ResearchForge / Calculators
← back to simulations
arXiv:synth-heat-1d · uncategorized

Synthetic 1D Heat Equation Benchmark

Test Harness

method: pdetier: T2fidelity 0.92arXiv abstract →read the article →

Abstract

A textbook 1D heat-equation case used to validate the PDE simulator path.

Extracted equations

  • du/dt = D * d^2 u / dx^2

Simulation outputs

plot /research-assets/synth-heat-1d/profile.pngplot /research-assets/synth-heat-1d/heatmap.png

Scalar outputs

u_max_final0.6105
u_min_final0.0000
u_mean_final0.3867
u_L2_initial0.7071
u_L2_final0.4317
dt_used1.010e-4
n_steps4.951e+3
energy_ratio0.6105

Paper claims vs. our run

The simulation demonstrates excellent agreement with the paper's claims. The L2 decay ratio matches the theoretical prediction to high precision (0.6105 vs. 0.610), and the solution exhibits the expected boundedness and non-negativity. The numerical setup (sin(π·x) IC, Dirichlet boundaries, heat equation with D=0.1) is correctly implemented, and the computed energy ratio validates the exponential decay law exp(-π²·D·t) at the specified parameters.

  • For sin(pi*x) IC on [0,1] with Dirichlet zeros, the L2 amplitude decays as exp(-pi^2 * D * t)
    partial
    The energy_ratio of 0.6105 at t=0.5 with D=0.1 matches the theoretical exp(-π²·0.1·0.5) ≈ 0.6105 very closely, confirming the decay law numerically, but we cannot directly verify the functional form exp(-π²*D*t) from scalar outputs alone.
  • At t=0.5 with D=0.1, the ratio ||u(t)||/||u(0)|| equals approximately 0.610
    supported
    The energy_ratio of 0.6104892677524105 matches the claimed value of 0.610 to three significant figures, and u_L2_final/u_L2_initial = 0.4317/0.7071 ≈ 0.6105, confirming the claim precisely.
  • The solution remains bounded and positive throughout the domain
    supported
    u_min_final = 0.0 and u_max_final = 0.610 confirm the solution is non-negative and bounded; positivity is maintained (minimum is zero, not negative).

Parameters

D0.1

Run notes

pde=heat; nx=200; dt=1.010e-04; steps=4951; boundary=dirichlet; IC='sin(pi*x)'