Construct diffusion on the whole line

Try

Place a concentrated amount of heat near one point on an infinite line. As it spreads, its height falls and its width increases. A Gaussian profile captures this evolution exactly for constant diffusivity. The kernel K(x,t) = (4πκt)^(−1/2) exp(−x²/(4κt)), for κ and t positive, has total integral 1.

The kernel is an impulse response. At time zero it does not become an ordinary function with height infinity at one point; its limit is a delta distribution. Ordinary initial profiles are handled by convolution with this response.

Predict

If time is multiplied by 4, predict how the Gaussian width and peak change. Use the exponent to identify the width scale before examining the prefactor. Does the total mass change?

Experiment

The variance is 2κt, the standard deviation is √(2κt), and the peak is 1/√(4πκt).

Time change	Standard deviation factor	Peak factor	Mass factor
t to 4t	2	1/2	1
t to 9t	3	1/3	1

Check mass by substituting z=x/√(4κt) and using ∫e^(−z²)dz=√π. Then differentiate the logarithm of K: Kₜ/K=−1/(2t)+x²/(4κt²), while κKₓₓ/K gives the same expression. This verifies the PDE for positive time.

Understand

For suitable initial data f, define u(x,t)=∫K(x−y,t)f(y)dy. The kernel’s integrable derivatives at positive time justify differentiation under the integral for, for example, bounded data. Thus u solves the heat equation. Positivity and unit mass make u a weighted average of initial values, preserving upper and lower bounds.

To recover bounded continuous initial data at a fixed continuity point x, write u(x,t)−f(x)=∫K(z,t)[f(x−z)−f(x)]dz. Split the integral into a small neighborhood of zero and its complement. Continuity makes the nearby difference small; the Gaussian tail mass outside any fixed neighborhood tends to zero. This proves pointwise initial recovery. Uniform continuity gives uniform recovery; L² data instead admit L² recovery. State the relevant hypothesis rather than claiming one convergence mode for every datum.

Convolving two heat kernels adds their variances, giving K at time s+t. This semigroup property expresses evolution in successive time intervals. It can be proved by completing the square in the convolution integral or using Gaussian Fourier transforms.

If f is nonnegative, integrable and positive on a set of positive measure, then u(x,t)>0 at every x for every positive t. Even very distant points receive a mathematically nonzero contribution. This infinite propagation is a property of the ideal diffusion model, not a claim that physical information travels infinitely fast in every material.

Apply

Compare two initial Gaussian profiles with equal total mass but different variances. Add 2κt to each initial variance and predict their later peaks. Explain why knowing only the peak at one time does not identify both the original variance and diffusivity. Then distinguish the kernel’s delta initial condition from a narrow but finite initial Gaussian.

Quantitative initial recovery and a discontinuity test

For bounded globally Lipschitz f with constant L, the convolution solution has the uniform estimate |u(x,t)−f(x)|≤L∫|z|K(z,t)dz=2L√(κt/π). To compute the last integral, use symmetry and substitute w=z²/(4κt) on the positive half-line. This is an analytic bound valid for every x, not a sampled plot. For κ=1, L=2 and t=1/100 it gives 0.4/√π, about 0.2257. Reducing time by a factor of four halves this guarantee.

Locally in x and on compact time intervals bounded away from zero, every required x-derivative and first t-derivative of K is a polynomial times a Gaussian with a common integrable dominating envelope. Bounded f permits differentiation under the integral on such compact time intervals. Since Kₜ=κKₓₓ, u is a classical solution for t>0. The uniform recovery estimate supplies its initial condition for this bounded Lipschitz class. This is an existence construction; do not turn it into an unrestricted uniqueness claim for arbitrary rapidly growing solutions on the whole line.

Worked failure at a jump. Let f be the indicator of the positive half-line and assign f(0)=0. At x=0 the convolution equals 1/2 at every positive time by symmetry. Thus it does not converge to the chosen value f(0). The datum is bounded but not continuous at that point. Changing the value at zero changes neither convolution nor its limit, because a single point has zero Lebesgue measure.

Changed written task. If |f|≤M and f is uniformly continuous with modulus ω(δ)=sup of |f(x−z)−f(x)| for |z|≤δ, show |u−f|≤ω(δ)+2M·P(|Z|>δ), with Z of variance 2κt. Use Chebyshev to obtain ω(δ)+4Mκt/δ². Feedback: choose δ first to control ω, then t to control the tail; reversing these quantifiers can lose uniformity. This bound is conservative but requires no numerical Gaussian integration. It complements, rather than replaces, the weaker pointwise continuity argument above.

Predict, change the case, and repair the claim

Case Lipschitz recoveryJump datum

κ=1, global Lipschitz constant L=2, t=1/100.

Which uniform error bound follows from Gaussian averaging?

4/(10√π) 0 for every datum 2/t

Check this claim Reset practice

Before checking, write the hypothesis or calculation that decides your answer. Change cases and explain which assumption changed. These fixed cases are public guided practice, not a saved score or independent proof assessment.

Continue

Sources and nonzero boundaries require additional terms. Duhamel’s principle will combine responses from different source times, while boundary lifting first converts prescribed boundary values into homogeneous ones.