4.2.1 Example 1 case one
Solve
\[ \left ( 1-x^{2}\right ) y^{\prime \prime }-2xy^{\prime }+6y=0 \]
Normalizing so that coefficient of
\(u^{\prime \prime }\) is one gives (assuming
\(x\neq 1\))
\begin{align} y^{\prime \prime }-2\frac {x}{\left ( 1-x^{2}\right ) }y^{\prime }+\frac {6}{\left ( 1-x^{2}\right ) }y & =0\nonumber \\ y^{\prime \prime }+ay^{\prime }\left ( x\right ) +by & =0 \tag {1}\end{align}
Hence
\begin{align*} a & =\frac {-2x}{\left ( 1-x^{2}\right ) }\\ b & =\frac {6}{\left ( 1-x^{2}\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 {-2x}{\left ( 1-x^{2}\right ) }\right ) ^{2}+\frac {1}{2}\frac {d}{dx}\left ( \frac {-2x}{\left ( 1-x^{2}\right ) }\right ) -\frac {6}{\left ( 1-x^{2}\right ) }\nonumber \\ & =\frac {6x^{2}-7}{\left ( x^{2}-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 {6x^{2}-7}{\left ( x^{2}-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
\(t_{1}=1,t_{2}=\left ( x^{2}-1\right ) \). Now
\(O\left ( \infty \right ) =\deg \left ( t\right ) -\deg \left ( s\right ) =4-2=2\). The
poles of
\(r\) are
\(x=1,-1\) each 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.
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 \(O\left ( \infty \right ) =2,t=\left ( x^{2}-1\right ) ^{2},t_{1}=1\) the above gives
\begin{align*} e_{fixed} & =\frac {1}{4}\left ( \min \left ( 2,2\right ) -4-3\left ( 0\right ) \right ) \\ & =\frac {1}{4}\left ( 2-4\right ) \\ & =-\frac {1}{2}\\ \theta _{fixed} & =\frac {1}{4}\left ( \frac {\frac {d}{dx}\left ( \left ( x^{2}-1\right ) ^{2}\right ) }{\left ( x^{2}-1\right ) ^{2}}+3\left ( 0\right ) \right ) \\ & =\frac {x}{x^{2}-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 \(t_{2}=x^{2}-1\). Label these poles \(c_{1},c_{2},\cdots ,c_{k_{2}}\). The zeros of \(t_{2}\) are \(\left \{ 1,-1\right \} \)
therefore \(c_{1}=1,c_{2}=-1\) and \(k_{2}=2\). 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*} \frac {6x^{2}-7}{\left ( x^{2}-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 {1}{4}\frac {1}{\left ( x-1\right ) ^{2}}-\frac {1}{4}\frac {1}{\left ( x+1\right ) ^{2}}\right ) +\left ( \frac {13}{4}\frac {1}{\left ( x-1\right ) }-\frac {13}{4}\frac {1}{x+1}\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 \(\frac {-1}{4}\). Hence \(b=\frac {-1}{4}\)
and \(e_{1}=\sqrt {1+4b}=\sqrt {1-4\frac {1}{4}}=0\). For \(c_{2}=-1\ \)looking at the above, we see that the coefficient of \(\frac {1}{\left ( x+1\right ) ^{2}}\) is \(\frac {-1}{4}\). Hence \(b=\frac {-1}{4}\) and \(e_{2}=\sqrt {1+4b}=\sqrt {1-4\frac {1}{4}}=0\).
Therefore
\begin{align*} e_{1} & =0\\ e_{2} & =0 \end{align*}
Now, \(\theta _{i}=\frac {e_{i}}{x-c_{i}}\). For \(c_{1}=1\) this gives \(\theta _{1}=\frac {e_{1}}{x-1}=0\) since \(e_{1}=0\). And \(\theta _{2}=\frac {e_{2}}{x-c_{2}}\). For \(c_{2}=-1\) this gives \(\theta _{2}=\frac {e_{2}}{x+1}=0\) since \(e_{2}=0\). Hence
\begin{align*} \theta _{1} & =0\\ \theta _{2} & =0 \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 \(4,6,8,\cdots ,M\) if any exist.
Since non exist here. This is skipped. This means
\[ M=2 \]
since
\(2\) is the number of poles of order
2.
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 \(O\left ( \infty \right ) =2\) then
\begin{align*} e_{0} & =\sqrt {1+4b}\\ \theta _{0} & =0 \end{align*}
Now we need to find \(b\). The Laurent series expansion of \(r\) at \(\infty \) is \(\frac {6}{x^{2}}+\frac {5}{x^{4}}+\frac {4}{x^{6}}+\cdots \). Hence \(b=6\). Therefore
\begin{align*} e_{0} & =\sqrt {1+4\left ( 6\right ) }\\ & =\sqrt {25}\\ & =5 \end{align*}
Now we have found all \(e_{i},\theta _{i}\). They are
\begin{align*} e & =\left \{ 5,0,0\right \} \\ \theta & =\left \{ 0,0,0\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 \(e_{fix},\theta _{fix}\). 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
\(M=2\) in this example (number of poles of order 2), then
these these are given by
\[ \Lambda =\begin {bmatrix} -\frac {1}{2} & -\frac {1}{2} & -\frac {1}{2}\\ -\frac {1}{2} & -\frac {1}{2} & +\frac {1}{2}\\ -\frac {1}{2} & +\frac {1}{2} & -\frac {1}{2}\\ -\frac {1}{2} & +\frac {1}{2} & +\frac {1}{2}\\ +\frac {1}{2} & -\frac {1}{2} & -\frac {1}{2}\\ +\frac {1}{2} & -\frac {1}{2} & +\frac {1}{2}\\ +\frac {1}{2} & +\frac {1}{2} & -\frac {1}{2}\\ +\frac {1}{2} & +\frac {1}{2} & +\frac {1}{2}\end {bmatrix} \]
We go over each row one at a time. Trying the first
row
\(s=\left \{ \frac {-1}{2},\frac {-1}{2},\frac {-1}{2}\right \} \) which means
\(s_{0}=\frac {-1}{2},s_{1}=\frac {-1}{2},s_{2}=-\frac {1}{2}\). Hence the first trial
\(d\) is (using Eq (7)) and recalling that
\(e_{fix}=-\frac {1}{2},\theta _{fixed}=\frac {x}{x^{2}-1}\)
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 ( -\frac {1}{2}\right ) +\left ( \frac {-1}{2}\right ) \left ( 5\right ) -\left ( s_{1}e_{1}+s_{2}e_{2}\right ) \\ & =\left ( 1\right ) \left ( -\frac {1}{2}\right ) +\left ( \frac {-1}{2}\right ) \left ( 5\right ) -\left ( \frac {-1}{2}\left ( 0\right ) -\frac {1}{2}\left ( 0\right ) \right ) \\ & =-3 \end{align*}
Since this is negative, then we skip this set \(s\). Now we try the second row of \(\Lambda \) which is \(s=\left \{ \frac {-1}{2},\frac {-1}{2},\frac {+1}{2}\right \} \).
Then above now gives
\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 ( -\frac {1}{2}\right ) +\left ( \frac {-1}{2}\right ) \left ( 5\right ) -\left ( s_{1}e_{1}+s_{2}e_{2}\right ) \\ & =\left ( 1\right ) \left ( -\frac {1}{2}\right ) +\left ( \frac {-1}{2}\right ) \left ( 5\right ) -\left ( \frac {-1}{2}\left ( 0\right ) +\frac {1}{2}\left ( 0\right ) \right ) \\ & =-3 \end{align*}
Since this is negative, then we skip this set \(s\). Now we try the third row of \(\Lambda \) which is \(s=\left \{ \frac {-1}{2},\frac {+1}{2},\frac {-1}{2}\right \} \). Then
above now gives
\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 ( -\frac {1}{2}\right ) +\left ( \frac {-1}{2}\right ) \left ( 5\right ) -\left ( s_{1}e_{1}+s_{2}e_{2}\right ) \\ & =\left ( 1\right ) \left ( -\frac {1}{2}\right ) +\left ( \frac {-1}{2}\right ) \left ( 5\right ) -\left ( \frac {+1}{2}\left ( 0\right ) -\frac {1}{2}\left ( 0\right ) \right ) \\ & =-3 \end{align*}
Since this is negative, then we skip this set \(s\). Now we try the row 4 of \(\Lambda \) which is \(s=\left \{ \frac {-1}{2},\frac {+1}{2},\frac {+1}{2}\right \} \). Then
above now gives
\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 ( -\frac {1}{2}\right ) +\left ( \frac {-1}{2}\right ) \left ( 5\right ) -\left ( s_{1}e_{1}+s_{2}e_{2}\right ) \\ & =\left ( 1\right ) \left ( -\frac {1}{2}\right ) +\left ( \frac {-1}{2}\right ) \left ( 5\right ) -\left ( \frac {+1}{2}\left ( 0\right ) +\frac {1}{2}\left ( 0\right ) \right ) \\ & =-3 \end{align*}
Since this is negative, then we skip this set \(s\). Now we try the row 5 of \(\Lambda \) which is \(s=\left \{ \frac {+1}{2},\frac {-1}{2},\frac {-1}{2}\right \} \). Then
above now gives
\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 ( -\frac {1}{2}\right ) +\left ( \frac {+1}{2}\right ) \left ( 5\right ) -\left ( s_{1}e_{1}+s_{2}e_{2}\right ) \\ & =\left ( 1\right ) \left ( -\frac {1}{2}\right ) +\left ( \frac {+1}{2}\right ) \left ( 5\right ) -\left ( \frac {+1}{2}\left ( 0\right ) +\frac {1}{2}\left ( 0\right ) \right ) \\ & =+2 \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 {x}{x^{2}-1}\right ) +\left ( s_{0}\theta _{0}+s_{1}\theta _{1}+s_{2}\theta _{2}\right ) \\ & =\frac {x}{x^{2}-1}+\left ( \frac {+1}{2}\theta _{0}-\frac {1}{2}\theta _{1}-\frac {1}{2}\theta _{2}\right ) \end{align*}
But all \(\theta _{i}=0\). Therefore
\[ \Theta =\frac {x}{x^{2}-1}\]
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 \(d=2\) and \(\Theta =\frac {x}{x^{2}-1}\) found from step 2 and also \(r=\frac {6x^{2}-7}{\left ( x^{2}-1\right ) ^{2}}\) 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}\\ & =x^{2}+a_{1}x+a_{0}\end{align*}
The goal is to solve for the \(a_{i}\) coefficient. Now depending on case number \(n\), we do the
following. Since we are in case \(n=1\) then
\begin{align*} p_{1} & =-p\\ & =-x^{2}-a_{1}x-a_{0}\\ p_{0} & =-p_{1}^{\prime }-\Theta p_{1}\\ & =-\left ( -x^{2}-a_{1}x-a_{0}\right ) ^{\prime }-\left ( \frac {x}{x^{2}-1}\right ) \left ( -x^{2}-a_{1}x-a_{0}\right ) \\ & =\frac {\left ( xa_{0}-a_{1}-2x+2x^{2}a_{1}+3x^{3}\right ) }{x^{2}-1}\\ p_{-1} & =-p_{0}^{\prime }-\Theta p_{0}-\left ( 1\right ) \left ( 1\right ) rp_{1}\\ & =-\left ( \frac {\left ( xa_{0}-a_{1}-2x+2x^{2}a_{1}+3x^{3}\right ) }{x^{2}-1}\right ) ^{\prime }-\left ( \frac {x}{x^{2}-1}\right ) \left ( \frac {\left ( xa_{0}-a_{1}-2x+2x^{2}a_{1}+3x^{3}\right ) }{x^{2}-1}\right ) -\frac {6x^{2}-7}{\left ( x^{2}-1\right ) ^{2}}\left ( -x^{2}-a_{1}x-a_{0}\right ) \\ & =2\frac {\left ( 2xa_{1}+3a_{0}+1\right ) }{x^{2}-1}\end{align*}
Now we try to solve for \(a_{i}\) using \(p_{-1}\left ( x\right ) =0\). This gives \(2xa_{1}+3a_{0}+1=0\) which gives \(a_{1}=0,a_{0}=-\frac {1}{3}\). Hence this implies
\begin{align*} p\left ( x\right ) & =x^{2}+a_{1}x+a_{0}\\ & =x^{2}-\frac {1}{3}\end{align*}
Since this is case \(n=1\) then
\begin{align*} \omega & =\frac {p^{\prime }}{p}+\Theta \\ & =\frac {2x}{x^{2}-\frac {1}{3}}+\frac {x}{x^{2}-1}\\ & =x\frac {9x^{2}-7}{3x^{4}-4x^{2}+1}\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 ( x\frac {9x^{2}-7}{3x^{4}-4x^{2}+1}\right ) +\left ( x\frac {9x^{2}-7}{3x^{4}-4x^{2}+1}\right ) ^{2} & =\frac {6x^{2}-7}{\left ( x^{2}-1\right ) ^{2}}\\ \frac {6x^{2}-7}{\left ( x^{2}-1\right ) ^{2}} & =\frac {6x^{2}-7}{\left ( x^{2}-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 \omega dx}\\ & =e^{\int \frac {x\left ( 9x^{2}-7\right ) }{3x^{4}-4x^{2}+1}dx}\\ & =e^{\frac {1}{2}\ln \left ( x^{2}-1\right ) +\ln \left ( x^{2}-\frac {1}{3}\right ) }\\ & =\sqrt {x^{2}-1}\left ( x^{2}-\frac {1}{3}\right ) \end{align*}
Hence first solution to the original ode is
\begin{align*} y & =ze^{\frac {-1}{2}\int adx}\\ & =\sqrt {x^{2}-1}\left ( x^{2}-\frac {1}{3}\right ) e^{\frac {-1}{2}\int \frac {-2x}{\left ( 1-x^{2}\right ) }dx}\\ & =\sqrt {x^{2}-1}\left ( x^{2}-\frac {1}{3}\right ) e^{\int \frac {x}{\left ( 1-x^{2}\right ) }dx}\\ & =\sqrt {x^{2}-1}\left ( x^{2}-\frac {1}{3}\right ) e^{\frac {-1}{2}\ln \left ( x^{2}-1\right ) }\\ & =\frac {\sqrt {x^{2}-1}\left ( x^{2}-\frac {1}{3}\right ) }{\sqrt {x^{2}-1}}\\ & =\left ( x^{2}-\frac {1}{3}\right ) \end{align*}