n-dimensional Fourier transform

The following are from Chapter 6 of Stein-Sharkarchi.

Calculus on $\mathbb{R}^n$¶

Integration¶

We use $\int_{\mathbb{R}} f(x) dx$ or $\int_{\mathbb R^n} f(x) dm(x)$ to denote the integral on $\mathbb{R}^n$ with the Lebesgue measure.

We’ll frequently consider integration on a hypersurface, in particular, the sphere. Let $B(0,r)$ be the ball of radius $r$ on $\mathbb R^n$, and $S(0,r)$ be its boundary, i.e. the $n$-sphere. The integral of restriction of $f$ on $S(0,r)$ on the $n$-sphere is denoted by $\int_{S(0,r)}f(y) d\sigma_r(y)$, where $d\sigma_r(y)$ is the area measure on the boundary of sphere.

The volume of ball and area of sphere is important quantity to know, we use $m(B(0,r))$ or $|B(0,r)|$ to denote the volume of ball radius $r$, and $|B^n|$ to denote the volume of unit ball in $\mathbb{R}^n$. Similarly, $|S^{d-1}|$ to denote the unit sphere in $\mathbb{R}^d$. For example, the volume of unit ball in $\mathbb{R}^3$ is $B^3$ and area of unit sphere is $S^2$.

In the exercise you’ll be asked to calculate the exact value of $|B^n|$ and $|S^{n-1}|$. Then we have $m(B(0,r)) = r^n|B^n|$ and $|S(0,r)| = r^{n-1}|S^{n-1}|$.

Fubini theorem¶

Integral on ball¶

Here $M_t$ means the average of $f$ on $S(0,t)$ given by $\int_{S(0,t)}f(y) \frac{d\sigma_t(y)}{|S(0,t)|}$. For example, on $\mathbb R^2$ we have $M_t(f) = \int_{S(0,t)}f \frac{d\sigma_t}{2\pi t} = \int_{0}^{2\pi}f(re^{it})\frac{dt}{2\pi}$.

Change of variable formula¶

Let $U$ be an open set, $\varphi:U\to V$ be a $\mathscr{C}^1$-diffeomorphism (i.e. φ is $\mathscr{C}^1$, one-to-one and onto, Jacobi matrix of φ invertible everywhere). Let $f:V\to \mathbb R$ be a function. Then $dx_1\wedge \cdots dx_n = \det (\varphi(y))dy_1\wedge \dots \wedge dy_n$, so $\int_{V}f(x) dx= \int_{U}f(\varphi(y))\det(\varphi(y))dy_1\wedge \dots dy_n$.

Example: Spherical coordinate¶

Let $\begin{pmatrix}x_1\\x_2\\x_3 \end{pmatrix}$ = $\begin{pmatrix} r\cos \varphi_1 \\ r\sin\varphi_1 \cos \varphi_2 \\ r\sin \varphi_1\sin\varphi_2 \end{pmatrix}$. Then we $dx_1\wedge dx_2 \wedge dx_3 = r^2 \sin\varphi_1 dr \wedge d\varphi_1 \wedge d\varphi_2$. Here $\varphi_1\in [-\pi/2,\pi/2], \varphi_2\in [-\pi,\pi]$.

More generally, on $\mathbb{R}^n$, let $\begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ \vdots \\ x_{n-1}\\x_n \end{pmatrix} = \begin{pmatrix} r\cos \varphi_1 \\ r \sin \varphi_1 \cos \varphi_2 \\ r\sin\varphi_1 \sin\varphi_2 \cos\varphi_3 \\ r\sin\varphi_1 \sin\varphi_2 \sin \varphi_3 \cos\varphi_4 \\ \vdots \\ r\sin\varphi_1 \cdots \sin\varphi_{n-2} \cos\varphi_{n-1} \\ r\sin\varphi_1 \dots\sin\varphi_{n-2} \sin\varphi_{n-1} \end{pmatrix}$.

Then $dx_1\wedge \cdots dx_n = r^{n-1}\sin\varphi_1\sin\varphi_2\cdot \sin \varphi_{n-2} dr\wedge d\varphi_1\wedge \cdots d\varphi_n$. Here $\varphi_1,\dots,\varphi_{n-2} \in [-\pi/2,\pi/2], \varphi_{n-1}\in [-\pi,\pi]$.

Differential¶

Let α be a multi-index. We use $x^{\alpha}$ to denote the monomial $x_1^{\alpha_1}\cdots x_n^{\alpha_n}$. We use $\partial_\alpha u$ to denote the operator $u\to \frac{\partial^{\alpha}u}{\partial x_1^{\alpha_1}\cdots \partial{x_n}^{\alpha_n}}$. For example, the Laplace operator $\Delta u = \frac{\partial^2}{\partial x_1^2}u + \cdots + \frac{\partial^2}{\partial x_n^2}u = \partial_{11}u + \cdots +\partial_{nn}u$. The order of derivative is given by $|\alpha| = \alpha_1+\cdots \alpha _n$.

Distribution theory¶

A test function on $\mathbb R^n$ is a smooth function with compact support on $\mathbb R^n$. With this, the distribution theory generalizes naturally to $\mathbb{R}^n$.

n-dimensional Fourier transform¶

Let $f:\mathbb{R}^n\to \mathbb{R}$ be a Schwartz function. The Fourier transform of $f$ is given by $\mathcal Ff(\xi) = \int_{\mathbb{R}^n} f(x)e^{-2\pi i \langle x,\xi\rangle}dx$, where $\langle x,\xi\rangle$ is the inner product given by $x_1\xi_1 + \cdots + x_n \xi_n$.

Basic properties¶

The following are direct generalization from one-dim case.

Derivative theorem¶

The Fubini theorem and one-dimensional case implies

• $\mathcal F(\partial_\alpha f)(\xi) = (2\pi i)^{|\alpha|} \xi^\alpha \mathcal Ff(\xi).$

• Conversely $(-2\pi i x)^{\alpha}f(x)$ ------> $\partial^{\alpha}\mathcal Ff(\xi)$

Shifting and scaling theorems¶

• $f(x+h)$ -----> $e^{2\pi i \langle x ,\xi\rangle }\mathcal Ff(\xi)$

• $f(x)e^{-2\pi i \langle x,h \rangle}$ -----> $\mathcal Ff(\xi + h)$

• $f(a x)$ -----> $|a|^{-n}\mathcal Ff(\frac{\xi}{a})$.

A new thing in high dimension is the linear transformation beocmes richer (much more than shifting and scaling).

Rotations¶

A rotation is a linear transformation which preserves length, i.e. a linear transformation $R:\mathbb{R}^n\to \mathbb{R}^n$ such that $|R(x)| = |x|$. We’ll also identify $R$ with its matrix and write $Rx$ for a rotation. By polarization identity, fixing norm is equivalent to fixing innner product. So we can also define rotation as a linear transformation which satisfies $\langle Rx,Ry \rangle = \langle x,y\rangle$ for every $x,y\in \mathbb R^n$.

The change of variable formula (apply to $\varphi(x) = Rx$) implies $\mathcal f(Rx)$ -----> $\mathcal Ff(R\xi)$.

Fourier transform of radial function¶

A radial function is a function $f:\mathbb{R}^n \to \mathbb{R}$, $x\mapsto f(x)$ that only depends on $|x|$. In other words, $f(x) = g(|x|)$ for some one-variable function $g$.

Wave equation¶

The one-dimensional wave equation is given by $u_{xx} = \frac{1}{c}u_{tt}$.

The wave equation is given by $\Delta u = \frac{1}{c^2}u_{tt}$. Here $u(x,t)$ is a function on $\mathbb{R}^n\times \mathbb R$. Note that here the $t$-variable, though represents time, is allowed to take negative value since we want to look at “backwards” wave as well.

We normalize so that $c = 1$. Take Fourier transform of wave equation on the spacial variable we get $-4\pi^2(s_1^2+\cdots+s_n^2) \hat{u}(s,t) = \widehat{\partial_{tt} u}(s,t)$. Here $\hat{u}(s,t):=\int_{\mathbb{R}^n} u(x,t)e^{-2\pi i \langle x,s \rangle} dm(x)$.

We expect that $\widehat{\partial_{tt}u} = \partial_{tt}\hat{u}$. This is justified by the “differential under integral sign” lemma, a consequence of dominated convergence theorem.

Lemma. If $\partial_t u(x,t)$ is continuous, then $\partial_t \int_{\mathbb R^n} u(x,t) e^{-2\pi i \langle x,\xi \rangle } dm(x) = \int_{\mathbb R^n} \partial_t u(x,t) e^{-2\pi i \langle \xi,x\rangle} dm(x)$.

So if $u$ is a solution of wave equation, then $\hat{u}(x,t)$ satisfies the ODE $-4\pi^2 \hat{u}(s,t) = \hat{u}_{tt}(s,t)$.

The ODE is can be easily solved (see here for a quick introduction of 2nd order ODE). Let $f(x) = u(x,0)$ and $g(x) = u_t(x,0)$ be the initial phase and speed. We have

It follows that $u$ is given by the inverse Fourier transform of the RHS.

Solving the wave equation¶

We’ll utalize our theory of distributions. For a distribution-free approach see Stein-Sharkarchi page 186-192.

1-dim case¶

We look at the one-dim case first using the fundamental solution approach. In the 1-dimensional case, $s\in \mathbb{R}$, so we know

And the second term is just our familiar sinc function. By scaling theorem $\mathcal F^{-1}(\frac{\sin(2\pi |s|t)}{2\pi |s|t} ) = \mathcal F^{-1}\mathrm{sinc}(2ts) = \frac{t}{2|t|}\Pi(\frac{x}{2t}) = \frac{1}{2}\chi_{[-t,t]}(x)$.

So by inverse Fourier transform to $(*)$ and apply convolution theorem we get

3-dim case¶

The 3 dimensional case turns out to be easier so we look at this case first. Observe that $\frac{\partial}{\partial t}(\frac{\sin(2\pi |s|t)}{2\pi|s|}) = \cos(2\pi|s|t)$ so the above argument can actually be simplified, and we see that the main problem is to calculate $\mathcal F^{-1}[\cos \xi]$ and $\mathcal F^{-1}[\mathrm{sinc}(\xi)]$ for $\xi\in \mathbb{R}^n$.

Lemma¶

Let σ be the area measure of $S(0,1)$ on $\mathbb{R}^3$. Then

Since the area of $S^2$ on $\mathbb R^3$ is $4\pi$, it also implies $\mathcal F(\overline \sigma)(\xi):=\mathcal F\frac{1}{4\pi}\sigma (\xi) = \frac{\sin(2\pi |\xi|)}{2\pi |\xi|}$. This is direct generalization of $\mathcal F \frac{1}{2}\chi_{[-1,1]}(s) = \frac{\sin 2\pi s}{2\pi s}$.

Proof. Note that the measure $d\sigma$ is rotation invariant, i.e. for every rotation $R$, $d\sigma(Rx) = d\sigma(x)$, it follows that $\mathcal F\sigma(\xi)$ is radial, consequently $\mathcal F\sigma(\xi) = \mathcal F\sigma\begin{pmatrix}|\xi|\\ 0 \\ 0 \end{pmatrix}$.

Since σ is compact support,

$\square$

By scaling theorem $\mathcal F^{-1}\frac{\sin (2\pi |\xi|t)}{2\pi |\xi|t} = \frac{1}{t^2}\frac{1}{4\pi} \sigma(\frac{x}{t}) = \frac{1}{4\pi t^2}\sigma_t(x) =: \bar{\sigma}_t$, here $\bar{\sigma}_t$ is the normalized area measure on $S(0,t)$, note that it is a probability measure.

For the cosine part, $\mathcal F^{-1}\cos(2\pi |\xi|t) = \mathcal F^{-1}\frac{\partial}{\partial t} \frac{\sin (2\pi |\xi|t)}{2\pi |\xi|} = \frac{\partial}{\partial t}\mathcal F^{-1}\frac{\sin(2\pi |\xi|t)}{2\pi |\xi|} =\frac{\partial}{\partial t}(t\bar{\sigma}_t)$. (The notation is a bit messy here but it will be clear when we apply it later, don’t worry.)

Another problem is about convolution with the area measure.

Lemma¶

Let $\sigma_t$ be the area measure on $S(0,t)$. Then for any $x$, let $(M_{t,x},f):=\frac{1}{4\pi t^2}\int_{S(x,t)}f(y)dS(y)$ be the average of $f$ on $\partial B(x,t)$. Then $(M_{t,x},f) = (f*\bar{\sigma}_t)(x)$. Note that here $f*\bar{\sigma}_t(x):=\int_{S(0,t)}f(x-y)\frac{d\sigma_t(y)}{4\pi t^2}$ is the convolution of the (compactly supported) probability measure.

Proof.

$\square$

Taking inverse Fourier transform to $(*)$ we get

Compare this with the one-dimensional case.

2-dim case¶

On Stein-Sharkarchi

Appendix: Speed of light¶

Differential forms on $\mathbb{R}^3$¶

The Maxwell equations¶

Let $C$ be a closed curve in $\mathbb{R}^3$ and $S$ be any surface patch surrounded by the curve. We assume everything are smooth for simplicity, though we mostly only need $\mathscr{C}^2$. Let $\mathbf{F}=(f_1,f_2,f_3)$ be a vector field. The line integral $\int_{C}F\cdot ds$ is given by the integral $\int_C f_1dx_1+f_2dx_2+f_3dx_3$. The surface integral $\int_S \mathbf{F}\cdot dS$ is given by $(-dS,f_1n_1+f_2n_2+f_3n_3)$ (see problem set 8) or the wiki page.

Take $\partial_t$ both side on (4) we get $\epsilon_0\mu_0 \partial_{tt}\mathfrak{e} = \partial_t d*\mathfrak{b} = d*\partial_t \mathfrak b = d*(-d*\mathfrak e) = -d*d*\mathfrak e$.

Laplace operator on differential forms¶

We need the computation from this note

• Let ω be a 2-form, then $\square \omega = d\delta\omega+\delta d\omega = d*d* \omega+*d*d\omega$. If ω is closed, then $\square\omega = d*d*\omega$ (which is the case of $\mathfrak e$ and $\mathfrak b$!)

• If $\omega = f(x,y,z)dx\wedge dy$, then $\square \omega = -(\partial_{xx}f+\partial_{yy}f+\partial_{zz}f)dx\wedge dy = -(\Delta f)dx\wedge dy$.

Combine the results we derive the wave equation:

The meaning of constant $c$ in the wave equation $\Delta f = \frac{1}{c^2}\partial_{tt}f$ is the speed of the wave. So $c = \frac{1}{\sqrt{\epsilon_0\mu_0}}$ is the speed of electronic wave in vaccum. So the Maxwel theory is not consistant with Newton mechanics because lightspeed is a constant, which leads to special relativity.