Alphabeta Math
CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-09
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Analytic heat data need not give a time-analytic germ

Statement refuted

Analytic initial data alone do not guarantee time-analytic solvability of ut=uxx. For u(0,x)=1/(1+x2), any analytic solution at zero would have ∂tku(0,0)=(−1)k(2k)! for every k, and its time Taylor series would have radius zero.

Facts & Assumptions

Given: The heat equation ut=uxx and initial function g(x)=1/(1+x2). An analytic solution is assumed temporarily to derive the forced Taylor coefficients and a contradiction.

[F2]

The geometric series for ratio -x squared converges to the reciprocal when its modulus is below one. (For ∣r∣<1, ∑k≥0rk=1/(1−r), and for ∣r∣≥1 the series diverges).

[F3]

The principal symbol tests whether a normal covector is characteristic. (Characteristic covectors, hypersurfaces, and noncharacteristic data).

Counterexample

1.1givenF1algebraF2

For ∣x∣<1, the geometric identity gives 1/(1+x2)=∑k≥0(−1)kx2k, hence the data are analytic with g(2k)(0)=(−1)k(2k)!. If u were analytic, repeated differentiation of its equation and F1 would give ∂tku=∂x2ku, beginning at k=0 and using ∂t∂x2ku=∂x2kut=∂x2k+2u at each step. Evaluation on the initial trace gives the asserted derivatives.

2.1step 1.1algebraF3∎

The time coefficient is ak=(−1)k(2k)!/k!. For every fixed t nonzero, ∣ak+1tk+1∣/∣aktk∣=2(2k+1)∣t∣→∞. Thus these terms eventually increase by a factor at least two and do not tend to zero. The time series diverges at every t nonzero and cannot represent an analytic germ. The PDE has total order two, with no u_tt term; its principal symbol is ξx2 and vanishes at dt. It therefore lies outside the noncharacteristic CK hypothesis.

Source notes

Gantumur, §4 Example 21, printed p. 11; Ageno §2.4.1, PDF p. 28, specifies the data 1/(1+x²).

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