I Apologize that this is a continuation of a question that I just asked. Anyway here is where I am:
Ok so I was trying to solve the Poisson's equation for a point charge with a Fourier transform to get the familiar equation.
This is what I did so far:
So ultimately I am trying to solve this in 3 dimensions but I am embarrassingly struggling with the 1-D solution right now.
$\frac{\partial^{2}}{\partial x^{2}} f(x) = \rho(x) $
I express f and ρ in terms of their Fourier transforms:
$f(x) = \frac{1}{ \sqrt{2 \pi}} \int_{-\infty}^{+\infty} f(\vec{k})e^{i \vec{k}\vec{x}}dk$
and
$\rho(x) = \frac{1}{ \sqrt{2 \pi}} \int_{-\infty}^{+\infty}\rho(\vec{k})e^{i \vec{k}\vec{x}} dk$
So from here I bring the derivative into the integral that is $f(x)$ and operate on the $e^{i \vec{k}\vec{x}}$ term:
$\frac{\partial^{2}}{\partial x^{2}} f(x) = \frac{1}{ \sqrt{2 \pi}} \int_{-\infty}^{+\infty} -k^{2} f(\vec{k})e^{i \vec{k}\vec{x}}dk$
I have:
$\frac{1}{ \sqrt{2 \pi}} \int_{-\infty}^{+\infty} -k^{2} f(\vec{k})e^{i \vec{k}\vec{x}}dk = \frac{1}{ \sqrt{2 \pi}} \int_{-\infty}^{+\infty}\rho(\vec{k})e^{i \vec{k}\vec{x}} dk$
And I am able to drop the integrals because the Fourier transform is unique.
$-k^{2}f(\vec{k}) = \rho(\vec{k})$
So Now I can solve for $f(x)$:
$f(\vec{k}) = \frac{\rho(\vec{k})}{-k^2}$
So now for a point charge I know that $\rho(x) = q \delta(x)$ which will leave me with the following result when i try to use Fourier transforms to transform $f(\vec{k})$ back to $f(\vec{x})$:
$f(\vec{x}) = \frac{-1}{2 \pi} \frac{q}{\epsilon_o} \int_{-\infty}^{+\infty} \frac{1}{k^2}e^{i \vec{k}\vec{x}}dk $
However I do not know how to integrate this to find the answer back in x-space. Have I went wrong somewhere or is their a certain trick to this integral?
Best Answer
With that note, let's begin by breaking the problem down into components, each of which is hopefully elementary.
STEP 1:
First, we know that the Fourier Transform of the Dirac Delta $\delta$ is
$$\begin{align} \mathscr{F}\{\delta\}(k)&=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\delta(x)e^{ikx}\,dx\\\\ &=\frac{1}{\sqrt{2\pi}} \end{align}$$
STEP 2:
Second, recall that the Fourier Transform has the property such that the Fourier Transform of the $n$'th order derivative, $D^nf$, of a function $f$, is given by
$$\bbox[5px,border:2px solid #C0A000]{\mathscr{F}\{D^nf\}(k)=(ik)^n\mathscr{F}\{f\}(k)} \tag 2$$
STEP 3:
Now, define the ramp function $r(x)$ as
$$r(x)= \begin{cases} x&,x\ge 0\\\\ 0&,x<0 \end{cases}$$
Note that the second derivative of the ramp function is $D^2r=\delta$. Using $(2)$ reveals that
$$\begin{align} \mathscr{F}\{r\}(k)&=\frac{1}{(ik)^2}\mathscr{F}\{D^2r\}(k)\\\\ &=\frac{1}{\sqrt{2\pi}}\frac{1}{(ik)^2} \end{align} \tag 3$$
STEP 4:
Taking the inverse Fourier Transform of $(3)$ and multiplying by $2\pi$ yields
$$\bbox[5px,border:2px solid #C0A000]{2\pi\,r(x)=\int_{-\infty}^{\infty}\frac{-1}{k^2}e^{-ikx}\,dk }\tag 4$$
STEP 5:
Finally, using $(4)$ we find that
$$\bbox[5px,border:2px solid #C0A000]{\frac{-1}{2\pi}\frac{q}{\epsilon_0}\int_{-\infty}^{\infty}\frac{1}{k^2}e^{ikx}\,dk=\frac{q}{\epsilon_0}r(x)}$$
We can use the preceding analysis to solve the more general one-dimensional Poisson Equation
$$f''(x)=\rho(x)$$
Taking Fourier Transforms yields
$$\hat f(k)=\frac{-1}{k^2}\hat \rho(k)$$
whereupon inversion reveals that
$$\begin{align} f(x)&=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\frac{-1}{k^2}\hat \rho(k)e^{-ikx}\,dk\\\\ &=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\frac{-1}{k^2} \left(\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\rho(x')e^{ikx'}\,dx'\right)e^{-ikx}\,dk\\\\ &=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\rho(x')\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\frac{-1}{k^2}e^{-ik(x-x')}\,dk\,dx'\\\\ &=\int_{-\infty}^{\infty}\rho(x')r(x-x')\,dx'\\\\ &=\int_{-\infty}^{x}(x-x')\,\rho(x')\,dx'\\\\ \end{align}$$
Thus, the general solution is
$$\bbox[5px,border:2px solid #C0A000]{f(x)=\int_{-\infty}^{x}(x-x')\,\rho(x')\,dx'}$$