General Solution to Chebyshev's Differential Equation/Lemma

Lemma for General Solution to Chebyshev's Differential Equation

Chebyshev's differential equation:

$\quad \ds \paren {1 - x^2} \frac {\d^2 y} {\d x^2} - x \frac {\d y} {\d x} + n^2 y = 0$

can be depressed down to:

$\ds \frac {\d^2 y} {\d \theta^2} + n^2 y = 0$

by substituting either:

$x = \sin \theta$

or:

$x = \cos \theta$

Proof 1

Let:

\(\ds x\)

\(=\)

\(\ds \cos \theta\)

\(\ds \leadsto \ \ \)

\(\ds \d x\)

\(=\)

\(\ds -\sin \theta \rd \theta\)

\(\ds \leadsto \ \ \)

\(\ds \frac {\d \theta} {\d x}\)

\(=\)

\(\ds -\frac 1 {\sin \theta}\)

\(\ds \leadsto \ \ \)

\(\ds \frac {\d^2 \theta} {\d x \d \theta}\)

\(=\)

\(\ds \frac {\cos \theta} {\sin^2 \theta}\)

Derivative of Cosecant Function

Then:

\(\ds 0\)

\(=\)

\(\ds \paren {1 - \cos^2 \theta} \frac {\d } {\d x} \paren {\frac {\d y} {\d \theta} \frac {\d \theta} {\d x} } - \cos \theta \frac {\d y} {\d \theta} \frac {\d \theta} {\d x} + n^2 y\)

substituting $x = \cos \theta$ and Chain Rule for Derivatives into Chebyshev's differential equation

\(\ds \)

\(=\)

\(\ds \paren {1 - \cos^2 \theta} \frac {\d \theta} {\d x} \map {\frac \d {\d \theta} } {\frac {\d y} {\d \theta} \frac {\d \theta} {\d x} } - \cos \theta \frac {\d y} {\d \theta} \frac {\d \theta} {\d x} + n^2 y\)

Chain Rule for Derivatives

\(\ds \)

\(=\)

\(\ds \paren {1 - \cos^2 \theta} \dfrac {\d \theta} {\d x} \paren {\frac {\d^2 y} {\d \theta^2} \dfrac {\d \theta} {\d x} + \frac {\d y} {\d \theta} \frac {\d^2 \theta} {\d x \d \theta} } - \cos \theta \frac {\d y} {\d \theta} \frac {\d \theta} {\d x} + n^2 y\)

Product Rule for Derivatives

\(\ds \)

\(=\)

\(\ds \paren {\sin^2 \theta} \paren {-\frac 1 {\sin \theta} } \paren {\frac {\d^2 y} {\d \theta^2} \paren {-\frac 1 {\sin \theta} } + \frac {\d y} {\d \theta} \frac {\cos \theta} {\sin^2 \theta} } - \cos \theta \frac {\d y} {\d \theta} \paren {-\frac 1 {\sin \theta} } + n^2 y\)

Sum of Squares of Sine and Cosine and substituting $\ds \frac {\d \theta} {\d x} = -\frac 1 {\sin \theta}$ and $\ds \frac {\d^2 \theta} {\d x \d \theta} = \frac {\cos \theta} {\sin^2 \theta}$

\(\ds \)

\(=\)

\(\ds \frac {\d^2 y} {\d \theta^2} - \frac {\d y} {\d \theta} \frac {\cos \theta } {\sin \theta} + \frac {\d y} {\d \theta} \frac {\cos \theta } {\sin \theta} + n^2 y\)

simplification: cancelling $\sin^2 \theta$ from top and bottom

\(\ds \leadsto \ \ \)

\(\ds 0\)

\(=\)

\(\ds \frac {\d^2 y} {\d \theta^2} + n^2 y\)

$\blacksquare$

Proof 2

Let:

\(\ds x\)

\(=\)

\(\ds \sin \theta\)

\(\ds \leadsto \ \ \)

\(\ds \d x\)

\(=\)

\(\ds \cos \theta \rd \theta\)

\(\ds \leadsto \ \ \)

\(\ds \frac {\d \theta} {\d x}\)

\(=\)

\(\ds \frac 1 {\cos \theta}\)

\(\ds \leadsto \ \ \)

\(\ds \frac {\d^2 \theta} {\d x \d \theta}\)

\(=\)

\(\ds \frac {\sin \theta} {\cos^2 \theta}\)

Derivative of Secant Function

Then:

\(\ds 0\)

\(=\)

\(\ds \paren {1 - \sin^2 \theta} \frac {\d } {\d x} \paren {\frac {\d y} {\d \theta} \frac {\d \theta} {\d x} } - \sin \theta \frac {\d y} {\d \theta} \frac {\d \theta} {\d x} + n^2 y\)

substituting $x = \sin \theta$ and Chain Rule for Derivatives into Chebyshev's differential equation

\(\ds \)

\(=\)

\(\ds \paren {1 - \sin^2 \theta} \frac {\d \theta} {\d x} \map {\frac \d {\d \theta} } {\frac {\d y} {\d \theta} \frac {\d \theta} {\d x} } - \sin \theta \frac {\d y} {\d \theta} \frac {\d \theta} {\d x} + n^2 y\)

Chain Rule for Derivatives

\(\ds \)

\(=\)

\(\ds \paren {1 - \sin^2 \theta} \dfrac {\d \theta} {\d x} \paren {\frac {\d^2 y} {\d \theta^2} \dfrac {\d \theta} {\d x} + \frac {\d y} {\d \theta} \frac {\d^2 \theta} {\d x \d \theta} } - \sin \theta \frac {\d y} {\d \theta} \frac {\d \theta} {\d x} + n^2 y\)

Product Rule for Derivatives

\(\ds \)

\(=\)

\(\ds \paren {\cos^2 \theta} \paren {\frac 1 {\cos \theta} } \paren {\frac {\d^2 y} {\d \theta^2} \paren {\frac 1 {\cos \theta} } + \frac {\d y} {\d \theta} \frac {\sin \theta} {\cos^2 \theta} } - \sin \theta \frac {\d y} {\d \theta} \paren {\frac 1 {\cos \theta} } + n^2 y\)

Sum of Squares of Sine and Cosine and substituting $\ds \frac {\d \theta} {\d x} = \frac 1 {\cos \theta}$ and $\ds \frac {\d^2 \theta} {\d x \d \theta} = \frac {\sin \theta} {\cos^2 \theta}$

\(\ds \)

\(=\)

\(\ds \frac {\d^2 y} {\d \theta^2} - \frac {\d y} {\d \theta} \frac {\sin \theta } {\cos \theta} + \frac {\d y} {\d \theta} \frac {\sin \theta } {\cos \theta} + n^2 y\)

simplification: cancelling $\cos^2 \theta$ from top and bottom

\(\ds \leadsto \ \ \)

\(\ds 0\)

\(=\)

\(\ds \frac {\d^2 y} {\d \theta^2} + n^2 y\)

$\blacksquare$