Legendre Polynomials

• The Legendre polynomials $P_\ell(x)$, sometimes called Legendre functions of the first kind, Legendre coefficients, or zonal harmonics are solutions to the Legendre differential equation.
• The Legendre polynomials satisfy the second-order differential equation.

  $$\boxed{(1-x^{2})\frac{\mathrm{d}^{2} y}{\mathrm{d} x^{2}}-2x\frac{\mathrm{d} y}{\mathrm{d} x}+\ell(\ell+1)y=0}$$

  where $y=P_\ell(x)$
• This equation has two regular singular points $x =\pm1$ where the leading coefficient $(1 - x^2)$ vanishes.
• Solutions of Legendre equations are Legendre polynomials

  $$\boxed{P_{\ell}(x)=\frac{1}{2^{\ell}\ell!}\left( \frac{\mathrm{d}}{\mathrm{d} x}\right)^{\ell}(x^2-1)^{\ell}}$$

• If $\ell$ ( $0\le \ell \le \infty$) is an integer, they are polynomials and make up an infinite set of functions of the variable x.
• We therefore have a function $P_0(x)$, another function $P_\ell(x)$, and an infinite number of additional functions belonging to the set of Legendre polynomials.
• Introduce a (generating) function $\Phi(x,h)$ of two variables, known as a generating function for the definition of the Legendre polynomials.

  $$\boxed{\Phi(x,h)=(1-2xh+h^2)^{-1/2}}$$

  • The first variable, x, is the same variable that appears as the argument of the Legendre polynomials.
  • The second variable, h, is an auxiliary variable with no particular meaning.

• Think of $\Phi$ as a function of a single variable h ( $\Phi=\Phi(h)$) and expand as a Taylor expansion in powers of h
  

  $$\Phi(h) = \Phi(0)+\frac{d\Phi}{dh}\bigg\rvert_{h=0}h+\frac{1}{2!}\frac{d^2\... ...gg\rvert_{h=0}h^2 +\frac{1}{3!}\frac{d^3\Phi}{dh^3}\bigg\rvert_{h=0}h^3+ \ldots$$

  $$= \sum_{\ell=0}^{\infty} \frac{1}{\ell!}\frac{d^\ell\Phi}{dh^\ell}\bigg\rvert_{h=0} h^\ell$$

  
  
• Restore the x-dependence of the generating function. This doesn't change the general appearance of the Taylor expansion but written as partial derivatives instead of total derivatives.

  $$\Phi(x,h)=\sum_{\ell=0}^{\infty} \frac{1}{\ell!}\frac{\partial^\e... ...partial h^\ell}\bigg\rvert_{h=0} h^\ell=\sum_{\ell=0}^{\infty} P_\ell(x) h^\ell$$

• Right hand side of this equation is the formal definition of the Legendre polynomials. They are identified as the coefficients in the Taylor expansion of the generating function about $h=0$.

• Let us use this equation to calculate the first few polynomials.
  • For $\ell=0$ we are instructed to take no derivatives, and to evaluate the generating function at $h=0$. This gives $P_0(x)=1$; the zeroth polynomial is actually a constant.
  • Moving on to $\ell=1$, we must differentiate $\Phi$ once with respect to h. Evaluating this at $h=0$ and dividing by $1!=1$ gives $P_1(x)=x$.
  • For $\ell=2$ we differentiate $\Phi$ twice. Evaluating this at $h=0$ and dividing by $2!=2$ produces $P_2(x)=1/2(3x^2-1)$. We can just keep going like this, and generate any number of polynomials.
• When $\ell$ is even, $P_\ell(x)$ contains only even powers of x, starting with $x^\ell$ and ending with $x^0$.
• When $\ell$ is odd, $P_\ell(x)$ contains only odd powers of x, starting with $x^\ell$ and ending with $x$.
• $P_\ell(x)$ is an even function of $x$ when $\ell$ is even, and an odd function of $x$ when $\ell$ is odd.

• The first few Legendre polynomials are given by
  

  $$P_0 = 1$$

  $$P_1 = x$$

  $$P_2 = \frac{1}{2}(3x^2-1)$$

  $$P_3 = \frac{1}{2}(5x^3-3x)$$

  $$P_4 = \frac{1}{8}(35x^4-30x^2+3)$$

  $$P_5 = \frac{1}{8}(63x^5-70x^3+15x)$$

  $$P_6 = \frac{1}{16}(231x^6-315x^4+105x^2-5)$$

  
  
• Recursion relation:

  $$\boxed{P_{\ell}(x)=\frac{1}{\ell}[(2\ell-1)xP_{\ell-1}(x)-(\ell-1)P_{\ell-2}(x)]}$$

Example py-file: The program to find first 6 Legendre polynomials: myLegendre.py

Figure 5.16: First 6 Legendre Polynomials $P_{\ell }(x)$ with Recursion Relation: $P_{\ell }(x)=\frac {1}{\ell }[(2\ell -1)xP_{\ell -1}(x)-(\ell -1)P_{\ell -2}(x)]$.

Figure 5.17: Plot of first 6 $P_{\ell }(x)$.