What happens at the bounce?
Calculus is great at describing smooth changes, but sometimes we deal with abrupt jumps - e.g. a ball hits the ground and bounces: in an idealized approximation, its velocity switches from downward to upward instantaneously. What is its acceleration, i.e. the derivative of velocity, at that instant?
As an ordinary derivative, acceleration is undefined at that instant. The usual answer invokes the Dirac delta “function”: informally, zero away from a single point and supposedly infinite at that point - clearly no ordinary function. Fourier analysis adds to the puzzle by suggesting that such an infinitely sharp spike can be assembled from perfectly smooth waves.
To make sense of this, we will explore hyperfunctions, introduced by Mikio Sato in the late 1950s. We will stray from the real line and venture into the complex plane. The complex plane is not strictly necessary to resolve these puzzles, but we like approaches that bring apparently different ideas together with a touch of mathematical elegance, so that is the route we shall take.
We will assume familiarity with calculus in the complex plane, including contour integration and the residue theorem. We will focus on the ideas, with some proofs along the way; more detailed proofs can be found elsewhere.
Functions as cliff jumps
Our bouncing ball’s velocity has the shape of a step. Choosing convenient units, we can write
$$ v(t)=H(t)-\frac{1}{2}, $$where \(H(t)\) is the ol’ good Heaviside step function, so the velocity switches from \(-1/2\) to \(+1/2\). Away from the collision, the derivative of this simplified velocity is zero, but at the collision, it does not exist.
A fox beside a grassy cliff
The ground of this landscape will be the complex plane. We represent “height” in the upper half-plane by \(F_+(z)\), and in the lower half-plane by \(F_-(z)\). Both functions have complex derivatives throughout their respective regions. Now, their values are complex rather than real numbers representing height, but the cliff gives us the right picture:
$$ f(x)=\lim_{\epsilon\to0^+} \left[F_+(x+i\epsilon)-F_-(x-i\epsilon)\right] $$We approach the cliff from both sides and take the difference. If the boundary values behave nicely, we recover an ordinary function, excellent! However! Even if the limit does not exist, we can keep the pair and call it a hyperfunction:
$$ \widehat f\equiv[F_+,F_-] $$Hyperfunctions are a generalization of functions. The operations on them, e.g. addition and differentiation, will be well defined even when cliff values are not. Before we move on to operations, as our first step, let us see how we could represent the Heaviside step function as a hyperfunction. The complex logarithm already contains a discrete jump as we do one full walk around the origin:
$$ \operatorname{Log}z=\ln|z|+i\arg z, \qquad -\pi\lt\arg z\lt\pi $$This gives us the idea to define the hyperfunction
$$ \widehat H= \left[ -\frac{\operatorname{Log}(-z)}{2\pi i}, -\frac{\operatorname{Log}(-z)}{2\pi i} \right] $$Let us verify that this hyperfunction’s cliff values reproduce the Heaviside step function. For \(x\lt0\), both sides approach the same value, so the difference equals zero. For \(x\gt0\), approaching from above and below gives
$$ \begin{aligned} f(x) &=\lim_{\epsilon\to0^+} \left[F_+(x+i\epsilon)-F_-(x-i\epsilon)\right]\\ &=-\frac{(\ln x-i\pi)-(\ln x+i\pi)}{2\pi i}\\ &=1. \end{aligned} $$As another example, let us see how a constant function \(c\) can be represented as a hyperfunction:
$$ \widehat c=[c,0] \qquad f(x)=\lim_{\epsilon\to0^+}(c-0)=c $$Things to do with hyperfunctions
We have seen that hyperfunctions can represent ordinary functions, but what else can we do with them? Let’s define some operations. We expect them to agree with the corresponding operations on ordinary functions whenever cliff values exist and behave nicely.
For example, we have hyperfunctions \(\widehat f=[F_+,F_-]\), \(\widehat g=[G_+,G_-]\), which have corresponding cliff values of \(f(x)\) and \(g(x)\). We define hyperfunction sum by:
$$ \widehat f+\widehat g \equiv [F_++G_+,\,F_-+G_-] $$Let us check the cliff values of \(\widehat f+\widehat g\):
$$ \begin{aligned} & \lim_{\epsilon\to0^+} \Bigl[ (F_+(x+i\epsilon)+G_+(x+i\epsilon)) -(F_-(x-i\epsilon)+G_-(x-i\epsilon)) \Bigr]\\ &= \lim_{\epsilon\to0^+} \Bigl[F_+(x+i\epsilon)-F_-(x-i\epsilon)\Bigr] + \lim_{\epsilon\to0^+} \Bigl[G_+(x+i\epsilon)-G_-(x-i\epsilon)\Bigr]\\ &=f(x)+g(x). \end{aligned} $$Thus, the hyperfunction sum has cliff value \(f(x)+g(x)\), as expected. However, hyperfunction addition still makes sense even when cliff values do not exist! We’ll skip the analogous checks for the other operations. Multiplication by an analytic function \(\phi\) means:
$$ \phi(x)\widehat f \equiv [\phi(z)F_+,\phi(z)F_-] $$Multiplication of two arbitrary hyperfunctions, however, is not in general defined. Differentiation means:
$$ \widehat f'\equiv[F_+',F_-'] $$For example, we can differentiate our logarithmic representation of the step:
$$ \frac{d}{dz}\operatorname{Log}(-z)=\frac{1}{z} $$Therefore,
$$ \widehat H'= \left[ -\frac{1}{2\pi iz}, -\frac{1}{2\pi iz} \right] $$Integration will further reveal what this specific derivative does. We will define integration by taking a clockwise contour that encircles the real line segment \([a,b]\). As the contour is brought down toward the real line, its upper part runs from \(a\) to \(b\), while its lower part runs back from \(b\) to \(a\).
$$ \int_a^b \widehat f(x)\,dx \equiv \oint F(z)\,dz = \lim_{\epsilon\to0^+} \int_a^b \left[ F_+(x+i\epsilon)-F_-(x-i\epsilon) \right]dx $$This form also helps to define the improper integral:
$$ \int_{-\infty}^{\infty}\widehat f(x)\,dx = \lim_{\epsilon\to0^+} \int_{-\infty}^{\infty} \left[ F_+(x+i\epsilon)-F_-(x-i\epsilon) \right]dx $$As promised, let us integrate \(\widehat H'\). Using the definitions and Cauchy’s integral formula:
$$ \int_a^b \phi(x)\widehat H'(x)\,dx = -\frac{1}{2\pi i} \oint \frac{\phi(z)}{z}\,dz = \phi(0) $$Checking for \(\phi(x)=1\):
$$ \int_a^b \widehat H'(x)\,dx = -\frac{1}{2\pi i} \oint \frac{1}{z}\,dz = 1 $$These are exactly the defining behaviors of the Dirac delta! The derivative of the Heaviside step function is the delta, and is represented in our complex-plane picture by a simple pole.
$$ \widehat H'=\widehat\delta $$The bouncing ball
Going back to our bouncing ball example, we can now take derivatives at the jump in peace. For a bouncing ball:
$$ \widehat v=\widehat H-\widehat{\frac{1}{2}} $$and therefore
$$ \frac{d\widehat v}{dt}=\widehat\delta(t) $$Take \(a\lt0\lt b\), so the interval contains the collision, then
$$ \int_a^b \frac{d\widehat v}{dt}\,dt =\int_a^b\widehat\delta(t)\,dt =1 $$Thus the integral gives exactly the change in velocity.
We used hats in the equations above to remind ourselves that we were working with hyperfunctions. Most of the time, they don’t add much, as when cliff values exist, they reproduce regular functions anyways. So we can leave the hats off and only remind ourselves of their true nature when we bump into discontinuities or deltas.
The delta as a sharp peak
Looking back, we used Cauchy’s integral formula directly to show the delta property. Let us now write the same integral in its limit form:
$$ \begin{aligned} \int_{\mathbb R}\phi(x)\widehat{\delta}(x)\,dx &=\lim_{\epsilon\to0^+} \int_{\mathbb R}\phi(x) \left[ -\frac{1}{2\pi i} \left(\frac{1}{x+i\epsilon}-\frac{1}{x-i\epsilon}\right) \right]dx\\ &=\lim_{\epsilon\to0^+} \int_{\mathbb R}\phi(x) \frac{\epsilon}{\pi(x^2+\epsilon^2)}\,dx\\ &=\phi(0). \end{aligned} $$The expression inside the integral deserves a name:
$$ \delta_\epsilon(x) \equiv -\frac{1}{2\pi i} \left(\frac{1}{x+i\epsilon}-\frac{1}{x-i\epsilon}\right) =\frac{\epsilon}{\pi(x^2+\epsilon^2)} $$For each \(\epsilon\gt0\), this is a smooth peak with area one. As \(\epsilon\to0^+\), it grows taller and narrower around \(0\), while tending to zero everywhere else. This is the ‘sharp peak’ picture of the delta.
$$ \lim_{\epsilon\to0^+}\int_{-\infty}^{\infty} \phi(x)\delta_\epsilon(x)\,dx=\phi(0) $$There is another way to rewrite \(\delta_\epsilon(x)\) using the following identity:
$$ \int_0^\infty e^{-sk}\,dk=\frac1s \qquad(\operatorname{Re}s\gt0) $$Using this for each term in the difference:
$$ -\frac{1}{2\pi i(y+i\epsilon)} =\frac{1}{2\pi}\int_0^\infty e^{iky}e^{-\epsilon k}\,dk, \qquad \frac{1}{2\pi i(y-i\epsilon)} =\frac{1}{2\pi}\int_{-\infty}^{0}e^{iky}e^{\epsilon k}\,dk $$Putting them together gives
$$ \delta_\epsilon(y) =\frac{1}{2\pi}\int_{\mathbb R} e^{iky}e^{-\epsilon|k|}\,dk $$Thus the peak is a continuous sum of smooth waves, where different waves are softened by the factor \(e^{-\epsilon|k|}\). As \(\epsilon\) shrinks, the sum sharpens toward the spike, giving a better picture of how a delta can emerge from smooth waves.
The emergence of the Fourier transform
Let us begin with the delta’s ability to recover a regular function \(f\), then use its wave representation from the previous section:
$$ \begin{aligned} f(x) &=\lim_{\epsilon\to0^+}\int_{\mathbb R}f(t)\delta_\epsilon(x-t)\,dt\\ &=\lim_{\epsilon\to0^+}\frac{1}{2\pi} \int_{\mathbb R}f(t) \left[\int_{\mathbb R}e^{ik(x-t)}e^{-\epsilon|k|}\,dk\right]dt\\ &=\lim_{\epsilon\to0^+}\frac{1}{2\pi} \int_{\mathbb R} \left[\int_{\mathbb R}f(t)e^{-ikt}\,dt\right] e^{ikx}e^{-\epsilon|k|}\,dk. \end{aligned} $$This motivates a definition, which we recognize as the Fourier transform of \(f\):
$$ \widetilde f(k)\equiv\int_{\mathbb R}f(t)e^{-ikt}\,dt $$So our expression becomes
$$ f(x)=\lim_{\epsilon\to0^+}\frac{1}{2\pi} \int_{\mathbb R}\widetilde f(k)e^{ikx}e^{-\epsilon|k|}\,dk $$When we can take the limit inside the integral, the damping factor tends to one, and we recover the familiar inverse Fourier transform:
$$ f(x)=\frac{1}{2\pi}\int_{\mathbb R}\widetilde f(k)e^{ikx}\,dk $$A way around
Even though we have glided over many details, the hyperfunction picture has allowed us to follow a very neat chain of ideas. A bouncing ball led us to a step, the step to a logarithm, and its derivative to a pole. Integrating around that pole revealed the Dirac delta, and expressing the delta through waves led us to the Fourier transform.
Sometimes the way past a difficulty is to give ourselves room to walk around it. Here, that extra room was the complex plane.