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  "body": "\n## Statement\n\nLet $t_0 \\in \\R$ and let $n \\ge 1$.\nLet $\\Lambda(t) = \\operatorname{diag}(\\lambda_1(t), \\dots, \\lambda_n(t))$ be a continuous [[diagonal matrix|044JNC7ZAA]] function on $[t_0, \\infty)$ with complex entries.\nLet $R(t)$ be a continuous complex $n \\times n$ matrix function on $[t_0, \\infty)$ with\n$$\\int_{t_0}^{\\infty} \\norm{R(t)} \\, dt \u003c \\infty.$$\nAssume the dichotomy condition: there is a constant $K$ such that for each pair $i \\ne j$ one of the following holds, where $D_{ij}(s, t) = \\int_s^t \\operatorname{Re}\\bigl(\\lambda_i(u) - \\lambda_j(u)\\bigr) \\, du$:\n\n1. $D_{ij}(t_0, t) \\to \\infty$ as $t \\to \\infty$, and $D_{ij}(s, t) \\ge -K$ for all $t \\ge s \\ge t_0$;\n2. $D_{ij}(s, t) \\le K$ for all $t \\ge s \\ge t_0$.\n\nThen the [[linear system|037ZNY7NAG]]\n$$x' = \\bigl(\\Lambda(t) + R(t)\\bigr) x$$\nhas a [[fundamental matrix|0H1GN6P9R5]] of the form\n$$X(t) = \\bigl(I + E(t)\\bigr) \\exp\\Bigl(\\int_{t_0}^{t} \\Lambda(s) \\, ds\\Bigr), \\qquad E(t) \\to 0 \\text{ as } t \\to \\infty.$$\nEquivalently, for each $k$ there is a solution $x_k(t) = \\bigl(e_k + o(1)\\bigr) \\exp\\bigl(\\int_{t_0}^t \\lambda_k(s) \\, ds\\bigr)$, where $e_k$ is the $k$-th standard basis vector.\n\n## Remarks\n\nThe proof writes the $k$-th solution as $x_k = \\exp\\bigl(\\int \\lambda_k\\bigr) z$ and turns the equation for $z$ into an integral equation.\nThe dichotomy condition splits the indices into those whose integrals are taken from $t_0$ to $t$ and those taken from $t$ to $\\infty$.\nWith this split the integral operator is a contraction on a far half-line $[T, \\infty)$, because $\\int_T^\\infty \\norm{R}$ is small.\nIts fixed point gives $z(t) \\to e_k$.\n\nExample: if $\\Lambda$ is a constant diagonal matrix, then each $D_{ij}(s, t)$ is $(t - s)$ times a constant, so the dichotomy condition holds automatically.\nHence for a constant diagonal $\\Lambda$ with distinct or equal eigenvalues and an integrable perturbation $R$, each solution of $x' = (\\Lambda + R) x$ behaves like $e^{\\lambda_k t}$ times a vector close to $e_k$.\n\nThe integrability of $R$ cannot be dropped.\nFor $n = 1$, $\\Lambda = 0$ and $R(t) = 1/t$ on $[1, \\infty)$, the solutions are $c t$, which do not tend to a constant multiple of $e^0 = 1$.\n\n## Explanations\n\n### L1 human\n\nMany linear systems look like a simple diagonal system plus a small disturbance.\nFor a diagonal system the solutions are explicit exponentials, one for each coordinate.\nLevinson's theorem says that if the disturbance is small in total, measured by the integral of its size, then the true solutions still look like these exponentials for large times.\nA mild condition on the growth rates is needed, so that the exponentials can be told apart.\n\n### L2 human\n\nThe theorem is the basic tool of asymptotic integration of linear systems.\nOne first brings a system into the form diagonal part plus integrable part, often by a change of variables that diagonalises the leading term.\nThen the theorem gives the leading behaviour of a fundamental matrix.\nThe WKB, or Liouville–Green, approximation for $y'' = Q(t) y$ can be justified in this way after a suitable substitution.\nHartman and Wintner proved a related result in which $R$ is only square integrable, at the cost of a stronger dichotomy condition and a correction term in the exponent.\n",
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