On the left linear Riemann problem in Clifford analysis
On the left linear Riemann problem in Clifford analysis
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DOI：
10.36045/bbms/1105652784
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发表时间：
1996
影响因子：
0.5
通讯作者：
S. Bernstein
中科院分区：
文献类型：
--
作者：
S. Bernstein
我们考虑一个类似于经典Riemann问题的左线性问题：Dau = 0 in R n n u += H（x）u + h（x）o n ju（x）j = O（jxj n 2 1）asjxj!一曰：为此，我们给出了扰动Dirac算子Da = D + a的一个Borel-Pompeiu公式和一些泛函理论结果.我们将黎曼问题转化为一个积分方程：Pau + HQau = h on;其中Pa = 1（I + Sa）a ndQa = I Pa：我们证明了借助于D + a的基本解构造的奇异积分算子Sa的本质部分正是与D相关联的奇异积分算子S：设Sa是单纯S且= Rn 1，则在1.H = P的假设下，He和H都是实值的、可测的且本质有界的;（1 + H（x））（1 + H（x））和H（x）H（x）对所有的x~2 R~n~1都是真实的数; 3. H的纯量部分H0对所有x~2 R~n~1都满足H0（x）>"> 0;黎曼问题在L~2; C（R~n~1）中唯一可解，
We consider a left-linear analogue to the classical Riemann problem: Dau =0 inR n n u + = H(x)u +h(x )o n ju(x)j = O(jxj n 2 1 )a sjxj!1: For this purpose, we state a Borel-Pompeiu formula for the disturbed Dirac operatorDa = D +a with a paravectora and some functiontheoretical results. We reformulate the Riemann problem as an integral equation: Pau +HQau = h on ; where Pa = 1 (I +Sa )a ndQa = I Pa: We demonstrate that the essential part of the singular integral operator Sa which is constructed by the aid of a fundamental solution of D +a is just the singular integral operator S associated toD: In case Sa is simplyS and = R n 1 , then under the assumptions 1.H= P He and allH are real-valued, measurable and essentially bounded; 2. (1 +H(x))(1 +H(x)) and H(x) H(x) are real numbers for all x2R n 1 ; 3. the scalar part H0 of H fulls H0(x) >"> 0 for all x2R n 1 ; the Riemann problem is uniquely solvable in L2;C(R n 1 ) and the successive approximation