Special Functions
Legendre Polynomials¶
\[
\begin{gather}
(1-x^2)y''-2xy'+\lambda y=0
\\\\
(1-x^2)y'' - 2xy' + n(n+1)y=0
\\\\
\big[(1-x^2)y'\big]+\lambda y = 0
\end{gather}
\]
solve the equation with \(y = \sum a_m x^m\), we can get the conclusion that
\[
\begin{gather}
\lambda_{n} = n(n+1), \qquad \text{for } n = 0, 1,2 , \dots
\\\\
a_{m+2} = \frac{m(m+1)-\lambda_{n}}{(m+2)(m+1)}a_m
\end{gather}
\]
thus \(a_{n+2} = a_{n+4} = \dots = 0\)
Normalization¶
-
Let the polynomial \(P_0(x)=1\), \(P_1(x) = x\)
-
Recurrence relation : (背)
\[\begin{align}
(n+1)P_{n+1} = (2n+1)x P_n - nP_{n-1}
\end{align}
\]
Fourier-Legendre Expansion¶
- \(f(x)\) is defined for \(-1 < x< 1\)
\[
\begin{gather}
f(x) = \sum_{n= 0}^{\infty}{c_{n}P_{n}(x)}
\\\\
c_{n} = \frac{(f, P_{n})}{(P_{n}, P_{n})} = \frac{\int_{-1}^{1}{f(x)}P_{n}(x)dx}{\int_{-1}^{1}{P^{2}_{n}(x)dx}}
\\\\
\int_{-1}^{1}{P^{2}_{n}(x)dx} = \frac{2}{2n+1}
\end{gather}
\]
Bessel Functions¶
- definition
\[
\begin{align}
x^2y'' + xy' + (x^2 - \nu^2)y = 0
\end{align}
\]
- solution ( for \(J_\nu\) and \(J_{-\nu}\) are independent)
- \(2\nu \notin ℕ\)
- \(2\nu\) is odd positive integer
\[
\begin{gather}
y(x) = c_1 J_\nu(x) + c_2J_{-\nu}(x)
\end{gather}
\]
- solution ( for \(J_\nu\) and \(J_{-\nu}\) are not independent)
- \(2\nu\) is even positive integer
\[
\begin{gather}
y(x) = c_1 J_\nu(x) + c_2Y_\nu(x)
\end{gather}
\]
- \(J_\nu\) is called a Bessel function of the 1st kind of order \(\nu\)
\[
\begin{gather}
J_\nu = \sum_{n=0}^{\infty}{\frac{(-1)^n}{2^{n+\nu}n!\ \Gamma(n+\nu+1)}x^{2n+\nu}}
\end{gather}
\]
- \(Y_\nu \rightarrow Y_{0}(0) = -\infty\)
Recurrence Relations¶
\[
\begin{gather}
\frac{d}{dx}(x^\nu J_\nu) = x^\nu J_{\nu-1}
\\\\
\frac{d}{dx}(x^{-\nu}J_\nu)=-x^{-\nu}J_{\nu+1}
\end{gather}
\]