Evolve a profile by decaying spatial modes

Try

A rod occupies [0,π], with its endpoints held at zero temperature relative to a reference level. Its interior evolves by uₜ = κuₓₓ, where κ is a positive diffusivity. Begin with two sine ripples of different frequencies. Predict which ripple will fade faster before calculating either rate.

The reference temperature can be subtracted from a physical temperature, so negative values here mean below reference. The model assumes constant diffusivity and no internal source. It is not a calibrated thermal simulation.

Predict

Take f(x) = 2sin x + sin(3x) and κ = 1. If the first coefficient is multiplied by e^(−t), should the third-frequency coefficient be multiplied by e^(−3t) or e^(−9t)? Use a second spatial derivative to decide.

Experiment

Live experiment: the same profile, two different equations

On 0≤x≤π, both endpoints stay at zero. Start with f(x)=sin(x)+a₂sin(2x)+a₃sin(3x). Predict which modes disappear fastest under heat flow, and why a wave can have small displacement without losing energy. All quantities use dimensionless teaching units.

Equation Heat: uₜ=κuₓₓWave: uₜₜ=c²uₓₓTime t 00.250.5123Diffusivity κ 0.512Second-mode coefficient a₂ -0.500.5Third-mode coefficient a₃ -0.2500.25Reset experiment

The dashed initial profile and solid current profile share fixed axes. At x=π/2, initial u=0.75 and current u=0.603753. These are evaluations of a three-term exact solution, not a spatial-grid PDE approximation. The drawing samples it at 241 positions.

Mode n	Initial aₙ	Current amplitude	Amplitude time derivative
1	1	0.606531	-0.606531
2	0.5	0.067668	-0.270671
3	0.25	0.002777	-0.024995

Mode n is multiplied by exp(−κn²t). Its time derivative is −κn² times its amplitude, matching κ times its second spatial derivative. The n² factor suppresses higher frequencies more rapidly. The solution satisfies the initial data at t=0 and both zero endpoint conditions.

Heat energy E=½∫u² dx: 0.292534. Initial E=1.030835. Its derivative is −κ∫uₓ² dx=-0.606743. This is a squared-amplitude energy, not total thermal mass.

Sine orthogonality gives ∫sin(nx)sin(mx) dx=0 for n≠m and π/2 for n=m over [0,π]. Thus the displayed energies come from exact modal sums; rounded numbers can conceal very small amplitudes. These smooth finite sums avoid convergence questions. Arbitrary initial data, infinite Fourier expansions and numerical stability require the separate arguments in the course.

Set t=0 to check the shared initial profile; then compare t=1 and t=3. Remove higher modes and explain the change in shape and decay. Switching equations compares models with different physical interpretations, not two solvers for one PDE. No animation runs; settings are temporary.

The widget compares finite sine expansions for heat and waves. Its controls are temporary exploration, not saved mastery evidence. Inspect the equation-specific energy labels: squared temperature and mechanical wave energy are different quantities, so their numerical values are not interchangeable physical measurements.

For the separate worked example above, the heat solution is 2e^(−t)sin x + e^(−9t)sin(3x).

Mode n	Initial coefficient	Decay rate κn²
1	2	1
3	1	9

Set a higher frequency in the widget, compare heat with wave evolution, and explain why rapid spatial oscillation produces rapid heat decay but not wave damping.

Understand

Separation u = X(x)T(t) gives X″ = −λX and T′ = −κλT after dividing in a region where the factors are nonzero; the resulting products can then be verified everywhere directly. Zero Dirichlet endpoints give Xₙ = sin(nx), λₙ = n². Projection of the initial data supplies coefficients aₙ = (2/π) times the integral of f(x)sin(nx) over [0,π]. The solution is Σaₙe^(−κn²t)sin(nx).

For finite sums, every differentiation and boundary check is immediate. For general L² data, Parseval and dominated convergence give u(t) approaching f in L² as t decreases to zero. At every time bounded away from zero, the exponential decay dominates polynomial powers of n, allowing differentiated series and smooth interior solutions. This does not imply that rough data become continuous at the initial instant or that incompatible endpoint values are attained continuously at the corners.

With insulated endpoints, cosine modes replace sines. The constant eigenfunction has eigenvalue zero and therefore does not decay. Its coefficient is the spatial average, conserved because the net boundary heat flux is zero. All positive modes decay, so the long-time state is that average rather than zero.

Apply

Build a fresh three-mode initial profile, calculate its three decay rates and verify both the PDE and zero endpoint values. Compare its L² energy at two times using coefficient squares, not a few selected point values. Then change to Neumann boundaries and explain which part survives forever. Keep exact series statements separate from any truncation used to draw the graph.

Continue

The next lesson proves decay and uniqueness without knowing any Fourier coefficients. Energy and maximum principles can establish structural facts even when an explicit eigenbasis is unavailable.