Three families of functions of complexity one Текст научной статьи по специальности «Математика»

УДК 517.55

Three Families of Functions of Complexity One

Valery K. Beloshapka*

Department of Mechanics and Mathematics Moscow State University GSP-1, Moscow, 119991

Russia

Received 17.06.2016, received in revised form 29.07.2016, accepted 24.08.2016 Three rare families of functions of analytic complexity one were studied. Main results are the description of linear differential equations with solutions of complexity one (Theorem 2), the description of L-pairs of complexity one (Theorem 55), the description of O (2)-simple functions (Theorem 7).

Keywords: rare family, analytic complexity. DOI: 10.17516/1997-1397-2016-9-4-416-426.

Introduction

The complexity of analytic functions of several variables has been studied in [1-5]. A method of measuring the complexity of an analytic function in two variables, possibly multivalued, is proposed in [3]. For any analytic function of two variables z(x,y) one can define its complexity N(z). It attains values 0, 1,..., to and is preserved under any analytic continuation. Functions of one variable have complexity N(z) = 0. Complexity one have functions z(x, y) of two variables if they have the form z = c(a(x) + b(y)), where a, b, c are nonconstant functions of one variable, and so on. In other words, for a function z of two variables we write N(z) = n if z can be represented in the form C(A(x,y) + B(x,y)), where C is a function of one variable, and the complexity of A and B is less than n, and there is no such representation with a smaller value of n. This produces an increasing system of classes of functions

ci0 c Ci1 c ci2....

If a function does not belong to any of these classes we write N(z) = to. Each of the above classes is defined by differential-algebraic relations. For example, Cl0 is defined by the condition z'x z'y = 0, and Cli by the condition

S(z) = zxzy (zxxy zy — zxyy zx) + zxy ((zx) zyy — (zy) zxx) = (1)

The differential polynomial S(z) is the numerator of the expression (\r\.(z'y/z'x))'£y.

1. Linear equations with constant coefficients

Consider the pair of functions (z1 = eax+by, z2 = epx+qy). If ab = pq = 0 then we have max(N(z1),N(z2)) = 0. If it is not so, then max(N(z1),N(z2)) = 1. What condition on (a, b, p, q) provides that the complexity of all linear combinations of z1 and z2 does not exceed one? The answer gives

* vkb@strogino.ru © Siberian Federal University. All rights reserved

Lemma 1. Let (ab,pq) = 0. The complexity of all linear combinations of zi and z2 does not exceed 1 only in three cases (1) p = a, (2) q = b, (3) aq = bp.

Proof. The condition (1) for z = tizi + t2z2 has the form

(b - q) (a - p) (qa - bp) (^(eax+byf abt22 - (epx+qyf pqti2) eax+byepx+qy ti t2 = 0.

So the lemma is proved. □

There is a curious corollary from this lemma. Consider a homogeneous linear equation with constant coefficients P(D)(z(x,y)) = 0 and let L be the space of its analytic solutions. The complexity N(L) of the space of solutions L is the maximum (finite or infinite) of the solutions' complexities.

Theorem 2. If N(L)< 1, then the equation P(D)(z(x,y)) = 0 has one of the forms:

(1) z'x - Az = 0, solutions have the form z = eAx b(y),

(2) z'y - Bz = 0, solutions have the form z = eBy a(x),

(3) kz'x + lz'y = 0, solutions have the form z = c(lx - ky),

(4) z'J.y = 0, solutions have the form z = a(x) + b(y).

Proof. Let x = {P(Ai,A2) = 0} be the characteristic set of this equation and let (zi = eax+by, z2 = epx+qy) be two solutions, i.e. (a,b), (p,q) G x. It follows from Lmma 1 that X belongs to a vertical line (case (1)) or to a horizontal line (case (2)), or to a line passing through the origin (case (3)). There is another case (case (4)) outside Lemma 1. In this case X is the coordinate cross and N(z1) = N(z2) = 0. The characteristic polynomials have one of the forms: in case (1) P(Xi, X2) = (Ai - A)ni, in case (2) P(\u\2) = (X2 - B)n2, in case (3) P(Ai, X2) = (kAi + lX2)n3, in case (4) P(Ai,A2) = (AiA2)n4. In all cases it is not difficult to solve these differential equations. The condition N(L)^ 1 is true only for ni = n2 = n3 = n4 = 1. The theorem is proved. □

Note that if the multiplicities (ni, n2 n3, n4) are arbitrary, then the complexities of the space of solutions are finite but greater than one.

2. L-pairs

A collection of functions forms a linear space if this collection is closed under addition and multiplication by a constant (complex numbers). Multiplication by a nonzero constant does not change the complexity of a function: N(Az(x, y)) = N(z(x, y)). This means that a nonzero function of complexity 1 generates a linear space lying in Cli. As for a sum of two functions, if N(zi(x,y)) and N(z2(x,y)) do not exceed n then N(zi(x,y) + z2(x,y)) < (n + 1). It can be shown that in 'general position' this inequality becomes the equality. There is a simple example: N(xy) = 1, N(x2) = 0, then N(xy + x2) = 2. But there exist exceptional pairs. For example N (xy) = 1, N (x + y) = 1 and N (ti (xy) + t2(x + y)) = 1 for any (tiM).

Definition. We call a pair of functions (zi(x, y), z2(x, y) an L-pair of complexity n if

N (ti (zi(x, y)+ t2z2(x, y)) < max(N (zi), N (z2)) = n for any (ti, t2).

Here we assume that zi and z2 have analytic germs at the same point. Lemma 1 then becomes a classification of L-pairs of a special form. Let us formulate several obvious statements.

Statement 3. Two functions (zi,z2) is an L-pair of complexity zero if and only if they are functions of the same argument x or y.

Statement 4. The property of being an L-pair is invariant under the action of

(1) the pseudo-group of transformations {(x ^ p(x), y ^ q(y))},

(2) the change {(x ^ y, y ^ x)},

(3) the affine group of transformations of (z1, z2)-plane.

The pseudo-group generated by the transformations (1), (2) and (3) we denote by G. The description of L-pairs is natural to give up to the G-action.

Now let us turn back to Lemma 1. If we assume only that N(zi + z2) < 1, we have the same description. Indeed, the condition (1) for z = z1 + z2 has the form

(b - q) (a - p) (qa - bp) ((eax+by)2 ab - (epx+qy)2pq^j eax+byepx+qy = 0,

and it is enough to reach the conclusion of Lemma 1. Taking this into account we modify the definition.

Definition. We call a pair (z1 (x,y), z2(x,y)) a pair of complexity n, if N(z1 (x,y) + z2(x,y)) ^ max(N (z1),N (z2)) = n.

We can strengthen Lemma 1 as follows.

Lemma 1'. Let (ab,pq) = 0. The pair (z1 = eax+by, z2 = epx+qy) is a pair of complexity one only in three cases (1) p = a, (2) q = b, (3) aq = bp.

Now we turn to the construction of an arbitrary L-pair of complexity one. Their description is given in the form of a list of cases that are specified and denoted in the course of exposition.

Let z1 and z2 be two functions of complexity not exceeding 1, that is z1 = c1 (a1(x) + b1(y)), z2 = c2(a2(x) + b2(y)). Assume also that max(N(z1),N(z2)) = 1, i.e one of the functions has complexity one, let it be z2. Then a2, b2, and r are non constant and locally invertible at a general point. Replace x by a-,1(x) and y by b--1(y). The condition takes the form

c(a(x) + b(y))+ t • r(x + y) e C/1 yt, r' = 0. (2)

Let the first term have complexity zero, this is Case (01). Then the first term is a function of one variable, denote it by a(x). From (1) for a(x) + t • r(x + y) we get

a1rr = 2 a1f22 - a2r^2, rr = r22.

By lower indices we denote orders of derivatives. If r2 =0 then r(x + y) = k • (x + y) + l and a(x) is arbitrary. This is Case (01.1). This pair is equivalent to (a(x), (x + y)). If r2 is not zero then from the second equation we have r(t) = p • emt + p. And from the first equation we have a(x) = a• emt + a. This pair is equivalent to (kx, xy). We call this Case (01.2)

Consider now Case (11) when both terms have complexity one. This means that a', b', cC , r' are nonconstant functions. From (1) for c(a(x) + b(y)) + t • r(x + y) we get

a12b1C3r12 - a1b12C3r12 - a12C2nr2 - a^2C2r12 + a2b1C2n2 + +b12C2r1 r2 - a^rr + 2 a1C1r22 - a2Cmr2 + b1C1r1r3 - 2 b1C1 r22 + b2Cmr2 = 0,

-afb1C1C3r12 - a13b1C22r12 - a1b13C1C3r12 + a1b13C22r12 - 2 a12b2C1C2r12 + 2 a2b12C1C2r12 -

-a12C12r1r3 + a12C12r22 +2 a1 b2C12r1 r2 - 2 a2^C12r1r2 + b12C12r1 r3 - b12C12r22 = 0, (3)

3 2 3 2 2 2 3 2 3 2 3 3

a1 b1 C1C3r1 - 2 a1 b1 C2 r1 - a1 b1 C1C3r1 +2 a1 b1 C2 r1 + a1 b1C1C2r2 - a1 b2C1C2n -

-a1b13C1C2r2 + a,2b^C^r! - a12b1 C12r3 + a12b2C12r2 + a1b12C12r3 - a2^2C12r2 = 0.

Eliminating c3 from the first and second equations and then from the first and third equations, we get two equations. Each of them is a quadratic form in (ci,c2) with a common factor aibiri (ai - bi)2. In our case this factor can be equal to zero only if ai - bi =0 (Case (11.1)). This pair has the form (c(x + y), r(x + y)).

Assume now ai - bi = 0. After dividing by the common factor we get

ai2bi022r12 + a1b12C22r12 - a12c1 c2T1 - 2 a1b1c1c2r1v2 + a1b2C1C2T12 + +a2b1c1c2r12 - bi2cic2rir2 + aiC{2T22 - a2ci2rir2 + bici2T22 - b2ci2rir2 = 0, 2ai2bi2c22 ri2 - 2ai2bi CIC2TIT2 + ai2b2cic2ri2 - 2a,ib2 CIC2TIT2 + +a22bi2cic2ri2 + 4aibic{2T22 - 2aib2ci2riT2 - 2a2bici2riT2 = 0.

(4)

After elimination of c2/ci we have

(ai - bi)3 aibiri6a2b2r2 (ai2bir2 - ai2b2ri - aibi2r2 + a2bi2ri) = 0. (5)

Consider all the possibilities separately.

Case (11.2). One of the functions a'' = 0 and b'' = 0 is linear, let it be b, then b(y) = k ■ y +1, where k = 0. Replace k ■ y + l by y and k ■ x - l by x, then r(t) becomes r(t/k). The condition (1) for c(a(x) + y) + t ■ r(x + y) takes the form

o O O Q Q Q Q

ai cic2r2 + ai cic3ri - 2 ai c2 ri - ai ci r3 - ai cic3ri + 2 ai2c22ri + aici2r3 - aicic2r2 - a,2ci2r2 + a2cic2ri = 0,

3 2 3 2 2 2 2 ,2 2 2 2 ,

ai cic3ri - ai c2 ri - ai ci rir3 + ai ci r2 - aicic3ri +

aic22ri2 - 2 a2ci2rir2 + 2 a2cic2ri2 + ci2rir3 - ci2r22 = 0,

2 i 2 2 i r> 2 2

-ai c2rir2 + ai c3ri - aicirir3 + 2 aicir2 - aic3ri -a2 ciri r2 + a2c2ri2 + cirir3 - 2 cir2" + c2rir2 = 0.

The expressions for c3 from each of these equations are fractions with the denominators

ai2ciri (ai - 1), aiciri2 (ai2 - 1) , airi2 (ai - 1) .

There are two possibilities for vanishing of one of the denominators: ai = 1 or ai = -1. In our case ai = bi, hence we have only the second possibility ai = -1, a(x) = -x + a. The condition (1) yields

-ci2r3 - cic3ri + 2 c22ri = 0, cirir3 - 2 cir22 + c3r2 = 0,

where c and r are functions of two independent variables x - y and x + y.

Separating the variables and solving the differential equations we arrive at Case (11.2.1) :

c(-x + y) = Yem(-x+y) + Y, r(x + y) = pe±m(x+y) + p. The pair then has the form (y/x, xy). If ai = ±1, we can eliminate c3 from (5) to get two quadratic form in (ci,c2):

(c2ri - cir2) (c2ai3ri + c2a,i2ri - ciai2r2 - ciair2 + a2ri) = 0, (c2ri - cir2) (2 c2ai2ri - 2 ciair2 + cari) = 0

with the common factor (c2ri - r2ci). If this factor is equal to zero (Case (11.2.2)), then we can separate the variables and, taking into account that the Jacobian of the change (t = a(x) + y; s = x + y) does not vanish, we see that both logarithmic derivatives are equal to the

same constant m. From this we get z\ = pem(a(x)+y) + p, z2 = pem(x+y) + p . The pair has the form (a(x)y,xy).

Otherwise, (Case (11.2.3)), dividing out the common factor and eliminating c2/c1 from two linear forms, we get a12a2r12 (a1 — 1) = 0. It vanishes only if a2 = 0, a1 is then the constant A. In this case Ac2/c1 = r2 /r1, and z1 = c(Ax + y) = Ye A (Ax+y), z2 = r(x + y) = pem(x+y). The pair has the form (xky, xy)

We see that Cases (11.2.1) and (11.2.3) are subcases of Case (11.2.2). Thus, in Case (11.2) the pair has the form (a(x)y, xy).

In Case (11.3) r2 = 0, i.e. r(x + y) = p(x + y) + p, where p = 0. By replacing x with px + p and y with py we obtain r(x + y) = x + y. The condition (1) for c(a(x) + b(y)) + (x + y) has the form

a13b12C1C3 — 2 a13b12 C22 — a12b13cc + 2 a12 b13 C22 — a13b2C1C2 + a2b13C1C2 = 0,

3 323 322 2

a1 b1C1C3 — a1 b1C2 — a^1 C1C3 + 0,^1 C2 — 2 a1 b2C1C2 +2 a2b1 C1C2 = 0,

a12b1C3 — a1b12 C3 — a^C2 + a^bc = 0.

By eliminating c3 and c2/c1, we get

(a1 — b1) (a12b2 — a2b12) = 0.

It may vanish only because of the second factor, therefore, separating the variables we get a2/a2 = b2/b2 = —m where m is a constant. Then

a(x) + b(y) = —(ln(mx + a) + ln(my + ¡3) + ln(n)), m

and three equations for c(t) are

C3 = mc2, C3C1 = c2, mc1c2 + C1C3 — 2c2 = 0.

Consequently, c(t) = Yemt + p, and the pair has the form (xy, x + y). Case (11.4)

a12br — a12 b2r1 — a1 b12r2 + a2b12r1 = 0. (6)

From this we get

r2 ai262 - a2b\

ri aibi (ai — bi)

(7)

r2

(the denominator is not zero). The condition that — is a function of x + y, namely the equality

r1

of its derivatives with respect to x and y, is

—a14b1 b3 + a14b22 + a13b12b3 — 2 a^bb'2 — a12a3b13 + 2 a1a22b13 + a1 a3b14 — a22b14 = 0 (8)

We can decrease the order of equation (8) twice. First, putting a1 = a'(x) = A, b1 = b'(y) = B. Second, introducing a2 = a''(x) = P(A), b2 = b''(y) = Q(B). In this notation we have a3 = a''' (x) = P'(A) P(A), b3 = b''' (y) = Q'(B) Q(B) and we can write (8) as

—A4B (dBG (B)) G (B) + A4 (G (B)f + A3B^ -BG (B)) G B) — —2 A3B (G (B))2 — A2 ^d[AF (A)^j F (A) B3 + 2 A (F (A))2 B3 +

dAF (A^j F (A) B4 — (F (A))2 B4 = 0-

After the substitution f (A) = y/F(A), g(B) = yjG(B) we previous equation becomes linear

-A4BB (B) + 2 A4g (B) + A3B2 dBg (B) - 4 A3Bg (B) -A2B3 A (A) + 4 Af (A) B3 + AB4-Af (A) - 2 f (A) B4 = 0.

From this we find -~Bg and write the condition of its independence from A:

- 2 - 2 - 2 - -

-A4B2 A f (A) + 2 A3B3-A2 f (A) - A2B4 ¿2 f A) + 6 A3B2 A A - W B3 -Af A +

+4 AB4-A f (A) + 2 A4g (B) - 12 A2B2f (A) + 16 Af (A) B3 - 6 f (A) B4 = 0. -A

Now we express g(B) and write the condition of its independence from A:

- 3 - 2 -

A3 d^ f (A) - 6 A2 ^ f (A) + 18 A A (A) - 24 f (A) = 0. By looking for solutions of the form f (A) = Am, we get the equation

m(m - 1)(m - 2) - 6m(m - 1) + 18m - 24= (m - 2)(m - 3)(m - 4).

Hence, a general solution to (9) is f (A) = liA4 + mi A3 + niA2. By eliminating f (A) from (9), we obtain g(B) = l2B4 + m2B3 + n2B2. Substituting these f (A) and g(B) in (9), we get li = l2, mi = m2, ni = n2. Finally, f (A) = lA4 + mA3 + nA2, g(B) = IB4 + mB3 + nB2. We see that a(x) = a'(x) and 3(y) = b'(y) satisfy the same differential equation

= \J la4 + ma3 + na2, -33 = \J l/34 + mfi3 + nft2. (9)

dx dy

Since a and b are not linear, we may assume that the constants l, m, and n are not zeros simultaneously. Thus, if l = n = 0 and m = 0 (Case (11.4.1)), then

f dt -2

t\Jmt y/t

Therefore

4 4 ~ —8

a' (x) = a(x) = —--—2, a(x) =----—— + C, a'' (x)

m(x + C)2' m(x + C) ' m(x + C )3

Analogously,

4 4 -8

b' (x) = ß(x) = -r—-2, b(y) =--+ D, b''(y)

Now, from (7) we get

m(y + D)2' m(y + D) ' m(y + D)3

r2 a2ib2 — a2b\

ri aibi(ai — bi)'

P

; + c + l

expression for r2/ri, we get

and then we have r(t) =--——. Computing c2/ci from any of (4) and substituting the

t + C + D

C2 aib2 — a2bi

ci aibi(ai — bi)

and

c(a(x) + b(y))

+ —

x+C + y+D

Thus, the pair has the form

xy

zi = —:—, ^2

x + y x + y

If l = 0 or n = 0 (Case (11.4.2)), then

dt —2 I V lt2 + mt + n

-- arctgh

( Vit2 + mt + n \

V Vit2 + n )'

J Wl-t2 + mt + n Vlt2 + n V Vlt2+ n

and we get a'(x) = a(x) and b'(y) = ¡¡(y) as inversion of the integrals, and a(x) and b(y) by one more integration. As in the previous case, from (7) we get r(t) and c(t) from any relation of (4).

Finally, we have the theorem.

Theorem 5. Let z1(x, y) and z2(x, y) is an L-pair of complexity one, then this pair up G-action has the form

For N(z1) = 0, N(z2) = 1

(01.1) z1 = a(x), z2 = x + y, a is arbitrary,

(01.2) z1 = x, z2 = xy, For N(z1) = N(z2) = 1

(11.1) z1 = c(x + y), z2 = r(x + y), where c and r are arbitrary,

(11.2) z1 = a(x)y, z2 = xy, a is arbitrary,

(11.3) z1 = xy, z2 = x + y,

xy 1

(11.4.1) z1 = , z2 = ——,

x + y x + y

(11.4.2) In this case there are no explicit expressions for the pair z1 = c(a(x) + b(y)), z2 = r(x + y). The four functions (a,b,c,r) are constructed as described above. In particular, they can be expressed by quadratures.

As shown above, all pairs in this list are L-pairs. In Cases (01.1), (01.2), (11.1), (11.2) it is obvious. In Case (11.3) we can also see it easily: z = xy + t(x + y) = (x + t)(y +1) — t2. In Case (11.4.1) it is not that clear. We need to check that

xy 1 t + xy - +1-=-e Cl1 for all t.

x + y x + y x + y After the change t by t2 we get

t2 + xy

z = -.

x+y

By replacing x with tx, y with ty, and z with t/z, we get

x+y

z = -.

1 + xy

Now, we replace x with th(x), y with th(y), and z with th(z) and use the addition formula

th(x + y)= th(x)+th(y) th(x + y)= 1 + th(x) th(y)

to get z = x + y. Since all the transformations here do not change complexity, this proves that the complexity of the original function is 1.

For Case (11.4.2) the author does not know a similar reasoning. The open question is what mysterious relations are behind that fact.

The set of pairs of complexity one is certainly wider than the set of L-pairs of complexity one. This is another open problem: to describe all pairs of complexity one.

1

1

3. O(2)-simplicity

The standard action of the O(2) on the (x, y)-plane is

g0 = ( x ^ cos(4>)x — sin($)y, y ^ sin($)x + cos($)y ) where $ G C. This action induces an action on functions

z(x, y) ^ g0(z)(x, y) = z(cos($)x — sin($)y, sin($)x + cos($)y). Denote t = tg($/2), then we have another form for this action

( 1 — t2 t 1 — t2 t gt = x —ï --k x — 2^-- y, y —ï -- y + 2--^x

J V 1+12 t2 + 1y' y t2 + 1 1+12

■

If N(z(x,y)) = n, then N(z(\x,\y)) = n also, therefore we can replace gt(x,y) with ht(x,y) = (1+ t2) gt(x,y).

If N(z) < n, then N(g$(z)) < n +1, and for arbitrary z and $ there is no reason to expect that N(g$(z)) < n. For example, let z = xy, then N(z) = 1. For S(ht(z)) we have

41 (x2 + y2) (t — 1)(t + 1) (t2 + 21 — 1)(t2 — 21 — 1) (t2 + 1)4 .

We see that N(ht(xy)) = 1 only for 9 values of t, namely t = 0, ±1, ±i, ±1 ±y/2. The corresponding functions are proportional to

xy, x2 — y2, (x ± iy)2.

For another values t the complexity N(ht(xy)) is equal to two.

Definition. A function z(x,y) is called O(2)-simple if N(gt(z)) ^ 1 for all t.

All linear functions are, of course, O(2)-simple. Now, we want to describe all O(2)-simple functions. It is clear that for such functions N(z) ^ 1, then z = c(a(x) + b(y)). If one of the functions (a, b, c) is constant, then N(z) = 0, and z depends on only one variable or a constant. Any such function is O(2)-simple (Case 0). Assume that N(z) = 1, i.e. a, b, c are not constant.

Statement 6. (1) z is O(2)-simple if and only if 5(gt(z)) = 0 for all (x,y,t).

(2) c(a(x) + b(y)) is O(2)-simple if and only if a(x) + b(y) is O(2)-simple. (3) z(x,y) is O(2)-

simple if and only if z(y, x) is O(2)-simple.

The proof is obvious.

Let a(x) + b(y) is O(2)-simple, then, in particular,

jt6(gt(a(x)+ b(y)))\t=o = 0, (10)

in index notation for derivatives we have

—a12a2b2 — a12bb + a12 b22 — a1a3b12 + a22b12 — a2^2b2 = 0. (11)

We can decrease the order of equation (11) twice. First, putting a1 = a (x) = A, b1 = b (y) = B. Second, introducing P(A) = a2 = a'' (x), Q(B) = b2 = b'' (y). In this notation we have a3 = a''' (x) = P'(A) P(A), b3 = b' '' (y) = Q'(B) Q(B) and we can write (11) as

—QA2P — BQ1QA2 + Q2A2 — B2AP1P + B2 P2 — B2 QP = 0 (12)

By differentiating (12) wit respect to A, we get

-2 QAP - QA2Pi - 2 ABQiQ + 2 Q2A + B2PiP - B2AP2P - B2AP2 - B2QP = 0. (13)

The relations (12) h (13) are a system of linear equations in Q(B) and Q'(B), its determinant is equal to

-BP A (A3Pi + B2APi - 2B2P) .

This determinant is identically equal to zero only if P(A) = 0 (Case 1). The solution to the system for Q(B) is

( =_B2(A2P2P + A2P2 - 3 APPi +2P2) Q( ) A3Pi + B2APi - 2 B2P .

The condition of independence Q from A is

-A3P3 PPi + A3P22P - 2 A3P2 Pi2 - AB2P3 PPi + +AB2P22P - 2 AB2P2 Pi2 + A2P2 PPi +4 A2Pi 3 + (14)

+2 B2P3 P2 + 3 B2P2 PPi +2 AP2 P2 - 10 APPi2 + 6 P2Pi =0,

which splits into two relations: terms free of B and terms with the factor B2. Eliminating P'''(A) from them, we get

P (APi - 2 P) (AP2 P - 2 APi2 + 3 Pi P) (A2PP2 + A2Pi 2 - 3 APPi +2 P2) = 0.

The case P = 0 ( Case 1) has been considered above. Now we turn to the remaining cases.

(APi - 2P) = 0 (Case 2), (AP2 P - 2 APi2 +3 Pi P) =0 (Case 3), (A2PP2 + A2Pi 2 - 3 APPi +2 P2) =0 (Case 4).

The solutions to the corresponding differential equations are

P(A) = 0 (Case 1), P(A) = CA2 (Case 2),

P (A) = ac+c (Case3),

P(A) = A yCTinjAY+C^ (Case 4).

To find Q(B) corresponding to P(A), we substitute these solutions in (13).

In Case 1 P(A) = 0, Q(B) = CB.

In Case 2 P(A) = CA2, Q(B) = -CB2.

In Case 3 P(A) = A2/(cA2 + d) and for Q(B) we have

-A6BQQi c3 + A6Q2c3 - 3 A4BQQi c2d - A6Qc2 - A4B2Qc2 + 3 A4Q2c2d -

-3 A2BQQi cd2 + B2A4c - 2 A4Qcd - 2 A2B2Qcd +3 A2Q2cd2 - (15)

-BQQi d3 - A2B2d - A2Qd2 - B2Qd2 + Q2d3 = 0,

which is a polynomial in A2 and splits into four differential equations of first order on Q(B) (the coefficients at 1, A2, A4, A6). These equations yield d = 0, and P(A) = Q(B) = C = const. In Case 4 we have P(A) = A ^J c ln(A) + d and

-2 QA^yOniAy+d - B2Ac - 2 ABQQi - 2 B2Qvicln(AJTd + 2 AQ2 =0.

The functions

^/c\n(Ây+d, A, A^Vcïn(Ây+d

are linearly independent, hence Q(B) = 0 and c = 0. So the answer in Case 4 coincides with the answer in Case 1 after replacing A ^ B.

Now we can return to equations in a(x) and b(y) and find the answers: In Case 1: P(A) = 0 means a''(x) = 0 and a(x) = a1x + a0, then Q(B) = CB means b''(y) = Cb'(y) and b(y) = ¡¡1eCy + ¡¡0. Then we write the O(2)-simplicity condition S(gt(z)) = 0 for a + b and see that it holds only for a1^1 = 0. The same goes in Case 4.

In Case 2: P(A) = CA2 means a''(x) = C(a'(x))2 and a(x) = — ln(a1x + a0)/C, then from Q(B) = CB2 we get b(y) = ln(p1y + ¡¡0)/C. Since

, , 1 , ( Piy + ¡30 )

a(x) + b(y) = — --- ,

C y aix + ao J it is enough to check the O(2)-simplicity condition only for

¡iy + ¡0

z = -.

aix + ao

It is easy to see that the condition S(gt(z)) = 0 holds.

In Case 3: P (A) = C means a''(x) = C and a(x) = Cx2 + a1x + a0, then from Q(B) = C we get b(y) = Cy2 + ¡1y + ¡¡0. We see that the O(2)-simplicity condition for a + b holds. Thus, we have the theorem.

Theorem 7. The complete list of O(2)-simple functions up to transformations ( z(x,y) ^ f (z(x, y)) and ( z(x,y) ^ z(y,x) ) is

¡i y + ¡0

z =

a\ x + ao 2 = (x2 + y2) + ax + ßy, z = ax + ßy.

Corollary 8. Any O(2)-simple function is a rational function, up to a transformation ( z(x, y) ^

f (z(x,y)) )■

The research was supported by the Russian Foundation for Basic Research, grants no. 14-01-00709-a and no■ 13-01-12417-ofi-m2■

References

[1] A.Ostrowski, Uber Dirichletsche Reihen und algebraische Differentialgleichungen, Math■ Z, 8(1920), no. 3, 241-298.

[2] A.G.Vitushkin, Hilbert's thirteenth problem and related questions, Russian Mathematical Surveys, 59(2004), no. 1, 11-25.

[3] V.K.Beloshapka, Analytic Complexity of Functions of Two Variables, Russian Journal of Mathematical Physics, 14(2007), no. 3, 243-249.

[4] V.K.Beloshapka, A seven-dimensional family of simple harmonic functions, Mathematical Notes, 98(2015), no. 6, 867-871.

[5] M.Stepanova, On Rational Functions of First-Class Complexity, Russian Journal of Mathematical Physics, 23(2016), no. 2, 251-256.

Три семейства функций сложности один

Валерий К. Белошапка

Механико-математический факультет МГУ Ленинские горы, 1, ГСП-1, Москва, 119991

Россия

В работе описаны некоторые семейства функций двух переменных аналитической сложности единица, обладающие некоторыми редкими свойствами. Во-первых, классифицированы линейные уравнения с постоянными коэффициентами,т.ч. все их аналитические решения имеют сложность не выше единицы (теорема 2). Во-вторых, классифицированы пары аналитических функций, таких что любая их линейная комбинация имеет сложность не выше единицы (теорема 5). В-третьих, дано явное описание функций, т.ч. их орбиты под действием группы O(2) состоят из функций сложности не выше единицы (теорема 7).

Ключевые слова: редкие семейства, аналитическая сложность.