5.3.2.2 Example 2
Let
\[ y^{\prime \prime }=\left ( \frac {2}{x^{2}}-1\right ) y \]
Therefore
\begin{align} r & =\frac {s}{t}\nonumber \\ & =\frac {2-x^{2}}{x^{2}} \tag {1}\end{align}
Step 1 In this we find all \(\left [ \sqrt {r}\right ] _{c}\) and associated \(\alpha _{c}^{\pm }\) for each pole. There is one pole at \(x=0\) of order 2. In
this case, from the description of the algorithm earlier, we write
\begin{align*} \left [ \sqrt {r}\right ] _{c} & =0\\ \alpha _{c}^{+} & =\frac {1}{2}+\frac {1}{2}\sqrt {1+4b}\\ \alpha _{c}^{-} & =\frac {1}{2}-\frac {1}{2}\sqrt {1+4b}\end{align*}
Where \(b\) is the coefficient of \(\frac {1}{\left ( x-0\right ) ^{2}}\) in the partial fraction decomposition of \(r\) which is \(\frac {2}{x^{2}}-1\). Hence \(b=2\).
Therefore
\begin{align*} \left [ \sqrt {r}\right ] _{c} & =0\\ \alpha _{c}^{+} & =\frac {1}{2}+\frac {1}{2}\sqrt {1+8}=\frac {1}{2}+\frac {3}{2}=2\\ \alpha _{c}^{-} & =\frac {1}{2}-\frac {1}{2}\sqrt {1+8}=\frac {1}{2}-\frac {3}{2}=-1 \end{align*}
We are done with all the poles. Now we consider \(O\left ( \infty \right ) \) which is \(\deg \left ( t\right ) -\deg \left ( s\right ) =2-2=0\). This falls in the case \(-2v\leq 0\). Hence
\[ v=0 \]
We need the Laurent series of
\(\sqrt {r}\) around
\(\infty \). Using the computer this is
\(i-\frac {i}{x^{2}}-\frac {1}{2x^{4}}i+\cdots \). Hence we need the
coefficient of
\(x^{0}\) in this series (
\(0\) because that is value of
\(v\)).
\[ \left [ \sqrt {r}\right ] _{\infty }=ix^{0}\]
Recall that
\(\left [ \sqrt {r}\right ] _{\infty }\) is the sum of terms
of
\(x^{j}\) for
\(0\leq j\leq v\) or for
\(j=0\) since
\(v=0\). Looking at the series above, we see that
\[ a=i \]
Which is the
coefficient of the term
\(x^{0}\). Now we need to find
\(\alpha _{\infty }^{\pm }\) associated with
\(\left [ \sqrt {r}\right ] _{\infty }\). For this we need
to first find
\(b\) which is the coefficient of
\(x^{v-1}=\frac {1}{x}\) in
\(r\) minus the coefficient of
\(x^{v-1}=\frac {1}{x}\) in
\(\left ( \left [ \sqrt {r}\right ] _{\infty }\right ) ^{2}\). But
\[ \left ( \left [ \sqrt {r}\right ] _{\infty }\right ) ^{2}=i^{2}=-1 \]
Hence the coefficient of
\(x^{-1}\) in
\(\left ( \left [ \sqrt {r}\right ] _{\infty }\right ) ^{2}\) is
\(0\). To find the coefficient of
\(x^{-1}\) in
\(r\) long division is
done
\begin{align*} r & =\frac {s}{t}\\ & =\frac {2-x^{2}}{x^{2}}\\ & =Q+\frac {R}{x^{2}}\end{align*}
Where \(Q\) is the quotient and \(R\) is the remainder. This gives
\[ r=-1+\frac {2}{x^{2}}\]
For the other case of
\(v=0\) then the
coefficient of
\(x^{-1}\) is found by looking up the coefficient in
\(R\) of
\(x\) to the degree of of
\(t\) then
subtracting one and dividing result by
\(lcoeff\left ( t\right ) \). But degree of
\(t\) is
\(2\). Therefore we want the
coefficient of
\(x^{2-1}\) or
\(x\) in
\(R\) which is zero. Hence
\(b=0-0=0\).
Now that we found \(a,b\), then from the above section describing the algorithm, we see in this
case that
\begin{align*} \alpha _{\infty }^{+} & =\frac {1}{2}\left ( \frac {b}{a}-v\right ) =\frac {1}{2}\left ( 0-0\right ) =0\\ \alpha _{\infty }^{-} & =\frac {1}{2}\left ( -\frac {b}{a}-v\right ) =\frac {1}{2}\left ( -0-0\right ) =0 \end{align*}
This completes step 1 of the solution. We have found \(\left [ \sqrt {r}\right ] _{c}\) and its associated \(\alpha _{c}^{\pm }\) and found \(\left [ \sqrt {r}\right ] _{\infty }\) and
its associated \(\alpha _{\infty }^{\pm }\). Now we go to step 2 which is to find the \(d^{\prime }s\).
step 2 Since we have a pole at zero, and we have one \(O\left ( \infty \right ) \), each with \(\pm \) signs, then we set up this
table to make it easier to work with. This implements
\[ d=\alpha _{\infty }^{\pm }-\sum _{i=1}^{1}\alpha _{c_{i}}^{\pm }\]
Therefore we obtain 4 possible
\(d\)
values.
| | | | |
| pole \(c\) |
\(\alpha _{c}\) value |
\(O\left ( \infty \right ) \) value |
\(d\) |
\(d\) value |
| | | | |
| \(x=0\) |
\(\alpha _{c}^{+}=2\) |
\(\alpha _{\infty }^{+}=0\) |
\(\alpha _{\infty }^{+}-\left ( \alpha _{c}^{+}\right ) =0-\left ( 2\right ) \) |
\(-2\) |
| | | | |
| \(x=0\) |
\(\alpha _{c}^{+}=2\) | \(\alpha _{\infty }^{-}=0\) | \(\alpha _{\infty }^{-}-\left ( \alpha _{c}^{+}\right ) =0-\left ( 2\right ) \) | \(-2\) |
| | | | |
| \(x=0\) | \(\alpha _{c}^{-}=-1\) | \(\alpha _{\infty }^{+}=0\) | \(\alpha _{\infty }^{+}-\left ( \alpha _{c}^{-}\right ) =0-\left ( -1\right ) \) | \(1\) |
| | | | |
| \(x=0\) |
\(\alpha _{c}^{-}=-1\) |
\(\alpha _{\infty }^{-}=0\) |
\(\alpha _{\infty }^{-}-\left ( \alpha _{c}^{-}\right ) =0-\left ( -1\right ) \) |
\(1\) |
| | | | |
We see from the above that we took each pole in this problem (there is only one pole here
at \(x=0\)) and its associated \(\alpha _{c}^{\pm }\) with each \(\alpha _{\infty }^{\pm }\) and generated all possible \(d\) values from all the
combinations. Hence we obtain 4 possible \(d\) values in this case. If we had 2 poles, then we
would have 8 possible \(d\) values. Hence the maximum possible \(d\) values we can get is \(2^{p+1}\) where \(p\)
is number of poles. Now we remove all negative \(d\) values. Hence the trial \(d\) values
remaining is
\[ d=\left \{ 1\right \} \]
There is one
\(d\) value to try. We can pick any one of the two values of
\(d\)
generated since there are both
\(d=1\). Both will give same solution. We generate
\(\omega \) using
\[ \omega =\left ( \sum _{c}s\left ( c\right ) \left [ \sqrt {r}\right ] _{c}+\frac {\alpha _{c}}{x-c}\right ) +s\left ( \infty \right ) \left [ \sqrt {r}\right ] _{\infty }\]
To
apply the above, we update the table above, now using only the first
\(d=1\) value in
the above table. (selecting the second
\(d=1\) row, will not change the final solution).
but we also add columns for
\(\left [ \sqrt {r}\right ] _{c},\left [ \sqrt {r}\right ] _{\infty }\) to make the computation easier. Here is the new
table
| | | | | | | | |
|
pole \(c\) |
\(\alpha _{c}\) value |
\(s\left ( c\right ) \) |
\(\left [ \sqrt {r}\right ] _{c}\) |
\(O\left ( \infty \right ) \) value |
\(s\left ( \infty \right ) \) |
\(\left [ \sqrt {r}\right ] _{\infty }\) |
\(d\) value |
| \(\omega \) value | | \(\left ( \sum _{c}s\left ( c\right ) \left [ \sqrt {r}\right ] _{c}+\frac {\alpha _{c}}{x-c}\right ) +s\left ( \infty \right ) \left [ \sqrt {r}\right ] _{\infty }\) |
|
| | | | | | | | |
| \(x=0\) | \(\alpha _{c}^{-}=-1\) | \(-\) | \(0\) | \(\alpha _{\infty }^{+}=0\) | \(+\) | \(i\) | \(1\) | \(\left ( -\left ( 0\right ) +\frac {-1}{x-0}\right ) +\left ( +\right ) \left ( i\right ) =\frac {-1}{x}+i\) |
| | | | | | | | |
The above gives one candidate \(\omega \) value to try. For this \(\omega \) we need to find polynomial \(P\) by
solving
\begin{equation} P^{\prime \prime }+2\omega P^{\prime }+\left ( \omega ^{\prime }+\omega ^{2}-r\right ) P=0 \tag {8}\end{equation}
If we are able to find
\(P\), then we stop and the ode
\(y^{\prime \prime }=ry\) is solved. If we try all
candidate
\(\omega \) and can not find
\(P\) then this case is not successful and we go to the next
case.
step 3 Now for each candidate \(\omega \) we solve the above Eq (8). Starting with \(\omega =\frac {-1}{x}+i\) associated
with first \(d=1\) in the table, then (8) becomes
\begin{align*} P^{\prime \prime }+2\left ( \frac {-1}{x}+i\right ) P^{\prime }+\left ( \left ( \frac {-1}{x}+i\right ) ^{\prime }+\left ( \frac {-1}{x}+i\right ) ^{2}-\left ( \frac {2}{x^{2}}-1\right ) \right ) P & =0\\ P^{\prime \prime }+2\left ( \frac {-1}{x}+i\right ) P^{\prime }+\left ( \frac {-2i}{x}\right ) P & =0 \end{align*}
This needs to be solved for \(P\). Since degree of \(p\left ( x\right ) \) is \(d=1\). Let \(p=x+a\). The above becomes
\begin{align*} 2\left ( \frac {-1}{x}+i\right ) +\left ( \frac {-2i}{x}\right ) \left ( x+a\right ) & =0\\ \frac {-2}{x}+2i-2i-\frac {2ia}{x} & =0\\ \frac {-2}{x}-\frac {2ia}{x} & =0 \end{align*}
Which means
\[ a=i \]
Hence we found the polynomial
\[ p\left ( x\right ) =x+i \]
Therefore the solution to
\(y^{\prime \prime }=ry\) is
\begin{align*} y & =pe^{\int \omega dx}\\ & =\left ( x+i\right ) e^{\int \frac {-1}{x}+i\ dx}\\ & =\left ( x+i\right ) \frac {1}{x}e^{ix}\\ & =\frac {x+i}{x}\left ( \cos x+i\sin x\right ) \end{align*}
The second solution can be found by reduction of order. The full general solution to \(y^{\prime \prime }=\left ( \frac {2}{x^{2}}-1\right ) y\)
is
\[ y\left ( x\right ) =\frac {c_{1}}{x}\left ( x\cos x-\sin x\right ) +\frac {c_{2}}{x}\left ( \cos x+x\sin x\right ) \]