Abstract. We consider the Dirac-type operators D + a, a is a paravector in the Clifford algebra. For this operator we state a Cauchy-Green formula in the spaces C

(1)

INSTITUTE OF MATHEMATICS POLISH ACADEMY OF SCIENCES

WARSZAWA 1996

FUNDAMENTAL SOLUTIONS FOR DIRAC-TYPE OPERATORS

S W A N H I L D B E R N S T E I N Freiberg University of Mining and Technology Faculty of Mathematics and Computer Science

Institute of Applied Mathematics I D-09596 Freiberg, Germany

Abstract. We consider the Dirac-type operators D + a, a is a paravector in the Clifford algebra. For this operator we state a Cauchy-Green formula in the spaces C

¹

(G) and W

p¹

(G).

Further, we consider the Cauchy problem for this operator.

1. Preliminaries. Dirac and Dirac-type operators are considered in many papers. A significant selection of papers is contained in the bibliography. Most of them consider the operators in the quaternionic algebra. We want to consider the operator in the Clifford algebra. Thus our considerations differ in some sense from the considerations in the quaternionic algebra. We consider stationary problems and obtain as the main result the Cauchy-Green formula. In the case of nonstationary problems we consider the Cauchy problem.

2. Introduction. Let (e

1

, . . . , e

m

) be an orthonormal basis of R

^m

, m ∈ N, then by C we denote the 2

^m

-dimensional Clifford algebra obtained from the generating relations e

j

e

k

+ e

k

e

j

= 2δ

jk

, j, k = 1, . . . , m. Thus the quaternionic case is not contained. An element of C is of the form a = P a

A

e

A

, a

A

∈ C, e

A

= e

a₁...a_k

= e

a₁

· . . . · e

a_k

for A = {a

1

, . . . , a

k

} with a

1

< . . . < a

k

, e

0

= 1 is the identity of C. We identify a vector x ∈ R

^m

with the element x = P

m

j=1

x

j

e

j

of the Clifford algebra. Let G be a bounded domain of R

^m

, with smooth boundary Γ. If F is a functionspace of complex-valued function, a function u = P u

A

e

A

is an element of F

_C

iff u

A

is a complex-valued function of the space F . We use the Sobolev-spaces W

_p^k

(G), 1 < p < ∞, k ∈ N, the space of continuously differentiable functions C

¹

(G) and the space L

2

[0, t; L

2

(R

^m

)] with the norm

1991 Mathematics Subject Classification: Primary 30G35, Secondary 47G10.

The paper is in final form and no version of it will be published elsewhere.

[159]

(2)

( R

T 0

||f ||

²_L

2(R^m)

dt)

^1/2

. We consider the Dirac operator D

_x

+ a =

m

X

j=1

e

_j

( ∂

∂x

j

+ a

_j

) + e

₀

a

₀

, m ∈ N,

where a = P

m

j=1

e

j

a

j

= a

0

e

0

+ a and D

x

= P

m j=1

e

j ∂

∂x_j

, then D

²_x

= ∇, ∇ being the Laplacian. We want to illustrate the Dirac-operator in the Pauli-algebra. The basic vectors are e

1

, e

2

, e

3

, e

²₁

= e

²₂

= e

²₃

= 1, the unit element of the algebra is e

0

. We have the bivectors e

23

= e

2

e

3

, e

31

= e

3

e

1

and e

12

= e

1

e

2

and the pseudoscalar e

123

= e

1

e

2

e

3

. We consider a function

u = u

₀

e

₀

+ u

₁

e

₁

+ u

₂

e

₂

+ u

₃

e

₃

+ u

₂₃

e

₂₃

+ u

₃₁

e

₃₁

+ u

₁₂

e

₁₂

+ u

₁₂₃

e

₁₂₃

.

The scalar part is u

0

, the pseudoscalar part u

123

, the vector u = (u

1

, u

2

, u

3

) and the bivector v = (u

23

, u

31

, u

12

). Then we obtain for the Dirac operator

Du =

3

X

i=1

e

i

∂

∂x

i

u = +( ∂

∂x

1

u

1

+ ∂

∂x

2

u

2

+ ∂

∂x

3

u

3

)e

0

+

+( ∂

∂x

3

u

₃₁

− ∂

∂x

2

u

₁₂

+ ∂

∂x

1

u

₀

)e

₁

+ ( ∂

∂x

1

u

₁₂

− ∂

∂x

3

u

₂₃

+ ∂

∂x

2

u

₀

)e

₂

+ +( ∂

∂x

₂

u

₃₁

− ∂

∂x

₁

u

₃₁

+ ∂

∂x

₃

u

₀

)e

₃

+ +( ∂

∂x

₂

u

₃

− ∂

∂x

₃

u

₂

+ ∂

∂x

₁

u

₁₂₃

)e

₂₃

+ ( ∂

∂x

₃

u

₁

− ∂

∂x

₁

u

₃

+ ∂

∂x

₂

u

₁₂₃

)e

₃₁

+ +( ∂

∂x

₁

u

₂

− ∂

∂x

₂

u

₁

+ ∂

∂x

₃

u

₁₂₃

)e

₁₂

+ ( ∂

∂x

₃

u

₂₃

− ∂

∂x

₂

u

₃₁

+ ∂

∂x

₃

u

₁₂

)e

₁₂₃

. This system is equivalent to another system

divu = − ∗ curl ∗ v +curl ∗ u − ∗grad ∗ u

123

− ∗ div ∗ v,

where ∗ is the Hodge-operator (multiplication with the pseudo scalar e

123

),

∗e

0

= e

123

, ∗e

123

= −e

0

; ∗e

2

= e

31

, ∗e

31

= −e

2

;

∗e

1

= e

23

, ∗e

23

= −e

1

; ∗e

3

= e

12

, ∗e

12

= −e

3

; and the operator div is defined for vector u

divu =

3

X

i=1

∂

∂x

i

u

i

,

and the operator curl is also defined for vectors u in the following way curlu = ( ∂

∂x

₂

u

3

− ∂

∂x

₃

u

2

)e

23

+ ( ∂

∂x

₃

u

1

− ∂

∂x

₁

u

3

)e

31

+ ( ∂

∂x

₁

u

2

− ∂

∂x

₂

u

1

)e

12

and the Hodge-operator transforms bivectors in vectors and vectors in bivectors. The

(3)

operator grad is defined for scalars u

0

gradu

₀

=

3

X

i=1

∂

∂x

i

u

₀

e

_i

and the Hodge-operator transforms scalars in pseudoscalars and pseudoscalars in scalars.

3. The Dirac-type operator D + a. Let m be a natural number and K

_a₀

(x) = K

_a₀

(|x|) = 1

π

^2/m

2

^m/2

( a

₀

|x|

m/2−1

K

^m

2−1

(a

₀

|x|))

where K denotes modified Bessel functions, the so-called MacDonald functions.

Lemma. The fundamental solution for ∇ − a

²0

is K

_a₀

. The proof is contained in [0rt].

Theorem. The fundamental solution for D

^x

+ a is

E

a

(x) = exp

−<a,x>

{(D

x

− a

0

)K

a₀

(x)} =

= exp

−<a,x>

{ 1 a

0

(2π)

^m/2

·

m

X

j=1

x

j

e

j

|x|

^m

(a

₀

|x|)

^m/2

K

_m/2

(a

₀

|x|)+

+ a

0

(2π)

^m/2

· 1

|x|

^m−2

(a

0

|x|)

^m/2−1

K

_m/2−1

(a

0

|x|)}, with ha, xi = P

m i=1

a

i

x

i

. P r o o f. We prove that E

_a

(x) is locally integrable and (D + a)E

_a

= 0 in R

^m

\ {0} and (D + a)E

a

= δ. On every compact subset of R

^m

the function e

−<a,x>

is bounded from above and positive,

K

a0

∼ 1

|x|

^m−2

, |x| → 0, K

a0

∼ e

^−|Ra⁰^|·|x|

|x|

^m/2−1/2

, |x| → ∞,

∂K

a₀

(x)

∂x

_j

∼ x

j

|x|

^m

, |x| → 0, ∂K

a₀

(x)

∂x

_j

∼ e

^−|Ra⁰^|·|x|

x

j

|x|

^m/2+1

, |x| → ∞,

Ra

0

denotes the real part of the complex number a

0

. Thus E

a

(x) is locally integrable in R

^m

. Next,

(D

_x

+ a)E

_a

(x) = (D

_x

+ a){e

−<a,x>

(D

_a

− a

₀

)K

_a₀

(x)} =

= −ae

−<a,x>

(D

_x

− a

₀

)K

_a₀

(x) + e

−<a,x>

D

_x

(D

_x

− a

₀

)K

_a₀

(x)+

+ae

−<a,x>

(D

x

− a

0

)K

a₀

(x) =

= e

−<a,x>

(D

x

+ a

0

)(D

x

− a

0

)K

a₀

(x) = e

−<a,x>

(∆ − a

²₀

)K

a₀

(x) = 0 in R

^m

\ {0}. On the other hand, let φ ∈ D(R

^m

):

((D + a)E

a

(x), φ(x)) = (e

−<a,x>

(∆ − a

²₀

)K

a₀

(x), φ(x)) =

= ((∆ − a

²₀

)K

a₀

(x), e

−<a,x>

φ(x)) = (δ, e

−<a,x>

φ(x)) =

= φ(0) = (δ, φ).

R e m a r k. The case a = a

₀

∈ C is discussed in [Xu] if e

²i

= −1 by using outer and

inner monogenics.

(4)

We introduce the integral operators T

_a

u :=

Z

G

E

_a

(x − y)u(y)dy, F

_a

u := − Z

Γ

E

_a

(x − y)n(y)u(y)dy, x 6∈ Γ.

Here, n(y) denotes the outward-normal at the point y ∈ Γ.

Theorem. The operator

T

_a

: W

_p,c^k

(G) → W

_p,C^k+1

(g), 1 < p < ∞, k = 0, 1, . . . is continuous.

P r o o f. T

_a

is a weakly singular integral operator with

|E

a

(x − y)| ≤ C

|x − y|

^m−1

and using [MP] we get thus the operator

T

a

: L

p,C

(G) → L

p,C

(g), 1 < p < ∞ is continuous. The rest is contained in the following lemmas.

Lemma. Let u ∈ L

^p,C

(G), 1 < p < ∞, k = 0, 1, . . .. Then

∂

∂x

k

T

a

u = Z

G

∂

∂x

k

E

a

(x − y)u(y)dy + u(x) A

m

Z

S1

y − x

|x − y| · y

k

− x

k

|x − y| dS

1

(y),

where the integral over S

1

is a constant only depending on the dimension m and k and A

m

=

^2π_Γ(^m/2m

2)

is the area of the unit sphere in R

^m

.

P r o o f. We have E

a

(x − y) = e

−<a,x−y>

(D

x

− a

0

)K

a0

(x − y), a

₀

K

_a₀

(x − y) ∼ 1

|x − y|

^m−2

, x → y, D

x

K

a₀

(x − y) = 1

A

_m

e

−<a,x−y>

x − y

|x − y| , x → y.

Now, we get

∂

∂x

k

Z

G

e

−<a,x−y>

(D

_x

− a

0

)K

_a₀

(x − y)u(y)dy = Z

G

∂

∂x

_k

{e

−<a,x−y>

(D

x

− a

0

)K

a₀

(x − y)}u(y)dy+

− Z

r==|x−y|

e

−<a,x−y>

(D

_x

− a

0

)K

_a₀

(x − y)u(y) cos(r, x

_k

)dS

= Z

G

∂

∂x

_k

{e

−<a,x−y>

(D

x

− a

0

)K

a₀

(x − y)}u(y)dy+

+ Z

=|x−y|

e

−<a,x−y>

a

0

K

a0

(x − y)u(y) cos(r, x

k

)dS

+

− Z

=|x−y|

e

−<a,x−y>

D

x

K

a₀

(x − y)u(y) cos(r, x

k

)dS

=

= Z

G

∂

∂x

k

{e

−<a,x−y>

(D

x

− a

0

)K

a₀

(x − y)}u(y)dy+

(5)

− Z

|x−y|=1

^m−1

e

−<a,x−(x+θ)>

a

0

K

a₀

(x − y)u(x + θ) cos(r, x

k

)dS

1

+

− Z

|x−y|=1

e

−<a,x−(x+θ)>

D

x

K

a0

(x − y)u(x + θ)

^m−1

cos(r, x

k

)dS

1

. We take → 0:

= Z

G

∂

∂x

k

E

a

(x − y)u(y)dy + u(x) A

m

Z

S₁

y − x

|x − y| · y

_k

− x

k

|x − y| dS

1

(x), e

−<a,x−(x+θ)>

= e

<a,θ>

→ 1).

Lemma. The operator

∂x^∂_k

T

a

: L

p,C

(g) → L

p,C

, 1 < p < ∞, is continuous.

P r o o f. From the Lemma above we get

∂

∂x

k

T

a

= Z

G

∂

∂x

k

E

a

(x − y)u(y)dy + C(k)u(x),

where C(k) is a constant only depending on k. Thus C(k)u ∈ L

_p,C

if u ∈ L

_p,C

. We consider the first term.

∂

∂x

k

E

a

(x − y) =

−aE

a

(x − y) − e

−<a,x−y>

a

0

(2π)

^m/2

· x

k

− y

k

|x − y|

^m

(a

0

|x − y|)

^m/2

K

_m/2

(a

0

|x − y|)+

e

−<a,x−y>

a

0

(2π)

^m/2

·

m

X

j=1

(x

j

− y

j

)(x

k

− y

k

)

|x − y|

^m

· (a

0

|x − y|)

^m/2−1

K

_m/2−1

(a

0

|x − y|)+

e

−<a,x−y>

1 a

₀

(2π)

^m/2

K

_m/2−

(a

0

|x − y|) ·

m

X

j=1

∂

∂x

k

( (x

j

− y

j

)e

j

|x − y| ).

We only have to consider the last part, because the other parts lead to weakly singular kernels. We prove that the last part creates a singular kernel of a Calderon-Zygmund operator. We choose k = 1 and use sphericals cordinates in the following way:

y

1

= x

1

+ r cos θ

1

, y

2

= x

2

+ r sin θ

1

cos θ

1

,

...,

y

_m−1

= x

_m−1

+ r sin θ

₁

sin θ

₂

. . . sin θ

_m−2

cos θ

_m−1

, y

_m

= x

_m

+ r sin θ

₁

sin θ

₂

. . . sin θ

_m−2

sin θ

_m−1

, r = |x − y|, θ

1

∈ [0, π], θ

i

∈ [−π, π], i = 2, 3, . . . , m − 1. We have

∂r

∂x

1

= x

1

− y

1

r = − cos θ

1

. Thus

− ∂

∂x

1

cos θ

1

= 1

r − (x

1

− y

1

)

²

r

³

1 r − cos

²

θ

1

r = sin

²

θ

1

r = sin θ

1

∂θ

1

∂x

1

and we obtain

∂θ

₁

∂x

1

= sin θ

₁

r

(6)

because of

θ

₁

∈ [0, π].

Further

∂

∂x

₁

( y

2

− x

2

r ) = ∂

∂x

₁

(sin θ

1

cos θ

2

)

= ∂

∂θ

₁

(sin θ

1

cos θ

2

) ∂θ

2

∂x

₁

= sin θ

1

r cos θ

1

cos θ

2

− sin θ

1

sin θ

2

∂θ

2

∂x

₁

cos θ

1

cos θ

2

∂θ

1

∂x

₁

− sin θ

1

sin θ

2

∂θ

2

∂x

₁

= sin θ

1

r cos θ

1

cos θ

2

− sin θ

1

sin θ

2

∂θ

2

∂x

₁

. On the other hand

∂

∂x

1

( y

₂

− x

2

r ) = (y

₂

− x

2

)

r · ( (y

₁

− x

1

) r ) · 1

r = sin θ

₁

r cos θ

1

cos θ

2

and we obtain

− sin θ

1

sin θ

₂

∂θ

₂

∂x

1

= 0 and thus

^∂θ_∂x²

1

= 0. Now, let

_∂x^∂θ^j

1

= 0, j = 2, ·, l − 1, l ≤ m − 1, then

∂

∂x

1

( y

₂

− x

₂

r ) = ∂

∂x

1

(sin θ

₁

· . . . · sin θ

_l−1

cos θ

_l

)

= ∂

∂θ

₁

(sin θ

₁

· . . . · sin θ

_l−1

cos θ

_l

) ∂θ

1

∂x

₁

+ ∂

∂θ

₁

(sin θ

₁

· . . . · sin θ

_l−1

cos θ

_l

) ∂θ

l

∂x

₁

= sin θ

1

r cos θ

₁

sin θ

₂

. . . sin θ

_l−1

cosθ

_l

− sin θ

₁

. . . sin θ

_l−1

sin θ

₁

∂θ

1

∂x

₁

. On the other hand

∂

∂x

1

( y

1

− x

1

r ) = (y

1

− x

1

)

r · ( (y

1

− x

1

) r ) · 1

r = sin θ

1

r cos θ

1

sin θ

2

. . . sin θ

l−1

cos θ

l

and we get

_∂x^∂θ^l

1

= 0, l = 2, . . . , m − 1. Though, we obtain

∂

∂x

1

( x

|x|

^m

) = ∂

∂x

1

(

m

X

j=1

(x

_j

− y

j

)

|x − y|

^m

e

_j

) = − ∂

∂x

1

(

m

X

j=1

(x

_j

− y

j

)

r · 1

r

^m

e

_j

) =

= (m − 1) r

^m

· ∂r

∂x

1

·

m

X

j=1

(x

_j

− y

_j

)

|x − y|

^m

e

_j

− 1 r

^m−1

∂

∂θ

1

(

m

X

j=1

(x

_j

− y

_j

)

|x − y|

^m

e

_j

) ∂θ

₁

∂x

1

=

= − 1 r

^m

{r ∂

∂θ

₁

(

m

X

j=1

(x

j

− y

j

)

|x − y|

^m

e

j

) ∂

∂x

₁

− (m − 1)(

m

X

j=1

(x

j

− y

j

)

|x − y|

^m

e

j

) ∂r

∂x

₁

}.

Let

φ(θ

1

, θ

2

, . . . , θ

m−1

) := −

m

X

j=1

(x

_j

− y

j

) r e

j

, then

∂

∂x

₁

( x

|x|

^m

) = 1 r

^m

{r ∂φ

∂θ

₁

∂θ

1

∂x

₁

− (m − 1)φ ∂r

∂x

₁

} = f (φ, r)

r

^m

.

(7)

We get

∂

∂x

1

E

a

(x − y) =

−aE

_a

(x − y) − e

−<a,x−y>

a

0

(2π)

^m/2

· x

k

− y

k

|x − y|

^m

(a

₀

|x − y|)

^m/2

K

_m/2

(a

₀

|x − y|)+

e

−<a,x−y>

a

0

(2π)

^m/2

·

m

X

j=1

(x

j

− y

j

)(x

k

− y

k

)

|x − y|

^m

· (a

0

|x − y|)

^m/2−1

K

_m/2−1

(a

0

|x − y|)+

e

−<a,x−y>

1 a

0

(2π)

^m/2

· (a

0

|x − y|)

^m/2

K

m/2

(a

0

|x − y|) · f (φ, |x − y|)

|x − y|

^m

. We have to prove that R

S₁

f (φ, 1)dS

1

= 0, where S

1

is the unit sphere in R

^m

with center in x, and that R

S1

|f (φ, 1)|

^p⁰

dS

1

= const, 1 < p

⁰

< ∞. Because of

dS

1

= sin

^m−2

θ

1

sin

^m−3

θ

2

. . . sin θ

m−2

dθ

1

dθ

2

. . . dθ

m−2

dθ

m−1

, we obtain

Z

S₁

f (φ, 1)dS

1

= Z

π

−π

dθ

m−1

Z

π 0

sin θ

m−2

dθ

m−2

. . . Z

π

0

sin

^m−3

θ

2

dθ

2

Z

π 0

f (φ, 1) sin

^m−2

θ

₁

dθ

₁

. We have

f (φ, 1) = C(a

0

){ ∂φ

∂θ

1

sin θ

1

+ (m − 1)φ cos θ

1

},

where C(a

₀

) is a constant only depending on a

₀

. Thus the inner integral is equal to C(a

0

)

Z

π 0

[(m − 1)φ cos θ

1

+ ∂φ

∂θ

₁

sin θ

1

sin

^m−2

θ

1

dθ

1

= C(a

0

)

Z

π 0

∂

∂θ

₁

[sin

^m−1

θ

1

· φ(θ

1

, . . . , θ

m−1

)]dθ

1

= 0 and thus

Z

S₁

f (φ, 1)dS

₁

= 0.

Furthermore, we have

∂φ

∂θ

₁

sin θ

1

= − sin

²

θ

1

e

1

+ sin θ

1

cos θ

1

cos θ

2

e

2

+ . . . + + sin θ

₁

cos θ

₁

sin θ

₂

. . . sin θ

_m−2

cos θ

_m−1

e

_m−1

+

+ sin θ

₁

cos θ

₁

sin θ

₂

. . . sin θ

_m−2

sin θ

_m−1

e

_m

and

φ cos θ

1

= cos

²

θ

1

e

1

+ sin θ

1

cos θ

1

cos θ

2

e

2

+ . . . + + sin θ

1

cos θ

1

sin θ

2

. . . sin θ

m−2

cos θ

m−1

e

m−1

+

+ sin θ

1

cos θ

1

sin θ

2

. . . sin θ

m−2

sin θ

m−1

e

m

(8)

and thus

∂φ

∂θ

1

sin θ

1

+ (m − 1)φ cos θ

1

= −1 · e

1

+ m · cos

²

θ

1

e

1

+

+m · sin θ

₁

cos θ

₁

cos θ

₂

e

₂

+ . . . + m · sin θ

₁

cos θ

₁

sin θ

₂

. . . sin θ

_m−2

sin θ

_m−1

e

_m

and

| ∂φ

∂θ

₁

sin θ

1

+ (m − 1)φ cos θ

1

| ≤ 1 + m

²

and we obtain finally

Z

S1

|f (φ, 1)|

^p⁰

dS

1

≤ (1 + m

²

)

^p⁰

· A

m

, 1 < p

⁰

< ∞.

If k 6= 1, we choose spherical coordinates such that the same situation arises.We obtain that R

G

∂

∂x_k

E

a

(x−y)u(y)dy consists of weakly singular integral operators and a Calderon- Zygmund operator (singular integral operator). This completes the proof.

Lemma. Let u ∈ C

C^d

(G) then we have

(D

_x

+ a)T

_a

u = u(x) in G 0 in R

^m

\ G.

This follows immediately from the construction of the operator T

a

.

An important connection between the operators D

x

+ a, T

a

and F

a

is given by the Cauchy-Green formula.

Theorem (Cauchy-Green formula). Let u ∈ C

C¹

(G) then we have F

a

u + T

a

(D

x

+ a)u = u(x) in G

0 in R

^m

\ G.

P r o o f. We have

(D

y

− a){e

−<a,x−y>

K

a₀

(x − y)u(y)} + e

−<a,x−y>

K

a₀

(x − y)au(y) =

= ae

−<a,x−y>

K

a₀

(x − y)u(y) − e

−<a,x−y>

D

x

K

a₀

(x − y)u(y)+

+e

−<a,x−y>

K

a₀

(x − y)D

y

u(y) − ae

−<a,x−y>

K

a₀

(x − y)u(y)+

+e

−<a,x−y>

K

a0

(x − y)au(y) =

−e

−<a,x−y>

(D

_x

+ a

₀

)K

_a₀

(x − y)u(y) + e

−<a,x−y>

K

_a₀

(x − y)(D

_y

+ a

₀

)u(y).

Let G

= {y ∈ G : |x − y| > } then Z

G

(D

y

− a){e

−<a,x−y>

K

a₀

(x − y)u(y)}dy =

− Z

G

e

−<a,x−y>

(D

_x

+ a

₀

)K

_a₀

(x − y)u(y)dy+

+ Z

G

e

−<a,x−y>

K

a₀

(x − y)(D

y

+ a)u(y)dya Z

G

e

−<a,x−y>

K

a₀

(x − y)u(y)dy =

− Z

G

e

−<a,x−y>

(D

_x

+ a

₀

)K

_a₀

(x − y)u(y)dy + Z

Γ

e

−<a,x−y>

K

_a₀

(x − y)n(y)u(y)dy+

− Z

S

e

−<a,x−y>

K

a₀

(x − y)n(y)u(y)dy − a Z

G

e

−<a,x−y>

K

a₀

(x − y)u(y)dy+

(9)

+ Z

G

e

−<a,x−y>

D

x

K

a₀

(x − y)u(y)dy = a Z

G

e

−<a,x−y>

K

a₀

(x − y)u(y)dy+

+ Z

Γ

e

−<a,x−y>

K

_a₀

(x − y)n(y)u(y)dy − Z

S

e

−<a,x−y>

K

_a₀

(x − y)n(y)u(y)dy.

Thus Z

G

e

−<a,x−y>

K

a₀

(x − y)(D

y

+ a)u(y)dy − Z

G

e

−<a,x−y>

(D

x

+ a

0

)K

a₀

(x − y)u(y)dy =

= Z

Γ

e

−<a,x−y>

K

a0

(x − y)n(y)u(y)dy − Z

S

e

−<a,x−y>

K

a0

(x − y)n(y)u(y)dy.

Now, tends to zero, than G

tends to G and the integral over S

tends to zero, thus

−T

_−a

u + Z

G

e

−<a,x−y>

K

a₀

(x − y)(D

y

+ a)u(y)dy = Z

Γ

e

−<a,x−y>

K

a₀

(x − y)n(y)u(y)dy.

Application of (D

x

− a) from the left leads to Z

Γ

e

−<a,x−y>

(D

_x

− a

0

)K

_a₀

(x − y)n(y)u(y)dy+

Z

G

e

−<a,x−y>

(D

x

− a

0

)K

a₀

(x − y)(D

y

+ a))u(y)dy = u(x) in G 0 in R

^m

\ G or

F

a

u + T

a

(D

x

+ a)u = u(x) in G 0 in R

^m

\ G . Because the operators

D + a : W

_p,C¹

(G) → L

_p,C

(G), 1 < p < ∞, T

a

: L

p,C

(G) → W

_p,C¹

(G), 1 < p < ∞ are continuous, the operator

F

a

: W

_p,C^1−1/p

(Γ) → W

_p,C¹

(G)

is also continuous. Thus we are able to extend the lemma and the Cauchy-Green formula:

Lemma. Let u ∈ W

p,C¹

(G), 1 < p < ∞, then we have (D

_x

+ a)T

_a

u = u(x) in G

0 in R

^m

\ G .

Theorem (Cauchy-Green formula). Let u ∈ W

p,C¹

(G), 1 < p < ∞, then we have F

a

u + T

a

(D

x

+ a)u = u(x) in G

0 in R

^m

\ G .

4. A note on elementary functions. If X is an arbitrary element of the Clifford- algebra C, then

e

^X

=

∞

X

n=0

X

ⁿ

n! = 1 + X + x

²

2! + X

³

3! + . . .

(10)

and

sinh x = e

^X

− e

^−X

2 =

∞

X

n=0

X

²ⁿ⁺¹

(2n + 1)! , cosh X = e

^X

+ e

^−X

2 =

∞

X

n=0

X

²ⁿ

(2n)! , thus

e

^X

= cosh X + sinh X.

Furthermore, we have sin X =

∞

X

n=0

(−1)

ⁿ

X

²ⁿ⁺¹

(2n + 1)! , cos X =

∞

X

n=0

(−1)

ⁿ

X

²ⁿ

(2n)! .

Lemma. If J X = XJ for all X and J

²

= −1, then we have cosh J X = cos X and sinh J X = J sin X and e

^{J X}

= cos X + J sin X.

R e m a r k. If we denote by I

m

= e

1

e

2

. . . e

m

the pseudoscalar of C the only possible J are the scalars ±i and the elements ±iI

_4p+1

and ±I

_4p+3

, where p = 0, 1, 2, . . ..

If a is a vector, i.e. a = P

m

j=1

a

_j

e

_j

, then

a

ⁿ

= |a|

ⁿ

, n = 2k

|a|

ⁿ⁻¹

a, n = 2k + 1 . In this case we obtain

e

^a

= cosh a + sinh a = cosh |a| + a

|a| sinh |a|

e

^{J a}

= e

^aJ

= cosh J a + sinh J a = cos a + J sin a = cos |a| + J a

|a| sinh |a|.

5. The Dirac-type operator

_∂t^∂

D. We consider the equation ( ∂

∂t D)E = δ(x) ⊗ δ(t)

in distributional sense. We use the partial Fourier transform to compute the fundamental solution. Thus

( ∂

∂t

E + D(iy) ˆ ˆ E) = l(y) ⊗ δ(t).

The solution of the problem is

E(t) = exp(−tD(iy)). ˆ

Theorem. The fundamental solution for the wave operator

_∂t^∂

+ D, t ≥ 0 is E(x, t) = ( ∂

∂t − D

x

)F

⁻¹

( sin |y|t

|y| ),

where F

⁻¹

(

^{sin |y|t}_|y|

) is the fundamental solution for the wave operator for t ≥ 0.

P r o o f. We have seen that the partial Fourier transform of the fundamental solution is

E(t) = exp(−tD(iy)). ˆ

(11)

We use the results of section 3 to get an explicit formula. We have exp(−tD(iy)) = exp(−i

m

X

j=1

e

j

y

j

t) = cos(|y|t) − i y

|y| sin(|y|t) = cos(|y|t) − D)iy)

|y| sin(|Y |t) = ( ∂

∂t − D(iy)) sin |y|t

|y| . Thus

E(x, t) = ( ∂

∂t − D

x

)F

⁻¹

sin |y|t

|y| .

R e m a r k. The fundamental solution for the wave operator depends on the dimension m. In general it is a distributional derivative of a measure.

A similar problem is the following operator

∂

∂t + β

₀

(D + a),

where β

₀

is a complex non-zero constant and a = a

₀

e

₀

+ P

m

j=1

a

_j

e

_j

= a

₀

e

₀

+ a.

Theorem. The fundamental solution of

∂

∂t + β

0

(D + a), for t ≥ 0 is

e

^−a⁰t<a,x>

· e

^β⁰

· ( ∂

∂t − D

x

)F

⁻¹

( sin |y|t

|y| ).

The proof is obvious.

6. The Cauchy problem for the operator

_∂t^∂

+D. An important problem for this Dirac-type operator is the Cauchy problem because this problem for hyperbolic operators is well-posed.

Theorem. The Cauchy problem for the Dirac-type operator

∂u

∂t + Du = 0, u(x, 0) = u

0

has a unique solution in D

⁰

for t ≥ 0

u(x, t) = F

⁻¹

(cos(|y|t))h∗, xiu

0

− D

x

F

⁻¹

( sin |y|t

|y| )h∗, xiu

0

,

where h∗, xi denotes the convolution only with respect to x. If u

₀

∈ L

2,C

(R

^m

) then (u, x) ∈ L

2,C

[0, T ; R

^m

].

P r o o f. The proof follows from the fact that cos(|y|t) and D(iy)

^{sin |y|t}_|y|

if t ≥ 0 are multipliers in L

_2,C

(R

^m

); see [DL].

To solve the inhomogeneous Cauchy problem we use again partial Fourier transform and obtain the problem

∂ ˆ u

∂t + D(iy)ˆ u = ˆ f , ˆ

u(0) = u

0

.

(12)

Now, we set ˆ u = e

^−D(iy)·t

v(t) and obtain a problem in v.

e

^−D(iy)·t

∂v

∂t = ˆ f , v(0) = ˆ u(0) and the solution is

v(t) = ˆ u

₀

+ Z

t

0

e

^−D(iy)·t

v(t) ˆ f (r)dr and thus

ˆ

u(t) = e

^−D(iy)·t

u ˆ

0

+ Z

t

0

e

−D(iy)(t−r)

f (r)dr = ˆ

= e

^−D(iy)·t

u ˆ

0

+ Y (t) · e

^−D(iy)·t

h∗, ti ˆ f (t), where Y (t) denotes the Heaviside function. To summarize we state

Theorem. The Cauchy problem

∂u

∂t + Du = f (x, t), u(x, 0) = u

0

has for u

₀

∈ L

2,C

(R

^m

) and f ∈ L

_2,C

[0, T ; L

_2,C

(R

^m

)] a unique solution u ∈ L

2,C

[0, T ; L

2,C

(R

^m

)]

and

ˆ

u(t) = e

^−D(iy)·t

u ˆ

0

+ Y (t) · e

^−D(iy)·t

h∗, ti ˆ f (t).

7. Examples of problems with Dirac-type operators. 1) A relativistic particle with spin 1/2 in an electromagnetic field with vector potential ~ A :

3

X

k=1

e

k

(i ∂

∂x

_t

+ b

k

) + e

0

m

0

with rest-mass m

0

and (b

1

, b

2

, b

3

) = ~b = −Q ~ A, where Q is the charge.

2) The Dirac ”Hamilton”-operator for a free particle H = e

0

− i X

j=1

e

j

∂

∂x

_j

and the corresponding Cauchy problem:

∂ψ

∂t + (

3

X

j=1

e

j

∂

∂x

j

+ ie

0

)ψ = 0, ψ(0) = ψ(0).

3) The equation of small perturbations for an irrotational perfect compressible gas:

∂p

∂t + ρ

₀

divv = 0, ρ

₀

∂v

∂t + gradp = 0 v(x, 0) = v

⁰

(x), p(x, 0) = p

⁰

(x), is equivalent to the system

( ∂

∂t + D) p

⁰

p

0

v

= 0,

(13)

p

⁰

p

₀

v

(x, 0) =

p

⁰

p

₀

v

⁰

. 4) Stationary Maxwell equations:

We set E(x, t) = E

0

(x)e

^iωt

B(x, t) = B

0

(x)e

^iωt

, where E(x, t) is the electrical field and B(x, t) is the magnetic inductivity and then

divE

0

= ρ

0

, iωE

0

− curlB

0

= −j

0

,

iωB

0

+ curlE

0

= 0, divB

0

= 0

in the domain G. If we set U = (0, E

0

, B

0

, 0), then the system above is equivalent to (

3

X

k=1

e

_k

∂

∂x

t

+ iωe

₀

)U = (ρ

₀

, −j

₀

0, 0).

5) Time-dependent Maxwell equations in vacuum divE

₀

− ρ = 0,

∂E

∂t − curlB + j = 0,

∂B

∂t + curlE = 0, divB = 0.

If we set V = (0, E, B, 0) the system above is equivalent to ( ∂

∂t +

3

X

k=1

e

k

∂

∂x

t

)V = (ρ, −j, 0, 0).

References

[Be1] S. B e r n s t e i n, Analytische Untersuchungen in unbeschr¨ ankten Gebieten mit Anwen- dungen auf quaternionische Operatortheorie und elliptische Randwertprobleme, PhD Thesis, Freiberg University of Mining and Technology, 1993.

[Be2] —, Cauchy-Green formulas in Clifford Analysis, to appear.

[BDS] F. B r a c k x, R. D e l a n g h e and F. S o m m e n, Clifford Analysis, Pitman, Boston- London-Melbourne, 1982.

[DL] R. D a u t r a y and J.-L. L i o n s, Mathematical Analysis and Numerical Methods for Sci- ence and Technology , vol. 1 and 5, Springer-Verlag, 1992.

[GS] K. G ¨ u r l e b e c k and W. S p r ¨ o ß i g, Quaternionic Analysis and Elliptic Boundary Value Problems, Akademie-Verlag, Berlin, 1989.

[Jan] B. J a n c e w i c z, Multivectors and Clifford Algebra in Electrodynamics, World Scientific Publ. Co. Pt;. Ltd., 1989.

[Kr1] V. V. K r a v c h e n k o, Integral representation of biquaternionic hyperholomorphic func-

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[Kr2] —, On the generalized holomorphic vectors, Abstracts of Republican Research Con- ference devoted to 200 birth of N.I. Lobachevskij, Odessa, Ukraine, 1992, Part 1, p.

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Vekua, Tbilisi 5(1) (1990), 106–109 (Russian).

[KS2] —, Helmholtz operator with a quaternionic wave number and associated function theory , Deformations of Mathematical Structures. Ed. by J. Lawrynowicz, Kluwer Academic Publishers, Dordrecht 1993, 101–128.

[MP] S. G. M i c h l i n and S. P r ¨ o ß d o r f, Singul¨ are Integraloperatoren, Akademie-Verlag, Berlin, 1980.

[Ob1] E. I. O b o l a s h v i l i, Space analogous of generalized analytic functions, Soobch. Akad.

Nauk Gruzin. SSR, 73,1 (1974), 21–24 (Russian).

[Ob2] —, Spatial generalized holomorphic vectors, Different. Uravneniya 11(1) (1975), 108–

115 (Russian).

[Ort] N. O r t n e r, Regularisierte Faltung von Distributionen. Teil 2: Eine Tabelle von Fun- damentall¨ osungen, Journal of Applied Mathematics and Physics (ZAMP), 31 (1980), 133–155.

[Xu1] Z. X u, Boundary value problems and function-theory for Spin-invariant differential operators, PhD Thesis, State University of Gent, 1989.

[Xu2] —, A function theory for the operator D → −λ, Complex Variables Theory Appl.,

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