4.1.2 Example 2 case one

Solve

\[ x\left ( x-1\right ) ^{2}y^{\prime \prime }-2y=0 \]

Normalizing so that coefficient of \(u^{\prime \prime }\) is one gives (assuming \(x\neq 1\) and \(x\neq 0\))

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

Hence

\begin{align*} a & =0\\ b & =-\frac {2}{x\left ( x-1\right ) ^{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 {2}{x\left ( x-1\right ) ^{2}} \tag {4}\end{align}

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

\begin{equation} z^{\prime \prime }=\frac {2}{x\left ( x-1\right ) ^{2}}z \tag {5}\end{equation}

Step 0 We need to find which case it is. \(r=\frac {s}{t}\). The free square factorization of \(t\) is \(t=t_{1}t_{2}^{2}\). Hence

\begin{equation} m=2 \tag {6}\end{equation}

And

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

Now \(O\left ( \infty \right ) =\deg \left ( t\right ) -\deg \left ( s\right ) =3-0=3\). The poles of \(r\) are \(x=0\) of order 1 and \(x=1\) of order 2.  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 all cases are possible. Hence \(L=\left \{ 1,2,4,6,12\right \} \). So \(n=1,n=2,n=4,n=6,n=12\) will be tried until one is successful. Starting with \(n=1\).

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 \(O\left ( \infty \right ) =3,t=x\left ( x-1\right ) ^{2},t_{1}=x\) the above gives

\begin{align*} e_{fixed} & =\frac {1}{4}\left ( \min \left ( 3,2\right ) -3-3\left ( 1\right ) \right ) \\ & =\frac {1}{4}\left ( 2-3-3\right ) \\ & =-1\\ \theta _{fixed} & =\frac {1}{4}\left ( \frac {\frac {d}{dx}\left ( x\left ( x-1\right ) ^{2}\right ) }{x\left ( x-1\right ) ^{2}}+3\left ( \frac {x^{\prime }}{x}\right ) \right ) \\ & =\frac {1}{4}\left ( \frac {3x^{2}-4x+1}{x\left ( x-1\right ) ^{2}}+\frac {3}{x}\right ) \\ & =\frac {1}{2x}\frac {3x-2}{x-1}\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 \(t_{2}=x-1\). Label these poles \(c_{1},c_{2},\cdots ,c_{k_{2}}\). The zeros of \(t_{2}\) are \(\left \{ 1\right \} \) therefore \(c_{1}=1\) and \(k_{2}=1\) since one zero. Hence

\[ M=1 \]

For each \(c_{i}\) then \(e_{i}=\sqrt {1+4b}\) where \(b\) is the coefficient of \(\frac {1}{\left ( x-c_{i}\right ) ^{2}}\) in the partial fraction expansion of \(r\) which is

\begin{align*} r & =\frac {2}{x\left ( x-1\right ) ^{2}}\\ & =\left ( \sum _{i}\frac {\alpha _{i}}{\left ( x-c_{i}\right ) ^{2}}\right ) +\left ( \sum _{j}\frac {\beta _{j}}{x-d_{j}}\right ) \\ & =\left ( \frac {2}{\left ( x-1\right ) ^{2}}\right ) +\left ( -\frac {2}{x-1}+\frac {2}{x}\right ) \end{align*}

Therefore for \(c_{1}=1\), looking at the above, we see that the coefficient of \(\frac {1}{\left ( x-1\right ) ^{2}}\) is \(2\). Hence \(b=2\) and \(e_{1}=\sqrt {1+4b}=\sqrt {1+8}=3\). Hence

\[ e_{1}=3 \]

Now, \(\theta _{i}=\frac {e_{i}}{x-c_{i}}\). For \(c_{1}=1\) this gives \(\theta _{1}=\frac {e_{1}}{x-1}=\frac {3}{x-1}\). Hence

\[ \theta _{1}=\frac {3}{x-1}\]

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 \(1\).

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 \(O\left ( \infty \right ) =3\) then

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

Now we have found all \(e_{i},\theta _{i}\). They are

\begin{align*} e & =\left \{ 1,3\right \} \\ \theta & =\left \{ 0,\frac {3}{x-1}\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 \(e_{fix},\theta _{fix}\). 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} \tag {7}\end{equation}

Where \(n\) is the case number. For case 1, it will be \(n=1\). For case 2 it will be \(n=2\). For case 3 it will be \(4\) and \(6\) and \(12.\) 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}

We need to first generate \(s\) sets. For \(n=1\) and since \(M=1\) in this example, then these these are given by

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

We go over each row one at a time. Trying the first row \(s=\left \{ \frac {-1}{2},\frac {-1}{2}\right \} \) which means \(s_{0}=\frac {-1}{2},s_{1}=\frac {-1}{2}\). Hence the first trial \(d\) is (using Eq (7)) and recalling that \(e_{fix}=-1,\theta _{fixed}=\frac {1}{2x}\frac {3x-2}{x-1}\) then

\begin{align*} d & =\left ( n\right ) \left ( e_{fix}\right ) +s_{0}e_{0}-\sum _{i=1}^{M}s_{i}e_{i}\\ & =\left ( 1\right ) \left ( -1\right ) +\left ( \frac {-1}{2}\right ) \left ( 1\right ) -\left ( s_{1}e_{1}\right ) \\ & =-1-\frac {1}{2}-\left ( \frac {-1}{2}\left ( 3\right ) \right ) \\ & =0 \end{align*}

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

\begin{align*} \Theta & =\left ( n\right ) \left ( \theta _{fix}\right ) +\sum _{i=0}^{M}s_{i}\theta _{i}\\ & =\left ( 1\right ) \left ( \frac {1}{2x}\frac {3x-2}{x-1}\right ) +\left ( s_{0}\theta _{0}+s_{1}\theta _{1}\right ) \\ & =\frac {1}{2x}\frac {3x-2}{x-1}+\left ( \frac {-1}{2}\theta _{0}-\frac {1}{2}\theta _{1}\right ) \\ & =\frac {1}{2x}\frac {3x-2}{x-1}-\frac {1}{2}\left ( 0\right ) -\frac {1}{2}\frac {3}{x-1}\end{align*}

Therefore

\[ \Theta =-\frac {1}{x\left ( x-1\right ) }\]

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 \(\Theta =-\frac {1}{x\left ( x-1\right ) }\) found from step 2 and also \(r=\frac {2}{x\left ( x-1\right ) ^{2}}\) 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 \(z^{\prime \prime }=rz\) is \(z=e^{\int \omega dx}\,\). 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 \\ & =0-\frac {1}{x\left ( x-1\right ) }\\ & =\frac {-1}{x\left ( x-1\right ) }\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 ( -\frac {1}{x\left ( x-1\right ) }\right ) +\left ( -\frac {1}{x\left ( x-1\right ) }\right ) ^{2} & =\frac {2}{x\left ( x-1\right ) ^{2}}\\ \frac {1}{x^{2}}\frac {2x-1}{\left ( x-1\right ) ^{2}}+\left ( -\frac {1}{x\left ( x-1\right ) }\right ) ^{2} & =\frac {2}{x\left ( x-1\right ) ^{2}}\\ \frac {2}{x\left ( x-1\right ) ^{2}} & =\frac {2}{x\left ( x-1\right ) ^{2}}\end{align*}

Verified. Since solution \(\omega \) is found and verified, then first solution to the ode is

\begin{align*} z & =e^{\int -\frac {1}{x\left ( x-1\right ) }dx}\\ & =e^{\ln x-\ln \left ( x-1\right ) }\\ & =\frac {x}{x-1}\end{align*}

Hence first solution to given ODE is

\begin{align*} y & =ze^{\frac {-1}{2}\int adx}\\ & =\frac {x}{x-1}e^{-\frac {1}{2}\int 0dx}\\ & =\frac {x}{x-1}\end{align*}