4.1.3 Example 3 case one

Solve

\begin{align} y^{\prime \prime }-x^{2}y^{\prime }-x^{2}y & =0\tag {1}\\ y^{\prime \prime }+ay^{\prime }\left ( x\right ) +by & =0\nonumber \end{align}

Hence

\begin{align*} a & =-x^{2}\\ b & =-x^{2}\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}{4}x^{4}-x+x^{2}\nonumber \\ & =\frac {x^{4}+4x^{2}-4x}{4} \tag {4}\end{align}

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

\begin{equation} z^{\prime \prime }=\left ( \frac {x^{4}+4x^{2}-4x}{4}\right ) z \tag {5}\end{equation}

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

\begin{align*} s & =x^{4}+4x^{2}-4x\\ t & =4 \end{align*}

The free square factorization of \(t\) is \(t=\left [ \left [ {}\right ] \right ] \). Hence

\begin{equation} m=0 \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 ) \\ & =0-4\\ & =-4 \end{align*}

There are no poles. Looking at the cases table giving up, 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 1 meets the necessary conditions. Hence \(L=\left [ 1\right ] \). So \(n=1.\)

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 \(O\left ( \infty \right ) =-4,t=4,t_{1}=4\) the above gives

\begin{align*} e_{fixed} & =\frac {1}{4}\left ( \min \left ( -4,2\right ) -0-3\left ( 0\right ) \right ) \\ & =\frac {1}{4}\left ( -4\right ) \\ & =-1\\ \theta _{fixed} & =\frac {1}{4}\left ( \frac {\frac {d}{dx}\left ( 4\right ) }{\left ( 4\right ) }+3\left ( \frac {\left ( 4\right ) ^{\prime }}{4}\right ) \right ) \\ & =0 \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 \(2\). These will be the zeros of \(t_{2}\) in the above square free factorization of \(t\). From above we found that \(t_{2}=1\). Label these poles \(c_{1},c_{2},\cdots ,c_{k_{2}}\). The zeros of \(t_{2}\) are \(\left \{ {}\right \} \). There are no zeros since constant. Therefore \(k_{2}=0\) since one zero. Hence

\[ M=0 \]

No \(e_{i},\theta _{i}\) are generated. i.e. \(e=\left \{ {}\right \} ,\theta =\left \{ {}\right \} \) so far.

Part (c)

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

Part(d)

Now we need to find \(e_{0},\theta _{0}\). If \(O\left ( \infty \right ) >2\) then \(e_{0}=1,\theta _{0}=0\). But if \(O\left ( \infty \right ) =2\) then \(\theta _{0}=0\) and \(e_{0}=\sqrt {1+4b}\) where \(b\) is the coefficient of \(\frac {1}{x^{2}}\) in the Laurent series expansion of \(r\) at \(\infty \). Since in this example \(O\left ( \infty \right ) =-4\) then none of these cases apply. We fall into the case that handles \(n=1\) only which is the current case which results in \(e_{0}=-2,\theta _{0}=2+x^{2}\). Hence

\begin{align*} e & =\left \{ -2\right \} \\ \theta & =\left \{ 2+x^{2}\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 \(e_{fix},\theta _{fix}\). Since \(n=1\) and \(M=0\) then we have \(\left ( n+1\right ) ^{M+1}=2^{1}=2\) sets \(s\) to try. These are given by

\[ \Lambda =\begin {bmatrix} -\frac {1}{2}\\ +\frac {1}{2}\end {bmatrix} \]

Now we generate trial \(d\) using

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

Since \(M=0\) 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}^{M}s_{i}\theta _{i} \tag {8}\end{equation}

Since \(M=0\). Hence the first trial \(d\) is (using Eq (7)) and recalling that \(e_{fix}=-1,\theta _{fixed}=0\) then

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

Since \(d\geq 0\), then we can use it. Using Eq (8) gives

\begin{align*} \Theta & =\left ( n\right ) \left ( \theta _{fix}\right ) +s_{0}\theta _{0}\\ & =0-\frac {1}{2}\left ( 2+x^{2}\right ) \\ & =-1-\frac {1}{2}x^{2}\end{align*}

Now that we have good trial \(d\) and \(\Theta \), then step 3 is called to generate \(\omega \) if possible.

Step 3

The input to this step is the integer \(d=0\) and \(\Theta =-1-\frac {1}{2}x^{2}\) found from step 2 and also \(r=\frac {x^{4}+4x^{2}-4x}{4}\) which comes from \(z^{\prime \prime }=rz\). This step is broken into these parts. First we find the \(p_{-1}\left ( x\right ) \) polynomial. If we are to solve for its coefficients, then next we build the minimal polynomial from the \(p_{i}\left ( x\right ) \) polynomials constructed during finding \(p_{-1}\left ( x\right ) \). The minimal polynomial \(p_{\min }\left ( x\right ) \) will be a function of \(\omega \). Next we solve for \(\omega \) from \(p_{\min }\left ( x\right ) =0\). If this is successful, then we have found \(\omega \) and the first solution to the ode \(y^{\prime \prime }=ry\) is \(e^{\int \omega dx}\,\). Below shows how this is done.

We start by forming a polynomial

\begin{align*} p\left ( x\right ) & =x^{d}+a_{d-1}x^{d-1}+\cdots +a_{0}\\ & =1 \end{align*}

Since this is case \(n=1\) then

\begin{align*} \omega & =\frac {p^{\prime }}{p}+\Theta \\ & =-1-\frac {1}{2}x^{2}\end{align*}

Before using this, we will verify it is correct. For case 1 the above should satisfy

\[ \omega ^{\prime }+\omega ^{2}=r \]

Let us see if this is the case or not.

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

It did not verify. This means this solution can not be used. If we try the next row in \(\Lambda \) we will find it gives negative \(d\). This means there is no Liouvillian solution. This is an example where even if we find \(d\geq 0\) we still can end up not finding a solution.