= Solution
Parameterize the initial line by $(s,2)$ and use characteristic time $\tau$. The <characteristic equations for a transport equation> integrate to
$$
x_1=se^{2\tau},\qquad x_2=2e^{4\tau},\qquad U=\frac{h}{1-h\tau},\qquad \tau<1/h.
$$
Here $dU/d\tau=U^2$ and $U(0)=h$. Since $\tau=\tfrac14\log(x_2/2)$, the solution is
$$
\boxed{u(x_1,x_2)=\frac{h}{1-\frac h4\log(x_2/2)},\qquad (x_1,x_2)\in\mathbb R\times(0,2e^{4/h}).}
$$
Direct differentiation gives $u_{x_1}=0$, $u_{x_2}=u^2/(4x_2)$, verifying the equation and the initial values. The map $(s,\tau)\mapsto(x_1,x_2)$ has determinant $8e^{6\tau}\ne0$, so every point of this strip is reached exactly once.
At the upper boundary $u\to+\infty$, a <finite-time blow-up of an ordinary differential equation>. At the lower boundary $u\to0$, but $u_{x_2}=u^2/(4x_2)$ diverges as $x_2\downarrow0$, so no $C^1$ continuation through that boundary is possible. Thus the displayed strip is the maximal connected domain containing the initial line on which this classical Cauchy solution exists.
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