Representations of Some Classes of Quaternionic Hyperholomorphic Functions

Abstract

In the algebra of complex quaternions $\mathbb {H(C)}$ we consider the left– and right–$\psi$–hyperholomorphic functions, and left–$\Lambda -\psi$–hyperholomorphic functions. We justify the transition in left– and right–$\psi$–hyperholomorphic functions to a simpler basis i.e., to the Cartan basis. Using Cartan’s basis we find the solution of Cauchy–Fueter equation. By the same method we find representations of left– and right–$\psi$–hyperholomorphic functions, and representation of left–$\Lambda -\psi$–hyperholomorphic functions.

1 Introduction

Our main object of interest is the set which is usually called the set of complex quaternions and which is traditionally denoted as $\mathbb {H(C)}$. It turns out to be an associative, non–commutative complex algebra generated by the elements 1, I, J, K such that the following multiplication rules hold:

$$\begin{aligned} {\begin{matrix} I^2 =J^2 &{} = K^2 = IJK=-1 , \\ I J = - JI = K , \quad JK &{} = -KJ = I , \quad KI = - IK = J, \end{matrix}} \end{aligned}$$

and the complex imaginary unit i commutes with I, J and K. For $\mathbb {H(C)}$ another name, the algebra of biquaternions, is used also.

Consider in $\mathbb {H(C)}$ another set $\{e_1,e_2,e_3,e_4\}$, which is Cartan’s basis [1] such that

$$\begin{aligned} e_1=\frac{1}{2}(1+i I), \quad e_2=\frac{1}{2}(1-i I), \quad e_3=\frac{1}{2}(iJ - K), \quad e_4=\frac{1}{2}(iJ +K) , \end{aligned}$$

(1.1)

where i is the complex imaginary unit. It is direct to check that we got a new basis.

The multiplication table can be represented as

(1.2)

The unit 1 can be decomposed as $1=e_1+e_2$ .

Note that the subalgebra with the basis $\{e_1,e_2\}$ is the algebra of bicomplex numbers $\mathbb{B}\mathbb{C}$ or Segre’s algebra of commutative quaternions (see, e.g., [2, 3]).

The following relations holds:

$$\begin{aligned} 1=e_1+e_2\,,\quad I=-ie_1+ie_2\,,\quad J=-ie_3-ie_4\,,\quad K=e_4-e_3\,. \end{aligned}$$

(1.3)

Of course, formulas (1.1) and (1.3) give the transition from one basis to the other.

Together with the Hamilton and Cartan bases, we will also consider the Pauli basis. It is a well-known fact that complex quaternions admit the matrix representations via the famous Pauli spin matrices:

$$ \sigma _0:= \left( \begin{array}{cc} 1 &{} 0 \\ 0 &{} 1 \\ \end{array} \right) , \,\, \sigma _1:= \left( \begin{array}{cc} 0 &{} 1 \\ 1 &{} 0 \\ \end{array} \right) , \,\, \sigma _2:= \left( \begin{array}{cc} 0 &{} -i \\ i &{} 0 \\ \end{array} \right) , \,\, \sigma _3:= \left( \begin{array}{cc} 1 &{} 0 \\ 0 &{} -1 \\ \end{array} \right) . $$

In the Pauli basis the multiplication table has the form:

$$ \sigma _1^2=\sigma _2^2=\sigma _3^2=\sigma _0\,, \quad \sigma _1\sigma _2\sigma _3=i\sigma _0\,, $$

$$ \sigma _1\sigma _2=-\sigma _2\sigma _1=i\sigma _3\,, \quad \sigma _2\sigma _3=-\sigma _3\sigma _2=i\sigma _1\,, \quad \sigma _1\sigma _3=-\sigma _3\sigma _1=i\sigma _2\,. $$

Formulas for the transition from the Pauli basis to the Cartan basis have the form

$$\begin{aligned} e_1=\frac{1}{2}(\sigma _0-\sigma _3), \,\, e_2=\frac{1}{2}(\sigma _0+\sigma _3), \,\, e_3=\frac{1}{2}(-\sigma _2-i\sigma _1), \,\, e_4=\frac{1}{2}(-\sigma _2+i\sigma _1). \end{aligned}$$

(1.4)

2 Classes of Hyperholomorphic Functions

Let $\psi _1,\psi _2,\psi _3,\psi _4$ be fixed elements in $\mathbb {H(C)}$ with the following representations in the Cartan’s basis:

$$\begin{aligned} \begin{aligned} \psi _1:=\sum \limits _{s=1}^4\alpha _se_s,\quad \alpha _s \in \mathbb {C},\qquad \psi _2:=\sum \limits _{s=1}^4\beta _se_s,\quad \beta _s \in \mathbb {C},\\ \psi _3:=\sum \limits _{s=1}^4\gamma _se_s,\quad \gamma _s \in \mathbb {C},\qquad \psi _4:=\sum \limits _{s=1}^4\delta _se_s,\quad \delta _s \in \mathbb {C}. \end{aligned}\nonumber \\ \end{aligned}$$

(2.1)

Consider a variable $z=z_1e_1+z_2e_2+z_3e_3+z_4e_4,\,z_s\in \mathbb {C},\,s=1,2,3,4$ and consider a function

$$f(z)=\sum \limits _{s=1}^4f_s(z_1,z_2,z_3,z_4)e_s,\quad f_s:\Omega \rightarrow \mathbb {H(C)},$$

where $\Omega$ is a domain in $\mathbb {C}^4$. Let components $f_s$, $s=1,2,3,4$, be holomorphic functions of four complex variables $z_1,z_2,z_3,z_4$ in $\Omega$.

Consider the operators

$$\begin{aligned} ^{\psi }D[f](z):= & {} \psi _1\frac{\partial f}{\partial z_1}+\psi _2\frac{\partial f}{\partial z_2}+\psi _3\frac{\partial f}{\partial z_3} +\psi _4\frac{\partial f}{\partial z_4}, \end{aligned}$$

(2.2)

$$\begin{aligned} D^\psi [f](z):= & {} \frac{\partial f}{\partial z_1}\,\psi _1+\frac{\partial f}{\partial z_2}\,\psi _2+\frac{\partial f}{\partial z_3}\,\psi _3 +\frac{\partial f}{\partial z_4}\,\psi _4. \end{aligned}$$

(2.3)

Definition 2.1

A function $f:\Omega \rightarrow \mathbb {H(C)}$, $\Omega \subset \mathbb {C}^4$, is called left–$\psi$–hyperholomorphic (or right–$\psi$–hyperholomorphic) if components $f_s$ are holomorphic functions of four complex variables $z_1,z_2,z_3,z_4$ in $\Omega$, and f satisfies the equation

$$\begin{aligned} ^{\psi }D[f](z)=0. \end{aligned}$$

(2.4)

(or $D^{\psi }[f](z)=0.$)

The class of $\psi$–hyperholomorphic functions in the real quaternions algebra is introduced for the first time by Shapiro and Vasilevski in the papers [4, 5]. Since then, these functions have attracted the attention of many researchers. K. Gürlebeck and his student H. M.Nguyen pay a special attention to the applications of $\psi$–hyperholomorphic functions. See, for example, the papers [6,7,8] and dissertation of Nguyen [9]. We note also that operators (2.2) and (2.2) are also called the weighted Dirac operators. Analysis and application of such operators are studied in papers [10, 11].

There are different generalizations of $\psi$–hyperholomorphic functions, which are being actively researched. Recently, generalizations to the case of fractional derivatives have become interesting. We will mark the works [12, 13].

Also, operators of a more general form than (2.2) are considered. Namely, in the paper [14], an operator of the following form is investigated:

$$\begin{aligned} ^{\psi }_\Lambda D[f]:=\Lambda f+\psi _1\frac{\partial f}{\partial z_1}+\psi _2\frac{\partial f}{\partial z_2}+\psi _3\frac{\partial f}{\partial z_3} +\psi _4\frac{\partial f}{\partial z_4}\,,\quad \Lambda \in \mathbb {H}(\mathbb {C}). \end{aligned}$$

(2.5)

Definition 2.2

A function $f:\Omega \rightarrow \mathbb {H(C)}$, $\Omega \subset \mathbb {C}^4$, is called left–$\Lambda -\psi$–hyperholomorphic if components $f_s$ are holomorphic functions of four complex variables $z_1,z_2,z_3,z_4$ in $\Omega$, and f satisfies the equation

$$\begin{aligned} ^{\psi }_\Lambda D[f](z)=0. \end{aligned}$$

(2.6)

In the paper [15] it is develop the theory of so-called $(\phi ,\psi )$–hyperholomorphic functions. Following a matrix approach, for such functions a generalized Borel–Pompeiu formula and the corresponding Plemelj-Sokhotski formulae are established. Research from paper [15] was continued in the papers [16,17,18,19].

At the same time, the problem of representation (or description in the explicit form) of $\psi$–hyperholomorphic and left–$\Lambda -\psi$–hyperholomorphic functions is open. This paper is devoted to solving this problem.

2.1 Examples

At first, we consider examples of left– and right–$\psi$–hyperholomorphic functions.

Example 1

Consider a domain $\Omega \subset \mathbb {C}^2 \simeq \mathbb{B}\mathbb{C}$ and consider a variable $\zeta =z_1e_1+z_2e_2$, and a function $f:\Omega \rightarrow \mathbb {H(C)}$ of the form

$$f=\sum \limits _{s=1}^4f_s(z_1,z_2)e_s,\quad f_s:\Omega \rightarrow \mathbb {C}.$$

This should be understood as follows. We identify $\mathbb {C}^2$ and $\mathbb{B}\mathbb{C}$ after which the set $\Omega$ in $\mathbb{B}\mathbb{C}$ becomes a subset in $\mathbb {H(C)}$, not in $\mathbb {C}^2$; next we consider some objects as being situated in $\mathbb {H(C)}$. In particular, the set $\Omega$ is situated in $\mathbb {H(C)}$. When saying that the domain of f is in $\mathbb {H(C)}$ we mean already the previous identifications. Hence we work with functions with both domains and ranges in $\mathbb {H(C)}$. Thus $\zeta$ is in a domain in $\mathbb {H(C)}$: we imbed everything in $\mathbb {H(C)}$.

With these agreements we introduce the following definitions.

A function $f:\Omega \rightarrow \mathbb {H(C)},\,\Omega \subset \mathbb{B}\mathbb{C},$ is called right-$\mathbb{B}\mathbb{C}$-hyperholomorphic if there exists an element of the algebra $\mathbb {H(C)}$, $f'_r(\zeta )$ such that

$$\begin{aligned} \lim \limits _{\varepsilon \rightarrow 0}\frac{f(\zeta +\varepsilon h)-f(\zeta )}{\varepsilon }= h\cdot f'_r(\zeta )\qquad \forall \,h\in \mathbb{B}\mathbb{C}. \end{aligned}$$

(2.7)

A function $f:\Omega \rightarrow \mathbb {H(C)},\,\Omega \subset \mathbb{B}\mathbb{C},$ is called left-$\mathbb{B}\mathbb{C}$-hyperholomorphic if there exists an element of the algebra $\mathbb {H(C)}$, $f'_l(\zeta )$ such that

$$\begin{aligned} \lim \limits _{\varepsilon \rightarrow 0}\frac{f(\zeta +\varepsilon h)-f(\zeta )}{\varepsilon }= f'_l(\zeta )\cdot h\qquad \forall \,h\in \mathbb{B}\mathbb{C}. \end{aligned}$$

(2.8)

Condition (2.7) implies

$$\begin{aligned} \frac{\partial f}{\partial z_1}=e_1f'_r(\zeta )\quad \text {for}\quad h=e_1 \end{aligned}$$

(2.9)

and

$$\begin{aligned} \frac{\partial f}{\partial z_2}=e_2f'_r(\zeta )\quad \text {for}\quad h=e_2. \end{aligned}$$

(2.10)

From (2.9) and (2.10) follows the analog of the Cauchy–Riemann condition

$$\begin{aligned} e_2\frac{\partial f}{\partial z_1}=e_1\frac{\partial f}{\partial z_2}. \end{aligned}$$

(2.11)

Analogously, from (2.8) follows

$$\begin{aligned} \frac{\partial f}{\partial z_1}\,e_2=\frac{\partial f}{\partial z_2}\,e_1. \end{aligned}$$

(2.12)

Thus, right– and left–$\mathbb {B}\mathbb {C}$–hyperholomorphic function generalize the concepts of holomorphic functions in the algebra $\mathbb {B}\mathbb {C}$ (see, e.g., [2, 3]).

It is easy to see that the set of right- and left–$\mathbb{B}\mathbb{C}$–hyperholomorphic functions is a subset of left–$\psi$–hyperholomorphic and right–$\psi$–hyperholomorphic function, respectively. Indeed, for $\zeta =z_1e_1+z_2e_2$ the equality (2.11) has the form of the equality (2.4) with $\psi _1=e_2$, $\psi _2=-e_1$, $\psi _3=\psi _4=0$. Analogously, right–$\mathbb{B}\mathbb{C}$–hyperholomorphic functions is a subset of a set of right–$\psi$–hyperholomorphic functions.

Another example of mappings from the domain in $\mathbb {R}^3$ into the algebra $\mathbb {H(C)}$, which are a particular case of left– and right–$\psi$–hyperholomorphic functions, is considered in [20, 21].

Example 2

In (2.4) we set $\psi _1=1,\,\psi _2=I,\,\psi _3=J,\,\psi _4=K$. In this case

$$\alpha _1=\alpha _2=1,\quad \alpha _3=\alpha _4=0,\quad \beta _1=-i,\quad \beta _2=i,\quad \beta _3=\beta _4=0,$$

$$\gamma _1=\gamma _2=0,\quad \gamma _3=-i,\quad \gamma _4=-i,\quad \delta _1=\delta _2=0,\quad \delta _3=-1,\quad \delta _4=1.$$

Then (2.4) takes the form

$$\frac{\partial f}{\partial z_1}+I\frac{\partial f}{\partial z_2}+J\frac{\partial f}{\partial z_3}+K\frac{\partial f}{\partial z_4}=0$$

that is well-known Cauchy–Fueter type equation (see, e.g., [22, 23]).

2.2 Main Property of Left– and Right–$\psi$–Hyperholomorphic functions

Theorem 2.3

Let a function f be left–$\psi$–hyperholomorphic (or right–$\psi$–hyperholomorphic) in some basis of the algebra $\mathbb {H(C)}$. Then in another basis of $\mathbb {H(C)}$ there exist a vector $\Psi :=(\Psi _1,\Psi _2,\Psi _3,\Psi _4)$, $\Psi _s\in \mathbb {H(C)}$, $s=1,2,3,4$, such that the function f is left–$\Psi$–hyperholomorphic (or right–$\Psi$–hyperholomorphic).

Proof

Let us prove the theorem for the case left–$\psi$–hyperholomorphic functions. Let $\{e_1,e_2,e_3,e_4\}$ be the Cartan basis in $\mathbb {H(C)}$ and let $\{i_1,i_2,i_3,i_4\}$ be another basis in $\mathbb {H(C)}$. It means that

$$ \begin{aligned}&e_1=k_1i_1+k_2i_2+k_3i_3+k_4i_4,\\&e_2=m_1i_1+m_2i_2+m_3i_3+m_4i_4,\\&e_3=n_1i_1+n_2i_2+n_3i_3+n_4i_4,\\&e_4=r_1i_1+r_2i_2+r_3i_3+r_4i_4, \end{aligned} $$

where $k_i,m_i,n_i,r_i,\,\,i=1,2,3,4,$ are complex numbers.

Consider the equation

$$\begin{aligned} ^\psi D[f](t):=\psi _1\frac{\partial f}{\partial t_1} + \psi _2\frac{\partial f}{\partial t_2}+ \psi _3\frac{\partial f}{\partial t_3}+\psi _4\frac{\partial f}{\partial t_4}=0, \end{aligned}$$

(2.13)

where $t:=t_1e_1+t_2e_2+t_3e_3+t_4e_4$,    $t_1,t_2,t_3,t_4\in \mathbb {C}$. In the variable t we passing to the basis $\{i_1,i_2,i_3,i_4\}$. Then

$$t=i_1(t_1k_1+t_2m_1+t_3n_1+t_4r_1)+i_2(t_1k_2+t_2m_2+t_3n_2+t_4r_2)$$

$$\qquad \quad +i_3(t_1k_3+t_2m_3+t_3n_3+t_4r_3)+i_4(t_1k_4+t_2m_4+t_3n_4+t_4r_4).$$

We set

$$\begin{aligned} \begin{aligned}&z_1:=t_1k_1+t_2m_1+t_3n_1+t_4r_1,\\&z_2:=t_1k_2+t_2m_2+t_3n_2+t_4r_2,\\&z_3:=t_1k_3+t_2m_3+t_3n_3+t_4r_3,\\&z_4:=t_1k_4+t_2m_4+t_3n_4+t_4r_4. \end{aligned}\nonumber \\ \end{aligned}$$

(2.14)

From equalities (2.14) we obtain

$$ \begin{aligned}&\frac{\partial f}{\partial t_1}=k_1\frac{\partial f}{\partial z_1}+k_2\frac{\partial f}{\partial z_2} +k_3\frac{\partial f}{\partial z_3}+k_4\frac{\partial f}{\partial z_4}\,,\\&\frac{\partial f}{\partial t_2}=m_1\frac{\partial f}{\partial z_1}+m_2\frac{\partial f}{\partial z_2} +m_3\frac{\partial f}{\partial z_3}+m_4\frac{\partial f}{\partial z_4}\,,\\&\frac{\partial f}{\partial t_3}=n_1\frac{\partial f}{\partial z_1}+n_2\frac{\partial f}{\partial z_2} +n_3\frac{\partial f}{\partial z_3}+n_4\frac{\partial f}{\partial z_4}\,,\\&\frac{\partial f}{\partial t_4}=r_1\frac{\partial f}{\partial z_1}+r_2\frac{\partial f}{\partial z_2}+r_3\frac{\partial f}{\partial z_3}+r_4\frac{\partial f}{\partial z_4}\,. \end{aligned} $$

Then Eq. (2.13) is equivalent to the following equation

$$^\psi D[f](t)=(\psi _1k_1+\psi _2m_1+\psi _3n_1+\psi _4r_1)\frac{\partial f}{\partial z_1}+(\psi _1k_2+\psi _2m_2+\psi _3n_2+\psi _4r_2)\frac{\partial f}{\partial z_2}$$

$$\begin{aligned} \qquad \quad +(\psi _1k_3+\psi _2m_3+\psi _3n_3+\psi _4r_3)\frac{\partial f}{\partial z_3}+(\psi _1k_4+\psi _2m_4+\psi _3n_4+\psi _4r_4)\frac{\partial f}{\partial z_4}. \nonumber \\ \end{aligned}$$

(2.15)

Using denotation (2.1), we have

$$\psi _1=\sum \limits _{s=1}^4\alpha _se_s=\sum \limits _{s=1}^4i_s(\alpha _1 k_s+\alpha _2 m_s+\alpha _3 n_s+\alpha _4 r_s), $$

$$\psi _2=\sum \limits _{s=1}^4\beta _se_s=\sum \limits _{s=1}^4i_s(\beta _1 k_s+\beta _2 m_s+\beta _3 n_s+\beta _4 r_s), $$

$$\psi _3=\sum \limits _{s=1}^4\gamma _se_s=\sum \limits _{s=1}^4i_s(\gamma _1 k_s+\gamma _2 m_s+\gamma _3 n_s+\gamma _4 r_s), $$

$$\psi _4=\sum \limits _{s=1}^4\delta _se_s=\sum \limits _{s=1}^4i_s(\delta _1 k_s+\delta _2 m_s+\delta _3 n_s+\delta _4 r_s), $$

From (2.15) we obtain

$$^\psi D[f](t)= \sum \limits _{s=1}^4i_s\Bigr [(\alpha _1 k_s+\alpha _2 m_s+\alpha _3 n_s+\alpha _4 r_s)k_1+(\beta _1 k_s+\beta _2 m_s+\beta _3 n_s+\beta _4 r_s)m_1$$

$$\qquad \quad + (\gamma _1 k_s+\gamma _2 m_s+\gamma _3 n_s+\gamma _4 r_s)n_1+ (\delta _1 k_s+\delta _2 m_s+\delta _3 n_s+\delta _4 r_s)r_1\Bigr ]\frac{\partial f}{\partial z_1} $$

$$\qquad \qquad +\sum \limits _{s=1}^4i_s\Bigr [(\alpha _1 k_s+\alpha _2 m_s+\alpha _3 n_s+\alpha _4 r_s)k_2+(\beta _1 k_s+\beta _2 m_s+\beta _3 n_s+\beta _4 r_s)m_2$$

$$\qquad \qquad + (\gamma _1 k_s+\gamma _2 m_s+\gamma _3 n_s+\gamma _4 r_s)n_2+ (\delta _1 k_s+\delta _2 m_s+\delta _3 n_s+\delta _4 r_s)r_2\Bigr ]\frac{\partial f}{\partial z_2} $$

$$\qquad \qquad +\sum \limits _{s=1}^4i_s\Bigr [(\alpha _1 k_s+\alpha _2 m_s+\alpha _3 n_s+\alpha _4 r_s)k_3+(\beta _1 k_s+\beta _2 m_s+\beta _3 n_s+\beta _4 r_s)m_3$$

$$\qquad \qquad + (\gamma _1 k_s+\gamma _2 m_s+\gamma _3 n_s+\gamma _4 r_s)n_3+ (\delta _1 k_s+\delta _2 m_s+\delta _3 n_s+\delta _4 r_s)r_3\Bigr ]\frac{\partial f}{\partial z_3} $$

$$\qquad \qquad +\sum \limits _{s=1}^4i_s\Bigr [(\alpha _1 k_s+\alpha _2 m_s+\alpha _3 n_s+\alpha _4 r_s)k_4+(\beta _1 k_s+\beta _2 m_s+\beta _3 n_s+\beta _4 r_s)m_4$$

$$+ (\gamma _1 k_s+\gamma _2 m_s+\gamma _3 n_s+\gamma _4 r_s)n_4+ (\delta _1 k_s+\delta _2 m_s+\delta _3 n_s+\delta _4 r_s)r_4\Bigr ]\frac{\partial f}{\partial z_4} $$

$$=:\Psi _1\frac{\partial f}{\partial z_1}+\Psi _2\frac{\partial f}{\partial z_2}+ \Psi _3\frac{\partial f}{\partial z_3}+\Psi _4\frac{\partial f}{\partial z_4}=0.$$

$\square$

Remark 2.4

In Clifford algebras, it is known that the equations

$$\begin{aligned} \frac{\partial f}{\partial t_0} + I\frac{\partial f}{\partial t_1}+ J\frac{\partial f}{\partial t_2}+K\frac{\partial f}{\partial t_3}=0 \end{aligned}$$

(2.16)

and $^\psi D[f](t)=0$ coincide, up to an orthogonal transformation. Note that, in essence, Theorem 2.3 is a similar statement, but formulated in other terms.

Remark 2.5

It follows from Theorem 2.3 that in future investigation it is enough to consider constants $\psi$ and function f in the simplest basis, i.e., in Cartan basis. The use of the Cartan basis is of principal importance, because in this basis the multiplication table has the simplest form. In addition, in the Cartan basis, Eqs. (2.4), (2.6) and (2.16) are reduced to systems of differential equations that we integrate in the explicit form. This is what we will do next.

3 Application to Solving Cauchy–Fueter Type Equation

Now, we will establish a connection between solutions of the equation

$$\begin{aligned} D[f](t):=\frac{\partial f}{\partial t_0} + I\frac{\partial f}{\partial t_1}+ J\frac{\partial f}{\partial t_2}+K\frac{\partial f}{\partial t_3}=0, \end{aligned}$$

(3.1)

where $t:=t_0+t_1I+t_2J+t_3K$,    $t_0,t_1,t_2,t_3\in \mathbb {C}$, and the solutions of Eq. (2.4). For this purpose, in t we passing to Cartan basis. We have

$$t=t_0(e_1+e_2)+t_1(-ie_1+ie_2)+t_2(-ie_3-ie_4)+t_3(e_4-e_3)$$

$$=(t_0-it_1)e_1+(t_0+it_1)e_2+(-it_2-t_3)e_3+(-it_2+t_3)e_4.$$

We set

$$\begin{aligned} z_1:=t_0-it_1,\quad z_2:=t_0+it_1,\quad z_3:=-it_2-t_3,\quad z_4:=-it_2+t_3. \end{aligned}$$

(3.2)

From equalities (3.2) we obtain

$$\frac{\partial f}{\partial t_0}=\frac{\partial f}{\partial z_1}+\frac{\partial f}{\partial z_2},\qquad \frac{\partial f}{\partial t_1}=-i\frac{\partial f}{\partial z_1}+i\frac{\partial f}{\partial z_2},$$

$$\frac{\partial f}{\partial t_2}=-i\frac{\partial f}{\partial z_3}-i\frac{\partial f}{\partial z_4},\qquad \frac{\partial f}{\partial t_3}=-\frac{\partial f}{\partial z_3}+\frac{\partial f}{\partial z_4}.$$

Then Eq. (3.1) is equivalent to the following equation

$$\begin{aligned} D[f]&=\frac{\partial f}{\partial z_1}+\frac{\partial f}{\partial z_2}-iI\frac{\partial f}{\partial z_1}+ iI\frac{\partial f}{\partial z_2}-iJ\frac{\partial f}{\partial z_3}-iJ\frac{\partial f}{\partial z_4}-K \frac{\partial f}{\partial z_3}+K\frac{\partial f}{\partial z_4} \\&=(1-iI)\frac{\partial f}{\partial z_1}+(1+iI)\frac{\partial f}{\partial z_2}+(-iJ-K)\frac{\partial f}{\partial z_3} +(-iJ+K)\frac{\partial f}{\partial z_4}\\&=2\left( e_2\frac{\partial f}{\partial z_1}+e_1\frac{\partial f}{\partial z_2}-e_4\frac{\partial f}{\partial z_3}- e_3\frac{\partial f}{\partial z_4}\right) =0. \end{aligned}$$

Thus, we proved the following theorem

Theorem 3.1

A function f of the variable $t=t_0+t_1I+t_2J+t_3K$ satisfies Eq. (3.1) if and only if the function f of the variable $z=z_1e_1+z_2e_2+z_3e_3+z_4e_4$ satisfies the equation

$$\begin{aligned} e_2\frac{\partial f}{\partial z_1}+e_1\frac{\partial f}{\partial z_2}-e_4\frac{\partial f}{\partial z_3}-e_3\frac{\partial f}{\partial z_4}=0, \end{aligned}$$

(3.3)

where z and t are related by equalities (3.2).

Now, we solve Eq. (3.3).

$$e_2\frac{\partial f}{\partial z_1}=e_2\left( \frac{\partial f_1}{\partial z_1}e_1+\frac{\partial f_2}{\partial z_1}e_2 +\frac{\partial f_3}{\partial z_1}e_3+\frac{\partial f_4}{\partial z_1}e_4\right) $$

$$\qquad =\frac{\partial f_2}{\partial z_1}e_2+\frac{\partial f_4}{\partial z_1}e_4\,,$$

$$e_1\frac{\partial f}{\partial z_2}=\frac{\partial f_1}{\partial z_2}e_1+\frac{\partial f_3}{\partial z_2}e_3,$$

$$e_4\frac{\partial f}{\partial z_3}=\frac{\partial f_1}{\partial z_3}e_4+\frac{\partial f_3}{\partial z_3}e_2,$$

$$e_3\frac{\partial f}{\partial z_4}=\frac{\partial f_2}{\partial z_4}e_3+\frac{\partial f_4}{\partial z_4}e_1.$$

Then Eq. (3.3) is equivalent to the system

$$\frac{\partial f_1}{\partial z_2}=\frac{\partial f_4}{\partial z_4},\qquad \frac{\partial f_2}{\partial z_1}=\frac{\partial f_3}{\partial z_3},$$

$$\frac{\partial f_3}{\partial z_2}=\frac{\partial f_2}{\partial z_4},\qquad \frac{\partial f_4}{\partial z_1}=\frac{\partial f_1}{\partial z_3}.$$

We have pair of systems

$$\begin{aligned} \frac{\partial f_1}{\partial z_2}=\frac{\partial f_4}{\partial z_4}\,,\qquad \frac{\partial f_1}{\partial z_3}=\frac{\partial f_4}{\partial z_1} \end{aligned}$$

(3.4)

and

$$\begin{aligned} \frac{\partial f_2}{\partial z_1}=\frac{\partial f_3}{\partial z_3}\,,\qquad \frac{\partial f_2}{\partial z_4}=\frac{\partial f_3}{\partial z_2}\,. \end{aligned}$$

(3.5)

In a simple connected domain $\Omega$, a solution of system (3.4) is an arbitrary holomorphic function

$$f_1=f_1(z_2,z_3)$$

and

$$f_4=z_4\frac{\partial f_1}{\partial z_2}+z_1\frac{\partial f_1}{\partial z_3}.$$

In a simple connected domain $\Omega$, a solution of system (3.5) is an arbitrary holomorphic function

$$f_2=f_2(z_1,z_4)$$

and

$$f_3=z_3\frac{\partial f_2}{\partial z_1}+z_2\frac{\partial f_2}{\partial z_4}.$$

Thus, we have the following solution of Eq. (3.3):

$$\begin{aligned} f(z)&=f_1(z_2,z_3)e_1+f_2(z_1,z_4)e_2 \nonumber \\&\quad +\left( z_3\frac{\partial f_2}{\partial z_1}+z_2\frac{\partial f_2}{\partial z_4}\right) e_3 +\left( z_4\frac{\partial f_1}{\partial z_2}+z_1\frac{\partial f_1}{\partial z_3}\right) e_4\,. \end{aligned}$$

(3.6)

Thus, accordingly to Theorem 3.1 we obtain

Theorem 3.2

In a simple connected domain, function (3.6), where $z_1,z_2,z_3,z_4$ are given by relations (3.2), satisfies Eq. (3.1).

Proposition 3.3

In a simple connected domain, function (3.6) satisfies the four-dimensional complex Laplace equation

$$\begin{aligned} \Delta _{\mathbb {C}^4}f:=\frac{\partial ^2 f}{\partial t_1^2}+\frac{\partial ^2 f}{\partial t_2^2} +\frac{\partial ^2 f}{\partial t_3^2}+\frac{\partial ^2 f}{\partial t_4^2}=0. \end{aligned}$$

(3.7)

About Eq. (3.7) and its relation to the Cauchy-Fueter equation see in [22].

4 Representation of Left–$\psi$–hyperholomorphic Functions in a Special Case

Now we will find a general solution of Eq. (2.4) for a special choice of parameters $\psi _1,\psi _2,\psi _3$ and $\psi _4$. For this purpose, we reduce Eq. (2.4) to a system of four PDEs. We have

$$\begin{aligned} \psi _1\frac{\partial f}{\partial z_1}&=\left( \alpha _1e_1+\alpha _2e_2+\alpha _3e_3+\alpha _4e_4\right) \left( \frac{\partial f_1}{\partial z_1}e_1+\frac{\partial f_2}{\partial z_1}e_2+\frac{\partial f_3}{\partial z_1}e_3+ \frac{\partial f_4}{\partial z_1}e_4\right) \\&=\frac{\partial f_1}{\partial z_1}\alpha _1e_1+\frac{\partial f_3}{\partial z_1}\alpha _1e_3+ \frac{\partial f_2}{\partial z_1}\alpha _2e_2+\frac{\partial f_4}{\partial z_1}\alpha _2e_4 \\&\quad +\frac{\partial f_2}{\partial z_1}\alpha _3e_3+\frac{\partial f_4}{\partial z_1}\alpha _3e_1+ \frac{\partial f_1}{\partial z_1}\alpha _4e_4+\frac{\partial f_3}{\partial z_1}\alpha _4e_2 \\&=\frac{\partial }{\partial z_1}\left( \alpha _1f_1+\alpha _3f_4\right) e_1+\frac{\partial }{\partial z_1}\left( \alpha _2f_2+\alpha _4f_3\right) e_2 \\&\quad +\frac{\partial }{\partial z_1}\left( \alpha _1f_3+\alpha _3f_2\right) e_3+\frac{\partial }{\partial z_1}\left( \alpha _2f_4+\alpha _4f_1\right) e_4. \end{aligned}$$

Similarly,

$$\begin{aligned} \psi _2\frac{\partial f}{\partial z_2}&=\frac{\partial }{\partial z_2}\left( \beta _1f_1+\beta _3f_4\right) e_1 +\frac{\partial }{\partial z_2}\left( \beta _2f_2+\beta _4f_3\right) e_2 \\&\quad +\frac{\partial }{\partial z_2}\left( \beta _1f_3+\beta _3f_2\right) e_3+\frac{\partial }{\partial z_2}\left( \beta _2f_4+\beta _4f_1\right) e_4, \\ \psi _3\frac{\partial f}{\partial z_3}&=\frac{\partial }{\partial z_3}\left( \gamma _1f_1+\gamma _3f_4\right) e_1+ \frac{\partial }{\partial z_3}\left( \gamma _2f_2+\gamma _4f_3\right) e_2 \\&\quad +\frac{\partial }{\partial z_3}\left( \gamma _1f_3+\gamma _3f_2\right) e_3+\frac{\partial }{\partial z_3}\left( \gamma _2f_4+\gamma _4f_1\right) e_4, \\ \psi _4\frac{\partial f}{\partial z_4}&=\frac{\partial }{\partial z_4}\left( \delta _1f_1+\delta _3f_4\right) e_1 +\frac{\partial }{\partial z_4}\left( \delta _2f_2+\delta _4f_3\right) e_2 \\&\quad +\frac{\partial }{\partial z_4}\left( \delta _1f_3+\delta _3f_2\right) e_3+\frac{\partial }{\partial z_4}\left( \delta _2f_4+\delta _4f_1\right) e_4. \end{aligned}$$

Then Eq. (2.4) is equivalent to the following system

$$\begin{aligned} \begin{aligned} \frac{\partial }{\partial z_1}(\alpha _1f_1+\alpha _3f_4)+\frac{\partial }{\partial z_2}(\beta _1f_1+\beta _3f_4) +\frac{\partial }{\partial z_3}(\gamma _1f_1+\gamma _3f_4)+\frac{\partial }{\partial z_4}(\delta _1f_1+\delta _3f_4)=0, \\ \frac{\partial }{\partial z_1}(\alpha _2f_2+\alpha _4f_3)+\frac{\partial }{\partial z_2}(\beta _2f_2+\beta _4f_3) +\frac{\partial }{\partial z_3}(\gamma _2f_2+\gamma _4f_3)+\frac{\partial }{\partial z_4}(\delta _2f_2+\delta _4f_3)=0, \\ \frac{\partial }{\partial z_1}(\alpha _1f_3+\alpha _3f_2)+\frac{\partial }{\partial z_2}(\beta _1f_3+\beta _3f_2) +\frac{\partial }{\partial z_3}(\gamma _1f_3+\gamma _3f_2)+\frac{\partial }{\partial z_4}(\delta _1f_3+\delta _3f_2)=0,\\ \frac{\partial }{\partial z_1}(\alpha _2f_4+\alpha _4f_1)+\frac{\partial }{\partial z_2}(\beta _2f_4+\beta _4f_1) +\frac{\partial }{\partial z_3}(\gamma _2f_4+\gamma _4f_1)+\frac{\partial }{\partial z_4}(\delta _2f_4+\delta _4f_1)=0. \end{aligned} \nonumber \\ \end{aligned}$$

(4.1)

Theorem 4.1

Let

$$\begin{aligned}{} & {} \psi _1=\alpha _1e_1+\alpha _2e_2+\alpha _3e_3+\alpha _4e_4,\qquad \alpha _1\alpha _2\ne \alpha _3\alpha _4\,, \nonumber \\{} & {} \psi _2=\lambda \alpha _1e_1+\mu \alpha _2e_2+\mu \alpha _3e_3+\lambda \alpha _4e_4,\nonumber \\{} & {} \psi _3=\theta \alpha _1e_1+\vartheta \alpha _2e_2+\vartheta \alpha _3e_3+\theta \alpha _4e_4,\nonumber \\{} & {} \psi _4=\nu \alpha _1e_1+\eta \alpha _2e_2+\eta \alpha _3e_3+\nu \alpha _4e_4, \end{aligned}$$

(4.2)

where $\alpha _1,\alpha _2,\alpha _3,\alpha _4,\lambda ,\mu ,\theta ,\vartheta ,\nu ,\eta$ are arbitrary complex numbers. Then every left–$\psi$–hyperholomorphic function is of the form

$$\begin{aligned} f(z)=f_1(\widetilde{\zeta }_2,\widetilde{\zeta }_3,\widetilde{\zeta }_4)e_1+f_2(\zeta _2,\zeta _3,\zeta _4)e_2+ f_3(\widetilde{\zeta }_2,\widetilde{\zeta }_3,\widetilde{\zeta }_4)e_3+f_4(\zeta _2,\zeta _3,\zeta _4)e_4, \end{aligned}$$

(4.3)

where

$$\begin{aligned} \begin{aligned}&\widetilde{\zeta }_2:=\lambda z_1-z_2,\quad \widetilde{\zeta }_3:=\theta z_1-z_3,\quad \widetilde{\zeta }_4:=\nu z_1-z_4,\\&\zeta _2:=\mu z_1-z_2,\quad \zeta _3:=\vartheta z_1-z_3,\quad \zeta _4:=\eta z_1-z_4, \end{aligned} \nonumber \\ \end{aligned}$$

(4.4)

and $f_1,f_2,f_3,f_4$ are arbitrary holomorphic functions of three their arguments.

Proof

For given parameters (4.2) the first equation of system (4.1) takes the form

$$\begin{aligned}{} & {} \frac{\partial }{\partial z_1}\left( \alpha _1f_1+\alpha _3f_4\right) +\frac{\partial }{\partial z_2}\left( \lambda \alpha _1f_1+\mu \alpha _3f_4\right) \nonumber \\{} & {} \quad +\frac{\partial }{\partial z_3}\left( \theta \alpha _1f_1+\vartheta \alpha _3f_4\right) +\frac{\partial }{\partial z_4}\left( \nu \alpha _1f_1+\eta \alpha _3f_4\right) =0. \end{aligned}$$

(4.5)

Similarly, for given parameters (4.2) the fourth equation of system (4.1) takes the form

$$\begin{aligned}{} & {} \frac{\partial }{\partial z_1}\left( \alpha _4f_1+\alpha _2f_4\right) +\frac{\partial }{\partial z_2}\left( \lambda \alpha _4f_1+\mu \alpha _2f_4\right) \nonumber \\{} & {} \quad +\frac{\partial }{\partial z_3}\left( \theta \alpha _4f_1+\vartheta \alpha _2f_4\right) +\frac{\partial }{\partial z_4}\left( \nu \alpha _4f_1+\eta \alpha _2f_4\right) =0. \end{aligned}$$

(4.6)

Consider the difference between Eq. (4.5) multiplied by $\alpha _2$ and Eq. (4.6) multiplied by $\alpha _3$. Then we obtain the following equation

$$\begin{aligned}{} & {} \frac{\partial }{\partial z_1}\Big (f_1(\alpha _1\alpha _2-\alpha _3\alpha _4)+f_4(\alpha _2\alpha _3-\alpha _2\alpha _3)\Big )\\{} & {} \quad +\frac{\partial }{\partial z_2}\Big (f_1(\lambda \alpha _1\alpha _2-\lambda \alpha _3\alpha _4)+f_4(\mu \alpha _2\alpha _3-\mu \alpha _2\alpha _3)\Big )\\{} & {} \quad +\frac{\partial }{\partial z_3}\Big (f_1(\theta \alpha _1\alpha _2-\theta \alpha _3\alpha _4)+f_4(\vartheta \alpha _2\alpha _3-\vartheta \alpha _2\alpha _3)\Big )\\{} & {} \quad +\frac{\partial }{\partial z_4}\Big (f_1(\nu \alpha _1\alpha _2-\nu \alpha _3\alpha _4)+f_4(\eta \alpha _2\alpha _3-\eta \alpha _2\alpha _3)\Big )=0. \end{aligned}$$

Thus, we obtain the equation

$$\begin{aligned} \frac{\partial f_1}{\partial z_1}+\lambda \frac{\partial f_1}{\partial z_2}+\theta \frac{\partial f_1}{\partial z_3}+\nu \frac{\partial f_1}{\partial z_4}=0. \end{aligned}$$

(4.7)

For Eq. (4.7) consider the characteristic equation

$$\begin{aligned} \frac{dz_1}{1}=\frac{dz_2}{\lambda }=\frac{dz_3}{\theta }=\frac{dz_4}{\nu }. \end{aligned}$$

(4.8)

The solutions of system (4.8) are the following integrals

$$c_2=\lambda z_1-z_2,\quad c_3=\theta z_1-z_3,\quad c_4=\nu z_1-z_4.$$

Therefore, the general solution of Eq. (4.7) has the form

$$f_1=f_1(\widetilde{\zeta }_2,\widetilde{\zeta }_3,\widetilde{\zeta }_4),$$

where $\widetilde{\zeta }_2,\widetilde{\zeta }_3,\widetilde{\zeta }_4$ are defined by equalities (4.4).

Note that polynomials (4.4) are similarly to the well-known Fueter’s polynomials [24].

Similarly, we obtain the representations for the components $f_2,f_3,f_4$. $\square$

Thus, formula (4.3) gives a representation of left–$\psi$–hyperholomorphic function for a special choice of $\psi$.

Remark 4.2

Using formulas (1.4) we can rewrite representation (4.3) in the Pauli basis:

$$\begin{aligned} f(z)= & {} \left( f_1(\widetilde{\zeta }_2,\widetilde{\zeta }_3,\widetilde{\zeta }_4)+f_2(\zeta _2,\zeta _3,\zeta _4)\right) \sigma _0+ \left( if_3(\widetilde{\zeta }_2,\widetilde{\zeta }_3,\widetilde{\zeta }_4)-if_4(\zeta _2,\zeta _3,\zeta _4)\right) \sigma _1 \\{} & {} +\left( -f_3(\widetilde{\zeta }_2,\widetilde{\zeta }_3,\widetilde{\zeta }_4)-f_4(\zeta _2,\zeta _3,\zeta _4)\right) \sigma _2+ \left( f_2(\zeta _2,\zeta _3,\zeta _4)-f_1(\widetilde{\zeta }_2,\widetilde{\zeta }_3,\widetilde{\zeta }_4)\right) \sigma _3\,. \end{aligned}$$

5 Representation of Right–$\psi$–hyperholomorphic Functions in a Special Case

In this section we will find a general solution of the equation

$$\begin{aligned} D^\psi [f](z)=\frac{\partial f}{\partial z_1}\,\psi _1+\frac{\partial f}{\partial z_2}\,\psi _2+\frac{\partial f}{\partial z_3}\,\psi _3 +\frac{\partial f}{\partial z_4}\,\psi _4=0 \end{aligned}$$

(5.1)

for a special choice of parameters $\psi _1,\psi _2,\psi _3$ and $\psi _4$. For this purpose, we reduce Eq. (5.1) to a system of four PDEs. We have

$$\begin{aligned} \frac{\partial f}{\partial z_1}\,\psi _1&= \left( \frac{\partial f_1}{\partial z_1}e_1+\frac{\partial f_2}{\partial z_1}e_2+\frac{\partial f_3}{\partial z_1}e_3+ \frac{\partial f_4}{\partial z_1}e_4\right) \left( \alpha _1e_1+\alpha _2e_2+\alpha _3e_3+\alpha _4e_4\right) \\&=\frac{\partial f_1}{\partial z_1}\alpha _1e_1+\frac{\partial f_1}{\partial z_1}\alpha _3e_3+ \frac{\partial f_2}{\partial z_1}\alpha _2e_2+\frac{\partial f_2}{\partial z_1}\alpha _4e_4\\&\quad +\frac{\partial f_3}{\partial z_1}\alpha _2e_3+\frac{\partial f_3}{\partial z_1}\alpha _4e_1+ \frac{\partial f_4}{\partial z_1}\alpha _1e_4+\frac{\partial f_4}{\partial z_1}\alpha _3e_2\\&=\frac{\partial }{\partial z_1}\left( \alpha _1f_1+\alpha _4f_3\right) e_1+\frac{\partial }{\partial z_1}\left( \alpha _2f_2+\alpha _3f_4\right) e_2\\&\quad +\frac{\partial }{\partial z_1}\left( \alpha _3f_1+\alpha _2f_3\right) e_3+\frac{\partial }{\partial z_1}\left( \alpha _4f_2+\alpha _1f_4\right) e_4. \end{aligned}$$

Similarly,

$$\begin{aligned} \frac{\partial f}{\partial z_2}\,\psi _2= & {} \frac{\partial }{\partial z_2}\left( \beta _1f_1+\beta _4f_3\right) e_1 +\frac{\partial }{\partial z_2}\left( \beta _2f_2+\beta _3f_4\right) e_2\\{} & {} +\frac{\partial }{\partial z_2}\left( \beta _3f_1+\beta _2f_3\right) e_3+\frac{\partial }{\partial z_2}\left( \beta _4f_2+\beta _1f_4\right) e_4,\\ \frac{\partial f}{\partial z_3}\,\psi _3= & {} \frac{\partial }{\partial z_3}\left( \gamma _1f_1+\gamma _4f_3\right) e_1+ \frac{\partial }{\partial z_3}\left( \gamma _2f_2+\gamma _3f_4\right) e_2\\{} & {} +\frac{\partial }{\partial z_3}\left( \gamma _3f_1+\gamma _2f_3\right) e_3+\frac{\partial }{\partial z_3}\left( \gamma _4f_2+\gamma _1f_4\right) e_4,\\ \frac{\partial f}{\partial z_4}\,\psi _4= & {} \frac{\partial }{\partial z_4}\left( \delta _1f_1+\delta _4f_3\right) e_1 +\frac{\partial }{\partial z_4}\left( \delta _2f_2+\delta _3f_4\right) e_2\\{} & {} +\frac{\partial }{\partial z_4}\left( \delta _3f_1+\delta _2f_3\right) e_3+\frac{\partial }{\partial z_4}\left( \delta _4f_2+\delta _1f_4\right) e_4. \end{aligned}$$

Then Eq. (5.1) is equivalent to the following system

$$\begin{aligned} \begin{aligned} \frac{\partial }{\partial z_1}(\alpha _1f_1+\alpha _4f_3)+\frac{\partial }{\partial z_2}(\beta _1f_1+\beta _4f_3) +\frac{\partial }{\partial z_3}(\gamma _1f_1+\gamma _4f_3)+\frac{\partial }{\partial z_4}(\delta _1f_1+\delta _4f_3)=0,\\ \frac{\partial }{\partial z_1}(\alpha _2f_2+\alpha _3f_4)+\frac{\partial }{\partial z_2}(\beta _2f_2+\beta _3f_4) +\frac{\partial }{\partial z_3}(\gamma _2f_2+\gamma _3f_4)+\frac{\partial }{\partial z_4}(\delta _2f_2+\delta _3f_4)=0, \\ \frac{\partial }{\partial z_1}(\alpha _3f_1+\alpha _2f_3)+\frac{\partial }{\partial z_2}(\beta _3f_1+\beta _2f_3) +\frac{\partial }{\partial z_3}(\gamma _3f_1+\gamma _2f_3)+\frac{\partial }{\partial z_4}(\delta _3f_1+\delta _2f_3)=0, \\ \frac{\partial }{\partial z_1}(\alpha _4f_2+\alpha _1f_4)+\frac{\partial }{\partial z_2}(\beta _4f_2+\beta _1f_4) +\frac{\partial }{\partial z_3}(\gamma _4f_2+\gamma _1f_4)+\frac{\partial }{\partial z_4}(\delta _4f_2+\delta _1f_4)=0. \end{aligned} \nonumber \\ \end{aligned}$$

(5.2)

Theorem 5.1

Let

$$\begin{aligned} \begin{aligned}&\psi _1=\alpha _1e_1+\alpha _2e_2+\alpha _3e_3+\alpha _4e_4,\qquad \alpha _1\alpha _2\ne \alpha _3\alpha _4\,,\\&\psi _2=\mu \alpha _1e_1+\lambda \alpha _2e_2+\mu \alpha _3e_3+\lambda \alpha _4e_4,\\&\psi _3=\vartheta \alpha _1e_1+\theta \alpha _2e_2+\vartheta \alpha _3e_3+\theta \alpha _4e_4,\\&\psi _4=\eta \alpha _1e_1+\nu \alpha _2e_2+\eta \alpha _3e_3+\nu \alpha _4e_4, \end{aligned}\nonumber \\ \end{aligned}$$

(5.3)

where $\alpha _1,\alpha _2,\alpha _3,\alpha _4,\lambda ,\mu ,\theta ,\vartheta ,\nu ,\eta$ are arbitrary complex numbers. Then every right–$\psi$–hyperholomorphic function is of the form

$$\begin{aligned} f(z)=f_1(\zeta _2,\zeta _3,\zeta _4)e_1+f_2(\widetilde{\zeta }_2,\widetilde{\zeta }_3,\widetilde{\zeta }_4)e_2+ f_3(\widetilde{\zeta }_2,\widetilde{\zeta }_3,\widetilde{\zeta }_4)e_3+f_4(\zeta _2,\zeta _3,\zeta _4)e_4, \end{aligned}$$

(5.4)

where $\zeta _2, \zeta _3, \zeta _4$, $\widetilde{\zeta }_2, \widetilde{\zeta }_3, \widetilde{\zeta }_4$ are defined by relations (4.4) and $f_1,f_2,f_3,f_4$ are arbitrary holomorphic functions of their three arguments.

Proof

For given parameters (5.3) the first equation of system (5.2) takes the form

$$\begin{aligned}&\frac{\partial }{\partial z_1}\left( \alpha _1f_1+\alpha _4f_3\right) +\frac{\partial }{\partial z_2}\left( \mu \alpha _1f_1+\lambda \alpha _4f_3\right) \nonumber \\&\quad +\frac{\partial }{\partial z_3}\left( \vartheta \alpha _1f_1+\theta \alpha _4f_3\right) +\frac{\partial }{\partial z_4}\left( \eta \alpha _1f_1+\nu \alpha _4f_3\right) =0. \end{aligned}$$

(5.5)

Similarly, for given parameters (5.3) the third equation of system (5.2) takes the form

$$\begin{aligned}{} & {} \frac{\partial }{\partial z_1}\left( \alpha _3f_1+\alpha _2f_3\right) +\frac{\partial }{\partial z_2}\left( \mu \alpha _3f_1+\lambda \alpha _2f_3\right) \nonumber \\{} & {} \quad +\frac{\partial }{\partial z_3}\left( \vartheta \alpha _3f_1+\theta \alpha _2f_3\right) +\frac{\partial }{\partial z_4}\left( \eta \alpha _3f_1+\nu \alpha _2f_3\right) =0. \end{aligned}$$

(5.6)

Consider the difference between Eq. (5.5) multiplied by $\alpha _2$ and Eq. (5.6) multiplied by $\alpha _4$. Then we obtain the following equation

$$\begin{aligned}{} & {} \frac{\partial }{\partial z_1}\Big (f_1(\alpha _1\alpha _2-\alpha _3\alpha _4)+f_3(\alpha _2\alpha _4-\alpha _2\alpha _4)\Big ) \\{} & {} \quad +\frac{\partial }{\partial z_2}\Big (\mu f_1(\alpha _1\alpha _2-\alpha _3\alpha _4)+\lambda f_3(\alpha _2\alpha _4-\alpha _2\alpha _4)\Big )\\{} & {} \quad +\frac{\partial }{\partial z_3}\Big (\vartheta f_1(\alpha _1\alpha _2-\alpha _3\alpha _4)+\theta f_3(\alpha _2\alpha _4-\alpha _2\alpha _4)\Big )\\{} & {} \quad +\frac{\partial }{\partial z_4}\Big (\eta f_1(\alpha _1\alpha _2-\alpha _3\alpha _4)+\nu f_3(\alpha _2\alpha _3-\alpha _2\alpha _3)\Big )=0. \end{aligned}$$

Thus, we obtain the equation

$$\begin{aligned} \frac{\partial f_1}{\partial z_1}+\mu \frac{\partial f_1}{\partial z_2}+\vartheta \frac{\partial f_1}{\partial z_3}+\eta \frac{\partial f_1}{\partial z_4}=0. \end{aligned}$$

(5.7)

For Eq. (5.7) consider the characteristic equation

$$\begin{aligned} \frac{dz_1}{1}=\frac{dz_2}{\mu }=\frac{dz_3}{\vartheta }=\frac{dz_4}{\eta }. \end{aligned}$$

(5.8)

The solutions of system (5.8) are the following integrals

$$c_2=\mu z_1-z_2,\quad c_3=\vartheta z_1-z_3,\quad c_4=\eta z_1-z_4.$$

be of the for Therefore, the general solution of Eq. (5.7) has the form

$$f_1=f_1(\zeta _2,\zeta _3,\zeta _4),$$

where $\zeta _2,\zeta _3,\zeta _4$ are defined by equalities (4.4).

Similarly, we obtain the representations for the components $f_2,f_3,f_4$. $\square$

Thus, formula (5.4) gives a representation of right–$\psi$–hyperholomorphic function for a special choice of $\psi$.

Comparing representations (4.3) and (5.4), we obtain the following statement.

Proposition 5.2

Let the conditions of Theorem 4.1 be satisfied. Then the function f is simultaneously left– and right–$\psi$–hyperholomorphic if it takes values on the set $\{h_3e_3+h_4e_4:h_3,h_4\in \mathbb {C}\}$.

Remark 5.3

Using formulas (1.4), we can rewrite representation (5.4) in the Pauli basis:

$$\begin{aligned} f(z)= & {} \left( f_1(\zeta _2,\zeta _3,\zeta _4)+f_2(\widetilde{\zeta }_2,\widetilde{\zeta }_3,\widetilde{\zeta }_4)\right) \sigma _0+ \left( if_3(\widetilde{\zeta }_2,\widetilde{\zeta }_3,\widetilde{\zeta }_4)-if_4(\zeta _2,\zeta _3,\zeta _4)\right) \sigma _1 \\{} & {} +\left( -f_3(\widetilde{\zeta }_2,\widetilde{\zeta }_3,\widetilde{\zeta }_4)-f_4(\zeta _2,\zeta _3,\zeta _4)\right) \sigma _2+ \left( f_2(\widetilde{\zeta }_2,\widetilde{\zeta }_3,\widetilde{\zeta }_4) -f_1(\zeta _2,\zeta _3,\zeta _4)\right) \sigma _3\,. \end{aligned}$$

6 Representation of Left–$\Lambda -\psi$–hyperholomorphic Functions in a Special Case

In this section we consider operator (2.5) and Eq. (2.6). Now we will find a representation of left–$\Lambda -\psi$–hyperholomorphic functions for a special choice of parameters $\psi _1,\psi _2,\psi _3$, $\psi _4$ and $\Lambda$. For this purpose, we reduce Eq. (2.6) to a system of four PDEs. Denote

$$\begin{aligned} \Lambda :=\Lambda _1e_1+\Lambda _2e_2+\Lambda _3e_3+\Lambda _4e_4. \end{aligned}$$

(6.1)

Using result of Sect. 4, we obtain that Eq. (2.6) is equivalent to the non-homogeneous system of PDEs:

$$\begin{aligned} \begin{array}{lr} \frac{\partial }{\partial z_1}(\alpha _1f_1+\alpha _3f_4)+\frac{\partial }{\partial z_2}(\beta _1f_1+\beta _3f_4) +\frac{\partial }{\partial z_3}(\gamma _1f_1+\gamma _3f_4)+\frac{\partial }{\partial z_4}(\delta _1f_1+\delta _3f_4) &{} \\ \hspace{50mm}=-\Lambda _1f_1-\Lambda _3f_4 ,&{}\\ \frac{\partial }{\partial z_1}(\alpha _2f_2+\alpha _4f_3)+\frac{\partial }{\partial z_2}(\beta _2f_2+\beta _4f_3) +\frac{\partial }{\partial z_3}(\gamma _2f_2+\gamma _4f_3)+\frac{\partial }{\partial z_4}(\delta _2f_2+\delta _4f_3) &{} \\ \hspace{50mm} =-\Lambda _2f_2-\Lambda _4f_3, &{}\\ \frac{\partial }{\partial z_1}(\alpha _1f_3+\alpha _3f_2)+\frac{\partial }{\partial z_2}(\beta _1f_3+\beta _3f_2) +\frac{\partial }{\partial z_3}(\gamma _1f_3+\gamma _3f_2)+\frac{\partial }{\partial z_4}(\delta _1f_3+\delta _3f_2) &{} \\ \hspace{50mm}=-\Lambda _1f_3-\Lambda _3f_2, &{}\\ \frac{\partial }{\partial z_1}(\alpha _2f_4+\alpha _4f_1)+\frac{\partial }{\partial z_2}(\beta _2f_4+\beta _4f_1) +\frac{\partial }{\partial z_3}(\gamma _2f_4+\gamma _4f_1)+\frac{\partial }{\partial z_4}(\delta _2f_4+\delta _4f_1) &{} \\ \hspace{50mm}=-\Lambda _4f_1-\Lambda _2f_4. &{}\\ \end{array} \end{aligned}$$

(6.2)

Theorem 6.1

Let

$$\begin{aligned} \begin{aligned}&\psi _1=\alpha _1e_1+\alpha _2e_2+\alpha _3e_3+\alpha _4e_4,\qquad \alpha _1\alpha _2\ne \alpha _3\alpha _4\,,\\&\psi _2=\lambda \alpha _1e_1+\mu \alpha _2e_2+\mu \alpha _3e_3+\lambda \alpha _4e_4,\\&\psi _3=\theta \alpha _1e_1+\vartheta \alpha _2e_2+\vartheta \alpha _3e_3+\theta \alpha _4e_4,\\&\psi _4=\nu \alpha _1e_1+\eta \alpha _2e_2+\eta \alpha _3e_3+\nu \alpha _4e_4, \end{aligned} \end{aligned}$$

(6.3)

where $\alpha _1,\alpha _2,\alpha _3,\alpha _4,\lambda ,\mu ,\theta ,\vartheta ,\nu ,\eta$ are arbitrary complex numbers. In additional, let $\Lambda$ be of the form (6.1) such that

$$\begin{aligned} \alpha _2\Lambda _3=\alpha _3\Lambda _2\,,\qquad \alpha _1\Lambda _4=\alpha _4\Lambda _1\,. \end{aligned}$$

(6.4)

Then every left–$\Lambda -\psi$–hyperholomorphic function is of the form

$$\begin{aligned} \begin{aligned} f(z)&=e_1\,\Phi _1(\widetilde{\zeta }_2,\widetilde{\zeta }_3,\widetilde{\zeta }_4)\cdot \exp \Big (\frac{\alpha _3\Lambda _4-\alpha _2\Lambda _1}{\alpha _1\alpha _2-\alpha _3\alpha _4}\,z_1 \Big )\\&\quad +e_2\,\Phi _2(\zeta _2,\zeta _3,\zeta _4)\cdot \exp \Big (\frac{\alpha _4\Lambda _3-\alpha _1\Lambda _2}{\alpha _1\alpha _2-\alpha _3\alpha _4}\,z_1\Big )\\&\quad +e_3\,\Phi _3(\widetilde{\zeta }_2,\widetilde{\zeta }_3,\widetilde{\zeta }_4)\cdot \exp \Big (\frac{\alpha _3\Lambda _4-\alpha _2\Lambda _1}{\alpha _1\alpha _2-\alpha _3\alpha _4}\,z_1\Big )\\&\quad +e_4\,\Phi _4(\zeta _2,\zeta _3,\zeta _4)\cdot \exp \Big (\frac{\alpha _4\Lambda _3-\alpha _1\Lambda _2}{\alpha _1\alpha _2-\alpha _3\alpha _4}\,z_1\Big ), \end{aligned} \nonumber \\ \end{aligned}$$

(6.5)

where $\widetilde{\zeta }_2,\widetilde{\zeta }_3, \widetilde{\zeta }_4, \zeta _2, \zeta _3, \zeta _4$ are defined by relations (4.4), and $\Phi _1,\Phi _2,\Phi _3,\Phi _4$ are arbitrary holomorphic functions of three complex variables.

Proof

For given parameters (6.3) the first equation of system (6.2) takes the form

$$\begin{aligned}{} & {} \frac{\partial }{\partial z_1}\left( \alpha _1f_1+\alpha _3f_4\right) +\frac{\partial }{\partial z_2}\left( \lambda \alpha _1f_1+\mu \alpha _3f_4\right) \nonumber \\{} & {} \quad +\frac{\partial }{\partial z_3}\left( \theta \alpha _1f_1+\vartheta \alpha _3f_4\right) +\frac{\partial }{\partial z_4}\left( \nu \alpha _1f_1+\eta \alpha _3f_4\right) =-\Lambda _1f_1-\Lambda _3f_4. \end{aligned}$$

(6.6)

Similarly, for given parameters (6.3) the fourth equation of system (6.2) takes the form

$$\begin{aligned}{} & {} \frac{\partial }{\partial z_1}\left( \alpha _4f_1+\alpha _2f_4\right) +\frac{\partial }{\partial z_2}\left( \lambda \alpha _4f_1+\mu \alpha _2f_4\right) \nonumber \\{} & {} \quad +\frac{\partial }{\partial z_3}\left( \theta \alpha _4f_1+\vartheta \alpha _2f_4\right) +\frac{\partial }{\partial z_4}\left( \nu \alpha _4f_1+\eta \alpha _2f_4\right) =-\Lambda _4f_1-\Lambda _2f_4 . \end{aligned}$$

(6.7)

Consider the difference between Eq. (6.6) multiplied by $\alpha _2$ and Eq. (6.7) multiplied by $\alpha _3$. Then, taking into account (6.4), we obtain the following equation

$$\begin{aligned}{} & {} \frac{\partial }{\partial z_1}\Big (f_1(\alpha _1\alpha _2-\alpha _3\alpha _4)+f_4(\alpha _2\alpha _3-\alpha _2\alpha _3)\Big )\\{} & {} \quad +\frac{\partial }{\partial z_2}\Big (f_1(\lambda \alpha _1\alpha _2-\lambda \alpha _3\alpha _4)+f_4(\mu \alpha _2\alpha _3-\mu \alpha _2\alpha _3)\Big )\\{} & {} \quad +\frac{\partial }{\partial z_3}\Big (f_1(\theta \alpha _1\alpha _2-\theta \alpha _3\alpha _4)+f_4(\vartheta \alpha _2\alpha _3-\vartheta \alpha _2\alpha _3)\Big )\\{} & {} \quad +\frac{\partial }{\partial z_4}\Big (f_1(\nu \alpha _1\alpha _2-\nu \alpha _3\alpha _4)+f_4(\eta \alpha _2\alpha _3-\eta \alpha _2\alpha _3)\Big )=(\Lambda _4\alpha _3-\Lambda _1\alpha _2)f_1. \end{aligned}$$

Thus, we obtain the equation

$$\begin{aligned} \frac{\partial f_1}{\partial z_1}+\lambda \frac{\partial f_1}{\partial z_2}+\theta \frac{\partial f_1}{\partial z_3}+\nu \frac{\partial f_1}{\partial z_4} =\frac{\Lambda _4\alpha _3-\Lambda _1\alpha _2}{\alpha _1\alpha _2-\alpha _3\alpha _4}\,f_1. \end{aligned}$$

(6.8)

For Eq. (6.8) consider the characteristic equation

$$\begin{aligned} \frac{dz_1}{1}=\frac{dz_2}{\lambda }=\frac{dz_3}{\theta }=\frac{dz_4}{\nu }=\frac{(\alpha _1\alpha _2-\alpha _3\alpha _4)\,d\,f_1}{(\Lambda _4\alpha _3-\Lambda _1\alpha _2)f_1}. \end{aligned}$$

(6.9)

The solutions of system (6.9) are the following integrals

$$c_2=\lambda z_1-z_2,\quad c_3=\theta z_1-z_3,\quad c_4=\nu z_1-z_4,$$

$$c_5=\ln f_1+\frac{\alpha _2\Lambda _1-\alpha _3\Lambda _4}{\alpha _1\alpha _2-\alpha _3\alpha _4}\,z_1. $$

Therefore, the general solution of Eq. (6.8) has the form

$$f_1=\Phi _1(\widetilde{\zeta }_2,\widetilde{\zeta }_3,\widetilde{\zeta }_4)\cdot \exp \Big (\frac{\alpha _3\Lambda _4-\alpha _2\Lambda _1}{\alpha _1\alpha _2-\alpha _3\alpha _4}\,z_1\Big ),$$

where $\widetilde{\zeta }_2,\widetilde{\zeta }_3,\widetilde{\zeta }_4$ are defined by equalities (4.4), and $\Phi _1$ is an arbitrary holomorphic function of three complex variables.

Similarly, we obtain the representations for the components $f_2,f_3,f_4$. $\square$

Thus, formula (6.5) gives a representation of left–$\Lambda -\psi$–hyperholomorphic function for a special choice of $\Lambda$ and $\psi$.

Remark 6.2

It is clear from the Eqs. (2.2) and (2.5) that for $\Lambda =0$ a set of left–$\Lambda -\psi$–hyperholomorphic functions coincide with a set of left–$\psi$–hyperholomorphic functions. This fact is also confirmed by Theorems 4.1 and 6.1, because representation (6.5) coincides with representation (4.3) for $\Lambda = 0$.

Remark 6.3

Using formulas (1.4), we can rewrite representation (6.5) in the Pauli basis:

$$\begin{aligned} f(z)= & {} \sigma _0\,\left( \Phi _1(\widetilde{\zeta }_2,\widetilde{\zeta }_3,\widetilde{\zeta }_4)+\Phi _2(\zeta _2,\zeta _3,\zeta _4)\right) \exp \Bigg (\frac{\alpha _3\Lambda _4-\alpha _2\Lambda _1}{\alpha _1\alpha _2-\alpha _3\alpha _4}\,z_1 \Bigg ) \\{} & {} +\sigma _1\, \left( i\Phi _3(\widetilde{\zeta }_2,\widetilde{\zeta }_3,\widetilde{\zeta }_4)-i\Phi _4(\zeta _2,\zeta _3,\zeta _4) \right) \exp \Bigg (\frac{\alpha _3\Lambda _4-\alpha _2\Lambda _1}{\alpha _1\alpha _2-\alpha _3\alpha _4}\,z_1 \Bigg ) \\{} & {} +\sigma _2\, \left( -\Phi _3(\widetilde{\zeta }_2,\widetilde{\zeta }_3,\widetilde{\zeta }_4)-\Phi _4(\zeta _2,\zeta _3,\zeta _4) \right) \exp \Bigg (\frac{\alpha _3\Lambda _4-\alpha _2\Lambda _1}{\alpha _1\alpha _2-\alpha _3\alpha _4}\,z_1 \Bigg ) \\{} & {} +\sigma _3\,\left( \Phi _2(\zeta _2,\zeta _3,\zeta _4)-\Phi _1(\widetilde{\zeta }_2,\widetilde{\zeta }_3,\widetilde{\zeta }_4)\right) \exp \Bigg (\frac{\alpha _3\Lambda _4-\alpha _2\Lambda _1}{\alpha _1\alpha _2-\alpha _3\alpha _4}\,z_1 \Bigg ). \end{aligned}$$