ote 9 To construct a potential that satisfies Laplace's equation and matches the undary conditions at more points, let u be represented by the potential due to a sum of n weighted point charges in the square's interior. As shown in the slide, we can think of the potential due to a sum of charges as a sum of Greens functions. Of course, we have to determine the weights on the n point charges and the weight on the i n charge is denoted hereby a 8.1.2 More points equations SLIDE 13 Source Strengths selec to give correct potent - testpoints a(x-x,￥-y G(x-x,x-Y Note 10 contd To determine a system of n equations for the n ais, consider selecting a set of test points, as shown in the slide above. Then, by superposition, for each test point ti, yt,, u(xr4,v,)=∑alog√(xt1-xo)2+(vt4-yo) Writing an equation like(6) for each test point yields the matrix equation or the slide The matrix A in the slide has some properties worth notin . A is dense, that is Ai, never equals zero. This is because every charge contributes to every potential If the test points and the charge points are ordered so that the ith test point is nearest the ith charge, then Ai. i will be larger than Ai i for all j Item 2 above seems to suggest that A is diagonally dominant, but this is not the case. Diagonal dominance requires that the absolute sum of the off-diagonal entries is smaller than the magnitude of the diagonal. The matrix above easil violates that conditionNote 9 ...contd To construct a potential that satisfies Laplace’s equation and matches the boundary conditions at more points, let u be represented by the potential due to a sum of n weighted point charges in the square’s interior. As shown in the slide, we can think of the potential due to a sum of charges as a sum of Green’s functions. Of course, we have to determine the weights on the n point charges, and the weight on the i th charge is denoted hereby αi. 8.1.2 "More points equations" Slide 13 ( ) x y 1 1 , ( ) x y 2 2 , ( ) x y n n , ( ) 1 1 x y t t , Source Strengths selec to give correct potent testpoints. ( )( ) ( )( ) ( ) ( ) 1 1 1 1 11 1 1 1 1 1 , ,, , ,, n n n n nn t t t nt n t t n t t t nt n t t Gx x y y Gx x y y x y Gx x y y Gx x y y x y α α    −− − − Ψ          =     −− − − Ψ       



Note 10 ...contd To determine a system of n equations for the n αi’s,consider selecting a set of n test points, as shown in the slide above. Then, by superposition, for each test point xti , yti , u(xti , yti ) = n i=1 αi log (xti − x0)2 + (yti − y0)2 = n i=1 αiG(xti − x0, yti − y0). (6) Writing an equation like (6) for each test point yields the matrix equation on the slide. The matrix A in the slide has some properties worth noting: • A is dense, that is Ai,j never equals zero. This is because every charge contributes to every potential. • If the test points and the charge points are ordered so that the i th test point is nearest the i th charge, thenAi,i will be larger than Ai,j for all j. Item 2 above seems to suggest that A is diagonally dominant, but this is not the case. Diagonal dominance requires that the absolute sum of the off-diagonal entries is smaller than the magnitude of the diagonal. The matrix above easily violates that condition. 9