### Author Topic: Web Bonus Problem to Week 8 (#2)  (Read 2095 times)

#### Victor Ivrii ##### Web Bonus Problem to Week 8 (#2)
« on: November 01, 2015, 05:12:04 PM »

#### Jeremy Li 2

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• Karma: 0 ##### Re: Web Bonus Problem to Week 8 (#2)
« Reply #1 on: December 01, 2015, 10:59:13 AM »
I'm really not sure if I'm doing this correctly but...

Since $f(\mathbf{x})$ depends only on $|\mathbf{x}|$, $f(\mathbf{x})$ is rotationally symmetric and so is $\hat{f}(\mathbf{k})$. Then we might as well rotate the vector $\mathbf{k}$ onto an arbitrary axis. So
\begin{equation}
\hat{f}(\mathbf{k})=\hat{f}(0,0,...,k)
\end{equation}

For concreteness, lets do this in 2D, with polar coordinates. Then, since we're free to choose the direction of $\mathbf{k}$, we let $\mathbf{k}=(k,0)$ in polar coordinates, and $\mathbf{x}=(r,\theta)$.

\begin{equation}
\hat{f}(\mathbf{k})=\frac{1}{2\pi}\iint_{\mathbb{R}^n} f(\mathbf{x}) e^{-i\mathbf{k}\cdot\mathbf{x}} d^2 \mathbf{x}=\frac{1}{2\pi}\iint_{\mathbb{R}^n} f(\mathbf{x}) e^{-ikx} d^2 \mathbf{x}
\end{equation}
(since in cartesian coordinates $k_x=k$ and $k_y=0$, we can simplify the dot product)

Doing a change of coordinates we get,
\begin{equation}
\hat{f}(\mathbf{k})=\frac{1}{2\pi}\int_0^\infty \int_0^{2\pi} f(\mathbf{x}) e^{-ikr\cos\theta} r dr d\theta
\end{equation}

Is this correct? This formula is very difficult to apply even for the simplest problems like (a). I'm guessing this is because I calculated $\mathbf{k}\cdot\mathbf{x}$ as the cartesian dot product, which in polar coordinates is $rk\cos(\theta-\phi)$. This style of fourier transform decomposes the circularly-symmetric function into complex plane waves. I tried this on a computer and got a Bessel function.

If you meant to do a multi-dimensional fourier transform such that the function is decomposed into complex radial waves and complex tangential waves, the FT is much simpler. This isn't the usual definition of a dot product though.
\begin{equation}
\hat{f}(\mathbf{k})=\frac{1}{2\pi}\int_0^\infty \int_0^{2\pi} f(\mathbf{x}) e^{-ikr} r dr d\theta
\end{equation}

I will assume that you want the latter and I will try doing it that way.
« Last Edit: December 02, 2015, 05:06:03 AM by Jeremy Li 2 »

#### Jeremy Li 2

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• Karma: 0 ##### Re: Web Bonus Problem to Week 8 (#2)
« Reply #2 on: December 02, 2015, 06:48:15 AM »
Note that $k$ not boldfaced means the magnitude of $\mathbf{k}$.

Since $f(\mathbf{x})$ depends only on $|\mathbf{x}|$, $f(\mathbf{x})$ is rotationally symmetric and so is $\hat{f}(\mathbf{k})$. Then we might as well rotate the vector $\mathbf{k}$ onto an arbitrary axis. So
\begin{equation}
\hat{f}(\mathbf{k})=\hat{f}(0,0,...,k)
\end{equation}

For concreteness, I'll do this in 2D, with polar coordinates. Then, since we're free to choose the direction of $\mathbf{k}$, we let $\mathbf{k}=(k,0)$ in polar coordinates, and $\mathbf{x}=(r,\theta)$.

Then
\begin{equation}
\hat{f}(\mathbf{k})=\frac{1}{2\pi}\int_0^\infty \int_0^{2\pi} f(\mathbf{x})e^{-ikr}rdrd\theta=\frac{1}{2\pi}\int_0^\infty \int_0^{2\pi} f(r)e^{-ikr}rdrd\theta
\end{equation}
Since $f$ depends on $r$ only. The equation falls into a 1d fourier transform.

We get the forumla
\begin{equation}
\hat{f}(\mathbf{k})=\int_0^{\infty} f(r)e^{-ikr}rdr
\end{equation}

a) It's a straightforward calculation:
\begin{equation}
\hat{f}(\mathbf{k})=\int_0^a r e^{-ikr}dr=\frac{-1+e^{-i k a} (1+i k a)}{k^2}
\end{equation}

b) We first calculate the fourier transform of $a$ alone, and then $|\mathbf{x}|$ alone and subtract the difference. The fourier transform of $a$ is just $a$ times eq 8. The fourier transform of $|\mathbf{x}|$ can be calculated using 5.2.3a, via the formula $\hat{f}(\mathbf{k})=i\hat{g}'(k)$ where $\hat{g}(k)$ is the fourier transform from part a, eq 8.
\begin{equation}
\int_0^a ar e^{-ikr}dr=\frac{-a +a e^{-i k a} (1+i k a)}{k^2}
\end{equation}

\begin{equation}
\int_0^a r^2 e^{-ikr}dr=
\frac{2i + e^{-ika} (2ka + i k^2 a^2 -2 i)}{k^3}
\end{equation}

Taking the difference, and simplifying (a lot of terms cancel)
\begin{equation}
\hat{f}(\mathbf{k})=\frac{-ka-2i +e^{-ika} (2 i-ka)}{k^3}
\end{equation}

c) and d) to follow...
« Last Edit: December 02, 2015, 09:08:02 AM by Jeremy Li 2 »

#### Jeremy Li 2

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• Karma: 0 ##### Re: Web Bonus Problem to Week 8 (#2)
« Reply #3 on: December 02, 2015, 09:04:46 AM »
For part c) we do essentially the same thing, calculating
\begin{equation}
\hat{f}(\mathbf{k})=a^2 \hat{g}(\mathbf{k})- i\hat{h}'(k)
\end{equation}

Where $\hat{g}(\mathbf{k})$ is equation 8 from post above, and $\hat{h}(k)$ is equation 10 from above. Like the last question, a lot of stuff cancels and we get the answer:

\begin{equation}
\hat{f}(\mathbf{k})=\frac{-k^2a^2-6+(6+6ika-2k^2a^2) e^{-ika}}{k^4}
\end{equation}
« Last Edit: December 02, 2015, 12:12:02 PM by Jeremy Li 2 »

#### Jeremy Li 2

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• Karma: 0 ##### Re: Web Bonus Problem to Week 8 (#2)
« Reply #4 on: December 02, 2015, 12:10:30 PM »
d)
We calculate

\begin{equation}
\hat{f}(\mathbf{k})=\int_0^a e^{-\alpha r} e^{-ikr} rdr=\frac{1-e^{-(\alpha+ik)a} (1+\alpha a +ika)}{(\alpha+ik)^2}
\end{equation}