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An eikonal kink and its monotone HJB grid updates

Statement

On T1, take A={−1,1}, b(a)=a, ℓ=0, so H(p)=∣p∣. Let dT1(x,0)=min⁡k∈Z∣x−k∣ and u(x,t)=max⁡{dT1(x,0)−t,0},0≤t≤1/4. This is the viscosity solution of ut+∣ux∣=0 with initial data u0(x)=dT1(x,0). On a grid with an even number N≥4 of nodes, h=1/N, and 0 a node, one step with 0<τ≤h gives both the positive-interpolation and upwind schemes value τ at x=0, whereas the exact solution there is 0. On the horizon T=1/4, choosing M=N/2 gives τ=T/M=h/2; along this even-grid family the one-step error is exactly τ and the positive-interpolation mesh condition is satisfied.

Facts & Assumptions

[F1]

For A={−1,1}, b(a)=a, ℓ=0, the model Hamiltonian is H(p)=max⁡a∈A(−ap)=∣p∣ (Periodic first-order controlled Hamilton–Jacobi Cauchy problem).

[F2]

The interpolation weights are nonnegative and reproduce linear functions on each grid cell; the upwind rows use the one-sided derivative in the velocity direction (Positive-interpolation semi-Lagrangian update for first-order HJB equations, CFL-monotone upwind update for first-order HJB equations).

[F3]

Under hk,max⁡2/τk→0, the semi-Lagrangian scheme covered by the main convergence theorem converges to its viscosity solution (Convergence of the positive-interpolation and upwind first-order HJB schemes).

Proof

1.1F1givenalgebra

Away from x=0,1/2 and the moving fronts dT1(x,0)=t, the function is either constant zero or has ut=−1 and ∣ux∣=1; it satisfies the PDE classically. Near a moving front on the increasing-distance side, put s=x−t in a local lift; on the other side put s=1−x−t. In either case u=max⁡(s,0). There is no C1 upper test at this convex kink: its directional slopes would have to be at least 1 in the positive s direction and at most 0 in the negative direction. Any C1 lower test has s-slope p∈[0,1]. Use a second coordinate tangent to the moving front: the profile is constant in that direction, so the test's tangential derivative is zero. Thus ϕt=−p and ϕx=p on the increasing-distance side, or ϕx=−p on the decreasing-distance side; in both cases its residual is −p+∣p∣=0. At the antipode x=1/2 for t<1/4, the local profile is 1/2−t−∣x−1/2∣. Its upper tests have spatial slope p∈[−1,1] and time derivative −1, giving residual −1+∣p∣≤0; no lower test exists at that concave cusp. Thus the sub- and supersolution tests hold everywhere for 0<t<1/4. The initial trace is u0.

2.1F2F3givenstep 1.1algebra∎

At x=0, the sampled data have U0=0 and U±1=h because u0(±h)=h. For either semi-Lagrangian action the footpoint is ±τ; linear interpolation between 0 and ±h returns τ when τ≤h, so the action minimum is τ. For the upwind update the forward difference is 1 and the backward difference is −1; multiplying by a=1 and a=−1, respectively, gives drift contribution 1 for both actions. The CFL condition is τ/h≤1, so its minimum update is also τ. Since u(0,τ)=0, both one-step errors equal τ. Taking τ=h/2 gives h2/τ=2h→0, and [F3] supplies convergence for the refined semi-Lagrangian family under its full hypotheses.

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