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Two-node Gauss collocation gives the exact endpoint for a nonstiff cubic IVP

Example

Assume Countable Choice for the Lebesgue moment data. On one step $[0, 1]$ , take the scalar IVP $y^{'} = t^{3}$ , $y (0) = 0$ . The field is independent of the state, so its state Lipschitz constant is zero; the example has no stiffness from a state-dependent fast mode.

Facts & Assumptions

Given: The two-node Gauss rule for Lebesgue measure on $[0, 1]$ and the IVP in the Example.

[F1]

The m-node Gaussian quadrature rule for a positive measure and Gaussian quadrature is exact through degree 2m−1 and has positive weights: the two-node Gauss rule integrates every polynomial of degree at most three exactly.

[F2]

Collocation tableau from distinct quadrature nodes: the collocation endpoint is $y (0) + \sum_{j = 1}^{2} b_{j} c_{j}^{3}$ for this step, and its stages are finite because the right-hand side is independent of stage state.

[F3]

Riemann and Lebesgue integrals agree for continuous data on [0,1] and The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ : the exact endpoint increment is $\int_{0}^{1} t^{3} d t = 1 / 4$ using either integral convention.

[F4]

Lebesgue measure on the unit interval supplies Gaussian moment data: Lebesgue measure on $[0, 1]$ supplies the required order-four two-node moment data under Countable Choice; these data also satisfy the order-three branch used for exactness.

[F5]

The m-node Gaussian quadrature rule for a positive measure and Collocation tableau from distinct quadrature nodes: Gaussian weights and collocation weights are integrals of the same cardinal polynomials, so [F3] identifies them.

[F6]

First-order systems, initial value problems, and solutions on intervals: the exact IVP solution satisfies $y^{'} = t^{3}$ and $y (0) = 0$ .

Verification

1.1F1F2F3F4F5F6givenalgebra

The integrand $s^{3}$ has degree three, so [F1] gives $\sum_{j = 1}^{2} w_{j} c_{j}^{3} = \int_{0}^{1} s^{3} d λ$ . The matching of weights in [F5] and agreement with the Riemann integral in [F3] give $\sum_{j = 1}^{2} b_{j} c_{j}^{3} = \int_{0}^{1} s^{3} d s$ . The stages solve $U_{i} = \sum_{j} Q_{i j} c_{j}^{3}$ and are well-defined finite real numbers because $f$ does not depend on $U$ . The rule data are valid by [F4] and the axiom premise in the Example.

2.1F2F3F6step 1.1algebra∎

The IVP identity [F6] and the scalar FTC [F3] give $y (s) = s^{4} / 4$ , hence $y (1) = 1 / 4$ . Step 1.1 gives the same collocation endpoint, so the one-step endpoint is exact despite the fact that the collocation polynomial need not equal the exact path at interior points.

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Sources

John A. Gubner, Gaussian Quadrature and the Eigenvalue Problem, §§2–5, 2020 (standard reference, not scraped)
Ernst Hairer, Syvert P. Nørsett, and Gerhard Wanner, Solving Ordinary Differential Equations I, 2nd ed., Chapter II §II.7 (standard reference, not scraped)