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An explicitly treated stiff decay mode is unstable for a large step

Example

Take the scalar decay equation $y^{'} = - κ y$ with $κ > 0$ and $y_{0} = 1$ . Treat the whole stiff term explicitly, so $a (y) = - κ y$ and $b (y) = 0$ . IMEX Euler gives amplification $1 - h κ$ ; for $h κ > 2$ its magnitude exceeds one although the exact solution $\exp (- κ t)$ decays. If instead this same term is implicit, the amplification is $(1 + h κ)^{- 1}$ , which lies in $(0, 1)$ for every $h κ > 0$ . This scalar comparison does not imply matrix stability for nonnormal operators.

Facts & Assumptions

Given: $κ > 0$ , $h > 0$ , and $y_{0} = 1$ .

[F1]

One-step IMEX Euler with an implicit branch and residual status: the exact-step equation is $Y = y + h a (y) + h b (Y)$ .

[F2]

The real exponential function and the number $e$ by a power series, The exponential function is smooth and $(\exp)^{'} = \exp$ , and The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ : $\exp (x)$ is differentiable with derivative $\exp (x)$ , so $y (t) = \exp (- κ t)$ solves $y^{'} = - κ y$ and $y (0) = 1$ .

[F3]

The exponential is positive and satisfies $\exp (- x) = 1 / \exp (x)$ and The exponential function is strictly increasing: $\exp (x) > 0$ and $x < 0$ implies $\exp (x) < \exp (0) = 1$ .

[F4]

Inverses of positives are positive, and reciprocation reverses order: if $0 < a < b$ , then $0 < b^{- 1} < a^{- 1}$ .

Verification

technique · direct

1.1givenF1F2F3algebra

With $a (y) = - κ y$ and $b = 0$ , [F1] gives $Y = (1 - h κ) y$ , so after $N$ steps the numerical value is $(1 - h κ)^{N}$ . If $h κ > 2$ , then $1 - h κ < - 1$ and $∣ 1 - h κ ∣^{N} > 1$ for every $N \geq 1$ . By [F2], $y (t) = \exp (- κ t)$ is the exact solution, and [F3] gives $∣ y (N h) ∣ = \exp (- κ N h) < 1$ .

1.2F1F4algebra

With $a = 0$ and $b (y) = - κ y$ , the equation in [F1] becomes $(1 + h κ) Y = y$ . Since $0 < 1 < 1 + h κ$ , [F4] gives $0 < (1 + h κ)^{- 1} < 1$ . Thus the implicit treatment damps every such scalar step, while the explicit treatment grows whenever $h κ > 2$ .

2.1F1algebra∎

At the boundary $h κ = 2$ , the explicit amplification is $- 1$ , whose magnitude is exactly one. Thus the strict instability condition is $h κ > 2$ ; equality is neutral for this scalar test equation.

Depends on

Used by

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Dependency tree · two levels

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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)