Form of Legendre Equation/Legendre Polynomials
The Legendre Equation is a second-order linear ODE that takes the form of
\begin{align}
(1-x^2)y^{\prime\prime} -2xy’ + l(l+1)y = 0 \tag{1}
\end{align}
where \( l \) is a given constant. \( x = -1, 1, \infty \) are the three regular singular points while other points are ordinary. We can resort to the series solution method (see this tutorial) to compute the two linearly independent solutions at the ordinary point \( x = 0 \). Substituting (1)-(3) in the aforementioned tutorial into (1), we obtain
\begin{align}
\begin{aligned} (1-x^2)\sum_{n=2}^{\infty} n(n-1) a_nx^{n-2} -2x \sum_{n=1}^{\infty} n a_nx^{n-1} \\
+ l(l+1)\sum_{n=0}^{\infty} a_n x^n \end{aligned} &= 0 \\
\begin{aligned} \sum_{n=2}^{\infty} n(n-1) a_nx^{n-2} -\sum_{n=2}^{\infty} n(n-1) a_nx^n \\ – 2\sum_{n=1}^{\infty} n a_nx^{n} + l(l+1)\sum_{n=0}^{\infty} a_n x^n \end{aligned} &= 0 \\
\begin{aligned} \sum_{n=0}^{\infty} (n+1)n a_{n+2}x^n -\sum_{n=2}^{\infty} n(n-1) a_nx^n \\ – 2\sum_{n=1}^{\infty} n a_nx^{n} + l(l+1)\sum_{n=0}^{\infty} a_n x^n \end{aligned} &= 0 \tag{2}
\end{align}
where we have shifted the index of the first summation by \(2\). The recurrence relation reads
\begin{align}
(n+1)na_{n+2} -n(n-1) a_n -2na_n + l(l+1)a_n = 0 \\
\Rightarrow a_{n+2} = \frac{n(n+1)-l(l+1)}{n(n+1)}a_n = -\frac{(l+(n+1))(l-n)}{n(n+1)}a_n \tag{3}
\end{align}
This is a two-term recurrence relationship with a stride of \( 2 \), so the two solutions cleanly separate into odd and even series:
\begin{align}
y &= \sum_{n=0}^{\infty} a_nx^n \\
&= (a_0 + a_2x^2 + a_4x^4 + \cdots) + (a_1x + a_3x^3 + a_5x^5 + \cdots) \\
&= a_0(1 -l(l+1)\frac{x^2}{2!} + (l-2)l(l+1)(l+3)\frac{x^4}{4!} -\cdots) \\
&\quad+ a_1(x -(l-1)(l+2)\frac{x^3}{3!} + (l-3)(l-1)(l+2)(l+4)\frac{x^5}{5!} -\cdots) \tag{4}
\end{align}
Notice that, for an integer \(l\), one of the series will break off according to (3):
\begin{align}
a_{l+2} = \frac{l(l+1)-l(l+1)}{n(n+1)} = 0 \tag{5}
\end{align}
which truncates to a polynomial of degree \(l\), known as the Legendre polynomial of order \(l\). This is particularly desired because these finite polynomials are bounded on the natural interval \( [-1,1] \). Adding on this, many Physics questions often require an integer value of \(l\) to work. The first few Legendre polynomials are then computed by (5):
\begin{align}
\begin{aligned}
P_0(x) &= 1 & & P_1(x) = x \\
P_2(x) &= \frac{1}{2}(3x^2 -1) & & P_3(x) = \frac{1}{2}(5x^3 -3x) \\
P_4(x) &= \frac{1}{8}(35x^4 -30x^2 + 3) & & P_5(x) = \frac{1}{8}(63x^5 -70x^3 +15x) \\
&\vdots\end{aligned} \tag{6}
\end{align}
The normalization factor in the front is made such that \( P_l(1) = 1\). Clearly, Legendre polynomials are either odd or even according to \(l\) (and hence \( P_l(-1) = (-1)^l\)). For the other infinite series that does not terminate, we denote it by \( Q_l(x) \) as the Legendre functions of the second kind, which diverges to infinity at the regular singular points \( x = -1, 1 \). (See exercise below)
Legendre Equation as a Sturm-Liouville Problem
Orthogonality
The Legendre Equation can be readily written in the Sturm-Liouville Form (see this tutorial):
\begin{align}
\frac{d}{dx}((1-x^2)y’) + l(l+1)y = 0 \tag{7}
\end{align}
with \( p(x) = 1-x^2 \). This means that the results derived for Sturm-Liouville Equations and Hermitian operators in this tutorial naturally apply to the Legendre equation. An important point to notice is that the form of the corresponding Sturm-Liouville operator \( \mathcal{L} = -\frac{d}{dx}[(1-x^2)\frac{d}{dx}] \) suggests that there will be the natural interval \( [-1, 1] \) where the boundary terms will automatically vanish as \( 1-x^2 = 0 \) so Hermicity holds, but a necessary caveat is that the eigenfunctions will also have to be bounded at \( x=-1,1\). Again, as will be shown by the exercise, only the Legendre polynomials at integer values of \( l \) are finite at \(x = -1,1\), and hence they will be the desired eigenfunctions in the context of the Sturm-Liouville theory.
An immediate consequence is the orthogonality between eigenfunctions (Theorem 2 of the mentioned tutorial), and hence Legendre polynomials of different \(l\) are also orthogonal with a unit weighting \( \rho = 1 \):
\begin{align}
\int_{-1}^{1} P_l(x) P_m(x) dx &= 0 &\text{if \( l \neq m \)} \tag{8}
\end{align}
Nevertheless, we can arrive at the same conclusion directly. \( y = P_l(x) \) satisfies (7), and let’s multiply it by \( P_m(x), l \neq m \) to get
\begin{align}
P_m(x)\frac{d}{dx}((1-x^2)P_l(x)’) + l(l+1)P_l(x)P_m(x) = 0 \tag{9}
\end{align}
Integrate it over \( x = [-1,1]\) by parts produces
\begin{align}
\int_{-1}^{1} P_m(x)\frac{d}{dx}((1-x^2)P_l'(x)) dx + \int_{-1}^{1} l(l+1)P_l(x)P_m(x) dx &= 0 \\
-\int_{-1}^{1} P_m'(x)(1-x^2)P_l'(x) dx + \int_{-1}^{1} l(l+1)P_l(x)P_m(x) dx &= 0 \tag{10}
\end{align}
Repeat the same process but with \( l, m \) interchanged and subtracting between them then leads to the desired result:
\begin{align}
\begin{aligned}
(-\int_{-1}^{1} P_m'(x)(1-x^2)P_l'(x) dx + \int_{-1}^{1} l(l+1)P_l(x)P_m(x) dx) \\
-(-\int_{-1}^{1} P_m'(x)(1-x^2)P_l'(x) dx + \int_{-1}^{1} m(m+1)P_l(x)P_m(x) dx ) \end{aligned} &= 0 \\
[l(l+1)-m(m+1)]\int_{-1}^{1} P_l(x)P_m(x) dx &= 0 \\
\Rightarrow \int_{-1}^{1} P_l(x)P_m(x) dx &= 0 \tag{11}
\end{align}
Rodrigues’ Formula
Before proving other properties, we will benefit from establishing the convenient Rodrigues’ Formula first:
\begin{align}
P_l(x) = \frac{1}{2^l l!} \frac{d^l}{dx^l} [(x^2-1)^l] \tag{12}
\end{align}
Let \( u = (x^2-1)^l \), then differentiate once, we have
\begin{align}
u’ &= 2lx(x^2-1)^{l-1} \\
(x^2-1)u’ &= 2lx(x^2-1)^{l} = 2lxu \\
(x^2-1)u’ -2lxu &= 0 \tag{13}
\end{align}
Further differentiate (13) by \( l + 1 \) times by Leibniz Product Rule:
\begin{align}
\begin{aligned}
((x^2-1)u^{(l+2)} + 2x(l+1)u^{(l+1)} + l(l+1)u^{(l)}) \\
-(2xlu^{(l+1)} + 2l(l+1)u^{(l)}) \end{aligned} &= 0 \\
(x^2-1)u^{(l+2)} + 2xu^{(l+1)} -l(l+1)u^{(l)} &= 0 \tag{14}
\end{align}
Comparing with the original form (1), we identify \( u^{(l)} = d^l/dx^l[(x^2-1)^l] \) as one solution of the Legendre equation with the integer order \(l\). Since it is a finite polynomial, it must coincide with \( P_l \) apart from a scaling factor. To find that, recall that the normalization requirement is \( P_l(1) = 1\). Computing the \(l\)-times derivative of \( u = (x^2-1)^l \), the only term that does not contain the factor \( (x^2-1) \) and will not vanish upon substituting \(x = 1\), is \( (2x)^l l! (x^2-1)^0 \). Hence \( u^{(l)}(1) = 2^l l! \) and we have this as the denominator in (12).
With (12), we can obtain the norm of the Legendre polynomials:
\begin{align}
\int_{-1}^1 [P_l(x)]^2 dx &= \frac{1}{2^{2l} (l!)^2} \int_{-1}^1 \frac{d^l}{dx^l} [(x^2-1)^l]\frac{d^l}{dx^l} [(x^2-1)^l] dx \\
&= \frac{(-1)^l}{2^{2l} (l!)^2} \int_{-1}^1 (x^2-1)^l\frac{d^{2l}}{dx^{2l}} [(x^2-1)^l] dx \\
&= \frac{(2l)!}{2^{2l} (l!)^2} \int_{-1}^1 (1-x^2)^l dx \tag{15}
\end{align}
where the second line is by repeating integration by parts \(l\) times and the boundary terms vanish. The integral can be computed recursively with integrating by parts as well:
\begin{align}
\int_{-1}^1 (1-x^2)^l dx &= -\int_{-1}^1 x d[(1-x^2)^l] \\
\int_{-1}^1 (1-x^2)^l dx &= \int_{-1}^1 2lx^2(1-x^2)^{l-1} dx \\
\int_{-1}^1 (1-x^2)^l dx &= \int_{-1}^1 2l(1-x^2)^{l-1} dx -\int_{-1}^1 2l(1-x^2)(1-x^2)^{l-1} dx \\
\int_{-1}^1 (1-x^2)^l dx &= \int_{-1}^1 2l(1-x^2)^{l-1} dx -\int_{-1}^1 2l(1-x^2)^l dx \\
(2l+1)\int_{-1}^1 (1-x^2)^l dx &= \int_{-1}^1 2l(1-x^2)^{l-1} dx \tag{16}
\end{align}
Therefore, putting this back to (15), we have
\begin{align}
\int_{-1}^1 [P_l(x)]^2 dx &= \frac{(2l)!}{2^{2l} (l!)^2} \int_{-1}^1 (1-x^2)^l dx \\
&= \frac{(2l)!}{2^{2l} (l!)^2} \frac{2l}{2l+1} \int_{-1}^1 (1-x^2)^{l-1} dx \\
&= \frac{(2l)!}{2^{2l} (l!)^2} \frac{2l}{2l+1} \frac{2l-2}{2l-1} \cdots \frac{2}{3} \int_{-1}^1 (1-x^2)^{0} dx \\
&= \frac{(2l)!}{2^{2l} (l!)^2} \frac{[(2l)(2l-2)\cdots(2)]^2}{(2l+1)!} (2) \\
&= \frac{(2l)!}{2^{2l} (l!)^2} \frac{2^{2l}(l!)^2}{(2l+1)!} (2) \\
&= \frac{2}{2l+1} \tag{17}
\end{align}
Generating Function
Another method to yield the Legendre polynomials is via a generating function, which has the form of
\begin{align}
\frac{1}{\sqrt{1-2xt+t^2}} = \sum_{n=0}^{\infty} P_n(x)t^n \tag{18}
\end{align}
i.e. the coefficient of this function for \( t^n \) will be the Legendre polynomial of order \(n\). To show this directly, we can expand it by using the Binomial Theorem twice and also triangular summation twice: (writing \( \Gamma(-1/2-n+1) = (-1/2-n)! \) for convenience)
\begin{align}
\frac{1}{\sqrt{1-2xt+t^2}} &= \sum_{n=0}^{\infty} \frac{(-\frac{1}{2})!}{n!(-\frac{1}{2}-n)!} (t^2-2xt)^{n} \\
&= \sum_{n=0}^{\infty} \frac{(-\frac{1}{2})!}{n!(-\frac{1}{2}-n)!} t^{n}(t-2x)^n \\
&= \sum_{n=0}^{\infty} \frac{(-\frac{1}{2})!}{n!(-\frac{1}{2}-n)!} t^{n} \sum_{m=0}^{n} \frac{n!}{m!(n-m)!} t^m (-2x)^{n-m} \\
&= \sum_{n=0}^{\infty} \sum_{m=0}^{n} \frac{(-\frac{1}{2})!}{(-\frac{1}{2}-n)!m!(n-m)!} t^{m+n} (-2x)^{n-m} \\
&= \sum_{m=0}^{\infty} \sum_{n=m}^{\infty} \frac{(-\frac{1}{2})!}{(-\frac{1}{2}-n)!m!(n-m)!} t^{m+n} (-2x)^{n-m} \\
&= \sum_{m=0}^{\infty} \sum_{n=2m}^{\infty} \frac{(-\frac{1}{2})!}{(-\frac{1}{2}-n+m)!m!(n-2m)!} t^{n} (-2x)^{n-2m} \\
&= \sum_{n=0}^{\infty} \sum_{m=0}^{\lfloor \frac{n}{2} \rfloor} \frac{(-\frac{1}{2})!}{(-\frac{1}{2}-n+m)!m!(n-2m)!} (-2x)^{n-2m} t^{n} \\
&= \sum_{n=0}^{\infty} \sum_{m=0}^{\lfloor \frac{n}{2} \rfloor} \frac{(-\frac{1}{2})(-\frac{3}{2})\cdots(-\frac{1}{2}-n+m-1)}{m!(n-2m)!} (-2x)^{n-2m} t^{n} \\
&= \sum_{n=0}^{\infty} \frac{1}{2^n} \sum_{m=0}^{\lfloor \frac{n}{2} \rfloor} (-1)^m\frac{(1)(3)\cdots(2n-2m-1)(2^{n-m})(n-m)!}{m!(n-m)!(n-2m)!} x^{n-2m} t^{n} \\
&= \sum_{n=0}^{\infty} (\frac{1}{2^n} \sum_{m=0}^{\lfloor \frac{n}{2} \rfloor}\frac{(-1)^m(2n-2m)!}{m!(n-m)!(n-2m)!} x^{n-2m}) t^{n} \tag{19}
\end{align}
The bracketed term coincides with the general form of the Legendre polynomial of degree \(n\).
Recurrence Relations
The generating function (18) can be used to derive several recurrence relations. First, if we differentiate (18) with respect to \( x \) and use (18) again, we have
\begin{align}
t(1-2xt+t^2)^{-\frac{3}{2}} &= \sum_{n=0}^{\infty} P_n'(x)t^n \\
t(1-2xt+t^2)^{-\frac{1}{2}} &= (1-2xt+t^2) \sum_{n=0}^{\infty} P_n'(x)t^n \\
t \sum_{n=0}^{\infty} P_n(x)t^n &= (1-2xt+t^2) \sum_{n=0}^{\infty} P_n'(x)t^n \\
\sum_{n=0}^{\infty} P_n(x)t^{n+1} &= \sum_{n=0}^{\infty} P_n'(x)t^n -2x\sum_{n=0}^{\infty} P_n'(x)t^{n+1} + \sum_{n=0}^{\infty} P_n'(x)t^{n+2} \\
\sum_{n=0}^{\infty} P_n(x)t^{n+1} &= \sum_{n=0}^{\infty} P_{n+1}'(x)t^{n+1} -2x\sum_{n=0}^{\infty} P_n'(x)t^{n+1} + \sum_{n=1}^{\infty} P_{n-1}'(x)t^{n+1} \tag{20}
\end{align}
So by comparing the coefficients for \( t^{n+1} \), we obtain
\begin{equation}
P_n(x) = P_{n+1}'(x) -2xP_n'(x) +P_{n-1}'(x) \tag{21}
\end{equation}
Other recurrence relations are listed in the reference Mathematical Methods for Physics and Engineering by Riley, Hobson, and Bence:
\begin{align}
\begin{aligned}
P’_{n+1} &= (n+1)P_n + xP_n’ \\
P’_{n-1} &= -nP_n + xP_n’ \\
(1-x^2) P’_n &= n(P_{n-1} -xP_n) \\
(2n+1) P_n &= P’_{n+1} -P’_{n-1} \\
(n+1) P_{n+1} &= (2n+1) xP_n -nP_{n-1}
\end{aligned} \tag{22}
\end{align}
Particularly, the last one of (22) allows us to compute \( P_n(x) \) recursively, starting from \( P_0(x) = 1, P_1(x) = x \) and so on.
Gram-Schmidt Orthogonalization
The final method we will discuss to generate the Legendre polynomials is via the Gram-Schmidt Orthogonalization. We will start from the standard polynomial basis \( \{1,x,x^2,x^3,\cdots \} \) over the interval \( [-1,1 ] \). (Its completeness follows from the Weierstrass Approximation Theorem.) Then, we iteratively compute the orthogonal part of each \( x^n \) with respect to the previously processed polynomials by subtracting the parallel projections onto them. Then the resulting polynomials will abide (8) and coincide with the Legendre polynomials. The steps for \( n = 0,1 \) are trivial, and so we will demonstrate starting from \( n=2 \):
\begin{align}
P_2(x) &= x^2 -\frac{\langle x^2, P_0 \rangle}{\lVert P_0 \rVert^2} P_0 -\frac{\langle x^2, P_1 \rangle}{\lVert P_1 \rVert^2} P_1 \\
&= x^2 -(\frac{1}{2}\int_{-1}^{1} x^2 dx) (1) -(\frac{3}{2}\int_{-1}^{1} (x^2)(x) dx) (x) \\
&= x^2 -\frac{1}{2}[\frac{x^3}{3}]_{-1}^{1} (1) -(0) \\
&= x^2 -\frac{1}{3} \tag{23}
\end{align}
in which we recall (17) for values of the norms, and any odd integral will vanish. This is consistent with (6) apart from only a multiplicative factor, and we can just normalize it according to the requirement \( P_n(1) = 1 \) again. Let’s also do the case for \( n = 3 \):
\begin{align}
P_3(x) &= x^3 -\frac{\langle x^3, P_0 \rangle}{\lVert P_0 \rVert^2} P_0 -\frac{\langle x^3, P_1 \rangle}{\lVert P_1 \rVert^2} P_1 -\frac{\langle x^3, P_2 \rangle}{\lVert P_2 \rVert^2} P_2 \\
&= x^3 -(\frac{1}{2}\int_{-1}^{1} x^3 dx) (1) -(\frac{3}{2}\int_{-1}^{1} (x^3)(x) dx) (x) \\
&\quad -(\frac{5}{2}\int_{-1}^{1} (x^3)(\frac{1}{2}(3x^2 -1)) dx) (\frac{1}{2}(3x^2 -1)) \\
&= x^3 -(0) -\frac{3}{2}[\frac{x^5}{5}]_{-1}^{1} x -(0) \\
&= x^3 -\frac{3}{5}x \tag{24}
\end{align}
Exercise
By a change of variable \( t = \frac{1}{2}(1-x) \), the regular singular point \( x=1 \) is sent to \( t = 0 \), and show that the Legendre equation (1) is transformed into a hypergeometric equation. Hence by analyzing the form of solutions near that point, argue that except for the Legendre polynomials, other solutions to (1) are unbounded at \( x=1 \) (and by symmetry, \(x=-1\)).
Answer
Following the instructions, \( x = 1-2t\), \( dy/dx = (dy/dt) (dt/dx) = (-1/2)dy/dt \) and similarly for \( d^2y/dx^2\) , the new equation is
\begin{align}
(1-(1-2t)^2)((-\frac{1}{2})^2y^{\prime\prime}) -2(1-2t)(-\frac{1}{2}y’) + l(l+1)y &= 0 \\
t(1-t)y^{\prime\prime} +(1-2t)y’ + l(l+1)y &= 0 \\
\end{align}
Comparing with (1) in this tutorial, this is a hypergeometric equation with \( a=-l, b=l+1, c=1 \). By the remark under (11) in that tutorial, when \( l \) is not an integer, the corresponding hypergeometric function will not converge at \( t = 1 \) and thus \( x = -1 \) since \( c = 1 = -l + (l+1) = a+b \), and this also implies the unboundedness at \( x = 1 \) by symmetry. When \( l \) is an integer, the hypergeometric function in the first solution will be a finite polynomial that unsurprisingly corresponds to the Legendre polynomials. The special discussion when \( c = l \) is an integer indicates that the second solution will contain a term consisting of the Legendre polynomial times a logarithm \( \ln t \). This logarithm then implies that this second solution will also be unbounded at \( t = 0 \) and \( x = 1 \).
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