Smooth data do not force an analytic solution

For x nonzero define polynomials recursively by $P_0(z)=1$ and $P_{m+1}(z)=-z^2{P}_m^{\mathrm{\prime }}(z)+2z^3{P}_m(z)$. F1–F3 show by successive differentiation that $g^{(m)}(x)=P_m(1\mathrm{/}x)\exp (-1\mathrm{/}{x}^2)$ off zero. Indeed $d(1\mathrm{/}x)\mathrm{/}dx=-1\mathrm{/}{x}^2$ and $d(-1\mathrm{/}{x}^2)\mathrm{/}dx=2\mathrm{/}{x}^3$, giving exactly that recurrence.

With $y=1\mathrm{/}{x}^2\to \mathrm{\infty }$, the absolute value of any polynomial in 1/x, and of that polynomial divided by x, is bounded by a constant times an integer power of y for y at least one. F4 makes both products with $\exp (-y)$ tend to zero. Starting with the continuity of g at zero, induction now gives $g^{(m)}(0)=0$: if the mth derivative equals the expression of step 1.1 off zero and is zero at zero, its difference quotient tends to zero, so the next derivative at zero exists and is zero. Its continuity follows from the same bound. Thus g is smooth and all its Taylor coefficients at zero vanish.

F5 gives g(x)>0 for x nonzero, arbitrarily close to zero. F6 therefore prevents g from being analytic at zero: its zero Taylor series could not equal those positive values. The function u(t,x)=g(x) is smooth, has u_t=0 and the required trace. If an analytic solution existed, substituting t=0 in its convergent two-variable series would give a convergent series for g with its derivative coefficients, contradicting the preceding conclusion. Hence no analytic germ has those data.