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Parabolic finite-difference methods — Examples

1 · Prerequisites

2 · Summary

The examples isolate the forward-Euler endpoint and instability above its CFL bound, contrast backward-Euler damping with Crank–Nicolson's weak damping of the highest mode, and compare the continuous and discrete maximum bounds on a smooth heat solution. Fourier conclusions are stated on periodic grids; the continuous and discrete boundary bounds use their own hypotheses.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

Forward Euler at and above the parabolic CFL threshold

Example

Take the one-dimensional periodic heat equation with κ>0 on an even N-point grid, and let μ=κΔt/h2. The alternating initial data Uj0=(−1)j keep their magnitude when μ=1/2 and grow geometrically when μ>1/2 under forward Euler.

Verification

Given: An even integer N≥2, periodic spacing h=L/N, Δt>0, and the forward-Euler scheme with diffusivity κ>0.

[F1] Fully discrete θ heat finite-difference scheme: the periodic update is Un+1−Un+ΔtAhperUn=0 when θ=0 and the source is zero.

[F2] Periodic heat-stencil eigenvalues and θ amplification factor: mode ℓ=N/2 has eigenvalue 4κ/h2 and multiplier 1−Δtλℓ.

[F3] Forward Euler heat CFL stability and sharp Nyquist instability: on even periodic grids μ≤1/2 gives nonexpansion, while μ>1/2 makes the Nyquist multiplier less than −1.

1.1F1F2given

The vector ϕj=(−1)j is the mode ℓ=N/2, so [F2] gives its multiplier g=1−4μ.

2.1F2F3step 1.1algebra∎

At μ=1/2, g=−1, and hence Ujn=(−1)nϕj has unchanged norm; [F3] shows all other Fourier modes are also nonexpansive. If μ>1/2, then g<−1, so ∥Un∥h=∣g∣n∥ϕ∥h grows geometrically.

Remarks

The endpoint is stable but not damping for the Nyquist mode. This example is periodic and does not assert a finite-boundary result.

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Backward Euler strongly damps high periodic modes

Example

Fix Δt>0 and consider even periodic one-dimensional grids with h=L/N→0. Backward Euler multiplies the Nyquist mode by 1/(1+4κΔt/h2), which tends to zero, while its weighted grid L2 norm never increases.

Verification

Given: κ>0, a fixed Δt>0, and an even periodic grid with spacing h=L/N.

[F1] Fully discrete θ heat finite-difference scheme: backward Euler is the θ update with θ=1 and zero source.

[F2] Periodic heat-stencil eigenvalues and θ amplification factor: the Nyquist eigenvalue is λ=4κ/h2, and its multiplier is gθ=(1−(1−θ)Δtλ)/(1+θΔtλ).

1.1F1F2givenalgebra

Setting θ=1 and λ=4κ/h2 in [F2] gives g=1/(1+4κΔt/h2).

2.1step 1.1algebra

As h→0 with κ,Δt>0 fixed, the denominator tends to infinity, so g→0. Thus this high-frequency mode is strongly damped.

3.1F2step 1.1algebra∎

For every mode λ≥0, the backward-Euler factor is 1/(1+Δtλ)∈(0,1], so the periodic Fourier basis gives nonincrease of the full weighted grid L2 norm.

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Crank–Nicolson is stable but does not strongly damp the highest mode

Example

Fix Δt>0 and consider even periodic one-dimensional grids with h=L/N→0. Crank–Nicolson is weighted-L2 stable, but its Nyquist multiplier tends to −1 and its magnitude tends to one, so the highest mode is not strongly damped.

Verification

Given: κ>0, a fixed Δt>0, and an even periodic grid with spacing h=L/N.

[F1] Fully discrete θ heat finite-difference scheme: Crank–Nicolson is the θ update with θ=1/2 and zero source.

[F2] Periodic heat-stencil eigenvalues and θ amplification factor: the Nyquist eigenvalue is λ=4κ/h2, and its multiplier is gθ=(1−(1−θ)Δtλ)/(1+θΔtλ).

1.1F1F2givenalgebra

Substituting θ=1/2 and λ=4κ/h2 into [F2] gives g=(1−2κΔt/h2)/(1+2κΔt/h2).

2.1step 1.1algebra

As h→0 with κ,Δt>0 fixed, the numerator and denominator have leading terms −2κΔt/h2 and 2κΔt/h2, so g→−1 and ∣g∣→1.

3.1F2step 1.1algebra∎

For any z=Δtλ≥0, ∣1−z/2∣≤1+z/2, so the factor in [F2] has magnitude at most one. Thus the mode is stable at each mesh while its high-frequency damping vanishes in the limit.

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Continuous and discrete heat maximum bounds on sampled data

Example

On Ω=(0,1)2, the function

u(x,y,t)=sin⁡(πx)sin⁡(πy)e−2π2t

solves the unit-diffusivity heat equation with zero Dirichlet data and initial values in [0,1]. It remains in [0,1] continuously. On an equal-spacing grid, forward Euler also remains in [0,1] provided 2Δt/h2≤1/2.

Verification

Given: T>0, N≥2, h=1/N, M≥1, Δt=T/M with 2Δt/h2≤1/2, and the exact sampled initial and zero Dirichlet boundary data for the forward-Euler heat scheme with κx=κy=1.

[F1] Fully discrete θ heat finite-difference scheme: the centered Dirichlet update imposes the specified boundary rows at every time level.

[F2] A boundary-aware discrete maximum principle for the θ scheme: for the zero-source Dirichlet θ scheme, if Δt(1−θ)(κx/hx2+κy/hy2)≤1/2, all grid values lie between the initial and boundary extrema.

1.1givenalgebra

Direct differentiation gives ut=−2π2u and Δu=−2π2u, so ut−Δu=0. Also u=0 on the four sides and u(x,y,0)=sin⁡(πx)sin⁡(πy)∈[0,1].

2.1givenstep 1.1algebra

On [0,1], 0≤sin⁡(πx),sin⁡(πy)≤1, and for t≥0 the factor e−2π2t lies in (0,1]. The explicit formula therefore gives 0≤u≤1 throughout the closed cylinder.

3.1F1F2givenalgebra∎

The sampled initial values lie in [0,1] and the boundary rows are zero. With θ=0 and κx=κy=1, the condition in [F2] is exactly 2Δt/h2≤1/2. Therefore [F2] gives 0≤Uijn≤1 at every grid node and time level.

Remarks

The continuous supplier is used only for unit diffusivity on a bounded rectangle with smooth exact data. The discrete conclusion follows from the separately proved finite-grid boundary condition.