Differential Equations, Vol. 41, No. 8, 2005, pp. 1061–1074. Translated from Differentsial’nye Uravneniya, Vol. 41, No. 8, 2005, pp. 1011–1023. c 2005 by Basov, Skitovich. Original Russian Text Copyright
ORDINARY DIFFERENTIAL EQUATIONS
A Generalized Normal Form and Formal Equivalence of Two-Dimensional Systems with Quadratic Zero Approximation: II V. V. Basov and A. V. Skitovich St. Petersburg State University, St. Petersburg, Russia Received June 16, 2003
The present paper is a continuation of [1], and hence we preserve the notation used there and continue the numbering of sections, formulas, theorems, remarks, assertions, corollaries, and examples. 7. SYSTEMS WITH ZERO APPROXIMATION OF THE FORM (αx21 + x1 x2 , x1 x2 ) Consider system (1) whose unperturbed part P (x) is irregular and can be reduced by a nondegenerate linear change of variables to the canonical form (25) (α, 1, 0)(0, 1, 0) with α 6= 0. One can always assume that |α| ≤ 1. Indeed, if |α| > 1 in (25), then the change of variables (α−1, 0) (α−1 − 1, 1) reduces it to the system (α−1, 1, 0) (0, 1, 0). Therefore, we shall assume that system (1) is already represented in the form x˙ 1 = αx21 + x1 x2 + X1 (x1 , x2 ) ,
x˙ 2 = x1 x2 + X2 (x1 , x2 )
(|α| ≤ 1,
α 6= 0),
(70)
i.e., that a1 = α, 2b1 , 2b2 = 1, and c1 , a2 , c2 = 0. Formally, if we set α = 0 in (25), then we obtain the canonical form (17) considered in Section 3. As usual, let the change of variables xi = yi + hi (y) (i = 1, 2) given by (2) reduce (70) to the system (3) y˙1 = αy12 + y1 y2 + Y1 (y1 , y2 ) , y˙ 2 = y1 y2 + Y2 (y1 , y2 ) , P∞ P p+1 (p+1) (p+1) (s,p+1−s) s p+1−s where Yi = p=2 Yi (y1 , y2 ) and Yi = s=0 Yi y1 y2 (i = 1, 2). By differentiating relation (2) with regard to (70) and (3) and by matching the coefficients of y1s y2p+1−s (0 ≤ s ≤ p + 1, p ≥ 2), we obtain system (5). Let us take into account notation (6) : (s,p−s) (s,p+1−s) (s,p+1−s) hpis = hi (0 ≤ s ≤ p) and hpis = 0 (s < 0, s > p); Yˆisp+1 = Y˜i − Yi (0 ≤ s ≤ p + 1), (s,p+1−s) p+1 ˆ ˜ Yis = 0 (s < 0 or s > p + 1), and the Yi are known. With regard to this notation, system (5) has the form (p − s + 1 + α(s − 3))hp1s−1 + (s − 1)hp1s − hp2s−1 = Yˆ1sp+1 , (p − s + α(s − 1))hp2s−1 + shp2s − hp1s = Yˆ2sp+1
(s = 0, . . . , p + 1; p ≥ 2).
(71)
Consider the equations with s = 0 in system (71). These are the equation p+1 −hp10 = Yˆ10 ,
(72)
p+1 which readily gives hp10 , and the equation −hp10 = Yˆ20 , which, together with (72), implies the first (p+1) relation for the coefficients of the polynomial Y : p+1 p+1 Yˆ10 − Yˆ20 =0
(p ≥ 2).
c 2005 Pleiades Publishing, Inc. 0012-2661/05/4108-1061
(73)
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p+1 and rewrite system (71) We introduce the vectors hpi = hpi0 , . . . , hpip and Yˆip = Yˆi1p+1 , . . . , Yˆip+1 p p p p ˆ without the two equations with s = 0 in the matrix form Υ h1 − h2 = Y1 , −Epp hp1 + ∆p hp2 = Yˆ2p , p p where Υp and ∆p are the bidiagonal matrices with entries υss = p − s + 1 + α(s − 3), υss+1 = s − 1, p p p δss = p − s + α(s − 1), and δss+1 = s and the matrix Ep has one nonzero diagonal, ess+1 = 1. The subscripts of all entries of the matrices vary from 1 to p + 1. By substituting hp2 from the first equation into the second equation, we obtain the system Θp hp1 = Y˘1p
(p ≥ 2),
(74)
where Θp = ∆p Υp − Epp is an upper tridiagonal square matrix and the vector Y˘1p is given by Y˘1p = ∆p Yˆ1p + Yˆ2p . Their entries have the form p cps = θss = (p − s + α(s − 1))(p − s + 1 + α(s − 3))
bps = aps =
θsp s+2 θsp s+1
= s2
(s = 1, . . . , p − 1),
ap1
=
p θ12
= (p − s)(2s − 1) − 1 + α 2s2 − 4s + 1
p+1 Y˘1sp = (p − s + α(s − 1))Yˆ1sp+1 + sYˆ1s+1 + Yˆ2sp+1
(s = 1, . . . , p + 1), (75)
= p − 2 − α, (s = 2, . . . , p),
(76) (s = 1, . . . , p + 1).
If det Θp = cp1 . . . cpp+1 6= 0, then system (74) is uniquely solvable for arbitrary right-hand sides and the presence of zero entries cps depends primarily on the value of α. In this connection, one should separately consider the case in which p = 2, since c23 = 0 for arbitrary α. For p = 2, the entries of the matrix Θp in system (74) acquire the form c21 = 2(1 − α), 2 c22 = α(1 − α), c23 = 0; a21 = −α, a22 = α − 1; b21 = 1. Therefore, the condition Y˘13 = 0, together with (76), implies the second relation 3 (2α − 1)Yˆ133 + Yˆ23 = 0,
(77)
and the component h212 in the last equation in system (74) is not constrained. By solving system (74), we obtain 2 α(1 − α)h211 = Y˘12 + (1 − α)h212 ,
2 2(1 − α)2 h210 = (1 − α)Y˘112 + Y˘12 ;
i.e., h210 is independent of h212 . By substituting h210 from (72) and Y˘1s2 from (76) into the last relation, we obtain the third relation for the coefficients of the homogeneous polynomial Y (3) : 3 3 3 2(1 − α)2 Yˆ10 + (1 − α)Yˆ113 + Yˆ12 + 2Yˆ133 + (1 − α)Yˆ213 + Yˆ22 = 0.
(78)
As a result, if p = 2, then the coefficients of the homogeneous polynomials Yi(3) should satisfy three relations (73), (77), and (78) for system (71) to be compatible. The component h212 remains arbitrary. Throughout the following, we assume that p ≥ 3. Let us find conditions under which Θp in (74) has zero eigenvalues cps , which, in view of (75), is equivalent to the existence of integer solutions s of the equation cps = (p − s + α(s − 1))(p − s + 1 + α(s − 3)) = 0
(1 ≤ s ≤ p + 1,
p ≥ 3).
(79)
Let α be a rational number; i.e., α = ±k/m, where k and m are coprime positive integers and k ≤ m, since 0 < |α| ≤ 1 in system (1). There exist integer solutions s of Eq. (79) in the specified range in the following three cases. 1. If α = −k/m, then p = lk + lm + 2 and s = lm + 3 (l = 1, 2, . . .). 2. If α = −k/m, then p = lk + lm + 1 and s = lm + 1 (l = 1, 2, . . .). 3. If α = 1/m (k = 1, m ≥ 3), then p = m and s = m + 1. DIFFERENTIAL EQUATIONS
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It is only in these cases that det Θp = 0 and one should impose additional conditions on the right-hand side to ensure the solvability of system (74). We split the set of pairs (α, p), where 0 < |α| ≤ 1 and p ≥ 3 is a positive integer, into four disjoint families {α, p}ν (ν = 0, 1, 2, 3) and introduce the corresponding constants sν as follows: {α, p}1 = {−k/m, lk + lm + 2}∞ k,l,m=1 and s1 = lm + 3; {α, p}2 = {−k/m, lk + lm + 1}∞ k,l,m=1 and s2 = lm + 1; ∞ {α, p}3 = {1/m, m}m=3 and s3 = m + 1 (k and m are coprime, and k ≤ m); n o S3 {α, p}0 = (α, p) 6∈ ν=1 {α, p}ν , α ∈ (−1, 0) ∪ (0, 1], p ≥ 3 and s0 = 0. We have α 6= −1 in the set {α, p}0 , since {−1, 2l + 2} ∈ {α, p}1 and {−1, 2l + 1} ∈ {α, p}2 . Since for ν = 1, 2, 3 and for each pair (α, p) ∈ {α, p}ν , the constant sν corresponds to the unique number s of the equation in system (74) such that cs = 0, it is natural to set s0 = 0 if Eq. (79) has no solutions, i.e., if (α, p) ∈ {α, p}0 . S3 If (α, p) ∈ ν=1 {α, p}ν , then the sν th equation in system (74) has the form 0 × hp1 sν −1 + apsν hp1 sν + bpsν hp1 sν +1 = Y˘1psν
(1 ≤ ν ≤ 6).
(80)
If sν = p + 1, which can happen only for ν = 3, then Eq. (80) is the last equation in system (74). Consequently, if (α, p) ∈ {α, p}3 , then Eq. (80) has the form 0 × h1m = Y˘1mm+1 , since hp1s = 0 for s > p. By using (76), from this equation, one can readily obtain the second resonance relation for the right-hand side of system (71) in case 3 : Yˆ2m+1 m+1 = 0
(m ≥ 3, (α, p) ∈ {α, p}3 ) .
(81)
Now let 0 ≤ sν ≤ p and hence ν = 0, 1, 2. We set s∗ν = p+1−sν (1 ≤ s∗ν ≤ p+1), s+ ν = sν +1, and p = sν − 1. We introduce the s∗ν × s∗ν matrix Θp+ obtained from the matrix Θ by deleting the first ∗ sν sν rows and columns. In view of (75), we have the upper-triangular tridiagonal matrix with main diagonal {cps+ , . . . , cpp+1 } and with first and second superdiagonals {aps+ , . . . , app } and {bps+ , . . . , bpp−1 }, ν ν ν p respectively (0 ≤ ν ≤ 2); moreover, det Θp+ 6= 0 and Θp+ s∗ p+1 = Θ , which is possible only for ν = 0. ν The last s∗ν equations in system (74) form the linear system s− ν
˘ p+ Θp+ hp+ s∗ 1 = Y1 ν
(p ≥ 3,
(74+ )
ν = 0, 1, 2),
p p ˘ p+ = Y˘ p + , . . . , Y˘1pp+1 . where hp+ 1 = h1sν , . . . , h1p and Y1 1s
p p+1 To solve system (74 ), we introduce the auxiliary upper triangular matrix Gp = gjs s,j=s+ ν whose entries are defined as follows: p p ∀s = s+ gsj =0 s+ gsp+ ,s = 0, gss = 1, ν ,...,p + 1 ν ≤ j < s , ν ν (82) p p p p p p gsj = − gs j−1 aj−1 + gs j−2 bj−2 /cj (s < j ≤ p + 1). ν
+
Since Θp+ in (74+ ) is a tridiagonal matrix, we have s∗ ν
o n p p p p p p p+1 p p Gp Θp+ = g b + g a + g c = diag c , . . . , c ∗ sν p+1 . s j−2 j−2 s j−1 j−1 s j j s,j=s+ s+ ν
ν
+
We multiply the left- and right-hand sides of system (74 ) by the matrix Gp on the left; then Pp+1 p ˘ p ˘p ˆ p+1 by (76), cps hp1 s−1 = j=s gsj Y1j (s = s+ ν , . . . , p + 1). By expressing the components Y1j via Yij we obtain p+1 X p p ˆ p+1 cps hp1 s−1 = (p − j + α(j − 1))gsj , (83) Y2j + (j − 1)gsp j−1 Yˆ1jp+1 + gsj j=s p + where s = s+ ν , . . . , p + 1 (1 ≤ sν ≤ p + 1), cs 6= 0, and the gsj are given by (82).
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Formula (83) permits readily writing out the second relation [in addition to (73)] on the coefficients of the homogeneous polynomial Y (p+1) for the case in which (α, p) ∈ {α, p}0 and hence ν = 0, sν = 0, s∗ν = 1, and the uniquely solvable system (74+ ) coincides with system (74). We set s = 1 in (83) and substitute hp10 from (72) into the left-hand side; then the abovementioned relation acquires the form p+1 u00 Yˆ10
p+1 X u0j Yˆ1jp+1 + vj0 Yˆ2jp+1 = 0 +
for
(α, p) ∈ {α, p}0 ,
(84)
j=1
u00
=
cp1 ,
u0j
= (p − j + α(j −
p 1))g1j
+ (j −
1)g1p j−1 ,
vj0
=
p g1j ,
p where the g1j are the entries given by (82) with cps , aps , and bps taken from (75). Throughout the following, we assume that ν = 1, 2, i.e., Eq. (80) is present. Using the entries of the matrix Θp from (75), we introduce the constants
a∗j−1 = apj−1 /cpj ,
b∗j−2 = bpj−2 /cpj
(sν + 1 ≤ j ≤ p + 1).
(85)
We substitute the entries hp1 sν and hp1 sν +1 found in (83) into the left-hand side of Eq. (80) and use the relations hp1s = 0 for s > p. Then on the right-hand side in Eq. (80), we replace Y˘1psν by its expression in (76); we obtain a∗sν
p+1 X
(p − j + α(j − 1))gspν +1 j + (j − 1)gspν +1 j−1 Yˆ1jp+1 + gspν +1 j Yˆ2jp+1
j=sν +1
+
b∗sν
p+1 X
(p − j + α(j − 1))gspν +2 j + (j − 1)gspν +2 j−1 Yˆ1jp+1 + gspν +2 j Yˆ2jp+1
j=sν +2
ˆ p+1 ˆ p+1 = (p − sν + α (sν − 1)) Yˆ1p+1 sν + sν Y1 sν +1 + Y2 sν . As a result, for cases 1 and 2, we also obtain the second relation p+1 X uνj Yˆ1jp+1 + vjν Yˆ2jp+1 = 0 j=sν uνsν
for (α, p) ∈ {α, p}ν ,
= − (p − sν + α (sν − 1)) ,
(86ν )
vsνν = −1;
uνsν +1 = (p − sν − 1 + αsν ) a∗sν − sν , uνj = (p − j + α(j − 1)) gspν +1 j a∗sν vjν = gspν +1 j a∗sν + gspν +2 j b∗sν
ν = 1, 2;
vsνν +1 = a∗sν ; + gspν +2 j b∗sν + (j − 1) gspν +1 j−1 a∗sν + gspν +2 j−1 b∗sν ,
(j = sν + 2, . . . , p + 1) ,
p where the gsj are defined in (82) and a∗j and b∗j are the constants given by (85). Now for ν = 1, 2, 3, one should consider the first sν − 1 equations in system (74) (s1 , s3 ≥ 4 and s2 ≥ 2), since their possible inconsistency would imply the presence of additional relations on the right-hand sides of system (74). p In cases 1–3, we introduce the matrix Θp− of size s− ν = sν − 1 obtained from Θ by deleting s− ν − the first column, the last p − s− ν columns, and the last p + 1 − sν rows. By (75), this matrix is a p p tridiagonal matrix with main diagonal {a1 , . . . , as− } and with sub- and superdiagonals {cp2 , . . . , cps− } ν ν and {bp1 , . . . , bps− −1 }, respectively (1 ≤ ν ≤ 3), where 1 ≤ s− ≤ p and cp2 , . . . , cps− 6= 0 by the ν ν ν definition of sν . p Consider the first s− ν equations in system (74). The component h10 occurs only in the first of p them, and h1 sν occurs only in the last equation. We transpose these components together with
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the corresponding factors to the right-hand sides; then the above-mentioned s− ν equations form the linear system Θp− hp− = Y˘1p− (p ≥ 3, ν = 1, 2, 3), (74− ) s− 1 ν
p p ˘ p− has the components where hp− 1 = (h11 , . . . , h1 s− ) and the vector Y1 ν
p− p Y˘11 = Y˘11 − cp1 hp10 ,
Y˘1sp− = Y˘1sp
2 ≤ s < s− ν ,
Y˘1p− = Y˘1ps− − bps− hp1 sν . s− ν
ν
ν
(87)
Moreover, the last component hp1 sν −1 of the vector hp− does not occur in any equations of sys1 tem (74) other than those occurring in system (74− ), since it has the zero coefficient in Eq. (80). The component hp10 is taken from (72). As a result, if det Θp− 6= 0, then system (74− ), as well as system (74+ ), is uniquely solvable; s− ν therefore, the only solvability condition for system (74) is given by relation (80) equivalent to (86ν ) with ν = 1, 2 and (81) with ν = 3. If, on the contrary, det Θp− = 0 for some ν, then there exists at s− ν least one more relation for the right-hand sides of system (71). To compute det Θp− and simplify system (74− ), we reduce it by the Gauss method to the system s− ν
(74− d )
p Θpd hp− 1 = Yd , p p − where Θpd is a bidiagonal s− ν × sν matrix with main diagonal d1 , . . . , ds− ν bp1 , . . . , bps− −1 . Then
and superdiagonal
ν
dp1 = ap1 , p p− Yd1 = Y˘11 ,
dps = aps − bps−1 cps /dps−1
s = 2, . . . , s− ν ,
p Yds = Y˘1sp− − Ydps−1 cps /dps−1
dp1 , . . . , dps− −1 6= 0, ν − s = 2, . . . , s− , s ≤ p , ν ν if
(88) (89)
where aps , bps , and cps are defined in (75) and Y˘1sp− is defined in (87). Lemma 1. Let a number p∗ (1 ≤ p∗ ≤ p) and a pair (α, p) satisfy the condition (p − s − 1 − α)(p − s + α(s − 2)) 6= 0
(s = 1, . . . , p∗ − 1) .
(90)
Then the numbers dp1 , . . . , dpp−1 defined by the recursion formula dp1 = p − 2 − α,
dps = aps − bps−1 cps /dps−1
aps = (p − s)(2s − 1) − 1 + α 2s2 − 4s + 1 , cps = (p − s + α(s − 1))(p − s + 1 + α(s − 3))
(s = 2, . . . , p∗ ) , bps−1 = (s − 1)2 , (s = 2, . . . , p∗ )
(88∗ )
are nonzero, and one has the closed-form expression dps = s(p − s − 1 − α)(p − s + α(s − 2))/(p − s − α)
(s = 1, . . . , p∗ ) .
(91)
Proof. If s = 1, then dp1 = p − α − 2 6= 0 in (91) by condition (90), which provides a basis of induction. Suppose that dps−1 = (s − 1)(p − s − α)(p − s + 1 + α(s − 3))/(p − s + 1 − α) DIFFERENTIAL EQUATIONS
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for 2 ≤ s ≤ p. Then dps−1 6= 0 by (90). Let us prove (91) : aps − bps−1 cps /dps−1
= (p − s)(2s − 1) − 1 + α 2s2 − 4s + 1 − (s − 1)(p − s + α(s − 1)) × (p − s + 1 − α)/(p − s − α) = (p − s − α) (p − s)(2s − 1) + α 2s2 − 4s + 1 − (s − 1)(p − s − 1 + α(s − 1)) . − (p − s)(s − 1) + α(s − 1)2 (p − s − α) = ((p − s − α)((p − s)s − 1 + αs(s − 2)) − ((p − s)s + αs(s − 2) − (p − s − α)))/(p − s − α),
and this expression exactly coincides with dps given by (91). The proof of the lemma is complete. We set p∗ = sν − for ν = 1, 2, 3; then 1 ≤ p∗ ≤ p, the first factor in (90) is always positive, and the second factor is nonzero since cp2 , . . . , cps− 6= 0 in the matrix Θp− of system (74− ). Consequently, s− ν ν by Lemma 1, formula (91) holds for dp1 , . . . , dps− given by (88) with ν = 1, 2, 3. ν
In case 2, we have (α, p, s2 ) = (−k/m, lk + lm + 1, lm + 1) and k, l, m ≥ 1. Therefore, s− 2 = lm p and ds = s(lk + lm − s + k/m)(lk + lm + 1 − s − (s − 2)k/m)/(lk + lm + 1 − s + k/m) > 0 for 1 ≤ s ≤ lm. p In case 3, we have (α, p, s3 ) = (m−1 , m, m + 1), m ≥ 3. Consequently, s− 3 = m and ds = s(m − s − 1 − 1/m)(m − s − (s − 2)/m)/(m − s − 1/m) 6= 0 for 1 ≤ s ≤ m. Therefore, the matrices Θps− and Θps− in system (74− ) have nonzero eigenvalues, and just as for 2 3 ν = 0, to provide the solvability of system (74), it suffices to satisfy one of relations (86ν ) for ν = 2 or (81) for ν = 3. In case 1, we have (α, p, s1 ) = (−k/m, lk +lm+2, lm+3) and k, l, m ≥ 1. Therefore, s− 1 = lm+2 and dps = s(lk + lm + 1 − s + k/m)(lk + lm + 2 − s − (s − 2)k/m)/(lk + lm + 2 − s + k/m) > 0 for 1 ≤ s ≤ lm + 1; dplm+2 = 0 for s = lm + 2. Consider system (74− d ) with ν = 1. Its last equation has the form 0 × hp1 s− = Ydps− p = lk + lm + 2, s− (92) 1 = lm + 2 1
1
and defines the second relation, which, together with relation (861 ), provides the consistency of system (74) and leaves the component hp1 s− arbitrary. 1
We express
Ydps− 1
via Yˆisp+1 . One can readily see that p Yds
s s X Y s−j ˘ p− = (−1) Y1j
cpµ dp µ=j+1 µ−1
j=1
s = 1, . . . , s− 1
(93)
in (89) for ν = 1. By using cps in (75) and dps in (91), in case 1, we introduce the constants −
s− 1 −j
βj1 = (−1)
−
Therefore, βj1 = (−1)s1 −j
s1 Y (p − µ + α(µ − 1))(p − µ + 1 − α) (µ − 1)(p − µ − α) µ=j+1
Qs−1 µ=j+1
j = 1, . . . , s− 1 −1 .
(94)
cpµ /dpµ−1 6= 0 (βs1− = 1, β01 = 0). 1
˘ p− We substitute the expansion (93) with s = s− 1 ≥ 3 into Eq. (92), replace Y1j by the component Y˘1jp according to (87) with ν = 1, and use the constants (94); then we obtain −
−cp1 β11 hp10 +
s1 X
βj1 Y˘1jp − b1s− hp1 s1 = 0. 1
j=1
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In the first term in this equation, we replace hp10 and cp1 in accordance with (72) and (75), respectively, in the second term, we expand Y˘1jp in accordance with (76), and in the third term, we replace hp1 s1 in accordance with (83) with s = s+ 1 = s1 + 1 ≤ p + 1 and with the use of notation (85) and (82). Then we obtain the equation (p − 1)(p −
p+1 2α)β11 Yˆ10
+
sX 1 −1
p+1 1 (p − j + α(j − 1))βj1 + (j − 1)βj−1 Yˆ1j + βj1 Yˆ2jp+1
j=1 ∗ + (s1 − 1) Yˆ1p+1 s1 − bs− 1
p+1 (95) X (p − j + α(j − 1))gsp+ j + (j − 1)gsp+ j−1 Yˆ1jp+1 + gsp+ j Yˆ2jp+1 1
j=s1 +1
1
1
for (α, p) ∈ {α, p}4 ,
=0
which gives the third relation for the right-hand sides of system (71) in case 1, where (α, p, s1 ) = (−k/m, lk + lm + 2, lm + 3). p+1 ˆ p+1 Furthermore, there are a fortiori no constraints on the components Yˆ20 , Y2 s1 , and hp1 s− occurring 1 in (71). Let us now state our results in terms of resonance equations. By replacing Yˆisp+1 in all above(s,p+1−s) (s,p+1−s) obtained relations by the difference Y˜i − Yi , for each p ≥ 2, one obtains a set of resonance equations of the form (7): p+1 X
(s,p+1−s)
apµs Y1
(s,p+1−s)
+ bpµs Y2
= c˜
(µ = 1, . . . , np ),
s=0
Pp+1 (s,p+1−s) (s,p+1−s) where c˜ = s=0 apµs Y˜1 and np = nαp is the number of resonance equations + bpµs Y˜2 for given α and p. Let us describe the factors apµs and bpµs (which are denoted by uνj and vjν in this section) and the function nαp for all p ≥ 2 and α 6= 0, |α| ≤ 1. For p = 2 and for each α, from (73), (77), and (78), we obtain the three equations (0,3)
Y1 2(1 − α)
2
(0,3) Y1
+ (1 −
(1,2) α)Y1
(0,3)
− Y2
+
(2,1) Y1
(3,0)
(2α − 1)Y1
= c˜,
+
(3,0) 2Y1
+ (1 −
(1,2) α)Y2
(3,0)
= c˜,
(2,1) Y2
= c˜.
− Y2 +
(96)
If p ≥ 3, then the number of resonance equations is equal to 2 or 3 depending on α. The first of them, obtained from (73), is universal: Y1(0,p+1) − Y2(0,p+1) = c˜
(p ≥ 3).
(97)
The second resonance equation also holds for each p ≥ 3, but its form depends on the value of the parameter α, since the set of pairs (α, p) has been divided into four disjoint sets {α, p}ν (ν = 0, 1, 2, 3), each having its own relation, namely, (84) for ν = 0, (86ν ) for ν = 1, 2, and (81) for ν = 3. By combining these relations, we obtain the second resonance equation p+1 X (j,p+1−j) (j,p+1−j) uνj Y1 = c˜ for (α, p) ∈ {α, p}ν + vjν Y2 j=0
where uνj = 0 and vjν = 0 (0 ≤ j < sν ) for ν = 1, 2, 3; DIFFERENTIAL EQUATIONS
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(ν = 0, 1, 2, 3),
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BASOV, SKITOVICH
p u00 = (p − 1)(p − 2α), v00 = 0, u01 = p − 1, v10 = 1, u0j = (p − j + α(j − 1))g1j + (j − 1)g1p j−1 , and p = g1j (2 ≤ j ≤ p + 1) for ν = 0 (s0 = 0); uνsν = − (p − sν + α (sν − 1)), vsνν = −1, uνsν +1 = (p − sν − 1 + αsν ) a∗sν − sν , vsνν +1 = a∗sν , uνj = (p − j + α(j − 1)) gspν +1 j a∗sν + gspν +2 j b∗sν + (j − 1) gspν +1 j−1 a∗sν + gspν +2 j−1 b∗sν , and vjν = gspν +1 j a∗sν + gspν +2 j b∗sν (sν + 2 ≤ j ≤ p + 1) for ν = 1, 2; 3 u3p+1 = 0 and vp+1 = 1 for ν = 3 (s3 = p + 1). p Here the elements gsj are defined in (82), and a∗j and b∗j are the constants given by (85). The third resonance equation, which follows from (95), appears only in case 1, where
vj0
(α, p, s1 ) = (−k/m, lk + lm + 2, lm + 3), and has the form p+1 X (j,p+1−j) ∗1 ˆ (j,p+1−j) ˆ u∗1 = c˜ for Y Y + v 1 2 j j
(α, p) ∈ {α, p}1 ,
(99)
j=0 1 ∗1 ∗1 1 1 ∗1 1 where u∗1 0 = (p − 1)(p − 2α)β1 , v0 = 0; uj = (p − j + α(j − 1))βj + (j − 1)βj−1 , and vj = βj (1 ≤ j < s1 ); p p + ∗1 ∗1 ∗1 ∗1 u∗1 s1 = s1 −1, vs1 = 0, us+ = p−s1 +αs1 < 0, vs+ = 1; uj = (p−j +α(j −1))gs+ j +(j −1)gs+ j−1 , 1
and vj∗1 = gsp+ j (s+ 1 < j ≤ p + 1).
1
1
1
1
p Here the constants βj1 are defined in (94), and the elements gsj are given by (82). Moreover, (s− ,p−s− )
1 by (92), there are no constraints for the coefficients h1 1 . α To state the results, we introduce two families, pl = {2 if α 6= −k/m; lk + lm + 2 if α = −k/m (l ≥ 0, 1 ≤ k ≤ m)} and {α}0 = [−1, 1]\{0, − k/m (1 ≤ k ≤ m), 1/m (m ≥ 3)}. Then pα0 = 2, and {α}0 contains the values of α not included in cases 1–3.
Theorem 10. 1. System (70) is formally equivalent to system (3) with zero approximation (p+1) P = (αy12 + y1 y2 , y1 y2 ) if, for each p ≥ 2, the coefficients of the homogeneous polynomials Yi in system (3) satisfy the np = {2 if p 6∈ pαl ; 3 if p ∈ pαl } resonance equations listed below. For p = 2, these are the three equations (96). If p ≥ 3, then the first equation is necessarily Eq. (97); further, depending on the parameter α (α 6= 0, |α| ≤ 1), one has the following equations : if α = −k/m, then one has Eq. (982 ) for p = lk + lm + 1, Eqs. (981 ) and (99) for p = lk + lm + 2, and Eq. (980 ) (1 ≤ k ≤ m, l ≥ 1) for p 6= lk + lm + 1, lk + lm + 2; if α = 1/m, then one has Eq. (983 ) for p = m and Eq. (980 ) (m ≥ 3) for p 6= m; if α ∈ {α}0 , then one has Eq. (980 ) for all p ≥ 3. (p+1)
2. Those coefficients of the homogeneous polynomials Yi which do not occur in the resonance equations (97), (98ν ), and (99) or are multiplied only by zero factors in these equations are nonresonance coefficients and can take arbitrary values. 3. In the change of variables (2) relating (70) and (3), the resonance coefficients h(lm+2,lk) 1 (l, k, m ≥ 1) have no constraints only for (α, p) ∈ {α, p}1 . Remark 3. We might not write out the resonance equations (96) with p = 2 separately, since this case is essentially contained in case 1; one should only extend the set {α, p}1 by adding the pairs (α, p) with l = 0. Then p = 2 and s1 = 3 for l = 0 and for arbitrary k and m. Therefore, it is natural that α has no constraints. In other words, setting {α, p}+ 1 = {α, p}1 ∪ {(α, 2), α ∈ [−1, 0) ∪ (0, 1]}, + + in (981 ) and (99), we assume that (α, p) ∈ {α, p}1 and denote these equations by (98+ 1 ) and (100 ). + + Then Eq. (981 ) with l = 0 coincides with the second equation in (96), and Eq. (100 ) coincides with the third equation. A coefficient in system (3) is referred to as a resonance coefficient if it occurs in at least one resonance equation with a nonzero factor. Therefore, the problem is to find out which factors uνj ∗1 and vjν in Eqs. (98ν ) and u∗1 j and vj in (99) are nonzero and which of them vanish. DIFFERENTIAL EQUATIONS
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Assertion 2. If j = 0, . . . , s− ν , then all factors multiplying Yi in (99) except for v0∗1 = 0 are nonzero.
(j,p+1−j)
1069
in (98ν ) are zero and those
1 Indeed, β11 , . . . , βs1− 6= 0 in (94). Therefore, in (99), we have u∗1 0 = (p − 1)(p − 2α)β1 6= 0 (since 1 1 α < 0) and u∗1 1 = (p − 1)β1 6= 0. Since, by (75) and (91), 1 βj−1 = −βj1 cpj /dpj−1 = −βj1 (p − j + α(j − 1))(p − j + 1 − α)/((j − 1)(p − j − α)) 1 in (94) for j = 2, . . . , s1 − 1, we have u∗1 j = βj (p − j + α(j − 1))/(p − j − α) 6= 0 (1 < j < s1 ) in (99).
Assertion 3. If j = sν , then the factors uνsν and vsνν in Eqs. (98ν ) with ν = 0, 1, 2, 3 and in Eq. (99) have the values u00 = (p − 1)(p − 2α) 6= 0 (p ≥ 3, α ≤ 1), 1 1 ulm+3 = 1 + 2k/m, vlm+3 = −1; u2lm+1 = 0,
2 vlm+1 = −1;
u3m+1 = 0,
3 vm+1 = −1;
u∗1 lm+3 = lm + 2,
∗1 vlm+3 =0
v00 = 0;
(k, l, m ≥ 1).
p Now let j = s+ ν , . . . , p + 1. Let us estimate the gsj given by (82). To this end, by analogy with (88), we introduce the recurrent set of numbers fjp . We set
fsp+ = −aps+ , ν
fsp = −aps −
ν
bps−1 cps p fs−1
if
By induction, one can show that
s = s+ ν + 1, . . . , p .
s+ ν +1 ≤ j ≤ p+1
p gsp+ j = gsp+ j−1 fj−1 /cpj ν
p fs−1 6= 0
ν
(100)
(101)
for gsp+ j in (82). ν
Consider the set of numbers fjp first for ν = 0, where we have (α, p) ∈ {α, p}0 . In this case, + sν = 1, and formula (100) differs from (88∗ ) only by the minus sign of apj . By using Lemma 1, one can readily show that fsp = −dps = −s(p − s − 1 − α)(p − s + α(s − 2))/(p − s − α)
(s = 1, . . . , p∗ ) ,
(102)
where p∗ (1 ≤ p∗ ≤ p) is chosen so as to satisfy condition (90). The second factor p − s + α(s − 2) in (90) is a factor in cps+1 in (75). But if ν = 0, then p c1 , . . . , cpp+1 6= 0, since these are the entries on the leading diagonal in the upper-triangular matrix p + Θp+ = Θp+ p+1 = Θ with nonzero determinant in system (74 ). Therefore, p − s + α(s − 2) 6= 0 s∗ 0 for s = 1, . . . , p. The first factor p − s − 1 − α in (90) can vanish only for α = 1 and s = p − 2. One should use the fact that (−1, p) 6∈ {α, p}0 . Therefore, if (α, p) ∈ {α, p}0 and α 6= 1, then condition (90) is valid for p∗ = p; moreover, fsp 6= 0 for s = 1, . . . , p in (102). If α = 1, then (1, p) ∈ {α, p}0 and p∗ = p − 2 for each p ≥ 3; i.e., by (102), p fsp 6= 0 for s = 1, . . . , p − 3, and fp−2 = 0. We substitute the expression for fsp given by (102) into (101) with ν = 0; then, by (75) and p p (82), we obtain g11 = 1 and g1j = −(j − 1)(p − j − α)(p − j + 1 − α)−1 (p − j + α(j − 1))−1 g1p j−1 6= 0 ∗ (2 ≤ j ≤ p + 1). p Consequently, in (980 ) with α 6= 1, we have vj0 = g1j 6= 0 and u0j = j − 1 − (j − 1)(p − j − α)(p − j + 1 − α)−1 g1p j−1 = (j − 1)(p − j + 1 − α)−1 g1p j−1 6= 0 DIFFERENTIAL EQUATIONS
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(2 ≤ j ≤ p + 1).
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vj0
BASOV, SKITOVICH
By arguing in a similar way, in (980 ) with α = 1 for p∗ = p − 2, we obtain u0j 6= 0 (2 ≤ j ≤ p − 1), 0 6= 0 (2 ≤ j ≤ p − 2), and g1p p−1 = vp−1 = 0. Further, vp0 = g1p p = −g1p p−2 bpp−2 /cpp 6= 0,
0 vp+1 = g1p p+1 = −g1p p app /cpp+1 6= 0
by (82). Then u0p = (p − 1)g1p p 6= 0 and u0p+1 = g1p p pcpp+1 − (p − 1)app /cpp+1 6= 0. By computing the first factors uνj and vjν with j ≥ s+ ν in (98ν ) with ν = 1, 2, we find that they can vanish; therefore, it is quite difficult to obtain some general results for them. The situation is clear for ν = 3 in (98ν ). To find the factors occurring in the third resonance equation (99), we estimate fsp in (100) in case 1 for ν = 1, sν = lm + 3, α = −k/m, and p = lk + lm + 2 ≥ 3. By induction, we show that fsp > ζsp = s(s + 1 − p − α(s − 2)) > 0 s = s+ (103) 1 ,...,p . If s = s+ 1 = lm + 4, then by (75), we have the inductive hypothesis fsp+ = −aps+ = 5lk + 4lm + 14 + 17k/m > ζsp+ = 2lk + 3lm + 12 + 8k/m > 0, 1
1
1
and since ζsp is an increasing function of s, we find that it is always positive. p p Suppose that fj−1 > ζj−1 for s+ 1 + 1 ≤ j ≤ p; then p (−aps − ζsp ) ζs−1 = (s − 1)(p − s + 1 + α(s − 1))(s − 1)(p − s + α(s − 3)) = (s − 1)2 ((p − s + α(s − 1))(p − s + 1 + α(s − 3)) − 2α) = bps−1 cps − 2α(s − 1)2 > bps−1 cps , p since α < 0. By using (100), the positivity of bps−1 , cps , and ζs−1 , the inductive hypothesis, and the p p p p p p last inequality, we obtain (103), since fs > −as −bs−1 cs /ζs−1 > ζs (s = s+ 1 , . . . , p). Moreover, cs > 0, p since cs+ = (2 + 4k/m)(1 + k/m) > 0 is also an increasing function of s in the above-mentioned 1 range. We use inequality (103) in (101) with ν = 1; then we obtain p gsp+ j > gsp+ j−1 ζj−1 /cpj > gsp+ j−1 (1 − j)/(p − j + α(j − 1)) 1
1
(104)
1
for j = s+ 1 + 1, . . . , p + 1. It follows from (104) that all gsp+ j > 0 since gsp+ s+ = 1; therefore, vj∗1 > 0 for j, s+ 1 < j ≤ p + 1, 1 1 1 p ∗1 in (99). By substituting gs+ j given by (104) into the formula for uj in (99), we obtain u∗1 j < 0 for 1 the same values of j. We have thereby proved the following assertion. + 0 0 0 Assertion 4. If j = s+ 0 , . . . , p + 1 (s0 = 1), then in (980 ), one has uj , vj 6= 0 but vp−1 = 0 for + ∗1 ∗1 α = 1. If j = s+ 1 , . . . , p + 1 (s1 = lm + 4), then all uj , vj 6= 0 in (99).
Theorem 10, together with Assertions 2–4, implies the following assertion. (s,p+1−s) Corollary 9. If (α, p) ∈ {α, p}+ are resonance coefficients, i.e., each of 1 , then all Yi (1,2) them occurs in at least one of Eqs. (97), (981 ), and (99) with nonzero factor except for Yi with (s,p+1−s) (s,p+1−s) α = 1 in (96). If (α, p) ∈ {α, p}2 , then Y1 , 1 ≤ s ≤ lm + 1, and Y2 (1 ≤ s ≤ lm) (0,p+1) are nonresonance coefficients. If (α, p) ∈ {α, p}3 , then all coefficients except for Y1 , Y2(0,p+1) , (p+1,0) (s,p+1−s) (p−1,2) and Y2 are nonresonance ones. If (α, p) ∈ {α, p}0 , then all Yi except for Y2 with α = 1 are resonance coefficients.
DIFFERENTIAL EQUATIONS
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Corollary 10. 1. If (α, p) 6∈ {α, p}1 , then each resonance set consists of two coefficients of (p+1) (p+1) (0,p+1) (0,p+1) the polynomials Y1 and Y2 , one of which is either Y1 or Y2 , and the other is any resonance coefficient in (98ν ) (ν 6= 1) other than the first one. If (α, p) ∈ {α, p}0 , then the (0,p+1) (0,p+1) resonance set can be formed by the pair Y1 , Y2 . 2. If (α, p) ∈ {α, p}1 , then each resonance set consists of three coefficients, the first of which (0,p+1) (0,p+1) is, just as above, either Y1 or Y2 , the second is any coefficient occurring in (981 ) with a (s,p+1−s) nonzero factor, and the third is any coefficient Yi (other than the second coefficient) in (99) − p with either 1 ≤ s ≤ s1 or s ≥ s1 such that det Υ 6= 0. (0,3) (0,3) (3,0) 3. If p = 2, then the resonance set contains either Y1 or Y2 , either Y1 (α 6= 1/2) or (3,0) (0,3) (1,2) (1,2) Y2 , and also any other coefficient except for Y1 , Y1 , and Y2 for α = 1. Corollary 11. System (3) with zero approximation (25) is a generalized normal form if for each (p+1) p ≥ 2, the polynomials Yi contain at most nαp nonzero coefficients that have arbitrary values and belong to one of the resonance sets described in Corollary 10. Theorem 11. An arbitrary system (70) can be reduced by a formal change of variables (2) to (p+1) the generalized normal form (3), where the coefficients of the homogeneous polynomials Yi in some resonance set described in Corollary 10 satisfy Eq. (96) with p = 2 and Eqs. (97)–(99) for any p ≥ 3, and the remaining coefficients vanish. Example 4. Suppose that α ∈ {α}0 and α 6= 1 in system (70); then, by Theorem 10, nαp = 2, the resonance equations (97) and (980 ) should be valid for p ≥ 3, and system (70) can be reduced to the generalized normal form (3), which has, for example, one of the following forms: y˙ 1 =
αy12
+ y1 y2 ,
y˙ 2 = y1 y2 +
(3,0) Y2 y13
∞ X (0,p+1) p+1 (1,p) Y2 + y2 + Y2 y1 y2p , p=2
y˙ 1 =
αy12
+ y1 y2 +
(1,2) Y1 y1 y22
+
∞ X
(0,p+1) p+1 Y1 y2 ,
p=2
y˙ 2 = y1 y2 +
∞ X
(0,p+1) p+1 y2 .
Y2
p=2
8. SYSTEMS WITH ZERO APPROXIMATION (±x1 x2 , x21 ) Consider system (1) whose unperturbed part P (x) belongs to the regular case and can be reduced by a linear nondegenerate change of variables to the canonical form (27) (0, −1, 0)(1, 0, 0) or (250 ) (0, 1, 0)(1, 0, 0) equivalent to the form (25) with α = −1 considered in Section 7. Therefore, we assume that system (1) has the form x˙ 1 = ±x1 x2 + X1 (x1 , x2 ) ,
x˙ 2 = x21 + X2 (x1 , x2 ) .
(105)
Then, by notation (6), system (5) acquires the form p+1 ∓hp10 = Yˆ10 ,
p+1 Yˆ20 = 0; p p p (p − s + 2)h1 s−2 ± (s − 1)h1s ∓ h2 s−1 = Yˆ1sp+1 , (p − s + 2)hp2 s−2 ± shp2s − 2hp1 s−1 = Yˆ2sp+1
(106) (s = 1, . . . , p + 1; p ≥ 2).
The first equation in (106) specifies hp10 , and the second equation defines the first resonance relation required for the solvability of system (5). Further, by substituting, as usual, hp2 s−1 from the first subsystem in (106) into the second one, we obtain a linear system for hp1 = hp10 , . . . , hp1p : Θp hp1 = Y˘1p DIFFERENTIAL EQUATIONS
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(107)
1072
BASOV, SKITOVICH
p+1 p where the tridiagonal square matrix Θp = {θsν }s,ν=1 and the vector Y˘1p = Y˘11 , . . . , Y˘1pp+1 have the following components: p θ11 = ±(p − 2), θss = ± 2p(s − 1) − 2s2 + 5s − 6 (s ≥ 2), θsp s+2 = s2 , θs s−2 = (p − s + 2)(p − s + 3); p p+1 p+1 ˆ p+1 ˆ p+1 Y˘11 = ± Yˆ12 + Yˆ21 , Y˘1sp = (p − s + 2)Yˆ1p+1 s−1 ± sY1 s+1 + Y2s
(s ≥ 2).
We reduce system (107) to an upper-triangular form. To this end, we subtract the first equation in system (106) from the first equation in (107) and then successively subtract the sth equation multiplied by ±(p − s + 1)/s from the (s + 2)nd equation (s ≥ 1). Then we obtain the linear system ±s(p − s)hp1 s−1 + s2 hp1 s+1 = Y¯1sp
(s = 1, . . . , p + 1),
(108)
p p p+1 ¯ p p ¯p where Y¯11 = Y˘11 − Yˆ10 , Y12 = Y˘12 , Y1s = Y˘1sp ∓ Y¯1p+1 s−2 (p − s + 3)/(s − 2) (s ≥ 3), and ±s(p − s) are the diagonal entries of the upper-triangular matrix in (108). Let s = 2j + l (l = 0, 1). System (108) splits into two systems:
±(2j + l)(p − 2j − l)hp1 2(j−1)+l+1 + (2j + l)2 hp1 2j+l+1 = Y¯1p2j+l
(l = 0, 1),
(109l )
Qj−1 l where 1 − l ≤ j ≤ (p + 1 − l)/2, and if βjν = (−1)l(j+ν) µ=ν (p − 2µ − l + 1)/(2µ + l), then, by P j p p+1 1 l ˘p passing to the direct formula, we obtain Y¯1p2j+l = lβj0 + ν=1 βjν Y˘11 Y1 2ν+l . − Yˆ10 Let p = 2r + k (r ≥ 1, k = 0, 1). If l 6= k, then system (109l ) is uniquely solvable for h2j+l−1 , and if l = k, then in the matrix of system (109k ), there appears a zero diagonal entry. If j = r, then we have 0 × hp1 p−1 = Y¯12r+k 2r+k . 2r+k By replacing Y¯1 2r+k in accordance with the formulas for systems (109k ) and (107), for each p = 2r + k ≥ 2, we obtain the second relation for the right-hand sides of system (106) : r X
k ˆ p+1 ˆ p+1 2(r − ν + 1)Yˆ1p+1 βrν ± (2ν + k) Y + Y 2ν+k−1 1 2ν+k+1 2 2ν+k
ν=1
(110)
p+1 p+1 p+1 1 − kβr0 = 0; Yˆ10 ∓ Yˆ12 ∓ Yˆ21
2r+k the component h2r+k occurs in the first equation in 1 2r+k−1 is not constrained. The component h10 p p+1 ˆ system (106), that is, ∓h10 = Y10 , and in the first equation (l = 1, j = 0) in system (1091 ), that p is, ±(p − 1)hp10 + hp12 = Y¯11 . We reduce system (1091 ) to a diagonal form; then its first equation becomes
±(p −
1)hp10
=
p−r X
gjp Y¯1 2j+1
−
p kgr−1 (p
−
2)2 hp1 p−1 ,
j=0
gjp
j (∓1)j Y 2ν − 1 = . 2j + 1 ν=1 p − 2ν − 1
(111)
Note that the gjp are obtained from the recursion formulas g0p = 1,
gjp = ∓gj−1 (2j − 1)2 ((2j + 1)(p − 2j − 1))−1 > 0
(j = 1, . . . , r − k)
describing the entries of the first row of an upper-triangular matrix that, being multiplied by the matrix of system (1091 ) on the right, gives a diagonal matrix. 2r+1 If p = 2r + 1, then the free component h2r+1 in (111) has a 1 2r permits one to ensure that h10 value satisfying the first equation in system (106). DIFFERENTIAL EQUATIONS
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1073
If p = 2r, i.e., k = 0, then h2r 1 2r−1 does not occur in Eq. (111), and for the consistency of Pr 2r+1 system (106), we obtain one more condition (2r − 1)Yˆ10 + j=0 gj2r Y¯12r2j+1 = 0, which, after the replacement of Y¯12r2j+1 by the term Yˆis2r+1 , implies the third resonance relation (2r − 1)Yˆ102r+1 +
r X
1 2r+1 2r+1 2r+1 −Yˆ10 gj2r βj0 ± Yˆ12 ± Yˆ21
j=0
+
j X
1 βjν
(2r − 2ν +
1)Yˆ12r+1 2ν
! 2r+1 2r+1 = 0. ± (2ν + 1)Yˆ1 2ν+2 + Y2 2ν+1
(112)
ν=1 p+1 The relations Yˆ20 = 0 in (106), (110), and (112) provide the resonance equations (0,p+1) Y2
= c˜,
r+1 X
(2j+k−1,2(r−j)+2) a2r+k Y1 j
+
j=1−k
r X
(2j+k,2(r−j)+1)
b2r+k Y2 j
= c˜,
(113)
j=1−k
1 k where p = 2r + k (r ≥ 1, k = 0, 1), a2r+1 = −βr0 , a2r+k = 2(r − j + 1)βrj ± (2j + k − 2)βrk j−1 6= 0 0 j 2r+k k (j = 1, . . . , r + 1), and bj = ±βrj (j = 1 − k, . . . , r), r X
(2j,2(r−j)+1)
u2r j Y1
(2j+1,2(r−j))
+ vj2r Y2
= c˜,
(114)
j=0
Pr Pr Pr 1 1 where p = 2r, u2r gν2r βν0 , u2r+k = (2r − 2j + 1) ν=j gν2r βνj ± (2j − 1) ν=j−1 gν2r βν1 j−1 j 0 = − ν=0 Pr 1 l (j ≥ 1), and vj2r = ± ν=j gν2r βrj (j ≥ 0). Moreover, the βjν are given by (109l ), the gjp are given (2r−1,1) by (111), and the h1 are arbitrary coefficients. Theorem 12. System (105) is formally equivalent to system (3) with zero approximation (p+1) P = (±y1 y2 , y12 ) if, for each p ≥ 2, the coefficients of the homogeneous polynomials Yi in system (3) satisfy the resonance equations (113) and (114). More precisely, if p = 2r, then (0,2r+1) (2j,2r−2j+1) (2j−1,2r−2j+2) (2j+1,2r−2j) Y2 satisfies (1131 ), Y2 and Y1 (j ≥ 1) satisfy (1132 ), and Y2 (2j,2r−2j+1) (0,2r+2) (2j+1,2r−2j+1) and Y1 (j ≥ 0) satisfy (114); if p = 2r + 1, then Y2 satisfies (1131 ), Y2 (2j,2r−2j+2) (2j,2r−2j+2) (2j+1,2r−2j+1) and Y1 (j ≥ 0) satisfy (1132 ), and Y2 (j ≥ 1) and Y1 (j ≥ 0) are nonresonance coefficients. In the change of variables (2) relating systems (105) and (3), the h(2r−1,1) 1 are resonance coefficients. (s,p+1−s)
Corollary 12. Each coefficient Yi equations (113) and (114).
in system (3) occurs in at most one of the resonance
We have shown that all coefficients occurring in (1132 ) are resonance coefficients, since they are 2r multiplied by nonzero factors. In general, some of the factors u2r j and vj given by (114) can be zero, which can readily be verified. For example, vr2r 6= 0; therefore, the Y2(2r+1,0) are resonance coefficients. Corollary 13. There may exist resonance sets only of the following types : if p = 2r, then (2j+1,p−2j) (2ν+1,p−2ν) (0,p+1) (2j+1,p−2j) (0,p+1) (2ν+1,p−2ν) (0,p+1) they include Y1 , Y1 , Y2 or Y1 , Y2 , Y2 or Y2 , (2j,p−2j+1) (2ν+1,p−2ν) (2j,p−2j+1) (0,p+1) (0,p+1) Y2 , and Y2 ; if p = 2r + 1, then they include Y1 , Y2 or Y2 , (2j+1,p−2j) and Y2 . Theorem 13. An arbitrary system (105) can be reduced by a formal change of variables to the generalized normal form (3), where three coefficients (for any even integer p ≥ 2) or two coefficients (p+1) (for any odd integer p ≥ 3) in the homogeneous polynomials Yi that belong to some resonance set described in Corollary 13 satisfy Eqs. (113) and (114) and the remaining coefficients vanish. DIFFERENTIAL EQUATIONS
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Example 5. Each system (105) is formally equivalent to the following generalized normal form with zero perturbation in the first equation: y˙ 1 = ±y1 y2 ,
y˙ 2 = y12 +
∞ X (0,p+1) p+1 (p,1) (p+1,0) p+1 Y2 , y2 + Y2 y1p y2 + σp Y2 y1 p=2
where σp = {1 if p = 2r; 0 if p = 2r + 1}. In conclusion, we note that general statements of the theory of equivalence of systems with zero characteristic numbers and the notion of generalized normal form based on the use of resonance equations were given in [2], where systems with the zero approximation (x2 , −x31 ) were investigated. The case (x32 , −x31 ) was considered in [3]. ACKNOWLEDGMENTS The work was financially supported by the Russian Foundation for Basic Research (project no. 02-01-00245). REFERENCES 1. Basov, V.V. and Skitovich, A.V., Differents. Uravn., 2003, vol. 39, no. 8, pp. 1016–1029. 2. Basov, V.V., Differents. Uravn., 2003, vol. 39, no. 2, pp. 154–170. 3. Basov, V.V., Differents. Uravn., 2004, vol. 40, no. 8, pp. 1011–1022.
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