Example 13 case one

4.2.13 Example 13 case one

Let

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

Hence

\begin{align*} a & =\frac {x}{1-x}\\ b & =-\frac {1}{1-x}\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}\left ( \frac {x}{1-x}\right ) ^{2}+\frac {1}{2}\frac {d}{dx}\left ( \frac {x}{1-x}\right ) -\left ( -\frac {1}{1-x}\right ) \nonumber \\ & =\frac {x^{2}-4x+6}{4\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 {x^{2}-4x+6}{4\left ( x-1\right ) ^{2}}z \tag {5}\end{equation}

Therefore

\begin{align} r & =\frac {s}{t}\nonumber \\ & =\frac {x^{2}-4x+6}{4\left ( x-1\right ) ^{2}}\nonumber \end{align}

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

\begin{align*} s & =x^{2}-4x+6\\ t & =4\left ( x-1\right ) ^{2}\end{align*}

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

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

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

\begin{align*} t_{1} & =1\\ t_{2} & =\left ( x-1\right ) \end{align*}

Now

\begin{align*} O\left ( \infty \right ) & =\deg \left ( t\right ) -\deg \left ( s\right ) \\ & =2-2\\ & =0 \end{align*}

There is one pole at \(x=1\) of order 2. Looking at the cases table


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 1,2 are possible. Hence \(L=\left [ 1,2\right ] \).

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 \(O\left ( \infty \right ) =0,t=4\left ( x-1\right ) ^{2},t_{1}=1\) the above gives

\begin{align*} e_{fixed} & =\frac {1}{4}\left ( \min \left ( 0,2\right ) -2-3\left ( 0\right ) \right ) \\ & =\frac {1}{4}\left ( 0-2\right ) \\ & =-\frac {1}{2}\\ \theta _{fixed} & =\frac {1}{4}\left ( \frac {\frac {d}{dx}\left ( 4\left ( x-1\right ) ^{2}\right ) }{4\left ( x-1\right ) ^{2}}+3\left ( 0\right ) \right ) \\ & =\frac {1}{2x-2}\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 \(t_{2}=\left ( x-1\right ) \). In other words, the number of poles of \(r\) that are of order \(2\). There is one pole of order 2. Hence \(k_{2}=1\). For each pole \(c_{i}\) 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 \(r\) and \(\theta _{i}=\frac {e_{i}}{x-c_{i}}\). The partial fraction expansion of \(r\) is

\[ \frac {x^{2}-4x+6}{4\left ( x-1\right ) ^{2}}=\frac {1}{4}+\frac {3}{4\left ( x-1\right ) ^{2}}-\frac {1}{2}\frac {1}{x-1}\]

The coeﬃcient of \(\frac {1}{\left ( x-1\right ) ^{2}}\) is \(b_{1}=\frac {3}{4}\) from looking at the above. Hence \(e_{1}=\sqrt {1+4b}=\sqrt {1+4\left ( \frac {3}{4}\right ) }=\allowbreak 2\) and \(\theta _{1}=\frac {e_{1}}{x-c_{1}}=\frac {\allowbreak 2}{x-1}\).Therefore the lists \(e,\theta \) are

\begin{align*} e & =\left \{ 2\right \} \\ \theta & =\left \{ \frac {2}{x-1}\right \} \end{align*}

Part (c)

This part applied only to case 1. It is used to generate \(e_{i},\theta _{i}\) for poles of \(r\) order \(4,6,8,\cdots ,k\) if any exist. There are none. This step is skipped.

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 \(O\left ( \infty \right ) =0\) here then none of these cases applies. For case 1 \(\left ( n=1\right ) \) we ﬁrst ﬁnd \(\left [ r\right ] _{\infty }\) the sum of terms \(x^{i}\) for \(i=-\frac {v}{2},\cdots 0\) where \(v\) is \(O\left ( \infty \right ) \) which is zero here. Hence \(v=0\). This sum of terms is from the Laurent series expansion of \(\sqrt {r}\) at \(x=\infty \) which is

\[ \left [ \sqrt {r}\right ] _{\infty }=\frac {1}{2}-\frac {1}{2x}+\frac {1}{x^{3}}+\cdots \]

We want only terms for \(0\leq i\leq v\) but \(v=0\). Therefore only the constant term. Hence

\[ \left [ \sqrt {r}\right ] _{\infty }=\frac {1}{2}\]

Then \(a\) is the coeﬃcient of \(x^{-\frac {v}{2}}=x^{0}\) or constant term. Hence

\[ a=\frac {1}{2}\]

And \(b\) is the coeﬃcient of \(x^{\frac {-v}{2}+1}=x\) in \(r-\left ( \left [ \sqrt {r}\right ] _{\infty }\right ) ^{2}\). This comes out to be

\[ b=-\frac {1}{2}\]

Therefore

\begin{align*} e_{0} & =\frac {b}{a}=\frac {-\frac {1}{2}}{\frac {1}{2}}=-1\\ \theta _{0} & =2\left [ \sqrt {r}\right ] _{\infty }=1 \end{align*}

Hence now we have

\begin{align*} e & =\left \{ -1,2\right \} \\ \theta & =\left \{ 1,\frac {2}{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 \(e_{fix},\theta _{fix}\).

Starting with \(n=1\). And since we have \(k_{2}=1\) then there are \(\left ( n+1\right ) ^{k_{2}+1}=2^{2}=4\) sets \(s\) to try. The ﬁrst set \(s\) is

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

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 \(k_{2}=1\) then the above becomes

\begin{equation} d=\left ( n\right ) \left ( e_{fix}\right ) +s_{0}e_{0}-s_{1}e_{1} \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 \(e_{fix}=-\frac {1}{2},\theta _{fixed}=\frac {1}{2x-2}\) gives using set \(\left \{ \frac {-1}{2},\frac {-1}{2}\right \} \)

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

This will work. The corresponding \(\Theta \) is from (8)

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

Let us ﬁnd all of the \(d\) and \(\Theta \) so to compare with the solution to same ode using original kovacic algorithm given earlier to see if we get same \(d^{\prime }s\). We try next set \(s=\left \{ \frac {-1}{2},\frac {+1}{2}\right \} \)

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

We skip this \(d\) since negative. Next is \(s=\left \{ \frac {+1}{2},\frac {-1}{2}\right \} \)

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

The corresponding \(\Theta \) is from (8)

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

The next set is \(\left \{ \frac {+1}{2},\frac {+1}{2}\right \} \)

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

OK, we have all \(d\) values. We now try the ones which are \(d\geq 0\) and these are \(d=0,d=1\). Let us use \(d=1\) case. Now that we have good trial \(d\) and \(\Theta \), then step 3 is called to generate \(P\left ( x\right ) \) if possible.

Step 3

The input to this step is the integer \(d=1\) and \(\Theta =-\frac {1}{2}\frac {x}{x-1}\) found from step 2 and also \(r=\frac {x^{2}-4x+6}{4\left ( x-1\right ) ^{2}}\) which comes from \(z^{\prime \prime }=rz\). Since degree \(d=1\), then let \(p\left ( x\right ) =x+a\). Solving for \(p\left ( x\right ) \) from

\[ P^{\prime \prime }+2\Theta P^{\prime }+\left ( \Theta ^{\prime }+\Theta ^{2}-r\right ) P=0 \]

gives \(p\left ( x\right ) =x\) as solution. Hence the solution is

\begin{align*} z & =P\left ( x\right ) e^{\int \Theta dx}\\ & =xe^{\int -\frac {1}{2}\frac {x}{x-1}dx}\\ & =x\frac {e^{-\frac {1}{2}x}}{\sqrt {x-1}}\end{align*}

Hence ﬁrst solution to given ODE is

\begin{align*} y_{1} & =x\frac {e^{-\frac {1}{2}x}}{\sqrt {x-1}}e^{\frac {-1}{2}\int adx}\\ & =x\frac {e^{-\frac {1}{2}x}}{\sqrt {x-1}}e^{\frac {-1}{2}\int \frac {x}{1-x}dx}\\ & =x \end{align*}

The second solution can be found by reduction of order.

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