# Airy Equation

1. Jun 28, 2011

### Ted123

1. The problem statement, all variables and given/known data

Find the general solution of Airy's equation $$f'' - zf=0$$ satisfying the initial conditions f(0)=1, f'(0)=0 as a power series expansion at z=0. Express the result in terms of the Gauss hypergeometric series.

3. The attempt at a solution

After subbing $$f(z)=\sum_{n=0}^{\infty} a_n z^n$$ into Airy's equation and manipulating the summations I get the recurrence relation: $$a_2 =0$$ $$a_{n+3} = \frac{a_{n}}{(n+2)(n+3)}\;,\;\;\;\forall \;\;n=0,1,2,3,...$$
Solving this:
$$a_{3k+2}=0\;\;\;\forall \;\;k=0,1,2,3,...$$ and solving separately for n=3k and n=3k+1, $$a_{3(k+1)} = \frac{a_{3k}}{(3k+3)(3k+2)} = \frac{a_{3k}}{9(k+1)(k+\frac{2}{3})}$$ $$a_{3k} = \frac{a_0}{9^k (1)_k (\frac{2}{3})_k}$$ Similarly for n=3k+1, $$a_{3k+1} = \frac{a_1}{9^k (1)_k (\frac{4}{3})_k}$$ so that the solution is $$f(z) = a_0 \left( \sum_{k=0}^{\infty} \frac{1}{9^k (1)_k (\frac{2}{3})_k} z^{3k} \right) + a_1 \left( \sum_{k=0}^{\infty} \frac{1}{9^k (1)_k (\frac{4}{3})_k} z^{3k+1} \right)$$

f(0)=a0, f'(0)=a1

Hence the solution to the initial value problem f(0)=1, f'(0)=0 is:

$$f(z) = \sum_{k=0}^{\infty} \frac{1}{9^k (1)_k (\frac{2}{3})_k} z^{3k} = \sum_{k=0}^{\infty} \frac{1}{9^k (\frac{2}{3})_k} \frac{z^{3k}}{k!}$$

How do I express this in terms of the Gauss hypergeometric series?:
[PLAIN]http://img200.imageshack.us/img200/5992/gauss.png [Broken]

Last edited by a moderator: May 5, 2017