Solution of Singular Integral Equations Involving Logarithmically Singular Kernel with an Application in a Water Wave Problem

A direct function theoretic method is employed to solve certain weakly singular integral equations arising in the study of scattering of surface water waves by vertical barriers with gaps. Such integral equations possess logarithmically singular kernel, and a direct function theoretic method is shown to produce their solutions involving singular integrals of similar types instead of the stronger Cauchy-type singular integrals used by previous workers. Two specific ranges of integration are examined in detail, which involve the following: Case i two disjoint finite intervals 0, a ∪ b, c and a, b, c being finite and Case ii a finite union of n disjoint intervals. The connection of such integral equations for Case i , with a particular water wave scattering problem, is explained clearly, and the important quantities of practical interest the reflection and transmission coefficients are determined numerically by using the solution of the associated weakly singular integral equation.


Introduction
The singular integral equation where B is a single or a finite union of finite intervals is encountered in many branches of mathematical physics cf.Lewin  We introduce a sectionally analytic function F z in the complex z x iy plane, cut along the segment −b, −c ∪ −a, 0 ∪ 0, a ∪ b, c on the real axis as given by If we use the following results, valid for the limiting values of Φ z log z ≡ log x iy : ln|x| ± πi for x < 0, 2.5 we find the following limiting values of F z :

2.6
Equations 2.6 give rise to the following Plemelj-type alternative formulae: where where B −c, −b ∪ −a, 0 .By using the relation 2.9 , the singular integral equation 2.2 can be expressed in the form of a Riemann-Hilbert problem as given by where

2.11
Again noting the relation 2.8 , we can write

2.13
Using the relation 2.13 in 2.4 , we get the solution of special Riemann-Hilbert problem 2.7 as

2.14
Utilizing the above idea, we can solve the Riemann-Hilbert problem 2.10 in the following manner. Let where F 0 z is the solution of the corresponding homogeneous problem 2.10 , that is, Thus, 2.10 becomes

2.17
This is a Riemann-Hilbert problem which is similar to 2.7 .Noting 2.7 , 2.12 , and 2.14 , the solution of this Riemann-Hilbert problem is given by where E z is an entire function.
We can choose with where

ISRN Applied Mathematics
Knowing F z from 2.18 , 2.19 , and 2.20 , we can find the limiting values F ± x of F z using the formula 2.5 .Substituting F ± x in 2.7 and noting the relation 2.8 , the solution of the integral equation 2.2 can be obtained as

2.24
We can determine the arbitrary constants A, B, C, D, and E by using the relation 2.24 .Thus, φ x can be determined completely by utilizing the relation 2.22 .It is observed that the solution as obtained by the relations 2.22 and 2.23 of 2.2 neither involves any strong singularity nor any arbitrary constant.

Case(ii)
Consider that

2.25
In this section, we proceed to solve the integral equation where In this case, we introduce the following sectionally analytic function Using the formula 2.5 , we get following limiting values of F z :

2.28
This gives the following Plemelj-type alternative formulae for the integral equation 2.26 : where

2.31
Hence, from the relation 2.30 , we get

2.32
Thus, noting the relation 2.32 , the equation 2.27 becomes

2.33
Utilizing the above idea, we can solve the Riemann-Hilbert problem 2.31 in the following form: where E 1 z is an entire function, and we can choose z , the solution of the homogeneous equation of 2.31 as

2.37
Using 2.35 in 2.34 , F 1 z is known.Noting the relation 2.5 , one can find the limiting values F ± 1 x of F 1 z .Using the limiting values F ± 1 x , we obtain, from 2.29 and 2.30 , the solution φ 1 x of the integral equation 2.26 as

2.39
We observe that the original integral equation 2.26 can be solved by using the above relations if and only if the following consistency conditions are satisfied:

2.40
From these 2n conditions, we find the arbitrary constant Π 2n 1 i 2 A i , and from the order condition, the arbitrary constant A 1 can be determined, and finally the solution of the singular integral equation 2.26 can be obtained from 2.38 .
In the next section, we give the genesis of the integral equation 2.2 with B 0, a ∪ b, c and use its solution to the problem of scattering of water waves by a partially immersed vertical barrier with gap.

The Detailed Analysis and the Analytical Solution
In this section, we show the connection of integral equation 2. in deep water occupying the position x 0, 0 < y < a, and b < y < c, y-axis being directed downwards and x-axis along the mean free surface.We assume the motion to be two dimensional, irrotational, and time harmonic.The velocity potential is represented by Re φ x, y exp −iσt , σ being circular frequency.Here, φ x, y satisfies the following boundary value problem: where K σ 2 /g, g being the acceleration of gravity, where R and T are reflection and transmission coefficients.A suitable representation of φ satisfying the equations 3.1 , 3.4 , and 3.5 is given by Then, using the condition 3.2 along with the condition of continuity of φ across the gap G : a, b ∪ c, ∞ , along x 0, we arrive at the following multiple integral equations:

3.10
The Relation 3.9 can be recast in the form

3.11
where D 1 and D 2 are two arbitrary constants.Then, applying d/dy K to the equations 3.10 and 3.11 , we obtain

3.14
It is observed from the relation 3.13 that we must choose C 1 0 for consistency.Setting where g y is an unknown function and using sine inversion formula, to the relations 3.12 and 3.15 , we obtain from 3.13

3.17
The solution of the integral equation 3.16 is given by the equations 2.22 , 2.23 , and 2.24 with f y given by relation 3.17 .Thus, knowing g t , we can obtain F k by using Fourier Sine inversion formulae to the relations 3.12 and 3.15 .Hence, A k can be obtained from equation 3.14 .Now, f y contains unknown constants D 1 , D 2 , and C 2 .To obtain the unknowns D 1 , D 2 , C 2 , and R, we substitute A k in 3.10 and 3.11 to get the following four equations:

3.19
and A, B, C, D, and E are arbitrary constants and Thus, the relations 3.18 can be solved to determine four unknowns D 1 , D 2 , C 2 , and R.This completes the description of the analytical solution of the water wave problem.

Numerical Results
The Numerical computation of |R| involves the solution of four equations 3.18 .It may be noted from the relation 3.19 that for the solution of 3.18 , one does not need to compute any integral with strong singularity.The integrals occurring in 3.18 are defined in Riemann sense.This is the main advantage in the present method as compared to the other methods based on strongly singular integral equation formulation available in the literature.Since there is no integral with strong singularity, so for numerical computation, we have used Mathematica 5.2 to evaluate various integrals in 3.18 and 3.19 and to solve the system of 3.18 .
In Table 1,   as observed in Figure 1.Thus, when the length of the lower part of the barrier increases, the peak value in |R| decreases.In Figure 3, |R| is plotted against Kb for a/b 0.25 and c/b 6.In this figure, it is observed that when the length of the gap is increased keeping the length of the lower part of the barrier the same, the sharp peak in the values of |R| occurs for 0.111 < Kb < 0.1111.In this case the peak value is much lower than the peak value observed in Figures 1 and 2 b .Thus, an increase in the length of the gap decreases the peak value in |R|.Thus, it is observed from Figures 1, 2, and 3 that for certain range of values of Kb, there occurs a sharp peak in |R| showing that |R| reaches a maximum value.In these three cases, the maximum value |R| is almost unity, that is, the incident wave experiences almost total reflection for certain range of small values of Kb.This type of behaviour of |R| is also observed by Banerjea in 9 .

Conclusion
A direct function theoretic method is applied here to solve the integral equations 2.2 and 2.26 without converting them to the integral equation with strong singularity.It may be noted here that the solution functions given in 2.22 and 2.38 do not involve any strong singularity.Also using the presently detailed solution of 2.2 , the problem of scattering of water waves by a partially immersed barrier with a gap is examined fully.

R
Figure 2

6 Figure 3 By
Figure 3 reflection coefficient |R| has been numerically evaluated for various values of wave number Kb for four cases, namely, a/b → 0, c/b 15; a/b 0.5, c/b 4; a/b 0.5, c/b 6; a/b 0.25, c/b 6.
we have presented the values of reflection coefficient |R A | computed from our result when a/b → 0 and c/b 15 and |R E |, the reflection coefficients corresponding to the problem of scattering by a submerged vertical barrier x 0, b < y < c c/b 15 as given in the literature cf.Banerjea and Dutta 6 for various values of wave number Kb.It is observed that the values reflection coefficient |R A | coincides with |R E | upto two or three places of decimal.This shows that in limiting case, the reflection coefficient |R A | computed from the present paper agrees with the exact value of |R E | as given in 6 .In Figures 1, 2 a , 2 b , and 3 f , |R| is depicted against Kb for various values of a/b and c/b 6.In each case, we have verified that |R| 2 |T| 2 1 for all values of Kb.In Figure 1, we have plotted |R| against various values of Kb when a/b 0.5 and c/b 4. For 0.15 < Kb < 0.25, it is observed that |R| rises steeply, becomes almost a unity, and then falls rapidly forming a sharp peak.This indicates that for this range of values of Kb, there occurs almost total reflection of wave energy.Also for 0.35 < Kb < 0.4, there occurs a small oscillation in |R|.In Figures 2 a and 2 b , |R| is depicted against Kb for a/b 0.5 and c/b 6.In Figure 2 a , it is observed that for very small values of Kb, that is, for Kb < 0.1, |R| increases at first and then decreases almost to zero indicating that most of the wave energy is transmitted.If the values of Kb are further increased, that is, Kb > 0.1, it is observed from Figure 2 b that |R| increases.For 0.13 < Kb < 0.15, |R| sharply increases almost to unity and then decreases sharply.The peak value in this case is lower than the peak value when c/b 4 1 , Ursell 2 , and Mandal and Kundu 3 .This type In this section, we present a method of solution of