4
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

We note here that all of the ensuing analysis uses notation that is interpreted in the sense of Distributions or Generalized Functions.

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

$$\begin{align} \mathscr{F}\{\delta\}(k)&=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\delta(x)e^{ikx}\,dx\\\\ &=\frac{1}{\sqrt{2\pi}} \end{align}$$

ASIDE:

This implies that the Inverse Fourier Transform of the constant function $1$ is the Dirac Delta $\delta(x)$. We can rewrite this relationship as $$\bbox[5px,border:2px solid #C0A000]{\int_{-\infty}^{\infty}e^{-ikx}\,dk=2\pi\,\delta(x)} \tag 1$$ where again the notation in $(1)$ is interpreted as a distribution.


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

$$\bbox[5px,border:2px solid #C0A000]{f(x)=\int_{-\infty}^{x}(x-x')\,\rho(x')\,dx'}$$

$\endgroup$
  • $\begingroup$ This presentation is book quality. The delta function is a magical little feature and is a transforms best friend. $\endgroup$ – Leucippus Sep 2 '15 at 3:43
  • $\begingroup$ @Leucippus Thanks as always!!!! The Dirac Delta is awesome. Did you know that Dirac's undergraduate degree was in Electrical Engineering. I wonder when he began first thinking about the Delta. Perhaps it was as an impulse to a circuits problem. $\endgroup$ – Mark Viola Sep 2 '15 at 3:46
  • $\begingroup$ That I did not. I recall that there was a historical presentation in his first edition quantum mechanics book. I'm sure somewhere it goes back to Heaviside and Laplace. $\endgroup$ – Leucippus Sep 2 '15 at 3:51
  • $\begingroup$ @Leucippus You are correct! Dirac coined the term in his seminal 1930 book.From here , it appears that it was not until Schwartz's work in 1945 that led to the formal development. $\endgroup$ – Mark Viola Sep 2 '15 at 4:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.