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Series Solution of Chebyshev Differential Equation 📂Odinary Differential Equations

Series Solution of Chebyshev Differential Equation

Definition

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

(1−x2)d2ydx2−xdydx+n2y=0 (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 xx 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 Tn(x)T_{n}(x).

Solution

(1−x2)y′′−xy′+λ2y=0 \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=a0+a1(x−x0)+a2(x−x0)2+⋯=∑n=0∞an(x−x0)n 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=0x=0, since the coefficient of y′′y^{\prime \prime} is (1−x2)∣x=0=1≠0(1-x^2)|_{x=0}=1\ne 0, let’s set it to x0=0x_{0}=0. Then,

y=a0+a1x+a2x2+⋯=∑n=0∞anxn \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 yy are finite. Now, to substitute into (1)\eqref{1}, let’s find y′y^{\prime} and y′′y^{\prime \prime}.

y′=a1+2a2x+3a3x2+⋯=∑n=1∞nanxn−1 y^{\prime}=a_{1}+2a_2x+3a_{3}x^2+\cdots=\sum \limits_{n=1}^\infty na_{n}x^{n-1}

y′′=2a2+3⋅2a3x+4⋅3a4x2+⋯=∑n=2n(n−1)anxn−2 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′,y′′y, y^{\prime}, y^{\prime \prime} into (1)\eqref{1} yields the following:

(1−x2)∑n=2∞n(n−1)anxn−2−x∑n=1∞nanxn−1+λ2∑n=0∞anxn=0 (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−x2)(1-x^2) gives:

∑n=2∞n(n−1)anxn−2−x2∑n=2∞n(n−1)anxn−2−x∑n=1∞nanxn−1+λ2∑n=0∞anxn=0 \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

  ⟹  ∑n=2∞n(n−1)anxn−2−∑n=2∞n(n−1)anxn−∑n=1∞nanxn+λ2∑n=0∞anxn=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 xx. While the rest are all expressed as xnx^n, only the first series is expressed as xn−2x^{n-2}, so substituting n+2n+2 instead of nn yields:

∑n=0∞(n+2)(n+1)an+2xn−∑n=2∞n(n−1)anxn−∑n=1∞nanxn+λ2∑n=0∞anxn=0 \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 x2x^2, we extract the term where n=0,1n=0,1 from the rest of the series, and group constant terms with constant terms, and first-order terms with first-order terms:

[2⋅1a2+λ2a0]+[3⋅2a3−a1+λ2a1]x+∑n=2∞[(n+2)(n+1)an+2−n(n−1)an−nan+λ2an]xn=0 \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 00.

2⋅1a2+λ2a0=0 2\cdot 1 a_2+\lambda^2 a_{0} = 0

3⋅2a3−a1+λ2a1=0 3\cdot 2 a_{3}-a_{1}+\lambda^2a_{1} =0

(n+2)(n+1)an+2−n(n−1)an−nan+λ2an=0 (n+2)(n+1)a_{n+2}-n(n-1)a_{n}-na_{n}+\lambda^2a_{n}=0

Organizing each gives:

a2=−λ22⋅1a0a3=−λ2−123⋅2a1an+2=−λ2−n2(n+2)(n+1)an \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 (5)\eqref{5}, we can determine all coefficients if we just know the values of a0a_{0} and a1a_{1}. From (3),(5)\eqref{3}, \eqref{5}, obtaining the coefficients of even-order terms:

a4=−λ2−224⋅3a2=λ2(λ2−22)4!a0a6=−λ2−426⋅5a4=−λ2(λ2−22)(λ2−42)6!a0⋮ \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,⋯ )n=2m (m=1,2,3,\cdots) then:

an=a2m=(−1)mλ2(λ2−22)⋯(λ2−(2m−2)2)(2m)!a0 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 (4),(5)\eqref{4}, \eqref{5}:

a5=−λ2−325⋅4a3=(λ2−12)(λ2−32)5!a1a7=−λ2−527⋅6a5=−(λ2−12)(λ2−32)(λ2−52)7!a1⋮ \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,⋯ )n=2m+1 (m=1,2,3,\cdots) then:

an=a2m+1=(−1)m(λ2−12)(λ2−32)⋯(λ2−(2m−1)2)(2m+1)!a1 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 (2)\eqref{2} to find the solution gives:

y=a0+a1x−λ22!a0x2−λ2−123!a1x3+λ2(λ2−22)4!a0x4+(λ2−12)(λ2−32)5!a1x5+⋯+(−1)mλ2(λ2−22)⋯(λ2−(2m−2)2)(2m)!a0x2m+(−1)m(λ2−12)(λ2−32)⋯(λ2−(2m−1)2)(2m+1)!a1x2m+1+⋯(m=1,2,3,⋯ ) \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 a0a_{0} and odd-order terms as a1a_{1} and organizing gives:

y=a0[1−λ22!x2+λ2(λ2−22)4!x4+∑m=3∞(−1)mλ2(λ2−22)⋯(λ2−(2m−2)2)(2m)!x2m+⋯ ]+a1[x−λ2−123!x3+(λ2−12)(λ2−32)5!x5+∑m=3∞(−1)m(λ2−12)(λ2−32)⋯(λ2−(2m−1)2)(2m+1)!x2m+1+⋯ ] \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 y0y_{0} and the second bracket y1y_{1}, then the general solution of the Chebyshev equation is as follows:

y=a0y0+a1y1 y=a_{0}y_{0}+a_{1}y_{1}

The two series y0y_{0} and y1y_{1} are known to converge within the range of ∣x∣<1|x|<1 according to the ratio test. Because of (5)\eqref{5}, an+2an=n2−λ2(n+2)(n+1)=n2−λ2n2+3n+2\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⁡n→∞n2−λ2n2+3n+2x2=x2<1 \lim \limits_{n \rightarrow \infty} \dfrac{n^2-\lambda^2}{n^2+3n+2}x^2=x^2<1

  ⟹  −1<x<1 \implies -1<x<1

However, in many problems, x=cos⁡θx=\cos \theta, λ\lambda appear in the form of non-negative integers, and we seek solutions that converge for all θ\theta. In other words, the goal is to find solutions that also converge at x=±1x=\pm 1. Fortunately, when λ\lambda is an integer, the desired solution exists, and depending on the value of λ\lambda, only one of the two solutions y0,y1y_{0}, y_{1} must exist. When λ\lambda is 00 or an even number, y1y_{1} diverges, and y0y_{0} becomes a finite-term polynomial with only even-order terms. When λ\lambda is odd, y0y_{0} diverges, and y1y_{1} becomes a finite-term polynomial with only odd-order terms. The summary is as follows:

Value of λ\lambday0y_{0}y1y_{1}Solution of the equation
00 or evenFinite-term polynomialDivergesy=a0y0y=a_{0}y_{0}
OddDivergesFinite-term polynomialy=a1y1y=a_{1}y_{1}

  • Case 1. When λ\lambda is 00 or even

    • When λ=0\lambda=0, from the second term, it takes λ2\lambda^2 as a factor, becoming all 00, thus y0=1y_{0}=1

    • When λ=2\lambda=2, from the fourth term, it takes (λ2−22)(\lambda^2-2^2) as a factor, becoming all 00, thus y0=1−x2y_{0}=1-x^2

    • When λ=4\lambda=4, from the sixth term, it takes (λ2−42)(\lambda^2-4^2) as a factor, becoming all 00, thus y0=1−8x2+8x4y_{0}=1-8x^2+8x^4

    And when λ=0\lambda=0, x=1x=1 diverges at y1=1+13!+1⋅325!+⋯y_{1}=1+\frac{1}{3!}+\frac{1\cdot3^2}{5!}+\cdots. The same applies to other even numbers. Therefore, when λ\lambda is 00 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 y0y_{0}. The opposite result is obtained when λ\lambda is odd.

  • Case 2. When λ\lambda is odd

    • When λ=1\lambda=1, from the third term, it takes (λ2−12)(\lambda^2-1^2) as a factor, becoming all 00, thus y1=xy_{1}=x

    • When λ=3\lambda=3, from the fifth term, it takes (λ2−32)(\lambda^2-3^2) as a factor, becoming all 00, thus y1=−3x+4x3y_{1}=-3x+4x^3

    • When λ=5\lambda=5, from the seventh term, it takes (λ2−52)(\lambda^2-5^2) as a factor, becoming all 00, thus y1=5x−20x3+16x5y_{1}=5x-20x^3+16x^5

    When λ=1\lambda=1, x2=1x^2=1 diverges at y0y_{0}, and the same applies to other odd numbers. Therefore, when λ\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 y1y_{1}.

It’s also noted that when λ\lambda is negative, it’s the same as when λ\lambda is positive, which can be seen by examining y0y_{0} and y1y_{1}. For instance, the case of λ=2\lambda=2 is the same as the case of λ=−2\lambda=-2, and the case of λ=1\lambda=1 is the same as the case of λ=−1\lambda=-1. Therefore, we only need to consider λ\lambda within the range of non-negative integers. Selecting a0a_{0} and a1a_{1} appropriately, when x=1x=1, the solution becomes y(x)=1y(x)=1, this is called the Chebyshev polynomial and is commonly denoted as Tn(x)T_{n}(x). The first few Chebyshev polynomials are as follows.

T0(x)=1T1(x)=xT2(x)=2x2−1T3(x)=4x3−3xT4(x)=8x4−8x2+1T5(x)=16x5−20x3+5x⋮ \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*}

See Also