Solutions of Robin Problems for Overdetermined Inhomogeneous Cauchy–Riemann Systems on the Unit...

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Compl. anal. oper. theory 4 (2010), 39–53 c 2008 Birkh¨auser Verlag Basel/Switzerland 1661-8254/010039-15, published online October 13, 2008 DOI 10.1007/s11785-008-0086-6 Complex Analysis and Operator Theory Solutions of Robin Problems for Overdetermined Inhomogeneous Cauchy–Riemann Systems on the Unit Polydisc Alip Mohammed and M. W. Wong Abstract. The inhomogeneous Robin condition with general coefficient for the overdetermined inhomogeneous Cauchy–Riemann system of equations on the polydisc is studied using Fourier analysis. It is shown that this problem for the case of nonholomorphic general coefficient, is actually a problem with essential singularity in the domain, but still well-posed under certain compatibility conditions. Under proper assumptions, the unique solution is given explicitly. Mathematics Subject Classification (2000). 32A10, 35C10, 35N10. Keywords. Robin problem, overdetermined inhomogeneous Cauchy–Riemann systems, essential singularities. 1. Introduction Let D n be the unit polydisc given by D n = z =(z 1 ,z 2 ,...,z n ) C n : |z k | < 1, 1 k n and let 0 D n be its essential boundary given by 0 D n = z =(z 1 ,z 2 ,...,z n ) C n : |z k | =1, 1 k n . We are interested in finding a function v in C 1 (D n ) such that ∂v z k = f k (z) , z D n , ∂v ∂ν ζ + α(ζ )v = γ 0 (ζ ) , ζ 0 D n , (1.1) This research is supported by the Natural Sciences and Engineering Research Council of Canada. Alip Mohammed is on leave from Department of Mathematics, Xinjiang University, Shenglilu 14, 830046 ¨ Ur¨ umqi, PR China.

Transcript of Solutions of Robin Problems for Overdetermined Inhomogeneous Cauchy–Riemann Systems on the Unit...

Compl. anal. oper. theory 4 (2010), 39–53c© 2008 Birkhauser Verlag Basel/Switzerland1661-8254/010039-15, published online October 13, 2008

DOI 10.1007/s11785-008-0086-6

Complex Analysisand Operator Theory

Solutions of Robin Problems forOverdetermined InhomogeneousCauchy–Riemann Systemson the Unit Polydisc

Alip Mohammed and M. W. Wong

Abstract. The inhomogeneous Robin condition with general coefficient for theoverdetermined inhomogeneous Cauchy–Riemann system of equations on thepolydisc is studied using Fourier analysis. It is shown that this problem for thecase of nonholomorphic general coefficient, is actually a problem with essentialsingularity in the domain, but still well-posed under certain compatibilityconditions. Under proper assumptions, the unique solution is given explicitly.

Mathematics Subject Classification (2000). 32A10, 35C10, 35N10.

Keywords. Robin problem, overdetermined inhomogeneous Cauchy–Riemannsystems, essential singularities.

1. Introduction

Let Dn be the unit polydisc given by

Dn =

{z = (z1, z2, . . . , zn) ∈ C

n : |zk| < 1, 1 ≤ k ≤ n}

and let ∂0Dn be its essential boundary given by

∂0Dn =

{z = (z1, z2, . . . , zn) ∈ C

n : |zk| = 1, 1 ≤ k ≤ n}

.

We are interested in finding a function v in C1(Dn) such that⎧⎨

∂v∂zk

= fk(z) , z ∈ Dn ,

∂v∂νζ

+ α(ζ)v = γ0(ζ) , ζ ∈ ∂0Dn ,

(1.1)

This research is supported by the Natural Sciences and Engineering Research Council of Canada.Alip Mohammed is on leave from Department of Mathematics, Xinjiang University, Shenglilu 14,830046 Urumqi, PR China.

40 A. Mohammed and M. W. Wong Comp.an.op.th.

where the given functions fk, 1 ≤ k ≤ n, are such that

∂fk

∂zl− ∂fl

∂zk= 0 , 1 ≤ k , l ≤ n, (1.2)

∂v∂νζ

denotes the outward normal derivative of v at the point ζ ∈ ∂0Dn, γ0 and α

are given Holder continuous functions on ∂0Dn.

The problem (1.1) is known as the Robin problem for the system of inhomo-geneous Cauchy–Riemann equations on the unit polydisc D

n.Let us recall some results in the case when n = 1. Since α is Holder continuous

on the boundary ∂D of the unit disc D, we can write

α(z) = α+(z) + α−(z) , z ∈ ∂D . (1.3)

In the preceding equation,

α+(z) =∞∑

k=1

α−kzk

and

α−(z) =m∑

k=0

αkz−k ,

where 0 ≤ m ≤ ∞, and the complex numbers α0, α±1, α±2, . . . , are the Fouriercoefficients of α.

If m = 0, then the function α is the boundary value of a holomorphic functionon D and to the problem (1.1) a unique explicit solution can be given [2] by justusing the real part of the boundary condition. This is then the Riemann–Hilbert–Poincare (RHP) problem. For the case when m = 0, the higher dimensional analogof the RHP problem for polydiscs has been investigated in [15].

If 0 < m < ∞, then the function α is the boundary value of a function, whichis holomorphic on the punctured disc D\{0} and at the origin there is a pole oforder m. In this case, the RHP problem has very different features from the casewhen m = 0 in [2]. The problem may be well posed. It may be just solvable andfor a unique solution, a finite number of pointwise boundary conditions have tobe imposed. And in the case of an “inconsistent” coefficient α, the problem is noteven solvable [2,17]. Through involved analysis and calculations, the exact numberand the exact form of finitely many additional pointwise boundary conditions aregiven in [17] to fix the solution uniquely.

If m = ∞, then the function α is the boundary value of a function, whichis holomorphic on D\{0} and at the origin there is an essential singularity. Inthis case, it is shown in [17] that only the Robin problem is well-posed, but theRHP problem cannot be well-posed unless infinitely many additional pointwiseboundary conditions are imposed.

In [2], the problem (1.1) is solved for the case when f = 0 and α is holomor-phic. It is also solved for the case when f = 0 and

α(z) = αmzm + α0 , z ∈ ∂D , (1.4)

Vol. 4 (2010) Cauchy–Riemann Systems on the Unit Polydisc 41

where αm and α0 are complex constants. In [17], the general problem (1.3) isreduced to the case when

α(z) = α−(z) =m∑

j=0

αjzj , z ∈ ∂D , (1.5)

where α0, α1, . . . , αm are complex constants. The two cases for m < ∞ and form = ∞ are then solved separately.

Arbitrary Holder continuous functions on ∂D can be treated as boundary val-ues of harmonic functions on D, but this is not the case for polydiscs D

n, n ≥ 2.This fact brings the compatibility conditions into consideration [3, 12–14] andproper notation is essential for the study of boundary value problems on poly-domains.

For the main results in this paper, we denote the set of all multi-indicesby Z

+n, which is simply the n-fold Cartesian product of the set Z+ of all nonneg-

ative integers. We assume as in [4, 16] that the boundary value α on the essentialboundary ∂0D

n satisfies the compatibility condition

α = α+ + α− , (1.6)

whereα+(ζ) =

κ∈Z+n

ακζκ ,

andα−(ζ) =

κ∈Z+n\{0}α−κζ−κ ,

for all ζ ∈ ∂0Dn. In other words, we assume that α is equal to the projection of

α onto the boundary value PBHn(α) of a harmonic function on Dn. Moreover, we

show in Section 4 that the function α in (1.6) can be reduced to the case in whichα = α−, i.e.,

α(ζ) = α−(ζ) =m∑

|κ|=0

α−κζ−κ , ζ ∈ ∂0Dn . (1.7)

Remark. To give some more insight into the nature of the compatibility condition,we let D

−n be the n-fold Cartesian product of the exterior of the unit disc D, i.e.,

D−n = (C − D) × (C − D) × · · · × (C − D)

︸ ︷︷ ︸n

.

Then we denote by BHn and BH−n the spaces of all boundary values of harmonicfunctions on, respectively, D

n and D−n. We also denote by ∂Hn and ∂H−n the

spaces of boundary values of holomorphic functions on, respectively, Dn and D

−n.Then

BHn = ∂Hn ⊕ ∂H−n

andBHn = BH−n .

42 A. Mohammed and M. W. Wong Comp.an.op.th.

In this paper we study the Robin problem for the case when m = ∞ in (1.7)for the unit polydisc D

n, n ≥ 2. The results in this paper are useful for the studyof RHP and Robin problems for the inhomogeneous pluriharmonic and pluriholo-morphic systems on polydiscs, which are higher dimensional analogs of Poissonand Bitsadze equations.

2. Homogenization

It is well-known that the general solution to the system of Cauchy-Riemann equa-tions (1.1) can be given by the Pompeiu formula to the effect that,

v(z) = u(z) + v0(z) , z ∈ Dn , (2.1)

where u is an arbitrary holomorphic function on Dn and

v0 =n∑

ν=1

(−1)ν+1∑

1≤k1<···<kν≤n

TkνTkν−1 . . . Tk1fk1ζk2

···ζkν. (2.2)

Each Tk in (2.2) is the Pompeiu operator given by

(Tkf)(z) = − 1π

Dk

f(ζ)dξkdηk

ζk − zk, z ∈ D

n ,

for 1 ≤ k ≤ n, where ζk = ξk + iηk. See [1, 20] in this connection.If u is a holomorphic function on the unit polydisc D

n, then

∂u

∂νζ=

n∑

j=1

ζj∂u

∂ζj, ζ ∈ ∂0D

n , (2.3)

where γ(ζ) = γ0(ζ)√

n. See [12–14]. Thus, by (2.1), the second equation in (1.1)becomes

n∑

j=1

ζj∂u

∂ζj+ α(ζ)u = γ(ζ) , ζ ∈ ∂0D

n , (2.4)

where

γ(ζ) = γ0(ζ) −[∂v0

∂νζ+ α(ζ)v0

], ζ ∈ ∂0D

n . (2.5)

So, to find v, it is enough to find the solution u to the Robin problem for fk =0, 1 ≤ k ≤ n, the given function α and the function γ just introduced. To this end,we need to assume that in addition to the compatibility condition (1.6) satisfiedby α, the boundary value γ in (2.4) satisfies the compatibility condition

γ = γ+ + γ− ,

whereγ+(ζ) =

κ∈Z+n

γκζκ

Vol. 4 (2010) Cauchy–Riemann Systems on the Unit Polydisc 43

andγ−(ζ) =

κ∈Z+n\{0}γ−κζ−κ

for all ζ ∈ ∂0Dn. So, the assumption on γ says that γ is equal to the projection

of γ onto the boundary value PBHn(γ) of a harmonic function on Dn.

3. The n-dimensional Fuchsian type equation

Let G be holomorphic on Dn and continuous on D

n ∪ ∂0Dn such that

Re G(ζ) = h(ζ) , ζ ∈ ∂0Dn , (3.1)

where h(ζ) = Re γ(ζ) for γ(ζ) in (2.4). Then a solution to the Schwarz prob-lem (3.1) is given by

G(z) =1

(2πi)n

∂0Dn

h(ζ)[2

11 − z/ζ

− 1]

ζ+ ih0 , z ∈ D

n , (3.2)

where h0 is an arbitrary real constant. Clearly h ∈ BHn.Let us write

G(z) =∞∑

|κ|=0

gκzκ , z ∈ Dn .

and assume for a moment that (1.7) is satisfied. Comparing (3.1) with (2.4), weget

Re

⎣n∑

j=1

ζj∂u

∂ζj+ α−(ζ)u

⎦ = Re G(ζ) , ζ ∈ ∂0Dn . (3.3)

Since the function u is holomorphic on Dn and the order ord(α−(z),0) of the

singularity of α− at 0 is equal to −m = −∞, the functionn∑

j=1

zj∂u

∂zj+ α−(z)u(z)

is holomorphic on Dn\{0} and has an essential singularity at 0, which is the n-

dimentional zero vector in Cn. Therefore from (3.3) we have

n∑

j=1

zj∂u

∂zj(z) + α−(z)u(z) = α∗

(1z

)− α∗(z) + G(z) , z ∈ D

n\{0} , (3.4)

where

α∗(z) =∞∑

|κ|=0

aκzκ ,

withord(α−(z),0

)= ord

(α∗(1/z),0

),

is a holomorphic function with arbitrary complex constants aκ, κ ∈ Z+n. For the

case when n = 1, see [1, 7, 11].

44 A. Mohammed and M. W. Wong Comp.an.op.th.

We are confronted with the problem of finding the two holomorphic func-tions u and α∗ from (3.4).

4. Reduction

In this section, we show that the Robin problem for the general coefficient α givenby (1.6) can be reduced to the case (1.7) when α = α−.

Theorem 4.1. The problem (2.4) for the general coefficient α given by (1.6) can bereduced to the case when α is given by (1.7).

Proof. Comparing (3.1) with (2.4) for the case (1.6), we have

Re

⎣n∑

j=1

zj∂u

∂zj(z) +

(α+(z) + α−(z)

)u(z)

⎦ = Re G(z) , z ∈ ∂0Dn . (4.1)

Since the function u is holomorphic on Dn, the function

n∑

j=1

zj∂u

∂zj(z) +

(α+(z) + α−(z)

)u(z)

is holomorphic on Dn\{0} and has an essential singularity at the origin. Therefore

from (4.1) we haven∑

j=1

zj∂u

∂zj(z) +

(α+(z) + α−(z)

)u(z) = α∗

(1z

)− α∗(z) + G(z) (4.2)

for all z ∈ Dn\{0}. Let the function α+ be decomposed as

α+(z) =n∑

i=1

ziαi(z) , z ∈ Dn ,

where α0 is a complex constant and αi is holomorphic on Dn. Transforming the

function u to w byw = ueF+

0 ,

where

F+0 (z) =

n∑

k=1

∫ zk

0

αk(0, . . . , 0, ζk, zk+1, . . . , zn)dζk , z ∈ Dn ,

as in [15], the equation (4.2) becomesn∑

j=1

zj∂w

∂zj(z) + α−(z)w(z) =

{

α∗(

1z

)− α∗(z) + G(z)

}

eF+0 (z) (4.3)

for all z ∈ Dn\{0}. We define the function E on D

n and represent it in terms ofits power series by

E(z) = eF+0 (z) =

∞∑

|κ|=0

eκzκ , z ∈ Dn .

Vol. 4 (2010) Cauchy–Riemann Systems on the Unit Polydisc 45

Then taking into account the fact that

α∗(

1z

)E(z) =

∞∑

|τ |=0

cτz−τ +∞∑

|τ |=1

⎧⎨

∞∑

|κ|=0

aκeκ+τ

⎫⎬

⎭zτ ,

where

cτ =∞∑

|κ|=0

aτ+κeκ ,

and writing

{G(z) − α∗(z)

}E(z) +

∞∑

|τ |=1

⎧⎨

∞∑

|κ|=0

aκeκ+τ

⎫⎬

⎭zτ +

∞∑

|t|=0

cτzτ =∞∑

|κ|=0

fκzκ ,

we can rewrite the equation (4.3) asn∑

j=1

zj∂w

∂zj(z)+α−(z)w(z) =

∞∑

|τ |=0

(cτz−τ−cτzτ )+∞∑

|κ|=0

fκzκ , z ∈ Dn\{0} . (4.4)

This completes the proof. �

5. Fixing free parameters

The equation (3.4) is almost hopeless to solve, due to essential singularities onboth sides of the equation at the origin. On the boundary ∂0D

n, however, we canobtain from (3.4) and (4.3)

n∑

j=1

ζj∂u

∂ζj(ζ) + α−(ζ)u(ζ) = G(ζ) − 2iIm α∗(ζ) , ζ ∈ ∂0D

n , (5.1)

andn∑

j=1

ζj∂w

∂ζj(ζ) + α−(ζ)w(ζ) =

(G(ζ) − 2iIm α∗(ζ)

)eF (ζ) , ζ ∈ ∂0D

n . (5.2)

Now we have to fix two holomorphic functions from one equation. Usually ahalf boundary condition (Schwarz boundary condition) has to be appended to fixone holomorphic function. So applying the Robin condition (2.4) on the boundary,by taking the real and the imaginary parts from both sides of the equation, bothholomorphic functions can be fixed uniquely up to constant free terms.

Applying (2.4) and taking (2.1) into account, the equation (5.1) becomes

γ∗(ζ) = ImG(ζ) − 2 Im α∗(ζ) , ζ ∈ ∂0Dn , (5.3)

where γ∗(ζ) = Im γ(ζ). For the function w in (5.2), the condition (5.3) means that

Im

⎧⎨

⎣n∑

j=1

ζj∂w

∂ζj(ζ) + α−(ζ)w(ζ)

⎦ e−F (ζ)

⎫⎬

⎭= γ∗(ζ) , ζ ∈ ∂0D

n . (5.4)

46 A. Mohammed and M. W. Wong Comp.an.op.th.

Theorem 5.1. The holomorphic function α∗ can be uniquely determined from equa-tions (5.1) and (5.2) up to one constant term.

Proof. For equations (5.1) and (5.2), using the boundary conditions (5.3) and (5.4),we have

Im α∗(ζ) = γ∗(ζ) , ζ ∈ ∂0Dn , (5.5)

where γ∗(ζ) = 12 (Im G(ζ) − γ∗(ζ)). Since this is a Schwarz problem for the holo-

morphic function α∗, we have for all z ∈ Dn,

α∗(z) =1

(2πi)n

∂0Dn

γ∗(ζ)[2

11 − z/ζ

− 1]

ζ+ ic∗0 =

∞∑

|κ|=0

γ∗κzκ , (5.6)

where c∗0 is an arbitrary real constant. Since

α∗(z) =∞∑

|κ|=0

aκzκ , z ∈ Dn ,

it follows thataκ = γ∗

κ , κ ∈ Z+n .

So, the complex constants aκ, κ ∈ Z+n, are uniquely determined. Only Im a0 = c∗0

remains free and will be determined in the sequel. This completes the proof. �

This means that the Robin boundary condition, in the case of essentiallysingular coefficient, for the system of Cauchy-Riemann equations is well posed, ifthe holomorphic function u can be determined uniquely and that will be the casein the sequel.

6. Inhomogeneous equation with essential singularity

Let the function α− be decomposed as

α−(z) = α0 +n∑

i=1

ziα−i (z)

where α0 is a complex constant and α−i is holomorphic in D

−n .Transforming the function u to w by

w = ueF−0 , (6.1)

where

F−0 (z) =

n∑

k=1

∫ zk

∞α−

k (∞, . . . ,∞, ζk, zk+1, . . . , zn)dζk , z ∈ D−n ,

the equation (3.4) becomesn∑

j=1

zj∂w

∂zj(z) + α0w(z) = F (z) , z ∈ D

n ∪ D−n , (6.2)

Vol. 4 (2010) Cauchy–Riemann Systems on the Unit Polydisc 47

where F = H∗ + G∗,

H∗(z) = H(z)e−F−0 (z) ,

H(z) =

{

α∗(

1z

)− α∗(z)

}

andG∗(z) = G(z)e−F−

0 (z) .

Write

H∗(z) =∞∑

|κ|=−∞hκzκ ,

G∗(z) =∞∑

|κ|=−∞g∗κzκ

and

F (z) =∞∑

|κ|=−∞fκzκ .

So, it follows from (6.2) that

w(z) =∞∑

|κ|=−∞wκzκ .

Now, we define the functions H±∗ , G±

∗ , F±∗ and w± by

H+∗ (z) =

∞∑

|κ|=0

hκzκ ,

G+∗ (z) =

∞∑

|κ|=0

g∗κzκ ,

F+(z) =∞∑

|κ|=0

fκzκ ,

w+(z) =∞∑

|κ|=0

wκzκ

for all z ∈ Dn, and

H−∗ (z) =

∞∑

|κ|=1

h−κz−κ ,

G−∗ (z) =

∞∑

|κ|=1

g∗−κz−κ ,

48 A. Mohammed and M. W. Wong Comp.an.op.th.

F−(z) =∞∑

|κ|=1

f−κz−κ ,

w−(z) =∞∑

|κ|=1

w−κz−κ

for all z ∈ D−n. Thus, the equation (6.2) becomes

n∑

j=1

zj∂w+

∂zj(z) + α0w

+(z) = F+(z) , z ∈ Dn , (6.3)

andn∑

j=1

zj∂w−

∂zj(z) + α0w

−(z) = F−(z) , z ∈ D−n , (6.4)

whereF+(z) = H+

∗ (z) + G∗+(z) , z ∈ D

n ,

andF−(z) = H−

∗ (z) + G∗−(z) , z ∈ D

−n .

Since equations (6.3) and (6.4) are different in character, we treat them separately.

7. The homogeneous problem

We assume first F (z) = 0 and then we consider the problem (6.2) with G0(z) = 0.

7.1. Nonsingular equation

Lemma 7.1. Let v be holomorphic on Dn and satisfy

n∑

l=1

zl∂v

∂z�(z) + α0v(z) = 0 , z ∈ D

n . (7.1)

If v is not identically zero, then α0 must be a nonpositive integer. If α0 > 0, then vmust be identically zero.

Proof. Let v be holomorphic on Ω = Dn and satisfy the equation (7.1). Applying

Green’s theorem, we getn∑

j=1

Ω

(|vzj |2 + |vzj|2)dΩz

=n∑

j=1

Ω

[∂

∂zj(vvzj

) +∂

∂zj(vvzj

)]

dΩz

=n∑

j=1

∂Ω

v(ζ)

(∂v

∂ζj

(ζ)

)

ρζj(ζ) +

(∂v

∂ζj(ζ))

ρζj(ζ)

⎦ dsζ

|(∇ρ)(ζ)|

Vol. 4 (2010) Cauchy–Riemann Systems on the Unit Polydisc 49

=n∑

j=1

∂Ω

v(ζ)

[(∂v

∂ζj(ζ))

ρζj(ζ)

]dsζ

|(∇ρ)(ζ)|

=∫

∂0Dn

v(ζ)

⎣n∑

j=1

ζj∂v

∂ζj(ζ)

⎦dsζ√n

= − α0√n

∂0Dn

|v(ζ)|2dsζ .

This means that if Reα0 = Re α0 > 0, then vzj= vzj

= 0, i.e., v is a constantfunction. Thus, v is identically zero when α0 = 0. Suppose that α0 = 0 and theholomorphic function v satisfying (7.1) has the form

v(z) =∑

|κ| ≥0

γκzκ , z ∈ Dn .

Then by (7.1), we have∑

|κ|≥0

γκ(|κ| + α0)zκ = 0 , z ∈ Dn .

Thus, if v is not identically zero on Dn, then there exists a nonnegative integer |κ|

such that α0 + |κ| = 0. So, α0 is a nonpositive integer. �

Lemma 7.2. Let α0 be a nonpositive integer and suppose that v satisfies (7.1).Then

v(z) =∑

|κ|=−α0

Cκzκ , z ∈ Dn ,

where Cκ, |κ| = −α0, are arbitrary complex constants.

Lemma 7.3. If α0 + |κ| = 0 for all |κ| ∈ Z+, then

v(z) =∑

|κ|≥0

|κ| + α0zκ , z ∈ D

n ,

is a solution ofn∑

l=1

zl∂v

∂zl(z) + α0v(z) = H+

∗ (z) , z ∈ Dn . (7.2)

Lemma 7.4. Let −α0 = k0 ∈ Z+. Assume that v is holomorphic on D

n satisfyingn∑

l=1

zl∂v

∂zl(z) + α0v(z) =

|κ|=k0

C∗κzκ , z ∈ D

n ,

where C∗κ, |κ| = k0, are arbitrary complex constants. Then

C∗κ = 0 , |κ| = k0 .

50 A. Mohammed and M. W. Wong Comp.an.op.th.

7.2. Singular equation

Lemma 7.5. Let w− be holomorphic on D−n and satisfy

n∑

j=1

zj∂w−

∂zj(z) + α0w

−(z) = 0 , z ∈ D−n . (7.3)

If w− is not identically zero, then α0 must be an integer.

Lemma 7.6. If α0 is a negative integer satisfying (7.3), then w− is identically zero.

We need some more lemmas. We omit the proofs, which are straightforward.

Lemma 7.7. Let α0 be nonnegative integer satisfying (7.3). Then

w−(z) = Cz−α0 , z ∈ D−n ,

where C is an arbitrary complex constant.

Lemma 7.8. If α0 ∈ N, then

w−(z) =∞∑

|κ|=1

h−κ

α0 − |κ|z−κ , z ∈ D

−n ,

is a solution ton∑

j=1

zj∂w−

∂zj(z) + α0w

−(z) = H−∗ (z) , z ∈ D

−n . (7.4)

Lemma 7.9. Suppose α0 − k0 = 0, k0 ∈ Z+ and w− is holomorphic on D

−n satis-fying

n∑

j=1

zj∂w−

∂zj(z) + α0w

−(z) = C0z−k0 , z ∈ D

−n , (7.5)

where C0 is a constant. Then C0 = 0 and hence

w−(z) = C0z−k0 , z ∈ D−n ,

where C0 is an arbitrary complex constant.

From the preceding lemmas, we have the following theorems.

Theorem 7.10. If α0 /∈ Z, then the homogeneous equation (3.4), i.e., G(z) = 0, z ∈D

n, has a nontrivial solution u given by

u(z) =

⎣∞∑

|κ|=0

α0 + |κ|zκ +

∞∑

|κ|=1

h−κ

α0 − |κ|z−κ

⎦ e−F0(z) , z ∈ Dn\{0} .

If α0 + k0 = 0 for some k0 ∈ Z+, then we have the following conclusions:

Vol. 4 (2010) Cauchy–Riemann Systems on the Unit Polydisc 51

1. If h−α0 = 0, then the homogeneous equation (3.4), i.e., G(z) = H+∗ (z) =

0, z ∈ Dn, has a nontrivial solution u given by

u(z) =

⎣Cz−α0 +∞∑

|κ|=1

h−κ

α0 − |κ|z−κ

⎦ e−F0(z) , z ∈ Dn\{0} ,

where C is an arbitrary complex constant.2. If h−α0 = 0, then the homogeneous equation (3.4), i.e., G(z) = 0, z ∈ D

n,has a nontrivial solution u given by

u(z) =

⎣Cz−α0 +∞∑

|κ|=0,|κ|�=−α0

α0 + |κ|zκ +

∞∑

|κ|=1

h−κ

α0 − |κ|z−κ

⎦ e−F0(z)

for all z ∈ Dn\{0}, where C is an arbitrary complex constant.

Theorem 7.11. If α0 ∈ N, then we have the following conclusions:1. If h−α0 = 0, then the equation (3.4) with G(z) = 0, z ∈ D

n, and H+∗ (z) =

0, z ∈ D−n, has a nontrivial solution u given by

u(z) =

⎣Cz−α0 +∞∑

|κ|=0

α0 + |κ|zκ

⎦ e−F0(z) , z ∈ Dn\{0} ,

where C is an arbitrary complex constant.2. If h−α0 = 0, then the equation (3.4) with

G(z) = 0 , z ∈ Dn ,

has a nontrivial solution u given by

u(z) =

⎣Cz−α0 +∞∑

|κ|=1,|κ|�=α0

h−κ

α0 − |κ|z−κ +

∞∑

|κ|=0

α0 + |κ|zκ

⎦ e−F0(z)

for all z ∈ Dn\{0}, where C is an arbitrary complex constant.

Remark. If e−F−0 (z) is replaced by e−(F−

0 (z)+F+0 (z)), then the above two theorems

are true for the equation (4.2) with G(z) = 0, z ∈ Dn.

8. The inhomogeneous problem

If we can find a special solution to the equation (6.2), then we have the generalsolution to the equation (3.4).

Theorem 8.1. If α0 /∈ Z, then the inhomogeneous equation (3.4) is solvable andthe solution u is

u(z) =

⎝∞∑

|κ|=0

hκ + g∗κα0 + |κ|z

κ +∞∑

|κ|=1

h−κ + g∗−κ

α0 − |κ| z−κ

⎠ e−F0(z) , z ∈ Dn\{0} .

52 A. Mohammed and M. W. Wong Comp.an.op.th.

Furthermore, we have the following conclusions:1. If −α0 ∈ Z

+, then the inhomogeneous equation (3.4) is solvable if and onlyif

h−α0 + g∗−α0= 0 .

If this condition is satisfied, then the solution u to the equation (3.4) is givenby

u(z) =

⎝Cz−α0 +∑

|κ|≥0,|κ|�=−α0

hκ + g∗κα0 + |κ|z

κ +∑

|κ|≥1

h−κ + g∗−κ

α0 − |κ| z−κ

⎠ e−F0(z)

for all z ∈ Dn\{0}, where C is an arbitrary complex constant.

2. If α0 ∈ N, then the inhomogeneous equation (3.4) is solvable if and only if

h−α0 + g∗−α0= 0 .

If this condition is satisfied, then the solution u to the equation (3.4) is givenby

u(z) =

⎝Cz−α0 +∑

|κ|≥1,|κ|�=α0

h−κ + g∗−κ

α0 − |κ| z−κ +∑

|κ|≥0

hκ + g∗κα0 + |κ|z

κ

⎠ e−F0(z)

for all z ∈ Dn\{0}, where C is an arbitrary complex constant.

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Alip Mohammed and M. W. WongDepartment of Mathematics and StatisticsYork University4700 Keele StreetToronto, Ontario M3J 1P3Canadae-mail: [email protected]

[email protected]

Submitted: April 4, 2008.

Accepted: August 31, 2008.