R E S E A R C H
Open Access
One class of generalized boundary value
problem for analytic functions
Pingrun Li
**Correspondence:
lipingrun@163.com
School of Mathematical Sciences, Qufu Normal University, Jingxuanxi Road 57, Qufu, Shandong 273165, P.R. China
Abstract
In this paper, a boundary value problem for analytic functions with two unknown functions on two parallel straight lines is studied, the general solutions in the different domains as well as the conditions of solvability are obtained in class{1}, and the behaviors of solutions are discussed atz=∞and in the different domains, respectively. Therefore, the classic Riemann boundary value problem is extended further.
Keywords: boundary value problem for analytic functions; index; canonical function; the function class{1}
1 Introduction and preliminaries
Many mathematicians have studied the boundary value problems of analytic functions and formed a perfect theoretical system; see [–]. The boundary value problem of an-alytic functions on an infinite straight line has been studied in the literature, and there has been a brief description of boundary value problems of analytic function with an un-known function on several parallel lines. In this paper, we will put forward the boundary value problems of analytic functions with two unknown functions on two parallel lines and a general method different from the one in classical boundary value theory. Moreover, we will give and discuss the general solution and solvability conditions, which will generalize the classical theory of boundary value problems of analytic functions.
Let us describe the definitions ofPlemeljformula and function class{}on an infinite straight line.
Definition . Assume thatω(x) is a continuous complex function on the real axisX. We say thatω(x)∈ ˆHif the following conditions hold:
() For any sufficiently large positive numberM,ω(x)satisfiesω(x)∈Hon[–M,M] (see [] for the definition ofH).
() |ω(x) –ω(x)| ≤A|x–x|, for any|xj|>M(j= , ) and some positive real
numberA.
Under condition (), we say thatω(x) satisfies the Hölder condition onN∞and denote
itω(x)∈H(N∞), whereN∞={x:|x|>M}is a neighborhood of∞.
Definition . Assume thatω(x) is continuous on (–∞,∞) and–∞∞|ω(x)|dx< +∞, then we say thatω(x)∈L(–∞,∞).
Definition . Ifω(x) satisfies: ()ω(x)∈ ˆH, ()ω(x)∈L(–∞,∞), then we say thatω(x)
belongs to the function class{}.
Definition . Assume thatω(x)∈ {}, then the integrals+(z) =√
π +∞
ω(t)eitzdtand
–(z) =√
π
–∞ω(t)eitzdtare called the left and right one-sided Fourier integral,
respec-tively.
Lemma .(see []) Ifω(z)∈H with respect to any finite part of some infinite domain D, andω(z)is analytic in any neighborhood of infinity,thenω(z)∈ ˆH.
Lemma .(see []) Ifω(t)belongs to the class{},then the left and right one-sided Fourier integrals defined in Definition.are analytic whenImz> andImz< ,respectively.
Lemma .(see []) Ifω(x)∈ ˆH,we have the Cauchy type integral(z) =πi–∞∞ωt–(tz)dt, z∈/(–∞,∞),then the following formula holds on the infinite straight line:
±(x) =±
ω(x) +(x), i.e.,
±(x) =±
ω(x) + πi
∞
–∞
ω(t) t–xdt.
2 Problem presentation
Now, we put forward the boundary value problem of analytic functions on two parallel lines.
Without loss of generality, we assume that the two lines are parallel to theX-axis (oth-erwise, we can translate them into this case by a linear transformation), and denote them byL,L, whereLjcan be expressed byζ =x+ilj(x∈(–∞,∞),l<lare real numbers)
and take the direction from left to right as the positive direction. LetL=L+L.
We want to get functions(z) and(z) such that(z) is analytic in{Imz>l} ∪ {Imz<
l},(z) is analytic in{z:l<Imz<l}, and we have the following boundary value
condi-tions: ⎧ ⎨ ⎩
+(ζ) =D
(ζ)–(ζ) +G(ζ), whenζ∈L,
–(ζ) =D(ζ)+(ζ) +G(ζ), whenζ ∈L,
(.)
whereLj:ζ=x+ilj(j= , ),x∈(–∞,∞).
Actually, (.) is a boundary value problem on two parallel straight linesImz=l,Imz=
lwith∞as a pole. Here+(ζ) is the boundary value of analytic function+(z) which
is analytic in{z:Imz>l}and belongs to the class{}onL,–(ζ) is the boundary value
of analytic function–(z) which is analytic in{z:Imz<l
}and belongs to the class{}
on L, and±(ζ) is the boundary value of analytic function (z) which is analytic in
{z:l<Imz<l}and belongs to the class{}onL,L, respectively. The functionsD(ζ)
andD(ζ) belong toHˆ onL,L, respectively. The functionsG(ζ) andG(ζ) belong to the
class{}onL,L, respectively. Hence, for the functions appearing in (.) the one-sided
limits exist whenx→ ∞onL,L.
It can be seen from (.) that the order of(z) is equal to that of(z) at infinity. There-fore, if the orders of(z) and(z) aremat infinity, then such a problem can be denoted asRm. Actually, problemRand problemR–are often discussed. On the problemR, both
assumed to be zero. Such a problemRis called regular ifDj(ζ) is not zero onL; otherwise,
it is called irregular or of exception type.
Remark . Since the positive direction ofLjis the direction from left to right, when the
observer moves from left to right onLj, the boundary values of left region ofLjis
posi-tive boundary value,i.e., the positive boundary value of(z) is the boundary value above L, and the negative boundary value of(z) is ones belowL. The positive or negative
boundary values of(z) can be defined in a similar way.
3 Resolution
We only consider problemRin this paper. Hence, we assume(∞) and(∞) are finite
and nonzero. For problemRm, similar arguments can be used. In this paper, we only
con-sider the normal case, that is,Dj(ζ) (j= , ) does not have zeroes and poles onLj. For the
irregular case, similar discussions also work (see Section . of Chapter in []). Equation (.) can be written as
⎧ ⎨ ⎩
+(ζ) =D(ζ)–(ζ) +G(ζ), whenζ∈L,
+(ζ) =D
(ζ)
–(ζ) –G(ζ)
D(ζ), whenζ ∈L.
(.)
In order to unify, letC(ζ) =G(ζ),C(ζ) =G(ζ)/D(ζ), and the above equation can be
transformed into ⎧
⎨ ⎩
+(ζ) =D(ζ)–(ζ) +C(ζ), whenζ∈L,
+(ζ) =
D(ζ)
–(ζ) –C
(ζ), whenζ∈L.
(.)
By putting κ=IndLD(ζ), κ=IndLD(ζ), andκ =
j=κj, we callκ as the index of
problem (.). Without loss of generality, we take three pointsz,z,zon theZplane such
thatl<Imz,l<Imz<l,Imz<l. Then we take the following piecewise function:
Y(z) =
⎧ ⎨ ⎩
e(z), Imz>l
,
(z–z z–z)
ke(z), Imz<l
,
Y(z) =
⎧ ⎨ ⎩
(z–z z–z)
ke(z), Imz>l
,
e(z), Imz<l
,
(.)
herej(z) (j= , ) is defined as follows:
j(z) =
√
π
+∞+ilj ilj
rj(t)eitzdt, whenlj<Imzj
and
j(z) =
√
π
ilj
–∞+ilj
rj(t)eitzdt, whenImzj<lj,
where
r(t) =
√
π
+∞+il
–∞+il
logD˜(τ)·e–iτtdτ, D˜(τ) =
τ–z
τ–z
κ
D(τ),
r(t) =
√
π
+∞+il
–∞+il
logD˜(τ)·e–iτtdτ, D˜(τ) =
τ–z
τ–z
κ
The function j(z) defined above is analytic on the complex plane except L andL.
The logarithmic function of the integrand has a certain analytic branch such that
logt–z
t–zj|t=∞= ; thenY
+
j (z) andYj–(z) are analytic in{z:Imz>lj}and{z:Imz<lj},
re-spectively. Moreover,
Y+ (t)
Y–(t)=
t–z
t–z –k
e+(t)––(t)=
t–z
t–z –k
exp
√
π
+∞+il
–∞+il
r(ζ)eitζdζ
,
owing to
Vr(t)
=√
π
+∞+il
–∞+il
r(ζ)eitζdζ,
by the representative ofr(t) as well as the relationship between Fourier transform and
inverse Fourier transform, we haveV[r(t)] =log[(tt––zz)kD(t)], therefore
Y+ (t) Y– (t) =
t–z
t–z –k
exp
log
t–z
t–z
k
D(t)
=D(t). (.)
Similarly, one has
Y+ (t)
Y–(t)=D
–
(t). (.)
Putting (.), (.) into (.), we can obtain
⎧ ⎨ ⎩
+(t)[Y+
(t)]–=–(t)[Y–(t)]–+C(t)[Y+(t)]–, t∈l,
+(t)[Y+(t)]–=–(t)[Y–(t)]––C(t)[Y+(t)]–, t∈l.
(.)
In (.), the first equality is multiplied by [Y+(t)]–, the second one is multiplied by
[Y–
(t)]–, then
⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
+(t)[Y+
(t)]–[Y+(t)]–
=–(t)[Y+
(t)]–[Y+(t)]–+C(t)[Y+(t)]–[Y+(t)]–, t∈l,
+(t)[Y+
(t)]–[Y–(t)]–
=–(t)[Y–
(t)]–[Y–(t)]––C(t)[Y+(t)]–[Y–(t)]–, t∈l,
(.)
denoting
F+(z) =√ π
+∞+il il
f(τ)eiτzdτ, F–(z) = –
√
π
il
–∞+il
f(τ)eiτzdτ, (.)
where
f(t) =
√
π
+∞+il
–∞+il
C(τ)
Y+(τ)Y+(τ)e
Using Lemma ., we know thatF+(z),F–
(z) are analytic inImz>l,Imz<l, respectively.
OnL, we have
F+(t) –F–(t) = C(t) Y+
(t)Y+(t)
.
Again we denote
F+(z) =√ π
+∞+il il
f(τ)eiτzdτ, F–(z) = –
√
π
il
–∞+il
f(τ)eiτzdτ, (.)
where
f(t) =
√
π
+∞+il
–∞+il
C(τ)
Y–(τ)Y+(τ)e
–iτtdτ.
Similarly,F+(z),F–(z) are analytic inImz>l,Imz<l, respectively. OnL, we obtain
F+(t) –F–(t) = C(t) Y–(t)Y+(t). Then (.) may be reduced to
⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
+(t)[Y+
(t)Y+(t)]––F+(t)
=–(t)[Y–
(t)Y+(t)]––F–(t), t∈l,
+(t)[Y–
(t)Y+(t)]–+F–(t)
=–(t)[Y–
(t)Y–(t)]–+F–(t), t∈l,
(.)
in the two sides of the first equation of (.), by addingF+(t); in the two sides of the second one by subtraction ofF–
(t), we have
⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
+(t)[Y+
(t)Y+(t)]––F+(t) +F+(t)
=–(t)[Y–
(t)Y+(t)]––F–(t) +F+(t), t∈l,
+(t)[Y–(t)Y+(t)]–+F–(t) –F–(t) =–(t)[Y–
(t)Y–(t)]–+F–(t) –F–(t), t∈l,
(.)
the left side of the first equation of (.) is denoted by M+
(t), the right side of one is
denoted byM–
(t); the left side of the second equation of (.) is denoted byM+(t), the
right side of this one is denoted byM–(t). Let
M(z) =
⎧ ⎨ ⎩
M+
(z), Imz>l,
M–(z), Imz<l,
(.)
where
() We firstly consider the solutions ofM+(z) andM–
(z), respectively inImz>l and
Imz<l.
Case:k≥. Since [Y+(z)]–, [Y+
(z)]–are analytic in{z:Imz>l}, [Y+(z)Y+(z)]–is analytic. It
fol-lows fromF+
j(z) (j= , ) is analytic thatM+(z) is analytic. Hence, thelimz→∞(Imz>l)M+(z)
exists. Suppose thatM+
(z) =H(z), then
+(z) =Y+(z)Y+ (z)
F+(z) –F+(z) +H(z)
, Imz>l, (.)
whereH(z) is analytic in{z:Imz>l}and thelimz→∞(Imz>l)H(z) exists.
When Imz<l, since –(z) is defined in {z :l<Imz<l}, z is a k-order pole of
[Y–
(z)Y+(z)]–(k> ). In order to ensure thatM–(z) is bounded at a polez, we can
mul-tiply by a factor (z–z)k. Thus, it has ak-order atz=∞,i.e., we have a polynomial withk
degree. Let (z–z)kM–(z) =pk(z), hence,M–(z) =
pk(z)
(z–z)k,i.e.,
–(z)Y–(z)Y+(z)––F–(z) +F+(z) = pk(z) (z–z)k
, Imz<l, (.)
wherepk(z) =C+C(z–z) +· · ·+Ck(z–z)kis a polynomial with degree no more thanκ.
Case:k< .
It follows from similar arguments as above thatM+(z) is analytic in{z:Imz>l}.
Because [Y–
(z)Y+(z)]–,F–(z) andF+(z) are analytic in{z:Imz<l}, so isM–(z).
More-over,M+
(t) =M–(t) onL. Hence,M(z) is holomorphic in the whole complex plane and
thelimz→∞M(z) exists. By the Liouville theorem and the principle of analytic
continua-tion, there is a constantCsuch thatM(z) =C, and one has
–(z) =Y–(z)Y+(z)F–(z) –F+(z) +C, Imz<l, (.)
–(z) =Y+(z)Y+(z)F+(z) –F+(z) +C, Imz>l. (.)
Noticing thatz=zis a pole ofY–(z)Y+(z) with order –k,–(z) has a singularity atz.
In order to ensure that–(z) is analytic in{z:Imz<l}(in fact, for solving–(z) in the
range thatImz<l, we should only consider the case thatl<Imz<l). For the case that
k= –, it is sufficient to eliminate the singularity by puttingC=F+
(z) –F–(z), then one
can define–(z) in the following way:
–(z) =Y–(z)Y+(z)F–(z) –F+(z) +F+(z) –F–(z)
. (.)
Whenκ≤–, such aCstill cannot eliminate the singularity of–(z) atz=z. But the
following condition should be satisfied:
F+(q)(z) –F–(q)(z) = , q= , , . . . , –κ– ,
i.e.,
il
–∞+il
τqf(τ)eiτzdτ+
+∞+il il
(.) is a solution of (.) if and only if (.) holds. Therefore,
ask> , –(z) =Y–(z)Y+(z)
F–(z) –F+(z) + pk(z) (z–z)k
, Imz<l, (.)
+(z) =Y+(z)Y+(z)F+(z) –F+(z) +H(z)
, Imz>l, (.)
ask< , –(z) =Y–(z)Y+(z)F–(z) –F+(z) +C, Imz<l, (.)
+(z) =Y+ (z)Y+(z)
F+
(z) –F+(z) +C
, Imz>l. (.)
However, the solvability conditions should be satisfied (.) fork< –. Similarly, we can define the following piecewise function:
M(z) =
⎧ ⎨ ⎩
M+(z), Imz>l,
M–
(z), Imz<l,
(.)
where
M+(z) =+(z)Y–(z)Y+(z)––F–(z) +F–(z), (.) M–(z) =–(z)Y–(z)Y–(z)––F–(z) +F–(z). (.)
() We secondly consider the solutions ofM+
(z) andM–(z), respectively, inImz>land
Imz<l.
Case:k≥.
SinceY–(z) andY–(z) are analytic in{z:Imz<l}, so is [Y–(z)Y–(z)]–. It follows from
F–
j(z) (j= , ) being analytic thatM–(z) is analytic. Hence, thelimz→∞(Imz<l)M–(z) exists.
Suppose thatM–
(z) =H(z), then
–(z) =Y–(z)Y–(z)F–(z) –F–(z) +H(z)
, (.)
whereH(z) is analytic in{z:Imz<l}and thelimz→∞(Imz<l)H(z) exists.
WhenImz>l,z=zis ak-order pole of [Y–(z)Y+(z)]–(k> ), thenz=zis ak-order
pole ofM+(z). By similar arguments to above, one has
+(z)Y–(z)Y+(z)––F–(z) +F+(z) = pk(z) (z–z)k
. (.)
Case:k< .
Since [Y–(z)Y–(z)]–has no singularity in{z:Imz<l},M–(z) is analytic in{z:Imz<l}.
Noticing that [Y–
(z)Y+(z)]–is analytic in{z:Imz>l}, one finds thatM+(z) is analytic
in{z:Imz>l}andM+(t) =M–(t) onL. Therefore,M(z) is holomorphic in the whole
complex plane and thelimz→∞M(z) exists. By similar arguments to (), there exists a
constantC, such thatM(z) =Cand
+(z) =Y–(z)Y+(z)F–(z) –F+(z) +C, Imz>l,
–(z) =Y+(z)Y+(z)F+(z) –F+(z) +C, Imz<l.
Moreover, (.) is also a necessary condition for solvability. Hence,
ask> , +(z) =Y–(z)Y+(z)
F–(z) –F+(z) + pk(z) (z–z)k
, (.)
–(z) =Y–(z)Y–(z)F–(z) –F–(z) +H(z)
, (.)
ask< , +(z) =Y–(z)Y+(z)F–(z) –F+(z) +C, (.)
–(z) =Y+(z)Y+(z)F+(z) –F+(z) +C. (.)
Collecting results, fork≥ andl<Imz<l, one has
(z) =Y–(z)Y+(z)
F–(z) –F+(z) + pk(z) (z–z)k
, (.)
and, fork< andl<Imz<l,
(z) =Y–(z)Y+(z)F–(z) –F+(z) +C, (.)
wherepk(z) andCas above. Forz∈ {z:Imz>l},
+(z) =Y+(z)Y+(z)F+(z) –F+(z) +H(z)
, (.)
whereH(z) is analytic in{z:Imz>l}ask≥ and thelimz→∞(Imz>l)H(z) exists. When
k< ,H(z)≡C(constant). Forz∈ {z:Imz<l},
–(z) =Y–(z)Y–(z)F–(z) –F–(z) +H(z)
, (.)
whereH(z) is analytic in{z:Imz<l}whenk≥ andlimz→∞(Imz<l)H(z) exists.H(z)≡
C(constant) whenk< .
Hence we get the solution of the boundary value problem (.).
Theorem . The boundary value problem(.)with two unknown functions(z)and
(z)on two parallel lines has a solution in{z:l<Imz<l}and{Imz>l} ∪ {Imz<l},
respectively.Moreover,the general solution can be expressed by(.)-(.),where Yj±(z) (j= , )is defined by(.)and Fj(z) (j= , )are defined by(.)and(.).Whenκ> –,
pk(z)is a polynomial withκorder,and whenκ≤–,the necessary conditions for solvability
still are(.).In all,the degree of freedom of the solution isκ+ .
4 Further discussion on solution and solvability conditions
In this section, we say more about the solution (.)-(.) of (.) and the solvability conditions.
() The case that the solution lies inImz>landImz<l.
As in (.) and (.),+(z) is analytic in{z:Imz>l}and–(z) is analytic in{z:
Imz<l}. No matter how we chooseκ, the boundary value problem (.) is solvable and
It can be seen from the expression of(z) thatzis a|κ|-order pole ofY–(z)Y+(z) when
κ< . In order to ensure (.) is solvable, one hasC=F+(z) –F–(z) whenk= –,i.e.,
C=√ π
il
–∞+il
f(τ)eiτzdτ+
√
π
+∞+il il
f(τ)eiτzdτ. (.)
Whenk< –, the following|κ|– conditions are required:
il
–∞+il
τqf(τ)eiτzdτ+
+∞+il il
τqf(τ)eiτzdτ= , q= , , . . . , –κ– . (.)
Then(z) is analytic in{z:l<Imz<l}and has a bounded solution. Whenκ> ,zis a
κ-order pole ofY–(z)Y+(z), and thereforeY–(z)Y+(z) pκ(z)
(z–z)κ is analytic in{z:l<Imz<l}.
Hence,(z) is analytic in{z:l<Imz<l}and(z) is a constant whilez=∞.
Forz∈ {z:l<Imz<l},(z) can be defined by (.), (.) ifD(z) andD– (z) are
not zero. Otherwise, ifz∗,z∗, . . . ,z∗nare common zero-points ofD(z) andD– (z) with the
orderss,s, . . . ,sn, respectively, then(j)(zq∗) = (≤q≤n, ≤j≤sq). Lets=
n q=sq.
Then the following solvability conditions must be augmented.
Asκ≥, the followingκ+ element equations with unknown numbersc,c, . . . ,cκ:
pκ(z) (z–z)κ
(j)
z=zq∗
= j!i
j
√ π
il
–∞+il
τjf(τ)eiτzdτ+
+∞+il il
τjf(τ)eiτzdτ
. (.)
Asκ< , the following condition is required:
il
–∞+il
τjf(τ)eiτzdτ+
+∞+il il
τjf(τ)eiτzdτ= (.)
(j= , , , . . . ,sq;q= , , . . . ,n), wherec,c, . . . ,cκare the coefficients ofpk(z).
() The case of solutions atz=∞.
In order to discuss the solution atz=∞, we denote
γ∞()=μ()∞+iν∞()= πi
logD(il+∞) –logD(il–∞)
,
γ∞()=μ()∞ +iν()∞ = πi
logD(il+∞) –logD(il–∞)
,
μ∞=μ()∞+μ()∞,
where the logarithm functionlogDj(τ) takes some certain continuous branch whenReζ>
orReζ< such that ≤μ∞< .
Ifz=∞is a common node, it follows fromFj(∞) = (j= , ) thatFj(ζ) = Fj∗(ζ)
|ζ|μ
∗
j ,μ
∗
j <μ
(j)
∞
andFj∗(ζ)∈Hnearz=∞. By the conditions of (.), one has(ζ)∈ {}. Therefore,(∞) exists and is finite. Denote F(ζ) =F(ζ) –F(ζ). Note that ≤μ∞< . If μ∞> , it is
clear thatY–
(ζ)Y+(ζ)F(ζ) =O(/|ζ|μ∞–ε) (whereεis a positive number sufficiently small
and|ζ|is large enough) andY–
(ζ)Y+(ζ)
pκ(ζ)
(ζ–z)κ =O(). Ifμ∞≤
, in order to ensure that
(ζ)∈ {}, the coefficientekofpk(z) should be taken as
ek=
√
π
il
–∞+il
f(τ)eiτzdτ–
√
π
+∞+il il
Forκ< , the condition (.) should hold andj= , , . . . ,|κ|.
Ifz=∞is a special node,i.e.,μ∞= , one can translate it into the case thatμ∞≤as a common node. For the rest, similar arguments can be used [].
As for the boundary value problem withnunknown functions onn(n> ) parallel lines, there is no essential difference for the solving method with the casen= . We will not elaborate.
5 Example
In this section we consider one important example in practice. In (.), suppose
D(ζ) =D(ζ) = , C(ζ) =
+ζ, C(ζ) =
+ζ,
L: ζ = , L: ζ=x+i (–∞<x< +∞).
Without loss of generality, we assume thatz=i,z=–i,z=i. Then we haveκ=κ=
and henceκ= . Therefore,γj(t) = ,j(t) = ,Yj(z) = (j= , ). In this case, by (.) and
(.), we obtain
f(t) =
√
π
+∞
–∞
+τe
–iτtdτ =
√
πe
–t
,
f(t) =
√
π
+∞+i
–∞+i
+τe
–iτtdτ=
√
πe
–√t
,
and then
F+(z) =√ π
+∞
f(τ)eiτzdτ=
i
√(z+i), (.)
F–(z) = –√ π
–∞
f(τ)eiτzdτ=
i
√(z–i). (.)
Similarly, we have
F+(z) = i
√(z+√i), F
– (z) =
i
√(z–√i). (.)
Then we obtain the solutions of (.):
(z) = i
√(z–i)– i
√(z+√i)+C, when <Im< ,
+(z) = i √(z+i)–
i
√(z+√i)+H(z), when Imz> , (.)
–(z) = i √(z–i)–
i
√(z–√i)+H(z), whenImz< , whereCis a constant,H(z) = ( –
√
)
+z,H(z) = .
Competing interests
Acknowledgements
The author expresses sincere thanks to the referee(s) for the careful and details reading of the manuscript and very helpful suggestions, which improved the manuscript substantially.
Received: 5 August 2014 Accepted: 4 February 2015 References
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