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  "body": "\n## Statement\n\nConsider a [[linear $k$-step method|BX4D85A79P]] $\\sum_{i=0}^k \\alpha_i y_{j+i} = h \\sum_{i=0}^k \\beta_i f_{j+i}$ with $\\alpha_k = 1$ and characteristic polynomials $\\rho(\\zeta) = \\sum_i \\alpha_i \\zeta^i$, $\\sigma(\\zeta) = \\sum_i \\beta_i \\zeta^i$, where $\\rho$ and $\\sigma$ have no common root.\nThe method has *order* $p$ if $\\rho(e^z) - z \\sigma(e^z) = O(z^{p+1})$ as $z \\to 0$.\nIt is *zero-stable* if every root of $\\rho$ has modulus at most $1$ and the roots of modulus $1$ are simple.\nIt is *A-stable* if, for every complex $\\lambda$ with $\\operatorname{Re} \\lambda \u003c 0$ and every $h \u003e 0$, every sequence produced by the method for $y' = \\lambda y$ tends to $0$.\n\n1. First Dahlquist barrier: a zero-stable linear $k$-step method has order $p \\le k + 2$ if $k$ is even, and $p \\le k + 1$ if $k$ is odd.\nIf it is moreover explicit, then $p \\le k$.\n2. Second Dahlquist barrier: an A-stable linear multistep method has order $p \\le 2$, and no explicit linear multistep method is A-stable.\n\n## Remarks\n\nFor the first barrier, the substitution $\\zeta = \\frac{1 + w}{1 - w}$ maps the left half plane $\\operatorname{Re} w \u003c 0$ onto the open unit disk.\nIt turns the order conditions into conditions on the power series of a rational function of $w$.\nZero-stability forces sign conditions on the coefficients that are incompatible with a higher order.\nFor the second barrier, A-stability means that $\\rho(\\zeta) - z \\sigma(\\zeta)$ has all its roots in the open unit disk whenever $\\operatorname{Re} z \u003c 0$.\nHence $\\operatorname{Re} \\frac{\\rho(\\zeta)}{\\sigma(\\zeta)} \\ge 0$ for $\\abs{\\zeta} \u003e 1$, and this positivity is incompatible with order greater than $2$.\nExample: Simpson's two-step method $y_{j+2} - y_j = \\frac{h}{3} (f_{j+2} + 4 f_{j+1} + f_j)$ has order $4 = k + 2$ and is zero-stable, so the first barrier is sharp for $k = 2$.\nThe trapezoidal rule $y_{j+1} - y_j = \\frac{h}{2} (f_{j+1} + f_j)$ has order $2$ and is A-stable, so the second barrier is sharp; among A-stable methods of order $2$ it has the smallest error constant.\nThe [[backward differentiation formulas|TDA94C2JYT]] of order $3$ to $6$ are zero-stable but not A-stable, as the second barrier requires.\n\n## Explanations\n\n### L1 human\n\nOne might hope to make a multistep method more accurate just by using more old values and choosing the weights cleverly.\nThe first barrier says that this hope fails beyond a fixed limit: past that order, the method becomes unstable and useless.\nThe second barrier says that methods that are safe for every decaying problem, however stiff, cannot be more than second-order accurate.\nSo accuracy and stability pull against each other, and the barriers say exactly where.\n\n### L2 human\n\nA $k$-step method can reach order $2k$, but such methods are not zero-stable for $k \\ge 3$.\nThe first barrier explains why practical codes use Adams methods instead, of order $k$ or $k + 1$ for $k$ steps.\nThe second barrier explains why stiff solvers use [[BDF methods|TDA94C2JYT]].\nAbove order $2$ these methods give up full A-stability, but they keep stability in a large sector of the left half plane.\nIt also motivates implicit Runge–Kutta methods, which escape the barrier because they are not linear multistep methods: the Gauss–Legendre methods are A-stable of every even order.\nBoth barriers were proved by Germund Dahlquist.\n",
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