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Exact exponential stepping captures nonnormal transient growth

Statement

Let $A = (\begin{array}{cc} - 1 & 2 e \\ 0 & - 1 \end{array}), g = 0, y (0) = (\begin{array}{c} 0 \\ 1 \end{array}) .$ Here nonnormal means $A A^{*} \neq A^{*} A$ . Compute the exact solution, show its norm grows above its initial value before decaying, and verify that exact-action exponential Euler and ETD2 reproduce it at every grid point.

Facts & Assumptions

Given: The finite matrix, zero nonlinear term, initial state, and methods of Exponential Euler and the two-stage ETD2 step maps.

[F1]

Exponential Euler and the two-stage ETD2 step maps: when $g = 0$ , both exact maps reduce to $y \mapsto e^{h A} y$ .

[F2]

The exponential series of a bounded operator: the finite-matrix exponential is defined by its convergent series and satisfies $e^{(r + s) A} = e^{r A} e^{s A}$ .

[F3]

The real exponential function and the number $e$ by a power series and The exponential series converges absolutely for every real argument: $e^{t}$ is defined by its convergent power series.

[F4]

The adjoint $T^{*} : W \to V$ is characterised by $⟨ T v, w ⟩_{W} = ⟨ v, T^{*} w ⟩_{V}$ , In orthonormal bases, the matrix of the adjoint is the conjugate transpose of the matrix, and Rectangular matrix multiplication and the identity matrix $I_{n}$ , including zero-sized shapes: $A^{*}$ is the conjugate transpose and products are the usual matrix products.

[F5]

The norm $∥ v ∥ = \sqrt{⟨ v, v ⟩}$ induced by a real or complex inner product and The standard formulas $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ and $\sum_{k < n} x_{k} \overline{y_{k}}$ on $C^{n}$ are inner products: $∥ v ∥_{2}^{2} = v^{*} v$ in standard complex coordinates.

[F6]

The exponential is positive and satisfies $\exp (- x) = 1 / \exp (x)$ : $e^{- t} = 1 / e^{t}$ and $e^{t} > 0$ for every real $t$ .

[F7]

Variation of constants for a finite-matrix semilinear system: when $g = 0$ , a solution satisfies $y (t) = e^{t A} y (0)$ .

[A1]

The exact IVP starts at the stated $y (0)$ and has $g = 0$ .

Verification

1.1givenF2F3F4inductionalgebra

Write $A = - I + N$ with $N = (\begin{array}{cc} 0 & 2 e \\ 0 & 0 \end{array})$ ; direct multiplication gives $N^{2} = 0$ [F4]. The formula holds for $n = 1$ since $A = - I + N$ . If $A^{n} = (- 1)^{n} I + n (- 1)^{n - 1} N$ , multiplying by $A = - I + N$ and using $N^{2} = 0$ gives $A^{n + 1} = (- 1)^{n + 1} I + (n + 1) (- 1)^{n} N$ , so induction proves it for every $n \geq 1$ . Substituting in the absolutely convergent matrix-exponential series [F2] gives $e^{t A} = e^{- t} I + \sum_{n \geq 1} t^{n} (- 1)^{n - 1} N / (n - 1)! = e^{- t} (I + t N)$ by the scalar series [F3].

2.1A1F3F4F5F6F7step 1.1algebra

By [F7] and step 1.1, the solution is $y (t) = e^{- t} (2 e t, 1)^{T}$ . Direct multiplication using [F4] gives $A A^{*} = (\begin{array}{cc} 1 + 4 e^{2} & - 2 e \\ - 2 e & 1 \end{array})$ and $A^{*} A = (\begin{array}{cc} 1 & - 2 e \\ - 2 e & 1 + 4 e^{2} \end{array})$ , which differ, so $A$ is nonnormal. By [F5] and [F6], its initial norm is $1$ , whereas $∥ y (1) ∥_{2} = e^{- 1} \sqrt{4 e^{2} + 1} > 2$ . For $t > 0$ , the exponential series gives $e^{t} \geq t^{2} / 2$ , and [F6] gives $e^{- t} = 1 / e^{t} \leq 2 / t^{2}$ ; hence $e^{- t} \to 0$ and $t e^{- t} \leq 2 / t \to 0$ , so $y (t) \to 0$ .

3.1F1F2step 1.1algebrainduction∎

By [F1], every exact-action step is multiplication by $e^{h A}$ . Repeated use of the semigroup identity in [F2] gives $(e^{h A})^{n} = e^{n h A}$ , so the method returns $y (t_{n})$ at every grid point. The transient growth is therefore present in the exact linear flow and is not removed by exponentiating this nonnormal matrix.

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Sources

Marlis Hochbruck and Alexander Ostermann, Exponential Integrators, Acta Numerica 19 (2010) (standard reference, not scraped)
Jitse Niesen and Will M. Wright, A Krylov Subspace Algorithm for Evaluating the Phi-Functions Appearing in Exponential Integrators, arXiv:0907.4631v2 (standard reference, not scraped)