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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/ii/paper-1/30e/b/solution

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Past exam of the mathematics course of the University of Cambridge / 2015 / ii / Paper 1 / 30E / b / Solution by

Codex  0 Created 2026-09-24 Updated 2026-10-06

Choose

u_{n} (x, y) = \frac{1}{n} cos (n x) cosh (n y), v_{n} (x, y) = - \frac{1}{n} sin (n x) sinh (n y) .

Direct differentiation gives

u_{n y} = v_{n x}

and

v_{n y} = - u_{n x}

. Both functions are smooth and

2 π

-periodic in

x

, with initial data

f_{n} (x) = n^{- 1} cos n x

,

v_{n} (x, 0) = 0

. Their squared norms are

\int_{0}^{2 π} ∣ f_{n} ∣^{2} d x = \frac{π}{n ^{2}} \to 0, \int_{0}^{2 π} ∣ u_{n} (x, y) ∣^{2} d x = \frac{π}{n ^{2}} cosh^{2} (n y) \to \infty

for every fixed

y \neq = 0

. Thus continuous dependence on the initial data fails in this $L^{2}$ topology, even on smooth analytic data. The Cauchy problem for a partial differential equation is ill-posed in the sense of Hadamard well-posedness; existence of analytic solutions from the preceding theorem does not supply stability for an elliptic Cauchy problem.

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