The differential equation $$(1-x^2)y''-xy'+p^2y=0$$ can be assumed to have series solution on the form $$y(x) = x^q \sum_0^\infty a_n x^n $$ with $q = 0, 1$ and recurrence relation $$a_{n+2} = \frac{(n+q)^2-p^2}{(n+q+1)(n+q+2)} a_n $$ Find the terminating polynomial series for $p = 2$.

First of all, I don't understand why q is not simply 0 in all cases, because the series expansion clearly is about x=0, which is an ordinary point of the equation and I don't see why to inwoke Fuch's theorem. However, approaching the question as suggested I think the series can end in two ways:

Case 1: $q = 0, n = 2$

This gives $a_2 = -2a_0$ hence $$T_2(x) = a_0 -2a_0x^2$$ or if we use the common normalisation $T(1) = 1 $ we get: $$T_2(x) = 2x^2-1$$ This far all good. However, I think there is also a second case:

Case 2: $q = 1, n = 1$

In this case we terminate with $a_3 = 0$ so we have a series: $$T'_2(x) = x^1 (a_1x) = a_1x^2 = x^2$$ if we normalise. I think this solution satisfies both the termination condition and the recurrence relation found for the series. However, upon substitution I notice that this is not a valid solution to the equation!

Why doesn't case $2$ work?

  • $\begingroup$ @Somos how does that follow from q = 1 and n = 1? $\endgroup$ – Jhonny Dec 21 '17 at 12:14

In Case $2$, for $p=2, q=1$ we are looking for a factor of $0$ in the numerator of the recurrence in order to get a polynomial solution. That factor is $(n+q-p)=(n-1)$ and this is zero when $n=1$. You are correct that the recursion terminates and the solution is $a_1x^2,$ but the recursion relation is incomplete. That is, the condition $0 = a_0 q(q-1)$ for even indexes and $0 = a_1 q(q+1)$ for odd indexes must also be satisfied as initial conditions. Thus $a_1=0$ and the solution equals $0$.

  • $\begingroup$ Where do we get the $0 = a_0q(q-1) = a_1q(q+1)$ condition from? I thought that odd and even coefficients were completely decoupled? $\endgroup$ – Jhonny Dec 22 '17 at 20:16

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