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) .

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