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The ARS(2,2,2) bivariate scalar stability function

Statement

For this stability calculation, extend the real tableau formulas in The ARS(2,2,2) additive tableau and its three stages to complex scalars by the real embedding $R ↪ C$ (equivalently, view $C$ as $R^{2}$ and let each $λ$ act by multiplication). Apply that extension to $y^{'} = λ_{a} y + λ_{b} y$ , with $z_{a} = h λ_{a}$ and $z_{b} = h λ_{b}$ in $C$ . If $1 - γ z_{b} \neq 0$ , the one-step amplification factor is $R (z_{a}, z_{b}) = \frac{1 + z_{a} + (1 - 2 γ) z_{b} + \frac{1}{2} z_{a}^{2} + (\frac{1}{2} - γ^{2}) z_{a} z_{b}}{(1 - γ z_{b})^{2}} .$ The exact scalar absolute-stability region for this tableau is $S = {(z_{a}, z_{b}) \in C^{2} : 1 - γ z_{b} \neq 0, ∣ 1 + z_{a} + (1 - 2 γ) z_{b} + \frac{1}{2} z_{a}^{2} + (\frac{1}{2} - γ^{2}) z_{a} z_{b} ∣ \leq ∣ 1 - γ z_{b} ∣^{2}} .$ On $z_{b} = 0$ this reduces to $∣ 1 + z_{a} + z_{a}^{2} / 2 ∣ \leq 1$ ; on $z_{a} = 0$ it reduces to $∣ 1 + (1 - 2 γ) z_{b} ∣ \leq ∣ 1 - γ z_{b} ∣^{2}$ . This is a scalar, two-variable region; it makes no matrix-nonnormality or PDE stability claim.

Facts & Assumptions

Given: The ARS(2,2,2) stages and weights, and the scalar test equation with the stated $z_{a}, z_{b}$ .

[F1]

The ARS(2,2,2) additive tableau and its three stages: $Y_{1} = y$ , $Y_{2} = y + γ h a (Y_{1}) + γ h b (Y_{2})$ , and $Y_{3} = y + h [δ a (Y_{1}) + (1 - δ) a (Y_{2})] + h [(1 - γ) b (Y_{2}) + γ b (Y_{3})]$ , with output $Y_{3}$ .

[F2]

The ARS(2,2,2) coefficients satisfy the coupled second-order identities: $γ (1 - δ) = 1 / 2$ , $γ (2 - γ) = 1 / 2$ , and $γ (1 - γ - δ) = 1 / 2 - γ^{2}$ .

[F3]

The complex numbers as $R [x] / (x^{2} + 1)$ , with the real embedding and imaginary unit $i$ and Real and imaginary parts, complex conjugation, and modulus: the scalar stability variables lie in $C$ and absolute stability is expressed using complex modulus.

Proof

1.1givenF1F2algebra

Substituting $a (y) = λ_{a} y$ and $b (y) = λ_{b} y$ into [F1] gives $Y_{2} / y = (1 + γ z_{a}) / (1 - γ z_{b})$ whenever the denominator is nonzero. The third stage then satisfies $\frac{Y_{3}}{y} = \frac{1 + δ z_{a} + [(1 - δ) z_{a} + (1 - γ) z_{b}] (Y_{2} / y)}{1 - γ z_{b}} .$ These formulas are linear coefficient identities, so they also define the amplification when $y = 0$ .

2.1step 1.1algebra

Substituting the stage-two ratio from step 1.1 gives numerator $(1 + δ z_{a}) (1 - γ z_{b}) + [(1 - δ) z_{a} + (1 - γ) z_{b}] (1 + γ z_{a})$ over denominator $(1 - γ z_{b})^{2}$ . Collecting terms yields $1 + z_{a} + (1 - 2 γ) z_{b} + γ (1 - δ) z_{a}^{2} + γ (1 - γ - δ) z_{a} z_{b}$ .

3.1F2step 2.1algebra

The identities in [F2] simplify the $z_{a}^{2}$ and $z_{a} z_{b}$ coefficients in step 2.1; the remaining $z_{b}$ coefficient is $1 - 2 γ$ . Substitution gives the displayed $R (z_{a}, z_{b})$ .

4.1F3step 3.1algebra∎

Repeated application to the scalar test equation multiplies its state by $R$ at each step, so nonincrease of the absolute value is exactly $∣ R ∣ \leq 1$ . Since $∣ (1 - γ z_{b})^{2} ∣ = ∣ 1 - γ z_{b} ∣^{2}$ , this is precisely the displayed set $S$ , with the pole excluded because the implicit stage equation is singular there. Setting either variable to zero gives the stated axis sections by substitution.

Depends on

Used by

ARS(2,2,2) amplification on a two-cell reaction-diffusion grid Example

Dependency tree · two levels

16 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Uri M. Ascher, Steven J. Ruuth, and Raymond J. Spiteri, Implicit-explicit Runge-Kutta methods for time-dependent partial differential equations (standard reference, not scraped)
Lorenzo Pareschi and Giovanni Russo, Implicit-explicit Runge-Kutta schemes and applications to hyperbolic systems with relaxation (standard reference, not scraped)