Legendre’s Differential Equation

Subsection 4.5.1 Legendre’s Differential Equation

The differential equation

\begin{equation} (1-x^{2})\frac{\,d^{2}y}{\,dx^{2}}-2x\frac{\,dy}{\,dx}+n(n+1)y=0\tag{4.5.1} \end{equation}

is known as Legendre’s differential equation. Such type of equation appears on solving Laplace’s equation ($\nabla^{2}y=0$) associated with spherical harmonics. In equation (4.5.1), $n$ is any number (integer or fraction). This equation can be solved in series of ascending or descending power of $x\text{.}$ It has no singularity at $x=0$ so its solution can be obtained as a power series developed about this point.

Let the series solution in descending powers of $x$ be

\begin{equation} y=\sum\limits_{k=0}^{\infty} a_{k}x^{m-k}\tag{4.5.2} \end{equation}

so that

\begin{equation*} \frac{\,dy}{\,dx}=\sum a_{k}(m-k)x^{m-k-1} \end{equation*}

and

\begin{equation*} \frac{\,d^{2}y}{\,dx^{2}}=\sum a_{k}(m-k)(m-k-1)x^{m-k-2} \end{equation*}

Substituting these in equation (4.5.1), we get-

\begin{equation*} (1-x^{2})\sum a_{k}(m-k)(m-k-1)x^{m-k-2}-2x\sum a_{k}(m-k)x^{m-k-1} \end{equation*}

\begin{equation*} +n(n+1)\sum a_{k}x^{m-k} =0 \end{equation*}

or,

\begin{equation*} \sum a_{k}(m-k)(m-k-1)x^{m-k-2} \end{equation*}

\begin{equation*} -\sum \{n(n+1)-(m-k)(m-k-1) -2(m-k)\}a_{k}x^{m-k}=0 \end{equation*}

or,

\begin{equation*} \sum a_{k}(m-k)(m-k-1)x^{m-k-2} \end{equation*}

\begin{equation} -\sum \{n(n+1) -(m-k)(m-k+1)\}a_{k}x^{m-k}=0\tag{4.5.3} \end{equation}

It is an identity and therefore, the coefficients of various poweers of $x$ must vanish. Now equating to zero, the coefficient of $x^{m}\text{,}$ the highest powers of $x\text{,}$ we have

\begin{equation*} a_{o}\{n(n+1)-m(m+1)\}=0 \end{equation*}

for $(k=0)\text{.}$ Since the coefficient of first term of the series can not be zero. Therefore $a_{o} \neq 0 \text{.}$ Hence the indicial equation is

\begin{equation*} n(n+1)-m(m+1) =0 \end{equation*}

or,

\begin{equation*} n^{2} +n-m^{2}-m=0 \end{equation*}

or,

\begin{equation*} n^{2}-m^{2}+(n-m)=0 \end{equation*}

or,

\begin{equation*} (n-m)(n+m+1) =0 \end{equation*}

or,

\begin{equation} m=n \quad \text{and}\quad m=-n-1. \tag{4.5.4} \end{equation}

Next equating the coefficient of $x^{m-1}$ to zero, we get -

\begin{equation*} a_{1}\left[n(n+1)-(m-1)m\right] =0 \end{equation*}

or,

\begin{equation*} a_{1}\left[n^{2}+n-m^{2}+m\right] =0 \end{equation*}

or,

\begin{equation*} a_{1}\left[(n+m)+(n+m)(n-m)\right] =0 \end{equation*}

or,

\begin{equation*} a_{1}(n+m)\{n-m+1\} =0 \end{equation*}

but,

\begin{equation*} (n+m)(n-m+1) \neq 0, \end{equation*}

for $m=n,-n-1\text{.}$ Hence,

\begin{equation} a_{1}=0\tag{4.5.5} \end{equation}

Again, equating the coefficient of general term $x^{m-k-2}$ to zero, we get the recurrence relation.

\begin{equation*} (m-k)(m-k-1)a_{k}+\{n(n+1)-(m-k-2)(m-k-1)\}a_{k+2}=0 \end{equation*}

now,

\begin{equation*} n(n+1)-(m-k-2)(m-k-1) \end{equation*}

\begin{equation*} =n^{2}+n-(m-k-1-1)(m-k-1) \end{equation*}

\begin{equation*} =n^{2}+n-(m-k-1)^{2}+(m-k-1) \end{equation*}

\begin{equation*} =(m-k+n-1)-\{(m-k-1)^{2}-n^{2}\} \end{equation*}

\begin{equation*} =(m-k+n-1)-(m-k-n-1)(m-k+n-1) \end{equation*}

\begin{equation*} =-(m-k+n-1)[m-k-n-2] \end{equation*}

or,

\begin{equation*} (m-k)(m-k-1)a_{k}-(m-k+n-1)(m-k-n-2)a_{k+2}=0 \end{equation*}

or,

\begin{equation} a_{k+2}=\frac{(m-k)(m-k-1)}{(m-k+n-1)(m-k-n-2)}a_{k}\tag{4.5.6} \end{equation}

Case I.

For, $m=n\text{,}$

\begin{equation*} a_{k+2} = \frac{(n-k)(n-k-1)}{(n-k+n-1)(n-k-n-2)}a_{k} \end{equation*}

\begin{equation*} = - \frac{(n-k)(n-k-1)}{(2n-k-1)(k+2)}a_{k} \end{equation*}

so that

\begin{equation*} a_{2}=-\frac{n(n-1)}{(2n-1)\cdot 2}a_{o}; \end{equation*}

\begin{equation*} a_{4}=\frac{(n-2)(n-3)}{(2n-3)\cdot 4}a_{2}=\frac{n(n-1)(n-2)(n-3)}{(2n-1)(2n-3)\cdot 4\cdot 2}a_{o} \end{equation*}

and so on. Also

\begin{equation*} a_{1}=a_{3}=a_{5}=\cdots =0. \end{equation*}

Hence the series (4.5.2) becomes

\begin{equation*} y=a_{o}\left[x^{n}-\frac{n(n-1)}{(2n-1)\cdot 2}x^{n-2}+\frac{n(n-1)(n-2)(n-3)}{(2n-1)(2n-3)\cdot 4\cdot 2}x^{n-4}-\cdots\right. \end{equation*}

\begin{equation*} +\left.(-1)^{r}\frac{n(n-1)(n-2)\cdots(n-2r+1)}{2\cdot 4\cdots 2r(2n-1)(2n-3)\cdots (2n-2r+1)}x^{n-2r}\right. \end{equation*}

\begin{equation} \left.\cdots+\cdots\right]\tag{4.5.7} \end{equation}

\begin{equation*} \therefore y=a_{o}x^{n}\left[1-\frac{n(n-1)}{(2n-1)\cdot 2}x^{-2}+\frac{n(n-1)(n-2)(n-3)}{(2n-1)(2n-3)\cdot 4\cdot 2}x^{-4}-\cdots\right. \end{equation*}

\begin{equation*} +\left.(-1)^{r}\frac{n(n-1)(n-2)\cdots(n-2r+1)}{2\cdot 4\cdots 2r(2n-1)(2n-3)\cdots (2n-2r+1)}x^{-2r}\right. \end{equation*}

\begin{equation} \left.\cdots +\cdots\right]\tag{4.5.8} \end{equation}

which is a solution of equation (4.5.1).

If we choose the arbitrary constant $a_{o}=\frac{1\cdot 3\cdot 5\cdots (2n-1)}{n!}$ where $n$ is a positive integer, then this solution of Legender’s equation is known as Legender’s Polynomials or Legender’s Function of First Kind, $P_{n}(x)\text{,}$ i.e.,

\begin{equation*} P_{n}(x) = \frac{1\cdot 3\cdot 5\cdots (2n-1)}{n!}\left[x^{n}-\frac{n(n-1)}{(2n-1)\cdot 2}x^{n-2} \right. \end{equation*}

\begin{equation*} \left. +\frac{n(n-1)(n-2)(n-3)}{(2n-1)(2n-3)\cdot 4\cdot 2}x^{n-4}-\cdots\right. \end{equation*}

\begin{equation*} +\left.(-1)^{r}\frac{n(n-1)(n-2)\cdots(n-2r+1)}{2\cdot 4\cdots 2r(2n-1)(2n-3)\cdots (2n-2r+1)}x^{n-2r}\right. \end{equation*}

\begin{equation} \left.\cdots +\cdots\right] \tag{4.5.9} \end{equation}

This is a terminating series. When $n$ is even, $n=2r$ and the last term is

\begin{equation*} (-1)^{\frac{n}{2}}\frac{n(n-1)(n-2)\cdots 3\cdot 2\cdot 1}{(2n-1)(2n-3)\cdots (n+1)\cdot 2\cdot 4\cdot 6\cdots n} \end{equation*}

when $n$ is odd, $n=2r+1$ and the last term is

\begin{equation*} (-1)^{\frac{n-1}{2}}\frac{n(n-1)(n-2)\cdots 3\cdot 2\cdot 1}{(2n-1)(2n-3)\cdots (n+2)\cdot 2\cdot 4\cdot 6\cdots (n-1)}x \end{equation*}

$P_{n}(x)$ is that solution of Legender’s equation which is equal to unity when $x=1\text{.}$

Case II.

For, $m=-n-1,$ we have from equation (4.5.6)

\begin{equation*} a_{k+2} = \frac{(n+k+1)(n+k+2)}{(k+2)(2n+k+3)}a_{k} \end{equation*}

so that

\begin{equation*} a_{2}=-\frac{(n+1)(n+2)}{2(2n+3)}a_{o}; \end{equation*}

\begin{equation*} a_{4}=\frac{(n+1)(n+2)(n+3)(n+4)}{2\cdot 4\cdot (2n+3)(2n+5)}a_{o} \end{equation*}

and so on. Also ,

\begin{equation*} a_{1}=a_{3}=a_{5}=\cdots =0. \end{equation*}

Hence,

\begin{equation*} y=a_{o}\left[x^{-n-1}+\frac{(n+1)(n+2)}{2(2n+3)}x^{-n-3} +\frac{(n+1)(n+2)(n+3)(n+4)}{2\cdot 4\cdot (2n+3)(2n+5)}x^{-n-5}+\right. \end{equation*}

\begin{equation} \left.\cdots+ \frac{(n+1)(n+2)(n+3)\cdots(n+2r)}{2\cdot 4\cdot 2r(2n+3)(2n+5)\cdots (2n+2r+1)}x^{-n-2r-1}+\cdots\right] \tag{4.5.10} \end{equation}

\begin{equation*} y=a_{o}x^{-n-1}\left[1+\frac{(n+1)(n+2)}{2(2n+3)}x^{-2}+\cdots \right. \end{equation*}

\begin{equation*} \left. + \frac{(n+1)(n+2)(n+3)\cdots(n+2r)}{2\cdot 4\cdot 2r(2n+3)(2n+5)\cdots (2n+2r+1)}x^{-2r}\right. \end{equation*}

\begin{equation} \left.\cdots +\cdots\right]\tag{4.5.11} \end{equation}

This gives an another solution of equation (4.5.1). If we choose, $a_{o}=\frac{n!}{1\cdot 3\cdot5 \cdots (2n+1)}$ where $n$ is a positive integer, then this solution of equation (4.5.1) is known as Legender’s Function of Second Kind, $Q_{n}(x)\text{,}$ i.e.,

\begin{equation*} Q_{n}(x) =\frac{n!}{1\cdot 3\cdot5 \cdots (2n+1)}\left[x^{-n-1}+\frac{(n+1)(n+2)}{2(2n+3)}x^{-n-3}+\cdots\right. \end{equation*}

\begin{equation} \left.\cdots+\frac{(n+1)(n+2)\cdots(n+2r)}{2\cdot 4\cdots2r\cdot(2n+3)\cdots(2n+2r+1)}x^{-n-2r-1}+\cdots\right]\tag{4.5.12} \end{equation}

This is a non - terminating series. The most general solution of Legender’s Equation is $y=AP_{n}(x)+BQ_{n}(x)$ where $A$ and $B$ are arbitrary constants.

Note:.

If the infinite series solution of a given differential equation is reduced into a finite series, then the solution is called the polynomial. "Polynomial is a finite series that consists of a finite number of terms. Each term in a polynomial is a product of a coefficient, a variable raised to a power, and there is a finite number of such terms in the polynomial expression. The degree of the polynomial determines the highest power of the variable present in the expression, and the number of terms is directly related to the degree". For a positive integer $n\text{,}$ equation (4.5.8) is a finite series (polynomial) and equation (4.5.10) is an infinite series. For a nigative integer $n\text{,}$ equation (4.5.8) is an infinite series and equation (4.5.11) is a polynomial. The general solution is a linear combination of a finite series and an infinite series.
If we want to solve the Legender’s equation in a series of ascending powers of $x\text{,}$ we may proceed by taking $y=\sum\limits_{k=0}^{\infty} a_{k}x^{m+k}\text{.}$
The expansion of $P_{n}(x)$ can also be written as

\begin{equation*} P_{n}(x) = \sum\limits_{r=0}^{N} (-1)^{r}\frac{1\cdot 3\cdot 5\cdots (2n-1)}{n!} \end{equation*}

\begin{equation*} \times\frac{n(n-1)\cdots(n-2r+1)}{2\cdot 4\cdots 2r\cdot(2n-1)(2n-3)\cdots (2n-2r+1)}x^{n-2r} \end{equation*}

\begin{equation*} =\sum\limits_{r=0}^{N} (-1)^{r}\frac{1\cdot 3\cdot 5\cdots (2n-2r-1)}{2^{r}\cdot r! (n-2r)!}x^{n-2r} \end{equation*}

\begin{equation*} =\sum\limits_{r=0}^{N} (-1)^{r}\frac{1\cdot 2\cdot 3\cdot 5\cdots (2n-2r)}{2^{r}\cdot r! (n-2r)!\cdot 2\cdot 4\cdots (2n-2r)}x^{n-2r} \end{equation*}

\begin{equation} =\sum\limits_{r=0}^{N} (-1)^{r}\frac{1(2n-2r)!}{2^{r}\cdot r! (n-r)!(n-2r)!}x^{n-2r}\tag{4.5.13} \end{equation}

where $N=\frac{n}{2}$ for $n$ even and $N=n-\frac{1}{2}$ for $n$ odd. In particular,

\begin{equation*} P_{n}(x) = \frac{1\cdot 3\cdot 5\cdots (2n-1)}{n!}\left[x^{n}-\frac{n(n-1)}{2(2n-1)}x^{n-2}\right. \end{equation*}

\begin{equation*} \left.+\frac{n(n-1)(n-2)(n-3)}{2\cdot 4\cdots (2n-1)(2n-3)}x^{n-4}-\cdots\right] \end{equation*}

\begin{equation*} =\frac{2n)!}{(2\cdot 4\cdot 6 \cdots 2n)n!}\left[x^{n}-\frac{n(n-1)}{2(2n-1)}x^{n-2} \right. \end{equation*}

\begin{equation*} \left.+\frac{n(n-1)(n-2)(n-3)}{2\cdot 4\cdots (2n-1)(2n-3)}x^{n-4}-\cdots\right] \end{equation*}

\begin{equation*} =\frac{2n)!}{(2^{n}n!n!}\left[x^{n}-\frac{n(n-1)}{2(2n-1)}x^{n-2}\right. \end{equation*}

\begin{equation*} \left.+\frac{n(n-1)(n-2)(n-3)}{2\cdot 4\cdots (2n-1)(2n-3)}x^{n-4}-\cdots\right] \end{equation*}

putting, $n=0,1,2,3,\cdots$ etc., we get - $P_{n}(x)=1\text{.}$

\begin{equation*} P_{1}(x)=\frac{2!}{2\cdot 1!\cdot1!}[x] =x, \end{equation*}

other terms are vanished.

\begin{equation*} P_{2}(x)=\frac{4!}{2^{2}\cdot 2!\cdot 2!}[x^{2}-\frac{1}{3}] = \frac{3}{2}(x^{2}-\frac{1}{3}) = \frac{3x^{2}-1}{2}, \end{equation*}

other terms are vanished.

\begin{equation*} P_{3}(x)=\frac{6!}{2^{3}\cdot 3!\cdot 3!}[x^{3}-\frac{3}{2}\frac{2}{5}x] = \frac{5x^{3}-3x}{2}, \end{equation*}

etc.
From expression (4.5.13), we deduce that

\begin{equation*} P_{n}(x) =\sum\limits_{r=0}^{N}\frac{ (-1)^{r} }{2^{n}r!(n-r)!}\frac{d^{n}}{dx^{n}}(x^{2n-2r}) \end{equation*}

\begin{equation*} =\frac{1}{2^{n}n!}\frac{d^{n}}{dx^{n}}\sum\limits_{r=0}^{N}(-1)^{r}\frac{n!}{r!(n-r)!}x^{2n-2r} \end{equation*}

\begin{equation*} \left[ \because \frac{d^{n}}{dx^{n}}x^{2n-2r} = \frac{(2n-2r)!}{(n-2r)!}x^{2n-2r}\right] \end{equation*}

Now extended the upper limit from $N$ to $n$ since the extra terms from $N+1$ to $n$ do not contribute anything because each term in the series

\begin{equation*} \sum\limits_{r=N+1}^{n} \frac{d^{n}}{dx^{n}}\frac{(-1)^{r}n!}{r!(n-r)!}x^{2n-2r} \end{equation*}

vanishes for $r$ and $n$ integrals. Also,

\begin{equation*} (x^{2}-1)^{n}= \sum\limits_{r=0}^{n} \frac{(-1)^{r}n!}{r!(n-r)!}x^{2(n-r)} \end{equation*}

from Binomial expansion of $(x^{2}-1)^{n}$

\begin{equation*} \therefore \quad P_{n}(x) = \frac{1}{2^{n}n!}\frac{d^{n}}{dx^{n}}(x^{2}-1)^{n} \end{equation*}

This is the Rodrigue’s Formula for Legender’s Polynomial. The plot for Legender’s polynomial is shown in Figure 4.5.1.

Subsubsection 4.5.1.1 Rodrigue’s Formula

\begin{equation*} P_{n}(x) = \frac{1}{2^{n}n!}\frac{d^{n}}{dx^{n}}(x^{2}-1)^{n} \end{equation*}

Proof.

Let

\begin{equation} v=(x^{2}-1)^{n}\tag{4.5.14} \end{equation}

then,

\begin{equation*} \frac{\,dv}{\,dx}=n(x^{2}-1)^{n-1}\cdot2x =\frac{n(x^{2}-1)^{n}}{(x^{2}-1)}\cdot 2x \end{equation*}

or,

\begin{equation*} (x^{2}-1)\frac{\,dv}{\,dx}=2n(x^{2}-1)^{n}x \end{equation*}

or,

\begin{equation} (x^{2}-1)\frac{\,dv}{\,dx} =2nvx\tag{4.5.15} \end{equation}

Now, differentiating equation (4.5.15), $(n+1)$ times using Leibnitz theorem, we get -

\begin{equation*} (x^{2}-1)\frac{d^{n+2}}{dx^{n+2}}v + \comb{n+1}{1} (2x)\frac{d^{n+1}}{dx^{n+1}}v+\comb{n+1}{2} (2)\frac{d^{n}}{dx^{n}}v \end{equation*}

\begin{equation*} =2n\left[x \frac{d^{n+1}}{dx^{n+1}}v+\comb{n+1}{1}(1)\frac{d^{n}}{dx^{n}}\right]. \end{equation*}

or,

\begin{equation*} (x^{2}-1)\frac{d^{n+2}}{dx^{n+2}}v +2x\left[\comb{n+1}{1} -n\right]\frac{d^{n+1}}{dx^{n+1}}v \end{equation*}

\begin{equation*} +2\left[\comb{n+1}{2} -n \comb{n+1}{1}\right]\frac{d^{n}}{dx^{n}}v=0 \end{equation*}

or,

\begin{equation} (x^{2}-1)\frac{d^{n+2}}{dx^{n+2}}v +2x \frac{d^{n+1}}{dx^{n+1}}v-n(n+1)\frac{d^{n}v}{dx^{n}}=0\tag{4.5.16} \end{equation}

If we put $\frac{d^{n}v}{dx^{n}}=y$ then equation (4.5.16) becomes

\begin{equation*} (x^{2}-1)\frac{d^{2}y}{dx^{2}} +2x \frac{dy}{dx}-n(n+1)y=0 \end{equation*}

or,

\begin{equation} (1-x^{2})\frac{d^{2}y}{dx^{2}} -2x \frac{dy}{dx}+n(n+1)y=0. \tag{4.5.17} \end{equation}

which is Legendre’s equation and $y= \frac{d^{n}v}{dx^{n}}$ is one of its solution.

\begin{equation} \therefore P_{n}(x)=C\frac{d^{n}v}{dx^{n}}\tag{4.5.18} \end{equation}

where C is a constant. But,

\begin{equation*} v=(x^{2}-1)^{n}=(x+1)^{n}(x-1)^{n}, \end{equation*}

so that

\begin{equation*} \frac{d^{n}v}{dx^{n}}=(x+1)^{n}\frac{d^{n}}{dx^{n}}(x-1)^{n}+\comb{n}{1}n(x+1)^{n-1}\frac{d^{n-1}}{dx^{n-1}}(x-1)^{n} \end{equation*}

\begin{equation*} +\cdots+(x-1)^{n}\frac{d^{n}}{dx^{n}}(x+1)^{n}. \end{equation*}

when $x=1\text{,}$

\begin{equation*} \frac{d^{n}v}{dx^{n}}=2^{n}n! \end{equation*}

All the other terms disappear as $(x-1)$ is a factor in every term except first.

\begin{equation*} \because \frac{d^{n}}{dx^{n}}(x-1)^{n}=n! ; \end{equation*}

\begin{equation*} \frac{d^{n-1}}{dx^{n-1}}(x-1)^{n}=n(x-1)\cdots 2(x-1); \end{equation*}

\begin{equation*} \frac{d^{n-2}}{dx^{n-2}}(x-1)^{n}=n(x-1)\cdots 3(x-1)^{2} \end{equation*}

Therefore, when $x=1\text{,}$ equation (4.5.18) gives

\begin{equation*} P_{n}(1) = C\cdot 2^{n}n!=1 \quad \therefore C =\frac{1}{2^{n}n!} \end{equation*}

Hence,

\begin{equation*} P_{n}(x) = \frac{1}{2^{n}n!}\frac{d^{n}v}{dx^{n}} \end{equation*}

from equation (4.5.18).

\begin{equation*} \therefore \quad P_{n}(x) = \frac{1}{2^{n}n!}\frac{d^{n}}{dx^{n}}(x^{2}-1)^{n} \end{equation*}

from equation (4.5.14). This is the Rodrigue’s Formula for Legendre Polynomial.

Subsubsection 4.5.1.2 Legendre Polynomials

The Rodrigue’s formula for Legendre’s Polynomials is

\begin{equation*} P_{n}(x) = \frac{1}{2^{n}n!}\frac{d^{n}}{dx^{n}}(x^{2}-1)^{n} \end{equation*}

If $n=0\text{,}$

\begin{equation*} P_{o}(x) = \frac{1}{2^{0}0!}=1 \end{equation*}

If $n=1\text{,}$

\begin{equation*} P_{1}(x) = \frac{1}{2^{1}1!}\frac{d}{dx}(x^{2}-1)=\frac{1}{2}(2x)=x. \end{equation*}

If $n=2\text{,}$

\begin{equation*} P_{2}(x) = \frac{1}{2^{2}2!}\frac{d^{2}}{dx^{2}}(x^{2}-1)^{2} \end{equation*}

\begin{equation*} =\frac{1}{8}\frac{d}{dx}[2(x^{2}-1)(2x)] = \frac{1}{2}[2(x^{2}\cdot 1+2x\cdot x] \end{equation*}

\begin{equation*} =\frac{1}{2}[3x^{2}-1] \end{equation*}

similarly,

\begin{equation*} P_{3}(x) = \frac{1}{2}(5x^{3}-3x) \end{equation*}

\begin{equation*} P_{4}(x) = \frac{1}{8}(35x^{4}-30x^{2}+3) \end{equation*}

\begin{equation*} P_{5}(x) = \frac{1}{8}(63x^{5}-70x^{2}+15x) \end{equation*}

\begin{equation*} \vdots \hspace{4cm} \vdots \end{equation*}

\begin{equation*} P_{n}(x) =\sum\limits_{r=0}^{N}(-1)^{r} \frac{(2n-2r)!}{2^{n}r!(n-r)!(n-2r)!}x^{n-2r}; \end{equation*}

where,

\begin{equation*} N = \quad \begin{cases} \frac{n}{2}, \quad \text{if n is even.}\\ \frac{1}{2}(n-1),\quad \text{if n is odd.} \end{cases} \end{equation*}

Note: we can evaluate $P_{n}(x)$ by expanding $(x^{2}-1)^{n}$ from Binomial theorem.

\begin{equation*} (x^{2}-1)^{n}=\sum\limits_{r=0}^{n}{\comb{n}{r}(x^{2})^{n-r}(-1)^{r}} \end{equation*}

\begin{equation*} = \sum\limits_{r=0}^{n}(-1)^{r}\frac{n!}{r!(n-r)!}(x^{2n-2r}) \end{equation*}

However,

\begin{equation*} P_{n}(x) = \frac{1}{2^{n}n!}\frac{d^{n}}{dx^{n}}(x^{2}-1)^{n}=\frac{1}{2^{n}n!}\sum\limits_{r=0}^{n}\frac{n!}{r!(n-r)!}\frac{d^{n}}{dx^{n}}(x^{2n-2r}) \end{equation*}

\begin{equation*} =\sum\limits_{r=0}^{n}(-1)^{r}\frac{(2n-2r)!}{2^{n}r!(n-r)!(n-2r)!}x^{n-2r} \end{equation*}

The last term contains either $x^{0}$ or $x^{1}\text{.}$

\begin{equation*} \therefore \quad n-2r = 0 \end{equation*}

or,

\begin{equation*} r=\frac{n}{2} \quad \text{(for n even)} \end{equation*}

or,

\begin{equation*} n-2r=1 \end{equation*}

or,

\begin{equation*} r =\frac{1}{2}(n-1) \quad \text{(for n odd)}. \end{equation*}

Python Code:.

import numpy as np

from scipy.special import eval_legendre

import matplotlib.pyplot as plt

from scipy.linalg import det

N = range(0, 6)

eval_legendre(N, 0)

X = np.linspace(-1, 1)

for n in range(0, 6):

y = eval_legendre(n, X)

plt.plot(X, y, label=r’$P_{}(x)$’.format(n))

plt.title("Legendre Polynomials")

plt.xlabel("x")

plt.ylabel(r’$P_n(x)$’)

plt.legend(loc=’lower right’)

plt.show()

Subsubsection 4.5.1.3 Generating Function for $P_{n}(x)$

Legendre Polynomial $P_{n}(x)$ is the coefficient of $z^{n}$ in $(1-2xz+z^{2})^{-\frac{1}{2}}\text{,}$ i.e.,

\begin{equation} (1-2xz+z^{2})^{-\frac{1}{2}} =\sum\limits_{n=0}^{\infty}P_{n}(x)z^{n} \tag{4.5.19} \end{equation}

Proof.

We have

\begin{equation*} (1-2xz+z^{2})^{-\frac{1}{2}} =\left[1-z(2x-z)\right]^{-\frac{1}{2}} \end{equation*}

\begin{equation*} =1+\frac{1}{2}z(2x-z)+\frac{(-\frac{1}{2})(-\frac{3}{2})}{2!}z^{2}(2x-z)^{2}+\cdots \end{equation*}

\begin{equation*} +\cdots \frac{(-\frac{1}{2})(-\frac{3}{2})(-\frac{5}{2})\cdots(-\frac{1}{2}-n+1)}{n!}(-z)^{n}(2x-z)^{n}+\cdots \end{equation*}

\begin{equation*} =1+\frac{1}{2}z(2x-z)+\frac{1\cdot 3}{2^{2}2!}z^{2}(2x-z)^{2}+ \end{equation*}

\begin{equation*} \cdots+\frac{1\cdot 3\cdot 5\cdots(2n-5)}{2^{n-2}(n-2)!}z^{n-2}(2x-z)^{n-2} \end{equation*}

\begin{equation*} + \frac{1\cdot 3\cdot 5\cdots(2n-3)}{2^{n-1}(n-1)!}z^{n-1}(2x-z)^{n-1} \end{equation*}

\begin{equation} +\frac{1\cdot 3\cdot 5\cdots(2n-1)}{2^{n}(n)!}z^{n}(2x-z)^{n}+\cdots\tag{4.5.20} \end{equation}

\begin{equation*} \left[\because (1-x)^{n}= 1-nx+\frac{n(n-1)}{2!}x^{2}-\frac{n(n-1)(n-2)}{3!}x^{2}+\cdots\right] \end{equation*}

\begin{equation*} \left[\because (x+y)^{n}= x^{n}+\comb{n}{1}x^{n-1}y+\comb{n}{2}x^{n-2}y^{2}+\cdots\right] \end{equation*}

But,

\begin{equation*} z^{n}(2x-z)^{n}=z^{n}(2x^{n}+\cdots \end{equation*}

\begin{equation*} z^{n-1}(2x-z)^{n-1}=z^{n-1}\left[(2x)^{n-1}-\comb{n-1}{1}(2x)^{n-2}z+\cdots\right] \end{equation*}

\begin{equation*} = z^{n-1}(2x)^{n-1}-(n-1)(2x)^{n-2}z^{n}+\cdots \end{equation*}

and

\begin{equation*} z^{n-2}(2x-z)^{n-2} \end{equation*}

\begin{equation*} =z^{n-2}\left[(2x)^{n-2}-\comb{n-2}{1}(2x)^{n-3}z+\cdots+\comb{n-2}{2}(2x)^{n-4}z^{2}+\cdots\right] \end{equation*}

\begin{equation*} =z^{n-2}(2x)^{n-2}-(n-2)(2x)^{n-3}z^{n-1}+\frac{(n-2)(n-3)}{2!}(2x)^{n-4}z^{n}+\cdots \end{equation*}

Hence, replacing by these values to collect the coefficient of $z^{n}\text{,}$ we get -

\begin{equation*} (1-2xz+z^{2})^{-\frac{1}{2}}=1+xz+\left(\frac{3}{2}x^{2}-\frac{1}{2}\right)z^{2}+\cdots \end{equation*}

\begin{equation*} \cdots+\left[\frac{1\cdot3\cdot5\cdots(2n-1)}{2^{n}n!}2^{n}x^{n}-\frac{1\cdot3\cdot5\cdots(2n-3)}{2^{n-1}(n-1)!}(n-1)(2x)^{n-2}\right. \end{equation*}

\begin{equation*} +\left.\frac{1\cdot3\cdot5\cdots(2n-5)(n-2)(n-3)}{2^{n-2}(n-2)!2!}2^{n-4}x^{n-4}\right]z^{n}+\cdots \end{equation*}

\begin{equation*} =1+xz+\frac{1}{2}(3x^{2}-1)z^{2}+\cdots \end{equation*}

\begin{equation*} +\frac{1\cdot3\cdot5\cdots(2n-1)}{n!}\left[x^{n}-\frac{n(n-1)}{2(2n-1)}x^{n-2}\right. \end{equation*}

\begin{equation*} \left. +\frac{n(n-1)(n-2)(n-3)}{2\cdot4(2n-1)(2n-3)}x^{n-4}+\cdots\right]z^{n}+\cdots \end{equation*}

\begin{equation*} =P_{o}(x)+P_{1}(x)z+P_{2}(x)z^{2}+\cdots+P_{n}(x)z^{n}+\cdots =\sum\limits_{n=0}^{\infty}P_{n}(x)z^{n} \end{equation*}

Thus the function $(1-2xz+z^{2})^{-\frac{1}{2}}$ is defined as generating function of $P_{n}(x) \text{.}$

Subsubsection 4.5.1.4 Some Important Results

\begin{equation*} P_{n}(1) =1 \end{equation*}

We know that $(1-2xz+z^{2})^{-\frac{1}{2}} $

\begin{equation*} =1+zP_{1}(x)+z^{2}P_{2}(x)+z^{3}P_{3}(x)+\cdots+z^{n}P_{n}(x)+\cdots \end{equation*}

substituting $'1'$ for $'x'$ in the above equation, we get -

\begin{equation*} (1-2xz+z^{2})^{-\frac{1}{2}} \end{equation*}

\begin{equation*} =1+zP_{1}(1)+z^{2}P_{2}(1)+z^{3}P_{3}(1)+\cdots+z^{n}P_{n}(1)+\cdots \end{equation*}

or,

\begin{equation*} \left[(1-z)^{2}\right]^{-\frac{1}{2}}=\sum\limits_{n=0}^{\infty}z^{n}P_{n}(1) \end{equation*}

or,

\begin{equation*} (1-z)^{-1}=\sum\limits_{n=0}^{\infty}z^{n}P_{n}(1) \end{equation*}

\begin{equation*} \sum\limits_{n=0}^{\infty}z^{n}P_{n}(1)=(1-z)^{-1} =1+z+z^{2}+z^{3}+\cdots+z^{n}+\cdots \end{equation*}

Equating the coefficients of $z^{n}$ on both sides, we get -

\begin{equation*} P_{n}(1) =1. \end{equation*}
\begin{equation*} P_{n}(-1) =(-1)^{n}P_{n} \end{equation*}

putting $x=-1$ in the generating function of $P_{n}(x)\text{,}$ we get -

\begin{equation*} (1+2z+z^{2})^{-\frac{1}{2}} =\sum\limits_{n=0}^{\infty}z^{n}P_{n}(-1)z^{n} \end{equation*}

or,

\begin{equation*} \sum\limits_{n=0}^{\infty}z^{n}P_{n}(-1)z^{n}=(1+z)^{-1}=\sum\limits_{n=0}^{\infty}(-1)^{n}z^{n} \end{equation*}

equating the coefficient of $z^{n}\text{,}$ we get -

\begin{equation*} P_{n}(-1)=(-1)^{n}, \end{equation*}

i.e.,

\begin{equation*} P_{n}(-1)=(-1)^{n}P_{n}(1) \end{equation*}
\begin{equation*} P_{n}(-x) =(-1)^{n} P_{n}(x) \end{equation*}

we have

\begin{equation*} (1-2xz+z^{2})^{-\frac{1}{2}} \end{equation*}

\begin{equation*} =[1-2(-x)(-z)+(-z)^{2}]^{-\frac{1}{2}}=\sum\limits_{n=0}^{\infty}P_{n}(-x)(-z)^{n} \end{equation*}

now replacing $x$ by $-x$ and $z$ by $-z\text{.}$

\begin{equation*} \sum\limits_{n=0}^{\infty}P_{n}(x)(z)^{n}=\sum\limits_{n=0}^{\infty}(-1)^{n}P_{n}(-x)(z)^{n} \end{equation*}

equating the coefficients of $z^{n}\text{,}$ we get -

\begin{equation*} P_{n}(x)=(-1)^{n}P_{n}(-x) \end{equation*}

\begin{equation*} P_{n}(-x)=(-1)^{n}P_{n}(x) \end{equation*}

Alternative: We know that

\begin{equation*} P_{n}(x)= \sum\limits_{r=0}^{N}(-1)^{r}\frac{(2n-2r)!}{2^{n}r!(n-r)!(n-2r)}(x)^{n-2r}; \end{equation*}

\begin{equation*} \begin{cases} \text{where}\quad N=\frac{n}{2} \quad \text{for n even}\\ \quad =\frac{n-1}{2} \quad \text{for n odd}\end{cases} \end{equation*}

Thus

\begin{equation*} P_{n}(-x) = \sum\limits_{r=0}^{N}(-1)^{r}\frac{(2n-2r)!}{2^{n}r!(n-r)!(n-2r)}(-1)^{n-2r}(x)^{n-2r} \end{equation*}

\begin{equation*} =(-1)^{n}\sum\limits_{r=0}^{N}(-1)^{r}\frac{(2n-2r)!}{2^{n}r!(n-r)!(n-2r)}(x)^{n-2r} \end{equation*}

\begin{equation*} =(-1)^{n}P_{n}(x) \end{equation*}

Subsubsection 4.5.1.5 Recurrence Relations for $P_{n}(x)$

Recurrence relations are the relationship among the polynomials of different orders and their derivatives of differential orders.

\begin{equation*} nP_{n}=(2n-1)xP_{n-1}-(n-1)P_{n-2}. \end{equation*}

Solution: We know that

\begin{equation*} (1-2xz+z^{2})^{-\frac{1}{2}}=\sum z^{n}[P_{n}(x)] \end{equation*}

Differentiating it with respect to $z\text{,}$ we get -

\begin{equation*} -\frac{1}{2}(1-2xz+z^{2})^{-\frac{3}{2}}(-2x+2z)=\sum n z^{n-1}P_{n}(x) \end{equation*}

or,

\begin{equation*} (1-2xz+z^{2})^{-\frac{1}{2}}(x-z)=(1-2xz+z^{2})\sum n z^{n-1}P_{n}(x) \end{equation*}

or,

\begin{equation*} (x-z)\sum z^{n}P_{n}(x)=(1-2xz+z^{2})\sum nz^{n-1}P_{n}(x) \end{equation*}

Equating the coefficient of $z^{n-1}$ on both sides, we get-

\begin{equation*} xP_{n-1}-P_{n-2}=nP_{n}-2x(n-1)P_{n-1}+(n-2)P_{n-2} \end{equation*}

or,

\begin{equation*} nP_{n}=(2n-1)xP_{n-1}-(n-1)P_{n-2} \end{equation*}

Note: If $n-1=m \text{,}$ then this result reduces to

\begin{equation*} (m+1)P_{m+1}(x)=(2M+1)xP_{m}(x)-mP_{m-1}(x). \end{equation*}
\begin{equation*} xP'_{n} - P'_{n-1}=n P_{n} \end{equation*}

Solution: We know that

\begin{equation} (1-2xz+z^{2})^{-\frac{1}{2}}=\sum z^{n}[P_{n}(x)]\tag{4.5.21} \end{equation}

Differentiating equation (4.5.21) with respect to $z\text{,}$ we get -

\begin{equation*} -\frac{1}{2}(1-2xz+z^{2})^{-\frac{3}{2}}(-2x+2z)=\sum n z^{n-1}P_{n}(x) \end{equation*}

or,

\begin{equation} (x-z)(1-2xz+z^{2})^{-\frac{3}{2}}=\sum nz^{n-1}P_{n}(x)\tag{4.5.22} \end{equation}

Differentiating equation (4.5.21) with respect to $x\text{,}$ we get -

\begin{equation*} -\frac{1}{2}(1-2xz+z^{2})^{-\frac{3}{2}}=\sum z^{n}P'_{n}(x) \end{equation*}

or,

\begin{equation} z(1-2xz+z^{2})^{-\frac{3}{2}}=\sum z^{n}P'_{n}(x)\tag{4.5.23} \end{equation}

Dividing equation (4.5.22) by equation (4.5.23), we get -

\begin{equation*} \frac{(x-z)}{z}=\frac{\sum nz^{n-1}P_{n}(x)}{\sum z^{n}P'_{n}(x)} \end{equation*}

or,

\begin{equation*} (x-z)\sum z^{n}P'_{n}(x)=\sum nz^{n}P_{n}(x) \end{equation*}

equating the coefficients of $z^{n} $ from both sides, we get -

\begin{equation*} xP'_{n}(x)-P'_{n-1}(x)=nP_{n}(x) \end{equation*}
\begin{equation*} P'_{n}-xP'_{n-1} = nP_{n-1}. \end{equation*}

solution: From relation I, we have -

\begin{equation*} nP_{n}=(2n-1)xP_{n-1}-(n-1)P_{n-2} \end{equation*}

Differentiating this with respect to $x\text{,}$ we get -

\begin{equation*} n P'_{n}=(2n-1)P_{n-1}+(2n-1)xP'_{n-1}-(n-1)P'_{n-2} \end{equation*}

or,

\begin{equation*} n\left[P'_{n}-xP'_{n-1}\right]-(n-1)\left[xP'_{n-1}-P'_{n-2}\right]=(2n-1)P_{n-1} \end{equation*}

or,

\begin{equation*} n\left[P'_{n}-xP'_{n-1}\right]-(n-1)\left[(n-1)P_{n-1}\right]=(2n-1)P_{n-1} \end{equation*}

from relation II. or,

\begin{equation*} n\left[P'_{n}-xP'_{n-1}\right]=\left[(n-1)^{2}+(2n-1)\right]P_{n-1} \end{equation*}

\begin{equation*} =\left[n^{2}-2n+1+2n-1\right]P_{n-1} = n^{2}P_{n-1} \end{equation*}

\begin{equation*} \therefore \quad P'_{n}-xP'_{n-1}=nP_{n-1}. \end{equation*}
\begin{equation*} P'_{n+1}-P'_{n-1}=(2n+1)P_{n}. \end{equation*}

Solution: From relation I., we have

\begin{equation*} nP_{n}=(2n-1)x P_{n-1}-(n-1)P_{n-2} \end{equation*}

replacing $n$ by (n+1), we get-

\begin{equation*} (n+1)P_{n+1}=(2n+1)xP_{n}-nP_{n-1} \end{equation*}

Differentiating this with respect to $x\text{,}$ we get -

\begin{equation*} (n+1)P'_{n+1}=(2n+1)P_{n}(2n+1)xP'_{n}-nP'_{n-1} \end{equation*}

\begin{equation*} =(2n+1)P_{n}+(2n+1)\left[nP_{n}+P'_{n-1}\right]-nP'_{n-1} \end{equation*}

from relation II. Or,

\begin{equation*} (n+1)P'_{n+1}-(n+1)P'_{n-1}=(2n+1)(1+n)P_{n} \end{equation*}

or,

\begin{equation*} P'_{n+1}-P'_{n-1}=(2n+1)P_{n} \end{equation*}
\begin{equation*} (x^{2}-1)P'_{n}=n\left[xP_{n}-P_{n-1}\right] \end{equation*}

Solution: From relation III, we have

\begin{equation} P'_{n}-xP'_{n-1}=nP_{n-1}\tag{4.5.24} \end{equation}

From relation II, we have

\begin{equation} xP'_{n}-P'_{n-1}=nP_{n}\tag{4.5.25} \end{equation}

multiplying equation (4.5.25) by $x$ and subtracting from equation (4.5.24), we get -

\begin{equation*} (1-x^{2})P'_{n}=n(P_{n-1}-xP_{n}) \end{equation*}
\begin{equation*} (x^{2}-1)P'_{n}=(n+1)\left[P_{n+1}-xP_{n}\right] \end{equation*}

Solution: From relation I, we have

\begin{equation*} nP_{n}=(2n-1)xP_{n-1}-(n-1)P_{n-2} \end{equation*}

Replacing $n$ by $n+1\text{,}$ we get -

\begin{equation*} (n+1)P_{n+1}=(2n+1)xP_{n}-nP_{n-1} \end{equation*}

or,

\begin{equation*} (n+1)\left[P_{n+1}-xP_{n}\right] =n\left[xP_{n}-P_{n-1}\right] \end{equation*}

or,

\begin{equation*} (n+1)\left[P_{n+1}-xP_{n}\right] +n\left[P_{n-1}-xP_{n}\right]=0 \end{equation*}

or,

\begin{equation*} (n+1)\left[P_{n+1}-xP_{n}\right] +(1-x^{2})P'_{n}=0 \end{equation*}

\begin{equation*} \therefore \quad (x^{2}-1)P'_{n}=(n+1)\left[P_{n+1}-xP_{n}\right] \end{equation*}

Subsubsection 4.5.1.6 Orthogonality of Legendre’s Polynomials

Legendre’s polynomiala are a set of orthogonal functions in the interval $(-1,1)\text{.}$

\begin{equation} \int\limits_{-1}^{+1}P_{m}(x)P_{n}(x)\,dx=\frac{2}{2n+1}\,\delta_{m,n};\tag{4.5.26} \end{equation}

where

\begin{equation*} \delta_{m,n}=\begin{cases} 0 \quad \text{if}\quad m \neq n \\ 1 \quad \text{if}\quad m=n \end{cases} \end{equation*}

here $\delta_{m,n} $ is a Kronecker delta symbol and $m$ and $n$ are positive integer.

Proof.

$P_{n}(x)$ is a solution of

\begin{equation} (1-x^{2})\frac{d^{2}y}{dx^{2}}-2x\frac{\,dy}{\,dx}+n(n+1) y =0\tag{4.5.27} \end{equation}

$P_{m}(x)$ is a solution of

\begin{equation} (1-x^{2})\frac{d^{2}z}{dx^{2}}-2x\frac{dz}{dx}+m(m+1) z =0 \tag{4.5.28} \end{equation}

Multiplying equation (4.5.27) by ’z’ and equation (4.5.28) by ’y’ and then subtracte, we get -

\begin{equation*} (1-x^{2})\left[z\frac{d^{2}y}{dx^{2}}-y\frac{d^{2}z}{dx^{2}}\right]-2x\left[z\frac{\,dy}{\,dx}-y\frac{\,dz}{\,dx}\right] \end{equation*}

\begin{equation*} +\left[n(n+1)-m(m+1)\right]y z =0 \end{equation*}

or,

\begin{equation*} \frac{\,d}{\,dx}\left[(1-x^{2})\left\{z\frac{\,dy}{\,dx}-y\frac{\,dz}{\,dx}\right\}\right]+(n-m)(n+m+1)yz =0 \end{equation*}

Integrating with respect to $x$ from -1 to +1, we get -

\begin{equation*} \left[(1-x^{2})\left(z\frac{dy}{dx}-y\frac{dz}{dx}\right)\right]^{1}_{-1}+(n-m)(n+m+1)\int\limits_{-1}^{+1}yz\,dx = 0 \end{equation*}

or,

\begin{equation} 0+(n-m)(n+m+1)\int\limits_{-1}^{+1}yz\,dx = 0\tag{4.5.29} \end{equation}

or,

\begin{equation*} \int\limits_{-1}^{+1}P_{m}(x)P_{n}(x)\,dx=0\quad \text{if}\quad m \neq n. \end{equation*}

since $P_{n}(x)$ satisfies equation (4.5.27) and $P_{m}(x)$ satisfies equation (4.5.28).

\begin{equation} \therefore \quad \int\limits_{-1}^{+1}P_{m}(x)P_{n}(x)\,dx=0\tag{4.5.30} \end{equation}

If $m=n\text{,}$ then equation (4.5.29) has zero on both its sides and hence we can not get the value of integral for $m=n$ from this equation. To calculate the integral in such a case we may proceed from generating function as $(1-2xz+z^{2})^{-\frac{1}{2}}=\sum z^{n}P_{n}(x)\text{.}$ Squaring both sides, we get -

\begin{equation*} (1-2xz+z^{2})^{-1}=\sum [z^{n}P_{n}(x)]^{2}=\sum[P_{n}(x)]^{2}z^{2n} \end{equation*}

Integrating with respect to $x$ from -1 to +1, we get -

\begin{equation*} \int\limits_{-1}^{+1}\frac{dx}{(1-2xz+z^{2})}=\sum\int\limits_{-1}^{+1}[P_{n}(x)]^{2}z^{2n}\,dx \end{equation*}

or,

\begin{equation*} \sum z^{2n} \int\limits_{-1}^{+1}[P_{n}(x)]^{2}\,dx=\int\limits_{-1}^{+1}\frac{dx}{(1-2xz+z^{2})}\times\frac{-2z}{-2z} \end{equation*}

\begin{equation*} =-\frac{1}{2z}\left[\log(1-2xz+z^{2})\right]_{-1}^{+1} =-\frac{1}{2z}\log\frac{1-2z+z^{2}}{1+2z+z^{2}} \end{equation*}

\begin{equation*} =-\frac{1}{2z}\log\left(\frac{1-z}{1+z}\right)^{2} =\frac{1}{z}\log\left(\frac{1+z}{1-z}\right) =\frac{1}{z}[\log(1+z)-\log(1-z)] \end{equation*}

\begin{equation*} =\frac{1}{z}\left[(z-\frac{z^{2}}{2}+\frac{z^{3}}{3}-\frac{z^{4}}{4}+\cdots)-(-z-\frac{z^{2}}{2}-\frac{z^{3}}{3}-\frac{z^{4}}{4}-\cdots)\right] \end{equation*}

\begin{equation*} =\frac{2}{z}\left[z+\frac{z^{3}}{3}+\frac{z^{5}}{5}+\cdots\right] \end{equation*}

\begin{equation*} =\frac{2}{z}\left[\sum\limits_{n=0}^{\infty}\frac{z^{2n+1}}{2n+1}\right]=2\sum \frac{z^{2n}}{2n+1} \end{equation*}

\begin{equation*} \therefore \quad \sum z^{2n}\int\limits_{-1}^{+1}[P_{n}(x)]^{2}\,dx=2\sum\frac{z^{2n}}{2n+1}. \end{equation*}

Equating the coefficient of $z^{2n}$ on both sides, we get -

\begin{equation} \int\limits_{-1}^{+1}[P_{n}(x)]^{2}\,dx=\frac{2}{2n+1}\tag{4.5.31} \end{equation}

Now, from equation (4.5.30) and equation (4.5.31), we have

\begin{equation*} \int\limits_{-1}^{+1}P_{m}(x)\cdot P_{n}(x)\,dx =\frac{2}{2n+1}\delta_{m,n}; \end{equation*}

where

\begin{equation*} \delta_{m,n}=\begin{cases} 0 \quad \text{for}\quad m \neq n \\ 1 \quad \text{for}\quad m=n \end{cases} \end{equation*}

Subsubsection 4.5.1.7 The Associated Legendre’s Polynomials

Laplace’s equation in spherical polar coordinates is given by $\nabla^{2}\psi=0\text{.}$ Or,

\begin{equation*} \frac{1}{r^{2}\sin\theta}\left[\sin\theta\frac{\partial}{\partial r}\left(r^{2}\frac{\partial\psi}{\partial r}\right)+\frac{\partial}{\partial \theta}\left(\sin\theta\frac{\partial\psi}{\partial \theta}\right)+\frac{1}{\sin\theta}\frac{\partial^{2}\psi}{\partial\phi^{2}}\right]=0 \end{equation*}

[see Laplacian in spherical curvilinear coordinates Subsection 1.6.6] or,

\begin{equation*} \frac{1}{r^{2}}\frac{\partial}{\partial r}\left(r^{2}\frac{\partial\psi}{\partial r}\right)+\frac{1}{r^{2}}\left[\frac{1}{\sin\theta}\frac{\partial}{\partial \theta}\left(\sin\theta\frac{\partial\psi}{\partial \theta}\right)+\frac{1}{\sin^{2}\theta}\frac{\partial^{2}\psi}{\partial\phi^{2}}\right]=0 \end{equation*}

or,

\begin{equation} \left[\nabla_{r}^{2}+\frac{1}{r^{2}}\nabla^{2}_{\theta,\phi}\right]\psi(r,\theta,\phi) =0\tag{4.5.32} \end{equation}

On solving the potential problems some solution yields to relate the Legendre’s Polynomials called associated Legendre’s polynomials. Let

\begin{equation} \psi(r,\theta,\phi)=R(r)Y(\theta,\phi)\tag{4.5.33} \end{equation}

be the solution of equation (4.5.32). Then

\begin{equation*} \nabla_{r}^{2}\left[R(r)Y(\theta,\phi)\right]+\frac{1}{r^{2}}\nabla^{2}_{\theta,\phi}\left[R(r)Y(\theta,\phi)\right]=0 \end{equation*}

or,

\begin{equation*} r^{2}\nabla_{r}^{2}\left[R(r)Y(\theta,\phi)\right]=-\nabla^{2}_{\theta,\phi}\left[R(r)Y(\theta,\phi)\right] \end{equation*}

Dividing by $R(r)Y(\theta,\phi)$ on both sides, we get -

\begin{equation} \frac{r^{2}}{R}\nabla_{r}^{2}R(r)=-\frac{1}{Y}\nabla^{2}_{\theta,\phi}Y(\theta,\phi) =\lambda \quad \text{(say)}\tag{4.5.34} \end{equation}

In equation (4.5.34) Left Hand Side (LHS) is a function of $r$ only and Right Hand Side (RHS) is a function of $(\theta,\phi)$ only hence such an equation must be equal to some constant, say $\lambda \text{.}$

\begin{equation} \nabla^{2}_{\theta,\phi}Y(\theta,\phi)+\lambda Y(\theta,\phi) =0\tag{4.5.35} \end{equation}

Assume

\begin{equation} Y(\theta,\phi) = \Theta(\theta)\Phi(\phi) \tag{4.5.36} \end{equation}

be the solution of equation (4.5.35), then

\begin{equation*} \nabla^{2}_{\theta,\phi}\Theta(\theta)\Phi(\phi)+\lambda \Theta(\theta)\Phi(\phi) =0 \end{equation*}

or,

\begin{equation*} \left[\frac{1}{\sin\theta}\frac{\partial}{\partial \theta}\left(\sin\theta\frac{\partial}{\partial \theta}\right)+\frac{1}{\sin^{2}\theta}\frac{\partial^{2}}{\partial \phi^{2}}\right]\Theta(\theta)\Phi(\phi)+\lambda \Theta(\theta)\Phi(\phi)=0 \end{equation*}

or,

\begin{equation*} \left[\nabla^{2}_{\theta}+\lambda\right]\Theta\cdot\Phi =-\frac{1}{\sin^{2}\theta}\nabla^{2}_{\phi}(\Theta\cdot\Phi) \end{equation*}

or,

\begin{equation*} \sin^{2}\theta\left[\nabla^{2}_{\theta}+\lambda\right]\Theta\cdot\Phi=-\nabla^{2}_{\phi}(\Theta\cdot\Phi) \end{equation*}

or,

\begin{equation} \sin^{2}\theta\frac{\nabla^{2}_{\theta}\Theta}{\Theta}+\sin^{2}\theta\lambda=-\frac{\nabla^{2}_{\phi}\Phi}{\Phi}=m^{2}\quad\text{(say)}\tag{4.5.37} \end{equation}

LHS is a function of $\theta$ only and RHS is a function of $\phi$ only which is possible only if they are equal to some constant. From equation (4.5.37), we have

\begin{equation*} \frac{\nabla^{2}_{\phi}\Phi}{\Phi}=-m^{2} \end{equation*}

or,

\begin{equation} \frac{\partial^{2}\Phi}{\partial\phi^{2}}+m^{2}\Phi =0\tag{4.5.38} \end{equation}

[Azimuthal equation]

\begin{equation} \therefore \quad \Phi=c_{1}e^{im\phi}+c_{2}e^{-im\phi}=A\cos m\phi+B\sin m\phi\tag{4.5.39} \end{equation}

From equation (4.5.37)

\begin{equation*} \sin^{2}\theta\frac{\nabla^{2}_{\theta}\Theta}{\Theta}+\lambda\sin^{2}\theta=m^{2} \end{equation*}

or,

\begin{equation*} \nabla^{2}_{\theta}\Theta+\left(\lambda-\frac{m^{2}}{\sin^{2}\theta}\right)=0 \end{equation*}

or,

\begin{equation} \frac{1}{\sin\theta}\frac{\partial}{\partial \theta}\left(\sin\theta\frac{\partial\Theta}{\partial \theta}\right)+\left(\lambda-\frac{m^{2}}{\sin^{2}\theta}\right)\Theta =0 \tag{4.5.40} \end{equation}

[Theta equation]

To solve equation (4.5.40), introduce a new variable $x=\cos\theta$ so that $\frac{\,dx}{\,d\theta}=-\sin\theta$ and $\sin^{2}\theta =1-x^{2}\text{.}$ Therefore, from equation (4.5.40), we have

\begin{equation*} -\frac{\,d\theta}{\,dx}\cdot\frac{\,d}{\,d\theta}\left(-\frac{\,dx}{\,d\theta}\frac{\,d\Theta}{\,d\theta}\right)+\left(\lambda-\frac{m^{2}}{1-x^{2}}\right)\Theta =0 \end{equation*}

or,

\begin{equation*} \frac{\,d}{\,dx}\left(\frac{\,dx}{\,d\theta}\cdot\frac{\,d\Theta}{\,dx}\cdot\frac{\,dx}{\,d\theta}\right)+\left(\lambda-\frac{m^{2}}{1-x^{2}}\right)\Theta =0 \end{equation*}

or,

\begin{equation*} \frac{\,d}{\,dx}\left(\sin^{2}\theta\frac{\,d\Theta}{\,dx}\right)+\left(\lambda-\frac{m^{2}}{1-x^{2}}\right)\Theta =0 \end{equation*}

or,

\begin{equation*} \frac{\,d}{\,dx}\left[(1-x^{2})\frac{\,d\Theta}{\,dx}\right]+\left(\lambda-\frac{m^{2}}{1-x^{2}}\right)\Theta =0 \end{equation*}

\begin{equation} (1-x^{2})\frac{d^{2}\Theta}{dx^{2}}-2x\frac{d\Theta}{dx}+\left(\lambda-\frac{m^{2}}{1-x^{2}}\right)\Theta =0\tag{4.5.41} \end{equation}

Equation (4.5.41) is known as associated Legendre’s differential equation. If $m=0\text{,}$ equation (4.5.41) becomes Legendre’s differential equation.

\begin{equation} (1-x^{2})\frac{d^{2}\Theta}{dx^{2}}-2x\frac{d\Theta}{dx}+\lambda\Theta =0 \tag{4.5.42} \end{equation}

If $\Theta=y$ and $\lambda=n(n+1) $ then equation (4.5.42) is Legendre’s differential equation in the cartesian system. From equation (4.5.34), we get -

\begin{equation*} \nabla^{2}_{r}R(r)=\frac{\lambda}{r^{2}}R(r) \end{equation*}

or,

\begin{equation*} \frac{1}{r^{2}}\frac{d}{dr}\left(r^{2}\frac{dR}{dr}\right)-\frac{\lambda}{r^{2}}R(r)=0 \end{equation*}

or,

\begin{equation} r^{2}\frac{d^{2}R}{dr^{2}}+2r\frac{dR}{dr}-\lambda R=0\tag{4.5.43} \end{equation}

[Radial Equation].

Prev Top Next

Mathematical Methods of Physics: For Undergraduate

Subsection 4.5.1 Legendre’s Differential Equation

Case I.

Case II.

Note:.

Subsubsection 4.5.1.1 Rodrigue’s Formula

Proof.

Subsubsection 4.5.1.2 Legendre Polynomials

Python Code:.

Subsubsection 4.5.1.3 Generating Function for \(P_{n}(x)\)

Proof.

Subsubsection 4.5.1.4 Some Important Results

Subsubsection 4.5.1.5 Recurrence Relations for \(P_{n}(x)\)

Subsubsection 4.5.1.6 Orthogonality of Legendre’s Polynomials

Proof.

Subsubsection 4.5.1.7 The Associated Legendre’s Polynomials