Basics and Flouqent's Theorem

Our derivation below can be found in greater detail and better form in many references [3,4,5], and our derivation follows the spirit of these. An equation such as Mathieu's equation,

$$\ddot{\eta}+(a-2q\cos(2\tau))\eta=0$$ (2.1)

is of a class of differential equations of the type [7],

$$L[y]=y''+p(t)y' + q(t)y = 0$$ (2.2)

Any two fundamental solutions to this equation, $y_1(t), y_2(t)$ , will satisfy the set of boundary value equations,

$$\left. \begin{array}{lll} c_1y_1(t_0) + c_2 y_2(t_0)&=&y_0 \\ c_1y'_1(t_0) + c_2y'_2(t_0)&=&y'_0 \end{array}\right\} Y{\bf c}={\bf y}$$

We thus require that the determinant of Y (called the Wronskian) is not equal to zero,

$$W(Y)=det(Y)=\left\vert\begin{array}{cc}y_1(t_0)&y_2(t_0)\ y'_1(t_0)&y'_2(t_0)\end{array}\right\vert \neq 0$$ (2.3)

The set of even/odd solutions:

$$\left.\begin{array}{l}y_1: y(t_0)=1, y'(t_0)=0\ y_2: y(t... ...}\right\} W(Y)=\left\vert\begin{array}{cc}1&0\ 0&1\end{array}\right\vert=1$$

Are thus fundamental sets of solutions. We may follow Floquet's theorem [3], which tells us that Mathieu's equation has at least one solution $\ni$

$$y(z):(y(z+\pi)=\sigma y(z))$$ (2.4)

The proof of this is outlined as follows.

Since Mathieu's equations will have an even ($w_1(\eta)$ ) and odd ($w_2(\eta)$ ) solution pair, these two functions may define any other solution, e.g. consider

$$w_1(\eta+\pi)=\alpha w_1(\eta)+ \beta w_2(\eta) \rightarrow w'_1(\eta+\pi)=\alpha w'_1(\eta)+ \beta w'_2(\eta) \ni$$

$$w_1(0)=w'_2(0)=1, w_1'(0)=w_2(0)=0 \Rightarrow w_1(\pi)=\alpha, w'_1(\pi)=\beta \ni$$

$$w_1(\eta+\pi)=w_1(\pi)w_1(\eta)+w'_1(\pi)w_2(\eta),$$   as well,$$w_2(\eta+\pi)=w_2(\pi)w_1(\eta)+w_2'(\pi)w_2(\eta)$$

Let

$$A = \left(\begin{array}{cc}w_1(\pi)&w'_1(\pi)\ w_2(\pi)&w'_2(\pi... ...\begin{array}{c}w_1(\eta)\ w_2(\eta)\end{array}\right)\ni Aw(\eta)=w(\eta+\pi)$$

According to Floquet's theorem, we thus require,

$$\left\vert A - \sigma I\right\vert = 0$$

an eigenvalue equation which can be satisfied with the proper value of $\sigma$ . We consider further that Mathieu's equation has a solution of the form $e^{\mu \eta}\phi(\eta)$

$$\sigma=e^{\mu \pi}, \phi(\eta)=e^{-\mu \eta}y(\eta) \ni \phi(\eta + \pi) = e^{-\mu(\eta+\pi)}y(\eta+\pi)=e^{-\mu \eta}y(\eta)=\phi(\eta)$$