 + 2 ln (y) x2y2 +2 xy2 − 1

dy 3 (ey)2 x4y2 − 7 eyx3y +3 x2 − y = − (137) dx x (ey)2 x4y2 − 3 eyx3y − x3ey + x2 − y − 1 dy 4 x3y6 +8 x4y3 +4 x5 − x4 + cos(y) = − (138) dx (y6 +2 xy3 + x2) sin (y) − 3 x4y2 +3 cos(y) y2

2. The following ﬁve 1ODEs have a non-elementary Liouvillian solution. We will refer to them as 1ODEs 6, 7, 8, 9, 10.

dy (2 x2 − y) (ln (x) y2 − x2y − 1) y = dx x (ln (x))2 y3 − ln (x) x2y2 − (ln (x))2 y2 + 2 ln (x) x2y − x4 − ln (x) y (139)

dy (ln (x))2 xy2 − 2 ln (x) x2y − ln (x) xy + ln (x) y2 + x3 = − (140) dx x (ln (x))2 y

39 dy (−e2 xx6y2 − 2 e2 xx5y2 + e2 xx4y2 + exx4y + 3 exx3y +1)e−x = dx x3 (exx3y − x − 1) (141)

dy (ln (y))2 xy4 − (ln (y))2 y4 − x2 ln (y) y2 + ln (y) xy2 + x3 = − dx xy (2 ln (y) + 1) (ln (y) y2 − x2 − x) (142)

dy y3 (−xy3 + ey + 1) = (143) dx eyx2y6 +3 x2y5 − 2 eyxy3 − 3 eyxy2 − 3 xy2 + ey

Applying the Solde command to each one of them, we have:

[> t0 := time(): solde[1] := Solde(_1ode[1],Deg=7); time()-t0; ln (x2 − y) ln (x) x2y + ln (x2 − y)+1 solde[1] := − = C1 ln (x) x2y +1 0.563

[> t0 := time(): solde[2] := Solde(_1ode[2],Deg=7); time()-t0;

xy2ex +1 solde[2] := − = C1 ex ln (x3 − y) xy2 + ln (x3 − y)+ y 1.078

[> t0 := time(): solde[3] := Solde(_1ode[3],Deg=5); time()-t0;

x 2 solde[3] := −x + ln (y) e xy2−1 = C1 0.209

[> t0 := time(): solde[4] := Solde(_1ode[4],Deg=5); time()-t0;

40 eyxy ln (x3ey + 1) − ln (x3ey +1)+1 solde[4] := − = C1 xyey − 1 0.625

[> t0 := time(): solde[5] := Solde(_1ode[5],Deg=7); time()-t0;

−iy4 − iyx +1 solde[5] := ln −2 eiyx4 + e2 iy +1 + = C1 y3 + x 3.687

[> t0 := time(): solde[6] := Solde(_1ode[6],Deg=5); time()-t0;

− − − 2 1 Ei 1, − (ln (x) y − x2) 1 y + e(ln(x)y x ) solde[6] := − = C1 y 0.203

[> t0 := time(): solde[7] := Solde(_1ode[7],Deg=7); time()-t0;

− −1 −1 solde[7] := Ei 1, (ln (x) y − x) 1 e(ln(x)y−x) + x e−(ln(x)y−x) = C1 0.218

[> t0 := time(): solde[8] := Solde(_1ode[8],Deg=9); time()-t0;

− −1 x 2 − 1 solde[8] := Ei 1, − exx2y − 1 x + e(e x y 1) x−1 = C1 0.687

[> t0 := time(): solde[9] := Solde(_1ode[9],Deg=6); time()-t0; x 2 x solde[9] := x eln (y) y − x + Ei 1, − = C1 ln (y) y2 − x 0.437

41 [> t0 := time(): solde[10] := Solde(_1ode[10],Deg=6); time()-t0;

− − − 3− 1 −1 3− 1 e (xy 1) Ei 1, − xy3 − 1 + ey e(xy 1) = C1 0.297 1ODE Time (sec.) Memory (MB) 1 (134) 0,672 21,07 2 (135) 1,078 35,60 3 (136) 0,209 4,18 4 (137) 0,625 33,18 5 (138) 3,687 28,47 6 (139) 0,329 17,41 7 (140) 0,218 16,18 8 (141) 0,687 21,26 9 (142) 0,437 29,86 10 (143) 0,297 18,42

Remark 4.1. Some comments: • When we start a Maple session, the simple loading of the basic packages that are activated with the opening of the program’s standard platform already uses 4.18 MB. So, when that number appears (in the table above), the meaning is basically ‘almost no memory expenditure’. • For the measurement of consumed CPU time, a similar problem occurs: the time measurement provided is ‘variable from session to session’. However, the number representing the time spent is always reasonably close to the average of the values obtained in different rounds. In the data placed in the table above, we avoid averaging over many sessions (which means that the last two digits are not significant) because we were focused on a more qualitative result. • Another piece of information (quite relevant for those who don’t fre- quently use the Maple command dsolve) is that the memory consumed in each one of the ten 1ODEs is relatively low. • Another interesting detail is that most of the time consumed (even in cases where the time consumed is very short) is generally used in determining the S-function, that is, in addition to validating our method, the previous result is also a reaffirmation of the S-function method.

42 • Finally, looking at the table below, we can see why these 1ODEs cause diﬃculties for the vast majority of methods. We only need to pay attention to the Lie symmetries or to the integrating factors of the 1ODEs’ set. We see that only 1ODEs 3, 5 and 10 have integrating factors in which the θ function is absent (and, even so, they are not simple).

1ODE Symmetry (ν) Integrating Factor 2 −1 (x2y ln(x)+1)2 e(x y ln(x)+1) 1 (134) − 2 −1 2 2 e(x y ln(x)+1) ((ln(x))2x4y2+ln(x)x4+x2y ln(x)+1) (x y ln(x)+1) −y y 2 x 2 xy2ex+1 2 x 2 (135) − (xy e +1) e e xy e +1 (ex)2x2y4+exx4y2−exxy3+2 xy2ex−x3+y+1 (xy2ex+1)2 −x x 2− 2 xy2−1 2− 3 (136) (xy 1) y e − e xy 1 −2 x4y2−x2y4+2 ln(y)x2y2+2 xy2−1 y (xy2−1)2 y −1 y −1 y− 2 (1−xye ) (xye −1) y 4 (137) (xye 1) e −e e eyx((ey)2x4y2−3 eyx3y−x3ey+x2−y−1) (xyey−1)2 3 −1 − 3 2 (−y −x) (y3+x) 1 5 (138) (y +x) e −e (sin(y)y6−3 x4y2+2 sin(y)xy3+3 cos(y)y2+sin(y)x2) (y3+x)2 2 −1 − − 2 2 2 (x −ln(x)y) (ln(x)y−x2) 1 6 (139) (ln(x)y x ) y e −e (ln(x))2y3−ln(x)x2y2−(ln(x))2y2+2 ln(x)x2y−x4−ln(x)y (ln(x)y−x2)2xy2 −1 (ln(x)y−x)2 −e−(ln(x)y−x) 7 (140) − e−(ln(x)y−x) 1 (ln(x))2 y (ln(x)y−x)2 x x 3 − x 3 − e x y x+1 e x y−x+1 exx2y−1 x 2 − 2 x 2 − 8 (141) e (e x y 1) −e e x y 1 x (exx3y−x−1) x2(exx2y−1)2 − x x ln(y)y2−x 2− 2 y y2−x 9 (142) e (ln(y)y x) −e ln( ) y (2 ln(y)+1) (ln(y)y2−x2−x) (ln(y)y2−x)2 x 3 −1 − −(xy −1) 3− 2 (xy3−1) 1 10 (143) e (xy 1) −e eyx2y6+3 x2y5−2 eyxy3−3 eyxy2−3 xy2+ey (xy3−1)2 Remark 4.2. For all 1ODEs in Kamke’s book [49] that obey the conditions of application of the algorithm our method does very well in all cases. As they are much simpler cases than those shown in the table, we decided not to include them in order not to increase the presentation unnecessarily.

4.2. Special features As we can see, in the ten 1ODEs shown in the section 4.1 we have that θ is a function of only one variable (x or y). In these cases, as shown in the section 2.4 (theorem 2.2), we can use an improvement that allows us to calculate only one polynomial instead of two (F astLs algorithm). However, due to some technical details, we decided to leave the ﬁrst part of the F astLs algorithm in the hands of the package user. In this section we will show you

43 how to use the commands in the package to achieve the solution in more diﬃcult cases. Consider the 1ODE given by:

2 3 x5y2 − 6 ln x x4y +3 ln x x3 − x4y + x3 + xy2 − y y dy y y = . 2 dx 4 2 3 x 2 x 3 2 x x y + x y − 2 ln y xy + ln y y + x − xy − y (144) After loading the package and the 1ODE (144), we can try (first) to find the S-function using the Sfunction command. However, if we type [> t0 := time(): S1 := op(Sfunction(_1ode,Deg=7)); time()-t0; we get 1 S1 := (145) y 11.812 after ≈ 12 seconds (and spending 215 MB of memory). However, if we type [> sol := Solde(_1ode,SF=S1); The output is sol := (146) which would probably leave us intrigued, because as the S-function is simple the method should have found the solution. What is actually happening is that the following invariant is found for the vector field associated χ: I = − ln(x)+ln(y)+z. Now, if we pay attention to the fact that the function θ present in 1ODE is θ = ln(x/y), making the substitution z → θ we will obtain I = − ln(x) + ln(y) + ln(x/y) = 0 that represents the identity θ = ln(x/y) (and therefore, as it is not an answer to what we seek, the program discards it). We can try using the parameter DegMNP which is, actually, a list with three integers [degM,degN,degP] which gives to the Sfunction command the degrees of the polynomials that are candidates for M0, N0 and P2. This, however, results in a very large consumption of resources and on a memory ‘overflow’. However, we can still succeed using the command Tr1ode to transform the 1ODE (144) into an 1ODE such that θ is a function of just one variable. We can do this by typing (for example)

44 [> _1odetr := Tr1ode(_1ode,TR=[x=x*y,y=y]); and obtaining a transformed 1ODE given by: dy −y 3 x5y6 − 6 x4 ln (x) y4 − x4y4 +3 x3 ln (x)2 y2 + x3y2 + xy2 − 1 = 6 6 5 4 5 4 4 2 2 3 4 2 2 2 2 2 dx 3x y −6x ln(x)y −2x y +3x ln(x) y −x y +2ln(x)x y +2x y −xln(x) (147) So, in the transformed 1ODE, we have that θ = ln(x). Now, we can apply again the command Sfunction (and, in this case, using the F astLs algorithm)

[> t0:=time(): s1tr := Sfunction(_1odetr[1],Deg=11); time()-t0; that returns us 3x5y6 −6x4zy4 −2x4y4 +3x3y2z2 −x2y4 +2zxy2 +2xy2−z2 s1tr := (148) y (x3y2 − 1) 2.375 i.e., the algorithm ﬁnds the S-function in less than 3 seconds (and consuming only 30 MB). At this point, we just have to type

[> soltr := Solde(_1odetr[1],ST=op(s1tr)); to obtain the general solution of the transformed 1ODE (147) in 0.3 seconds:

−xy2ln(x3y2 −1)−xy2ln(y)+ln(x) ln(x3y2 −1)+ln(x) ln(y)−1 soltr := = C1 xy2 − ln (x) (149) and

[> Itr := Solde(_1odetr[1],ST=op(s1tr),II); to obtain the ﬁrst integral of the transformed vector ﬁeld:

xy2ln(x3y2 −1)+xy2ln(y)− z ln(x3y2 −1)− z ln(y)+1 Itr := , z = ln(x) −xy2 + z (150) The command Tr1ode returns (as second output) the inverse transformation. So, by typing

45 [> itr := _1odetr[2]: [> Ior := subs(itr,Itr[1]); we obtain the ﬁrst integral of the vector ﬁeld associated with the 1ODE (144):

x3 x3 xy ln y − 1 + xy ln (y) − z ln y − 1 − z ln (y)+1 Ior := (151) z − yx We can check this with [> X := Xi(_1ode): [> X(Ior); 0 Finally, doing the substitution z → ln(x/y), we ﬁnd the general solution of the original 1ODE (144): [> DX := Dx(_1ode): [> sol := subs(z=ln(x/y),Ior=_C1); [> DX(Ior);

x3 x x3 x xy ln y − 1 + xy ln (y) − ln( y ) ln y − 1 − ln( y ) ln (y)+1 sol := = C1 x ln( y ) − yx (152) 0

4.3. Integrability analisys In addition to the features shown above, we will present in this section a very useful way to apply the LeapS1ode package: the Par parameter. When we are studying an ODE presenting parameters, we can use the parameter Par in the entries of the Sfunction command. Thus, if the command was unable to ﬁnd a S-function for generic parameter values, the entry Par = {α,β, ...} tells the command to try to ﬁnd a S-function for some relationship between them, i.e., the method brings us the possibility of an analysis of the integrability region of the 1ODE’s parameters.

To see a real case of using the package in the search for integrability regions in terms of Liouvillian functions (closed solutions), let’s take as an

46 example an 1ODE obtained from the study of 2ODEs that model nonlinear oscillation phenomena. In our case, the 1ODE we are going to present appears in the reduction of a Levinson-Smith equation11 [50]

x¨ + F (x, x˙)x ˙ + G(x)=0, (153) a 2ODE that is a generalization of the Li´enard equation 12 [52]

x¨ + F (x)x ˙ + G(x)=0. (154)

As 2ODE (153) does not explicitly depend on time, we can do the following coordinate transformation to reduce its order: x = x, x˙ = y, (155)  dy  x¨ = dx y. The application of this transformation to the 2ODE (153) leads to the 1ODE given by dy F (x, y) y + G(x) = − . (156) dx y We can now, from very generic F and G functions, try to obtain integrable cases in terms of Liouvilian functions (i.e., closed form solutions). Let’s analyze the following case:

dy −fx2yex − ex2ex − xexyb − y2c − dx − ya = (157) dx y Applying the Sfunction command and using the parameter Par as assigned below

[> _1ode := diff(y(x),x)= \cdots: [> ss1 := [Sfunction(_1ode,Par={b,d,e},DegMNP=[6,4,5])]: [> for i from 3 by 3 to nops(ss1) do for j to nops(ss1[i]) do

11The Levinson-Smith equation has applications in astrophysics [51]. 12There is a vast literature on the use of this equation to model various phenomena: ranging from electrical circuits, study of heartbeat, neuron activity, chemical kinetics to turbulence phenomena [53, 54, 55, 56].

47 print([i,j],______,traperror(factor(subs(ss1[i][j], ss1[i-2]))),subs(ss1[i][j],[‘b=‘,b,‘c=‘,c,‘d=‘,d,‘e=‘,e,‘f=‘,f])); trys[i,j]:=traperror(factor(subs(ss1[i][j],ss1[i-1][1]/ss1[i-1][2]))); print(trys[i,j]); end do; end do; we obtain13 . . cz [3, 4], , − [‘b = ‘, b, ‘c = ‘, c, ‘d = ‘, 0, ‘e = ‘, 0, ‘f = ‘, f] cy + a (cfx2z + bcxz + c2y + ac + cy + cz + a) z cy + a . . Using the determined region on 1ODE (157) we ﬁnally obtain [> new1ode := subs(ss1[3][4],_1ode); d −fx2y (x) ex − xexy (x) b − (y (x))2 c − y (x) a y (x)= dx y (x) (158) that can be solved with [> s1 := ss1[1]; [> new1ode := subs(ss1[3][4],_1ode); [> Solde(new1ode,SF=s1): cz − cy + a d −fx2y (x) ex − xexy (x) b − (y (x))2 c − y (x) a y (x)= dx y (x) Rearranging the solution we have

f ex (c + 1)2 x2 − 2 (c + 1) x +2 b ex ((c + 1) x − 1) a y = − 3 − 2 − + C1 (c + 1) (c + 1) c !

13We omit the full output because it is too long.

48 Replacing y with z and x with y, we will obtain the ﬁrst integral of the Levinson-Smith equation (for the parameter regions we ﬁnd):

f ey (c + 1)2 y2 − 2 (c + 1) y +2 b ey ((c + 1) y − 1) a I = z + 3 + 2 + (c + 1) (c + 1) c !

5. Conclusion

In this work we developed a new method to solve 1ODEs presenting an elementary function. Our procedure is applicable to 1ODEs that have a general Liouvillian solution and, in addition, it is restricted to 1ODEs that can be expressed in the following form: dy M(x, y, θ) y′ = = φ(x, y, θ)= (159) dx N(x, y, θ) where φ is a rational function of (x, y, θ) and θ is an elementary function of (x, y) (M and N are co-polynomials in (x, y, θ)). The method is based on a connection between an 1ODE with an elementary function θ and a polynomial vector field in three variables. The main point is that, if this 1ODE has a Liouvillian general solution, we can associate a rational 2ODE with the vector field and, therefore, use the S-function method (see [3]) to find a first integral for the vector field and (using the simple substitution z → θ) find the general solution of the 1ODE. One might consider that exchanging the problem of solving a 1ODE for the problem of finding a first integral of a 2ODE may not seem like a good idea at first. However, there are two facts that make this exchange result in a very efficient procedure: first, the 1ODE has an elementary function whereas the 2ODE is rational; second, the S-function method is effective in cases where the search for first integrals of 2ODEs by procedures such as the Lie symmetries method and Darboux approaches are particularly problematic. The method we developed first (Lsolv algorithm) was (in a first assess- ment) reasonably efficient compared to the more canonical alternative that would be to try a DPS approach or the Lie symmetry method. However, the appearance of an unexpected result, increased our work by far exceeding our expectations. These result (expressed in the theorems 2.2 and 2.3) allowed

49 the creation of the F astLS algorithm that, for a very comprehensive range of 1EDOs (the class LS - see the deﬁnition 2.2, performed very well. The algorithms Lsolv and F astLS were implemented in a computational package (LeapS1ode) on the Maple symbolic computing platform. The package, in addition to solving 1ODEs using the methods mentioned above, presents commands that materialize all stages of the process. The commands also make use of several parameters that help both in the search for solutions as well as in the research in physics and mathematics. Finally, due to the nature of the algorithm search process, we can perform an analysis of integrability regions for 1ODEs that present parameters (see the example in section 4.3).

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