A smooth nonanalytic solution of a first-order PDE

Smoothness. The map $x\mapsto x_1$ is linear, hence smooth, and $\beta$ is smooth by [F2]; the composition $u=\beta \circ (x\mapsto x_1)$ is therefore smooth on ${\mathbb{R}}^n$, with ${\mathrm{\partial }}_{x_2}u(x)={\beta }^{\mathrm{\prime }}(x_1)\cdot 0=0$ and more generally $D^{\alpha }u(x)={\beta }^{(\mathrm{\mid }\alpha \mathrm{\mid })}(x_1)$ if $\alpha ={\alpha }_1{e}_1$ and $D^{\alpha }u(x)=0$ whenever $\alpha$ has a nonzero entry outside the first coordinate.

A first-order PDE. Since $u(x)$ depends on $x$ only through the first coordinate, ${\mathrm{\partial }}_{x_2}u\equiv 0$ on ${\mathbb{R}}^n$; thus $u$ solves the first-order linear equation ${\mathrm{\partial }}_{x_2}u=0$, which is not the Laplace equation.

Vanishing Taylor coefficients. Let $\alpha$ be any multi-index. If ${\alpha }_1=\mathrm{\mid }\alpha \mathrm{\mid }$ then $D^{\alpha }u(0)={\beta }^{(\mathrm{\mid }\alpha \mathrm{\mid })}(0)=0$ by [F2]; otherwise $D^{\alpha }u\equiv 0$ by step 1.1 and again $D^{\alpha }u(0)=0$. Hence every coefficient $c_{\alpha }=D^{\alpha }u(0)\mathrm{/}\alpha !$ of the Taylor expansion of $u$ at the origin vanishes, so the only candidate series is the zero series.

Failure of the representation. For every $\delta >0$ and every $0<x_1<\delta$ we have $u(x_1{e}_1)=\beta (x_1)=\exp (-1\mathrm{/}{x}_1)>0$ by [F1], while the candidate series of step 2.2 sums to $0$; hence no neighbourhood of the origin carries a power-series representation of $u$. By [F3], $u$ is not real analytic at the origin.

Steps 1.1, 2.1 and 3.1 exhibit a $C^{\mathrm{\infty }}$ function that solves a PDE and is not real analytic at a point, so smoothness of a solution does not imply real analyticity for general partial differential equations; the harmonic conclusion of the companion page uses the Laplace equation, not smoothness alone.