Abstract
Purpose
This paper investigates a different method to evaluate different real improper integrals and also to obtain the solutions of various types of Cauchy-type singular integral equations of the first kind.
Methods
Methods using the analysis of functions of real variables only are reviewed and utilized for the above purpose. These methods clearly demonstrate that details of complex function theory which are normally employed in handling such integral equations for their solutions can be avoided altogether. Also, some approximate methods of solution of such integral equations are developed.
Results
The solutions of real singular integral equations over different intervals such as (−1,1); (a, b); (0, a) ∪ (b, c); (−1, k) ∪ (k, 1); (−∞, b); (a, + ∞); (−∞, + ∞); infinite intervals with a gap are obtained by using the proposed methods.
Conclusion
The proposed methods are new and each has its own structure.
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Background
Real singular integral equations involving Cauchy-type singularities arise (see[1–10]) in a natural way in handling a large class of mixed boundary value problems of mathematical physics, especially when two-dimensional problems are encountered. The integrals occurring in these integral equations are in fact improper and their evaluations in most cases can be rendered by using the theory of functions of complex variables involving the application of Cauchy’s residue theorem. It is desirable, as is always felt, to avoid the use of complex function theory to evaluate real integrals because the details can be more involved analytically speaking than what is actually necessary for being able to use the final results in practical problems. It is with this idea in mind, in the present paper, that we have first reviewed the problems of evaluation of several real improper integrals (see[11]) by the help of the theory of functions of real variables only whilst the application of complex function theory is also demonstrated for these problems for comparison as well as for realizing the major differences of the analysis involved.
Thus, by developing the feeling that complex function theory can be avoided for problems involving improper real integrals, we have next taken up some known real singular integral equations involving Cauchy-type kernels and have presented the real variable method of solution of these equations.
The plan of this paper is as follows: In the ‘Evaluation of real improper integrals’ section, we consider some problems of real improper integrals and their solutions. In the ‘Solution of a Cauchy-type singular integral equation of the first kind’ section, a Cauchy-type singular integral equation of the first kind is considered for its solution in the intervals (−1,1) and (a, b). We consider a Cauchy-type singular integral equation of the first kind over two disjoint intervals, (0,a)∪(b,c) and (−1,−k)∪(k,1) and its solution in the ‘Solution of a Cauchy-type singular integral equation of the first kind over an interval with a gap’ section. In the ‘Verification of the solutions for homogeneous Cauchy-type singular integral equation of the first kind’ section, we verify the solution of the homogeneous problem obtained in the previous section. We then we determine the approximate solution of Cauchy-type singular integral equations in the intervals (0,1), (−1,−k)∪(k,1) and then (0,a)∪(b,c) in the ‘Approximate solution of singular integral equations of the Cauchy type’ section. Finally, in the ‘Solutions of Cauchy-type singular integral equations over semi-infinite and infinite intervals’ and ‘Solution of Cauchy-type singular integral equations of the first kind over infinite intervals with a gap’ sections, we derive the solutions of singular integral equations of the Cauchy type, involving semi-infinite as well as infinite intervals, as special limiting cases and show that the final results agree with the known ones.
Results and discussion
Evaluation of real improper integrals
In this section, we consider the problems of evaluation of certain special real improper integrals and their solutions by using the complex variable method as well as the real variable method.
Problem 1
Evaluate
Solution (using complex analysis):
Let. Using Cauchy’s residue theorem, we obtain
where Γis the closed contour consisting of the upper half of the large circle |z| = R and the real axis from -R to R which avoids the origin, with a semicircular indentation of radius r.
Then, letting R → ∞ and r → 0, we get
giving
Solution (without using complex analysis):
We write
which matches with Equation 4.
Problem 2
Evaluate
Solution (using complex analysis):
Using Cauchy’s residue theorem, we obtain
where and Γ is the same contour as was used for problem 1.
Again, letting R → ∞ and r → 0, we get
giving
Thus, we find that
Solution (without using complex analysis):
We write
which agrees with Equation 10.
Problem 3
Evaluate
Solution (using complex analysis):
where
We use Cauchy’s residue theorem and get
where and Γ is a contour comprising of a circular indentation of radius r at the origin along with the two parts of the positive real axis, one lying above and the other lying below, as well as a large circle of radius R.
Then, letting R → ∞ and r → 0, we obtain
where
We then find that
Solution (without using complex analysis):
We write
where.
Equation 18 agrees with Equation 16.
Problem 4
Evaluate
Solution: We write
where
Solution (using complex analysis):
Applying Cauchy’s residue theorem, we first get
where Γ is the contour lying above the real axis, with a small semicircular indentation of radius h and a large semicircular arc of radius R, giving
Then, we find that
Solution (without using complex analysis):
We can write
Then, using Equation 20, we get
which agrees with Equation 24.
Problem 5
Evaluate
Solution (using complex analysis):
Cauchy’s residue theorem gives
where Γ is the contour lying at the right side of the y-axis with a large semicircular arc of radius R, giving
But J = 2iI, which gives
Solution (without using complex analysis):
We write
which matches with Equation 29.
A different approach:
We write
which matches with Equations 29 and 31.
Problem 6
Evaluate
Solution (using complex analysis):
We write
where Γ is the unit circle around the origin.
Then, applying Cauchy’s residue theorem, we get
Solution (without using complex analysis):
We can write Equation 33 as
and this agrees with Equation 35.
Problem 7
Evaluate
Solution:
We can write Equation 37 as
Solution (using complex analysis):
Applying Cauchy’s residue theorem, involving the contour Γcomprising of the real axis and a large semicircular arc lying above the real axis, we obtain
Solution (without using complex analysis):
We can write Equation 38 as
and this matches with Equation 39.
Problem 8
Evaluate
Solution:
I(x) can be written as
where
This ψ(t) can be written as
where with.
From Equations 42 and 44, I(x) can be written as
where
Evaluation of I n (x) (using complex analysis):
from which we obtain
Evaluation of I n (x) (without using complex analysis):
I n (x) can also be written as
where
The values of I n (x) can then be determined easily which are the same as the ones obtained in Equation 48.
Substituting the values of I n (x)(n = 0,1,2,…) in Equation 45, the approximate value of I(x) can be written as
The exact value of I(x) is given by (see[2])
In Table1, we find that the exact values and approximate values of I(x) are nearly equal.
Evaluation of some real singular integrals by real variable method
Here, we consider some problems involving singular integrals and their solutions by real variable method. These results are useful to evaluate some other integral equations in the succeeding sections.
Problem 1
Evaluate
Solution: Setting, we get
Case 1: When x > 1, we obtain
Case 2: When x < 0, we obtain
So, we get
Problem 2
Evaluate
Solution: Setting, we get
Case 1: When x > 1, we obtain
Case 2: When x < 0, we obtain
So, we get
Problem 3
Evaluate
Solution: Setting, we get
Case 1: When x > c (i.e., x − c > 0 and x − a > 0), we obtain
Case 2: When x < a (i.e., a − x > 0, c − x > 0), we obtain
So, we obtain
Problem 4
Evaluate
Solution: Setting, we get
Case 1: When x > c (i.e., x − c > 0 and x − a > 0), we get
Case 2: When x < a (i.e., a − x > 0, c − x > 0), we get
So, we obtain
Solution of a Cauchy-type singular integral equation of the first kind
Here, we consider a Cauchy-type singular integral equation of the first kind in the interval (-1,1) and obtain its solution by real variable method. Then, we have generalized the result for the interval (a,b).
Problem 1
Solve the singular integral equation of the first kind
Solution: Set
Get: t − x = 2(u−ξ).
Hence, the given integral equation (Equation 74) becomes
where
Now, we set
Then, we get, from Equation 76,
Now, we set (see[12])
Then, Equations 79 and 80 give
Now, we have the following results:
Result 1:
Result 2: Let
Now,
Again, we get
So, we get
Now, let us assume that
We have
We then get
Hence, the assumption in Equation 87 holds good for all r = 1,2,…, by induction.
Thus, by using Equation 87, we get, from Equation 81,
Hence, we obtain, by the Fourier series method,
This, then, gives, from Equation 80,
Now, we have the following result (using a limiting procedure of the type explained above):
Thus, Equations 92 and 93 give
Hence, putting back (see Equation 78):, we obtain, from Equation 94,
where c = (1/4)a0 = an arbitrary constant.
Note that Equation 95 is the well-known form of the solution of the integral equation (Equation 76), obtainable by using the theory of Riemann-Hilbert problems involving functions of a complex variable.
Finally, substituting, we obtain
which is the well-known form of the solution of the given integral equation (Equation 74) where c0 = 2c is an arbitrary constant.
Now, we consider the integral equation given by Equation 74 for its solution in the interval (a, b). So, we have the following problem to solve:
Problem 2
Solve the singular integral equation
Solution: Set
Get: t − x = (b−a)(u−ξ).
Hence, the given integral equation (Equation 97) becomes
where
By the help of Equations 76 and 95, the solution of the integral equation (Equation 99) can be written as
where c = (1/4)a0 = an arbitrary constant.
Now, substituting,, in Equation 101, we get
which is the well-known form of the solution of the given integral equation (Equation 74), where c0 = c(b−a) is an arbitrary constant.
Solution of a Cauchy-type singular integral equation of the first kind over an interval with a gap
Here, we consider a Cauchy-type singular integral equation of the first kind over two disjoint intervals (0,a)∪(b c) and obtain its solution by real variable method. We also find, here, its solution, applicable to the intervals (−1,−k)∪(k,1) as considered by Tricomi[13].
Problem: Solve the singular integral equation of the first kind, involving a finite interval with a gap, as given by
Solution: We shall use the known solution of the following singular integral equation:
which is given by
where c0 is an arbitrary constant.
Setting
we obtain, from Equations 104 and 106, that the solution of the singular integral equation
is given by
where D0 is an arbitrary constant.
Now, we shall first solve the homogeneous integral equation (Equation 103) as follows:
Consider the homogeneous integral equation (Equation 103), as given by
Let us assume that there exist two functions: ψ0(t) and f0(x), such that
and
where is an unknown function.
Then, using Equations 109, 110, and 111, we obtain an integral equation which is given by
which possesses the solution (see Equations 107 and 108)
where E0 is an arbitrary constant.
Now, ψ0(x) = 0, for x ∈ (a,b) (see Equation 110), gives an integral equation for the unknown function, as given by (see Equation 113)
with its solution, as given by the equation (see Equations 105 and 108)
where.
Thus, using Equations 113 and 115, as well as Equation 110, we obtain the solution of the homogeneous equation (Equation 109), as given by
Now, we can evaluate the integral in Equation 116 for x ∉ (a b) (see[2]) and obtain
Hence, by using Equations 116 and 117, we obtain
which is the solution of the homogeneous equation (Equation 118), where, are two arbitrary constants.
Now, we solve the inhomogeneous equation (Equation 103) (for a particular solution):
We define
and
where is an unknown function.
Then, Equation 103 can be expressed as
with the particular solution
Then, ψ(x) = 0, for x ∈ (a,b), gives the integral equation (see Equations 120, 121, and 123):
Now the singular integral equation (Equation 124) possesses the particular solution as given by
Using Equation 121, Equation 123 can be written as
From the Equations 119 and 127, the general solution of the singular integral equation (Equation 103) can be obtained as
where (note that N0 and are two arbitrary constants).
Now, we set
and get (note that the integral equation (Equation 103) transforms to a new equation (see[13]))
which exactly matches with the result obtained by Tricomi[13], where are two arbitrary constants.
Verification of the solutions for homogeneous Cauchy-type singular integral equation of the first kind
In this section, we verify the solutions obtained in the ‘Solution of Cauchy-type singular integral equations of the first kind over an interval with a gap’ section for the homogeneous equation involving a Cauchy-type singular integral associated with the disjoint intervals (0,a)∪(b,c) and (−1,−k)∪(k,1).
Problem 1
Prove that
is a solution of the homogeneous integral equation
Solution: Substituting Equation 131 in the left-hand side of Equation 132, we get
For evaluation of the integrals, we will consider two cases: Case I: x ∈ (0,a) and Case II: x ∈ (b,c).
For t ∈ (0,a)∪(b,c), take
For x ∈ (0,a)∪(b,c), take
Case I: When x ∈ (0, a )
Case II: When x ∈ ( b , c )
Hence, the problem gets solved. ▇
Problem 2
Prove that
is a solution of the homogeneous integral equation
Solution: Substituting Equation 136 in the left-hand side of equation 137, we get
Putting τ2 = u and ξ2 = v, we get
Now, substituting, we get
This solves the problem. ▇
Approximate solution of singular integral equations of the Cauchy type
Here, we find an approximate solution of singular integral equations of the Cauchy type, involving the intervals (0,1), (−1,−k)∪(k,1), and (0,a)∪(b,c).
Problem 1
Solve
Solution: Let
where a n (n = 0,1,2,…) and f n (n = 0,1,2,…) are real constants.
Substituting Equation 141 in Equation 140, we get
For n = 0,1,2,3, we obtain
Equating the coefficients of xn(n = 0,1,2,3) from both sides, we get
Example: Let f(x) = 1 (i.e., f0 = 1,f1 = f2 = 0).
From Equations 1, we obtain
Hence,
with a0 as an arbitrary constant.
A different approach
From Equations 97 and 102, the solution of Equation 140 can be written as
For f(x) = 1,
Substituting Equation 148 in Equation 147, we get
which matches exactly with Equation 146, where is an arbitrary constant.
Problem 2
Find an approximate solution of the integral equation
Solution: Assume
is a solution of the integral equation (Equation 150) and also assume that
where a n (n = 0,1,2,…) and g n (n = 0,1,2,…) are real constants.
Substituting Equations 151 and 152 in Equation 150, we get
Now, consider the left side (LS) of Equation 153 and take k, l = 0,1 and τ2 = u ξ2 = v, and obtain (see[2])
Then, from Equations 153 and 154, we get
Equating the coefficients of ξn(n = 0,1,2,3), we obtain
In particular, taking g(ξ) = 1 (i.e., g0 = 1,g1 = g2 = g3 = 0), we get
Hence,
is the special case g = 1, where a0 and a1 are two arbitrary constants.
Matching with the closed form solution (Equation 130): Using Equation 130, the solution of Equation 150 can be written as
where
Now Q can be written as
In particular, taking g(ξ) = 1 (g(−ξ)=1)
From Equations 158 and 160, we then get
which exactly matches with Equation 157, where.
Note that A1 and B are two arbitrary constants.
We can also find an approximate solution for the non-homogeneous integral equation (Equation 150) involving the disjoint intervals (0,a)∪(b,c)
Substituting,,, and ϕ(τ) = ϕ(t) in Equation 157, we get
where and are two arbitrary constants.
Solutions of Cauchy-type singular integral equations over semi-infinite and infinite intervals
Here, we consider a Cauchy-type singular integral equation of the first kind in the intervals (−∞,b), (a,∞), and (−∞,∞) and obtain the solutions by a limiting process applied to the solutions of similar equations associated with finite intervals.
Problem 1
Solve the singular integral equation
Solution: We know that the general solution of the singular integral equation
is given by
where c0 is an arbitrary constant.
Taking limit as a → −∞, keeping x fixed, we get
which is the general solution of the integral equation (Equation 163), where c1 is an arbitrary constant as given by.
Problem 2
Solve the singular integral equation
Solution: Taking limit as b → ∞, keeping x fixed, in Equation 165, we get
which is the general solution of the integral equation (Equation 167), where c2 is an arbitrary constant as given by.
Problem 3
Solve the singular integral equation
Solution: We know that the general solution of the singular integral equation
is given by
where c3 is an arbitrary constant.
Taking limit as a → ∞, keeping x fixed, we get
where is an arbitrary constant.
Now, for consistency, we must have c4 = 0, since ϕ(x) must tend to 0 as x → ∞. Hence, from Equation 172, we get
which is the general solution of the integral equation (Equation 169). Equations 169 and 173 exactly match with the results obtained in[14], known as Hilbert’s pair of formulae.
Solution of Cauchy-type singular integral equations of the first kind over infinite intervals with a gap
Here, we consider a Cauchy-type singular integral equation of the first kind over two disjoint intervals (−∞,b1)∪(c1,d1); (a1,b1)∪(c1,∞); and (−∞,−k)∪(k,∞) and obtain its solution by limiting process applied to the known solution for equations involving two disjoint finite intervals.
Problem 1
Solve the singular integral equation of the first kind, involving a semi-infinite interval with a gap, as given by
Solution: We know that the solution of the integral equation
is given by
where N0 and H0 are two arbitrary constants.
Now, we put
in Equations 175 and 176 and find that the solution of the integral equation
is given by
where and are arbitrary constants.
Then, taking limit as a1 → −∞, keeping x fixed, we get
which is the general solution of the integral equation (Equation 174), where is an arbitrary constant.
Problem 2
Solve the singular integral equation of the first kind, involving a semi-infinite interval with a gap, as given by
Solution: Taking limit as d1→∞, keeping x fixed, in Equation 179, we obtain
which is the general solution of the integral equation (Equation 181), where is an arbitrary constant.
Problem 3
Solve the singular integral equation of the first kind, involving an infinite interval with a gap, as given by
Solution: Taking limit as a1→−∞, d1→∞, b1→−k, and c1→k, keeping x fixed, in Equation 179, we obtain
which is the required solution of the integral equation (183), whereis an arbitrary constant.
Conclusions
Methods involving evaluation of real improper integrals by the use of the ideas of the theory of functions of real variables only are highlighted, and solutions of certain real singular integral equations of the Cauchy type have been re-derived. Particular examples have been worked out in detail, and the final results have been compared with the known ones.
Methods
The principal methods used in the present work involve application of the theory of functions of real variables only to analyze and solve singular integral equations involving real valued functions everywhere. As a result, the presently developed methods are free from other complicated methods used earlier by various workers to determine the solutions of such integral equations.
Endnotes
Methods of solution of singular integral equations involve, generally speaking, details of complex function theory needing to analyze new types of boundary value problems of the Riemann Hilbert type. In this paper, we have demonstrated, in a systematic manner, that if our concern is to determine solutions of singular integral equations in which the unknown function, the kernel, as well as the forcing term are all functions of real variables, then methods based on the theory of functions of real variables only help in finding out the all wanted real valued solution function. The methods developed in this paper replaces the detailed use of complex function theory and the final forms of the solutions are expected to be useful for direct application to practical problems.
Authors’ information
AC received his M.Sc. and Ph.D. in Applied Mathematics from Calcutta University, Kolkata, India and retired as a professor in the Department of Mathematics, Indian Institute of Science, Bangalore, India. He is currently a NASI-Senior Scientists Platinum Jubilee Fellow in the same institute. His areas of research are diffraction theory, elasticity, fluid mechanics, integral equations, and boundary value problems. SCM received his M.Sc. from Sambalpur University, Odisha, India and Ph.D. from Indian Institute of Technology Guwahati, Guwahati, India in Mathematics. He is an assistant professor in the Department of Mathematics, Indian Institute of Technology Ropar, India. His main areas of interest are mathematical modeling of water wave phenomena, scattering by very large floating structures (VLFS), and integral equation.
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Acknowledgements
A. Chakrabarti thanks the National Academy of Sciences India (NASI) for awarding him a Senior Scientists Platinum Jubilee Fellowship, and S.C. Martha is grateful to both the National Board for Higher Mathematics, India, and the Indian Institute of Science, Bangalore, India, for providing a Post Doctoral Fellowship and necessary infrastructure, respectively, for carrying out research when this paper work was initiated. He also thanks the Institute of Technology Ropar India for awarding the ISIRD grant and for providing all the necessary facilities.
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AC and SCM contributed equally to this work. Both authors read and approved the final manuscript.
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Chakrabarti, A., Martha, S.C. Methods of solution of singular integral equations. Math Sci 6, 15 (2012). https://doi.org/10.1186/2251-7456-6-15
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DOI: https://doi.org/10.1186/2251-7456-6-15