※ 引述《jmKevin (谢谢)》之铭言： : [91 中山光电] : 题目 : Find the function f(x,y) satisfying the Laplace equation : d^2 f d^2 f : ------- + ------- = 0 , for x^2 + y^2 = a , a > 0 : d^2 x d^2 y : and the boundary condition f(x,y) = x^3 for x^2 + y^2 = a 换柱座标,let x=rcosθ,y=rsinθ f(x,y)=f(r,θ) , 又f(r,θ)=f(r,θ+2π) 令 f(r,θ) = a0 + ΣQn cosnθ + Wnsinnθ ------------------------------------------------------------ 这中间带回Laplace equation,可推出Qn,Wn 是r^n + r^(-n) 的函数 Qn = An*r^n + Bn*r^(-n) , Wn = Cn*r^n + Dn*r^(-n) f(0,θ)有限 Bn = Dn = 0 B.C. f(r,θ) = r^3*cos^3θ = a0 + ΣQn cosnθ + Wnsinnθ note: cos3θ=4cos^3θ-3cosθ => cos^3θ = 1/4[3cosθ+cos3θ] f(a,θ)= a^3*1/4[3cosθ+cos3θ] = a0 + Σ[An*a^n ]cosnθ+[Cn*a^n]sinnθ 由函数正交性知 A1 , A3 =\=0 , otherAn=Cn=0 π a0 =1/2π ∫ a^3*1/4[3cosθ+cos3θ dθ , a0 = 0 -π π A1*a = 1/π∫ 3/4 a^3* cos^2θdθ , A1 = 3a^2 / 4 -π π A3*a^3 = 1/π∫ 1/4 a^3* cos^2 3θdθ ,A3 = 1/4 -π 3 1 f(r,θ) = ___ a^2 r cosθ + ___ r^3 cos3θ 4 4 : 3 1 : 答案 : f = --- a r cosθ + --- r^3 cosθ : 4 4 : 不知道该如何把边界条件换成柱座标，麻烦板上高手帮忙指导一下，谢谢 --

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