One-step stability after removing the exact linear flow

Given: The finite matrix, interval, methods, Lipschitz constant, and tube in the Statement.

Exponential Euler and the two-stage ETD2 step maps: ${\mathrm{\Psi }}_h^E$ and ${\mathrm{\Psi }}_h^2$ have the displayed exponential Euler and ETD2 formulas, with the predictor $a_h(t,y)={\mathrm{\Psi }}_h^E(t,y)$.

Integral representations of the first two phi functions: ${\phi }_1(X)={\int }_0^1{e}^{(1-\theta )X}d\theta$ and ${\phi }_2(X)={\int }_0^1\theta e^{(1-\theta )X}d\theta$.

A bounded linear operator between normed spaces, The operator norm as the least bound and as the unit-sphere or unit-ball supremum, Every finite-dimensional normed space is Banach, and The exponential series of a bounded operator: finite-matrix exponentials are continuous, differentiable and satisfy $e^{(r+q)A}=e^{rA}{e}^{qA}$.

The norm $\parallel v\parallel =\sqrt{⟨v,v⟩}$ induced by a real or complex inner product, The standard formulas $⟨x,y⟩={\sum }_{k<n}{x}_k{y}_k$ on ${\mathbb{R}}^n$ and ${\sum }_{k<n}{x}_k\overline{y_k}$ on ${\mathbb{C}}^n$ are inner products, and The operator norm is zero on the zero domain and otherwise is max_{||v||=1} ||Tv||: $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_2$ is the Euclidean vector norm and the induced operator norm.

The operator norm is submultiplicative and satisfies ||T^*T|| = ||T||^2 and For $a\le b$ and $f:[a,b]\to {\mathbb{R}}^m$ integrable when $a<b$, $\parallel {\int }_a^{b}f{\parallel }_2\le {\int }_a^b\parallel f{\parallel }_2$; for $a<b$, $\parallel f{\parallel }_2$ is integrable: products obey $\mathrm{\parallel }ST{\mathrm{\parallel }}_2\le \mathrm{\parallel }S{\mathrm{\parallel }}_2\mathrm{\parallel }T{\mathrm{\parallel }}_2$, and vector integrals obey $\mathrm{\parallel }\int v{\mathrm{\parallel }}_2\le \int \mathrm{\parallel }v{\mathrm{\parallel }}_2$.

The exponential series of a bounded operator: for finite $A$ on the finite-dimensional Banach space ${\mathbb{C}}^d$, $\mathrm{\parallel }{e}^{\pm rA}{\mathrm{\parallel }}_2\le e^{r\mathrm{\parallel }A{\mathrm{\parallel }}_2}$ for $0\le r\le T$. For $0\le r\le T$, nonnegativity of $\mathrm{\parallel }A{\mathrm{\parallel }}_2$ gives $r\mathrm{\parallel }A{\mathrm{\parallel }}_2\le T\mathrm{\parallel }A{\mathrm{\parallel }}_2$, so scalar exponential monotonicity (The exponential function is strictly increasing) gives $e^{r\mathrm{\parallel }A{\mathrm{\parallel }}_2}\le e^{T\mathrm{\parallel }A{\mathrm{\parallel }}_2}$. Both sets contain $\mathrm{\parallel }{I}_d{\mathrm{\parallel }}_2$ at $r=0$ and are therefore bounded above by $e^{T\mathrm{\parallel }A{\mathrm{\parallel }}_2}$; Dedekind completeness: the least-upper-bound property supplies finite suprema.

$\mathbb{C}$ is the real coordinate plane, with coordinate arithmetic, The standard formulas $⟨x,y⟩={\sum }_{k<n}{x}_k{y}_k$ on ${\mathbb{R}}^n$ and ${\sum }_{k<n}{x}_k\overline{y_k}$ on ${\mathbb{C}}^n$ are inner products, and The norm $\parallel v\parallel =\sqrt{⟨v,v⟩}$ induced by a real or complex inner product: for $v=(v_0,\dots ,v_{d-1})$, ${\mathrm{\Phi }}_d(v):=(\mathrm{Re}{v}_0,\mathrm{Im}{v}_0,\dots ,\mathrm{Re}{v}_{d-1},\mathrm{Im}{v}_{d-1})$ is real-linear and satisfies $\mathrm{\parallel }{\mathrm{\Phi }}_d(v){\mathrm{\parallel }}_{{\mathbb{R}}^{2d}}^2={\sum }_{j<d}\mathrm{\mid }{v}_j{\mathrm{\mid }}^2=\mathrm{\parallel }v{\mathrm{\parallel }}_2^2$. Thus the real vector-integral estimate [F5] applies to the complex-valued integrands with the same norms.

On the stated tube, $\mathrm{\parallel }g(r,u)-g(r,v){\mathrm{\parallel }}_2\le L\mathrm{\parallel }u-v{\mathrm{\parallel }}_2$, and all transformed inputs and ETD2 predictor stages considered in the Statement remain admissible.