4.2.11 Example 11 case one
Solve
\begin{align} x^{2}\left ( x^{2}+x+1\right ) y^{\prime \prime }-x\left ( -2x^{2}-4x+1\right ) y^{\prime }+y & =0\tag {1}\\ y^{\prime \prime }-\frac {x\left ( -2x^{2}-4x+1\right ) }{x^{2}\left ( x^{2}+x+1\right ) }y^{\prime }+\frac {1}{x^{2}\left ( x^{2}+x+1\right ) }y & =0\nonumber \\ y^{\prime \prime }+ay^{\prime }\left ( x\right ) +by & =0\nonumber \end{align}
Hence
\begin{align*} a & =-\frac {x\left ( -2x^{2}-4x+1\right ) }{x^{2}\left ( x^{2}+x+1\right ) }\\ b & =\frac {1}{x^{2}\left ( x^{2}+x+1\right ) }\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\left ( -2x^{2}-4x+1\right ) }{x^{2}\left ( x^{2}+x+1\right ) }\right ) ^{2}+\frac {1}{2}\frac {d}{dx}\left ( -\frac {x\left ( -2x^{2}-4x+1\right ) }{x^{2}\left ( x^{2}+x+1\right ) }\right ) -\frac {1}{x^{2}\left ( x^{2}+x+1\right ) }\nonumber \\ & =\frac {10x^{2}-8x-1}{4x^{2}\left ( x^{2}+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 {10x^{2}-8x-1}{4x^{2}\left ( x^{2}+x+1\right ) ^{2}}z \tag {5}\end{equation}
Step 0 We need to
find which case it is.
\(r=\frac {s}{t}\) where
\begin{align*} s & =10x^{2}-8x-1\\ t & =4x^{2}\left ( x^{2}+x+1\right ) ^{2}\\ & =\left ( x^{3}+x^{2}+x\right ) ^{2}\end{align*}
The free square factorization of \(t\) is \(t=\left [ 1,x^{3}+x^{2}+x\right ] \). Hence
\begin{equation} m=2 \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} & =x^{3}+x^{2}+x \end{align*}
Now
\begin{align*} O\left ( \infty \right ) & =\deg \left ( t\right ) -\deg \left ( s\right ) \\ & =6-2\\ & =4 \end{align*}
There are poles of order 2. 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 all cases are possible. Hence \(L=\left [ 1,2,4,6,12\right ] \).
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 \(O\left ( \infty \right ) =2,t=\left ( x^{3}+x^{2}+x\right ) ^{2},t_{1}=1\) the above gives
\begin{align*} e_{fixed} & =\frac {1}{4}\left ( \min \left ( 6,2\right ) -6-3\left ( 0\right ) \right ) \\ & =\frac {1}{4}\left ( 2-6\right ) \\ & =-1\\ \theta _{fixed} & =\frac {1}{4}\left ( \frac {\frac {d}{dx}\left ( \left ( x^{3}+x^{2}+x\right ) ^{2}\right ) }{\left ( x^{3}+x^{2}+x\right ) ^{2}}+3\left ( 0\right ) \right ) \\ & =\frac {1}{2}\frac {3x^{2}+2x+1}{x^{3}+x^{2}+x}\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 \(t_{2}=x^{3}+x^{2}+x\). In other words, the
number of poles of \(r\) that are of order \(2\). There are three poles of order \(2.\) Hence \(k_{2}=3\). These poles
at \(x=\left \{ 0,-\frac {1}{2}\pm \frac {1}{2}i\sqrt {3}\right \} \). The coefficient of \(\frac {1}{\left ( x-c_{1}\right ) ^{2}}\) where \(c_{1}\) is first pole is \(b_{1}=-\frac {1}{4}\). Hence \(e_{1}=\sqrt {1+4b}=\sqrt {1+4\left ( -\frac {1}{4}\right ) }=0\) and \(\theta _{1}=\frac {e_{1}}{x-c_{1}}=0\). The coefficient of \(\frac {1}{\left ( x-c_{2}\right ) ^{2}}\) where \(c_{2}\) is
second pole is \(b_{2}=\frac {9i\sqrt {3}+2}{3\left ( -1+i\sqrt {3}\right ) ^{2}}\). Hence \(e_{2}=\sqrt {1+4b}=\frac {\sqrt {3}\sqrt {2+30i\sqrt {3}}}{3\left ( -1+i\sqrt {3}\right ) }\) and \(\theta _{2}=\frac {e_{2}}{x-c_{2}}=\frac {-2\sqrt {3}\sqrt {2+30i\sqrt {3}}}{3\left ( -1+i\sqrt {3}\right ) \left ( i\sqrt {3}-2x-1\right ) }\). The coefficient of \(\frac {1}{\left ( x-c_{3}\right ) ^{2}}\) where \(c_{3}\) is the third pole is \(b_{3}=\frac {-9i\sqrt {3}+2}{3\left ( 1+i\sqrt {3}\right ) ^{2}}\). Hence \(e_{3}=\sqrt {1+4b}=\frac {\sqrt {3}\sqrt {2-30i\sqrt {3}}}{3\left ( 1+i\sqrt {3}\right ) }\) and \(\theta _{3}=\frac {e_{3}}{x-c_{3}}=\frac {2\sqrt {3}\sqrt {2-30i\sqrt {3}}}{3\left ( 1+i\sqrt {3}\right ) \left ( i\sqrt {3}+2x+1\right ) }\).
Hence
\begin{align*} e & =\left \{ 0,\frac {\sqrt {3}\sqrt {2+30i\sqrt {3}}}{3\left ( -1+i\sqrt {3}\right ) },\frac {\sqrt {3}\sqrt {2-30i\sqrt {3}}}{3\left ( 1+i\sqrt {3}\right ) }\right \} \\ \theta & =\left \{ 0,\frac {-2\sqrt {3}\sqrt {2+30i\sqrt {3}}}{3\left ( -1+i\sqrt {3}\right ) \left ( i\sqrt {3}-2x-1\right ) },\frac {2\sqrt {3}\sqrt {2-30i\sqrt {3}}}{3\left ( 1+i\sqrt {3}\right ) \left ( i\sqrt {3}+2x+1\right ) }\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 \(4,6,8\) if any exist.
Since only order 2 pole exist, then this is skipped. Hence \(k_{2}\) stays \(3\).
Part(d)
Now we need to find \(e_{0},\theta _{0}\). Since this is case \(O\left ( \infty \right ) =4>2\) and since there are no poles or order \(4,6,8,\cdots \) then we do
not need to handle case \(n=1\). Instead we use
\begin{align*} e_{0} & =1\\ \theta _{0} & =0 \end{align*}
Hence now we have
\begin{align*} e & =\left \{ 1,0,\frac {\sqrt {3}\sqrt {2+30i\sqrt {3}}}{3\left ( -1+i\sqrt {3}\right ) },\frac {\sqrt {3}\sqrt {2-30i\sqrt {3}}}{3\left ( 1+i\sqrt {3}\right ) }\right \} \\ \theta & =\left \{ 0,0,\frac {-2\sqrt {3}\sqrt {2+30i\sqrt {3}}}{3\left ( -1+i\sqrt {3}\right ) \left ( i\sqrt {3}-2x-1\right ) },\frac {2\sqrt {3}\sqrt {2-30i\sqrt {3}}}{3\left ( 1+i\sqrt {3}\right ) \left ( i\sqrt {3}+2x+1\right ) }\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 \(k_{2}=3\) then there are \(\left ( n+1\right ) ^{k_{2}+1}=2^{4}=16\) sets \(s\) to try. The first set \(s\) is
\[ s=\left \{ \frac {-n}{2},\frac {-n}{2},\frac {-n}{2},\frac {-n}{2}\right \} =\left \{ \frac {-1}{2},\frac {-1}{2},\frac {-1}{2},\frac {-1}{2}\right \} \]
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
\(k_{2}=3\) then the above becomes
\begin{equation} d=\left ( n\right ) \left ( e_{fix}\right ) +s_{0}e_{0}-s_{1}e_{1}-s_{2}e_{2}-s_{3}e_{3} \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
\(e_{fix}=-1,\theta _{fixed}=\frac {1}{2}\frac {3x^{2}+2x+1}{x^{3}+x^{2}+x}\) gives
\begin{align*} d & =\left ( 1\right ) \left ( -1\right ) +\left ( \frac {-1}{2}\right ) \left ( +1\right ) -\left ( \frac {-1}{2}\right ) \left ( 0\right ) -\left ( \frac {-1}{2}\right ) \left ( \frac {\sqrt {3}\sqrt {2+30i\sqrt {3}}}{3\left ( -1+i\sqrt {3}\right ) }\right ) -\left ( \frac {-1}{2}\right ) \left ( \frac {\sqrt {3}\sqrt {2-30i\sqrt {3}}}{3\left ( 1+i\sqrt {3}\right ) }\right ) \\ & =-\frac {7}{6}i\sqrt {3}-\frac {3}{2}\end{align*}
Since not an integer, we try next set \(s=\left \{ \frac {-1}{2},\frac {-1}{2},\frac {-1}{2},\frac {+1}{2}\right \} \). If we continue this process we will find that set \(s=\left \{ \frac {1}{2},\frac {-1}{2},\frac {-1}{2},\frac {+1}{2}\right \} \)
works and generates
\begin{align*} d & =\left ( 1\right ) \left ( -1\right ) +\left ( \frac {1}{2}\right ) \left ( +1\right ) -\left ( \frac {-1}{2}\right ) \left ( 0\right ) -\left ( \frac {-1}{2}\right ) \left ( \frac {\sqrt {3}\sqrt {2+30i\sqrt {3}}}{3\left ( -1+i\sqrt {3}\right ) }\right ) -\left ( \frac {1}{2}\right ) \left ( \frac {\sqrt {3}\sqrt {2-30i\sqrt {3}}}{3\left ( 1+i\sqrt {3}\right ) }\right ) \\ & =0 \end{align*}
We can use this \(d.\ \)From Eq (8)
\begin{align*} \Theta & =\left ( n\right ) \left ( \theta _{fix}\right ) +s_{0}\theta _{0}+s_{1}\theta _{1}+s_{2}\theta _{2}+s_{3}\theta _{3}\\ & =\left ( 1\right ) \left ( \frac {1}{2}\frac {3x^{2}+2x+1}{x^{3}+x^{2}+x}\right ) +\left ( \frac {-1}{2}\right ) \left ( 0\right ) +\left ( \frac {-1}{2}\right ) \left ( 0\right ) +\left ( \frac {-1}{2}\right ) \left ( \frac {-2\sqrt {3}\sqrt {2+30i\sqrt {3}}}{3\left ( -1+i\sqrt {3}\right ) \left ( i\sqrt {3}-2x-1\right ) }\right ) +\left ( \frac {1}{2}\right ) \left ( \frac {2\sqrt {3}\sqrt {2-30i\sqrt {3}}}{3\left ( 1+i\sqrt {3}\right ) \left ( i\sqrt {3}+2x+1\right ) }\right ) \\ & =\frac {1}{2x}\frac {2x^{2}-2x+1}{x^{2}+x+1}\end{align*}
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 \(d=0\) and \(\Theta =\frac {1}{2x}\frac {2x^{2}-2x+1}{x^{2}+x+1}\) found from step 2 and also \(r=\frac {10x^{2}-8x-1}{4x^{2}\left ( x^{2}+x+1\right ) ^{2}}\). Since degree \(d=0\), then
let \(p\left ( x\right ) =1\). A constant. We need to verify
\begin{align*} P^{\prime \prime }+2\Theta P^{\prime }+\left ( \Theta ^{\prime }+\Theta ^{2}-r\right ) P & =0\\ \Theta ^{\prime }+\Theta ^{2}-r & =0 \end{align*}
Substituting gives
\begin{align*} \frac {d}{dx}\left ( \frac {1}{2x}\frac {2x^{2}-2x+1}{x^{2}+x+1}\right ) +\left ( \frac {1}{2x}\frac {2x^{2}-2x+1}{x^{2}+x+1}\right ) ^{2}-\frac {10x^{2}-8x-1}{4x^{2}\left ( x^{2}+x+1\right ) ^{2}} & =0\\ 0 & =0 \end{align*}
Verified. The solution is
\begin{align*} z & =P\left ( x\right ) e^{\int \Theta dx}\\ & =e^{\int \frac {1}{2x}\frac {2x^{2}-2x+1}{x^{2}+x+1}dx}\\ & =\left ( x^{2}+x+1\right ) ^{\frac {1}{4}}\sqrt {x}e^{-\frac {7}{6}\sqrt {3}\arctan \left ( \frac {\left ( 2x+1\right ) \sqrt {3}}{3}\right ) }\end{align*}
Hence first solution to given ODE is
\begin{align*} y_{1} & =ze^{\frac {-1}{2}\int adx}\\ & =\left ( x^{2}+x+1\right ) ^{\frac {1}{4}}\sqrt {x}e^{-\frac {7}{6}\sqrt {3}\arctan \left ( \frac {\left ( 2x+1\right ) \sqrt {3}}{3}\right ) }e^{\frac {-1}{2}\int \left ( \frac {1}{x^{2}\left ( x^{2}+x+1\right ) }\right ) dx}\\ & =\frac {xe^{-\frac {7}{3}\sqrt {3}\arctan \left ( \frac {\left ( 2x+1\right ) \sqrt {3}}{3}\right ) }}{\sqrt {x^{2}+x+4}}\end{align*}
Second solution \(y_{2}\) can now be find by reduction of order.