Find the Green's function of the differential operator for periodic boundaries.

Question

On the space of functions $f(x)$ where $x \in (-\pi,\pi]$ satisfying $\int^{\pi}_{-\pi} f(x)dx = 0$ and periodic boundary conditions, find the Green's function for the operator $-\frac{d^2}{dx^2}$.

I know that to solve this problem, homogenous boundary conditions are needed for the operator to be nonsingular, but I am having trouble figuring out how to enforce the boundary conditions. Here is what I do know:

Knowing the properties of Green's function, we can say,

$-\frac{d^2}{dx^2}G(x,\xi) = \delta(x - \xi)$

When $x \neq \xi \rightarrow -\frac{d^2}{dx^2}G(x,\xi) = 0$

From here I am wondering if I would enforce the boundary conditions at both ends of the "crease"; when $x\in (-\pi,\xi]$ and $x\in [\xi,\pi]$. However, I am a little bit unsure what to do with that noninclusive boundary condition.

I have attempted to tackle this problem in a different way by expanding $G(x,\xi)$ as a Fourier series

$G(x,\xi) = \sum_{n = -\infty}^{\infty} g_n e^{in(x-\xi)}$

Then applying the operator:

$-\frac{d^2}{dx^2} \sum_{n = -\infty}^{\infty} g_n e^{in(x-\xi)} = \sum_{n = -\infty}^{\infty} g_n n^2 e^{in(x-\xi)} = \delta(x-\xi)$

Then when compared to the delta function Fourier expansion:

$\delta(x-\xi) = \frac{1}{2\pi}\sum_{n = -\infty}^{\infty} e^{in(x-\xi)}$

We see that $g_n = \frac{1}{2\pi n^2}$. Thus,

$G(x,\xi) = \frac{1}{2\pi} \sum_{n = -\infty}^{\infty} \frac{1}{n^2} e^{in(x-\xi)}$

From here, I don't really know what to do with this infinite series.

So, between these two methods I would really appreciate some guidance on how best to proceed with either method I started this problem with. If something is unclear, please let me know and I will do my best to explain. Thanks in advance!

Answer 1

It's not possible to enforce periodic boundary conditions on the solution to the differential equation $$ -\frac{d^2}{dx^2}G(x,\xi)=\delta(x-\xi)\qquad(-\pi<x,\xi<\pi). \tag{1} $$ Indeed, if $G(-\pi,\xi)=G(\pi,\xi)$ and $G_x(-\pi,\xi)=G_x(\pi,\xi)$, then $$ \int_{-\pi}^{\pi}-\frac{d^2}{dx^2}G(x,\xi)\,dx=-G_x(\pi,\xi)+G_x(-\pi,\xi)=0, \tag{2} $$ whereas $$ \int_{-\pi}^{\pi}\delta(x-\xi)=1. \tag{3} $$ Such impossibility appears in disguise in your solution of the problem using Fourier series: the term corresponding to $n=0$ in $$ G(x,\xi)=\frac{1}{2\pi} \sum_{n = -\infty}^{\infty} \frac{1}{n^2}\, e^{in(x-\xi)} \tag{4} $$ is ill-defined. However, if we remove that term from $(4)$, we obtain the Green's function of the operator $\hat{H}=-\frac{d^2}{dx^2}$ in the space $\mathcal{P}$ of $2\pi$-periodic functions satisfying the condition $\int_{-\pi}^{\pi}f(x)\,dx=0$. Indeed, if we use the method of Fourier series to solve the ODE $$ -y''(x)=f(x), \tag{5} $$ subject to the boundary conditions $y(-\pi)=y(\pi)$ and $y'(-\pi)=y'(\pi)$, we get $$ -\frac{d^2}{dx^2}\underbrace{\sum_{n\in\mathbb{Z}}y_ne^{inx}}_{=\,y(x)} =\sum_{n\in\mathbb{Z}}n^2y_ne^{inx} =\underbrace{\sum_{n\in\mathbb{Z}}f_ne^{inx}}_{=\,f(x)} $$ $$ \implies n^2y_n=f_n=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(\xi)e^{-in\xi}\,d\xi. \tag{6} $$ For $n\neq 0$, Eq. $(6)$ implies $y_n=\frac{1}{n^2}f_n$. On the other hand, Eq. $(6)$ does not determine $y_0$, since it yields $$ 0y_0=f_0=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(\xi)\,d\xi=0\qquad(f\in\mathcal{P}). \tag{7} $$ To determine $y_0$, we use the condition that $y$ is also in $\mathcal{P}$: $$ y_0=\frac{1}{2\pi}\int_{-\pi}^{\pi}y(\xi)\,d\xi=0. \tag{8} $$ Therefore, we can write the solution to $(5)$ as \begin{align} y(x)&=\sum_{n\in\mathbb{Z}^*}\frac{1}{n^2}f_ne^{inx} \\ &=\sum_{n\in\mathbb{Z}^*}\frac{1}{2\pi n^2}\int_{-\pi}^{\pi}f(\xi)e^{in(x-\xi)}\,d\xi \\ &=\int_{-\pi}^{\pi}\tilde{G}(x,\xi)f(\xi)\,d\xi, \tag{9} \end{align} where $$ \tilde{G}(x,\xi)=\frac{1}{2\pi}\sum_{n\in\mathbb{Z}^*}\frac{1}{n^2}\,e^{in(x-\xi)}, \tag{10} $$ thus confirming that $(4)$ without the term corresponding to $n=0$ is the Green's function of $\hat{H}=-\frac{d^2}{dx^2}$ in the space of functions $\mathcal{P}$.

As a final remark, we notice that $\tilde{G}(x,\xi)$ can be rewritten as$^{(*)}$ $$ \tilde{G}(x,\xi)=\frac{1}{\pi}\sum_{n=1}^{\infty}\frac{1}{n^2}\cos(x-\xi)=\frac{\pi}{6}-\frac{1}{2}|x-\xi|+\frac{1}{4\pi}(x-\xi)^2. \tag{11} $$ Differentiating the above expression twice with respect to $x$, we obtain a variant of Eq. $(1)$: $$ -\frac{d^2}{dx^2}\tilde{G}(x,\xi)=\delta(x-\xi)-\frac{1}{2\pi}. \tag{12} $$ The extra term on the RHS of $(12)$ makes it consistent with Eq. $(2)$: $$ \int_{-\pi}^{\pi}\left[\delta(x-\xi)-\frac{1}{2\pi}\right]dx=1-1=0. \tag{13} $$

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$^{(*)}$ See Explicit form of $\sum_{m=1}^{\infty}\frac{\cos(mk)}{m^2}$ for $k \in [-\pi,\pi]$.