Heat Kernel
y = ∫_(x − 12)^(x + 12) sign(sin(u)) · exp(−(x − u)²/(0.2 + mod(t, 20))) du/sqrt(π · (0.2 + mod(t, 20)))
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What it draws
The curve is an integral, computed at every point: y = ∫ sign(sin u) · e^(−(x − u)²/(0.2 + τ)) du / √(π·(0.2 + τ)), with τ = t mod 20. sign(sin u) is a square wave of period 2π ≈ 6.28, at +1 for half a period and −1 for the other half. The rest is a Gaussian centred on x with unit area, so y is a weighted average of the square wave around x. That is a convolution, and the width of the blur grows with the clock.
At τ = 0 the kernel is only √0.2 ≈ 0.45 wide, and the square wave shows with its corners softened. The blur then widens as √(0.2 + τ) and the wave melts, until at t = 20 seconds at Speed 1 the modulo restarts the cycle. The limits x ± 12 stand in for ±∞, the kernel being negligible beyond them.
The heat equation
A Gaussian whose variance grows in step with time is the heat kernel, the fundamental solution of the heat equation ∂u/∂τ = D · ∂²u/∂x², which turns an initial temperature into a later one by convolution. Here the initial profile is the square wave, and the diffusivity is D = 1/4, because the kernel's denominator 4Dτ is 0.2 + τ.
The square wave is a Fourier series, (4/π)·(sin x + sin 3x/3 + sin 5x/5 + …), and the heat equation damps each harmonic sin kx by e^(−k²(0.2 + τ)/4). The high harmonics are the corners, and they go first. The third harmonic is 22% of the fundamental at the start and 3% one second in, when the curve is already nearly a sine of height 0.91. The fundamental halves every 4·ln 2 ≈ 2.77 seconds: 0.34 at t = 5, under 0.1 by t = 10.
History
Joseph Fourier wrote down the heat equation and solved it with trigonometric series in a memoir of 1807, which the Paris Academy's referees, Lagrange among them, declined to publish. His book of 1822, Théorie analytique de la chaleur, set out the method in full: any profile is a sum of sines, and here each sine cools on its own.
Try
- Start from a sine, integral(sin(u)·exp(−(x − u)²/(0.2 + mod(t, 20))), u, x − 12, x + 12)/sqrt(π·(0.2 + mod(t, 20))): one harmonic only shrinks, never changes shape, its height exactly e^(−(0.2 + τ)/4).
- Double the frequency, sign(sin(2u)) in place of sign(sin(u)): the crests are half as far apart and the wave melts four times as fast, since the decay rate goes as k².
- Change both 0.2 to 2: the cycle starts already smoothed, a near sine of height ≈ 0.77.
- Change both 20 to 5: the reset comes while the crest still stands at ≈ 0.34.