# Legendre Differential Equation(#4) : A friendly introduction

When you are working with Spherical harmonics, then the Legendre Differential Equation does not appear in its natural form i.e

$(1-x^2)y^{''} -2xy^{'} + l(l+1)y = 0$

Instead, it appears in this form:

$\frac{d^2 y}{d\theta^2} + \frac{cos\theta}{sin\theta} \frac{dy}{d\theta} + l(l+1)y = 0$

It seems daunting but the above is the same as the LDE. We can arrive at it by taking $x = cos(\theta)$ and proceeding as follows:

$\frac{dy}{dx} = \frac{dy}{d(cos\theta)} = \frac{-1}{sin\theta}\frac{dy}{d \theta}$

$\frac{d^2 y}{dx^2} = \frac{d}{d(cos\theta)}\left( \frac{-1}{sin\theta}\frac{dy}{d \theta} \right) = \frac{-1}{sin\theta}\frac{d}{d \theta}\left( \frac{-1}{sin\theta}\frac{dy}{d \theta} \right)$

Now, applying chain rule, we obtain that

$\frac{d^2 y}{dx^2} = \frac{-1}{sin\theta} \left( \frac{-1}{sin\theta} \frac{d^2 y}{d\theta^2} - \frac{cos\theta}{sin^2 \theta} \frac{dy}{d\theta} \right)$

Now simplifying the above expression, we obtain that:

$\frac{d^2 y}{dx^2} = \frac{1}{sin^2\theta} \left( \frac{d^2 y}{d\theta^2} + \frac{cos\theta}{sin\theta} \frac{dy}{d\theta} \right)$

Plugging in the values of $\frac{dy}{dx}$ and $\frac{d^2 y}{dx^2}$ into the Legendre Differential Equation,

$(1-x^2)y^{''} -2xy^{'} + l(l+1)y = 0$

$(1-cos^2 \theta)y^{''} -2cos\theta y^{'} + l(l+1)y = 0$

$\frac{1- cos^2 \theta}{sin^2 \theta} \left( \frac{d^2 y}{d\theta^2} + \frac{cos\theta}{sin\theta}\frac{dy}{d\theta} \right) +\frac{2cos\theta}{sin\theta} \frac{dy}{d\theta} + l(l+1)y = 0$

Now if we do some algebra and simplify the trigonometric identities, we will arrive at the following expression for the Legendre Differential Equation:

$\frac{d^2 y}{d\theta^2} + \frac{cos\theta}{sin\theta} \frac{dy}{d\theta} + l(l+1)y = 0$

If we take the solution for the LDE as $f(x)$, then the solution to the LDE in the above form is merely $f(cos\theta)$.