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.

<hr /> <p><input type="button" value="Hide details" onclick="document.getElementById('p2_div').innerHTML=document.getElementById('blank').value;"><br /> A second-order linear differential equation with periodic coefficients has the form,</p> $$C_1(x)\frac{d^2\psi(x)}{dx^2} + C_2(x)\frac{d\psi(x)}{dx} + C_3(x)\psi(x) = 0, $$ <p>where the coefficients are all periodic with the same period $a$, $C_1(x +a)=C_1(x)$, $C_2(x +a)=C_2(x)$, $C_3(x +a)=C_3(x)$.</p> <p><b>Transformation into Hill's Form</b><br /> This equation can be transformed to Hill's equation</p> $$\frac{d^2y(x)}{dx^2} + Q(x)y(x) = 0, $$ <p>by dividing through by $C_1(x)$ and then making the transformation,</p> $$y(x) = e^{\frac{1}{2}A(x)}\psi(x), \hspace{0.3cm} \mathrm{with} \hspace{0.3cm} \frac{dA(x)}{dx} = \frac{C_2(x)}{C_1(x)}, $$ <p>and</p> $$Q(x) = -\frac{1}{2}\frac{d}{dx}\left(\frac{C_2}{C_1}\right)-\frac{1}{4}\left(\frac{C_2}{C_1}\right)^2 + \left(\frac{C_3}{C_1}\right).$$ <p>To check this, differentiate $y$,</p> $$\frac{dy}{dx} = \frac{1}{2}\frac{dA}{dx}e^{\frac{1}{2}A}\psi+e^{\frac{1}{2}A}\frac{d\psi}{dx},$$ $$\frac{d^2y}{dx^2} = \frac{1}{2}\frac{d^2A}{dx^2}e^{\frac{1}{2}A}\psi+\frac{1}{2}\frac{dA}{dx}\left(\frac{1}{2}\frac{dA}{dx}e^{\frac{1}{2}A}\psi+e^{\frac{1}{2}A}\frac{d\psi}{dx}\right)+ \frac{1}{2}\frac{dA}{dx}e^{\frac{1}{2}A}\frac{d\psi}{dx}+e^{\frac{1}{2}A}\frac{d^2\psi}{dx^2}.$$ $$\frac{d^2y}{dx^2} = \frac{1}{2}\frac{d^2A}{dx^2}e^{\frac{1}{2}A}\psi+\frac{1}{4}\left(\frac{dA}{dx}\right)^2e^{\frac{1}{2}A}\psi+\frac{dA}{dx}e^{\frac{1}{2}A}\frac{d\psi}{dx}+e^{\frac{1}{2}A}\frac{d^2\psi}{dx^2}.$$ <p>Substituting the expressions for $\frac{d^2y}{dx^2}$, $Q(x)$, and $\frac{dA(x)}{dx}$ into Hill's equation yields,</p> $$\left(\frac{C_2}{C_1}\right)\frac{d\psi}{dx}+\frac{d^2\psi}{dx^2} + \left(\frac{C_3}{C_1}\right)\psi =0.$$ <hr /> 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 $\lambda_{\pm}=e^{ik_{\pm}a}$,

$$\text{T}e^{ik_{\pm}x}u_{k_{\pm}}(x)=e^{ik_{\pm}(x+a)}u_{k_{\pm}}(x+a)=e^{ik_{\pm}a}e^{ik_{\pm}x}u_{k_{\pm}}(x).$$

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.

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• Lineare Differentialgleichungen mit periodischen Koeffizienten (in german)
  http://www.mathematik.uni-dortmund.de/lsi/kaballo/DGL/Kap12.pdf.
• On a class of Hill's equations having explicit solutions
  https://www.sciencedirect.com/science/article/pii/S0893965913001419.
• V. Pašić, Differential equations with periodic coefficients, 2001, http://frontslobode.org/vedad/pdf/diss.pdf.
• W. Magnus, S. Winkler, Hill's Equation, Dover Publications Inc., 1979.