= Solution
For a regular defining function $\phi$, the hypersurface is <non-characteristic> when the characteristic <vector> has nonzero normal component:
$$
\boxed{x_2\phi_{x_1}-x_1\phi_{x_2}+a\phi_{x_3}\ne0\quad\text{on }S_\phi.}
$$
For $\phi=x_3$ this is precisely $a\ne0$. The defining function must also have nonzero gradient so that it actually defines a hypersurface.
The <method of characteristics> gives $\dot x_1=x_2$, $\dot x_2=-x_1$, $\dot x_3=a$, $\dot u=u$. Starting at $(\xi_1,\xi_2,0)$, the first two coordinates rotate clockwise through characteristic time $s$ and $x_3=as$. Solving back for the initial point gives
$$
\boxed{u(x_1,x_2,x_3)=e^{x_3/a}f\left(x_1\cos\frac{x_3}a-x_2\sin\frac{x_3}a,\ x_1\sin\frac{x_3}a+x_2\cos\frac{x_3}a\right).}
$$
It meets the data and is $C^1$ when $f$ is $C^1$.
For the rotationally invariant data, $u=(x_1^2+x_2^2)e^{x_3/a}$. As $a\to0^+$ it tends to zero for fixed $x_3<0$, diverges to positive infinity for fixed $x_3>0$ and nonzero radius, and retains the initial value on $x_3=0$. It is identically zero on the vertical axis. \b[There is no regular limit across the initial plane.] At $a=0$ that plane is characteristic and the data would need to satisfy $x_2f_{x_1}-x_1f_{x_2}=f$; the chosen radial data violate this except at the origin, explaining the singular limit.
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