Questions: The Heat Equation and Diffusion Problems

Question 1

A rod's initial temperature profile is a sharp spike near the center, with both endpoints held at 0°C. After a long time, what does the temperature distribution look like?

• A. The spike migrates toward one endpoint and stays there
• B. The temperature oscillates symmetrically around zero, never fully settling
• C. The temperature smoothly decays to zero everywhere as all modes die out
• D. The spike broadens but retains its shape indefinitely

Answer: (C)

With zero-temperature boundary conditions, the steady-state solution is u = 0 everywhere. Every term in the Fourier series solution carries an exponential time factor e^(−k(nπ/L)²t) that decays to zero. The spike, however sharp, is composed of many Fourier modes — each one decays exponentially, so the entire profile smoothly flattens to zero. There is no mode that sustains itself; all decay. Option D is the classic misconception: diffusion does spread the spike, but it does not preserve it — the peak amplitude also continuously shrinks.

Question 2

In the heat equation solution u(x,t) = Σ bₙ sin(nπx/L) e^(−k(nπ/L)²t), why does a fine-grained initial temperature pattern smooth out faster than a broad one?

• A. Fine-grained patterns have larger Fourier coefficients bₙ, so they dominate initially
• B. Fine-grained patterns correspond to higher-n modes, whose decay rate k(nπ/L)² grows as n², making them decay much faster
• C. Fine-grained patterns create larger temperature gradients that drive faster conduction
• D. The diffusion constant k is larger for high-spatial-frequency components

Answer: (B)

The decay rate of each mode is k(nπ/L)², which grows as n² — the decay rate of the n=10 mode is 100 times that of the n=1 mode. Fine-grained patterns are dominated by large-n terms, so they decay rapidly. Broad, smooth patterns are dominated by n=1 and n=2 terms, which decay slowly. This is the mathematical reason why diffusion is a smoothing process: it preferentially destroys spatial detail. Option C is physically intuitive but does not explain the mechanism — the n² scaling in the exponent is the precise cause.

Question 3

The heat equation ∂u/∂t = k∂²u/∂x² is time-reversible: if u(x,t) solves it, then u(x,−t) also solves it.

• T. True
• F. False

Answer: False

This is false. Substituting t → −t gives ∂u/∂(−t) = k∂²u/∂x², i.e., −∂u/∂t = k∂²u/∂x², which is a different equation (heat flow in reverse). Physical diffusion irreversibly spreads heat from hotter to colder regions — you cannot unscramble a smoothed temperature profile by 'running it backward.' This asymmetry reflects the second law of thermodynamics. The wave equation, by contrast, IS time-reversible; this is a key distinction between parabolic (heat) and hyperbolic (wave) PDEs.

Question 4

If a rod has insulated endpoints (no heat escapes), the appropriate Fourier series for the solution uses sine functions.

• T. True
• F. False

Answer: False

Insulated endpoints impose the Neumann condition ∂u/∂x = 0 at x = 0 and x = L, meaning zero flux (no heat flow out). Cosine functions satisfy this condition because their derivatives at 0 and L are zero. Sine functions satisfy Dirichlet conditions (u = 0 at the endpoints, corresponding to endpoints held at zero temperature). The choice between sine and cosine series is dictated by the physical boundary conditions — a key link between the PDE, its boundary conditions, and the Fourier representation.

Question 5

Explain why the solution method 'separation of variables' works for the heat equation, and what the resulting Fourier coefficients represent physically.

Model answer: Separation of variables assumes u(x,t) = X(x)T(t), substituting into the PDE and dividing produces X''/X = T'/(kT) = −λ (a constant, since both sides depend on different variables). This separates into two ODEs. The spatial ODE with boundary conditions yields a discrete set of eigenfunctions X_n(x) (e.g., sines for zero-endpoint conditions) — the natural spatial 'modes' of the rod. The temporal ODE gives exponential decay T_n(t) = e^(−kλ_n t). The Fourier coefficients bₙ represent the initial amplitude of each spatial mode — how much of the initial temperature profile f(x) 'projects onto' each eigenfunction. Each mode evolves independently, decaying at its own rate.

The key physical picture: any initial condition can be decomposed into spatial modes (via the Fourier representation). Once decomposed, each mode evolves independently and exponentially decays. This decomposition is possible because the heat equation is linear — modes do not interact. The Fourier coefficients bₙ are found by matching u(x,0) = f(x) to the Fourier series, which is exactly the Fourier analysis technique from the prerequisite course.