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One-step Lipschitz bound for classical RK4 on an admitted tube

Statement

Let $f : [0, T] \times U \to R^{m}$ be uniformly $L$ -Lipschitz in its state variable on a set containing every stage generated by the classical RK4 step of The explicit Euler, midpoint, and classical fourth-order Runge–Kutta tableaus from any two admitted starting states $u, v$ at the same base time $t$ . Assume all these stage evaluations are defined, and let $Ψ_{h} (t, \cdot)$ be the resulting one-step map for $0 < h \leq h_{0}$ . Then $∣ Ψ_{h} (t, u) - Ψ_{h} (t, v) ∣ \leq (1 + L_{*} h) ∣ u - v ∣, L_{*} = L e^{L h_{0}} .$

Facts & Assumptions

Given: The explicit RK4 stage equations, a time-dependent vector field uniformly Lipschitz in state with constant $L$ , admitted stages at common stage times, and $0 < h \leq h_{0}$ .

[F1]

An explicit tableau has stages $Y_{i} = u + h \sum_{j < i} a_{i j} K_{j}$ and output $u + h \sum_{i} b_{i} K_{i}$ (Explicit Runge–Kutta tableaus, stage maps, and failure status).

[F2]

The classical RK4 coefficients are $(1 / 2, 1 / 2, 1)$ in the stage rows and $(1 / 6, 1 / 3, 1 / 3, 1 / 6)$ in the output (The explicit Euler, midpoint, and classical fourth-order Runge–Kutta tableaus).

[F3]

The Euclidean norm satisfies the triangle inequality and is homogeneous (Cauchy-Schwarz $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥_{2} ∥ y ∥_{2}$ with its equality case, the triangle inequality for $∥ \cdot ∥_{2}$ , the parallelogram law and polarisation, The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ).

Proof

1.1F1F2F3givenalgebra

Put $δ = ∣ u - v ∣$ and let $d_{i} = ∣ K_{i} (t, u) - K_{i} (t, v) ∣$ . The stage times are identical for the two starting states. The stage equations [F1]-[F2] and the uniform state-Lipschitz assumption give, in order, $d_{1} \leq L δ, d_{2} \leq L (1 + h L / 2) δ, d_{3} \leq L (1 + h L / 2 + h^{2} L^{2} / 4) δ,$ and $d_{4} \leq L (1 + h L + h^{2} L^{2} / 2 + h^{3} L^{3} / 4) δ .$ Each bound follows by substituting the preceding stage bound into $∣ f (t_{i}, Y_{i} (u)) - f (t_{i}, Y_{i} (v)) ∣ \leq L ∣ Y_{i} (u) - Y_{i} (v) ∣$ .

2.1F1F2F3step 1.1algebra∎

The output formula in [F1]-[F2] gives $∣ Ψ_{h} (t, u) - Ψ_{h} (t, v) ∣ \leq (1 + h L + \frac{1}{2} (h L)^{2} + \frac{1}{6} (h L)^{3} + \frac{1}{24} (h L)^{4}) δ \leq e^{h L} δ .$ For $h \leq h_{0}$ , $e^{h L} - 1 = \int_{0}^{h L} e^{s} d s \leq h L e^{L h_{0}}$ , so $e^{h L} \leq 1 + L_{*} h$ . This proves the claim.

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Sources

Tobias Embree, Lecture Notes on Numerical Analysis, Sections 5.2.3-5.3 (standard reference, not scraped)
MIT OpenCourseWare 18.330, Introduction to Numerical Analysis, Chapter 5 (standard reference, not scraped)