Series Solution of Chebyshev Differential Equation
Definition
The following differential equation is referred to as the Chebyshev Differential Equation:
$$ (1-x^2)\dfrac{d^2 y}{dx^2} -x\dfrac{dy}{dx}+n^2 y=0 $$
Description
It’s a form that includes the independent variable $x$ in the coefficient, and assuming that the solution is in the form of a power series, it can be solved. The solution to the Chebyshev equation is called the Chebyshev polynomial, often denoted as $T_{n}(x)$.
Solution
$$ \begin{equation} (1-x^2)y^{\prime \prime} -xy^{\prime}+\lambda^2 y=0 \label{1} \end{equation} $$
Let’s assume the solution to the given Chebyshev differential equation is as follows:
$$ y=a_{0}+a_{1}(x-x_{0})+a_2(x-x_{0})^2+\cdots=\sum \limits_{n=0}^\infty a_{n}(x-x_{0})^n $$
When it is $x=0$, since the coefficient of $y^{\prime \prime}$ is $(1-x^2)|_{x=0}=1\ne 0$, let’s set it to $x_{0}=0$. Then,
$$ \begin{equation} y=a_{0}+a_{1}x+a_2x^2+\cdots=\sum \limits_{n=0}^\infty a_{n}x^n \label{2} \end{equation} $$
We start solving by assuming a series solution, but at the end of the solution, we find that the terms of $y$ are finite. Now, to substitute into $\eqref{1}$, let’s find $y^{\prime}$ and $y^{\prime \prime}$.
$$ y^{\prime}=a_{1}+2a_2x+3a_{3}x^2+\cdots=\sum \limits_{n=1}^\infty na_{n}x^{n-1} $$
$$ y^{\prime \prime}=2a_2+3\cdot 2a_{3}x+4\cdot 3 a_{4}x^2 +\cdots = \sum \limits_{n=2} n(n-1)a_{n}x^{n-2} $$
Substituting $y, y^{\prime}, y^{\prime \prime}$ into $\eqref{1}$ yields the following:
$$ (1-x^2)\sum \limits_{n=2}^\infty n(n-1)a_{n}x^{n-2} -x\sum \limits_{n=1}^\infty na_{n}x^{n-1}+\lambda^2 \sum \limits_{n=0}^\infty a_{n}x^n=0 $$
Expanding and rearranging the first term’s coefficient $(1-x^2)$ gives:
$$ \sum \limits_{n=2}^\infty n(n-1)a_{n}x^{n-2} -x^2\sum \limits_{n=2}^\infty n(n-1)a_{n}x^{n-2} -x\sum \limits_{n=1}^\infty na_{n}x^{n-1}+\lambda^2 \sum \limits_{n=0}^\infty a_{n}x^n = 0 $$
$$ \implies \sum \limits_{n=2} ^\infty n(n-1)a_{n}x^{n-2} -\sum \limits_{n=2}^\infty n(n-1)a_{n}x^{n} -\sum \limits_{n=1}^\infty na_{n}x^{n}+\lambda^2 \sum \limits_{n=0}^\infty a_{n}x^n = 0 $$
The key here is to match the order of $x$. While the rest are all expressed as $x^n$, only the first series is expressed as $x^{n-2}$, so substituting $n+2$ instead of $n$ yields:
$$ \sum \limits_{n=0} ^\infty (n+2)(n+1)a_{n+2}x^{n} -\sum \limits_{n=2}^\infty n(n-1)a_{n}x^{n} -\sum \limits_{n=1}^\infty na_{n}x^{n}+\lambda^2 \sum \limits_{n=0}^\infty a_{n}x^n=0 $$
Since the second series starts from the term $x^2$, we extract the term where $n=0,1$ from the rest of the series, and group constant terms with constant terms, and first-order terms with first-order terms:
$$ \left[ 2\cdot 1 a_2+\lambda^2 a_{0} \right]+\left[ 3\cdot 2 a_{3}-a_{1}+\lambda^2a_{1} \right]x \\ + \sum \limits_{n=2}^\infty \left[ (n+2)(n+1)a_{n+2}-n(n-1)a_{n}-na_{n}+\lambda^2a_{n} \right] x^n=0 $$
For this equation to hold, all coefficients must be $0$.
$$ 2\cdot 1 a_2+\lambda^2 a_{0} = 0 $$
$$ 3\cdot 2 a_{3}-a_{1}+\lambda^2a_{1} =0 $$
$$ (n+2)(n+1)a_{n+2}-n(n-1)a_{n}-na_{n}+\lambda^2a_{n}=0 $$
Organizing each gives:
$$ \begin{align} a_2 &= -\dfrac{\lambda^2}{2 \cdot 1}a_{0} \label{3} \\ a_{3} &=-\dfrac{\lambda^2-1^2}{3\cdot 2} a_{1} \label{4} \\ a_{n+2} &= -\dfrac{\lambda^2-n^2}{(n+2)(n+1)}a_{n} \label{5} \end{align} $$
Having obtained the recurrence relation $\eqref{5}$, we can determine all coefficients if we just know the values of $a_{0}$ and $a_{1}$. From $\eqref{3}, \eqref{5}$, obtaining the coefficients of even-order terms:
$$ \begin{align*} a_{4} &= -\dfrac{\lambda^2-2^2}{4\cdot 3}a_2=\dfrac{\lambda^2(\lambda^2-2^2)}{4!}a_{0} \\ a_{6} &= -\dfrac{\lambda^2-4^2}{6\cdot 5}a_{4}= -\dfrac{\lambda^2(\lambda^2-2^2)(\lambda^2-4^2)}{6!}a_{0} \\ &\vdots \end{align*} $$
If we say $n=2m (m=1,2,3,\cdots)$ then:
$$ a_{n}=a_{2m}=(-1)^m \dfrac{\lambda^2(\lambda^2-2^2)\cdots(\lambda^2-(2m-2)^2)}{(2m)!}a_{0} $$
Similarly, obtaining the coefficients of odd-order terms from $\eqref{4}, \eqref{5}$:
$$ \begin{align*} a_{5} &= -\dfrac{\lambda^2-3^2}{5\cdot 4}a_{3}=\dfrac{(\lambda^2-1^2)(\lambda^2-3^2)}{5!}a_{1} \\ a_{7} &= -\dfrac{\lambda^2-5^2}{7\cdot 6 }a_{5}=-\dfrac{(\lambda^2-1^2)(\lambda^2-3^2)(\lambda^2-5^2)}{7!}a_{1} \\ &\vdots \end{align*} $$
If we say $n=2m+1 (m=1,2,3,\cdots)$ then:
$$ a_{n}=a_{2m+1}=(-1)^m\dfrac{(\lambda^2-1^2)(\lambda^2-3^2) \cdots (\lambda^2-(2m-1)^2)}{(2m+1)!}a_{1} $$
Substituting the obtained coefficients into $\eqref{2}$ to find the solution gives:
$$ \begin{align*} y = &a_{0}+a_{1}x -\dfrac{\lambda^2}{2!}a_{0}x^2-\dfrac{\lambda^2-1^2}{3!} a_{1}x^3 + \dfrac{\lambda^2(\lambda^2-2^2)}{4!}a_{0}x^4 \\ &+\dfrac{(\lambda^2-1^2)(\lambda^2-3^2)}{5!}a_{1}x^5+ \cdots +(-1)^m \dfrac{\lambda^2(\lambda^2-2^2)\cdots(\lambda^2-(2m-2)^2)}{(2m)!}a_{0}x^{2m} \\ &+(-1)^m\dfrac{(\lambda^2-1^2)(\lambda^2-3^2) \cdots (\lambda^2-(2m-1)^2)}{(2m+1)!}a_{1}x^{2m+1}+\cdots\quad(m=1,2,3,\cdots) \end{align*} $$
Here, grouping even-order terms as $a_{0}$ and odd-order terms as $a_{1}$ and organizing gives:
$$ \begin{align*} y&=a_{0}\left[1-\dfrac{\lambda^2}{2!}x^2+\dfrac{\lambda^2(\lambda^2-2^2)}{4!}x^4+\sum \limits_{m=3}^\infty (-1)^m \dfrac{\lambda^2(\lambda^2-2^2)\cdots(\lambda^2-(2m-2)^2)}{(2m)!} x^{2m} + \cdots \right] \\ & + a_{1}\left[x-\dfrac{\lambda^2-1^2}{3!}x^3+\dfrac{(\lambda^2-1^2)(\lambda^2-3^2)}{5!}x^5+\sum \limits_{m=3}^\infty (-1)^m\dfrac{(\lambda^2-1^2)(\lambda^2-3^2) \cdots (\lambda^2-(2m-1)^2)}{(2m+1)!} x^{2m+1} + \cdots\right] \end{align*} $$
If we call the first bracket $y_{0}$ and the second bracket $y_{1}$, then the general solution of the Chebyshev equation is as follows:
$$ y=a_{0}y_{0}+a_{1}y_{1} $$
The two series $y_{0}$ and $y_{1}$ are known to converge within the range of $|x|<1$ according to the ratio test. Because of $\eqref{5}$, $\dfrac{a_{n+2}}{a_{n}}=\dfrac{n^2-\lambda^2}{(n+2)(n+1)}=\dfrac{n^2-\lambda^2}{n^2+3n+2}$, so if we use the ratio test:
$$ \lim \limits_{n \rightarrow \infty} \dfrac{n^2-\lambda^2}{n^2+3n+2}x^2=x^2<1 $$
$$ \implies -1<x<1 $$
However, in many problems, $x=\cos \theta$, $\lambda$ appear in the form of non-negative integers, and we seek solutions that converge for all $\theta$. In other words, the goal is to find solutions that also converge at $x=\pm 1$. Fortunately, when $\lambda$ is an integer, the desired solution exists, and depending on the value of $\lambda$, only one of the two solutions $y_{0}, y_{1}$ must exist. When $\lambda$ is $0$ or an even number, $y_{1}$ diverges, and $y_{0}$ becomes a finite-term polynomial with only even-order terms. When $\lambda$ is odd, $y_{0}$ diverges, and $y_{1}$ becomes a finite-term polynomial with only odd-order terms. The summary is as follows:
| Value of $\lambda$ | $y_{0}$ | $y_{1}$ | Solution of the equation |
|---|---|---|---|
| $0$ or even | Finite-term polynomial | Diverges | $y=a_{0}y_{0}$ |
| Odd | Diverges | Finite-term polynomial | $y=a_{1}y_{1}$ |
Case 1. When $\lambda$ is $0$ or even
When $\lambda=0$, from the second term, it takes $\lambda^2$ as a factor, becoming all $0$, thus $y_{0}=1$
When $\lambda=2$, from the fourth term, it takes $(\lambda^2-2^2)$ as a factor, becoming all $0$, thus $y_{0}=1-x^2$
When $\lambda=4$, from the sixth term, it takes $(\lambda^2-4^2)$ as a factor, becoming all $0$, thus $y_{0}=1-8x^2+8x^4$
And when $\lambda=0$, $x=1$ diverges at $y_{1}=1+\frac{1}{3!}+\frac{1\cdot3^2}{5!}+\cdots$. The same applies to other even numbers. Therefore, when $\lambda$ is $0$ or an even number, the solution becomes a finite-term polynomial with only even-order terms. That is, we obtain a form of the solution that only remains up to a specific term of the series $y_{0}$. The opposite result is obtained when $\lambda$ is odd.
Case 2. When $\lambda$ is odd
When $\lambda=1$, from the third term, it takes $(\lambda^2-1^2)$ as a factor, becoming all $0$, thus $y_{1}=x$
When $\lambda=3$, from the fifth term, it takes $(\lambda^2-3^2)$ as a factor, becoming all $0$, thus $y_{1}=-3x+4x^3$
When $\lambda=5$, from the seventh term, it takes $(\lambda^2-5^2)$ as a factor, becoming all $0$, thus $y_{1}=5x-20x^3+16x^5$
When $\lambda=1$, $x^2=1$ diverges at $y_{0}$, and the same applies to other odd numbers. Therefore, when $\lambda$ is odd, the solution becomes a finite-term polynomial with only odd-order terms. That is, we obtain a form of the solution that only remains up to a specific term of the series $y_{1}$.
It’s also noted that when $\lambda$ is negative, it’s the same as when $\lambda$ is positive, which can be seen by examining $y_{0}$ and $y_{1}$. For instance, the case of $\lambda=2$ is the same as the case of $\lambda=-2$, and the case of $\lambda=1$ is the same as the case of $\lambda=-1$. Therefore, we only need to consider $\lambda$ within the range of non-negative integers. Selecting $a_{0}$ and $a_{1}$ appropriately, when $x=1$, the solution becomes $y(x)=1$, this is called the Chebyshev polynomial and is commonly denoted as $T_{n}(x)$. The first few Chebyshev polynomials are as follows.
$$ \begin{align*} T_{0}(x) &= 1 \\ T_{1}(x) &= x \\ T_2(x) &= 2x^2-1 \\ T_{3}(x) &= 4x^3-3x \\ T_{4}(x) &= 8x^4-8x^2+1 \\ T_{5}(x) &= 16x^5-20x^3+5x \\ \vdots & \end{align*} $$