# Download e-book for iPad: A Projection Transformation Method for Nearly Singular by Ken Hayami

By Ken Hayami

In 3 dimensional boundary point research, computation of integrals is a crucial element because it governs the accuracy of the research and likewise since it often takes the foremost a part of the CPU time. The integrals which ensure the effect matrices, the inner box and its gradients comprise (nearly) singular kernels of order lIr a (0:= 1,2,3,4,.··) the place r is the gap among the resource aspect and the combination element at the boundary aspect. For planar components, analytical integration might be attainable 1,2,6. despite the fact that, it truly is turning into more and more very important in functional boundary aspect codes to exploit curved parts, comparable to the isoparametric components, to version common curved surfaces. considering analytical integration isn't attainable for common isoparametric curved components, one has to depend on numerical integration. while the gap d among the resource element and the aspect over which the mixing is played is adequately huge in comparison to the aspect dimension (d> 1), the traditional Gauss-Legendre quadrature formulation 1,3 works successfully. even though, whilst the resource is admittedly at the aspect (d=O), the kernel 1I~ turns into singular and the simple software of the Gauss-Legendre quadrature formulation breaks down. those integrals might be known as singular integrals. Singular integrals happen while calculating the diagonals of the impression matrices.

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For discontinuous elements 1, Ci = 1 / 2 for i = 1-N, and calculating the singular integrals Hii directly are reported to give more accurate results 15. 68) where u* = 1/(47l"r) and G ii = L k x = x. 71) which contribute to the non-diagonal element Hij and Gij. It will be shown in Chapter 3 that the integrals hee k1 , gee k1 (k:;t: l) are not singular for three dimensional potential problems in the sense that the integral kernels are of order 0(1) or O(r) , where r = Ix -xekl, since e1(xe k) = 0 for k:;t: I.

44) is solved for the boundary values u and q, the potential u(xs) at an internal point Xs can be calculated by U (xs ) n =i i. (1J 1 ,1J 2) x! 55) dTJ 1 dTJ 2 ,and n is the unit outward normal vector of the boundary element Se at x E Se. 4. 13) when Xs is a node shared by two or more elements as shown in Fig. 8 involves the calculation of the solid angle w subtended by the region V at Xs on S. In order to avoid this, one may use the row sum elimination method in many cases. For the three dimensional potential problem defined in the interior region, consider the equipotential solution u(x)== 1 to the original Laplace equation b.

D. 6) are only weakly singular. They can be calculated efficiently using polar coordinates on the plane tangent at Xs, as will be shown by numerical experiments in Chapter 7. For the case when e' = e but k:;t: l, it is shown in the following that the numerators of the integral kernels also become zero at the source point xe' , so that the integrals gee kl and hee kl have a even weaker singularity compare to gee" and hee". 42) and let the source xek be x(O, 0). 2) or 0('71 '7:J in the neighborhood of xek = x(O, 0) and will not include a constant term.