Expansions of solutions to the equation P₁² by algorithms of power geometry
Algorithms of Power Geometry allow to find all power expansions of solutions to ordinary differential equations of a rather general type. Among these, there are Painlev´e equations and their generalizations. In the article we demonstrate how to find by these algorithms all power expansions of soluti...
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Інститут прикладної математики і механіки НАН України
2009
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| Цитувати: | Expansions of solutions to the equation P₁² by algorithms of power geometry / A.D. Bruno, N.A. Kudryashov // Український математичний вісник. — 2009. — Т. 6, № 3. — С. 311-337. — Бібліогр.: 48 назв. — англ. |
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irk-123456789-1243622017-09-25T03:02:56Z Expansions of solutions to the equation P₁² by algorithms of power geometry Bruno, A.D. Kudryashov, N.A. Algorithms of Power Geometry allow to find all power expansions of solutions to ordinary differential equations of a rather general type. Among these, there are Painlev´e equations and their generalizations. In the article we demonstrate how to find by these algorithms all power expansions of solutions to the equation P₁² at the points z = 0 and z = ∞. Two levels of the exponential additions to the expansions of solutions near z = ∞ are computed. We also describe an algorithm of computation of a basis of a minimal lattice containing a given set. 2009 Article Expansions of solutions to the equation P₁² by algorithms of power geometry / A.D. Bruno, N.A. Kudryashov // Український математичний вісник. — 2009. — Т. 6, № 3. — С. 311-337. — Бібліогр.: 48 назв. — англ. 1810-3200 http://dspace.nbuv.gov.ua/handle/123456789/124362 2000 MSC. 34E05, 41A58, 41A60. en Український математичний вісник Інститут прикладної математики і механіки НАН України |
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Algorithms of Power Geometry allow to find all power expansions of solutions to ordinary differential equations of a rather general type. Among these, there are Painlev´e equations and their generalizations. In the article we demonstrate how to find by these algorithms all power expansions of solutions to the equation P₁² at the points z = 0 and z = ∞. Two levels of the exponential additions to the expansions of solutions near z = ∞ are computed. We also describe an algorithm of computation of a basis of a minimal lattice containing a given set. |
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Article |
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Bruno, A.D. Kudryashov, N.A. |
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Bruno, A.D. Kudryashov, N.A. Expansions of solutions to the equation P₁² by algorithms of power geometry Український математичний вісник |
| author_facet |
Bruno, A.D. Kudryashov, N.A. |
| author_sort |
Bruno, A.D. |
| title |
Expansions of solutions to the equation P₁² by algorithms of power geometry |
| title_short |
Expansions of solutions to the equation P₁² by algorithms of power geometry |
| title_full |
Expansions of solutions to the equation P₁² by algorithms of power geometry |
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Expansions of solutions to the equation P₁² by algorithms of power geometry |
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Expansions of solutions to the equation P₁² by algorithms of power geometry |
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expansions of solutions to the equation p₁² by algorithms of power geometry |
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Інститут прикладної математики і механіки НАН України |
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2009 |
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http://dspace.nbuv.gov.ua/handle/123456789/124362 |
| citation_txt |
Expansions of solutions to the equation P₁² by algorithms of power geometry / A.D. Bruno, N.A. Kudryashov // Український математичний вісник. — 2009. — Т. 6, № 3. — С. 311-337. — Бібліогр.: 48 назв. — англ. |
| series |
Український математичний вісник |
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AT brunoad expansionsofsolutionstotheequationp12byalgorithmsofpowergeometry AT kudryashovna expansionsofsolutionstotheequationp12byalgorithmsofpowergeometry |
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2025-07-09T01:19:18Z |
| last_indexed |
2025-07-09T01:19:18Z |
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1837130277649907712 |
| fulltext |
Український математичний вiсник
Том 6 (2009), № 3, 311 – 337
Expansions of solutions to the equation P 2
1 by
algorithms of power geometry
Alexander D. Bruno, Nikolai A. Kudryashov
(Presented by I. A. Lukovsky)
Abstract. Algorithms of Power Geometry allow to find all power ex-
pansions of solutions to ordinary differential equations of a rather general
type. Among these, there are Painlevé equations and their generaliza-
tions. In the article we demonstrate how to find by these algorithms all
power expansions of solutions to the equation P
2
1 at the points z = 0
and z = ∞. Two levels of the exponential additions to the expansions
of solutions near z = ∞ are computed. We also describe an algorithm
of computation of a basis of a minimal lattice containing a given set.
2000 MSC. 34E05, 41A58, 41A60.
Key words and phrases. Polygon of an equation, asymptotic form of
solution, power expansion of solution, exponential addition of the first
and the second level.
1. Introduction
In [1] a hierarchy of the first Painlevé equation was suggested. It can
be described by the relation
Ln[w] = z, (1.1)
where Ln is the Lenard’s operator determined by the relation [2]
d
dz
Ln+1 = Ln
zzz − 4wLn
z − 2wzL
n, L0[w] = −1
2
. (1.2)
Assuming that n = 0 in (1.2), we have L1[w] = w. In the case n = 2,
using (1.1), we obtain the first Painlevé equation [3]
wzz − 3w2 = z. (1.3)
Received 26.06.2009
The work was partly supported by RFBR, grant 08-01-00082.
ISSN 1810 – 3200. c© Iнститут математики НАН України
312 Expansions of solutions...
If n = 3 in (1.1), we obtain the fourth-order equation [1, 2]
wzzzz − 10w wzz − 5w2
z + 10w3 = z. (1.4)
Using n = 4, we obtain the sixth-order equation from (1.1)
wzzzzzz − 14wwzzzz − 28 wz wzzz − 21 w2
zz + 70ww2
z − 35 w4 = z. (1.5)
In general, equation (1.1) has order 2(n − 1).
Equation (1.4) is used in describing the waves on water [4, 5] and in
the Hénon–Heiles model that characterizes the behavior of a star in the
mean field of galaxy [6–8].
In papers [9–23], it was shown that equation (1.4) has properties that
are typical for the Painlevé equations P1 ÷ P6. Equation (1.4) belongs
to the class of exactly solvable equations, since it has the Lax pair and
many other typical properties of the exactly solvable equations. It does
not have the first integral in the polynomial form. It has the Painlevé
property [41]. Equation (1.4) seems to determine new transcendental
functions just as equations P1 ÷ P6, although the rigorous proof of the
irreducibility of equation (1.4) remains an open problem. Thereupon the
study of all the asymptotic forms and asymptotic expansions of solutions
to equation (1.4) is the important stage of the analysis of this equation,
as they can indirectly confirm the irreducibility of equation (1.4).
Recently developed methods of Power Geometry [24,25] allow to ob-
tain algorithmically asymptotic expansions of solutions for a wide class
of ordinary differential equations (see also [26–29]). Further development
of these methods see in [30, 31]. Applications of these methods for ob-
taining asymptotic expansions of solutions to the Painlevé equations see
in [32–40].
We remark that algorithms in the articles [25,30,31] allow to find all
power expansions as well as power-logarithmic, complicated and exotic
ones. In particular, this made it possible to find all local and asymptotic
expansions of solutions to the sixth Painlevé equation [40].
The power expansion of a function is an expansion over simpliest
functions (power functions). It allows to clarify local properties of the
expanded function. The Taylor and Laurent series are not for nothing
used in Analysis. Certainly, expansions over other functions such as ra-
tional or elliptic ones are possible, but such expansions are not so conve-
nient for local analysis of the expanded function and they are used not
so frequently.
The main goal of this article is to demonstrate the application of these
methods to the simplest equation, which is different from the Painlevé
equations, i.e. to the equation (1.4), that is often denoted as P 2
1 .
A. D. Bruno, N. A. Kudryashov 313
Let us find all the power expansions for the solution to equation (1.4)
in the form of
w(z) = cr zr +
∑
cs zs, s ∈ K; (1.6)
as z → 0, then ω
def
= − 1, s > r; and as z → ∞, then ω
def
= 1, s < r.
We use notions and notation from [25] or [26–29], but here we have
no place to repeat them.
2. The general properties of equation (1.4)
Let us write the fourth-order equation (1.4) in the form
f(z, w)
def
= wzzzz − 10wwzz − 5w2
z + 10w3 − z = 0. (2.1)
Monomials of equation (2.1) have vectorial power exponents
M1 = (−4, 1), M2 = (−2, 2), M3 = (−2, 2),
M4 = (0, 3), M5 = (1, 0).
The support of the equation consists of four points
Q1 = M1, Q2 = M4, Q3 = M5 and Q4 = M2 = M3. (2.1′)
Their convex hull Γ is the triangle (Fig. 1).
Figure 1
314 Expansions of solutions...
This triangle has vertexes Qj (j = 1, 2, 3) and edges Γ
(1)
1 = [Q3, Q1],
Γ
(1)
2 = [Q1, Q2], Γ
(1)
3 = [Q2, Q3].
Outward normal vectors Nj (j = 1, 2, 3) of edges Γ
(1)
j (j = 1, 2, 3) are
N1 = (−1,−5), N2 = (−1, 2), N3 = (3, 1). (2.2)
The normal cones U
(1)
j of edges Γ
(1)
j are rays
U
(1)
j = µNj , µ > 0, j = 1, 2, 3. (2.3)
They and the normal cones U
(0)
j of vertices Γ
(0)
j = Qj (j = 1, 2, 3) are
represented in Fig. 2.
Figure 2
If the support (2.1′) of equation (2.1) is shifted by vector −Q3, then
it is situated at the lattice Z, formed by vectors
Q1 − Q3 = (−5, 1), Q4 − Q3 = (−3, 2). (2.4)
We choose the basis of the lattice as
B1 = (−5, 1), B2 = (−3, 2). (2.5)
A. D. Bruno, N. A. Kudryashov 315
To each face Γ
(d)
j there corresponds the truncated equation f̂
(d)
j (z, w) =
0 where the sum f̂
(d)
j (z, w) contains only such monomials of the sum
f(z, w) in (2.1) whose vectorial power exponents Mk belong to the face
Γ
(d)
j . If expansion (1.6) is a solution to equation (2.1), then its first term
w = crw
r is a solution of the truncated equation f̂
(d)
j (z, w) = 0, if vector
ω(1, r) ∈ U
(d)
j [25]. Thus, to each face Γ
(d)
j there corresponds a set of
families of expansions (1.6) of solutions to equation (2.1).
Let us study solutions corresponding to the faces Γ
(d)
j , d = 0, 1; j =
1, 2, 3 taking into account the truncated equations corresponding to ver-
tices Γ
(0)
j (j = 1, 2, 3)
f̂
(0)
1
def
= wzzzz = 0, (2.6)
f̂
(0)
2
def
= 10w3 = 0, (2.7)
f̂
(0)
3
def
= −z = 0. (2.8)
and truncated equations corresponding to edges Γ
(1)
j (j = 1, 2, 3)
f̂
(1)
1
def
= wzzzz − z = 0, (2.9)
f̂
(1)
2
def
= wzzzz − 10 w wzz − 5w2
z + 10w3 = 0, (2.10)
f̂
(1)
3
def
= 10w3 − z = 0. (2.11)
Note that the truncated equations (2.7) and (2.8) are algebraic ones.
According to [25] they do not have non-trivial solutions.
3. Solutions, corresponding to the vertex Q1
The vertex Q1 = (−4, 1) corresponds to the truncated equation (2.6).
Let us find the truncated solutions
w = crz
r, cr 6= 0 (3.1)
with ω(1, r) ∈ U
(0)
1 .
Since p1 < 0 in the cone U
(0)
1 , then ω = −1, z → 0, and the expan-
sions are by ascending powers of z. The dimension of the face d = 0,
therefore
g(z, w) = w4 w−1 wzzzz. (3.2)
We obtain the characteristic polynomial
χ(r)
def
= g(z, zr) = r(r − 1)(r − 2)(r − 3). (3.3)
316 Expansions of solutions...
Its roots are
r1 = 0, r2 = 1, r3 = 2, r4 = 3 (3.4)
Let us explore all these roots. According to Fig. 2, all vectors ω(1, ri)
belong to the normal cone U
(0)
1 (i = 1, 2, 3, 4).
For the root r1 = 0, we obtain the family F (1)
1 1 of truncated solutions
y = c0, where c0 6= 0 is arbitrary constant. The first variation of equation
(2.6)
δf̂
(0)
1
δw
=
d4
dz4
(3.5)
gives the operator
L(z) =
d4
dz4
6= 0. (3.6)
Its characteristic polynomial is
ν(k) = z4−kL(z) zk = k(k − 1)(k − 2)(k − 3). (3.7)
Equation
ν(k) = 0 (3.8)
has four roots
k1 = 0, k2 = 1, k3 = 2, k4 = 3. (3.9)
Since ω = −1 and r = 0, then the cone of the problem is
K = {k > 0}. (3.10)
It contains the critical numbers k2 = 1, k3 = 2, and k4 = 3. Expansions
(1.6) for the solutions corresponding to the truncated solution (3.1) take
the form
w = c0 + c1z + c2z
2 + c3z
3 +
∞
∑
k=4
ckz
k, (3.11)
where all the coefficients are constants, c0 6= 0, c1, c2, c3 are arbitrary
ones, and ck (k ≥ 4) are uniquely determined. Denote this family as
G(0)
1 1. Expansion (3.11) with eight terms is
w(z) = c0 + c1 z + c2 z2 + c3 z3 +
(5
6
c0c2 −
5
12
c0
3 +
5
24
c1
2
)
z4
+
( 1
120
α − 1
4
c1c0
2 +
1
2
c0c3 +
1
3
c1c2
)
z5
+
( 7
36
c2c0
2 − 5
36
c0
4 − 1
72
c0c1
2 +
1
9
c2
2 +
1
4
c1c3
)
z6
A. D. Bruno, N. A. Kudryashov 317
+
( 1
36
c1
3 +
1
504
c0 +
1
12
c3c0
2 +
1
6
c0c1c2 −
5
36
c1c0
3 +
1
6
c2c3
)
z7 + · · ·
(3.12)
Similarly, for each root rj+1 = j, j = 1, 2, 3, the cone of the problem
is K = {k > j}, the truncated solution F (0)
1 (j + 1): w = cjz
j and the
expansion is (3.11), (3.12) with ci = 0, i < j, cj 6= 0, and cj , . . . , c3 are
arbitrary constants. Denote this family as G(0)
1 (j + 1). For j = 2, the
expansion (3.11) with eight terms is
w(z) = c2z
2 + c3z
3 +
1
120
z5 +
1
9
c2
2z6 +
1
6
c2c3z
7 +
1
16
c3
2z8
+
1
1134
c2 z9 +
( 5
648
c2
3 +
41
60480
c3
)
z10 + · · · (3.13)
The expansions of solutions converge for sufficiently small |z|. The
existence and analyticity of expansions (3.11), (3.12), (3.13) follows from
the Cauchy theorem.
4. Solutions corresponding to the edge Γ
(1)
1
The edge Γ
(1)
1 corresponds to the truncated equation
f̂
(1)
1 (z, y)
def
= yzzzz − z = 0. (4.1)
The normal cone is
U
(1)
1 = {−µ(1, 5), µ > 0}. (4.2)
Therefore ω = −1, i.e. z → 0, and r = 5. We are looking for power
solutions to (4.1) in the form
w = c5z
5.
For c5 we have equation 120c5 − 1 = 0, i.e.
c5 =
1
120
. (4.3)
The only power solution is
F (1)
2 1 : w =
z5
120
. (4.4)
Compute the critical numbers. The first variation of (2.9) is
δf̂
(1)
1
δw
=
d4
dz4
. (4.5)
318 Expansions of solutions...
We obtain the eigenvalues (3.9)
k1 = 0, k2 = 1, k3 = 2, k4 = 3. (4.6)
The cone of the problem
K = {k > 5}
does not contain them. Solution (4.4) corresponds to two vectorial power
exponents Q̃1 = (0, 1), Q̃2 = (5, 0). Their difference B = Q̃1 − Q̃2 =
(−5, 1) equals the vector Q1 − Q3. So the solution (4.4) corresponds
to the lattice Z which consists of the points Q = (q1, q2) = k(−3, 2) +
m(−5, 1) = (−3k − 5l, 2k + l), where k and l are integers. These points
lie on the line q2 = −1, if l = −1 − 2k. In this case q1 = 5 + 7k. Since
the cone of the problem here is K = {k > 5}, then the support K of the
solutions expansion takes the form
K = {5 + 7n, n ∈ N}. (4.7)
Then the expansion of the solution can be written as
w(z) = z5
(
1
120
+
∞
∑
m=1
c5+7m z7m
)
. (4.8)
Expansion (4.8) with three terms takes the form
w(z) =
z5
120
(
1 +
13 z7
57024
+
2851 z14
79569008640
+ · · ·
)
(4.9)
and coincides with expansion (3.13) with c2 = c3 = 0. It does not have
exponential additional terms. Equation (4.1) does not give non-power
asymptotics as well.
5. Solutions corresponding to the edge Γ
(1)
2
The edge Γ
(1)
2 corresponds to the truncated equation
f̂
(1)
2 (z, w)
def
= wzzzz − 10 w wzz − 5 w2 + 10w3 = 0. (5.1)
The normal cone is
U
(1)
2 = {−µ(1,−2), µ > 0}. (5.2)
Therefore ω = −1, i.e. z → 0, and r = −2. Hence the solution to
equation (5.1) must be found in the form
w = c−2z
−2. (5.3)
A. D. Bruno, N. A. Kudryashov 319
For c−2 we have the determining equation
c2
−2 − 8 c−2 + 12 = 0. (5.4)
Consequently we obtain
c
(1)
−2 = 2, c
(2)
−2 = 6. (5.5)
The truncated solutions are
F (1)
2 1 : w = 2z−2, (5.6)
F (1)
2 2 : w = 6z−2. (5.7)
Let us compute the corresponding critical numbers. The first variation is
δf̂
(1)
2
δw
=
d4
dz4
− 10wzz − 10w
d2
dz2
− 10 wz
d
dz
+ 30w2. (5.8)
Substituting the solution (5.6), it produces operator
L(1)(z) =
d4
dz4
− 20
z2
d2
dz2
+
40
z3
d
dz
, (5.9)
which corresponds to the characteristic polynomial
ν(k) = k4 − 6k3 − 9k2 + 54k. (5.10)
Equation
ν(k) = 0 (5.11)
has the roots
k1 = −3, k2 = 0, k3 = 3, k4 = 6. (5.12)
For the solution (5.7), the variation (5.8) gives the operator
L(2)(z) =
d4
dz4
+
60
z2
d2
dz2
− 120
z3
d
dz
+
720
z4
, (5.13)
which corresponds to the characteristic polynomial
ν(k) = k4 − 6k3 − 49k2 + 174k + 720 (5.14)
with roots
k1 = −5, k2 = −3, k3 = 6, k4 = 8. (5.15)
The cone of the problem here is
K = {k > −2}. (5.16)
320 Expansions of solutions...
Therefore the truncated solution (5.6) has three critical numbers k2 = 0,
k3 = 3, k4 = 6; and there are two critical numbers k3 = 6, k4 = 8 for the
truncated solution (5.7) in the cone of the problem.
Let us consider expansions of solutions beginning with (5.6). Similarly
to Section 4, here the shifted support of the truncated solution (5.6)
belongs to the lattice Z. Hence, similarly to (4.7) we have
K = {−2 + 7n, n ∈ N}. (5.17)
The sets K(0), K(0, 3), and K(0, 3, 6) are
K(0) = {−2 + 7n + 2m, n, m ∈ N ∪ {0}, n + m > 0}
= {0, 4, 6, 5, 7, 8, . . . }, (5.18)
K(0, 3) = {−2 + 7n + 2m + 5k, n, m, k ∈ N ∪ {0}, m + n + k > 0}
= {0, 2, 3, 4, 5, 6, 7, 8, . . . }, (5.19)
K(0, 3, 6) = {−2 + 7n + 2m + 5k + 8l, n, m, k, l ∈ N ∪ {0},
m + n + k + l > 0} = {0, 2, 3, 4, 5, 6, 7, 8, . . . }. (5.20)
Here the expansion of the solution to the equation (2.1) can be written
as
w(z) =
2
z2
+
∑
n,m,k,l
c5+7n+2m+2k+8l z
5+7n+2m+2k+8l. (5.21)
Denote this family as G(1)
2 1. The critical number 0 does not belong to the
set K, so the compatibility condition for c0 holds automatically, and c0 is
an arbitrary constant. The critical number 3 also does not belong to set
K(0), therefore the compatibility condition for c3 holds as well, and c3
is an arbitrary constant. But the critical number 6 belong to sets K(0)
and K(0, 3), so it is necessary to verify that the compatibility condition
for c6 holds and that c6 is an arbitrary constant. The calculation shows
that here the condition holds and c6 is an arbitrary constant as well.
The three-parameter power expansion of solutions corresponding to the
truncated solution (5.6) takes the form
w(z) =
2
z2
+ c0 −
3
2
c0
2z2 + c3z
3 − 5
2
c0
3z4
+
(3
4
c0c3 −
1
80
)
z5 + c6z
6 − 1
280
c0 z7
+
(153
352
c0
5 +
9
44
c0c6 +
9
176
c3
2
)
z8 +
( 19
12096
c0
2 − 5
16
c0
3c3
)
z9
A. D. Bruno, N. A. Kudryashov 321
+
( 25
104
c0
6 − 29
29120
c3 −
3
26
c0
2c6 +
3
52
c0c3
2
)
z10 + · · · (5.22)
The support of the power expansion corresponding to the truncated
solution (5.7), is determined by the sets
K(6) = {−2 + 7n + 8m, n, m ∈ N ∪ {0}, m + n > 0}
= {5, 6, 12, 14, 20, 21, 22, 27, 28, 29, 30, 34, 35, 36, 37, 38, 41, . . . }, (5.23)
K(6, 8) = {−2 + 7n + 8m + 10k, n, m, k ∈ N ∪ {0}, m + n + k > 0}
= {5, 6, 8, 12, 13, 14, 15, 16, 18, 19, 20, 21, . . . }. (5.24)
The expansion of solution to the equation can be written as
w(z) =
6
z2
+
∑
n,m,k
c5+7n+8m+10k z5+7n+8m+10k. (5.25)
Denote this family as G(1)
2 2. The critical numbers 6 and 8 do not belong
to the set K, and the number 8 does not belong to the set K(6). For
numbers 6 and 8 the compatibility conditions holds automatically, there-
fore the coefficients c6 and c8 are arbitrary constants. The two-parameter
expansion of solution corresponding to the truncated solution (5.7) is
w(z) =
6
z2
+
1
240
z5 + c6z
6 + c8z
8 +
29
70502400
z12+
+
11
60480
c6z
13 +
25
1292
c6
2z14 +
1
6804
c8z
15 + · · · (5.26)
According to [25, §§7,5], the expansions of solutions (5.25) and (5.26)
do not have exponential additions and equation (5.1) does not give non-
power asymptotics.
In Sections 3–5, we obtained all power expansions of solutions to
the equation (1.4) in the neighborhood of the origin z = 0. All these
expansions were Taylor or Laurent series. According to [25], they all are
convergent in some (punctured) neighborhood of the origin. Similarly,
we can obtain all power expansions of solutions at an arbitrary point
z = z0 6= 0. For this we need to introduce a new independent variable
z̃ = z − z0. Then the equation (1.4) would take the form
w(4) − 10ww′′ − 5w′2 + 10w3 − z̃ − z0 = 0. (5.27)
It differs from the equation (2.1) only by the term −z0, which corresponds
to the new point Q5 = (0, 0) of the support. This point also is a vertex
322 Expansions of solutions...
of a new polygon; other its vertices are Q1, Q2, Q3. Using the described
technique, we can obtain expansions of solutions to the equation (5.27)
as z̃ → 0. These expansions are analogous to the already found ones,
and for them, the consistency conditions for the critical power exponents
are satisfied identically with respect to z0. The solutions to the equation
(1.4) do not have movable singularities, i.e. the equation (1.4) possesses
the Painlevé property [41].
6. Solutions corresponding to the edge Γ
(1)
3
The edge Γ
(1)
3 corresponds to the truncated equation
f̂
(1)
3 (z, w)
def
= 10w3 − z = 0. (6.1)
It has three power solutions
F (1)
3 1 : w = ϕ(1)(z) = c
(1)
1/3 z1/3, c
(1)
1/3 =
( 1
10
)1/3
; (6.2)
F (1)
3 2 : w = ϕ(2)(z) = c
(2)
1/3 z1/3, c
(2)
1/3 =
(1
2
+ i
√
3
)( 1
10
)1/3
; (6.3)
F (1)
3 3 : w = ϕ(3)(z) = c
(3)
1/3 z1/3, c
(3)
1/3 =
(1
2
− i
√
3
)( 1
10
)1/3
. (6.4)
The shifted support of the truncated solutions (6.2)–(6.4) gives the vector
B =
(1
3
,−1
)
, (6.5)
which equals the third part of the vector Q2 − Q3.
Let us find a basis of the lattice generated by vectors B1, B2 from
(2.5) and vector B. According to the algorithm of Appendix, we compute
determinants D1 = |B1B2|, D2 = |B1B|, D3 = |B2B| :
D1 =
∣
∣
∣
∣
−5 1
−3 2
∣
∣
∣
∣
= −7, D2 =
∣
∣
∣
∣
−5 1
1/3 −1
∣
∣
∣
∣
=
14
3
,
D3 =
∣
∣
∣
∣
−3 2
1/3 −1
∣
∣
∣
∣
=
7
3
.
Since |D3| is the minimum of absolute values and other determinants
are its multiplies, then vectors B2 and B form a basis of the lattice
corresponding to solutions (6.2)–(6.4). Points of the lattice have the
form
Q = (q1, q2) = l(−3, 2) + m(1/3,−1) = (−3l + m/3, 2l − m),
A. D. Bruno, N. A. Kudryashov 323
where l, m are integer numbers. On the line q2 = −1 we have 2l−m = −1.
Hence, m = 2l+1 and q1 = (2l+1−9l)/3. So the support of the solution
is
K =
{
k =
1 − 7n
3
, n ∈ N
}
= {−2,−11/3, . . .}, (6.6)
and the expansions of solutions take the form
G(1)
3 l : w = ϕ(l)(z) = c
(l)
1/3z
1/3 +
∞
∑
n=1
c
(l)
(1−7n)/3 z(1−7n)/3. (6.7)
Here c
(l)
1/3 are given in (6.2)–(6.4); coefficients c
(l)
(1−7n)/3 are computed
sequentially. The expansion of the solution with five terms is
ϕ(l)(z) = c1/3 z1/3 − 1
18
z−2 − 7
108
1
c1/3
z−13/3
− 4199
17496
c1/3
−2 z−20/3 − 28006583
23514624
1
c1/3
3
z−9 + · · · (6.8)
The obtained expansions seem to diverge [25, §7].
7. Exponential additions of the first level
Let us find the exponential additions [25, §7] to solutions (6.7). We
look for the solutions in the form
w = ϕ(l)(z) + u(l), l = 1, 2, 3.
The truncated equation for the addition u(l) is
M(1)
l (z)u(l) = 0, (7.1)
where M(1)
l (z) is the first variation of f(z, w) from (2.1) at the solution
w = ϕ(l)(z). As we have
δf
δw
=
d4
dz4
− 10wzz − 10w
d2
dz2
− 10wz
d
dz
+ 30w2, (7.2)
then
M(1)
l (z) =
d4
dz4
− 10ϕ(l)
zz − 10ϕ(l) d2
dz2
− 10ϕ(l)
z
d
dz
+ 30ϕ(l)2. (7.3)
Equation (7.1) takes the form
324 Expansions of solutions...
d4u(l)
dz4
− 10ϕ(l)
zzu(l) − 10ϕ(l)
d2u(l)
dz2
− 10ϕ(l)
z
du(l)
dz
+ 30ϕ(l)2u(l) = 0,
l = 1, 2, 3. (7.4)
Denote
ζ(l) =
d lnu(l)
dz
, (7.5)
then from (7.5) we have
du(l)
dz
= ζ(l)u(l),
d2u(l)
dz2
= ζ(l)
z u(l) + ζ(l)2u(l),
d3u(l)
dz3
= ζ(l)
zz u(l) + 3ζ(l)ζ(l)
z u(l) + ζ(l)3u(l),
d4u(l)
dz4
= ζ(l)
zzzu
(l) + 4ζ(l)ζ(l)
zz u(l) + 3ζ(l)
z
2
u(l) + 6ζ(l)2ζ(l)
z u(l) + ζ(l)4u(l).
Substituting the derivatives
du(l)
dz
,
d2u(l)
dz2
,
d4u(l)
dz4
into equation (7.4) we obtain the equation in the form
u(l)[ζ(l)
zzz + 4ζ(l)ζ(l)
zz + 3ζ(l)
z
2
+ 6ζ(l)2ζ(l)
z + ζ(l)4 − 10ϕ(l)
zz
− 10ϕ(l)ζ(l)
z − 10ϕ(l)ζ(l)2 − 10ϕ(l)
z ζ(l) + 30ϕ(l)2] = 0. (7.6)
Let us find the power expansions for solutions to the equation (7.6).
The support of the equation (7.6), divided by u(l), consists of points
Q1 = (−3, 1), Q2 = (−2, 2), Q3 = (−1, 3),
Q4 = (0, 4), Q5 =
(1
3
, 2
)
, Q6 =
(2
3
, 0
)
, Q7 =
(
− 2
3
, 1
)
,
Q8 =
(
− 5
3
, 0
)
, Q8+n =
(
− 5 + 7n
3
, 0
)
,
Q9+m =
(
− 2 + 7m
3
, 1
)
,
Q10+k =
(1 − 7k
3
, 2
)
, Q11+k =
(12 − 14l
3
, 0
)
, m, n, k, l ∈ N.
(7.7)
The closure of the convex hull of the points of the support of the equation
(7.6) is the band in Fig. 3.
A. D. Bruno, N. A. Kudryashov 325
Figure 3
The boundary of the band contains the edges Γ
(1)
j (j = 1, 2, 3) with the
normal vectors N1 = (6, 1), N2 = (0,−1), N3 = (0, 1). We must consider
the edge Γ
(1)
1 only. This edge corresponds to the truncated equation
h
(1)
1 (z, ζ)
def
= ζ4 − 10ϕ̂(l)ζ2 + 30ϕ̂(l)2 = 0, (7.8)
where ϕ̂(l) = c
(l)
1/3z
1/3 from (6.2)–(6.4). Wherefrom we have
ζ2 =
(
5 + (−1)m−1i
√
5
)
ϕ̂(l), m = 1, 2. (7.9)
We obtain twelve solutions to the equation (7.8)
ζ(l,m,k) = g
(l,m,k)
1/6 z1/6, l = 1, 2, 3; m, k = 1, 2, (7.10)
where
g
(1,m,k)
1/6 =
( 1
10
)1/3
(−1)k−1
(
5 + (−1)m−1i
√
5
)1/2
, m, k = 1, 2, (7.11)
g
(2,m,k)
1/6 =
(1
2
+i
√
3
)( 1
10
)1/3
(−1)k−1
(
5+(−1)m−1i
√
5
)1/2
, m, k = 1, 2,
(7.12)
g
(3,m,k)
1/6 =
(1
2
−i
√
3
)( 1
10
)1/3
(−1)k−1
(
5+(−1)m−1i
√
5
)1/2
, m, k = 1, 2.
(7.13)
326 Expansions of solutions...
The truncated equation (7.8) is an algebraic one, so it gives no critical
numbers. Let us compute the support of the expansion for solution to
the equation (7.6). The shifted support of equation (7.6) is contained in
the lattice generated by vectors B̃1 = (7/3, 0) , B̃2 = (1, 1). The shifted
support of solutions (7.10) gives the vector B̃3 = (−1/6, 1).
According to Appendix, we compute the determinants |B̃1B̃2|= 7/3=
|B̃1B̃3|, |B̃2B̃3| = 7/6. Hence, the vectors B̃2, B̃3 or B̃2, B̃4
def
= B̃2−B̃3 =
(7/6, 0) form a basis of the lattice containing B̃1, B̃2, B̃3. The points of
this lattice can be written as
Q = (q1, q2) = k(1, 1) + m
(7
6
, 0
)
=
(
k +
7m
6
, k
)
.
On the line q2 = −1 we have k = −1, and so q1 = −1 + 7m/6. Since
the cone of the problem here is K = {k < 1/6}, then the support K of
expansions is
K =
{1 − 7n
6
, n ∈ N
}
. (7.14)
The expansion of the solution to the equation (7.6) takes the form
ζ(l,m,k) = g
(l,m,k)
1/6 z1/6 +
∑
n
g
(l,m,k)
(1−7n)/6 z(1−7n)/6,
l = 1, 2, 3; m = 1, 2; k = 1, 2. (7.15)
Coefficients g
(l,m,k)
1/6 are determined by expressions (7.11), (7.12), and
(7.13). The expansion of the solution with four terms takes the form
ζ(l,m,k) = g1/6z
1/6 − 1
4
z−1 − 7
288
(
30 g1/6
2 − 7 102/3
)
g1/6
(
2 g1/6
2 − 102/3
)z−13/6
− 49
1728
30 g1/6
4 − 6 102/3g1/6
2 + 35 3
√
10
g1/6
2
(
5 3
√
10 − 2 102/3 g1/6
2 + 2 g1/6
4
)z−10/3 + · · · (7.16)
In view of (7.5) we can find the additions u(l,m,k)(z). We have
u(l,m,k)(z) = C exp
∫
ζ(l,m,k)(z) dz. (7.17)
Wherefrom we obtain
A. D. Bruno, N. A. Kudryashov 327
u(l,m,k)(z)
= C1 z−1/4 exp
[
6
7
g
(l,m,k)
1/6 z7/6 +
∞
∑
n=2
6
7(1 − n)
g
(l,m,k)
(1−7n)/6z
7(1−n)/6
]
,
l = 1, 2, 3; m = 1, 2; k = 1, 2. (7.18)
Here C1 and further C2 and C3 are arbitrary constants. The addition
u(l,m,k)(z) near z → ∞ is exponentially small in those sectors of the
complex plane z where
Re
[
g
(l,m,k)
1/6 z7/6
]
< 0. (7.19)
Thus for each of three expansions G(1)
3 l we obtain four one-parameter
families of additions G(1)
3 lG(1)
1 mk, where l = 1, 2, 3, m = 1, 2 and k = 1, 2.
8. Exponential additions of the second level
Let us find exponential additions of the second level v(p), i.e. the
additions to ζ(l,m,k)(z). The truncated equation for the addition v(p) is
M(2)
p (z)v(p) = 0, (8.1)
where the operator M(2)
p is the first variation of the square brackets in
(7.6). Equation (8.1) for v = v(p) after substitution d ln v/dz = ξ takes
the form
ξzz + 3 ξ ξz + ξ3 + 4ζzz + 4ξz ζ + 4ξ2 ζ + 6 ξ ζz + 12ζζz
+ 6 ξ ζ2 + 4 ζ3 − 10 ϕ(l)ξ − 20 ζ ϕ(l) − 10 ϕ(l)
z = 0. (8.2)
Monomials of the equation (8.2) have the following power exponents
M1 = (−2, 1), M2 = (−1, 2), M3 = (0, 3), M4 =
(
− 11
6
, 0
)
,
M5 =
(
− 5
6
, 1
)
, M6 =
(1
6
, 2
)
, M7 =
(
− 5
6
, 1
)
,
M8 =
(
− 2
3
, 0
)
, M9 =
(1
3
, 1
)
, M10 =
(1
2
, 0
)
,
M11 =
(1
3
, 1
)
, M12 =
(1
2
, 0
)
, M13 =
(
− 2
3
, 0
)
, . . .
(8.3)
The support of the equation (8.2) consists of points of the set (8.3). Its
convex hull forms the band, which is similar to the band represented
328 Expansions of solutions...
in Fig. 3. We must examine the edge Γ
(1)
1 , passing through the points
Q1 = (1/2, 0), Q2 = (1/3, 1), Q3 = (0, 3).
The truncated equation corresponding to this edge is
ξ3 + 4 ξ2 ζ + 6 ξ ζ2 + 4 ζ3 − 20 ζ ϕ(l) − 10 ξ ϕ(l) = 0. (8.4)
The basis of the lattice corresponding to the support of the equation (8.2)
is
B1 = (1, 1), B2 =
(7
6
, 0
)
.
The solutions to the equation (8.4) take the form
ξ(l,m,k,p) = r
(l,m,k,p)
1/6 z1/6, m, k = 1, 2; l = 1, 2, 3; p = 1, 2, 3, (8.5)
where r = r
(l,m,k,p)
1/6 , p = 1, 2, 3 are the roots of the equation
5 r3 + 4 r2 g
(l,m,k)
1/6 +
(
6 g
(l,m,k)2
1/6 − 10 c
(l)
1/3
)
r
+ 4 g
(l,m,k)
1/6
3
− 20 g
(l,m,k)
1/6 c
(l)
1/3 = 0. (8.6)
Equation (8.6) has the roots
r
(l,m,k,1)
1/6 = −2 g
(l,m,k)
1/6 ,
r
(l,m,k,2)
1/6 = −g
(l,m,k)
1/6 +
(
10 c
(l)
1/3 − g
(l,m,k)
1/6
)1/2
,
r
(l,m,k,3)
1/6 = −g
(l,m,k)
1/6 −
(
10 c
(l)
1/3 − g
(l,m,k)
1/6
)1/2
.
(8.7)
The support K of expansions for the solution coincides with (7.14). The
expansion of solution for ξ(l,m,k,p) takes the form
ξ(l,m,k,p) = r
(l,m,k,p)
1/6 z1/6 +
∞
∑
n=1
r
(l,m,k,p)
(1−7n)/6 z(1−7n)/6,
l = 1, 2, 3; m = 1, 2; k = 1, 2; p = 1, 2, 3. (8.8)
The expansion of the solution with three terms is
ξ(l,m,k,p) = r1/6z
1/6 +
1
6
z−1 + (−30 g4
1/6 − 9 102/3g2
1/6
+ 150 g2
1/6c1/3 − 35 c1/3102/3 − 30 r2
1/6g
2
1/6 + 7 r2
1/6102/3
− 60 g3
1/6r1/6 + 6 r1/6g1/6102/3)(102/3 − 2 g2
1/6)
−1
× (6 g2
1/6 − 10 c1/3 + 3 r2
1/6 + 8 g1/6r1/6)
−1g−1
1/6z
− 13
6 . (8.9)
A. D. Bruno, N. A. Kudryashov 329
The exponential addition v(l,m,k,p)(z) to ζ(l,m,k)(z) is
v(l,m,k,p)(z)
= C2 z1/6 exp
[
6
7
r
(l,m,k,p)
1/6 z7/6 +
∞
∑
n=2
6
7(1 − n)
r
(l,m,k,p)
(1−7n)/6z
7(1−n)/6
]
,
l = 1, 2, 3; m = 1, 2; k = 1, 2; p = 1, 2, 3. (8.10)
Solutions v(l,m,k,p)(z) seem to diverge as well.
Thus for each one-parameter family of additions G(1)
3 lG(1)
1 mk of the
first level we obtain 3 families of additions G(1)
3 lG(1)
1 mkG(1)
1 p, where p =
1, 2, 3, of the second level.
9. Exponential additions of the third level
Similarly we obtain that the exponential addition y(s,p,l,m,k)(z) to the
log v(l,m,k,p)(z) is
y(l,m,k,p,s)(z)
= C3 z1/6 exp
[
6
7
q
(l,m,k,p,s)
1/6 z7/6 +
∞
∑
n=2
6
7(1 − n)
q
(l,m,k,p,s)
(1−7n)/6 z7(1−n)/6
]
l = 1, 2, 3; m = 1, 2; k = 1, 2; p = 1, 2, 3; s = 1, 2, (9.1)
where
q
(l,m,k,p,s)
1/6 = −3
2
r
(l,m,k,p)
1/6 − 2 g
(l,m,k)
1/6
+ (−1)s−1
(9
4
r
(l,m,k,p)
1/6
2
− 2 r
(l,m,k,p)
1/6 g
(l,m,k)
1/6 − 2 g
(l,m,k)
1/6
2
+ 10 c
(l)
1/3
)1/2
,
l = 1, 2, 3; m, k = 1, 2; p = 1, 2, 3; s = 1, 2. (9.2)
Thus we have found three levels of exponential additions to the ex-
pansions of solutions to the equation near the point z = ∞. Solution
w(z) with exponential additions as z → ∞ has the expansion
w(z) = c
(l)
1/3z
1/3 − 1
18z2
+
∞
∑
n=2
c
(l)
(1−7n)/3 z(1−7n)/3
+ C1 z−1/4 exp
{
F1(z) + C2
∫
z1/6 exp
[
F2(z)
+ C3
∫
(z1/6 exp F3(z)) dz
]
dz
}
, (9.3)
330 Expansions of solutions...
where c
(l)
1/3 can be computed by formulas (6.2), (6.3), and (6.4); F1(z) =
F
(l,m,k)
1 (z), F2(z) = F
(l,m,k,p)
2 (z), and F3(z) = F
(l,m,k,p,s)
3 (z), (l = 1, 2, 3;
m, k = 1, 2; p = 1, 2, 3; s = 1, 2) are
F
(l,m,k)
1 (z) =
6
7
g
(l,m,k)
1/6 z7/6 +
∞
∑
n=2
6
7(1 − n)
g
(l,m,k)
(1−7n)/6z
7(1−n)/6, (9.4)
F
(p,l,m,k)
2 (z) =
6
7
r
(l,m,k,p)
1/6 z7/6 +
∞
∑
n=2
6
7(1 − n)
r
(l,m,k,p)
(1−7n)/6z
7(1−n)/6, (9.5)
F
(l,m,k,p,s)
3 (z) =
6
7
q
(l,m,k,p,s)
1/6 z7/6 +
∞
∑
n=2
6
7(1 − n)
q
(l,m,k,p,s)
(1−7n)/6 z7(1−n)/6.
(9.6)
Coefficients g
(l,m,k)
1/6 , r
(l,m,k,p)
1/6 , and q
(l,m,k,p,s)
1/6 are defined by formulas
(7.11), (7.12), (7.13), (8.7), and (9.2). Other coefficients are computed
sequentially.
Here the additions of the first level F
(l,m,k)
1 (z) exist only in those
sectors of the complex plane, where the condition (7.19) holds, i.e.
Re
[
g
(l,m,k)
1/6 z7/6
]
< 0. (9.7)
Additions of the second level F
(l,m,k,p)
2 (z) exist there, where the condition
Re
[
r
(l,m,k,p)
1/6 z7/6
]
< 0 (9.8)
holds together with (9.7). Additions of the third level F
(l,m,k,p,s)
3 (z) exist
there, where the condition
Re
[
q
(l,m,k,p,s)
1/6 z7/6
]
< 0 (9.9)
holds together with (9.7) and (9.8).
The inequality
Re[a x] < 0
means that
π
4
− arg a < arg x <
3π
4
− arg a.
Hence the inequalities (9.7) are always satisfied in the open half-plane of
the complex plane x = z7/6. We obtain all 12 cases. The inequalities
(9.7) and (9.8) are satisfied simultaneously in some part of this plane if
arg r
(l,m,k,p)
1/6 6= −arg g
(l,m,k)
1/6 .
A. D. Bruno, N. A. Kudryashov 331
According to (8.7), this inequality is satisfied for p = 2, 3, since
10 6= (−1)k−1(5 + (−1)m−1i
√
5)1/2.
We obtain 24 cases of two inequalities (9.7) and (9.8) being satisfied
simultaneously. The three inequalities (9.7)–(9.9) are never satisfied si-
multaneously, which is demonstrated by the computation of the complex
numbers g1/6, r1/6, q1/6. Hence additions of the third level (9.6) are
absent.
According to [25, §7], the expansions (6.8) diverge, since the exponen-
tial additions are present. The exponential additions themselves describe
the Stokes phenomenon.
Since the truncated equation (6.1) is algebraic, then it has not non-
power solutions and does not give non-power asymptotics.
Note that the asymptotic study of P 2
1 has rather long history. In [42],
asymptotics were obtained corresponding to the first terms in the expan-
tions (6.7), (6.8). In [43], a one-parameter asymptotics were obtained
corresponding to the first terms of the exponential additions of the first
level (7.16), (7.18). In [44], the following equation was suggested
X = TU −
[1
6
U3 +
1
24
(U ′2 + 2UU ′′) +
1
24
U (4)
]
, (9.10)
which differs from the equation (1.4) by the additional term TU contain-
ing a new parameter T . For solutions to this equation, in [44], the first
term of the asymptotics was written, which coincides with the first terms
in expansions (6.7), (6.8) for T = 0. A more accurate asymptotics of
solutions to the equation (9.10) was written in [45], but for T = 0, it
coincides with already mentioned.
Finally, in [46], the equation (9.10) was cited, but there was con-
sidered a more complicated equation of the fourth order containing two
parameters. There the existence of asymptotics depending on four pa-
rameters was shown for solutions of this equation. It is possible that
two new parameters in the asymptotics of this solution correspond to
arbitrary constants in the exponential additions of the first and the sec-
ond levels. But this is a subject of a separate study, since algorithms of
Power Geometry are applicable to this equation as well and allow to ob-
tain asymptotic expansions and exponential additions of different levels.
In [42–46] there are only asymptotic forms of solutions, they do not con-
tain asymptotic expansions of solutions, but they contain global results:
the existence and uniqueness of a remarkable solution of equation (9.10)
which has the asymptotic behaviour ±(6|x|)1/3 as x approaches ±∞.
332 Expansions of solutions...
10. Summary of the results and discussion
We obtained the following expansions for the solutions to the fourth-
order equation (2.1).
Near point z = 0:
1. The four-parameter (with arbitrary constants c0, c1, c2 and c3)
family G(0)
1 1 of expansions (3.11), (3.12).
2. The three-parameter (with arbitrary constants c1, c2 and c3) family
G(0)
1 2 of expansions (3.11), (3.12) with c0 = 0.
3. Two-parameter (with arbitrary constants c2 and c3) family G(0)
1 3 of
expansions (3.11), (3.13) with c0 = c1 = 0.
4. One-parameter (with arbitrary constant c3) family G(0)
1 4 of expan-
sions (3.11), (3.13) with c0 = c1 = c2 = 0.
5. Family G(1)
1 1 of one expansion (4.8).
6. Three-parameter (with arbitrary constants c0, c3 and c6) family
G(1)
2 1 of expansions (5.21).
7. Two-parameter (with arbitrary constants c6 and c8) family G(1)
2 2 of
expansions (5.25).
All listed expansions converge for sufficiently small |z|, z 6= 0.
Near point z = ∞:
8. Three expansions G(1)
3 l (l = 1, 2, 3) described by formulas (6.7),
(6.2)–(6.4). For each of these expansions we found four exponential
additions G(1)
3 lG1
1mk (m, k = 1, 2) of the first level expressed by
the formula (7.18). For them we also computed two exponential
additions G(1)
3 lG(1)
1 mkG(1)
1 p (p = 2, 3) of the second level expressed
by the formula (8.10).
Theorem. All power expansions of solutions to the equation (1.4) at the
points z = 0 and z = ∞ are exhausted by the enumerated families 1–8.
The existence and analyticity of expansions described in items 1–5
follows from the Cauchy theorem. Families G(1)
2 1 and G(1)
2 2 were first
found in the paper [13]. However the structure of supports of expansions
G(1)
2 1 and G(2)
2 2 was not discussed earlier. The other families of expansions
of solutions and their exponential additions of two levels are found for
the first time.
A. D. Bruno, N. A. Kudryashov 333
Asymptotic expansions allow us to find asymptotic forms with any
accuracy. New precise asymptotic forms of solutions to the equation
(1.4) can be applied to study the waves on water [4, 5] and to study
the behaviour of a star [6–8], but such studies are subjects for separate
papers.
11. Appendix.
The computation of the basis of a lattice
Let there be a set S of points Q1, . . . , Qm in the plane R
2 with the
origin among them. Our aim is to compute the basis B1, B2 of the
minimal lattice Z that contains all the points of the set S. The minimality
of the lattice Z means that there is no other lattice Z1 ⊂ Z and Z1 6= Z
which also contains the set S. The computation is divided into three
steps.
Step 1. Let Qm = 0, and the others Qj 6= 0. For all pairs of vectors
Qj , Qk, 1 ≤ j, k < m, j 6= k compose the determinants
det(QjQk)
def
= ∆jk. (11.1)
Among pairs with ∆jk 6= 0 we find one with |∆jk| = min |∆jk| 6= 0
for all j, k = 1, . . . , m − 1. If there are several such pairs, then we take
any one of them. Suppose for the sake of simplicity that it is the pair
Q1, Q2. Other points Q3, . . . , Qm−1 are arbitrary ordered.
Step 2. Let us find the basis of the lattice generated by the vectors
Q1, Q2, Q3. Let Q3 = aQ1 + bQ2, where a and b are rational. Denote
the integral part of the real number a as [a] and the fractional part as
{a}, i.e. {a} = a − [a]. Denote Q′
3 = {a}Q1 + {b}Q2. Suppose that
min |det(Q′
3Qi)| for i = 1, 2 is attained at i = 1. Then we take Q1
and Q′
3 as the basis vectors and use them to express Q2, i.e. we get
Q2 = a1Q1 + bQ′
3. Replace the vector Q2 by Q′
2 = {a1}Q1 + {b1}Q′
3.
Among the three vectors Q1, Q
′
2, Q
′
3 we find the pair with the minimal
modulus of the determinant. Using this pair we expand the third vector,
take its fractional part and so on. At some step l we obtain that the
fractional part of the third vector equals zero. The latest pair of vectors
Q
(l)
2 , Q
(l)
3 gives the basis of the minimal lattice containing the points
Q1, Q2, Q3.
Step 3. For the vectors Q
(l)
2 , Q
(l)
3 , Q4 we repeat Step 2 and obtain
vectors Q̃3, Q̃4 and so on. After looking through all Qj , j ≤ m − 1, we
obtain the pair of vectors Q∗
m−2, Q∗
m−1 which is the basis of the minimal
lattice containing the set S.
334 Expansions of solutions...
Remark. An analogous algorithm allows to find the basis of the minimal
lattice in R
n containing a given finite set S. If n = 1 it is the Euclid
algorithm.
Example. Let us consider the equation (2.1). Its support consists of
four points (2.1′). Shift them by the vector −Q3 = −(1, 0). We obtain
Q′
1 = (−5, 1), Q′
2 = (−1, 3), Q′
3 = 0, Q′
4 = (−3, 2).
For vectors Q′
1, Q′
2, Q′
4 we compute the pairwise determinants
∆14 =
∣
∣
∣
∣
−5 1
−3 2
∣
∣
∣
∣
= −7, ∆12 =
∣
∣
∣
∣
−5 1
−1 3
∣
∣
∣
∣
= −14,
∆42 =
∣
∣
∣
∣
−3 2
−1 3
∣
∣
∣
∣
= −7.
(11.2)
Thus we can use the vectors Q′
1, Q′
4 or Q′
2, Q′
4 as the initial pair.
Let us take Q′
1, Q′
4 for example. We are looking for the expansion Q′
2 =
aQ′
1 + bQ′
4 = a(−5, 1) + b(−3, 2); i.e. we solve the linear system of
equations
−5a − 3b = −1,
a + 2b = 3.
(11.3)
We obtain a = −1, b = 2. Since {a} = {b} = 0, then the vectors
B1 = Q′
1 and B2 = Q′
4 generate the basis of the lattice of shifted support
of the equation (2.1).
The preliminary versions of that article are preprints [47,48].
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Contact information
Alexander
Dmitrievich Bruno
Keldysh Institute of Applied Mathematics,
Miusskaja Sq. 4,
Moscow 125047
Russia
E-Mail: abruno@keldysh.ru
Nikolai Alekseevich
Kudryashov
Moscow Engineering and Physics Institute,
31 Kashirskoe Shosse,
Moscow 115409
Russia
E-Mail: kudryashov@mephi.ru
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