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

Examples:

1) Consider the Chebyshev differential equation \((1-x^2)y''(x)-xy'(x)+\alpha^2y(x)=0\). Then \(x=\pm1\) are regular singular points of the differential equation. Now we have:

\[p_0=\lim\limits_{x\to\pm1}(x\mp1)p(x)=\lim\limits_{x\to\pm1}(x\mp1)\frac{-x}{1-x^2}=\lim\limits_{x\to\pm1}\frac{x}{x\pm1}=\frac{1}{2} \quad\text{and}\quad q_0=\lim\limits_{x\to\pm1}(x-1)^2q(x)=\lim\limits_{x\to\pm1}(x-1)^2\frac{\alpha^2}{1-x^2}=0.\]

So for both \(x=1\) and \(x=-1\) the indicial equation is equal to \(r(r-1)+\frac{1}{2}r=0\) or equivalently \(r(r-\frac{1}{2})=0\). So the indices are: \(r_1=\frac{1}{2}\) and \(r_2=0\). So for \(x\in(-1,1)\) there exist solutions in the form \(y(x)=\displaystyle\sum_{n=0}^{\infty}c_n(1\pm x)^n\). Now suppose that:

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

Substitution then gives:

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

Now we sue that \(x=1-(1-x)\) and that \(1-x^2=(1-x)(1+x)=(1-x)(2-(1-x))=2(1-x)-(1-x)^2\):

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

or equivalently

\[2\sum_{n=0}^{\infty}(n+1)na_{n+1}(1-x)^n-\sum_{n=0}^{\infty}n(n-1)a_n(1-x)^n+\sum_{n=0}^{\infty}(n+1)a_{n+1}(1-x)^n -\sum_{n=0}^{\infty}na_n(1-x)^n+\alpha^2\sum_{n=0}^{\infty}a_n(1-x)^n=0.\]

This implies that \((n+1)(2n+1)a_{n+1}-(n^2-\alpha^2)a_n=0\) for \(n=0,1,2,\ldots\). Hence:

\begin{align*} a_n&=\frac{(n-1-\alpha)(n-1+\alpha)}{n(2n-1)}a_{n-1}=\frac{(n-1-\alpha)(n-1+\alpha)}{n(2n-1)}\cdot\frac{(n-2-\alpha)(n-2+\alpha)}{(n-1)(2n-3)}a_{n-2}\\[2.5mm] &=\frac{(-\alpha)(1-\alpha)\cdots(n-1-\alpha)\alpha(1+\alpha)\cdots(n-1+\alpha)}{n!\,1\cdot3\cdots(2n-1)}a_0,\quad n=1,2,3,\ldots. \end{align*}

Hence: \(y(x)=\displaystyle 1+\sum_{n=1}^{\infty}\frac{(-\alpha)(1-\alpha)\cdots(n-1-\alpha)\alpha(1+\alpha)\cdots(n-1+\alpha)}{n!\,1\cdot3\cdots(2n-1)}(1-x)^n\) is a solution.

Note that for \(\alpha\in\mathbb{N}\) these reduce to polynomila solutions \(T_{\alpha}(x)\). We have: \(T_0(x)=1\), \(T_1(x)=1-(1-x)=x\),

\[T_2(x)=1-4(1-x)+2(1-x)^2=2x^2-1,\quad T_3(x)=1-9(1-x)+12(1-x)^2-4(1-x)^3=4x^3-3x,\quad\ldots.\]

For \(x\in(-1,1)\) there also exist solutions in the form \(y(x)=\displaystyle(1\pm x)^{\frac{1}{2}}\sum_{n=0}^{\infty}c_n(1\pm x)^n=\sum_{n=0}^{\infty}c_n(1\pm x)^{n+\frac{1}{2}}\). Now suppose that:

\[y(x)=\sum_{n=0}^{\infty}b_n(1-x)^{n+\frac{1}{2}}\quad\Longrightarrow\quad y'(x)=-\sum_{n=0}^{\infty}(n+\tfrac{1}{2})b_n(1-x)^{n-\frac{1}{2}} \quad\Longrightarrow\quad y''(x)=\sum_{n=0}^{\infty}(n^2-\tfrac{1}{4})b_n(1-x)^{n-\frac{3}{2}}.\]

Substitution then gives:

\[(1-x^2)\sum_{n=0}^{\infty}(n^2-\tfrac{1}{4})b_n(1-x)^{n-\frac{3}{2}}+x\sum_{n=0}^{\infty}(n+\tfrac{1}{2})b_n(1-x)^{n-\frac{1}{2}} +\alpha^2\sum_{n=0}^{\infty}b_n(1-x)^{n+\frac{1}{2}}=0.\]

Now again we use that \(x=1-(1-x)\) and that \(1-x^2=(1-x)(1+x)=(1-x)(2-(1-x))=2(1-x)-(1-x)^2\):

\begin{align*} &2\sum_{n=0}^{\infty}(n^2-\tfrac{1}{4})b_n(1-x)^{n-\frac{1}{2}}-\sum_{n=0}^{\infty}(n^2-\tfrac{1}{4})b_n(1-x)^{n+\frac{1}{2}} +\sum_{n=0}^{\infty}(n+\tfrac{1}{2})b_n(1-x)^{n-\frac{1}{2}}\\[2.5mm] &{}\hspace{25mm}{}-\sum_{n=0}^{\infty}(n+\tfrac{1}{2})b_n(1-x)^{n+\frac{1}{2}}+\alpha^2\sum_{n=0}^{\infty}b_n(1-x)^{n+\frac{1}{2}}=0. \end{align*}

Note that the coefficient of \((1-x)^{-\frac{1}{2}}\) equals \(2\left(-\frac{1}{4}\right)b_0+\frac{1}{2}b_0=0\). Further we have

\begin{align*} &2\sum_{n=0}^{\infty}(n^2+2n+\tfrac{3}{4})b_{n+1}(1-x)^{n+\frac{1}{2}}-\sum_{n=0}^{\infty}(n^2-\tfrac{1}{4})b_n(1-x)^{n+\frac{1}{2}} +\sum_{n=0}^{\infty}(n+\tfrac{3}{2})b_{n+1}(1-x)^{n+\frac{1}{2}}\\[2.5mm] &{}\hspace{25mm}{}-\sum_{n=0}^{\infty}(n+\tfrac{1}{2})b_n(1-x)^{n+\frac{1}{2}}+\alpha^2\sum_{n=0}^{\infty}b_n(1-x)^{n+\frac{1}{2}}=0. \end{align*}

This implies that \((2n^2+5n+3)b_{n+1}-(n^2+n+\frac{1}{4}-\alpha^2)b_n\) for \(n=0,1,2,\ldots\). Hence:

\begin{align*} b_n&=\frac{(n-\frac{1}{2}-\alpha)(n-\frac{1}{2}+\alpha)}{n(2n+1)}b_{n-1}=\frac{(n-\frac{1}{2}-\alpha)(n-\frac{1}{2}+\alpha)}{n(2n+1)}\cdot\frac{(n-\frac{3}{2}-\alpha)(n-\frac{3}{2}+\alpha)}{(n-1)(2n-1)}b_{n-2}\\[2.5mm] &=\frac{(\frac{1}{2}-\alpha)(\frac{3}{2}-\alpha)\cdots(n-\frac{1}{2}-\alpha)(\frac{1}{2}+\alpha)(\frac{3}{2}+\alpha)\cdots(n-\frac{1}{2}+\alpha)}{n!\,3\cdot5\cdots(2n+1)}b_0,\quad n=1,2,3,\ldots. \end{align*}

Hence: \(\displaystyle y(x)=(1-x)^{\frac{1}{2}}+(1-x)^{\frac{1}{2}}\sum_{n=1}^{\infty}\frac{(\frac{1}{2}-\alpha)(\frac{3}{2}-\alpha)\cdots(n-\frac{1}{2}-\alpha)(\frac{1}{2}+\alpha)(\frac{3}{2}+\alpha)\cdots(n-\frac{1}{2}+\alpha)}{n!\,3\cdot5\cdots(2n+1)}(1-x)^n\) is also a solution.

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

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

Substitution then gives:

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

Now again we use that \(x=1-(1-x)\) and that \(1-x^2=(1-x)(1+x)=(1-x)(2-(1-x))=2(1-x)-(1-x)^2\):

\begin{align*} &2\sum_{n=0}^{\infty}(n+r)(n+r-1)c_n(1-x)^{n+r-1}-\sum_{n=0}^{\infty}(n+r)(n+r-1)c_n(1-x)^{n+r} +2\sum_{n=0}^{\infty}(n+r)c_n(1-x)^{n+r-1}\\[2.5mm] &{}\hspace{25mm}{}-2\sum_{n=0}^{\infty}(n+r)c_n(1-x)^{n+r}+\alpha(\alpha+1)\sum_{n=0}^{\infty}c_n(1-x)^{n+r}=0. \end{align*}

Now the coefficient of \((1-x)^{r-1}\) is: \(2r(r-1)c_0+2rc_0=2r^2c_0\). This implies that \(r=0\) (twice). Further we have:

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

or equivalently

\begin{align*} &2\sum_{n=0}^{\infty}(n+1)nc_{n+1}(1-x)^n-\sum_{n=0}^{\infty}n(n-1)c_n(1-x)^n+2\sum_{n=0}^{\infty}(n+1)c_{n+1}(1-x)^n\\[2.5mm] &{}\hspace{25mm}{}-2\sum_{n=0}^{\infty}nc_n(1-x)^n+\alpha(\alpha+1)\sum_{n=0}^{\infty}c_n(1-x)^n=0. \end{align*}

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

\begin{align*} c_n&=\frac{(n-1-\alpha)(n+\alpha)}{2n^2}c_{n-1}=\frac{(n-1-\alpha)(n+\alpha)}{2n^2}\cdot\frac{(n-2-\alpha)(n-1+\alpha)}{2(n-1)^2}c_{n-2}\\[2.5mm] &=\frac{(-\alpha)(1-\alpha)\cdots(n-2-\alpha)(1+\alpha)(2+\alpha)\cdots(n-1+\alpha)}{2^n(n!)^2}c_0,\quad n=1,2,3,\ldots. \end{align*}

Hence: \(y(x)=1+\displaystyle\sum_{n=1}^{\infty}\frac{(-\alpha)(1-\alpha)\cdots(n-2-\alpha)(1+\alpha)(2+\alpha)\cdots(n-1+\alpha)}{2^n(n!)^2}(1-x)^n\) is a solution.

Note that for \(\alpha\in\mathbb{N}\) these reduce to polynomila solutions \(P_{\alpha}(x)\). We have: \(P_0(x)=1\), \(P_1(x)=1-(1-x)=x\),

\[P_2(x)=1-3(1-x)+\frac{3}{2}(1-x)^2=\frac{3}{2}x^2-\frac{1}{2},\quad P_3(x)=1-5(1-x)+\frac{15}{2}(1-x)^2-\frac{5}{2}(1-x)^3=\frac{5}{2}x^3-\frac{3}{2}x,\quad\ldots.\]

3) Consider the Laguerre differential equation \(xy''(x)+(1-x)y'(x)+\alpha y(x)=0\). Then \(x=0\) is a regular singular point of the differential equation. So we look for solutions in the form of a generalized power series about \(x=0\). So suppose that

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

Substitution then gives:

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

or equivalently

\[\sum_{n=0}^{\infty}(n+r)(n+r-1)c_nx^{n+r-1}+\sum_{n=0}^{\infty}(n+r)c_nx^{n+r-1}-\sum_{n=0}^{\infty}(n+r)c_nx^{n+r}+\alpha\sum_{n=0}^{\infty}c_nx^{n+r}=0.\]

The coefficient of \(x^{r-1}\) is \(r(r-1)c_0+rc_0\). This implies that \(r=0\) (twice). Further we have:

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

or equivalently

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

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

\[c_n=\frac{n-1-\alpha}{n^2}c_{n-1}=\frac{n-1-\alpha}{n^2}\cdot\frac{n-2-\alpha}{(n-1)^2}c_{n-2} =\frac{(-\alpha)(1-\alpha)\cdots(n-1-\alpha)}{(n!)^2}c_0,\quad n=1,2,3,\ldots.\]

Hence: \(y(x)=1+\displaystyle\sum_{n=1}^{\infty}\frac{(-\alpha)(1-\alpha)\cdots(n-1-\alpha)}{(n!)^2}x^n\) is a solution.

Note that for \(\alpha\in\mathbb{N}\) these reduce to polynomial solutions \(L_{\alpha}(x)\). We have: \(L_0(x)=1\), \(L_1(x)=1-x\),

\[L_2(x)=1-2x+\frac{1}{2}x^2,\quad L_3(x)=1-3x+\frac{3}{2}x^2-\frac{1}{6}x^3,\quad L_4(x)=1-4x+3x^2-\frac{2}{3}x^3+\frac{1}{24}x^4,\quad\ldots.\]

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Last modified on May 13, 2021