[...]... best A simple one is to solve for un,m in Equation 1. 1.64 Assuming ∆r = ∆z, we obtain 1 4 i +1 i +1 i +1 i i ui n +1, m + un 1, m + un,m +1 + un,m 1 + un +1, m − un 1, m /(2n) , (1. 1.69) where n = 1, 2, , N − 1 and m = 1, 2, , M − 1 Similarly, ui +1 = n,m ui +1 = 0,m 1 4 i +1 2ui + ui 1, m 0,m +1 + u0,m 1 , (1. 1.70) where m = 1, 2, , M − 1 Here, we denote the value of un,m during the ith iteration with the... Equation 1. 1 .14 become d 2 U (k, y) − k 2 U (k, y) = 0, dy 2 0 < y < h, (1. 1 .16 ) with U (k, h) = 0 Solving Equation 1. 1 .16 and inverting the transform, we find that u(x, y) = 2 π ∞ A(k) 0 sinh[k(h − y)] cos(kx) dk sinh(kh) (1. 1 .17 ) Substituting Equation 1. 1 .17 into Equation 1. 1 .13 , we obtain − 2 π ∞ 0 ≤ x < 1, k coth(kh)A(k) cos(kx) dk = 1/ h, 0 and 2 π ∞ A(k) cos(kx) dk = 0, 0 1 < x < ∞ (1. 1 .18 ) (1. 1 .19 )... to solve these mixed boundary value problems 1. 1 EXAMPLES OF MIXED BOUNDARY VALUE PROBLEMS Before we plunge into the details of how to solve a mixed boundary value problem, let us examine the origins of these problems and the challenges to their solution 1 © 2008 by Taylor & Francis Group, LLC 2 Mixed Boundary Value Problems • Example 1. 1 .1: Separation of variables Mixed boundary value problems arise... (1. 1 .11 ) subject to the boundary conditions ux (0, y) = 0 and lim u(x, y) → 0, x→∞ uy (x, 0) = 1/ h, u(x, 0) = 0, 0 < y < h, 0 ≤ x < 1, 1 < x < ∞, (1. 1 .12 ) (1. 1 .13 ) and u(x, h) = 0, 0 ≤ x < ∞ (1. 1 .14 ) The interesting aspect of this problem is the boundary condition given by Equation 1. 1 .13 It changes from a Neumann condition to a Dirichlet condition along the boundary x = 1 To solve this boundary value. .. L L and (1. 1.38) ∞ G(k, y|ξ, η) = 2 −ikξ sin(nπη/L) sin(nπy/L) e L k 2 + n2 π 2 /L2 n =1 (1. 1.39) Inverting the Fourier transforms in Equation 1. 1.39 term by term, we obtain g(x, y|ξ, η) = = ∞ nπ nπη nπy 1 exp − |x − ξ| sin sin n L L L n =1 1 π 1 2π (1. 1.40) ∞ nπ 1 exp − |x − ξ| n L n =1 × cos nπ(y − η) nπ(y + η) − cos L L (1. 1. 41) Because ∞ qn cos(nα) = − ln n n =1 1 − 2q cos(α) + q 2 , (1. 1.42) provided... along 1 < (t) < 1, (t) = 0 Consider next, the fractional linear transformation s= αt + β , γt + δ (1. 1. 51) 5 Laporte, O., and R G Fowler, 19 66: Resistance of a plasma slab between juxtaposed disk electrodes Phys Rev., 14 8, 17 0 17 5 © 2008 by Taylor & Francis Group, LLC 10 Mixed Boundary Value Problems Figure 1. 1.2: Same as Figure 1. 1 .1 except that we have the additional mapping given by Equation 1. 1. 51. .. the boundary condition given by Equation 1. 1.4, we obtain ∞ An cos n − n− 1 2 n =1 and ∞ An cos n − 1 2 1 2 x = 1, x = 0, 0 ≤ x < c, c < x ≤ π (1. 1.9) (1. 1 .10 ) n =1 Both Equations 1. 1.9 and 1. 1 .10 have the form of a Fourier series except that there are two of them! Clearly the challenge raised by the boundary condition along y = 0 is the solution of this dual Fourier cosine series given by Equation 1. 1.9... 1. 1. 51 with c/a = 1 or k = 0.430 by Equation 1. 1.55 If D = 1, α = 2.325, β = −9.344, γ = 1, and δ = 6. 018 Figure 1. 1.3: Same as Figure 1. 1.2 except that we have the additional mapping s = sn(ζ, k), where K = 1. 779 and K = 1. 918 © 2008 by Taylor & Francis Group, LLC Overview 11 1. 1.50, becomes ∂2u ∂2u + 2 = 0, ∂ξ 2 ∂η −K < ξ < K, 0