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Undamped nonlinear RKC2 has fixed-stage order two on an admitted tube

Statement

Fix $s \geq 2$ and $T > 0$ . Let $f (t, y)$ be $C^{2}$ on an open time-state domain, and let the exact IVP solution $y$ be defined on $[0, T]$ . Assume its derivatives through order three and the derivatives of $f$ through order two are bounded on a compact convex tube $K$ contained in the domain. Assume the stage recursion of Second-order nonlinear stabilized Runge–Kutta–Chebyshev step is admitted from every starting state in $K$ for sufficiently small uniform steps and that every stage and output remains in $K$ . Assume all line segments used in Taylor expansions stay in $K$ , and that on each state slice of $K$ , $f$ is Lipschitz in its state argument with one common constant $L_{f}$ . Evaluation failures and domain failures are absent. The global-error claim additionally assumes the exact and numerical trajectories stay in $K$ .

Then the nonlinear RKC2 method has a uniform one-step defect bound $C_{s} h^{3}$ and one-step stability bound $(1 + L_{s} h)$ on that tube, for constants $C_{s}, L_{s}$ independent of the step index and $h$ (but allowed to depend on fixed $s$ and the tube bounds).

For $h = T / N$ , start the method at $y (0)$ and suppose its numerical stages and states remain in the same tube for all $N$ steps. Then the global grid error is at most $\max_{0 \leq n \leq N} ∥ y_{n} - y (t_{n}) ∥_{2} \leq {\begin{cases} C_{s} \frac{\exp (L_{s} T) - 1}{L_{s}} h^{2}, & L_{s} > 0, \\ C_{s} T h^{2}, & L_{s} = 0. \end{cases}$ The statement is fixed-stage and fixed-time. It gives no uniform-in- $s$ accuracy, nonlinear stability, positivity, or stiffness-uniform accuracy.

Facts & Assumptions

Given: A fixed stage count $s \geq 2$ , a $C^{2}$ time-dependent vector field and exact solution on a fixed finite interval, the common compact convex tube and bounded-derivative hypotheses, state-Lipschitz constant $L_{f}$ , admitted method stages, and exact stage evaluations.

[F1]

Second-order nonlinear stabilized Runge–Kutta–Chebyshev step: $b_{1} = 1 / 4$ , $w_{1} = 3 / (s^{2} - 1)$ , $Y_{1} = y + b_{1} w_{1} h f (t, y)$ , and for $j \geq 2$ , $Y_{j} = (1 - μ_{j} - ν_{j}) y + μ_{j} Y_{j - 1} + ν_{j} Y_{j - 2} + m_{j} h f (t + c_{j - 1} h, Y_{j - 1}) + g_{j} h f (t, y)$ with the stated coefficients.

[F2]

RKC stability polynomials and certified negative-real intervals: the scalar linear specialization has $R_{2, s} (z) = 1 + z + z^{2} / 2 + O (z^{3})$ .

[F3]

One-step ODE methods, local defect, uniform order, and step stability: local defect is the exact-start method output minus exact next state; order two requires a uniform $O (h^{3})$ defect, and one-step stability has the form $1 + L h$ .

[F4]

Multivariable Taylor formula with a Lagrange remainder along a line segment and $C^{k}$ maps and multi-index derivative notation in Euclidean space: a $C^{2}$ vector field has a first-order Taylor expansion on each admitted segment with a uniformly bounded quadratic remainder; apply the scalar theorem to each component.

[F5]

Chain sum product and composition rules for Banach derivatives and Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ : differentiating the ODE along its exact solution gives $y^{''} (t) = \partial_{t} f (t, y (t)) + D_{y} f (t, y (t)) f (t, y (t))$ .

[F6]

Global error from uniform one-step defect and stability: on a common compact tube, initial error $C_{0} h^{p}$ , local defect $C h^{p + 1}$ , and one-step bound $1 + L h$ imply the stated fixed-time global estimate.

[F7]

The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ and 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 state norm is Euclidean and obeys the triangle inequality.

[F8]

The standard list $e : n \to F^{n}$ with $e_{i} (i) = 1_{F}$ and $e_{i} (j) = 0_{F}$ for $j \neq i$ is an ordered basis of $F^{n}$ ; hence $\dim_{F} F^{n} = n$ , and $F^{0}$ is the zero space with basis $\emptyset$ and dimension $0$ and Every finite-dimensional normed space is Banach: the standard coordinate lists are finite bases of $R$ and $R^{d}$ , so these Euclidean spaces are real Banach spaces. This supplies the Banach-space hypotheses when [F5] is applied to the ODE solution and vector field.

[A1]

All exact states, numerical states, stage inputs, and Taylor segments used below lie in the stated compact convex tube, where the derivative and state-Lipschitz bounds are uniform.

Proof

1.1F1algebrainduction

Apply the recurrence of [F1] to the time component of the augmented field $(t, y)^{'} = (1, f (t, y))$ . Its stage abscissae satisfy $c_{0} = 0$ , $c_{1} = b_{1} w_{1}$ , and for $j \geq 2$ , $c_{j} = μ_{j} c_{j - 1} + ν_{j} c_{j - 2} + 2 b_{j} w_{1},$ because $m_{j} + g_{j} = m_{j} b_{j - 1} = 2 b_{j} w_{1}$ . At $j = 2$ , $μ_{2} = 2$ , $ν_{2} = - 1$ , and $2 b_{2} w_{1} = w_{1} / 2$ , so $c_{2} = 2 c_{1} + w_{1} / 2 = w_{1}$ . At $j = 3$ (when $s \geq 3$ ), $b_{3} = 8 / 27$ , $μ_{3} = 64 / 27$ , $ν_{3} = - 32 / 27$ , and $2 b_{3} w_{1} = 16 w_{1} / 27$ , so $c_{3} = 8 w_{1} / 3$ . For every $j \geq 4$ , the previous two values are $c_{j - 1} = ((j - 1)^{2} - 1) / (s^{2} - 1)$ and $c_{j - 2} = ((j - 2)^{2} - 1) / (s^{2} - 1)$ . Using $w_{1} = 3 / (s^{2} - 1)$ and $3 b_{k} k^{2} = k^{2} - 1$ for $k = j, j - 1, j - 2$ gives $μ_{j} c_{j - 1} + ν_{j} c_{j - 2} + 2 b_{j} w_{1} = \frac{b_{j}}{s^{2} - 1} (6 (j - 1)^{2} - 3 (j - 2)^{2} + 6) = \frac{3 b_{j} j^{2}}{s^{2} - 1} = \frac{j^{2} - 1}{s^{2} - 1} .$ Thus induction gives $c_{j} = b_{j} w_{1} j^{2} = (j^{2} - 1) / (s^{2} - 1)$ for $j \geq 2$ , so $c_{s} = 1$ . The actual nonautonomous stage times are the declared $t + c_{j} h$ .

2.1A1F1F4step 1.1inductionalgebra

For fixed $s$ , stage increments $Z_{j} : = Y_{j} - y$ are uniformly $O (h)$ . The base is $Z_{0} = 0$ , $Z_{1} = b_{1} w_{1} h f (t, y)$ ; for $j \geq 2$ , the coefficient sum $(1 - μ_{j} - ν_{j}) + μ_{j} + ν_{j} = 1$ rewrites [F1] as $Z_{j} = μ_{j} Z_{j - 1} + ν_{j} Z_{j - 2} + m_{j} h f (t + c_{j - 1} h, Y_{j - 1}) + g_{j} h f (t, y)$ . Boundedness of $f$ , the fixed finite number of stages, and induction give a uniform $O (h)$ bound. Refine the same induction: $Z_{1} = c_{1} h f (t, y)$ , and Taylor to first order gives $f (t + c_{j - 1} h, Y_{j - 1}) = f (t, y) + O (h)$ , so the recurrence for its $h$ coefficient is $c_{j} = μ_{j} c_{j - 1} + ν_{j} c_{j - 2} + m_{j} + g_{j}$ , the abscissa recurrence in step 1.1. Thus $Z_{j} = c_{j} h f (t, y) + O (h^{2})$ for every stage.

3.1A1F1F4step 2.1inductionalgebra

Write $q : = \partial_{t} f (t, y) + D_{y} f (t, y) f (t, y)$ . From step 2.1 and [F4], uniformly at every stage, $f (t + c_{j - 1} h, Y_{j - 1}) = f (t, y) + c_{j - 1} h q + O (h^{2}) .$ A second induction in [F1] therefore gives $Y_{j} = y + c_{j} h f (t, y) + d_{j} h^{2} q + O (h^{3}),$ where $d_{0} = d_{1} = 0$ and $d_{j} = μ_{j} d_{j - 1} + ν_{j} d_{j - 2} + m_{j} c_{j - 1}$ . The remainders are uniform because $s$ is fixed and the derivatives are bounded on $K$ .

3.2A1F1F3F7step 2.1algebrainduction

For stability, fix two admitted initial states $u, v$ in the same convex state slice and let $D_{0} = ∥ u - v ∥_{2}$ . The finite recurrence [F1], the triangle inequality, and the state-Lipschitz assumption imply by induction a finite constant $M_{j}$ with $∥ Y_{j} (u) - Y_{j} (v) ∥_{2} \leq M_{j} D_{0}$ for every $j \leq s$ and $h \leq h_{0}$ : take $M_{0} = 1$ , $M_{1} = 1 + b_{1} w_{1} h_{0} L_{f}$ , and for $j \geq 2$ take $M_{j} = ∣ 1 - μ_{j} - ν_{j} ∣ + ∣ μ_{j} ∣ M_{j - 1} + ∣ ν_{j} ∣ M_{j - 2} + ∣ m_{j} ∣ h_{0} L_{f} M_{j - 1} + ∣ g_{j} ∣ h_{0} L_{f} .$ Writing $Z_{j} (y) = Y_{j} (y) - y$ , the same recurrence after the constant base-state coefficients cancel gives $∥ Z_{j} (u) - Z_{j} (v) ∥_{2} \leq K_{j} h D_{0}$ : use $K_{0} = 0$ , $K_{1} = b_{1} w_{1} L_{f}$ , and $K_{j} = ∣ μ_{j} ∣ K_{j - 1} + ∣ ν_{j} ∣ K_{j - 2} + ∣ m_{j} ∣ L_{f} M_{j - 1} + ∣ g_{j} ∣ L_{f} .$ All constants are finite because $s$ is fixed. Hence $∥ Y_{s} (u) - Y_{s} (v) ∥_{2} \leq ∥ u - v ∥_{2} + ∥ Z_{s} (u) - Z_{s} (v) ∥_{2} \leq (1 + K_{s} h) ∥ u - v ∥_{2}$ . This proves one-step stability with $L_{s} = K_{s}$ .

4.1A1F1F2F4F5F8step 3.1algebra

For the scalar linear equation $y^{'} = y$ at starting value $1$ , the coefficient of $h^{2}$ in the same recurrence is exactly $d_{s}$ . By [F2], the scalar stability polynomial is $1 + z + z^{2} / 2 + O (z^{3})$ , so $d_{s} = 1 / 2$ . Since [F1] and the recurrence defining $d_{j}$ are independent of the choice of $f$ , step 3.1 yields $Y_{s} = y + h f (t, y) + \frac{h^{2}}{2} (\partial_{t} f (t, y) + D_{y} f (t, y) f (t, y)) + O (h^{3}) .$ By [F5], whose Banach-space hypotheses hold by [F8], the exact solution has the same expansion through degree two, with a uniform $O (h^{3})$ remainder on the tube. The method-minus-exact defect is therefore bounded by $C_{s} h^{3}$ .

5.1A1F3F6step 4.1step 3.2∎

Apply [F6] with $p = 2$ , $C_{0} = 0$ , $C = C_{s}$ , and $L = L_{s}$ , using the local-defect and step-stability bounds from steps 4.1 and 3.2. It gives exactly the two global-error cases in the Statement.

Remarks

The imported global theorem is conditional: tube admissibility, local defect, and one-step stability are proved here for fixed $s$ . Its use does not establish a uniform error constant as the stage count changes.

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Sources

Ibrahim Almuslimani and Gilles Vilmart, Explicit Stabilized Integrators for Stiff Optimal Control Problems, §2.2.2 (standard reference, not scraped)
B.P. Sommeijer, L.F. Shampine, and J.G. Verwer, RKC: An explicit solver for parabolic PDEs, §2 (standard reference, not scraped)