Electronic Journal of Differential Equations, Vol. 2024 (2024), No. 71, pp. 1–12.

ISSN: 1072-6691. URL: https://ejde.math.txstate.edu, https://ejde.math.unt.edu

DOI: 10.58997/ejde.2024.71

CONSTRUCTION OF SOLUTIONS TO PDES USING

HOLOMORPHIC FUNCTIONS OF SEVERAL VARIABLES

VITALII SHPAKIVSKYI

Abstract. The work develops Humbert’s method of solving PDEs. It applies

method for constructing solutions of the three-dimensional Laplace, Helmholtz,
and Poisson equations in the form of components of holomorphic functions of

several variables (complex and hypercomplex).

1. Introduction

In 1929, Humbert [3] proposed an original idea for constructing solutions of the
two-dimensional Laplace equation

∆2U(x, y) :=
∂2U

∂x2
+
∂2U

∂y2
= 0. (1.1)

Namely, introducing the change of variables

u = x+ iy,

v = x− iy,
(1.2)

it is easy to verify that in the new variables u, v equation (1.1) takes the form

∂2U

∂u∂v
= 0. (1.3)

The general solution of equation (1.3) is the function U = f(u)+g(v), where f and
g are ”arbitrary” functions of their arguments from a certain class of functions.

The goal of Humbert was to generalize equation (1.3) to the spatial case. Thus,
he considers the equation

∂3U

∂u∂v∂w
= 0. (1.4)

By introducing the change of variables

u = x+ y + z,

v = x+ jy + j2z,

w = x+ j2y + jz,

(1.5)

2020 Mathematics Subject Classification. 32A10, 30G35, 35A25.
Key words and phrases. Holomorphic functions of several complex variables;

holomorphic functions of several hypercomplex variables; harmonic algebra; elliptic PDE.
©2024. This work is licensed under a CC BY 4.0 license.

Submitted September 12, 2024. Published November 11, 2024.

1



2 V. SHPAKIVSKYI EJDE-2024/71

where {1, j, j2} is a basis of a commutative associative algebra such that j3 = 1,
Humbert arrives to the equation

∂3U

∂x3
+
∂3U

∂y3
+
∂3U

∂z3
− 3

∂3U

∂x∂y∂z
= 0, (1.6)

which he calls the spatial generalization of equation (1.1). Equation (1.6) is usually
called Humbert’s equation.

Note that (1.4) has the general solution

U = F (u, v) +G(v, w) +H(w, u),

where F,G,H are “arbitrary” functions of their arguments from a certain class of
functions. In particular, a solution of (1.6) is the function

U0 = lnu+ ln v + lnw = lnuvw.

Since uvw = x3 + y3 + z3 − 3xyz, it follows that U0 = ln(x3 + y3 + z3 − 3xyz).
The dissertation by Devisme [2] is devoted to the study of the properties of

solutions (1.6). Also, (1.6) and the function theory of the hypercomplex variable v =
x+jy+j2z were considered by Roşculeţ [12]. He constructed a theory of monogenic
functions similar to the theory of analytic functions of a complex variable.

Our goal is to develop Humbert’s idea and make it suitable for a predetermined
PDEs. The main one the problem here is to choose a suitable algebra (in our case,
commutative and associative) and the appropriate substitution of type (1.2), (1.5).
The basic principles of algebra selection and suitable replacement of variables will
be demonstrated on the example of the three-dimensional Laplace, Helmholtz, and
Poisson equations.

2. Change of variables for a differential equation in the real case

For simplicity, we study the equation in the three variables x, y, z. In the domain
Ω ⊂ R3 we consider the differential equation

F
(
x, y, z,

∂f

∂x
,
∂f

∂y
,
∂f

∂z
, . . . ,

∂nf

∂xα∂yβ∂zγ

)
= 0, α+ β + γ = n. (2.1)

In this article, we use the notation ∂f
∂x = fx,

∂2f
∂x∂y = fxy etc.

One of the methods for solving equation (2.1) is as follows:

(1) for the given equation, we find such a change of variables so that the new
equation has a simpler form in the new coordinates;

(2) we solve the obtained equation in the new coordinates;
(3) we return to the solutions of given equation (2.1).

Humbert used this approach in his reasoning, but, unlike the classical method,
he used the change of variables with values in a commutative associative algebra.

For equation (2.1) we define the change of variables

φ = φ(x, y, z),

ψ = ψ(x, y, z),

η = η(x, y, z),

(2.2)

where the real functions φ,ψ, η are defined in a domain Ω and are continuously
differentiable in it with respect to x, y, z a sufficient number of times. Assume also
that Q is a domain and a function f : Q → R, where Q := {(φ,ψ, η) ∈ R3 :



EJDE-2024/71 SOLUTIONS TO PDES USING HOLOMORPHIC FUNCTIONS 3

(x, y, z) ∈ Ω}, is continuously differentiable in Q with respect to φ,ψ, η a sufficient
number of times.

Now, with the assumptions on a real function f , let us write the first few formulas
for the transition from the variables x, y, z to the variables φ,ψ, η.

fx = fφφx + fψψx + fηηx , (2.3)

fxx = fφφφ
2
x + fψψψ

2
x + fηηη

2
x + 2fφψφxψx + 2fφηφxηx

+ 2fψηψxηx + fφφxx + fψψxx + fηηxx ,
(2.4)

fxy = fφφφxφy + fψψψxψy + fηηηxηy + fφψ(φyψx + φxψy)

+ fφη(φyηx + φxηy) + fψη(ψyηx + ψxηy) + fφφxy + fψψxy + fηηxy ,

fxxx = fφφφφ
3
x + fψψψψ

3
x + fηηηη

3
x + 3fφψψφxψ

2
x + 3fφηηφxη

2
x + 3fφφψφ

2
xψx

+ 6fφψηφxψxηx + 3fφφηφ
2
xηx + 3fψηηψxη

2
x + 3fψψηψ

2
xηx + fφφφxxφx

+ fψψψxxψx + fηηηxxηx + fφψ(φxψxx + φxxψx) + fφη(φxηxx + φxxηx)

+ fψη(ψxηxx + ψxxηx) + fφφxx + fψψxx + fηηxx ,

fxxy = fφφφφ
2
xφy + fψψψψ

2
xψy + fηηηη

2
xηy + fφψψ(ψ

2
xφy + 2φxψxψy)

+ fφηη(η
2
xφy + 2φxηxηy) + fφφψ(φ

2
xψy + 2φxψxφy) + fφφη(φ

2
xηy

+ 2φxφyηx) + fφψη(2ψxηxφy + 2φxηxψy + 2φxψxηy)

+ fψηη(η
2
xψy + 2ψxηxηy) + fψψη(ψ

2
xηy + 2ψxηxψy)

+ fφφφxxφy + fψψψxxψy + fηηηxxηy

+ fφψ(ψxxφy + φxxψy) + fφη(ηxxφy + φxxηy)

+ fψη(ηxxψy + ψxxηy) + fφφyy + fψψyy + fηηyy ,

fxyz = fφφφφxφyφz + fψψψψxψyψz + fηηηηxηyηz

+ fφψψ(ψxψyφz + φyψxψz + φxψyψz) + fφηη(ηxηyφz + φyηxηz + φxηyηz)

+ fφφψ(φzφyψx + φxψyφz + φxφyψz) + fφφη(φyφzηx

+ φxηyφz + φxφyηz) + fφψη(φzψyηx + ψxηyφz + φyηxψz

+ φxηyψz + φyψxηz + φxψyηz) + fψψη(ψyψzηx + ψxηyψz + ψxψyηz)

+ fψηη(ηxηyψz + ψyηxηz + ψxηyηz) + fφφφxyφz + fψψψxyψz

+ fηηηxyηz + fφψ(ψxyφz + φxyψz) + fφη(ηxyφz + φxyηz)

+ fψη(ηxyψz + ψxyηz) + fφφzz + fψψzz + fηηzz .

Part I

Solutions of Laplace, Helmholtz, and Poisson equations in the form of
components of holomorphic functions of several complex variables.



4 V. SHPAKIVSKYI EJDE-2024/71

3. change of variables in a differential equation in the complex case

Now let functions (2.2) be complex-valued: φ,ψ, η : Ω → C, Ω ⊂ R3, and
the functions φ,ψ, η are continuously differentiable with respect to x, y, z in Ω a
sufficient number of times. Suppose also that the function f : Q → C, where
Q := {(φ,ψ, η) ∈ C3 : (x, y, z) ∈ Ω}. Now note that for the holomorphic function f
of three complex variables φ,ψ, η the formulas for the transition from the derivatives
by x, y, z to the derivatives by φ,ψ, η, which are given in section 2, remain valid.

By dividing the increment of the holomorphic function

∆f(φ,ψ, η) = fφ∆φ+ fψ∆ψ + fη∆η + o
(√

|∆φ|2 + |∆ψ|2 + |∆η|2
)

on ∆x and passing to the limit as ∆x → 0, we obtain equality (2.3). Taking into
that the holomorphic function has derivatives of all orders by all variables, which
are again holomorphic functions, we get the rest of the transition formulas. The
class of functions f : Q → C, holomorphic in the domain Q ⊂ C3, is denoted by
HC(Q).

4. Solutions of the Laplace equation in the form of components of
holomorphic functions of three complex variables

We consider the Laplace equation

∆3f(x, y, z) :=
∂2f

∂x2
+
∂2f

∂y2
+
∂2f

∂z2
= 0 (4.1)

in a domain Ω ⊂ R3. For this, in equation (4.1) we will make twice continuously
differentiable change of variables (2.2). In this case, equation (4.1) will turn into
a certain equation that we will solve in the class HC(Q), where Q := {(φ,ψ, η) ∈
C3 : (x, y, z) ∈ Ω}.

Let us transition to the new coordinates (2.2), using the transition formulas
(2.4). We have

∆3f(x, y, z)

= fφφ(φ
2
x + φ2

y + φ2
z) + fψψ(ψ

2
x + ψ2

y + ψ2
z) + fηη(η

2
x + η2y + η2z)

+ 2fφψ(φxψx + φyψy + φzψz) + 2fφη(φxηx + φyηy + φzηz)

+ 2fψη(ψxηx + ψyηy + ψzηz) + fφ(φxx + φyy + φzz)

+ fψ(ψxx + ψyy + ψzz) + fη(ηxx + ηyy + ηzz) = 0.

(4.2)

Now we need to choose the functions φ,ψ, η so that equation (4.2) turns into an
equation that can be solved explicitly. Or, much better, let the equation turn into
an identity. For this purpose, let us consider the linear change of variables

φ = a1x+ b1y + c1z,

ψ = a2x+ b2y + c2z,

η = a3x+ b3y + c3z.

(4.3)

In this case, equation (4.2) takes the form

∆3f(x, y, z) = fφφ(a
2
1 + b21 + c21) + fψψ(a

2
2 + b22 + c22) + fηη(a

2
3 + b23 + c23)

+ 2fφψ(a1a2 + b1b2 + c1c2) + 2fφη(a1a3 + b1b3 + c1c3)

+ 2fψη(a2a3 + b2b3 + c2c3) = 0.

(4.4)



EJDE-2024/71 SOLUTIONS TO PDES USING HOLOMORPHIC FUNCTIONS 5

For equation (4.4) to turn into an identity, it is sufficient that the substitution
coefficients (4.3) satisfy the system of equations

a21 + b21 + c21 = 0,

a22 + b22 + c22 = 0,

a23 + b23 + c23 = 0,

a1a2 + b1b2 + c1c2 = 0,

a1a3 + b1b3 + c1c3 = 0,

a2a3 + b2b3 + c2c3 = 0.

(4.5)

Note that the system of equations (4.5) does not have non-trivial solutions on the
set of real numbers. But this system has solutions on the set of complex numbers.
Thus, we have proved the following theorem.

Theorem 4.1. In the class HC(Q) equation (4.1) is satisfied by the function

f(x, y, z) = F (a1x+ b1y + c1z, a2x+ b2y + c2z, a3x+ b3y + c3z), (4.6)

where F is an arbitrary holomorphic function of three complex variables in Q =
{(φ,ψ, η) ∈ C3 : (x, y, z) ∈ Ω}, φ,ψ, η are defined by equalities (4.3), and the coef-
ficients ak, bk, ck at k = 1, 2, 3, are arbitrary complex numbers that satisfy system
(4.5).

Note that the real and imaginary part of function (4.6) satisfies equation (4.1)
in the domain Ω.

5. Solutions of the Helmholtz equation in the form of components
of holomorphic functions of three complex variables

Consider the three-dimensional Helmholtz equation

(∆3 + λ2)f(x, y, z) =
∂2f

∂x2
+
∂2f

∂y2
+
∂2f

∂z2
+ λ2f = 0, λ ∈ C (5.1)

in a domain Ω ⊂ R3. For this, in equation (5.1), we will replace variables (4.3). In
this case, equation (5.1) will turn into a certain equation that we will solve in the
class HC(Q), where Q := {(φ,ψ, η) ∈ C3 : (x, y, z) ∈ Ω}.

Under such assumptions in equation (5.1), we will move to new coordinates (4.3).
Using equality (4.4), we have

∆3f(x, y, z) = fφφ(a
2
1 + b21 + c21) + fψψ(a

2
2 + b22 + c22) + fηη(a

2
3 + b23 + c23)

+ 2fφψ(a1a2 + b1b2 + c1c2) + 2fφη(a1a3 + b1b3 + c1c3)

+ 2fψη(a2a3 + b2b3 + c2c3) + λ2f = 0.

(5.2)

On the coefficients ak, bk, ck, k = 1, 2, 3, we impose the following conditions

a21 + b21 + c21 = λ2,

a22 + b22 + c22 = 0,

a23 + b23 + c23 = 0,

a1a2 + b1b2 + c1c2 = 0,

a1a3 + b1b3 + c1c3 = 0,

a2a3 + b2b3 + c2c3 = 0.

(5.3)



6 V. SHPAKIVSKYI EJDE-2024/71

In this case, equation (5.2) will turn into the equation

fφφ(φ,ψ, η) + f(φ,ψ, η) = 0. (5.4)

The general solution of equation (5.4) is the function

f(φ,ψ, η) = (α cosφ+ β sinφ)G(ψ, η),

where α, β are arbitrary complex constants, and G(ψ, η) is an arbitrary holomorphic
function in Q2 := {(ψ, η) ∈ C2 : (x, y, z) ∈ Ω}. Thus, we have proved the following
theorem.

Theorem 5.1. In the class HC(Q) equation (5.1) is satisfied by the function

f(x, y, z) =
(
α cos(a1x+ b1y + c1z) + β sin(a1x+ b1y + c1z)

)
××G(a2x+ b2y + c2z, a3x+ b3y + c3z),

(5.5)

where G is an arbitrary holomorphic function in the domain Q2 := {(ψ, η) ∈ C2 :
(x, y, z) ∈ Ω}, ψ, η are defined by equalities (4.3), coefficients α, β are arbitrary com-
plex numbers, and numbers ak, bk, ck for k = 1, 2, 3, are arbitrary complex numbers,
which satisfy the system (5.3).

Note that the system of equations (5.3) does not have non-trivial solutions on the
set of real numbers, but this system has solutions on the set of complex numbers.

Note that the real and imaginary parts of function (5.5) satisfy equation (5.1)
in the domain Ω.

6. Solutions of Poisson equation in the form of components of
holomorphic functions of three complex variables

Consider the three-dimensional Poisson equation

∂2f

∂x2
+
∂2f

∂y2
+
∂2f

∂z2
= g(x, y, z) (6.1)

in a domain Ω ⊂ R3. For this, in equation (6.1), we will replace variables (4.3). In
this case, equation (6.1) will turn into a certain equation that we will solve in the
class HC(Q), where Q := {(φ,ψ, η) ∈ C3 : (x, y, z) ∈ Ω}. The conditions for the
function g will be given below.

Under such assumptions in equation (6.1), we will move to new coordinates (4.3).
Using equality (4.4), equation (6.1) takes the form

fφφ(a
2
1 + b21 + c21) + fψψ(a

2
2 + b22 + c22) + fηη(a

2
3 + b23 + c23)

+ 2fφψ(a1a2 + b1b2 + c1c2) + 2fφη(a1a3 + b1b3 + c1c3)

+ 2fψη(a2a3 + b2b3 + c2c3) = g(φ,ψ, η).

(6.2)

On the coefficients ak, bk, ck, k = 1, 2, 3, we impose the following conditions

a21 + b21 + c21 = 1,

a22 + b22 + c22 = 0,

a23 + b23 + c23 = 0,

a1a2 + b1b2 + c1c2 = 0,

a1a3 + b1b3 + c1c3 = 0,

a2a3 + b2b3 + c2c3 = 0.

(6.3)



EJDE-2024/71 SOLUTIONS TO PDES USING HOLOMORPHIC FUNCTIONS 7

In this case, equation (6.2) turns into the equation

fφφ(φ,ψ, η) = g(φ,ψ, η). (6.4)

The general solution of equation (6.4) is the function

f(φ,ψ, η) =

∫∫
g(φ,ψ, η)dφdφ+ φh1(ψ, η) + h2(ψ, η),

where h1, h2 are arbitrary holomorphic functions in the domain Q2 := {(ψ, η) ∈
C2 : (x, y, z) ∈ Ω}. Thus, we have proved the following theorem.

Theorem 6.1. Let in equation (6.1) the function g(φ,ψ, η) for some set of param-
eters (6.3) be a holomorphic in the domain Q := {(ψ,ψ, η) ∈ C3 : (x, y, z) ∈ Ω}.
Then in the class HC(Q) equation (6.1) is satisfied by the function

f(φ,ψ, η) =

∫∫
g(φ,ψ, η)dφdφ+ φh1(ψ, η) + h2(ψ, η), (6.5)

where h1, h2 are arbitrary holomorphic functions in the domain Q2 := {(ψ, η) ∈
C2 : (x, y, z) ∈ Ω}, φ,ψ, η are defined by equalities (4.3), and coefficients ak, bk, ck,
k = 1, 2, 3, are determined by equalities (6.3).

Note that the system of equations (6.3) does not have non-trivial solutions on the
set of real numbers, but this system has solutions on the set of complex numbers.
Also note that the real and imaginary parts of function (6.5) satisfy equation (6.1)
in the domain Ω.

Part 2

Solutions of the Laplace equation in the form of components of holomor-
phic functions of three hypercomplex variables.

7. Holomorphic functions of three hypercomplex variables

Now let functions (2.2) take values in some commutative associative algebra A
over the field R or C, i.e. φ,ψ, η : Ω → A, Ω ⊂ R3. Let the functions φ,ψ, η be
continuously differentiable in x, y, z in Ω a sufficient number of times. Suppose that
the function f takes values in the algebra A: f : D → A, where D := {(φ,ψ, η) ∈
A3 : (x, y, z) ∈ Ω}. Let us ask the question: under what conditions for the function
f do the formulas for the transition from the derivatives by x, y, z to the derivatives
by φ,ψ, η, which are given in section 2? For this, first of all, it is necessary to explain
how we will understand the derivative.

Definition 7.1. We say that the function f : D → A, D ⊂ A3, of three hyper-
complex variables φ,ψ, η ∈ A has a partial derivative with respect to the variable
φ in the domain D, if f is differentiable with respect to this variable in the sense of
Lorch [6] at each point of the domain D, i.e. if there exists an element of algebra
fφ such that for an arbitrary ε > 0 there exists δ > 0 such that for all h1 ∈ A with
∥h1∥ < δ the inequality

∥f(φ+ h1, ψ, η)− f(φ,ψ, η)− h1fφ(φ,ψ, η)∥ ≤ ∥h1∥ε
holds. The element fφ is called the Lorch partial derivative of the function f with
respect to the variable φ.

Similar to the complex case [13, p. 270], we introduce the following definition.



8 V. SHPAKIVSKYI EJDE-2024/71

Definition 7.2. The function f : D → A, D ⊂ A3, of three hypercomplex vari-
ables φ,ψ, η ∈ A is called holomorphic in the domain D, if in this domain there
exist continuous Lorch partial derivatives fφ, fψ, fη and in D the increment of the
function ∆f := f(φ+∆φ,ψ +∆ψ, η +∆η)− f(φ,ψ, η) is given in the form

∆f = fφ∆φ+ fψ∆ψ + fη∆η + o
(√

∥∆φ∥2 + ∥∆ψ∥2 + ∥∆η∥2
)
, (7.1)

where o(·) → 0 as ∆φ→ 0,∆ψ → 0,∆η → 0.

Remark 7.3. In equality (7.1) we pass to the limit as ∆φ→ 0,∆ψ → 0,∆η → 0.
It is obvious that in this case ∆f → 0. And this means that every holomorphic
function in the domain D ⊂ A3 is continuous in this domain.

It is obvious that every function holomorphic of three hypercomplex variables is
differentiable in the sense of Lorch with respect to each variable separately.

Note that all elementary functions of the three variables φ,ψ, η ∈ A are holo-
morphic.

Now we divide equality (7.1) by ∆x and pass to the limit as ∆x → 0, we
obtain equality (2.3). To obtain PDEs faster, without resorting to the study of
the holomorphic functions theory, let us additionally assume that the functions fφ,
fψ, fη are also holomorphic in D. Under this assumption, from equality (2.3) we
immediately obtain equality (2.4).

The class of functions f : D → A, holomorphic in the domain D ⊂ A3 together
with Lorch partial derivatives fφ, fψ, fη, denote by H2

A(D).

8. Solutions of the Laplace equation in the form of components of
holomorphic functions of three hypercomplex variables

In this section, we look for solutions of equation (4.1) in a domain Ω ⊂ R3 using
the change of variables (4.3) under the assumption that the coefficients ak, bk, ck at
k = 1, 2, 3, are elements of some three-dimensional commutative associative algebra
over the field C. In this case, equation (4.1) will turn into a certain equation that
we will solve in class H2

A(D), where D := {(φ,ψ, η) ∈ A3 : (x, y, z) ∈ Ω}.
So, in system (4.5) we put a1 = c2 = b3 = e1, b1 = a2 = c3 = e2, c1 = b2 = a3 =

e3, where e1, e2, e3 are basis elements of some algebra. That is,

φ = xe1 + ye2 + ze3 ,

ψ = xe2 + ye3 + ze1 ,

η = xe3 + ye1 + ze2 .

(8.1)

Then the first three equations of system (4.5) are equivalent to one equation – this
is the condition of harmonicity of vectors (see [4, 10]):

e21 + e22 + e23 = 0.

If we consider e1 = 1 as a unit of algebra, then the fourth, fifth and sixth equations
of system (4.5) are equivalent to one condition:

e2e3 = −e2 − e3. (8.2)

Our problem was reduced to finding a three-dimensional (though not necessarily
three-dimensional) harmonic algebra with multiplication condition (8.2).



EJDE-2024/71 SOLUTIONS TO PDES USING HOLOMORPHIC FUNCTIONS 9

In [1], a multiplication table of all harmonic bases for e1 = 1 in three-dimensional
algebras was obtained:

e22 =
(
p(p+ t)∓m

√
−1− (p+ t)2

)
e1 +

(
m±

√
−1− (p+ t)2

)
e2 + pe3,

e2e3 =
(
−m(p+ t)∓ p

√
−1− (p+ t)2

)
e1 + te2 +me3,

e23 =
(
− 1− p(p+ t)±m

√
−1− (p+ t)2

)
e1

+
(
−m∓

√
−1− (p+ t)2

)
e2 − pe3 ,

(8.3)

where m, t, p are arbitrary complex numbers, and in double signs ∓, ±, either
upper or lower signs are taken simultaneously. Thus, condition (8.2) leads us to the
equalities t = m = −1 and to the equation

(p− 1)∓ p
√
−1− (p− 1)2 = 0.

The solutions of the above equation are numbers p = 1
2 ±

√
3
2 i. By substituting the

obtained values of p and t = m = −1, we obtain two families of harmonic bases.
The first family of harmonic bases is multiplied by the following multiplication
rules:

e22 = (−1

2
±

√
3

2
i)e1 + (−1

2
±

√
3

2
i)e2 + (

1

2
±

√
3

2
i)e3,

e2e3 = −e2 − e3,

e23 = (−1

2
∓

√
3

2
i)e1 + (

1

2
∓

√
3

2
i)e2 + (−1

2
∓

√
3

2
i)e3 ,

(8.4)

and the second family multiplies as follows:

e22 = (−3

2
∓

√
3

2
i)e1 + (−3

2
∓

√
3

2
i)e2 + (

1

2
±

√
3

2
i)e3,

e2e3 = −e2 − e3,

e23 = (
1

2
±

√
3

2
i)e1 + (

3

2
±

√
3

2
i)e2 + (−1

2
∓

√
3

2
i)e3 ,

(8.5)

where in double signs ∓, ±, either upper or lower signs are taken simultaneously.
Thus, we have proved the following theorem.

Theorem 8.1. In the class H2
A(D), equation (4.1) is satisfied by the function

f(x, y, z) = F (xe1 + ye2 + ze3, xe2 + ye3 + ze1, xe3 + ye1 + ze2), (8.6)

where F is an arbitrary holomorphic function in D = {(φ,ψ, η) ∈ A3 : (x, y, z) ∈
Ω}, φ,ψ, η are defined by equalities (8.1), and the vectors e1, e2, e3 are defined by
equalities (8.4) and (8.5).

If we consider that function (8.6) depends only of one variable xe1+ye2+ze3, then
the solution f = F (xe1+ye2+ze3) will be satisfy equation (4.1) due to the condition
of harmonicity of vectors e1, e2, e3. Solutions of the form f = F (xe1 + ye2 + ze3)
are studied in works [7, 11, 9, 8].

Since the Laplace equation (4.1) is linear, all real-valued components of function
(8.6) are solutions of equation (4.1).



10 V. SHPAKIVSKYI EJDE-2024/71

9. Example

Let us show how some harmonic functions are presented in form (8.6). For this,
consider, for example, homogeneous linearly independent harmonic polynomials of
the third degree. It is known that there are 7 of them. Let us denote

U1 := y3 − 3z2y, U2 := z3 − 3y2z, U3 := x3 − 3y2x, U4 := x3 − 3z2x,

U5 := y3 − 3x2y, U6 := z3 − 3x2z, U7 := xyz.

In the notation (8.1) we have:

U1 = Re
{ 1√

3

(
− 2ie3

1 + i
√
3
+ 1

)
η3
}
,

U2 = Re
{ 1√

3

(
− 2ie3

1 + i
√
3
− i+ 1−

√
3
)
φ3

}
,

U3 = Re
{ 1√

3

(
− 2ie3

1 + i
√
3
+ 1

)
φ3

}
,

U4 = Re
{ 1√

3

(
− 2ie3

1 + i
√
3
− i+ 1−

√
3
)
η3
}
,

U5 = Re
{ 1√

3

(
− 2ie3

1 + i
√
3
− i+ 1−

√
3
)
ψ3

}
,

U6 = Re
{ 1√

3

(
− 2ie3

1 + i
√
3
+ 1

)
ψ3

}
.

10. Necessary and sufficient conditions for the holomorphicity of a
function in an algebra

Theorem 10.1. For the function f : D → A to be holomorphic in the domain
D ⊂ A3 it is necessary and sufficient that the function f be a differentiable function
of three real variables in the domain Ω := {(x, y, z) ∈ R3 : (φ,ψ, η) ∈ D}, where
x, y, z are related to φ,ψ, η by equalities (8.1), and that the following conditions be
fulfilled:

fx = fφe1 + fψe2 + fηe3 ,

fy = fφe2 + fψe3 + fηe1 ,

fz = fφe3 + fψe1 + fηe2 .

(10.1)

Proof. Necessity. Let the function f be holomorphic. Then the equality (7.1) will
be rewritten in the form

∆f = fφ(∆xe1 +∆ye2 +∆ze3) + fψ(∆xe2 +∆ye3 +∆ze1)

+ fη(∆xe3 +∆ye1 +∆ze2) + o
(√

∥∆φ∥2 + ∥∆ψ∥2 + ∥∆η∥2
)

= (fφe1 + fψe2 + fηe3)∆x+ (fφe2 + fψe3 + fηe1)∆y

+ (fφe3 + fψe1 + fηe2)∆z + o
(√

3 ·
√

(∆x)2 + (∆y)2 + (∆z)2
)
.

(10.2)

Equality (10.2) means that the function f is differentiable as a function of three
real variables x, y, z, and the conditions (10.1) are fulfilled.

Sufficiency. Since the function f is differentiable then the equality

∆f = fx∆x+ fy∆y + fz∆z + o
(√

3 ·
√

(∆x)2 + (∆y)2 + (∆z)2
)

(10.3)



EJDE-2024/71 SOLUTIONS TO PDES USING HOLOMORPHIC FUNCTIONS 11

holds. We will substitute relation (10.1) into equality (10.3), and we have

∆f = (fφe1 + fψe2 + fηe3)∆x+ (fφe2 + fψe3 + fηe1)∆y

+ (fφe3 + fψe1 + fηe2)∆z + o
(√

3 ·
√
(∆x)2 + (∆y)2 + (∆z)2

)
= fφ∆φ+ fψ∆ψ + fη∆η + o

(√
∥∆φ∥2 + ∥∆ψ∥2 + ∥∆η∥2

)
.

□

Equalities (10.1) are analogs of the Cauchy-Riemann conditions. Let us trans-
form these equalities. We multiply the first equality with (10.1) by e1 and add to
the second equality multiplied by e2 and add to the third equality multiplied by e3.
At the same time we get equality

fxe1 + fye2 + fze3 = 0.

Similarly, multiplying equalities (10.1) respectively by e2, e3, e1 and adding the
resulting expressions, we will have

fxe2 + fye3 + fze1 = 0.

Finally, multiplying equalities (10.1) respectively by e3, e1, e2 and adding the re-
sulting expressions, we obtain

fxe3 + fye1 + fze2 = 0.

So the consequence of conditions (10.1) is the conditions

fxe1 + fye2 + fze3 = 0,

fxe2 + fye3 + fze1 = 0,

fxe3 + fye1 + fze2 = 0.

(10.4)

That is, the holomorphic function of three hypercomplex variables φ,ψ, η necessar-
ily is hyperholomorphic (see, for example, [5]) for each variable separately.

Acknowledgments. This work was supported by a grant from the Simons Foun-
dation (1030291,V.S.Sh.).

References

[1] Beckh-Widmanstetter, H. A. v.; Laßt sich die Eigenschaft der analytischen Funktionen einer
gemeinen komplexen veranderlichen, Potentiale als Bestandteile zu liefern, auf Zahlsysteme

mit drei einheiten verallgemeinern? Monatshefte für Mathematik und Physik, 23 (1912),

257–260.
[2] Devisme, J.; Sur l’équation de M. Pierre Humbert, Theses de l’entre-deux-guerres, 148

(1933), 108 p.

[3] Humbert, P.; Sur une généralisation de l’équation de Laplace, Journal de Mathématiques
Pures et Appliquées, Serie 9, 8 (1929), 145–159.

[4] Ketchum, P. W.; Analytic Functions of Hypercomplex Variables, Trans. Amer. Math. Soc.,
30 (1928), 641–667.

[5] Kravchenko, V. V.; Shapiro, M. V.; Integral representations for spetial models of mathemat-

ical physics. Pitman Research Notes in Mathematics Series 351, Addison Wesley Longman,
1996, 247 p.

[6] Lorch, E. R.; The theory of analytic function in normed abelin vector rings, Trans. Amer.

Math. Soc., 54 (1943), 414–425.
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Laplace equation, Ukr. Math. J., 55 (9) (2003), 1551–1557.



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[8] Plaksa, S. A.; Pukhtaevich, R. P.; Constructive description of monogenic functions in a three-

dimensional harmonic algebra with one-dimensional radical, Ukr. Math. J., 65 (5) (2013),

740–751.
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commutative algebra, An. St. Univ. Ovidius Constanta, 22 (1) (2014), 221–235.

[10] Plaksa, S. A.; Shpakivskyi, V. S.; Monogenic Functions in Spaces with Commutative Multi-
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siuni, Studii şi Cercetări Matematice, 5, 3–4 (1954), 361–401.

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(in Russian).

Vitalii Shpakivskyi

Department of Complex Analysis and Potential Theory, Institute of Mathematics of
the National Academy of Sciences of Ukraine, 3, Tereschenkivska st., Kyiv-4, 01024,

Ukraine

Email address: shpakivskyi86@gmail.com


	1. Introduction
	2. Change of variables for a differential equation in the real case
	Part I
	Solutions of Laplace, Helmholtz, and Poisson equations in the form of components of holomorphic functions of several complex variables

	3. change of variables in a differential equation in the complex case
	4. Solutions of the Laplace equation in the form of components of holomorphic functions of three complex variables
	5. Solutions of the Helmholtz equation in the form of components of holomorphic functions of three complex variables
	6. Solutions of Poisson equation in the form of components of holomorphic functions of three complex variables
	Part 2
	Solutions of the Laplace equation in the form of components of holomorphic functions of three hypercomplex variables

	7. Holomorphic functions of three hypercomplex variables
	8. Solutions of the Laplace equation in the form of components of holomorphic functions of three hypercomplex variables
	9. Example
	10. Necessary and sufficient conditions for the holomorphicity of a function in an algebra
	Acknowledgments

	References