General Solution to Chebyshev's Differential Equation

From ProofWiki
Jump to navigation Jump to search

Theorem

Consider Chebyshev's differential equation:

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

where $n \in \N$.


The general solution to $(1)$ is given by:

$y = \begin {cases} A \map {T_n} x + B \sqrt {1 - x^2} \, \map {U_{n - 1} } x + C \map \cos {n \arcsin x} + D \map \sin {n \arcsin x} & : n = 1, 2, 3, \ldots \\ \\ A \arccos x + B \arcsin x + C & : n = 0 \end {cases}$

where:

$\map {T_n} x$ denotes the Chebyshev polynomial of the first kind of order $n$
$\map {U_n} x$ denotes the Chebyshev polynomial of the second kind of order $n$
$\size x < 1$


Proof

First a Lemma:

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:

$x = \cos \theta$

$\Box$


Let $n = 0$.

In our Lemma, we assumed that either $x = \sin \theta$ or $x = \cos \theta$.

Assuming $x = \cos \theta$, then:

\(\ds \leadsto \ \ \) \(\ds 0\) \(=\) \(\ds \frac {\d^2 y} {\d \theta^2}\)
\(\ds \leadsto \ \ \) \(\ds y\) \(=\) \(\ds A \theta + D\) Primitive of Constant twice
\(\ds \leadsto \ \ \) \(\ds y\) \(=\) \(\ds A \arccos x + D\) substituting $x = \cos \theta$, that is, $\theta = \arccos x$

Assuming $x = \sin \theta$, then:

\(\ds \leadsto \ \ \) \(\ds 0\) \(=\) \(\ds \frac {\d^2 y} {\d \theta^2}\)
\(\ds \leadsto \ \ \) \(\ds y\) \(=\) \(\ds B \theta + E\) Primitive of Constant twice
\(\ds \leadsto \ \ \) \(\ds y\) \(=\) \(\ds B \arcsin x + E\) substituting $x = \sin \theta$, that is, $\theta = \arcsin x$


From Linear Combination of Solutions to Homogeneous Linear 2nd Order ODE, we have that a linear combination of particular solutions to a homogeneous linear second order ODE is also a particular solution to that ODE.

Therefore:

\(\ds y\) \(=\) \(\ds \paren {A \arccos x + D} + \paren {B \arcsin x + E}\)
\(\ds \leadsto \ \ \) \(\ds y\) \(=\) \(\ds A \arccos x + B \arcsin x + C\) Linear Combination of Solutions to Homogeneous Linear 2nd Order ODE

$\Box$

In our Lemma, we assumed that either $x = \sin \theta$ or $x = \cos \theta$.

We note here that since $x = \sin \theta$ or $x = \cos \theta$, then from Real Sine Function is Bounded and Real Cosine Function is Bounded:

$\size x \le 1$

If $n = 0$ and $x = \pm 1$, then Chebyshev's differential equation reduces to:

$\pm \dfrac {\d y} {\d x} = 0$

From Derivative of Arccosine Function, we have:

$\dfrac {\d y} {\d x} = -\dfrac A {\sqrt {\paren {1 - x^2} } }$

From Derivative of Arcsine Function, we have:

$\dfrac {\d y} {\d x} = \dfrac B {\sqrt {\paren {1 - x^2} } }$

The derivative is not defined at $x = \pm 1$ under either assumption, therefore:

$\size x < 1$

$\Box$


Now let $n > 0$.

Assuming $x = \cos \theta$.

Then:

\(\ds 0\) \(=\) \(\ds \frac {\d^2 y} {\d \theta^2} + n^2 y\) Lemma: using $x = \cos \theta$
\(\ds \leadsto \ \ \) \(\ds y\) \(=\) \(\ds A \map \cos {n \theta} + B \map \sin {n \theta}\) Linear Second Order ODE: $y' ' + k^2 y = 0$
\(\ds \) \(=\) \(\ds A \map \cos {n \arccos x} + B \map \sin {n \arccos x}\) substituting $x = \cos \theta$, that is, $\theta = \arccos x$
\(\ds \) \(=\) \(\ds A \map \cos {n \arccos x} + B \map \sin {\arccos x} \, \dfrac {\map \sin {n \arccos x} } {\map \sin {\arccos x} }\) multiplying top and bottom by $\map \sin {\arccos x}$ and rearranging
\(\ds \) \(=\) \(\ds A \map \cos {n \arccos x} + B \sqrt {1 - \paren {\map \cos {\arccos x} }^2 } \, \dfrac {\map \sin {n \arccos x} } {\map \sin {\arccos x} }\) Sum of Squares of Sine and Cosine
\(\ds \leadsto \ \ \) \(\ds y\) \(=\) \(\ds A \map {T_n} x + B \sqrt {1 - x^2} \, \map {U_{n - 1} } x\) Definition of Chebyshev Polynomial of the First Kind and Definition of Chebyshev Polynomial of the Second Kind

Assuming $x = \sin \theta$.

Then:

\(\ds 0\) \(=\) \(\ds \frac {\d^2 y} {\d \theta^2} + n^2 y\) Lemma: using $x = \sin \theta$
\(\ds \leadsto \ \ \) \(\ds y\) \(=\) \(\ds C \map \cos {n \theta} + D \map \sin {n \theta}\) Linear Second Order ODE: $y' ' + k^2 y = 0$
\(\ds \) \(=\) \(\ds C \map \cos {n \arcsin x} + D \map \sin {n \arcsin x}\) substituting $x = \sin \theta$, that is, $\theta = \arcsin x$
\(\ds \leadsto \ \ \) \(\ds y\) \(=\) \(\ds C \map \cos {n \arcsin x} + D \map \sin {n \arcsin x}\)

Once again, from Linear Combination of Solutions to Homogeneous Linear 2nd Order ODE, we have that a linear combination of particular solutions to a homogeneous linear second order ODE is also a particular solution to that ODE.

Therefore:

\(\ds y\) \(=\) \(\ds \paren {A \map {T_n} x + B \sqrt {1 - x^2} \, \map {U_{n - 1} } x } + \paren {C \map \cos {n \arcsin x} + D \map \sin {n \arcsin x} }\)
\(\ds \leadsto \ \ \) \(\ds y\) \(=\) \(\ds A \map {T_n} x + B \sqrt {1 - x^2} \, \map {U_{n - 1} } x + C \map \cos {n \arcsin x} + D \map \sin {n \arcsin x}\) Linear Combination of Solutions to Homogeneous Linear 2nd Order ODE


$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=General_Solution_to_Chebyshev%27s_Differential_Equation&oldid=745885"