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H λ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq1_HTML.gif-regular vector functions and their boundary value problems

Boundary Value Problems20122012:75

DOI: 10.1186/1687-2770-2012-75

Received: 22 February 2012

Accepted: 2 July 2012

Published: 19 July 2012

Abstract

Let D = ( λ + t 2 z ¯ 2 z λ t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq2_HTML.gif, where λ is a positive real constant. In this paper, by using the methods from quaternion calculus, we investigate the H λ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq1_HTML.gif-regular vector functions, that is, the complex vector solutions Ψ ( t , z ) = ( ψ 1 ( t , z ) ψ 2 ( t , z ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq3_HTML.gif of the equation D Ψ = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq4_HTML.gif, and work out a systematic theory analogous to quaternionic regular functions. Differing from that, the component functions of quaternionic regular functions are harmonic, the component functions of H λ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq1_HTML.gif-regular functions satisfy the modified Helmholtz equation, that is ( λ 2 Δ ) ψ i = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq5_HTML.gif, i = 1 , 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq6_HTML.gif. We give out a distribution solution of the inhomogeneous equation D u = f https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq7_HTML.gif and study some properties of the solution. Moreover, we discuss some boundary value problems for H λ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq1_HTML.gif-regular functions and solutions of equation D u = f https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq7_HTML.gif.

MSC:30G35, 35J05.

Keywords

quaternion calculus H λ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq1_HTML.gif-regular vector function modified Helmholtz equation Riemann-Hilbert type boundary value problem
It is well known that the theories of holomorphic functions of one complex variable and regular functions of quaternion as well as Clifford calculus are closely connected with the theory of harmonic functions, i.e., their component functions are all harmonic. But side by side with the Laplace operator is the Helmholtz operator and modified Helmholtz operator
λ = λ 2 ± , λ ( R ) 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equa_HTML.gif

which play an important role and are often met in application. In recent years, it has been considered that by replacing the harmonic function with the solutions of Helmholtz equation and modified Helmholtz equation, the theory of regular functions is naturally generalized in quaternion calculus and Clifford calculus. The theory has been well developed and has been applied to the research of some partial differential equations such as Helmholtz equation, Klein-Cordon equation, and Schroding equation. The corresponding results can be found in [13, 511, 1315].

Let H ( R ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq8_HTML.gif and H ( C ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq9_HTML.gif denote the real and complex quaternion space respectively. Their basis elements 1, i, j, k satisfy the following relations: i 2 = j 2 = k 2 = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq10_HTML.gif, i j = j i = k https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq11_HTML.gif, j k = k j = i https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq12_HTML.gif.

In [2, 3], the authors introduced a differential operator of first order D λ = λ + i x 1 + j x 2 + k x 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq13_HTML.gif, where λ is a positive real constant. It is easy to see that
D λ D λ = λ 2 + Δ , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equb_HTML.gif
where λ 2 + Δ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq14_HTML.gif is namely the 3-dimensional Helmholtz operator. A quaternion function theory associated with the operator was established which involved the Pompeiu formula corresponding to D λ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq15_HTML.gif, the Cauchy integral formula for solutions of equation D λ u = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq16_HTML.gif, the Plemelj formula of Cauchy type integral and the theory of operator T λ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq17_HTML.gif. By using these results, the authors investigated the Dirichlet boundary problems for Helmholtz equation
( λ 2 + Δ ) u = f . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equc_HTML.gif

Since the operator λ 2 Δ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq18_HTML.gif can not be factorized into the product of two differential operators of first order in H ( R ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq19_HTML.gif, the quaternion function theory about modified Helmholtz equation was developed in complex quaternion space H ( C ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq9_HTML.gif, namely the operator λ 2 + Δ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq14_HTML.gif, λ C https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq20_HTML.gif and some related equations were directly investigated by H ( C ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq9_HTML.gif. However, different from H ( R ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq19_HTML.gif, H ( C ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq9_HTML.gif is a Euclidean 8-space; and since there exists a set of zero divisors in H ( C ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq21_HTML.gif, a non-zero complex quaternion is not necessarily invertible. There exist many differences between the two theories.

In this article, we shall use the quasi-quaternion space introduced in [18, 19] and transform the modified Helmholtz operator into matric form ( λ 2 ) e 0 = D D https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq22_HTML.gif. By using the quaternion technique, we obtain a systematic theory about the H λ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq23_HTML.gif-regular vector functions, that is, the complex vector solutions Ψ ( t , z ) = ( ψ 1 ( t , z ) ψ 2 ( t , z ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq24_HTML.gif of the equation D Ψ = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq4_HTML.gif, analogous to the quaternion regular function. Because the H λ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq1_HTML.gif-regular vector functions are two-dimensional complex vector functions, this is more similar to the case of H ( R ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq8_HTML.gif.

For applications of partial differential equations, the research of boundary value problems is very important. How should appropriate boundary data be chosen for the Helmholtz equation or modified Helmholtz equation of first order? So far, there have been very few research works on the aspect. In this article, we introduce and investigate some Riemann-Hilbert type boundary value problems for H λ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq1_HTML.gif-regular vector functions and solutions of the equation D u = f https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq7_HTML.gif, obtain general solutions and solvable conditions respectively in different cases.

1 Some notations and definitions

Denote
e 0 = ( 1 0 0 1 ) , e 1 = ( 1 0 0 1 ) , e 2 = ( 0 1 1 0 ) , e 3 = ( 0 i i 0 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equd_HTML.gif
It is easy to see that
e 1 2 = e 2 2 = e 3 2 = e 0 , e 1 e 2 = e 2 e 1 = i e 3 , e 2 e 3 = e 3 e 2 = i e 1 , e 3 e 1 = e 1 e 3 = i e 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Eque_HTML.gif

Henceforth we shall abbreviate e 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq25_HTML.gif to 1.

Introduce the three-dimensional modified Helmholtz operator D = λ + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq26_HTML.gif of first order, where = t e 1 + x e 2 + y e 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq27_HTML.gif, λ is a positive real constant. Define D = λ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq28_HTML.gif, then D D = D D = ( λ 2 ) e 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq29_HTML.gif, where is the three-dimensional Laplace operator. The matrix forms of D, D https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq30_HTML.gif are
D = ( λ + t 2 z ¯ 2 z λ t ) , D = ( λ t 2 z ¯ 2 z λ + t ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equf_HTML.gif
where
z ¯ = 1 2 ( x + i y ) , z = 1 2 ( x i y ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equg_HTML.gif
and then
D D = D D = ( λ 2 Δ 0 0 λ 2 Δ ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equh_HTML.gif
Let Ω be a region in R × C https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq31_HTML.gif which identifies with R 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq32_HTML.gif. Ψ ( t , z ) = ( ψ 1 ( t , z ) ψ 2 ( t , z ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq33_HTML.gif is a complex vector function defined in Ω. If Ψ ( t , z ) C 1 ( Ω ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq34_HTML.gif and satisfies the equation
D Ψ = 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equ1_HTML.gif
(1)

then Ψ ( t , z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq35_HTML.gif will be called H λ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq1_HTML.gif-regular vector function in Ω.

2 Pompeiu formula and Cauchy integral formula of H λ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq23_HTML.gif-regular vector function

Let Ω be a bounded domain in R × C https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq36_HTML.gif with piecewise smooth boundary S. U ( t , z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq37_HTML.gif, V ( t , z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq38_HTML.gif are two-dimensional complex vector functions defined in Ω and U ( t , z ) , V ( t , z ) C 1 ( Ω ) C ( Ω ¯ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq39_HTML.gif. By the divergence theorem
Ω [ ( U ) V + U ( V ) ] d σ = Ω [ x 1 ( U e 1 V ) + x 2 ( U e 2 V ) + x 3 ( U e 3 V ) ] d σ = S U V d S , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equ2_HTML.gif
(2)
where = e 1 cos α 1 + e 2 cos α 2 + e 3 cos α 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq40_HTML.gif, ( cos α 1 , cos α 2 , cos α 3 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq41_HTML.gif denotes the unit outward normal to the surface S. From the equality (2), we have
Ω [ U ( D V ) ( U D ) V ] d σ = S U V d S . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equ3_HTML.gif
(3)
It is easy to show that 1 4 π r e λ r https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq42_HTML.gif, r = ( t 2 + x 2 + y 2 ) 1 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq43_HTML.gif, is a fundamental solution of the modified Helmholtz operator λ 2 Δ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq18_HTML.gif. When r 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq44_HTML.gif, ( λ 2 Δ ) ( 1 4 π r e λ r ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq45_HTML.gif. We write
E ( t , z ) = D ( 1 4 π r e λ r ) = 1 4 π [ λ r ( λ r 2 + 1 r 3 ) ( t e 1 + x e 2 + y e 3 ) ] e λ r . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equi_HTML.gif
Suppose u ( t , z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq46_HTML.gif is a complex vector function defined in Ω and u ( t , z ) C 1 ( Ω ) C ( Ω ¯ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq47_HTML.gif. Let p 0 = ( t 0 , z 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq48_HTML.gif be a fixed point in Ω and B ε ( p 0 ) = { | p p 0 | < ε } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq49_HTML.gif be an open ball whose center is p 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq50_HTML.gif, and the radius ε is so small that B ε ( p 0 ) ¯ Ω https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq51_HTML.gif. Write Ω ε = Ω B ε ( p 0 ) ¯ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq52_HTML.gif. Using the formula (3) in Ω ε https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq53_HTML.gif and replacing U, V by E ( p p 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq54_HTML.gif, u ( p ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq55_HTML.gif respectively, we have
Ω ε E ( p p 0 ) D u d σ = S E ( p p 0 ) u ( p ) d S B ε ( p 0 ) E ( p p 0 ) u ( p ) d S . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equ4_HTML.gif
(4)
Where
B ε ( p 0 ) E ( p p 0 ) u ( p ) d S = I 1 + I 2 + I 3 , I 1 = λ e λ ε 4 π ε 2 B ε ( p 0 ) [ ( t t 0 ) e 1 + ( x x 0 ) e 2 + ( y y 0 ) e 3 ] u ( p ) d S , I 2 = λ e λ ε 4 π ε B ε ( p 0 ) u ( p ) d S , I 3 = e λ ε 4 π ε 2 B ε ( p 0 ) u ( p ) d S . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equj_HTML.gif
It is easy to show that
lim ε 0 I 1 = 0 , lim ε 0 I 2 = 0 , lim ε 0 I 3 = u ( p 0 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equk_HTML.gif

Then letting ε tend to zero in (4), we obtain the following Pompeiu formula corresponding to the operator D.

Theorem 1 Let Ω be a bounded domain in R × C https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq31_HTML.gif with piecewise smooth boundary S. If u ( t , z ) = u ( p ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq56_HTML.gif is a complex vector function defined in Ω and u ( t , z ) C 1 ( Ω ) C ( Ω ¯ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq47_HTML.gif, then
u ( t , z ) = S E ( p p ) u ( p ) d S p + Ω E ( p p ) D p u d σ p , ( t , z ) Ω . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equ5_HTML.gif
(5)

By applying Theorem 1, we can deduce the following Cauchy integral formula of the H λ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq1_HTML.gif-regular vector function.

Theorem 2 If a complex vector function Ψ ( t , z ) C 1 ( Ω ) C ( Ω ¯ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq57_HTML.gif and satisfies the equation D Ψ = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq58_HTML.gif in Ω, then
Ψ ( t , z ) = S E ( p p ) u ( p ) d S p , ( t , z ) Ω , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equ6_HTML.gif
(6)
and if ( t , z ) ¯ Ω https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq59_HTML.gif, then
S E ( p p ) u ( p ) d S p = 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equ7_HTML.gif
(7)

Proof The formula (6) follows directly from the Pompeiu formula (5) and the equality (7) can easily be derived from (3). □

3 Cauchy type integral and Plemelj formula

Let φ ( t , z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq60_HTML.gif be a complex vector function defined on a closed smooth surface S in R × C https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq31_HTML.gif, φ ( t , z ) C α ( S ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq61_HTML.gif, 0 < α < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq62_HTML.gif. Denote
Ψ ( t , z ) = S E ( p p ) φ ( p ) d S p , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equ8_HTML.gif
(8)

and call Φ ( t , z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq63_HTML.gif the Cauchy type integral with respect to the operator D. In the following, we shall simply call it the Cauchy type integral. In addition, φ ( t , z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq60_HTML.gif is called the density function of Ψ ( t , z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq35_HTML.gif.

For arbitrary p = ( t , z ) ¯ S https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq64_HTML.gif, there exists a neighborhood B ρ ( p ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq65_HTML.gif of p which does not intersect with S. In B ρ ( p ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq65_HTML.gif,
D Ψ ( t , z ) = ( λ + p ) S ( λ + p ) ( 1 4 π | p p | e λ | p p | ) φ ( p ) d S p = ( λ + p ) S ( λ p ) ( 1 4 π | p p | e λ | p p | ) φ ( p ) d S p = S ( λ 2 p ) ( 1 4 π | p p | e λ | p p | ) φ ( p ) d S p = 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equl_HTML.gif

Consequently, Ψ ( t , z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq35_HTML.gif is H λ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq1_HTML.gif-regular in the exterior of S. In addition, it is easy to see that Ψ ( t , z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq35_HTML.gif converges to 0 as ( t , z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq66_HTML.gif.

When ( t , z ) S https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq67_HTML.gif, we provide that the integral on the right-hand side of (8) represents Cauchy’s principal value.

Lemma 1 Let Ω be a bounded domain in R × C https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq31_HTML.gif with smooth boundary S. If p S https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq68_HTML.gif, in the sense of Cauchy’s principal value, then
S E ( p p ) d S p = 1 2 e 0 + λ Ω E ( p p ) d σ p . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equ9_HTML.gif
(9)
Proof Let B ε ( p ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq69_HTML.gif be an open ball with the radius ε and the center p, write the component of B ε ( p ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq70_HTML.gif lying in the exterior of Ω as Γ. Then x is an interior point of the region inclosed by the closed surface S = ( S ( S B ε ( p ) ) ) Γ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq71_HTML.gif. By the Pompeiu formula (5), we have
e 0 = ( S ( S B ε ( p ) ) + Γ ) E ( p p ) d S p + λ Ω B ε ( p ) E ( p p ) d σ p . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equ10_HTML.gif
(10)
Similarly to the proof of Theorem 1, we can derive
lim ε 0 Γ E ( p p ) d S p = 1 2 e 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equm_HTML.gif

Letting ε 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq72_HTML.gif in (10), it follows that (9) holds. □

By using Lemma 1, we can obtain the following Plemelj formula of the Cauchy type integral (8).

Theorem 3 Write the domain Ω as Ω + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq73_HTML.gif and the complementary domain of Ω ¯ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq74_HTML.gif as Ω https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq75_HTML.gif. When p tends to p 0 ( S ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq76_HTML.gif from Ω + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq73_HTML.gif and Ω https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq75_HTML.gif respectively, the limits of the Cauchy type integral (8) exist, which will be written as Ψ + ( p 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq77_HTML.gif and Ψ ( p 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq78_HTML.gif respectively, and
{ Ψ + ( p 0 ) = S E ( p p 0 ) φ ( p ) d S p + 1 2 φ ( p 0 ) , Ψ ( p 0 ) = S E ( p p 0 ) φ ( p ) d S p 1 2 φ ( p 0 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equ11_HTML.gif
(11)
The above formula can be rewritten as
{ Ψ + ( p 0 ) Ψ ( p 0 ) = φ ( p 0 ) , Ψ + ( p 0 ) + Ψ ( p 0 ) = 2 S E ( p p 0 ) φ ( p ) d S p . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equ12_HTML.gif
(12)
Proof Since φ ( p ) C α ( S ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq79_HTML.gif, 0 < α < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq62_HTML.gif, therefore the improper integral S E ( p p 0 ) ( φ ( p ) φ ( p 0 ) ) d S p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq80_HTML.gif is convergent. By Lemma 1, we have
S E ( p p 0 ) φ ( p ) d S p = S E ( p p 0 ) ( φ ( p ) φ ( p 0 ) ) d S p 1 2 φ ( p 0 ) + λ Ω E ( p p 0 ) d σ y φ ( x 0 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equn_HTML.gif
The Cauchy type integral (8) can be written in the following form:
S E ( p p ) φ ( p ) d S p = S E ( p p ) ( φ ( p ) φ ( p 0 ) ) d S p S E ( p p ) d S p φ ( p 0 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equ13_HTML.gif
(13)
By the Pompeiu formula, we obtain
S E ( p p ) d S p = { e 0 + λ Ω E ( p p ) d σ p , p Ω + , λ Ω E ( p p ) d σ p , p Ω . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equo_HTML.gif
When p ( ¯ S ) p 0 ( S ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq81_HTML.gif, using the method similar to one complex variable [12, 13], we can show that
lim p p 0 S E ( p p ) ( φ ( p ) φ ( p 0 ) ) d S p = S E ( p p 0 ) ( φ ( p ) φ ( p 0 ) ) d S p . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equp_HTML.gif
Moreover, by using the Hölder inequality, it is easy to show that
lim p p 0 Ω E ( p p ) d σ p = Ω E ( p p 0 ) d σ p . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equq_HTML.gif
Thus letting p tend to p 0 ( S ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq76_HTML.gif from Ω + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq73_HTML.gif and Ω https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq75_HTML.gif respectively in (13), we get
Ψ + ( p 0 ) = S E ( p p 0 ) ( φ ( p ) φ ( p 0 ) ) d S p + φ ( p 0 ) λ Ω E ( p p 0 ) d σ p φ ( p 0 ) Ψ + ( p 0 ) = S E ( p p 0 ) φ ( p ) d S p + 1 2 φ ( p 0 ) , Ψ ( p 0 ) = S E ( p p 0 ) ( φ ( p ) φ ( p 0 ) ) d S p λ Ω E ( p p 0 ) d σ p φ ( p 0 ) Ψ ( p 0 ) = S E ( p p 0 ) φ ( p ) d S p 1 2 φ ( p 0 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equr_HTML.gif

This is (11), and (12) is easily deduced from (11). □

The following result follows directly from Theorem 3.

Corollary 1 Let Ω be a bounded domain in R × C https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq31_HTML.gif whose boundary is a closed smooth surface S. φ ( t , z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq60_HTML.gif is a complex vector function defined on the surface S, and φ ( t , z ) C α ( S ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq61_HTML.gif, 0 < α < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq62_HTML.gif. Then the Cauchy type integral (8) whose density function is φ ( t , z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq60_HTML.gif is a Cauchy integral if and only if ( t 0 , z 0 ) S https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq82_HTML.gif,
Ψ ( t 0 , z 0 ) = 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equs_HTML.gif

4 Operator T Ω f https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq83_HTML.gif

Let f ( t , z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq84_HTML.gif be a complex vector function defined in a bounded domain Ω of R × C https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq31_HTML.gif. Denote
T Ω f = Ω E ( p p ) f ( p ) d σ p , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equ14_HTML.gif
(14)
where
E ( p p ) = 1 4 π [ λ | p p | ( λ | p p | 2 + 1 | p p | 3 ) ( ( t t ) e 1 + ( x x ) e 2 + ( y y ) e 3 ) ] e λ | p p | . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equt_HTML.gif
In this section, we shall get that if f ( t , z ) L 1 ( Ω ¯ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq85_HTML.gif, then T Ω f https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq83_HTML.gif is a distribution solution of the inhomogeneous equation
D u = f , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equ15_HTML.gif
(15)

and shall discuss some properties of the operator T Ω f https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq83_HTML.gif.

Similarly to the quaternion calculus [3, 17], we can obtain the following results.

Theorem 4 If f ( t , z ) L 1 ( Ω ¯ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq85_HTML.gif, then T Ω f https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq83_HTML.gif exists for all ( t , z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq86_HTML.gif in the exterior of Ω. Beside T Ω f https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq83_HTML.gif is H λ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq1_HTML.gif-regular in the exterior of Ω and
T Ω f = 0 , ( t , z ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equu_HTML.gif

Theorem 5 Let f ( t , z ) L 1 ( Ω ¯ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq85_HTML.gif, then T Ω f https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq83_HTML.gif exists almost everywhere on R × C https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq36_HTML.gif and belongs to L p ( Ω ¯ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq87_HTML.gif, 1 p < 3 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq88_HTML.gif, where Ω https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq89_HTML.gif denotes any bounded domain in R × C https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq31_HTML.gif.

For complex vector functions f ( t , z ) = ( f 1 ( t , z ) f 2 ( t , z ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq90_HTML.gif, φ ( x ) = ( φ 1 ( t , z ) φ 2 ( t , z ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq91_HTML.gif given on Ω, define
( f , φ ) = Ω ( f 1 ¯ φ 1 + f 2 ¯ φ 2 ) d σ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equv_HTML.gif

When f ( t , z ) L 1 ( Ω ¯ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq85_HTML.gif, φ ( t , z ) C 0 ( Ω ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq92_HTML.gif, it is easy to show that f ( φ ) = ( f , φ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq93_HTML.gif is a distribution on C 0 ( Ω ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq94_HTML.gif.

Theorem 6 Let f ( t , z ) L 1 ( Ω ¯ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq85_HTML.gif. Then for any φ ( t , z ) = ( φ 1 ( t , z ) φ 2 ( t , z ) ) C 0 ( Ω ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq95_HTML.gif,
( T Ω f , D φ ) = ( f , φ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equw_HTML.gif

holds.

Proof From the equality (2), we get
Ω [ U ( D V ) ( U D ) V ] d σ = S U V d S . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equx_HTML.gif
In the above equality replacing U, V by E ( p p ) = D ( 1 4 π | p p | e λ | p p | ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq96_HTML.gif, u ( p ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq55_HTML.gif respectively, by using the method analogous to the proof of Pompeiu formula (5), we can derive the Pompeiu formula corresponding to the operator D https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq30_HTML.gif, i.e., if u ( t , z ) C 1 ( Ω ) C ( Ω ¯ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq47_HTML.gif, then
u ( t , z ) = S E ( p p ) u ( p ) d S p + Ω E ( p p ) D p u d σ p , ( t , z ) Ω . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equ16_HTML.gif
(16)
Thus for any φ ( t , z ) C 0 ( Ω ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq97_HTML.gif,
φ ( t , z ) = T Ω D φ = Ω E ( p p ) D p φ d σ p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equy_HTML.gif

holds.

In accordance with Theorem 5, T Ω f L 1 ( Ω ¯ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq98_HTML.gif. Thereby by the Fubini theorem,
( T Ω f , D φ ) = ( f , T Ω ( D φ ) ) = ( f , φ ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equz_HTML.gif

the desired result follows. □

Let complex vector functions f , g L 1 ( Ω ¯ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq99_HTML.gif. If for any φ ( t , z ) C 0 ( Ω ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq97_HTML.gif,
( g , D φ ) = ( f , φ ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equaa_HTML.gif

then f is called a generalized derivative corresponding to the operator D of g. The derivative is denoted by f = ( g ) D https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq100_HTML.gif. From Theorem 6 and the definition, ( T Ω f ) D = f https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq101_HTML.gif.

Theorem 7 If a complex vector function g C 1 ( Ω ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq102_HTML.gif and satisfies the equation D g = f https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq103_HTML.gif, then
( g ) D = f . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equab_HTML.gif

This shows that if the complex vector function g is a classical solution of the equation (15), then it is also a distributional solution of the equation.

Proof It follows by the definition and the divergence theorem. □

Let a = ( a 1 , a 2 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq104_HTML.gif be a complex vector. The model of a is defined
| a | = ( | a 1 | 2 + | a 2 | 2 ) 1 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equac_HTML.gif
It is easy to show that
| a ( t e 1 + x e 2 + y e 3 ) | = | ( a 1 , a 2 ) ( t z z ¯ t ) | = | ( a 1 t + a 2 z ¯ , a 1 z a 2 t ) | = [ ( | a 1 | 2 + | a 2 | 2 ) ( t 2 + | z | 2 ) ] 1 2 = | a | | t e 1 + x e 2 + y e 3 | . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equad_HTML.gif

By using similar methods to those used when proving the Hölder continuous of the operator T in quaternion calculus [16, 17], we can prove the following theorem.

Theorem 8 Let Ω be a bounded domain in R × C https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq31_HTML.gif, the complex vector function f ( t , z ) L p ( Ω ¯ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq105_HTML.gif, p > 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq106_HTML.gif.
  1. (a)
    For any ζ R × C https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq107_HTML.gif,
    | T Ω f ( ζ ) | M ( p , Ω ) f L p , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equ17_HTML.gif
    (17)
     
where M ( p , Ω ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq108_HTML.gif is a positive real constant depending only on p, Ω.
  1. (b)
    g ( ζ ) = T Ω f ( ζ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq109_HTML.gif satisfies
    | g ( ζ 1 ) g ( ζ 2 ) | M ( p ) f L p | ζ 1 ζ 2 | α , α = p 3 p , ζ 1 , ζ 2 R × C , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equ18_HTML.gif
    (18)
     

where M ( p ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq110_HTML.gif is a positive real constant depending only on p.

The inequalities (17) and (18) imply that T Ω f https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq83_HTML.gif is a compact mapping from L p ( Ω ¯ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq111_HTML.gif, p > 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq106_HTML.gif into C α ( Ω ¯ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq112_HTML.gif, α = p 3 p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq113_HTML.gif, and
T Ω f C α ( Ω ¯ ) M f L p , p > 3 , α = p 3 p . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equ19_HTML.gif
(19)
Proof (a) From the definition (14) of T Ω f https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq83_HTML.gif,
| T Ω f | λ 2 π Ω e λ | ζ ζ | | ζ ζ | | f ( ζ ) | d σ ζ + 1 4 π Ω e λ | ζ ζ | | ζ ζ | 2 | f ( ζ ) | d σ ζ = I 1 + I 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equae_HTML.gif
Since
e λ | ζ ζ | 1 λ | ζ ζ | , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equaf_HTML.gif
we have by Hölder’s inequality
I 1 1 2 π ( Ω 1 | ζ ζ | 2 q d σ ζ ) 1 q f L p , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equag_HTML.gif

where 1 p + 1 q = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq114_HTML.gif. By hypothesis p > 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq106_HTML.gif, we have q < 3 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq115_HTML.gif. Let d = sup ζ , ζ Ω ¯ | ζ ζ | https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq116_HTML.gif, d 1 = dist ( ζ , Ω ¯ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq117_HTML.gif, namely d, d 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq118_HTML.gif denote the diameter of a bounded domain Ω and the distance between ζ and Ω ¯ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq74_HTML.gif respectively.

If ζ Ω ¯ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq119_HTML.gif,
( Ω 1 | ζ ζ | 2 q d σ ζ ) 1 q ( | ζ ζ | d 1 | ζ ζ | 2 q d σ ζ ) 1 q = ( 4 π d 3 2 q 3 2 q ) 1 q . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equah_HTML.gif
If ζ ¯ Ω ¯ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq120_HTML.gif,
( Ω 1 | ζ ζ | 2 q d σ ζ ) 1 q ( d 1 | ζ ζ | d 1 + d 1 | ζ ζ | 2 q d σ ζ ) 1 q = ( 4 π d 3 2 q 3 2 q ) 1 q [ ( 1 + d 1 d ) 3 2 q ( d 1 d ) 3 2 q ] 1 q ( 4 π d 3 2 q 3 2 q ) 1 q . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equai_HTML.gif
The last inequality is immediate from
0 < ( 1 + d 1 d ) 3 2 q ( d 1 d ) 3 2 q 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equaj_HTML.gif

In fact, from 0 < β = 3 2 q < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq121_HTML.gif, it is easy to see that the real function μ ( x ) = ( 1 + x ) β x β https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq122_HTML.gif is a monotone decreasing function in [ 0 , + ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq123_HTML.gif and μ ( 0 ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq124_HTML.gif, so that 0 < μ ( x ) 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq125_HTML.gif.

Let
M 1 ( p , Ω ) = 1 2 π ( 4 π d 3 2 q 3 2 q ) 1 q . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equak_HTML.gif
Hence we obtain
I 1 M 1 ( p , Ω ) f L p . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equ20_HTML.gif
(20)
Noting
e λ | ζ ζ | 1 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equal_HTML.gif
thus
I 2 = 1 4 π Ω e λ | ζ ζ | | ζ ζ | 2 | f ( ζ ) | d σ ζ 1 4 π ( Ω 1 | ζ ζ | 2 q d σ ζ ) 1 q f L p 1 4 π ( 4 π d 3 2 q 3 2 q ) 1 q f L p , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equam_HTML.gif
i.e.,
I 2 1 2 M 1 ( p , Ω ) f L p . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equ21_HTML.gif
(21)
The inequality (17) follows immediately from (20) and (21).
  1. (b)
    Without loss of generality, we may take | ζ 1 ζ 2 | < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq126_HTML.gif. We write
    g ( ζ ) = λ 4 π Ω e λ | ζ ζ | | ζ ζ | f ( ζ ) d σ ζ λ 4 π Ω ( ζ ζ ) e λ | ζ ζ | | ζ ζ | 2 f ( ζ ) d σ ζ 1 4 π Ω ( ζ ζ ) e λ | ζ ζ | | ζ ζ | 3 f ( ζ ) d σ ζ = g 1 ( ζ ) g 2 ( ζ ) g 3 ( ζ ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equan_HTML.gif
     
where
ζ ζ = ( t t ) e 1 + ( x x ) e 2 + ( y y ) e 3 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equao_HTML.gif

It is easy to see that ( ζ ζ ) 2 = | ζ ζ | 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq127_HTML.gif.

We have
| g 1 ( ζ 1 ) g 1 ( ζ 2 ) | = λ 4 π | Ω ( | ζ ζ 2 | | ζ ζ 1 | ) e λ | ζ ζ 1 | + | ζ ζ 1 | ( e λ | ζ ζ 1 | e λ | ζ ζ 2 | ) | ζ ζ 1 | | ζ ζ 2 | f ( ζ ) d σ ζ | λ 4 π Ω | ζ 1 ζ 2 | ( 1 + λ | ζ ζ 1 | ) | ζ ζ 1 | | ζ ζ 2 | | f ( ζ ) | d σ ζ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equap_HTML.gif
Here we use the estimates
e λ | ζ ζ 1 | 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equaq_HTML.gif
and
| e λ | ζ ζ 1 | e λ | ζ ζ 2 | | λ | | ζ ζ 2 | | ζ ζ 1 | | λ | ζ 1 ζ 2 | . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equar_HTML.gif
We get by Hölder’s inequality
Ω 1 | ζ ζ 1 | | ζ ζ 2 | | f ( ζ ) | d σ ζ ( Ω 1 | ζ ζ 1 | q | ζ ζ 2 | q d σ ζ ) 1 q f L p , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equas_HTML.gif
and
Ω 1 | ζ ζ 2 | | f ( ζ ) | d σ ζ ( Ω 1 | ζ ζ 2 | q d σ ζ ) 1 q f L p . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equat_HTML.gif
Using the inequality
Ω 1 | ζ ζ 1 | α | ζ ζ 2 | β d σ ζ { c 1 | ζ 1 ζ 2 | 3 α β , 0 < α , β < 3 , α + β > 3 , c 2 , 0 α , β < 3 , α + β < 3 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equau_HTML.gif
where c 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq128_HTML.gif, c 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq129_HTML.gif are positive real constants, and noting q < 3 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq115_HTML.gif, we then obtain
| g 1 ( ζ 1 ) g 1 ( ζ 2 ) | M 1 ( p ) f L p | ζ 1 ζ 2 | . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equ22_HTML.gif
(22)
By simple computation we have
| g 2 ( ζ 1 ) g 2 ( ζ 2 ) | = λ 4 π | Ω ( ζ 1 ζ 2 ) e λ | ζ ζ 1 | + ( ζ ζ 1 ) ( e λ | ζ ζ 1 | e λ | ζ ζ 2 | ) ( ζ ζ 1 ) ( ζ ζ 2 ) f ( ζ ) d σ ζ | , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equav_HTML.gif
and
| g 3 ( ζ 1 ) g 3 ( ζ 2 ) | = 1 4 π | Ω { [ | ζ ζ 2 | ( ζ 1 ζ 2 ) + ( ζ ζ 1 ) ( | ζ ζ 2 | | ζ ζ 1 | ) ] e λ | ζ ζ 1 | | ζ ζ 1 | | ζ ζ 2 | ( ζ ζ 1 ) ( ζ ζ 2 ) + ( ζ ζ 1 ) ( e λ | ζ ζ 1 | e λ | ζ ζ 2 | ) | ζ ζ 2 | ( ζ ζ 1 ) ( ζ ζ 2 ) } f ( ζ ) d σ ζ | . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equaw_HTML.gif
By using a similar method, we can obtain
| g 2 ( ζ 1 ) g 2 ( ζ 2 ) | M 2 ( p ) f L p | ζ 1 ζ 2 | https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equ23_HTML.gif
(23)
and
| g 3 ( ζ 1 ) g 3 ( ζ 2 ) | M 3 ( p ) f L p | ζ 1 ζ 2 | α . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equ24_HTML.gif
(24)

The required estimate then follows by combining the resulting inequalities. □

5 Some boundary value problems for H λ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq1_HTML.gif-regular functions

It is well known that the Dirichlet problem for analytic functions w ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq130_HTML.gif in a bounded domain of the complex plane, boundary value of which is a given complex value function, is overdetermined, thereby being unsolvable in general. In the theory of boundary value problems for analytic functions, the boundary condition is replaced by Re w = r ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq131_HTML.gif, and a more general problem is the so-called Riemann-Hilbert problem with boundary condition Re λ ( t ) ¯ w = r ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq132_HTML.gif. Analogously to this, the Dirichlet problem for H λ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq1_HTML.gif-regular functions, boundary value of which is a given complex value vector function, is also overdetermined, and we have therefore to consider new boundary conditions. In this section, we introduce and discuss some Riemann-Hilbert type boundary value problems for H λ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq1_HTML.gif-regular vector functions.

Let Ω be a bounded domain with smooth boundary S in R × C https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq31_HTML.gif, 0 Ω https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq133_HTML.gif. S satisfies the exterior sphere condition, that is, for every point ζ S https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq134_HTML.gif, there exists a ball B satisfying B ¯ Ω ¯ = ζ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq135_HTML.gif. Ω 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq136_HTML.gif denotes the transversal domain of Ω on the plane t = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq137_HTML.gif, its boundary L = Ω 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq138_HTML.gif is a closed smooth curve and the projection of every point of Ω on the plane t = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq137_HTML.gif is in Ω 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq136_HTML.gif. We consider the following boundary value problems:

Find a continuous solution u ( t , z ) = ( u 1 ( t , z ) u 2 ( t , z ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq139_HTML.gif of the equation
D u = f https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equ25_HTML.gif
(25)
in Ω ¯ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq74_HTML.gif, satisfying the boundary conditions
u 1 | S = φ , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equ26_HTML.gif
(26)
Re λ ( τ ) ¯ u 2 ( 0 , τ ) = r ( τ ) , τ L , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equ27_HTML.gif
(27)

where φ is a given complex value function on S, φ C ( S ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq140_HTML.gif, λ ( τ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq141_HTML.gif is a given complex value function on L, λ ( τ ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq142_HTML.gif, τ L https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq143_HTML.gif. r ( τ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq144_HTML.gif is a given real value function on L, λ ( τ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq141_HTML.gif, r ( τ ) C α https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq145_HTML.gif, 0 < α < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq62_HTML.gif. This problem is called problem H of the equation (25), and κ = 1 2 π L arg λ ( τ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq146_HTML.gif is called index of the problem H.

When κ = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq147_HTML.gif, if u satisfies the condition
Im u 2 ( 0 , 0 ) = a https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equ28_HTML.gif
(28)

besides the above boundary conditions, where a is a real constant, then the problem is called problem D.

In particular, when f = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq148_HTML.gif in the equation (25), the above problems are namely the problem H and problem D for the H λ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq1_HTML.gif-regular vector functions.

Lemma 2 Suppose complex value functions g 1 ( t , z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq149_HTML.gif, g 2 ( t , z ) C ( Ω ¯ ) C 1 ( Ω ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq150_HTML.gif. If g 1 ( t , z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq149_HTML.gif, g 2 ( t , z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq151_HTML.gif satisfy compatible condition
( λ t ) g 1 = z ¯ g 2 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equ29_HTML.gif
(29)
then the following overdetermined system with respect to u ( t , z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq46_HTML.gif
{ z ¯ u = g 1 ( t , z ) , ( λ t ) u = g 2 ( t , z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equ30_HTML.gif
(30)
has the general solution
u ( t , z ) = T Ω 0 g 1 + Φ ( z ) e λ t F ( t , z ) e λ t , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equ31_HTML.gif
(31)
here Φ ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq152_HTML.gif is any analytic function in Ω 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq136_HTML.gif,
T Ω 0 g 1 = 1 π Ω 0 g 1 ( t , ζ ) ζ z d σ ζ , F ( t , z ) = 0 t e λ ξ ( 1 2 π i Γ g 2 ( ξ , ζ ) ζ z d ζ ) d ξ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equax_HTML.gif

Proof Noting the compatible condition and that F ( t , z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq153_HTML.gif is an analytic function with respect to z, using the Pompeiu formula [12], it is not difficult to verify by direct calculation that u ( t , z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq46_HTML.gif expressed by (31) is the general solution of the system (30). □

As a special case of Theorem 6.13 in [4], we can derive the following result.

Lemma 3 If φ is continuous on S, then the Dirichlet problem with the boundary condition
w | S = φ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equay_HTML.gif

for the equation ( λ 2 ) w = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq154_HTML.gif in Ω has a unique solution w ( t , z ) C ( Ω ¯ ) C α 2 ( Ω ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq155_HTML.gif.

Similarly to harmonic function, we have the following result.

Lemma 4 For the Dirichlet problem of the equation ( λ 2 ) w = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq154_HTML.gif in Ω, Green functions G ( p , p ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq156_HTML.gif exist such that the solutions of the problem can be represented by G ( p , p ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq156_HTML.gif, namely we have
w ( p ) = S w ( p ) G ν d S p . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equ32_HTML.gif
(32)

These Green functions G ( p , p ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq156_HTML.gif are unique.

Proof Suppose functions w , v C 1 ( Ω ¯ ) C 2 ( Ω ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq157_HTML.gif. By Green’s second identity
Ω ( w v v w ) d σ = S ( w v ν v w ν ) d S , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equaz_HTML.gif
where ν denotes the unit outward normal to the surface S, we obtain
Ω { v [ ( λ 2 ) w w ( λ 2 ) v ] } d σ = S ( w v ν v w ν ) d S . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equ33_HTML.gif
(33)
Let p be a fixed point in Ω and B ε ( p ) = { | p p | < ε } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq158_HTML.gif be an open ball whose radius ε is so small that B ε ( p ) ¯ Ω https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq159_HTML.gif. Write Ω ε = Ω B ε ( p ) ¯ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq160_HTML.gif. Replacing v by 1 4 π e λ | p p | | p p | https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq161_HTML.gif, using the formula (33) in Ω ε https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq53_HTML.gif and letting ε tend to zero, similarly to the proof of Theorem 1, we can derive
w ( p ) = 1 4 π S [ w ν ( e λ | p p | | p p | ) ( e λ | p p | | p p | ) w ν ] d S + 1 4 π Ω e λ | p p | | p p | ( λ 2 ) w d σ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equba_HTML.gif
Thus when w C 1 ( Ω ¯ ) C 2 ( Ω ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq162_HTML.gif and satisfies the equation ( λ 2 ) w = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq154_HTML.gif,
w ( p ) = 1 4 π S [ w ν ( e λ | p p | | p p | ) ( e λ | p p | | p p | ) w ν ] d S . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equ34_HTML.gif
(34)
For a given p in Ω, find g ( p , p ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq163_HTML.gif which satisfies the equation ( λ 2 ) g = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq164_HTML.gif in Ω and the boundary condition g ( p , p ) = 1 4 π e λ | p p | | p p | https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq165_HTML.gif on S. By virtue of Lemma 3, this g ( p , p ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq163_HTML.gif is existential and unique. Write G ( p , p ) = 1 4 π e λ | p p | | p p | g ( p , p ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq166_HTML.gif. When w satisfies the equation ( λ 2 ) w = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq154_HTML.gif in Ω, from (33) we derive
S ( w g ν g w ν ) d S = 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equbb_HTML.gif
Subtracting this from (34), we get
w ( p ) = S w ( p ) G ν d S p . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equbc_HTML.gif

A simple approximation argument shows that this formula continues to hold for w C ( Ω ¯ ) C 2 ( Ω ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq167_HTML.gif. □

With the aid of the methods of conformal mapping and standardizing boundary condition from complex analysis (see [12, 13]), we can map conformally Ω 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq136_HTML.gif into the unit disk on the plane t = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq137_HTML.gif, and transform λ ( τ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq141_HTML.gif in the condition (33) into τ κ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq168_HTML.gif. Hence without loss of generality, we shall directly suppose that Ω 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq136_HTML.gif is the unit disk B 1 ( 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq169_HTML.gif on the plane t = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq137_HTML.gif and replace (27) by the following condition
Re τ ¯ κ u 2 ( 0 , τ ) = r ( τ ) , τ L = B 1 ( 0 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equ35_HTML.gif
(35)

Using these results, we can discuss the solvability of the problem H and the problem D for the H λ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq1_HTML.gif-regular vector functions and the equation D u = f https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq170_HTML.gif.

Theorem 9 (1) If the index κ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq171_HTML.gif, the problem H for the H λ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq1_HTML.gif-regular vector functions in Ω is solvable. The problem has the general solutions Ψ ( t , z ) = ( ψ 1 ( t , z ) ψ 2 ( t , z ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq33_HTML.gif, with
ψ 1 ( t , z ) = S φ ( p ) G ν d S p , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equ36_HTML.gif
(36)
ψ 2 ( t , z ) = T B 1 ( 0 ) g 1 + Φ ( z ) e λ t F ( t , z ) e λ t , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equ37_HTML.gif
(37)
where
g 1 = 1 2 ( λ + t ) ψ 1 , g 2 ( t , z ) = 2 z ψ 1 , Φ ( z ) = 1 π B 1 ( 0 ) z 2 κ + 1 g 1 ( 0 , ζ ) ¯ 1 ζ ¯ z d σ ζ + z κ 2 π i L r ( τ ) τ + z τ z d τ τ + m = 0 2 κ c m z m , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equbd_HTML.gif
here c m https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq172_HTML.gif, m = 0 , , κ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq173_HTML.gif are arbitrary complex constants, satisfying
c 2 κ m + c m ¯ = 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Eqube_HTML.gif
  1. (2)
    If the index κ < 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq174_HTML.gif, the problem H for the H λ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq1_HTML.gif-regular vector functions in Ω is solvable if and only if the function r ( τ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq144_HTML.gif in the boundary conditions (27) satisfies the following conditions
    https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equ38_HTML.gif
    (38)
     
When the conditions (38) hold, the solution then has the same expression as (1), except that
Φ ( z ) = 1 π B 1 ( 0 ) ζ ¯ 2 κ 1 g 1 ( 0 , ζ ) ¯ 1 ζ ¯ z d σ ζ + 1 π i L r ( τ ) τ κ ( ζ z ) d τ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equbf_HTML.gif
Proof If Ψ ( t , z ) = ( ψ 1 ( t , z ) ψ 2 ( t , z ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq3_HTML.gif is a H λ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq1_HTML.gif-regular vector function, then Ψ ( t , z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq35_HTML.gif satisfies the equation D Ψ = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq58_HTML.gif which is equivalent to
{ z ¯ ψ 2 = 1 2 ( λ + t ) ψ 1 = g 1 ( t , z ) , ( λ t ) ψ 2 = 2 z ψ 1 = g 2 ( t , z ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equ39_HTML.gif
(39)
From Lemma 4, the function ψ 1 ( t , z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq175_HTML.gif expressed in (36) is the unique solution of the Dirichlet problem with the boundary condition (26) for the equation ( λ 2 ) w = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq176_HTML.gif in Ω, so that g 1 ( t , z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq149_HTML.gif, g 2 ( t , z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq151_HTML.gif satisfy the compatible condition of Lemma 2
( λ t ) g 1 = z ¯ g 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equ40_HTML.gif
(40)
Consequently, ψ 2 ( t , z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq177_HTML.gif can be given by the formula (37). Furthermore, ψ 2 ( t , z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq177_HTML.gif expressed in (37) satisfies the boundary condition (35) if and only if the analytic function Φ ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq152_HTML.gif satisfies the following boundary condition
Re τ ¯ κ Φ ( τ ) = r ( τ ) Re τ ¯ κ T B 1 ( 0 ) g 1 ( 0 , τ ) , τ L . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equ41_HTML.gif
(41)

By means of the results about the Riemann-Hilbert boundary value problem for analytic function in the unit disk [13], we can derive the solvable conditions and the expression of solutions. □

Corollary 2 The problem D for the H λ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq1_HTML.gif-regular vector functions in Ω has a unique solution, and the solution is Ψ ( t , z ) = ( ψ 1 ( t , z ) ψ 2 ( t , z ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq178_HTML.gif which ψ 1 ( t , z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq175_HTML.gif is given by (36) and ψ 2 ( t , z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq177_HTML.gif expressed as (37) where
Φ ( z ) = 1 π B 1 ( 0 ) z g 1 ( 0 , ζ ) ¯ 1 ζ ¯ z d σ ζ + 1 2 π i L r ( τ ) τ + z τ z d τ τ + i ( a Im T B 1 ( 0 ) g 1 ( 0 , z ) ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equbg_HTML.gif

Proof The result follows immediately from Theorem 9 and the results of the Dirichlet boundary value problem for analytic function in the unit disk. □

Since the solution u of the equation D u = f https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq7_HTML.gif can be expressed as u = Ψ + T Ω f https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq179_HTML.gif, where Ψ is any H λ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq1_HTML.gif-regular vector functions in Ω, if f L p ( Ω ¯ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq180_HTML.gif, p > 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq106_HTML.gif, then T Ω f C α ( Ω ¯ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq181_HTML.gif, therefore the problem H of the equation D u = f https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq7_HTML.gif in Ω can be transformed into the problem H of the H λ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq1_HTML.gif-regular vector function Ψ ( t , z ) = ( ψ 1 ( t , z ) ψ 2 ( t , z ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq33_HTML.gif in Ω with the following boundary conditions
ψ 1 ( t , z ) = φ 1 ( t , z ) = φ ( t , z ) T Ω 1 f , ( t , z ) S , Re τ ¯ κ ψ 2 ( 0 , τ ) = r 1 ( τ ) = r ( τ ) Re τ ¯ κ T Ω 2 f , τ L , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equbh_HTML.gif
where
T Ω 1 f = 1 4 π Ω { λ e λ | ζ ζ | | ζ ζ | f 1 ( λ | ζ ζ | 2 + 1 | ζ ζ | 3 ) e λ | ζ ζ | [ ( t t ) f 1 + ( z z ) f 2 ] } d σ ζ , T Ω 2 f = 1 4 π Ω { λ e λ | ζ ζ | | ζ ζ | f 2 ( λ | ζ ζ | 2 + 1 | ζ ζ | 3 ) e λ | ζ ζ | [ ( z ¯ z ¯ ) f 1 ( t t ) f 2 ] } d σ ζ , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_Equbi_HTML.gif

namely T Ω f = ( T Ω 1 f T Ω 2 f ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq182_HTML.gif. Using Theorem 10, we obtain the following result about the problem H for the equation D u = f https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq7_HTML.gif in Ω.

Theorem 10 Let f L p ( Ω ¯ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq180_HTML.gif, p > 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq106_HTML.gif.

(a) If the index κ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq171_HTML.gif, the problem H for the equation D u = f https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq7_HTML.gif in Ω has the solution u ( t , z ) = Ψ ( t , z ) + T Ω f https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq183_HTML.gif, where the H λ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq1_HTML.gif-regular vector function Ψ ( t , z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq35_HTML.gif is expressed as (a) of Theorem  9 with φ 1 ( t , z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq184_HTML.gif, r 1 ( τ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq185_HTML.gif replacing φ ( t , z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq60_HTML.gif, r ( τ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq144_HTML.gif respectively.

(b) If the index κ < 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq174_HTML.gif, replacing φ ( t , z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq60_HTML.gif by φ 1 ( t , z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq184_HTML.gif, the problem H for the equation D u = f https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq7_HTML.gif in Ω is solvable if and only if the function r 1 ( τ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq185_HTML.gif satisfies the conditions (38). When the conditions (38) hold, the problem then has the solution u ( t , z ) = Ψ ( t , z ) + T Ω f https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq186_HTML.gif, where the H λ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq1_HTML.gif-regular vector function Ψ ( t , z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq35_HTML.gif is expressed as (b) of Theorem  9 with φ 1 ( t , z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq184_HTML.gif, r 1 ( τ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq185_HTML.gif replacing φ ( t , z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq60_HTML.gif, r ( τ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq144_HTML.gif respectively.

In the same way, we can obtain the result about the problem D for the equation D u = f https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq7_HTML.gif in Ω.

Corollary 3 Suppose that f L p ( Ω ¯ ) , p > 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq187_HTML.gif. The problem D for the equation D u = f https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq7_HTML.gif in Ω has a unique solution u ( t , z ) = Ψ ( t , z ) + T Ω f https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq186_HTML.gif, where the H λ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq1_HTML.gif-regular vector function Ψ ( t , z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq35_HTML.gif is expressed as Corollary  2 with φ 1 ( t , z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq184_HTML.gif, r 1 ( τ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq185_HTML.gif and a 1 = a Im T Ω 2 f https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq188_HTML.gif replacing φ ( t , z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq60_HTML.gif, r ( τ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-75/MediaObjects/13661_2012_Article_175_IEq144_HTML.gif and a respectively.

Declarations

Acknowledgement

This work is supported by National Natural Science Foundation of China (61173121), the Foundation of Doctor Education of China (20095134110001), and the Key Project Foundation of the Education Department of Sichuan Province of China (12ZA136). The authors would like to thank the referee for helpful comments and suggestions.

Authors’ Affiliations

(1)
Department of Mathematics, Sichuan Normal University

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