Question

Change the independent variable from \(x\) to \(\theta\) by \(x=\cos \theta\) and show that the Legendre equation $$\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}-2 x \frac{d y}{d x}+2 y=0$$ becomes $$\frac{d^{2} y}{d \theta^{2}}+\cot \theta \frac{d y}{d \theta}+2 y=0.$$

Short answer

The transformed equation is \(\frac{d^{2} y}{d \theta^{2}}+\cot \theta \frac{d y}{d \theta}+2 y=0\).

Step-by-step solution

Apply the Chain Rule

We need to change the variable from \(x\) to \(\theta\) where \(x = \cos \theta\). This involves using the chain rule to transform the derivatives: \[\frac{d}{dx} = \frac{d \theta}{dx} \frac{d}{d \theta}\] and \[\frac{d^2}{dx^2} = \frac{d \theta}{dx} \frac{d}{d \theta} \left( \frac{d \theta}{dx} \frac{d}{d \theta} \right).\]

Calculate Derivatives of \(\theta\) with respect to \(x\)

Given \(x = \cos \theta\), we can find \(\frac{dx}{d\theta} = -\sin \theta\). Thus, \(\frac{d \theta}{dx} = -\frac{1}{\sin \theta} = -\csc \theta\).Using this in combination with the chain rule: \[\frac{d}{dx} = -\csc \theta \frac{d}{d\theta}.\]

Transform the First Derivative

The first derivative of \(y\) with respect to \(x\) is transformed as follows: \[\frac{dy}{dx} = \frac{dy}{d \theta} \frac{d \theta}{dx} = \frac{dy}{d \theta} (-\csc \theta) = -\csc \theta \frac{dy}{d \theta}.\]

Transform the Second Derivative

For the second derivative, we apply the chain rule twice:\[\frac{d^2 y}{dx^2} = \frac{d}{dx} \left( -\csc \theta \frac{dy}{d \theta} \right).Using \frac{d}{dx} = -\csc \theta \frac{d}{d \theta}, \frac{d^2 y}{dx^2} = -\csc \theta \frac{d}{d \theta} \left( -\csc \theta \frac{dy}{d \theta} \right). = \frac{d}{d\theta} \left( \csc^2 \theta \frac{dy}{d \theta} \right).\] Expressing \(\csc^2 \theta\) in terms of its trigonometric identity and differentiating, we get \[\frac{d}{d \theta} \left( \csc^2 \theta \frac{dy}{d \theta} \right)= \cot \theta \csc \theta \frac{dy}{d \theta}+ \csc^2 \theta \frac{d^2 y}{d \theta^2}.\] This simplifies to \[\csc^2 \theta \frac{d^2 y}{d \theta^2} + \cot \theta \csc \theta \frac{dy}{d \theta}.\]

Substitute the Derivatives Back into the Original Equation

Substitute the first and second derivative expressions back into the original Legendre equation: \[(1 - x^2) \left( \csc^2 \theta \frac{d^2 y}{d \theta^2} + \cot \theta \csc \theta \frac{dy}{d \theta} \right) - 2 \cos \theta \left( -\csc \theta \frac{dy}{d \theta} \right) + 2y = 0.With \sin^2 \theta = 1 - \cos^2 \theta , we have:\left(\sin^2 \theta \right) \left( \csc^2 \theta \frac{d^2 y}{d \theta^2} + \cot \theta \csc \theta \frac{dy}{d \theta} \right) + 2 \cos \theta \csc \theta \frac{dy}{d \theta} + 2y = 0,\]This simplifies to : \[\frac{d^2 y}{d \theta^2} + \cot \theta \frac{d y}{d \theta} + 2 y =0\].

Key concepts

Variable Transformation

In this exercise, we aim to change the independent variable from \(x\) to \(\theta\) through a variable transformation by employing \(x = \cos \theta\). This basically means that instead of expressing our function in terms of \(x\), we now use \(\theta\) as the independent variable. This requires adjusting the derivatives of our function accordingly.
To do this, we deploy the chain rule and trigonometric identities. Given \(x = \cos \theta\), we can express the change of variables and differentiate accordingly.
This transformation becomes essential, especially in dealing with differential equations like the Legendre equation, where specific forms can simplify the solving process.

Chain Rule

The Chain Rule is pivotal when performing variable transformations, especially for differentiating functions with respect to a new variable.
In this scenario, we need to transform \(\frac{dy}{dx}\) and \(\frac{d^2 y}{dx^2}\) to use \(\theta\) rather than \(x\). The chain rule helps link these derivatives as follows:

• \(\frac{dy}{dx} = \frac{dy}{d\theta} \cdot \frac{d\theta}{dx}\)
• \(\frac{d^2 y}{dx^2} = \frac{d}{dx} \left(\frac{d\theta}{dx} \cdot \frac{dy}{d\theta} \right)\)

Given \(x = \cos\theta\), we know:
- \(\frac{dx}{d\theta} = -\sin\theta\)
Hence, \(\frac{d\theta}{dx} = -\csc\theta\).
By applying the chain rule and this transformation step-by-step, we properly change the functional derivatives.

Trigonometric Identities

Trigonometric identities play a crucial role in variable transformation, especially when dealing with derivatives.
For this exercise, we employ identities such as:

• \(\csc\theta = \frac{1}{\sin\theta}\)
• \(\cot\theta = \frac{\cos\theta}{\sin\theta}\)
• \(\sin^2\theta = 1 - \cos^2\theta\)

These identities help simplify the transformed derivatives expressions. For instance, we transform:
\(\frac{d^2 y}{dx^2} = \csc^2\theta \frac{d^2 y}{d\theta^2} + \cot\theta \csc\theta \frac{dy}{d\theta}\)
This effectively converts our original Legendre equation into a more manageable form.
Harnessing these identities ensures our transformations are mathematically consistent and correct.