Hyperbolic function theory in R

Hyperbolic function theory in R3 Sirkka-Liisa Eriksson-Bique University of Joensuu Department of Mathematics P.O.Box 111 FIN-80101 Joensuu, Finland November 22, 2004 Abstract We consider the Poincare model of a hyperbolic geometry in R3 (ie., 2 2 +dt2 the metric is ds2 = dx +dy ) and Möbius transformations. We t2 study also an old problem how to extend one variable complex function theory to higher dimensions. We denote by H the real associative algebra of quaternions generated by e1 ,e2 satisfying e21 = e22 = 1 and e1 e2 = e2 e1 . In 1992 Leutwiler noticed that xm (x 2 H) is a conjugate gradient of a harmonic function with respect to a hyperbolic metric. We study an extension of solutions of Leutwiler, called hyperholomorphic. Hyperholomorphic functions satisfy the equation @f @f tDf (x; y; t) e2 y f (x; y; t) = 0; where Df = @f @x + e1 @y + e2 @t and 0

e2 y f = e2 f 2f e2 ( 0 is the main involution). Note that e2 y f is a quaternionic derivation operation. We prove a version of Cauchy theorem and Cauchy formula for hyperholomorphic functions

1

Preliminaries

Let H be the real associative algebra of quaternions generated by e1 , e2 satisfying e21 = e22 = 1; e1 e2 = e2 e1 : Set e12 = e1 e2 : 1

Any element in H may be written as x = x0 + x1 e1 + x2 e2 + x12 e12 for x0 ; x1 ; x2 ; x12 2 R. The elements x = x0 + x1 e1 + x2 e2 for x0 ; x1 ; x2 2 R are called paravectors in H. The space R3 is identi…ed with the set of paravectors. The complex number …eld C is identi…ed with the set fx0 + x1 e1 : x0 ; x1 2 Rg. The main involution 0 : H ! H is the isomorphism de…ned by q 0 = x0

x1 e1

x2 e2 + x12 e12

for x0 ; x1 ; x2 ; x12 2 R. Hence we have (ab)0 = a0 b0 : The second involution : H ! H, called reversion, is the anti-isomorphism de…ned by q = x0 + x1 e1 + x2 e2

x12 e12 ; (ab) = b a

for x0 ; x1 ; x2 ; x12 2 R: The conjugation

: H ! H, is de…ned by

q = (q )0 = x0

x1 e1

x2 e2

x12 e12

for x0 ; x1 ; x2 ; x12 2 R. Note that, ab = ba: The involutionb: H ! H is de…ned by qb = x0 + x1 e1

x2 e2

x12 e12 :

Useful properties of involutions are the following ze2 = e2 z 0 = e2 z; z 2 C, q 0 e2 = e2 q^; q 2 H. The projecting operators P : H ! C and Q : H ! C are de…ned by P (z0 + z1 e2 ) = z0 ; Q (z0 + z1 e2 ) = z1 2

(1)

and for z 2 C. Note that

P 2 x = P x; Q2 x = 0; P (Qx) = Qx; Q (P x) = 0:

for any x 2 H. If x 2 H, then

x

1

=

where jxj2 = xx = x20 + x21 + x22 + x212 : The relation

x ; jxj2

(x1 e1 + x2 e2 + x12 e12 )2m = ( 1)m x21 + x22 + x212

m

(2)

is useful. The operators P and Q satisfy the following product rules: Lemma 1 If a and b belong to H, then P (ab) = P aP b Qa (Qb)0 ; Q (ab) = aQb + (Qa) b0 : Proof. Write a = P a + (Qa) e2 and b = P b + (Qb) e2 . Applying (1) we obtain ab = (P a) (P b) + (Qa) e2 (Qb) e2 + (P a) (Qb) e2 + (Qa) e2 (P b) = (P a) (P b)

(Qa) (Qb)0 + (P a) (Qb) + (Qa) (P b)0 e2 :

Hence Q (ab) = (P a) (Qb) + (Qa) (P b)0 = (a = aQb

Qae2 ) (Qb) + (Qa) (b

Qbe2 )0

Qa (Qb)0 e2 + Qab0 + Qa (Qb)0 e2 = aQb + (Qa) b0 :

The following rules are easily proved P (a0 ) = (P a)0 ; 0

Q (a ) =

(Qa) :

We abbreviate (P a)0 = P 0 a; (Qa)0 = Q0 a: 3

(3) 0

(4)

Lemma 2 If w 2 H then Qw =

e2 w0

Pw =

we2 2

;

we2 + e2 w0 e2 : 2

Proof. Setting w = P w + (Qw) e2 we infer w0 = P 0 w (1) we obtain e2 w0

we2 = e2 P 0 w = 2Qw:

e2 (Q0 w) e2

(Q0 w) e2 . Using

P we2 + Qw

Let w be a paravector and a 2 H. The exterior product of w and a is wa + a0 w : w^a= 2 The contraction of w and a is wy a =

a0 w

wa 2

:

We have the identity wa = w ^ a + wy a: Our de…nitions of the exterior product and contraction for a vector has been introduced by M. Riesz 1958. Lounesto in 1993 pointed out that this contraction is a derivation for both Cli¤ord and exterior product. The contraction operator is a generalization of the Q -operator, since en y a = Q0 a: We use the notation P f for the function from (P f ) (x) = P (f (x)) and similarly for Qf . It is clear that @ (P f ) @f =P @xi @xi @ (Qf ) @f =Q @xi @xi 4

; :

into H de…ned by

We use the following standard notations for x = x0 +x1 e1 +x2 e2 +x12 e12 2 H and a multi-index = ( 0 ; 1 ; 2 ; 12 ) 2 N4 x = x0 0 :::x1212 ; ! = 0 !::: 12 !; j j = 0 + ::: + m m! = ; 0 !::: 12 !

12 ;

if

j j = m:

A multi-index = ( 0 ; :::; 12 ) 2 N4 is called even if all denote "i = ("i0 ; ::; "i12 ) for i = 0; 1; 2; 3;where "ij = ij .

i

are even. We

Theorem 3 Let x = x0 + x1 e1 + x2 e2 + x12 e12 be a quaternion and m 2 N. Then X m xm = c( )x ; j j=m

where the coe¢ cients c (:) are given by 8 m 0 2 > > m 0 "0 > 0 > 2 > ( 1) 2 ; > m 0 > < ( 0 "0 ) m 0 1 c( ) = 2 > ( " m 0 0 "i ) 0 > > 2 2 > ( 1) m > 0 ) > ( > 0 "0 : 0; otherwise.

0 "0

even;

1

ei ;

"i even,

0 "0

Proof. Let x = x0 e0 + x1 e1 + x2 e2 + x12 e12 . Since x0 commutes with all ei , we infer that m X m m x = x0 0 (x1 e1 + x2 e2 + x12 e12 )m 0 . 0 =0

0

Applying (2) and multinomial theorem we infer that

(x1 e1 + x2 e2 + x12 e12 )m 0 m 0 m X 0 2 n 2 2 = ( 1) x21 1 x22 2 x12 ; j j=

when m

0

= ( 1)

0

2

is even, and

(x1 e1 + x2 e2 + x12 e12 )m m

m

0 2

0

1

(x1 e1 + x2 e2 + x12 e12 ) j j=

5

X m

0 2

m

0

2 1

1

x21 1 x22 2 x2123

when m

m

is odd. Since

(m)

; the result follows. ( 00 ) The following binomial theorem is easily obtained. 0

0

=

m

Theorem 4 If x = x0 + x1 e1 + x2 e2 + x12 e12 and y = y0 + y1 e1 + y2 e2 + y12 e12 are paravectors, then X

(x + y)m =

m ;

j j+j j=m

c( + )x y :

Proof. Using Theorem 3 we infer (x + y)m =

X

m

c ( ) (x + y) .

(5)

j j=m

Set

= ( 0 ; :::;

12 ).

Substituting

(x + y) = (x0 + y0 ) 0 ::: (x12 + y12 ) 12 X 0 1 n = ::: x y i+ i= i i=0;::;3

=

X

i+ i= i i=0;1;2;12

0

n

1

0 !::: 3 !x

y ; 0 !:: 12 ! 0 !:: 12 !

in (5) we establish the assertion. Applying Theorem 3 it is simple to di¤erentiate xm . Theorem 5 If x = x0 + x1 e1 + x2 e2 + x12 e12 and @xm+j j = @x1 1 @x2 2 @x1212

m+j j ;m

!

X

=(

m

1;

2;

3 ),

then

c( + )x

j j=m

for any natural number m.

2

Hyperholomorphic functions

The Dirac operator (also called generalized Cauchy-Riemann operator) is de…ned by @f @f @f Df = + e1 + e2 @x0 @x1 @x2 6

for a mapping f : ! H; whose domain of the de…nition is an open subset of R3 and whose components are partially di¤erentiable. The operator D is de…ned by @f @f @f Df = e1 e2 : @x0 @x1 @x2 If Df = 0 the function f is called (left) monogenic (or regular). The modi…ed Dirac operator Mk is de…ned by (Mk f ) (x) = (Df ) (x) +

k 0 Qf x2

and the operator M by M k f (x) = Df (x)

k 0 Q f: x2

It is easy to see that Mk f + M k f = Df + Df = 2

@f . @x0

De…nition 6 Let R3 be open. If f 2 C 2 ( ) and Mk f (x) = 0 for any x 2 n fx j x2 = 0g, the function f is called k-hyperholomorphic in . If f is paravector valued k hyperholomorphic in ; the function f is called an Hk -solution. The functions that are 1-hyperholomorphic are called hyperholomorphic. The H-solutions in R3 were introduced by H. Leutwiler in 1992 in ([21]) . They are notably studied in Rn by H. Leutwiler ([22], [23], [24]), Th. Hemp‡ing ([16]), J. Cnops ([5]), P. Cerejeiras[4]and S.-L. Eriksson-Bique ( [7], [10], [11]). In R3 the hyperholomorphic functions are researched by W. Hengartner and H. Leutwiler ([19]) in and in Rn by H. Leutwiler and S.-L. Eriksson-Bique ([13], [14], [15]). Example 7 (Fueter construction) ([21]) Let f = u + iv be holomorphic in an open set C. We de…ne the mapping : R3 ! C by (x0 ; x1 ; x2 ) = x0 + x1 i and the mapping

: R4 ! C by

q (x0 ; x1 ; x2 ) = x0 + i x21 + x22 : 7

Then the function f is hyperholomorphic in the set fx j (x) 2 g. Moree over the function f de…ned by fe(x) = u

x1 e1 + x2 e2 (x) + p 2 v x1 + x22

(x)

is hyperholomorphic in the set fx j (x) 2 g . The following Lemma is useful.

Lemma 8 Let be an open subset of R3 . If the components of a function f : ! H are partially di¤erentiable, then @ (Qf )0 P (Df ) = D1 (P f ) ; @x2 @ (P f )0 ; Q (Df ) = D1 (Qf ) + @x2

where D1 f =

@f @f + e1 : @x0 @x1

Proof. Decomposing f = P f + (Qf ) e2 we obtain Df = D (P f ) + D ((Qf ) e2 ) @ (P f ) @ (Qf ) = D1 (P f ) + e2 + D1 (Qf ) e2 + e2 e2 : @x2 @x3 Applying (1) we deduce Df = D1 (P f )

@ (Qf )0 @ (P f )0 + D1 (Qf ) + @x2 @x2

e2 ;

completing the proof. Lemma 9 Let f : ! H be twice continuously di¤erentiable on an open 3 subset of R . Then k @P f x2 @x2 k @Qf k + 4 (Qf ) + 2 Qf e2 x2 @x2 x2 0 @f e2 f e2 + f = 4f k +k ; x2 @x2 2x22

M k Mk f = Mk M k f = 4 (P f )

where 4 is the Laplacian in R3 . 8

Proof. Using Lemma 8 we obtain k Q (Q0 f ) x2 @P 0 f = D1 Qf + : @x2

QMk f = QDf +

Hence we have M k Mk f = DDf + kD

Q0 f x2

k x2

(D1 Qf )0 +

@P f @x2

:

Since D

Q0 f x2

2 1 X @Q0 f e2 Q0 f e2 + x2 i=1 @xi x22

1 @Q0 f = x2 @x0

(D2 Qf )0 Qf + 2 e2 = x2 x2

1 @Qf e2 ; x2 @x2

we have M k Mk f = 4P f

k @P f + 4Qf x2 @x2

Hence we have Mk M k f = 4f

k

k @Qf Qf +k 2 x2 @x2 x2

e2 :

@f Qf e2 +k 2 : x2 @x2 x2

Using Lemma 2 we conclude Mk M k f = 4f

k

@f e2 f 0 e2 + f +k : x2 @x2 2x22

@f From the de…nitions we note that Mk f + M k f = 2 @x . Hence we obtain 0

@f @Mk f =2 @x0 @x0 = Mk + Mk (Mk f ) = M k Mk f + Mk2 f;

Mk M f + Mk2 f = 2Mk

which implies Mk M k f = Mk Mk f , completing the proof De…nition 10 Let f : ! H be twice continuously di¤erentiable on an open subset of R3 . If M k Mk f = 0 for any x2 6= 0, then f is called k-hyperbolic harmonic. If f is 1-hyperbolic harmonic, then f is called hyperbolic harmonic. 9

Proposition 11 If f : ! H is k-hyperholomorphic, then P f and Qf e2 are k hyperbolic harmonic and the function h : ! R de…ned by ( Qf (x) ; if x2 6= 0; x2 h (x) = @Qf (x) ; if x2 = 0; @x2 is (k

2)-hyperbolic harmonic on

.

Proof. If x2 6= 0, then 4h

k

2 @h 4Qf = x2 @x2 x2

Using Lemma 9 we note that 4h Corollary 12 Let f : f is real analytic.

k 2 @h x2 @x2

k@Qf kQf + 3 : 2 x2 @x2 x2 = 0.

! H be a hyperholomorphic function

R3 , then

The equation k @g =0 x2 @x2 is the Laplace-Beltrami equation associated with the hyperbolic metric 4g

ds2 = x2 2k dx20 + dx21 + dx22 : Proposition 13 The function xm is hyperholomorphic for all m 2 Z. This result may be proved using the following results. Lemma 14 An element x 2 H is a paravector if and only if 2 X

ei xei =

x0 :

(6)

i=0

Proof. It is easy to see that the equation (6) holds for all ei with i = 0; 1; 2. Using the linearity we infer that it holds for all paravectors. Conversely, calculate 2 X ei e12 ei = e12 : i=0

Hence comparing the components of the left and right side of the equality (6) we note that the equality (6) implies that x has to be a paravector. 10

Theorem 15 Let be an open subset of R3 and f : ! H be hyperholomorphic. Then the product f (x) x is hyperholomorphic if and only if f is an H-solution. Proof. Assume that f : by Lemma 1 we have

! H and f (x) x are hyperholomorphic. Then

M1 (f x) = (Df ) x +

2 X

ei f ei +

i=0

= (Df ) x +

Q0 (f x) x3

X Q0 f x + f0 + ei f ei : x3 i=0 2

Hence we obtain 0

M1 (f x) = (M1 f ) x + f +

2 X

ei f ei :

(7)

i=0

Using the assumptions we infer 0 = M1 (f x) =

2 X

ei f ei + f 0 :

i=0

By virtue of Lemma 14 we …nd out that f is paravector valued. The converse statement is proved similarly. Theorem 16 Let be an open subset of R3 and F : ! H be hyperholomorphic. Then the function f (x) = F (x) x 1 is hyperholomorphic in nf0g if and only if it is paravector valued. Proof. Assume that F : ! H and f (x) = F (x) x 1 are hyperholomorphic. Since F (x) = (F (x) x 1 ) x we obtain from the preceding theorem that (F (x) x 1 ) is paravector valued. On the other hand, assume that f is paravector valued and F is hyperholomorphic. Then by Lemma 1 and Lemma 14, we conclude 0 = M1 F = (M1 f ) x +

2 X

ej f ej + f 0 = (M1 f ) x

j=0

and therefore f is also hyperholomorphic. Corollary 17 The function x

m

is an H-solution. 11

Proposition 18 The set of hyperholomorphic functions in an open subset of R3 is a right complex vector space. Proof. Let z 2 C. Using (1) we notice that M1 (f z) = (Df ) z +

Q0 (f z) = (M1 f ) z; x2

implying the assertion. Lemma 19 Let be an open subset of R3 and f : @f Then @x is hyperholomorphic for l = 0; 1: l

! H be k hyperholomorphic.

Proposition 20 Let be an open subset of R3 and f : ! C complex monogenic (and therefore Qf = 0). Then f is k-hyperholomorphic. Moreover the function F de…ned by F (x) = xk2 f (x) e2 is k-hyperholomorphic. Proposition 21 Let be an open subset of R3 and f : morphic. Then f is holomorphic on \ C.

! H be hyperholo-

Theorem 22 Let be an open subset of R3 and f : ! H be a mapping with continuous partial derivatives. The equation Mk f = 0 is equivalent with the system of equations D1 (P f ) D1 (Qf ) + where D1 =

@Q0 f Q0 f + k @x2 x2 @P 0 (f ) = 0; @x2

= 0;

(8)

P1

@f i=0 ei @xi :

The corresponding system for monogenic functions is the following. Theorem 23 Let be an open subset of R3 and f : ! H be a mapping with continuous partial derivatives. The equation Df = 0 is equivalent with the system of equations D1 (P f ) D1 (Qf ) +

12

@Q0 f @x2 @P 0 f @x2

= 0; = 0:

Theorem 24 Let be an open subset of R3 and f : ! H three times continuously di¤erentiable. Then f is k-hyperholomorphic (resp. monogenic) if and only if locally f = DH for some complex valued k hyperbolic harmonic (resp. harmonic) function. Proof. Assume that a mapping H : B (x; r) ! C is hyperbolic harmonic and f = DH. Since DD = 4 we have x2 Df = x2 4 H. The equality @H = Q0 f follows from @x2 Qf e2 = QDH =

e2

@H : @x2

Hence Mk f = 0 and therefore f is k-hyperholomorphic. Conversely assume that f : ! H is k-hyperholomorphic. Set P f = f0 + f1 e1 . Let B (a; r) be a ball in R3 centered at a = (a0 ; a1 ; a2 ) satisfying B (a; r) . De…ne a mapping p : R3 ! R2 by p (x0 ; x1 ; x2 ) = (x0 ; x1 ). Let si : p (B (a; r) \ fx j x2 = a2 g) ! R be a three times continuously di¤erentiable solution of the Poisson equation 4si (p (x)) = fi (p (x) ; a2 ) for i = 0; 1 and a ball B ((a0 ; a1 ; a2 ) ; r). Set s = s0 + s1 e1 and de…ne a mapping H : B (a; r) ! C by Z x2 H (x) = Q0 f (~ x; t) dt + D1 s (p (x)) . a2

Then we have by Proposition 22 Z x2 0 DH (x) = e2 Q f (x) D1 Q0 f (p (x) ; t) dt + D1 D1 s (p (x)) Z ax2 2 @P f = Qf (x) e2 + (p (x) ; t) dt + (P f ) (p (x) ; a2 ) @x2 a2 = f (x) : Since the image space of H is C we note that Qf e2 =

e2

@H : @x2

Using the assumption f is k-hyperholomorphic we obtain 0 = x2 Df + kQ0 f = x2 DDH

k

@H : @x2

Hence the mapping H is hyperbolic harmonic, completing the proof. 13

Theorem 25 A mapping f is k-hyperholomorphic on a domain and only if there exist locally Hk -solutions gi such that

R3 if

f = g0 + g1 e1 : Proof. Assume that f is k-hyperholomorphic on a ball B (a; r) R3 . Applying the preceding Theorem we …nd a k-hyperbolic harmonic mapping H from B (a; r) into C satisfying f = DH. Denote H = h0 + h1 e1 for real k-hyperbolic harmonic functions hi . Then the mapping gi = Dhi is para vector valued and therefore an Hk -solution. Clearly we have f = g0 + g1 e1 . If f is paravector valued the following result holds. Proposition 26 Let solution, then

be an open subset of R3 . If f :

! H is an Hk -

@f = D (Re f ) : @x0 Proof. Assume that f : valued, we have

! H is an Hk -solution. Since f is paravector

2D (Re f ) = Df + Df 0 = 2

@f @x0

Df + (Df )0 :

Since f is an Hk -solution, we obtain 2D (Re f ) = 2

Q0 f @f +k @x0 x2

k

Qf @f =2 ; x2 @x0

completing the proof. Corollary 27 If f = xn+1 we obtain (n + 1) xn = D Re xn+1 : A relation between harmonic and hyperholomorphic functions is given next. Theorem 28 Let be an open subset of R3 . Let h : ! C be k-hyperbolic harmonic. Then there exists locally a k +2-hyperholomorphic function f with values in H satisfying Qf (x) = x2 h (x) :

14

Proof. Let h : ! C be k-hyperbolic harmonic. Assume that a 2 and B (a; r) is a ball satisfying B (a; r) . De…ne a mapping p : R3 ! R2 by p (x0 ; x1 ; x2 ) = (x0 ; x1 ). We denote D1 =

1 X

ei

i=0

@ : @xi

Find a mapping from B (a; r) \ fx j x2 = a2 g into C satisfying D1 g (p (x)) = a2

@h (p (x) ; a2 ) @x2

(k + 1) h (p (x) ; a2 ) :

Set Qf (x) = x2 h0 (x) and P f (x) =

Z

x2

tD1 h (p (x) ; t) dt + g (p (x)) :

a2

We verify that the function f = P f + Qf e2 satis…es the equations of (8). Using the de…nition of P f and Qf we obtain @P 0 f = @x2

0

x2 D1 h (x) =

0

D1 Q0 f (x) =

D1 Qf:

Hence the second equation of (8) holds. Since h is k-hyperbolic harmonic, the mapping Q0 f = x2 h satis…es the equation 4Q0 f

(k + 2)

and therefore

@h =0 @x2

@h @ 2 Q0 f 41 Q f = (k + 2) : @x2 @x22 Substituting this into the term D1 P f of the left side of the …rst equation of the system (8) and using the equality Q0 f = x2 h we infer Z x2 D1 P f (x) = 41 Q0 f dt + D1 g (p (x)) a Z 2x2 @h @ 2 Q0 f = (k + 2) (p (x) ; t) + (p (x) ; t) dt + D1 g (p (x)) @x2 @x22 a2 @Q0 f @Q0 f = (k + 2) h (x) + (k + 2) h (p (x) ; a2 ) + (x) (p (x) ; a2 ) @x2 @x2 @h + a2 (p (x) ; a2 ) (k + 1) h (p (x) ; a2 ) @x2 @Q0 f = (k + 2) h (x) + (x) : @x2 0

15

Hence we have x 2 D1 P f

@Q0 f @x2

+ (k + 2) Q0 f =

(k + 2) x2 h (x) + (k + 2) Q0 f = 0:

Theorem 29 Let f : ! H be a k-hyperholomorphic function and be an open subset of R3 . Then 4f is k 2-hyperholomorphic in . Conversely, if g : ! H is k 2-hyperholomorphic in there exists locally a k-hyperholomorphic function f with values in H such that 4 (f ) = g: Proof. De…ne a function h : ( h (x) =

! C by Q0 f (x) ; x2 0 @Q f (x) ; @x2

if x2 6= 0;

if x2 = 0:

Then h is k 2-hyperbolic harmonic by Proposition 11 and the equation Df + kh (x) = 0 implies the property 4f =

k Dh:

Since h is k 2-hyperbolic the function k Dh is k 2-hyperholomorphic. Conversely, assume that g : ! H is k 2-hyperholomorphic. Then there exists locally a k 2-hyperbolic harmonic function H with values in C satisfying kDH = g: The functions H being complex valued k 2-hyperbolic harmonic yields that there exists a k-hyperholomorphic functions f such that Q0 f = x2 H: Hence we have Df + k

Q0 f = 0; x2

which implies 4f = DDf =

kDH = g:

Using the previous result inductively we obtain a correspondence between k + 2s-hyperholomorphic functions and k-hyperholomorphic functions.

16

Theorem 30 Let be an open subset of R3 . Let f : ! H be k-hyperholomorphic. Then there exists locally a k + 2s-hyperholomorphic function g : ! H s such that f = 4 g. Conversely, if a mapping g : ! H is k + 2ss hyperholomorphic, then f = 4 g is k-hyperholomorphic: Proof. Set g0 = f . Applying the previous theorem we …nd locally k +2jhyperholomorphic functions gj satisfying the properties gj 1 = 4gj . Then the function gs is k + 2s-hyperholomorphic and has the desired property f = 4s gs . Conversely, assume that a function g : ! H is k+2s-hyperholomorphic. Then we have D 4s g = 4s Dg =

Q0 g x2

(k + 2s) 4s

(9)

:

Since s

s

4 Qg = x2 4

Qg x2

+ 2s

@ 4s

Qg x2

1

@x2

;

we infer from (9) s

s

0

x2 D 4 g + k 4 Q g = Using (9) we obtain 0 = x2 4

s

2sx2 4 Qg x2

Q0 g x2

(k + 2s

+ 2ks

2)

@

@ 4s

Qg x2

@x2

@g x2 4 g + 2 @x , we deduce inductively 2

j

0 = x2 4

Qg x2

(k + 2s

2j)

@ 4j

1

1

@x2

Qg x2

:

. Since 4 (x2 g) = Qg x2

@x2

for any j = 1; :::; s: Consequently, we have x2 D 4s g + k 4s Q0 g = 0:

Corollary 31 Let be an open subset of R3 . Let g : ! H be monogenic. Then there exists locally a 2s-hyperholomorphic function f such that g = 4s f . Conversely, if f is 2s-hyperholomorphic, then 4s g is monogenic.

17

Theorem 32 Let be an open subset of R3 . Let g : continuously di¤erentiable solution of the equation 4g

k

! H be in…nitely

@g g + k 2 = 0; x2 @x2 x2

(10)

then the function M k (ge2 ) =

@g 0 @x2

is k-hyperholomorphic in

k

g0 + D1 ge2 = x2

@g 0 + x2 4 g 0 + D1 ge2 @x2

:

Proof. Using the proof of Proposition 11 we note that the function h de…ned by ( g(x) ; if x2 6= 0; x2 h (x) = @g (x) ; if x2 = 0; @x2 is k

2-hyperbolic harmonic. Using Lemma 9 we infer 1 @P f (ge2 ) x2 @x2 k k @Qf (ge2 ) + 4 (Q (ge2 )) + 2 Qf (ge2 ) e2 x2 @x2 x2 k @g k + 2 g e2 = 0: = 4g x2 @x2 x2

Mk Mk g = 4 (P (ge2 ))

Hence we have Mk g is hyperholomorphic. Theorem 33 Let Then the function g (x) =

be an open subset of R3 . Let f :

! H be k-hyperholomorphic.

@f Qf (x) @f (x) e2 + k = (x) e2 @x2 x2 @x2

Df 0

is hyperholomorphic in . Moreover, if f is left and right k-hyperholomorphic, then 1 X @f @f Qf (x) (x) e2 = ei k : @x2 @xi x2 i=0 Proof. Assume that f : ! H be k-hyperholomorphic and a 2 . Then there exists a complex-valued k-hyperbolic harmonic function h : B (a; r) ! 18

C for some r > 0 such that f = Dh on B (a; r). This implies that Qf = Q Dh = Q D2 h + Q e2 =

@h @x2

@h0 : @x2

@h satis…es the equation (10) and applying the preceding Hence the function @x 2 result we infer that the function

Mk

@h e2 @x2

@h0 =D kx2 @x2 Qf (x) @f = (x) e2 + k @x2 x2 @h e2 @x2

1

is k-hyperholomorphic. Polyharmonic functions can be presented in terms of hyperholomorphic functions. Theorem 34 Let be an open subset of R3 . If a function H : ! C satis…es 4k H = 0, then there exist locally 2i-hyperholomorphic functions P i2 gij : ! H (j = 1; 2) such that ki=1 gi1 + @g e = H: @x2 2

Proof. Let k = 1 and assume that a mapping H : ! C satis…es 4H = 0. Applying Theorem 28 we may …nd a 2-hyperholomorphic mapping g12 such that 2Qg12 = x2 H. By virtue of Theorem 33 we deduce that the 0 12 function g11 = @g e Dg12 is 2-hyperholomorphic. Moreover we have @x2 2 g11

@g12 e2 = @x2

0 Dg12 =2

Qg12 = H: x2

Hence the assertion holds for k = 1. Next assume that the assertion holds for some k. Assume that a mapping H : ! C satis…es 4k+1 H = 0. Using the …rst part of the proof we may 12 …nd 2-hyperholomorphic functions f1j such that f11 + @f e = 4k H. Ap@x2 2 plying Theorem 30 we may …nd 2 (k + 1)-hyperholomorphic functions g(k+1)j satisfying 4k g(k+1)j = f1j for j = 1; 2. Hence we have 4k g(k+1)j +

@g(k+1)2 e2 @x2

H

= 0:

Applying the induction hypothesis we obtain the assertion. 19

Lastly we compare hyperholomorphic functions with 4k -monogenic functions, that is functions satisfying D 4k f = 0. In complex Cli¤ord algebras they were considered by J. Ryan in [27]. In real Cli¤ord algebras they are studied by G. Laville and I. Ramadano¤ in [20]. They have also been researched by M. Sce in [28]. Theorem 35 If a function f : ! H satisfy D 4k f = 0, then there exist locally 2i-hyperholomorphic functions gij : ! H (j = 1; :::; 4) such that k X

gi1 + gi2 e2 +

i=1

@ (gi3 + gi4 e2 ) @x2

= f:

Proof. Assume that D 4k f = 0. Then 4k f is monogenic and by Corollary 31 there exists a 2k-hyperholomorphic function such that 4k f = 4k g. Hence 4k (f g) = 0 and f g = H1 + H2 e2 ; where H1 and H2 are C-valued functions satisfying 4k Hi = 0. By virtue of the preceding theorem there exist locally 2i-hyperholomorphic functions hij : ! H (j = 1; 2; 3; 4) Pk Pk @hi2 @hi4 such that = H1 and = H2 . We i=1 hi3 + @x2 e2 i=1 hi1 + @x2 e2 compute that f =g+

k X

hi1 + hi3 e2 +

i=1

@ ( hi4 + hi2 e2 ) ; @x2

yielding to the desired expression. Theorem 36 If h is real hyperbolic harmonic in B (0; r) then there exists a hyperholomorphic function f such that Re f = h. Moreover this function is given by Z 1

Dh (sx) x ds + h (0) .

f (x) =

0

Proof. First we compute that f is hyperholomorphic. Denote sx, s 2 R. Then Z 1 M1 f (x) = M1 Dh (sx) x ds:

x

(s) =

0

Since h is real the function x ! M 1 h (sx) = Dh (sx) is paravector-valued. Hence by (7) we have Z 1 M1 f (x) = M1 Dh (sx) x ds 0

20

and further we note that M1 f (x) =

Z

1

s M1 Dh (sx) x ds = 0:

0

Moreover we deduce Z 1 Dh (sx) x ds + h (0) Re 0

Lemma 37 If f :

Z

1

X

@h (sx) ds + h (0) @xi 0 Z 1 @ (h x ) = (s) ds + h (0) @s 0 = h (x) : =

xi

! H is twice continuously di¤erentiable, then 4 (xf ) = 2Df + x4f .

Proof. Note …rst that D (xf ) = 3f +

2 X i=0

ei x

@f : @xi

Since 4 = DD, we infer 4 (xf ) = 3Df +

2 X 2 X j=0 i=0

= 2Df + x4f .

Lemma 38 If f :

XX @f @f ej ei ej + ej ei x @xi j=0 i=0 @xi @xj 2

2

! H is twice continuously di¤erentiable, then Mk M k (xf ) = 2Mk f + xMk M k f:

Proof. We just calculate Mk M k (xf ) = 4 (xf )

k

@ (xf ) Q (xf ) e2 +k : x2 @x2 x22

The product rule of Q leads to Q (xf ) = xQf + x2 f 0 : 21

Applying the preceding Lemma we obtain M M f (x) = 2 (Df ) (x) + x4f

k

e2 f xQf e2 f 0 e2 +k + k x2 x22 x2 f 0 e2 e2 f Q0 f 2k +k : x2 x2

x@f x2 @x2

= 2 (M f ) (x) + xMk M k f

k

Hence the result holds by Lemma 2. Theorem 39 The set of hyperbolic harmonic functions is left and right complex vector space. Theorem 40 Let f : ! H be twice continuously di¤erentiable. Then f is k-hyperholomorphic if and only if f and xf are k-hyperbolic harmonic functions. Theorem 41 If fn : ! H are hyperholomorphic functions and the sequence (fn )n is converging locally uniformly, then the limit function lim fn is hyperholomorphic. P xn Example 42 The exponential function ex = 1 n=0 n! is hyperholomorphic.

3

Homogenous hyperholomorphic polynomials

De…nition 43 The homogeneous polynomials Lm are de…ned for any k 2 N0 and a non-negative integer m by Lkm =

1 @ k xm+k : ! @xk1

Theorem 44 The set fLm j j j mg is a basis of the right C-module of homogeneous hyperholomorphic polynomials of degree m. Proof. See [10]. Theorem 45 Assume that f 2 C 1 ( ) is hyperholomorphic in a neighborhood of a point x = (x0 ; x1 ; 0). Then there exist constants bk ( ) 2 C such that 0 1 1 k X X @ f= Lk bk ( )A k=0

j j=0

22

Proof. Assume that f 2 C 1 ( ) is hyperholomorphic in a neighborhood of a point x = (x0 ; x1 ; 0). If T (y) = y + a for a 2 R3 with the last coordinate a2 = 0 then f T is also hyperholomorphic. Hence we may assume that x = 0. Since f is hyperholomorphic, f is real analytic and therefore admits the presentation X f (y) = a( )y : 2N30

in some neighborhood Br (0). Applying M we obtain 0 1 1 X X 0 = M f (y) = M@ a( )y A: k=0

Since M

P

j j=k

a( )y

j j=k

is a homogeneous polynomial of degree k we infer 0 1 X M@ a ( ) y A = 0: j j=k

P

This implies that j j=k a ( ) y is hyperholomorphic. Applying Theorem 44 we obtain the result. Theorem 46 The dimension of right complex vector space generated by hyperholomorphic homogeneous polynomials of degree m is m + 1. P Proof. Denote p (x) = j j=m a ( ) x . Then we have X P p (x) = Pa( )x ; j j=m

Qp (x) =

X

Qa ( ) x :

j j=m

Using Theorem 22 we infer that M p = 0 is equivalent with the following system 0 x2 D1 (P p (x)) @Q@xp(x) + Q0 p (x) = 0; 2 D1 (Qp (x)) +

@P 0 (p) @x2

= 0:

Hence we obtain the system 1 X X

i ei P a (

)x

"i +"2

j j=m i=0

X

0 2Q a (

)x +

j j=m

X

1 X

i ei Qa (

j j=m i=0

Q0 a ( ) x = 0;

j j=m

)x

"i

+

X

j j=m

23

X iP

0

a( )x

"2

= 0:

This is equivalent with 1 X X

(

i

+ 1) ei P a ( + "i ) x

+"2

+ (1

2)

j j=m 1 i=0 1 X X

(

X

Q0 a ( ) x = 0;

j j=m

i + 1) ei Qa ( + "i ) x +

j j=m 1 i=0

X

(

2

+ 1) P 0 a ( + "2 ) x = 0;

j j=m 1

from which we obtain the system 1 X

(

i

0 2Q a (

+ "2 ) = 0; if j j = m

1;

+ 1) P 0 a ( + "2 ) = 0; if j j = m

1;

+ 1) ei P a ( + "i ) +

i=0

1 X i=0

(

i

+ 1) ei Qa ( + "i ) + (

2

Qa ( ) = 0, if

2

= 0 and j j = m:

Hence we may choose freely in C the variables Qa ( ) with 2 = 1 and they determine all the variables P a ( ) with 2 2 even and Qa ( ) with 2 odd. Using the second equation we infer that the property Qa ( ) = 0, if 2 = 0 and j j = m, implies that P a ( ) = 0, if 2 = 1. If j j = m 1 and 2 = 1, then the …rst equation implies that Q0 a ( + "2 ) = 0 and further Qa ( ) = 0 for any with j j = m and 2 even. If now j j = m 1 and 2 = 0, then we can solve P a ( + "1 ) as follows ! 1 X e1 ( i + 1) ei P a ( + "i ) : P a ( + "1 ) = 1+1 i=0 This means that the free variables P a ( ) with j j = m and 2 = 1 = 0 determine all other variables P a ( ) with j j = m and 2 = 0. Consequently the dimension is m + 1: Theorem 47 Denote by Mm the set of hyperholomorphic homogeneous polynomials of degree m. Then then the operator M 1 maps Mm+1 onto Mm . Proof. We note that the operator M 1 maps Mm+1 into Mm , since MMp = MMp = 0 for any hyperholomorphic p. Using the linear algebra we know that dim M 1 (Mm+1 ) = dim Mm+1 = ker M1 = dim Mm+1 24

ker M 1 :

Assume that p 2 ker M 1 . Then M 1 p = M1 p = 0 and therefore 2

@p = M 1 p + M1 p = 0: @x0

Hence p is a hyperholomorphic homogenous polynomial of degree m + 1 independent of x0 . By virtue of the previous proof we notice that then p P = j j=m+1 a ( ) x is determined by the coe¢ cients a ( ) with 2 N30 j j j = m + 1;

2

0

= 0 and

2

=1 :

Hence we have dim ker M1 = 1; which implies dim M 1 (Mm+1 ) = m + 2

1 = dim Mm :

This means that the operator M 1 maps Mm+1 onto Mm :

4

Möbius transformations and isometries for the upper half space model

In C Möbius transformations has the form T (z) =

az + b ; cz + d

where a; b; c; d 2 C and ad bc 6= 0: The group of Möbius transformations is generated by all similarities and the inverse z 1 transformation. Möbius transformation’s are conformal (angle-preserving and sense-preserving). It is also well known that any Möbius transformation in C can be written as a product of elementary maps: z z z z

!z+a (a 2 C) : ! a ( 2 R) : ! ei' z (' 2 R) : z 1 : ! z = jzj2

The transformation Ta z =

z 1

25

a az

translation dilatation rotation re‡ection

2

2

maps the unit disc jzj < 1 onto itself. Harmonic functions @@xf2 + @@yf2 = 0 are preserved by Möbius transformations. Moreover the Poisson formula may be obtained using Ta 1 v (a) = 2 where the Poisson kernel

Z

2

v ei

0

1 jaj2 j1 azei j2

1 j1

jaj2 d ; azei j2

is the Jacobian of the transformation Ta .

c (R3 )is the group generated by all De…nition 48 The full Möbius group M similarities and the inverse transformation Fx =

(x0 ; x1 ; x2 ) : jxj2

b 3 = R3 [ f1g. The Möbius group M R3 is the subin the extended space R + 3 c c (R3 )(that group of M (R ) containing sense preserving transformations of M c (R3 ) and j@T j 0;where j@T j is the Jacobian). is T 2 M c (R3 ) there exists a positive number Lemma 49 For any T 2 M that @T (x) = O (3) ; (x)

(x) such

where @T (x) is the derivative matrix of T . In other words all Möbius transformations are conformal. Of special interest is the subgroup M (R3+ ) of those Möbius transformations which map the upper half space bijectively onto itself. Möbius transformations are the composition of the folowing mappings translations dilatations orthogonal maps inversion

x x x I:x

! x+c ! x ! Ux x ! kxk 2

(c 2 R3 ) ( 2 R) (U x; U y) = (x; y); U 1 = U T I(0) = 1; I(1) = 0:

Subgroup M (R3+ ) of all those maps in M (R3 ) which the upper half space R3+ bijectively onto itself. Möbius transformations can be presented as T (x) = (ax + b) (cx + d) 26

1

where a; b; c; d 2 H. More speci…cally Möbius transformations in R3 are transformations T mapping paravectors to paravectors. The hyperbolic metric in R3+ = f(x; y; t) 2 R3 ; t > 0g is induced by Z dx2 + dy 2 + dt2 2 ds = ; l( ) = ds: t2 Denote hyperbolic distance by dh . The geodesics are half-circles or half-lines in R3+ , which are orthogonal to the x-y-plane. The angles in this model coincide with the euclidean ones. Lemma 50 If P = (x0 ; x1 ; x2 ) 2 R3+ , Q = (y0 ; y1 ; y2 ) 2 R3+ , then (P; Q) =

(x0

y0 )2 + (x1 y1 )2 + x22 + y22 2x2 y2

is a point-pair invariant under M (R3+ ) i.e. ( P; Q) = (P; Q);

2 M (R3+ ):

Lemma 51 M (R3+ ) is ”double transitive”, i.e. to every P; Q; P 0 ; Q0 2 R3+ with dh (P; Q) = dh (P 0 ; Q0 ) there is a 2 M (R3+ ) such that (P ) = P 0 and (Q) = Q0 . Lemma 52 The distance formula dh (P; Q) for P = (x0 ; x1 ; x2 ) 2 R3+ , Q = (y0 ; y1 ; y2 ) 2 R3+ is the following dh (P; Q) = arcosh (P; Q): Proof. Indeed, if P = (0; 0; 1) and Q = (0; 0; ) with Z dt dh (P; Q) = = ln t 1 1+ 2 1 1 (P; Q) = = ( + ): 2 2

> 1, then

We obtain cosh dh (P; Q) = cosh ln

1 = (eln + e 2

ln

)

1 1 = ( + ) = (P; Q): 2 Using the previous lemma we obtain the general distance formula for any points in the upper half space. 27

Theorem 53 The group Iso+ (R3+ ) of sense- preserving isometries of the upper half space model agrees with the Möbius group M (R3+ ) and hence is isomorphic to P SL(2; C): . Lemma 54 The n.e. sphere with n.e. center a = a0 + e1 a1 + e2 a2 2 R3+ and n.e. radius Rh , i.e. the set fz 2 R3+ j dh (z; a) = Rh g is an euclidean sphere with euclidean center b = a0 + e1 a1 + e2 a2 cosh Rh and euclidean radius q Re = a2 cosh Rh2

1 = a2 sinh Rh :

Moreover, if dh (z; a) = Rh , then z z

a Rh = tanh : b a 2

Proof. We consider the n.e. sphere with n.e. center a = a0 +e1 a1 +e2 a2 2 R3+ and n.e. radius Rk : Let z = x0 + e1 x1 + e2 x2 and dh (z; a) = Rh . Then (z; a) =

(x0

a0 )2 + (x1 a1 )2 + x22 + a22 . 2x2 a2

Denote (z; a) = A. Then A = cosh Rh Then

(x0 (x0

a0 )2 + (x1 a1 )2 + x22 + a22 = 2x2 a2 A a0 )2 + (x1 a1 )2 + (x2 a2 )2 = 2x2 a2 (A 1) a0 )2 + (x1 a1 )2 + (x2 + a2 )2 = 2x2 a2 (A + 1) :

(x0

a0 )2 + (x1

(x0

Hence and

jz jz

a1 )2 + (x2

Aa2 )2 = A2

1 a22

aj2 A 1 cosh Rh 1 Rh = = = tanh2 : 2 b aj A+1 cosh Rh + 1 2 28

5

Laplace-Beltrami operator R3 has the form

Riemannian metric on an open set ds2 =

2 X

gij dxi dxj ;

i;j=0

where (gij ) denotes a family of positive de…nite matrices, which are second order covariant tensors, i.e., if x = (x0 ; x1 ; x2 ) are the old coordinates and x = (x0 ; x1 ; x2 ) = h (x0 ; x1 ; x2 ) the new coordinates in terms of the change xi = hi (x0 ; x1 ; x2 ) then ! X @xi @xj 0 0 t ghk = h 1 (ghk )n n h 1 : (g ij )3 3 = @x @x h k h;k 3 3

We consider the conformal metric ds2 =

2 X

gii dx2i ;

i=0

where gij =

2

(x)

ij :

The mapping h : R3 ! R3 is conformal when h0 (x) 2 O (3) jh0 (x)j where O (3) is the set orthogonal transformations in R3 . Möbius transformations are conformal. The determinant g = det(gik ) de…nes the volume: if K is compact Z p V (K) = g dx0 dx1 dx2 : K

The divergence of a (contravariant) vector …eld ! v = (v0 ; v1 ; v2 ) on by 2 1 X @ p ! ( g vi ): div v = p g m=0 @xi 29

is given

The (contravariant) gradient of a function f : matrix (g lm ) of (gik ) as follows grad f = (

2 X

g

0m

m=0

! R is de…ned by the inverse

2 X @f @f ; : ::; g 2m ) @xm @xm m=0

The Laplace-Beltrami operator is the combination of these two operators = div (grad f )

LB f

2 1 X @ p ik @f ): =p ( gg @xk g i;k=0 @xi

If we compare the Euclidean and the hyperbolic case we obtain the following table. Euclidean case Hyperbolic case (conformal) P2 @vi P @vi 3 P2 @ ! div v = i=0 @xi divh ! v = 2i=0 @x + i=0 @xi vi i @f @f @f @f @f 1 @f grad f = ( @x0 ; @x1 ; @x2 ) gradh f = 2 ( @x0 ; @x1 ; @x ) P2 @ 2 f P2 @2 @f 1 1 f = i=0 @x2 LB f = 2 ( f + i=0 @xi @xi ) i

6

Integral formulas

We recall an important integral formula. Lemma 55 Let R3 be open and f; g : ! H continuously di¤erentiable functions. Let K be a compact set with smooth boundary K and = 0 + 1 e1 + 2 e2 its outer unit normal …eld. Then Z Z f gd = (Dr f g + f Dl g) dx @K

K

where dx denotes the volume element in R3 and d the surface element. Proof. See [3, 9.2 Proposition, p. 52]. Lemma 56 Let R3 be open and f; g : ! H continuously di¤erentiable functions. Let K be a compact set with smooth boundary K and = + e + e its outer unit normal …eld. Then 0 1 1 2 2 Z Z f gd = Dr f g + f Dl g dx @K

K

where dx denotes the volume element in R3 and d the surface element. 30

Proof. Using the preceding Lemma we obtain Z Z Z f gd = g fd = Dr gf + gDl f dx @K @K K Z Z Dr f g + f Dl g dx: Dr gf + gDl f dx = = K

K

Dirac operator and the modi…ed Dirac operator are related as follows. Lemma 57 Let f be twice continuously di¤erentiable, then P 0f Dl k k+1 e2 ; x2 Pf Dr k k+1 e2 ; x2 f P (Mk f ) = xk2 P D ; xk2 Q (Mk f ) = Q (Df ) ; Qf e2 Mk Qf e2 D = : k x2 xk2 f xk2 f xk2

Mk f = k x2 M rf = kk x2

(11) (12) (13) (14) (15)

Proof. We just compute 0

f xk2

D

Df e2 f Df Qf = k k k+1 = k + k k+1 x2 x2 x2 x2 0 Mk f P f e2 = k k k+1 ; x2 x2

k

P 0 f e2 xk+1 2

which implies the …rst equality. If we take P from the both sides of this equality we obtain the third equality. The fourth equality follows directly from the de…nition of Mk . The last equality follows from Qf e2 xk2

D Let K

=

D (Qf e2 ) xk2

k

e2 Qf e2 Mk (Qf e2 ) = : k+1 xk2 x2

be an open subset of R3 n fx2 = 0g and K is an 3-chain satisfying . De…ne a 2-form by d

k

=

dx1 ^ dx2

e1 dx0 ^ dx2 + e2 dx0 ^ dx1 xk2

and a 3-form by dmk =

1 dx0 ^ dx1 ^ dx2 : xk2

31

Lemma 58 If f 2 C 1 ( ; H), then Z Z b (d k + d 0 ) f + (d bk d b0 ) f = 2 (P (d k f ) + Q (d 0 f ) e2 ) @K Z@K =2 (P (Mk f ) dmk + Q (Mk f ) e2 dm0 ) K Z d Mk f + M = k f dmk K Z d + Mk f M k f dm0 : K

Proof. Recall …rst that Z Z gd 0 f = ((Dr g) f + gDl f ) dm0 ; @K

(16)

K

see [3, 9.2 Proposition, p. 52].Using this and the previous Lemma we obtain Z Z Z f Mk f P 0 f e2 d kf = Dl dm0 = k k+1 dm0 k xk2 x2 @K K K x2 Z 0 P f e2 = Mk f k dmk x2 K Z Z d 0f = Dl f dm0 : @K

K

Hence we have Z

Z

P (d k f ) =

Z@K

Q (d 0 f ) =

@K

Hence Z Z P (d k f ) + @K

Q (d 0 f ) e2 =

@K

P (Mk f ) dmk

ZK

Q (Mk f ) dm0 :

K

Z

P (Mk f ) dmk +

K

Z

Q (Mk f ) e2 dm0 :

K

Using the equalities (2) we obtain the last equality. Theorem 59 A function f 2 C 1 ( ; H) is k-hyperholomorphic in an open subset R3 n fx2 = 0g if and only if the property Z Z P (d k f ) + Q (d 0 f ) e2 = 0 @K

@K

32

or equivalently Z

@K

(d

k

+ d 0 ) f + (d bk

holds for any 3-chain K satisfying K

:

d b0 ) fb = 0

Lemma 60 Let be an open subset of R3 n fx2 = 0g and K an 3-chain satisfying K . If f; g 2 C 1 ( ; H), then Z Z kP ge2 gd k f = Mkr g P f + gMk f + (Dr g) Qf e2 dmk : x 2 @K K Proof. Using (16) and Lemma 57 we obtain Z Z Z g Qf e2 gd k f = d 0P f + gd 0 k k x x2 @K @K Z@K 2 g g Qf e2 Dr gQf e2 = Dr P f + k Dl P f + gDr + dm0 k k x2 x2 x2 xk2 K Z g Mk (Qf e2 ) Dr gQf e2 g = Dr P f + k Dl P f + g + dm0 k x2 x2 xk2 xk2 K Z kP ge2 = Mkr g P f + gMk f + Dr gQf e2 dmk : x2 K Theorem 61 Let be an open subset of R3 n fx2 = 0g and K an 3-chain . If f; g 2 C 1 ( ; H) and Dr g 2 C, then satisfying K Z Z P (gd k f ) = P (Mkr gP f + gMk f ) dmk @K

K

Proof. Using (16) and Lemma 57 we obtain Z Z Z Qf e2 g gd k f = d 0P f + gd 0 k k x x2 @K @K Z@K 2 g g Mk (Qf e2 ) Dr gQf e2 = Dr P f + k Dl P f + g + k x2 x2 xk2 xk2 K Z kP gP 0 f e2 = Mkr gP f + gMk f + Dr gQf e2 dmk : x2 K

dm0

Hereby we used that Dl (P f ) = Mkl (P f ). Since Dr g 2 C by the assumption we conclude the result. 33

Lemma 62 Let be an open subset of R3 n fx3 = 0g and K a 3-chain satisfying K . If f; g 2 C 1 ( ; H)and f is k-hyperholomorphic, then Z Z Q (gd f ) = Q (Dr gf + gDl f ) dm @K K Z =: Q (Dr g) f 0 + gQ (Dl f ) + Dr gQf + Qg (Dl f )0 Z K = Q M r k g f 0 + M r k gQf dm: K

Lemma 63 The function x2 g (x) = 2y2

(x

y) yj jx

jx

1

(x ybj

jx

yb) yj jx

1

(x y) 1 (x yb) 1 1 x = x2 e2 = 3 2D jx yj jx ybj 2 y2

!

ybj Z

1

s2

1

jx yj jx y bj

s2

!

ds

is left and right hyperholomorphic on R3 n fy; ybg for each y with y2 6= 0. (x y) 1 jx yj

Proof. Note …rst that

(x jx

2y2 Dg = x2

Since x

yb = x

yb) (x y) yj jx ybj3

1

are monogenic. Hence

(x y) (x yb) 1 + jx yj3 jx ybj

(x y) 1 (x yb) 1 + 2y2 e2 e2 : jx yj jx ybj y + 2y2 e2 and x

2y2 Dg =

Since

(x yb) 1 jx ybj

and

jx

y=x

x2 yj jx

3

ybj

yb

2y2 e2 we infer

x2 (2y2 e2 (x y)) jx yj3 jx ybj3

x2 2y2 e2 x b y x2 + + jx yj3 jx ybj3 jx yj3 jx ybj (x y) 1 (x yb) 1 + 2y2 e2 e2 : jx yj jx ybj 1

1

jx

2

yj

jx

2

ybj

=

34

jx

4x2 y2 yj2 jx

ybj2

!

and (x

yb) = e2 (x

y) e2 = e2 (b x 2y2 Dg =

4x22 y2

4x2 y2 e2 x

2y2 Dg +

3

yb

3

jx yj jx ybj 4x22 y2 + 4x2 y22 = jx yj3 jx ybj3

Q0 g x2

2x2 e2 ) we obtain

2x2 y2 e2 (x jx

+

Hence we have

yb

y)

yj3 jx jx jx

2x2 y2 e2 x ybj2

2y2 yj3 jx 2y2 yj3 jx

ybj2

ybj2

yb

:

4x22 y2 + 4x2 y22 2y2 3 3 jx yj jx ybj jx yj3 jx ybj y2 x2 (y2 + x2 ) + + 3 jx yj jx ybj jx yj jx ybj3 =

1 (y2 + x2 ) + jx 1ybj2 4x22 y2 + 4x2 y22 jx yj2 = + jx yj jx ybj jx yj3 jx ybj3 2 2 4x y2 + 4x2 y2 4x2 (y2 + x2 ) y2 = 2 = 0: jx yj3 jx ybj3

Leutwiler in [21] stated that g is a fundamental H-solution. Note that the point pair-invariance of the hyperbolic distance yields to jx yj j' (x) ' (y)j = jx ybj [ ' (x) ' (y)

for each Möbius transformation mapping the upper half space onto itself.

Theorem 64 Let be an open subset of R3+ and K a 3-chain satisfying K . If f is hyperholomorphic in and y 2 K , then ! Z y22 (x y) 1 (x yb) 1 P f (y) = P x2 e2 d 1 (x) f (x) jx yj jx ybj @K ! Z y22 (x y) 1 (x yb) 1 = P e2 d 0 (x) f (x) jx yj jx ybj @K Z y22 (x y) 1 (x yb) 1 = e2 d 0 (x) f (x) 2 @K jx yj jx ybj Z y22 (b x yb) 1 (b x y) 1 e2 d b0 (x) fb(x) : 2 @K jx yj jb x yj 35

Proof. Using Theorem 61 we obtain ! Z (x y) 1 (x yb) 1 P x2 e2 d 1f jx yj jx ybj @K Z Z = P (g (x) d 1 f ) + P (g (x) d 1 f ) @(KnBr (y))

=

Z

@Br (y)

P (g (x) d 1 f ) = P

@Br (y)

Z

(x y) 1 (x yb) 1 e2 d 0f jx yj jx ybj

@Br (y)

!

1

: 1

The preceding Lemma implies that the function g (x) = x2 (xjx y)yj e2 (xjx yb)ybj is paravector valued. Thus we have g (x) = g (x) = x2 Hence Z

P

@K

Z

=P

(x y) 1 (x yb) 1 x2 e2 d jx yj jx ybj

1

!

(x) f (x)

(x yb) 1 (x y) 1 e2 d jx ybj jx yj

@Br (y)

Z

=P

(x yb) 1 (x y) 1 e2 : jx ybj jx yj

(x yb) 1 (x y) e2 jx ybj r

@Br (y)

1

0

!

(x) f (x)

(x

y) r

!

f (x) dS (x) ;

where S is the usual surface measure of the ball Br (y).When r ! 0, we obtain the result. The Q-part of the hyperholomorphic functions satis…es the following result. Theorem 65 Let be an open subset of R3+ and K a 3-chain satisfying K . If f is hyperholomorphic in and y 2 K, then @

Qf (y) y2

@y2

=

y2 2

Z

@K

P

y

D 1 x2

(x jx

y) (x e2 yj3 jx

yb) ybj3

d

1

(x) f 0 (x) :

Proof. Assume that f is hyperholomorphic in . Using Lemma 23 we obtain y @ Qfy2(y) D1 (P 0 f ) : = y2 @y2 Hence by the previous theorem we conclude the result. 36

Proposition 66 If f be k-hyperholomorphic, then the function hyperholomorphic.

f e2 xk2

is

k-

Proof. We just compute D

Since Q

f e2 xk2

=

Pf xk2

f e2 xk2

(Df ) e2 e2 f e2 k k+1 k x2 x2 P 0f (Mk f ) e2 = + k : xk2 xk+1 2 =

we obtain the result.

Corollary 67 If f be a hyperholomorphic, then the function x2 1 is

@f Qf e2 + = @x2 x22

x2 1

@f e2 f 0 e2 + f + @x2 2x22

1-hyperholomorphic. Proof. Using [11, Theorem 11] we infer that @ f x2 1 @x2

is

Q f x2 1 e2 x2

1-hyperholomorphic. Since Q f x2 1 e2 = x2 1 P f , we conclude the result

Proposition 68 If f :

! C is

1-hyperbolic harmonic, then the function

hf (u0 ; u1 ; u2 ; u3 ) = f

u0 ; u1 ;

q u22 + u23

is harmonic . Proposition 69 If f : gf (u0 ; u1 ; u2 ; u3 ) = P f

! H is u0 ; u1 ;

q

1-hyperholomorphic, then the function u22

+

u23

is monogenic in H. 37

u2 + u3 e1 +p 2 Qf u2 + u23

u0 ; u1 ;

q

u22 + u23 e2

Proof. Assume that If f : ! H is 1-hyperholomorphic. Then there exists locally a 1-hyperbolic harmonic g such that f = Dg, Using the previous result the function q hg (u0 ; u1 ; u2 ; u3 ) = g u0 ; u1 ; u22 + u23 is harmonic. Hence the function Since

@hg @u0

g e1 @h @u1

g e2 @h @u2

g e12 @h is monogenic. @u3

@g u @hg p 2 = 2 @u2 @x2 u2 + u23 @hg @g u p 3 = @u3 @x2 u22 + u23

and f = Dg we obtain the result. Proposition 70 If f :

! H is hyperholomorphic, then the function q q u2 + u3 e1 4Qf u ; u ; gf (u0 ; u1 ; u2 ; u3 ) = 4P f u0 ; u1 ; u22 + u23 + p 2 u22 + u23 e2 0 1 2 u2 + u3

is monogenic

Proof. If f is hyperholomorphic, then 4f is we may apply the previous result. Theorem 71 If f : 1

f (x0 ) =

2

2

Z

@C

1-hyperholomorphic and

! H satis…es D 4 f = 0, then (x

x0 ) 4

1

4f (x) +

1 2 jx

x0 j2

! (x x0 ) 1 f (x) d : Df (x) + jx x0 j2

Proof. Let Br (a) be a ball with center a and radius r, contained in C. We …rst prove that Z (x a) 1 (x a) 1 1 Dl f gd = 0: (17) f f 4f + 2 4 jx aj 2 jx aj2 @(CnBr (a)) From Lemma 55 and the fact that Z

@(CnBr (a))

x 1 jxj2

(x a) 1 f (x) d = jx aj2 38

is monogenic we infer that Z

CnBr (a)

(x a) 1 Dl f (x) dx: jx aj2

Using the preceding Lemma 55 and the hypothesis D4f = 0 we conclude that R R (x a) 1 1 4f (x) d = D (x a) 1 4f (x) dx 4 4 r @(CnBr (a)) CnBr (a) R (x) = CnBr (a) 2jx4f aj 2 dx; since Dl x R

1

1

= Dr x 1

@(CnBr (a)) 2jx aj

2

=

2 . jxj2

Dl f (x) d

From Lemma 56 we deduce = =

R

R

CnBr (a) CnBr (a)

1 D 2 r

jx

(x a) jx aj2

Thus (17) holds true. Since the outside normal (x) =

x a , r

R

@C

=

R

aj 1

2

Dl f +

Dl f (x) +

4f (x) 2jx aj2

4f (x) 2jx aj2

dx dx:

on the sphere @Br (a) is

we obtain from (17) (x a) 4

@Br (a)

1

4f (x) + 2jx 1 aj2 Dl f (x) + (xjx R R (x a)Dl f 4f d + d + 3 4r 2r @Br (a) @Br (a)

a) 1 aj2 f (x) d r3

f (x) d :

R R When r ! 0, then limr!0 @Br (a) d4r = 0 and also limr!0 @Br (a) x2r3a d = 0 . R Since in addition limr!0 @Br (a) fr(x) = 2 2 f (a), the desired result follows. 3 d

Remark. Similar types of representation formulas have already been established by G. Laville and I. Ramadano¤ in ([20]), L. Pernas in ([26]), and J. Ryan ([27]). Corollary 72 Let be an open subset of R4 , C a compact set with smooth boundary @C and = 0 + 1 e1 + 2 e2 + 3 e12 its outer unit normal …eld. If f : ! H is harmonic, then for any a 2 C f (a) =

1 2

2

R

@C

(x a) 1 jx aj2

f (x) +

1 2jx aj2

Df d :

Proof. Just use 4f = 0:

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