Lecture 24: Laplace s Equation
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1 Introductory lecture notes on Prtil Differentil Equtions - c Anthony Peirce. Not to e copied, used, or revised without explicit written permission from the copyright owner. 1 Lecture 24: Lplce s Eqution (Compiled 4 August 217 In this lecture we strt our study of Lplce s eqution, which represents the stedy stte of field tht depends on two or more independent vriles, which re typiclly sptil. We demonstrte the decomposition of the inhomogeneous Dirichlet Boundry vlue prolem for the Lplcin on rectngulr domin into sequence of four oundry vlue prolems ech hving only one oundry segment tht hs inhomogeneous oundry conditions nd the reminder of the oundry is suject to homogeneous oundry conditions. These ltter prolems cn then e solved y seprtion of vriles. Key Concepts: Lplce s eqution; Stedy Stte oundry vlue prolems in two or more dimensions; Linerity; Decomposition of complex oundry vlue prolem into suprolems Reference Section: Boyce nd Di Prim Section Lplce s Eqution 24.1 Summry of the equtions we hve studied thus fr In this course we hve studied the solution of the second order liner PDE. u t = α 2 u Het eqution: Prolic T = α 2 X 2 Dispersion Reltion σ = α 2 k 2 2 u t 2 = c 2 u Wve eqution: Hyperolic T 2 c 2 X 2 = A Dispersion Reltion σ = ±ick (24.1 u = Lplce s eqution: Elliptic X 2 + Y 2 = A Dispersion Reltion σ = ±k Importnt: (1 These equtions re second order ecuse they hve t most 2nd prtil derivtives. (2 These equtions re ll liner so tht liner comintion of solutions is gin solution Stedy stte solutions in higher dimensions Lplce s Eqution rises s stedy stte prolem for the Het or Wve Equtions tht do not vry with time so tht u t = = 2 u t 2. 2D: u = 2 u x u =. (24.2 y2
2 2 3D: No initil conditions required. Only oundry conditions. The Lplcin in Polr Coordintes: u = 2 u r u r r u r 2 θ 2 =. Physicl prolems in which Lplce s eqution rises 2D Stedy-Stte Het Conduction, Sttic Deflection of Memrne, Electrosttic Potentil. u = 2 u x u y u =. (24.3 z Lplce s Eqution in two dimensions u t = α 2 (u xx + u yy u(x, y, t inside domin D. (24.4 Stedy-Stte Solution stisfies: u = u xx + u yy = (x, y D (24.5 BC: u prescried on D. (24.6 We consider domins D tht re rectngulr, circulr, pizz slices Rectngulr Domins Consider solving the Lplce s eqution on rectngulr domin (see figure 4 suject to inhomogeneous Dirichlet Boundry Conditions u = u xx + u yy = (24.7 BC: u(x, = f 1 (x, u(, y = g 2 (y, u(x, = f 2 (x, u(, y = g 1 (y (24.8 Figure 1. Inhomogeneous Dirichlet Boundry conditions on rectngulr domin s prescried in (24.8
3 Lplce s Eqution 3 Ide for solution - divide nd conquer We wnt to use seprtion of vriles so we need homogeneous oundry conditions. Since the eqution is liner we cn rek the prolem into simpler prolems which do hve sufficient homogeneous BC nd use superposition to otin the solution to (24.8. Pictorilly: Figure 2. Decomposition of the inhomogeneous Dirichlet Boundry vlue prolem for the Lplcin on rectngulr domin s prescried in (24.8 into sequence of four oundry vlue prolems ech hving only one oundry segment tht hs inhomogeneous oundry conditions nd the reminder of the oundry is suject to homogeneous oundry conditions 24.4 Solution to Prolem (1A y Seprtion of Vriles Figure 3. Boundry vlue prolem for su-solution u A (x, y (1A u xx + u yy = (24.9 u(, y = = u(, y = u(x, ; u(x, = f 1 (x. (24.1 Let u(x, y = X(xY (y. (24.11 X (xy (y + X(xY (y = (24.12 X (x X(x = Y (y Y (y = const = ±λ2 (24.13
4 4 λ 2 : X + λ 2 X = Y λ 2 Y = X = A cos λx + B sin λx Y = C cosh λx + D sinh λx X( = = X( Y ( =... Y ( = Becuse sin nd cos hve n # of rel roots the choice λ 2 is good for BC s for Prolems (A nd (C. +λ 2 : X λ 2 X = Y + λ 2 Y = X = A cosh(λx + B sinh(λx Y = C cos(λy + D sin(λy X( =... X( =... Y ( = = Y (. (24.14 Agin ecuse sin nd cos hve n # of rel roots the choice +λ 2 is good for BC s for Prolems (B nd (D. Bck to Solving (1A: X( = A = (24.15 X( = B sin(λ = λ n = nπ n = 1, 2,... X n (x = sin ( nπx. (24.16 u(x, = X(xY ( = Y ( = (24.17 Y ( = C cosh(λ + D sinh(λ = c = D tn h(λ (24.18 Y (y = D tn h(λ cosh(λy + D sinh(λy (24.19 { } sinh(λy cosh(λ cosh(λy sinh(λ = D (24.2 cosh(λ D = cosh(λ sinh λ(y = D sinh λ(y. (24.21 Note: We could sve ourselves the time y uilding the BC y( = directly into the solution y letting Y n (y = D sinh λ n (y (24.22 directly. ( nπ Now the functions: u n (x, y = sin sinh (y n = 1, 2,... stisfy ll the homogeneous BC of Prolem (1A. In order to mtch the BC u(x, = f 1 (x we need to superimpose ll these solutions. ( nπ u(x, y = B n sin sinh (y (24.23 { ( } nπ f 1 (x = u(x, = B n sinh sin (24.24 }{{} n where ( nπ B n sinh = n = 2 f 1 (x sin dx. (24.25
5 Therefore Specific Exmple Let f L (x = 1 = Therefore u(x, y = 1 u(x, y = Lplce s Eqution 5 2 where B n = sinh ( nπ n sin. n = 2 nπ 2 nπ ( nπ B n sinh (y sin ; f 1 (x sin dx (24.26 [ 1 + ( 1 n+1 ] ( nπ = B n sinh. (24.27 [1 + ( 1 n+1 ] sinh ( sin nπ ( nπ sinh (y. ( Solution to Prolem (1B y Seprtion of Vriles Figure 4. Boundry vlue prolem for su-solution u A (x, y Let u = u xx + u yy = (24.29 = u(x, = u(x, = u(, y; u(, y = g 2 (y (24.3 u(x, y = X(xY (y (24.31 X (x X(x = Y (y Y (y = ±λ2. (24.32 Since we hve homogeneous BC t y = nd y = we wnt the function Y (y to ehve like sines nd cosines. X λ 2 X = Y + λ 2 Y = X = c 1 cosh λx + c 2 sinh λx Y = A cos(λx + B sin(λx (24.33
6 6 u(x, = X(xY ( = Y ( = Y ( = A = (24.34 u(x, = X(xY ( = Y ( = Y = B sin(λ =, λ n = nπ n = 1, 2,... (24.35 ( nπy Y n = sin u(, y = X(Y (y = ( X( = c 1 =. nπx Therefore X n (x = c 2 sinh. ( nπy Therefore u n (x, y = sin sinh stisfy the homogeneous BC. ( nπy Therefore u(x, y = c n sinh sin. Now to stisfy the inhomogeneous BC ( nπ ( nπy g 2 (y = u(, y = c n sinh sin }{{} n where Summrizing: u(x, y = c n sinh c n sinh sin (24.36 ( nπ = 2 ( nπy g 2 (y sin dy. (24.37 ( nπy ; c n = 2 sinh ( nπ g 2 (y sin ( nπy dy. (24.38
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