Re: [理工] [工數]-pde
※ 引述《sptadao (咪咪)》之銘言：
: ▽^2 U(x,y)=0 for x^2+y^2<4
: U(x,y)=x^2-y^2 for x^2+y^2=4
: 請問這題如何解
let x=rcosθ , y=rsinθ
▽^2 U(r,θ) = 0 , B.C U(r=2,θ) = 4cos^2θ - 4sin^2θ
let U = a0 + Σan*cos(nθ) + bn*sin(nθ)
1 d dU d^2U
___ ___(r___) + ______ = 0
r dr dr r^2dθ^2
d^2U dU d^2U
r^2____ + r ____ + ____ = 0
dr^2 dr dθ^2
r^2[a0'' + Σan''*cos(nθ) + bn''*sin(nθ)]
+ r[a0' + Σan'*cos(nθ) + bn'*sin(nθ)] + -n^2[Σan*cos(nθ) + bn*sin(nθ)]=0
r^2a0''+ra0' = 0
r^2an''+r*an'-n^2an = 0
r^2bn''+r*bn'-n^2bn = 0
a0 = c1+c2lnr
an = A*r^(n) + B*r^(-n)
bn = C*r^(n) + D*r^(-n)
U = c1+c2lnr + Σ[A*r^(n) + B*r^(-n)]*cos(nθ) + [C*r^(n) + D*r^(-n)]*sin(nθ)
B.C r=0有限 , B=D=c2 = 0
U = c1 + Σ An*r^(n)*cos(nθ) + Cn*r^(n)*sin(nθ)
U(2,θ) = c1 + Σ An*2^(n)*cos(nθ) + Cn*2^(n)*sin(nθ) = 4cos^2θ - 4sin^2θ
n=1
= 4*cos2θ
c1 = Cn = 0 , A_2=1 ,An(n≠2) = 0
U = r^2cos(2θ) = x^2-y^2
......是不是哪裡怪怪= =
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02/04 14:22, , 1F
02/04 14:22, 1F
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