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  "body": "\n## Statement\n\nLet $A$ be a real $d \\times d$ matrix, let $N \\colon \\R^d \\to \\R^d$ be continuous, and consider the [[initial value problem|0FXDW992DS]] $y' = A y + N(y)$, $y(t_0) = y_0$.\nPut $\\varphi_1(z) = (e^z - 1)/z$ for $z \\ne 0$ and $\\varphi_1(0) = 1$, so that $\\varphi_1(hA) = \\sum_{k \\ge 0} \\frac{(hA)^k}{(k+1)!}$.\n\n1. The exact solution satisfies the variation-of-constants formula\n$$y(t_0 + h) = e^{hA} y_0 + \\int_0^h e^{(h - s)A} N(y(t_0 + s))\\, ds,$$\nwhere $e^{hA}$ is the [[matrix exponential|020M04FN1H]].\n2. The exponential Euler method is the step\n$$y_{n+1} = e^{hA} y_n + h\\, \\varphi_1(hA)\\, N(y_n).$$\nIf $N$ is constant, it gives the exact solution at every step, for every $h \u003e 0$.\n\n## Remarks\n\nClause 1 follows by differentiating $e^{-tA} y(t)$.\nFor clause 2, if $N \\equiv b$, the integral in clause 1 equals $\\int_0^h e^{(h-s)A} b\\, ds = h \\varphi_1(hA) b$, which is the method.\nIn general, the exponential Euler method replaces $N(y(t_0 + s))$ by $N(y_n)$ in the integral; it has order $1$ when $N$ is Lipschitz and the solution is smooth.\nExample: for the scalar equation $y' = -100 y + 1$ the method reproduces $y(t) = 1/100 + (y_0 - 1/100) e^{-100 t}$ exactly, whatever the step size.\nHigher-order exponential Runge–Kutta and Rosenbrock-type methods use the functions $\\varphi_k(z) = \\sum_{j \\ge 0} z^j / (j + k)!$.\n\n## Explanations\n\n### L1 human\n\nMany equations split into a simple linear part, which may be very fast, and a slower remainder.\nAn exponential integrator solves the linear part exactly with a matrix exponential and approximates only the remainder.\nSo the fast part does not force tiny time steps.\nWhen the remainder is constant, the method is exact.\n\n### L2 human\n\nExponential integrators are a tool for [[stiff equations|RR5FCG92PN]] where the stiffness sits in a known linear operator, such as the discretised Laplacian of a semilinear parabolic equation.\nThey are explicit in $N$ and do not need the solution of nonlinear equations.\nThey need products of $\\varphi_k(hA)$ with vectors, computed with Krylov methods, Padé approximants or contour integrals.\nTheir error analysis for parabolic problems uses that the bounds do not depend on the norm of $A$.\n",
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