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✓ 12 results · all verified · 7 also independently AI-judged

Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 5 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Imex and operator splitting methods

1 · Prerequisites

2 · Summary

This page studies finite-dimensional autonomous additive systems $y^{'} = a (y) + b (y)$ . Under the stated common-tube and branch hypotheses for sufficiently small steps, IMEX Euler evaluates $a$ explicitly and solves the $b$ equation on a selected local branch. Its local defect is second order in the step, its exact-root map has a one-step Lipschitz bound, and its fixed-time global error is first order. An unscaled per-step update residual of order $h^{2}$ preserves that first-order global bound.

The fixed ARS(2,2,2) pair is defined by its explicit and diagonally implicit tableaus and three stages. The local proof expands both component fields and verifies the coupled colored second-order terms directly. Under uniform selected-branch and smooth-tube hypotheses for sufficiently small steps, its two implicit stages use separate branch conditions; a stage-two residual of order $h^{2}$ and final-stage residual of order $h^{3}$ give an order-two global error. The scalar stability proposition gives the exact two-variable region and makes no matrix or PDE stability claim.

For exact finite-dimensional split flows, Lie composition has global order one and symmetric Strang composition has global order two for sufficiently small steps under common-tube hypotheses. Approximate subflow errors have their own per-step budgets and require a stability bound for the composed approximate map. The companion page supplies noncommuting matrix, finite-matrix Schrödinger, two-cell reaction-diffusion, stiff scalar, and boundary-forcing examples.

The boundary example is a fully specified diagonal sequence-space heat model. It proves the generator domain, identifies constant Dirichlet-incompatible forcing outside that domain, and computes a two-sided $h^{5 / 4}$ Strang error at fixed time. This is not a convergence theorem for general unbounded operators or PDEs.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Additive autonomous ODE partitions

Definition

Let $d \geq 1$ , let $U \subseteq R^{d}$ be open, and let $a, b : U \to R^{d}$ . The additive autonomous ODE partition with components $a$ and $b$ is the initial-value problem $y^{'} (t) = a (y (t)) + b (y (t)), y (t_{0}) = y_{0} \in U,$ in the sense of First-order systems, initial value problems, and solutions on intervals. The term autonomous means that neither component has a separate time argument. A numerical method may evaluate $a$ explicitly and solve an equation involving $b$ implicitly; this naming convention does not assert that either exact subflow exists or that an implicit stage has a solution.

DefinitionDefinition: Literature-sourcedProof: Not applicableOpen item page →

Diagonal Dirichlet sine-mode heat model with constant forcing

Definition

Let $I = {k \in N : k \geq 1}$ and $H = ℓ^{2} (I, R)$ , with square-summability as in Square-summable families on an arbitrary index set and the space $ℓ^{2} (I)$ . All modal coordinates below are indexed by $k \in I$ . Define the diagonal operator $A$ by $D (A) = {u \in H : \sum_{k \geq 1} k^{4} ∣ u_{k} ∣^{2} < \infty}, (A u)_{k} = - k^{2} u_{k} .$ Let $g \in H$ be the fixed sequence $g_{k} = 1 / k$ for odd $k$ and $g_{k} = 0$ for even $k$ ; membership follows from The p-series for a real exponent p converges exactly when p is greater than one with exponent $2$ . The split evolution is $u^{'} = A u + g$ . For $t \geq 0$ , define the exact diagonal subflow $E_{t} u = (\exp (- k^{2} t) u_{k})_{k \geq 1}$ and the exact translation subflow $B_{t} u = u + t g$ . By The exponential is positive and satisfies $\exp (- x) = 1 / \exp (x)$ and The exponential function is strictly increasing, $0 < \exp (- k^{2} t) \leq 1$ , so each $E_{t}$ maps $H$ to itself; each $B_{t}$ does so because $g \in H$ . The associated Strang modal step is $S_{h} = E_{h / 2} \circ B_{h} \circ E_{h / 2}, h > 0,$ where the rightmost map is applied first. Each coordinate of $E_{t}$ solves $u_{k}^{'} = - k^{2} u_{k}$ , and each coordinate of $B_{t}$ solves $u_{k}^{'} = g_{k}$ . This is a precisely defined sequence-space model; it records one Dirichlet sine-mode boundary mismatch and makes no general convergence assertion for unbounded PDE operators.

AlgorithmAlgorithm: Literature-sourcedProof: Not applicableOpen item page →

One-step IMEX Euler with an implicit branch and residual status

Definition

For the additive ODE partition Additive autonomous ODE partitions, take a current state $y \in U$ , a step $h > 0$ , a closed ball $\overline{B} (c, R) \subset U$ centered at $c : = y + h a (y)$ , bounds $M_{b}, L_{b} \geq 0$ for $∥ b ∥_{2}$ and its Lipschitz constant on that ball, a tolerance $τ \geq 0$ , and a solve routine that returns either a candidate $z \in \overline{B} (c, R)$ or a failure status. The procedure is:

If the solve routine is unavailable, return solver-unavailable.
If $h M_{b} > R$ or $h L_{b} \geq 1$ , return branch-not-certified without claiming a root.
Call the solve routine on $z = c + h b (z)$ .
If it returns failure, return that failure status. If it returns $z \notin \overline{B} (c, R)$ , return out-of-branch-ball.
Compute the update residual $r : = z - c - h b (z)$ in the state units of Implicit residuals and their update and derivative units. If $∥ r ∥_{2} > τ$ , return residual-too-large with $z$ and $r$ ; otherwise return success with $z$ and $r$ .

When the routine returns an exact root, $r = 0$ and $z$ is the IMEX Euler step $Y = y + h a (y) + h b (Y)$ . A success status records only the requested residual check; the forward error conclusion is supplied separately by the residual theorem. This procedure uses one explicit evaluation of $a$ and one implicit solve.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Lie and Strang compositions of exact split flows

Definition

Let $a, b : U \to R^{d}$ be $C^{3}$ fields on an open set, and restrict attention to points and time intervals for which the unique local solutions of $x^{'} = a (x)$ and $x^{'} = b (x)$ remain in $U$ . Denote their exact solution maps by $Φ_{a}^{t}$ and $Φ_{b}^{t}$ . The Lie split map applies $a$ and then $b$ : $L_{h} : = Φ_{b}^{h} \circ Φ_{a}^{h} .$ The Strang split map applies half of $a$ , then $b$ , then half of $a$ : $S_{h} : = Φ_{a}^{h / 2} \circ Φ_{b}^{h} \circ Φ_{a}^{h / 2} .$ Define the vector-field Lie bracket convention $[a, b] (y) : = D b (y) a (y) - D a (y) b (y) .$ This definition is finite-dimensional. It asserts neither that an unbounded operator has a common invariant domain nor that its commutators are bounded.

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passOpen item page →

The diagonal heat semigroup has the stated generator domain

Statement

Let $H$ , $A$ , and $E_{t}$ be as in Diagonal Dirichlet sine-mode heat model with constant forcing. The maps $E_{t}$ for $t \geq 0$ satisfy $E_{0} = I$ and $E_{t + s} = E_{t} E_{s}$ , are contractions in $ℓ^{2}$ , and are strongly continuous at $t = 0$ . Define the generator domain to consist of those $u \in H$ for which the right limit $\lim_{t ↓ 0} \frac{E_{t} u - u}{t}$ exists in $H$ . This domain is exactly $D (A) = {u : \sum_{k \geq 1} k^{4} ∣ u_{k} ∣^{2} < \infty}$ , and the limit equals $A u = (- k^{2} u_{k})_{k \geq 1}$ .

Facts & Assumptions

Given: The sequence space, diagonal operator, and maps in the Statement.

[F1]

Diagonal Dirichlet sine-mode heat model with constant forcing and Square-summable families on an arbitrary index set and the space $ℓ^{2} (I)$ : $H$ consists of real families $u = (u_{k})_{k \geq 1}$ with $\sum_{k \geq 1} ∣ u_{k} ∣^{2} < \infty$ , and $(E_{t} u)_{k} = \exp (- k^{2} t) u_{k}$ for every $k \geq 1$ .

[F2]

The exponential addition formula $\exp (x + y) = \exp (x) \exp (y)$ : $\exp (x + y) = \exp (x) \exp (y)$ and $\exp (0) = 1$ .

[F3]

The exponential is positive and satisfies $\exp (- x) = 1 / \exp (x)$ and The exponential function is strictly increasing: $\exp (x) > 0$ , and $x \leq y$ implies $\exp (x) \leq \exp (y)$ .

[F4]

The exponential function is smooth and $(\exp)^{'} = \exp$ , The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ , and The mean value theorem, as the case $g (x) = x$ of Cauchy's: for $f$ continuous on $[a, b]$ with $a < b$ and differentiable on $(a, b)$ there is $c \in (a, b)$ with $f (b) - f (a) = f^{'} (c) (b - a)$ : the scalar exponential is differentiable with derivative $\exp (x)$ , its compositions obey the chain rule, and a differentiable scalar function on an interval obeys the mean-value theorem.

Proof

1.1F1F2F3algebra

Coordinatewise, [F2] gives $\exp (- k^{2} (t + s)) = \exp (- k^{2} t) \exp (- k^{2} s)$ and $E_{0} u = u$ , so $E_{t + s} = E_{t} E_{s}$ . If $t \geq 0$ , [F3] gives $0 < \exp (- k^{2} t) \leq 1$ ; hence $\sum_{k} ∣ (E_{t} u)_{k} ∣^{2} \leq \sum_{k} ∣ u_{k} ∣^{2}$ , proving that $E_{t}$ maps $H$ to itself and is a contraction.

1.2F1F3F4algebra

At $x = 0$ the bound $0 \leq 1 - \exp (- x) \leq x$ is equality; for $x > 0$ , [F4] applied to $s \mapsto \exp (- s)$ on $[0, x]$ gives $1 - \exp (- x) = x \exp (- ξ)$ for some $ξ \in (0, x)$ , and [F3] bounds $\exp (- ξ)$ by $1$ . Thus the bound holds for all $x \geq 0$ . For $u \in D (A)$ , the $k$ th coordinate of $(E_{t} u - u) / t$ tends to $- k^{2} u_{k}$ by the chain rule and exponential derivative in [F4]. Its difference from that limit has absolute value at most $2 k^{2} ∣ u_{k} ∣$ , by the same bound. The squared tail of these differences is therefore at most $4 \sum_{k > K} k^{4} ∣ u_{k} ∣^{2}$ , which tends to zero; on every finite head the coordinate limits give norm convergence. Hence the right limit exists in $H$ and equals $A u$ .

2.1F1F3F4step 1.1algebrachoose

Fix $u \in H$ and $ε > 0$ . By convergence of its square sum, choose $K$ so that $\sum_{k > K} ∣ u_{k} ∣^{2} < ε^{2} / 4$ . The squared norm of the tail of $E_{t} u - u$ is at most this tail because $∣ \exp (- k^{2} t) - 1 ∣ \leq 1$ by [F3]; for the finitely many indices $k \leq K$ , continuity of the exponential at zero from [F4] makes their squared sum less than $ε^{2} / 4$ for all sufficiently small $t > 0$ . Thus $∥ E_{t} u - u ∥_{2} < ε$ , proving strong continuity at zero.

3.1F1F4step 1.2algebra∎

Conversely, suppose $(E_{t} u - u) / t$ converges in $H$ to $v$ . For each $k$ , coordinate evaluation is continuous since $∣ w_{k} ∣ \leq ∥ w ∥_{2}$ ; therefore its limit is $v_{k} = - k^{2} u_{k}$ , using [F4]. Since $v \in H$ , $\sum_{k} k^{4} ∣ u_{k} ∣^{2} = \sum_{k} ∣ v_{k} ∣^{2} < \infty$ , so $u \in D (A)$ . Together with step 1.2 this identifies the generator domain and its value.

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

A local IMEX Euler branch from contraction

Statement

Let $d \geq 1$ , $c \in R^{d}$ , $R > 0$ , $h > 0$ , and let $b$ be defined on the closed Euclidean ball $\overline{B} (c, R)$ . Suppose $∥ b (z) ∥_{2} \leq M$ and $∥ b (z) - b (w) ∥_{2} \leq L ∥ z - w ∥_{2}$ there, with $h M \leq R$ and $h L < 1$ . Then the IMEX Euler equation $Y = c + h b (Y)$ has exactly one solution in $\overline{B} (c, R)$ . Every $\tilde{Y}$ in that ball satisfies $∥ \tilde{Y} - Y ∥_{2} \leq \frac{∥ \tilde{Y} - c - h b (\tilde{Y}) ∥_{2}}{1 - h L} .$ In the algorithm One-step IMEX Euler with an implicit branch and residual status, $c = y + h a (y)$ ; this is a local branch result and gives no root outside the selected ball.

Facts & Assumptions

Given: $d \geq 1$ , $c, R, h, b, M, L$ and the bounds in the Statement.

[F1]

One-step IMEX Euler with an implicit branch and residual status: the implicit update is $Y = c + h b (Y)$ and its unscaled update residual is $\tilde{Y} - c - h b (\tilde{Y})$ .

[F2]

$R$ and $R^{n}$ for $n \geq 1$ with the Euclidean metric are complete, componentwise from the Cauchy criterion in $R$ : $R^{d}$ with the Euclidean metric is complete.

[F3]

Complete metric space: every Cauchy sequence converges in the space and Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric: a metric space is complete when every Cauchy sequence converges to a point in that space.

[F4]

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 norm satisfies $∥ u + v ∥_{2} \leq ∥ u ∥_{2} + ∥ v ∥_{2}$ .

[F5]

Lipschitz map, $α$ -Hölder map for rational $0 < α \leq 1$ , and contraction: a map is a contraction with constant $q$ when its output distance is at most $q$ times its input distance, where $0 \leq q < 1$ .

[F6]

A contraction of a nonempty complete metric space into itself has exactly one fixed point, the limit of the iterates from any starting point: a contraction from a nonempty complete metric space into itself has a unique fixed point.

Proof

1.1givenF2F3F4construct

The ball is nonempty because it contains $c$ . If $(z_{n})$ is Cauchy in the ball, it is Cauchy in $R^{d}$ , so [F2] gives a limit $z \in R^{d}$ . By [F4], $∥ z - c ∥_{2} \leq ∥ z - z_{n} ∥_{2} + ∥ z_{n} - c ∥_{2} \leq ∥ z - z_{n} ∥_{2} + R$ ; letting $n \to \infty$ gives $∥ z - c ∥_{2} \leq R$ , so $z$ remains in the ball. Thus the ball is a nonempty complete metric space by [F3].

1.2givenF5algebra

Define $T (z) : = c + h b (z)$ . For every $z$ in the ball, $∥ T (z) - c ∥_{2} = h ∥ b (z) ∥_{2} \leq h M \leq R$ , so $T$ maps the ball into itself; for $z, w$ in the ball, $∥ T (z) - T (w) ∥_{2} \leq h L ∥ z - w ∥_{2}$ , and $h L < 1$ makes $T$ a contraction as in [F5].

2.1step 1.1step 1.2F6

Applying [F6] to the complete ball and the contraction from step 1.2 gives a unique fixed point $Y$ , which is exactly the unique solution of $Y = c + h b (Y)$ in the ball.

3.1F1F4step 1.2step 2.1algebra∎

For any $\tilde{Y}$ in the ball, [F4] and step 2.1 give $∥ \tilde{Y} - Y ∥_{2} \leq ∥ \tilde{Y} - T (\tilde{Y}) ∥_{2} + ∥ T (\tilde{Y}) - T (Y) ∥_{2} \leq ∥ \tilde{Y} - c - h b (\tilde{Y}) ∥_{2} + h L ∥ \tilde{Y} - Y ∥_{2}$ . Since $1 - h L > 0$ , rearranging proves the residual bound.

DefinitionDefinition: Literature-sourcedProof: Not applicableOpen item page →

The ARS(2,2,2) additive tableau and its three stages

Definition

For the autonomous split problem $y^{'} = a (y) + b (y)$ of Additive autonomous ODE partitions, set $γ : = 1 - \frac{1}{\sqrt{2}}, δ : = - \frac{1}{\sqrt{2}} .$ The explicit tableau and the lower-triangular implicit tableau, using the separate stage conventions of Explicit Runge–Kutta tableaus, stage maps, and failure status and Implicit Runge–Kutta stage equations and update, are $\tilde{A} = (\begin{array}{ccc} 0 & 0 & 0 \\ γ & 0 & 0 \\ δ & 1 - δ & 0 \end{array}), \tilde{b} = (δ, 1 - δ, 0), \tilde{c} = (0, γ, 1),$ $A = (\begin{array}{ccc} 0 & 0 & 0 \\ 0 & γ & 0 \\ 0 & 1 - γ & γ \end{array}), b = (0, 1 - γ, γ), c = (0, γ, 1) .$ For a step $h > 0$ from $y$ , the method is defined by $Y_{1} = y, Y_{2} = y + γ h a (Y_{1}) + γ h b (Y_{2}),$ $Y_{3} = y + h [δ a (Y_{1}) + (1 - δ) a (Y_{2})] + h [(1 - γ) b (Y_{2}) + γ b (Y_{3})], y^{+} = Y_{3} .$ Thus there are two implicit solves per step, one for each nonzero diagonal implicit stage. The unscaled stage-equation residual uses the update-residual convention of Implicit residuals and their update and derivative units. The tableaus and stages alone assert neither branch existence nor stability.

LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

Lie and Strang splitting errors have commutator expansions

Statement

Let $a, b \in C^{3} (U, R^{d})$ on an open set and let $K_{0} \subset K \subset U$ , where $K_{0}$ is compact and $K$ is compact and convex. Assume the exact $a$ - and $b$ -subflows, their Lie and Strang compositions, and the exact flow $Φ_{a + b}^{h}$ from $K_{0}$ are defined and remain in $K$ for all sufficiently small $h > 0$ , and the fields and derivatives through order three are bounded on $K$ . Uniformly for $y \in K_{0}$ , $L_{h} (y) - Φ_{a + b}^{h} (y) = \frac{h^{2}}{2} [a, b] (y) + O (h^{3}),$ and the Strang expansion, pointwise in $y$ , is $S_{h} (y) - Φ_{a + b}^{h} (y) = - \frac{h^{3}}{24} [a, [a, b]] (y) - \frac{h^{3}}{12} [b, [a, b]] (y) + o (h^{3}) .$ The Strang defect is also uniformly $O (h^{3})$ on $K_{0}$ . The maps and bracket use Lie and Strang compositions of exact split flows; the claim is finite-dimensional.

Facts & Assumptions

Given: The fields, exact subflows, compact tube, and bounded derivatives in the Statement.

[F1]

Lie and Strang compositions of exact split flows: $L_{h} = Φ_{b}^{h} \circ Φ_{a}^{h}$ , $S_{h} = Φ_{a}^{h / 2} \circ Φ_{b}^{h} \circ Φ_{a}^{h / 2}$ , and $[a, b] = D b a - D a b$ .

[F2]

The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ : differentiating a composition gives $D (g \circ f) = D g (f) \circ D f$ .

[F3]

The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ : an integrable derivative satisfies $\int_{α}^{β} g^{'} = g (β) - g (α)$ .

[F4]

Continuous mixed partials of order $k$ are invariant under permutations: for a $C^{2}$ scalar function, the two Hessian arguments are symmetric.

[A1]

Each flow is defined on the stated tube, and its time derivatives through order three are bounded there; these follow from the $C^{3}$ fields and the compact-tube hypotheses.

Proof

1.1A1F2F3algebra

For a $C^{3}$ field $c$ and a smooth scalar coordinate function $q (y) = y_{j}$ , define $L_{c} q : = D q c$ . By [F2], differentiating $q (Φ_{c}^{t} (y))$ successively in $t$ gives $L_{c} q$ , $L_{c}^{2} q$ , and $L_{c}^{3} q$ , evaluated along the flow. Three uses of [F3] give the expansion through degree two with integral remainder $\frac{1}{2} \int_{0}^{h} (h - s)^{2} (L_{c}^{3} q) (Φ_{c}^{s} (y)) d s$ ; [A1] bounds this remainder uniformly by $C h^{3}$ and continuity at $s = 0$ gives cubic coefficient $\frac{1}{6} L_{c}^{3} q (y)$ with pointwise remainder $o (h^{3})$ . The same reasoning applies successively to the finite flow compositions.

2.1F1step 1.1algebra

For a map $M$ , define its pullback by $M^{*} q : = q \circ M$ , and for a vector field $c$ put $L_{c} q : = D q c$ . Set $X = L_{a}$ and $Y = L_{b}$ . The pullback of a composition reverses composition order, so [F1] and step 1.1 give, through degree two, $L_{h}^{*} q = q + h (X + Y) q + h^{2} (\frac{1}{2} X^{2} + X Y + \frac{1}{2} Y^{2}) q + O (h^{3}) .$ The exact total-flow pullback is $q + h (X + Y) q + \frac{1}{2} h^{2} (X + Y)^{2} q + O (h^{3})$ . Subtracting leaves $\frac{1}{2} h^{2} (X Y - Y X) q + O (h^{3})$ .

3.1F1F4step 2.1algebra

For any $C^{2}$ scalar $q$ , the chain rule gives $[L_{a}, L_{b}] q = D^{2} q (a, b) + D q D b a - D^{2} q (b, a) - D q D a b = D q (D b a - D a b),$ where [F4] cancels the Hessian terms. Thus $[X, Y] q = L_{[a, b]} q$ . Applying this to each coordinate function in step 2.1 gives the stated Lie map expansion.

3.2F1step 1.1step 2.1algebra

Pullback reversal and [F1] give $S_{h}^{*} = Φ_{a}^{h / 2 *} Φ_{b}^{h *} Φ_{a}^{h / 2 *}$ . Expanding its three factors by step 1.1 gives the cubic part $\frac{1}{6} X^{3} + \frac{1}{6} Y^{3} + \frac{1}{8} X^{2} Y + \frac{1}{4} X Y X + \frac{1}{8} Y X^{2} + \frac{1}{4} X Y^{2} + \frac{1}{4} Y^{2} X .$ Multiplying the three degree-three Taylor polynomials gives these coefficients: the $X^{3}$ term is $1 / 48 + 1 / 48 + 1 / 16 + 1 / 16 = 1 / 6$ , and the $Y^{3}$ term is $1 / 6$ . The mixed coefficients are $X^{2} Y : (1 / 8) (1) = 1 / 8$ , $X Y X : (1 / 2) (1) (1 / 2) = 1 / 4$ , $Y X^{2} : (1) (1 / 8) = 1 / 8$ , $X Y^{2} : (1 / 2) (1 / 2) = 1 / 4$ , and $Y^{2} X : (1 / 2) (1 / 2) = 1 / 4$ ; $Y X Y$ is absent because neither outer factor contains $Y$ . The exact total-flow cubic part is $\frac{1}{6} (X + Y)^{3}$ , whose eight words each have coefficient $1 / 6$ . Degrees zero through two agree, so subtraction gives $- \frac{1}{24} (X^{2} Y - 2 X Y X + Y X^{2}) - \frac{1}{12} (2 Y X Y - Y^{2} X - X Y^{2}) + o (h^{3}) .$

4.1F1step 1.1step 3.1step 3.2algebra∎

Step 3.1 also applies with the second field equal to $[a, b]$ , which is $C^{2}$ under the $C^{3}$ hypotheses. Hence $[X, [X, Y]] = L_{[a, [a, b]]}$ and $[Y, [X, Y]] = L_{[b, [a, b]]}$ . Applying step 3.2 to each coordinate yields the stated Strang coefficient and pointwise $o (h^{3})$ remainder. The uniform $O (h^{3})$ defect follows already from the uniform degree-two remainder in step 1.1 and the equality of all coefficients through degree two.

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passOpen item page →

Constant Dirichlet forcing is square summable but outside the heat-generator domain

Statement

For the constant forcing $f (x) : = π / 4$ on $(0, π)$ , define its Fourier sine coefficient by ${\hat{f}}_{k} : = \frac{2}{π} \int_{0}^{π} f (x) \sin (k x) d x .$ Then ${\hat{f}}_{k} = g_{k}$ from Diagonal Dirichlet sine-mode heat model with constant forcing. Thus $g \in H$ but $g$ is outside the generator domain of $A$ . The constant forcing has nonzero endpoint value, while every Dirichlet sine mode vanishes at both endpoints.

Facts & Assumptions

Given: The modal space, generator domain, forcing coefficients, and Dirichlet sine modes in the statement.

[F1]

Diagonal Dirichlet sine-mode heat model with constant forcing and The diagonal heat semigroup has the stated generator domain: $H = ℓ^{2} ({k \in N : k \geq 1}, R)$ , and the generator of $E_{t}$ has domain $D (A) = {u : \sum_{k \geq 1} k^{4} ∣ u_{k} ∣^{2} < \infty}$ ; $g_{k} = 1 / k$ for odd $k$ and $0$ for even $k$ .

[F2]

The p-series for a real exponent p converges exactly when p is greater than one: $\sum_{k \geq 1} k^{- p}$ converges exactly when $p > 1$ .

[F3]

The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ : if $G^{'} = q$ and $q$ is integrable, then $\int_{a}^{b} q = G (b) - G (a)$ .

[F4]

The derivatives of sine and cosine are cosine and minus sine: $\frac{d}{d x} \cos (k x) = - k \sin (k x)$ .

[F5]

Quarter-turn values and shifts by pi/2 and pi: $\sin (k π) = 0$ and $\cos (k π) = (- 1)^{k}$ for each integer $k$ .

[F6]

Pi as twice the smallest positive zero of cosine: $π$ is strictly positive.

Proof

1.1givenF1F3F4F5algebra

By [F4], an antiderivative of $\sin (k x)$ is $- \cos (k x) / k$ ; applying [F3] and [F5] gives ${\hat{f}}_{k} = \frac{2}{π} \cdot \frac{π}{4} \cdot \frac{1 - (- 1)^{k}}{k} = \frac{1 - (- 1)^{k}}{2 k} .$ This is $1 / k$ for odd $k$ and $0$ for even $k$ , exactly the sequence in [F1].

2.1F1F2step 1.1algebra

The squared coefficients sum to $\sum_{j \geq 0} (2 j + 1)^{- 2} \leq \sum_{k \geq 1} k^{- 2} < \infty$ by [F2], so $g \in H$ .

2.2F1step 1.1algebra

The generator-domain sum is $\sum_{k \geq 1} k^{4} ∣ g_{k} ∣^{2} = \sum_{j \geq 0} (2 j + 1)^{2}$ , whose partial sums are at least the number of terms and hence unbounded. Therefore $g \notin D (A)$ by [F1].

3.1givenF5F6algebra∎

By [F6], the constant forcing $f = π / 4$ is nonzero. By [F5], every sine mode $\sin (k x)$ vanishes at both $x = 0$ and $x = π$ . Hence the forcing is incompatible with the homogeneous Dirichlet endpoint values. This modal observation supplies no general PDE error theorem.

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IMEX Euler has a first-order local defect and a one-step stability bound

Statement

Let $d \geq 1$ and let $F = a + b$ on an open set $U \subseteq R^{d}$ as in Additive autonomous ODE partitions. On a common compact convex tube $T \subset U$ , suppose $a, b$ are bounded and Lipschitz with constants $M_{a}, M_{b}, L_{a}, L_{b}$ . For every $u \in T$ and sufficiently small $h > 0$ , assume there is a closed ball $\overline{B} (c_{u}, R_{u}) \subset T$ with $R_{u} > 0$ , centered at $c_{u} : = u + h a (u)$ , and with $h M_{b} \leq R_{u}$ ; the branch lemma then gives a unique root $Ψ_{h} (u) = c_{u} + h b (Ψ_{h} (u))$ in that ball when $h L_{b} \leq 1 / 2$ . Require that the exact solution segment from each reference state $y (t) \in T$ stays in $T$ and that its endpoint $y (t + h)$ lies in $\overline{B} (c_{y (t)}, R_{y (t)})$ . Then, uniformly on the tube, $∥ Ψ_{h} (y (t)) - y (t + h) ∥_{2} \leq C h^{2}, ∥ Ψ_{h} (u) - Ψ_{h} (v) ∥_{2} \leq (1 + L h) ∥ u - v ∥_{2},$ where one may take $L = 2 (L_{a} + L_{b})$ and $C$ depends only on the stated tube bounds and branch margins. The local defect is therefore $O (h^{2})$ , and the exact-root method is first order locally.

Facts & Assumptions

Given: The additive IVP, a common tube with uniform field bounds, the positive-radius branch balls and branch hypotheses for every starting state in that tube, exact path segments and endpoints admitted in their branch balls, and $h L_{b} \leq 1 / 2$ .

[F1]

One-step IMEX Euler with an implicit branch and residual status: the exact update is $Y = y + h a (y) + h b (Y)$ .

[F2]

A local IMEX Euler branch from contraction: if $∥ b ∥ \leq M_{b}$ , $b$ is $L_{b}$ -Lipschitz on the selected ball, $h M_{b} \leq R$ , and $h L_{b} < 1$ , the exact branch exists uniquely and satisfies the residual-to-root bound.

[F3]

Along each exact solution segment in $T$ , $s \mapsto F (y (s))$ is continuous because $F$ is Lipschitz on $T$ and $y$ is continuous. Each coordinate is therefore integrable on the compact segment (A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion), so the vector derivative is integrable; the vector-valued fundamental theorem of calculus (If $f : [a, b] \to R^{m}$ is differentiable with integrable $f^{'}$ then $\int_{a}^{b} f^{'} = f (b) - f (a)$ ; and a bounded derivative makes $f$ Lipschitz) gives $y (t + s) - y (t) = \int_{0}^{s} F (y (t + r)) d r$ .

[F4]

For $a \leq b$ and $f : [a, b] \to R^{m}$ integrable when $a < b$ , $∥ \int_{a}^{b} f ∥_{2} \leq \int_{a}^{b} ∥ f ∥_{2}$ ; for $a < b$ , $∥ f ∥_{2}$ is integrable: for an integrable vector function on an increasing interval, $∥ \int f ∥_{2} \leq \int ∥ f ∥_{2}$ .

[A1]

$F = a + b$ is bounded by $M_{F} : = M_{a} + M_{b}$ and Lipschitz with $L_{F} : = L_{a} + L_{b}$ on the tube; the exact path and its endpoints remain in the stated convex tube, and each selected root belongs to its stated branch ball in that tube.

Proof

1.1A1F3F4algebra

By [F3], for $0 \leq s \leq h$ , $ϕ_{s} - y = \int_{0}^{s} F (ϕ_{r}) d r$ , so [F4] and [A1] give $∥ ϕ_{s} - y ∥_{2} \leq M_{F} s$ . Subtracting $h F (y)$ from the integral at $h$ and applying [F4] again yields $∥ ϕ_{h} - y - h F (y) ∥_{2} \leq \int_{0}^{h} L_{F} ∥ ϕ_{s} - y ∥_{2} d s \leq \frac{1}{2} L_{F} M_{F} h^{2}$ .

2.1A1F1F2step 1.1algebra

Set $c = y + h a (y)$ and evaluate the implicit residual of the exact endpoint: $r_{ϕ} : = ϕ_{h} - c - h b (ϕ_{h}) = (ϕ_{h} - y - h F (y)) - h (b (ϕ_{h}) - b (y))$ . By [A1] and step 1.1, $∥ r_{ϕ} ∥_{2} \leq M_{F} (\frac{1}{2} L_{F} + L_{b}) h^{2}$ . The exact endpoint and the branch root lie in the same admitted branch ball, so [F2] gives $∥ Ψ_{h} (y) - ϕ_{h} ∥_{2} \leq 2 M_{F} (\frac{1}{2} L_{F} + L_{b}) h^{2}$ .

3.1A1F1algebra∎

For two inputs $u, v$ with selected roots $Y_{u}, Y_{v}$ , subtracting their equations in [F1] and using the Lipschitz bounds gives $(1 - h L_{b}) ∥ Y_{u} - Y_{v} ∥_{2} \leq (1 + h L_{a}) ∥ u - v ∥_{2}$ . Since $h L_{b} \leq 1 / 2$ , $(1 + h L_{a}) / (1 - h L_{b}) = 1 + h (L_{a} + L_{b}) / (1 - h L_{b}) \leq 1 + 2 (L_{a} + L_{b}) h$ . This is the stated one-step stability bound.

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The ARS(2,2,2) coefficients satisfy the coupled second-order identities

Statement

For $γ = 1 - 1 / \sqrt{2}$ and $δ = - 1 / \sqrt{2}$ in The ARS(2,2,2) additive tableau and its three stages, one has $γ > 0$ and $δ = 1 - \frac{1}{2 γ}, γ (1 - δ) = \frac{1}{2}, γ (2 - γ) = \frac{1}{2}, γ (1 - γ - δ) = \frac{1}{2} - γ^{2} .$

Facts & Assumptions

Given: The coefficient definitions in the statement.

[F1]

The ARS(2,2,2) additive tableau and its three stages: $γ = 1 - 1 / \sqrt{2}$ and $δ = - 1 / \sqrt{2}$ .

[F2]

Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ : $r : = \sqrt{2}$ is nonnegative and $r^{2} = 2$ .

[F3]

Squaring is monotone on the nonnegatives: squaring is strictly increasing on nonnegative reals.

[F4]

Inverses of positives are positive, and reciprocation reverses order: if $0 < a < b$ , then $0 < b^{- 1} < a^{- 1}$ .

Proof

1.1F2F3F4algebra

Let $r = \sqrt{2}$ . If $r \leq 1$ , then $0 \leq r \leq 1$ and [F3] gives $r^{2} \leq 1$ , contradicting $r^{2} = 2$ from [F2]; hence $r > 1$ . Applying [F4] to $0 < 1 < r$ gives $0 < 1 / r < 1$ , so $γ = 1 - 1 / r > 0$ .

2.1F1F2step 1.1algebra

By [F1] and $γ = (r - 1) / r$ from step 1.1, $1 - 1 / (2 γ) = 1 - r / (2 (r - 1)) = (r - 2) / (2 (r - 1))$ . Since $r^{2} = 2$ , $r (r - 2) = 2 - 2 r = - 2 (r - 1)$ ; dividing by the nonzero $2 r (r - 1)$ gives $(r - 2) / (2 (r - 1)) = - 1 / r = δ$ .

2.2F1F2step 1.1algebra

Using $δ = - 1 / r$ and $γ = (r - 1) / r$ , $γ (1 - δ) = ((r - 1) / r) (1 + 1 / r) = (r^{2} - 1) / r^{2} = 1 / 2$ , and $γ (2 - γ) = 1 - (1 - γ)^{2} = 1 - 1 / r^{2} = 1 / 2$ .

3.1step 2.1step 2.2algebra∎

The identity in step 2.1 gives $γ (1 - γ - δ) = γ (1 / (2 γ) - γ) = 1 / 2 - γ^{2}$ . This proves all coefficient identities.

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The two implicit ARS stages have selected branches and residual bounds

Statement

Let $γ = 1 - 1 / \sqrt{2}$ as in The ARS(2,2,2) additive tableau and its three stages. For each of the two implicit stages, let $c_{j}$ be its center after the previous stages have been selected. Suppose $b$ is bounded by $M$ and $L$ -Lipschitz on a closed ball $\overline{B} (c_{j}, R_{j}) \subset U$ with $R_{j} > 0$ , and $∣ γ ∣ h M \leq R_{j}$ , $∣ γ ∣ h L < 1$ . Then stage $Y_{j} = c_{j} + γ h b (Y_{j})$ has a unique root in that ball. Every candidate ${\tilde{Y}}_{j}$ in the same ball with unscaled stage-equation residual $r_{j}$ satisfies $∥ {\tilde{Y}}_{j} - Y_{j} ∥_{2} \leq \frac{∥ r_{j} ∥_{2}}{1 - ∣ γ ∣ h L} .$ The claim is local to these stage balls; it asserts no remote-root or large-step branch.

Facts & Assumptions

Given: The ARS(2,2,2) coefficients, $h > 0$ , and for each sequential stage its center, positive-radius closed ball, bounds $M, L$ , and the two small-step inequalities.

[F1]

The ARS(2,2,2) additive tableau and its three stages: $Y_{2} = c_{2} + γ h b (Y_{2})$ with $c_{2} = y + γ h a (y)$ ; after selecting $Y_{2}$ , $Y_{3} = c_{3} + γ h b (Y_{3})$ with $c_{3} = y + h [δ a (y) + (1 - δ) a (Y_{2}) + (1 - γ) b (Y_{2})]$ .

[F2]

Small-step implicit Runge–Kutta stages have a unique branch and controlled residual error: on an invariant product ball, if $h ∥ A ∥_{\infty} M \leq R$ and $h ∥ A ∥_{\infty} L < 1$ , the implicit stages have a unique fixed point and the unscaled residual bounds the stage error by the reciprocal contraction margin.

[A1]

The two balls are admitted in $U$ , $b$ has the stated bounds on each, and $∣ γ ∣ h M \leq R_{j}$ , $∣ γ ∣ h L < 1$ for $j = 2, 3$ .

Proof

1.1A1F1F2algebra

For stage two in [F1], apply [F2] to the one-stage matrix $A = [γ]$ , the one-block function $F_{1} = b$ , and center $c_{2}$ ; its matrix row norm is $∥ A ∥_{\infty} = ∣ γ ∣$ , so [A1] gives the invariant-ball and contraction hypotheses. Hence the second stage root exists uniquely, and its residual bound is the one in [F2], with contraction factor $∣ γ ∣ h L$ .

2.1A1F1F2step 1.1

With that selected $Y_{2}$ , the third-stage center $c_{3}$ in [F1] is fixed. Apply [F2] again to the one-stage matrix $[γ]$ , function $b$ , and ball centered at $c_{3}$ ; [A1] verifies the same hypotheses, giving a unique $Y_{3}$ and its residual bound with contraction factor $∣ γ ∣ h L$ .

3.1step 1.1step 2.1∎

The stages are solved in order, so step 1.1 determines the center used in step 2.1. Both roots are unique in their selected balls, and the residual estimates hold for every candidate in those balls; no statement is made outside them.

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Lie and Strang splitting have global orders one and two, with separate subflow errors

Statement

Let $a, b \in C^{3} (U, R^{d})$ define the additive IVP of Additive autonomous ODE partitions and have exact subflows on a common compact convex tube over $[0, T]$ . Assume $a, b$ are Lipschitz there with constants $L_{a}, L_{b}$ , and the exact solution, all split stages, and all numerical states remain in that tube with positive margin. For all sufficiently small $h = T / N$ and exact initial data, the exact Lie and Strang split maps of Lie and Strang compositions of exact split flows satisfy $\max_{0 \leq n \leq N} ∥ y_{n}^{L i e} - y (t_{n}) ∥_{2} \leq C_{1} h, \max_{0 \leq n \leq N} ∥ y_{n}^{S t r a n g} - y (t_{n}) ∥_{2} \leq C_{2} h^{2} .$ The constants may depend on the tube, the fields, and $T$ . For approximate subflows, suppose $∥ {\hat{Φ}}_{c}^{s} (u) - Φ_{c}^{s} (u) ∥_{2} \leq ϵ_{c} (s)$ for $c \in {a, b}$ and every admitted starting state $u$ . If all approximate compositions stay in the tube, their composed one-step maps are Lipschitz with constant $1 + \hat{L} h$ for some $\hat{L} \geq 0$ , and $ϵ_{b} (h) + \exp (L_{b} h) ϵ_{a} (h) = O (h^{2})$ for Lie and $ϵ_{a} (h / 2) + \exp (L_{a} h / 2) ϵ_{b} (h) + \exp ((L_{a} / 2 + L_{b}) h) ϵ_{a} (h / 2) = O (h^{3})$ for Strang, then the same respective global orders hold. These additional conditions keep subflow-solver errors separate from splitting errors.

Facts & Assumptions

Given: The finite-dimensional fields, interval, exact and numerical paths, and tube hypotheses in the Statement.

[F1]

Lie and Strang splitting errors have commutator expansions: the exact Lie split has a uniform one-step defect $O (h^{2})$ and the exact Strang split has a uniform one-step defect $O (h^{3})$ on the stated tube.

[F2]

Lie and Strang compositions of exact split flows: the Lie and Strang maps are the compositions $Φ_{b}^{h} \circ Φ_{a}^{h}$ and $Φ_{a}^{h / 2} \circ Φ_{b}^{h} \circ Φ_{a}^{h / 2}$ .

[F3]

A first-order initial value problem is equivalent to its Volterra integral equation: a solution of $x^{'} = c (x)$ satisfies $x (t) = x (0) + \int_{0}^{t} c (x (s)) d s$ .

[F4]

For $a \leq b$ and $f : [a, b] \to R^{m}$ integrable when $a < b$ , $∥ \int_{a}^{b} f ∥_{2} \leq \int_{a}^{b} ∥ f ∥_{2}$ ; for $a < b$ , $∥ f ∥_{2}$ is integrable: $∥ \int_{0}^{t} g (s) d s ∥_{2} \leq \int_{0}^{t} ∥ g (s) ∥_{2} d s$ for continuous vector functions and $t \geq 0$ .

[F5]

Gronwall's integral inequality with variable and constant coefficients: if $u (t) \leq A + K \int_{0}^{t} u (s) d s$ with continuous $u \geq 0$ and constants $A, K \geq 0$ , then $u (t) \leq A \exp (K t)$ .

[F6]

The exponential function is smooth and $(\exp)^{'} = \exp$ and The exponential is positive and satisfies $\exp (- x) = 1 / \exp (x)$ : the real exponential is positive and has derivative $\exp (x)$ .

[F7]

The exponential function is strictly increasing and The mean value theorem, as the case $g (x) = x$ of Cauchy's: for $f$ continuous on $[a, b]$ with $a < b$ and differentiable on $(a, b)$ there is $c \in (a, b)$ with $f (b) - f (a) = f^{'} (c) (b - a)$ : on a bounded interval, the exponential mean-value estimate gives $\exp (K h) \leq 1 + K \exp (K h_{0}) h$ for $0 \leq h \leq h_{0}$ and $K \geq 0$ .

[F8]

Global error from uniform one-step defect and stability: a uniform one-step defect $C h^{p + 1}$ and one-step Lipschitz bound $1 + L h$ on a common compact tube imply fixed-time global error $O (h^{p})$ for exact initial data.

[A1]

Exact and approximate subflow stages are defined and remain in the tube; for the approximate case the pointwise subflow errors and the composed-map stability bounds stated in the Statement hold uniformly in the step and starting state.

Proof

1.1A1F2F3F4F5algebra

Let $K = L_{a} + L_{b}$ . For either field $c \in {a, b}$ , let $x (t)$ and $z (t)$ start at $u$ and $v$ . By [F3] and [F4], $∥ x (t) - z (t) ∥_{2} \leq ∥ u - v ∥_{2} + L_{c} \int_{0}^{t} ∥ x (s) - z (s) ∥_{2} d s .$ The constant case of [F5] yields $∥ Φ_{c}^{t} (u) - Φ_{c}^{t} (v) ∥_{2} \leq \exp (L_{c} t) ∥ u - v ∥_{2}$ . Composing these bounds according to [F2] shows both exact split maps are Lipschitz with factor $\exp (K h)$ .

2.1F6F7step 1.1algebra

Fix $h_{0} > 0$ within the admitted step range. For $0 \leq h \leq h_{0}$ , [F6] and the mean value theorem give $\exp (K h) - 1 = K h \exp (ξ) \leq K h \exp (K h_{0})$ for some $ξ \in [0, K h]$ , using monotonicity from [F7]. Thus both exact split maps satisfy a one-step bound $1 + L h$ , with $L = K \exp (K h_{0})$ (and $L = 0$ when $K = 0$ ).

2.2A1F2step 1.1algebra

Fix a starting state $u$ in the tube. For Lie, insert the exact $b$ flow between the two compositions and use its Lipschitz bound from step 1.1: $∥ {\hat{Φ}}_{b}^{h} ({\hat{Φ}}_{a}^{h} (u)) - Φ_{b}^{h} (Φ_{a}^{h} (u)) ∥_{2} \leq ϵ_{b} (h) + \exp (L_{b} h) ϵ_{a} (h) .$ For Strang, inserting the exact outer $a$ and middle $b$ maps in sequence gives $∥ {\hat{Φ}}_{a}^{h / 2} ({\hat{Φ}}_{b}^{h} ({\hat{Φ}}_{a}^{h / 2} (u))) - Φ_{a}^{h / 2} (Φ_{b}^{h} (Φ_{a}^{h / 2} (u))) ∥_{2} \leq ϵ_{a} (h / 2) + \exp (L_{a} h / 2) ϵ_{b} (h) + \exp ((L_{a} / 2 + L_{b}) h) ϵ_{a} (h / 2) .$ The inequalities hold uniformly over all starting states in the tube.

3.1F1F8step 2.1

The local defects are uniformly $O (h^{2})$ for Lie and $O (h^{3})$ for Strang by [F1]. Combine these defects with the stability bounds from step 2.1 and apply [F8] with $p = 1$ and $p = 2$ , respectively. The exact-root Lie and Strang global estimates follow.

4.1A1F1F8step 2.2∎

Under the stated $O (h^{2})$ Lie and $O (h^{3})$ Strang perturbation budgets, step 2.2 and [F1] give the same respective local-defect orders for the approximate compositions. Their one-step Lipschitz constants are assumed to be $1 + \hat{L} h$ in [A1], so [F8] applies with $p = 1$ and $p = 2$ and proves the approximate-subflow global bounds. This does not assert that local subflow accuracy alone implies stability.

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IMEX Euler is first-order convergent on a common compact tube

Statement

Let $T > 0$ , $N \geq 1$ , $h = T / N$ , and let $y (t)$ solve the additive IVP of Additive autonomous ODE partitions on $[0, T]$ . Assume the exact path, all numerical states, and every selected implicit branch remain in a common compact convex tube with positive margin in the state domain. Suppose the uniform hypotheses of IMEX Euler has a first-order local defect and a one-step stability bound hold on this tube and use the exact-root IMEX Euler map with exact initial value $y_{0} = y (0)$ . For all sufficiently small $h = T / N$ satisfying these hypotheses, there is a constant $C_{T}$ , independent of $h$ and the step index, such that $\max_{0 \leq n \leq N} ∥ y_{n} - y (t_{n}) ∥_{2} \leq C_{T} h .$ This is a fixed-time first-order result for the selected local branch; it makes no assertion about other implicit roots.

Facts & Assumptions

Given: The finite grid, exact solution, exact-root IMEX Euler map, common tube, and branch hypotheses in the Statement.

[F1]

IMEX Euler has a first-order local defect and a one-step stability bound: on the common tube, the exact-start defect is bounded by $C h^{2}$ and the selected-root map is Lipschitz with constant $1 + L h$ .

[F2]

Global error from uniform one-step defect and stability: for a one-step map on a common compact tube, uniform local defect $C h^{p + 1}$ , one-step Lipschitz constant $1 + L h$ , and initial error at most $C_{0} h^{p}$ imply a fixed-time global error bound of order $h^{p}$ .

[A1]

The exact path, numerical states, and all implicit stages stay in the tube, so the uniform constants and map hypotheses of [F1] hold at every step.

Proof

1.1F1A1algebra

By [F1] and [A1], for each $n < N$ the exact-start local defect satisfies $∥ Ψ_{h} (y (t_{n})) - y (t_{n + 1}) ∥_{2} \leq C h^{2}$ , and for all compared states in the tube $∥ Ψ_{h} (u) - Ψ_{h} (v) ∥_{2} \leq (1 + L h) ∥ u - v ∥_{2}$ . The exact initial value gives zero initial error, which is at most $0 \cdot h$ .

2.1F2A1step 1.1∎

Apply [F2] with $p = 1$ , $C_{0} = 0$ , and the constants from step 1.1. All hypotheses are satisfied by [A1], so its fixed-time estimate gives $\max_{n \leq N} ∥ y_{n} - y (t_{n}) ∥_{2} \leq C_{T} h$ for a constant independent of $h$ and $n$ .

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IMEX Euler preserves first-order convergence under per-step residuals of order h squared

Statement

Let $T > 0$ , $N \geq 1$ , $h = T / N$ , and let the exact solution and exact-root IMEX Euler map satisfy the common-tube hypotheses of IMEX Euler has a first-order local defect and a one-step stability bound. Set ${\tilde{y}}_{0} : = y_{0}$ . At step $n$ , suppose the solver returns ${\tilde{y}}_{n + 1}$ in the selected branch ball centered at $c_{n} = {\tilde{y}}_{n} + h a ({\tilde{y}}_{n})$ , with unscaled update residual $r_{n} : = {\tilde{y}}_{n + 1} - c_{n} - h b ({\tilde{y}}_{n + 1}), ∥ r_{n} ∥_{2} \leq C_{r} h^{2} .$ Assume $h L_{b} \leq 1 / 2$ in every branch ball, and that the exact path, numerical states, candidates, and exact branches remain in the common tube. For all sufficiently small $h = T / N$ satisfying these hypotheses, if $∥ y_{0} - y (0) ∥_{2} \leq C_{0} h$ , then $\max_{0 \leq n \leq N} ∥ {\tilde{y}}_{n} - y (t_{n}) ∥_{2} \leq C_{T} h,$ where $C_{T}$ is independent of $h$ and the step index. The residual is in update units; a derivative-scaled residual bound must be multiplied by $h$ before applying this statement.

Facts & Assumptions

Given: The grid, additive IVP, residual candidates, branch balls, common tube, and bounds in the Statement.

[F1]

IMEX Euler has a first-order local defect and a one-step stability bound: the exact-root map has local defect at most $C_{ℓ} h^{2}$ and is Lipschitz with constant $1 + L h$ on the tube.

[F2]

A local IMEX Euler branch from contraction: on a branch ball with $h L_{b} < 1$ , the unscaled update residual gives $∥ {\tilde{y}}_{n + 1} - y_{n + 1}^{*} ∥_{2} \leq ∥ r_{n} ∥_{2} / (1 - h L_{b})$ for its unique exact root $y_{n + 1}^{*}$ .

[F3]

Finite geometric error bound from a one-step recurrence: if $e_{n + 1} \leq (1 + L h) e_{n} + C h^{p + 1}$ for $h = T / N$ and $e_{0} \leq C_{0} h^{p}$ , then $\max_{n} e_{n} \leq C_{T} h^{p}$ with a constant independent of $h$ and $n$ .

[F4]

Implicit residuals and their update and derivative units: the update residual has the same units as the unknown; the derivative residual is the update residual divided by $h$ .

[A1]

In each step's closed branch ball $\overline{B} (c_{n}, R_{n}) \subset U$ , $R_{n} > 0$ , $∥ b ∥_{2} \leq M_{b}$ , $b$ is $L_{b}$ -Lipschitz, $h M_{b} \leq R_{n}$ , and $h L_{b} \leq 1 / 2$ ; both the unique exact root and returned candidate lie in that ball. The uniform tube constants in [F1] hold.

Proof

1.1givenF2A1algebra

Since $h L_{b} \leq 1 / 2$ , [F2] gives $∥ {\tilde{y}}_{n + 1} - Ψ_{h} ({\tilde{y}}_{n}) ∥_{2} \leq 2 C_{r} h^{2}$ , where $Ψ_{h} ({\tilde{y}}_{n}) = y_{n + 1}^{*}$ is the exact-root update from the current numerical state. The bound is valid at every step by [A1].

2.1F1step 1.1algebra

By the triangle inequality, [F1], and step 1.1, $e_{n + 1} : = ∥ {\tilde{y}}_{n + 1} - y (t_{n + 1}) ∥_{2} \leq (1 + L h) e_{n} + (C_{ℓ} + 2 C_{r}) h^{2} .$ Here the first term compares the exact-root map at ${\tilde{y}}_{n}$ with that map at $y (t_{n})$ ; the local-defect term compares the latter output with the exact next state.

3.1F3F4step 2.1given∎

Apply [F3] to this recurrence with $p = 1$ , $C = C_{ℓ} + 2 C_{r}$ , and the initial error from the Statement. This proves the displayed $O (h)$ global bound. By [F4], if a solver instead reports the derivative-scaled residual $r_{n} / h$ , the assumed update-residual budget is obtained by multiplying that bound by $h$ .

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The ARS(2,2,2) pair is second order under distinct stage residual tolerances

Statement

Use the tableau and stage equations in The ARS(2,2,2) additive tableau and its three stages, with $F = a + b$ . Let $a, b \in C^{2} (U, R^{d})$ on an open set, and suppose the exact solution path and, for every starting state in the tube, all exact stages, approximate candidates, and joining line segments remain in a common compact convex tube on which $a, b$ and their first two derivatives are bounded and $a, b$ are Lipschitz with constants $L_{a}, L_{b}$ . For every starting state in the tube and each implicit stage, assume the selected branch hypotheses of The two implicit ARS stages have selected branches and residual bounds hold with uniform bounds and $∣ γ ∣ h L_{b} \leq 1 / 2$ . All one-step conclusions below are for sufficiently small $h > 0$ , and the global conclusion is for $h = T / N$ in that same range.

For exact stage roots, the exact-start one-step defect is at most $C h^{3}$ , and the exact-root one-step map is Lipschitz with constant $1 + L h$ on the tube, for constants independent of sufficiently small $h$ . If computed stages have unscaled update residuals $r_{2}, r_{3}$ satisfying $∥ r_{2} ∥_{2} \leq C_{2} h^{2}, ∥ r_{3} ∥_{2} \leq C_{3} h^{3},$ then their output differs from the exact-root output at the same input by at most $C_{a l g} h^{3}$ . Consequently, for $h = T / N$ and initial error $∥ y_{0} - y (0) ∥_{2} \leq C_{0} h^{2}$ , the computed method satisfies $\max_{0 \leq n \leq N} ∥ y_{n} - y (t_{n}) ∥_{2} \leq C_{T} h^{2} .$ The second-stage residual has the weaker $O (h^{2})$ budget because its stage error is multiplied by $h$ in the last stage; the final-stage residual must be $O (h^{3})$ . All residuals are in update units.

Facts & Assumptions

Given: The split IVP, ARS tableau and stages, tube, branch balls, residuals, and initial error in the Statement.

[F1]

The ARS(2,2,2) additive tableau and its three stages: $Y_{2} = y + γ h a (y) + γ h b (Y_{2})$ and $Y_{3} = y + h [δ a (y) + (1 - δ) a (Y_{2}) + (1 - γ) b (Y_{2}) + γ b (Y_{3})], y^{+} = Y_{3} .$

[F2]

The two implicit ARS stages have selected branches and residual bounds: on each admitted stage ball, the implicit root is unique and an unscaled stage residual is bounded in forward error by the reciprocal contraction margin.

[F3]

Implicit residuals and their update and derivative units: the unscaled stage residual is measured in the units of the unknown stage; a derivative-scaled residual is divided by $h$ .

[F4]

A first-order initial value problem is equivalent to its Volterra integral equation and For $a \leq b$ and $f : [a, b] \to R^{m}$ integrable when $a < b$ , $∥ \int_{a}^{b} f ∥_{2} \leq \int_{a}^{b} ∥ f ∥_{2}$ ; for $a < b$ , $∥ f ∥_{2}$ is integrable: an IVP solution satisfies its integral equation, and the norm of a vector integral is bounded by the integral of the norm.

[F5]

Multivariable Taylor formula with a Lagrange remainder along a line segment: on an admitted line segment, a $C^{2}$ field has its first-order Taylor polynomial with a remainder bounded by a constant times the squared segment length.

[F6]

The ARS(2,2,2) coefficients satisfy the coupled second-order identities: $γ > 0$ , $γ (1 - δ) = 1 / 2$ , and $γ (2 - γ) = 1 / 2$ .

[F7]

Finite geometric error bound from a one-step recurrence: a recurrence $e_{n + 1} \leq (1 + L h) e_{n} + C h^{p + 1}$ with $h = T / N$ and $e_{0} \leq C_{0} h^{p}$ gives $\max_{n} e_{n} = O (h^{p})$ .

[A1]

The exact path and all exact and computed stages remain in the stated convex tube; for each stage the closed ball is contained in $U$ , $∥ b ∥_{2} \leq M_{b}$ , $b$ is $L_{b}$ -Lipschitz, $∣ γ ∣ h M_{b} \leq R_{j}$ , and $∣ γ ∣ h L_{b} \leq 1 / 2$ .

Proof

1.1A1F4F5algebra

Let $ϕ_{s}$ be the exact solution starting at $y$ . By [F4], $ϕ_{s} - y = \int_{0}^{s} F (ϕ_{r}) d r$ . Boundedness gives $∥ ϕ_{r} - y ∥ = O (r)$ , and Lipschitz continuity then gives $∥ ϕ_{s} - y - s F (y) ∥ \leq \int_{0}^{s} L_{F} ∥ ϕ_{r} - y ∥ d r = O (s^{2})$ , where $L_{F} = L_{a} + L_{b}$ . Taylor's formula [F5] gives $F (ϕ_{s}) = F (y) + D F (y) (ϕ_{s} - y) + O (s^{2})$ , hence $F (ϕ_{s}) = F (y) + s D F (y) F (y) + O (s^{2})$ . Integrating this identity from $0$ to $h$ yields $ϕ_{h} = y + h F (y) + \frac{1}{2} h^{2} D F (y) F (y) + O (h^{3}) .$

1.2A1F1F6algebra

For two inputs $u, v$ with exact selected second stages $Y_{2} (u), Y_{2} (v)$ , subtract the first equation in [F1] and use the Lipschitz bounds. Since $γ > 0$ from [F6] and $γ h L_{b} \leq 1 / 2$ , this gives $∥ Y_{2} (u) - Y_{2} (v) ∥_{2} \leq \frac{1 + γ h L_{a}}{1 - γ h L_{b}} ∥ u - v ∥_{2} \leq (1 + 2 γ (L_{a} + L_{b}) h) ∥ u - v ∥_{2} .$

1.3A1F1F3F6algebra

Let $e_{2} = ∥ {\tilde{Y}}_{2} - Y_{2} ∥_{2}$ . By $γ > 0$ from [F6] and $∣ γ ∣ h L_{b} \leq 1 / 2$ from [A1], subtracting the second-stage equations and using its residual gives $(1 - γ h L_{b}) e_{2} \leq ∥ r_{2} ∥_{2}$ , hence $e_{2} \leq 2 ∥ r_{2} ∥_{2}$ . For the final stage, subtract the equations with the same initial input; the terms depending on $Y_{2}$ are bounded by $h K_{s} e_{2}$ , where $K_{s} = ∣ 1 - δ ∣ L_{a} + ∣ 1 - γ ∣ L_{b}$ . The final-stage residual therefore gives $∥ {\tilde{Y}}_{3} - Y_{3} ∥_{2} \leq 2 ∥ r_{3} ∥_{2} + 4 K_{s} h ∥ r_{2} ∥_{2} .$ This is at most $C_{a l g} h^{3}$ under the two stated residual budgets.

2.1A1F1F2F5step 1.1algebra

By [F2] and [A1], the selected second-stage root exists in its branch ball. Its equation in [F1] and boundedness first give $Y_{2} - y = O (h)$ . Lipschitz continuity gives $b (Y_{2}) = b (y) + O (h)$ , so the same equation gives $Y_{2} - y = γ h F (y) + O (h^{2})$ . Applying [F5] to $b (Y_{2})$ then gives $b (Y_{2}) = b (y) + γ h D b (y) F (y) + O (h^{2})$ , and substitution into [F1] yields $Y_{2} = y + γ h F (y) + γ^{2} h^{2} D b (y) F (y) + O (h^{3}) .$

2.2A1F1step 1.2algebra

Let $K_{2} = 2 γ (L_{a} + L_{b})$ and $K_{3} = ∣ δ ∣ L_{a} + (∣ 1 - δ ∣ L_{a} + ∣ 1 - γ ∣ L_{b}) (1 + K_{2} h_{0})$ for $h \leq h_{0}$ , where $h_{0}$ is any fixed upper bound on the small steps. The third-stage centers in [F1] then differ by at most $(1 + K_{3} h) ∥ u - v ∥_{2}$ . Applying the same subtraction to the third-stage equation and using $(1 - γ h L_{b})^{- 1} \leq 1 + 2 γ h L_{b}$ gives $∥ Ψ_{h} (u) - Ψ_{h} (v) ∥_{2} \leq (1 + L h) ∥ u - v ∥_{2}$ for $L : = K_{3} + 2 γ L_{b} + 2 γ L_{b} K_{3} h_{0}$ , which is independent of $h$ and the step index.

3.1A1F1F2F5step 2.1algebra

By [F2] and [A1], the selected third-stage root exists after $Y_{2}$ is selected. Since $Y_{2} - y = O (h)$ and the stage equations give $Y_{3} - y = O (h)$ , Lipschitz continuity and the weight sums $δ + (1 - δ) = 1$ and $(1 - γ) + γ = 1$ imply $Y_{3} - y = h F (y) + O (h^{2})$ . Taylor's formula applied to the last-stage terms gives $a (Y_{2}) = a (y) + γ h D a (y) F (y) + O (h^{2}), b (Y_{2}) = b (y) + γ h D b (y) F (y) + O (h^{2}), b (Y_{3}) = b (y) + h D b (y) F (y) + O (h^{2}) .$ Substituting these expansions into [F1] and collecting the coefficients gives $Y_{3} = y + h F (y) + h^{2} {γ (1 - δ) D a (y) F (y) + γ (2 - γ) D b (y) F (y)} + O (h^{3}) .$

4.1F6step 1.1step 3.1algebra

The identities [F6] make both second-order coefficients in step 3.1 equal to $1 / 2$ , so its expansion agrees with the exact expansion in step 1.1. Thus $∥ Ψ_{h} (y) - ϕ_{h} ∥_{2} \leq C h^{3}$ uniformly on the tube.

5.1A1F3F7step 1.3step 2.2step 4.1algebra∎

Put $e_{n} = ∥ y_{n} - y (t_{n}) ∥_{2}$ . Compare the computed output first with the exact-root ARS output at $y_{n}$ , then with that map at $y (t_{n})$ , then with the exact next state. Step 1.3 bounds the first difference, step 2.2 bounds the second, and step 4.1 bounds the third, giving $e_{n + 1} \leq (1 + L h) e_{n} + (C + C_{a l g}) h^{3} .$ Apply [F7] with $p = 2$ and the initial error from the Statement to obtain the claimed fixed-time $O (h^{2})$ bound. By [F3], a derivative-scaled residual must be multiplied by $h$ before it is tested against either stage budget.

PropositionStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

The ARS(2,2,2) bivariate scalar stability function

Statement

For this stability calculation, extend the real tableau formulas in The ARS(2,2,2) additive tableau and its three stages to complex scalars by the real embedding $R ↪ C$ (equivalently, view $C$ as $R^{2}$ and let each $λ$ act by multiplication). Apply that extension to $y^{'} = λ_{a} y + λ_{b} y$ , with $z_{a} = h λ_{a}$ and $z_{b} = h λ_{b}$ in $C$ . If $1 - γ z_{b} \neq 0$ , the one-step amplification factor is $R (z_{a}, z_{b}) = \frac{1 + z_{a} + (1 - 2 γ) z_{b} + \frac{1}{2} z_{a}^{2} + (\frac{1}{2} - γ^{2}) z_{a} z_{b}}{(1 - γ z_{b})^{2}} .$ The exact scalar absolute-stability region for this tableau is $S = {(z_{a}, z_{b}) \in C^{2} : 1 - γ z_{b} \neq 0, ∣ 1 + z_{a} + (1 - 2 γ) z_{b} + \frac{1}{2} z_{a}^{2} + (\frac{1}{2} - γ^{2}) z_{a} z_{b} ∣ \leq ∣ 1 - γ z_{b} ∣^{2}} .$ On $z_{b} = 0$ this reduces to $∣ 1 + z_{a} + z_{a}^{2} / 2 ∣ \leq 1$ ; on $z_{a} = 0$ it reduces to $∣ 1 + (1 - 2 γ) z_{b} ∣ \leq ∣ 1 - γ z_{b} ∣^{2}$ . This is a scalar, two-variable region; it makes no matrix-nonnormality or PDE stability claim.

Facts & Assumptions

Given: The ARS(2,2,2) stages and weights, and the scalar test equation with the stated $z_{a}, z_{b}$ .

[F1]

The ARS(2,2,2) additive tableau and its three stages: $Y_{1} = y$ , $Y_{2} = y + γ h a (Y_{1}) + γ h b (Y_{2})$ , and $Y_{3} = y + h [δ a (Y_{1}) + (1 - δ) a (Y_{2})] + h [(1 - γ) b (Y_{2}) + γ b (Y_{3})]$ , with output $Y_{3}$ .

[F2]

The ARS(2,2,2) coefficients satisfy the coupled second-order identities: $γ (1 - δ) = 1 / 2$ , $γ (2 - γ) = 1 / 2$ , and $γ (1 - γ - δ) = 1 / 2 - γ^{2}$ .

[F3]

The complex numbers as $R [x] / (x^{2} + 1)$ , with the real embedding and imaginary unit $i$ and Real and imaginary parts, complex conjugation, and modulus: the scalar stability variables lie in $C$ and absolute stability is expressed using complex modulus.

Proof

1.1givenF1F2algebra

Substituting $a (y) = λ_{a} y$ and $b (y) = λ_{b} y$ into [F1] gives $Y_{2} / y = (1 + γ z_{a}) / (1 - γ z_{b})$ whenever the denominator is nonzero. The third stage then satisfies $\frac{Y_{3}}{y} = \frac{1 + δ z_{a} + [(1 - δ) z_{a} + (1 - γ) z_{b}] (Y_{2} / y)}{1 - γ z_{b}} .$ These formulas are linear coefficient identities, so they also define the amplification when $y = 0$ .

2.1step 1.1algebra

Substituting the stage-two ratio from step 1.1 gives numerator $(1 + δ z_{a}) (1 - γ z_{b}) + [(1 - δ) z_{a} + (1 - γ) z_{b}] (1 + γ z_{a})$ over denominator $(1 - γ z_{b})^{2}$ . Collecting terms yields $1 + z_{a} + (1 - 2 γ) z_{b} + γ (1 - δ) z_{a}^{2} + γ (1 - γ - δ) z_{a} z_{b}$ .

3.1F2step 2.1algebra

The identities in [F2] simplify the $z_{a}^{2}$ and $z_{a} z_{b}$ coefficients in step 2.1; the remaining $z_{b}$ coefficient is $1 - 2 γ$ . Substitution gives the displayed $R (z_{a}, z_{b})$ .

4.1F3step 3.1algebra∎

Repeated application to the scalar test equation multiplies its state by $R$ at each step, so nonincrease of the absolute value is exactly $∣ R ∣ \leq 1$ . Since $∣ (1 - γ z_{b})^{2} ∣ = ∣ 1 - γ z_{b} ∣^{2}$ , this is precisely the displayed set $S$ , with the pole excluded because the implicit stage equation is singular there. Setting either variable to zero gives the stated axis sections by substitution.

5 · Examples, counterexamples and false statements

None yet.

Imex and operator splitting methods

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

Additive autonomous ODE partitions

Definition

Diagonal Dirichlet sine-mode heat model with constant forcing

Definition

One-step IMEX Euler with an implicit branch and residual status

Definition

Lie and Strang compositions of exact split flows

Definition

The diagonal heat semigroup has the stated generator domain

Statement

Facts & Assumptions

Proof

A local IMEX Euler branch from contraction

Statement

Facts & Assumptions

Proof

The ARS(2,2,2) additive tableau and its three stages

Definition

Lie and Strang splitting errors have commutator expansions

Statement

Facts & Assumptions

Proof

Constant Dirichlet forcing is square summable but outside the heat-generator domain

Statement

Facts & Assumptions

Proof

IMEX Euler has a first-order local defect and a one-step stability bound

Statement

Facts & Assumptions

Proof

The ARS(2,2,2) coefficients satisfy the coupled second-order identities

Statement

Facts & Assumptions

Proof

The two implicit ARS stages have selected branches and residual bounds

Statement

Facts & Assumptions

Proof

Lie and Strang splitting have global orders one and two, with separate subflow errors

Statement

Facts & Assumptions

Proof

IMEX Euler is first-order convergent on a common compact tube

Statement

Facts & Assumptions

Proof

IMEX Euler preserves first-order convergence under per-step residuals of order h squared

Statement

Facts & Assumptions

Proof

The ARS(2,2,2) pair is second order under distinct stage residual tolerances

Statement

Facts & Assumptions

Proof

The ARS(2,2,2) bivariate scalar stability function

Statement

Facts & Assumptions

Proof

5 · Examples, counterexamples and false statements

Sources