Hill's Method solution

With Floquent's theorem we assume a series solution, due to G. W. Hill,

$$w=e^{\mu \eta}\phi(\eta)=e^{\mu \eta}\sum^{\infty}_{r=-\infty}c_{2r}e^{2ri\eta}=\sum^{\infty}_{r=-\infty}c_{2r}e^{(\mu + 2ri)\eta}$$ (2.5)

When we put this into Mathieu's equation,

$$\sum^{\infty}_{r=-\infty}c_{2r}\left((\mu+2ir)^2 + a -2q(\frac{e^{2i\eta}+e^{-2i\eta}}{2})\right)e^{(\mu + 2ri)\eta}=0$$

matching terms in power of r, we get the equation

$$-qc_{2r-2} + \left((\mu + 2ir)^2 + a\right)c_{2r} - qc_{2r+2}=0$$ (2.6)

Multiplying through by $-1=i^2$ , and then dividing by the middle term,

$$\frac{q}{(2r-\mu i)^2 - a}c_{2r-2} + c_{2r} + \frac{q}{\left(2r-\mu i\right)^2-a}c_{2r+2}=0$$ (2.7)

We now define

$$\gamma_{2r}=\frac{q}{(2r-\mu i)^2 - a}$$

That these coefficents, $c_{i}$ have non-trivial solutions requires the infinite determinant $\Delta$ to vanish for noninfinite r:

$$\Delta(i\mu)=\left\vert\begin{array}{ccccc}\cdots&&&&\ \gamma_{-... ...}&1&\gamma_{0}&\\ &&\gamma_2&1&\gamma_2\\ &&&&\cdots \end{array}\right\vert=0$$ (2.8)

But of course, this is not a simple object to understand and solve. We can approach this problem from a rather clever angle introduced by E. T. Whittaker.

Consider the function

$$\lambda=\frac{1}{\cos\pi i \mu - \cos \pi \sqrt{a}}$$

Like our determinant, $\lambda$ has a simple pole at $a=(2r-i\mu)^2$ , so that the function

$$\zeta = \Delta(i\mu)-\kappa \lambda$$

has no singularities if $\kappa$ is chosen properly and is bound at infinity, where $\Delta(i\mu)=1$ since the $\gamma$ functions all vanish and the diagonal term is all that remains, and $\lambda = 0$ since $\cosh(x)$ limits to zero as x tends towards infinity.

$$\mho=\Delta(i\mu)-\kappa \lambda \rightarrow 1-0$$

By Liouville's theorem (of complex calculus), since this limits to a constant, it is a constant always, so we have

$$\kappa = \frac{\Delta(i\mu)-1}{\lambda}$$

Next we consider the $\mu=0$ case and find,

$$\kappa = (\Delta(0)-1)(1-\cos \pi \sqrt{a}) \rightarrow \frac{\Delta(i\mu)-1}{\lambda}= (\Delta(0)-1)(1-\cos \pi \sqrt{a})$$

Next we suppose that $\mu$ is chosen to satisfy our requirement that the determinant vanish. We thus have

$$\cos \pi i \mu - \cos \pi \sqrt{a} = (1-\Delta(0))(1-\cos \pi \sq... ...rrow i\mu=\frac{1}{\pi}\cos^{-1}\left(1-\Delta(0)(1-\cos \pi \sqrt{a})\right)$$

Recall that our solution took the form,

$$w=e^{\mu \eta}\phi(\eta)$$

This solution will be unbounded unless $\mu \in \Im$ , in which case we have

$$\mu=\frac{1}{\pi}\cos^{-1}\left(1-\Delta(0)(1-\cosh \pi \sqrt{a})\right)$$ (2.9)

We can easily encode this result, say,

if(a>=0){  mu=acos( 1 - (d[100])*(1-cos(pi*sqrt(a)))) / (pi);}
if(a<0){   mu=acos( 1 - (d[100])*(1-cosh(pi*sqrt(fabs(a))))) / (pi);}
if (mu != mu){mu=0.000000;}  //If mu=nan then make it zero

But first we must calculate $\Delta (0)$ . This task has been made exceedingly simple by the recent work of J. E. Sträng [5] who has found an efficient recursion formula.