Differential equations – Series solutions of linear differential equations – Regular points

Theorem: If \(x_0\) is a regular point of the differential equation

\[y''(x)+p(x)y'(x)+q(x)y(x)=0,\]

then both \(p(x)\) and \(q(x)\) are analytic at \(x=x_0\). That means that there exist positive constants \(R_1\) and \(R_2\) such that

\[p(x)=\sum_{n=0}^{\infty}p_0(x-x_0)^n,\quad|x-x_0| < R_1\;\quad\text{and}\quad\;q(x)=\sum_{n=0}^{\infty}q_n(x-x_0)^n,\quad|x-x_0| < R_2.\]

Then there exist solutions \(y(x)\) of the differential equation that are analytic at \(x=x_0\). In fact, if \(R=\min\{R_1,R_2\}\) then the general solution can be written as

\[y(x)=\sum_{n=0}^{\infty}c_n(x-x_0)^n,\quad|x-x_0| < R.\]

Moreover \(c_0=y(x_0)\) and \(c_1=y'(x_0)\) can be chosen arbitrarily and we have that (y(x)=c_0y_1(x)+c_1y_2(x)\) where \(y_1(x)\) en \(y_2(x)\) are linearly independent solutions.

We have already seen some examples in the course Calculus. Here we will consider some other examples.

Examples:

1) Consider the Airy differential equation \(y''(x)-xy(x)=0\). Then \(x=0\) is a regular point of the differential equation. So there exist solutions in the form of a power series about \(x=0\). So suppose that

\[y(x)=\sum_{n=0}^{\infty}c_nx^n\quad\Longrightarrow\quad y'(x)=\sum_{n=1}^{\infty}nc_nx^{n-1}\quad\Longrightarrow\quad y''(x)=\sum_{n=2}^{\infty}n(n-1)c_nx^{n-2}.\]

Substitution then gives:

\[\sum_{n=2}^{\infty}n(n-1)c_nx^{n-2}-x\sum_{n=0}^{\infty}c_nx^n=0\quad\Longleftrightarrow\quad \sum_{n=0}^{\infty}(n+2)(n+1)c_{n+2}x^n-\sum_{n=1}^{\infty}c_{n-1}x^n=0.\]

This implies that \(2c_2=0\) and \((n+2)(n+1)c_{n+2}-c_{n-1}=0\) for \(n=1,2,3,\ldots\). Hence: \(c_2=0\) and \(c_{n+1}=\displaystyle\frac{c_{n-1}}{(n+1)(n+2)}\) for \(n=1,2,3,\ldots\). Since \(c_2=0\) this implies that \(c_{3n+2}=0\) for \(n=0,1,2,\ldots\). Further we have for \(n=1,2,3,\ldots\) that

\[c_{3n}=\frac{c_{3n-3}}{(3n-1)(3n)}=\frac{c_{3n-6}}{(3n-4)(3n-3)(3n-1)(3n)}=\cdots=\frac{c_0}{2\cdot3\cdot5\cdot6\cdots(3n-1)(3n)}\]

and

\[c_{3n+1}=\frac{c_{3n-2}}{(3n)(3n+1)}=\frac{c_{3n-5}}{(3n-3)(3n-2)(3n)(3n+1)}=\cdots=\frac{c_1}{3\cdot4\cdot6\cdot7\cdots(3n)(3n+1)}.\]

So the general solution is \(y(x)=c_0y_1(x)+c_1y_2(x)\) with

\[y_1(x)=1+\sum_{n=1}^{\infty}\frac{x^{3n}}{2\cdot3\cdot5\cdot6\cdots(3n-1)(3n)}\quad\text{and}\quad y_2(x)=x+\sum_{n=1}^{\infty}\frac{x^{3n+1}}{3\cdot4\cdot6\cdot7\cdots(3n)(3n+1)}.\]

2) Consider the Hermite differential equation \(y''(x)-2xy'(x)+2\lambda y(x)=0\). Then \(x=0\) is a regular point of the differential equation. So there exist solution in the form of a power series about \(x=0\). So suppose that

\[y(x)=\sum_{n=0}^{\infty}c_nx^n\quad\Longrightarrow\quad y'(x)=\sum_{n=1}^{\infty}nc_nx^{n-1}\quad\Longrightarrow\quad y''(x)=\sum_{n=2}^{\infty}n(n-1)c_nx^{n-2}.\]

Substitution then gvives:

\[\sum_{n=2}^{\infty}n(n-1)c_nx^{n-2}-2x\sum_{n=1}^{\infty}nc_nx^{n-1}+2\lambda\sum_{n=0}^{\infty}c_nx^n=0 \quad\Longleftrightarrow\quad\sum_{n=0}^{\infty}(n+2)(n+1)c_{n+2}x^n-2\sum_{n=0}^{\infty}nc_nx^n+2\lambda\sum_{n=0}^{\infty}c_nx^n=0.\]

This implies: \((n+2)(n+1)c_{n+2}-(2n-2\lambda)c_n=0\) for \(n=0,1,2,\ldots\). Hence: \(\displaystyle c_{n+2}=\frac{2(n-\lambda)}{(n+2)(n+1)}c_n\) for \(n=0,1,2,\ldots\). Hence:

\[c_{2n}=\frac{2(2n-2-\lambda)}{(2n)(2n-1)}c_{2n-2}=\frac{2(2n-2-\lambda)2(2n-4-\lambda)}{(2n)(2n-1)(2n-2)(2n-3)}c_{2n-4} =\cdots=\frac{2^n(-\lambda)(2-\lambda)\cdots(2n-2-\lambda)}{(2n)!}c_0\]

and

\[c_{2n+1}=\frac{2(2n-1-\lambda)}{(2n+1)(2n)}c_{2n-1}=\frac{2(2n-1-\lambda)2(2n-3-\lambda)}{(2n+1)(2n)(2n-1)(2n-2)}c_{2n-3} =\cdots=\frac{2^n(1-\lambda)(3-\lambda)\cdots(2n-1-\lambda)}{(2n+1)!}c_1\]

for \(n=1,2,3,\ldots\). So the general solution is \(y(x)=c_0y_1(x)+c_1y_2(x)\) with

\[y_1(x)=1+\sum_{n=1}^{\infty}\frac{2^n(-\lambda)(2-\lambda)\cdots(2n-2-\lambda)}{(2n)!}x^{2n}\quad\text{and}\quad y_2(x)=x+\sum_{n=1}^{\infty}\frac{2^n(1-\lambda)(3-\lambda)\cdots(2n-1-\lambda)}{(2n+1)!}x^{2n+1}.\]

Note that for \(\lambda\in\mathbb{N}\) one of the power series solutions breaks down and reduces to a polynomial (solution). These solutions \(H_{\lambda}(x)\) are called Hermite polynomials: \(H_0(x)=1\), \(H_1(x)=x\),

\[H_2(x)=1-2x^2,\quad H_3(x)=x-\frac{2}{3}x^3,\quad H_4(x)=1-4x^2+\frac{4}{3}x^4,\quad H_5(x)=x-\frac{4}{3}x^3+\frac{4}{15}x^5,\quad\ldots.\]

3) Consider the Chebyshev differential equation \((1-x^2)y''(x)-xy'(x)+\alpha^2y(x)=0\). Then \(x=0\) is a regular point of the differential equation. So for \(x\in(-1,1)\) there exist solutions in the form of a power series about \(x=0\). So suppose that

\[y(x)=\sum_{n=0}^{\infty}c_nx^n\quad\Longrightarrow\quad y'(x)=\sum_{n=1}^{\infty}nc_nx^{n-1}\quad\Longrightarrow\quad y''(x)=\sum_{n=2}^{\infty}n(n-1)c_nx^{n-2}.\]

Substitution then gives:

\[(1-x^2)\sum_{n=2}^{\infty}n(n-1)c_nx^{n-2}-x\sum_{n=1}^{\infty}nc_nx^{n-1}+\alpha^2\sum_{n=0}^{\infty}c_nx^n=0\]

or equivalently

\[\sum_{n=0}^{\infty}(n+2)(n+1)c_{n+2}x^n-\sum_{n=0}^{\infty}n(n-1)c_nx^n-\sum_{n=0}^{\infty}nc_nx^n+\alpha^2\sum_{n=0}^{\infty}c_nx^n=0.\]

This implies that \((n+2)(n+1)c_{n+2}-\{n(n-1)+n-\alpha^2\}a_n=0\) or equivalently \(c_{n+2}=\displaystyle\frac{n^2-\alpha^2}{(n+2)(n+1)}c_n\) for \(n=0,1,2,\ldots\). Hence:

\[c_{2n}=\frac{(2n-2)^2-\alpha^2}{(2n)(2n-1)}c_{2n-2}=\cdots=\frac{(-\alpha^2)(4-\alpha^2)\cdots((2n-2)^2-\alpha^2)}{(2n)!}c_0\]

and

\[c_{2n+1}=\frac{(2n-1)^2-\alpha^2}{(2n+1)(2n)}c_{2n-1}=\cdots=\frac{(1-\alpha^2)(9-\alpha^2)\cdots((2n-1)^2-\alpha)}{(2n+1)!}c_1\]

for \(n=1,2,3,\ldots\). So the general solution is \(y(x)=c_0y_1(x)+c_1y_2(x)\) with

\[y_1(x)=1+\sum_{n=1}^{\infty}\frac{(-\alpha^2)(4-\alpha^2)\cdots((2n-2)^2-\alpha^2)}{(2n)!}x^{2n}\quad\text{and}\quad y_2(x)=x+\sum_{n=1}^{\infty}\frac{(1-\alpha^2)(9-\alpha^2)\cdots((2n-1)^2-\alpha)}{(2n+1)!}x^{2n+1}.\]

Note that for \(\alpha\in\mathbb{N}\) one of the power series solutions breaks down and reduces to a polynomial (solution). These solutions \(T_{\alpha}(x)\) are called Chebyshev polynomials: \(T_0(x)=1\), \(T_1(x)=x\),

\[T_2(x)=1-2x^2,\quad T_3(x)=x-\frac{4}{3}x^3,\quad T_4(x)=1-8x^2+8x^4,\quad T_5(x)=x-4x^3+\frac{16}{5}x^5,\quad\ldots.\]

4) Consider the Legendre differential equation \((1-x^2)y''(x)-2xy'(x)+\alpha(\alpha+1)\). Then \(x=0\) is a regular point of the differential equation. So for \(x\in(-1,1)\) there exist solutions in the form of a power series about \(x=0\). So suppose that

\[y(x)=\sum_{n=0}^{\infty}c_nx^n\quad\Longrightarrow\quad y'(x)=\sum_{n=1}^{\infty}nc_nx^{n-1}\quad\Longrightarrow\quad y''(x)=\sum_{n=2}^{\infty}n(n-1)c_nx^{n-2}.\]

Substitution then gives:

\[(1-x^2)\sum_{n=2}^{\infty}n(n-1)c_nx^{n-2}-2x\sum_{n=1}^{\infty}nc_nx^{n-1}+\alpha(\alpha+1)\sum_{n=0}^{\infty}c_nx^n=0\]

or equivalently

\[\sum_{n=0}^{\infty}(n+2)(n+1)c_{n+2}x^n-\sum_{n=0}^{\infty}n(n-1)c_nx^n-2\sum_{n=0}^{\infty}nc_nx^n+\alpha(\alpha+1)\sum_{n=0}^{\infty}c_nx^n=0.\]

This implies that \((n+2)(n+1)c_{n+2}-\{n(n-1)+2n-\alpha(\alpha+1)\}a_n=0\) or equivalently \(c_{n+2}=\displaystyle\frac{(n-\alpha)(n+\alpha+1)}{(n+2)(n+1)}c_n\) for \(n=0,1,2,\ldots\). Similar as above we now obtain:

\[c_{2n}=\frac{(-\alpha)(2-\alpha)\cdots(2n-2-\alpha)(\alpha+1)(\alpha+3)\cdots(\alpha+2n-1)}{(2n)!}c_0\]

and

\[c_{2n+1}=\frac{(1-\alpha)(3-\alpha)\cdots(2n-1-\alpha)(\alpha+2)(\alpha+4)\cdots(\alpha+2n)}{(2n+1)!}c_1\]

for \(n=1,2,3,\ldots\). So the general solution is \(y(x)=c_0y_1(x)+c_1y_2(x)\) with

\[y_1(x)=1+\sum_{n=1}^{\infty}\frac{(-\alpha)(2-\alpha)\cdots(2n-2-\alpha)(\alpha+1)(\alpha+3)\cdots(\alpha+2n-1)}{(2n)!}x^{2n}\]

and

\[y_2(x)=x+\sum_{n=1}^{\infty}\frac{(1-\alpha)(3-\alpha)\cdots(2n-1-\alpha)(\alpha+2)(\alpha+4)\cdots(\alpha+2n)}{(2n+1)!}x^{2n+1}.\]

Note that for \(\alpha\in\mathbb{N}\) one of the power series solutions breaks down and reduces to a polynomial (solution). These solutions \(P_{\alpha}(x)\) are called Legendre polynomials: \(P_0(x)=1\), \(P_1(x)=x\),

\[P_2(x)=1-3x^2,\quad P_3(x)=x-\frac{5}{3}x^3,\quad P_4(x)=1-10x^2+\frac{35}{3}x^4,\quad P_5(x)=x-\frac{14}{3}x^3+\frac{21}{5}x^5,\quad\ldots.\]

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Last modified on May 13, 2021