Complex Analysis and Operator Theory (2020) 14:53 Complex Analysis https://doi.org/10.1007/s11785-020-01010-5 and Operator Theory

Dirac Operators with Gradient Potentials and Related Monogenic Functions

Longfei Gu1 · Daowei Ma2

Received: 25 June 2019 / Accepted: 28 May 2020 © Springer Nature Switzerland AG 2020

Abstract We investigate some properties of solutions to Dirac operators with gradient potentials. Solutions to Dirac operators with gradient potentials are called monogenic functions with respect to the potential functions. We establish a Borel–Pompeiu formula, and obtain Cauchy integral formula and mean value formula about such functions. Based on integral formulas, we prove some geometric properties of monogenic functions with respect to the potential functions and construct some integral transforms. The boundedness of integral transforms in Hölder space is given. We prove the Plemelj– Sokhotski formula and the Painlevé theorem. As applications, we firstly prove that a kind of Riemann–Hilbert problem for monogenic functions with respect to the poten- tial functions is solvable. Explicit representation formula of the solution is also given. We then establish solvability conditions of a kind of general Riemann–Hilbert prob- lem for monogenic functions with respect to the potential functions. Finally, we give some examples about the above Riemann–Hilbert problem.

Keywords Dirac operators · Gradient potentials · Clifford analysis

Mathematics Subject Classification 30G35

Communicated by Irene Sabadini, Michael Shapiro and Daniele Struppa.

This article is part of the topical collection “Higher Dimensional Geometric Function Theory and Hypercomplex Analysis” edited by Irene Sabadini, Michael Shapiro and Daniele Struppa. This work was partially supported by NSF of China (Grant Nos. 11401287, 11771195) and AMEP of Linyi University.

B Longfei Gu [email protected] Daowei Ma [email protected]

1 Department of Mathematics, Linyi University, Linyi 276005, Shandong, People’s Republic of China 2 Department of Mathematics, Wichita State University, Wichita, KS 67260-0033, USA

0123456789().: V,-vol 53 Page 2 of 19 L. Gu, D. Ma

1 Introduction

In books, [4,10], R. Delanghe, F, Brackx, F. Sommen and V. Soucek use Clifford algebra to establish a function theory for the Dirac operator. This theory is also called Clifford analysis. Clifford analysis is a higher dimensional function theory offering a refinement of classical harmonic analysis and generalize classical complex analysis in plane to a higher dimension in a natural and elegant way. The theory is centered around the concept of monogenic functions, i.e., null solutions of the Dirac operator. Under the framework of Clifford analysis, in [1–3,13,15,16,18–20,30], many interesting results about boundary value problems for monogenic functions and poly-monogenic func- tions, singular integral equations, and partial differential equations in Clifford analysis and quaternionic analysis are presented. These results in higher dimensions are sim- ilar to classical boundary value problems for analytic functions and singular integral equations in complex plane (see, [5,6,21,26]). In [31], Xu obtain the general solution of the equation (D − λ)w = 0 in SO(n)-invariant open subsets  of Rn, whereby D is the Dirac operator and λ is a complex constant, and construct the fundamental solution of D − λ and integral representation formulas. From then on, many scholars in mathematics and physics have researched theories of integral representation for- mulas about the equation (D − λ)kw = 0 where k is a positive integer and λ is a real, complex or quaternionic constant by Clifford analytic methods. Moreover, based on integral representation formulas, some Dirichlet, Riemann–Hilbert boundary value problems are considered. We refer to [12,15,17,18,25,28,29,31]. However, to the best of our knowledge, about function theories for the operator D − λ(x) where λ(x) is a real functions in Rn have not been considered. In this paper, we mainly study a func- tion theory for the operator D − λ(x) and associate Riemann–Hilbert problems in the universal Clifford algebra Cl(Vn,n). The main motivations is that a large number of interesting physical applications (see [3,22–24]), for instance, Maxwell’s equations in physics, which are the fundamental equations of electromagnetism, can be expressed in terms of Clifford analysis as (D − λ(x))[u]=0. Its boundary value problems reduce to the Riemann–Hilbert problems. This article is organized as follows: In Sect. 2, we recall some basic facts about uni- versal Clifford algebras and Clifford analysis which will be needed in the article. In the section, we also introduce the Dirac operators with gradient potentials. In Sect. 3,we establish some integral formulas, including Borel–Pompeiu formula, Cauchy integral formula, and the mean value formula. Using the Cauchy integral formula, we investi- gate some geometric properties of the “monogenic functions” related. In Sect. 4,we discuss some integral transforms in Hölder space. Moreover we obtain Plemelj– for- mula and Painlevé theorem. Finally, in Sects. 5 and 6, two kinds of Riemann–Hilbert problems for such monogenic functions are considered. Dirac Operators with Gradient Potentials and Related … Page 3 of 19 53

2 Preliminaries

Let A := R(e1,...,en) denote the free R-algebra with n indeterminants {e1,...,en}. Let J be the two-sided ideal in A generated by the elements

{ 2 − , = ,..., ; 2 + , = + ,..., ; + , ≤ < ≤ }. ei 1 i 1 s ei 1 i s 1 n ei e j e j ei 1 i j n

The quotient algebra Cl(Vn,s) := A/J is called the Clifford algebra with parameters n, s. Without risk of ambiguity, we take the usual practice of using the same symbol to denote an indeterminant ei in A and its equivalent class in A/J. Therefore, e1,...,en considered as elements of A/J have the following relations: ⎧ 2 = , = ,..., , ⎨ ei 1 i 1 s 2 =− , = + ,..., , (2.1) ⎩ ei 1 i s 1 n ei e j + e j ei = 0, i = j.

Set

:= ··· , ≤ < ···< ≤ . el1...lr el1 elr while 1 l1 lr n

For more information on Cl(Vn,s), we refer to [4,10]. In this article, we only consider s = n. Thus Cl(Vn,n) is a real linear non-commutative algebra. An involution is defined by ⎧ ( )( ( )+ ) ⎨ = (− ) n A n A 3 , ∈ P , eA 1 2 eA if A N (2.2) ⎩ λ = λAeA, if λ = λAeA, A∈P N A∈P N where

{eA, A ={l1,...,lr }∈P N, 1 ≤ l1 < ···< lr ≤ n}, n(A) is the cardinal number of the set A, N stands for the set {1, 2,...,n} and P N denotes the family of all order-preserving subsets of N in the above way. The Cl(Vn,n)- value n-1-differential form

n σ = (− )i−1 N d 1 ei dxi i=1 is exact, where

N = ∧···∧ ∧ ∧···∧ . dxi dx1 dxi−1 dxi+1 dxn 53 Page 4 of 19 L. Gu, D. Ma

If dS stands for the classical surface element and

n n = ei ni , i=1 where ni is the i-th component of the outward pointing normal, which yields the Clifford-valued surface element dσ can be written as

dσ = ndS. (2.3)  λ λ=( |λ |2) 1 R (λ) The norm of is defined by A∈P N A 2 .If e denotes the scalar portion of λ ∈ Cl(Vn,n), then it follows  2 2 Re(λλ) = Re(λλ) = |λA| =λ . A∈P N

n Suppose  be an open bounded non-empty subset of R (n ≥ 3). We introduce = n ∂ . =   the Dirac operator D i=1 ei ∂x In particular, we obtain that DD where n i is the Laplacian over R . A function f :  → Cl(Vn,n) is said to be left monogenic if it satisfies the equation D[ f ](x) = 0 for each x ∈ . A similar definition can be given for right monogenic functions. Denote

z j = x j − x1e1e j , j = 2, 3,...,n and

1  V ,..., (x) = z ···z l1 l p p! l1 l p π(lr ,...,l p)

p where (l1,...,l p) ∈{1, 2,...,n} , the sum is taken over all permutations with rep- ( ,..., ) ( ) = = etition of the sequence l1 l p . In particular we define Vl1,...,l p x 1forp 0 ( ) = < and Vl1,...,l p x 0forp 0. In what follows, M(r, f ) denotes the maximum modulus of a monogenic function f in a ball centered at 0, with radius r. Based on the maximum principle holds for monogenic functions, we obtain

M(r, f ) = max { f (x)}, x=r which is well defined and strictly monotonic increasing whenever f is not constant. ( , ) M(r, f ) Moreover, M r f is a continuous function about r. When the expression rm remains bounded for r →∞, f is a monogenic polynomial. As the general version of Liouville’s theorem in Clifford analysis, we have the following result: Dirac Operators with Gradient Potentials and Related … Page 5 of 19 53

n M(r, f ) Lemma 2.1 [7,16] Let D[ f ]=0 in R and lim inf m = L < ∞,m∈ N\{0}. r→∞ r Then

m  ( ) = ( ) . f x Vl1,...,l p x Cl1,...,l p (2.4) p=0 (l1,...,l p)

Elementary properties of the Dirac operators and left monogenic functions can be found in References [4,8–10,15,27,32,34]. n In this article, assume that B : R → R is a bounded continuous differentiable 1 n ∂B function, i.e., C -function and b := = e . We now consider the following i 1 i ∂xi inhomogeneous Dirac operator:

Db(x)[ f ](x) = D[ f ](x) + b(x) f (x).

1 Definition 2.2 A C -function f :  → Cl(Vn,n) is said to be (left) monogenic with respect to the potential B if Db(x)[ f ](x) = 0 for all x ∈ . The function f (x) is right monogenic with respect to the potential B if [ f (x)]Db(x) =[f (x)]D + f (x)b(x) = 0. Remark 2.3 The function b(x) appearing in the previous definition is called a gradient potential. For this reason, the inhomogeneous Dirac operator Db(x) is said to be the Dirac operator with gradient potential b(x).

We shall denote the right Cl(Vn,n) module of monogenic functions defined on the domain  by M(, Cl(Vn,n)), while we denote the right Cl(Vn,n) module of mono- genic functions with respect to the potential B defined over  by MB(, Cl(Vn,n)).

Proposition 2.4 The modules M(, Cl(Vn,n)) and MB(, Cl(Vn,n)) are canonically isomorphic.

Proof If f (x) ∈ M(, Cl(Vn,n)), then D[ f ](x) = 0in. It is easy to check that

−B −B(x) Db(x)[ fe ](x) = b(x) f (x)e n ∂ ( ) ∂B( ) f x −B(x) x −B(x) + ei [ e − e f (x)] ∂xi ∂xi i=1 = 0.

Conversely, if u(x) ∈ MB(, Cl(Vn,n)), then we consider

n ∂ ( ) ∂B( ) B u x B(x) x B(x) D[ue ](x) = ei [ e + u(x)e ] ∂xi ∂xi i=1 B(x) = Db(x)[u](x)e = 0.

The proof is finished. 53 Page 6 of 19 L. Gu, D. Ma

In view of Proposition 2.4 and Lemma 2.1, we directly get the following result: n M(r, f ) Theorem 2.5 Let Db(x)[ f ](x) = 0 in R and lim inf m = L < ∞,m∈ N\{0}. r→∞ r Then m  ( ) = −B(x) ( ) . f x e Vl1,...,l p x Cl1,...,l p (2.5) p=0 (l1,...,l p)

3 Some Integral Representation Formulas in Clifford Analysis

Throughout this work, suppose  is a bounded, open set of Rn with a Lyapunov boundary ∂, denote + = , − = Rn\. By Stokes’ Theorem in Clifford analysis in [4,8,9], we can prove the following lemma. 1 Lemma 3.1 Let f , g ∈ C (, Cl(Vn,n)) ∩ C(, Cl(Vn,n)). Then   

fdσ g = [ f ]Db(x)gdV + fD−b(x)[g]dV ∂  

= [ f ]Db(x)gdV + fD−b(x)[g]dV.   where dV denotes the Lebesgue volume measure. Denote − 1 y x B(y)−B(x) B( , ) = , K y x n e (3.1) ωn y − x   = n = n ω where y i=1 yi ei , x i=1 xi ei and n denotes the surface area of the unit sphere in Rn. 1 Theorem 3.2 If f ∈ C (, Cl(Vn,n)) ∩ C(, Cl(Vn,n)), then  

KB(y, x)dσy f (y) − KB(y, x)Db(y)[ f ](y)dV ∂  (3.2) ( ), ∈  = f x x 0, x ∈ Rn\ where KB(y, x) is as in (3.1). n Proof Let x ∈ R \. It is obvious that [KB(y, x)]D−b(y) = 0, y ∈ ∂.By Lemma 3.1, we get   

KB(y, x)Db(y)[ f ](y)dV =− [KB(y, x)]D−b(y)u(y)dV + KB(y, x)dσy f (y).   ∂ Dirac Operators with Gradient Potentials and Related … Page 7 of 19 53

Then the left hand side of the stated formula (3.2) apparently equals to zero. Now take x ∈  and r > 0 such that B(x, r) ⊂ . Invoking the previous case, we then write  

KB(y, x)dσy f (y) = KB(y, x)Db(y)[ f ](y)dV. (3.3) ∂(\B(x,r)) \B(x,r)

Denote 

F(r)  KB(y, x)Db(y)[ f ](y)dV, (3.4) \B(x,r) applying spherical coordinates, 0 < r1 < r2, we find  1 F(r ) − F(r )  D ( )[ f ](y)dV 2 1 y − xn−1 b y ( , )\ ( , ) Bx r2 B x r1 r2  Db(y)[ f ](y)drdω n S r1

 (r2 − r1).

Hence, we obtain that  

lim KB(y, x)Db(y)[ f ](y)dV = KB(y, x)Db(y)[ f ](y)dV. r→0 \B(x,r) 

For the left hand side of (3.3), we can write the following form  

KB(y, x)dσy f (y) − KB(y, x)dσy f (y), (3.5) ∂ ∂ B(x,r) we denote 

(x)  KB(y, x)dσy f (y), ∂ B(x,r) applying the Lebesgue differential theorem, we have

lim (x) = f (x). (3.6) r→0

In view of (3.3)–(3.6), the result follows. The proof is done. Using Theorem 3.2, this implies the following conclusion. 53 Page 8 of 19 L. Gu, D. Ma

1 Theorem 3.3 If f (x) ∈ C (, Cl(Vn,n)) C(, Cl(Vn,n)) and Db(x)[ f ](x) = 0 in , then  f (x), x ∈ , KB(y, x)dσ f (y) = y 0, x ∈ Rn\, ∂ where KB(y, x) is as in (3.1).

Remark 3.4 In case of B ≡ C, where C is a constant. Theorems 3.2–3.3 are the classical Borel–Pompeiu formula and Cauchy integral formula in Clifford analysis respectively.

Corollary 3.5 If f is monogenic with respect to the potential B in  and x ∈ , then  −B(x) ne B(y) ( ) = ( ) n e f y dV f x (3.7) ωnr B(x,r) and  −B(x) e B( ) e y f (y)dS = f (x) n−1 (3.8) ωnr ∂B(x,r) for each r > 0 such that B(x, r) ⊂ .

Proof In view of Theorem 3.3, we can obtain the results.

Combining integral representation formulas with the following lemma, we now study a geometric property for the monogenic functions with respect to the potential B.

n Lemma 3.6 Suppose that x1, x2 and y ∈ R , x1, x2 = y. Then we have

− y − x y − x n 2 x − x   1 − 2 ≤ 1 2 , (3.9) y − x n y − x n y − x n−i−1y − x i+1 1 2 i=0 1 2   = n = n where x i=1 xi ei and y i=1 yi ei .

n Proof For x1, x2, y ∈ R and x1, x2 = y,wehave − −  y x1 − y x2 I n n y − x1 y − x2 ( − )[ − n−2( − ) − ( − ) − n−2]( − ) = y x1 y x2 y x2 y x1 y x1 y x2 , n n y − x1 y − x2 Dirac Operators with Gradient Potentials and Related … Page 9 of 19 53 further obtain that

y − x n−2(y − x ) − (y − x )y − x n−2 I ≤ 2 2 1 1 n−1 n−1 y − x1 y − x2 ( − )( − n−2 − − n−2) + ( − ) − n−2 = y x2 y x2 y x1 x1 x2 y x1 n−1 n−1 y − x1 y − x2 n−2  −  ≤ x1 x2 . y − x n−i−1y − x i+1 i=0 1 2

The proof is finished.

Remark 3.7 In view of some properties of universal Clifford algebra, we prove the above lemma which improves the results Lemma 2.3 in [33] and Proposition 3.63 in [14]. Our proof is much more direct and simpler than the proof of Lemma 4.22 in [11], which is called Hile’s Lemma and plays an important role in studying singular integral operators in Clifford analysis.

n Theorem 3.8 Suppose B(0, 1) is an open unit ball in R ,f: B(0, 1) → Cl(Vn,n) is a monogenic function with respect to the potential B(x), and  f (x)≤1, ∀x ∈ B(0, 1) and f (0) = 0. Then ∀x ∈ B(0, 1) we have

2n−1x, B(x) ≡ C,  f (x)≤ C x, B(x) = C, where C is a real constant function on Rn, C = √2δ , δ = 2n−1δ+1− (2n−1δ+1)2−4δ e2M max{1, m},M= max |B(x)| and m = max b(x). x≤1 x≤1

Proof For B(x) = C, by Theorem 3.3, ∀x ∈ B(0, r),0< r < 1,



f (x) = KB(y, x)dσy f (y). ∂B(0,r)

In view of f (0) = 0, we have



f (x) = f (x) − f (0) = (KB(y, x) − KB(y, 0))dσy f (y). ∂B(0,r) 53 Page 10 of 19 L. Gu, D. Ma

Using Lemma 3.6, mean value theorem in several variables and  f (x)≤1, ∀x ∈ B(0, 1), we get

 n− e2M 2 1  f (x)≤x ( )dS + me2M x 1+i n−i−1 ωn ∂B( , ) y y − x 0 r i=0  2M  −  n−2  − i = e 1 y x ( y x ) + 2M   x n i dS me x ωn ∂B( , ) y − x y y 0 r i=0  2M 2 − 2 n−2 +  ≤ e r x 1 ( (r x )i ) + 2M   x n dS me x ωnr ∂B( , ) y − x r −x r 0 r i=0 − e2M n 2 r +x = ( ( )i )x+me2M x. (3.10) r −x r i=0

When r → 1in(3.10), it follows that

2n−1 − 1  f (x)≤[e2M + me2M ]x 1 −x (3.11) 2n−1 − 1 ≤ δ[ + 1]x. 1 −x

For 0 < x < 1, we have

1  f (x)≤1 = x. (3.12) x

( ) := δ 2n−1−1 + δ ( ) := 1 , Denote f x 1−x and g x x it is easy to check that

2δ sup min{ f (x), g(x)}= . (3.13) − − 0≤x<1 2n 1δ + 1 − (2n 1δ + 1)2 − 4δ

Basedon(3.10)–(3.13), the result follows. For B(x) ≡ C, we use a similar method of proof in the above and prove the result. The proof is finished.

n Theorem 3.9 Let B(0, r1) be an open ball with center origin and radius r1 in R , f : B(0, r1) → Cl(Vn,n) is a monogenic function with respect to the potential B(x) and  f (x)≤r2, ∀x ∈ B(0, r1) and f (0) = 0. Then ∀x ∈ B(0, r1) we have

2n−1 r2 x, B(x) ≡ C,  f (x)≤ r1 C r2 x, B(x) = C, r1 where C is a positive constant mentioned in Theorem 3.8. Dirac Operators with Gradient Potentials and Related … Page 11 of 19 53

Proof Let f (r y) F(y) := 1 r2 for y ∈ B(0, 1). We intend to apply Theorem 3.8 to F(y). In view of properties of f (·), we note that F(y) is a monogenic function with respect to the potential B(x) in B(0, 1), F(0) = 0 and

 f (r y) F(y)= 1 ≤ 1. r2 In light of Theorem 3.8,itfollowsthat

 ( ) n−1 , B( ) ≡ ,  ( )= f r1y ≤ 2 y x C F y  , B( ) = , (3.14) r2 C y x C

Denote x := r1y , then x ∈ B(0, r1) and use (3.14), the result holds.

Remark 3.10 When r1 = r2 = 1, Theorem 3.9 is just Theorem 3.8.

˛ 4 Some Integral Transforms in H (@Ä, Cl(Vn,n))

α The space H (∂, Cl(Vn,n)) consists of all Hölder continuous functions with values in Cl(Vn,n) on ∂ (the Hölder exponent is α,0<α≤ 1), which the norm

 f (α,∂) =f ∞ +f α (4.1)

 f (x1)− f (x2) is finite, where  f ∞ := sup  f (x),  f α := sup α . Obviously, x1−x2 x∈∂ x1,x2∈∂ x =x α 1 2 the space of functions H (∂, Cl(Vn,n)) is a Banach space. Next, we introduce the following integral transforms: 

C∂[ f ]:=2 KB(y, x)dσy f (y), x ∈ , (4.2) ∂

S∂[ f ]:=2 KB(y, x)dσy f (y), x ∈ ∂. (4.3) ∂ where KB(y, x) is as in (3.1). Theorem 4.1 Let  be a bounded, convex domain in Rn with a Lyapunov boundary α α ∂. Then the integral transform S∂ : H (∂, Cl(Vn,n)) → H (∂, Cl(Vn,n)) defined by (4.3) is bounded, i.e.,  S∂[ f ](α,∂) ≤ C f (α,∂), 53 Page 12 of 19 L. Gu, D. Ma where C is a nonnegative constant, depending on the dimension n.

α Proof For f (x) ∈ H (∂, Cl(Vn,n)), using the fact ∂ is of dimension n − 1 and convex domain, the weak singularity of kernel function KB(y, x),wehave

 2 beM 1 S∂[ f ](x)≤ dS f ∞ n−2 ωn y − x ∂ (4.4)

≤ C1 f ∞, where M = max |B(x)| and b = max b(x). x∈ x∈ For x1, x2 ∈ ∂, it is enough to consider the case of x1 − x2 being sufficiently small. We denote x1 − x2=η and take η ∂ Lx := {y ∈ ∂ :y − x ≤ }, 1 1 3 η ∂ Lx := {y ∈ ∂ :y − x ≤ }. 2 2 3

∂ ∂ =∅ ( , ) It is clearly that Lx1 Lx2 . Using the weak singularity of KB y x again, we then have

S∂[ f ](x ) − S∂[ f ](x ) 1 2 

≤2 KB(y, x1)dσy f (y)+2 KB(y, x1)dσy f (y) ∂\∂ L ∂ L x1 x1

+2 KB(y, x2)dσy f (y)+2 KB(y, x2)dσy f (y) ∂\∂ ∂ Lx2 Lx2   2 beM 1 2 beM 1 ≤ dS f (α,∂) + dS f (α,∂) n−2 n−2 ωn y − x1 ωn y − x1 ∂\∂ ∂ Lx1 Lx1  2 beM 1 + dS f (α,∂) n−2 ωn y − x2 ∂\∂ Lx2  2 beM 1 + dS f (α,∂) := I . n−2 ωn y − x2 ∂ Lx2 Dirac Operators with Gradient Potentials and Related … Page 13 of 19 53

In view of ∂ being compact, there exist constants δ1 and δ2 independent of x1, x2 and are comparable to η such that

   η M δ M δ M 2be 1 2be 2 4be 3 I ≤{ dr + dr + dr}Cn f (α,∂) ω η ω η ω n 3 n 3 n 0 1−α α ≤ Cnη  f (α,∂)x1 − x2 α ≤ Cn f (α,∂)x1 − x2 , where Cn is a nonnegative constant, only depending on the dimension n and not necessarily the same one. Thus we have

α S∂[ f ](x1) − S∂[ f ](x2)≤Cn f (α,∂)x1 − x2 , (4.5) it follows from (4.4) and (4.5) that

 S∂[ f ](α,∂) ≤ C f (α,∂),

 where C = max{Cn, C1}. The proof is done.

Theorem 4.2 Suppose that  is an open nonempty bounded subset of Rn with oriented α Lyapunov boundary ∂, and f ∈ H (∂, Cl(Vn,n)), 0 <α≤ 1. Then

f (x0) 1 lim C∂[ f ](x) = + S∂[ f ](x0), x→x0∈∂ 2 2 x∈ f (x0) 1 lim C∂[ f ](x) =− + S∂[ f ](x0). x→x0∈∂ 2 2 x∈Rn\

α 1 n Proof For x1, x2 ∈ ∂, since f ∈ H (∂, Cl(Vn,n)) and B(y) ∈ C (R , R),we have

B(x1) B(x2) α  f (x1)e − f (x2)e ≤Cx1 − x2 , where C is a positive constant independent of x1 and x2. In view of the Plemelj– Sokhotski formula for functions with parameter in Clifford analysis that can be found in [15], the result follows.

For simplicity, we denote

± f (x) = lim f (y), y→x∈∂ y∈± where + =  and − = Rn\. 53 Page 14 of 19 L. Gu, D. Ma

Theorem 4.3 (Painlevé theorem) Suppose that f is a monogenic function with respect to the potential B in Rn\∂, and satisfies the following conditions:

f +(x) = f −(x), x ∈ ∂, + − α f (x) = f (x) ∈ H (∂, Cl(Vn,n)), 0 <α≤ 1.

n Then Db(x)[ f ](x) = 0 in R .

+ − Proof Let f (x) = f (x) = f (x), x ∈ ∂. For any x0 ∈ ∂, there exists a constant r > 0 such that  ⊂ B(x0, r) where B(x0, r) is a ball with the center at x0 and radius r. In view of Theorem 3.3, we get 

f (x1) = KB(y, x1)dσy f (y), x1 ∈ , (4.6) ∂ 

f (x2) = KB(y, x2)dσy f (y), x2 ∈ B(x0, r)\. (4.7)

∂∪∂B(x0,r)

Using Theorem 4.2 to (4.6) and (4.7), we obtain that

+ f (x0) 1 f (x ) = f (x ) = + S∂[ f ](x ), (4.8) 0 0 2 2 0 − f (x0) 1 f (x ) = f (x ) = + S∂∪∂B( , )[ f ](x ). (4.9) 0 0 2 2 x0 r 0

From (4.8) and (4.9), we have 

f (x0) = KB(y, x0)dσy f (y), ∂

[ ]( ) = ∈  therefore Db(x0) f x0 0. Due to arbitrary x0 , the result follows.

5 Riemann–Hilbert Problems for the Dirac Operators with Gradient Potentials B

We consider in this section the Riemann–Hilbert problems: ⎧ n ⎨⎪ Db(x)[ f ]=0, in R \∂, + − f (x) = f (x)A + g(x), x ∈ ∂, (5.1) ⎩⎪ M(r, f ) lim inf m = L < ∞, m ∈ N\{0}, r→∞ r

α where A is an invertible Clifford constant and g(x) ∈ H (∂, Cl(Vn,n)) 0 <α≤ 1. We give the explicit representation formula of the solution for (5.1). Dirac Operators with Gradient Potentials and Related … Page 15 of 19 53

Theorem 5.1 The Riemann–Hilbert problem (5.1) is solvable and the solution is given by ⎧  ⎪ ⎪ KB(y, x)dσ g(y) ⎪ y ⎪ ⎪∂ ⎪ m  ⎪ + −B(x) ( ) , ∈ +, ⎨ e Vl1,...,l p x Cl1,...,l p x p=0 (l ,...,l ) f (x) =  1 p (5.2) ⎪ − ⎪ KB(y, x)dσ g(y)A 1 ⎪ y ⎪ ⎪∂ ⎪ m  ⎪ + −B(x) ( ) −1, ∈ −. ⎩ e Vl1,...,l p x Cl1,...,l p A x p=0 (l1,...,l p)

Proof Let

1, x ∈ +, (x) = A−1, x ∈ −.

Furthermore, we have

+ − 1, x ∈  , 1(x) = A, x ∈ −.

The transmission condition

f +(x) = f −(x)A + g(x), can be changed into

+ − + − − − − + f (x)[ 1] (x) = f (x)[ 1] (x) + g(x)[ 1] (x), (5.3) and if we denote  n u(x) = KB(y, x)dσyg(y), x ∈ R \∂, ∂

n then Db(x)[u](x) = 0, x ∈ R \∂, and u(∞) = 0. According to Plemelj–Sokhotski formula, we get

+ − − + u (x) − u (x) = g(x)[ 1] (x), x ∈ ∂. (5.4)

Combining (5.3) with (5.4), we obtain that

[ f −1 − u]+(x) =[f −1 − u]−(x), x ∈ ∂. 53 Page 16 of 19 L. Gu, D. Ma

In light of Theorem 4.3, we thus deduce that F(x) := f −1 − u is monogenic with respect to the potential B in Rn, and

M(r, F) lim inf < ∞, r→∞ r m In view of Theorem 2.5, the proof is done.

Remark 5.2 According to Proposition 2.4, we can change the Riemann–Hilbert prob- lem (5.1) into a classical Riemann–Hilbert problem for monogenic functions and also obtain the above result.

In the following, let  be a bounded, convex domain of Rn with a Lyapunov boundary ∂, we consider the following Riemann–Hilbert problem which 1 and A are replaced by Clifford value functions in transmission conditions in (5.1). ⎧ n ⎨ Db(x)[ f ]=0, in R \∂, +( ) ( ) = −( ) ( ) + ( ), ∈ ∂, ⎩ f x B x f x A x g x x (5.5)  f (∞)=0,

α where A(x), B(x) and g(x) ∈ H (∂, Cl(Vn,n)) 0 <α<1. We establish solvability conditions of the Riemann–Hilbert problem (5.1). α Theorem 5.3 Suppose A(x),B(x) and g(x) ∈ H (∂, Cl(Vn,n)), 0 <α<1. More- over A(x) and B(x) satisfy the following condition:  A − B(α,∂)(1 + C) +1 − B(α,∂) < 1, (5.6) where C is a positive constant mentioned in Theorem 4.1. Then there exists a unique solution to (5.5).

Proof The solution to this Riemann–Hilbert problem (5.5) can be written in the form 

f (x) = KB(y, x)dσyu(y), (5.7) ∂ where u(y) is a Hölder continuous function to be determined on ∂. By Theorem 4.2, we obtain that

+ u(x) 1 f (x) = + S∂[u](x), (5.8) 2 2 − u(x) 1 f (x) =− + S∂[u](x). (5.9) 2 2 We then combine (5.8), (5.9) and the transmission conditions in (5.5) to reduce the Riemann–Hilbert problem (5.5) to an equivalent singular integral equation,

u(x) ={S∂[u](x) − u(x)}(A(x) − B(x)) + u(x)(1 − B(x)) + 2g(x). Dirac Operators with Gradient Potentials and Related … Page 17 of 19 53

Set

(Mu)(x) ={S∂[u](x) − u(x)}(A(x) − B(x)) (5.10) + u(x)(1 − B(x)) + 2g(x).

α For any u1, u2 ∈ H (∂, Cl(Vn,n)),

 Mu1 − Mu2(α,∂) ≤ (1 + C)A − B(α,∂)u1 − u2(α,∂) (5.11) +1 − B(α,∂)u1 − u2(α,∂).

α From (5.11), the mapping M is a strict contraction from H (∂, Cl(Vn,n)) into itself. Then the mapping M has a unique fixed point. Hence, there exists a unique solution to (5.5). The proof is finished.

6 Some Examples

Example 6.1 Consider the following boundary value problem of Riemann–Hilbert type: ⎧ n ⎪ D  2 [ f ]=0, in R \∂, ⎨⎪ − x −xe 2 +( ) = −( ) + ( ), ∈ ∂, ⎪ f x f x A g x x (6.1) ⎩⎪ M(r, f ) lim inf m = L < ∞, r→∞ r

α where A is an invertible Clifford constant and g(x) ∈ H (∂, Cl(Vn,n)) 0 <α≤ 1.

− x2 − x2 Since B(x) = e 2 ,wehaveb(x) =−xe 2 . According to Theorem 5.1,we claim that the Riemann type boundary value problem (6.1) is solvable and the solutions can be written as ⎧   2  2 ⎪ − y − x ⎪ 1 y−x e 2 −e 2 σ ( ) ⎪ ω  − n e d yg y ⎪ n y x ⎪ ⎪ ∂ ⎪  2 ⎪ − x m  ⎪ + −e 2 ( ) , ∈ +, ⎨ e Vl1,...,l p x Cl1,...,l p x = ( ,..., ) ( ) =  p 0 l1 l p f x  2  2 ⎪ − − y − x ⎪ 1 y x e 2 −e 2 σ ( ) −1 ⎪ ω  − n e d yg y A ⎪ n y x ⎪ ⎪ ∂ ⎪  2 ⎪ − x m  ⎪ + −e 2 ( ) −1, ∈ −. ⎩⎪ e Vl1,...,l p x Cl1,...,l p A x p=0 (l1,...,l p) 53 Page 18 of 19 L. Gu, D. Ma

Example 6.2 More generally, let us consider the following Riemann–Hilbert problem: ⎧ n ⎨ Db(x)[ f ]=0, in R \∂, +( ) = −( ) ( ) + ( ), ∈ ∂, ⎩ f x f x A x g x x (6.2)  f (∞)=0,

α where A(x) and g(x) ∈ H (∂, Cl(Vn,n)) 0 <α<1. Moreover A(x) satisfies the following condition:

 A − 1(α,∂)(1 + C)<1, (6.3) where C is a positive constant mentioned in Theorem 4.1. Using Theorem 5.3,the Riemann–Hilbert problem (6.2) exists a unique solution.

Acknowledgements The authors are very grateful to the reviewers for their suggestions.

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