 Solution

For a regular defining function $\varphi$, the hypersurface is non-characteristic when the characteristic vector has nonzero normal component:For $\varphi =x_3$ this is precisely $a\ne 0$. 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 givesIt 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 $a\to 0^{+}$ 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. There is no regular limit across the initial plane. At $a=0$ that plane is characteristic and the data would need to satisfy $x_2{f}_{x_1}-x_1{f}_{x_2}=f$; the chosen radial data violate this except at the origin, explaining the singular limit.