Chebyshev Differential Equations and Chebyshev Polynomials 📂Odinary Differential Equations

Chebyshev Differential Equations and Chebyshev Polynomials

Definition

The following differential equation is referred to as the Chebyshev Differential Equation.

$\begin{equation} (1-x^2)\dfrac{d^2 y}{dx^2} -x\dfrac{dy}{dx}+n^2 y=0 \label{def1} \end{equation}$

The solution to the Chebyshev differential equation is known as Chebyshev polynomials, commonly denoted by $T_{n}(x)$ . The general term of $T_{n}(x)$ is as follows:

When $n$ is even
$1-\dfrac{\lambda^2}{2!}x^2+\dfrac{\lambda^2(\lambda^2-2^2)}{4!}x^4+\sum \limits_{m=3}^\infty (-1)^m \dfrac{\lambda^2(\lambda^2-2^2)\cdots(\lambda^2-(2m-2)^2)}{(2m)!} x^{2m}$
When $n$ is odd

$x-\dfrac{\lambda^2-1^2}{3!}x^3+\dfrac{(\lambda^2-1^2)(\lambda^2-3^2)}{5!}x^5+\sum \limits_{m=3}^\infty (-1)^m\dfrac{(\lambda^2-1^2)(\lambda^2-3^2) \cdots (\lambda^2-(2m-1)^2)}{(2m+1)!} x^{2m+1}$

Especially, the first few polynomials are as follows:

$\begin{align*} T_{0}(x) &= 1 \\ T_{1}(x) &= x \\ T_2(x) &= 2x^2-1 \\ T_{3}(x) &= 4x^3-3x \\ \vdots & \end{align*}$

Theorem

For Chebyshev polynomials $T_{n}$ , the following equation holds:

$T_{n}(\cos t)= \cos (nt)$

Description

Since $T_{n}(x)$ was a $n$ th degree polynomial with respect to $x$ , $T_{n}(\cos t)$ is a polynomial with respect to $\cos t$ . Thus, Chebyshev polynomials can also be understood as the expansion of $\cos (nt)$ into a $n$ th degree polynomial with respect to $\cos t$ .

When $n=2,\ 3$ , to see if it fits well

$\cos 2t=\cos ^2 t-1=T_2(\cos t)\iff T_2(x)=x^2-1$

$\cos 3t=4\cos ^3 t-3\cos t=T_{3}(\cos t) \iff T_{3}(x)=4x^3-3x$

Also, since $x=\cos t$ results in $\arccos x=t$ , substituting into the above equation yields

$T_{n}(x)=\cos(n\arccos x) \quad \text{or} \quad T_{n}(x) = \cos (n\cos^{-1}x)$

Proof

Strategy: Demonstrate that by substituting $x=\cos t$ , $y=\cos (nt)$ becomes a solution to the Chebyshev differential equation.

When it’s given $x=\cos t$ ,

$dx=-\sin t dt \quad \implies \quad \dfrac{dt}{dx}=-\dfrac{1}{\sin t}$

Therefore, $y^{\prime}$ is as follows.

$\dfrac{dy}{dx}=\dfrac{dy}{dt} \dfrac{dt}{dx}=-\dfrac{1}{\sin t}\dfrac{dy}{dt}$

And $y^{\prime \prime}$ is as follows.

$\begin{align*} \dfrac{d^2 y}{dx^2} &= \dfrac{d}{dx} \left(\dfrac{dy}{dx}\right) \\ &= \dfrac{d}{dt}\left(\dfrac{dy}{dx} \right) \dfrac{dt}{dx} \\ &=\dfrac{d}{dt}\left(\dfrac{1}{\sin t}\dfrac{dy}{dt}\right) \dfrac{1}{\sin t} \\ &= \dfrac{1}{\sin t} \left( \dfrac{-\cos t}{\sin^2 t}\dfrac{dy}{dt}+\dfrac{1}{\sin t}\dfrac{d^2y}{dt^2} \right) \\ &= \dfrac{1}{\sin ^2 t} \left( \dfrac{-\cos t}{\sin t}\dfrac{dy}{dt}+ \dfrac{d^2y}{dt^2} \right) \end{align*}$

Substituting the above into $\eqref{def1}$ results in the following.

$(1-\cos ^2t)\dfrac{1}{\sin ^2 t} \left( \dfrac{-\cos t}{\sin t}\dfrac{dy}{dt}+ \dfrac{d^2y}{dt^2} \right)-\cos t \left( -\dfrac{1}{\sin t} \right)\dfrac{dy}{dt} +n^2y=0$

Sorting it out yields

$y^{\prime \prime}+n^2y=0$

Thus $T_{n}(\cos t)$ is a solution to the above differential equation. However, the equation is a very simple second-order differential equation, and its general solution is $y=C_{1}\cos (nt) + C_2\sin (nt)$ . Therefore,

$T_{n}(\cos t)=\cos (nt)$

■