3 Handling initial conditions for series solution

In series solutions, initial conditions must be specified at same location as the expansion point. For ordinary point, this is the easy case, as solution can always be written as

\begin{align*} y\left ( x\right ) & =c_{1}\left ( \cdots \right ) \\ y\left ( x\right ) & =y\left ( x_{0}\right ) \left ( \cdots \right ) \end{align*}

Where \(\cdots \) is the series found using power series or taylor expansion. For second order it becomes

\begin{align*} y\left ( x\right ) & =c_{1}\left ( \cdots \right ) +c_{2}\left ( \cdots \right ) \\ & =y\left ( x_{0}\right ) \left ( \cdots \right ) +y^{\prime }\left ( x_{0}\right ) \left ( \cdots \right ) \end{align*}

Where \(x_{0}\) above is the expansion point, typically zero. So if we are given IC \(y\left ( 0\right ) =y_{0},y^{\prime }\left ( 0\right ) =y_{0}\) we just plugin in these values in the above. But for regular point, this is not the case. If the solution found was

\[ y\left ( x\right ) =c_{1}\left ( \cdots \right ) +c_{2}\left ( \cdots \right ) \]

And we are given IC \(y\left ( 0\right ) =y_{0},y^{\prime }\left ( 0\right ) =y_{0}\), we can not replace \(c_{1}\) by \(y\left ( 0\right ) \) and replace \(c_{2}\) by \(y^{\prime }\left ( 0\right ) \) as with ordinary point.

We have to set up two equations and solve for \(c_{1},c_{2}\) as we do with normal non series solutions. Only issue is that if we find one part of the above solution not defined at \(x_{0}\) then this part is removed. For an example, lets say we obtained this solution for regular singular point

\begin{align*} y & =c_{1}y_{1}+c_{2}y_{2}\\ & =c_{1}\left ( 1-\frac {1}{6}x^{2}+\frac {x^{4}}{120}+\cdots \right ) +c_{2}\left ( \frac {1}{x}-\frac {1}{2}x+\frac {1}{24}x^{3}+\cdots \right ) \end{align*}

And the initial conditions were \(y\left ( 0\right ) =1,y^{\prime }\left ( 0\right ) =0\). The at \(x=0\) we have

\begin{align*} 1 & =c_{1}\lim _{x\rightarrow 0}\left ( 1-\frac {1}{6}x^{2}+\frac {x^{4}}{120}+\cdots \right ) +\lim _{x\rightarrow 0}c_{2}\left ( \frac {1}{x}-\frac {1}{2}x+\frac {1}{24}x^{3}+\cdots \right ) \\ & =c_{1}+c_{2}\lim _{x\rightarrow 0}\frac {1}{x}\end{align*}

Since the second basis solution diverges, then we need to have \(c_{2}=0\) since\(\ \lim _{x\rightarrow 0}\frac {1}{x}\) is undefined. This gives \(c_{1}=1\) and the solution now becomes

\[ y=1-\frac {1}{6}x^{2}+\frac {x^{4}}{120}+\cdots \]

This needs to be verified that it satisfies both initial conditions, which it does. If both basis solution diverges at the IC given, then no solution exist.

The following is another example. The ode \(x^{2}y^{\prime \prime }=xy^{\prime }+\left ( x^{2}-4\right ) y=x^{3}\) has the series solution

\begin{align*} y & =c_{1}y_{1}+c_{2}y_{2}+y_{p}\\ & =c_{1}\left ( x^{2}-\frac {1}{12}x^{4}+\frac {1}{384}x^{6}+\cdots \right ) +c_{2}\left ( \ln \left ( x\right ) \left ( 9x^{2}-\frac {3}{4}x^{4}+\cdots \right ) -\left ( \frac {144}{x^{2}}-36+\frac {1}{2}x^{4}+\cdots \right ) \right ) +\left ( \frac {1}{5}x^{3}-\frac {1}{105}x^{2}+\cdots \right ) \end{align*}

Given IC \(y\left ( 0\right ) =0,y^{\prime }\left ( 0\right ) =1\), then using \(y\left ( 0\right ) =0\) gives

\begin{align*} 0 & =c_{1}\lim _{x\rightarrow 0}\left ( x^{2}-\frac {1}{12}x^{4}+\frac {1}{384}x^{6}+\cdots \right ) +c_{2}\lim _{x\rightarrow 0}\left ( \ln \left ( x\right ) \left ( 9x^{2}-\frac {3}{4}x^{4}+\cdots \right ) -\left ( \frac {144}{x^{2}}-36+\frac {1}{2}x^{4}+\cdots \right ) \right ) +\lim _{x\rightarrow 0}\left ( \frac {1}{5}x^{3}-\frac {1}{105}x^{5}+\cdots \right ) \\ & =c_{2}\lim _{x\rightarrow 0}\left ( \ln \left ( x\right ) \left ( 9x^{2}-\frac {3}{4}x^{4}+\cdots \right ) -\left ( \frac {144}{x^{2}}-36+\frac {1}{2}x^{4}+\cdots \right ) \right ) \end{align*}

Therefore we set \(c_{2}=0\) since the limit is not defined. This makes the solution now as

\[ y=c_{1}\left ( x^{2}-\frac {1}{12}x^{4}+\frac {1}{384}x^{6}+\cdots \right ) +\left ( \frac {1}{5}x^{3}-\frac {1}{105}x^{5}+\cdots \right ) \]

Taking derivative gives

\[ y^{\prime }=c_{1}\left ( 2x-\frac {4}{12}x^{4}+\frac {6}{384}x^{5}+\cdots \right ) +\left ( \frac {3}{5}x^{2}-\frac {5}{105}x^{4}+\cdots \right ) \]

Using \(y^{\prime }\left ( 0\right ) =1\) gives

\begin{align*} 1 & =c_{1}\lim _{x\rightarrow 0}\left ( 2x-\frac {4}{12}x^{4}+\frac {6}{384}x^{5}+\cdots \right ) +\lim _{x\rightarrow 0}\left ( \frac {3}{5}x^{2}-\frac {5}{105}x^{4}+\cdots \right ) \\ & =0 \end{align*}

Which gives \(1=0\). Therefore there is no solution.

To handle IC for series solution, we start with general solution found \(y=c_{1}y_{1}+c_{2}y_{2}+y_{p}\) and apply the above step by step in order to determine what happens. Let us change the IC for the above example to \(y\left ( 0\right ) =0,y^{\prime }\left ( 0\right ) =0\) and see what happens. We found that applying \(y\left ( 0\right ) =0\) gives

\[ y=c_{1}\left ( x^{2}-\frac {1}{12}x^{4}+\frac {1}{384}x^{6}+\cdots \right ) +\left ( \frac {1}{5}x^{3}-\frac {1}{105}x^{5}+\cdots \right ) \]

Taking derivative gives

\[ y^{\prime }=c_{1}\left ( 2x-\frac {4}{12}x^{4}+\frac {6}{384}x^{5}+\cdots \right ) +\left ( \frac {3}{5}x^{2}-\frac {5}{105}x^{4}+\cdots \right ) \]

Using \(y^{\prime }\left ( 0\right ) =0\) gives

\begin{align*} 0 & =c_{1}\lim _{x\rightarrow 0}\left ( 2x-\frac {4}{12}x^{4}+\frac {6}{384}x^{5}+\cdots \right ) +\lim _{x\rightarrow 0}\left ( \frac {3}{5}x^{2}-\frac {5}{105}x^{4}+\cdots \right ) \\ 0 & =0c_{1}\end{align*}

Which means \(c_{1}\) is arbitrary. This means the final solution remains

\[ y=c_{1}\left ( x^{2}-\frac {1}{12}x^{4}+\frac {1}{384}x^{6}+\cdots \right ) +\left ( \frac {1}{5}x^{3}-\frac {1}{105}x^{5}+\cdots \right ) \]

Even though we had two initial conditions, the final solution still has one arbitrary constant. This happens quite often when the the expansion point is singular as in this example.