5.4.2.8 Example 8 case two

Solve

\begin{align} y^{\prime \prime } & =\frac {1}{x^{3}}y\tag {1}\\ y^{\prime \prime }+ay^{\prime }\left ( x\right ) +by & =0\nonumber \end{align}

Hence

\begin{align*} a & =0\\ b & =-\frac {1}{x^{3}}\end{align*}

It is first transformed to the following ode by eliminating the first derivative

\begin{equation} z^{\prime \prime }=rz \tag {2}\end{equation}

Using what is known as the Liouville transformation given by

\begin{equation} y=ze^{\frac {-1}{2}\int adx} \tag {3}\end{equation}

Where it can be found that \(r\) in (2) is given by

\begin{align} r & =\frac {1}{4}a^{2}+\frac {1}{2}a^{\prime }-b\nonumber \\ & =\frac {1}{x^{3}} \tag {4}\end{align}

Hence the DE we will solve using Kovacic algorithm is Eq (2) which is

\begin{equation} z^{\prime \prime }=\frac {1}{x^{3}}z \tag {5}\end{equation}

Step 0 We need to find which case it is. \(r=\frac {s}{t}\) where

\begin{align*} s & =1\\ t & =x^{3}\end{align*}

The free square factorization of \(t\) is \(t=\left [ 1,1,x\right ] \). Hence

\begin{equation} m=3 \tag {6}\end{equation}

Since \(m\) is number of elements in the free square factorization. in this special case we set

\begin{align*} t_{1} & =1\\ t_{2} & =1 \end{align*}

Now

\begin{align*} O\left ( \infty \right ) & =\deg \left ( t\right ) -\deg \left ( s\right ) \\ & =3-0\\ & =3 \end{align*}

There is pole \(x=0\) of order 3. Looking at the cases table, reproduced here

case	allowed pole order for \(r=\frac {s}{t}\)	allowed \(O\left ( \infty \right ) \) order	\(L\)
1	\(\left \{ 0,1,2,4,6,8,\cdots \right \} \)	\(\left \{ \cdots ,-8,-6,-4,-2,0,2,3,4,5,6,7,\cdots \right \} \)	\(\left [ 1\right ] \)
2	\(\left \{ 2,3,5,7,9,\cdots \right \} \)	no condition	\(\left [ 2\right ] \)
3	\(\left \{ 1,2\right \} \)	\(\left \{ 2,3,4,5,6,7,\cdots \right \} \)	\(\left [ 4,6,12\right ] \)

Shows that only case 2 is possible (since odd pole is only allowed in case 2). Hence \(L=\left [ 2\right ] \).

Step 1

This step has 4 parts (a,b,c,d).

part (a) Here the fixed parts \(e_{fixed},\theta _{fixed}\) are calculated using

\begin{align*} e_{fixed} & =\frac {1}{4}\left ( \min \left ( O\left ( \infty \right ) ,2\right ) -\deg \left ( t\right ) -3\deg \left ( t_{1}\right ) \right ) \\ \theta _{fixed} & =\frac {1}{4}\left ( \frac {t^{\prime }}{t}+3\frac {t_{1}^{\prime }}{t_{1}}\right ) \end{align*}

Using \(O\left ( \infty \right ) =1,t=x^{3},t_{1}=1\) the above gives

\begin{align*} e_{fixed} & =\frac {1}{4}\left ( \min \left ( 3,2\right ) -3-3\left ( 0\right ) \right ) \\ & =\frac {1}{4}\left ( 2-3\right ) \\ & =-\frac {1}{4}\\ \theta _{fixed} & =\frac {1}{4}\left ( \frac {\frac {d}{dx}\left ( x^{3}\right ) }{x^{3}}+3\left ( 0\right ) \right ) \\ & =\frac {3}{4x}\end{align*}

part (b)

Here the values \(e_{i},\theta _{i}\) are found for \(i=1\cdots k_{2}\) where \(k_{2}\) is the number of roots of \(t_{2}\). In other words, the number of poles of \(r\) that are of order \(2\). There are no poles of order 2. Hence \(k_{2}=0\).

Part (c)

This part applied only to case 1. It is used to generate \(e_{i},\theta _{i}\) for poles of \(r\) order \(4,6,8,\cdots \) if any exist. Since only case 2 exist in this example. This is skipped. Hence \(k_{2}\) stays \(0\).

Part(d)

Now we need to find \(e_{0},\theta _{0}\). If \(O\left ( \infty \right ) >2\) then \(e_{0}=1,\theta _{0}=0\). Hence

\begin{align*} e_{0} & =1\\ \theta _{0} & =0 \end{align*}

Hence now we have

\begin{align*} e & =\left \{ 1\right \} \\ \theta & =\left \{ 0\right \} \end{align*}

The above are arranged such that \(e_{0}\) is the first entry. Same for \(\theta \). This to keep the same notation as in the paper. The above complete step 1, which is to generates the candidate \(e^{\prime }s\) and \(\theta ^{\prime }s\). In step 2, these are used to generate trials \(d\) and \(\theta \) and find from them \(P\left ( x\right ) \) polynomial if possible.

Step 2

In this step, we now have all the \(e_{i},\theta _{i}\) values found above in addition to \(e_{fix},\theta _{fix}\).

Starting with \(n=2\).  Since case 2 only applies here. And since we have \(k_{2}=0\) then there are \(\left ( n+1\right ) ^{k_{2}+1}=3\) sets \(s\) to try. The first set \(s\) is

\[ s=\left \{ \frac {-n}{2}\right \} =\left \{ -1\right \} \]

Now we generate trial \(d\) using

\begin{equation} d=\left ( n\right ) \left ( e_{fix}\right ) +s_{0}e_{0}-\sum _{i=1}^{k_{2}}s_{i}e_{i}\nonumber \end{equation}

Since \(k_{2}=0\) then the above becomes

\begin{equation} d=\left ( n\right ) \left ( e_{fix}\right ) +s_{0}e_{0} \tag {7}\end{equation}

If \(d\geq 0\) then we go and find a trial \(\Theta \). We need to have both \(d,\Theta \) to go to the next step.  \(\Theta \) is found using

\begin{equation} \Theta =\left ( n\right ) \left ( \theta _{fix}\right ) +\sum _{i=0}^{k_{2}}s_{i}\theta _{i} \tag {8}\end{equation}

Hence the first trial \(d\) is (using Eq (7)) and recalling that \(e_{fix}=-\frac {1}{4},\theta _{fixed}=\frac {3}{4x}\) gives

\begin{align*} d & =\left ( 2\right ) \left ( -\frac {1}{4}\right ) +\left ( -1\right ) \left ( 1\right ) \\ & =-\frac {3}{2}\end{align*}

Since negative then we can not use it. Now we try the next set \(s=\left \{ 0\right \} \). Then Eq(7) gives

\begin{align*} d & =\left ( 2\right ) \left ( -\frac {1}{4}\right ) +\left ( 0\right ) \left ( 1\right ) \\ & =-\frac {1}{2}\end{align*}

Since negative then we can not use it. Now we try the last set \(s=\left \{ +1\right \} \). Then Eq(7) gives

\begin{align*} d & =\left ( 2\right ) \left ( -\frac {1}{4}\right ) +\left ( +1\right ) \left ( 1\right ) \\ & =\frac {1}{2}\end{align*}

Since not an integer, then we can not use it. We are run out of sets \(s\) to try. Therefore there is no Liouvillian solution.