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SSPRK(3,3) is third order and preserves conditional convex bounds

Statement

Assume the smoothness and compact-tube hypotheses of Sufficient explicit Runge–Kutta order conditions through order four. The step in The three-stage third-order SSP Runge–Kutta step has uniform local order at least three.

For an autonomous field, take an initial state y∈C, where C is a convex subset of the state domain, and let Eh(u)=u+hf(u) be the forward-Euler map. If Eh(C)⊆C and all displayed stages are admitted, then every SSPRK(3,3) stage and the output lie in C. If Φ:C→R is convex, meaning Φ(θu+(1−θ)v)≤θΦ(u)+(1−θ)Φ(v) for u,v∈C and 0≤θ≤1, and Φ(Eh(u))≤Φ(u) for every u∈C, then Φ does not increase at any of the three stages or at the output.

For a nonautonomous field, assume the initial state y∈C and that the forward-Euler maps at the actual stage start times t, t+h, and t+h/2 preserve C and satisfy the stated Φ inequality on all admitted stage states. These conditions are premises of the theorem; the method alone does not assert positivity, total-variation diminishing, or any universal invariant-region property.

Facts & Assumptions

Given: The SSPRK(3,3) recurrence, an initial state y∈C, the convex state set, the exact forward-Euler preservation hypotheses, and the smoothness/tube assumptions for the order claim.

[F1]

The three-stage third-order SSP Runge–Kutta step: Y1=Et,h(y), Y2=34y+14Et+h,h(Y1), and y+=13y+23Et+h/2,h(Y2), where Eτ,h(u)=u+hf(τ,u); in the autonomous case Eτ,h=Eh.

[F2]

A convex subset of Rm contains every line segment between two of its points: a convex set contains θu+(1−θ)v whenever it contains u,v and 0≤θ≤1.

[F3]

Explicit Runge–Kutta tableaus, stage maps, and failure status: the tableau has row-sum abscissae and the method output is the weighted sum of its stage slopes.

[F4]

Sufficient explicit Runge–Kutta order conditions through order four: the first four explicit tableau order equations through order three imply uniform local order at least three under the stated tube hypotheses.

[A1]

C is convex, and the stated functional Φ is convex on C.

[A2]

The relevant forward-Euler maps send every admitted point of C into C and do not increase Φ.

Proof

1.1F1F3algebra

Rewriting the recurrences in [F1] as explicit RK stages gives rows of A equal to (0,0,0), (1,0,0), and (1/4,1/4,0), with c=(0,1,1/2) and b=(1/6,1/6,2/3). Thus ∑bi=1, ∑bici=1/2, ∑bici2=1/3, and ∑bi(Ac)i=1/6.

1.2givenA1A2F1F2construct

In the autonomous case Y1=Eh(y)∈C. Since Y1∈C, [A2] gives Eh(Y1)∈C; [A1] and [F2] then give Y2=34y+14Eh(Y1)∈C. Applying [A2] to Y2 gives Eh(Y2)∈C, and [A1] gives y+=13y+23Eh(Y2)∈C.

2.1F4step 1.1

The order equations in step 1.1 are exactly the order-one through order-three hypotheses of [F4], so the stated tube assumptions imply uniform local order at least three.

2.2givenA1A2F1step 1.2algebra

Convexity in [A1] gives Φ(Y1)≤Φ(y), Φ(Y2)≤34Φ(y)+14Φ(Eh(Y1))≤Φ(y), and Φ(y+)≤13Φ(y)+23Φ(Eh(Y2))≤Φ(y). Each last inequality uses [A2] at the relevant stage.

3.1givenA1A2F1F2step 1.2step 2.2∎

For a nonautonomous field the three formulas in [F1] use exactly Et,h at y, Et+h,h at Y1, and Et+h/2,h at Y2. Applying [A2] at those three start times and repeating the convex-combination calculations in steps 1.2–2.2 proves invariance and nonincrease.

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