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Omitting the second-stage correction destroys cubic local accuracy

Statement

On $y^{'} = y$ from $y = 1$ , with $J = 1$ , $\tilde{T} = 0$ , and $α = 1 - 1 / \sqrt{2}$ , consider the incorrect variant that retains $k_{1} = (1 - α h)^{- 1} y$ and the midpoint evaluation $F_{1} = f (y + h k_{1} / 2)$ but replaces the corrected second stage by $(1 - α h) ℓ_{2} = F_{1}$ and returns $y + h ℓ_{2}$ . Its one-step expansion is $y_{+} = y + h y + (α + \frac{1}{2}) h^{2} y + O (h^{3}),$ whereas the exact flow is $\exp (h) = 1 + h + \frac{1}{2} h^{2} + O (h^{3})$ . Since $α > 0$ , the altered method has a nonzero $O (h^{2})$ local defect and fails the cubic local-error condition for order two. The altered formula does not satisfy the proper method's stage equations.

Facts & Assumptions

Given: $f (y) = y$ , initial $y = 1$ , $J = 1$ , $h > 0$ small, and $α = 1 - 1 / \sqrt{2}$ .

[F1]

Two-stage second-order Rosenbrock W-method with shared factorization: the correct stages use the second right-hand side $F_{1} - k_{1}$ and then set $k_{2} = q_{2} + k_{1}$ .

[F2]

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: a differentiable vector curve with integrable derivative has endpoint difference equal to the integral of its derivative.

[F3]

The exponential is the unique solution of $y^{'} = y$ with $y (0) = 1$ : any differentiable solution on $R$ of $Y^{'} = Y$ , $Y (0) = 1$ equals $\exp$ .

[F4]

The exponential function is smooth and $(\exp)^{'} = \exp$ : every derivative of $\exp$ equals $\exp$ .

[F5]

The exponential function is strictly increasing: $\exp$ is increasing on $R$ .

[F6]

The real exponential function and the number $e$ by a power series: $\exp (s) = \sum_{n = 0}^{\infty} s^{n} / n!$ , so $\exp (0) = 1$ .

[F7]

A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion: a continuous real function on a compact interval is integrable.

Proof

Given: The data in the Statement.

1.1givenalgebra

Define the altered variant as stated. Since $W = 1 - α h$ , its first stage is $k_{1} = y / (1 - α h)$ . The midpoint value is $F_{1} = y + (h / 2) k_{1}$ , and the altered second solve gives $ℓ_{2} = F_{1} / (1 - α h)$ .

1.2givenF3F4F6

Define $Y (s) = \exp (s)$ for $s \in R$ . By [F4], $Y^{'} = Y$ , and [F6] gives $Y (0) = 1$ ; thus this is an exact solution of the IVP (unique on $R$ by [F3]).

2.1step 1.1algebra

Put $r = (1 - α h)^{- 1}$ . The exact identity $(1 - x)^{- 1} = 1 + x + x^{2} + x^{3} / (1 - x)$ shows $r = 1 + α h + α^{2} h^{2} + O (h^{3})$ ; for $0 < h \leq 1 / (2 α)$ the remainder is bounded by $2 α^{3} h^{3}$ . Squaring this finite expansion gives $r^{2} = 1 + 2 α h + 3 α^{2} h^{2} + O (h^{3})$ , with a bounded remainder on the same interval. Since $F_{1} = y + (h / 2) k_{1} = y (1 + h r / 2)$ , the altered second stage is $ℓ_{2} = F_{1} r = y (r + h r^{2} / 2) = y (1 + (α + 1 / 2) h + (α^{2} + α) h^{2} + O (h^{3}))$ . Multiplying by $h$ and adding $y$ gives the displayed one-step expansion.

2.2step 1.2F4F5

By [F4], $Y^{'''} (s) = \exp (s)$ ; since $0 \leq s \leq h \leq 1$ , [F5] gives $Y^{'''} (s) \leq \exp (1)$ .

3.1F2F4F7step 2.1step 2.2algebra

The derivatives of $Y$ are continuous by [F4] and integrable on each compact subinterval by [F7]. Three applications of [F2] give $\exp (h) = 1 + h + h^{2} / 2 + R$ , where $R = \int_{0}^{h} (h - s)^{2} Y^{'''} (s) / 2 d s$ and $∣ R ∣ \leq \exp (1) h^{3} / 6$ . Subtracting this expansion from the altered update in step 2.1 leaves $α h^{2} + O (h^{3})$ , which is not bounded by $C h^{3}$ for any fixed $C$ as $h ↓ 0$ because $α > 0$ . Thus the altered scheme fails the cubic local-error condition for order two.

4.1F1step 3.1∎

The proper method's second right-hand side and corrected output are exactly the two terms changed in this variant by [F1]. Step 3.1 proves the altered formula has no cubic local defect, and this item makes no global-rate claim for it.

Finite check

For $h = 0.01$ and $y = 1$ , the leading local-defect coefficient is $α h^{2} \approx 2.9289 \times 10^{- 5}$ ; the exact Taylor coefficient at $h^{2}$ is $1 / 2$ , while the altered coefficient is $1 / 2 + α$ .

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Sources

WIAS Berlin, Numerical Methods for Stiff Ordinary Differential Equations, §2.4 Rosenbrock Methods (standard reference, not scraped)
Daniel Kressner, Advanced Numerical Analysis, §3.5 Rosenbrock methods (standard reference, not scraped)