Regular singular point using Frobenius series method.

ode internal name "first_order_ode_series_method_regular_singular_point"

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

Example 1

\[ y^{\prime }+2xy=\sqrt {x}\]

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*}

The (homogenous) ode becomes

\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*}

Reindex so all powers on \(x\) are the lowest gives

\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}

For \(n=0\,\), Eq(1) gives

\[ ra_{0}x^{r-1}=0 \]

Hence \(r=0\) since \(a_{0}\neq 0\). Therefore the balance equation is

\[ mc_{0}x^{m-1}=\sqrt {x}\]

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*}

For \(n\geq 2\) the recurrence relation is from (1)

\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}

Or for \(r=0\) the above simplifies to

\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

\[ a_{2}=-1 \]

For \(n=3\)

\[ a_{3}=-\frac {2}{3}a_{1}=0 \]

For \(n=4\)

\[ a_{4}=-\frac {1}{2}a_{2}=\frac {1}{2}\]

For \(n=5,7,\cdots \) and all odd \(n\) then \(a_{n}=0\). For \(n=6\)

\[ a_{6}=-\frac {1}{3}a_{4}=-\frac {1}{6}\]

And so on. Hence (using \(a_{0}=1\))

\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

\[ ra_{0}x^{r-1}=x^{\frac {1}{2}}\]

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

\[ mc_{0}x^{m-1}=x^{\frac {1}{2}}\]

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

\[ \left ( n+r\right ) a_{n}+2a_{n-2}=0 \]

Hence it becomes

\[ \left ( n+m\right ) c_{n}+2c_{n-2}=0 \]

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

\[ c_{1}=0 \]

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*}

Hence for \(n=2\)

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

For \(n=3\)

\[ c_{3}=\frac {-2c_{1}}{\left ( 3+\frac {3}{2}\right ) }=0 \]

For \(n=4\)

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

And so on. Hence

\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*}

Hence the final solution is

\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*}

Example 2

\[ y^{\prime }+2xy=\frac {1}{x}\]

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

\[ ra_{0}x^{r-1}=\frac {1}{x}\]

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

\[ mc_{0}x^{m-1}=x^{-1}\]

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.

Example 3

\[ y^{\prime }=\frac {1}{x}\]

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*}

The (homogenous) ode becomes

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

For \(n=0\)

\[ ra_{0}x^{r-1}=0 \]

Hence \(r=0\) since \(a_{0}\neq 0\). Therefore the ode satisfies

\[ y^{\prime }=ra_{0}x^{r-1}\]

Eq (1) becomes

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

Therefore for all \(n\geq 1\) we have \(a_{n}=0\). Hence

\[ y_{h}=a_{0}\]

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

\[ ra_{0}x^{r-1}=\frac {1}{x}\]

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

\[ mc_{0}x^{m-1}=x^{-1}\]

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 \(y\left ( x\right ) =c+\ln x\).

Example 4

\[ y^{\prime }=\frac {1}{x^{2}}\]

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*}

The (homogenous) ode becomes

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

For \(n=0\)

\[ ra_{0}x^{r-1}=0 \]

Hence \(r=0\) since \(a_{0}\neq 0\). Therefore the balance equation is

\[ ra_{0}x^{r-1}=\frac {1}{x^{2}}\]

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}

Eq (1) becomes, where \(r=0\) now

\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

\[ y_{h}=c_{1}\left ( a_{0}+O\left ( x\right ) \right ) \]

Letting \(a_{0}=1\) the above becomes

\[ y_{h}=c_{1}\left ( 1+O\left ( x\right ) \right ) \]

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

\[ mc_{0}x^{m-1}=x^{-2}\]

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

\[ y_{p}=x^{m}\sum _{n=0}^{\infty }c_{0}x^{n}\]

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*}

Hence the solution is

\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*}

If we to ignore the big \(O\), the above becomes

\[ y=c_{1}-\frac {1}{x}\]

To verify, we see that \(y^{\prime }=\frac {1}{x^{2}}\).

Example 5

\[ y^{\prime }+\frac {y}{x}=0 \]

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}

The ode becomes

\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}

For \(n=0\)

\[ \left ( r+1\right ) a_{0}=0 \]

Hence \(r=-1\) since \(a_{0}\neq 0\). Eq (1) becomes, where \(r=-1\) now

\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*}

Letting \(a_{0}=1\) the above becomes

\[ y=c_{1}\left ( x^{-1}+O\left ( x\right ) \right ) \]