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  "body": "\n## Statement\n\nLet $X$ be a [[Banach space|06GZ956657]] and $T \u003e 0$.\nLet $A$ be a linear operator in $X$ such that the [[initial value problem|075VBY4K4X]] $u'(t) = A u(t)$, $u(0) = u_0$, is [[well-posed|W09B9NZ9B5]].\nThis means there are bounded linear solution operators $E(t)$, $t \\ge 0$, with $\\sup_{0 \\le t \\le T} \\norm{E(t)} \u003c \\infty$.\nMoreover $u(t) = E(t) u_0$ is a genuine solution for $u_0$ in a dense subspace $D \\subseteq X$.\nLet $C(k)$, $0 \u003c k \\le k_0$, be bounded linear operators on $X$ defining the one-step [[finite difference scheme|X7HYEBA5SA]] $u^{n+1} = C(k) u^n$ with time step $k$.\nAssume the scheme is consistent: for every $u_0 \\in D$,\n$$\\lim_{k \\to 0} \\sup_{0 \\le t \\le T} \\frac{\\norm{C(k) u(t) - u(t + k)}}{k} = 0, \\qquad u(t) = E(t) u_0.$$\nThen the following are equivalent:\n\n1. The scheme is stable: there are constants $K$ and $k_1 \u003e 0$ with $\\norm{C(k)^n} \\le K$ whenever $0 \u003c k \\le k_1$ and $0 \\le nk \\le T$, where $\\norm{\\cdot}$ is the [[operator norm|06EKW1AKXF]].\n2. The scheme is convergent: for every $u_0 \\in X$, every $t \\in [0, T]$, and all sequences $k_j \\to 0$ and integers $n_j$ with $n_j k_j \\to t$, one has $C(k_j)^{n_j} u_0 \\to E(t) u_0$ in $X$.\n\n## Remarks\n\nStability implies convergence by a telescoping sum: $C(k)^n u_0 - E(nk) u_0 = \\sum_{m=0}^{n-1} C(k)^{n-1-m} \\left( C(k) E(mk) - E((m+1)k) \\right) u_0$.\nEach factor $C(k)^{n-1-m}$ is bounded by $K$, and the $n$ consistency errors are each $o(k)$.\nDensity of $D$ and the uniform bounds then cover all $u_0$.\nConvergence implies stability by the uniform boundedness principle, since the orbits $C(k)^n u_0$ stay bounded for each $u_0$.\nExample: for $u_t = u_{xx}$ on $\\R$ in the maximum norm, the explicit scheme $u_j^{n+1} = u_j^n + r(u_{j+1}^n - 2u_j^n + u_{j-1}^n)$ with fixed $r = k/h^2 \\le \\tfrac12$ satisfies $\\norm{C(k)} \\le 1$, so it converges.\nWith $r \u003e \\tfrac12$ the mode $u_j^0 = (-1)^j$ is multiplied by $1 - 4r \u003c -1$ at each step, so the scheme is unstable and does not converge.\nThe theorem is for linear problems; for nonlinear schemes stability and consistency do not in general characterize convergence.\n\n## Explanations\n\n### L1 human\n\nA numerical method for a time-dependent equation is consistent if each single step makes only a small error.\nIt is stable if errors do not blow up when many steps are taken.\nThe theorem says that, for linear problems that are well-posed, a consistent method gives the right answer as the steps shrink exactly when it is stable.\nSo the hard question \"does the method converge?\" reduces to the easier question \"is it stable?\".\n\n### L2 human\n\nIn practice consistency is checked by Taylor expansion, and stability by a norm bound on the step operator $C(k)$, often through von Neumann's Fourier analysis in $L^2$.\nFor explicit schemes for hyperbolic equations, stability leads to the [[Courant–Friedrichs–Lewy condition|FQVVNYV6N8]] on $k/h$.\nThe result is due to Lax and Richtmyer, in their 1956 paper on the stability of linear finite difference equations.\nThe same structure, \"consistency plus stability gives convergence\", reappears for boundary value problems and for finite element methods, where Céa's lemma plays the role of stability.\n",
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