The Legendre Equation

2019-12-16 Sin-iu 發表迴響

Definition 1

(the Legendre equation)
The Legendre equation of order $\alpha$ is the second-order linear differential equation
$\displaystyle (1-x^2)y'' - 2xy' + \alpha(\alpha+1)y = 0,$
where $\alpha > -1.$

The only singular points of the Legendre equation are at +1 and -1, so it has two linearly independent solutions that can be expressed as power series in powers of $x$ with radius of convergence at least 1.

The substitution $y = \sum c_m x^m$ in the Legendre equation leads to the recurrence relation

$\displaystyle c_{m+2} = -\frac{(\alpha - m)(\alpha + m +1)}{(m+1)(m+2)}c_m$

Proof
Let $\displaystyle y = \sum^\infty_{m=0} c_m x^m$ , so that
$\displaystyle y' = \sum^\infty_{m=1} mc_m x^{m-1}, \quad y'' = \sum^\infty_{m=2} m(m-1)c_m x^{m-2}$ .

$\displaystyle (1-x^2)\sum^\infty_{m=2} m(m-1)c_m x^{m-2} - 2x\sum^\infty_{m=1} mc_m x^{m-1} + \alpha(\alpha+1)\sum^\infty_{m=0} c_m x^m = 0 \\ \sum^\infty_{m=2} m(m-1)c_m x^{m-2}-\sum^\infty_{m=2} m(m-1)c_m x^m \\ \phantom{\sum^\infty_{m=2} m(m-1)} - \sum^\infty_{m=1} 2mc_m x^m + \sum^\infty_{m=0} \alpha(\alpha+1)c_m x^m = 0\\ \sum^\infty_{m=0} (m+2)(m+1)c_{m+2} x^{m}-\sum^\infty_{m=2} m(m-1)c_m x^m \\ \phantom{\sum^\infty_{m=2} m(m-1)} - \sum^\infty_{m=1} 2mc_m x^m + \sum^\infty_{m=0} \alpha(\alpha+1)c_m x^m = 0 \\ \sum^\infty_{m=2}\Big[(m+2)(m+1)c_{m+2}-m(m-1)c_m - 2mc_m + \alpha(\alpha+1)c_m \Big]x^m \\ \phantom{\sum^\infty_{m=2} }+2c_2x^0 + 6c_3x^1 - 2c_1x^1 + \alpha(\alpha+1)c_0 x^0 + \alpha(\alpha+1)c_1 x^1 = 0\\ \sum^\infty_{m=2}\Big\{(m+2)(m+1)c_{m+2}-[\alpha(\alpha+1)-m(m+1)]c_m \Big\}x^m \\ \phantom{\sum^\infty_{m=2} }+[2c_2+\alpha(\alpha+1)c_0]x^0 + [6c_3 - 2c_1+ \alpha(\alpha+1)c_1] x^1 = 0 \\ \sum^\infty_{m=2}\Big[(m+2)(m+1)c_{m+2}-(\alpha-m)(\alpha+m+1)c_m \Big]x^m \\ \phantom{\sum^\infty_{m=2} }+[2c_2+\alpha(\alpha+1)c_0]x^0 + [6c_3 - 2c_1+ \alpha(\alpha+1)c_1] x^1 = 0$

The identity principle requires that

the coefficient of $x^0$ satisfies $\displaystyle 2c_2+\alpha(\alpha+1)c_0 = 0 \Rightarrow c_2 = -\frac{\alpha(\alpha+1)}{2!}c_0$ ;
the coefficient of $x^1$ satisfies $\displaystyle 6c_3 - 2c_1+ \alpha(\alpha+1)c_1 = 0 \Rightarrow c_3 = -\frac{(\alpha-1)(\alpha+2)}{3!}c_1$ ;
the coefficients of $x^m$ for $m = 2,3,\ldots$ satisfy $\displaystyle (m+2)(m+1)c_{m+2}-(\alpha-m)(\alpha+m+1)c_m = 0$ , which yields the recurrence relation

$\displaystyle c_{m+2} = -\frac{(\alpha - m)(\alpha + m +1)}{(m+1)(m+2)}c_m$ .

Actually the recurrence relation is applicable for $m = 0,1,2,\ldots$ .

微分方程

Intro to DE / Lecture Note 12

2019-12-11 Sin-iu 發表迴響

Consider a second order linear homogeneous differential equation

$\displaystyle a_2(x)y''+a_1(x)y'+a_0(x)y = 0.$

As defined in the last lecture, if $\displaystyle \frac{a_1(x)}{a_2(x)}$ and $\displaystyle \frac{a_0(x)}{a_2(x)}$ are analytic at $x_0$ , then $x_0$ is an ordinary point; otherwise, $x_0$ is an singular point (singularity).

Theorem 1

If $x_0$ is an ordinary point of Equation (1), then there exist two linearly independent solutions of the form $\displaystyle \sum_{n=0}^\infty c_n(x-x_0)^n$ , where the series converges on $|x-x_0| < R$ for some $R>0$ .

Note
For convenience, it’s often to let $x_0 = 0$ . When $x_0 = 0$ is not an ordinary point, then make a shift!

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2019-12-05 Sin-iu 發表迴響

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