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Constructing Solutions of Cauchy Type Integral Equations by Using Four Kinds of Basis

  • PARTIAL DIFFERENTIAL EQUATIONS
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Abstract

We have four kinds of solutions for Cauchy type integral equations: by expanding on known functions and using the Maclaurin series, we can convert these four kinds of solutions into linear combinations of some elements that are basis of these solutions. Using these bases gives exact solutions for polynomials and, for some other functions, a high-accuracy approximate solution.

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RESEARCH DATA POLICY AND DATA AVAILABILITY STATEMENTS

The datasets generated during or analyzed during the current study are available from the corresponding author on reasonable request.

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The authors declare that no funds, grants, or other support were received during the preparation of this manuscript.

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Authors and Affiliations

  1. Department of Mathematics, Shahid Rajaee Teacher Training University, Tehran, Iran

    M. Yaghobifar & F. Hosseini Shekarabi

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  1. M. Yaghobifar
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  2. F. Hosseini Shekarabi
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Correspondence to M. Yaghobifar or F. Hosseini Shekarabi.

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APPENDIX

APPENDIX

List of some \({{I}_{n}}(x)\) are as below:

$${{I}_{0}}(x) = 0.$$
$${{I}_{1}}(x) = - \pi .$$
$${{I}_{2}}(x) = - \pi x.$$
$${{I}_{3}}(x) = - \pi \left[ {{{x}^{2}} + \frac{1}{4}\left( {\begin{array}{*{20}{c}} 2 \\ 1 \end{array}} \right)} \right].$$
$${{I}_{4}}(x) = - \pi \left[ {{{x}^{3}} + \frac{1}{4}\left( {\begin{array}{*{20}{c}} 2 \\ 1 \end{array}} \right)x} \right].$$
$${{I}_{5}}(x) = - \pi \left[ {{{x}^{4}} + \frac{1}{4}\left( {\begin{array}{*{20}{c}} 2 \\ 1 \end{array}} \right){{x}^{2}} + \frac{1}{{{{4}^{2}}}}\left( {\begin{array}{*{20}{c}} 4 \\ 2 \end{array}} \right)} \right].$$
$${{I}_{6}}(x) = - \pi \left[ {{{x}^{5}} + \frac{1}{4}\left( {\begin{array}{*{20}{c}} 2 \\ 1 \end{array}} \right){{x}^{3}} + \frac{1}{{{{4}^{2}}}}\left( {\begin{array}{*{20}{c}} 4 \\ 2 \end{array}} \right)x} \right].$$
$${{I}_{7}}(x) = - \pi \left[ {{{x}^{6}} + \frac{1}{4}\left( {\begin{array}{*{20}{c}} 2 \\ 1 \end{array}} \right){{x}^{4}} + \frac{1}{{{{4}^{2}}}}\left( {\begin{array}{*{20}{c}} 4 \\ 2 \end{array}} \right){{x}^{2}} + \frac{1}{{{{4}^{3}}}}\left( {\begin{array}{*{20}{c}} 6 \\ 3 \end{array}} \right)} \right].$$
$${{I}_{8}}(x) = - \pi \left[ {{{x}^{7}} + \frac{1}{4}\left( {\begin{array}{*{20}{c}} 2 \\ 1 \end{array}} \right){{x}^{5}} + \frac{1}{{{{4}^{2}}}}\left( {\begin{array}{*{20}{c}} 4 \\ 2 \end{array}} \right){{x}^{3}} + \frac{1}{{{{4}^{3}}}}\left( {\begin{array}{*{20}{c}} 6 \\ 3 \end{array}} \right)x} \right].$$
$${{I}_{9}}(x) = - \pi \left[ {{{x}^{8}} + \frac{1}{4}\left( {\begin{array}{*{20}{c}} 2 \\ 1 \end{array}} \right){{x}^{6}} + \frac{1}{{{{4}^{2}}}}\left( {\begin{array}{*{20}{c}} 4 \\ 2 \end{array}} \right){{x}^{4}} + \frac{1}{{{{4}^{3}}}}\left( {\begin{array}{*{20}{c}} 6 \\ 3 \end{array}} \right){{x}^{2}} + \frac{1}{{{{4}^{4}}}}\left( {\begin{array}{*{20}{c}} 8 \\ 4 \end{array}} \right)} \right].$$
$${{I}_{{10}}}(x) = - \pi \left[ {{{x}^{9}} + \frac{1}{4}\left( {\begin{array}{*{20}{c}} 2 \\ 1 \end{array}} \right){{x}^{7}} + \frac{1}{{{{4}^{2}}}}\left( {\begin{array}{*{20}{c}} 4 \\ 2 \end{array}} \right){{x}^{5}} + \frac{1}{{{{4}^{3}}}}\left( {\begin{array}{*{20}{c}} 6 \\ 3 \end{array}} \right){{x}^{3}} + \frac{1}{{{{4}^{4}}}}\left( {\begin{array}{*{20}{c}} 8 \\ 4 \end{array}} \right)x} \right].$$

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Yaghobifar, M., Shekarabi, F.H. Constructing Solutions of Cauchy Type Integral Equations by Using Four Kinds of Basis. Comput. Math. and Math. Phys. 63, 1671–1680 (2023). https://doi.org/10.1134/S0965542523090142

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  • DOI: https://doi.org/10.1134/S0965542523090142

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