4.1.10 Example 10 case two
Solve
\begin{align} \left ( x^{2}+2\right ) y^{\prime \prime }+3xy^{\prime }-y & =0\tag {1}\\ y^{\prime \prime }+\frac {3x}{\left ( x^{2}+2\right ) }y^{\prime }-\frac {1}{\left ( x^{2}+2\right ) }y & =0\nonumber \\ y^{\prime \prime }+ay^{\prime }\left ( x\right ) +by & =0\nonumber \end{align}
Hence
\begin{align*} a & =\frac {3x}{\left ( x^{2}+2\right ) }\\ b & =-\frac {1}{\left ( x^{2}+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 {3x}{\left ( x^{2}+2\right ) }\right ) ^{2}+\frac {1}{2}\frac {d}{dx}\left ( \frac {3x}{\left ( x^{2}+2\right ) }\right ) +\frac {1}{\left ( x^{2}+2\right ) }\nonumber \\ & =\frac {7x^{2}+20}{4\left ( x^{2}+2\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 {7x^{2}+20}{4\left ( x^{2}+2\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 & =7x^{2}+20\\ t & =4\left ( x^{2}+2\right ) ^{2}=16+16x^{2}+4x^{4}\end{align*}
The free square factorization of \(t\) is \(t=\left [ 1,\left ( x^{2}+2\right ) \right ] \). 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} & =\left ( x^{2}+2\right ) \end{align*}
Now
\begin{align*} O\left ( \infty \right ) & =\deg \left ( t\right ) -\deg \left ( s\right ) \\ & =4-2\\ & =2 \end{align*}
There is pole \(x=\pm i\sqrt {2}\) 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 \(O\left ( \infty \right ) =2,t=4\left ( x^{2}+2\right ) ^{2},t_{1}=1\) 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 ( 4\left ( x^{2}+2\right ) ^{2}\right ) }{4\left ( x^{2}+2\right ) ^{2}}+3\left ( 0\right ) \right ) \\ & =\frac {x}{x^{2}+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 \(t_{2}=\left ( x^{2}+2\right ) \). In other words, the number of poles of \(r\) that are of order \(2\). There are two poles of order \(2.\) Hence \(k_{2}=2\). These poles at \(x=\pm i\sqrt {2}\). The coefficient of \(\frac {1}{\left ( x-c_{1}\right ) ^{2}}\) where \(c_{1}\) is first pole is \(b_{1}=-\frac {3}{16}\). Hence \(e_{1}=\sqrt {1+4b}=\sqrt {1+4\left ( -\frac {3}{16}\right ) }=\frac {1}{2}\) and \(\theta _{1}=\frac {e_{1}}{x-c_{1}}=\frac {1}{2\left ( x-i\sqrt {2}\right ) }\). The coefficient of \(\frac {1}{\left ( x-c_{2}\right ) ^{2}}\) where \(c_{2}\) is second pole is \(b_{2}=-\frac {3}{16}\). Hence \(e_{2}=\sqrt {1+4b}=\sqrt {1+4\left ( -\frac {3}{16}\right ) }=\frac {1}{2}\) and \(\theta _{2}=\frac {e_{1}}{x-c_{2}}=\frac {1}{2\left ( x+i\sqrt {2}\right ) }\) Hence
\begin{align*} e & =\left \{ \frac {1}{2},\frac {1}{2}\right \} \\ \theta & =\left \{ \frac {1}{2\left ( x-i\sqrt {2}\right ) },\frac {1}{2\left ( x+i\sqrt {2}\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 \(2\).
Part(d)
Now we need to find \(e_{0},\theta _{0}\). Since this is case \(O\left ( \infty \right ) =2\), then \(e_{0}=\sqrt {1+4b}\) where \(b=\frac {lcoeff\left ( s\right ) }{lcoeff\left ( t\right ) }\) where \(\,lcoeff\left ( s\right ) \) is leading coefficient of \(s=7x^{2}+20\) which is \(7\) and \(lcoeff\left ( t\right ) \) is leading coefficient of \(t=16+16x^{2}+4x^{4}\) which is \(4\). Hence \(b=\frac {7}{4}\). Therefore
\begin{align*} e_{0} & =\sqrt {1+4b}=\sqrt {1+4\left ( \frac {7}{4}\right ) }=2\sqrt {2}\\ \theta _{0} & =0 \end{align*}
Hence now we have
\begin{align*} e & =\left \{ 2\sqrt {2},\frac {1}{2},\frac {1}{2}\right \} \\ \theta & =\left \{ 0,\frac {1}{2\left ( x-i\sqrt {2}\right ) },\frac {1}{2\left ( x+i\sqrt {2}\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 \(k_{2}=2\) then there are \(\left ( n+1\right ) ^{k_{2}+1}=2^{3}=8\) sets \(s\) to try. The first set \(s\) is
\[ s=\left \{ \frac {-n}{2},\frac {-n}{2},\frac {-n}{2}\right \} =\left \{ \frac {-1}{2},\frac {-1}{2},\frac {-1}{2}\right \} \]
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 \(k_{2}=2\) 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} \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 \(e_{fix}=-\frac {1}{2},\theta _{fixed}=\frac {x}{x^{2}+2}\) gives
\begin{align*} d & =\left ( 1\right ) \left ( -\frac {1}{2}\right ) +\left ( \frac {-1}{2}\right ) \left ( 2\sqrt {2}\right ) -\left ( \frac {-1}{2}\right ) \left ( \frac {1}{2}\right ) -\left ( \frac {-1}{2}\right ) \left ( \frac {1}{2}\right ) \\ & =-\sqrt {2}\end{align*}
Since not an integer, we try next set \(s=\left \{ \frac {-1}{2},\frac {-1}{2},\frac {+1}{2}\right \} \) and now Eq (7) gives
\begin{align*} d & =\left ( 1\right ) \left ( -\frac {1}{2}\right ) +\left ( \frac {-1}{2}\right ) \left ( 2\sqrt {2}\right ) -\left ( \frac {-1}{2}\right ) \left ( \frac {1}{2}\right ) -\left ( \frac {+1}{2}\right ) \left ( \frac {1}{2}\right ) \\ & =-\sqrt {2}-\frac {1}{2}\end{align*}
Since not an integer, we try next set \(s=\left \{ \frac {-1}{2},\frac {+1}{2},\frac {-1}{2}\right \} \). If we continue this way we will find that all sets \(s\) will fail to generate a \(d\geq 0\). Hence case one did not work. Now we go to case 2 \(\left ( n=2\right ) \).
Starting with \(n=2\). And since we have \(k_{2}=2\) then there are \(\left ( n+1\right ) ^{k_{2}+1}=3^{3}\) sets \(s\) to try. The first set \(s\) is
\[ s=\left \{ \frac {-n}{2},\frac {-n}{2},\frac {-n}{2}\right \} =\left \{ -1,-1,-1\right \} \]
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 \(k_{2}=2\) 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} \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 \(e_{fix}=-\frac {1}{2},\theta _{fixed}=\frac {x}{x^{2}+2}\) gives
\begin{align*} d & =\left ( 2\right ) \left ( -\frac {1}{2}\right ) +\left ( -1\right ) \left ( 2\sqrt {2}\right ) -\left ( -1\right ) \left ( \frac {1}{2}\right ) -\left ( -1\right ) \left ( \frac {1}{2}\right ) \\ & =-2\sqrt {2}\end{align*}
Since not an integer, we try next set \(s=\left \{ -1,-1,+1\right \} \). If we continue this way we will find that set \(s=\left \{ 0,-1,-1\right \} \) works.
\begin{align*} d & =\left ( 2\right ) \left ( -\frac {1}{2}\right ) +\left ( 0\right ) \left ( 2\sqrt {2}\right ) -\left ( -1\right ) \left ( \frac {1}{2}\right ) -\left ( -1\right ) \left ( \frac {1}{2}\right ) \\ & =0 \end{align*}
We can use this \(d.\ \)From Eq (8)
\begin{align*} \Theta & =\left ( 2\right ) \left ( \frac {1}{2x}\right ) +s_{0}\theta _{0}+s_{1}\theta _{1}+s_{2}\theta _{2}\\ & =\left ( 2\right ) \left ( \frac {x}{x^{2}+2}\right ) \left ( 0\right ) \left ( 0\right ) -\left ( -1\right ) \left ( \frac {1}{2\left ( x-i\sqrt {2}\right ) }\right ) -\left ( -1\right ) \left ( \frac {1}{2\left ( x+i\sqrt {2}\right ) }\right ) \\ & =\frac {x}{x^{2}+2}\end{align*}
Since this is case 2 (\(n=2\)) then we need to first find \(P\left ( x\right ) \). The degree is \(d=0\). Hence constant. Say \(P=1\). But we need to verify this is valid. Setting up the equation
\[ P^{\prime \prime \prime }+3\Theta P^{\prime \prime }+\left ( 3\Theta ^{2}+3\Theta ^{\prime }-4r\right ) P^{\prime }+\left ( \Theta ^{\prime \prime }+3\Theta \Theta ^{\prime }+\Theta ^{3}-4r\Theta -2r^{\prime }\right ) P=0 \]
Which simplifies to (since \(P=1\))
\[ \Theta ^{\prime \prime }+3\Theta \Theta ^{\prime }+\Theta ^{3}-4r\Theta -2r^{\prime }=0 \]
Using \(\Theta =\frac {x}{x^{2}+2},r=\frac {7x^{2}+20}{4\left ( x^{2}+2\right ) ^{2}}\) the above reduces to
\[ 0=0 \]
Hence \(P\left ( x\right ) =1\) can be used. Now let
\begin{align*} \phi & =\Theta +\frac {P^{\prime }}{P}\\ & =\frac {x}{x^{2}+2}\end{align*}
We now need to solve for \(\omega \) from (notice that original Kovacic paper has \(+\) and not \(-\) after first term in the following equation. The \(+\) is from Smith paper. It seems to have been a typo in original paper as this version gives the correct solution).
\begin{align*} \omega ^{2}-\phi \omega +\left ( \frac {1}{2}\phi ^{\prime }+\frac {1}{2}\phi ^{2}-r\right ) & =0\\ \omega ^{2}-\left ( \frac {x}{x^{2}+2}\right ) +\omega \left ( \frac {1}{2}\left ( \frac {x}{x^{2}+2}\right ) ^{\prime }+\frac {1}{2}\left ( \frac {x}{x^{2}+2}\right ) ^{2}-\frac {7x^{2}+20}{4\left ( x^{2}+2\right ) ^{2}}\right ) & =0\\ \frac {4\omega ^{2}x^{4}-4\omega x^{3}+\left ( 16\omega ^{2}-7\right ) x^{2}-8x\omega +16\omega ^{2}-16}{4\left ( x^{2}+2\right ) ^{2}} & =0\\ 4\omega ^{2}x^{4}-4\omega x^{3}+\left ( 16\omega ^{2}-7\right ) x^{2}-8x\omega +16\omega ^{2}-16 & =0 \end{align*}
Solving for \(\omega \) (and picking first root) gives
\[ \omega =\frac {x+2\sqrt {2x^{2}+4}}{2\left ( x^{2}+2\right ) }\]
Before using this, we verify it satisfies \(\omega ^{\prime }+\omega ^{2}=r\)
\begin{align*} \frac {d}{dx}\left ( \frac {x+2\sqrt {2x^{2}+4}}{2\left ( x^{2}+2\right ) }\right ) +\left ( \frac {x+2\sqrt {2x^{2}+4}}{2\left ( x^{2}+2\right ) }\right ) ^{2} & =\frac {7x^{2}+20}{4\left ( x^{2}+2\right ) ^{2}}\\ \frac {7x^{2}+20}{4x^{4}+16x^{2}+16} & =\frac {7x^{2}+20}{4\left ( x^{2}+2\right ) ^{2}}\end{align*}
Verified OK. Hence solution is
\begin{align*} z & =e^{\int \omega dx}\\ & =e^{\int \frac {x+2\sqrt {2x^{2}+4}}{2\left ( x^{2}+2\right ) }dx}\\ & =\left ( x^{2}+2\right ) ^{\frac {1}{4}}e^{\sqrt {2}\operatorname {arcsinh}\left ( \frac {\sqrt {2}}{2}x\right ) }\end{align*}
Hence first solution to given ODE is
\begin{align*} y_{1} & =ze^{\frac {-1}{2}\int adx}\\ & =\left ( x^{2}+2\right ) ^{\frac {1}{4}}e^{\sqrt {2}\operatorname {arcsinh}\left ( \frac {\sqrt {2}}{2}x\right ) }e^{-\frac {1}{2}\int \frac {3x}{\left ( x^{2}+2\right ) }dx}\\ & =\frac {e^{\sqrt {2}\operatorname {arcsinh}\left ( \frac {\sqrt {2}}{2}x\right ) }}{\sqrt {\left ( x^{2}+2\right ) }}\end{align*}
Second solution \(y_{2}\) can now be find by reduction of order.