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  "body": "\n## Definition\n\nLet $\\tau \u003e 0$ and let $f \\colon \\R \\times \\R^n \\times \\R^n \\to \\R^n$ be continuous.\nConsider the [[delay differential equation|39PCDRDB01]] $x'(t) = f(t, x(t), x(t - \\tau))$ for $t \\ge 0$, with a continuous history $x(\\theta) = \\phi(\\theta)$ for $\\theta \\in [-\\tau, 0]$.\nThe *method of steps* constructs the solution on the intervals $I_j = [j\\tau, (j+1)\\tau]$, $j = 0, 1, 2, \\ldots$, in order.\n\n1. On $I_0$, $x(t - \\tau) = \\phi(t - \\tau)$ is known, so $x$ solves the [[initial value problem|0FXDW992DS]] $x'(t) = f(t, x(t), \\phi(t - \\tau))$, $x(0) = \\phi(0)$.\n2. Suppose $x$ is known on $[-\\tau, j\\tau]$.\n   On $I_j$ the term $x(t - \\tau)$ is then known, and $x$ solves $x'(t) = f(t, x(t), x(t - \\tau))$ with the value $x(j\\tau)$ from the previous step.\n\nLet $f$ be locally Lipschitz in its second argument.\nThen each step has a unique local solution, so the solution is unique as long as it exists.\n\n## Remarks\n\nExample: $x'(t) = -x(t - 1)$ with $\\phi \\equiv 1$.\nOn $[0, 1]$: $x' = -1$ and $x(0) = 1$, so $x(t) = 1 - t$.\nOn $[1, 2]$: $x'(t) = -(1 - (t - 1)) = t - 2$ and $x(1) = 0$, so $x(t) = (t - 1)(t - 3)/2$.\nCheck: the derivative of $(t^2 - 4t + 3)/2$ is $t - 2$, and its value at $t = 1$ is $0$.\n\nNon-example: the method does not apply directly to $x'(t) = \\int_{t - 1}^{t} x(s) \\, ds$.\nThere the right side at time $t$ involves $x$ on all of $[t - 1, t]$, including times close to $t$ that are not yet known.\nSo no step reduces it to an ordinary equation with known data.\n\nThe solution is in general less smooth at the points $j\\tau$; for the example, $x'$ jumps at $t = 0$ from $0$ to $-1$, and smoothness increases with each step.\n\n## Explanations\n\n### L1 human\n\nSolve one delay-length at a time, using the part already found as the past for the next part.\n",
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