## Trends in Classical Analysis, Geometric Function Theory, and Geometry of Conformal Invariants

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Arif A. M. Yunus, Ali H. M. Murid, Mohamed M. S. Nasser, "Conformal Mapping of Unbounded Multiply Connected Regions onto Canonical Slit Regions", *Abstract and Applied Analysis*, vol. 2012, Article ID 293765, 29 pages, 2012. https://doi.org/10.1155/2012/293765

# Conformal Mapping of Unbounded Multiply Connected Regions onto Canonical Slit Regions

**Academic Editor:**Matti Vuorinen

#### Abstract

We present a boundary integral equation method for conformal mapping of unbounded multiply connected regions onto five types of canonical slit regions. For each canonical region, three linear boundary integral equations are constructed from a boundary relationship satisfied by an analytic function on an unbounded multiply connected region. The integral equations are uniquely solvable. The kernels involved in these integral equations are the modified Neumann kernels and the adjoint generalized Neumann kernels.

#### 1. Introduction

In this paper, we present a new method for numerical conformal mapping of unbounded multiply connected regions onto five types of canonical slit regions. A canonical region in conformal mapping is known as a set of finitely connected regions such that each finitely connected nondegenerate region is conformally equivalent to a region in . With regard to conformal mapping of multiply connected regions, there are several types of canonical regions as listed in [1–4]. The five types of canonical slit regions are disk with concentric circular slits , annulus with concentric circular slits , circular slit regions , radial slit regions , and parallel slit regions . One major setback in conformal mapping is that only for certain regions exact conformal maps are known. One way to deal with this limitation is by numerical computation. Trefethen [5] has discussed several methods for computing conformal mapping numerically. Amano [6] and DeLillo et al. [7] have successfully map unbounded regions onto circular and radial slit regions. Boundary integral equation related to a boundary relationship satisfied by a function which is analytic in a simply or doubly connected region bounded by closed smooth Jordan curves has been given by Murid [8] and Murid and Razali [9]. Special realizations of this integral equation are the integral equations related to the Szegö kernel, Bergmann kernel, Riemann map, and Ahlfors map. The kernels arise in these integral equations are the Neumann kernel and the Kerzman-Stein kernel. Murid and Hu [10] managed to construct a boundary integral equation for numerical conformal mapping of a bounded multiply connected region onto a unit disk with slits. However, the integral equation involves unknown radii which lead to a system of nonlinear algebraic equations upon discretization of the integral equation. Nasser [11–13] produced another technique for numerical conformal mapping of bounded and unbounded multiply connected regions by expressing the mapping function in terms of the solution of a uniquely solvable Riemann-Hilbert problem. This uniquely solvable Riemann Hilbert problem can be solved by means of boundary integral equation with the generalized Neumann kernel. Recently, Sangawi et al. [14–17] have constructed new linear boundary integral equations for conformal mapping of bounded multiply region onto canonical slit regions, which improves the work of Murid and Hu [10] where in [10], the system of algebraic equations are nonlinear. In this paper, we extend the work of [14–17] for numerical conformal mapping of unbounded multiply connected regions onto all five types of canonical slit regions. These boundary integral equations are constructed from a boundary relationship satisfied by an analytic function on an unbounded multiply connected region.

The plan of this paper is as follows: Section 2 presents some auxiliary material. Section 3 presents a boundary integral equation related to a boundary relationship. In Sections 4–8, we present the derivations for numerical conformal mapping for all five types of canonical regions. In Section 9, we give some examples to illustrate the effectiveness of our method. Finally, Section 10 presents a short conclusion.

#### 2. Auxiliary Material

Let be an unbounded multiply connected region of connectivity . The boundary consists of smooth Jordan curves , and will be denoted by . The boundaries are assumed in clockwise orientation (see Figure 1). The curve is parameterized by -periodic twice continuously differentiable complex function with nonvanishing first derivative, that is, The total parameter domain is the disjoint union of intervals . We define a parameterization of the whole boundary on by

Let be the conformal mapping function that maps onto , where represents any canonical region mentioned above, is a prescribed point located inside , is a prescribed point inside and is a prescribed point located in . In this paper, we determine the mapping function by computing the derivatives of the mapping function and two real functions on , that is, the unknown function and a piecewise constant real function . Let be the space of all real Hölder continuous -periodic functions and be the subspace of which contains the piecewise real constant functions . The piecewise real constant function can be written as briefly written as . Let be a complex continuously differentiable -periodic function for all . We define two real kernels formed with as [18] The kernel is known as the generalized Neumann kernel formed with complex-valued functions and . The kernel is continuous with The kernel has a cotangent singularity where the kernel is continuous with The adjoint function of is defined by The generalized Neumann kernel and the real kernel formed with are defined by Then, where is the adjoint kernel of the generalized Neumann kernel . We define the Fredholm integral operators by Integral operators , , and are defined in a similar way. Throughout this paper, we will assume the functions and are given by It is known that is not an eigenvalue of the kernel and is an eigenvalue of the kernel with multiplicity [18]. The eigenfunctions of corresponding to the eigenvalue are , where We also define an integral operator by (see [14]) The following theorem gives us a method for calculating the piecewise constant real function in connection with conformal mapping later. This theorem can be proved by using the approach as in [19, Theorem 5].

Theorem 2.1. * Let , , and such that
**
are the boundary values of a function analytic in . Then the function has each element given by
**
where is the unique solution of the integral equation
*

#### 3. The Homogeneous Boundary **Relationship**

Suppose we are given a function which is analytic in , continuous on , Hölder continuous on and is finite. The boundary is assumed to be a smooth Jordan curve. The unit tangent to at the point will be denoted by . Suppose further that satisfies the exterior homogeneous boundary relationship where and are complex-valued functions with the following properties: (1) is analytic in and does not have zeroes on , (2), is finite,(3), . Note that the boundary relationship (3.1) also has the following equivalent form: Under these assumptions, an integral equation for can be constructed by means of the following theorem.

Theorem 3.1. *If the function satisfies the exterior homogeneous boundary relationship (3.1), then
**
where
*

* Proof. * Consider the integral ,
Since the boundary is in clockwise orientation and is analytic in , then by [20, p. 2] we have
Now, let the integral be defined as
Using the boundary relationship (3.1) and the fact that and does not contain zeroes, then by [20, p. 2] we obtain
Next, by taking with further arrangement yields
Then multiplying (3.9) with and , subsequently yields (3.3).

Theorem 3.2. *The kernel is continuous with
*

* Proof. * Let the kernel be written as
where
Notice that is the classical Neumann kernel with . Now for , as we take the limit we have,
Hence, by combining and , we obtain

Note that when , the kernel reduces to the classical Neumann kernel.

We define the Fredholm integral operator by Hence, (3.3) becomes The solvability of the integral equation (3.16) depends on the possibility of being an eigenvalue of the kernel . For the numerical examples considered in this paper, is always an eigenvalue of the kernel . Although there is no theoretical proof yet, numerical evidence suggests that is an eigenvalue of . If the multiplicity of the eigenvalue is , then one need to add conditions to the integral equation to ensure the integral equation is uniquely solvable.

#### 4. Exterior Unit Disk with Circular Slits

The canonical region is the exterior unit disk along with arcs of circles. We assume that maps the curve onto the unit circle , the curve , where , onto circular slit on the circle , where are unknown real constants. The circular slits are traversed twice. The boundary values of the mapping function are given by where represents the boundary correspondence function and . By differentiating (4.1) with respect to and dividing the result obtained by its modulus, we have Using the fact that unit tangent and , it can be shown that Boundary relationship (4.3) is useful for computing the boundary values of provided , and are all known. By taking logarithmic derivative on (4.1), we obtain The mapping function can be uniquely determined by assuming Thus, the mapping function can be expressed as [12] where is an analytic function and . By taking logarithm on both sides of (4.6), we obtain Hence (4.7) satisfies boundary values (2.15) in Theorem 2.1 with , Hence, the values of can be calculated by To find , we began by differentiating (4.7) and comparing with (4.4) which yields In view of and letting , where is analytic function in , (4.10) becomes By [18, Theorem 2(c)], we obtain which implies However, this integral equation is not uniquely solvable according to [18, Theorem 12]. To overcome this, since the image of the curve is clockwise oriented and the images of the curves are slits so we have and , which implies By adding this condition to (4.12), the unknown function is the unique solution of the integral equation Next, the presence of in (4.3) can be eliminated by squaring both sides of (4.3), that is, Upon comparing (4.15) with (3.2), this leads to a choice of , , , , . Hence, satisfies the integral equation (3.16) with Numerical evidence shows that is an eigenvalue of of multiplicity , which means one needs to add conditions. Since is assumed to be single-valued, it is also required that the unknown mapping function satisfies [4] that is, Then, is the unique solution of the following integral equation By obtaining , the derivatives of the mapping function, can be found using By obtaining the values of , and , the boundary value of can be calculated by using (4.3).

#### 5. Annulus with Circular Slits

The canonical region consists of an annulus centered at the origin together with circular arcs. We assume that maps the curve onto the unit circle , the curve onto the circle and onto circular slit , where . The slit are traversed twice. The boundary values of the mapping function are given by where represents the boundary correspondence function and is a piecewise real constant function. By using the same reasoning as in Section 4, we get The mapping function can be uniquely determined by assuming . Thus, the mapping function can be expressed as [12] By taking logarithm on both sides of (5.4), we obtain Hence, (5.5) satisfies boundary values (2.15) in Theorem 2.1 with =1, By obtaining , the values of can be computed by By differentiating (5.5) and comparing with (5.3) yields In view of and , where is analytic in , (5.8) is equivalent to By [18, Theorem 2(c)], we obtain Note that the image of the curve is counterclockwise oriented, is clockwise oriented and the images of the curves are slits so we have , and , which implies Hence, the unknown function is the unique solution of the integral equation Next, the presence of in (5.2) can be eliminated by squaring both sides of the equation, that is, Comparing (5.13) with (3.2) leads to a choice of , , , , . Hence, satisfies the integral equation (3.16) with Numerical evidence shows that is an eigenvalue of of multiplicity , which implies one needs to add conditions. Since is assumed to be single-valued, hence it is also required that the unknown mapping function satisfies [4] that is, Since we assume the mapping function can be uniquely determined by , hence by [20] If we define the Fredholm operator as then is the unique solution of the following integral equation: By obtaining , the derivatives of the mapping function, can be obtained by

#### 6. Circular Slits

The canonical region consists of slits along the circle where and are unknown real constants. The boundary values of the mapping function are given by where represents the boundary correspondence function and . By using the same reasoning as in Section 4, we get The mapping function can be uniquely determined by assuming , and . Thus, the mapping function can be expressed as [12] By taking logarithm on both sides of (6.4), we obtain Hence, (6.5) satisfies boundary values in Theorem 2.1 with , By obtaining , the values of can be obtained by By differentiating (6.5) and comparing with (6.3) yields In view of , and where is analytic in and is analytic in , we rewrite (6.8) as Let . Then by [18, Theorems 2(c) and 2(d)], we obtain Subtracting (6.11) from (6.10), we get Since the images of the curve are slits, we have . Thus Hence the unknown function is the unique solution of the integral equation where By squaring both sides of (6.2) and dividing the result by , we obtain Upon comparing (6.16) with (3.2) leads to a choice of , , , , . Hence, satisfies the integral equation (3.16) with Numerical evidence shows that is an eigenvalue of of multiplicity , thus one needs to add conditions. Since is assumed to be single-valued, it is also required that the unknown mapping function satisfies [4] that is, Then, is the unique solution of the integral equation By obtaining , the derivatives of the mapping function, can be obtained by

#### 7. Radial Slits

The canonical region consists of slits along segments of the rays with arg. Then, the boundary values of the mapping function are given by where the boundary correspondence function now becomes real constant function and is an unknown function. By taking logarithmic derivative on (7.1), we obtain It can be shown that the mapping function can be determined using The mapping function can be uniquely determined by assuming , and . Thus, the mapping function can be expressed as [12] By taking logarithm on both sides of (7.4) and multiplying the result by , we obtain Hence, (7.5) satisfies boundary values in Theorem 2.1 with , By obtaining , one can obtain the values of by Then, by differentiating (7.5) and comparing with (7.2) yields In view of , and where is analytic in and is analytic in , we rewrite (7.8) as Let . Then by [18, Theorems 2(c) and 2(d)], we obtain Adding these equations, we get Since the images of the curve are slits, we have . Therefore Hence, the unknown function is the unique solution of the integral equation where The boundary relationship (7.3) can be rewritten as Squaring both sides of (7.15) yields Upon comparing (7.16) with (3.1) leads to a choice of , , , , . Hence, satisfies the integral equation (3.16) with Numerical evidence shows that is an eigenvalue of of multiplicity , which suggests one needs to add conditions. Since is assumed to be single-valued, hence it is also required that the unknown mapping function satisfies [4] that is, Then, is the solution of the following integral equation By obtaining , the derivatives of the mapping function can be found using

#### 8. Parallel Slits

The canonical region consists of a parallel straight slits on the -plane. Let , then the boundary values of the mapping function are given by where is the angle of intersection between the lines and the real axis. is a piecewise real constant function and is an unknown function. It can be shown that (8.1) can be written as The mapping function can be uniquely determined by assuming and . Thus, the mapping function can be expressed as [12] where is an analytic function with . By multiplying both sides of (8.3) with , we obtain Hence, (8.4) satisfies the boundary values in Theorem 2.1 with , Differentiating (8.4) and comparing the result with (8.2) yield In view of , and , where is analytic in and is analytic in , we rewrite (8.6) as Assuming , then by [18, Theorems 2(c) and 2(d)], we obtain These two equations yields Note that, the images of the curves are slits, so we have , which implies Hence, the unknown function is the unique solution of the integral equation where From (8.1), we introduce an analytic function such that where . By differentiating (8.13) with respect to , we obtain Let be an analytic function such that it has the following representation where . From (8.13)–(8.15) it can be shown that, the function can be rewritten as By squaring both sides of (8.16), the sign is eliminated, that is, Comparing (8.17) with (3.2) leads to a choice of , , , , . Hence, satisfies the integral equation (3.16) with Numerical evidence shows that is an eigenvalue of of multiplicity , which suggests one needs to add conditions. Since , hence we can have additional conditions for the integral equation above as in the following: