Partial Differential Equations
Partial Differential Equations MATH 467
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This 8 page Class Notes was uploaded by Vernie McCullough on Thursday October 15, 2015. The Class Notes belongs to MATH 467 at Millersville University of Pennsylvania taught by Staff in Fall. Since its upload, it has received 40 views. For similar materials see /class/223531/math-467-millersville-university-of-pennsylvania in Mathematics (M) at Millersville University of Pennsylvania.
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Date Created: 10/15/15
Laplace s Equation on a Sphere de Laplace Pierre Simon Domenic Scorzetti Millersville University dascorzemarauderrnillersvilleedu April 25 2006 Problem Outline A spherical shell ha an inner radius p l and an outer radius p 2 The temperatures of the 39nner ou er s are 39 n b 6 1007 200056 f06 50 300056 a State the boundary value problem for the steadystate temperature Within the shell b Find the steadystate temperature inside the shell 0 Plot the steadystate temperature inside the shell for several slices parallel to the shell s equator Problem CrossSection 1W lilmerest a Problem Statement Finding the steady state temperature will require the use of the Laplacian operator Au 2 0 As described above the solution will be for the region bounded by the two surfaces Also note the functional dependence only on 6 in this case used as the angle from the sphere s north pole and the absence of azimuthal angle b 2 Problem Solution The spherical Laplacian with only a 6 component looks like this 1 a 2 an 1 a an 1 Au p sm6 p2 0p 0p p2 sing 06 06 Since Laplace s equation is linear and homogeneous I will begin solving by separation of variables This assumes up t9 RpTt9 which is then substituted into 1 1 a sz jT isin6T R 0 p2 6 p2 sin 6 66 The primed variables indicate necessary regular derivatives I continue the process by Z 39 Z multiplying through by L26 2 Sinz 9 3amp2 v Sing iltsin6T39 0 R 0p T 66 QDependent Portion of u To nd this part I begin by dividing equation 2 by sinze and rearranging the terms 16 2 1 3 F50 RTsin666sm6T Since the lefthand side only depends on p and the righthand side depends solely on 6 then both sides must equal a constant For convenience say this constant has the form mm l with m e N U 0 meaning m is a natural number and can also be zero Under these assumptions I am left with an ordinary differential equation for the LHS 1 6 p2le mm 1 Applying the product rule szwsz39 mm 1 4 sz 2pR mm 1R 0 Equation 4 is Euler s Equation see 1 with a solution which looks like 5 for m e N U 0 s Rltpgt Cp quot Dp quot 1 9Dependent Portion of u Referring back to equation 3 I now isolate the 6 component and set it equal to the same constant o sin mm 1 Once again apply the product iulc Tir116T sin6T cos6 mm 1 TTH ggfnsg mml 0 Multiply through by 1 and T 7 T TCTO6 mm 1T 0 To further simplify the calculations I put 7 into another form through a change of variables and by letting w cos 9 see 1 This will require the chain rule dT dT dw dT s1nl9 d6 dw d6 dw d2T d dT d dT 2 s1nl9 ch92 d6 d6 d6 dw 8 d2T cos6d T s1 6d2T dw d 2 dw dw2 d6 d T dT 2 d2T cos6 sm 6 dw w2 Substitute 8 and 9 into 7 2 sin2 Hd T cos d T CO56 sin j T w 1 T0 dwz dw sing ltmltm gtgt 2 sin2 Hd T cos d T cos d T mm 1T 0 dwl dw dw 2 sin2 ed T 2cos6d T mm 1T 0 dw2 dw Recall the trigonometric property sin2 9 cos2 9 l to make the following substitution 2 1 cos2 6d T 2cos6d T mm 1T 0 Also recall thatw cos 9 2 10 1 w2 2wZ Vmm 1T 0 W Note that equation 10 is a Legendre polynomial which has solutions of the form 11 Tmw with m e NU0 see 2 Sec 7103 m 0 T0w 1 T0cos6 1 m I T1w w T1cos9 cose m 2 T2w 053w21 T2cos9 053cos291 General Solution This is found by combining equations 5 and l 1 inside an in nite series 6 00 C m D m 1 T 6 12 um mZO mp mp mcos However notice my original problem consists of only a nite sum of constants and cosines so I can nd an exact solution with p l and p 2 per equation 12 Additionally I only need to sum through m 1 since there are no squared cosine terms to account for p1 OO ul6 6 100 20cos6 mzzocm DmTm cos 100 20cos6 C0 DOTOcos6C1D1T1cos6 By substitution from the solution table above 13 100 2000s6 C0 DOC1D1cos6 o2 00 m m l u26f06503000s6 2 C 2 D 2 T cos mzo m m m 50 30cos6 C020 D021T0cos C121 D 2 1 1T1cos6 l l l 50 3000s6 C0 EDOTOcos62C1ZD1T1cos6 By substitution from the solution table above 1 l 14 503000s6 C0 EDO2C1ZD1cos6 Using equations 13 and 14 I will match up coefficients and find C0 D0 C1 and D1 and use these in my exact solution by plugging them into equation 12 100CODO 20C1D1 1 1 502CO EDO 30 2C1ZD1 For clarity the details of the coefficient calculations are in Appendix A These calculations yield C020 C1220 D0 100 D140 Here I will construct the final form of the steadystate temperature using equation 12 and my tables of Legendre solutions and coefficients up6 CO30 D0p1T0cos6Clp1D 1p11T1cos6 up6 0100p120p1 40p2cos6 15 up6 20p 4 gicos6 p k p Equation 15 represents the steadystate temperature for the spherical shell c Plot the Solution For my nlminn nlm r a the L 4 www of interest from p 1 and p 2 In L L 39 439 below L L 1 1 1 L I C L y You Lrl I I I I I I I L straight instead of curved Lastly J A L Lu uum m outer References 1 Dr Robert Buchanan Laplace s Equation on a Sphere lthttpbanachmillersvilleeduNbobmath467sa1nplepr0jectpdfgt 2 Richard Haberman Applied Partial Differential Equations with Fourier Series anal Boundary Value Problems 43911 Edition Pearson Prentice Hall Upper Saddle River NJ 2004 Appendix A C0 100 D09 C0 0 50100 D0D0 1 50100 D0 2 loo 50 2 D0 100 C1 20 D1 a C1 20 30 2 20 D1D1 30 40 211 io1 30 40 1D1 4 70 1D1 4 D 40 1
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