In this section we discuss RHBVPs with variable coefficients for monogenic functions. In this case we will restrict ourselves to functions which have axial symmetry and are defined over axial domains of \(\mathbb{R}^{4}\). We first solve the RHBVP in \(\mathbb{R}^{4}\) by transforming it into a RHBVP for complex analytic functions. In the case of boundary values belonging to a Hölder space we represent its solution in terms of an explicit integral representation formula. As a corollary we obtain the solution to the corresponding Schwarz problem in quaternion analysis.
A lemma
Let us start with the following lemma.
Lemma 3.1
Let
\(A(D,\mathbb{C})\)
denote the set of analytic functions defined in
\(D\subset\mathbb {C_{+}}\). Then there exists an injective mapping
ψ
between
\(A (D, \mathbb{C})/\operatorname{ker} \varphi\)
and
\(M (\Omega,\mathbb{H})\), where
φ
is a surjective mapping from
\(A(D,\mathbb{C})\)
onto
\(M(\Omega ,\mathbb{H})\)
given by
$$\begin{aligned}& A(D,\mathbb{C})\rightarrow M(\Omega,\mathbb {H}), \\& f=u(x_{0},y)+iv(x_{0},y)\mapsto\phi =A(x_{0},r) + \underline{\omega} B(x_{0},r), \end{aligned}$$
where
$$A(x_{0},r)+\underline{\omega} B(x_{0},r )=\Delta \bigl(u(x_{0},r)+\underline{\omega} v (x_{0},r)\bigr), $$
\(x=x_{0}+r\underline{\omega} \in\mathbb{R}^{4}\), and
D
is the projection of Ω onto the
\((x_{0},r )\)plane.
In fact the surjective map φ is coming from Fueter’s theorem [14, 27, 28].
Proof
Let the map ψ be given by
$$\begin{aligned}& A(D,\mathbb{C})/\operatorname{ker} \varphi\rightarrow M(\Omega,\mathbb{H}), \\& f+\hat{g}\mapsto\phi, \end{aligned}$$
where \(\operatorname{ker} \varphi=\{f:\varphi(f)=0,f\in A (D,\mathbb{C})\}\), \(f=u(x_{0},r)+iv (x_{0},r)\), \(\phi=A(x_{0},r)+\underline{\omega} B (x_{0},r)\), \(\hat{g}=u_{1}(x_{0},r)+iv_{1} (x_{0},r)\), and \(\hat{g}\in\operatorname{ker} \varphi\). Then the map ψ is an injective mapping between \(A (D,\mathbb{C})/\operatorname{ker} \varphi\) and \(M (\Omega,\mathbb{H})\), where \(u_{1}\), \(v_{1}\) are defined similarly to u, v, respectively, and \(M(\Omega,\mathbb{H} )\) is the same as that in Section 2. The result follows. □
RHBVPs for monogenic functions
The idea behind this lemma allows us to link RHBVPs for axially monogenic functions with RHBVPs for analytic functions over the complex plane. Let us start the discussion with the following problem.
Problem I
Find a function \(\phi\in C^{1}(\Omega,\mathbb{H})\) of axial type which satisfies the condition
$$ (\star)\quad \left \{ \textstyle\begin{array}{l} \mathcal{D}\phi(x)=0,\quad x\in\Omega, \\ \operatorname{Re}\{\lambda(t)\phi(t)\}=g(t),\quad t\in \partial\Omega, \end{array}\displaystyle \right . $$
where \(g:\partial\Omega\rightarrow\mathbb{R}\) and λ is a \(\mathbb{R}\)valued function defined on ∂Ω.
Let us first transform the above problem to a RHBVP for analytic functions. The result is given in the following theorem.
Theorem 3.1
The RiemannHilbert boundary value problem (⋆) is equivalent to the following problem: Find a complexvalued analytic function
h, such that
$$ \left \{ \textstyle\begin{array}{l} \partial_{\bar{z}}h(z)=0,\quad z\in D\subset\mathbb {C}_{+}, \\ \operatorname{Re}\{\lambda(z)h(z)\}=\frac{r}{2}g(z),\quad z\in \partial D\subset\mathbb{C}_{+}, \end{array}\displaystyle \right . $$
where
\(z=x_{0}+ir\), \(\partial_{\bar{z}}=\frac{1}{2}(\partial _{x_{0}}+i\partial_{r})\), \(h=\partial_{r}f\), f, and
r
given as in Lemma
3.1, D
is a domain of
\(\mathbb{C}_{+}\)
with boundary
∂D
given as the projection of
\(\Omega\subset\mathbb{R}^{4}\)
into the complex plane, and
\(\lambda, g:\partial D\rightarrow\mathbb {R}\)
are both scalarvalued functions.
Proof
From Lemma 3.1 we have complexvalued functions \(f=u+iv\in A(D,\mathbb{C})\) and \(\hat {g}=u_{1}+iv_{1}\in A(D,\mathbb{C})\) with \(\hat {g}\in\operatorname{ker} \varphi\), such that \(\psi(f+\hat {g})=\phi\). Hence, \(\varphi\circ\psi^{1}(\phi)=f\), where \(\psi^{1}\) is the inverse of ψ on its range and \(\varphi\circ \psi^{1}\) denotes the composition of φ and \(\psi^{1}\). Adapting the boundary condition we see that the RiemannHilbert boundary value problem (⋆) is equivalent to the case
$$ (\divideontimes)\quad \left \{ \textstyle\begin{array}{l} \partial_{\bar{z}}f(z)=0,\quad z\in D\subset\mathbb {C}_{+}, \\ \operatorname{Re}\{\lambda(t)\Delta u\}(t)=g(t),\quad t\in \partial D\subset\mathbb{C}_{+}. \end{array}\displaystyle \right . $$
Using the Laplacian operator \(\Delta=\partial ^{2}_{x^{2}_{0}}+\partial^{2}_{r^{2}}+\frac{2}{r}\partial_{r}+\frac {1}{r^{2}}\partial_{\omega}\frac{1}{r^{2}}\partial^{2}_{\omega ^{2}}\), we have \(\Delta u=\frac{2}{r}\partial_{r}u\).
Now, let \(h=\partial_{r}f\) then we have \(\partial_{\bar {z}}h(z)=0\), \(z\in D\) since f is analytic in D.
Hence, our boundary value problem (⋇) reduces to the case
$$ \left \{ \textstyle\begin{array}{l} \partial_{\bar{z}}h(z)=0,\quad z\in D\subset\mathbb {C}_{+}, \\ \operatorname{Re}\{\frac{2}{r}\lambda(z)h(z)\}=g(z),\quad z\in \partial D\subset\mathbb{C}_{+}. \end{array}\displaystyle \right . $$
□
The above theorem allows us to study the solvability of our original RHBVP by studying the equivalent RHBVP over the complex plane.
Theorem 3.2
Suppose
\(g\in H^{\mu}(\partial \Omega,\mathbb{R})\), and
\(D=\{z:za<1\}\subset \mathbb{C}_{+}\)
with
\(z=x_{0}+ir\), \(a=a_{0}+ia_{1}\in\mathbb{C}_{+}\)
being the corresponding domain of
\(\mathbb{C}_{+}\)
with boundary
∂D.

(i)
If
\(\lambda\in H^{\mu}(\partial\Omega,\mathbb{R})\), and
\(\lambda\neq0\)
for arbitrary
\(z\in\partial D\)
then the RiemannHilbert boundary value problem (⋆) is solvable and its solution is given by
$$ \phi(x)=\Delta\bigl(\operatorname{Re}(f) \bigl(x_{0},\vert \underline{x}\vert \bigr)+\underline{\omega}\operatorname{Im}(f) \bigl(x_{0},\vert \underline{x}\vert \bigr)\bigr),\quad x\in\Omega, $$
where
\(\operatorname{Re}(f)\)
and
\(\operatorname{Im}(f)\)
denote the real and imaginary part of the complexvalued function
f, respectively. f
itself is given by
$$ f(z)=\frac{1}{2\pi} \int^{z}_{a} \int_{\partial D}\frac{\tilde {g}(\zeta)}{\zeta\xi}\, d\zeta\, d\xi+\sum ^{+\infty }_{n=0}t_{n}(za)^{n},\quad z\in D, $$
with
\(\tilde{g}=\lambda^{1}\frac{r}{2}g\), \(t_{n}\in\mathbb{C}\).

(ii)
Suppose
\(\lambda=\Pi\hat{\lambda}\)
with
\(\Pi(x)=\Pi ^{m}_{i=1}(x\hat{\alpha}_{i})^{\nu_{i}}\), \(\hat{\alpha}_{i}\in\partial\Omega\)
and
\(\nu_{i}\in\mathbb {N}\). Furthermore, if
\(\hat{\lambda}\in H^{\mu}(\partial \Omega,\mathbb{R})\)
and
\(\hat{\lambda}\neq0\)
for arbitrary
\(x\in\partial\Omega\)
then the RiemannHilbert boundary value problem (⋆) is solvable, and its solution is given again by
$$ \phi(x)=\Delta\bigl(\operatorname{Re}(f) \bigl(x_{0},\vert \underline{x}\vert \bigr)+\underline{\omega}\operatorname{Im}(f) \bigl(x_{0},\vert \underline{x}\vert \bigr)\bigr),\quad x\in\Omega, $$
where
f
is given by
$$ f(z)=\frac{1}{2\pi} \int^{z}_{a} \int_{\partial D}\frac{1}{\Pi(\xi )}\frac{\hat{g}(\zeta)}{\zeta\xi}\, d\zeta\, d\xi+ \sum^{+\infty}_{n=0} \int^{z}_{a}\frac{l_{n}(\xia)^{n}}{\Pi(\xi)}\, d\xi, \quad z\in D, $$
with
\(\hat{g}=\hat{\lambda}^{1}\frac{r}{2}g\), \(l_{n}\in \mathbb{C}\).
Proof
Since \(D=\{z:za<1\}\subset\mathbb {C}_{+}\) is the projection of Ω into \((x_{0},r)\)plane from Theorem 3.1 we see that the problem (⋆) is equivalent to the problem
$$ ({\divideontimes\divideontimes})\quad \left \{ \textstyle\begin{array}{l} \partial_{\bar{z}}h(z)=0,\quad z\in D, \\ \operatorname{Re}\{\lambda(z)h(z)\}=\frac{r}{2}g(z), \quad z\in \partial D. \end{array}\displaystyle \right . $$
Consider now the function
$$ H(z)=\left \{ \textstyle\begin{array}{l@{\quad}l} h(z),&z\in D\subset\mathbb{C}_{+}, \\ h_{*}(z)=\bar{h}(1/z),&z\in\mathbb{C}_{+}\backslash D\cup\partial D, \end{array}\displaystyle \right . $$
and let us study each case separately.
For the case (i) we can proceed in the following way. Since \(\lambda,g\in H^{\mu}(\partial\Omega,\mathbb{R})\) with \(\lambda\neq0\) for arbitrary \(x\in\partial\Omega\) we can recast our problem as the following Riemann boundary value problem:
$$ (\aleph)\quad \left \{ \textstyle\begin{array}{l} \partial_{\bar{z}}H(z) = 0, \quad z\in\mathbb {C}_{+}\backslash\partial D, \\ H^{+}(z)=H^{}(z)+\lambda^{1}(z)\frac{r}{2}g(z),\quad z\in\partial D, \end{array}\displaystyle \right . $$
where \(H^{\pm}\) are the boundary values of H on ∂D from D and \(\mathbb{C}_{+}\backslash(D\cup\partial D)\), respectively.
Because of \(\lambda,g\in H^{\mu}(\partial D,\mathbb {R})\) we have \(\tilde{g}\triangleq\lambda^{1}\frac {r}{2}g=\frac{1}{4i}\lambda^{1}(z\bar{z})g\in H^{\mu }(\partial D,\mathbb{R})\). Therefore, the solution to the boundary value problem (ℵ) is given by
$$ H(z)=\frac{1}{2\pi i} \int_{\partial D}\frac{\tilde{g}(\zeta )}{\zetaz}\, d\zeta+\sum ^{+\infty}_{n=0}l_{n}(za)^{n}, \quad z\in D, $$
where \(l_{n}\in\mathbb{C}\).
Hence, the unique solution to the boundary value problem (⋇⋇) is expressed by
$$ f(z)=\frac{1}{2\pi} \int^{z}_{a} \int_{\partial D}\frac{\tilde {g}(\zeta)}{\zeta\xi}\,d\zeta \,d\xi+\sum ^{+\infty }_{n=0}t_{n}(za)^{n}, \quad z\in D, $$
(2)
where \(t_{n}\in\mathbb{C}\).
We end up with the solution to the RiemannHilbert boundary value problem (⋆) in the form
$$ \phi(x)=\Delta\bigl(\operatorname{Re}(f) \bigl(x_{0},\vert \underline{x}\vert \bigr) +\underline{\omega}\operatorname{Im}(f) \bigl(x_{0},\vert \underline{x}\vert \bigr)\bigr),\quad x\in\Omega. $$
In the second case (ii) we map again Ω to D and \(\lambda= \Pi\hat{\lambda}\) with \(\Pi(x)=\Pi^{m}_{i=1}(x\hat{\alpha}_{i})^{\nu_{i}}\) where \(\hat{\alpha}_{i}\in\partial\Omega\) becomes \(\lambda (z)=\Pi(z)\hat{\lambda}(z)\), \(z\in\partial D\), where \(\Pi(z)=\Pi^{m}_{i=1}(z\alpha_{i})^{\nu_{i}}\), \(\alpha _{i}\in\partial D\) and \(\hat{\lambda}\neq0\), \(z\in\partial D\). By transforming the boundary value problem in the same way as in the first case we arrive at the problem
$$ \left \{ \textstyle\begin{array}{l} \partial_{\bar{z}}\widetilde{H}(z) = 0,\quad z\in \mathbb{C}_{+}\backslash\partial D, \\ \widetilde{H}^{+}(z)=\widetilde{H}^{}(z)+\hat{\lambda }^{1}(z)\frac{r}{2}g(z), \quad z\in\partial D, \end{array}\displaystyle \right . $$
where \(\widetilde{H}=\Pi H\), \(z\in\mathbb{C}_{+}\backslash\partial D\), and \(\widetilde{H}^{\pm}\), \(z\in\partial D\) are the boundary values of H̃ on ∂D from D and \(\mathbb {C}_{+}\backslash( D\cup\partial D)\), respectively.
Since \(\hat{\lambda}\in H^{\mu}(\partial D,\mathbb {R})\), we have \(\hat{g}\triangleq\hat{\lambda }^{1}\frac{r}{2}g=\frac{1}{4i}\hat{\lambda}^{1}(z\bar {z})g \in H^{\mu}(\partial D,\mathbb{R})\). This allows us to write the solution to the boundary value problem (⋇⋇) in the form
$$ f(z)=\frac{1}{2\pi} \int^{z}_{a} \int_{\partial D}\frac{1}{\Pi(\xi )}\frac{\hat{g}(\zeta)}{\zeta\xi}\,d\zeta \,d\xi+\sum ^{+\infty}_{n=0} \int^{z}_{a}\frac{l_{n}(\xia)^{n}}{\Pi(\xi)}\,d\xi, \quad z\in D, $$
(3)
where \(l_{n}\in\mathbb{C}\).
Therefore, the solution to the RiemannHilbert boundary value problem (⋆) is given by
$$ \phi(x)=\Delta\bigl(\operatorname{Re}(f) \bigl(x_{0},\vert \underline{x}\vert \bigr)+\underline{\omega}\operatorname{Im}(f) \bigl(x_{0},\vert \underline{x}\vert \bigr)\bigr), \quad x\in \Omega. $$
This finishes the proof. □
As a special case of Problem I, we can consider the following Schwarz problem.
Schwarz problem
Find a function \(\phi\in C^{1}(\Omega,\mathbb{H})\), which satisfies the system
$$ (\sharp)\quad \left \{ \textstyle\begin{array}{l} \mathcal{D}\phi(x)=0,\quad x\in\Omega, \\ \operatorname{Re}\{\phi(t)\}=g(t),\quad t\in \partial\Omega, \end{array}\displaystyle \right . $$
where \(g:\partial\Omega\rightarrow\mathbb{R}\) is a \(\mathbb {R}\)valued function defined on ∂Ω.
From Theorem 3.2 we can deduce the following theorem.
Theorem 3.3
If
\(g\in H^{\mu}(\partial \Omega,\mathbb{R})\)
then the Schwarz problem (♯) is solvable, and its solution is given by
$$ \phi(x)=\Delta\bigl(\operatorname{Re}(f) \bigl(x_{0},\vert \underline{x}\vert \bigr)+\underline{\omega}\operatorname{Im}(f) \bigl(x_{0},\vert \underline{x}\vert \bigr)\bigr), \quad x\in \Omega, $$
where
f
is given by
$$ f(z)=\frac{1}{2\pi} \int^{z}_{a} \int_{\partial D}\frac{\tilde {g}_{1}(\zeta)}{\zeta\xi}\,d\zeta \,d\xi+\sum ^{+\infty }_{n=0}t_{n}(za)^{n}, \quad z\in D, $$
with
\(\tilde{g}_{1}=\frac{r}{2}g\), \(t_{n}\in\mathbb{C}\).
Remark 3
Theorem 3.2 actually gives us a method to solve the RHBVPs with variable coefficients for axially monogenic functions in \(\mathbb{R}^{4}\) by transferring it to the study of the corresponding RHBVP for analytic functions over the complex plane.
When the space dimension is 2, then trivially RHBVP (⋆) reduces to the RHBVP for analytic functions on the complex plane [1, 3, 5]. Moreover, when the coefficient λ is a constant equal to 1 for all \(x\in\partial\Omega\) our RHBVP (♯) is just the Schwarz problem for analytic functions over the complex plane [1, 2, 8].
Remark 4
In fact, Theorem 3.3 gives us explicit solutions to the Schwarz problem for a special type of Vekua system defined over the complex plane [13, 14]. As far as we know, in general it is not easy to get explicit but nonformal analytic solutions of this system.
RHBVPs for nullsolutions to \((\mathcal {D}\alpha)\phi=0\), \(\alpha\in\mathbb{R}\)
The above approach can be adapted and generalized to other cases. In the following we will discuss RHBVPs with variable coefficients for functions of axial type defined over an axial domain of \(\mathbb {R}^{4}\) which are nullsolutions to the equation \((\mathcal {D}\alpha)\phi=0\), \(\alpha\in\mathbb{R}\). We begin with the following extension of Problem I.
Problem II
Find a function \(\phi\in\mathcal {C}^{1}(\Omega,\mathbb{H})\) of axial type which satisfies the condition
$$ ({\star\star})\quad \left \{ \textstyle\begin{array}{l} (\mathcal{D}\alpha)\phi=0,\quad\alpha \in\mathbb{R}, x\in\Omega, \\ \operatorname{Re}\{\lambda(t)\phi(t)\}=g(t),\quad t\in \partial\Omega, \end{array}\displaystyle \right . $$
where α is understood as αI, with I being the identity operator, \(g:\partial\Omega\rightarrow\mathbb{R}\), and λ is a \(\mathbb{R}\)valued function defined on ∂Ω.
Theorem 3.4
Suppose
\(g\in H^{\mu}(\partial \Omega,\mathbb{R})\), and
\(D=\{z:za<1\}\subset \mathbb{C}_{+}\)
with
\(z=x_{0}+ir\), \(a=a_{0}+ia_{1}\in\mathbb{C}_{+}\)
being the corresponding domain of
\(\mathbb{C}_{+}\)
with boundary
∂D.

(i)
If
\(\lambda\in H^{\mu}(\partial\Omega,\mathbb{R})\), and
\(\lambda\neq0\)
for arbitrary
\(z\in\partial D\)
then the RiemannHilbert boundary value problem (⋆) is solvable and its solution is given by
$$ \phi(x)= e^{\alpha x_{0}}\Delta\bigl(\operatorname{Re}(f) \bigl(x_{0}, \vert \underline{x}\vert \bigr)+\underline{\omega}\operatorname{Im}(f) \bigl(x_{0},\vert \underline{x}\vert \bigr)\bigr),\quad x\in\Omega, $$
where
\(\operatorname{Re}(f)\)
and
\(\operatorname{Im}(f)\)
denote the real and imaginary part of the complexvalued function
f, respectively. f
itself is given by
$$ f(z)=\frac{1}{2\pi} \int^{z}_{a} \int_{\partial D}\frac{\widetilde {e^{\alpha x_{0}}g}(\zeta)}{\zeta\xi}\,d\zeta \,d\xi+\sum ^{+\infty}_{n=0}t_{n}(za)^{n},\quad z\in D, $$
with
\(\tilde{g} =\lambda^{1}\frac{r}{2}g\), \(t_{n}\in\mathbb {C}\).

(ii)
Suppose
\(\lambda=\Pi\hat{\lambda}\)
with
\(\Pi(x)=\Pi ^{m}_{i=1}(x\hat{\alpha}_{i})^{\nu_{i}}\), \(\hat{\alpha}_{i}\in\partial\Omega\), and
\(\nu_{i}\in\mathbb {N}\). Furthermore, if
\(\hat{\lambda}\in H^{\mu}(\partial \Omega,\mathbb{R})\)
and
\(\hat{\lambda}\neq0\)
for arbitrary
\(x\in\partial\Omega\)
then the RiemannHilbert boundary value problem (⋆) is solvable, and its solution is given again by
$$ \phi(x) = e^{\alpha x_{0}}\Delta\bigl(\operatorname{Re}(f) \bigl(x_{0}, \vert \underline{x}\vert \bigr)+\underline{\omega}\operatorname{Im}(f) \bigl(x_{0},\vert \underline{x}\vert \bigr)\bigr), \quad x\in \Omega, $$
where
f
is given by
$$ f(z)=\frac{1}{2\pi} \int^{z}_{a} \int_{\partial D}\frac{1}{\Pi(\xi )}\frac{\widehat{e^{\alpha x_{0}} g}(\zeta)}{\zeta\xi}\,d\zeta \,d\xi + \sum^{+\infty}_{n=0} \int^{z}_{a}\frac{l_{n}(\xi a)^{n}}{\Pi(\xi)}\,d\xi, \quad z\in D, $$
with
\(\hat{g}=\hat{\lambda}^{1}\frac{r}{2}g\), \(l_{n}\in \mathbb{C}\).
Proof
Since \((\mathcal{D}\alpha)\phi= \mathcal{D}(e^{\alpha x_{0}}\phi)\), \(\alpha\in\mathbb {R}\), then
$$ (\mathcal{D}\alpha)\phi=0 \quad \mbox{is equivalent to}\quad \mathcal{D} \bigl(e^{\alpha x_{0}}\phi\bigr)=0,\quad \alpha\in \mathbb{R}. $$
(4)
Therefore, problem (⋆⋆) is equivalent to the case
$$ \left \{ \textstyle\begin{array}{l} \mathcal{D}(e^{\alpha x_{0}}\phi) = 0,\quad\alpha\in\mathbb{R}, x\in\Omega, \\ \operatorname{Re}\{\lambda(t)e^{\alpha x_{0}}\phi(t)\} = e^{\alpha x_{0}}g(t),\quad t\in \partial\Omega, \end{array}\displaystyle \right . $$
where \(g:\partial\Omega\rightarrow\mathbb{R}\) and λ is a \(\mathbb{R}\)valued function defined on ∂Ω.
Noting that \(D=\{z:za<1\}\subset\mathbb{C}_{+}\) with \(z=x_{0}+ir\), \(a=a_{0}+ia_{1}\in\mathbb{C}_{+}\) is the projection of Ω onto the \((x_{0},r)\)plane, associating with \(g\in H^{\mu}(\partial\Omega,\mathbb{R})\), we have \(e^{\alpha x_{0}}g\in H^{\mu}(\partial\Omega,\mathbb{R})\).
We apply Theorem 3.2.
(i) If \(\lambda\in H^{\mu}(\partial\Omega,\mathbb{R})\), and \(\lambda\neq0\) for arbitrary \(z\in\partial D\) then the RiemannHilbert boundary value problem (⋆) is solvable and its solution is given by
$$ \phi(x)= e^{\alpha x_{0}}\Delta\bigl(\operatorname{Re}(f) \bigl(x_{0}, \vert \underline{x}\vert \bigr)+\underline{\omega}\operatorname{Im}(f) \bigl(x_{0},\vert \underline{x}\vert \bigr)\bigr), \quad x\in \Omega, $$
where \(\operatorname{Re}(f)\) and \(\operatorname{Im}(f)\) denote the real and imaginary part of the complexvalued function f, respectively. f itself is given by
$$ f(z)=\frac{1}{2\pi} \int^{z}_{a} \int_{\partial D}\frac{\widetilde {e^{\alpha x_{0}}g}(\zeta)}{\zeta\xi}\,d\zeta \,d\xi+\sum ^{+\infty}_{n=0}t_{n}(za)^{n},\quad z\in D, $$
with \(\tilde{g} =\lambda^{1}\frac{r}{2}g\), \(t_{n}\in\mathbb {C}\).
(ii) Suppose \(\lambda=\Pi\hat{\lambda}\) with \(\Pi(x)=\Pi ^{m}_{i=1}(x\hat{\alpha}_{i})^{\nu_{i}}\), \(\hat{\alpha}_{i}\in\partial\Omega\), and \(\nu_{i}\in\mathbb {N}\). Furthermore, if \(\hat{\lambda}\in H^{\mu}(\partial \Omega,\mathbb{R})\) and \(\hat{\lambda}\neq0\) for arbitrary \(x\in\partial\Omega\) then the RiemannHilbert boundary value problem (⋆) is solvable, and its solution is given again by
$$ \phi(x) = e^{\alpha x_{0}}\Delta\bigl(\operatorname{Re}(f) \bigl(x_{0}, \vert \underline{x}\vert \bigr)+\underline{\omega}\operatorname{Im}(f) \bigl(x_{0},\vert \underline{x}\vert \bigr)\bigr), \quad x\in \Omega, $$
where f is given by
$$ f(z)=\frac{1}{2\pi} \int^{z}_{a} \int_{\partial D}\frac{1}{\Pi(\xi )}\frac{\widehat{e^{\alpha x_{0}} g}(\zeta)}{\zeta\xi}\,d\zeta \,d\xi + \sum^{+\infty}_{n=0} \int^{z}_{a}\frac{l_{n}(\xi a)^{n}}{\Pi(\xi)}\,d\xi, \quad z\in D, $$
with \(\hat{g}=\hat{\lambda}^{1}\frac{r}{2}g\), \(l_{n}\in \mathbb{C}\). The result follows. □
Remark 5
When the space dimension is 2, Problem II reduces to the RiemannHilbert boundary value problem for metaanalytic functions defined over the complex plane [9, 10].
Similar to the Schwarz problem in the previous subsection, let \(\lambda=1\) in Problem II, we can take it in account in the following problem.
Problem III
Find a function \(\phi\in C^{1}(\Omega,\mathbb{H})\) which satisfies the system
$$ \left \{ \textstyle\begin{array}{l} (\mathcal{D}\alpha)\phi=0,\quad \alpha \in\mathbb{R},x\in\Omega, \\ \operatorname{Re}\{\phi(t)\}=g(t),\quad t\in \partial\Omega, \end{array}\displaystyle \right . $$
where α is understood as αI, with I being the identity operator, and \(g:\partial\Omega\rightarrow\mathbb{R}\) is a \(\mathbb{R}\)valued function defined on ∂Ω.
Thanks to Theorem 3.4 we can derive the following theorem.
Theorem 3.5
If
\(g\in H^{\mu}(\partial \Omega,\mathbb{R})\)
then Problem
III
is solvable, and its solution is given by
$$ \phi(x) = e^{\alpha x_{0}} \Delta\bigl(\operatorname{Re}(f) \bigl(x_{0},\vert \underline{x}\vert \bigr)+\underline{\omega} \operatorname{Im}(f) \bigl(x_{0},\vert \underline{x}\vert \bigr) \bigr),\quad x\in\Omega, $$
where
f
is given by
$$ f(z)=\frac{1}{2\pi} \int^{z}_{a} \int_{\partial D}\frac{\widetilde {e^{\alpha x_{0}} g}_{1}(\zeta)}{\zeta\xi}\,d\zeta \,d\xi+\sum ^{+\infty}_{n=0}t_{n}(za)^{n}, \quad z\in D, $$
with
\(\tilde{g}_{1}=\frac{r}{2}g\), \(t_{n}\in\mathbb{C}\).
Remark 6
If the boundary data are of Hölder class, we consider the RiemannHilbert boundary value problems I, II, III with variable coefficients for monogenic functions of axial type in \(\mathbb{R}^{4}\) and nullsolutions to \((\mathcal{D}\alpha )\phi=0\), \(\alpha\in\mathbb{R}\). Note that Problem II reduces to Problem I when α equals 0 while Problem III reduces to Problem II when λ equals 1. Following the same argument, all results can be extended to the RiemannHilbert boundary value problems of monogenic functions of axial type in \(\mathbb{R}^{4}\), with boundary data belonging to \(L_{p}\) (\(1< p<+\infty\)).