Conformal Mapping of Unbounded Multiply Connected Regions onto Canonical Slit Regions

and Applied Analysis 3 ΓM Γ2 Γ1


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 S such that each finitely connected nondegenerate region is conformally equivalent to a region in S. 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 The total parameter domain J is the disjoint union of m intervals J 1 , . . ., J m .We define a parameterization η of the whole boundary Γ on J by

2.2
Let Φ z be the conformal mapping function that maps Ω − onto U − , where U − represents any canonical region mentioned above, z 1 is a prescribed point located inside Γ 1 , z m is a prescribed point inside Γ m and β is a prescribed point located in Ω − .In this paper, we determine the mapping function Φ z by computing the derivatives of the mapping function Φ η t and two real functions on J, that is, the unknown function ϕ t and a piecewise constant real function R t .Let H be the space of all real H ölder continuous 2π-periodic functions and S be the subspace of H which contains the piecewise real constant functions R t .The piecewise real constant function R t can be written as briefly written as R t R 1 , . . ., R m .Let A t be a complex continuously differentiable 2πperiodic function for all t ∈ J.We define two real kernels formed with A as 18

2.11
Integral operators M * , N, and M are defined in a similar way.Throughout this paper, we will assume the functions A and A are given by

2.13
We also define an integral operator J by see 14 χ j s χ j t μ t dt.

2.14
The following theorem gives us a method for calculating the piecewise constant real function h t in connection with conformal mapping later.This theorem can be proved by using the approach as in 19, Theorem 5 .where ρ t is the unique solution of the integral equation 2.17

3.4
Proof.Consider the integral I 1 η t , Since the boundary is in clockwise orientation and D is analytic in Ω − , then by 20, p. 2 we have Now, let the integral I 2 η t be defined as

3.7
Using the boundary relationship 3.1 and the fact that T η t |dt| dt and P η t does not contain zeroes, then by 20, p. 2 we obtain Next, by taking I 2 η t − I 1 η t with further arrangement yields 3.9 Then multiplying 3.9 with T η t and |η t |, subsequently yields 3.

3.16
The solvability of the integral equation 3.16 depends on the possibility of λ −1 being an eigenvalue of the kernel K s, t .For the numerical examples considered in this paper, λ −1 is always an eigenvalue of the kernel K s, t .Although there is no theoretical proof yet, numerical evidence suggests that λ −1 is an eigenvalue of K s, t .If the multiplicity of the eigenvalue λ −1 is m, then one need to add m conditions to the integral equation to ensure the integral equation is uniquely solvable.

Exterior Unit Disk with Circular Slits
The canonical region U d is the exterior unit disk along with m − 1 arcs of circles.We assume that w Φ z maps the curve Γ 1 onto the unit circle |w| 1, the curve Γ j , where j 2, 3, . . ., m, onto circular slit on the circle |w| R j , where R 2 , R 3 , . . ., R m are unknown real constants.The circular slits are traversed twice.The boundary values of the mapping function Φ are given by The mapping function Φ z can be uniquely determined by assuming Thus, the mapping function can be expressed as 12 where F z is an analytic function and F ∞ 0. By taking logarithm on both sides of 4.6 , we obtain Hence 4.7 satisfies boundary values 2.15 in Theorem 2.1 with A t 1, Hence, the values of R j can be calculated by R j e h j −h 1 for j 1, 2, . . ., m. 4.9 To find θ t , we began by differentiating 4.7 and comparing with 4.4 which yields In view of A η t and letting f z By 18, Theorem 2 c , we obtain I − N θ 0 which implies Abstract and Applied Analysis However, this integral equation is not uniquely solvable according to 18, Theorem 12 .To overcome this, since the image of the curve Γ 1 is clockwise oriented and the images of the curves Γ j , j 2, 3, . . ., m are slits so we have θ 1 2π − θ 1 0 −2π and θ j 2π − θ j 0 0, which implies Jθ t h t −1, 0, . . ., 0 .

4.13
By adding this condition to 4.12 , the unknown function θ t is the unique solution of the integral equation Next, the presence of sign θ t in 4.3 can be eliminated by squaring both sides of 4.3 , that is,

4.15
Upon comparing 4.15 with 3.2 , this leads to a choice of P η t Numerical evidence shows that λ −1 is an eigenvalue of K s, t of multiplicity m, which means one needs to add m conditions.Since Φ z is assumed to be single-valued, it is also required that the unknown mapping function Φ z satisfies 4 Then, φ t is the unique solution of the following integral equation

4.20
By obtaining φ t , the derivatives of the mapping function, Φ t can be found using By obtaining the values of R t , θ t and Φ η t , the boundary value of Φ η t can be calculated by using 4.3 .

Annulus with Circular Slits
The canonical region U a consists of an annulus centered at the origin together with m − 2 circular arcs.We assume that Φ z maps the curve Γ

5.3
The mapping function Φ z can be uniquely determined by assuming c Φ ∞ > 0. Thus, the mapping function can be expressed as 12

5.4
By taking logarithm on both sides of 5.4 , we obtain Note that the image of the curve Γ 1 is counterclockwise oriented, Γ m is clockwise oriented and the images of the curves Γ j , j 2, 3, . . ., m − 1 are slits so we have −2π and θ j 2π − θ j 0 0, which implies Jθ t h t 1, 0, . . ., −1 .

5.11
Hence, the unknown function θ t is the unique solution of the integral equation Next, the presence of sign θ t in 5.2 can be eliminated by squaring both sides of the equation, that is,

5.15
Numerical evidence shows that λ −1 is an eigenvalue of K s, t of multiplicity m 1, which implies one needs to add m 1 conditions.Since Φ z is assumed to be single-valued, hence it is also required that the unknown mapping function Φ z satisfies 4 that is, Jφ 0. 5.17 Since we assume the mapping function Φ z can be uniquely determined by c Φ ∞ > 0, hence by 20

5.18
If we define the Fredholm operator G as then φ t is the unique solution of the following integral equation:

5.20
By obtaining φ η t , the derivatives of the mapping function, Φ t can be obtained by Φ t φ η t η t .5.21

Circular Slits
The canonical region U c consists of m slits along the circle |Φ| R j where j 1, 2, . . ., m and R 1 , R

Radial Slits
The canonical region U r consists of m slits along m segments of the rays with arg Φ θ j , j 1, 2, . . ., m.Then, the boundary values of the mapping function Φ are given by Φ η t r t e iθ t e R t e iθ t , 7.1 where the boundary correspondence function θ t θ 1 , θ 2 , . . ., θ m now becomes real constant function and R t is an unknown function.By taking logarithmic derivative on 7.1 , we obtain It can be shown that the mapping function Φ z can be determined using The mapping function Φ z can be uniquely determined by assuming Φ β 0, Φ ∞ ∞ and Φ ∞ 1.Thus, the mapping function Φ z can be expressed as 12

7.4
By taking logarithm on both sides of 7.4 and multiplying the result by −i, we obtain

7.12
Hence, the unknown function R t is the unique solution of the integral equation Numerical evidence shows that λ −1 is an eigenvalue of K s, t of multiplicity m, which suggests one needs to add m conditions.Since Φ z is assumed to be single-valued, hence it is also required that the unknown mapping function Φ z satisfies 4 Then, φ t is the solution of the following integral equation

7.21
By obtaining φ η t , the derivatives of the mapping function Φ t can be found using

Parallel Slits
The canonical region U p consists of a m parallel straight slits on the w-plane.Let B e i π/2−θ , then the boundary values of the mapping function Φ are given by where θ is the angle of intersection between the lines Re BΦ R j and the real axis.R t R 1 t , R 2 t , . . ., R M t is a piecewise real constant function and δ t is an unknown function.It can be shown that 8.1 can be written as The mapping function Φ z can be uniquely determined by assuming Φ ∞ ∞ and lim z → ∞ Φ z − z 0. Thus, the mapping function Φ can be expressed as 12 where F z is an analytic function with F ∞ 0. By multiplying both sides of 8.

8.10
Hence, the unknown function δ t is the unique solution of the integral equation Numerical evidence shows that λ −1 is an eigenvalue of K s, t of multiplicity m, which suggests one needs to add m conditions.Since e Bη j t ϑ η j t | 2π 0 0, hence we can have m additional conditions for the integral equation above as in the following: e Bη j t ϑ j η j t dt 0, 2π 0 e Bη j t ϑ j η j t Bϑ η j t η t dt 0, 2π 0 e Bη j t σ η j t η t dt 0, 2π 0 e Bη j t φ η j t dt 0, q 1, 2, . . ., m.

8.20
We define Fredholm operator L as Lμ s J j e Bη j t μ t dt.

8.21
Then, φ t is the solution of the following integral equation:

Numerical Examples
Since the boundaries Γ j are parameterized by η t which are 2π-periodic functions, the reliable method to solve the integral equations are by means of Nystr öm method with trapezoidal rule 21 .Each boundary will be discretized by n number of equidistant points.The resulting linear systems are then solved by using Gaussian elimination.For numerical examples, we choose regions with connectivities one, two, three and four.For the region

9.2
For this example, we compare the error for each boundary value between our method and the exact mapping function.See Table 1 for Error Norm of ||w j − w j || ∞ .
Example 9.2.Consider an unbounded region Ω − bounded by a circle and an ellipse

9.3
Figure 2 shows the region and its five canonical images by using our proposed method.
Example 9.3.Consider an unbounded region Ω − bounded by 3 circles This example has also been considered in 6, 12 .Figure 3 shows the regions and its five canonical images by using our proposed method.See Table 3 for numerical comparison between our parameter values see Table 2 those in 12 .Note that our method has considered exterior unit disk with slits as a canonical region while 12 has considered interior    involves θ t , which varies with the number of collocation points, n.However, this does not have any effect on the accuracy of the method.
Example 9.4.Consider an unbounded region Ω − of 4-connectivity with boundaries Figure 6 shows the region and its five canonical images by using our proposed method.

Conclusion
In this paper, we have constructed a unified method for numerical conformal mapping of unbounded multiply connected regions onto canonical slit regions.The advantage of this method is that the integral equations are all linear which overcomes the nonlinearity problem encountered in 10 .From the numerical experiments, we can conclude that our method works on any finite connectivity with high accuracy.By computing the boundary values of the mapping function, the exterior points will be calculated by means of Cauchy's integral formula.

Figure 2 :
Figure 2: The original region Ω − and its canonical images with θ p π/3 for parallel slits.

Figure 4 :
Figure 4: Condition numbers of the matrices for our method for U d , U a , U c , U r and U p .

Figure 5 :
Figure 5: Condition number of the matrices for generalized Neumann kernel G.N.K , adjoint generalized Neumann kernel Adj.G.N. K , charge simulation for circular slits C.S. circular and charge simulation for radial slits C.S. radial .
The kernel N s, t is known as the generalized Neumann kernel formed with complex-valued functions A and η.The kernel N s, t is continuous with The generalized Neumann kernel N s, t and the real kernel M formed with A are defined by * s, t N t, s is the adjoint kernel of the generalized Neumann kernel N s, t .We define the Fredholm integral operators N * by N * υ t J N * t, s υ s ds, t ∈ J.
Suppose further that D η t satisfies the exterior homogeneous boundary relationship Suppose we are given a function D z which is analytic in Ω − , continuous on Ω − ∪ Γ, Hölder continuous on Γ and D ∞ is finite.The boundary Γ j is assumed to be a smooth Jordan curve.The unit tangent to Γ at the point η t ∈ Γ will be denoted by T η t η t /|η t |.where c t and P are complex-valued functions with the following properties: 3 .
By differentiating 4.1 with respect to t and dividing the result obtained by its modulus, we have Boundary relationship 4.3 is useful for computing the boundary values of Φ z provided θ t , R t , and Φ η t are all known.By taking logarithmic derivative on 4.1 , we obtain 4.1 where θ t represents the boundary correspondence function and R t 1, R 2 , . . ., R m .
1 onto the unit circle |Φ| 1, the curve Γ m onto the circle |Φ| R m and Γ j onto circular slit |Φ| R j , where j 2, 3, . . ., m − 1.The slit are traversed twice.The boundary values of the mapping function Φ are given by where θ t represents the boundary correspondence function and R t 1, R 2 , . . ., R m is a piecewise real constant function.By using the same reasoning as in Section 4, we get 2 , . . ., R m are unknown real constants.The boundary values of the mapping function Φ 1 , ln R 2 , ..., ln R m , γ t − ln η t − β .*θt−ϕ t Mψ t , 6.10I − N * ϕ t Mψ t .By squaring both sides of 6.2 and dividing the result by η t − β 2 , we obtainΦ η t 2 η t − β 2 η t η t .6.18Numerical evidence shows that λ −1 is an eigenvalue of K s, t of multiplicity m, thus one needs to add m conditions.Since Φ z is assumed to be single-valued, it is also required that the unknown mapping function Φ z satisfies 4

Table 1 :
12ror norm w j − w j ∞ for Example 9.1.We choose the special point β 2.5 1.5i.The exact mapping function for U c , U r , and U p are given respectively by12

Table 2 :
The values for approximated parameters in Example 9.3 with n 256.

Table 4 :
Error norm max 1≤j≤3 w j − w j ∞ of our method with 12 for Example 9.3.