Numerical Solution of Stoke's First Problem

Question

$f''+2 \eta f-4mf=0$ where $f=1$ at $\eta=0$, $f \to 0$ as $\eta \to \infty$. The case of $m=0$ arises in the analysis of motion of fluid above a suddenly accelerated flat plate.

Answer 1

The general solution of your differential equation is $$ f(\eta) = a \text{Ai}(2^{1/3}(2m-\eta)) + b \text{Bi}(2^{1/3}(2m-\eta))$$ where Ai and Bi are Airy functions, and $a$ and $b$ are arbitrary constants. Now both $\text{Ai}(t)$ and $\text{Bi}(t)$ go to $0$ as $t \to -\infty$: according to Wolfram Alpha, $$\eqalign{\text{Ai}(-t) &= \frac{t^{-1/4}}{\sqrt{\pi}} \sin(2 t^{3/2}/3+\pi/4) + O(t^{-7/4})\cr \text{Bi}(-t) &= \frac{t^{-1/4}}{\sqrt{\pi}} \cos(2 t^{3/2}/3+\pi/4) + O(t^{-7/4})}$$ So the boundary condition at $\infty$ does not restrict the solution. You need another boundary condition to determine a solution.