Next: Sträng's recursion formula for Up: Mathieu's Equation, solution, and Previous: Basics and Flouqent's Theorem

## Hill's Method solution

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

 (2.5)

When we put this into Mathieu's equation,

matching terms in power of r, we get the equation

 (2.6)

Multiplying through by , and then dividing by the middle term,

 (2.7)

We now define

That these coefficents, have non-trivial solutions requires the infinite determinant to vanish for noninfinite r:

 (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

Like our determinant, has a simple pole at , so that the function

has no singularities if is chosen properly and is bound at infinity, where since the functions all vanish and the diagonal term is all that remains, and since limits to zero as x tends towards infinity.

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

Next we consider the case and find,

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

Recall that our solution took the form,

This solution will be unbounded unless , in which case we have

 (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 . This task has been made exceedingly simple by the recent work of J. E. Sträng [5] who has found an efficient recursion formula.

Next: Sträng's recursion formula for Up: Mathieu's Equation, solution, and Previous: Basics and Flouqent's Theorem
tim jones 2008-07-07