第十四章习题课-武汉大学数学物理方法

WuhanUniversity武汉大学物理科学与技术学院MethodsinMathematicalPhysics第十四章勒让德多项式LegendrepolynomialWuhanUniversity第十四章勒让德多项式习题课™本章主要内容：™例题分析：一、有关特殊函数性质二、在球坐标中的解0=ΔuWuhanUniversity()()()⎯⎯⎯⎯⎯→⎯=ΔϕΦθΘ=rRuu令0.1()())()(012102xpxyyllyxyxml=→=++′−′′−=↓()()()()∑∑∑∑∞=∞==+−=∞==+−=+=++=00,0)1(000)1(coscos)(cos)(sincoslllllarmlllllmllmmlllllmlmlprcprdrcprdrcmBmAuθθθϕϕ™本章主要内容()ϕ+ϕ=ϕΦ→=Φ+Φ′′mBmAmmmmsincos02)1(2)(0)1(2+−+=→=+−′+′′llllrdrcrRRllRrRr()()())(01121)(222xpxyyxmllyxyxml=→=⎥⎦⎤⎢⎣⎡−−++′−′′−{第十四章习题课WuhanUniversity())1(1,21102=+−∑∞=ttxpttxlll()()()())3(12.211xpxpxpllll−+′−′=+()()()()())2(0121.111=++−+−+xplxpxlxpllll()())6(,...,2,1,0,,12211=+=∫−lkldxxpxpklklδ{()()()()∫∑−∞=+==110212,dxxpxflCxpCxflllll母函数关系式递推公式()()()()()正交性广义傅氏展开∑⎥⎦⎤⎢⎣⎡=−−−−−=202!2!!2!221lnnllnlxnlnlnnlxp(),10=xp(),1xxp=(),1321)(22−=xxp1)1(≡lpllp)1()1(−≡−™本章主要内容:)(.2xpl第十四章习题课WuhanUniversity()()()xpxxpmlmml)(221−=()()()()klmkmllmlmldxxpxpδ122!!11+−+=∫−()(),0∑∞==lmlmlxpCxf()()()()∫−++−=11212!!dxxpxflmlmlCmlml™本章主要内容:)(.3xpml第十四章习题课WuhanUniversity第十四章习题课™例题分析)()()()1(112xlpxxplxpxlll−−=′−、证明：()()[]()())1(112xpllxpxdxdll+=′−一、有关特殊函数性质∫xdx1:)1(()()[]()()dxxpllxpxxll∫+=′−1211()()()()dxxpxplllllx][121111−+′−′++=∫()()()()())]1(1[1211111−−+++−−++=llllpxppxplll()()()()()]11[1211xplxplllll−++−++=()()()()()()]112[1211xplxlpxxpllllll−−+−−++=证明：WuhanUniversity™例题分析)()()()1(112xlpxxplxpxlll−−=′−、证明：()()[]()()0112=++′−xpllxpxdxdll()()()()xpxpxpllll1112−+′−′=+1)1(≡lp一、有关特殊函数性质()()()()()012111=++−+−+xplxpxlxpllll知识点：第十四章习题课WuhanUniversity第十四章习题课™例题分析?)(211=∫−dxxpl、1)1(≡lp00≠=l，()()()()xpxpxpllll1112−+′−′=+()(),...2,1,0,,12211=+=∫−lkldxxpxpklklδllp)1()1(−≡−一、有关特殊函数性质1)(0=xp()()∫∫−−+−′−′+=111111][121)(dxxPxPldxxplll()()()()]1111[1211111−+−−−+=−−++llllppppl1)(0=xpdxxpxpdxxpll)()()(11011∫∫−−=法二：⎩⎨⎧=≠=0,200ll，知识点：WuhanUniversity()()()()()∑⎥⎦⎤⎢⎣⎡=−−−−−=202!2!!2!221lnnllnlxnlnlnnlxPWuhanUniversity思考：?)(10=∫dxxpl⎩⎨⎧=≠=∫0,10,0)(102mmdxxpm=∫+dxxpm)(1012222)!(2)!2()1()0(mmpmmm−=222122])!1[(2)]!1(2[)1()0(++−=+++mmpmmm2211012])![(2)!2()1(121)(mmmdxxpmmm−−−−=∫一、有关特殊函数性质:2)1ml=:12)2+=ml2221012])!1[(2)!22()1(121)(++−+=++∫mmmdxxpmmm)]0()0([341222+−+=mmppmdxxpxpmmm)]()([1)12(2121022′−′++∫+WuhanUniversity第十四章习题课展开。按、将)()(3xpxxfl=()∑∞==0lllxpCx()∫−+=11212dxxpxlCll012=+ncdxxpdxxpnn)()(10121012∫∫+−=－一、有关特殊函数性质()()∑∞==0lllxPCxf()()∫−+=11212dxxPxflCll:12)1+=nl:2)2nl=()()()()xPxPxPllll1112−+′−′=+dxxxpncnn)(221221022∫⋅+⋅=∫+′−′+=−[phone])()()14(dxnxpxpxnnn222221])!1[(2)!22()1(121])![(2)!2()1(121++−+−−−=+−nnnnnnnnnn2211012])![(2)!2()1(121)(mmmdxxpmmm−−−−=∫2221012])!1[(2)!22()1(121)(++−+=++∫mmmdxxpmmmWuhanUniversity展开。按、将)(21)(32xptxtxfl+−=一、有关特殊函数性质™例题分析：222121)(txttxtxf+−+−=())1(1,21102=+−∑∞=ttxPttxlll()∑∞=+−=02]21[llltxptxt()()∑∑∞=+∞=+−+=01022][lllllllxxptxptt()()()()()012111=++−+−+xPlxPxlxPllll()()()∑∑∞=−++∞=+++++−+=011102]12121[2][llllllllxpllxplltxptt()()∑∑∞=∞==+++−=++10111]12[121[kkklklllxpkktxpllt()∑∞=−=0]12[kkkxpkkt()()∑∑∞−=+∞=−=−+++=+120111]321[]12[kkkllkllxpkktxpllt()∑∞=+++=02]321[kkkxpkkt()∑∞=+−−+=+−=022]1232[21)(llllxpltlttxtxf答：WuhanUniversity二、在球坐标中的解0=Δu解：⎩⎨⎧===Δ==02,02,0uuuarauarar⎪⎪⎩⎪⎪⎨⎧=+=+∑∑∞=+−∞=+−00)1(0)1()(cos])2()2([0)(cos][llllllllllllupadacpadacθθ⋅aa2•1、设有一内半径为外半径为的均匀球壳，其内外表面的温度分布分别保持为零和，试求球壳的稳定温度分布。0uaa2法一：()∑∞=+−+=0)1(cos)(llllllprdrcuθ第十四章习题课WuhanUniversity第十四章习题课二、在球坐标中的解0=Δu解：⎪⎩⎪⎨⎧=+=+000002101uadcadc⋅aa2•法一：()∑∞=+−+=0)1(cos)(llllllprdrcuθ00)2()2(0)1()1(≠⎩⎨⎧=+=++−+−ladacadacllllllll0,0≠==ldcll;2,20000auduc−==())1(2cos)1(2000rauprauu−=−=θWuhanUniversity第十四章习题课二、在球坐标中的解0=Δu解：法二：;2,20201aucuc−==→),(),,(ruru=ϕθ)(1),,(22drdurdrdrru=Δ→ϕθ⎪⎩⎪⎨⎧=====0222,00)(1uuudrdurdrdrararrccru1)(21+=→()∑∞=+−+=0)1(cos)(llllllprdrcuθ())1(2cos)1(2000rauprauu−=−=θWuhanUniversity二、在球坐标中的解0=Δu2、设有一半径为的介质球，介电常数为，在与球心的距离为的地方放有一点电荷，求介质球内、外的电位。aε)(abbq04πε解：设球内外电位分别为和且ivevθcos2221rbbrqvve−++=）（有限1,00⎩⎨⎧→=Δ=riivarv）（有限2,011⎩⎨⎧→=Δ∞→rvarv）（3⎪⎩⎪⎨⎧∂∂=∂∂=====areariarearirvrvvvε（设真空中介电常数为1。）⋅rdbaθq04πεM第十四章习题课WuhanUniversity第十四章习题课WuhanUniversity则令),()(),(θθΘ=rRru)4()(cos)1(0θlllliprcv∑∞==→),(cos)2(0)1(1θllllprdv∑∞=+−=→)5(cos2)(cos220)1(θθrbbrqprdvlllle−++=∑∞=+−于是∑∞==−+=−+∴0222))((cos1cos2)(111cos21lllbrpbbrbrbrbbrbrθθθ时：当二、在球坐标中的解0=Δu())1(1,21102=+−∑∞=ttxPttxlllWuhanUniversity第十四章习题课)(cos)()(cos00)1(θθlllllllepbrbqprdv∑∑∞=∞=+−+=∴:br当)7()(cos)()(cos)(cos00)1(0θθθlllllllllllpbabqpadpac∑∑∑∞=∞=+−∞=+=)8()(cos)()(cos)1()(cos010)2(01θθθεlllllllllllpbablbqpaldplac∑∑∑∞=−∞=+−∞=−++−=⎪⎩⎪⎨⎧++−=+=→−+−−+−12)2(1)1()()1([)(llllllllllbabqlaldlacbabqadacε,]1)1[()12(1++++=llbllqcε112]1)1[()1(++++−−=lllbllaqdεε二、在球坐标中的解0=Δu)4()(cos0θ=∑∞=lllliprcvWuhanUniversity)(cos)(1)1()12(0θ++ε+=∑∞=lllipbrllbqv)(cos)(1)1()1(cos212022θ++ε−ε−θ−+=+∞=∑lllepbrallaqrbbrqv二、在球坐标中的解0=Δu）（有限1,00⎩⎨⎧→=Δ=riivarv）（有限2,011⎩⎨⎧→=Δ∞→rvarv）（3⎪⎩⎪⎨⎧∂∂=∂∂=====areariarearirvrvvvε第十四章习题课WuhanUniversity2、有一均匀球体，球心在原点，在球面上的温度为ϕθθcossin)cos31(+==aru试在稳定状态下求球内的温度分布。解：⎩⎨⎧+==Δ=ϕθθcossin)cos31(,0aruaru法一：∑∑∞==+=00)(cos)sincos(),,(llmmlmlmllpmBmArruθϕϕϕθϕθϕθθϕϕcos2sin23cossin)(cos)sincos(00+=+∑∑∞==llmmllmllmlpmaBmaA)(coscos)(coscos1211θϕθϕpp+=θθθθ2sin23)(cos,sin)(cos