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A dissipative scalar heat mode with Parareal correction

Example

For the dissipative scalar mode $\dot{y} = - y$ , take macro interval length $H = 1 / 2$ , two forward-Euler microsteps of length $h = 1 / 4$ for the fine propagator, and one backward-Euler step of length $H$ for the coarse propagator. Starting from $x_{0} = 1$ , after the first Parareal correction the node at $t = 1$ is $11 / 36$ , while the sequential fine value is $81 / 256$ . The coarse initial value there is $4 / 9$ , so this correction reduces the error relative to the sequential fine trajectory from $295 / 2304$ to $25 / 2304$ . This finite comparison makes no error claim about the exact ODE solution.

Facts & Assumptions

Given: Forward Euler and backward Euler on the stated macro grid, with the Parareal recurrence.

[F1]

The explicit Euler, midpoint, and classical fourth-order Runge–Kutta tableaus: Forward Euler has $s = 1$ , $A = (0)$ , $b = (1)$ , and $c = (0)$ .

[F2]

Backward Euler step equation and algebraic residual: a backward-Euler endpoint satisfies $y_{n + 1} = y_{n} + h_{n} f (t_{n} + h_{n}, y_{n + 1})$ .

[F3]

Finite-budget Parareal correction sweeps: the correction is $U_{n + 1}^{k + 1} = G_{n} (U_{n}^{k + 1}) + F_{n} (U_{n}^{k}) - G_{n} (U_{n}^{k})$ .

[A1]

$N = 3$ , $x_{0} = 1$ , $H = 1 / 2$ , and the fine map uses two forward-Euler microsteps of length $1 / 4$ for $\dot{y} = - y$ .

Verification

1.1givenF1F2A1algebra

By [F1], one fine microstep maps $x$ to $x - \frac{1}{4} x = \frac{3}{4} x$ , so the two-step fine macro map is $F (x) = \frac{9}{16} x$ . By [F2], the coarse backward-Euler equation is $z = x - \frac{1}{2} z$ , whose unique solution is $G (x) = \frac{2}{3} x$ .

2.1givenstep 1.1algebra

The sequential fine values are $x_{1} = \frac{9}{16}$ and $x_{2} = \frac{81}{256}$ , while coarse initialization gives $U_{1}^{0} = \frac{2}{3}$ and $U_{2}^{0} = \frac{4}{9}$ .

2.2givenF3step 1.1algebra

The first correction has $U_{1}^{1} = F (1) = \frac{9}{16}$ . Using [F3], $U_{2}^{1} = G (\frac{9}{16}) + (F - G) (\frac{2}{3}) = \frac{3}{8} + (\frac{9}{16} - \frac{2}{3}) \frac{2}{3} = \frac{11}{36}$ .

3.1givenstep 2.1step 2.2algebra∎

Direct subtraction gives $∣ U_{2}^{0} - x_{2} ∣ = ∣ \frac{4}{9} - \frac{81}{256} ∣ = \frac{295}{2304}$ and $∣ U_{2}^{1} - x_{2} ∣ = ∣ \frac{11}{36} - \frac{81}{256} ∣ = \frac{25}{2304}$ . Thus the error decreases for these parameters; the calculation does not assert monotone decrease for other propagators.

Depends on

Used by

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Dependency tree · two levels

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Sources

Martin J. Gander, Time Parallel Time Integration, Chapter 2: Methods Based on Multiple Shooting (standard reference, not scraped)
R. D. Falgout, S. Friedhoff, Tz. V. Kolev, S. P. MacLachlan, and J. B. Schroder, Parallel Time Integration with Multigrid (standard reference, not scraped)