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  "body": "\n## Counterexample\n\nConsider the Cauchy problem for the [[Laplace equation|S8RG6N1PRT]] in the half-plane $y \u003e 0$:\n$$\\partial_x^2 u + \\partial_y^2 u = 0 \\ \\text{ for } y \u003e 0, \\qquad u(x, 0) = 0, \\qquad \\partial_y u(x, 0) = g(x).$$\nFor an integer $n \\ge 1$, take the datum $g_n(x) = \\frac{1}{n} \\sin(n x)$.\nThe function\n$$u_n(x, y) = \\frac{1}{n^2} \\sinh(n y) \\sin(n x)$$\nsolves the problem.\nIndeed $\\partial_x^2 u_n = -n^2 u_n$ and $\\partial_y^2 u_n = n^2 u_n$, so the Laplace equation holds.\nAlso $u_n(x, 0) = 0$, since $\\sinh 0 = 0$, and $\\partial_y u_n(x, 0) = \\frac{1}{n} \\cosh(0) \\sin(n x) = g_n(x)$.\nThe data tend to $0$ uniformly: $\\sup_x \\abs{g_n(x)} = 1/n \\to 0$.\nFor the zero datum $g = 0$ the solution is $u = 0$.\nBut for every fixed $y \u003e 0$, $\\sup_x \\abs{u_n(x, y)} = \\sinh(n y) / n^2 \\to \\infty$.\nSo the solution does not depend continuously on the data, and the problem is not [[well posed|W09B9NZ9B5]].\n\n## Remarks\n\nUniqueness holds here, by [[Holmgren's uniqueness theorem|1VVKF0Y19B]], so the failure is only in the continuous dependence.\nThe same holds with any norm on the data that involves finitely many derivatives, since the $k$-th derivative of $g_n$ has size $n^{k-1}$, while $\\sinh(n y)$ grows exponentially in $n$.\nThe example also explains why the [[Cauchy–Kovalevskaya theorem|97HMV9BT2W]], which does give a solution for analytic data, is not a well-posedness result.\nThe natural problems for the Laplace equation are boundary value problems such as the Dirichlet problem, which are well posed.\n\n## Explanations\n\n### L1 human\n\nFor the Laplace equation one may not prescribe both the value and the slope of the solution along a line.\nData that are tiny, but oscillate fast, produce solutions that become enormous a short distance away.\nSo arbitrarily small errors in the data can cause arbitrarily large errors in the solution.\n",
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