Heat kernels and semigroup smoothing
Derive diffusion's averaging structure while distinguishing whole-space kernels, periodic modes, and initial traces.
On this page
The question for today
What you’ll understand
- Use the Gaussian kernel to solve a whole-space heat problem.
- Explain mass preservation, instantaneous spreading, and positive-time smoothing.
- State the norm and data assumptions behind a semigroup continuity claim.
Diffusion smooths sharp features by averaging neighboring values, but the average is neither over a fixed radius nor over a fixed time step. The heat kernel describes exactly how the averaging scale grows with elapsed time. Its convolution law organizes evolution into a semigroup. That structure proves useful estimates while also revealing what changes when the spatial domain has boundaries, periodicity, or an initial datum too rough for a classical time-zero derivative.
A Gaussian gives the whole-space evolution
On Euclidean d-space, consider uₜ=κΔu with constant κ>0 and initial data u₀ in L¹. The Gaussian kernel displayed below has total integral one, is positive everywhere for t>0, and solves the heat equation away from its initial singularity. Convolving it with u₀ defines a smooth solution at every positive time. The initial condition is attained in L¹ as t decreases to zero; an arbitrary integrable datum need not have a pointwise value or a classical derivative at every initial point.
If u₀ is nonnegative and not zero as an L¹ function, positivity of the kernel makes u(x,t)>0 at every spatial point for every positive time. This is infinite propagation speed, even though the distant values can be extremely small. It is a mathematical feature of the diffusion model, not a statement that every underlying microscopic mechanism transmits signals infinitely fast. Integrating the convolution preserves the total mass when u₀ is integrable. For signed data, absolute integrability justifies the corresponding exchange of integrals.
Composition of averages explains the semigroup law
Convolution of two centered Gaussians adds their variances, so evolving for time s and then time t equals evolving for time s+t. Writing S(t)u₀=Gκ(t)*u₀ gives S(t)S(s)=S(t+s), with S(0) the identity. Positivity and unit kernel mass imply contraction in each Lᵖ norm for 1≤p≤∞. Strong continuity at zero holds on Lᵖ for p<∞. It does not hold in the norm of all L∞ functions; a jump can remain a fixed essential-supremum distance from any smooth approximation near the jump.
The kernel also gives quantitative improvement between norms. Its supremum times the initial L¹ norm bounds the positive-time supremum of the solution. Derivatives can be placed on the Gaussian to obtain further estimates, with factors reflecting inverse powers of the diffusion length. These bounds become singular as t approaches zero, which is consistent with smoothing rough data instantly without pretending the rough initial datum was already smooth. They provide a priori control on the constructed solution, not backward stability for reconstructing an earlier state.
Domain and boundary conditions choose a different semigroup
The same Gaussian cannot simply be used unchanged on every domain. Periodic heat flow uses a periodized kernel or Fourier series. Homogeneous Dirichlet flow loses heat through an absorbing boundary, whereas homogeneous Neumann flow preserves mass in a bounded insulated region. Their eigenfunctions, long-time limits, and possible zero modes differ. An expression that solves the differential equation in the interior can still fail the prescribed boundary condition, so the spatial operator and its domain belong in the definition of the evolution.
Smoothing is strongly directional in time. A Fourier component with frequency magnitude ξ is multiplied by exp(−κ|ξ|²t), strongly suppressing fine scales. Reversing the multiplier amplifies measurement errors exponentially at high frequency, making the backward heat problem unstable in ordinary data norms. This does not contradict forward uniqueness. It separates the existence and continuous dependence of the forward evolution from the very different inverse task of recovering information that diffusion has nearly erased.
Work it through
A Gaussian initial profile broadens exactly
In one dimension with κ=1/2, start from u₀(x)=1/√(4π)exp(−x²/4). Find the profile and variance at t=2.
The initial profile is a centered unit-mass Gaussian with variance 2. The heat kernel at elapsed time t has variance 2κt=t.
Convolution adds these variances, giving variance 2+t and profile 1/√[4π(1+t/2)]exp(−x²/[4(1+t/2)]).
At t=2 the profile is 1/√(8π)exp(−x²/8), with variance 4. Its peak is the initial peak divided by √2, while total mass remains one.
The width grows and the peak falls without any loss of whole-space mass. Variance, not standard deviation, adds linearly under this convolution.
The living notebook
Explore the model
Interactive lab · heat
Compare the damping of coarse and fine modes
Reset, set n=1 and κ=0.5, then Step once to t=0.5. Predict amplitude exp(−0.25). For an independent run, Reset, set n=2 and κ=0.5, then Step once: predict exp(−1). Compare the amplitude readouts. This instrument is an exact periodic sine mode, not a Gaussian packet or a nonnegative compact initial bump.
- Current amplitude
- 1
- Spatial L² norm
- 1.772
- Decay rate · κn²
- 3.6 /s
The exact periodic heat-equation mode is u(x,t) = exp(−κn²t) sin(nx), on [0,2π]. Playback advances physical time. The initial sine is shown separately, and the unnormalized L² norm is √π exp(−κn²t). This analytic instrument does not run a finite-difference solver or test discretization stability. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forFine spatial frequencies decay faster, providing the spectral mechanism behind heat smoothing.
Pause & reason
Check your understanding
Your turn
Try it yourself
For u₀ equal to the indicator of [−1,1] on the real line and κ>0, find its conserved mass and two useful upper bounds on ‖u(t)‖∞. Does the solution stay compactly supported?
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Use both the initial supremum and the initial L¹ norm in the two semigroup estimates.
Compare with the solution
The mass is 2. Supremum contraction gives ‖u(t)‖∞≤1, and the L¹-to-L∞ estimate gives ‖u(t)‖∞≤2/√(4πκt)=1/√(πκt). Thus the smaller of these bounds is valid. The nonnegative datum is nonzero and the Gaussian is positive everywhere, so u(x,t)>0 for all real x at every positive time. Its tails are small but its support is no longer compact.
Keep the idea
Take it with you
- A Gaussian semigroup gives forward smoothing and norm estimates with explicit data assumptions.
- Mass behavior and zero modes depend on the domain and boundary conditions.
- Forward uniqueness does not make backward heat reconstruction stable.
Read further
Original sources and deeper explanations to follow at your own pace.
Optional progress, saved in this browser when available. Revisit any lesson at any time.
Lesson not marked complete.