2010
DOI: 10.2140/jomms.2010.5.679
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Influence of different integral kernels on the solutions of boundary integral equations in plane elasticity

Abstract: A modified integral kernel is introduced for boundary integral equations (BIE). The formulation for the modified kernel is based on a representation in pure deformable form of the fundamental solution of concentrated forces. It is found that the modified kernel can be applied to any case, even if the loadings on the contour are not in equilibrium in an exterior boundary value problem. The influence of different integral kernels on solutions of BIE, particularly in the Neumann problem and Dirichlet problem, are… Show more

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Cited by 4 publications
(2 citation statements)
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“…Concerning the results in [3,7], it is worth noting that these papers present the results for a complex potential. This complex potential leads to the following Green tensor, which is dierent from (5) [6,8]: Circle of radius 1 ρ 0 = 1 ρ 1 = ρ 2 = e 1/2κ [3] Ellipse with half axis a and b such that a + b = 2 ρ 0 = 1 ρ 1 = e (1−m)/2κ ; ρ 2 = e (1+m)/2κ with m = (a − b)/(a + b) [3] Segment of length 4 ρ 0 = 1 ρ 1 = 1; ρ 2 = e 1/κ [27] Hypotrochoid, approximating an equilateral triangle, image of the unit circle by z = (ζ + 1/3ζ 2 ) ρ 0 = 1 ρ 1 = ρ 2 = e −1/18κ 2 +1/2κ [5] Hypotrochoid, approximating a square, image of the unit circle by z = (ζ − 1/6ζ 3 ) ρ 0 = 1 ρ 1 = ρ 2 = e −1/36κ 2 +1/2κ [5] Ellipse with s 1 = −900m/(225 − 4m); s 2 = −4m/(225 − 4m) [5] These results from [3,7] have been processed to correspond with the Green tensor dened previously in (5), corresponding to the factor e 1/2κ in the last column of Table 2.…”
Section: B Inequalities For the Logarithmic Capacitymentioning
confidence: 99%
“…Concerning the results in [3,7], it is worth noting that these papers present the results for a complex potential. This complex potential leads to the following Green tensor, which is dierent from (5) [6,8]: Circle of radius 1 ρ 0 = 1 ρ 1 = ρ 2 = e 1/2κ [3] Ellipse with half axis a and b such that a + b = 2 ρ 0 = 1 ρ 1 = e (1−m)/2κ ; ρ 2 = e (1+m)/2κ with m = (a − b)/(a + b) [3] Segment of length 4 ρ 0 = 1 ρ 1 = 1; ρ 2 = e 1/κ [27] Hypotrochoid, approximating an equilateral triangle, image of the unit circle by z = (ζ + 1/3ζ 2 ) ρ 0 = 1 ρ 1 = ρ 2 = e −1/18κ 2 +1/2κ [5] Hypotrochoid, approximating a square, image of the unit circle by z = (ζ − 1/6ζ 3 ) ρ 0 = 1 ρ 1 = ρ 2 = e −1/36κ 2 +1/2κ [5] Ellipse with s 1 = −900m/(225 − 4m); s 2 = −4m/(225 − 4m) [5] These results from [3,7] have been processed to correspond with the Green tensor dened previously in (5), corresponding to the factor e 1/2κ in the last column of Table 2.…”
Section: B Inequalities For the Logarithmic Capacitymentioning
confidence: 99%
“…The null field method of Chen and the study of Chen in [13][14][15] seem to be more focused on the degenerate scale problems. For the circular domains with circular holes by the MFEs, our efforts are paid not only for the degenerate scales in [18,19], but also for the numerical algorithms themselves in [20] and this paper.…”
Section: Introductionmentioning
confidence: 99%