The two-stage ETD2 method has a third-order local defect

Given: The exact solution, forcing curve, Lipschitz tube, and bounds in the Statement.

Variation of constants for a finite-matrix semilinear system: the exact endpoint is the exponential-linear term plus ${\int }_0^h{e}^{(h-s)A}q(t+s)ds$.

Exponential Euler and the two-stage ETD2 step maps: the ETD2 predictor is $a_h(t,y)={\mathrm{\Psi }}_h^E(t,y)$ and its correction uses ${\phi }_1-{\phi }_2$ at $q(t)$ and ${\phi }_2$ at $g(t+h,a_h)$.

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

Exponential Euler has a second-order local defect: the exact-start predictor obeys $\mathrm{\parallel }{a}_h(t,y(t))-y(t+h){\mathrm{\parallel }}_2\le C_E{h}^2$.

If $f:[a,b]\to {\mathbb{R}}^m$ is differentiable with integrable $f^{\mathrm{\prime }}$ then ${\int }_a^b{f}^{\mathrm{\prime }}=f(b)-f(a)$; and a bounded derivative makes $f$ Lipschitz 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: Taylor integral remainders and their norms are bounded componentwise.

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||: vector and induced matrix norms are the Euclidean norms used in the bounds.

The second derivative bound $Q_2$, state Lipschitz bound $L$, and semigroup bound $K_{+}$ hold on the complete path and predictor tube.

A bounded linear operator between normed spaces, The operator norm is zero on the zero domain and otherwise is max_{||v||=1} ||Tv||, Every finite-dimensional normed space is Banach, and The exponential series of a bounded operator: for finite $A$ on ${\mathbb{C}}^d$, $\mathrm{\parallel }{e}^{sA}{\mathrm{\parallel }}_2\le e^{\mathrm{\mid }s\mathrm{\mid }\mathrm{\parallel }A{\mathrm{\parallel }}_2}$. For $0\le s\le T$, nonnegativity of $\mathrm{\parallel }A{\mathrm{\parallel }}_2$ gives $s\mathrm{\parallel }A{\mathrm{\parallel }}_2\le T\mathrm{\parallel }A{\mathrm{\parallel }}_2$, so scalar exponential monotonicity (The exponential function is strictly increasing) gives $e^{s\mathrm{\parallel }A{\mathrm{\parallel }}_2}\le e^{T\mathrm{\parallel }A{\mathrm{\parallel }}_2}$. The set defining $K_{+}$ contains $\mathrm{\parallel }{I}_d{\mathrm{\parallel }}_2$ at $s=0$ and is therefore bounded above by $e^{T\mathrm{\parallel }A{\mathrm{\parallel }}_2}$; completeness of $\mathbb{R}$ (Dedekind completeness: the least-upper-bound property) supplies its finite supremum.

$\mathbb{C}$ 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 $\mathrm{\parallel }{\mathrm{\Phi }}_d(v){\mathrm{\parallel }}_{{\mathbb{R}}^{2d}}^2={\sum }_{j<d}\mathrm{\mid }{v}_j{\mathrm{\mid }}^2=\mathrm{\parallel }v{\mathrm{\parallel }}_2^2$. It preserves real-time difference quotients and their limits, so the real vector-valued FTC and integral norm bound in [F4] apply to the complex forcing curve with the same norms.