0
$\begingroup$

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?

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

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 $0$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.