Lecture 25: More Rectangular Domains: Neumann Problems, mixed BC, and semiinfinite strip problems


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1 Introductory lecture notes on Prtil ifferentil Equtions  y Anthony Peirce UBC 1 Lecture 5: More Rectngulr omins: Neumnn Prolems, mixed BC, nd semiinfinite strip prolems Compiled 6 Novemer 13 In this lecture we Proceed with the solution of Lplce s equtions on rectngulr domins with Neumnn, mixed oundry conditions, nd on regions which comprise semiinfinite strip. Key Concepts: Lplce s eqution; Rectngulr domins; The Neumnn Prolem; Mixed BC nd semiinfinite strip prolems. Reference Section: Boyce nd i Prim Section More Rectngulr omins with mixed BC nd semiinfinite strip prolems 5.1 The Neumnn Prolem on rectngle  only flux oundry conditions Exmple 5.1 The Neumnn Prolem: Figure 1. Inhomogeneous Neumnn Boundry conditions on rectngulr domin s prescried in?? u xx + u yy =, < x < < y < 5.1 u x, y = u x, y = fy 5. u y x, = = u y x,. 5.3 Let ux, y = XxY y. X x Xx = Y y Y y = λ 5.4
2 Y y + λ Y y = Y = = Y } Y = A λy + B sin λy Y = Aλ sin λy + Bλ λy 5.5 Y = λb = λ = or B =. 5.6 Y = Aλ sin λ = λ n = nπ/ n =, 1,... nπy Y n =, Y = X n λ X n = 5.8 X n = 5.9 n = : X =, X = c x + X = c X = c =. Choose = 1: X = 1 n 1 X n = c n hλ n x + n λ n x X n = c n λ λ n x + n λ hλ n x X n = λ n n = 5.1 Choose c n = 1: X n = hλ n x. Thus u n x, y = X n Y n = hλ n x λ n y u x, y = X Y = 1 } stisfy homog. BC ux, y = A + A n h. 5.1 Now fy = u x, y. u x x, y = u x, y = nπ A n { A n nπ nπ } This is like Fourier Cosine Series for fy ut without the constnt term = fy Recll fy = + n, n = fy dy Thus the expnsion 5.14 is consistent only if =. For this to e true we require tht if fy dy then there is no solution to the oundry vlue prolem 1. fy dy = 5.16
3 Lplce s Eqution 3 Note 1 If fy dy there is net flux into the domin through the right hnd oundry nd, since the other oundries re insulted, there cn e no stedy solution the temperture will continully chnge with time. If fy dy = there is no net flux through the oundry nd stedy stte cn exist. i.e. It is possile tht u xx + u yy = u t =. If fy dy = then A n nπ nπ = fy dy nd A n = nπ nπ u x, y = A + fy A n h dy n L where A is undetermined. ux, y is sid to e known up to n ritrry constnt If u x, y is the stedy stte of Het Eqution u t = u xx + u yy with ux, y, = u x, y then u u t dx dy = u dx dy = ds =. 5. n t u dx dy = u dx dy = const for ll time = u x, y dx dy. 5.1 Now u x, ydxdy = A re = Which the condition tht determines A. u x, y dx Rectngulr domins with mixed BC Exmple 5. Insulting BC long two sides nd specified tempertures on the others: u = u xx + u yy = 5.3 = u x, y = u x, y = ux, 5.4 ux, = fx. 5.5
4 4 Figure. Mixed Boundry conditions on rectngulr domin s prescried in 5.4 Let ux, y = XxY y. X X = Y Y = ±λ. 5.6 Since we hve homogeneous BC on X = = X choose λ. 1 X + λ X = X = = X. Xx = A λx + B sin λx X = Bλ = B = X x = Aλ sinλx + Bλ λx X = Aλ sinλ = 5.7 re eigenfunctions nd eigenvlues. λ n = nπ/ n =, 1,,... X n x = λ n : Y λ Y = nd Y = Y n y = A u n x, y = n. Thus stisfy homogeneous BC. λ = : In this cse the OE for Y is: Y = Y y = c 1 y + c 5.3 Y = c = Y y = y 5.31 nd u x, y = y 1 stisfies the homogeneous BC.
5 Lplce s Eqution 5 ux, y = c y + ux, = c c = c = 1 ux, y = c y + c n nπ + c n fx dx; c n nπ = fx dx; c n = nπ c n 5.3 = fx 5.33 fx dx 5.34 fx dx Semiinfinite strip prolems Exmple 5.3 A Semiinfinite strip with specified tempertures: Figure 3. iriclet Boundry conditions on semiinfinite strip s prescried in 5.39 Let ux, t = XxT t nd plug into 1?: u xx + u yy = < x <, < y < 5.37 u, y = = u, y 5.38 ux, = fx ux, y s y 5.39 X x Xx = Y y Y y = λ since we hve homogeneous BC on X. 5.4
6 6 1 X + λ X = X = = X } λ n = nπ/ n = 1,,... X n = sin 5.41 Y λ Y = Y y = Ae λy + Be λy. Since ux, y s y we require B =. u n x, y = e λny sin 5.4 stisfy the homogeneous BC nd the BC t. Thus ux, y = nπ c n e y sin fx = ux, = c n sin c n = fx sin dx Exmple 5.4 Semiinfinite strip with inhomogeneous BC: Figure 4. iriclet Boundry conditions on semiinfinite strip s prescried in 5.47 u xx + u yy = < x <, < y < 5.45 u, y = A, B = u, y 5.46 ux, = fx ux, y s y 5.47 Look for function vx for which v = nd which stisfies the inhomogeneous BC. v = αx + β v = A = β v = α + A = B B A vx = x + A.
7 Lplce s Eqution 7 Now let ux, y = vx + wx, y. Thus w stisfies the sme BVP s does u in Eg. 3 ove. where = u xx + u yy = v xx +w xx + v yy +w yy w = 5.48 A = u, y = v + w, y w, y = 5.49 B = u, y = v + w, y w, y = 5.5 fx = ux, = vx + wx, wx, = fx vx ux, y = B Ax/ + A + d n = {fx vx} sin nπ d n e y sin 5.5 dx. 5.53
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