📂Functions

Orthogonality of Legendre Polynomials

Theorem

In the interval $[-1,\ 1]$, Legendre polynomials form an orthogonal set.

$$\int_{-1}^{1} P_{l}(x)P_{m}(x) dx =\frac{2}{2l+1}\delta_{lm} \quad (l, m = 0, 1, 2, \dots)$$

Proof

Case 1: $l \ne m$

Legendre Differential Equation

The following differential equation is called the Legendre differential equation.

$$\dfrac{d}{d x}\left[ (1-x)^{2} \dfrac{d y}{d x} \right] +l(l+1)y = 0$$

Since Legendre polynomials are solutions to the Legendre differential equation, they satisfy the above equation. Substituting $P_{l}$ and $P_{m}$ into the equation, we get,

$$\begin{align*} \dfrac{d}{dx} \left[ (1-x^2)P^{\prime}_{l}(x) \right] + l(l+1)P_{l}(x) &= 0 \\ \dfrac{d}{dx} \left[ (1-x^2)P^{\prime}_{m}(x) \right] + m(m+1)P_{m}(x) &= 0 \end{align*}$$

Multiplying both equations by $P_{m}(x)$ and $P_{l}(x)$ respectively and subtracting, we obtain the following.

$$\begin{equation} P_{m}\dfrac{d}{dx}\left[ (1-x^2)P^{\prime}_{l} \right] - P_{l}\dfrac{d}{dx}[(1-x^2)P^{\prime}_{m}] + [l(l+1)-m(m+1)]P_{l}P_{m} = 0 \end{equation}$$

Meanwhile, the following equation holds.

$$\begin{align*} & \dfrac{d}{dx}[(1-x^2)(P_{m}P^{\prime}_{l}-P_{l}P^{\prime}_{m})] \\ &= \dfrac{d}{dx}[{\color{blue}(1-x^2)P^{\prime}_{l}}P_{m}] -\dfrac{d}{dx}[{\color{blue}(1-x^2)P^{\prime}_{m}}P_{l}] \end{align*}$$

Thinking of the blue-painted part as one function and expanding the equation with the product rule gives,

$$\begin{align*} & \dfrac{d}{dx}[(1-x^2)P^{\prime}_{l}]P_{m}+(1-x^2)P^{\prime}_{l}P^{\prime}_{m}-\dfrac{d}{dx}[(1-x^2)P^{\prime}_{m}]P_{l}-(1-x^2)P^{\prime}_{m}P^{\prime}_{l} \\ &= \dfrac{d}{dx}[(1-x^2)P^{\prime}_{l}]P_{m}-\dfrac{d}{dx}[(1-x^2)P^{\prime}_{m}]P_{l} \end{align*}$$

This is the same as the first two terms of $(1)$. Thus, $(1)$ can be organized as follows.

$$\dfrac{d}{dx}[(1-x^2)(P_{m}P^{\prime}_{l}-P_{l}P^{\prime}_{m})]+ [l(l+1)-m(m+1)]P_{l}P_{m}=0$$

Integrating both sides over the interval $[-1, 1]$ yields,

$$(1-x^2)(P_{m}P^{\prime}_{l}-P_{l}P^{\prime}_{m})\Big|_{-1}^{1} +[l(l+1)-m(m+1)]\int_{-1}^{1}P_{l}(x)P_{m}(x)dx=0$$

The first term is $(1-x^2)\Big|_{x = \pm 1}=0$, so it becomes $0$. Due to the condition of $l, m$, the constant in front of the integration in the second term absolutely cannot be $0$. Therefore, we get the following.

$$\int_{-1}^{1}P_{l}(x)P_{m}(x)dx=0$$

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Case 2: $l = m$

Recurrence relation of Legendre polynomials

$$lP_{l}(x) = xP^{\prime}_{l}(x) - P^{\prime}_{l-1}(x)$$

Multiplying both sides of the above equation by $P_{l}(x)$ and integrating gives,

$$\begin{equation} l\int_{-1}^{1}[P_{l}(x)]^{2} dx= \int_{-1}^{1}xP_{l}(x)P^{\prime}_{l}(x)dx -\int_{-1}^{1} P_{l}(x)P^{\prime}_{l-1}(x)dx. \end{equation}$$

Here, $P^{\prime}_{l-1}(x)$ is a polynomial of degree $l-2$, and since Legendre polynomials are orthogonal to polynomials of lower degree, the last term on the right-hand side is $0$. The first term on the right-hand side can be solved with integration by parts.

$$\begin{align} \int_{-1}^{1}xP_{l}(x)P^{\prime}_{l}(x)dx &= \int_{-1}^{1}\frac{ x}{2}[2P_{l}(x)P^{\prime}_{l}(x)]dx \nonumber \\ &= \frac{ x}{2}[P_{l}(x)]^{2}\bigg|_{-1}^{1}-\frac{1}{2}\int_{-1}^{1}[P_{l}(x)]^{2}dx \nonumber \\ &= 1-\frac{1}{2}\int_{-1}^{1}[P_{l}(x)]^{2}dx. \end{align}$$

The third equation is valid due to $P_{l}(1)=1$. Therefore, substituting $(3)$ into $(2)$ gives,

$$\begin{align*} && l\int_{-1}^{1}[P_{l}(x)]^{2} dx &= 1-\frac{1}{2}\int_{-1}^{1}[P_{l}(x)]^{2}dx \\ \implies && \int_{-1}^{1}[P_{l}(x)]^{2} dx &= \frac{2}{2l+1} \end{align*}$$

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