【转载请注明出处】http://www.cnblogs.com/mashiqi
2017/06/16
适合于自己的关于Jacobi-Anger expansion的推导方法，这里记下来，方便以后查阅。
现记住下面四个关系式：
egin{align*}
& (1)~ |x-y|=|x| -hat{x} cdot y + mathcal{O}left(frac{1}{|x|}
ight), ~|x| o +infty. \
& (2)~ sum_{m=-n}^{n} Y_n^m(hat{x})overline{Y_n^m(hat{y})} = frac{2n+1}{4pi} P_n(cos heta). \
& (3)~ Phi (x,y) riangleq frac{e^{ik|x-y|}}{4pi|x-y|} = ik sum_{n=-infty}^{infty}sum_{m=-n}^{n} h_n^{(1)}(k|x|)Y_n^m(hat{x}) j_n(k|y|)overline{Y_n^m(hat{y})}, forall~ |x| > |y|. \
&(4)~ h_n^{(1)}(t) = frac{1}{i^{n+1}t} e^{it} left{1 + mathcal{O}left(frac{1}{t}
ight)
ight}, ~t o +infty.
end{align*}
于是当$|x|$充分大时，我们可以得到
egin{align*}
frac{e^{ik|x-y|}}{4pi|x-y|} & = frac{e^{ik|x|}}{4pi|x|} left{ e^{-ikhat{x} cdot y} + mathcal{O}left(frac{1}{|x|}
ight)
ight} \
& = ik sum_{n=-infty}^{infty}sum_{m=-n}^{n} h_n^{(1)}(k|x|)Y_n^m(hat{x}) j_n(k|y|)overline{Y_n^m(hat{y})} \
& = ik sum_{n=-infty}^{infty} left{ j_n(k|y|)h_n^{(1)}(k|x|) left[ sum_{m=-n}^{n} Y_n^m(hat{x}) overline{Y_n^m(hat{y})}
ight]
ight} \
& = ik sum_{n=-infty}^{infty} left{ j_n(k|y|)h_n^{(1)}(k|x|) frac{2n+1}{4pi} P_n(cos heta)
ight} \
& = ik sum_{n=-infty}^{infty} frac{2n+1}{4pi} j_n(k|y|) P_n(cos heta) h_n^{(1)}(k|x|) \
& = ik sum_{n=-infty}^{infty} frac{2n+1}{4pi} j_n(k|y|) P_n(cos heta) frac{e^{ik|x|}}{i^{n+1}k|x|} left{1 + mathcal{O}left(frac{1}{|x|}
ight)
ight} \
& = frac{e^{ik|x|}}{4pi |x|} sum_{n=-infty}^{infty} frac{2n+1}{i^n} j_n(k|y|) P_n(cos heta) left{1 + mathcal{O}left(frac{1}{|x|}
ight)
ight} \
& = frac{e^{ik|x|}}{4pi |x|} left{ sum_{n=-infty}^{infty} frac{2n+1}{i^n} j_n(k|y|) P_n(cos heta) + mathcal{O}left(frac{1}{|x|}
ight)
ight}.
end{align*}
于是$$e^{-ikhat{x} cdot y} = sum_{n=-infty}^{infty} frac{2n+1}{i^n} j_n(k|y|) P_n(cos heta).$$将$hat{x}$换做$-d$，$y$换做$x$，可得：
egin{align*}
e^{ikd cdot x} & = sum_{n=-infty}^{infty} frac{2n+1}{i^n} j_n(k|x|) P_n(cos(pi- heta)) \
& = sum_{n=-infty}^{infty} frac{2n+1}{i^n} j_n(k|x|) (-1)^n P_n(cos heta) \
& = sum_{n=-infty}^{infty} i^n(2n+1) j_n(k|x|) P_n(cos heta).
end{align*}