Regular singular point using Frobenius series method.

3.3.25.5.1 Example 1
3.3.25.5.2 Example 2
3.3.25.5.3 Example 3
3.3.25.5.4 Example 4
3.3.25.5.5 Example 5

expansion point is a regular singular point. Standard power series. The ode must be linear in \(y^{\prime }\) and \(y\) at this time.

Expansion is around \(x=0\). The (homogenous) ode has the form \(y^{\prime }+p\left ( x\right ) y=0\). We see that \(p\left ( x\right ) \) is analytic at \(x=0\). However the RHS has no series expansion at \(x=0\) (not analytic there). Therefore we must use Frobenius series in this case. Let

\begin{align*} y & =\sum _{n=0}^{\infty }a_{n}x^{n+r}\\ y^{\prime } & =\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}\end{align*}

\begin{align*} \sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+2x\sum _{n=0}^{\infty }a_{n}x^{n+r} & =0\\ \sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+\sum _{n=0}^{\infty }2a_{n}x^{n+r+1} & =0 \end{align*}

\begin{equation} \sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+\sum _{n=2}^{\infty }2a_{n-2}x^{n+r-1}=0 \tag {1}\end{equation}

Where \(r\) is replaced my \(m\) and \(a_{n}\) is replaced by \(c_{n}\). The above will used below to ﬁnd \(y_{p}\). For \(n=1\), Eq(1) gives

\begin{align*} \left ( 1+r\right ) a_{1}x^{r} & =0\\ a_{1} & =0 \end{align*}

\begin{align} \left ( n+r\right ) a_{n}+2a_{n-2} & =0\nonumber \\ a_{n} & =-\frac {2a_{n-2}}{\left ( n+r\right ) } \tag {2}\end{align}

\begin{equation} a_{n}=-\frac {2}{n}a_{n-2} \tag {2A}\end{equation}

Eq (2A) is what is used to ﬁnd all \(a_{n}\) for For \(n\geq 2\). Hence for \(n=2\) and remembering that \(a_{0}=1\) gives

\begin{align*} y_{h} & =c_{1}\sum _{n=0}^{\infty }a_{n}x^{n+r}\\ & =c_{1}\sum _{n=0}^{\infty }a_{n}x^{n}\\ & =c_{1}\left ( a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+\cdots \right ) \\ & =c_{1}\left ( 1-x^{2}+\frac {1}{2}x^{4}-\frac {1}{6}x^{6}+\cdots \right ) \end{align*}

Now we need to ﬁnd \(y_{p}\) using the balance equation. From above we found that

Renaming \(a\) to \(c\) and \(r\) as \(m\) so not to confuse terms used for \(y_{h}\), the above becomes

Hence \(m-1=\frac {1}{2}\) or \(m=\frac {3}{2}\). Therefore \(mc_{0}=1\) or \(c_{0}=\frac {2}{3}\). Now we can ﬁnd the series for \(y_{p}\) using

\begin{align*} y_{p} & =\sum _{n=0}^{\infty }c_{n}x^{n+m}\\ & =x^{\frac {3}{2}}\sum _{n=0}^{\infty }c_{n}x^{n}\end{align*}

To ﬁnd \(c_{m}\) we use the same recurrence relation found for \(y_{h}\) but change \(r\) to \(m\) and \(a\) to \(c\). From above we found

The above is valid for \(n\geq 2\). For \(n=0\) we have found \(c_{0}\) already. For \(c_{1}\) using the above \(ra_{1}=0\) hence it becomes \(mc_{1}=0\) which implies

since \(m\neq 0\). Now we are ready to ﬁnd few \(c_{n}\) terms. The above recurrence relation becomes for \(m=\frac {3}{2}\)

\begin{align*} \left ( n+\frac {3}{2}\right ) c_{n}+2c_{n-2} & =0\\ c_{n} & =\frac {-2c_{n-2}}{\left ( n+\frac {3}{2}\right ) }\end{align*}

\[ c_{2}=\frac {-2c_{0}}{\left ( 2+\frac {3}{2}\right ) }=\frac {-2\left ( \frac {2}{3}\right ) }{\left ( 2+\frac {3}{2}\right ) }=-\frac {8}{21}\]

\[ c_{4}=\frac {-2c_{2}}{\left ( 4+\frac {3}{2}\right ) }=\frac {-2\left ( -\frac {8}{21}\right ) }{\left ( 4+\frac {3}{2}\right ) }=\frac {32}{231}\]

\begin{align*} y_{p} & =x^{\frac {3}{2}}\sum _{n=0}^{\infty }c_{n}x^{n}\\ & =x^{\frac {3}{2}}\left ( c_{0}+c_{1}x+c_{2}x^{2}+\cdots \right ) \end{align*}

\begin{align*} y & =y_{h}+y_{p}\\ & =c_{1}\left ( 1-x^{2}+\frac {1}{2}x^{4}-\frac {1}{6}x^{6}+\cdots \right ) +x^{\frac {3}{2}}\left ( \frac {2}{3}+-\frac {8}{21}x^{2}+\frac {32}{231}x^{4}-\cdots \right ) \end{align*}

Expansion is around \(x=0\). The (homogenous) ode has the form \(y^{\prime }+p\left ( x\right ) y=0\). We see that \(p\left ( x\right ) \) is deﬁned as is at \(x=0\). However the RHS has no series expansion at \(x=0\). Therefore we must use Frobenius series. This is the same ode as example 1. So we go straight to ﬁnd \(y_{p}\) as \(y_{h}\) is the same. Now we need to ﬁnd \(y_{p}\) using the balance equation. From above we found that

Renaming \(a\) to \(c\) and \(r\) as \(m\) so not to confuse terms used for \(y_{h}\), the above becomes

Hence \(m-1=-1\) or \(m=0\). Therefore \(mc_{0}=1\). But since \(m=0\) then no solution for \(c_{0}\). Hence it is not possible to ﬁnd series solution. This is an example where the balance equation fails and so we have to use asymptotic expansion to ﬁnd solution, which is not supported now.

Expansion is around \(x=0\). The (homogenous) ode has the form \(y^{\prime }+p\left ( x\right ) y=0\). We see that \(p\left ( x\right ) =0\) is analytic at \(x=0\). However the RHS has no series expansion at \(x=0\) (not analytic there). Therefore we must use Frobenius series in this case. Let

\begin{align*} y & =\sum _{n=0}^{\infty }a_{n}x^{n+r}\\ y^{\prime } & =\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}\end{align*}

\begin{equation} \sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}=0 \tag {1}\end{equation}

\begin{align} \sum _{n=0}^{\infty }na_{n}x^{n-1} & =0\nonumber \\ na_{n}x^{n-1} & =0 \tag {2}\end{align}

Now we need to ﬁnd \(y_{p}\) using the balance equation. From above we found that

Changing \(r\) to \(m\) and \(a_{0}\) to \(c_{0}\) so not to confuse notation gives

Hence \(m-1=-1\) or \(m=0\). Therefore there is no solution for \(c_{0}\). Unable to ﬁnd \(y_{p}\) therefore no series solution exists. Asymptotic methods are needed to solve this. Mathematica AsymptoticDSolveValue gives the solution as \(y\left ( x\right ) =c+\ln x\).

\begin{align*} y & =\sum _{n=0}^{\infty }a_{n}x^{n+r}\\ y^{\prime } & =\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}\end{align*}

\begin{equation} \sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}=0 \tag {1}\end{equation}

Or by changing \(r\) to \(m\) and \(a_{0}\) to \(c_{0}\) so not to confuse notation with \(y_{h}\) gives

\begin{equation} mc_{0}x^{m-1}=x^{-2} \tag {2}\end{equation}

\begin{align} \sum _{n=0}^{\infty }na_{n}x^{n-1} & =0\nonumber \\ na_{n}x^{n-1} & =0 \tag {2}\end{align}

\(n=0\) is not used since that was used to ﬁnd \(r\). Therefore we start from \(n=1\). For all \(n\geq 1\) we see from (2) that \(a_{n}=0\). Hence

Now we need to ﬁnd \(y_{p}\) using the balance equation. From (2) above we found that

To balance, we need \(m-1=-2\) or \(m=-1\) and \(mc_{0}=1\) or \(c_{0}=-1\). Therefore

Where \(c_{0}=-1\) and all \(c_{n}\) for \(n\geq 1\) are found using the recurrence relation from ﬁnding \(y_{h}\). But from above we found that all \(a_{n}=0\) for \(n\geq 1\). Hence \(c_{n}=0\) also for \(n\geq 1\). Therefore

\begin{align*} y_{p} & =x^{m}c_{0}\\ & =\frac {-1}{x}+O\left ( x^{2}\right ) \end{align*}

\begin{align*} y & =y_{h}+y_{p}\\ & =c_{1}\left ( 1+O\left ( x^{2}\right ) \right ) +\left ( \frac {-1}{x}+O\left ( x^{2}\right ) \right ) \end{align*}

Expansion is around \(x=0\). The (homogenous) ode has the form \(y^{\prime }+p\left ( x\right ) y=0\). We see that \(p\left ( x\right ) =\frac {1}{x}\) is not analytic at \(x=0\) but \(\lim _{x\rightarrow 0}xp\left ( x\right ) =0\) is analytic. Therefore we must use Frobenius series in this case. Let

\begin{align} y & =\sum _{n=0}^{\infty }a_{n}x^{n+r}\tag {A}\\ y^{\prime } & =\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}\nonumber \end{align}

\begin{align} \sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+\frac {1}{x}\sum _{n=0}^{\infty }a_{n}x^{n+r} & =0\nonumber \\ \sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+\sum _{n=0}^{\infty }a_{n}x^{n+r-1} & =0\nonumber \\ \sum _{n=0}^{\infty }\left ( \left ( n+r\right ) a_{n}+a_{n}\right ) x^{n+r-1} & =0\nonumber \\ \sum _{n=0}^{\infty }\left ( n+r+1\right ) a_{n}x^{n+r-1} & =0 \tag {1}\end{align}

\begin{align} \sum _{n=0}^{\infty }na_{n}x^{n} & =0\nonumber \\ na_{n}x^{n-1} & =0 \tag {2}\end{align}

\(n=0\) is not used since that was used to ﬁnd \(r\). Therefore we start from \(n=1\). For \(n=1\) the above gives \(a_{1}=0\) and same for all \(n\geq 1\). Hence from Eq (A), since \(y=\sum _{n=0}^{\infty }a_{n}x^{n+r}\) then (note: When there is only one \(\sum \) term left in (1) as in this case, then this means there is no recurrence relation and all \(a_{n}=0\) for \(n>0\)).

\begin{align*} y & =c_{1}\left ( \sum _{n=0}^{\infty }a_{n}x^{n+r}\right ) \\ & =c_{1}\left ( \sum _{n=0}^{\infty }a_{n}x^{n-1}\right ) \\ & =c_{1}\left ( a_{0}x^{-1}+0+0+\cdots +O\left ( x\right ) \right ) \end{align*}

3.3.25.5 Regular singular point using Frobenius series method.