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On linear equations with general polynomial solutions
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On linear equations with general polynomial solutions A. Laradji Department of Mathematics & Statistics, King Fahd University of Petroleum & Minerals, Dhahran 31261, Saudi Arabia Received 8 September 2016; revised 22 September 2017
Abstract We provide necessary and sufficient conditions for which an nth-order linear differential equation has a general polynomial solution. We also give necessary conditions that can directly be ascertained from the coefficient functions of the equation. © 2018 Published by Elsevier Inc. MSC: 34A05
1. Introduction Consider the equation p0 y + p1 y + · · · + pn−1 y (n−1) + pn y (n) = 0
(1)
where the pk are functions (of a single variable x) continuous on some real interval in which pn does not vanish. If this equation has n linearly independent polynomial solutions yi (1 ≤ i ≤ n), then an application of Cramer’s rule to the system p0 pn−1 (n−1) (n) yi + · · · + y = −yi (1 ≤ i ≤ n) pn pn i E-mail address: [email protected]. https://doi.org/10.1016/j.jde.2018.01.002 0022-0396/© 2018 Published by Elsevier Inc.
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pk is a rational function. We will therefore assume, without loss of generality, pn that the coefficients pk in (1) are polynomials with no (non-constant) common factor and that pn is monic. Let K be the smallest integer k for which pk in (1) is not the zero polynomial. Clearly, (1) has n linearly independent polynomial solutions if and only if the equation shows that each
pK y + pK+1 y + · · · + pn−1 y (n−K−1) + pn y (n−K) = 0 has n − K linearly independent polynomial solutions. We can therefore assume that p0 in (1) is not the zero polynomial. For notational convenience, we will write each ph in (1) in “exponen phk k tial” form: ph = x with ph = 0 if h > n. If ph = 0, we denote its leading coefficient k≥0 k! by γh (with γn = 1). Determining conditions for which (1) has a fundamental set of polynomial solutions is a problem that has been discussed in several papers for various subclasses of (1). In [3], Calogero provided conditions for a wide class of second-order linear differential equations (with an arbitrary number of free parameters) to have general polynomial solutions. See also Calogero [4] in connection with a certain class of solvable N-body problems, Calogero [5] on the generalized hypergeometric equation, Calogero and Yi [6] concerning Jacobi polynomials and where para Jacobi polynomials are introduced, and Bagchi, Grandati and Quesne [2] where these polynomials are applied to the trigonometric Darboux–Pöschl–Teller potential. The main objective in this note is to give conditions, which do not seem to be known, for which (1) and its nonhomogeneous counterpart have general polynomial solutions. In Proposition 2, we provide necessary conditions that can quickly be ascertained from the leading coefficients and degrees of the polynomials pk . Propositions 4 and 5, while perhaps computationally more demanding, provide necessary and sufficient conditions. We end this note with a systematic way to construct nth-order linear equations containing an arbitrary number of parameters and having general polynomial solutions (cf. [3]). 2. Results We will need the following lemma. Lemma 1. Let r1 < · · · < rn be a sequence of nonnegative integers and y1 , . . . , yn be monic polynomials with respective degrees d1 < · · · < dn . Consider the generalized Wronskian rj y 1 , . . . , yn W , i.e. the determinant of the n ×n matrix whose (i, j )th-element is yi . Then, eir1 , . . . , rn y 1 , . . . , yn is the zero polynomial or it has degree ni=1 (di − ri ) and positive leading ther W r1 , . . . , rn di y1 , . . . , yn y 1 , . . . , yn . Furthermore, if W = 0, then W =0 coefficient det r1 , . . . , rn s1 , . . . , sn rj 1≤i,j ≤n for any sequence s1 < · · · < sn of nonnegative integers satisfying ri ≤ si for all i. y 1 , . . . , yn Proof. Clearly, the degree of W does not exceed the sum ni=1 (di − ri ) of the r1 , . . . , rn n y 1 , . . . , yn , and the coefficient c of x i=1 (di −ri ) is degrees of the diagonal polynomials in W r1 , . . . , rn
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the same as that of W
x d1 , . . . , x dn r1 , . . . , rn
3
(since the remaining terms of the yi have lower order). Ap n di rj !. plying elementary operations on this determinant, we obtain c = det rj 1≤i,j ≤n j =1 y 1 , . . . , yn By [7], c is nonnegative and c > 0 iff ri ≤ di for each i. This implies deg W = r1 , . . . , rn n (di − ri ) iff ri ≤ di for each i. Suppose ri > di for some i. Then, in the determinant i=1 y1 , . . . , y n W , the only possibly nonzero entries in the first i rows are the first (i − 1) entries, r1 , . . . , rn y 1 , . . . , yn = 0. Suppose now that s1 < and so these i rows are linearly dependent and W r1 , . . . , rn y 1 , . . . , yn = 0, · · · < sn is a sequence of nonnegative integers satisfying ri ≤ si for all i. If W s1 , . . . , sn di , which is positive by the foregoing argument. then its leading coefficient is det sj 1≤i,j ≤n di y1 , . . . , yn > 0. This implies W Hence di ≥ si ≥ ri for all i, and det = 0. 2 r1 , . . . , rn rj 1≤i,j ≤n Proposition 2. Suppose (1) (where the pk are polynomials with no common factor, p0 is not identically zero and pn is monic) has n linearly independent polynomial solutions. Then, for 0 ≤ k ≤ n, (i) (ii) (iii) (iv) (v)
pk = 0 1 + deg pk = deg pk+1 γk and γk+1 are integers with opposite signs deg pn ≤ −γn−1 (−1)n γ0 ≥ n!
Proof. Let y1 , . . . , yn be linearly independent polynomial solutions of the differential equation. We may clearly assume that the yi are monic with respective degrees di where d1 < · · · < dn (the degrees di are then uniquely determined by the n-dimensional solution space of polynoincreasing mials). Let rk1 , . . . , rkn denote the strictly sequence consisting of the elements of y 1 , . . . , yn {0, 1, 2, . . . , n}\{k} and let Wk := W be the n × n determinant defined in the rk1 , . . . , rkn above lemma. A straightforward application of Cramer’s rule to the system (2) with appropriate interchanges of columns, shows that for 0 ≤ k ≤ n pk Wn = (−1)n−k pn Wk .
(3)
By the above lemma, if Wk = 0, then its leading coefficient ck = det
di rkj
positive. Clearly, in this case, deg Wk =
n
i=1 (di
− rki ) =
n
i=1 di
+k−
n (n + 1) . 2
(1 ≤ i, j ≤ n) is
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If pk = 0, then deg pk + deg Wn = deg pn +
n
i=1 di
+k−
n (n + 1) 2
and in particular deg pk − k = deg pn − n.
(4)
This shows that if pk and pk+1 are nonzero polynomials, then deg pk+1 = 1 + deg pk . Suppose next that k < n and pk+1 = 0. Clearly, rk+1,j ≤ rkj for each j . Hence, by the lemma, Wk = 0 when Wk+1 = 0, i.e. pk = 0. This proves (i) and (ii). (k) To prove (iii), fix i in {1, . . . n}. For 0 ≤ k ≤ n, if pk yi is a nonzero polynomial, then it has degree deg pk − k + di and leading coefficient γk (di )k , where, for any number a, (a)k := a (a − 1) · · · (a − k + 1), (a)0 := 1. Hence, by Eq. (4), we obtain for each i n
(di )k γk = 0
k=0
i.e. the polynomial f (x) =
n
(x)k γk
k=0
which has degree n, has the n distinct positive integer roots d1 , . . . , dn (note that none of the di is zero since, otherwise, (1) would have a constant as a solution, contradicting the assumption that p0 = 0). From the fact that (x − d1 ) · · · (x − dn ) = nk=0 (x)k γk , we obtain that γk is a Z-linear combination of γk+1 , . . . , γn , d1 , . . . , dn for 0 ≤ k < n. Since γn = 1 and the di are integers, we infer (by induction) that each γk is an integer. Part (iii) of the proposition now follows from Eq. (3). x − rj where the rj are the (not necessarily distinct) complex Suppose next that pn = j
roots of pn . By Eq. (3), each rj is a root of Wn (otherwise it would be a common root for all n the pk ), and so pn divides Wn . Also, the coefficient − di of x n−1 in the polynomial f (x) can i=1
n n (n − 1) n (n − 1) easily shown to be γn−1 − di − , so that −γn−1 = = deg Wn ≥ deg pn . 2 2 i=1 This proves (iv). Finally, (v) follows from the fact that the strictly increasing sequence of degrees di of polyn nomial solutions of (1) satisfies n! ≤ di = (−1)n γ0 . 2 i=1
Remark 3. Suppose (1) has a fundamental set of polynomial solutions. n n (i) With the above notation and using the fact that (x − di ) = (x)k γk , one can easily i=1
k=0
obtain bounds, similar to Proposition 2 (v), involving other leading coefficients γk . Bounds can also be obtained for dn , the highest degree possible for a polynomial solution of (1); for example,
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|γ0 | . Note also that if |γ0 | is a prime power p α , then the degrees of all (n − 1)! √
1 + 8α + 1 n (n − 1) ≤ α, i.e. n ≤ . polynomial solutions of (1) are also powers of p and 2 2 pn−1 pn−1 Wn−1 dWn − pn dx is a (ii) By Eq. (3), =− . It is easy to see that = Wn−1 , so that e pn Wn dx pn−1 polynomial of the same degree as Wn , i.e. its degree is −γn−1 . This also means that = pn Aj for some negative integers Aj and roots rj of pn . In particular, all the roots of pn that j x − rj are not roots of pn−1 must be simple. |γ0 |1/n ≤ dn ≤
Note the converse of Proposition 2 does not hold: the equation x 2 y − 3xy + 2y = 0 that √ √ has x 2− 2 , x 2+ 2 as a fundamental set of solutions although it satisfies all the conditions of Proposition 2, as well as the condition in Remark 3 (ii). We next give a necessary and sufficient condition for (1) to have polynomial solutions only. n By the above proposition, we will assume that the polynomial (x)k γk has n distinct roots k=0
d1 , . . . , dn and that deg ph = h + d (0 ≤ h ≤ n) where d := deg p0 > 0 (so that phk = 0 if k > phk k x with ph = 0 if h > n. h + d). Recall that ph has leading coefficient γh and that ph = k≥0 k! For each positive integer N , let AN be the (N + d + 1) × (N + 1) matrix with (i, j )th entry i−1 i−1 pk+j −i,k . k k=0 Proposition 4. Equation (1) has a fundamental set of polynomial solutions if and only if d1 , . . . , dn are positive integers and rankAds = rankAds (1 ≤ s ≤ n), where Ads is the matrix obtained from Ads by deleting its last column. In this case: (i) d1 , . . . , dn are the degrees of n linearly independent solutions of (1); (ii) if Cj is the (j + 1)st column of Ads , then the coefficients of a monic solution q0 + q1 x + · · · + qds −1 x ds −1 + x ds of (1) satisfy Cds = −q0 C0 − · · · − qds −1 Cds −1 . Proof. Suppose that (1) has a polynomial solution of positive degree N , q = ql = 0 if l > N ). Clearly, N is a root of the polynomial h≥0
ph
n
(x)k γk . Direct substitution gives
k=0
qh+l l x = 0, i.e. l≥0 l! k + l h≥0 k≥0 l≥0
k
phk qh+l
x k+l = 0. (k + l)!
Relabeling indices gives N+d
u xu = 0. phk qu+h−k u! k
u=0 h≥0 0≤k≤u
ql l x say (with l≥0 l!
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u phk qu+h−k = 0, which gives the following system in the unh≥0 0≤k≤u k knowns ql (with the convention that phk = 0 if h < 0) We therefore have
N u pl+k−u,k ql = 0 (0 ≤ u ≤ N + d) . k
(5)
l=0 0≤k≤u
i −1 Let AN be the (N + d + 1) × (N + 1) coefficient matrix pk+j −i,k of the system. k k=0 It is easy to show that this system has a solution (q0 , . . . , qN ) with qN = 0 if and only if the rank of AN is equal to the rank of the matrix AN obtained from AN by deleting its last column (see [1, Lemma 1]). If, conversely, rankAN = rankAN , then the system (5) has a solution (q0 , . . . , qN ) with qN = 0, and so, clearly, (1) will have a polynomial solution of degree N . Finally, (i) follows from the proof of Proposition 2 (iii), and (ii) follows from the fact that Cds is a linear combination of C0 , . . . , Cds −1 . 2 i−1
We next turn to the non-homogeneous case. Consider the equation p0 y + p1 y + · · · + pn−1 y (n−1) + pn y (n) = f
(6)
where the pi and f are continuous on some interval I in which pn and f do not vanish. Let pk gk = (0 ≤ k ≤ n). We then obtain the equivalent equation on I f g0 y + g1 y + · · · + gn−1 y (n−1) + gn y (n) = 1. The general solution of (6) is y = z + C1 y1 + · · · + Cn yn , where z is a solution of (6), the yi are linearly independent solutions of the associated homogeneous equation p0 y + p1 y + · · · + pn−1 y (n−1) + pn y (n) = 0
(7)
and the Ci are arbitrary constants. Consider the system g0 yi + g1 yi + · · · + gn−1 yi(n−1) + gn yi(n) = 0 (1 ≤ i ≤ n) g0 z + g1 z + · · · + gn−1 z(n−1) + gn z(n) = 1 in the unknowns gk (0 ≤ k ≤ n). If (6) has a general polynomial solution, then, clearly, each yi and z are polynomials, and since the yi and z are linearly independent (the yi form a fundamental set of solutions for (7) and clearly z cannot be a linear combination of the yi ), we can apply Cramer’s rule to infer that each gk is a rational function. We will therefore assume in the sequel that the pk and f in (6) are polynomials with no (non-constant) common factor. If (6) has a general polynomial solution, then so too does (7), and therefore, when pn is monic, the necessary conditions in Proposition 2 hold in the nonhomogeneous case as well. Combining the next result with Proposition 4 (which concerns homogeneous equations) provides a necessary and sufficient rank condition for (6) to have a general polynomial solution.
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Proposition 5. Equation (6) has a general polynomial solution if and only if the homogeneous (n + 1)st-order equation n fp0 − f p0 y + fpk−1 + fpk − f pk y (k) + fpn y (n+1) = 0
(8)
k=1
has a general polynomial solution. n p d p0 i (i) Proof. Observe first that (8) is the equation y+ y = 0. Suppose that (6) dx f i=1 f has general polynomial solution y = z + C1 y1 + · · · + Cn yn , where z is a solution of (6), the yi are linearly independent solutions of (7) and the Ci are arbitrary constants. It is then easy to verify that z, z + yi (1 ≤ i ≤ n) are (n + 1) linearly independent solutions of (8). This proves “only the if” part. For the converse, suppose (8) has a fundamental set of polynomial solutions wj 0≤j ≤n . n p n p d p0 p0 i (i) i (i) Then, for 0 ≤ j ≤ n, wj + wj = 0, so that wj + wj = Aj for some dx f f f f i=1 i=1 constant Aj . If Aj = 0 for each j , then the wj would be (n + 1) linearly independent solutions of the nth-order equation (7), which is impossible. We can therefore assume that at least one of w0 is a solution of (6). For 1 ≤ j ≤ n, let the Aj , A0 say, is not equal to 0 so that A0 wj /Aj − w0 /A0 if Aj = 0 yj = if Aj = 0. wj f2
Then it is easy to see that y = y0 + quired.
2
n j =1
Cj yj is a general polynomial solution of (6), as re-
Remark 6. If (7) has a general polynomial solution and f is a polynomial, then it does not necessarily follow that (6) has also a general polynomial solution: For x > 0, the equation y − xy = x has general solution y = Cx − x ln x. In [3], Calogero gives a family of homogeneous second-order linear equations containing an arbitrary number of parameters and having general polynomial solutions. We end this note with a systematic way of constructing equations with such property. Start with any nth-order homogedy d ny neous equation, say P0 (x) y + P1 (x) + · · · + Pn (x) n = 0, where the Pj are polynomials, dx dx and suppose it has a general polynomial solution y = C1 y1 (x) + · · · + Cn yn (x). Since the polynomials yk (x) are linearly independent, we can clearly assume that they have distinct degrees αk (1 ≤ k ≤ n). Let x = ϕ(t) = a0 + a1 t + · · · + aN t N be an arbitrary polynomial of degree dj y N ≥ 1. Then, using Faà di Bruno’s formula (see [8], for example), we can express each dx j diϕ diy (1 ≤ j ≤ n) in terms of the derivatives i and i (1 ≤ i ≤ j ) and obtain an equation of the dt dt dy d ny form Q0 (t) y + Q1 (t) + · · · + Qn (t) n = 0 containing the constants a0 , . . . , aN as paramdt dt eters. This equation clearly has polynomial solutions y1 (ϕ(t)) , . . . , yn (ϕ(t)), which are linearly
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independent since they have distinct degrees αk N (1 ≤ k ≤ n), and has therefore a general polynomial solution. dy d 2y As an illustration, let f (x) y + g (x) + h (x) 2 = 0 have general polynomial solution dx dx y = C1 y1 (x) + C2 y2 (x). Then, putting x = ϕ(t) where ϕ is a polynomial of degree ≥ 1, we 2 2 d y d 2 y d 2 ϕ dy dϕ dy dϕ dy . The differential equation then becomes = and 2 = 2 + have dt dt dx dt dt dx dt dx 2
dϕ f (ϕ(t)) dt
3
dϕ y + g (ϕ(t)) dt
2
d 2ϕ − h (ϕ(t)) 2 dt
dϕ d 2 y dy = 0. + h (ϕ(t)) dt dt dt 2
(9)
Clearly, (9) contains the coefficients of ϕ as parameters and has polynomial solutions y1 (ϕ(t)) and y2 (ϕ(t)). A specific example is given below. Example. The Cauchy–Euler equation 2y − 2xy + x 2 y = 0 has solution y = C1 x + C2 x 2 . Putting x = a0 + a1 t + · · · + aN t N , we obtain the equation F (t)y + G (t)
dy d 2y + H (t) 2 = 0 dt dt
where F (t) = 2
N
3 kak t k−1
k=1
G (t) = −2
N
ak t
k
k=0
H (t) =
N k=0
2 kak t
k−1
−
k=1
2 ak t
N
k
N
N
2 ak t
k
k=0
N
k (k − 1) ak t
k−2
k=2
kak t
k−1
k=1
and whose general solution is y = C1
N k k=0 ak t
+ C2
N k k=0 ak t
2
.
Acknowledgment The author gratefully acknowledges the support of King Fahd University of Petroleum and Minerals. References [1] H. Azad, A. Laradji, T. Mustafa, Polynomial solutions of certain differential equations arising in physics, Math. Methods Appl. Sci. 36 (2013) 1615–1624. [2] B. Bagchi, Y. Grandati, C. Quesne, Rational extensions of the trigonometric Darboux–Pöschl–Teller potential based on para-Jacobi polynomials, J. Math. Phys. 56 (2015) 062103. [3] F. Calogero, A linear second-order ODE with only polynomial solutions, J. Differential Equations 255 (2013) 2130–2135.
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[4] F. Calogero, A solvable many-body problem, its equilibria, and a second-order ordinary differential equation whose general solution is polynomial, J. Math. Phys. 54 (2013) 012703. [5] F. Calogero, Can the generalized hypergeometric equation feature several independent polynomial solutions?, J. Phys. A 47 (2014) 045205. [6] F. Calogero, G. Yi, Can the general solution of the second-order ODE characterizing Jacobi polynomials be polynomial?, J. Phys. A 45 (2012) 095206. [7] I. Gessel, G. Viennot, Binomial determinants, paths, and hook length formulae, Adv. Math. 58 (1985) 300–321. [8] S. Roman, The formula of Faà di Bruno, Amer. Math. Monthly 87 (1980) 805–809.
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Available online at www.sciencedirect.com
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On linear equations with general polynomial solutions A. Laradji Department of Mathematics & Statistics, King Fahd University of Petroleum & Minerals, Dhahran 31261, Saudi Arabia Received 8 September 2016; revised 22 September 2017
Abstract We provide necessary and sufficient conditions for which an nth-order linear differential equation has a general polynomial solution. We also give necessary conditions that can directly be ascertained from the coefficient functions of the equation. © 2018 Published by Elsevier Inc. MSC: 34A05
1. Introduction Consider the equation p0 y + p1 y + · · · + pn−1 y (n−1) + pn y (n) = 0
(1)
where the pk are functions (of a single variable x) continuous on some real interval in which pn does not vanish. If this equation has n linearly independent polynomial solutions yi (1 ≤ i ≤ n), then an application of Cramer’s rule to the system p0 pn−1 (n−1) (n) yi + · · · + y = −yi (1 ≤ i ≤ n) pn pn i E-mail address: [email protected]. https://doi.org/10.1016/j.jde.2018.01.002 0022-0396/© 2018 Published by Elsevier Inc.
(2)
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pk is a rational function. We will therefore assume, without loss of generality, pn that the coefficients pk in (1) are polynomials with no (non-constant) common factor and that pn is monic. Let K be the smallest integer k for which pk in (1) is not the zero polynomial. Clearly, (1) has n linearly independent polynomial solutions if and only if the equation shows that each
pK y + pK+1 y + · · · + pn−1 y (n−K−1) + pn y (n−K) = 0 has n − K linearly independent polynomial solutions. We can therefore assume that p0 in (1) is not the zero polynomial. For notational convenience, we will write each ph in (1) in “exponen phk k tial” form: ph = x with ph = 0 if h > n. If ph = 0, we denote its leading coefficient k≥0 k! by γh (with γn = 1). Determining conditions for which (1) has a fundamental set of polynomial solutions is a problem that has been discussed in several papers for various subclasses of (1). In [3], Calogero provided conditions for a wide class of second-order linear differential equations (with an arbitrary number of free parameters) to have general polynomial solutions. See also Calogero [4] in connection with a certain class of solvable N-body problems, Calogero [5] on the generalized hypergeometric equation, Calogero and Yi [6] concerning Jacobi polynomials and where para Jacobi polynomials are introduced, and Bagchi, Grandati and Quesne [2] where these polynomials are applied to the trigonometric Darboux–Pöschl–Teller potential. The main objective in this note is to give conditions, which do not seem to be known, for which (1) and its nonhomogeneous counterpart have general polynomial solutions. In Proposition 2, we provide necessary conditions that can quickly be ascertained from the leading coefficients and degrees of the polynomials pk . Propositions 4 and 5, while perhaps computationally more demanding, provide necessary and sufficient conditions. We end this note with a systematic way to construct nth-order linear equations containing an arbitrary number of parameters and having general polynomial solutions (cf. [3]). 2. Results We will need the following lemma. Lemma 1. Let r1 < · · · < rn be a sequence of nonnegative integers and y1 , . . . , yn be monic polynomials with respective degrees d1 < · · · < dn . Consider the generalized Wronskian rj y 1 , . . . , yn W , i.e. the determinant of the n ×n matrix whose (i, j )th-element is yi . Then, eir1 , . . . , rn y 1 , . . . , yn is the zero polynomial or it has degree ni=1 (di − ri ) and positive leading ther W r1 , . . . , rn di y1 , . . . , yn y 1 , . . . , yn . Furthermore, if W = 0, then W =0 coefficient det r1 , . . . , rn s1 , . . . , sn rj 1≤i,j ≤n for any sequence s1 < · · · < sn of nonnegative integers satisfying ri ≤ si for all i. y 1 , . . . , yn Proof. Clearly, the degree of W does not exceed the sum ni=1 (di − ri ) of the r1 , . . . , rn n y 1 , . . . , yn , and the coefficient c of x i=1 (di −ri ) is degrees of the diagonal polynomials in W r1 , . . . , rn
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the same as that of W
x d1 , . . . , x dn r1 , . . . , rn
3
(since the remaining terms of the yi have lower order). Ap n di rj !. plying elementary operations on this determinant, we obtain c = det rj 1≤i,j ≤n j =1 y 1 , . . . , yn By [7], c is nonnegative and c > 0 iff ri ≤ di for each i. This implies deg W = r1 , . . . , rn n (di − ri ) iff ri ≤ di for each i. Suppose ri > di for some i. Then, in the determinant i=1 y1 , . . . , y n W , the only possibly nonzero entries in the first i rows are the first (i − 1) entries, r1 , . . . , rn y 1 , . . . , yn = 0. Suppose now that s1 < and so these i rows are linearly dependent and W r1 , . . . , rn y 1 , . . . , yn = 0, · · · < sn is a sequence of nonnegative integers satisfying ri ≤ si for all i. If W s1 , . . . , sn di , which is positive by the foregoing argument. then its leading coefficient is det sj 1≤i,j ≤n di y1 , . . . , yn > 0. This implies W Hence di ≥ si ≥ ri for all i, and det = 0. 2 r1 , . . . , rn rj 1≤i,j ≤n Proposition 2. Suppose (1) (where the pk are polynomials with no common factor, p0 is not identically zero and pn is monic) has n linearly independent polynomial solutions. Then, for 0 ≤ k ≤ n, (i) (ii) (iii) (iv) (v)
pk = 0 1 + deg pk = deg pk+1 γk and γk+1 are integers with opposite signs deg pn ≤ −γn−1 (−1)n γ0 ≥ n!
Proof. Let y1 , . . . , yn be linearly independent polynomial solutions of the differential equation. We may clearly assume that the yi are monic with respective degrees di where d1 < · · · < dn (the degrees di are then uniquely determined by the n-dimensional solution space of polynoincreasing mials). Let rk1 , . . . , rkn denote the strictly sequence consisting of the elements of y 1 , . . . , yn {0, 1, 2, . . . , n}\{k} and let Wk := W be the n × n determinant defined in the rk1 , . . . , rkn above lemma. A straightforward application of Cramer’s rule to the system (2) with appropriate interchanges of columns, shows that for 0 ≤ k ≤ n pk Wn = (−1)n−k pn Wk .
(3)
By the above lemma, if Wk = 0, then its leading coefficient ck = det
di rkj
positive. Clearly, in this case, deg Wk =
n
i=1 (di
− rki ) =
n
i=1 di
+k−
n (n + 1) . 2
(1 ≤ i, j ≤ n) is
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If pk = 0, then deg pk + deg Wn = deg pn +
n
i=1 di
+k−
n (n + 1) 2
and in particular deg pk − k = deg pn − n.
(4)
This shows that if pk and pk+1 are nonzero polynomials, then deg pk+1 = 1 + deg pk . Suppose next that k < n and pk+1 = 0. Clearly, rk+1,j ≤ rkj for each j . Hence, by the lemma, Wk = 0 when Wk+1 = 0, i.e. pk = 0. This proves (i) and (ii). (k) To prove (iii), fix i in {1, . . . n}. For 0 ≤ k ≤ n, if pk yi is a nonzero polynomial, then it has degree deg pk − k + di and leading coefficient γk (di )k , where, for any number a, (a)k := a (a − 1) · · · (a − k + 1), (a)0 := 1. Hence, by Eq. (4), we obtain for each i n
(di )k γk = 0
k=0
i.e. the polynomial f (x) =
n
(x)k γk
k=0
which has degree n, has the n distinct positive integer roots d1 , . . . , dn (note that none of the di is zero since, otherwise, (1) would have a constant as a solution, contradicting the assumption that p0 = 0). From the fact that (x − d1 ) · · · (x − dn ) = nk=0 (x)k γk , we obtain that γk is a Z-linear combination of γk+1 , . . . , γn , d1 , . . . , dn for 0 ≤ k < n. Since γn = 1 and the di are integers, we infer (by induction) that each γk is an integer. Part (iii) of the proposition now follows from Eq. (3). x − rj where the rj are the (not necessarily distinct) complex Suppose next that pn = j
roots of pn . By Eq. (3), each rj is a root of Wn (otherwise it would be a common root for all n the pk ), and so pn divides Wn . Also, the coefficient − di of x n−1 in the polynomial f (x) can i=1
n n (n − 1) n (n − 1) easily shown to be γn−1 − di − , so that −γn−1 = = deg Wn ≥ deg pn . 2 2 i=1 This proves (iv). Finally, (v) follows from the fact that the strictly increasing sequence of degrees di of polyn nomial solutions of (1) satisfies n! ≤ di = (−1)n γ0 . 2 i=1
Remark 3. Suppose (1) has a fundamental set of polynomial solutions. n n (i) With the above notation and using the fact that (x − di ) = (x)k γk , one can easily i=1
k=0
obtain bounds, similar to Proposition 2 (v), involving other leading coefficients γk . Bounds can also be obtained for dn , the highest degree possible for a polynomial solution of (1); for example,
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|γ0 | . Note also that if |γ0 | is a prime power p α , then the degrees of all (n − 1)! √
1 + 8α + 1 n (n − 1) ≤ α, i.e. n ≤ . polynomial solutions of (1) are also powers of p and 2 2 pn−1 pn−1 Wn−1 dWn − pn dx is a (ii) By Eq. (3), =− . It is easy to see that = Wn−1 , so that e pn Wn dx pn−1 polynomial of the same degree as Wn , i.e. its degree is −γn−1 . This also means that = pn Aj for some negative integers Aj and roots rj of pn . In particular, all the roots of pn that j x − rj are not roots of pn−1 must be simple. |γ0 |1/n ≤ dn ≤
Note the converse of Proposition 2 does not hold: the equation x 2 y − 3xy + 2y = 0 that √ √ has x 2− 2 , x 2+ 2 as a fundamental set of solutions although it satisfies all the conditions of Proposition 2, as well as the condition in Remark 3 (ii). We next give a necessary and sufficient condition for (1) to have polynomial solutions only. n By the above proposition, we will assume that the polynomial (x)k γk has n distinct roots k=0
d1 , . . . , dn and that deg ph = h + d (0 ≤ h ≤ n) where d := deg p0 > 0 (so that phk = 0 if k > phk k x with ph = 0 if h > n. h + d). Recall that ph has leading coefficient γh and that ph = k≥0 k! For each positive integer N , let AN be the (N + d + 1) × (N + 1) matrix with (i, j )th entry i−1 i−1 pk+j −i,k . k k=0 Proposition 4. Equation (1) has a fundamental set of polynomial solutions if and only if d1 , . . . , dn are positive integers and rankAds = rankAds (1 ≤ s ≤ n), where Ads is the matrix obtained from Ads by deleting its last column. In this case: (i) d1 , . . . , dn are the degrees of n linearly independent solutions of (1); (ii) if Cj is the (j + 1)st column of Ads , then the coefficients of a monic solution q0 + q1 x + · · · + qds −1 x ds −1 + x ds of (1) satisfy Cds = −q0 C0 − · · · − qds −1 Cds −1 . Proof. Suppose that (1) has a polynomial solution of positive degree N , q = ql = 0 if l > N ). Clearly, N is a root of the polynomial h≥0
ph
n
(x)k γk . Direct substitution gives
k=0
qh+l l x = 0, i.e. l≥0 l! k + l h≥0 k≥0 l≥0
k
phk qh+l
x k+l = 0. (k + l)!
Relabeling indices gives N+d
u xu = 0. phk qu+h−k u! k
u=0 h≥0 0≤k≤u
ql l x say (with l≥0 l!
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u phk qu+h−k = 0, which gives the following system in the unh≥0 0≤k≤u k knowns ql (with the convention that phk = 0 if h < 0) We therefore have
N u pl+k−u,k ql = 0 (0 ≤ u ≤ N + d) . k
(5)
l=0 0≤k≤u
i −1 Let AN be the (N + d + 1) × (N + 1) coefficient matrix pk+j −i,k of the system. k k=0 It is easy to show that this system has a solution (q0 , . . . , qN ) with qN = 0 if and only if the rank of AN is equal to the rank of the matrix AN obtained from AN by deleting its last column (see [1, Lemma 1]). If, conversely, rankAN = rankAN , then the system (5) has a solution (q0 , . . . , qN ) with qN = 0, and so, clearly, (1) will have a polynomial solution of degree N . Finally, (i) follows from the proof of Proposition 2 (iii), and (ii) follows from the fact that Cds is a linear combination of C0 , . . . , Cds −1 . 2 i−1
We next turn to the non-homogeneous case. Consider the equation p0 y + p1 y + · · · + pn−1 y (n−1) + pn y (n) = f
(6)
where the pi and f are continuous on some interval I in which pn and f do not vanish. Let pk gk = (0 ≤ k ≤ n). We then obtain the equivalent equation on I f g0 y + g1 y + · · · + gn−1 y (n−1) + gn y (n) = 1. The general solution of (6) is y = z + C1 y1 + · · · + Cn yn , where z is a solution of (6), the yi are linearly independent solutions of the associated homogeneous equation p0 y + p1 y + · · · + pn−1 y (n−1) + pn y (n) = 0
(7)
and the Ci are arbitrary constants. Consider the system g0 yi + g1 yi + · · · + gn−1 yi(n−1) + gn yi(n) = 0 (1 ≤ i ≤ n) g0 z + g1 z + · · · + gn−1 z(n−1) + gn z(n) = 1 in the unknowns gk (0 ≤ k ≤ n). If (6) has a general polynomial solution, then, clearly, each yi and z are polynomials, and since the yi and z are linearly independent (the yi form a fundamental set of solutions for (7) and clearly z cannot be a linear combination of the yi ), we can apply Cramer’s rule to infer that each gk is a rational function. We will therefore assume in the sequel that the pk and f in (6) are polynomials with no (non-constant) common factor. If (6) has a general polynomial solution, then so too does (7), and therefore, when pn is monic, the necessary conditions in Proposition 2 hold in the nonhomogeneous case as well. Combining the next result with Proposition 4 (which concerns homogeneous equations) provides a necessary and sufficient rank condition for (6) to have a general polynomial solution.
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Proposition 5. Equation (6) has a general polynomial solution if and only if the homogeneous (n + 1)st-order equation n fp0 − f p0 y + fpk−1 + fpk − f pk y (k) + fpn y (n+1) = 0
(8)
k=1
has a general polynomial solution. n p d p0 i (i) Proof. Observe first that (8) is the equation y+ y = 0. Suppose that (6) dx f i=1 f has general polynomial solution y = z + C1 y1 + · · · + Cn yn , where z is a solution of (6), the yi are linearly independent solutions of (7) and the Ci are arbitrary constants. It is then easy to verify that z, z + yi (1 ≤ i ≤ n) are (n + 1) linearly independent solutions of (8). This proves “only the if” part. For the converse, suppose (8) has a fundamental set of polynomial solutions wj 0≤j ≤n . n p n p d p0 p0 i (i) i (i) Then, for 0 ≤ j ≤ n, wj + wj = 0, so that wj + wj = Aj for some dx f f f f i=1 i=1 constant Aj . If Aj = 0 for each j , then the wj would be (n + 1) linearly independent solutions of the nth-order equation (7), which is impossible. We can therefore assume that at least one of w0 is a solution of (6). For 1 ≤ j ≤ n, let the Aj , A0 say, is not equal to 0 so that A0 wj /Aj − w0 /A0 if Aj = 0 yj = if Aj = 0. wj f2
Then it is easy to see that y = y0 + quired.
2
n j =1
Cj yj is a general polynomial solution of (6), as re-
Remark 6. If (7) has a general polynomial solution and f is a polynomial, then it does not necessarily follow that (6) has also a general polynomial solution: For x > 0, the equation y − xy = x has general solution y = Cx − x ln x. In [3], Calogero gives a family of homogeneous second-order linear equations containing an arbitrary number of parameters and having general polynomial solutions. We end this note with a systematic way of constructing equations with such property. Start with any nth-order homogedy d ny neous equation, say P0 (x) y + P1 (x) + · · · + Pn (x) n = 0, where the Pj are polynomials, dx dx and suppose it has a general polynomial solution y = C1 y1 (x) + · · · + Cn yn (x). Since the polynomials yk (x) are linearly independent, we can clearly assume that they have distinct degrees αk (1 ≤ k ≤ n). Let x = ϕ(t) = a0 + a1 t + · · · + aN t N be an arbitrary polynomial of degree dj y N ≥ 1. Then, using Faà di Bruno’s formula (see [8], for example), we can express each dx j diϕ diy (1 ≤ j ≤ n) in terms of the derivatives i and i (1 ≤ i ≤ j ) and obtain an equation of the dt dt dy d ny form Q0 (t) y + Q1 (t) + · · · + Qn (t) n = 0 containing the constants a0 , . . . , aN as paramdt dt eters. This equation clearly has polynomial solutions y1 (ϕ(t)) , . . . , yn (ϕ(t)), which are linearly
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independent since they have distinct degrees αk N (1 ≤ k ≤ n), and has therefore a general polynomial solution. dy d 2y As an illustration, let f (x) y + g (x) + h (x) 2 = 0 have general polynomial solution dx dx y = C1 y1 (x) + C2 y2 (x). Then, putting x = ϕ(t) where ϕ is a polynomial of degree ≥ 1, we 2 2 d y d 2 y d 2 ϕ dy dϕ dy dϕ dy . The differential equation then becomes = and 2 = 2 + have dt dt dx dt dt dx dt dx 2
dϕ f (ϕ(t)) dt
3
dϕ y + g (ϕ(t)) dt
2
d 2ϕ − h (ϕ(t)) 2 dt
dϕ d 2 y dy = 0. + h (ϕ(t)) dt dt dt 2
(9)
Clearly, (9) contains the coefficients of ϕ as parameters and has polynomial solutions y1 (ϕ(t)) and y2 (ϕ(t)). A specific example is given below. Example. The Cauchy–Euler equation 2y − 2xy + x 2 y = 0 has solution y = C1 x + C2 x 2 . Putting x = a0 + a1 t + · · · + aN t N , we obtain the equation F (t)y + G (t)
dy d 2y + H (t) 2 = 0 dt dt
where F (t) = 2
N
3 kak t k−1
k=1
G (t) = −2
N
ak t
k
k=0
H (t) =
N k=0
2 kak t
k−1
−
k=1
2 ak t
N
k
N
N
2 ak t
k
k=0
N
k (k − 1) ak t
k−2
k=2
kak t
k−1
k=1
and whose general solution is y = C1
N k k=0 ak t
+ C2
N k k=0 ak t
2
.
Acknowledgment The author gratefully acknowledges the support of King Fahd University of Petroleum and Minerals. References [1] H. Azad, A. Laradji, T. Mustafa, Polynomial solutions of certain differential equations arising in physics, Math. Methods Appl. Sci. 36 (2013) 1615–1624. [2] B. Bagchi, Y. Grandati, C. Quesne, Rational extensions of the trigonometric Darboux–Pöschl–Teller potential based on para-Jacobi polynomials, J. Math. Phys. 56 (2015) 062103. [3] F. Calogero, A linear second-order ODE with only polynomial solutions, J. Differential Equations 255 (2013) 2130–2135.
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[4] F. Calogero, A solvable many-body problem, its equilibria, and a second-order ordinary differential equation whose general solution is polynomial, J. Math. Phys. 54 (2013) 012703. [5] F. Calogero, Can the generalized hypergeometric equation feature several independent polynomial solutions?, J. Phys. A 47 (2014) 045205. [6] F. Calogero, G. Yi, Can the general solution of the second-order ODE characterizing Jacobi polynomials be polynomial?, J. Phys. A 45 (2012) 095206. [7] I. Gessel, G. Viennot, Binomial determinants, paths, and hook length formulae, Adv. Math. 58 (1985) 300–321. [8] S. Roman, The formula of Faà di Bruno, Amer. Math. Monthly 87 (1980) 805–809.