Hasegawa(1969) 5번 식 해설

$$p=A{e}^{i(\omega t-kx)}=A{e}^{i\omega }{e}^{-ikx}=A{e}^{i\omega t}{e}^{-ikr\cos \theta }$$

$$(x=r\cos \theta )$$

 

이번 페이지에서 다룰 내용은 Fourier-Legendre series를 다룰 예정입니다.

목표는 아래 두 식을 유도해내는 것이에요.

$$f(x)=\sum _{n=0}^{\mathrm{\infty }}{a}_n{P}_n$$

$$a_m=\frac{2m+1}{2}{\int }_{-1}^{1}f(x)P_m(x)dx$$

 

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Step 1: Legendre Polynomial

To derive above equations, we need 'Legendre polynomial'

$$P_n(x)=\sum _{m=0}^M(-1{)}^m\frac{(2n-2m)!}{2^{n}m!(n-m)!(n-2m)!}{x}^{n-2m}$$

 

Similarily, we can express $f(x)$ with Legendre polynomial

$$f(x)=\sum _{n=0}^{\mathrm{\infty }}{a}_n{P}_n(x)$$

 

And coefficient $a_n$ can be derived by using orthogonality

(We derived/studied 'orthogonality' before.)

$${\int }_{-1}^{1}r(x)P_m(x)P_n(x)=0(m\ne n)$$

 

$$f(x)=\sum _{n=0}^{\mathrm{\infty }}{a}_n{P}_n(x)$$

$$r(x)f(x)P_m(x)=\sum _{n=0}^{\mathrm{\infty }}{a}_n{P}_n(x)r(x)P_m(x)$$

$${\int }_{-1}^{1}r(x)f(x)P_m(x)dx={\int }_{-1}^1\sum _{n=0}^{\mathrm{\infty }}{a}_{n}r(x)P_n(x)P_m(x)dx\cdots (1)$$

 

According to orthogonality under \(m\neq n)

$${\int }_{-1}^1{P}_n(x)P_m(x)dx=0$$

 

So, equation (1) will be...

$${\int }_{-1}^{1}r(x)f(x)P_m(x)dx=a_n{\int }_{-1}^{1}r(x)P_m(x)P_n(x)dx(m=n)$$

$$a_n=\frac{{\int }_{-1}^{1}r(x)f(x)P_m(x)dx}{{\int }_{-1}^{1}r(x)P_m^2(x)dx}$$

 

Step 2: Rodrigues formula

We derived/studied 'Rodrigues formula' before.

$$P_n(x)=\frac{1}{2^{n}n!}\frac{d^n}{d{x}^n}(x^2-1{)}^n$$

 

Let's put this equation in our integral form.

$${\int }_{-1}^1{P}_n^2(x)dx=\frac{1}{2^{2n}(n!{)}^2}{\int }_{-1}^1\frac{d^n}{d{x}^n}(x^2-1{)}^n\frac{d^n}{d{x}^n}(x^2-1{)}^{n}dx$$

 

As it is too complex to solve at once, use integration by parts.

 

 

 

Reference

Advanced Engineering Mathematics p.178 SEC 5.2 Legendre's equation. Legendre polynomials