Math 2.0

Partial differential equations Existence theory Counterexample

Hadamard's example of ill-posedness

Builds on Well-posed problemLaplace's equation

Counterexample

Consider the Cauchy problem for the Laplace equation in the half-plane y>0y > 0: ∂x2u+∂y2u=0  for y>0,u(x,0)=0,∂yu(x,0)=g(x).\partial_x^2 u + \partial_y^2 u = 0 \ \text{ for } y > 0, \qquad u(x, 0) = 0, \qquad \partial_y u(x, 0) = g(x). For an integer n≥1n \ge 1, take the datum gn(x)=1nsin⁡(nx)g_n(x) = \frac{1}{n} \sin(n x). The function un(x,y)=1n2sinh⁡(ny)sin⁡(nx)u_n(x, y) = \frac{1}{n^2} \sinh(n y) \sin(n x) solves the problem. Indeed ∂x2un=−n2un\partial_x^2 u_n = -n^2 u_n and ∂y2un=n2un\partial_y^2 u_n = n^2 u_n, so the Laplace equation holds. Also un(x,0)=0u_n(x, 0) = 0, since sinh⁡0=0\sinh 0 = 0, and ∂yun(x,0)=1ncosh⁡(0)sin⁡(nx)=gn(x)\partial_y u_n(x, 0) = \frac{1}{n} \cosh(0) \sin(n x) = g_n(x). The data tend to 00 uniformly: sup⁡x∣gn(x)∣=1/n→0\sup_x \abs{g_n(x)} = 1/n \to 0. For the zero datum g=0g = 0 the solution is u=0u = 0. But for every fixed y>0y > 0, sup⁡x∣un(x,y)∣=sinh⁡(ny)/n2→∞\sup_x \abs{u_n(x, y)} = \sinh(n y) / n^2 \to \infty. So the solution does not depend continuously on the data, and the problem is not well posed.

Remarks

Uniqueness holds here, by Holmgren's uniqueness theorem, so the failure is only in the continuous dependence. The same holds with any norm on the data that involves finitely many derivatives, since the kk-th derivative of gng_n has size nk−1n^{k-1}, while sinh⁡(ny)\sinh(n y) grows exponentially in nn. The example also explains why the Cauchy–Kovalevskaya theorem, which does give a solution for analytic data, is not a well-posedness result. The natural problems for the Laplace equation are boundary value problems such as the Dirichlet problem, which are well posed.

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  1. Laplace's equation concept
  2. Well-posed problem concept
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Further reading

  1. Web https://en.wikipedia.org/wiki/Well-posed_problem
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MSC35R25, 35J05
UsesWell-posed problem, Laplace's equation

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