This paper deals with the initial value problem of the type ∂w ∂t =L t, x, w, ∂w ∂x i 1 where t is the time, L is a linear first order operator matrix-type in Quaternionic Analysis and
Trang 1© 2011 Springer Basel AG
0188-7009/030591-15
published online January 4, 2011
DOI 10.1007/s00006-010-0272-2
Differential Operators Associated to the Cauchy-Fueter Operator in Quaternion
Algebra
Nguyen
Thanh Van
Abstract. This paper deals with the initial value problem of the type
∂w
∂t =L
t, x, w, ∂w ∂x
i
(1)
where t is the time, L is a linear first order operator (matrix-type)
in Quaternionic Analysis and ϕ is a regular function taking values in
the Quaternionic Algebra The article proves necessary and sufficient conditions on the coefficients of operatorL under which L is associated
to the Cauchy-Fueter operator of Quarternionic Analysis
This criterion makes it possible to construct the operatorL for
which the initial problem (1), (2) is solvable for an arbitrary initial regular functionϕ and the solution is also regular for each t.
Mathematics Subject Classification (2010).35B45; 35F10; 47H10
Keywords.Initial value problem, associated space, interior estimate
1 Preliminaries and Notations
LetH be a Quaternion algebra with the basis formed by e0, e1, e2, e3 where
e0= 1, e3= e1e2= e12.
Suppose that Ω is a bounded domain ofR4 A function f defines in Ω and takes values in the Quaternionic AlgebraH which can be presented as
f =
3
j=0
f j e j ,
where f j (x) are real-valued functions.
We introduce the Cauchy-Fueter operator
D =
3
k=0
e k ∂x ∂ k
Advances in Applied Cliff ord Algebras
Trang 2592 T.V Nguyen Adv Appl Cliff ord Algebras
Definition 1. A function f ∈ C1(Ω, H) is said to be regular in Ω if f satisfies
Df = 0.
2 Necessary and Sufficient Conditions for Associated Pairs
Suppose that f = 3
j=0 f j e j is a twice continuously differentiable function with respect to the space-like x0, x1, x2, x3 Now assume that f is regular.
This means that Df = 0 It is easy to verify that the condition Df = 0 is
equivalent to
3
i=0
A i
∂f
∂x i = 0,
where
A0=
⎡
⎢
⎣
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
⎤
⎥
⎦ , A1=
⎡
⎢
⎣
0 −1 0 0
⎤
⎥
⎦ , A2=
⎡
⎢
⎣
0 −1 0 0
⎤
⎥
⎦
A3=
⎡
⎢
⎣
0 0 −1 0
⎤
⎥
⎦ , ∂x ∂f i =
⎛
⎜
⎜
⎝
∂f0
∂x i
∂f1
∂x i
∂f2
∂x i
∂f3
∂x i
⎞
⎟
⎟
⎠.
We define an operator as follows,
f =
3
i=0
A i ∂x ∂f
It is clear that Df = 0 if and only if f = 0 Next, we identify the function
f with f :=
⎛
⎜
⎝
f0
f1
f2
f3
⎞
⎟
⎠ and introduce a differential operator L as follows,
Lf =
3
j=0
B j ∂x ∂f j
where B j = [b (j) αβ ], C = [c αβ ], K =
⎛
⎜
⎝
d0
d1
d2
d3
⎞
⎟
⎠, b (j) αβ , c αβ , d α , (α, β = 0, 1, 2, 3) are
real-valued functions which are supposed to depend at least continuously on
the time t and the space-like x0, x1, x2, x3 A pair of operators , L is said
to be associated (see [9]) if f = 0 implies (Lf ) = 0 (for each t in case the coefficients of L depend on t) Now we formulate necessary and sufficient conditions on the coefficients of operator L under which L is associated to
Trang 3Vol 21 (2011) Diff erential Associated Operators and Their Applications 593
the operator (in other words, L is associated to the Cauchy-Fueter operator
of Quaternionic Analysis) Assume that the functions b (j)
αβ , c αβ , d α (j, α, β =
0, 1, 2, 3) are continuously differentiable with respect to the space-like variable
x0, x1, x2, x3 and differentiable on t.
Put
P j = [p (j) αβ ] = A j B j , j = 0, 1, 2, 3 (5)
Q ij = [q αβ (ij) ] = A i B j + A j B i , 0≤ i < j ≤ 3 (6)
R j = [r αβ (j)] =
3
i=0
A i ∂B ∂x j
i + A j C, j = 0, 1, 2, 3, α, β = 0, 1, 2, 3. (7)
Then we get the following theorem
Theorem 1. The operator L is associated to the operator if and only if the following conditions are satisfied
i) The functions h (α) = 3
i=0
c iα e i , α = 0, 1, 2, 3, and g =
3
i=0
d i e i are regu-lar.
ii)
⎧
⎪
⎪
⎪
⎪
⎪
⎪
r(1)
i0 = r(0)i1 , r(2)i0 = r i2(0), r i0(3)= r i3(0)
r(1)
i1 =−r(0)i0 , r(2)
i1 =−r i3(0), r(3)
i1 = r(0)i2
r(1)
i2) = r(0)i3 , r(2)i2 =−r(0)i0 , r(3)
i2 =−r(0)i1
r(1)
i3 =−r(0)i2 , r(2)
i3 = r(0)i1 , r(3)i3 =−r(0)i0
i = 0, 1, 2, 3.
iii)
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
q(01)
i0 = p(0)i1 − p(1)i1 , q(02)
i0 = p(0)i2 − p(2)i2 , q(03)
i0 = p(0)i3 − p(3)i3
q(12)
i0 =−p(1)i3 + p(2)
i3 , q(13)i0 = p(1)i2 − p(3)i2 , q(23)
i0 =−p(2)i1 + p(3)
i1
q(01)
i1 =−p(0)i0 + p(1)
i0 , q(02)i1 =−p(0)i3 + p(2)
i3 , q(03)i1 = p(0)i2 − p(3)i2
q(12)
i1 =−p(1)i2 + p(2)
i2 , q(13)i1 =−p(1)i3 + p(3)
i3 , q(23)i1 = p(2)i0 − p(3)i0
q(01)
i2 = p(0)i3 − p(1)i3 , q(02)
i2 =−p(0)i0 + p(2)
i0 , q i2(03)=−p(0)i1 + p(3)
i1
q(12)
i2 = p(1)i1 − p(2)i1 , q(13)
i2 =−p(1)i0 + p(3)
i0 , q i2(23)=−p(2)i3 + p(3)
i3
q(01)
i3 =−p(0)i2 + p(1)
i2 , q(02)i3 = p(0)i1 − p(2)i1 , q(03)
i3 =−p(0)i0 + p(3)
i0
q(12)
i3 = p(1)i0 − p(2)i0 , q(13)
i3 = p(1)i1 − p(3)i1 , q(23)
i3 = p(2)i2 − p(3)i2
i = 0, 1, 2, 3.
Proof We get
(Lf ) =
3
i=0
A i ∂(Lf ) ∂x
i
=
3
i=0
A i
∂
∂x i
⎛
⎝3
j=0
B j
∂f
∂x j + Cf + K
⎞
⎠
Trang 4594 T.V Nguyen Adv Appl Cliff ord Algebras
=
3
i=0
A i
∂
∂x i
⎛
⎝3
j=0
B j
∂f
∂x j
⎞
⎠ +3
i=0
A i
∂(Cf )
∂x i
+ 3
i=0
A i
∂K
∂x i
=
3
i=0
3
j=0
A i B j
∂2f
∂x i ∂x j +
3
i=0
3
j=0
A i
∂B j
∂x i
∂f
∂x j
+
3
i=0
A i
∂C
∂x i
f +
3
i=0
A i C
∂f
∂x i +
3
i=0
A i
∂K
∂x i
=
3
i=0
A i B i ∂
2f
∂x2
i
0≤i<j≤3 (A i B j + A j B i) ∂
2f
+
3
j=0
3
i=0
A i ∂B ∂x j
i + A j C
∂f
∂x j +
3
i=0
A i ∂x ∂C i
f +
3
i=0
A i ∂K ∂x
i .
By (5), (6) and (7), then (8) can be rewritten as
l(Lf ) =
3
i=0 2P i ∂
2f
∂x2
i
0≤i<j≤3
Q ij ∂
2f
∂x i ∂x j +
3
j=0
R j ∂x ∂f j
+
3
i=0
A i ∂x ∂C i
f +
3
i=0
A i ∂K ∂x
Write
M =
3
i=0
P i ∂
2f
∂x2
i
0≤i<j≤3
Q ij ∂
2f
∂x i ∂x j =
⎛
⎜
⎝
m0
m1
m2
m3
⎞
⎟
⎠
N =
3
j=0
R j ∂x ∂f
j =
⎛
⎜
⎝
n0
n1
n2
n3
⎞
⎟
⎠
S =
3
i=0
A i ∂x ∂C i
f, T =
3
i=0
A i ∂K ∂x
i .
Then we obtain
We get
m i = p(0)i0 ∂
2f
0
∂x2 + p(0)i1
∂2f 1
∂x2 + p(0)i2
∂2f 2
∂x2 + p(0)i3
∂2f 3
∂x2
+ p(1)
i0
∂2f
0
∂x2 + p(1)i1
∂2f 1
∂x2 + p(1)i2
∂2f 2
∂x2 + p(1)i3
∂2f 3
∂x2
+ p(2)
i0
∂2f
0
∂x2 + p(2)i1
∂2f 1
∂x2 + p(2)i2
∂2f 2
∂x2 + p(2)i3
∂2f 3
∂x2
Trang 5Vol 21 (2011) Diff erential Associated Operators and Their Applications 595
+ p(3)
i0
∂2f
0
∂x2 + p(3)i1
∂2f 1
∂x2 + p(3)i2
∂2f 2
∂x2 + p(3)i3
∂2f 3
∂x2
+ q(01)
i0
∂2f 0
∂x0∂x1 + q
(01)
i1
∂2f 1
∂x0∂x1 + q
(01)
i2
∂2f 2
∂x0∂x1 + q
(01)
i3
∂2f 3
∂x0∂x1
+ q(02)
i0
∂2f0
∂x0∂x2
+ q(02)
i1
∂2f1
∂x0∂x2
+ q(02)
i2
∂2f2
∂x0∂x2
+ q(02)
i3
∂2f3
∂x0∂x2
+ q(03)
i0
∂2f 0
∂x0∂x3 + q
(03)
i1
∂2f 1
∂x0∂x3 + q
(03)
i2
∂2f 2
∂x0∂x3 + q
(03)
i3
∂2f 3
∂x0∂x3
+ q(12)
i0
∂2f 0
∂x1∂x2 + q
(12)
i1
∂2f 1
∂x1∂x2 + q
(12)
i2
∂2f 2
∂x1∂x2 + q
(12)
i3
∂2f 3
∂x1∂x2
+ q(13)
i0
∂2f 0
∂x1∂x3 + q
(13)
i1
∂2f 1
∂x1∂x3 + q
(13)
i2
∂2f 2
∂x1∂x3 + q
(13)
i3
∂2f 3
∂x1∂x3
+ q(23)
i0
∂2f0
∂x2∂x3
+ q(23)
i1
∂2f1
∂x2∂x3
+ q(23)
i2
∂2f2
∂x2∂x3
+ q(23)
i3
∂2f3
∂x2∂x3
. (11)
Similarly, one gets
n i = r(0)i0 ∂x ∂f0
0
+ r(0)
i1
∂f1
∂x0
+ r(0)
i2
∂f2
∂x0
+ r(0)
i3
∂f3
∂x0
+ r(1)
i0
∂f0
∂x1+ r
(1)
i1
∂f1
∂x1+ r
(1)
i2
∂f2
∂x1+ r
(1)
i3
∂f3
∂x1
+ r(2)
i0
∂f0
∂x2+ r
(2)
i1
∂f1
∂x2+ r
(2)
i2
∂f2
∂x2+ +r
(2)
i3
∂f3
∂x2
+ r(3)
i0
∂f0
∂x3+ r
(3)
i1
∂f1
∂x3+ r
(3)
i2
∂f2
∂x3+ +r
(3)
i3
∂f3
Suppose that f is regular function, then
⎧
⎪
⎪
⎪
⎪
∂f0
∂x0 − ∂f1
∂x1 − ∂f2
∂x2 − ∂f3
∂x3 = 0
∂f0
∂x1 +∂f1
∂x0 − ∂f2
∂x3 +∂f3
∂x2 = 0
∂f0
∂x2 +∂f1
∂x3 +∂f2
∂x0 − ∂f3
∂x1 = 0
∂f0
∂x3 − ∂f1
∂x2 +∂f2
∂x1 +∂f3
∂x0 = 0.
(13)
It followa from (13), that
⎧
⎪
⎪
⎪
⎪
∂f0
∂x0 = ∂f1
∂x1 +∂f2
∂x2+ ∂f3
∂x3
∂f1
∂x0 =− ∂f0
∂x1 +∂f2
∂x3 − ∂f3
∂x2
∂f2
∂x0 =− ∂f0
∂x2 − ∂f1
∂x3 +∂f3
∂x1
∂f3
∂x0 =− ∂f0
∂x3 +∂f1
∂x2 − ∂f2
∂x1
(14)
and
Trang 6596 T.V Nguyen Adv Appl Cliff ord Algebras
⎧
⎪
⎪
⎪
⎪
∂2f0
∂x2 = ∂2f1
∂x0∂x1 + ∂2f2
∂x0∂x2 + ∂2f3
∂x0∂x3
∂2f0
∂x2 =− ∂2f1
∂x0∂x1 − ∂2f3
∂x1∂x2 + ∂2f2
∂x1∂x3
∂2f0
∂x2 =− ∂2f1
∂x2∂x3 − ∂2f2
∂x0∂x2 + ∂2f3
∂x1∂x2.
∂2f0
∂x2 = ∂2f1
∂x2∂x3 − ∂2f2
∂x1∂x3 − ∂2f3
∂x0∂x3.
(15)
and similar expression for the other ∂2f i
∂x2
j , i = 1, 2, 3; j = 0, 1, 2, 3.
Hence we get 3 remaining systems having the form of (15) Thus, one has a total of 12 equations Substituting above 12 equations into (11), and after a calculation, we obtain
m i=
−p(0)i1 + p(1)
i1 + q i0(01)
∂2f0
∂x0∂x1
+
−p(0)i2 + p(2)
i2 + q(02)i0
∂2f0
∂x0∂x2 +
−p(0)i3 + p(3)
i3 + q i0(03)
∂2f 0
∂x0∂x3 +
p(1)
i3 − p(2)i3 + q(12)
i0
∂2f 0
∂x1∂x2 +
−p(1)i2 + p(3)
i2 + q i0(13)
∂2f 0
∂x1∂x3 +
p(2)
i1 − p(3)i1 + q(23)
i0
∂2f 0
∂x2∂x3 +
p(0)
i0 − p(1)i0 + q(01)
i1
∂2f 1
∂x0∂x1 +
p(0)
i3 − p(2)i3 + q(02)
i1
∂2f 1
∂x0∂x2 +
−p(0)i2 + p(3)
i2 + q i1(03)
∂2f1
∂x0∂x3
+
p(1)
i2 − p(2)i2 + q(12)
i1
∂2f1
∂x1∂x2 +
p(1)
i3 − p(3)i3 + q(13)
i1
∂2f 1
∂x1∂x3 +
−p(2)i0 + p(3)
i0 + q i1(23)
∂2f 1
∂x2∂x3 +
−p(0)i3 + p(1)
i3 + q i2(01)
∂2f 2
∂x0∂x1 +
p(0)
i0 − p(2)i0 + q(02)
i2
∂2f 2
∂x0∂x2 +
p(0)
i1 − p(3)i1 + q(03)
i2
∂2f 2
∂x0∂x3 +
−p(1)i1 + p(2)
i1 + q i2(12)
∂2f 2
∂x1∂x2 +
p(1)
i0 − p(3)i0 + q(13)
i2
∂2f2
∂x1∂x3
+
p(2)
i3 − p(3)i3 + q(23)
i2
∂2f2
∂x2∂x3 +
p(0)
i2 − p(1)i2 + q(01)
i3
∂2f 3
∂x0∂x1 +
−p(0)i1 + p(2)
i1 + q i3(02)
∂2f 3
∂x0∂x2 +
p(0)
i0 − p(3)i0 + q(03)
i3
∂2f 3
∂x0∂x3 +
−p(1)i0 + p(2)
i0 + q i3(12)
∂2f 3
∂x1∂x2 +
−p(1)i1 + p(3)
i1 + q i3(13)
∂2f3
∂x1∂x3 +
−p(2)i2 + p(3)
i2 + q i3(23)
∂2f3
∂x2∂x3. (16) Analogously, substituting the relation (14) into (12), one gets
n i= (−r(0)i1 + r(1)
i0 )
∂f0
∂x1+ (−r(0)i2 + r(2)
i0 )
∂f0
∂x2 + (−r i3(0)+ r(3)
i0 )
∂f0
∂x3
+ (r(0)
i0 + r(1)i1 )
∂f1
∂x1 + (r
(0)
i3 + r(2)i1 )
∂f1
∂x2 + (−r i2(0)+ r(3)
i1 )
∂f1
∂x3
Trang 7Vol 21 (2011) Diff erential Associated Operators and Their Applications 597
+ (−r(0)i3 + r(1)
i2 )
∂f2
∂x1 + (r
(0)
i0 + r(2)i2 )
∂f2
∂x2 + (r
(0)
i1 + r(3)i2 )
∂f2
∂x3
+ (r(0)
i2 + r(1)i3 )
∂f3
∂x1 + (−r i1(0)+ r(2)
i3 )
∂f3
∂x2 + (r
(0)
i0 + r(3)i3 )
∂f3
∂x3. (17)
(*)Sufficient condition
Suppose that the conditions (i), (ii) and (iii) of the Theorem are satisfied From the relation (i), it follows that S = T = 0 Because of (ii) it leads
to n i = 0, i = 0, 1, 2, 3 Using the condition (iii) it implies m i = 0, i =
0, 1, 2, 3 This means that M = N = 0.
Hence l(Lf ) = M + N + S + T = 0 for all regular functions f
The sufficient condition is proved
(*)Necessary condition
Assume that a (l, L) is an associated pair, i.e., if lf = 0, then l(Lf ) = 0 We
will choose 38 regular functions as follows
First, choose f(1) = 0, then (10) passes into T Because of l(Lf ) = 0,
then T = 0 This means that g = 3
i=0
d i e i is a regular function Thus the term T can be omitted in (10) Next, we choose f(2)is arbitrary Quaternionic
constant, f(2)= 0 For this choice (10) implies S = 0 Since f(2) is arbitrary, then 2
i=0
A i ∂x ∂C i = 0 In other words h (α)=
3
i=0
c iα e i , α = 0, 1, 2, 3 are regular functions Hence S vanished in (10) Now, choose f(3)= x0+ x1e1, then (10)
leads to N = 0, so n i = 0, i = 0, 1, 2, 3 But in fact n i = r i0(0) + r(1)i1 . Therefore, we get r(1)
i1 =−r(0)i0 Note that the equality is the same as the condition 4th of the relation (i).
By similar method, choose
f(4)= x
1− x0e1, f(5)= x0e2+ x1e3, f(6)= x1e2− x0e3,
f(7)= x
0+ x2e2, f(8)= x0e1− x2e3, f(9)= x2− x0e2,
f(10)= x
2e1+ x0e3, f(11)= x0+ x3e3, f(12)= x0e1+ x3e2,
f(13)=−x3e1+ x0e2, f(14)=−x3+ x0e3
and substitute these functions into (10) we obtain N = 0 for all f (i) , i =
4, , 14 From this, we have remaining equalities which are contained in the condition (ii) Hence N can be omitted in (10).
Now we choose f(15) = (x2− x2) + 2x
0x1e1 and replace f in (10) by
f(15), it follows that M = 0.
This means
m i =−p(0)i1 + p(1)
i1 + q i0(01)= 0, i = 0, 1, 2, 3.
The equality leads to
q(01)
Trang 8598 T.V Nguyen Adv Appl Cliff ord Algebras
Note that (18) is the same as the first condition of (iii) Similarly, choose
f(16)=−2x0x1+ (x20− x2
1)e1 f(17)= (x20− x2
1)e2+ 2x0x1e3
f(18)=−2x0x1e2+ (x20− x2
1)e3 f(19) = (x20− x2
2) + 2x0x2e2
f(20)= (x2
0− x2
2)e1− 2x0x2e3 f(21)=−2x0x2+ (x20− x2
2)e2
f(22)= 2x
0x2e1+ (x20− x2
2)e3, f(23) = (x21− x2
2)− 2x1x2e3
f(24)= (x2
1− x2
2)e1− 2x1x2e2 f(25)= 2x1x2e1+ (x21− x2
2)e2
f(26)= 2x
1x2+ (x21− x2
2)e3, f(27)= (x20− x2
3) + 2x0x3e3
f(28)= (x2− x2)e
1+ 2x0x3e2, f(29) =−2x0x3e1+ (x2− x2)e
2
f(30)=−2x0x3+ (x20− x2
3)e3 f(31)= (x21− x2
3) + 2x1x3e2
f(32)= (x2
1− x2
3)e1− 2x1x3e3 f(33)=−2x1x3+ (x21− x2
3)e2
f(34)= 2x
1x3e1+ (x21− x2
3)e3 f(35)= (x22− x2
3)− 2x2x3e1
f(36)= 2x
2x3+ (x22− x2
3)e1 f(37)= (x22− x2
3)e2− 2x2x3e3
f(38)= 2x
2x3e2+ (x22− x2
3)e3
and substitute f = f (j) , j = 16, , 38 into (10) one obtains M = 0 By similar arguments we get all remaining equalities of the condition (iii) This
completes the proof of necessary condition
Remark 1. We can see that the conditions (ii) and (iii) of Theorem 1 can
be written as follows
R i = R0A i , i = 1, 2, 3
Q ij = (P i − P j )A j A i , 0≤ i < j ≤ 3.
So we get
l(Lf ) =
3
i=0
P i ∂
2f
∂x2
i
0≤i<j≤3 (P i − P j )A j A i ∂
2f
∂x i ∂x j +
3
i=0
R0A i ∂x ∂f
i . Note that A0 = E, A2i = −E and A i A j + A j A i = 0, i = j, i, j ∈ {1, 2, 3}.
Then we easily obtain
l(Lf ) = (P0∂x ∂
0 − P1A1∂x ∂
1− P2A2∂x ∂
2− P3A3∂x ∂
3+ R0)(
3
i=0
A i ∂x ∂f
i)
= V (lf ),
with V = P0∂x ∂
0− P1A1∂x ∂
1 − P2A2∂x ∂
2 − P3A3∂x ∂
3 + R0,
and P0, P1, P2, P3, R0 are given in (5) and (7).
Therefore one gets the following theorem
Trang 9Vol 21 (2011) Diff erential Associated Operators and Their Applications 599
Theorem 2. The operator L is associated to the operator if and only if
lL = V l,
where V = P0∂x ∂
0 − P1A1∂x ∂
1 − P2A2∂x ∂
2 − P3A3∂x ∂
3
+ R0.
3 Example
First, we choose c αβ as the arbitrary real-constants, g =
3
i=0
d i e i is arbitrary regular function and choose elements b(0)
αβ , α, β = 0, 1, 2, 3 of the matrix B0
as follows
b(0)
00 =
1
2[−(γ − c00)x0− c10x1− c20x2− c30x3] + δ(0)00
b(0)
01 =
1
2[c01x0+ (γ − c11)x1− c21x2− c31x3] + δ01(0)
b(0)
02 =
1
2[c02x0− c12x1+ (γ − c22)x2− c32x3] + δ02(0)
b(0)
03 =
1
2[c03x0− c13x1− c23x2+ (γ − c33)x3] + δ03(0)
b(0)
10 =
1
2[c10x0− (γ − c00)x1+ c30x2− c20x3] + δ10(0)
b(0)
11 =
1
2[−(γ − c11)x0+ c01x1+ c31x2− c21x3] + δ(0)11
b(0)
12 =
1
2[c12x0+ c02x1+ c32x2+ (γ − c22)x3] + δ12(0)
b(0)
13 =
1
2[c13x0+ c03x1− (γ − c33)x2− c23x3] + δ13(0)
b(0)
20 =
1
2[c20x0− c30x1− (γ − c00)x2+ c10x3] + δ20(0)
b(0)
21 =
1
2[c21x0− c31x1+ c01x2− (γ − c11)x3] + δ21(0)
b(0)
22 =
1
2[−(γ − c22)x0− c32x1+ c02x2+ c12x3] + δ(0)22
b(0)
23 =
1
2[c23x0+ (γ − c33)x1+ c03x2+ c13x3] + δ23(0)
b(0)
30 =
1
2[c30x0+ c20x1− c10x2− (γ − c00)x3] + δ30(0)
b(0)
31 =
1
2[c31x0+ c21x1+ (γ − c11)x2+ c01x3] + δ31(0)
b(0)
32 =
1
2[c32x0− (γ − c22)x1− c12x2+ c02x3] + δ32(0)
b(0)
33 =
1
2[−(γ − c33)x0+ c23x1− c13x2+ c03x3] + δ(0)33,
where γ, δ(0)
αβ , α, β = 0, 1, 2, 3 are arbitrary real-constants.
Trang 10600 T.V Nguyen Adv Appl Cliff ord Algebras
Second, choose
⎧
⎪
⎪
B1=−A1B0
B2=−A2B0
B3=−A3B0. Then it is easy to verify that all the conditions of Theorem 1 are satisfied By
this way one obtains a class of differential operators L which are associated
to the Cauchy-Fueter of Quaternionic Analysis
Remark 2. If b (j)
αβ , c αβ , d αβ and f α , α, β = 0, 1, 2, 3 do not depend on x3, then
we obtain the necessary and sufficient conditions under which the operator L
is associated to the Cauchy-Riemann Operator in Quaternionic Analysis (see [8]) Using Theorem 1, we can construct a class of the differential operators
L which are associated to the Cauchy-Riemann Operator as follows.
First, we take g =3
i=0 d i e i arbitrary regular function and
c00= s1x1+ s2x2+ γ00
c10=−2s1x0−13s0x2+ γ10
c20=−2s2x0+13s0x1+ γ20
c30=−23s0x0− s2x1+ s1x2+ γ30
c01= s0x2+ γ01
c11= 3s1x1+ 3s2x2+ γ11
c21= 3s2x1− 3s1x2+ γ21
c31= s0x1+ γ31
c02=−s0x1+ γ02
c12=−3s2x1+ 3s1x2+ γ12
c22= 3s1x1+ 3s2x2+ γ22
c32= s0x2+ γ32
c03=−2
3s0x0+ s2x1− s1x2+ γ03
c13=−2s2x0−1
3s0x1+ γ13
c23= 2s1x0−1
3s0x2+ γ23
c33= s1x1+ s2x2+ γ33,
where s0, s1, s2, γ ij are arbitrary real-constants.
The elements of the matrix B0 are given by