Journal of Mathematical Sciences, Vol. 126, No. 5, 2005
GENERALIZED NORMAL FORM AND THE FORMAL EQUIVALENCE OF SYSTEMS WITH ZERO APPROXIMATION (x32 , −x31 ) V. V. Basov
UDC 517.925
Abstract. We consider formal systems of differential equations of the form ∞ ∞ (p) (p) Y1 (y1 , y2 ), y˙ 2 = −y13 + Y2 (y1 , y2 ), y˙ 1 = y23 + p=4
p=4
(p) Yi
where are homogeneous polynomials of order p. Such systems are obtained from initial systems of the same form by using formal invertible changes of variables xi = yi + hi (y1 , y2 ) (i = 1, 2). For any p ≥ 4, we explicitly write np = {5, if p = 4r + 1; 4, if p = 4r + 1} linear resonant equations. The initial system (p) is formally equivalent to the above system if the coefficients of the polynomials Yi satisfy the specified resonant equations.
This paper continues [1, 2]. In [1, Part I], we introduced the notions of resonance equations and generalized normal form for the system [κ]
x˙ i = Pγi (x) + Xi (x)
(i = 1, . . . , n),
[κ]
where the zero approximation Pγ is a quasihomogeneous polynomial of degree κ with weight γ = [κ] (γ1 , . . . , γn ), i.e., the generalized order of Pγi (x) equals κ + γi , and the lowest generalized orders of the series Xi (x) are not lower than κ + γi + 1. In [1, Part II] and [2], the results obtained were applied to the cases where an initial system is twodimensional and has the zero approximation (x2 , −x31 ) or a quadratic polynomial. In this paper, we consider an initial system in the case where n = 2, the variables x1 and x2 have first order, i.e., γ = (1, 1), the degree κ = 2, and a quasihomogeneous first-approximation polynomial [2] P(1,1) = (x32 , −x31 ) is an ordinary polynomial of third order. Thus, the initial system has the form x˙ 1 = x32 + X1 (x), where Xi =
∞
(p+2) Xi (x),
x˙ 2 = −x31 + X2 (x),
(r) Xi
=
p=2
r
(s,r−s) s r−s x1 x2
Xi
(1)
(i = 1, 2).
s=0
Let xi = yi + hi (y)
(i = 1, 2),
(2)
be a formal or convergent at zero transformation, where hi =
∞
(p)
hi (y).
p=2
It transforms system (1) into the system y˙ 1 = y23 + Y1 (y),
y˙ 2 = −y13 + Y2 (y),
(3)
Translated from Sovremennaya Matematika i Ee Prilozheniya (Contemporary Mathematics and Its Applications), Vol. 8, Suzdal Conference–2, 2003.
1392
c 2005 Springer Science+Business Media, Inc. 1072–3374/05/1265–1392
where Yi =
∞
(p+2)
Yi
(y)
p=2
and, as usual, (r)
Zi Then the homogeneous
=
r
(s,r−s) s r−s y1 y 2 .
Zi
s=0 (p) (p+2) satisfy the conditions polynomials hi and Yi (p) (p) ∂h1 3 ∂h1 3 (p) (p+2) y2 − y1 − 3h2 y22 = Y1 , ∂y1 ∂y2 (p) (p) ∂h2 3 ∂h2 3 (p) (p+2) y2 − y1 + 3h1 y12 = Y2 , ∂y1 ∂y2
where (p+2) (p+2) (p+2) Yi = Yi (y) − Yi (y),
P = (3h22 y2 + h32 , −3h21 y1 − h31 ), ∂hi ∂hi (p+2) (p+2) Yi = Xi (y + h) + Pi (h) − Y1 − Y2 . ∂y1 ∂y2 (p+2) (r) (y) are known since they depend only on hν and Obviously, the homogeneous polynomials Yi (r+2) , where 2 ≤ r ≤ p − 1 (p ≥ 2, ν = 1, 2). Yν Equating the coefficients of y1s y2p+2−s (s = 0, 1, . . . , p + 2) in the first equation and the coefficients of y1s+2 y2p−s (s = −2, −1, . . . , p) in the second equation; thus, we obtain the following system of 2(p + 3) equations with 2(p + 1) unknowns: (s+1,p−s−1)
− (p − s + 3)h1
(s+3,p−s−3)
− (p − s + 1)h2
(s + 1)h1 (s + 3)h2
(s−3,p−s+3) (s−1,p−s+1)
(s +
− (p − s +
1)hp2s−1
+
3hp1s
=
(s,p−s+2) = Y1 ,
(s,p−s)
= Y2
+ 3h1
Rewrite it in a more convenient form: p (s + 1)hp1s+1 − (p − s + 3)hp1s−3 − 3hp2s = Y1s 3)hp2s+3
(s,p−s)
− 3h2
p Y2s
(s+2,p−s)
.
(s = 0, 1, . . . , p + 2), (s = −2, −1, . . . , p),
(4)
where (s,p−s)
hp1s = h1 p Y1s
=
,
(s,p−s+2) Y1 ,
(s,p−s)
hp2s = h2 p Y2s
=
,
(s+2,p−s) Y2 ,
hpis
= 0 for s < 0 or s > p (i = 1, 2). and, by definition, We emphasize the following symmetry property of the linear system (4): this system does not change p p p p , and Y2s by −hp2p−s , −hp1p−s , Y2p−s , and Y1p−s , respectively. if we replace hp1s , hp2s , Y1s In other words, the (2p + 6) × (2p + 2)-matrix A defining the left-hand side of (4) satisfies the condition D2p+6 A D2p+2 = −A, where
0 0 Dl = · 0 1
0 0 · 1 0
... ... ··· ... ...
0 1 · 0 0
1 0 · . 0 0 (l×l) 1393
The multiplication of the matrix Dl by a vector of dimension l is the permutation of the components of this vector in inverse order. Therefore, if a vector a has dimension 2p + 6, then a ⊥ Im A ⇔ D2p+6 a ⊥ Im A.
(5)
Indeed, let a ⊥ Im A, i.e., (a, Ax) = 0 for any vector x of dimension 2p + 2. We consider another vector y of the same dimension and set x = D2p+2 y. Then y = D2p+2 x and (D2p+6 a, Ay) = (a, D2p+6 AD2p+2 x) = (a, −Ax) = 0, i.e., D2p+6 a ⊥ Im A. We eliminate four redundant equations from system (4). We take the equations with s = p + 1, p + 2 from the first moiety of (4) and the equations with s = −2, −1 from the second moiety of (4): p , −2hp1p−2 = Y1p+1 p hp21 = Y2,−2 ,
p −hp1p−1 = Y1p+2 , p 2hp22 = Y2,−1 .
(6)
We rewrite the remainder 2p + 2 equations of system (4) in the matrix form: Γp hp1 − 3hp2 = Y1p ,
3hp1 + ∆p hp2 = Y2p ,
(7)
where hpi = (hpi0 , . . ., hpip ), Yip = (Yi0p , . . . , Yipp ), Γp and ∆p are two-diagonal matrices of dimension (p + 1)× (p + 1) with the following diagonal elements: γss+1 = s + 1, δss+3 = s + 3,
γss−3 = s − 3 − p; δss−1 = s − 1 − p
(subscripts vary from 0 to p). The first equation of system (7) allows one to find the vector hp2 : 3hp2 = Γp hp1 − Y1p
p or 3hp2s = (s + 1)hp1s+1 + (s − 3 − p)hp1s−3 − Y1s .
(8)
p p and 3hp22 = 2hp13 − Y12 (h213 = 0). Substituting these In particular, we have in (8) 3hp21 = 2hp12 − Y11 p components of the vector h2 into the third and fourth equations of system (6), we obtain four additional equations for the components of hp1 : p p + Y11 , 2hp12 = 3Y2,−2 p −2hp1p−2 = Y1p+1 ,
p p 6hp13 = 3Y2,−1 + 2Y12 , p −hp1p−1 = Y1p+2 .
(9)
Eliminating hp2 from (8) and the second equation of system (7), we obtain the system Θp hp1 = Y˘2p ,
(10)
where Θp = ∆p Γp + 9E is a three-diagonal (p + 1) × (p + 1)-matrix and Y˘2 = ∆Y1p + 3Y2p . It is easy to verify that θss = δss+3 γs+3s + δss−1 γs−1s + 9, θss+4 = δss+3 γs+3s+4 , θss−4 = δss−1 γs−1s−4 , p p p p = δss+3 Y1s+3 + δss−1 Y1s−1 + 3Y2s , Y˘2s
where all subscripts vary from 0 to p. 1394
Let
0 0 0 bp0 0 −ap0 p p 0 0 0 0 b −a 1 1 0 0 0 0 0 −ap2 0 0 0 0 0 −ap3 Θp = p p c 0 0 0 −a4 0 0 p p 0 0 0 0 −a c 1 5 · · · · · · 0 0 0 0 0 0 then aps
··· ··· ··· ··· ··· ··· ··· ···
0 0 0 0 ; 0 0 · −app
(p − s)(s + 3) + (p − s + 1)s − 9 for s = 0, . . . , p − 3, = (p − s + 1)s − 9 for s = p − 2, p − 1, p,
bps−4 = (s − 1)s, cps−4 = (s − 1 − p)(s − 4 − p) for s = 4, . . . , p, p p p p Y˘2s = (s + 3)Y1s+3 + (s − 1 − p)Y1s−1 + 3Y2s
(11)
(Y1lp = 0 for l < 0, l > p).
Note that all components of the vector hp1 that comprise Eqs. (9) are distinct if 3 < p − 2, i.e., p ≥ 6. For such p, we can obtain general formulas for the resonance equation. Initial values of p that equal 2, 3, 4, and 5, will be considered separately. Let p = 2. Then system (10) has the form 2 −a2s h21s = Y˘2s
or 2 2 ((s − 3)s + 9)h21s = (s − 3)Y1s−1 + 3Y2s
(s = 0, 1, 2).
Thus, we have 2 3h210 = Y20 ,
2 2 7h211 = −2Y10 + 3Y21 ,
2 2 7h212 = −Y11 + 3Y22 ,
det Θ2 = 0.
Substituting these values of h21s into (9) with p = 2, we obtain four resonance constraints for 10 components of the vectors Yi2 : 2 2 + 3Y2,−1 = 0, 2Y12 2 2 2 −2Y10 + 7Y14 + 3Y21 = 0, 2 2 + 2Y20 = 0, 3Y13
(122 )
2 2 2 3Y11 + 7Y2,−2 − 2Y22 = 0;
any component is a term of exactly one of Eqs. (122 ). Let p = 3. Then system (10) takes the form 3 , −a3s h31s = Y˘2s
or 3 6h311 = −3Y10 + 3Y21 , 3 5h312 = −2Y11 + 3Y22 , 3 6h313 = −Y12 + 3Y23 , 3 3 0 · h310 = 3Y13 + 3Y20 .
The last equation is an additional equation, which appears because the diagonal matrix Θ3 has zero eigenvalue. The component h310 has no restrictions. 1395
Substituting these values of h31s into (9) with p = 3, we obtain five resonance constraints for 12 components of the vectors Yi3 : 3 3 3 + Y14 + Y21 = 0, −Y10 3 3 3 −2Y11 + 5Y15 + 3Y22 = 0, 3 3 3 + Y2,−1 − Y23 = 0, Y12
(123 )
3 3 3 3Y11 + 5Y2,−2 − 2Y22 = 0, 3 3 Y13 + Y20 = 0.
If p = 4, the four relations (9) contain only the components h412 and h413 and the matrix Θ4 becomes three-diagonal. Since a40 = 0, system (10) with p = 4 is reduced to the upper-triangular system 4 −a40 h410 + b40 h414 = Y˘20 , 4 −a4s h41s = Y˘2s (s = 1, 2, 3), 4 −d44 h414 = Yd4 ,
where
4 4 4 = Y˘24 + (c40 /a40 )Y˘20 . d44 = a44 − b40 c40 /a40 , Yd4 4 4 It is easy to see that a = (3, 7, −3, −3, −5) and d4 = 21. Therefore, det Θ4 = 0 and the vector h41 is uniquely determined by system (10). In particular, for s = 2, 3 we have 4 4 + 3Y22 , 3h412 = −3Y11
4 4 3h413 = −2Y12 + 3Y23 .
Substituting this into (9), we obtain four resonance constraints for eight components of Yis4 (the total number of components is 14): 4 4 4 + Y15 + 2Y22 = 0, −2Y11 4 4 4 + 3Y2,−2 − 2Y22 = 0, 3Y11 4 4 2Y12 + Y2,−1 4 4 −2Y12 + 3Y16
− +
4 2Y23 4 3Y23
= 0,
(124 )
= 0.
For p = 5, relations (9) and system (10) take the form 5 5 2h512 = 3Y2,−2 + Y11 , 5 5 6h513 = 3Y2,−1 + 2Y12 , 5 −2h513 = Y16 ,
−a5s h51s + b5s+4 h51s+4
5 −h514 = Y17 ; 5 = Y˘2s (s = 0, 1, 2, 3),
5 −d5s h51s = Yds
where
d5s = a5s − b5s−4 c5s−4 /a5s−4 ,
(s = 4, 5),
5 5 5 Yds = Y˘2s + (c5s−4 /a5s−4 )Y˘2s−4 .
It is easy to see that a5 = (6, 12, 14, 0, −1, −4), d54 = −21, and d55 = −32/3. System (10) for s = 2, 3, and 4 implies that 5 5 5 + 5Y15 + 3Y22 , −14h512 = −4Y11 5 5 0 · h513 = −Y12 + Y23 , 5 5 5 21h514 = 3Y13 + 5Y20 + 3Y24 .
We have obtained that a53 = 0 but the component h513 is not arbitrary because it is a term of two relations in (9); we must eliminate it from these two relations. 1396
Thus, 12 components of the vectors Yi5 (its total number is 16) satisfy the following four resonance constraints: 5 5 5 5 + 5Y1,5 + 21Y2,−2 + 3Y22 = 0, 3Y11 5 5 5 2Y12 + 3Y16 + 3Y2,−1 = 0,
(125 )
5 5 5 5 3Y13 + 21Y17 + 5Y20 + 3Y24 = 0, 5 5 −Y12 + Y23 = 0.
Note that the symmetry property (5) is clearly salient in formulas (12τ ) (τ = 2, 3, 4, 5). The third and fourth equations in (12τ ) can be obtained from the first and second equations respectively by replacing p 5 = Y 5 , which Yisp by Y3−i,p−s . The symmetry does not break in (125 ) since the incomplete equation Y12 23 appears since a53 = 0, transforms into the equation 5 5 5 3Y16 + 3Y2,−1 + 2Y23 = 0.
Moreover, the unique unpaired fifth equation in (123 ) is invariant under property (5). Now we consider system (10) for p ≥ 6. It can be shown by using induction in rows that it is equivalent to the system (13) Θpd hp1 = Ydp , where
0 0 0 bp0 0 −dp0 p p 0 0 0 0 b −d 1 1 p 0 0 0 0 0 −d2 0 0 0 0 0 −dp3 Θpd = p 0 0 0 0 0 −d4 p 0 0 0 0 0 −d 5 · · · · · · 0 0 0 0 0 0
··· ··· ··· ··· ··· ··· ··· ···
0 0 0 0 0 0 · −dpp
if dp0 , . . . , dpp−4 = 0. In this case, dps = aps , Y p = Y˘ p
ds dps p Yds
= =
2s (0 ≤ s ≤ 3); p as − bps−4 cps−4 /dps−4 , p p Y˘2s + Yds−4 cps−4 /dps−4
(14) (4 ≤ s ≤ p).
Explicit formulas for diagonal elements dps of the matrix Θpd and for eigenvalues of the matrix Θp can be obtained only for particular values of s. Therefore, if we want to know whether elements dps vanish, we must construct majorizing nonpositive and minorizing nonnegative functions for them. It is convenient to represent the elements aps in (11) for s = 0, . . . , p − 3 in the form aps = (2(p − s) + 1)s + 3(p − s − 3), bps−4 cps−4 = (s − 1)s(p − s + 1)(p − s + 4). p = 0. Using We introduce the functions ζsp = (p − s − 3)(s + 3). Then ζsp > 0 for 0 ≤ s ≤ p − 4 and ζp−3 p p induction, we prove that ds ≥ ζs for 0 ≤ s ≤ p − 3. Since aps − ζsp = (p − s + 4)s for p ≥ 6 and s ≤ p − 3, we have dp0 = ζ0p and dpj > ζjp (j = 1, 2, 3); the base of induction is proved. p = (p − s + 1)(s − 1) Let s = 4, . . . , p − 3. Assume that dpj ≥ ζjp for 4 ≤ j ≤ s − 1. Taking the relation ζs−4 and the induction inference into account, we obtain p = ζsp . dps ≥ aps − bps−4 cps−4 /ζs−4
1397
Considering the initial values of the eigenvalues, we specify the estimate of dps . Obviously, p dp4l = ζ4l
and dp4l+j 4r+3 =0 d4r
p ζ4l+j
(0 ≤ l ≤ (p − 3)/4),
dps
> for j = 1, 2, 3. Therefore, > 0 for p ≥ 6 and 0 ≤ s ≤ p − 3 except for the elements p (r ≥ 1), and d4l = (p − 4l − 3)(4l + 3) for 0 ≤ l ≤ (p − 3)/4. We introduce the function ηsp = ζsp + 3p + 3 = (p − s)(s + 6) − 6. Then ηsp > 0 for 0 ≤ s ≤ p − 1 and ηpp = −6. We prove that dps ≤ ηsp for 0 ≤ s ≤ p − 3. Since aps − ηsp = (p − s + 1)(s − 3) for p ≥ 6 and s ≤ p − 3, we have dpj ≤ ηjp (j = 0, 1, 2, 3); the base of induction is proved. Let s = 4, . . . , p − 3. We assume that dpj ≤ ηjp for 4 ≤ j ≤ s − 1. Then p , dps ≤ aps − bps−4 cps−4 /ηs−4
since dps−4 > 0 for s ≤ p (see above). The required inequality dps ≤ ηsp will hold if p aps − bps−4 cps−4 /ηs−4 ≤ ηsp ,
or p ≤ bps−4 cps−4 , (aps − ηsp )ηs−4
or (s − 3)((p − s + 4) × (s + 2) − 6) ≤ (s − 1)s(p − s + 4); the last inequality always holds. We estimate dpp−j from above for j = 2, 1, 0. Since p 0 ≤ dpp−j−4 ≤ ηp−j−4 ,
we have p ; dpp−j = app−j − bpp−j−4 cpp−j−4 /dpp−j−4 ≤ app−j − bpp−j−4 cpp−j−4 /ηp−j−4
in this case, app−j = (j + 1)(p − j) − 9, bpp−j−4 cpp−j−4 = (p − j − 1)(p − j)(j + 1)(j + 4), p = (j + 4)(p − j + 2) − 6. ηp−j−4
Therefore, p+1 3p + 1 < 0, dpp−1 ≤ −9 < 0, p−1 5p − 1 These arguments prove the following lemma. dpp−2 ≤ −3
dpp ≤ −3
5p + 3 < 0. 2p + 1
Lemma 1. For any p ≥ 6, diagonal elements dps of the matrix Θpd of system (13) are nonzero for all s = 0, . . . , p, except for the cases where p = 4r + 3, s = 4r (r ≥ 1), i.e., we have d4r+3 = 0. In this case, 4r (p − s − 3)(s + 3) ≤ dps ≤ (p − s − 3)(s + 3) + 3p + 3 for s = 0, . . . , p − 3; if s = 4l, then for l = 0, . . . , [(p − 3)/4].
dp4l = (p − 4l − 3)(4l + 3)
Now we obtain the restrictions for Yisp (these quantities are terms on the right-hand sides of (7) and (9) for p ≥ 6). Relations (9) contain four different components of the vector hp1 : hp12 , hp13 , hp1p−2 , and hp1p−1 . We express them from system (10), which is equivalent to (7); this yields four resonance constraints. If p = 4r +3, then we see that in system (10) dpp−3 = 0; this yields an additional constraint for the right-hand sides of (7). 1398
p To find hp1p−2 and hp1p−1 (and for p = 4r + 3, immediately Ydp−3 ), we represent the number p ≥ 6 in one of the following forms:
(r ≥ 1,
p = sjr + k sj0
sjl
j = 0, 1, 2, 3,
sjl = 4l + j
k = 1, 2, 3),
(l is integer).
(15)
sjl+1
= j, + 3 = − 1; p = 4r + j + k and r ≥ 2 if j = 0 and k = 1. In this case, if k = 3, then Then we must consider only the case where j = 0. We introduce two sets of constants depending, in particular, on dps (see (14)): αlpj = sjl + 3 −
dpj sl
(p − sjl − 3)(p − sjl ) = 0 dpj
βlpj =
, p − sjl
(0 ≤ l < r);
sl
dpj
(16)
r = 0 (jk = 03), αr4r+3 0 = 4r + 3. k We use system (13) obtained from (10). The (p − k)th equation of this system has the form
αrpj = −
s
p −dpp−k hp1p−k = Ydp−k
for k = 1, 2, 3 or, by (15),
−dpj hp
= Y pj
sr 1sjr
(k = 1, 2, 3).
dsr
(17)
We express the quantities Y p j in (17) immediately by the components of the vectors Yip of system (7). dsr We prove by induction that Y p j = (sjl + 3)Y p j
1sl +3
dsl
+
l−1
(ανpj Y p j
1sν +3
ν=0
l−1
+ 3Y p j ) 2sν
+ 3Y p j 2sl
l−1
βµpj + (j − 1 − p)Y p j
1s−1 +3
µ=ν p Y1,s
where p and j are as in (15), k = 1, 2, 3, 0 ≤ l ≤ r, and Indeed, for l = 0, (14) and (11) (and also (18)) imply
(18)
βµpj ,
µ=0
= 0 (s < 0, s > p).
p p + 3Y2jp + (j − 1 − p)Y1j−1 , Ydjp = Y˘2jp = (j + 3)Y1j+3
and the base of induction is proved. Assume that formula (18), where 0 ≤ l ≤ r − 1, holds. We find Y p j
dsl+1
into account that βlpj = cpj /dpj . We have sl
sl
Y pj
dsl+1
= (sjl+1 + 3)Y p j
1sl+1 +3
+3Y p j βlpj 2sl
from (14), (11), and (17), taking
+
l−1 ν=0
= Y˘ p j
2sl+1
+ 3Y p j
2sl+1
(ανpj Y p j 1s +3 ν
+
+ Y p j βlpj dsl
+ ((sjl + 3 − p) + (sjl + 3)βlpj )Y p j
3Y p j ) 2sν
1sl +3
l
βµpj
+ (j − 1 −
p p)Y1j−1
µ=ν
l
βµpj ;
µ=0
this is exactly formula (18) with l replaced by l + 1, since the factor of
Y pj 1sl +3
in the third term of the
right-hand side equals αlpj βlpj . Let k = 1, 2; then for l = r, the first term on the right-hand side of (18) vanishes since sjl + 3 > p, and (18) has the form Y pj dsr
=
3Y p j 2sr
+
r−1 ν=0
(ανpj Y p j 1s +3 ν
+
3Y p j ) 2sν
r−1
µ=ν
βµpj
+ (j − p −
p 1)Y1j−1
r−1
βµpj ;
µ=0
1399
p note that for j = 0, the last term vanishes because Y1,−1 = 0. p Substituting the quantities h1p−k into Eqs. (17) from the last two of Eqs. (9), −khp j = Y p j , we 1sr 1sr +3 obtain for any p ≥ 6 in (15) two resonance constraints:
Y pj
dp /(−k) 1sr +3 sjr
+ Y pj = 0 dsr
or r ν=0
ανpj Y p j 1s +3 ν
where j = 1, 2, 3, k = 1, 2; p and and k = 1, 2, then Eqs. (19jk ) do
+
3Y p j 2sν
r−1
βµpj
− (4r + k +
1)Y1pj−1
µ=ν
sjν
r−1
βµpj = 0,
(19jk )
µ=0
ανpj
are from (15), and and βµpj are not contain the components Y p j . 2s−1
from (16). In this case, if j = 2, 3
Formulas (19jk ) without the last term are valid for j = 0, but in this case they can be simplified since the quantities αlpj and βlpj given by the recurrence formulas (16) can be easily calculated for j = 0. Let p = 4r + k (j = 0, k = 1, 2). By Lemma 1, dp4l = (p − 4l − 3)(4l + 3) for 0 ≤ l ≤ (p − 3)/4. Therefore, 4r+k can be found by (14). By (11), we have l ≤ r − 1 and d4r+k 4(r−1) = (k + 1)(4r − 1) and d4r 4r+k = (k + 1)4r − 9, a4r
4r+k b4r+k 4r−4 c4r−4 = (4r − 1)4r(k + 1)(k + 4)
4r+k = −3(4r + 3) (k = 1, 2). and, therefore, d4r
Substituting p = 4r + k into (16), we find ανp0 =
3(4ν + 3) 4(r − µ) + k for ν = 0, . . . , r and βµp0 = 4(r − ν) + k 4µ + 3
for µ = 0, . . . , r − 1 (k = 1, 2). Obviously, these formulas hold for k = 3 and ν < r. Thus, for p = 4r + k and j = 0, formulas (19jk ) take the form r−1 r
4(r − µ) + k 4ν + 3 4r+k 4r+k Y1 4ν+3 + Y2 4ν =0 4(r − ν) + k 4µ + 3 µ=ν
(k = 1, 2);
(190k )
ν=0
in addition, r ≥ 2 for k = 1 and r ≥ 1 for k = 2. By Lemma 1, systems (13) and (10) are uniquely solvable for any p ≥ 6 except for the case where p = 4r + 3 (r ≥ 1), i.e., where j = 0 and k = 3. In this case, the matrix Θpd in system (13) has the last diagonal element d4r+3 = 0; this imposes an 4r p additional condition on Yd4r . The fourth equation from the end of system (13) is such that 0 · h4r+3 14r = 4r+3 4r+3 Yd4r ; therefore, the coefficient h14r is arbitrary. In (18), we set p = 4r + 3 (j = 0, k = 3) and s0ν = 4ν. Then the last term on the right-hand side of (18) vanishes; the first term is such that for ν = r, we have s0r + 3 = 4r + 3. Moreover, we have explicit p = 0, we obtain for any formulas for ανp0 and βµp0 . Substituting the found quantities into the equation Yd4r p = 4r + 3 (r ≥ 1) the relation r−1 r
4(r − µ) + 3 4ν + 3 4r+3 4r+3 Y1 4ν+3 + Y2 4ν = 0. (1903 ) 4(r − ν) + 3 4µ + 3 µ=ν ν=0
Obviously, this formula coincides with (190k ), where k = 3. Moreover, we have proved that the 4r+3 components h14r in system (7) have no restrictions. It remains to find the components hp12 and hp13 in system (7), substitute them in the first two of Eqs. (9), and obtain the last two resonance constraints for any p ≥ 6. Unfortunately, the right-hand sides of system (13) become cumbersome after reduction to the diagonal form. Another way is the reduction of system (10) to the system Θpe hp1 = Yep with upper-triangular, twodiagonal matrix Θpe with the principal diagonal −ep0 , . . . , −epp and the fixed fourth subdiagonal cp0 , . . . , cpp−4 , 1400
if the elements epp , . . . , ep1 are nonzero. For p as in (15), where j = 2, 3 and k = 0, 1, 2, 3, explicit formulas for epj can be written only for k = 3 (otherwise, even the separation of ep4l+j from zero is difficult). sl
The quantities hp1j (j = 2, 3) can be found from system (10) by using the Kramer formulas, but this is not necessary: missing formulas can be easily obtained by using property (5). p p by Y2p−s and vice versa, we Premultiplying the vector Y (p+2) by the matrix Dp+6 , i.e., replacing Y1s 0k obtain from (19 ) for k = 1, 2 the relation r−1 r
4(r − µ) + k 4ν + 3 4r+k + Y Y24r+k = 0; (200k ) 4(r−ν)+k−3 1 4(r−ν)+k 4(r − ν) + k 4µ + 3 µ=ν ν=0
formulas (19jk ) for p and sjν satisfying (15), k = 1, 2, and j = 1, 2, 3 imply r r−1
pj p r−1 p p pj βµ − (4r + k + 1)Y2 4r+k+1 βµpj = 0. αν Y24(r−ν)+k−3 + 3Y14(r−ν)+k µ=ν
ν=0
(20jk )
µ=0
p Since the components Y1s are defined for s ≤ p + 2, Eqs. (20jk ) do not contain Y1p4r+k+4 for k = 1, 2 and j = 2, 3. It is easy to see that the resonance equation (1903 ) is invariant under such transformations since r−ν
µ=ν
4(r − µ) + 3 =1 4µ + 3
for any ν ≤ r/2. For Eqs. (19) and (20), we introduce universal constants by using (16), (14), and (11). We set b0k rν
=
r−1
µ=ν
4(r − µ) + k = 0, 4µ + 3
a0k rν =
4ν + 3 b0k = 0 (ν = r, . . . , 0) 4(r − ν) + k rν
for j = 0 and k = 1, 2, 3; bjk rr bjk rν =
r−1
µ=ν
ajk rr
= 1,
(p − sjµ − 3)(p − sjµ ) = 0, −dpj
dpj s
= − r = 0, 3k
ajk rν =
sjν + 3 −
sµ
bjk r,−1
= 0,
ajk r,−1
dpj
sν
p − sjν
bjk rν 3
(0 ≤ ν ≤ r),
(21)
(4r + k + 1)bjk r0 =− = 0 3
for j = 1, 2, 3 and k = 1, 2, where p = 4r + j + k (r ≥ 1), sjl = 4l + j, dpj = (2(p − j) + 1)j + 3(p − j − 3), s0
dpj = (2(p − sjl ) + 1)sjl + 3(p − sjl − 3) − sl
1 (sjl p dj sl−1
dpj = dpp−k = (k + 1)(p − k) − 9 − sr
− 1)sjl (p − sjl + 1)(p − sjl + 4) (0 < l < r), 1
(k dpj sr−1
+ 1)(k + 4)(sjr − 1)sjr .
(s,p−s+2) p = Y1 Thus, we have obtained linear equations (12τ ) (τ = 2, 3, 4, 5) for the components Y1s (s+2,p−s) p (0 ≤ s ≤ p + 2) and the components Y2s = Y2 (−2 ≤ s ≤ p) of system (4) for p = 2, 3, 4, 5; for jk p = 4r + j + k ≥ 6, we have Eqs. (19 ) and (20jk ) (j = 0, 1, 2, 3 and k = 1, 2). Moreover, we have Eq. (1903 ).
1401
According to the notation for the coefficients of system (3), we have (l,ν) 2 ∂hi (l,ν) (l,ν) (l,ν) (l,ν) Yi = Yi − Yi , Yi = Xi (y + h) + Pi (h) − Yj , ∂yj j=1
(l,ν)
(i = 1, 2) are known constants containing the coefficients of the series h and Y and the where Yi coefficients of the series X. In what follows, we represent an integer number p ≥ 2 in the form (r ≥ 0,
p = 4r + τ
2 ≤ τ ≤ 5).
(22)
We rewrite Eqs. (12), (19), and (20) immediately for the coefficients of system (3) assuming that in the (l,ν) (l,ν) c, the constant c always equals F (Yi ). obtained linear equations of type F (Yi ) = If r = 0 in (22), then the seventeen equations (12τ ) for the coefficients of the homogeneous polynomials (τ +2) (τ +2) Y1 and Y2 take the following forms: (3,1)
3Y1 (0,4) −2Y1
+
(0,5)
−Y1
(1,4)
−2Y1
(4,0) 7Y1
(2,2)
+ 2Y2 +
(4,1)
+ Y1
(5,0)
+ 5Y1
(1,5)
−2Y1
(2,4)
−2Y1
(2,5)
2Y1
(5,1)
+ Y1
(6,0)
+ 3Y1
(6,1)
+ 3Y1
=
= c,Y1
(4,1)
= c,3Y1
(2,3)
= c;
(4,2)
= c,3Y1
(5,1)
= c,2Y1
+ Y2
+ 2Y2 + 3Y2
(1,6)
+ 3Y2
(1,3)
+ 3Y2
(1,3) c,3Y1
(3,2)
+ Y2
+ 3Y2
(3,2)
Y1
(3,1) 3Y2
(2,2)
= c,2Y1
(2,3)
+
(1,4)
+ Y2
(1,4)
(3,4)
3Y1
(1,6)
3Y1
(7,0)
+ 21Y1
(5,2)
− 2Y2
(0,6)
− 2Y2
(1,5)
(6,1)
+ 5Y1
(0,5)
+ Y2
3Y1
−
(5,0)
+ 3Y2
(2,4)
(4,0) 2Y2
− Y2
+ 5Y2
(1,5)
= c,
(0,4) 7Y2
= c,
(1,6) (2,5)
+ 5Y2
(0,7)
+ 21Y2
= c,
(4,1)
= c,
(4,2)
= c,
(5,1)
− 2Y2
+ 3Y2
= c;
(2320 )
= c;
(5,2)
= c,
(6,1)
= c,
(4,3)
= c.
+ 2Y2 + 3Y2 + 3Y2
(2330 )
(2340 )
(2350 )
Now let r ≥ 1 in (22). For τ = 2, 3, 4, 5, we consecutively write (in terms of homogeneous polynomials (4r+τ +2) Yi ) those of Eqs. (19jk ) and (20jk ) in which j + k = τ or j + k = 1, where τ = 5; for this, we use constants (21). For τ = 2, we obtain four equations (1911 ), (2011 ), (2002 ), and (1902 ): r+1
(4ν,4(r−ν)+4)
a11 rν−1 Y1
+
ν=0 r
(4ν+1,4(r−ν)+3)
b11 r r−ν Y1
ν=0 r
(4ν+2,4(r−ν)+2)
b02 r r−ν Y1
ν=0
1402
(4ν+3,4(r−ν)+1)
a02 rν Y1
(4ν+3,4(r−ν)+1)
b11 rν Y2
ν=0 r+1
+
+
ν=0
r
r
+
(4ν,4(r−ν)+4)
a11 r r−ν Y2
ν=0 r
ν=0
= c, (232r )
(4ν+1,4(r−ν)+3)
a02 r r−ν Y2
ν=0 r
= c,
(4ν+2,4(r−ν)+2)
b02 rν Y2
= c,
= c.
For τ = 3, we have five equations (1912 ), (1921 ), (2021 ), (2012 ), and (1903 ): r+1
(4ν,4(r−ν)+5)
a12 rν−1 Y1
ν=0 r+1
+
(4ν+1,4(r−ν)+4)
a21 rν−1 Y1
ν=0 r+1
(4ν+1,4(r−ν)+4)
b21 r r−ν Y1
ν=0 r
(4ν+2,4(r−ν)+3)
b12 r r−ν Y1
(4ν+3,4(r−ν)+2)
a03 rν Y1
(4ν+3,4(r−ν)+2)
b12 rν Y2
ν=0 r+1
+
+
+
ν=0
r
r
+
ν=0
ν=0 r+1
= c,
(4ν,4(r−ν)+5)
= c,
(4ν,4(r−ν)+5)
= c,
b21 rν−1 Y2
a21 r r−ν Y2
ν=0 r+1
(4ν+1,4(r−ν)+4)
a12 r r−ν Y2
ν=0 r
(4ν+2,4(r−ν)+3)
b03 rν Y2
(233r )
= c,
= c.
ν=0
For τ = 4, we have four equations (1922 ), (2031 ), (1931 ), and (2022 ): r+1
(4ν+1,4(r−ν)+5)
a22 rν−1 Y1
+
r+1
ν=0
ν=0
r+1
r+1
(4ν+1,4(r−ν)+5)
b31 r r−ν Y1
ν=0 r+1
(4ν+2,4(r−ν)+4)
a31 rν−1 Y1
ν=0 r+1
(4ν+2,4(r−ν)+4)
b22 r r−ν Y1
+
+
+
ν=0
(4ν,4(r−ν)+6)
= c,
(4ν,4(r−ν)+6)
= c,
b22 rν−1 Y2
a31 r r−ν Y2
ν=0 r+1
(4ν+1,4(r−ν)+5)
= c,
ν=0 r+1
(4ν+1,4(r−ν)+5)
= c.
(234r )
b31 rν−1 Y2
a22 r r−ν Y2
ν=0
For τ = 5, we have Eqs. (1932 ) and (2032 ) and also Eqs. (2001 ) and (1901 ), in which r should be replaced by ρ = r + 1 since p = 4r + 5 = 4ρ + 1: r+1 ν=0 r+1
(4ν+2,4(r−ν)+5) a32 rν−1 Y1 (4ν+2,4(r−ν)+5)
b32 r r−ν Y1
ν=0 ρ
+
+
(4ν+1,4(ρ−ν)+2)
b01 ρρ−ν Y1
ν=0 ρ ν=0
r+1 ν=0 r+1
(4ν+1,4(r−ν)+6)
= c,
(4ν+1,4(r−ν)+6)
= c,
b32 rν−1 Y2
a32 r r−ν Y2
ν=0 ρ
+
(235r ) (4ν,4(ρ−ν)+3)
a01 ρρ−ν Y2
= c,
ν=0 (4ν+3,4(ρ−ν)) a01 ρν Y1
+
ρ
(4ν+2,4(ρ−ν)+1)
b01 ρν Y2
= c.
ν=0
We set nτ = {5 for τ = 3; 4 for τ = 2, 4, 5}. Denote by (23τrµ ) the µth equation in system (23τr ), where µ = 1, . . . , nτ . Thus, we have proved the following theorem. Theorem 1. System (1) is formally equivalent to sustem (3) (i.e., there exists a formal change of variables (2), which tranforms (1) into (3)) if and only if for any p ≥ 2 in (22), the coefficients of the 1403
homogeneous polynomial Y (p+2) of system (3) satisfy nτ linear resonance equations (23τr ) whose left-hand sides are determined by the zero approximation of system (1) or (3). To state some consequences of this theorem, we represent the coefficient vector of the polynomial Y (p+2) (p = 3r + τ ) (obviously, this vector has 2(4r + τ + 3) components) as the sum of four nonintersecting vectors Yrτ l (l = 0, 1, 2, 3) with the following components: (4ν,4(r−ν)+τ +2)
Yrτ 0 : Y1 Yrτ 1 : Yrτ 2 : Yrτ 3 :
,
(4ν+3,4(r−ν)+τ −1)
Y2
;
(4ν+1,4(r−ν)+τ +1) (4ν,4(r−ν)+τ +2) Y1 , Y2 ; (4ν+2,4(r−ν)+τ ) (4ν+1,4(r−ν)+τ +1) Y1 , Y2 ; (4ν+3,4(r−ν)+τ −1) (4ν+2,4(r−ν)+τ ) Y1 , Y2 ,
where ν varies such that all indices are nonnegative. Corollary 1. For any r ≥ 0, the seventeen resonance equations (23r ) can be divided into two groups (a) and (b) as follows. Group (a) consists of nine equations: (232rµ ) (µ = 1, 2, 3, 4), (233r1 ), (233r4 ), (233r5 ), (235r3 ), and (235r4 ); each of these equations contains its own set of coefficients of system (3), namely, Yr2 µ−1 , Yr30 , Yr32 , Yr33 , Yr51 , and Yr53 , respectively. Group (b) consists of four pairs: (233r2 ) and (233r3 ), (234r1 ) and (234r2 ), (234r3 ) and (234r4 ), and (235r1 ) and (235r2 ); each of these pairs contains its own set of coefficients of system (3), namely, Yr31 , Yr41 , Yr42 , and Yr52 , respectively. Corollary 2. For any r ≥ 0, the following quantities do not have restrictions: the coefficient vectors Yr40 and Yr43 of the polynomial Y (4r+6) , the coefficient vector Yr50 of the polynomial Y (4r+7) of system (3), and (4r,3) (4r+3) of h1 in transformation (2). the coefficient h1 A coefficient of system (3) comprising at least one of Eqs. (23) with nonzero factor is said to be resonant; otherwise, it is said to be nonresonant. Contrarily, a coefficient of transformation (2) is said to be resonant if it has no restrictions. Thus, Corollary 2 specifies all resonant coefficients of transformation (2) and all nonresonant coefficient vectors of system (3). Unfortunately, we cannot assert that other coefficients of system (3) are resonant. A coefficient really comprises some equation (23) if its factor—a number a or b with indices defined jk pj jk by (21)—is nonzero. By (16), all these factors are nonzero except for bjk r,−1 and, perhaps, arν = αν brν /3, where p = 4r + j + k, r ≥ 1, j = 1, 2, 3, k = 1, 2, and 0 ≤ ν < r, since some constants ανpj = sjν + 3 − dpj /(p − sjν ) in (16) may vanish because the quantities dps are defined by recurrence formulas (14) sν and (11). For example, we calculate α0pj (see (16)). For any r ≥ 1, j > 0, and k = 1, 2 we have = a4r+k+j = (4r + k)(j + 3) + (4r + k + 1)j − 9; dpj = d4r+k+j j j s0
this implies α0pj = 9 − j(4r + k + 1)/(4r + k) > 0. The quantities α1pj , α2pj etc. can be found similarly. Thus, all factors of the coefficients of the polynomials Y (4r+τ +2) in Eqs. (232r3 ), (232r4 ), (233r5 ), (235r3 ), and (235r4 ) are nonzero since j = 0. On the other hand, by Corollary 1, all coefficients of system (3), which comprise the paired resonant equations of group (b), are also resonant since for ν > 0, they have jk the nonzero factor bjk rν−1 in one of these paired equations and for ν = 0, where br,−1 = 0, their factors
ajrrk are nonzero. As a result, we obtain the following consequence of Theorem 1, which generalizes the previous two corollaries. Corollary 3. For any r ≥ 0, all coefficients comprising the vectors Yr40 , Yr43 , and Yr50 are nonresonant, i.e., 3/16 of the total number of coefficients of the perturbed part of system (3). Without additional 1404
(4ν,4(r−ν)+4)
(4ν,4(r−ν)+4)
(4ν,4(r−ν)+5)
calculation, we cannot assert that the coefficients Y1 in Yr20 , Y2 in Yr21 , Y1 (4ν+1,4(r−ν)+4) in Yr30 , and Y2 in Yr32 are resonant. Other coefficients of the polynomials Y (4r+τ +2) (τ = 2, 3, 4, 5) in system (3) are resonant. Remark 1. The second equations in pairs of group (b) obtained from the first equations by using property (5) do not coincide with the first equations (unlike Eq. (233r5 ), which is invariant under property (5)) since only one of the paired equations really contains a coefficient of the series Y1 with ν = 0 and only another equation contains a coefficient of the series Y2 with ν = r + 1. We present some general definitions from [1] for the case of systems with zero approximations that are treated in this paper. (s ,p−sm +2) We fix p (see (22)). To any nτ coefficients Yimm of the homogeneous polynomial Y (4r+τ +2) , where m = 1, . . . , nτ , 0 ≤ sm ≤ p + 2, and 1 ≤ im ≤ 2, we put into correspondence a matrix Υp = (s ,p−sm +2) p p τ {υµm }nµ,m=1 whose element υµm vanishes if Yimm does not comprise Eq. (23τrµ ) or equals the (s ,p−s +2)
m corresponding factor of Yimm in (23τrµ ). The set of nτ coefficients of the homogeneous polynomial Y (4r+τ +2) is said to be resonant if det Υ4r+τ = 0. Thus, resonant equations (23τr ) are uniquely solvable with respect to any resonant set. System (3) is a generalized normal form if for any p ≥ 2, all coefficients of the polynomial Y (p+2) vanish except for the coefficients of some resonant set; these coefficients may take arbitrary values.
Corollary 4. System (3) is a generalized normal form if for any r ≥ 0, four of its homogeneous polynomials Y (4r+τ +2) (2 ≤ τ ≤ 5) contain not greater than seventeen nonzero coefficients that compose a resonant set. Among them, there are coefficients of nine vectors Yr20 , Yr21 , Yr22 , Yr23 , Yr30 , Yr32 , Yr33 , Yr51 , and Yr53 (one coefficient of each of these vectors), each of which comprises its own resonant equation (23r ) of group (a) with a nonzero factor, and coefficients of four vectors Yr31 , Yr41 , Yr42 , and Yr52 (two coefficients of each of these vectors). Factors of coefficients in (233r2 ) and (233r3 ), (234r1 ) and (234r2 ), (234r3 ) and (234r4 ), and (235r1 ) and (235r2 ) of group (b) form a nondegenerate (2 × 2)-matrix. Values of coefficients of this resonant set are arbitrary. The following theorem is obvious. Theorem 2. An arbitrary system (1) can be transformed into a generalized normal form (3) by using formal transformation (2). In the obtained generalized normal form, for any p ≥ 2 (see (22)), all coefficients of the generalized polynomial Y (p+2) vanish except for nτ coefficients of one of the resonant sets, which satisfy the resonant equations (23τr ). Example 1. Theorem 2 and Corollary 4 imply that any system (1) is formally equivalent to the following generalized normal form: ∞ (0,4r+4) 4 (2,4r+2) 2 2 (0,4r+5) 5 (1,4r+4) 3 Y1 y2 + Y1 y1 y2 + Y1 y2 + Y1 y1 y24 y˙ 1 = +y2 + r=0
+
(1,4r+5) Y1 y1 y25
y˙ 2 = −y13 +
∞
(2,4r+4) 2 4 y1 y2
+ Y1
(0,4r+4) 4 Y2 y2
+
(1,4r+6)
+ Y1
(1,4r+3) Y2 y1 y23
(2,4r+5) 2 5 y1 y2
y1 y26 + Y1
+
(0,4r+5) 5 Y2 y2
+
y24r ,
r=0
+
(2,4r+3) 2 3 Y2 y1 y2
+
(0,4r+6) 6 Y2 y2
+
(1,4r+5) Y2 y1 y25
+
(24)
(1,4r+4) Y2 y1 y24
(1,4r+6) Y2 y1 y26
+
(2,4r+5) 2 5 Y2 y1 y2
y24r .
Indeed, it is easy to see that nτ coefficients comprising system (24) for any r ≥ 0 and τ = 2, 3, 4, 5, form a resonant set and, therefore, their values can be uniquely obtained from the system of nτ resonant equations (23τr ) (of course, up to the choice of resonant coefficients of transformation (2), which are terms of constants c). 1405
(1,4r+4)
For example, factors of coefficients Y1 erate matrix since b21 r,−1
(0,4r+5)
in Eqs. (233r2 ) and (233r3 ) form a nondegen-
and Y2
21 a21 r,−1 br,−1 b21 a21 rr rr = 0. Therefore, for τ = 3, these coefficients belong to a resonant set; similarly, for τ = 5, the (2,4r+5)
(1,4r+6)
(1,4r+5)
and Y2 also belong to a resonant set. Two pairs of coefficients Y1 with coefficients Y1 (0,4r+6) (2,4r+4) (1,4r+5) 4 Y2 and Y1 with Y2 , comprising Eqs. (23r ), form a resonant set for τ = 4. For any r ≥ 0, generalized normal form (24) contains not greater than seventeen nonzero terms; the variable y1 comprising these terms has degree not greater than two. (2,4r+3) Note that for r ≥ 0, the coefficient Y2 in system (24) is chosen from the unique resonant equa3 tion (23r5 ). The reason for the appearance of this resonant equation is not that the number of variables in the linear system (4) connecting the coefficients of the series X, h, and Y is less than the number of equations, but that the principal minor of its matrix (or, equivalently, the determinant of the matrix Θp of the linear system (10)) vanishes. Example 2. Similarly to Example 1, we can assert that system (1) is formally equivalent to the following generalized normal form: y˙ 1 = y23 , y˙ 2 = −y13 + + +
∞
(0,4r+4) 4 y2
Y2
r=0 (0,4r+5) 5 Y2 y2 (0,4r+6) 6 Y2 y2
+ Y2
(0,4r+7) 7 y2
+ Y2
+ Y2
(1,4r+3)
+ Y2
(2,4r+2) 2 2 y1 y2
y1 y23 + Y2
(1,4r+4)
y1 y24 + Y2
+ Y2
(1,4r+5)
y1 y25 +
+
(1,4r+6)
y1 y26 + Y2
+ Y2
(2,4r+3) 2 3 y1 y2 (4,4r+2) 4 2 Y2 y1 y2 (2,4r+5) 2 5 y1 y2
(3,4r+2) 3 2 y1 y2 (5,4r+1) 5 Y2 y1 y 2 (5,4r+2) 5 2 y1 y2
+ Y2
(0,4r+5)
The coefficients of paired resonant equations, for example, Y2 set since their factors (see (233r )) form the matrix 21 br,−1 b21 4r+5 r0 . = Υ a21 a21 rr rr−1
(3,4r+1) 3 y1 y 2
+ Y2
(4,4r+1) 4 y1 y 2
+ Y2
(25)
y24r . (4,4r+1)
and Y2
, comprise a resonant
4r+5 = 0 since, by (21), the factors b21 and a21 are nonzero Although the quantity a21 rr rr−1 is unknown, det Υ r0 and b21 r,−1 = 0. Therefore, for any r ≥ 0, the first equation of system (25) has no perturbations and the right-hand side of the second equation contains not greater than seventeen nonzero terms whose orders are 4r + τ + 2 (2 ≤ τ ≤ 5); these terms contain y1 in powers not greater than five.
REFERENCES 1. V. V. Basov, “Generalized normal forms and formal equivalence of systems of differential equations with zero eigenvalues,” Differ. Equations (in press). 2. V. V. Basov and A. V. Skitovich, “Generalized normal forms and formal equivalence of two-dimensional systems with zero quadratic approximations,” Differ. Equations (in press). V. V. Basov St.-Petersbourg State University E-mail:
[email protected]
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