Rational Solutions of Riccati-like Partial Differential Equations
MM Research Preprints, 166–187 No. 18, Dec. 1999. Beijing
Rational Solutions of Riccati-like Partial Differential Equations1) Ziming Li MMRC, Institute of Systems Science Academia Sinica, Beijing 100080, China Email: [email protected] Fritz Schwarz GMD, Institut SCAI 53754 Sankt Augustin, Germany Email: [email protected] Abstract. When factoring linear partial differential systems with a finite-dimensional solution space or analyzing symmetries of nonlinear ode’s, we need to look for rational solutions of certain nonlinear pde’s. The nonlinear pde’s are called Riccati-like because they arise in a similar way as Riccati ode’s. In this paper we describe the structure of rational solutions of a Riccati-like system, and an algorithm for computing them. The algorithm is also applicable to finding all rational solutions of Lie’s system {∂x u + u2 + a1 u+a2 v+a3 , ∂y u+uv+b1 u+b2 v+b3 , ∂x v+uv+c1 u+c2 v+c3 , ∂y v+v 2 +d1 u+d2 v+d3 }, where a1 . . . d3 are rational functions of x and y.
1. Introduction Riccati’s equation is one of the first examples of a differential equation that has been considered extensively in the literature, shortly after Leibniz and Newton had introduced the concept of the derivative of a function at the end of the 17th century. Riccati equations occur in many problems of mathematical physics and pure mathematics. A good survey is given in the book by Reid (1972). Of particular importance is the relation between Riccati’s equation and a linear ode y 00 + ay 0 + by = 0 with a, b ∈ C(x), where C is the field of the complex numbers. For example, solutions Rh with the property that the quotient p = h0 /h ∈ C(x) may be represented as h = exp ( p dx) if p satisfies the first-order Riccati equation p0 +p2 +ap+b = 0. Equivalently, this linear ode allows the first-order right factor y 0 −qy over C(x) if q obeys the same equation as p. In general, finding the first-order right rational factors of a linear homogeneous ode is equivalent to finding the rational solutions of its associated Riccati equation (see e.g. Singer 1991). It turns out that this correspondence carries over to systems of linear homogeneous partial differential equations with a finite-dimensional solution space. Systems of this kind occur 1)
The work is done when the first author visited GMD from September, 1997 to December, 2000
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in Lie’s symmetry theory for solving nonlinear ode’s and related equivalence problems. For example, Lie studied the coherent nonlinear system {∂x u + u2 + a1 u + a2 v + a3 , ∂y u + uv + b1 u + b2 v + b3 , ∂x v + uv + c1 u + c2 v + c3 , ∂y v + v 2 + d1 u + d2 v + d3 }
(1)
for the first time in connection with the symmetry analysis of second-order ode’s with projective symmetry group (see, e.g. Lie 1873, page 365). It is suggested therefore to call (??) Lie’s (2) system. We will show that Lie’s system may be transformed to the Riccati-like system R3 given in Example ??. In this paper the following problem will be considered. Given a set L of linear homogeneous pde’s in one unknown function z(x, y) whose coefficients are in C(x, y) and whose R solution space is of finite dimension, find a solution of L in the form exp ( u dx + v dy), where u and v are in C(x, y) with ∂y u = ∂x v. In other words, we want to find a linear differential ideal over C(x, y) with one-dimensional solution space containing L. As shown in Section ??, this problem is equivalent to finding a rational solution of the Riccati-like system associated with L. We describe an algorithm, implemented in the ALLTYPE system Schwarz (1998b), that computes all rational solutions of an associated Riccati-like system. The algorithm is also applicable to finding rational solutions of Lie’s system and hyperexponential solutions of linear homogeneous pde’s with finite-dimensional solution space in several unknowns. The paper is organized as follows. Section ?? contains necessary preliminaries. Section ?? presents the main results: the structure of rational solutions of a Riccati-like system and an algorithm for computing them. Applications are given in Section ??. 2. Preliminaries We denote by K the field of rational functions C(x, y), and by ∂x and ∂y the usual partial differential operators acting on K. Let Ω be the field of of all complex meromorphic functions in the space of complex variables of x and y. Note that both Ω and C have the same constant field C. In this paper we are concerned with solutions in Ω. An element a of Ω is called an x-constant (resp. y-constant) if ∂x a = 0 (resp. ∂y a = 0). An x-derivative (resp. y-derivative) of a means ∂xk a (resp. ∂yk a) for some nonnegative integer k. By a system we mean a finite subset of some differential polynomial ring over K. Basic notions related to differential polynomials are used such as: rankings, leaders, autoreduced sets, differential remainders, and characteristic sets for differential ideals. For their definitions, the reader is referred to Ritt (1950), Rosenfeld (1959), and Kolchin (1973). The differential ideal generated by a subset P of some differential polynomial ring is denoted by [P]. In the rest of this section, we describe three useful notions: the elimination property of linear differential ideals with finite linear dimension, integrable pairs, and coherent orthonomic systems, 2.1. Linear systems with finite linear dimension Let z be a differential indeterminate over K and fix an orderly ranking on the differential polynomial ring K{z}. We denote by L the K-linear space consisting of all linear homogeneous polynomials in K{z}.
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Given a linear system L ⊂ L, the linear dimension of [L] in K{z} is the codimension of L ∩ [L] in L (Kolchin 1973, page 151). The general solution of L depends on a finite number of unspecified constants if and only if the linear dimension of [L] is finite (Kolchin 1973, page 152). To check if the linear dimension of [L] is finite, we compute a coherent autoreduced set A (Rosenfeld, 1959) such that [L] = [A]. This computation can be done by various methods such as: Janet bases (Janet 1920, Schwarz 1998a) the characteristic set method (Wu 1989, Li and Wang 1999), and Gr¨obner bases for differential operators (Kandri-Rody and Weispfennig) . The linear dimension of [L] is finite if and only if an x-derivative and a y-derivative of z appear as leaders of some elements of A. We denote by Lx (resp. Ly ) the subset of L consisting of differential polynomials involving only x-derivatives (resp. y-derivatives) of z. The following elimination property is wellknown. Lemma 2.1 Let L be a finite subset of L. If [L] is of finite linear dimension, then there is an algorithm for computing two nonzero elements [L] ∩ Lx and [L] ∩ Ly , respectively. Proof. Compute a coherent autoreduced set A in L such that [L] = [A]. Assume that A is not {1}, for, otherwise, [L] is trivial. A finite base B of the K-linear space V = L/(L ∩ [L]) consists of all differential monomials ∂xi ∂yj z that cannot be reduced w.r.t. A. Using A, we can express any element of V as a K-linear combination of elements of B. Hence, we can find smallest integer n such that z, ∂x z, . . . , ∂xn z are K-linearly dependent, that is, we can find a0 , a1 , . . . , an−1 ∈ K such that Lx = a0 z + a1 ∂x z + · · · + an−1 ∂xn−1 z + ∂xn z ∈ [L]. Similarly, we can find a nonzero linear differential polynomial in [L] ∩ Ly .
2
2.2. Integrable pairs To describe the structure of rational solutions in Section ??, we define a pair of rational functions (f, g) ∈ K × K to be integrable if ∂y f = ∂x g. For an integrable pair (f, g), the R expression H = exp ( f dx + g dy) defines an element of Ω unique up to a nonzero multiplicative constant. We regard H as an element of Ω when the value of the multiplicative constant is irrelevant to our discussion. Note that ∂x H/H = f and ∂y H/H = g. Two integrable pairs (f, g) and (p, q) are said to be equivalent if there exists a nonzero h in K such that f − p = ∂x h/h and g − q = ∂y h/h. Write (f, g) ∼ (p, q) when they are equivalent. R R Since (f, g) ∼ (p, q) if and only if the ratio of exp ( f dx + g dy) and exp ( p dx + q dy) is a rational function, ∼ is an equivalence relation on the set of integrable pairs. It is easy to verify that, for integrable pairs (f1 , g1 ), (p1 , q1 ), (f2 , g2 ), (p2 , q2 ) and integers m, n, (f1 , g1 ) ∼ (p1 , q1 ) and (f2 , g2 ) ∼ (p2 , q2 ) implies (mf1 + nf2 , mg1 + ng2 ) ∼ (mp1 + np2 , mq1 + nq2 ). An element a in Ω is said to be a hyperexponential if ∂x a/a and ∂y a/a belong to K. If a is a hyperexponential, (∂x a/a, ∂y a/a) is an integrableR pair. Conversely, for an integrable pair (f, g), we may construct a hyperexponential exp ( f dx + g dy) in Ω. For an integrable R pair (f, g), define E (f,g) to be the C-linear space {h exp ( f dx + g dy) | h ∈ K } . A direct calculations shows that E (f,g) consists of all hyperexponentials a such that (∂x a/a, ∂y a/a) ∼ (f, g). The following lemma is used to group rational solutions of a Riccati-like system. Lemma 2.2 If (f1 , g1 ), (f2 , g2 ), . . . , (fn , gn ) are mutually inequivalent integrable pairs, the sum of C-linear spaces E (f1 ,g1 ) , E (f2 ,g2 ) , . . . , E (fn ,gn ) is direct.
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Proof. We proceed by induction. For n = 2, if E (f1 ,g1 ) ∩ E (f2 ,g2 ) contains a nonzero element a, (∂x a/a, ∂y a/a) ∼ (fi , gi ), for i = 1, 2. Hence, (f1 , g1 ) ∼ (f2 , g2 ), a contradiction. The sum of E (f1 ,g1 ) and E (f2 ,g2 ) is direct. Assume that the result is proved for lower values of n. If the sum of E (f1 ,g1 ) , E (f2 ,g2 ) , . . . , E (fn ,gn ) is not direct, there are nonzero z1 ∈ E (f2 ,g2 ) , z2 ∈ E (f2 ,g2 ) , . . . , zn ∈ E (fn ,gn ) which are C-linearly dependent. By a possible rearrangement of indexes, we have zn = c1 z1 + c2 z2 + · · · + cn−1 zn−1
(2)
for some c1 , c2 , . . . , cn−1 ∈ C. Since z1 , z2 , . . . , zn−1 are linearly independent over C by the induction hypothesis, Theorem 1 in Kolchin (1973, page 86) implies that there exist derivatives θ1 , θ2 , . . . , θn−1 such that W = det(θi zj ) is nonzero, where 1 ≤ i ≤ n − 1 and 1 ≤ j ≤ n−1. Since the zi ’s are hyperexponentials, there exists rij ∈ K such that θj zi = rji zi . Applying θ1 , θ2 , . . . , θn−1 to (??) then yields a linear system
r11 r21 · rn−1,1
r12 r22 · rn−1,2
··· ··· ··· ···
r1,n−1 r2,n−1 · rn−1,n−1
c1 z1 c2 z2 .. . cn−1 zn−1
=
r1n zn r2n zn .. . rn,n−1 zn
whose coefficient matrix (rji ) is of full rank because W 6= 0. Solving this system, we get ci zi = si zn , where si ∈ K. Since ck 6= 0 for some k with 1 ≤ k ≤ n − 1, zk /zn is rational, so (fk , gk ) ∼ (fn , gn ), a contradiction. 2 2.3. Coherent orthonomic systems A system P in a differential polynomial ring D is orthonomic if it is an autoreduced set and any element of P is linear w.r.t. its leader and monic, or, equivalently, both initial and separant of any element in P is one. A linear autoreduced set can be considered as an orthonomic system. System (??) is orthonomic w.r.t. any orderly ranking on u and v. Systems studied in this paper are always orthonomic with respect to certain orderly ranking on their differential indeterminates and derivatives. Let P be an orthonomic system in D. We recall the notions of ∆-polynomials and coherence in Rosenfeld (1959) for P. Note that these two notions are originally defined for autoreduced sets and the name “∆-polynomial” is due to Boulier et. al (1995). Let P1 and P2 belong to P with the respective leaders θ1 z and θ2 z, where z is a differential indeterminate, and θ1 , θ2 are derivatives. Then there exist derivatives φ1 and φ2 with minimal orders φ1 θ1 = φ2 θ2 . The ∆-polynomial of P1 and P2 , denoted by ∆(P1 , P2 ), is φ1 P1 − φ2 P2 . Note that ∆(P1 , P2 ) is well-defined provided the leaders of P1 and P2 are derivatives of the same indeterminate. The system P is coherent if, for every such pair P1 , P2 in P, ∆(P1 , P2 ) can be written as a D-linear combination of derivatives of elements of P, in which each derivative has its leader lower than φ1 θ1 z (in a preselected ranking). To study orthonomic systems, we need to make sure that they cannot be formally reduced to non-orthonomic or trivial ones. Corollary 3 in Chapter I of Rosenfeld (1959) asserts that an orthonomic system P is a characteristic set of [P] if and only if P is coherent. Hence, P cannot be formally reduced any further if it is reduced. By the same corollary, one can easily show
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Lemma 2.3 An orthonomic system P is coherent if and only if all ∆-polynomials (possibly) formed by elements of P have zero as their differential remainders w.r.t. P. Hence, we are able to decide algorithmically whether an orthonomic system is coherent. 3. Rational solutions of associated Riccati-like systems Given a linear differential system L ⊂ L with finite linear dimension, we want to compute its hyperexponential solutions. The substitution: Z
z ← exp
u dx + v dy
where ∂y u = ∂x v
(3)
transforms L to a nonlinear system in K{u, v}. The union of this system and {∂y u − ∂x v} is called the Riccati-like system associated with L, or, simply, an associated Riccati-like system. Conversely, the substitution: u ← ∂x z/z, v ← ∂y z/z transforms an associated Riccati-like system in K{u, v} to a system in L. As in the ordinary case, computing hyperexponential solutions of L is equivalent to computing rational solutions of its associated Riccati-like system. Example 3.1 Let L2 = {∂x2 z + a1 ∂x z + a2 z, ∂y z + b1 ∂x z + b2 z} be coherent. Then the linear dimension of L is 2. The Riccati-like system R2 associated with L is {∂x u + u2 + a1 u + (1) a2 , v + b1 u + b2 , ∂y u − ∂x v}. Coherent linear systems with dimension 3 may be either L3 (2) equal to {∂x3 z + a1 ∂x2 z + a2 ∂x z + a3 z, ∂y z + b1 ∂x2 z + b2 ∂x z + b3 z}, or L3 equal to {∂x2 z + a1 ∂x z + a2 ∂y z + a3 z, ∂y ∂x z + b1 ∂x z + b2 ∂y z + b3 z, ∂y2 z + c1 ∂x z + c2 ∂y z + c3 z}. (1)
Their respective associated Riccati-like systems are R3 equal to {∂x2 u + 3u∂x u + u3 + a1 (∂x u + u2 ) + a2 u + a3 , v + b1 (∂x u + u2 ) + b2 u + b3 , ∂y u − ∂x v} (2)
and R3 equal to {∂x u + u2 + a1 u + a2 v + a3 , ∂y u + uv + b1 u + b2 v + b3 , ∂y v + v 2 + c1 u + c2 v + c3 , ∂y u − ∂x v}. In the rest of this section, let L be a system in L with finite linear dimension d. The Riccatilike system associated with L is denoted by R. The set of rational solutions of R is denoted by S. All elements of S are integrable pairs, because ∂y u−∂x v is contained in R. We describe the structure of S and present an algorithm for computing S. 3.1. Standard representations of rational solutions To describe rational solutions of Riccati ode’s and Riccati-like systems precisely, we introduce some notation. Let F be a subfield of K. By an F -linearly independent set, we mean a finite subset of C[x, y] whose elements are F -linearly independent. Let a, b be in K. Let A = {a1 , . . . , ak } and B = {b1 , . . . , bl } be C(y) and C(x)-independent sets, respectively. We define P k \ ∂x ci ai i=1 a XA = a + Pk | c1 , . . . , ck are x-constants K i=1 ci ai
Rational Solutions of Riccati-like Systems
and YBb =
b+
∂y
P
l i=1 ci bi
Pl
i=1 ci bi
171
| c1 , . . . , cl are y-constants
\
K.
If, moreover, (a, b) is integrable and H = {h1 , . . . , hm } is a C-linearly independent set, we define P P ∂x ( m ∂y ( m (a,b) i=1 ci hi ) i=1 ci hi ) SH = a + Pm , b + Pm | c1 , . . . , cm ∈ C . i=1 ci hi i=1 ci hi Lemma 3.1 Let r1 , r2 , . . . , rk belong to C(y)[x] and c1 , c2 , . . . , ck be x-constants of Ω. If f=
c1 ∂x r1 + c2 ∂x r2 + · · · + ck ∂x rk c1 r1 + c2 r2 + · · · + ck rk
belongs to K, then there exist s1 , s2 , . . . , sk ∈ C[x, y] and d1 , d2 , . . . , dk ∈ C[y] such that f=
d1 ∂x s1 + d2 ∂x s2 + · · · + dk ∂x sk . d1 s1 + d2 s2 + · · · + dk sk
Proof. Assume that f = f1 /f2 , where f1 , f2 ∈ C[x, y]. Equating the coefficients of the like powers of x in (c1 ∂x r1 + c2 ∂x r2 + · · · + ck ∂x rk )f2 = (c1 r1 + c2 r2 + · · · + ck rk )f1 , we find that c1 , c2 , . . . , ck satisfy a linear system over C(y). Hence, c1 , c2 , . . . , ck can be chosen as elements in C(y). Let h be the common denominator of all the ci and the ri . Then h is in C[y]. We derive that f=
h(c1 ∂x r1 + c2 ∂x r2 + · · · + ck ∂x rk ) d1 ∂x s1 + d2 ∂x s2 + · · · + dk ∂x sk = h(c1 r1 + c2 r2 + · · · + ck rk ) d1 s1 + d2 s2 + · · · + dk sk
where d1 , d2 , . . . , dk ∈ C[y] and s1 , s2 , . . . , sk ∈ C[x, y]. 2 a b By Lemma ?? the ci ’s in XA and YB can be chosen as elements in C[y] and C[x], respectively. The next theorem describes the structure of S. Theorem 3.2 There exist mutually inequivalent integrable pairs (a1 , b1 ), . . . , (am , bm ), and (a ,b ) C-linearly independent sets H1 , . . . , Hm such that S is the (disjoint) union of SHii i , i = 1, Pm . . . , m. Moreover, i=1 |Hi | is no greater than d. Proof. Let V be the solution space of L and (f1 , g1 ), . . . , (fm , gm ) in S, none of which is equivalent to the other. Then E (f1 ,g1 ) ∩ V , . . . , E (fm ,gm ) ∩ V are all nontrivial, and form a direct sum Lemma ??. Thus, m is no greater than d. We may further assume that there are only m equivalence classes (w.r.t. ∼) in S. Let (f, g) be one of the (fi , gi )’s. Assume that the intersection of V and E (f,g) is of dimension k over RC. Then there exist C-linearly independent rational functions r1 , . . . , rk R such that r1 exp ( f dx + g dy) , . . . , rk exp ( f dx + g dy) form a basis for E (f,g) ∩ V . Let S(f,g) be the subset of S consisting of integrable pairs equivalent to (f, g). Then
S(f,g) =
c1 ∂x r1 + · · · + ck ∂x rk c1 ∂y r1 + · · · + ck ∂y rk , g+ f+ c1 r1 + · · · + ck rk c1 r1 + · · · + ck rk
| c1 , . . . , ck ∈ C .
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Without loss of generality, we may assume that rj = hj /h where h, hj ∈ C[x, y], for j = 1, (a,b) . . . , k. Letting (a, b) = (f − ∂x h/h, g − ∂y h/h) and H = {h1 , . . . , hk }, we find S(f,g) = SH . Since S is the (disjoint) union of S(fi ,gi ) , i = 1, . . . , m, there exist integrable pairs (ai , bi ), with (ai , bi ) ∼ (fi , gi ), and C-linearly independent sets Hi such that S is the (disjoint) union Pm (ai ,bi ) (a ,b ) i i of SHi , i = 1, . . . , m. Moreover, i=1 |Hi | ≤ d, because dim E ∩ V = |Hi |. 2 n
o
(a ,b )
We call the set SHii i | i = 1, . . . , m a standard representation of S. Theorem ?? can be specialized to describe the rational solutions of an ordinary Riccati equation. More precisely, define two elements f and g of K are equivalent w.r.t. x if f − g = ∂x h/h for some h ∈ K, and denote this relation by ∼x . Likewise, we define the relation ∼y on K. Both ∼x and ∼y are equivalence relations. Remark that, for f, g, p, q ∈ K, f ∼x p and g ∼y q do not imply (f, g) ∼ (p, q), because (f, g) and (p, q) may not be integrable. Let Rx be an ordinary Riccati equation w.r.t. ∂x associated with a dx th order linear ordinary differential equation Lx whose coefficient field is K. Denote by Sx the set of rational solutions of Rx in K. Along the same line, one can show Corollary 3.3 There exists a1 , . . . , am in K, none of which is equivalent to the other w.r.t. ai , x, and C(y)-linearly independent sets A1 , . . . , Am such that Sx is the (disjoint) union of XA i Pm i = 1, . . . , m. Moreover, i=1 |Ai | is no greater than dx . ai | i = 1, . . . , m} a standard representation of Sx . We also call the set {XA i Algorithms for computing Sx are found in Singer (1991) and Bronstein (1992). Actually, their algorithms can compute all solutions of Rx in C(y)(x). But we only need Sx , a subset bi of K. Their algorithms output a finite list of sets XH whose union is Sx . We modify the i output to give a standard representation of Sx . If none of the bi ’s is equivalent to the other, then we are done. Otherwise, the next lemma is applied.
Lemma 3.4 If b1 and b2 are equivalent w.r.t. x, we can compute b ∈ K and a C(y)-linearly b1 b2 b ⊂S . independent set H such that XH ∪ XH ⊂ XH x 1 2 Proof. Let Hi be {hi1 , . . . , hiki } for k = 1, 2. Assume that b1 = b2 + ∂x g/g for some g ∈ K. Then c11 ∂x (h11 g) + · · · + c1k1 (∂x h1k1 g) b1 XH1 = b2 + | c11 , . . . c1k1 ∈ C(y) . c11 (h11 g) + · · · + c1k1 (h1k1 g) From the set G = {h11 g, . . . , h1k1 g, h21 , . . . , h2k2 }, we pick up a maximally C(y)-linearly independent set hk h1 ,..., | h, h1 , . . . , hk , ∈ C(y)[x] B= h h b , which contains both X b1 and Setting b = b2 − ∂x h/h and H = {h1 , . . . , hk }, we obtain XH H1 b2 XH because each element of G is a C (y)-linear combination of some elements of B. Note 2 that, for all c1 , . . . , ck ∈ C(y),
Z
(c1 h1 + · · · + ck hk ) exp
h1 hk b dx = c1 + · · · + ck h h
b is contained in S . which is a solution of Lx . Hence, XH x
Z
exp
b2 dx , 2
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Example 3.2 The ordinary Riccati equation x2 (∂x u + u2 ) + xu − 1 = 0 has rational solutions u = 1/x and −1/x + 2cx/(1 + cx2 ), where c is an x-constant. The set of rational solutions −1
1
x x are the union of X{1} and X{1,x 2 } . Note that (1/x) ∼x (−1/x) because
1 −1 2 ∂x x2 − = = 2 . x x x x −1
1
−1
1
x x x x Applying Lemma ?? to X{1} and X{1,x 2 } , we find that X{1} is contained in X{1,x2 } .
3.2. Computing standard representations Our idea for computing a standard representation of S consists of four steps. First, compute two nonzero ode’s Lx (z) ∈ [L] ∩ Lx and Ly (z) ∈ [L] ∩ Ly (with lowest order). Second, translate Lx (z) to the Riccati ode Rx (u), and Ly (z) to Ry (v) by (??) and compute respective standard representations of rational solutions of the Riccati ode’s in K. In the rest of this section, we denote by X and Y the standard representations of Rx (u) and Ry (v), respectively. Third, from X and Y, construct a finite number of mutually inequivalent integrable pairs such that an element of S is equivalent to such a pair. At last, for each pair obtained from the third step, compute elements of S that are equivalent to it. The first step is carried out by Lemma ??, and the second by known algorithms. Before describing the last two steps, we remind the reader of a useful identity
∂y
∂x a a
= ∂x
∂y a a
for all a ∈ Ω.
(4)
a ∈ X and Y b ∈ Y such that Lemma 3.5 If (f, g) belongs to S, then there exist unique XA B a × Y b. all rational solutions of R equivalent to (f, g) are contained in XA B
Proof. Since (f, g) ∈ S, f and g are rational solutions of Rx (u) and Ry (v), respectively. a ∈ X and unique Y b ∈ Y such that (f, g) is in X a × Y b . If (p, q) There then exist unique XA B A B a and q ∈ Y b by Corollary ??. is in S and equivalent to (f, g), p ∈ XA 2 B a b a b Lemma ?? reduces our task to computing S ∩ XA × YB , for all XA ∈ X and YB ∈ Y. Now, we search for elements in X × Y that have possibly nonempty intersection with S. We a × Y b of X × Y a candidate if (a, b) is integrable, and try to transform call an element XA B other elements in X × Y to candidates by the following lemma. Lemma 3.6 Let a and b belong to K. There exist two polynomials p, q ∈ C[x, y] such that (a + ∂x p/p, b + ∂y q/q) is integrable if and only if ∂x ∂y (log z) = ∂y a − ∂x b
(5)
has a solution in K. Proof. Let r be ∂y a − ∂x b. Assume that there exist such p and q. Then
r = ∂x
∂y q q
− ∂y
∂x p p
= ∂x
∂y q q
= ∂x (∂y log q − ∂y log p) = ∂x ∂y log
− ∂x q . p
∂y p p
(by (??))
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Conversely, if z = q/p is a solution of (??), reversing the above calculation shows that
a+
∂x p ∂y q , b+ p q
2
is integrable. Now, we present an algorithm for computing such p and q.
Algorithm IntegrablePair (Find an integrable pair). Given a, b in K, the algorithm finds p, q in C[x, y] such that (a + ∂x p/p, b + ∂y q/q) is integrable, or determines that no such p and q exist. I1. [Initialize.] Set r ← ∂y a − ∂x b. If r = 0, set p ← 1, q ← 1 and exit. I2. [Hermite’s reduction.] Apply Hermite’s reduction (w.r.t. x) to get f, h ∈ K such that r = ∂x f + h
(6)
If h is nonzero, the algorithm terminates; no such p and q exist. I3. [Partial fraction.] Compute the irreducible partial fraction decomposition of f w.r.t. y over its coefficient field. If the decomposition is X ∂y qi X ∂y pj mi − nj +g (7) qi pj i j where pi , qj ∈ C[x, y], mi , nj ∈ Z+ ∪ {0}, and g ∈ C(y), set p ← Otherwise, no such p and q exist. 2
Q
n
j
pj j , q ←
Q
mi i qi .
R
Step I1 is clear. If h 6= 0, then r dx is not rational by Hermite’s reduction (Geddes, et. al 1992, Bronstein 1997). Thus, (??) has no rational solution, and such p and q do not exist by Lemma ??. Suppose now h = 0. Equations (??) and (??) imply that ∂y z =f +w (8) z where w is an x-constant. Assume that the irreducible partial fraction decomposition of f w.r.t. y is X ∂y qj X X sk ∂y pi f= nj − mi + r, + (9) qj pi tk j i k |
{z G
}
|
{z g
}
where pi , qi , sk , tk , r ∈ C[x, y], pi , qi , tk are irreducible over C(x)[y] and relatively prime to each other. If z is rational, then the partial decomposition of ∂y z/z is of form G, so that g must be an x-constant because of (??), (??) and the uniqueness of partial fraction decomposition. Suppose now that g belongs to C(y). We compute ∂x p ∂y q − ∂x = r + ∂y − ∂x ∂y p q ∂y p ∂y q (??) = r + ∂x − p q ∂y p ∂y q (??) (??) = ∂x f + − = ∂x g = 0. p q IntegrablePair then returns p and q, as desired.
∂x p a+ p
∂y q b+ q
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Example 3.3 Given a=
y 2 x3 + x3 + xy 2 + y − x2 y − x yx3 + y 2 x4 − x5 y − x4
and b =
1 , x−y
IntegrablePair proceeds as follows. I1. r = 1/(1 + xy)2 . I2. f = −1/(xy 2 + y), h = 0. I3. f = x/(1 + xy) − 1/y, p = 1, q = 1 + xy.
Hence, a, b +
x 1+xy
is integrable.
Example 3.4 Apply IntegrablePair to a=
y 3 + xy 2 + 2y − x2 y − x xy 3 + y 2 − x2 y 2 − xy
and b =
1 . x−y
We find that r=−
x2 y 2 + 2xy + 1 − y 2 x2 y 4 + 2xy 3 + x2 y 4 + y 2
and f = −
x2 y + x + y . y 2 (xy + 1)
In step I3 f decomposes into 1 x x − − . 1 + xy y y 2 Since g = −x/y 2 is not in C(y), no such p and q exist. a × Y b ”, we mean that we are given a, b ∈ K, a C(y)-linearly In what follows, by “given XA B independent set A, and a C(x)-linearly independent set B. a × Y b contains no rational solutions of R if IntegrablePair confirms that The set XA B no p, q ∈ C[x, y] are such that (a + ∂x p/p, b + ∂y q/q) is integrable, because a rational solution of R must be integrable. Otherwise, we construct a candidate described below. a × Y b , the algorithm finds Algorithm Candidate (Find a solution candidate). Given XA B f g f g a × Y b , or determines that XF × YG such that (f, g) is an integrable pair and XF × YG = XA B a × Y b contains no integrable pairs. XA B
C1. [construct f and g.]
If IntegrablePair(a, b) returns p, q ∈ C[x, y], set f ←a−
∂x q ∂y p , g ←b− . q p
a × Y b contains no integrable pairs. Otherwise, the algorithm terminates; XA B
C2. [construct F and G.]
Set F ← {qh | h ∈ A}, G ← {ph | h ∈ B}. 2
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In Step C1 we set the integrable pair (f, g) to be (a − ∂x q/q, b − ∂y p/p) instead of (a + ∂x p/p, b + ∂y q/q), because the former construction makes step C2 simpler. Equation (??) a × Y b , we implies that (f, g) is integrable if (a + ∂x p/p, b + ∂y q/q) is. To see XFf × YGg = XA B P let α = s∈A cs s, where the cs ’s are in C[y], not all zero. Since a + ∂x α/α = f + ∂x (qα)/qα, a . Likewise, Y g = Y b . The algorithm Candidate is correct. XFf = XA B G Example 3.5 Let a and b be the same as those in Example ??. Let A = {1, x} and B = {1}. IntegrablePair(a, b) returns p = 1 and q = 1 + xy. Candidate then returns f = a − y/(1 + xy), g = b, F = {1 + xy, x(1 + xy)} and G = B. Applying the above algorithm to each member of X × Y, we obtain a set of disjoint candidates o n C = XFf11 × YGg11 , . . . , XFfkk × YGgkk
such that S is contained in XFf11 × YGg11 can belong to only one member in C.
S
···
S
XFfkk × YGgkk , and a rational solution of R
Lemma 3.7 Let XFf × YGg be one of the sets in C. Let ex = max degx p p∈F
and
ey = max degy q. q∈G
If an integrable pair (a, b) belongs to XFf × YGg , then there exists a polynomial h ∈ C[x, y] with degx h ≤ ex and degy h ≤ ey such that
∂x h ∂y h , g+ . h h
(a, b) = f +
(10)
Proof. Let (a, b) = (f + ∂x s/s, g + ∂y t/t) , where s and t are, respectively, C(y)- and C(x)linear combinations of elements in F and G. Since (a, b) and (f, g) are integrable pairs, so is (∂x s/s, ∂y t/t) . The function Z
h = exp
∂x s ∂y t dx + dy s t
is well-defined and has the property ∂x h/h = ∂x s/s and ∂y h/h = ∂y t/t. It follows that s = ah and t = bh
(11)
for some x-constant a and y-constant b. Consequently, sb = ta. Let α be an element of C such that t(α, y) 6= 0. Then a = s(α, y)b(α)/t(α, y), which implies that a belongs to C(y), and, therefore, h is in C(y)[x] by (??). Likewise, h is in C(x)[y]. Hence, h is in C[x, y]. Accordingly, degx h = degx s ≤ ex and degy h = degy t ≤ ey by (??). 2 f g Suppose that (a, b) is a rational solution of R contained in XF × YG . By Lemma ?? there exists h ∈ C[x, y] such that (??) holds. Moreover, respective degree bounds for h in x and y are known. The next algorithm PolynomialPart determines h. Algorithm PolynomialPart (Find polynomial part). Given an associated Riccati-like system R, and a candidate XFf × YGg , the algorithm computes a C-linearly independent set (f,g) H such that the set of rational solutions of R equivalent to (f, g) is equal to SH . If H is empty, then such rational solutions do not exist.
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177
P1. [bound degrees.] Set ex ← maxp∈F degx p, ey ← maxq∈G degy q. P2. [ex = ey = 0.] If both ex and ey are zero, check whether (f, g) satisfies all the equations in R; if the answer is affirmative, set H ← {1}, otherwise, set H ← ∅; the algorithm terminates. Pe
y i j x P3. [form a linear algebraic system.] Set h ← ei=0 j=0 cij x y , where the cij are unspecified constants. Substitute a + ∂x h/h and b + ∂y h/h for u and v in each equation of R, respectively. Set L be the result, which is a linear homogeneous algebraic system in the cij ’s.
P
P4. [compute H.] Calculate a basis B for the solution space of L. If B consists of only zero vector, then set H ← ∅. Otherwise, set H to be the set consisting of polynomials corresponding to vectors of B. 2 In Step P3 substituting u ← a+∂x h/h, v ← b+∂y h/h into R is equivalent to substituting z ← h exp (f dx + g dy) into L; the latter yields a linear homogeneous system in the unspecified constants cij ’s, because exp (f dx + g dy) is a nonzero overall factor. Thus, L obtained in step P3 is a linear homogeneous algebraic system. The correctness of PolynomialPart then follows from Lemma ??. Example 3.6 Determine the rational solutions of )
(
R=
2y 2 − 2y 2 2 2 v + 2 , ∂y u + uv, ∂y v + v 2 + v, ∂y u − ∂x v . ∂x u + u − u + 2 x x x y−1 2
−1
y y−1 x 0 in S1 = X{1, x x2 } × Y{y−1, 1} and S2 = X{1, x} × Y{1, y} , respectively. Applying PolynomialPart to S1 yields
P1. ex = 2 and ey = 1. P2. Skipped. P3. h = c0 + c1 x + c2 y + c3 x2 + c4 xy + c5 x2 y, u ← ∂x h/h, v ← −1/(2y) + ∂y h/h, L = {c0 , c4 + c1 , c5 + c3 }. P4. H = {y, x − xy, x2 − x2 y}. −1 y−1 0 The rational solutions in X{1, x x2 } × Y{y−1, 1} are
c2 (1 − y) + 2c3 x(1 − y) 1 c1 − c2 x − c3 x2 , − + c1 y + c2 (x − xy) + c3 (x2 − x2 y) y − 1 c1 y + c2 (x − xy) + c3 (x2 − x2 y) where c1 , c2 , c3 ∈ C. Applying PolynomialPart to S2 yields P1. ex = 1 and ey = 1. P2. Skipped.
!
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P3. h = c0 + c1 x + c2 y + c3 xy, u ← y + ∂x h/h, v ← x + ∂y h/h, L = {c0 , c1 , c2 , c3 }. P4. H = ∅. Now, we present the main algorithm. Algorithm RationalSolution (Find rational solutions of an associated Riccati-like system). Given an associated Riccati-like system R, the algorithm computes a standard representation of all rational solutions of R. R1. [Compute a coherent autoreduced set.] Transform R to the linear system L by the substitution u ← ∂x z/z, v ← ∂y z/z. Compute a coherent autoreduced set A such that [A] = [L]. If A = {1}, the algorithm terminates; no solution exists. R2. [Eliminate.] Use A and Lemma ?? to compute two linear ode’s Lx (z) (w.r.t. x) and Ly (z) (w.r.t. y). Transform Lx (z) and Ly (z) to their associated Riccati ode’s Rx (u) and Ry (v), respectively. R3. [Find rational solutions of the Riccati ode’s.] Compute respective standard representations X and Y of rational solutions of Rx (u) and Ry (v). If either X or Y is empty, the algorithm terminates; no rational solution exists. R4. [Construct candidates.] Apply Candidateoto each member of X × Y to get a set of n f1 candidates C = XF1 × YGg11 , . . . , XFfkk × YGgkk . If C is empty, the algorithm terminates; no rational solution exists. R5. [Compute polynomial parts.] Apply PolynomialPart to each member of C and collect all nonempty results to construct a standard representation of S. 2 A few words need to be said about the correctness of RationalSolution. The set S is the set of rational solutions of the Riccati-like system associated with A because [L] = [A]. The set S is contained in the union of members of X × Y by step R3, and in the union of members in C by step R4. Hence, RationalSolution returns a standard representation of S in step R5. Step R1 is a necessary preparation for step R2, because the operations given in Lemma ?? are based on a characteristic set (Janet basis) of [L]. Step R1 may be omitted if Rx (u) and Ry (v) can be easily read off from R. If the coefficients of equations in R belong to Q(x, y), step R3 may require to introduce a finite algebraic extension of Q. Nevertheless, steps R4 and R5 can be carried out. A few examples illustrate how RationalSolution works. Example 3.7 Find rational solutions of the Riccati-like system (
R=
)
2 2y 2 − 2y 2 2 ∂x u + u − u + v + 2 , ∂y u + uv, ∂y v + v 2 + v, ∂y u − ∂x v . x x2 x y−1 2
Apply RationalSolution to get n
R1. L = A = ∂x2 z − x2 ∂x z +
2y 2 −2y ∂y z x2
+
2 z, x2
∂x ∂y z, ∂y2 z +
2 y−1 ∂y z
o
.
Rational Solutions of Riccati-like Systems
179
R2. Lx (z) = ∂x3 z, Ly (z) = ∂y2 z + 2∂y z/(y − 1), Rx (u) = ∂x2 u + 3u∂x u + u3 , Ry (v) = ∂y v + v 2 + 2v/(y − 1). n
o
a | a = 0, A = {1, x, x2 } , Y = Y b | b = −1/(y − 1), B = {y − 1, 1} . R3. X = XA B
n
a ×Yb R4. C = XA B
o
R5. A standard representation of S is ple ??).
n
(a,b)
SH
| H = {y, x − xy, x2 − x2 y}
o
(see Exam-
In other words, the hyperexponential solutions of L are
c1 y + c2 (x − xy) + c3 (x2 − x2 y) exp
1 dy , y−1
Z
0 dx −
where c1 , c2 , c3 are arbitrary elements of C. Example 3.8 Compute rational solutions of the Riccati-like system R =
n
6x2 −6xy+y 2 (∂x u + u2 ), x2 (2x−y) o 2y−3x 2 ) + y−2x v . (∂ v + v y 2 x(y−x) x (y−x)
∂x2 u + 3u∂x u + u3 +
∂y2 v + 3v∂y v + v 3 +
We skip steps R1 and R2 because the first and second equations in oR can be used n n as Rx (u) o a1 a2 and Ry (v), respectively. In step R3 we compute X = XA , X and Y = YBb11 , YBb22 , A2 1 where a1 = 0, A1 = {1, x}, a2 = y/x2 , A2 = {1}, b1 = 0, B1 = {1}, b2 = −1/x, and a1 a2 B2 = {1, y 2 }. Step R4 finds two candidates XA × YBb11 and XA × YBb22 . Step R5 finds a 1 2 standard representation of S as the union of (0,0)
S{1,x} = and (a ,b ) S{1,2y22}
=
c2 , 0 | c1 , c2 ∈ C c1 + c2 x
y 1 2c4 y ,− + x2 x c3 + c4 y 2
| c3 , c4 ∈ C .
The linear system with which R is associated, is (
)
6x2 − 6xy + y 2 2 2y − 3x 2 y − 2x ∂x3 z + ∂x z, ∂y3 z + ∂y z + 2 ∂y z . 2 x (2x − y) x(y − x) x (y − x)
Its hyperexponential solutions are c1 +c2 x and (c3 +c4 y) exp are arbitrary elements of C.
R
y x2
dx −
1 x
dy where c1 , c2 , c3 , c4
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Z. Li and F. Schwarz
4. Applications In this section we apply the algorithm RationalSolution to finding rational solutions of Lie’s system and hyperexponential solutions of linear homogeneous differential systems with finite linear dimension in several unknowns. Lie’s system (??) occurred originally in his investigation of certain second-order ode’s that were based on its symmetries (Lie 1873). It turned out to be almost as ubiquitous as the Riccati ode’s, e.g. in decomposing systems of linear pde’s into smaller components. To extend the applicability of the algorithm, we consider the following more general orthonomic system {P1 = ∂x u + a0 u2 + a1 u + a2 v + a3 , P2 = ∂y u + b0 uv + b1 u + b2 v + b3 , P3 = ∂x v + c0 uv + c1 u + c2 v + c3 , P4 = ∂y v + d0 v 2 + d1 u + d2 v + d3 }
(12)
in K{u, v} w.r.t. the ranking 1 < v < u < ∂y v < ∂x v < ∂y u < ∂x u < · · ·. Moreover, we assume that both a0 and d0 are nonzero. The differential remainders of ∆(P1 , P2 ) and ∆(P3 , P4 ) are, respectively, R12 = b0 (c0 − a0 )vu2 + a2 (b0 − d0 )v 2 + terms involving u2 , uv, u, v, 1 and R34 = c0 (d0 − b0 )v 2 u + d1 (a0 − c0 )u2 + terms involving v 2 , uv, u, v, 1. Observe that R12 and R34 are in K[u, v]. Hence, if either R12 or R34 is nonzero, all solutions of (??) would be solutions of some polynomials in K[u, v]. We will not consider this degenerating case. Lemma 4.1 If (??) is coherent, either a0 = c0 and b0 = d0 , or b0 = c0 = a2 = d1 = 0. Proof. If (??) is coherent, R12 = R34 = 0 by Lemma ??. Hence b0 (c0 − a0 ) = a2 (b0 − d0 ) = c0 (d0 − b0 ) = d1 (a0 − c0 ) = 0. Since a0 d0 = 6 0 in (??), either a0 = c0 and b0 = d0 , or b0 = c0 = a2 = d1 = 0. 2 Lemma ?? splits (??) into two systems
and
{F1 = ∂x u + a0 u2 + a1 u + a2 v + a3 , F2 = ∂y u + d0 uv + b1 u + b2 v + b3 , F3 = ∂x v + a0 uv + c1 u + c2 v + c3 , F4 = ∂y v + d0 v 2 + d1 u + d2 v + d3 }
(13)
{G1 = ∂x u + a0 u2 + a1 u + a3 , G2 = ∂y u + b1 u + b2 v + b3 , G3 = ∂x v + c1 u + c2 v + c3 , G4 = ∂y v + d0 v 2 + d2 v + d3 .}
(14)
Note that Lie’s system is a special case of (??). Clearly, the coherence of (??) implies the coherence of (??) and (??). We solve (??) and (??) separately. Theorem 4.2 If (??) is coherent, then the substitution 1 1 ∂x (a0 d0 ) 1 1 ∂y (a0 d0 ) u← U− a1 + c2 − , v← V − b1 + d 2 − a0 3a0 a0 d0 d0 3d0 a0 d0
(2)
transforms (??) into an associated Riccati-like system (in U and V ) of type R3 ple ??
(15) in Exam-
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181
Proof. Normalizing (??) by the substitution u←
u ¯ v¯ , v← , a0 d0
(16)
we transform (??) into the coherent system {F¯1 = ∂x u ¯+u ¯2 + a ¯1 u ¯+a ¯2 v¯ + a ¯3 , F¯2 = ∂y u ¯+u ¯v¯ + ¯b1 u ¯ + ¯b2 v¯ + ¯b3 , 2 ¯ ¯ ¯ F3 = ∂x v¯ + u ¯v¯ + c¯1 u ¯ + c¯2 v¯ + c¯3 , F4 = ∂y v¯ + v¯ + d1 u ¯ + d¯2 v¯ + d¯3 , }
(17)
where a ¯1 , . . . , d¯3 ∈ K. Since the differential remainders of ∆(F¯1 , F¯2 ) and ∆(F¯3 , F¯4 ) are, respectively, ¯ 12 = (¯ ¯ R c1 − ¯b1 )¯ u2 + (¯ c2 − ¯b2 )¯ uv¯ + (¯ c3 + ¯b2 c¯1 + ∂y a ¯1 − ∂x¯b1 − 2¯b3 − a ¯2 d¯1 ) u |
{z
}
p
+ terms involving v¯ and 1, and ¯ 34 = (¯ R c1 − ¯b1 )¯ uv¯ + (¯ c2 − ¯b2 )¯ v 2 + (2¯ c3 − ¯b2 c¯1 − ¯b3 + a ¯2 d¯1 − ∂x d¯2 + ∂y c¯2 ) v¯ |
{z
}
q
+ terms involving u ¯ and 1, we deduce ¯b1 = c¯1 , ¯b2 = c¯2 and p = q = 0. It follows that p + q = 3¯ c3 + ∂y (¯ a1 + c¯2 ) − 3¯b3 − ∂x (¯b1 + d¯2 ) = 0.
(18)
Applying the substitution u ¯ ← U − s, v¯ ← V − t,
where s = 13 (¯ a1 + c¯2 ) and t = 31 (¯b1 + d¯2 )
(19)
to (??), we get {f1 = ∂x U + U 2 + A1 U + A2 V + A3 , f2 = ∂y U + U V + B1 U + B2 V + B3 , f3 = ∂x V + U V + C1 U + C2 V + C3 , f4 = ∂y V + V 2 + D1 U + D2 V + D3 }
(20)
where A1 , . . . , D3 ∈ K. Since ¯b1 = c¯1 and ¯b2 = c¯2 , B1 = C1 and B2 = C2 . We compute B3 − C3 = (−∂y s − ¯b1 s − ¯b2 t + st + ¯b3 ) − (−∂x t − c¯1 s − c¯2 t + st + c¯3 ) = ¯b3 + ∂x t − c¯3 − ∂y s (since ¯b1 = c¯1 and ¯b2 = c¯2 ) 1 ¯ 3b3 + ∂x (¯b1 + d¯2 ) − 3¯ c3 − ∂y (¯ a1 + c¯2 ) 3 = 0 (by (??)).
=
(by (??))
Therefore, f2 − f3 = ∂y U − ∂x V . This implies that {f1 , f2 , f2 − f3 , f4 } is the associated (2) Riccati-like system R3 , so is (??). Since a ¯1 = a1 − ∂x a0 /a0 , c¯2 = c2 − ∂x d0 /d0 , ¯b1 = b1 − ∂y a0 /a0 , d¯2 = d2 − ∂x d0 /d0 , substitution (??) is the result of the composition of (??) and (??).
2
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Example 4.1 Consider the coherent Lie’s system y + x2 4y(y + x2 ) 6yx2 + x4 + 4y 2 u + v + = 0, x(y − x2 ) x2 (y − x2 ) x2 (y 2 − 2yx2 + x4 ) 2y + x2 2(y + x2 ) 1 u − v − = 0, ∂y u + uv + y − x2 x(y − x2 ) x(y 2 − 2yx2 + x4 ) 2y + x2 1 x2 − 2y u − ∂x v + uv + v + = 0, y − x2 x(y − x2 ) x(y 2 − 2yx2 + x4 ) 2 4 v+ 2 = 0. ∂y v + v 2 + 2 y−x y − 2yx2 + x4 ∂x u + u2 +
By (??) we apply the transformation u←U+
y , 3x(y − x2 )
v←V −
5 3(y − x2 )
to get 4y(y + x2 ) 9x4 + 6yx2 − 23y 2 5y + 3x2 U + V + = 0, 3x(y − x2 ) x2 (y − x2 ) 9x2 (y 2 − 2yx2 + x4 ) 2 5y + 3x2 2(5y − 3x2 ) ∂y U + U V − U − V + = 0, 3(y − x2 ) 3x(y − x2 ) 9x(y 2 − 2yx2 + x4 ) 2 5y + 3x2 2(5y − 3x2 ) ∂x V + U V − U − V + = 0, 3(y − x2 ) 3x(y − x2 ) 9x(y 2 − 2yx2 + x4 ) 2 2 ∂y V + V 2 + V − = 0, 2 2 3(y − x ) 9(y − 2yx2 + x4 ) ∂x U + U 2 +
(2)
which is equivalent to the Riccati-like system R3 in Example ??, because the difference between the second and third equations is ∂y U − ∂x V . Apply RationalSolution to this system to find the rational solutions: U=
−1 2x 2c2 x −1 c1 + + , V = + , 2 2 2 3x 3(y − x ) c1 y + c2 x 3(y − x ) c1 y + c2 x2
c1 , c2 ∈ C.
Hence, the rational solutions of the original system are u=
x 2c2 x −2 c1 + , v= + , 2 2 2 y−x c1 y + c2 x y−x c1 y + c2 x2
c1 , c2 ∈ C.
We turn our attention to (??). Theorem 4.3 If (??) is coherent, it decouples into two individual coherent systems {∂x u + a0 u2 + a1 u + a3 , ∂y u + b1 u + b3 } for u and v, respectively.
{∂y v + d0 v 2 + d2 v + d3 , ∂x v + c2 v + c3 }
(21)
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183
Proof. The differential remainders of ∆(G1 , G2 ) and ∆(G3 , G4 ) are, respectively R12 = −2b2 a0 uv + terms involving u2 , u. v and 1 and R34 = 2c1 d0 uv + terms involving v 2 , u. v and 1 2
Since (??) is coherent, b2 = c1 = 0. The theorem follows. The following system also appears frequently in symmetry analysis. Theorem 4.4 Let a1 , . . . , b3 ∈ K and a1 b1 6= 0. The first-order Riccati-like system {F = ∂x z + a1 z 2 + a2 z + a3 , G = ∂y z + b1 z 2 + b2 z + b3 }
(22)
is coherent if and only if its general solution depends on a single constant. If (??) has a rational solution, one of the following alternatives applies. 1. The general solution is rational and has the form 1 ∂y r 1 ∂x r +p= +p a1 r + c b1 r + c with p, r ∈ K and c ∈ C ∪ {∞}. 2. There are at most two special rational solutions not involving unspecified constants. Proof. Since the differential remainder of ∆(F, G) is an algebraic polynomial in K[z] of degree no greater than 2, (??) has at most two solutions if it is not coherent. Assume that (??) is coherent. We show that it is the system R2 in disguise (see Example ??). The substitution 1 ∂x a1 1 z ← u− a2 − (23) a1 2a1 a1 transforms (??) into the coherent system {f = ∂x u + u2 + A3 , g = ∂y u + B1 u2 + B2 u + B3 },
(24)
where A3 , B1 , B2 , B3 ∈ K. Since the differential remainder of ∆(f, g) is zero, B2 = −∂x B1 , ∂x B2 = 2A3 B1 − 2B3 , which implies 1 1 g − B1 f = ∂y u − B1 ∂x u − (∂x B1 )u + ∂x2 B1 = ∂y u − ∂x B1 u − ∂x B1 . 2 2
Hence, (??) is equivalent to the system
2
∂x u + u + A3 , ∂y u − ∂x
1 B1 u − ∂x B1 2
.
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Set v = B1 w − 12 ∂x B1 . The above system becomes 1 ∂x u + u + A3 , v − B1 u + ∂x B1 , ∂y u − ∂x v 2
2
(25)
which is of type R2 in Example (??). It follows from (??) and the definition of v that (u, v) is a solution of (??) if and only if z given in (??) is a solution of (??). Since (??) is associated with a coherent linear system with linear dimension two, the general solution of (??) can be written as c1 ∂x s1 + c2 ∂x s2 c1 ∂y s1 + c2 ∂y s2 (u, v) = , , c1 s1 + c2 s2 c1 s1 + c2 s2 where s1 and s2 are in some differential extension F of K, linearly independent over the constant field of F, and c1 and c2 are in the same constant field. Therefore, (??) implies that the general solution of (??) can be written as 1 c1 ∂x s1 + c2 ∂x s2 1 ∂x a1 − a2 − . a1 c1 s1 + c2 s2 2a1 a1
z=
Setting c = c1 /c2 , we prove that the general solution of (??) depends on one constant. We now consider the rational solutions of (??). By (??) and the second equation in (??), z is a rational solution of (??) if and only if (u, v) is a rational solution of (??). By Theorem ?? (??) has either infinitely many rational solutions or at most two inequivalent rational solutions The former case corresponds to the first alternative, and the latter to the second. Assume that (??) has infinitely many rational solutions. Then, by Theorem ??,
(u, v) =
c1 ∂x h1 + c2 ∂x h2 c1 ∂y h1 + c2 ∂y h2 + a, +b c1 h1 + c2 h2 c1 h1 + c2 h2
where h1 , h2 , a, b ∈ K and c1 , c2 ∈ C. Setting r = h2 /h1 , c = c1 /c2 , f = ∂x h1 /h1 + a and g = ∂y h1 /h1 + b, we get ∂x r ∂y r (u, v) = + f, +g . c+r c+r Transformation (??) implies that z=
1 ∂x r +p a1 c + r
for some p ∈ K. Substituting ∂y r/(c + r) + g for v in the second equation of (??) yields z=
1 ∂y r +q a1 B1 c + r
for some q ∈ K. Hence, p = q because we may set c = ∞, i.e., c2 = 0. The theorem is then proved by noticing that a1 B1 = b1 . 2 According to the proof of Theorem ??, the rational solutions of (??) can be computed by the algorithm RationalSolution. We may also proceed as follows. Compute the rational solutions of F . If there are only a finite number of solutions, we need only check if they satisfy G. Otherwise, the rational solutions of F are given by ∂x r +f C +r
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where r, f ∈ K and C ∈ C(y). Substituting this expression for z in G yields H = ∂y C + B1 C 2 + B2 C + B3 , for some B1 , B2 , B3 ∈ K. Collecting coefficients of H w.r.t. the powers of x yields a system in C(y){C} consisting possibly of first-order Riccati ode’s, first-order linear ode’s and algebraic equations, whose rational solutions can be easily found. Example 4.2 Compute the rational solutions of {∂x z + z 2 , ∂y z + (1 − x2 − 2xy − y 2 )z 2 + (2x + 2y)z − 1 = 0.} The rational solutions of the first equation are 1/(C(y) + x), where C(y) is an x-constant. Substituting the expression into the second equation yields ∂y C(y) + C(y)2 − 2yC(y) + y 2 − 1 = 0, so that C(y) = y + 1/(y + c), where c is a constant. This system has the rational solutions z=
1 x+y+
1 y+c
=
xy+y 2 +1 x+y xy+y 2 +1 + x+y
∂x c
+
1 . x+y
At last, the following problem is considered. Let D be the differential polynomial ring K{z1 , . . . , zn }. Given a linear system L ⊂ D with finite linear dimension, find all hyperexponential solutions of L. By general elimination procedures, we compute a linear characteristic set Li for [L] ∩ K{zi }, for i = 1, . . . , n. Since each [Li ] is also of finite linear dimension, the algorithm RationalSolution computes a standard representation Si of the rational solutions of the Riccati-like system associated with Li . Assume that n o (f ,g ) Si = SHijij ij | j = 1, . . . , mi .
The hyperexponential solutions of Li are then expressed as Ei = Vij
Smi
j=1 Vij ,
where
Z X = ch h exp fij dx + gij dy | ch ∈ C . h∈Hij
The problem is thus reduced to computing hyperexponential solutions of L contained in V1j1 × · · · × Vnjn for 1 ≤ j1 ≤ m1 , . . . , 1 ≤ jn ≤ mn . Substituting
X
ch h exp
Z
fiji dx + giji dy
h∈Hiji
for zi in L yields a linear algebraic system A in the unspecified constants c’s. Notice that the coefficients of A may be hyperexponential. Nevertheless, the constant solutions of A gives us the hyperexponential solutions of L in V1j1 × · · · × Vnjn .
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Example 4.3 Consider the system L { x2 ∂x z2 − xy∂x z1 + yz1 , x2 ∂x2 z1 − x∂x z1 + z1 , y∂y z1 − x∂x z1 + z1 , xy∂y z2 + xy∂x z1 − xz2 + yz1 }. By elimination we get L1 = {y∂y z1 − x∂x z1 + z1 , x2 ∂x2 z1 − x∂x z1 + z1 } and L2 = {y∂y z2 − xy 2 ∂x z2 − z2 , x2 ∂x3 z2 + 3x∂x2 z2 + ∂x z2 }. By the algorithm RationalSolution we find that respective hyperexponential solutions of L1 and L2 are c1 x and c2 y, where c1 , c2 ∈ C. Substituting c1 x for z1 and c2 y for z2 into L yields the linear system {c1 = 0}. Hence, the hyperexponential solutions of L are (0, c2 y), where c2 ∈ C. References Boulier, F., Lazard, D., Ollivier, F., Petitot, M. (1995). Representation for the radical of a finitely generated differential ideal. In: Levelt, A.H.M. (ed.): Proc. Int. Symp. on Symbolic and Algebraic Computation, Montreal, Canada, ACM Press, 158–166. Bronstein, M. (1992). Linear ordinary differential equations: breaking through the order 2 barrier. In: Wang. P. (ed.): Proc. Int. Symp. on Symbolic and Algebraic Computation, Berkeley, ACM Press, 42–48. Bronstein, M. (1997). Symbolic Integration I: Transcendental Functions. Springer. Geddes, K., Czapor, S., Labahn, G. (1992). Algorithms for Computer Algebra. Kluwer Academic Publishers. Janet, M. (1920). Les syst´ems d’´equations aux d´eriv´ees partielles. Journal de math´ematiques 83, 65–123. Kandru-Rody, A. Weispfenning, V. (1990). Non-commutative Gr¨obner Bases in algebras of solvable type. J. Symb. Comput. 9, 1–26. Kolchin, E. (1973). Differential Algebra and Algebraic Groups. New York: Academic Press. Li, Z., Wang, D.-m. (1999) Coherent, regular and simple systems in zero decompositions of partial differential systems. Syst. Sci. Math. Sci. 12 Suppl., 43–60. Lie, S. Klassifikation und Integration von gew¨ohnlichen Differentialgleichungen zwischen x, y, die eine Gruppe von Transformationen gestatten III. Archiv for Mathematik VIII, 371–458 (Gesammelte Abhandlungen V, 362–427). Reid, W. T. (1972). Riccati Differential Equations. Academic Press, New York and London. Ritt, J. F. (1950). Differential Algebra. New York: Amer. Math. Soc. Rosenfeld, A. (1959). Specializations in differential algebra. Trans. Amer. Math. Soc. 90, 394–407. Schwarz, F. (1998a). Janet bases for symmetry groups. In: Buchberger, B. and Winkler, F. (eds.): Gr¨ obner Bases and Applications, London Math. Soc. Lecture Notes Series 251. Cambridge: Cambridge University Press, 221–234. Schwarz, F. (1998b). ALLTYPES: An ALgebraic Language and Type System. In: Calmet J. and Plaza J. (eds.): Artificial Intelligence and Symbolic Computation, Lecture Notes in Artificial Intelligence 1476. Springer, 270 – 283.
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Singer, M. (1991). Liouillian solutions of linear differential equations with Liouillian coefficients. J. Symb. Comput. 11, 251–273. Wu, W.-t. (1989). On the foundation of algebraic differential geometry. Syst. Sci. Math. Sci. 2, 289–312.
Rational Solutions of Riccati-like Partial Differential Equations1) Ziming Li MMRC, Institute of Systems Science Academia Sinica, Beijing 100080, China Email: [email protected] Fritz Schwarz GMD, Institut SCAI 53754 Sankt Augustin, Germany Email: [email protected] Abstract. When factoring linear partial differential systems with a finite-dimensional solution space or analyzing symmetries of nonlinear ode’s, we need to look for rational solutions of certain nonlinear pde’s. The nonlinear pde’s are called Riccati-like because they arise in a similar way as Riccati ode’s. In this paper we describe the structure of rational solutions of a Riccati-like system, and an algorithm for computing them. The algorithm is also applicable to finding all rational solutions of Lie’s system {∂x u + u2 + a1 u+a2 v+a3 , ∂y u+uv+b1 u+b2 v+b3 , ∂x v+uv+c1 u+c2 v+c3 , ∂y v+v 2 +d1 u+d2 v+d3 }, where a1 . . . d3 are rational functions of x and y.
1. Introduction Riccati’s equation is one of the first examples of a differential equation that has been considered extensively in the literature, shortly after Leibniz and Newton had introduced the concept of the derivative of a function at the end of the 17th century. Riccati equations occur in many problems of mathematical physics and pure mathematics. A good survey is given in the book by Reid (1972). Of particular importance is the relation between Riccati’s equation and a linear ode y 00 + ay 0 + by = 0 with a, b ∈ C(x), where C is the field of the complex numbers. For example, solutions Rh with the property that the quotient p = h0 /h ∈ C(x) may be represented as h = exp ( p dx) if p satisfies the first-order Riccati equation p0 +p2 +ap+b = 0. Equivalently, this linear ode allows the first-order right factor y 0 −qy over C(x) if q obeys the same equation as p. In general, finding the first-order right rational factors of a linear homogeneous ode is equivalent to finding the rational solutions of its associated Riccati equation (see e.g. Singer 1991). It turns out that this correspondence carries over to systems of linear homogeneous partial differential equations with a finite-dimensional solution space. Systems of this kind occur 1)
The work is done when the first author visited GMD from September, 1997 to December, 2000
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in Lie’s symmetry theory for solving nonlinear ode’s and related equivalence problems. For example, Lie studied the coherent nonlinear system {∂x u + u2 + a1 u + a2 v + a3 , ∂y u + uv + b1 u + b2 v + b3 , ∂x v + uv + c1 u + c2 v + c3 , ∂y v + v 2 + d1 u + d2 v + d3 }
(1)
for the first time in connection with the symmetry analysis of second-order ode’s with projective symmetry group (see, e.g. Lie 1873, page 365). It is suggested therefore to call (??) Lie’s (2) system. We will show that Lie’s system may be transformed to the Riccati-like system R3 given in Example ??. In this paper the following problem will be considered. Given a set L of linear homogeneous pde’s in one unknown function z(x, y) whose coefficients are in C(x, y) and whose R solution space is of finite dimension, find a solution of L in the form exp ( u dx + v dy), where u and v are in C(x, y) with ∂y u = ∂x v. In other words, we want to find a linear differential ideal over C(x, y) with one-dimensional solution space containing L. As shown in Section ??, this problem is equivalent to finding a rational solution of the Riccati-like system associated with L. We describe an algorithm, implemented in the ALLTYPE system Schwarz (1998b), that computes all rational solutions of an associated Riccati-like system. The algorithm is also applicable to finding rational solutions of Lie’s system and hyperexponential solutions of linear homogeneous pde’s with finite-dimensional solution space in several unknowns. The paper is organized as follows. Section ?? contains necessary preliminaries. Section ?? presents the main results: the structure of rational solutions of a Riccati-like system and an algorithm for computing them. Applications are given in Section ??. 2. Preliminaries We denote by K the field of rational functions C(x, y), and by ∂x and ∂y the usual partial differential operators acting on K. Let Ω be the field of of all complex meromorphic functions in the space of complex variables of x and y. Note that both Ω and C have the same constant field C. In this paper we are concerned with solutions in Ω. An element a of Ω is called an x-constant (resp. y-constant) if ∂x a = 0 (resp. ∂y a = 0). An x-derivative (resp. y-derivative) of a means ∂xk a (resp. ∂yk a) for some nonnegative integer k. By a system we mean a finite subset of some differential polynomial ring over K. Basic notions related to differential polynomials are used such as: rankings, leaders, autoreduced sets, differential remainders, and characteristic sets for differential ideals. For their definitions, the reader is referred to Ritt (1950), Rosenfeld (1959), and Kolchin (1973). The differential ideal generated by a subset P of some differential polynomial ring is denoted by [P]. In the rest of this section, we describe three useful notions: the elimination property of linear differential ideals with finite linear dimension, integrable pairs, and coherent orthonomic systems, 2.1. Linear systems with finite linear dimension Let z be a differential indeterminate over K and fix an orderly ranking on the differential polynomial ring K{z}. We denote by L the K-linear space consisting of all linear homogeneous polynomials in K{z}.
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Given a linear system L ⊂ L, the linear dimension of [L] in K{z} is the codimension of L ∩ [L] in L (Kolchin 1973, page 151). The general solution of L depends on a finite number of unspecified constants if and only if the linear dimension of [L] is finite (Kolchin 1973, page 152). To check if the linear dimension of [L] is finite, we compute a coherent autoreduced set A (Rosenfeld, 1959) such that [L] = [A]. This computation can be done by various methods such as: Janet bases (Janet 1920, Schwarz 1998a) the characteristic set method (Wu 1989, Li and Wang 1999), and Gr¨obner bases for differential operators (Kandri-Rody and Weispfennig) . The linear dimension of [L] is finite if and only if an x-derivative and a y-derivative of z appear as leaders of some elements of A. We denote by Lx (resp. Ly ) the subset of L consisting of differential polynomials involving only x-derivatives (resp. y-derivatives) of z. The following elimination property is wellknown. Lemma 2.1 Let L be a finite subset of L. If [L] is of finite linear dimension, then there is an algorithm for computing two nonzero elements [L] ∩ Lx and [L] ∩ Ly , respectively. Proof. Compute a coherent autoreduced set A in L such that [L] = [A]. Assume that A is not {1}, for, otherwise, [L] is trivial. A finite base B of the K-linear space V = L/(L ∩ [L]) consists of all differential monomials ∂xi ∂yj z that cannot be reduced w.r.t. A. Using A, we can express any element of V as a K-linear combination of elements of B. Hence, we can find smallest integer n such that z, ∂x z, . . . , ∂xn z are K-linearly dependent, that is, we can find a0 , a1 , . . . , an−1 ∈ K such that Lx = a0 z + a1 ∂x z + · · · + an−1 ∂xn−1 z + ∂xn z ∈ [L]. Similarly, we can find a nonzero linear differential polynomial in [L] ∩ Ly .
2
2.2. Integrable pairs To describe the structure of rational solutions in Section ??, we define a pair of rational functions (f, g) ∈ K × K to be integrable if ∂y f = ∂x g. For an integrable pair (f, g), the R expression H = exp ( f dx + g dy) defines an element of Ω unique up to a nonzero multiplicative constant. We regard H as an element of Ω when the value of the multiplicative constant is irrelevant to our discussion. Note that ∂x H/H = f and ∂y H/H = g. Two integrable pairs (f, g) and (p, q) are said to be equivalent if there exists a nonzero h in K such that f − p = ∂x h/h and g − q = ∂y h/h. Write (f, g) ∼ (p, q) when they are equivalent. R R Since (f, g) ∼ (p, q) if and only if the ratio of exp ( f dx + g dy) and exp ( p dx + q dy) is a rational function, ∼ is an equivalence relation on the set of integrable pairs. It is easy to verify that, for integrable pairs (f1 , g1 ), (p1 , q1 ), (f2 , g2 ), (p2 , q2 ) and integers m, n, (f1 , g1 ) ∼ (p1 , q1 ) and (f2 , g2 ) ∼ (p2 , q2 ) implies (mf1 + nf2 , mg1 + ng2 ) ∼ (mp1 + np2 , mq1 + nq2 ). An element a in Ω is said to be a hyperexponential if ∂x a/a and ∂y a/a belong to K. If a is a hyperexponential, (∂x a/a, ∂y a/a) is an integrableR pair. Conversely, for an integrable pair (f, g), we may construct a hyperexponential exp ( f dx + g dy) in Ω. For an integrable R pair (f, g), define E (f,g) to be the C-linear space {h exp ( f dx + g dy) | h ∈ K } . A direct calculations shows that E (f,g) consists of all hyperexponentials a such that (∂x a/a, ∂y a/a) ∼ (f, g). The following lemma is used to group rational solutions of a Riccati-like system. Lemma 2.2 If (f1 , g1 ), (f2 , g2 ), . . . , (fn , gn ) are mutually inequivalent integrable pairs, the sum of C-linear spaces E (f1 ,g1 ) , E (f2 ,g2 ) , . . . , E (fn ,gn ) is direct.
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Proof. We proceed by induction. For n = 2, if E (f1 ,g1 ) ∩ E (f2 ,g2 ) contains a nonzero element a, (∂x a/a, ∂y a/a) ∼ (fi , gi ), for i = 1, 2. Hence, (f1 , g1 ) ∼ (f2 , g2 ), a contradiction. The sum of E (f1 ,g1 ) and E (f2 ,g2 ) is direct. Assume that the result is proved for lower values of n. If the sum of E (f1 ,g1 ) , E (f2 ,g2 ) , . . . , E (fn ,gn ) is not direct, there are nonzero z1 ∈ E (f2 ,g2 ) , z2 ∈ E (f2 ,g2 ) , . . . , zn ∈ E (fn ,gn ) which are C-linearly dependent. By a possible rearrangement of indexes, we have zn = c1 z1 + c2 z2 + · · · + cn−1 zn−1
(2)
for some c1 , c2 , . . . , cn−1 ∈ C. Since z1 , z2 , . . . , zn−1 are linearly independent over C by the induction hypothesis, Theorem 1 in Kolchin (1973, page 86) implies that there exist derivatives θ1 , θ2 , . . . , θn−1 such that W = det(θi zj ) is nonzero, where 1 ≤ i ≤ n − 1 and 1 ≤ j ≤ n−1. Since the zi ’s are hyperexponentials, there exists rij ∈ K such that θj zi = rji zi . Applying θ1 , θ2 , . . . , θn−1 to (??) then yields a linear system
r11 r21 · rn−1,1
r12 r22 · rn−1,2
··· ··· ··· ···
r1,n−1 r2,n−1 · rn−1,n−1
c1 z1 c2 z2 .. . cn−1 zn−1
=
r1n zn r2n zn .. . rn,n−1 zn
whose coefficient matrix (rji ) is of full rank because W 6= 0. Solving this system, we get ci zi = si zn , where si ∈ K. Since ck 6= 0 for some k with 1 ≤ k ≤ n − 1, zk /zn is rational, so (fk , gk ) ∼ (fn , gn ), a contradiction. 2 2.3. Coherent orthonomic systems A system P in a differential polynomial ring D is orthonomic if it is an autoreduced set and any element of P is linear w.r.t. its leader and monic, or, equivalently, both initial and separant of any element in P is one. A linear autoreduced set can be considered as an orthonomic system. System (??) is orthonomic w.r.t. any orderly ranking on u and v. Systems studied in this paper are always orthonomic with respect to certain orderly ranking on their differential indeterminates and derivatives. Let P be an orthonomic system in D. We recall the notions of ∆-polynomials and coherence in Rosenfeld (1959) for P. Note that these two notions are originally defined for autoreduced sets and the name “∆-polynomial” is due to Boulier et. al (1995). Let P1 and P2 belong to P with the respective leaders θ1 z and θ2 z, where z is a differential indeterminate, and θ1 , θ2 are derivatives. Then there exist derivatives φ1 and φ2 with minimal orders φ1 θ1 = φ2 θ2 . The ∆-polynomial of P1 and P2 , denoted by ∆(P1 , P2 ), is φ1 P1 − φ2 P2 . Note that ∆(P1 , P2 ) is well-defined provided the leaders of P1 and P2 are derivatives of the same indeterminate. The system P is coherent if, for every such pair P1 , P2 in P, ∆(P1 , P2 ) can be written as a D-linear combination of derivatives of elements of P, in which each derivative has its leader lower than φ1 θ1 z (in a preselected ranking). To study orthonomic systems, we need to make sure that they cannot be formally reduced to non-orthonomic or trivial ones. Corollary 3 in Chapter I of Rosenfeld (1959) asserts that an orthonomic system P is a characteristic set of [P] if and only if P is coherent. Hence, P cannot be formally reduced any further if it is reduced. By the same corollary, one can easily show
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Lemma 2.3 An orthonomic system P is coherent if and only if all ∆-polynomials (possibly) formed by elements of P have zero as their differential remainders w.r.t. P. Hence, we are able to decide algorithmically whether an orthonomic system is coherent. 3. Rational solutions of associated Riccati-like systems Given a linear differential system L ⊂ L with finite linear dimension, we want to compute its hyperexponential solutions. The substitution: Z
z ← exp
u dx + v dy
where ∂y u = ∂x v
(3)
transforms L to a nonlinear system in K{u, v}. The union of this system and {∂y u − ∂x v} is called the Riccati-like system associated with L, or, simply, an associated Riccati-like system. Conversely, the substitution: u ← ∂x z/z, v ← ∂y z/z transforms an associated Riccati-like system in K{u, v} to a system in L. As in the ordinary case, computing hyperexponential solutions of L is equivalent to computing rational solutions of its associated Riccati-like system. Example 3.1 Let L2 = {∂x2 z + a1 ∂x z + a2 z, ∂y z + b1 ∂x z + b2 z} be coherent. Then the linear dimension of L is 2. The Riccati-like system R2 associated with L is {∂x u + u2 + a1 u + (1) a2 , v + b1 u + b2 , ∂y u − ∂x v}. Coherent linear systems with dimension 3 may be either L3 (2) equal to {∂x3 z + a1 ∂x2 z + a2 ∂x z + a3 z, ∂y z + b1 ∂x2 z + b2 ∂x z + b3 z}, or L3 equal to {∂x2 z + a1 ∂x z + a2 ∂y z + a3 z, ∂y ∂x z + b1 ∂x z + b2 ∂y z + b3 z, ∂y2 z + c1 ∂x z + c2 ∂y z + c3 z}. (1)
Their respective associated Riccati-like systems are R3 equal to {∂x2 u + 3u∂x u + u3 + a1 (∂x u + u2 ) + a2 u + a3 , v + b1 (∂x u + u2 ) + b2 u + b3 , ∂y u − ∂x v} (2)
and R3 equal to {∂x u + u2 + a1 u + a2 v + a3 , ∂y u + uv + b1 u + b2 v + b3 , ∂y v + v 2 + c1 u + c2 v + c3 , ∂y u − ∂x v}. In the rest of this section, let L be a system in L with finite linear dimension d. The Riccatilike system associated with L is denoted by R. The set of rational solutions of R is denoted by S. All elements of S are integrable pairs, because ∂y u−∂x v is contained in R. We describe the structure of S and present an algorithm for computing S. 3.1. Standard representations of rational solutions To describe rational solutions of Riccati ode’s and Riccati-like systems precisely, we introduce some notation. Let F be a subfield of K. By an F -linearly independent set, we mean a finite subset of C[x, y] whose elements are F -linearly independent. Let a, b be in K. Let A = {a1 , . . . , ak } and B = {b1 , . . . , bl } be C(y) and C(x)-independent sets, respectively. We define P k \ ∂x ci ai i=1 a XA = a + Pk | c1 , . . . , ck are x-constants K i=1 ci ai
Rational Solutions of Riccati-like Systems
and YBb =
b+
∂y
P
l i=1 ci bi
Pl
i=1 ci bi
171
| c1 , . . . , cl are y-constants
\
K.
If, moreover, (a, b) is integrable and H = {h1 , . . . , hm } is a C-linearly independent set, we define P P ∂x ( m ∂y ( m (a,b) i=1 ci hi ) i=1 ci hi ) SH = a + Pm , b + Pm | c1 , . . . , cm ∈ C . i=1 ci hi i=1 ci hi Lemma 3.1 Let r1 , r2 , . . . , rk belong to C(y)[x] and c1 , c2 , . . . , ck be x-constants of Ω. If f=
c1 ∂x r1 + c2 ∂x r2 + · · · + ck ∂x rk c1 r1 + c2 r2 + · · · + ck rk
belongs to K, then there exist s1 , s2 , . . . , sk ∈ C[x, y] and d1 , d2 , . . . , dk ∈ C[y] such that f=
d1 ∂x s1 + d2 ∂x s2 + · · · + dk ∂x sk . d1 s1 + d2 s2 + · · · + dk sk
Proof. Assume that f = f1 /f2 , where f1 , f2 ∈ C[x, y]. Equating the coefficients of the like powers of x in (c1 ∂x r1 + c2 ∂x r2 + · · · + ck ∂x rk )f2 = (c1 r1 + c2 r2 + · · · + ck rk )f1 , we find that c1 , c2 , . . . , ck satisfy a linear system over C(y). Hence, c1 , c2 , . . . , ck can be chosen as elements in C(y). Let h be the common denominator of all the ci and the ri . Then h is in C[y]. We derive that f=
h(c1 ∂x r1 + c2 ∂x r2 + · · · + ck ∂x rk ) d1 ∂x s1 + d2 ∂x s2 + · · · + dk ∂x sk = h(c1 r1 + c2 r2 + · · · + ck rk ) d1 s1 + d2 s2 + · · · + dk sk
where d1 , d2 , . . . , dk ∈ C[y] and s1 , s2 , . . . , sk ∈ C[x, y]. 2 a b By Lemma ?? the ci ’s in XA and YB can be chosen as elements in C[y] and C[x], respectively. The next theorem describes the structure of S. Theorem 3.2 There exist mutually inequivalent integrable pairs (a1 , b1 ), . . . , (am , bm ), and (a ,b ) C-linearly independent sets H1 , . . . , Hm such that S is the (disjoint) union of SHii i , i = 1, Pm . . . , m. Moreover, i=1 |Hi | is no greater than d. Proof. Let V be the solution space of L and (f1 , g1 ), . . . , (fm , gm ) in S, none of which is equivalent to the other. Then E (f1 ,g1 ) ∩ V , . . . , E (fm ,gm ) ∩ V are all nontrivial, and form a direct sum Lemma ??. Thus, m is no greater than d. We may further assume that there are only m equivalence classes (w.r.t. ∼) in S. Let (f, g) be one of the (fi , gi )’s. Assume that the intersection of V and E (f,g) is of dimension k over RC. Then there exist C-linearly independent rational functions r1 , . . . , rk R such that r1 exp ( f dx + g dy) , . . . , rk exp ( f dx + g dy) form a basis for E (f,g) ∩ V . Let S(f,g) be the subset of S consisting of integrable pairs equivalent to (f, g). Then
S(f,g) =
c1 ∂x r1 + · · · + ck ∂x rk c1 ∂y r1 + · · · + ck ∂y rk , g+ f+ c1 r1 + · · · + ck rk c1 r1 + · · · + ck rk
| c1 , . . . , ck ∈ C .
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Without loss of generality, we may assume that rj = hj /h where h, hj ∈ C[x, y], for j = 1, (a,b) . . . , k. Letting (a, b) = (f − ∂x h/h, g − ∂y h/h) and H = {h1 , . . . , hk }, we find S(f,g) = SH . Since S is the (disjoint) union of S(fi ,gi ) , i = 1, . . . , m, there exist integrable pairs (ai , bi ), with (ai , bi ) ∼ (fi , gi ), and C-linearly independent sets Hi such that S is the (disjoint) union Pm (ai ,bi ) (a ,b ) i i of SHi , i = 1, . . . , m. Moreover, i=1 |Hi | ≤ d, because dim E ∩ V = |Hi |. 2 n
o
(a ,b )
We call the set SHii i | i = 1, . . . , m a standard representation of S. Theorem ?? can be specialized to describe the rational solutions of an ordinary Riccati equation. More precisely, define two elements f and g of K are equivalent w.r.t. x if f − g = ∂x h/h for some h ∈ K, and denote this relation by ∼x . Likewise, we define the relation ∼y on K. Both ∼x and ∼y are equivalence relations. Remark that, for f, g, p, q ∈ K, f ∼x p and g ∼y q do not imply (f, g) ∼ (p, q), because (f, g) and (p, q) may not be integrable. Let Rx be an ordinary Riccati equation w.r.t. ∂x associated with a dx th order linear ordinary differential equation Lx whose coefficient field is K. Denote by Sx the set of rational solutions of Rx in K. Along the same line, one can show Corollary 3.3 There exists a1 , . . . , am in K, none of which is equivalent to the other w.r.t. ai , x, and C(y)-linearly independent sets A1 , . . . , Am such that Sx is the (disjoint) union of XA i Pm i = 1, . . . , m. Moreover, i=1 |Ai | is no greater than dx . ai | i = 1, . . . , m} a standard representation of Sx . We also call the set {XA i Algorithms for computing Sx are found in Singer (1991) and Bronstein (1992). Actually, their algorithms can compute all solutions of Rx in C(y)(x). But we only need Sx , a subset bi of K. Their algorithms output a finite list of sets XH whose union is Sx . We modify the i output to give a standard representation of Sx . If none of the bi ’s is equivalent to the other, then we are done. Otherwise, the next lemma is applied.
Lemma 3.4 If b1 and b2 are equivalent w.r.t. x, we can compute b ∈ K and a C(y)-linearly b1 b2 b ⊂S . independent set H such that XH ∪ XH ⊂ XH x 1 2 Proof. Let Hi be {hi1 , . . . , hiki } for k = 1, 2. Assume that b1 = b2 + ∂x g/g for some g ∈ K. Then c11 ∂x (h11 g) + · · · + c1k1 (∂x h1k1 g) b1 XH1 = b2 + | c11 , . . . c1k1 ∈ C(y) . c11 (h11 g) + · · · + c1k1 (h1k1 g) From the set G = {h11 g, . . . , h1k1 g, h21 , . . . , h2k2 }, we pick up a maximally C(y)-linearly independent set hk h1 ,..., | h, h1 , . . . , hk , ∈ C(y)[x] B= h h b , which contains both X b1 and Setting b = b2 − ∂x h/h and H = {h1 , . . . , hk }, we obtain XH H1 b2 XH because each element of G is a C (y)-linear combination of some elements of B. Note 2 that, for all c1 , . . . , ck ∈ C(y),
Z
(c1 h1 + · · · + ck hk ) exp
h1 hk b dx = c1 + · · · + ck h h
b is contained in S . which is a solution of Lx . Hence, XH x
Z
exp
b2 dx , 2
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Example 3.2 The ordinary Riccati equation x2 (∂x u + u2 ) + xu − 1 = 0 has rational solutions u = 1/x and −1/x + 2cx/(1 + cx2 ), where c is an x-constant. The set of rational solutions −1
1
x x are the union of X{1} and X{1,x 2 } . Note that (1/x) ∼x (−1/x) because
1 −1 2 ∂x x2 − = = 2 . x x x x −1
1
−1
1
x x x x Applying Lemma ?? to X{1} and X{1,x 2 } , we find that X{1} is contained in X{1,x2 } .
3.2. Computing standard representations Our idea for computing a standard representation of S consists of four steps. First, compute two nonzero ode’s Lx (z) ∈ [L] ∩ Lx and Ly (z) ∈ [L] ∩ Ly (with lowest order). Second, translate Lx (z) to the Riccati ode Rx (u), and Ly (z) to Ry (v) by (??) and compute respective standard representations of rational solutions of the Riccati ode’s in K. In the rest of this section, we denote by X and Y the standard representations of Rx (u) and Ry (v), respectively. Third, from X and Y, construct a finite number of mutually inequivalent integrable pairs such that an element of S is equivalent to such a pair. At last, for each pair obtained from the third step, compute elements of S that are equivalent to it. The first step is carried out by Lemma ??, and the second by known algorithms. Before describing the last two steps, we remind the reader of a useful identity
∂y
∂x a a
= ∂x
∂y a a
for all a ∈ Ω.
(4)
a ∈ X and Y b ∈ Y such that Lemma 3.5 If (f, g) belongs to S, then there exist unique XA B a × Y b. all rational solutions of R equivalent to (f, g) are contained in XA B
Proof. Since (f, g) ∈ S, f and g are rational solutions of Rx (u) and Ry (v), respectively. a ∈ X and unique Y b ∈ Y such that (f, g) is in X a × Y b . If (p, q) There then exist unique XA B A B a and q ∈ Y b by Corollary ??. is in S and equivalent to (f, g), p ∈ XA 2 B a b a b Lemma ?? reduces our task to computing S ∩ XA × YB , for all XA ∈ X and YB ∈ Y. Now, we search for elements in X × Y that have possibly nonempty intersection with S. We a × Y b of X × Y a candidate if (a, b) is integrable, and try to transform call an element XA B other elements in X × Y to candidates by the following lemma. Lemma 3.6 Let a and b belong to K. There exist two polynomials p, q ∈ C[x, y] such that (a + ∂x p/p, b + ∂y q/q) is integrable if and only if ∂x ∂y (log z) = ∂y a − ∂x b
(5)
has a solution in K. Proof. Let r be ∂y a − ∂x b. Assume that there exist such p and q. Then
r = ∂x
∂y q q
− ∂y
∂x p p
= ∂x
∂y q q
= ∂x (∂y log q − ∂y log p) = ∂x ∂y log
− ∂x q . p
∂y p p
(by (??))
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Conversely, if z = q/p is a solution of (??), reversing the above calculation shows that
a+
∂x p ∂y q , b+ p q
2
is integrable. Now, we present an algorithm for computing such p and q.
Algorithm IntegrablePair (Find an integrable pair). Given a, b in K, the algorithm finds p, q in C[x, y] such that (a + ∂x p/p, b + ∂y q/q) is integrable, or determines that no such p and q exist. I1. [Initialize.] Set r ← ∂y a − ∂x b. If r = 0, set p ← 1, q ← 1 and exit. I2. [Hermite’s reduction.] Apply Hermite’s reduction (w.r.t. x) to get f, h ∈ K such that r = ∂x f + h
(6)
If h is nonzero, the algorithm terminates; no such p and q exist. I3. [Partial fraction.] Compute the irreducible partial fraction decomposition of f w.r.t. y over its coefficient field. If the decomposition is X ∂y qi X ∂y pj mi − nj +g (7) qi pj i j where pi , qj ∈ C[x, y], mi , nj ∈ Z+ ∪ {0}, and g ∈ C(y), set p ← Otherwise, no such p and q exist. 2
Q
n
j
pj j , q ←
Q
mi i qi .
R
Step I1 is clear. If h 6= 0, then r dx is not rational by Hermite’s reduction (Geddes, et. al 1992, Bronstein 1997). Thus, (??) has no rational solution, and such p and q do not exist by Lemma ??. Suppose now h = 0. Equations (??) and (??) imply that ∂y z =f +w (8) z where w is an x-constant. Assume that the irreducible partial fraction decomposition of f w.r.t. y is X ∂y qj X X sk ∂y pi f= nj − mi + r, + (9) qj pi tk j i k |
{z G
}
|
{z g
}
where pi , qi , sk , tk , r ∈ C[x, y], pi , qi , tk are irreducible over C(x)[y] and relatively prime to each other. If z is rational, then the partial decomposition of ∂y z/z is of form G, so that g must be an x-constant because of (??), (??) and the uniqueness of partial fraction decomposition. Suppose now that g belongs to C(y). We compute ∂x p ∂y q − ∂x = r + ∂y − ∂x ∂y p q ∂y p ∂y q (??) = r + ∂x − p q ∂y p ∂y q (??) (??) = ∂x f + − = ∂x g = 0. p q IntegrablePair then returns p and q, as desired.
∂x p a+ p
∂y q b+ q
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Example 3.3 Given a=
y 2 x3 + x3 + xy 2 + y − x2 y − x yx3 + y 2 x4 − x5 y − x4
and b =
1 , x−y
IntegrablePair proceeds as follows. I1. r = 1/(1 + xy)2 . I2. f = −1/(xy 2 + y), h = 0. I3. f = x/(1 + xy) − 1/y, p = 1, q = 1 + xy.
Hence, a, b +
x 1+xy
is integrable.
Example 3.4 Apply IntegrablePair to a=
y 3 + xy 2 + 2y − x2 y − x xy 3 + y 2 − x2 y 2 − xy
and b =
1 . x−y
We find that r=−
x2 y 2 + 2xy + 1 − y 2 x2 y 4 + 2xy 3 + x2 y 4 + y 2
and f = −
x2 y + x + y . y 2 (xy + 1)
In step I3 f decomposes into 1 x x − − . 1 + xy y y 2 Since g = −x/y 2 is not in C(y), no such p and q exist. a × Y b ”, we mean that we are given a, b ∈ K, a C(y)-linearly In what follows, by “given XA B independent set A, and a C(x)-linearly independent set B. a × Y b contains no rational solutions of R if IntegrablePair confirms that The set XA B no p, q ∈ C[x, y] are such that (a + ∂x p/p, b + ∂y q/q) is integrable, because a rational solution of R must be integrable. Otherwise, we construct a candidate described below. a × Y b , the algorithm finds Algorithm Candidate (Find a solution candidate). Given XA B f g f g a × Y b , or determines that XF × YG such that (f, g) is an integrable pair and XF × YG = XA B a × Y b contains no integrable pairs. XA B
C1. [construct f and g.]
If IntegrablePair(a, b) returns p, q ∈ C[x, y], set f ←a−
∂x q ∂y p , g ←b− . q p
a × Y b contains no integrable pairs. Otherwise, the algorithm terminates; XA B
C2. [construct F and G.]
Set F ← {qh | h ∈ A}, G ← {ph | h ∈ B}. 2
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In Step C1 we set the integrable pair (f, g) to be (a − ∂x q/q, b − ∂y p/p) instead of (a + ∂x p/p, b + ∂y q/q), because the former construction makes step C2 simpler. Equation (??) a × Y b , we implies that (f, g) is integrable if (a + ∂x p/p, b + ∂y q/q) is. To see XFf × YGg = XA B P let α = s∈A cs s, where the cs ’s are in C[y], not all zero. Since a + ∂x α/α = f + ∂x (qα)/qα, a . Likewise, Y g = Y b . The algorithm Candidate is correct. XFf = XA B G Example 3.5 Let a and b be the same as those in Example ??. Let A = {1, x} and B = {1}. IntegrablePair(a, b) returns p = 1 and q = 1 + xy. Candidate then returns f = a − y/(1 + xy), g = b, F = {1 + xy, x(1 + xy)} and G = B. Applying the above algorithm to each member of X × Y, we obtain a set of disjoint candidates o n C = XFf11 × YGg11 , . . . , XFfkk × YGgkk
such that S is contained in XFf11 × YGg11 can belong to only one member in C.
S
···
S
XFfkk × YGgkk , and a rational solution of R
Lemma 3.7 Let XFf × YGg be one of the sets in C. Let ex = max degx p p∈F
and
ey = max degy q. q∈G
If an integrable pair (a, b) belongs to XFf × YGg , then there exists a polynomial h ∈ C[x, y] with degx h ≤ ex and degy h ≤ ey such that
∂x h ∂y h , g+ . h h
(a, b) = f +
(10)
Proof. Let (a, b) = (f + ∂x s/s, g + ∂y t/t) , where s and t are, respectively, C(y)- and C(x)linear combinations of elements in F and G. Since (a, b) and (f, g) are integrable pairs, so is (∂x s/s, ∂y t/t) . The function Z
h = exp
∂x s ∂y t dx + dy s t
is well-defined and has the property ∂x h/h = ∂x s/s and ∂y h/h = ∂y t/t. It follows that s = ah and t = bh
(11)
for some x-constant a and y-constant b. Consequently, sb = ta. Let α be an element of C such that t(α, y) 6= 0. Then a = s(α, y)b(α)/t(α, y), which implies that a belongs to C(y), and, therefore, h is in C(y)[x] by (??). Likewise, h is in C(x)[y]. Hence, h is in C[x, y]. Accordingly, degx h = degx s ≤ ex and degy h = degy t ≤ ey by (??). 2 f g Suppose that (a, b) is a rational solution of R contained in XF × YG . By Lemma ?? there exists h ∈ C[x, y] such that (??) holds. Moreover, respective degree bounds for h in x and y are known. The next algorithm PolynomialPart determines h. Algorithm PolynomialPart (Find polynomial part). Given an associated Riccati-like system R, and a candidate XFf × YGg , the algorithm computes a C-linearly independent set (f,g) H such that the set of rational solutions of R equivalent to (f, g) is equal to SH . If H is empty, then such rational solutions do not exist.
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177
P1. [bound degrees.] Set ex ← maxp∈F degx p, ey ← maxq∈G degy q. P2. [ex = ey = 0.] If both ex and ey are zero, check whether (f, g) satisfies all the equations in R; if the answer is affirmative, set H ← {1}, otherwise, set H ← ∅; the algorithm terminates. Pe
y i j x P3. [form a linear algebraic system.] Set h ← ei=0 j=0 cij x y , where the cij are unspecified constants. Substitute a + ∂x h/h and b + ∂y h/h for u and v in each equation of R, respectively. Set L be the result, which is a linear homogeneous algebraic system in the cij ’s.
P
P4. [compute H.] Calculate a basis B for the solution space of L. If B consists of only zero vector, then set H ← ∅. Otherwise, set H to be the set consisting of polynomials corresponding to vectors of B. 2 In Step P3 substituting u ← a+∂x h/h, v ← b+∂y h/h into R is equivalent to substituting z ← h exp (f dx + g dy) into L; the latter yields a linear homogeneous system in the unspecified constants cij ’s, because exp (f dx + g dy) is a nonzero overall factor. Thus, L obtained in step P3 is a linear homogeneous algebraic system. The correctness of PolynomialPart then follows from Lemma ??. Example 3.6 Determine the rational solutions of )
(
R=
2y 2 − 2y 2 2 2 v + 2 , ∂y u + uv, ∂y v + v 2 + v, ∂y u − ∂x v . ∂x u + u − u + 2 x x x y−1 2
−1
y y−1 x 0 in S1 = X{1, x x2 } × Y{y−1, 1} and S2 = X{1, x} × Y{1, y} , respectively. Applying PolynomialPart to S1 yields
P1. ex = 2 and ey = 1. P2. Skipped. P3. h = c0 + c1 x + c2 y + c3 x2 + c4 xy + c5 x2 y, u ← ∂x h/h, v ← −1/(2y) + ∂y h/h, L = {c0 , c4 + c1 , c5 + c3 }. P4. H = {y, x − xy, x2 − x2 y}. −1 y−1 0 The rational solutions in X{1, x x2 } × Y{y−1, 1} are
c2 (1 − y) + 2c3 x(1 − y) 1 c1 − c2 x − c3 x2 , − + c1 y + c2 (x − xy) + c3 (x2 − x2 y) y − 1 c1 y + c2 (x − xy) + c3 (x2 − x2 y) where c1 , c2 , c3 ∈ C. Applying PolynomialPart to S2 yields P1. ex = 1 and ey = 1. P2. Skipped.
!
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P3. h = c0 + c1 x + c2 y + c3 xy, u ← y + ∂x h/h, v ← x + ∂y h/h, L = {c0 , c1 , c2 , c3 }. P4. H = ∅. Now, we present the main algorithm. Algorithm RationalSolution (Find rational solutions of an associated Riccati-like system). Given an associated Riccati-like system R, the algorithm computes a standard representation of all rational solutions of R. R1. [Compute a coherent autoreduced set.] Transform R to the linear system L by the substitution u ← ∂x z/z, v ← ∂y z/z. Compute a coherent autoreduced set A such that [A] = [L]. If A = {1}, the algorithm terminates; no solution exists. R2. [Eliminate.] Use A and Lemma ?? to compute two linear ode’s Lx (z) (w.r.t. x) and Ly (z) (w.r.t. y). Transform Lx (z) and Ly (z) to their associated Riccati ode’s Rx (u) and Ry (v), respectively. R3. [Find rational solutions of the Riccati ode’s.] Compute respective standard representations X and Y of rational solutions of Rx (u) and Ry (v). If either X or Y is empty, the algorithm terminates; no rational solution exists. R4. [Construct candidates.] Apply Candidateoto each member of X × Y to get a set of n f1 candidates C = XF1 × YGg11 , . . . , XFfkk × YGgkk . If C is empty, the algorithm terminates; no rational solution exists. R5. [Compute polynomial parts.] Apply PolynomialPart to each member of C and collect all nonempty results to construct a standard representation of S. 2 A few words need to be said about the correctness of RationalSolution. The set S is the set of rational solutions of the Riccati-like system associated with A because [L] = [A]. The set S is contained in the union of members of X × Y by step R3, and in the union of members in C by step R4. Hence, RationalSolution returns a standard representation of S in step R5. Step R1 is a necessary preparation for step R2, because the operations given in Lemma ?? are based on a characteristic set (Janet basis) of [L]. Step R1 may be omitted if Rx (u) and Ry (v) can be easily read off from R. If the coefficients of equations in R belong to Q(x, y), step R3 may require to introduce a finite algebraic extension of Q. Nevertheless, steps R4 and R5 can be carried out. A few examples illustrate how RationalSolution works. Example 3.7 Find rational solutions of the Riccati-like system (
R=
)
2 2y 2 − 2y 2 2 ∂x u + u − u + v + 2 , ∂y u + uv, ∂y v + v 2 + v, ∂y u − ∂x v . x x2 x y−1 2
Apply RationalSolution to get n
R1. L = A = ∂x2 z − x2 ∂x z +
2y 2 −2y ∂y z x2
+
2 z, x2
∂x ∂y z, ∂y2 z +
2 y−1 ∂y z
o
.
Rational Solutions of Riccati-like Systems
179
R2. Lx (z) = ∂x3 z, Ly (z) = ∂y2 z + 2∂y z/(y − 1), Rx (u) = ∂x2 u + 3u∂x u + u3 , Ry (v) = ∂y v + v 2 + 2v/(y − 1). n
o
a | a = 0, A = {1, x, x2 } , Y = Y b | b = −1/(y − 1), B = {y − 1, 1} . R3. X = XA B
n
a ×Yb R4. C = XA B
o
R5. A standard representation of S is ple ??).
n
(a,b)
SH
| H = {y, x − xy, x2 − x2 y}
o
(see Exam-
In other words, the hyperexponential solutions of L are
c1 y + c2 (x − xy) + c3 (x2 − x2 y) exp
1 dy , y−1
Z
0 dx −
where c1 , c2 , c3 are arbitrary elements of C. Example 3.8 Compute rational solutions of the Riccati-like system R =
n
6x2 −6xy+y 2 (∂x u + u2 ), x2 (2x−y) o 2y−3x 2 ) + y−2x v . (∂ v + v y 2 x(y−x) x (y−x)
∂x2 u + 3u∂x u + u3 +
∂y2 v + 3v∂y v + v 3 +
We skip steps R1 and R2 because the first and second equations in oR can be used n n as Rx (u) o a1 a2 and Ry (v), respectively. In step R3 we compute X = XA , X and Y = YBb11 , YBb22 , A2 1 where a1 = 0, A1 = {1, x}, a2 = y/x2 , A2 = {1}, b1 = 0, B1 = {1}, b2 = −1/x, and a1 a2 B2 = {1, y 2 }. Step R4 finds two candidates XA × YBb11 and XA × YBb22 . Step R5 finds a 1 2 standard representation of S as the union of (0,0)
S{1,x} = and (a ,b ) S{1,2y22}
=
c2 , 0 | c1 , c2 ∈ C c1 + c2 x
y 1 2c4 y ,− + x2 x c3 + c4 y 2
| c3 , c4 ∈ C .
The linear system with which R is associated, is (
)
6x2 − 6xy + y 2 2 2y − 3x 2 y − 2x ∂x3 z + ∂x z, ∂y3 z + ∂y z + 2 ∂y z . 2 x (2x − y) x(y − x) x (y − x)
Its hyperexponential solutions are c1 +c2 x and (c3 +c4 y) exp are arbitrary elements of C.
R
y x2
dx −
1 x
dy where c1 , c2 , c3 , c4
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Z. Li and F. Schwarz
4. Applications In this section we apply the algorithm RationalSolution to finding rational solutions of Lie’s system and hyperexponential solutions of linear homogeneous differential systems with finite linear dimension in several unknowns. Lie’s system (??) occurred originally in his investigation of certain second-order ode’s that were based on its symmetries (Lie 1873). It turned out to be almost as ubiquitous as the Riccati ode’s, e.g. in decomposing systems of linear pde’s into smaller components. To extend the applicability of the algorithm, we consider the following more general orthonomic system {P1 = ∂x u + a0 u2 + a1 u + a2 v + a3 , P2 = ∂y u + b0 uv + b1 u + b2 v + b3 , P3 = ∂x v + c0 uv + c1 u + c2 v + c3 , P4 = ∂y v + d0 v 2 + d1 u + d2 v + d3 }
(12)
in K{u, v} w.r.t. the ranking 1 < v < u < ∂y v < ∂x v < ∂y u < ∂x u < · · ·. Moreover, we assume that both a0 and d0 are nonzero. The differential remainders of ∆(P1 , P2 ) and ∆(P3 , P4 ) are, respectively, R12 = b0 (c0 − a0 )vu2 + a2 (b0 − d0 )v 2 + terms involving u2 , uv, u, v, 1 and R34 = c0 (d0 − b0 )v 2 u + d1 (a0 − c0 )u2 + terms involving v 2 , uv, u, v, 1. Observe that R12 and R34 are in K[u, v]. Hence, if either R12 or R34 is nonzero, all solutions of (??) would be solutions of some polynomials in K[u, v]. We will not consider this degenerating case. Lemma 4.1 If (??) is coherent, either a0 = c0 and b0 = d0 , or b0 = c0 = a2 = d1 = 0. Proof. If (??) is coherent, R12 = R34 = 0 by Lemma ??. Hence b0 (c0 − a0 ) = a2 (b0 − d0 ) = c0 (d0 − b0 ) = d1 (a0 − c0 ) = 0. Since a0 d0 = 6 0 in (??), either a0 = c0 and b0 = d0 , or b0 = c0 = a2 = d1 = 0. 2 Lemma ?? splits (??) into two systems
and
{F1 = ∂x u + a0 u2 + a1 u + a2 v + a3 , F2 = ∂y u + d0 uv + b1 u + b2 v + b3 , F3 = ∂x v + a0 uv + c1 u + c2 v + c3 , F4 = ∂y v + d0 v 2 + d1 u + d2 v + d3 }
(13)
{G1 = ∂x u + a0 u2 + a1 u + a3 , G2 = ∂y u + b1 u + b2 v + b3 , G3 = ∂x v + c1 u + c2 v + c3 , G4 = ∂y v + d0 v 2 + d2 v + d3 .}
(14)
Note that Lie’s system is a special case of (??). Clearly, the coherence of (??) implies the coherence of (??) and (??). We solve (??) and (??) separately. Theorem 4.2 If (??) is coherent, then the substitution 1 1 ∂x (a0 d0 ) 1 1 ∂y (a0 d0 ) u← U− a1 + c2 − , v← V − b1 + d 2 − a0 3a0 a0 d0 d0 3d0 a0 d0
(2)
transforms (??) into an associated Riccati-like system (in U and V ) of type R3 ple ??
(15) in Exam-
Rational Solutions of Riccati-like Systems
181
Proof. Normalizing (??) by the substitution u←
u ¯ v¯ , v← , a0 d0
(16)
we transform (??) into the coherent system {F¯1 = ∂x u ¯+u ¯2 + a ¯1 u ¯+a ¯2 v¯ + a ¯3 , F¯2 = ∂y u ¯+u ¯v¯ + ¯b1 u ¯ + ¯b2 v¯ + ¯b3 , 2 ¯ ¯ ¯ F3 = ∂x v¯ + u ¯v¯ + c¯1 u ¯ + c¯2 v¯ + c¯3 , F4 = ∂y v¯ + v¯ + d1 u ¯ + d¯2 v¯ + d¯3 , }
(17)
where a ¯1 , . . . , d¯3 ∈ K. Since the differential remainders of ∆(F¯1 , F¯2 ) and ∆(F¯3 , F¯4 ) are, respectively, ¯ 12 = (¯ ¯ R c1 − ¯b1 )¯ u2 + (¯ c2 − ¯b2 )¯ uv¯ + (¯ c3 + ¯b2 c¯1 + ∂y a ¯1 − ∂x¯b1 − 2¯b3 − a ¯2 d¯1 ) u |
{z
}
p
+ terms involving v¯ and 1, and ¯ 34 = (¯ R c1 − ¯b1 )¯ uv¯ + (¯ c2 − ¯b2 )¯ v 2 + (2¯ c3 − ¯b2 c¯1 − ¯b3 + a ¯2 d¯1 − ∂x d¯2 + ∂y c¯2 ) v¯ |
{z
}
q
+ terms involving u ¯ and 1, we deduce ¯b1 = c¯1 , ¯b2 = c¯2 and p = q = 0. It follows that p + q = 3¯ c3 + ∂y (¯ a1 + c¯2 ) − 3¯b3 − ∂x (¯b1 + d¯2 ) = 0.
(18)
Applying the substitution u ¯ ← U − s, v¯ ← V − t,
where s = 13 (¯ a1 + c¯2 ) and t = 31 (¯b1 + d¯2 )
(19)
to (??), we get {f1 = ∂x U + U 2 + A1 U + A2 V + A3 , f2 = ∂y U + U V + B1 U + B2 V + B3 , f3 = ∂x V + U V + C1 U + C2 V + C3 , f4 = ∂y V + V 2 + D1 U + D2 V + D3 }
(20)
where A1 , . . . , D3 ∈ K. Since ¯b1 = c¯1 and ¯b2 = c¯2 , B1 = C1 and B2 = C2 . We compute B3 − C3 = (−∂y s − ¯b1 s − ¯b2 t + st + ¯b3 ) − (−∂x t − c¯1 s − c¯2 t + st + c¯3 ) = ¯b3 + ∂x t − c¯3 − ∂y s (since ¯b1 = c¯1 and ¯b2 = c¯2 ) 1 ¯ 3b3 + ∂x (¯b1 + d¯2 ) − 3¯ c3 − ∂y (¯ a1 + c¯2 ) 3 = 0 (by (??)).
=
(by (??))
Therefore, f2 − f3 = ∂y U − ∂x V . This implies that {f1 , f2 , f2 − f3 , f4 } is the associated (2) Riccati-like system R3 , so is (??). Since a ¯1 = a1 − ∂x a0 /a0 , c¯2 = c2 − ∂x d0 /d0 , ¯b1 = b1 − ∂y a0 /a0 , d¯2 = d2 − ∂x d0 /d0 , substitution (??) is the result of the composition of (??) and (??).
2
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Example 4.1 Consider the coherent Lie’s system y + x2 4y(y + x2 ) 6yx2 + x4 + 4y 2 u + v + = 0, x(y − x2 ) x2 (y − x2 ) x2 (y 2 − 2yx2 + x4 ) 2y + x2 2(y + x2 ) 1 u − v − = 0, ∂y u + uv + y − x2 x(y − x2 ) x(y 2 − 2yx2 + x4 ) 2y + x2 1 x2 − 2y u − ∂x v + uv + v + = 0, y − x2 x(y − x2 ) x(y 2 − 2yx2 + x4 ) 2 4 v+ 2 = 0. ∂y v + v 2 + 2 y−x y − 2yx2 + x4 ∂x u + u2 +
By (??) we apply the transformation u←U+
y , 3x(y − x2 )
v←V −
5 3(y − x2 )
to get 4y(y + x2 ) 9x4 + 6yx2 − 23y 2 5y + 3x2 U + V + = 0, 3x(y − x2 ) x2 (y − x2 ) 9x2 (y 2 − 2yx2 + x4 ) 2 5y + 3x2 2(5y − 3x2 ) ∂y U + U V − U − V + = 0, 3(y − x2 ) 3x(y − x2 ) 9x(y 2 − 2yx2 + x4 ) 2 5y + 3x2 2(5y − 3x2 ) ∂x V + U V − U − V + = 0, 3(y − x2 ) 3x(y − x2 ) 9x(y 2 − 2yx2 + x4 ) 2 2 ∂y V + V 2 + V − = 0, 2 2 3(y − x ) 9(y − 2yx2 + x4 ) ∂x U + U 2 +
(2)
which is equivalent to the Riccati-like system R3 in Example ??, because the difference between the second and third equations is ∂y U − ∂x V . Apply RationalSolution to this system to find the rational solutions: U=
−1 2x 2c2 x −1 c1 + + , V = + , 2 2 2 3x 3(y − x ) c1 y + c2 x 3(y − x ) c1 y + c2 x2
c1 , c2 ∈ C.
Hence, the rational solutions of the original system are u=
x 2c2 x −2 c1 + , v= + , 2 2 2 y−x c1 y + c2 x y−x c1 y + c2 x2
c1 , c2 ∈ C.
We turn our attention to (??). Theorem 4.3 If (??) is coherent, it decouples into two individual coherent systems {∂x u + a0 u2 + a1 u + a3 , ∂y u + b1 u + b3 } for u and v, respectively.
{∂y v + d0 v 2 + d2 v + d3 , ∂x v + c2 v + c3 }
(21)
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Proof. The differential remainders of ∆(G1 , G2 ) and ∆(G3 , G4 ) are, respectively R12 = −2b2 a0 uv + terms involving u2 , u. v and 1 and R34 = 2c1 d0 uv + terms involving v 2 , u. v and 1 2
Since (??) is coherent, b2 = c1 = 0. The theorem follows. The following system also appears frequently in symmetry analysis. Theorem 4.4 Let a1 , . . . , b3 ∈ K and a1 b1 6= 0. The first-order Riccati-like system {F = ∂x z + a1 z 2 + a2 z + a3 , G = ∂y z + b1 z 2 + b2 z + b3 }
(22)
is coherent if and only if its general solution depends on a single constant. If (??) has a rational solution, one of the following alternatives applies. 1. The general solution is rational and has the form 1 ∂y r 1 ∂x r +p= +p a1 r + c b1 r + c with p, r ∈ K and c ∈ C ∪ {∞}. 2. There are at most two special rational solutions not involving unspecified constants. Proof. Since the differential remainder of ∆(F, G) is an algebraic polynomial in K[z] of degree no greater than 2, (??) has at most two solutions if it is not coherent. Assume that (??) is coherent. We show that it is the system R2 in disguise (see Example ??). The substitution 1 ∂x a1 1 z ← u− a2 − (23) a1 2a1 a1 transforms (??) into the coherent system {f = ∂x u + u2 + A3 , g = ∂y u + B1 u2 + B2 u + B3 },
(24)
where A3 , B1 , B2 , B3 ∈ K. Since the differential remainder of ∆(f, g) is zero, B2 = −∂x B1 , ∂x B2 = 2A3 B1 − 2B3 , which implies 1 1 g − B1 f = ∂y u − B1 ∂x u − (∂x B1 )u + ∂x2 B1 = ∂y u − ∂x B1 u − ∂x B1 . 2 2
Hence, (??) is equivalent to the system
2
∂x u + u + A3 , ∂y u − ∂x
1 B1 u − ∂x B1 2
.
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Set v = B1 w − 12 ∂x B1 . The above system becomes 1 ∂x u + u + A3 , v − B1 u + ∂x B1 , ∂y u − ∂x v 2
2
(25)
which is of type R2 in Example (??). It follows from (??) and the definition of v that (u, v) is a solution of (??) if and only if z given in (??) is a solution of (??). Since (??) is associated with a coherent linear system with linear dimension two, the general solution of (??) can be written as c1 ∂x s1 + c2 ∂x s2 c1 ∂y s1 + c2 ∂y s2 (u, v) = , , c1 s1 + c2 s2 c1 s1 + c2 s2 where s1 and s2 are in some differential extension F of K, linearly independent over the constant field of F, and c1 and c2 are in the same constant field. Therefore, (??) implies that the general solution of (??) can be written as 1 c1 ∂x s1 + c2 ∂x s2 1 ∂x a1 − a2 − . a1 c1 s1 + c2 s2 2a1 a1
z=
Setting c = c1 /c2 , we prove that the general solution of (??) depends on one constant. We now consider the rational solutions of (??). By (??) and the second equation in (??), z is a rational solution of (??) if and only if (u, v) is a rational solution of (??). By Theorem ?? (??) has either infinitely many rational solutions or at most two inequivalent rational solutions The former case corresponds to the first alternative, and the latter to the second. Assume that (??) has infinitely many rational solutions. Then, by Theorem ??,
(u, v) =
c1 ∂x h1 + c2 ∂x h2 c1 ∂y h1 + c2 ∂y h2 + a, +b c1 h1 + c2 h2 c1 h1 + c2 h2
where h1 , h2 , a, b ∈ K and c1 , c2 ∈ C. Setting r = h2 /h1 , c = c1 /c2 , f = ∂x h1 /h1 + a and g = ∂y h1 /h1 + b, we get ∂x r ∂y r (u, v) = + f, +g . c+r c+r Transformation (??) implies that z=
1 ∂x r +p a1 c + r
for some p ∈ K. Substituting ∂y r/(c + r) + g for v in the second equation of (??) yields z=
1 ∂y r +q a1 B1 c + r
for some q ∈ K. Hence, p = q because we may set c = ∞, i.e., c2 = 0. The theorem is then proved by noticing that a1 B1 = b1 . 2 According to the proof of Theorem ??, the rational solutions of (??) can be computed by the algorithm RationalSolution. We may also proceed as follows. Compute the rational solutions of F . If there are only a finite number of solutions, we need only check if they satisfy G. Otherwise, the rational solutions of F are given by ∂x r +f C +r
Rational Solutions of Riccati-like Systems
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where r, f ∈ K and C ∈ C(y). Substituting this expression for z in G yields H = ∂y C + B1 C 2 + B2 C + B3 , for some B1 , B2 , B3 ∈ K. Collecting coefficients of H w.r.t. the powers of x yields a system in C(y){C} consisting possibly of first-order Riccati ode’s, first-order linear ode’s and algebraic equations, whose rational solutions can be easily found. Example 4.2 Compute the rational solutions of {∂x z + z 2 , ∂y z + (1 − x2 − 2xy − y 2 )z 2 + (2x + 2y)z − 1 = 0.} The rational solutions of the first equation are 1/(C(y) + x), where C(y) is an x-constant. Substituting the expression into the second equation yields ∂y C(y) + C(y)2 − 2yC(y) + y 2 − 1 = 0, so that C(y) = y + 1/(y + c), where c is a constant. This system has the rational solutions z=
1 x+y+
1 y+c
=
xy+y 2 +1 x+y xy+y 2 +1 + x+y
∂x c
+
1 . x+y
At last, the following problem is considered. Let D be the differential polynomial ring K{z1 , . . . , zn }. Given a linear system L ⊂ D with finite linear dimension, find all hyperexponential solutions of L. By general elimination procedures, we compute a linear characteristic set Li for [L] ∩ K{zi }, for i = 1, . . . , n. Since each [Li ] is also of finite linear dimension, the algorithm RationalSolution computes a standard representation Si of the rational solutions of the Riccati-like system associated with Li . Assume that n o (f ,g ) Si = SHijij ij | j = 1, . . . , mi .
The hyperexponential solutions of Li are then expressed as Ei = Vij
Smi
j=1 Vij ,
where
Z X = ch h exp fij dx + gij dy | ch ∈ C . h∈Hij
The problem is thus reduced to computing hyperexponential solutions of L contained in V1j1 × · · · × Vnjn for 1 ≤ j1 ≤ m1 , . . . , 1 ≤ jn ≤ mn . Substituting
X
ch h exp
Z
fiji dx + giji dy
h∈Hiji
for zi in L yields a linear algebraic system A in the unspecified constants c’s. Notice that the coefficients of A may be hyperexponential. Nevertheless, the constant solutions of A gives us the hyperexponential solutions of L in V1j1 × · · · × Vnjn .
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Example 4.3 Consider the system L { x2 ∂x z2 − xy∂x z1 + yz1 , x2 ∂x2 z1 − x∂x z1 + z1 , y∂y z1 − x∂x z1 + z1 , xy∂y z2 + xy∂x z1 − xz2 + yz1 }. By elimination we get L1 = {y∂y z1 − x∂x z1 + z1 , x2 ∂x2 z1 − x∂x z1 + z1 } and L2 = {y∂y z2 − xy 2 ∂x z2 − z2 , x2 ∂x3 z2 + 3x∂x2 z2 + ∂x z2 }. By the algorithm RationalSolution we find that respective hyperexponential solutions of L1 and L2 are c1 x and c2 y, where c1 , c2 ∈ C. Substituting c1 x for z1 and c2 y for z2 into L yields the linear system {c1 = 0}. Hence, the hyperexponential solutions of L are (0, c2 y), where c2 ∈ C. References Boulier, F., Lazard, D., Ollivier, F., Petitot, M. (1995). Representation for the radical of a finitely generated differential ideal. In: Levelt, A.H.M. (ed.): Proc. Int. Symp. on Symbolic and Algebraic Computation, Montreal, Canada, ACM Press, 158–166. Bronstein, M. (1992). Linear ordinary differential equations: breaking through the order 2 barrier. In: Wang. P. (ed.): Proc. Int. Symp. on Symbolic and Algebraic Computation, Berkeley, ACM Press, 42–48. Bronstein, M. (1997). Symbolic Integration I: Transcendental Functions. Springer. Geddes, K., Czapor, S., Labahn, G. (1992). Algorithms for Computer Algebra. Kluwer Academic Publishers. Janet, M. (1920). Les syst´ems d’´equations aux d´eriv´ees partielles. Journal de math´ematiques 83, 65–123. Kandru-Rody, A. Weispfenning, V. (1990). Non-commutative Gr¨obner Bases in algebras of solvable type. J. Symb. Comput. 9, 1–26. Kolchin, E. (1973). Differential Algebra and Algebraic Groups. New York: Academic Press. Li, Z., Wang, D.-m. (1999) Coherent, regular and simple systems in zero decompositions of partial differential systems. Syst. Sci. Math. Sci. 12 Suppl., 43–60. Lie, S. Klassifikation und Integration von gew¨ohnlichen Differentialgleichungen zwischen x, y, die eine Gruppe von Transformationen gestatten III. Archiv for Mathematik VIII, 371–458 (Gesammelte Abhandlungen V, 362–427). Reid, W. T. (1972). Riccati Differential Equations. Academic Press, New York and London. Ritt, J. F. (1950). Differential Algebra. New York: Amer. Math. Soc. Rosenfeld, A. (1959). Specializations in differential algebra. Trans. Amer. Math. Soc. 90, 394–407. Schwarz, F. (1998a). Janet bases for symmetry groups. In: Buchberger, B. and Winkler, F. (eds.): Gr¨ obner Bases and Applications, London Math. Soc. Lecture Notes Series 251. Cambridge: Cambridge University Press, 221–234. Schwarz, F. (1998b). ALLTYPES: An ALgebraic Language and Type System. In: Calmet J. and Plaza J. (eds.): Artificial Intelligence and Symbolic Computation, Lecture Notes in Artificial Intelligence 1476. Springer, 270 – 283.
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Singer, M. (1991). Liouillian solutions of linear differential equations with Liouillian coefficients. J. Symb. Comput. 11, 251–273. Wu, W.-t. (1989). On the foundation of algebraic differential geometry. Syst. Sci. Math. Sci. 2, 289–312.
