PHY.K02UF Molecular and Solid State Physics

Linear second-order differential equations with periodic coefficients

Hill's equation is a second-order differential equation with a periodic coefficient,

$$ \frac{d^2y(x)}{dx^2} + Q(x)y(x) = 0.$$

Here $Q(x)$ has a periodicity $a$, $Q(x + a) = Q(x)$.

Many problems can be put into the form of Hill's equation such as, an electron moving in a one-dimensional periodic potential, a child on a swing, Mathieu's differential equation, and the equation for the spatial part of the normal modes for waves moving in a one-dimensional perioidic medium.

Any second-order differential equation with periodic coefficients can be put in the form of Hill's equation.

Solutions to Hill's equation
There are two linearly independent solution to Hill's equation since it is a second-order differential equation. Because of the periodic symmetry, the solutions can be found that are simultaneously eigenfunctions of the translation operator $\textbf{T}$ with eigenvalue $\lambda$, $\textbf{T}y(x)=y(x+a)=\lambda y(x).$ These two solutions have the form,

$$y_+(x) = e^{ik_{+}x}u_{k_+}(x),\qquad\text{and}\qquad y_{-}(x) = e^{ik_{-}x}u_{k_{-}}(x).$$

where $u_{k_{\pm}}(x)$ is a periodic function with period $a$. Notice that functions of the form $e^{ik_{\pm}x}u_{k_{\pm}}(x)$ are eigenfunctions of the translation operator with eigenvalue $\lambda_{\pm}=e^{ik_{\pm}a}$,


Floquet theory shows that the two solutions $y_{\pm}(x)$ can be constructed from two linearly independent solutions that are specified by the boundary conditions,

$$y_1(0)=1, \, \left.\frac{dy_1}{dx}\right|_{x=0}=0 \hspace{0.3cm} \text{and} \hspace{0.3cm} y_2(0)=0, \, \left.\frac{dy_2}{dx}\right|_{x=0}=1. $$

The values of these two solutions are known at $x=0$ and have to be calculated at $x=a$. For some problems, like the Kronig-Penney model, it is possible to calculate $y_1(a)$ and $y_2(a)$ analytically but for most problems they are calculated numerically as described on the page on Bloch waves in 1-D. Once values of the functions $y_1(x)$ and $y_2(x)$ and their derivatives are known at $x=a$, the eigenvalues of the translation operator can be determined from the characteristic equation,

$$\lambda^2 - \alpha\lambda + 1 = 0, \quad \text{where}\quad\alpha = y_1(a) - \left. \frac{dy_2}{dx}\right|_{x=a}.$$

The characteristic equation has two roots, $\lambda_+$ and $\lambda_-$. If $|\alpha| < 2$, $y_+(x)$ and $y_-(x)$ are complex conjugates of each other and have the form,

$$y_+(x) = e^{ik_{+}x}u_{k_+}(x),\qquad\text{and}\qquad y_{-}(x) = e^{ik_{-}x}u_{k_{-}}(x),$$

where $k_+$ is purely real and $k_+ = -k_-$. If $|\alpha| > 2$, $k_+$ and $k_-$ have a complex component and one solution grows exponentially while the other decays exponentially.