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By Baracco L., Zaitsev D., Zampieri G.

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21 log r q where r =  , x2 +  , y 2 (singular at the point ;  2 ). 11), using the property of the Dirac delta Z ur ! , the domain integral has been replaced by a point value. 11) becomes Z @u Z @! u ;  + u @n d, = @n ! 13) @ @ This equation contains only boundary integrals (and no domain integrals as in Finite Elements) and is referred to as a boundary integral equation. It relates the value of u at some point inside @u the solution domain to integral expressions involving u and @n over the boundary of the solution domain.

19) are then known. 19) to find the solution at any point P 2 . Thus we solve for the boundary data first, and find the volume data as a separate step. 19) only involves surface integrals, as opposed to volume integrals in a finite element formulation, the overall size of the problem has been reduced by one dimension (from volumes to surfaces). , problems with large domains). Also the effort required to produce a volume mesh of a complex three-dimensional object is far greater than that required to produce a mesh of the surface.

Is a weight function but what should this be chosen to be? For a Galerkin formulation choose ! , weight function is one of the basis functions used to approximate the dependent variable. 32) , where the stiffness matrix is Emn where m = 1; : : : ; 4 and n = 1; : : : ; 4 and Fm is the (element) load vector. , F = kx where k is the stiffness of spring and F is the force/load. This yields the system of equations Emn un = Fm . 6). 6: Considering heat flow in a unit square. 7 A SSEMBLE G LOBAL E QUATIONS 37 The first component E11 is calculated as E11 = k Z1 Z1 1 , y2 + 1 , x2 dxdy 0 0 = 23 k and similarly for the other components of the matrix.

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A Burns-Krantz type theorem for domains with corners by Baracco L., Zaitsev D., Zampieri G.


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