I've run into an issue a couple times and have not really stopped to consider what it is that is happening. The general procedure where this has appeared is the following:
- I have a differential equation that has some poles in it. For example:
$$ y''(x) A(x) + y'(x) B(x) + y(x) C(x) =0 $$
where $A(x)$ has a singular behavior at $x=0$.
- Use Frobenius method to find power series solutions $y$. I may find something like:
$$ y = 1/x + \mathcal{O}(x)^0 $$ and
$$ y = 1 + \mathcal{O}(x) $$
for instance.
- This suggests to me that I would have 2 linearly independent solutions. So, I should be able to combine them to the form of:
$$ y = \alpha_0 \frac{1}{x} + \alpha_1 + h_1(\alpha_0,\alpha_1) x + h_2(\alpha_0,\alpha_1) x^2 + \cdots $$
So I take the solution: $$ y = \alpha_0 \frac{1}{x} + \alpha_1 + \alpha_2 x + \alpha_3 x^2 + \cdots $$ insert this into the differential equation, take a Taylor series (with the expectation that I would get a set of equations defining $\alpha_2,\alpha_3$ in terms of $\alpha_0,\alpha_1$. However, I will often get an equation where $\alpha_0=0$ and the rest of the coefficients depend only on $\alpha_1$.
A resolution to this has been to assume that there is also a logarithmic series of the form:
$$ y = \alpha_0 \frac{1}{x} + \alpha_1 + \alpha_2 x + \alpha_3 x^2 + \log(x)(\alpha_4 + x \alpha_5 +\cdots) $$
then, in the power series expansion there are terms that connect $\alpha_0$ and $\alpha_4$ and $\alpha_4$ to $\alpha_1$ and $\alpha_2$. Thus, the solution truly has 2 free parameters.
My question is summed up into a couple parts:
Why/how does a solution like $1/x$ appear from frobenius method if it seems to immediately become degenerate?
Is there any good source that discusses the meaning of these degenerate solutions and what they may imply about the differential equation?
Is there any motivation to adding a logarithmic series beyond getting the correct number of free parameters?
Note: My actual experience with this kind of equation appearing is a 4th order DE, where there are also derivatives of $A$ and $B$. I have left this detail out of the main text as I think it is besides the point, but here in case it is. As well, I am unsure if I have used the correct terminology or included enough information. If I need to add more detail please let me know and I will add.
2nd order Example
I've played around and found an example of a 2nd order DE that displays this feature. Consider the DE: $$ A(x) y''(x) + B(x) y'(x) + C(x) y(x) =0, $$ with: $$ A(x) = x^2(1+3x), \ \ \ B(x) = x(1+x), \ \ \ C(x) = -(1+x) $$ Using Frobenius method we would find that the Asymptotic solutions begin at orders: $$ y_1 = \frac{c_1}{x} + c_2 + c_3 x + \mathcal{O}(x)^2, $$ and $$ y_2 = \frac{f_1} x + \mathcal{O}(x)^2. $$ If we put the first series expression into the differential equation and take a series of that, enforcing that each order goes to zero, the first couple of orders would take the form of: $$ (4 c_1 - c_2) - c_2 x + 3 c_4 x^2 + \mathcal{O}(x)^2 $$ from which we see that $c_2=0$ then $c_1=0$ and $c_4=0$ leaving only $c_3$ as a free parameter and so $y_1$ becomes degenerate with $y_2$.