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One class of generalized boundary value problem for analytic functions
Boundary Value Problems volumeÂ 2015, ArticleÂ number:Â 40 (2015)
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 at \(z=\infty\) and in the different domains, respectively. Therefore, the classic Riemann boundary value problem is extended further.
1 Introduction and preliminaries
Many mathematicians have studied the boundary value problems of analytic functions and formed a perfect theoretical system; see [1â€“7]. The boundary value problem of analytic 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 unknown 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 of Plemelj formula and function class \(\{1\}\) on an infinite straight line.
Definition 1.1
Assume that \(\omega(x)\) is a continuous complex function on the real axis X. We say that \(\omega(x)\in\hat{H}\) if the following conditions hold:

(1)
For any sufficiently large positive number M, \(\omega(x)\) satisfies \(\omega(x)\in H\) on \([M,M]\) (see [8] for the definition of H).

(2)
\(\omega(x_{1})\omega(x_{2})\leq A\frac{1}{x_{1}}\frac{1}{x_{2}}\), for any \(x_{j}>M\) (\(j=1,2\)) and some positive real numberÂ A.
Under condition (2), we say that \(\omega(x)\) satisfies the HÃ¶lder condition on \(N_{\infty}\) and denote it \(\omega(x)\in H(N_{\infty})\), where \(N_{\infty}=\{x: x>M\}\) is a neighborhood of âˆž.
Definition 1.2
Assume that \(\omega(x)\) is continuous on \((\infty, \infty)\) and \(\int_{\infty}^{\infty}\omega(x)\, dx<+\infty\), then we say that \(\omega(x)\in L_{1}(\infty,\infty)\).
Definition 1.3
If \(\omega(x)\) satisfies: (1) \(\omega(x)\in\hat {H}\), (2) \(\omega(x)\in L_{1}(\infty,\infty)\), then we say that \(\omega (x)\) belongs to the function class \(\{1\}\).
Definition 1.4
Assume that \(\omega(x)\in\{1\}\), then the integrals \(\Omega^{+}(z)=\frac{1}{\sqrt{2\pi}}\int_{0}^{+\infty}\omega (t)e^{itz}\, dt\) and \(\Omega^{}(z)=\frac{1}{\sqrt{2\pi}}\int_{\infty }^{0}\omega(t)e^{itz}\, dt\) are called the left and right onesided Fourier integral, respectively.
Lemma 1.1
(see [1])
If \(\omega(z)\in H\) with respect to any finite part of some infinite domain D, and \(\omega(z)\) is analytic in any neighborhood of infinity, then \(\omega(z)\in\hat{H}\).
Lemma 1.2
(see [2])
If \(\omega(t)\) belongs to the class \(\{1\} \), then the left and right onesided Fourier integrals defined in DefinitionÂ 1.4 are analytic when \(\operatorname{Im} z>0\) and \(\operatorname{Im} z<0\), respectively.
Lemma 1.3
(see [8])
If \(\omega(x)\in\hat{H}\), we have the Cauchy type integral \(\Omega(z)=\frac{1}{2\pi i}\int_{\infty}^{\infty }\frac{\omega(t)}{tz}\, dt\), \(z\notin(\infty,\infty)\), then the following formula holds on the infinite straight line:
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 the Xaxis (otherwise, we can translate them into this case by a linear transformation), and denote them by \(L_{1}\), \(L_{2}\), where \(L_{j}\) can be expressed by \(\zeta=x+il_{j}\) (\(x\in(\infty,\infty)\), \(l_{2}< l_{1}\) are real numbers) and take the direction from left to right as the positive direction. Let \(L=L_{1}+L_{2}\).
We want to get functions \(\Phi(z)\) and \(\Psi(z)\) such that \(\Phi(z)\) is analytic in \(\{\operatorname{Im} z>l_{1}\} \cup\{\operatorname{Im} z< l_{2}\}\), \(\Psi(z)\) is analytic in \(\{z: l_{2} < \operatorname{Im} z < l_{1}\}\), and we have the following boundary value conditions:
where \(L_{j}\): \(\zeta=x+il_{j}\) (\(j=1,2\)), \(x\in(\infty, \infty)\).
Actually, (2.1) is a boundary value problem on two parallel straight lines \(\operatorname{Im} z=l_{1}\), \(\operatorname{Im} z=l_{2}\) with âˆž as a pole. Here \(\Phi^{+}(\zeta )\) is the boundary value of analytic function \(\Phi^{+}(z)\) which is analytic in \(\{z: \operatorname{Im} z> l_{1}\}\) and belongs to the class \(\{1\}\) on \(L_{1}\), \(\Phi^{}(\zeta)\) is the boundary value of analytic function \(\Phi ^{}(z)\) which is analytic in \(\{z: \operatorname{Im} z< l_{2}\}\) and belongs to the class \(\{1\}\) on \(L_{2}\), and \(\Psi^{\pm}(\zeta)\) is the boundary value of analytic function \(\Psi(z)\) which is analytic in \(\{z: l_{2} <\operatorname{Im} z < l_{1}\} \) and belongs to the class \(\{1\}\) on \(L_{1}\), \(L_{2}\), respectively. The functions \(D_{1}(\zeta)\) and \(D_{2}(\zeta)\) belong to \(\hat{H}\) on \(L_{1}\), \(L_{2}\), respectively. The functions \(G_{1}(\zeta)\) and \(G_{2}(\zeta)\) belong to the class \(\{1\}\) on \(L_{1}\), \(L_{2}\), respectively. Hence, for the functions appearing in (2.1) the onesided limits exist when \(x\rightarrow\infty\) on \(L_{1}\), \(L_{2}\).
It can be seen from (2.1) that the order of \(\Phi(z)\) is equal to that of \(\Psi(z)\) at infinity. Therefore, if the orders of \(\Phi(z)\) and \(\Psi(z)\) are m at infinity, then such a problem can be denoted asÂ \(R_{m}\). Actually, problem \(R_{0}\) and problem \(R_{1}\) are often discussed. On the problem \(R_{0}\), both \(\Phi(\infty)\) and \(\Psi(\infty)\) are supposed to be finite and nonzero. On \(R_{1}\), both \(\Phi(\infty)\) and \(\Psi(\infty)\) are assumed to be zero. Such a problem R is called regular if \(D_{j}(\zeta)\) is not zero on L; otherwise, it is called irregular or of exception type.
Remark 2.1
Since the positive direction of \(L_{j}\) is the direction from left to right, when the observer moves from left to right on \(L_{j}\), the boundary values of left region of \(L_{j}\) is positive boundary value, i.e., the positive boundary value of \(\Phi(z)\) is the boundary value above \(L_{1}\), and the negative boundary value of \(\Phi(z)\) is ones below \(L_{2}\). The positive or negative boundary values of \(\Psi(z)\) can be defined in a similar way.
3 Resolution
We only consider problem \(R_{0}\) in this paper. Hence, we assume \(\Phi (\infty)\) and \(\Psi(\infty)\) are finite and nonzero. For problem \(R_{m}\), similar arguments can be used. In this paper, we only consider the normal case, that is, \(D_{j}(\zeta)\) (\(j=1,2\)) does not have zeroes and poles on \(L_{j}\). For the irregular case, similar discussions also work (see SectionÂ 2.5 of ChapterÂ 2 in [1]). Equation (2.1) can be written as
In order to unify, let \(C_{1}(\zeta)=G_{1}(\zeta)\), \(C_{2}(\zeta)=G_{2}(\zeta )/D_{2}(\zeta)\), and the above equation can be transformed into
By putting \(\kappa_{1}=\operatorname{Ind}_{L_{1}}D_{1}(\zeta)\), \(\kappa_{2}=\operatorname{Ind}_{L_{2}}D_{2}(\zeta )\), and \(\kappa=\sum_{j=1}^{2}\kappa_{j}\), we call Îº as the index of problem (2.1). Without loss of generality, we take three points \(z_{0}\), \(z_{1}\), \(z_{2}\) on the Z plane such that \(l_{1}< \operatorname{Im} z_{1}\), \(l_{2}<\operatorname{Im} z_{0}<l_{1}\), \(\operatorname{Im} z_{2}< l_{2}\). Then we take the following piecewise function:
here \(\Omega_{j}(z)\) (\(j=1,2\)) is defined as follows:
and
where
The function \(\Omega_{j}(z)\) defined above is analytic on the complex plane except \(L_{1}\) and \(L_{2}\). The logarithmic function of the integrand has a certain analytic branch such that \(\log\frac {tz_{0}}{tz_{j}}_{t=\infty}=0\); then \(Y_{j}^{+}(z)\) and \(Y_{j}^{}(z)\) are analytic in \(\{z: \operatorname{Im} z>l_{j}\}\) and \(\{z: \operatorname{Im} z< l_{j}\}\), respectively. Moreover,
owing to
by the representative of \(r_{1}(t)\) as well as the relationship between Fourier transform and inverse Fourier transform, we have \(V[r_{1}(t)]=\log [(\frac{tz_{0}}{tz_{1}})^{k_{1}}D_{1}(t)]\), therefore
Similarly, one has
Putting (3.4), (3.5) into (3.2), we can obtain
In (3.6), the first equality is multiplied by \([Y_{2}^{+}(t)]^{1}\), the second one is multiplied by \([Y_{1}^{}(t)]^{1}\), then
denoting
where
Using LemmaÂ 1.2, we know that \(F_{1}^{+}(z)\), \(F_{1}^{}(z)\) are analytic in \(\operatorname{Im} z>l_{1}\), \(\operatorname{Im} z< l_{1}\), respectively. On \(L_{1}\), we have
Again we denote
where
Similarly, \(F_{2}^{+}(z)\), \(F_{2}^{}(z)\) are analytic in \(\operatorname{Im} z>l_{2}\), \(\operatorname{Im} z< l_{2}\), respectively. On \(L_{2}\), we obtain
Then (3.7) may be reduced to
in the two sides of the first equation of (3.10), by adding \(F_{2}^{+}(t)\); in the two sides of the second one by subtraction of \(F_{1}^{}(t)\), we have
the left side of the first equation of (3.11) is denoted by \(M_{1}^{+}(t)\), the right side of one is denoted by \(M_{1}^{}(t)\); the left side of the second equation of (3.11) is denoted by \(M_{2}^{+}(t)\), the right side of this one is denoted by \(M_{2}^{}(t)\). Let
where
(1) We firstly consider the solutions of \(M_{1}^{+}(z)\) and \(M_{1}^{}(z)\), respectively in \(\operatorname{Im} z>l_{1}\) and \(\operatorname{Im} z< l_{1}\).
Case: \(k\geq0\).
Since \([Y_{1}^{+}(z)]^{1}\), \([Y_{2}^{+}(z)]^{1}\) are analytic in \(\{z: \operatorname{Im} z>l_{1}\}\), \([Y_{1}^{+}(z)Y_{2}^{+}(z)]^{1}\) is analytic. It follows from \(F_{j}^{+}(z)\) (\(j=1,2\)) is analytic that \(M_{1}^{+}(z)\) is analytic. Hence, the \(\lim_{z\rightarrow\infty\ (\operatorname{Im} z>l_{1})}M_{1}^{+}(z)\) exists. Suppose that \(M_{1}^{+}(z)=H_{1}(z)\), then
where \(H_{1}(z)\) is analytic in \(\{z: \operatorname{Im} z>l_{1}\}\) and the \(\lim_{z\rightarrow\infty\ (\operatorname{Im} z>l_{1})}H_{1}(z)\) exists.
When \(\operatorname{Im} z< l_{1}\), since \(\Psi^{}(z)\) is defined in \(\{z: l_{2}<\operatorname{Im} z<l_{1}\}\), \(z_{0}\) is a korder pole of \([Y_{1}^{}(z)Y_{2}^{+}(z)]^{1}\) (\(k>0\)). In order to ensure that \(M_{1}^{}(z)\) is bounded at a pole \(z_{0}\), we can multiply by a factor \((zz_{0})^{k}\). Thus, it has a korder at \(z=\infty\), i.e., we have a polynomial with k degree. Let \((zz_{0})^{k}M_{1}^{}(z)=p_{k}(z)\), hence, \(M_{1}^{}(z)=\frac{p_{k}(z)}{(zz_{0})^{k}}\), i.e.,
where \(p_{k}(z)=C_{0}+C_{1}(zz_{0})+\cdots+C_{k}(zz_{0})^{k}\) is a polynomial with degree no more thanÂ Îº.
Case: \(k<0\).
It follows from similar arguments as above that \(M_{1}^{+}(z)\) is analytic in \(\{z: \operatorname{Im} z>l_{1}\}\).
Because \([Y_{1}^{}(z)Y_{2}^{+}(z)]^{1}\), \(F_{1}^{}(z)\) and \(F_{2}^{+}(z)\) are analytic in \(\{z: \operatorname{Im} z< l_{1}\}\), so is \(M_{1}^{}(z)\). Moreover, \(M_{1}^{+}(t)=M_{1}^{}(t)\) on \(L_{1}\). Hence, \(M_{1}(z)\) is holomorphic in the whole complex plane and the \(\lim_{z\rightarrow\infty} M_{1}(z)\) exists. By the Liouville theorem and the principle of analytic continuation, there is a constant C such that \(M_{1}(z)=C\), and one has
Noticing that \(z=z_{0}\) is a pole of \(Y_{1}^{}(z)Y_{2}^{+}(z)\) with order âˆ’k, \(\Psi^{}(z)\) has a singularity at \(z_{0}\). In order to ensure that \(\Psi ^{}(z)\) is analytic in \(\{z: \operatorname{Im} z< l_{1}\}\) (in fact, for solving \(\Psi ^{}(z)\) in the range that \(\operatorname{Im} z< l_{1}\), we should only consider the case that \(l_{2}< \operatorname{Im} z<l_{1}\)). For the case that \(k=1\), it is sufficient to eliminate the singularity by putting \(C=F_{2}^{+}(z_{0})F_{1}^{}(z_{0})\), then one can define \(\Psi^{}(z)\) in the following way:
When \(\kappa\leq2\), such a C still cannot eliminate the singularity of \(\Psi^{}(z)\) at \(z=z_{0}\). But the following condition should be satisfied:
i.e.,
(3.15) is a solution of (2.1) if and only if (3.18) holds. Therefore,
However, the solvability conditions should be satisfied (3.18) for \(k<1\). Similarly, we can define the following piecewise function:
where
(2) We secondly consider the solutions of \(M_{2}^{+}(z)\) and \(M_{2}^{}(z)\), respectively, in \(\operatorname{Im} z>l_{2}\) and \(\operatorname{Im} z< l_{2}\).
Case: \(k \geq0\).
Since \(Y_{1}^{}(z)\) and \(Y_{2}^{}(z)\) are analytic in \(\{z: \operatorname{Im} z< l_{2}\}\), so is \([Y_{1}^{}(z)Y_{2}^{}(z)]^{1}\). It follows from \(F_{j}^{}(z)\) (\(j=1,2\)) being analytic that \(M_{2}^{}(z)\) is analytic. Hence, the \(\lim_{z\rightarrow \infty\ (\operatorname{Im} z< l_{2})}M_{2}^{}(z)\) exists. Suppose that \(M_{2}^{}(z)=H_{2}(z)\), then
where \(H_{2}(z)\) is analytic in \(\{z: \operatorname{Im} z< l_{2}\}\) and the \(\lim_{z\rightarrow\infty\ (\operatorname{Im} z< l_{2})}H_{2}(z)\) exists.
When \(\operatorname{Im} z>l_{2}\), \(z=z_{0}\) is a korder pole of \([Y_{1}^{}(z)Y_{2}^{+}(z)]^{1}\) (\(k>0\)), then \(z=z_{0}\) is a korder pole of \(M_{2}^{+}(z)\). By similar arguments to above, one has
Case: \(k<0\).
Since \([Y_{1}^{}(z)Y_{2}^{}(z)]^{1}\) has no singularity in \(\{z: \operatorname{Im} z< l_{2}\} \), \(M_{2}^{}(z)\) is analytic in \(\{z: \operatorname{Im} z< l_{2}\}\). Noticing that \([Y_{1}^{}(z)Y_{2}^{+}(z)]^{1}\) is analytic in \(\{z: \operatorname{Im} z>l_{2}\}\), one finds that \(M_{2}^{+}(z)\) is analytic in \(\{z: \operatorname{Im} z>l_{2}\}\) and \(M_{2}^{+}(t)=M_{2}^{}(t)\) on \(L_{2}\). Therefore, \(M_{2}(z)\) is holomorphic in the whole complex plane and the \(\lim_{z\rightarrow\infty}M_{2}(z)\) exists. By similar arguments to (1), there exists a constant C, such that \(M_{2}(z)=C\) and
Moreover, (3.18) is also a necessary condition for solvability. Hence,
Collecting results, for \(k\geq0\) and \(l_{2}<\operatorname{Im} z< l_{1}\), one has
and, for \(k< 0 \) and \(l_{2}< \operatorname{Im} z<l_{1}\),
where \(p_{k}(z)\) and C as above. For \(z\in\{z: \operatorname{Im} z>l_{1}\}\),
where \(H_{1}(z)\) is analytic in \(\{z: \operatorname{Im} z>l_{1}\}\) as \(k\geq0\) and the \(\lim_{z\rightarrow\infty\ (\operatorname{Im} z>l_{1})}H_{1}(z)\) exists. When \(k<0\), \(H_{1}(z)\equiv C\) (constant). For \(z\in\{z: \operatorname{Im} z< l_{2}\}\),
where \(H_{2}(z)\) is analytic in \(\{z: \operatorname{Im} z< l_{2}\}\) when \(k\geq0\) and \(\lim_{z\rightarrow\infty\ (\operatorname{Im} z< l_{2})}H_{2}(z)\) exists. \(H_{2}(z)\equiv C\) (constant) when \(k<0\).
Hence we get the solution of the boundary value problem (2.1).
Theorem 3.1
The boundary value problem (3.1) with two unknown functions \(\Psi(z)\) and \(\Phi(z)\) on two parallel lines has a solution in \(\{z: l_{2}< \operatorname{Im} z<l_{1}\}\) and \(\{\operatorname{Im} z>l_{1}\}\cup\{\operatorname{Im} z< l_{2}\}\), respectively. Moreover, the general solution can be expressed by (3.33)(3.36), where \(Y_{j}^{\pm }(z)\) (\(j=1,2\)) is defined by (3.3) and \(F_{j}(z)\) (\(j=1,2\)) are defined by (3.8) and (3.9). When \(\kappa>1\), \(p_{k}(z)\) is a polynomial with Îº order, and when \(\kappa\leq1\), the necessary conditions for solvability still are (3.18). In all, the degree of freedom of the solution is \(\kappa+1\).
4 Further discussion on solution and solvability conditions
In this section, we say more about the solution (3.33)(3.36) of (2.1) and the solvability conditions.
(1) The case that the solution lies in \(\operatorname{Im} z>l_{1}\) and \(\operatorname{Im} z< l_{2}\).
As in (3.27) and (3.28), \(\Phi^{+}(z)\) is analytic in \(\{z: \operatorname{Im} z> l_{1}\}\) and \(\Phi^{}(z)\) is analytic in \(\{z: \operatorname{Im} z< l_{2}\}\). No matter how we choose Îº, the boundary value problem (2.1) is solvable and its solution can be expressed by (3.35)(3.36).
(2) The case that the solution lies in \(l_{2}<\operatorname{Im} z<l_{1}\).
It can be seen from the expression of \(\Psi(z)\) that \(z_{0}\) is a \(\kappa \)order pole of \(Y_{1}^{}(z)Y_{2}^{+}(z)\) when \(\kappa<0\). In order to ensure (2.1) is solvable, one has \(C=F_{2}^{+}(z_{0})F_{1}^{}(z_{0})\) when \(k=1\), i.e.,
When \(k<1\), the following \(\kappa1\) conditions are required:
Then \(\Psi(z)\) is analytic in \(\{z: l_{2}<\operatorname{Im} z<l_{1}\}\) and has a bounded solution. When \(\kappa>0\), \(z_{0}\) is a Îºorder pole of \(Y_{1}^{}(z)Y_{2}^{+}(z)\), and therefore \(Y_{1}^{}(z)Y_{2}^{+}(z)\frac{p_{\kappa }(z)}{(zz_{0})^{\kappa}}\) is analytic in \(\{z: l_{2}<\operatorname{Im} z< l_{1}\}\). Hence, \(\Psi(z)\) is analytic in \(\{z: l_{2}<\operatorname{Im} z<l_{1}\}\) and \(\Psi(z)\) is a constant while \(z=\infty\).
For \(z\in\{z: l_{2}< \operatorname{Im} z< l_{1}\}\), \(\Psi(z)\) can be defined by (3.33), (3.34) if \(D_{1}(z)\) and \(D_{2}^{1}(z)\) are not zero. Otherwise, if \(z_{1}^{*},z_{2}^{*},\ldots,z_{n}^{*}\) are common zeropoints of \(D_{1}(z)\) and \(D_{2}^{1}(z)\) with the orders \(s_{1},s_{2},\ldots,s_{n}\), respectively, then \(\Psi^{(j)}(z_{q}^{*})=0\) (\(1\leq q\leq n\), \(1\leq j\leq s_{q}\)). Let \(s=\sum_{q=1}^{n} s_{q}\). Then the following solvability conditions must be augmented.
As \(\kappa\geq0\), the following \(\kappa+1\) element equations with unknown numbers \(c_{0},c_{1},\ldots,c_{\kappa}\):
As \(\kappa<0\), the following condition is required:
(\(j=0,1,2,\ldots,s_{q}\); \(q=1,2,\ldots,n\)), where \(c_{0},c_{1},\ldots,c_{\kappa }\) are the coefficients of \(p_{k}(z)\).
(3) The case of solutions at \(z=\infty\).
In order to discuss the solution at \(z=\infty\), we denote
where the logarithm function \(\log D_{j}(\tau)\) takes some certain continuous branch when \(\operatorname{Re} \zeta>0\) or \(\operatorname{Re} \zeta<0\) such that \(0\leq\mu _{\infty}<1\).
If \(z=\infty\) is a common node, it follows from \(F_{j}(\infty)=0\) (\(j=1,2\)) that \(F_{j}(\zeta)=\frac{F_{j}^{*}(\zeta)}{\zeta^{\mu_{j}^{*}}}\), \(\mu_{j}^{*}< \mu _{\infty}^{(j)}\) and \(F_{j}^{*}(\zeta)\in H\) near \(z=\infty\). By the conditions of (2.1), one has \(\Psi(\zeta)\in\{1\}\). Therefore, \(\Psi (\infty)\) exists and is finite. Denote \(F(\zeta)=F_{1}(\zeta)F_{2}(\zeta )\). Note that \(0\leq\mu_{\infty}<1\). If \(\mu_{\infty}>\frac{1}{2}\), it is clear that \(Y_{1}^{}(\zeta)Y_{2}^{+}(\zeta)F(\zeta)=O(1/\zeta^{\mu_{\infty }\varepsilon})\) (where Îµ is a positive number sufficiently small and \(\zeta\) is large enough) and \(Y_{1}^{}(\zeta)Y_{2}^{+}(\zeta)\frac {p_{\kappa}(\zeta)}{(\zetaz_{0})^{\kappa}}=O(1)\). If \(\mu_{\infty}\leq \frac{1}{2}\), in order to ensure that \(\Psi(\zeta)\in\{1\}\), the coefficient \(e_{k}\) of \(p_{k}(z)\) should be taken as
For \(\kappa<0\), the condition (4.2) should hold and \(j=1,2,\ldots ,\kappa\).
If \(z=\infty\) is a special node, i.e., \(\mu_{\infty}=0\), one can translate it into the case that \(\mu_{\infty}\leq\frac{1}{2}\) as a common node. For the rest, similar arguments can be used [9].
As for the boundary value problem with n unknown functions on n (\(n>2\)) parallel lines, there is no essential difference for the solving method with the case \(n=2\). We will not elaborate.
5 Example
In this section we consider one important example in practice. In (3.2), suppose
Without loss of generality, we assume that \(z_{1}=\frac{3i}{2}\), \(z_{2}=\frac{i}{2}\), \(z_{0}=\frac{i}{2}\). Then we have \(\kappa_{1}=\kappa_{2}=0\) and hence \(\kappa=0\). Therefore, \(\gamma_{j}(t)=0\), \(\Omega_{j}(t)=0 \), \(Y_{j}(z)=1\) (\(j=1,2\)). In this case, by (3.8) and (3.9), we obtain
and then
Similarly, we have
Then we obtain the solutions of (3.2):
where C is a constant, \(H_{1}(z)=(1\frac{\sqrt{2}}{2})\frac {1}{1+z^{2}}\), \(H_{2}(z)=0\).
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Li, P. One class of generalized boundary value problem for analytic functions. Bound Value Probl 2015, 40 (2015). https://doi.org/10.1186/s1366101503010
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DOI: https://doi.org/10.1186/s1366101503010
Keywords
 boundary value problem for analytic functions
 index
 canonical function
 the function class \(\{1\}\)