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Adaptive ODE solvers and error estimation — Examples

1 · Prerequisites

2 · Summary

Summary

These examples isolate three limits of an embedded controller: the leading estimator term can vanish, a fast-decay step can be rejected down to the minimum-step safeguard, and a rounded implementation can accept while its exact endpoint error exceeds the requested tolerance. They illustrate the stated method and do not supply global adaptive accuracy or a general stiffness diagnosis.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

A tolerance below roundoff can pass with a zero embedded difference

Example

Use the toy binary format whose nonzero representable numbers are $\pm m 2^{e}$ with integer $e$ and $m \in {4, 5, 6, 7}$ (three significant bits), rounded to nearest with ties to an even significand. The next larger representable value after $8 = 4 \cdot 2$ is $10 = 5 \cdot 2$ , so their midpoint $9$ rounds to $8$ because $4$ is even and $5$ is odd. Apply a rounded Euler–Heun implementation to $y^{'} = 1$ , $y (0) = 8$ on $[0, 1]$ , with one step $h = 1$ , $h_{\min} = h_{\max} = h_{i n i t} = 1$ , attempt budget $N = 1$ , $a t o l = 1 / 4$ , $r t o l = 0$ , safety $0.9$ , and scale bounds $0.2$ and $5$ . Both computed endpoints round to $8$ ; the stored difference is zero and the step is accepted, although the exact endpoint is $9$ and the actual error is $1$ .

This example concerns the rounded realization described in Finite-budget adaptive Runge–Kutta controller. It does not contradict its exact-arithmetic specification and shows why that acceptance test is not a certificate.

Facts & Assumptions

Given: The toy format and rounding rule in the Example, field $f (y) = 1$ , initial state $8$ , and controller tolerance parameters.

[F1]

Euler–Heun embedded Runge–Kutta pair: both pair members use $K_{1} = f (y_{n})$ and $K_{2} = f (y_{n} + h K_{1})$ , with Euler and Heun weighted endpoints.

[F2]

Finite-budget adaptive Runge–Kutta controller: a rounded realization uses stored high and low outputs in the difference and accepts when the scaled difference $Q \leq 1$ .

[F3]

Local orders and leading estimator term for the Euler–Heun pair: for this pair the low member is order $p = 1$ , which is the controller exponent input.

Verification

1.1givenF1F2F3algebra

The two stages evaluate to $K_{1} = K_{2} = 1$ ; each exact endpoint formula is $8 + 1 = 9$ . By the specified tie rule, the Euler endpoint rounds to $8$ and the Heun endpoint also rounds to $8$ . Hence the stored difference is $0$ and the scaled error is $Q = 0 / (1 / 4) = 0 \leq 1$ , so the rounded controller accepts.

2.1givenstep 1.1algebra∎

The exact solution is $y (t) = 8 + t$ , so $y (1) = 9$ . The accepted stored endpoint is $8$ , giving absolute error $∣ 9 - 8 ∣ = 1 > 1 / 4 = a t o l$ . Thus the embedded difference can be zero while the actual error exceeds the requested tolerance.

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

A fast-decay test problem is rejected at the minimum step

Example

Run the Euler–Heun controller on the scalar fast-decay equation $y^{'} = - 1000 y$ , $y (0) = 1$ , with $T = 0.004$ , $h_{\min} = 0.002$ , $h_{\max} = h_{0} = 0.004$ , $a t o l = 10^{- 4}$ , $r t o l = 0$ , and attempt budget $N = 2$ , $p = 1$ , safety $σ = 0.9$ , and scale factors $γ_{\min} = 0.2$ , $γ_{\max} = 5$ . The computed pair difference rejects both attempted steps, so the procedure stops at its minimum-step safeguard. This scalar fast-decay equation is used as a stiffness illustration: the calculation verifies repeated rejection of this explicit trial only, and gives no general stiffness diagnosis.

Facts & Assumptions

Given: The scalar field $f (y) = - 1000 y$ , starting value $1$ , trial sizes $0.004$ and $0.002$ , and controller parameters in the Example.

[F1]

Euler–Heun embedded Runge–Kutta pair: $K_{1} = f (y_{n})$ , $K_{2} = f (y_{n} + h K_{1})$ , $Y_{E} = y_{n} + h K_{1}$ , and $Y_{H} = y_{n} + \frac{h}{2} (K_{1} + K_{2})$ .

[F2]

Finite-budget adaptive Runge–Kutta controller: reject if $Q > 1$ , scale by $\max (γ_{\min}, \min (γ_{\max}, σ Q^{- 1 / (p + 1)}))$ , clamp the next proposal to the step bounds, and return minimum-step-rejected for a rejection at $h_{\min}$ .

[F3]

The bounded adaptive Runge–Kutta controller terminates: this finite-budget controller reaches a terminal status after finitely many attempts.

[F4]

Local orders and leading estimator term for the Euler–Heun pair: the Euler–Heun pair has low order $p = 1$ .

Verification

1.1givenF1F2F4algebra

At $h = 0.004$ , $K_{1} = - 1000$ , the Euler stage value is $- 3$ , and $K_{2} = 3000$ . Thus $Y_{E} = - 3$ , $Y_{H} = 5$ , the difference magnitude is $8$ , and $Q = 8 / 10^{- 4} = 80000 > 1$ . The proposed scale is clamped to $0.2$ , so the next step is $\max (0.002, 0.2 \cdot 0.004) = 0.002$ .

2.1F1F2F3F4step 1.1algebra∎

At $h = 0.002$ , $K_{1} = - 1000$ , the Euler stage value is $- 1$ , and $K_{2} = 1000$ . Thus $Y_{E} = - 1$ , $Y_{H} = 1$ , the difference magnitude is $2$ , and $Q = 20000 > 1$ . This rejection occurs at $h_{\min}$ , so [F2] returns minimum-step-rejected; finite termination is also covered by [F3].

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

The leading Euler–Heun estimator coefficient can vanish

Example

Apply the Euler–Heun pair to $y^{'} = t^{2}$ , $y (0) = 0$ , on one step of size $h > 0$ . The low-order defect begins at order $h^{3}$ , not $h^{2}$ , so the embedded difference is not asymptotically equal to the low-order defect in ratio.

Facts & Assumptions

Given: The field $f (t, y) = t^{2}$ , initial time and value $t_{0} = y_{0} = 0$ , and $h > 0$ .

[F1]

Euler–Heun embedded Runge–Kutta pair: $K_{1} = f (0, 0)$ , $K_{2} = f (h, h K_{1})$ , $Y_{E} = h K_{1}$ , and $Y_{H} = \frac{h}{2} (K_{1} + K_{2})$ .

[F2]

Local orders and leading estimator term for the Euler–Heun pair: the local error expansions compare each one-step value with the exact endpoint.

[F3]

A nonzero leading defect makes the embedded difference asymptotically match the low-order defect: a ratio-one conclusion requires a nonzero leading low-defect coefficient.

Verification

1.1givenF1F2algebra

Here $K_{1} = 0$ and $K_{2} = h^{2}$ , so $Y_{E} = 0$ and $Y_{H} = h^{3} / 2$ . Since $y (h) = \int_{0}^{h} t^{2} d t = h^{3} / 3$ , the Euler defect is $- h^{3} / 3$ , the Heun defect is $h^{3} / 6$ , and the embedded estimator is $∣ Y_{H} - Y_{E} ∣ = h^{3} / 2$ .

2.1F1F3step 1.1algebra∎

The Euler defect has zero coefficient at order $h^{2}$ and the estimator-to-Euler-defect ratio is $(h^{3} / 2) / (h^{3} / 3) = 3 / 2$ for every $h > 0$ . This concrete case shows why [F3] requires a nonzero leading coefficient before asserting ratio one.

Adaptive ODE solvers and error estimation — Examples

1 · Prerequisites

2 · Summary

Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

A tolerance below roundoff can pass with a zero embedded difference

Example

Facts & Assumptions

Verification

A fast-decay test problem is rejected at the minimum step

Example

Facts & Assumptions

Verification

The leading Euler–Heun estimator coefficient can vanish

Example

Facts & Assumptions

Verification

Sources