Example 4 case one

4.2.4 Example 4 case one

Solve

\begin{align} \left ( x^{3}+1\right ) y^{\prime \prime }+7x^{2}y^{\prime }+9xy & =0\tag {1}\\ y^{\prime \prime }+ay^{\prime }\left ( x\right ) +by & =0\nonumber \end{align}

Hence

\begin{align*} a & =\frac {7x^{2}}{\left ( x^{3}+1\right ) }\\ b & =\frac {9x}{\left ( x^{3}+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 {7x^{2}}{\left ( x^{3}+1\right ) }\right ) ^{2}+\frac {1}{2}\left ( \frac {d}{dx}\left ( \frac {7x^{2}}{\left ( x^{3}+1\right ) }\right ) \right ) -\frac {9x}{\left ( x^{3}+1\right ) }\nonumber \\ & =\frac {1}{4}\left ( \frac {7x^{2}}{\left ( x^{3}+1\right ) }\right ) ^{2}+\frac {1}{2}\left ( \frac {-7x\left ( x^{3}-2\right ) }{\left ( x^{3}+1\right ) ^{2}}\right ) -\frac {9x}{\left ( x^{3}+1\right ) }\nonumber \\ & =\frac {-x\left ( x^{3}+8\right ) }{4\left ( x^{3}+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\left ( x^{3}+8\right ) }{4\left ( x^{3}+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 & =-x\left ( x^{3}+8\right ) \\ t & =4\left ( x^{3}+1\right ) ^{2}\end{align*}

The free square factorization of \(t\) is \(t=\left [ 1,x^{3}+1\right ] \). 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}+1 \end{align*}

Now

\begin{align*} O\left ( \infty \right ) & =\deg \left ( t\right ) -\deg \left ( s\right ) \\ & =6-4\\ & =2 \end{align*}

There are three poles 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 three 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 \(O\left ( \infty \right ) =2,t=4\left ( x^{3}+1\right ) ^{2},t_{1}=1\) the above gives

\begin{align*} e_{fixed} & =\frac {1}{4}\left ( \min \left ( 2,2\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 ( 4\left ( x^{3}+1\right ) ^{2}\right ) }{4\left ( x^{3}+1\right ) ^{2}}+3\left ( 0\right ) \right ) \\ & =\frac {3}{2}\frac {x^{2}}{x^{3}+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 \(2\).

\[ r=-\frac {x\left ( x^{3}+8\right ) }{4\left ( x^{3}+1\right ) ^{2}}\]

These will be the zeros of \(t_{2}\) in the above square free factorization of \(t\). From above we found that

\[ t_{2}=x^{3}+1 \]

Label these zeros of \(t_{2}\) as \(c_{1},c_{2},\cdots ,c_{k_{2}}\). The zeros of \(t_{2}\) are \(\left \{ -1,\left ( -1\right ) ^{\frac {1}{3}},-\left ( -1\right ) ^{\frac {1}{3}}\right \} =\left \{ -1,\frac {1}{2}-\frac {1}{2}i\sqrt {3},\frac {1}{2}+\frac {1}{2}i\sqrt {3}\right \} \). Therefore \(k_{2}=3\). Hence

\[ M=3 \]

Now we iterate over each zero \(c_{i}\) times ﬁnding \(e_{i}\) and \(\theta _{i}\) from each. These are found to be (following formula in paper) to be

\begin{align*} b_{1} & =\frac {7}{36}\\ e_{1} & =\frac {4}{3}\\ b_{2} & =\frac {7}{36}\\ e_{2} & =\frac {4}{3}\\ b_{3} & =\frac {7}{36}\\ e_{3} & =\frac {4}{3}\end{align*}

And

\begin{align*} \theta _{1} & =\frac {4}{3\left ( x+1\right ) }\\ \theta _{2} & =\frac {8}{3\left ( i\sqrt {3}+2x-1\right ) }\\ \theta _{3} & =\frac {-8}{3\left ( i\sqrt {3}-2x+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 \(4,6,8,\cdots ,M\) if any exist. Since non exist here. This is skipped. Hence \(M\) stays \(3\).

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 this case applies. \(b=\frac {lcoeff\left ( s\right ) }{lcoeff\left ( t\right ) }\) where \(lcoeff\) gives the leading coeﬃcient. Since \(s=-x\left ( x^{3}+8\right ) =-x^{4}-8x\) then \(lcoeff\left ( s\right ) \) \(=-1\). And since \(t=4\left ( x^{3}+1\right ) ^{2}=4x^{6}+8x^{3}+4\) then \(lcoeff\left ( t\right ) =4\). Therefore \(b=\frac {-1}{4}\) and therefore

\begin{align*} e_{0} & =\sqrt {1+4b}\\ & =0 \end{align*}

Hence now we have

\begin{align*} e & =\left \{ 0,\frac {4}{3},\frac {4}{3},\frac {4}{3}\right \} \\ \theta & =\left \{ 0,\frac {4}{3\left ( x+1\right ) },\frac {8}{3\left ( i\sqrt {3}+2x-1\right ) },\frac {-8}{3\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 \(M=3\) then there are \(\left ( n+1\right ) ^{M+1}=2^{4}=16\) 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}^{M}s_{i}e_{i}\nonumber \end{equation}

Since \(M=3\) 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}^{M}s_{i}\theta _{i} \tag {8}\end{equation}

Hence the ﬁrst trial \(d\) is (using Eq (7)) and recalling that \(e_{fix}=-1,\theta _{fixed}=\frac {3}{2}\frac {x^{2}}{x^{3}+1}\) gives

\begin{align*} d & =\left ( 1\right ) \left ( -1\right ) +\left ( \frac {-1}{2}\right ) \left ( 0\right ) -\left ( \frac {-1}{2}\right ) \left ( \frac {4}{3}\right ) -\left ( -\frac {1}{2}\right ) \left ( \frac {4}{3}\right ) -\left ( -\frac {1}{2}\right ) \left ( \frac {4}{3}\right ) \\ & =1 \end{align*}

Since \(d\geq 0\), then we can use it. Using Eq (8) gives (using \(M=3\))

\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 {3}{2}\frac {x^{2}}{x^{3}+1}\right ) +\left ( \frac {-1}{2}\right ) \left ( 0\right ) +\left ( \frac {-1}{2}\right ) \left ( \frac {4}{3\left ( x+1\right ) }\right ) +\left ( \frac {-1}{2}\right ) \left ( \frac {8}{3\left ( i\sqrt {3}+2x-1\right ) }\right ) +\left ( \frac {-1}{2}\right ) \left ( \frac {-8}{3\left ( i\sqrt {3}-2x+1\right ) }\right ) \\ & =\left ( \frac {3}{2}\frac {x^{2}}{x^{3}+1}\right ) -\frac {2}{3\left ( x+1\right ) }-\frac {4}{3\left ( i\sqrt {3}+2x-1\right ) }+\frac {4}{3\left ( i\sqrt {3}-2x+1\right ) }\\ & =\frac {2x^{2}}{\left ( x+1\right ) \left ( i\sqrt {3}+2x-1\right ) \left ( i\sqrt {3}-2x+1\right ) }\\ & =\frac {-x^{2}}{2x^{3}+2}\end{align*}

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 \(d=1\) and \(\Theta =\frac {-x^{2}}{2x^{3}+2}\) found from step 2 and also \(r=-\frac {x\left ( x^{3}+8\right ) }{4\left ( x^{3}+1\right ) ^{2}}\) 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 \(y^{\prime \prime }=ry\) is \(e^{\int \omega dx}\,\). 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 \end{align*}

Since this is case \(n=1\) then

\begin{align*} \omega & =\frac {p^{\prime }}{p}+\Theta \\ & =\frac {1}{x}-\frac {x^{2}}{2x^{3}+2}\\ & =\frac {x^{3}+2}{2x\left ( x^{3}+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 {x^{3}+2}{2x\left ( x^{3}+1\right ) }\right ) +\left ( \frac {x^{3}+2}{2x\left ( x^{3}+1\right ) }\right ) ^{2} & =-\frac {x\left ( x^{3}+8\right ) }{4\left ( x^{3}+1\right ) ^{2}}\\ -\frac {\left ( x^{6}+6x^{3}+2\right ) }{2x^{2}\left ( x^{3}+1\right ) ^{2}}+\frac {\left ( x^{3}+2\right ) ^{2}}{4x^{2}\left ( x^{3}+1\right ) ^{2}} & =-\frac {x\left ( x^{3}+8\right ) }{4\left ( x^{3}+1\right ) ^{2}}\\ -\frac {x\left ( x^{3}+8\right ) }{4\left ( x^{3}+1\right ) ^{2}} & =-\frac {x\left ( x^{3}+8\right ) }{4\left ( x^{3}+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^{3}+2}{2x\left ( x^{3}+1\right ) }dx}\\ & =\frac {x}{\sqrt [6]{x^{3}+1}}\end{align*}

Hence ﬁrst solution to given ODE is

\begin{align*} y & =ze^{\frac {-1}{2}\int adx}\\ & =\frac {x}{\sqrt [6]{x^{3}+1}}e^{-\frac {1}{2}\int \frac {7x^{2}}{\left ( x^{3}+1\right ) }dx}\\ & =\frac {x}{\sqrt [6]{x^{3}+1}}\frac {1}{\left ( x^{3}+1\right ) ^{\frac {7}{6}}}\\ & =\frac {x}{\left ( x^{3}+1\right ) ^{\frac {4}{3}}}\end{align*}

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