ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)

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A solution with only H1 time regularity loses CN second order

Example

In the scalar evolution triple $V = H = V^{*} = R$ , use $a (x, y) = x y$ , $T = 1$ , exact solution $u (t) = t^{3 / 4}$ , and load $f (t) = u^{'} (t) + u (t)$ . Apply Crank–Nicolson with exact interval-average load, uniform step $τ = 1 / N$ , and $U^{0} = u (0) = 0$ . Then the first nodal error is $e^{1} = \frac{τ^{7 / 4}}{14 (1 + τ / 2)} .$ Thus the maximum nodal error is not $O (τ^{2})$ although the exact solution belongs to $H^{1} (0, 1)$ and the load belongs to $L^{2} (0, 1)$ , as assumed by the method. The $H^{2}$ time regularity in the Crank–Nicolson theorem is a real condition.

Facts & Assumptions

Given: $V = H = V^{*} = R$ with the identity embeddings, the form $a (x, y) = x y$ , the function $u (t) = t^{3 / 4}$ , its load $f = u^{'} + u$ , and the Crank–Nicolson algorithm with interval-average forcing.

[F1]

Crank–Nicolson uses the average of the endpoint states in the autonomous form and the interval-average load (Crank–Nicolson and cG(1) for an autonomous parabolic form).

[F2]

A form is bounded and coercive when $∣ a (x, y) ∣ \leq M ∥ x ∥ ∥ y ∥$ and $a (x, x) \geq α ∥ x ∥^{2}$ (Bounded, coercive and symmetric sesquilinear forms).

[F3]

An evolution triple uses dense continuous embeddings and identifies $H$ with its dual in the canonical way (Evolution triples and the parabolic energy space).

Verification

1.1F2F3givenalgebra

Here $a (x, y) = x y$ has $M = α = 1$ . The derivative is $u^{'} (t) = \frac{3}{4} t^{- 1 / 4}$ and $\int_{0}^{1} ∣ u^{'} (t) ∣^{2} d t = \frac{9}{16} \int_{0}^{1} t^{- 1 / 2} d t = \frac{9}{8} < \infty$ ; also $u \in L^{2} (0, 1)$ , so $u \in H^{1} (0, 1)$ and $f = u^{'} + u \in L^{2} (0, 1)$ . But $u^{''} (t) = - \frac{3}{16} t^{- 5 / 4}$ has nonintegrable square near zero, so $u \notin H^{2} (0, 1)$ .

1.2F1givenalgebra

The exact equation integrated over $(0, τ)$ is $\frac{u (τ) - u (0)}{τ} + \frac{1}{τ} \int_{0}^{τ} u (t) d t = \frac{1}{τ} \int_{0}^{τ} f (t) d t .$ The first CN step from [F1] is $\frac{U^{1} - U^{0}}{τ} + \frac{U^{1} + U^{0}}{2} = \frac{1}{τ} \int_{0}^{τ} f (t) d t$ . Subtract, use $U^{0} = u (0) = 0$ , and set $e^{1} = U^{1} - u (τ)$ . This gives $e^{1} (\frac{1}{τ} + \frac{1}{2}) = \frac{1}{τ} \int_{0}^{τ} t^{3 / 4} d t - \frac{τ^{3 / 4}}{2} = (\frac{4}{7} - \frac{1}{2}) τ^{3 / 4} = \frac{1}{14} τ^{3 / 4} .$

2.1step 1.2algebra∎

Solving the scalar equation in step 1.2 yields $e^{1} = τ^{7 / 4} / (14 (1 + τ / 2))$ . Along $τ = 1 / N \to 0$ , $∣ e^{1} ∣ / τ^{2} = τ^{- 1 / 4} / (14 (1 + τ / 2)) \to \infty$ ; the maximum nodal error is at least $∣ e^{1} ∣$ , so it cannot be $O (τ^{2})$ .

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