The Boundary Element Method for the Helmholtz Equation ... - FEI VÅ B

73Now we consider the situation αs x ≠ t x and α ≠ 0. For K we haveK = u 2 (1 + α 2 )x (1 + α 2 ) u 2 x + (αsx−tx)2 − (αs1+α 2 x − t x )with a nonzero limit of the denominator. Thus, the function u 2 x ln |K| behaves like u2x ln |cu 2 x| → 0 for ux → 0. (4.28)Finally, let us assume the case of αs x ≠ t x and α = 0. For K we haveK =u 2 xu 2 x + t 2 x + t x.For t x ≥ 0 the limit is similar to (4.28). For t x < 0 we haveK =u 2 xu 2 x + t 2 x + t xu 2 x + t 2 x − t xu 2 x + t 2 x − t x= u 2 x + t 2 x − t xand the term u2x ln |K| vanishes for ux → 0.