A Robin boundary value problem in the upper half plane for the Bitsadze equation

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Doctopic: Complex Analysis

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J. Math. Anal. Appl. ••• (••••) •••–•••

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Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa

A Robin boundary value problem in the upper half plane for the Bitsadze equation Yankis R. Linares a , Carmen J. Vanegas b,∗ a

Universidad Militar Bolivariana de Venezuela, Fuerte Tiuna, Caracas, Venezuela Departamento de Matemáticas Puras y Aplicadas, Universidad Simón Bolívar, Caracas 1080-A, Venezuela b

a r t i c l e

i n f o

Article history: Received 20 December 2012 Available online xxxx Submitted by T. Witelski Keywords: Bitsadze equation Robin boundary condition

a b s t r a c t In this article we give the solvability conditions and an integral representation of the solution of a Robin problem for the Bitsadze equation in the upper half plane. In order to do that, we use classical results of complex analysis and carry out the composition of two Robin problems for the Cauchy Riemann operator. © 2014 Elsevier Inc. All rights reserved.

1. Introduction There are two basic second order differential operators in complex analysis, the Laplace operator ∂z ∂z¯ and the Bitsadze operator ∂z¯2 = ∂z¯∂z¯, where the complex partial differential operators ∂z and ∂z¯ are defined by the real partial differential operators ∂x and ∂y as ∂z = 12 (∂x − i∂y ) and ∂z¯ = 12 (∂x + i∂y ), respectively. Formally they are deduced by considering z = x + iy and z¯ = x − iy with x, y ∈ R, as independent variables using the chain rule of differentiation. The second operator leads to the model equation of second order called Bitsadze equation which corresponds to the case n = 2 of the polyanalytic equation ∂z¯n w = 0. The Dirichlet and the Neumann problems are basic boundary value problems in complex analysis. The Dirichlet problem is related to the Riemann jump problem. It is well known that a Dirichlet problem is the problem of finding a function which solves a specified partial differential equation in the interior of a given region that takes prescribed values on the boundary of the region. In the particular case of Laplace’s equation, the existence of a unique solution can be justified on the basis of the physical argument: any charge distribution on the boundary should, by the laws of electrostatics, determine an electrical potential as solution. The classical methods of potential theory allow the Dirichlet problem to be solved directly in

* Corresponding author. E-mail addresses: [email protected] (Y.R. Linares), [email protected] (C.J. Vanegas). http://dx.doi.org/10.1016/j.jmaa.2014.04.056 0022-247X/© 2014 Elsevier Inc. All rights reserved.

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terms of integral operators, for which the standard theory of compact and Fredholm operators is applicable. Dirichlet problems are typical for elliptic partial differential equations and potential theory. The Neumann problem is connected to the Dirichlet problem. The Neumann or flux boundary condition is a type of boundary condition imposed on a differential equation which specifies the values that the derivative of a solution takes on the boundary of the domain. For example, in problems of heat diffusion, an insulation type flux condition is a condition where the normal derivative is zero. Because the flux is proportional to the temperature gradient, the insulation condition requires that no heat flows across the boundary. Using Neumann functions (Green’s functions of second type) for certain partial differential operators, we can give an explicit representation of the solution to the Neumann problem for the equation determined by that operator. Building Neumann functions for partial differential operators could be a pretty difficult problem. A linear combination of Dirichlet and Neumann conditions yields the Robin condition on the boundary: ∂ν w + αw = γ, ν being the outer unit normal to the boundary and α a given function defined on the boundary. Robin boundary conditions are also called impedance boundary conditions, from their application in electromagnetic problems. They are commonly used in solving Sturm–Liouville problems which appear in many contexts in science and engineering. The aim of this paper is to solve a Robin boundary value problem for the Bitsadze equation in the upper half plane. The purpose of studying this equation lies in the importance of its applications. The Bitsadze equation arises in many areas including structural mechanics, electrostatics, magneto statics, power electromagnetic, conductive media, heat transfer and diffusion, see [12,15,4] and references therein. The mathematical formulation of our problem is as follows. Find a function w satisfying ∂z ∂z w = 0 in H, w − ∂y w = γ, ∂z w + ∂y ∂z w = γ1 on R, ¨ 1 dξ dη w(0) + = c1 , for ζ = ξ + iη, ∂z¯w(i) = c, ∂ζ¯w(ζ) π ζ H

where H denotes the upper half plane and R its boundary (the x-axis), γ, γ1 ∈ L2 (R, C) ∩ C(R, C), w ∈ Lp,2 (H, C) ∩ C 2 (H, C), p > 2 and c, c1 ∈ C (the complex plane). Here Lp,2 (H, C) means the space of ¯ 1 , C), |z|−2 f ( 1 ) ∈ Lp (H ¯ 1 , C), complex-valued functions f defined in H satisfying the properties f (z) ∈ Lp (H z 2 ¯ where H1 = {z ∈ C : |z|  1, Im(z) > 0} and C (H, C) the space of complex-valued functions f defined ¯ 1 , C), f ( 1 ) ∈ C 2 (H ¯ 1 , C). in H satisfying the properties f (z) ∈ C 2 (H z We observe two Robin conditions on the x-axis, one for the function w and the other one for its partial derivative with respect to z¯. They are combining the behavior of the function w (∂z w) and the normal derivative of the function ∂y w (∂y ∂z w) over the x-axis. The expressions in the third line represent the normalization conditions and they are introduced in order to get the uniqueness of the solution. The problem is overdetermined. Therefore we have to look for solvability conditions. In this work we give the solvability conditions and an integral representation of the solution of this problem. As far as we know the Robin problem has not been considered for higher-order model equations in the case of unbounded domains. Hence our results are the first in that direction. We have been able to get our explicit formulae by using classical results of complex analysis and carrying out the composition of two Robin problems for the Cauchy Riemann operator ∂z¯. We have organized this paper in such a way that in Section 2 we give the basic mathematical tools and known results related to our research. To avoid extending the length of the article we will not give any proof of such results. Anyone interested in the background will have to consult the provided references. In Section 3 we introduce a general Robin problem for the nonhomogeneous Cauchy Riemann equation ∂z¯w = f and derive from it two particular Robin problems which play a crucial role in the process of

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obtaining the solution of our main problem. To do that we will use the Dirichlet problem in the upper half plane [8] and some identities the proofs of which are obtained very easily. These identities are grouped in Lemma 1. In Section 4, which is the main section of this paper, we will obtain the solution of the Robin problem for the Bitsadze equation. Following the idea given by Begehr in [2], we reduce the problem into two boundary value problems for first order complex differential equations. In other words, taking wz¯ = u the problem is split into two: a Robin problem for the Cauchy Riemann equation uz¯ = 0 in H,

on R,

u + ∂y u = γ1

u(i) = c

and a Robin problem for the inhomogeneous Cauchy Riemann equation wz¯ = u

in H,

w − ∂y w = γ

1 w(0) + π

on R,

¨ ∂ζ¯w(ζ) H

dξ dη = c1 . ζ

These boundary problems are those solved in Section 3. Replacing the solution of the first problem into the solvability condition and the solution of the second one, we get a function which becomes a solution for the main problem. The proof of this fact is far from being obvious. Finally, we summarize and conclude in Section 5. 2. Preliminaries Some basic tools for solving boundary value problems for partial differential equations in complex analysis are Gauss’ Theorem and Cauchy–Pompeiu Integral Formula [17]. The generalization of these formulae to the upper half plane is as follows Theorem 1 (The Gauss theorem for the upper half plane). (See [8].) Let H := {z : Im(z) > 0} be the upper half plane, w ∈ W 1,1 (H, C) ∩ C(H, C), HR := {z : 0 < Im(z), |z| < R} and M (R, w) := max|z|=R,0θπ {|w(z)| : z = Reiθ }. Assuming RM (R, w) tends to zero as R tends to infinity, we have 1 2i

¨

ˆ∞ w(t) dt = −∞

∂z¯w(z) dx dy

and

H

1 − 2i

¨

ˆ∞ w(t) dt = −∞

∂z w(z) dx dy. H

Here and from now on W m,p stands for the Sobolev space of functions f ∈ Lp (H) for which the derivatives Dν f , ν a multi-index, up to order m also belong to Lp (H). Theorem 2 (Cauchy–Pompeiu representations for the upper half plane). (See [8].) Suppose w : H → C satisfies |w(x)|  C|x|− for |x| > k, > 0 and ∂z¯w ∈ L1 (H, C). If limR→∞ M (R, w) = 0, then the Cauchy–Pompeiu Integral Formulae for the upper half plane are given by 1 w(z) = 2πi

ˆ∞ w(t) −∞

1 w(z) = − 2πi

ˆ∞ −∞

1 1 dt − t−z π

¨ ∂ζ¯w(ζ) 0
1 1 dt − w(t) t − z¯ π

1 dξ dη, ζ −z

¨ ∂ζ w(ζ) 0
1 dξ dη. ζ −z

Now we give some nonelementary results which will be used as properties of some singular integral operators.

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In the upper half plane H, for f ∈ Lp (H, C), p > 1 the Pompeiu operator T f is defined by T f (z) = −

1 π

¨ f (ζ) H

dξ dη ζ −z

and satisfies (in the distributional sense) [18] ∂z¯T f (z) = f (z),

1 ∂z T f (z) = Πf (z) = − π

¨ f (ζ) H

dξ dη , (ζ − z)2

the integral is a Cauchy principal value integral. For a bounded domain we have Theorem 3. (See [18].) Let D ⊂ C be a bounded domain with smooth boundary ∂D and f ∈ C 1 (D, C). Then Πf (z) = T (fz )(z) −

1 2πi

ˆ ∂D

f (ζ) ¯ dζ, ζ −z

where 1 T f (z) := − π

¨ D

dξ dη , f (ζ) ζ −z

and

1 Πf (z) := − π

¨ f (ζ) D

dξ dη , (ζ − z)2

Πf is singular integral to be taken in the Cauchy principal sense. On the upper half plane this theorem becomes Theorem 4. Suppose p > 1, and 0 < α < 1. If f ∈ W 1,p (H, C) ∩ C 1,α (H, C), such that limR→∞ M (R, f ) = 0, then 1 Πf (z) = T (fz )(z) − 2πi

ˆ∞ −∞

f (t) dt. t−z

(1)

Here C 1,α (H; C) denotes the space of complex valued functions f defined and Hölder continuously differen¯ 1 , C), f ( 1 ) ∈ C 1,α (H ¯ 1 , C). tiable in H, satisfying the properties f (z) ∈ C 1,α (H z Proof. Suppose R > 0 and consider the domain HR = {z ∈ C : |z| < R, Im(z) > 0}. Then by Theorem 3, we have −

1 π

¨ f (ζ) HR

dξ dη 1 =− (ζ − z)2 π

¨ fζ (ζ) HR

1 dξ dη − ζ −z 2πi

ˆ f (ζ) ∂HR

dζ¯ . ζ −z

Since 1 − 2πi

ˆ ∂HR

1 dζ¯ =− f (ζ) ζ −z 2πi

ˆR −R

1 dt − f (t) t−z 2πi

ˆπ 0

  −Rie−iθ dθ f Reiθ Reiθ − z

(2)

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and    1 ˆπ   1 −iθ  1 −Rie dθ   , f Reiθ    RM (R, f ) iθ  2πi  Re − z 2 R − |z| 0

we obtain identity (1) after taking limit when R → ∞ in (2). 2 Definition 1. (See [13].) Let C be a smooth curve. A certain direction on C will be defined as positive. If C is a smooth open curve, around each point ζ ∈ C and different from the end points of C, a circle may be drawn with radius so small that it is divided by C into two parts, lying respectively to the right and left of C, when viewed in the positive direction of C. Accordingly they will be regarded as the left and right neighbourhoods of ζ. If C is a smooth closed curve and is the boundary of some connected region, the positive direction of C will be selected so that, if C is described in this direction, that region is always on the left or on the right. That part of the plane which lies on the left will be denoted by D+ and that to the right by D− . Now we recall the well known Plemelj–Sokhotzki Formula [9]. Theorem 5 (Plemelj–Sokhotzki Formula). Let C be a closed or open smooth curve and ϕ a Hölder continuous function defined on C. Then the Cauchy type integral 1 φ(z) = 2πi

ˆ C

ϕ(τ ) dτ τ −z

(3)

has the limiting values φ+ (ζ), φ− (ζ) at all points of the curve C not coinciding with its ends, on approaching the curve from left or from the right along an arbitrary path, and these limiting values are expressed by: 1 1 PV φ+ (ζ) = ϕ(ζ) + 2 2πi

ˆ

ϕ(τ ) dτ, τ −ζ

C

where PV

´

ϕ(τ ) C τ −ζ

1 1 φ− (ζ) = − ϕ(ζ) + PV 2 2πi

ˆ C

ϕ(τ ) dτ, τ −ζ

dτ denotes a Cauchy principal value integral.

Theorem 6. Suppose f ∈ W 1,p (H; C) ∩ C 1,α (H; C), p > 1 and 0 < α < 1. Then for z ∈ H 1 lim z→t 2πi

ˆ∞ −∞

f (t) 1 f (τ ) dτ = + PV τ −z 2 2πi

ˆ∞

−∞

f (τ ) dτ τ −t

(4)

and 1 f (t) − PV (Πf ) (t) = lim Πf (z) = − z→t 2 2πi

ˆ∞

+

−∞

Here the integral PV ˆ∞ PV −∞

´∞

f (τ ) −∞ τ −t

f (τ ) dτ + T (fz )(t). τ −t

dτ means a Cauchy principal value integral i.e.,

f (τ ) dτ = lim PV N →+∞ τ −t

ˆN

−N

  ˆt− ˆN f (τ ) f (τ ) f (τ ) dτ = lim lim dτ + dτ N →+∞ →0 τ −t τ −t τ −t −N

t+

(5)

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Proof. 1 lim z→t 2πi

ˆ∞ −∞

1 f (τ ) dτ = PV τ −z 2πi

ˆ∞

−∞

1 f (τ ) dτ + τ −t 2πi

ˆ S

f (ρ) dρ, ρ−t

where S is a semi-circle above the real axis centered at x = t and radius . Since f is a continuous function, then the second term on the right-hand side equals πif (t), when we take → 0. Therefore it follows identity (4). By Theorem 4 and using the last identity we obtain (5). 2 In order to define the Poisson kernel for the upper half plane we use the following classic theorem Theorem 7. (See [7].) Let f be an analytic function in the upper half plane H, such that |f (x + iy)|p , p  1, is integrable for each y > 0 and  ˆ∞

  f (x + iy)p dx

 p1

−∞

is bounded. Then 1 f (z) = 2πi

ˆ∞ −∞

f (t) dt, t−z

where Im(z) = y > 0,

and the integral vanishes for y < 0. Consequently, if h ∈ Lp (R, C) and 1 2πi

ˆ∞ −∞

h(t) dt = 0, t−z

where Im(z) = y < 0,

then the function 1 f˜(z) = 2πi

ˆ∞

h(t) dt, t−z

−∞

where Im(z) = y > 0

is analytic in the upper half plane H, |f˜(x + iy)|p is integrable for each y > 0 and  ˆ∞

  f˜(x + iy)p dx

 p1

−∞

is bounded. Moreover, for t0 ∈ R lim f˜(z) = lim

z→t0

After this theorem, if h ∈ Lp (R, C) and 1 lim z→t0 π

ˆ∞ −∞

z→t0

1 2πi

y 1 h(t) dt = lim z→t0 2πi (x − t)2 + y 2

1 2πi

´∞

ˆ∞ −∞

h(t) −∞ t−z

ˆ∞  −∞

h(t) dt = h(t0 ). t−z

dt = 0, we get

 ˆ∞ 1 1 1 h(t) − h(t) dt = lim dt = h(t0 ), z→t0 2πi t−z t − z¯ t−z −∞

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where Im(z) = y > 0, t ∈ R, z ∈ H. We call the function

1 1 y 1 1 − = K(z, t) = 2πi t − z t − z¯ π |t − z|2 the Poisson Kernel for the upper half plane. 3. Robin problem for the inhomogeneous Cauchy–Riemann equation in the upper half plane In order to solve the Robin problem under consideration in this section, we need some identities which will be proved in the next lemma. This will be done now to simplify calculations in the solution of the Robin problem. Lemma 1. Let O(H) be the set of analytic functions in H and α ∈ O(H) ∩ C(H; C) such that for some δ > 0, (1 + |z|)δ α(z) is bounded. Suppose for p > 2, f ∈ Lp,2 (H, C) ∩ C 1 (R, C) ∩ L2 (R, C). Then for z ∈ H we have i. ii. iii. iv. v. vi.

´ ∞ −iα(t)T f (t) ˜ 1 dt = −iα(z)T f (z) − πi H f (ζ)α(ζ) 2πi −∞ t−z ζ−z ´ ∞ −iα(t)T f (t) ˜ f (ζ)α(ζ) 1 i dt = − dξ dη. t−¯ z π ζ−¯ z −∞ H ´2πi ∞ dt = iπ. −∞´ t−z ´∞ ´∞ ∞ 1 1 dτ dt 1 dt 2πi ´−∞ 2πi PV −∞ f (τ ) τ −t t−z = 4πi −∞ f (t) t−z . ∞ 1 dt 2πi ´−∞ −T (fζ )(t) t−z = 0. ∞ 1 dt z ). 2πi −∞ −T (fζ )(t) t−¯ z = T (fζ )(¯

dξ dη.

Proof. For z ∈ H we have i. 1 2πi

ˆ∞ −∞

i −iα(t)T f (t) dt = − t−z π i =− π =−

i π

¨ H

¨ H

¨ H



1 f (ζ) 2πi 

ˆ∞ −∞

f (ζ) 1 ζ − z 2πi

α(t) dt dξ dη (t − ζ)(t − z)

ˆ∞ −∞

1 α(t) dt − t−ζ 2πi

ˆ∞ −∞

α(t) dt dξ dη t−z

 f (ζ) α(ζ) − α(z) dξ dη ζ −z

i = −iα(z)T f (z) − π

¨

f (ζ)α(ζ) dξ dη. ζ −z

H

ii. 1 2πi

ˆ∞ −∞

i −iα(t)T f (t) dt = − t − z¯ π i =− π =−

i π

¨ H

¨ H

¨ H



1 f (ζ) 2πi 

ˆ∞ −∞

f (ζ) 1 ζ − z¯ 2πi

α(t) dt dξ dη (t − ζ)(t − z¯)

ˆ∞ −∞

1 α(t) dt − t−ζ 2πi

f (ζ)α(ζ) dξ dη. ζ − z¯

ˆ∞ −∞

α(t) dt dξ dη t − z¯

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iii. If z = x + iy then ˆ∞ −∞

dt = t−z

ˆ∞

t − z¯ dt = |t − z|2

−∞



ˆ∞ −∞

t−x dt + i (t − x)2 + y 2

1 (a − x) + y = lim ln 2 a→+∞ (a + x)2 + y 2 2

2

ˆ∞

−∞



y dt (t − x)2 + y 2

a−x arctan y

+ i lim

a→+∞





−a − x − arctan y



= iπ.

iv. First we observe ˆ∞ PV −∞

dτ = lim PV τ − t N →+∞

ˆN

−N

=

lim ln

N →+∞

  ˆt− ˆN dτ dτ dτ = lim lim + τ − t N →+∞ →0 τ −t τ −t −N

t+

N −t = 0. N +t

Therefore 1 2πi

ˆ∞ −∞

1 PV 2πi

ˆ∞

−∞

−1 dτ dt = f (τ ) PV τ −tt−z 4π 2

  ˆ∞ ˆ∞ f (τ ) 1 dt PV + dτ τ −z τ −t t−z

ˆ∞

−∞

=

−1 PV 4π 2

−∞

ˆ∞

−∞

1 f (τ ) (iπ) dτ = τ −z 4πi

−∞

ˆ∞ f (t) −∞

dt . t−z

v. 1 2πi

ˆ∞ −∞

1 −T (fζ )(t) dt = − t−z π 1 =− π =−

1 π

¨ H

¨ H

¨ H



1 fζ (ζ) 2πi fζ (ζ) 1 ζ − z 2πi

ˆ∞ −∞

dt dξ dη (t − ζ)(t − z)

 ˆ∞



−∞

ˆ∞

dt − t−ζ

−∞



dt dξ dη t−z

fζ (ζ) 1 1 − dξ dη = 0. ζ −z 2 2

vi. 1 2πi

ˆ∞ −∞

1 −T (fζ )(t) dt = − t − z¯ π 1 =− π =−

1 π

¨ H

¨ H

¨ H



1 fζ (ζ) 2πi fζ (ζ) 1 ζ − z¯ 2πi 

ˆ∞ −∞

dt dξ dη (t − ζ)(t − z¯)

 ˆ∞ −∞



dt − t−ζ

ˆ∞

−∞

dt dξ dη t − z¯

fζ (ζ) 1 1 + dξ dη = T (fζ )(¯ z ). ζ − z¯ 2 2

2

Now we recall the Dirichlet problem for the Cauchy–Riemann equation in the upper half plane (see [8]), which will be also used in the proof of the Robin boundary value problem for the inhomogeneous Cauchy– Riemann equation.

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Consider γ ∈ L2 (R, C) ∩ C(R, C). Find w such that 

in H, on R.

∂z w = 0 w=γ

This problem has a unique solution if and only if for z ∈ H 1 2πi

ˆ∞ −∞

γ(t) dt = 0. t − z¯

The solution is 1 w(z) = 2πi

ˆ∞ −∞

γ(t) dt. t−z

Now we state the problem to solve. Robin problem. Let O(H) be the set of analytic functions in H. Consider α ∈ O(H) ∩ C(H; C) such that for some δ > 0, (1 + |z|)δ α(z) is bounded, γ ∈ L2 (R, C) ∩ C(R, C) and f ∈ Lp,2 (H, C) ∩ C 1 (H, C) ∩ L2 (R, C) for p > 2. Under these hypotheses we want to find a function w satisfying ⎧ in H, ⎨ ∂z w = f on R, αw − ∂y w = γ ⎩ w(i) = c + T f (i), where c ∈ C. Theorem 8. The above Robin problem is uniquely solvable if and only if for z ∈ H 1 2πi

ˆ∞ −∞



 dt i − iγ(t) + f (t) t − z¯ π

¨ H

α(ζ)f (ζ) dξ dη + T (fζ )(¯ z ) = 0. ζ − z¯

(6)

The solution is given by

w(z) = ce

−

´z

i − π

i

iα(ζ) dζ

¨

1 + 2πi ˆz

f (ζ) H

i



ˆ∞





ˆz

iγ(t) + 2f (t)

−∞

i ´u

α(ζ) − α(u)

e

z

e

´u z

iα(ζ) dζ

t−u

du dt

iα(ζ) dζ

ζ −u

du dξ dη + T f (z).

(7)

Proof. Suppose w is the solution of the Robin problem. Consider w(z) = ϕ(z) + T f (z) where ϕz¯ = 0. Then    α(z)w(z) − ∂ y w(z) = α(z) − i(∂z − ∂z¯) ϕ(z) + T f (z) = α(z)ϕ(z) + α(z)T f (z) − iϕ (z) − iΠf (z) + if (z). Taking limit when z → t, t ∈ R, we get: α(t)w(t) − ∂ y w(t) = α(t)ϕ(t) + α(t)T f (t) − iϕ (t) − i(Πf )+ (t) + if (t) = γ(t), where

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Doctopic: Complex Analysis

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Y.R. Linares, C.J. Vanegas / J. Math. Anal. Appl. ••• (••••) •••–•••

10

ˆ∞

1 1 PV (Πf ) (t) = − f (t) − 2 2πi

f (τ ) dτ + T (fζ )(t). τ −t

+

−∞

Then, 1 3 PV ϕ (t) + iα(t)ϕ(t) = iγ(t) − iα(t)T f (t) + f (t) + 2 2πi 

ˆ∞

−∞

f (τ ) dτ − T (fζ )(t) := γ˜ (t). τ −t

Since ϕ (z) + iα(z)ϕ(z) is analytic in H with boundary value γ˜ , then we have a Dirichlet problem for an analytic function on H. Hence for z ∈ H ˆ∞

1 ϕ (z) + iα(z)ϕ(z) = 2πi 

−∞

γ˜ (t) dt t−z

if and only if 1 2πi

ˆ∞ −∞

γ˜ (t) dt = 0. t − z¯

Using the identities of Lemma 1 we obtain the solvability conditions and integral representation of ϕ (z) + iα(z)ϕ(z), for z ∈ H: 1 2πi

ˆ∞



iγ(t) + f (t)

−∞

 dt i − t − z¯ π

¨ H

α(ζ)f (ζ) dξ dη + T (fζ )(¯ z ) = 0, ζ − z¯

and 1 ϕ (z) + iα(z)ϕ(z) = 2πi

ˆ∞



−∞

  dt i − iα(z)T f (z) − iγ(t) + 2f (t) t−z π

¨ H

α(ζ)f (ζ) dξ dη. ζ −z

Solving this differential equation we have

ϕ(z) = ce

−

−

´z i

i π

iα(ζ) dζ

1 + 2πi ˆz

¨ f (ζ) H

ˆ∞

  iγ(t) + 2f (t)

−∞

ˆz

e

´u z

t−u

i ´u

 e α(ζ) − α(u)

i

z

iα(ζ) dζ

du dt

iα(ζ) dζ

du dξ dη.

ζ −u

Therefore w(z) = ce

−

− and ∂z¯w = f (z).

´z

i π

i

iα(ζ) dζ

1 + 2πi ˆz

¨ f (ζ) H

i



ˆ∞





ˆz

iγ(t) + 2f (t)

−∞

i ´u

e

α(ζ) − α(u)

z

e

´u z

iα(ζ) dζ

t−u

du dt

iα(ζ) dζ

ζ −u

du dξ dη + T f (z)

JID:YJMAA

AID:18481 /FLA

Doctopic: Complex Analysis

[m3L; v 1.133; Prn:2/05/2014; 14:49] P.11 (1-18)

Y.R. Linares, C.J. Vanegas / J. Math. Anal. Appl. ••• (••••) •••–•••

11

Now suppose the solvability condition (6) is valid. Then multiplying it by i and summing the result to 1 α(z)w(z) − ∂ y w(z) = 2πi

ˆ∞

2i γ(t) dt − t−z 2πi

−∞

¨

1 − π

ˆ∞ −∞

f (t) dt t−z

f (ζ)α(ζ) dξ dη − iΠf (z) + if (z) ζ −z

H

we obtain 1 α(z)w(z) − ∂ y w(z) = π

ˆ∞ −∞

−

1 π



ˆ∞



i y dt − γ(t) − if (t) 2 |t − z| 2πi 

¨ f (ζ)α(ζ) H



−∞

f (t) dt t−z

1 1 − dξ dη − iΠf (z) + if (z) + iT (fζ )(¯ z ). ζ −z ζ − z¯

Taking limit when z → t0 and using the Plemelj–Sokhotzki Formula we get   1 lim α(z)w(z) − ∂ y w(z) = lim z→t0 z→t0 π

+

ˆ∞



γ(t) − if (t)



−∞

i y if (t0 ) − PV dt − |t − z|2 2 2πi

ˆ∞

−∞

if (t0 ) i + PV 2 2πi

ˆ∞

−∞

f (τ ) dτ τ − t0

f (τ ) dτ − iT (fζ )(t0 ) + if (t0 ) + iT (fζ )(t0 ) τ − t0

and since 1 lim z→t0 π

ˆ∞





γ(t) − if (t)

−∞

y dt = γ(t0 ) − if (t0 ), |t − z|2

it follows   lim α(z)w(z) − ∂y w(z)

z→t0

i if (t0 ) = γ(t0 ) − if (t0 ) − − PV 2 2πi

ˆ∞

−∞

+

i if (t0 ) + PV 2 2πi

ˆ∞

−∞

f (τ ) dτ τ − t0

f (τ ) dτ − iT (fζ )(t0 ) + if (t0 ) + iT (fζ )(t0 ) = γ(t0 ). τ − t0

2

Corollary 1. The Robin problem for the Cauchy–Riemann equation ⎧ ⎨ ∂z w = 0 αw − ∂y w = γ ⎩ w(i) = c,

in H, on R,

for α ∈ O(H) ∩ C(H; C) such that for some δ > 0, (1 + |z|)δ α(z) is bounded, γ ∈ L2 (R, C) ∩ C(R, C) and c ∈ C, is uniquely solvable if and only if

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AID:18481 /FLA

Doctopic: Complex Analysis

[m3L; v 1.133; Prn:2/05/2014; 14:49] P.12 (1-18)

Y.R. Linares, C.J. Vanegas / J. Math. Anal. Appl. ••• (••••) •••–•••

1 2πi

ˆ∞ γ(t) −∞

dt =0 t − z¯

(8)

for z ∈ H. The solution is given by w(z) = ce

−

´z i

i + π

iα(ζ) dζ

ˆ∞

ˆz γ(t)

−∞

e

´u z

iα(ζ) dζ

i

Im(u) du dt. |t − u|2

(9)

Remark 1. Corollary 1 appears as Theorem 16 in Section 3.5 in [8]. Taking α(z) = 1 and the normalization condition at zero in Theorem 8, we obtain the following particular case. Proposition 1. The Robin problem for the inhomogeneous Cauchy–Riemann equation ⎧ ⎨ ∂z¯w = f in H, w − ∂y w = γ on R, ⎩ w(0) = c + T f (0) (i.e. w(0) +

1 π

˜

f (ζ) ζ

H

c∈C

dξ dη = c),

¯ C) ∩ L2 (R, C), p > 2 and γ ∈ L2 (R, C) ∩ C(R, C) is uniquely solvable if for given f ∈ Lp,2 (H, C) ∩ C 1 (H, and only if for z ∈ H 1 2πi

ˆ∞ −∞

  dt + iT f (¯ z ) + T (fζ )(¯ iγ(t) + f (t) z ) = 0. t − z¯

(10)

The solution is given by

w(z) = ce

−iz

1 + 2πi

ˆ∞



z−t ˆ

 γ(t) − 2if (t) ei(t−z)

−∞

−t

eiζ dζ dt + T f (z). iζ

(11)

Proof. (10) follows from (6) when we take α = 1. From (7) with α = 1 and observing the normalization condition at zero, we obtain

w(z) = ce

−

´z 0

i dζ

= ce−iz +

1 + 2πi

1 2πi

ˆ∞

ˆ∞





ˆz

iγ(t) + 2f (t)

−∞

0

  iγ(t) + 2f (t) e−iz

−∞

ˆz 0

´u

e z i dζ du dt + T f (z) t−u eiu du dt + T f (z). t−u

Making the change u = t + ζ, the last integral becomes 1 2πi

ˆ∞



 iγ(t) + 2f (t) e−iz

−∞

1 = 2πi

z−t ˆ

−t

ˆ∞ −∞



ei(ζ+t) dζ dt −ζ

 γ(t) − 2if (t) ei(t−z)

z−t ˆ

−t

eiζ dζ dt. iζ

2

JID:YJMAA

AID:18481 /FLA

Doctopic: Complex Analysis

[m3L; v 1.133; Prn:2/05/2014; 14:49] P.13 (1-18)

Y.R. Linares, C.J. Vanegas / J. Math. Anal. Appl. ••• (••••) •••–•••

13

Remark 2. Corollary 1 and Proposition 1 appear in [8] as theorems independently from Theorem 8. However here they are deduced from Theorem 8. 4. Main result A Robin problem for the Bitsadze equation is set in the following theorem. Theorem 9. Let γ, γ1 ∈ L2 (R, C) ∩ C(R, C), w ∈ Lp,2 (H, C) ∩ C 2 (H, C) ∩ L2 (R, C), p > 2 and c, c1 ∈ C. The Robin boundary value problem for the Bitsadze equation ⎧ ∂ ∂ w = 0 in H, z z ⎪ ⎪ ⎨ w − ∂y w = γ on R, ∂z w + ∂y ∂z w = γ1 on R, ⎪ ⎪ ˜ ⎩ w(0) + π1 H ∂ζ¯w(ζ) dξζdη = c1 ,

∂z¯w(i) = c

is uniquely solvable if and only if for z ∈ H 1 2πi

ˆ∞ γ1 (t) −∞

dt =0 t − z¯

(12)

and 1 2πi

 ˆ∞  ˆt dt it −i(τ −t) + iT u(¯ z ) + T (uζ )(¯ iγ(t) + c2 e − iγ1 (τ )e dτ z ) = 0, t − z¯

−∞

(13)

0

´∞ ´z Im(s) where u(z) = ce1+iz − πi −∞ γ1 (t) i ei(z−s) |t−s| 2 ds dt and c2 = u(0). The solution is given by w(z) = c1 e−iz + T u(z)   z−t ˆ∞  ˆt ˆ 1 eiζ it −i(τ −t) i(t−z) dζ dt. dτ e γ(t) − 2i c2 e − iγ1 (τ )e + 2πi iζ −∞

(14)

−t

0

Proof. Taking wz¯ = u the problem is divided in two parts uz¯ = 0 in H, wz¯ = u,

in H,

u + ∂y u = γ1

w − ∂y w = γ

on R,

on R,

u(i) = c, ¨ 1 dξ dη = c1 . w(0) + ∂ζ¯w(ζ) π ζ

(15) (16)

H

Note that since u ∈ C 1 (H, C) then u ∈ C 1 (H1 , C) and u( z1 ) ∈ C 1 (H1 , C). This imply that u ∈ Lp (H1 , C) and z −2 u( z1 ) ∈ Lp (H1 , C) which in turn means that u ∈ Lp,2 (H, C), p > 2. On the other hand, as |u|2  |γ1 |2 on R and γ1 ∈ L2 (R, C), then u ∈ L2 (R, C). Therefore the function u ∈ Lp,2 (H, C) ∩ C 1 (H, C) ∩ L2 (R, C), p > 2. According to Corollary 1, the problem (15) is uniquely solvable if and only if for z ∈ H 1 2πi

ˆ∞ γ1 (t) −∞

dt =0 t − z¯

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AID:18481 /FLA

Doctopic: Complex Analysis

[m3L; v 1.133; Prn:2/05/2014; 14:49] P.14 (1-18)

Y.R. Linares, C.J. Vanegas / J. Math. Anal. Appl. ••• (••••) •••–•••

14

and the solution is 1+iz

u(z) = ce

i − π

ˆ∞

ˆz ei(z−s)

γ1 (t) −∞

i

Im(s) ds dt. |t − s|2

(17)

On the other hand, by the Proposition 1, the problem (16) is uniquely solvable if and only if for z ∈ H 1 2πi

ˆ∞ −∞

  dt + iT u(¯ z ) + T (uζ )(¯ z) = 0 iγ(t) + u(t) t − z¯

(18)

and its solution is given by

w(z) = c1 e

−iz

1 + 2πi

ˆ∞



 γ(t) − 2iu(t) ei(t−z)

−∞

z−t ˆ

−t

1 eiζ dζ dt − iζ π

¨ H

u(ζ) dξ dη. ζ −z

(19)

Observe that u + ∂y u = γ1 if and only if u (t) − iu(t) = −iγ1 (t). Therefore ˆt u(t) = c2 e − it

iγ1 (τ )e−i(τ −t) dτ.

(20)

0

Substituting (20) into (18) and (19) we get 1 2πi

ˆ∞ 

 iγ(t) +

ˆt

c2 eit −

−∞

 iγ1 (τ )e−i(τ −t) dτ

0

dt + iT u(¯ z ) + T (uζ )(¯ z) = 0 t − z¯

and w(z) = c1 e−iz + T u(z)   z−t ˆ∞  ˆt ˆ 1 eiζ it −i(τ −t) i(t−z) + dζ dt, dτ e γ(t) − 2i c2 e − iγ1 (τ )e 2πi iζ −∞

−t

0

respectively. Now we will prove that if (12) and (13) are satisfied, then (14) is the solution. First we write

 ∂z¯w(z) + ∂y ∂z¯w(z) = u(z) + ∂y u(z) = u(z) + i u (z) − ∂z¯u(z) = u(z) + iu (z). Since 

1+iz

u (z) = ice

i − π

ˆ∞ −∞

y 1 γ1 (t) dt + 2 |t − z| π

ˆ∞

ˆz ei(z−s)

γ1 (t) −∞

i

then 1 iu (z) = −u(z) + π 

ˆ∞ γ1 (t) −∞

y dt. |t − z|2

Im(s) ds dt, |t − s|2

(21)

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AID:18481 /FLA

Doctopic: Complex Analysis

[m3L; v 1.133; Prn:2/05/2014; 14:49] P.15 (1-18)

Y.R. Linares, C.J. Vanegas / J. Math. Anal. Appl. ••• (••••) •••–•••

15

Therefore 1 u(z) + iu (z) = π

ˆ∞



γ1 (t) −∞

y dt. |t − z|2

(22)

Substituting (22) in (21) we have 1 ∂z¯w(z) + ∂y ∂z¯w(z) = π

ˆ∞ γ1 (t) −∞

y dt. |t − z|2

Now taking limit when z → t0 ∈ R, we obtain 

ˆ∞



1 lim ∂z¯w(z) + ∂y ∂z¯w(z) = lim z→t0 z→t0 π

γ1 (t) −∞

y dt = γ1 (t0 ). |t − z|2

From the last equality it follows u(t) + iu (t) = γ1 (t), then the boundary value of u(z) is given by ˆt u(t) = c2 e − it

iγ1 (τ )e−i(τ −t) dτ,

0

hence (13) and (14) can be written in the following form 1 2πi

ˆ∞



iγ(t) + u(t)

−∞

 dt + iT u(¯ z ) + T (uζ )(¯ z) = 0 t − z¯

and w(z) = c1 e

−iz

1 + 2πi

ˆ∞

  γ(t) − 2iu(t) ei(t−z)

−∞

z−t ˆ

−t

eiζ dζ dt + T u(z), iζ

respectively. Now we will prove w − ∂y w = γ on R.   w(z) − ∂y w(z) = w(z) − i ∂z w(z) − ∂z¯w(z) .

(23)

We calculate ∂z w and ∂z¯w. ∂z w(z) = −ic1 e

+

−iz

1 2πi

i − 2πi

ˆ∞

ˆ∞

 γ(t) − 2iu(t) ei(t−z)

−∞

−t



γ(t) − 2iu(t)

−∞

z−t ˆ

eiζ dζ dt iζ

dt + Πu(z) i(z − t)

and ∂z¯w(z) = u(z). Substituting the last two equalities in (23) we have: 1 w(z) − ∂y w(z) = T u(z) + 2πi

ˆ∞ −∞

 dt − iΠu(z) + iu(z). γ(t) − 2iu(t) t−z

(24)

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AID:18481 /FLA

Doctopic: Complex Analysis

[m3L; v 1.133; Prn:2/05/2014; 14:49] P.16 (1-18)

Y.R. Linares, C.J. Vanegas / J. Math. Anal. Appl. ••• (••••) •••–•••

Multiplying (13) by i we obtain 1 2πi

ˆ∞ −∞

  dt − T u(¯ z ) + iT (uζ )(¯ −γ(t) + iu(t) z) = 0 t − z¯

(25)

then, adding (25) to (24) we get 1 w(z) − ∂y w(z) = iu(z) + π

ˆ∞

−∞



y i γ(t) − iu(t) dt − 2 |t − z| 2πi

ˆ∞ u(t) −∞

dt t−z

+ T u(z) − T u(¯ z ) − iΠu(z) + iT (uζ )(¯ z ).

(26)

Taking limit when z → t0 ∈ R in (26) we have 



1 lim w(z) − ∂y w(z) = iu(t0 ) + lim z→t0 z→t0 π − lim

z→t0

i 2πi

ˆ∞

 γ(t) − iu(t)

−∞

ˆ∞ u(t) −∞

y dt |t − z|2

dt + lim T u(z) − lim T u(¯ z) z→t0 t − z z→t0

− i lim Πu(z) + i lim T (uζ )(¯ z ). z→t0

z→t0

(27)

Applying Theorem 6 it follows 1 lim z→t0 2πi

ˆ∞ −∞

u(t0 ) 1 u(τ ) dτ = + PV τ −z 2 2πi

ˆ∞

−∞

1 u(t0 ) (Πu) (t0 ) = lim Πu(z) = − − PV z→t0 2 2πi

ˆ∞

+

−∞

u(τ ) dτ , τ − t0

u(τ ) dτ + T (uζ )(t0 ) τ − t0

(28)

(29)

and 1 lim z→t0 π

ˆ∞

 γ(t) − iu(t)

−∞

y dt = γ(t0 ) − iu(t0 ). |t − z|2

(30)

Substituting (28), (29) and (30) in (27) we have   lim w(z) − ∂y w(z) = γ1 (t0 ).

z→t0

Finally it is easy to check w(0) = c1 + T u(0) and since ∂z¯w(z) = u(z) then w(0) = c1 + T ∂z¯w(0) = ˜ c1 − π1 H ∂ζ¯w(ζ) dξζdη and also ∂z¯w(i) = u(i) = c. 2 Remark 3. The Robin problem for the inhomogeneous Bitsadze equation involves other iterated integrals that make the calculation for the solvability conditions and integral representation of the solution longer and more difficult. Having solved the inhomogeneous problem, one can show the existence and uniqueness of the solution of the Robin problem for polyanalytic equations applying the method used in the main theorem and mathematical induction. To present the integral representations of the solution and solvability conditions

JID:YJMAA

AID:18481 /FLA

Doctopic: Complex Analysis

[m3L; v 1.133; Prn:2/05/2014; 14:49] P.17 (1-18)

Y.R. Linares, C.J. Vanegas / J. Math. Anal. Appl. ••• (••••) •••–•••

17

requires a laborious calculation of iterated integrals and kernel functions. In [3] the Robin problem for the inhomogeneous polyanalytic equation of arbitrary order is investigated in the unit disc. 5. Concluding remarks In this paper we have introduced and solved the Robin boundary value problem for the Bitsadze equation in an unbounded domain. We have given an integral representation for the solution expressing it through the Robin boundary values of the solution and of its derivative with respect to z¯. Assuming these values as given, the resulting function in general fails to satisfy the respective boundary conditions. Therefore necessary and sufficient solvability conditions are described for this problem. In order to get unique solution two normalization conditions are introduced. This work is also concerned with the continue development of basic boundary value problems for complex partial differential equations restricted to model equations [1] which have a lot of applications in physics and engineering, see [9,11,14,18,19,5,16,10]. After this work we can deal with other types of boundary problems. For instance, mixed boundary value problems for the upper half plane, first quarter or cracked unbounded domains. These problems can be investigated for the model equation studied here or for other models such as the Laplace and Helmholtz equations. Also they can be treated in different ways. We can follow the composition method used here or construct special kernel functions [19]. On the other side, in some cases integral representation formulae for the solutions provide the tool to deal with more general partial differential equations whose leading term is a model operator, reducing them to some singular integral equations to which the Fredholm theory applies [18,1]. Algorithms for the fast and accurate computation can be presented to solve the singular integrals that arise in the solutions [6]. Acknowledgments The first version of this paper was done when the first author was working at Universidad Simón Bolívar. References [1] Ü. Aksoy, O. Çelebi, A survey on boundary value problems for complex partial differential equations, Adv. Dyn. Syst. Appl. 5 (2) (2010) 133–158. [2] H. Begehr, Boundary value problems in complex analysis I, Bol. Asoc. Mat. Venez. XII (2) (2005) 65–85; H. Begehr, Boundary value problems in complex analysis II, Bol. Asoc. Mat. Venez. XII (2) (2005) 217–250. [3] H. Begehr, G. Harutjunjan, Robin boundary value problem for the Cauchy–Riemann operator, Complex Var. Theory Appl. 50 (15) (2005) 1125–1136. [4] A.V. Bitsadze, Boundary Value Problems of Second Order Elliptic Equations, North-Holland Publishing Company, Amsterdam, 1968. [5] S. Ciulli, S. Ispas, M.K. Pidcock, A. Stroian, On a mixed Neumann–Robin boundary value problem in electrical impedance tomography, Z. Angew. Math. Mech. 80 (10) (2000) 681–696. [6] P. Daripa, A fast algorithm to solve nonhomogeneous Cauchy–Riemann equations in the complex plane, SIAM J. Sci. Statist. Comput. 13 (6) (1992) 1418–1432. [7] P.L. Duren, Theory of H P Spaces, Academic Press, New York, 1970. [8] E. Gaertner, Basic complex boundary value problems in the upper half plane, PhD thesis, FU Berlin, 2006, www.diss. fu-berlin.de, 2006. [9] F.D. Gakhov, Boundary Value Problems, Pergamon Press, Oxford, 1966. [10] D. Medkova, P. Krutitskii, Neumann and Robin problems in a cracked domain with jump conditions on cracks, J. Math. Anal. Appl. 301 (2005) 99–114. [11] E. Meister, F.-O. Speck, A contribution to the quarter-plane problem in diffraction theory, J. Math. Anal. Appl. 130 (1988) 223–236. [12] C. Miranda, Partial Differential of Elliptic Type, Springer-Verlag, New York, 1970. [13] N.I. Muskhelishvili, Singular Integral Equations, Noordhoff International Publishing, Groningen, 1953 (English translation). [14] N.I. Muskhelishvili, Some Basic Problems of the Mathematical Theory of Elasticity, Noordhoff International Publishing, The Netherlands, 2010.

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Doctopic: Complex Analysis

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Y.R. Linares, C.J. Vanegas / J. Math. Anal. Appl. ••• (••••) •••–•••

[15] M. Schechter, L. Bers, F. John, Partial Differential Equations, Interscience, New York, 1964. [16] A. Soldatov, On elliptic boundary value problems on upper half-plane, Complex Var. Elliptic Equ. 50 (7–11) (10 June– 15 September 2005) 719–731. [17] W. Tutschke, H. Vasudena, An Introduction to Complex Analysis: Classical and Modern Approaches, Chapman and Hall, 2004. [18] I.N. Vekua, Generalized Analytic Functions, Pergamon Press, Oxford, 1962. [19] I.N. Vekua, New Methods for Solving Elliptic Equations, North-Holland Publ. Co., John Wiley, Amsterdam, New York, 1967.