On conjugate pseudo-harmonic functions
We prove the following theorem. Let U be a pseudo-harmonic function on a surface M². For a real valued continuous function V : M² → R to be a conjugate pseudo-harmonic function of U on M² it is necessary and sufficient that V is open on level sets of U.
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Інститут математики НАН України
2009
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| Цитувати: | On conjugate pseudo-harmonic functions / Ye. Polulyakh // Збірник праць Інституту математики НАН України. — 2009. — Т. 6, № 2. — С. 505-517. — Бібліогр.: 4 назв. — англ. |
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Digital Library of Periodicals of National Academy of Sciences of Ukraine| _version_ | 1860088236452347904 |
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| author | Polulyakh, Ye. |
| author_facet | Polulyakh, Ye. |
| citation_txt | On conjugate pseudo-harmonic functions / Ye. Polulyakh // Збірник праць Інституту математики НАН України. — 2009. — Т. 6, № 2. — С. 505-517. — Бібліогр.: 4 назв. — англ. |
| collection | DSpace DC |
| description | We prove the following theorem. Let U be a pseudo-harmonic function on a surface M². For a real valued continuous function V : M² → R to be a conjugate pseudo-harmonic function of U on M² it is necessary and sufficient that V is open on level sets of U.
|
| first_indexed | 2025-12-07T17:21:20Z |
| format | Article |
| fulltext |
Çáiðíèê ïðàöüIí-òó ìàòåìàòèêè ÍÀÍ Óêðà¨íè2009, ò.6, �2, 505-517UDC 517.54, 517.57Yevgen PolulyakhInstitute of mathemati
s of NAS of UkraineE-mail: polulyah�imath.kiev.uaOn
onjugate pseudo-harmoni
fun
tions
We prove the following theorem. Let U be a pseudo-harmonic function
on a surface M2. For a real valued continuous function V : M2 → R to
be a conjugate pseudo-harmonic function of U on M2 it is necessary and
sufficient that V is open on level sets of U .
Keywords: a pseudo-harmonic function, a conjugate, a surface, an interior
transformation
Let M2 be a surface, i.e. a 2-dimensional and separable mani-
fold, U : M2 → R be a real-valued function on M2. Denote also
by
D = {(x, y) ∈ R2 |x2 + y2 < 1}
the open unit disk in the plane.
Definition 1 (see [1,2]). A function U is called pseudo-harmonic
in a point p ∈ M2 if there exist a neighbourhood N of p on M2
and a homeomorphism T : D → N such that T (0, 0) = p and a
function
u = U ◦ T : D → R2
is harmonic and not identically constant.
A neighbourhood N is called simple neighbourhood of p.
We can even choose N and T from previous definition to comply
with the equality
u(z) = U ◦ T (z) = Re zn + U(p) , z = x+ iy ∈ D ,
for a certain n = n(p) ∈ N (see [2]).
c© Yevgen Polulyakh, 2009
506 Yevgen Polulyakh
Definition 2 (see [1,2]). A function U is called pseudo-harmonic
on M2 if it is pseudo-harmonic in each point p ∈M2.
Let U : M2 → R be a pseudo-harmonic function on M2 and
V : M2 → R be a real valued function.
Definition 3 (see [1]). A function V is called a conjugate pseudo-
harmonic function of U in a point p ∈ M2 if there exist a neigh-
bourhood N of p on M2 and a homeomorphism T : D → N such
that T (0, 0) = p and
u = U ◦ T : D → R2 and v = V ◦ T : D → R2
are conjugate harmonic functions.
We can choose N and T from previous definition in such way
that
u(z) = U ◦ T (z) = Re zn + U(p) ,
v(z) = V ◦ T (z) = Im zn + V (p) , z = x+ iy ∈ D ,
for a certain n = n(p) ∈ N (see [2]).
Definition 4 (see [1]). A function V is called a conjugate pseudo-
harmonic function of U on M2 if it is a conjugate pseudo-harmonic
function of U in every p ∈M2.
Definition 5. Let U and V be continuous real valued functions
on a surface M2. We say that V is open on level sets of U if for
every c ∈ U(M2) a mapping
V |U−1(c) : U−1(c) → R
is open on the space U−1(c) in the topology induced from M2.
Theorem 1. Let U be a pseudo-harmonic function on M2. For
a real valued continuous function V : M2 → R to be a conjugate
pseudo-harmonic function of U on M2 it is necessary and suffi-
cient that V is open on level sets of U .
Let us remind following definition.
On conjugate pseudo-harmonic functions 507
Definition 6 (see [3]). A mapping G : M2
1 → M2
2 of a surface
M2
1 to a surface M2
2 is called interior if it complies with conditions:
1) G is open, i. e. an image of any open subset of M2
1 is
open in M2
2 ;
2) for every p ∈M2
2 its full preimage G−1(p) does not contain
any nondegenerate continuum (closed connected subset of
M2
1 ).
In order to prove theorem 1 we need following
Lemma 1. Let U be a pseudo-harmonic function on M2. Let a
real valued continuous function V be open on level sets of U .
Then the mapping F : M2 → C,
F (p) = U(p) + iV (p) , p ∈M2
is interior.
First we will verify one auxiliary statement. Denote I = [0, 1],
I̊ = (0, 1) = I \ {0, 1}.
Proposition 1. In the condition of Lemma 1 the following state-
ment holds true.
Let γ : I →M2 be a simple continuous curve and γ(I) ⊆ U−1(c)
for a certain c ∈ R. If the set γ(I̊) is open in U−1(c) in the
topology induced from M2, then the function V ◦ γ : I → R is
strictly monotone.
Proof. Suppose that contrary to the statement of Proposition the
equality V ◦γ(τ1) = V ◦γ(τ2) is valid for certain τ1, τ2 ∈ I, τ1 < τ2.
Since the function V ◦ γ is continuous and a set [τ1, τ2] is com-
pact, then following values
d1 = min
t∈[τ1,τ2]
V ◦ γ(t) ,
d2 = max
t∈[τ1,τ2]
V ◦ γ(t) ,
are well defined. Let us fix s1, s2 ∈ [τ1, τ2] such that di = V ◦γ(si),
i = 1, 2.
508 Yevgen Polulyakh
We designate W = (τ1, τ2). It is obviously the open subset of I̊.
Let us consider first the case d1 = d2. It is clear that
[τ1, τ2] ⊆ (V ◦ γ)−1(d1)
in this case. So the open subset γ(W ) of the level set U−1(c) is
mapped by V onto a one-point set {d1} which is not open in R
and V is not open on level sets of U .
Assume now that d1 6= d2. Since V ◦ γ(τ1) = V ◦ γ(τ2) due to
our previous supposition, then either s1 or s2 is contained in W .
Let s1 ∈ W (the case s2 ∈ W is considered similarly). Then
V ◦ γ(W ) ⊆ [d1,+∞) and the open subset γ(W ) of the level set
U−1(c) can not be mapped by V to an open subset of R since its
image containes the frontier point d1 = V ◦ γ(s1). So, in this case
V is not open on level sets of U .
The contradiction obtained shows that our initial supposition
is false and the function V ◦ γ is strictly monotone on I. �
Proof of Lemma 1. Let p ∈M2 and Q be an open neighbourhood
of p.
We are going to show that the set F (Q) containes a neigbour-
hood of F (p). At the same time we shall show that p is an isolated
point of a level set F−1(F (p)).
Without loss of generality we can assume that U(p) = V (p) = 0.
Let N be a simple neighbourhood of p and T : D → N be a
homeomorphism such that for a certain n ∈ N the folloving equal-
ity holds true u(z) = U ◦T (z) = Re zn, z ∈ D (see Definition 1 and
the subsequent remark). It is clear that without losing generality
we can regard that N is small enough to be contained in Q.
Observe that for an arbitrary level set Γ of U an intersection
Γ ∩ T (D) = Γ ∩ N is open in Γ. Consequently, since T is home-
omorphism then a mapping v = V ◦ T : D → R is open on level
sets of u = U ◦ T : D → R (see Definition 5).
Let us consider two possibilities.
Case 1. Zero is a regular point of the smooth function u =
U ◦ T , i. e. n = 1 and u(z) = Re z, z ∈ D.
On conjugate pseudo-harmonic functions 509
In this case
u−1(u(0)) = u−1(U(p)) = T−1(U−1(U(p))) = {0} × (−1, 1).
According to Proposition 1 the function v is strictly monotone on
every segment which is contained in this interval, so it is strictly
monotone on {0} × (−1, 1). Consequently, for points z1 = 0− i/2
and z2 = 0+i/2 the following inequality holds true v(z1)·v(z2) < 0.
Let us note that from previous it follows that V is monotone
on the arc β = T ({0} × (−1, 1)) = U−1(U(p)) ∩ N . And since
F−1(F (p))∩N ⊂ β then F−1(F (p))∩N = {p} and p is an isolated
point of its level set F−1(F (p)).
Let d1 = v(z1) < 0 and d2 = v(z2) > 0 (The case d1 > 0 and
d2 < 0 is considered similarly). Denote
ε =
1
2
min(|d1|, |d2|) > 0 .
Function v is continuous, so there exists δ > 0 such that following
implications are fulfilled
|z − z1| < δ ⇒ |v(z) − d1| < ε ,
|z − z2| < δ ⇒ |v(z) − d2| < ε .
Let us examine a neighbourhood W = (−δ, δ) × (−1/2, 1/2) of
0, which is depicted on Figure 13. It can be easily seen that for
every x ∈ (−δ, δ) following relations are valid
u(x+ iy) = x , y ∈ (−ε, ε) ,
v(x− i/2) < v(z1) + ε < −2ε+ ε = −ε ,
v(x+ i/2) > v(z2) − ε > 2ε− ε = ε .
From two last lines and from the continuity of v on a segment
{x} × [−1/2, 1/2] it follows that v({x} × [−1/2, 1/2]) ⊇ (−ε, ε).
Therefore
F ◦ T ({x} × [−1/2, 1/2]) ⊇ {x} × (−ε, ε) , x ∈ (−δ, δ) .
Since T (W ) ⊆ N ⊆ Q by the choise of N , then
0 = F (p) ∈ (−δ, δ) × (−ε, ε) ⊆ F ◦ T (W ) ⊆ F (Q) .
510 Yevgen Polulyakh
Figure 13
Case 2. Zero is a saddle point of u = U ◦T , i. e. u(z) = Re zn,
z ∈ D for a certain n > 1.
In this case
u−1(u(0)) = T−1(U−1(U(p))) = {0} ∪
2n−1⋃
k=0
γk ,
where γk = {z ∈ D | z = a · exp(πi(k − 1/2)/n), a ∈ (0, 1)},
k = 1, . . . , 2n − 1.
As above, applying Proposition 1 we conclude that function
v = V ◦ T
is strictly monotone on each arc γk, k = 1, . . . , 2n − 1. Since v is
continuous and 0 is a boundary point for each γk, then
v(z) 6= v(0)
for all z ∈ ⋃
k γk. Therefore, 0 = (F ◦ T )−1(F ◦ T (0)) and
F−1(F (p)) ∩ N = {p}, i. e. p is the isolated point if its level
set F−1(F (p).
On conjugate pseudo-harmonic functions 511
Let us designate by
Rk =
{
z ∈ D
∣∣ z = aeiϕ, a ∈ [0, 1), ϕ ∈
[
π(k−1/2)
2 , π(k+1/2)
2
]}
,
k = 0, . . . , 2n − 1
sectors on which disk D is divided by the level set u−1(u(0)).
We also denote
Dl = {z ∈ D | Re z ≤ 0} ,
Dr = {z ∈ D | Re z ≥ 0} .
Consider map Φ : D → D given by the formula Φ(z) = zn,
z ∈ D. It is easy to see that for every k ∈ {0, . . . , 2n−1} depending
on its parity sector Rk is mapped homeomorphically by Φ either
onto Dl or onto Dr. Let a mapping Φk : Rk → Dr is given by
relation
Φk =
{
Φ|Rk
, if k = 2m,
Inv ◦ Φ|Rk
, if k = 2m+ 1 ,
k = 0, . . . , 2n− 1 ,
where Inv : D → D is defined by formula Inv(z) = −z, z ∈ D.
Evidently, all Φk are homeomorphisms.
We consider now inverse mappings ϕk = Φ−1
k : Dr → D,
k = 0, . . . , 2n − 1. By construction all of these mappings are
embeddings. Moreover, it is easy to see that
uk(z) = u ◦ ϕk(z) =
{
Re z , when k = 2m,
−Re z , when k = 2m+ 1 .
Let us fix k ∈ {0, . . . , 2n− 1}. It is clear that ϕk homeomorphi-
cally maps a domain
D̊r = {z ∈ D | Re z > 0}
onto a domain
R̊k =
{
z ∈ D
∣∣ z = aeiϕ, a ∈ (0, 1), ϕ ∈
(
π(k−1/2)
2 , π(k+1/2)
2
)}
,
so with the help of argument similar to the observation preceding
to case 1 we conclude that the mapping v̊k = v ◦ ϕk|D̊r
: D̊r → R
512 Yevgen Polulyakh
is open on level sets of the function ůk = u ◦ ϕk|D̊r
: D̊r → R.
As above, applying Proposition 1 we conclude that function v̊k is
strictly monotone on each arc
αc = ů−1
k (̊uk(c+ 0i)) = {z ∈ D̊r | Re z = c} , c ∈ (0, 1) .
We already know that the function v is strictly monotone on the
arcs γk and γs, where s ≡ k+1 (mod 2n). Therefore the function
vk = v ◦ ϕk : Dr → R is strictly monotone on the arcs
α− = ϕ−1
k (γk) = {z ∈ Dr | Re z = 0 and Im z < 0} ,
α+ = ϕ−1
k (γs) = {z ∈ Dr | Re z = 0 and Im z > 0} .
Let us verify that vk is strictly monotone on the arc
α0 = α− ∪ {0} ∪ α+ = u−1
k (uk(0)) = {z ∈ Dr | Re z = 0} .
Since vk(0) = v(0) = V (p) = 0 according to our initial assump-
tions and 0 is the boundary point both for α− and α+, then vk is
of fixed sign on each of these two arcs.
So we have two possibilities:
• either vk has the same sign on α− and α+, then vk|α0 has
a local extremum in 0;
• or vk has different signs on α− and α+, then vk is strictly
monotone on α0.
Suppose that vk has the same sign on α− and α+.
We will assume that vk is negative both on α− and α+. The
case when vk is positive on α− and α+ is considered similarly.
Denote z1 = 0 − i/2 ∈ α−, z2 = 0 + i/2 ∈ α+. Let
ε̂ = 1
2 min(|vk(z1)|, |vk(z2)|) > 0 .
From the continuity of vk it follows that there exists δ̂ > 0 to
comply with the following implications
(1)
|z − z1| < δ̂ ⇒ |vk(z) − vk(z1)| < ε̂ ,
|z − z2| < δ̂ ⇒ |vk(z) − vk(z2)| < ε̂ ,
|z| = |z − 0| < δ̂ ⇒ |vk(z) − vk(0)| = |vk(z)| < ε̂ .
On conjugate pseudo-harmonic functions 513
Let c ∈ (0, δ̂). Then the point w0 = c + i0 is situated on the
curve αc between points w1 = c− i/2 and w2 = c+ i/2. It follows
from (1) that vk(w1) < −ε̂, vk(w2) < −ε̂ and vk(w0) ∈ (−ε̂, 0).
But these three correlations can not hold true simultaneously since
vk is strictly monotone on αc as we already know.
The contradiction obtained shows us that vk has different signs
on α− and α+. So, vk is strictly monotone on α0.
Now, repeating argument from case 1 we find such εk > 0 and
δk > 0 that the set
Ŵk = [0, δk) ×
(
−1
2 ,
1
2
)
meets the relations
(2)
F ◦ T ◦ ϕk(Ŵk) ⊇ [0, δk) × (−εk, εk) , if k = 2m,
F ◦ T ◦ ϕk(Ŵk) ⊇ (−δk, 0] × (−εk, εk) , if k = 2m+ 1 .
Let us denote Wk = ϕk(Ŵk),
W =
2n−1⋃
k=0
Wk , δ = min
k=0,...,2n−1
δk > 0 , ε = min
k=0,...,2n−1
εk > 0 .
Figure 14
514 Yevgen Polulyakh
It is easy to show that W is an open neighbourhood of 0 in D.
From (2) and from our initial assumptions it follows that
F (Q) ⊇ F (N) ⊇ F ◦ T (W ) ⊇ (−δ, δ) × (−ε, ε) .
So, we have proved that for an arbitrary point p ∈ M2 and
its open neighbourhood Q a set F (Q) contains a neigbourhood of
F (p). Hence the mapping F : M2 → C is open.
At the same time we have shown that an arbitrary p ∈ M2 is
an isolated point of its level set F−1(F (p)). It is easy to see now
that any level set F−1(F (p)) can not contain a nondegenerate
continuum.
Consequently, the map F is interior. �
Proof of Theorem 1. Necessity. Let U, V : M2 → R be conjugate
pseudoharmonic functions on M2 (see Definitions 3 and 4).
Obviously, V is continuous onM2. Suppose that contrary to the
statement of Theorem there exists such c ∈ R that V is not open on
the level set Γc = U−1(c) ⊂ M2, i. e. a map Vc = V |Γc : Γc → R
is not open on Γc in the topology induced from M2.
Let us verify that Vc has a local extremum in some p ∈ Γc.
Note that the space Γc is locally arcwise connected, i. e. for
every point a ∈ Γc and its open neighbourhood Q there exists a
neighbourhood Q̂ ⊆ Q of a such that every two points b1, b2 ∈ Q̂
can be connected by a continuous curve in Q. This is a straight-
forward corollary of the remark subsequent to Definition 1.
Since the map Vc is not open by our supposition, then there
exists an open subset O of Γc such that its image R = Vc(O) is
not open in R. Therefore there is a point d ∈ R \ IntR. Fix
p ∈ V −1
c (d) ∩O.
Let us show that p is a point of local extremum of Vc. Fix a
neighbourhood Ô ⊆ O of p such that every two points b1, b2 ∈ Ô
can be connected by a continuous curve βb1,b2 : I → Γc which
meets relations β(0) = b1, β(1) = b2 and β(I) ⊆ O. It is clear that
an image of a path-connected set under a continuous mapping is
On conjugate pseudo-harmonic functions 515
path-connected, therefore following inclusions are valid
(
Vc(b1), Vc(b2)
)
⊂ Vc(I) if Vc(b1) < Vc(b2) ,(
Vc(b2), Vc(b1)
)
⊂ Vc(I) if Vc(b2) > Vc(b1) .
Evidently, p is not an interior point of Vc(Ô) since it is not the
interior point of Vc(O) by construction and Vc(Ô) ⊆ Vc(O). Then
there does not exist a pair of points b1, b2 ∈ Ô such that
Vc(b1) < Vc(p) < Vc(b2)
and either V (b) ≤ V (p) for all b ∈ Ô or V (b) ≥ V (p) for all b ∈ Ô,
i. e. p is the point of local extremum of Vc.
Now, since V is the conjugate pseudo-harmonic function of U
in the point p (see Definition 3), we can take by definition a neigh-
bourhood N of p in M2 and a homeomorphism T : D → N such
that a map f : D → C
f(z) = u(z) + iv(z) , z ∈ D
is holomorphic on D. Here
u = U ◦ T : D → R
and
v = V ◦ T : D → R.
It is clear that without loss of generality we can choose N so
small that either V (b) = Vc(b) ≤ Vc(p) = V (p) for every b ∈ N∩Γc
or V (b) ≥ V (p) for all b ∈ N ∩ Γc.
Let for definiteness p is the local maximum of Vc and
V (b) ≤ V (p)
for every b ∈ N ∩Γc. The case when p is the local minimum of Vc
is considered similarly.
On one hand it follows from what we said above that
(
{U(p)} × (V (p),+∞)
)
∩ f(D) = ∅
516 Yevgen Polulyakh
since u−1(U(p)) = T−1(Γc ∩ N) and v(z) = V (T (z)) ≤ V (p) for
all z ∈ T−1(Γc ∩N) by construction. Therefore a point
U(p) + iV (p) = f(T−1(p))
is not the interior point of a set f(D).
On the other hand it is known that the holomorphic map f is
open, so the point f(T−1(p)) must be the interior point of the
domain f(D).
The contradiction obtained shows that our initial assumption is
false and V is open on level sets of U .
Sufficiency. Let U be a pseudo-harmonic function on M2 and
a continuous function V : M2 → R be open on level sets of U .
From Lemma 1 it follows that the mapping F : M2 → C, F (p) =
U(p) + iV (p), p ∈M2 is interior.
Let p ∈M2 and N is a simple neighbourhood of p in M2. Then
there exists a homeomorphism T : D → N . It is straightforward
that for the open set N a mapping FN = F |N : N → C is in-
terior and its composition FN ◦ T = F ◦ T : D → C with the
homeomorfism T is also an interior mapping.
Now from Stoilov theorem it follows that there exists a complex
structure on D such that the mapping F ◦T is holomorphic in this
complex structure (see [3]). But from the uniformization theorem
(see [4]) it follows that a simple-connected domain has a unique
complex structure. So the mapping F ◦ T is holomorphic on D in
the standard complex structure. Thus the functions
u = Re(F ◦ T ) = U ◦ T
and
v = Im(F ◦ T ) = V ◦ T
are conjugate harmonic functions on D. Consequently, V is a
conjugate pseudo-harmonic function of U in the point p.
From arbitrariness in the choise of p ∈ M2 it follows that V is
a conjugate pseudo-harmonic function of U on M2. �
On conjugate pseudo-harmonic functions 517
Corollary 1. Let U, V : M2 → R be conjugate pseudoharmonic
functions on M2.
Then there exists a complex structure on M2 with respect to
which U and V are conjugate harmonic functions on M2.
Proof. This statement follows from Theorem 1, Lemma 1 and the
Stoilov theorem which says that there exists a complex structure
on M2 such that the interior mapping F (p) = U(p) + iV (p), p ∈
M2 is holomorphic in this complex structure (see [3]). �
References
[1] Tôki Y., A topological characterization of pseudo-harmonic functions, Os-
aka Math. Journ. — 1951 — V.3, N 1. — P. 101–122.
[2] Morse M., Topological methods in the theory of functions of a complex
variable. — Princeton, 1947.
[3] Ñ. Ñòîèëîâ, Ëåêöèè î òîïîëîãè÷åñêèõ ïðíöèïàõ òåîðèè àíàëèòè-÷åñêèõ �óíêöèé. � Ì.: Íàóêà, 1964. � 228 ñ.
[4] Forster O., Lectures on Riemann Surfaces. // Springer Graduate Texts
in Math. — 1981. — V. 81.
|
| id | nasplib_isofts_kiev_ua-123456789-6330 |
| institution | Digital Library of Periodicals of National Academy of Sciences of Ukraine |
| issn | 1815-2910 |
| language | English |
| last_indexed | 2025-12-07T17:21:20Z |
| publishDate | 2009 |
| publisher | Інститут математики НАН України |
| record_format | dspace |
| spelling | Polulyakh, Ye. 2010-02-23T14:52:26Z 2010-02-23T14:52:26Z 2009 On conjugate pseudo-harmonic functions / Ye. Polulyakh // Збірник праць Інституту математики НАН України. — 2009. — Т. 6, № 2. — С. 505-517. — Бібліогр.: 4 назв. — англ. 1815-2910 https://nasplib.isofts.kiev.ua/handle/123456789/6330 517.54,517.57 We prove the following theorem. Let U be a pseudo-harmonic function on a surface M². For a real valued continuous function V : M² → R to be a conjugate pseudo-harmonic function of U on M² it is necessary and sufficient that V is open on level sets of U. en Інститут математики НАН України Геометрія, топологія та їх застосування On conjugate pseudo-harmonic functions Article published earlier |
| spellingShingle | On conjugate pseudo-harmonic functions Polulyakh, Ye. Геометрія, топологія та їх застосування |
| title | On conjugate pseudo-harmonic functions |
| title_full | On conjugate pseudo-harmonic functions |
| title_fullStr | On conjugate pseudo-harmonic functions |
| title_full_unstemmed | On conjugate pseudo-harmonic functions |
| title_short | On conjugate pseudo-harmonic functions |
| title_sort | on conjugate pseudo-harmonic functions |
| topic | Геометрія, топологія та їх застосування |
| topic_facet | Геометрія, топологія та їх застосування |
| url | https://nasplib.isofts.kiev.ua/handle/123456789/6330 |
| work_keys_str_mv | AT polulyakhye onconjugatepseudoharmonicfunctions |