How to solve the following Cauchy problem. Solve the following Cauchy problem for a first order PDE:
$$(2x_1 + x_2)u_{x_1} + (x_2 + 1)u_{x_2} = u^2, \ \  u(x_1, 1) = x_1^2 + 1, \ \ x_1 \ge 0, x_2 \ge 1$$
and find an implicit conldition over $x_1$ and $x_2$ under which this Cauchy problem has a bounded solution.
Attempt at the problem:
Our characteristic ODEs are:
\begin{align}
\frac{dx_1}{dt} &= 2x_1 +x_2  &x_1(s, 0) =& s\\
 \frac{dx_2}{dt} &= x_2 + 1 &x_2(s, 0) = 1 \\
\frac{dx_3}{dt} &= x_3^2   &x_3(s, 0) = s^2 + 1 \\
\end{align}
which have solutions:
\begin{align}
x_1(s, t) &= (s+\frac{3}{2})e^{2t} - 2 e^t+\frac{1}{2} \\
x_2(s, t) &= 2e^t - 1 \\
x_3(s, t) &= \frac{1}{\frac{1}{s^2 + 1} - t}\\
\end{align}
Which implies:
\begin{align}
s &= \frac{4x_1 + 4x_2 + 2}{(x_2 + 1)^2} - \frac{3}{2} \\
t &= \ln\frac{x_2+1}{2}
\end{align}
Substituting $s,t$ into $x_3$ gives a complex expression that seems wrong? Is this the correct way to go about solving this problem?

Consider writing your system as follows
$$\left( 2x+t \right){{u}_{x}}+\left( t+1 \right){{u}_{t}}={{u}^{2}}$$   where   $$u\left( x,1 \right)={{x}^{2}}+1,\,\,t\ge 1$$ 
Let $u\left( x,t \right)=u\left( x\left( s \right),t\left( s \right) \right)\Rightarrow \frac{du}{ds}=\frac{\partial u}{\partial x}\frac{dx}{ds}+\frac{\partial u}{\partial t}\frac{dt}{ds}$  .  
Equating ‘coefficients’ against ${{u}_{t}}$implies
$$\frac{dt}{ds}=t+1\Rightarrow \ln \left( \frac{t\left( s \right)+1}{t\left( 0 \right)+1} \right)=s$$  where $t\left( 0 \right)=1$, so $t\left( s \right)=2{{e}^{s}}-1$.
We also have by the same process
$$\frac{dx}{ds}=2x+t=2x+2{{e}^{s}}-1$$
Or
$$\frac{dx}{ds}-2x=2{{e}^{s}}-1$$
which we can write as 
$$\frac{d}{ds}\left( x{{e}^{-2s}} \right)=\left( 2{{e}^{s}}-1 \right){{e}^{-2s}}$$
Integrating
$$x\left( s \right){{e}^{-2s}}-x\left( 0 \right)=\frac{1}{2}{{e}^{-2s}}-2{{e}^{-s}}+\frac{3}{2}$$
Or
$$x\left( 0 \right)=x\left( s \right){{e}^{-2s}}+2{{e}^{-s}}-\frac{1}{2}{{e}^{-2s}}-\frac{3}{2}$$
which is constant along the characteristics.  Now we also have
$$\frac{du}{ds}={{u}^{2}}\Rightarrow \frac{1}{u\left( x\left( 0 \right),t\left( 0 \right) \right)}-\frac{1}{u\left( x\left( s \right),t\left( s \right) \right)}=s$$
Or
$$u\left( x\left( s \right),t\left( s \right) \right)=\frac{u\left( x\left( 0 \right),t\left( 0 \right) \right)}{1-su\left( x\left( 0 \right),t\left( 0 \right) \right)}$$ 
Note the initial condition being $u\left( x\left( 0 \right),t\left( 0 \right) \right)=u\left( x\left( 0 \right),1 \right)=x{{\left( 0 \right)}^{2}}+1$, therefore
$$u\left( x\left( s \right),t\left( s \right) \right)=\frac{x{{\left( 0 \right)}^{2}}+1}{1-s\left( x{{\left( 0 \right)}^{2}}+1 \right)}$$ 
Now from $t\left( s \right)=2{{e}^{s}}-1\Rightarrow s=\ln \left( \frac{1}{2}\left( t+1 \right) \right)$.  we have
$$x\left( 0 \right)=\frac{2\left( 2x-1 \right)+4\left( t+1 \right)}{{{\left( t+1 \right)}^{2}}}-\frac{3}{2}$$ 
Substituting we obtain finally
$$u\left( x,t \right)=\frac{1+{{\left( \frac{2\left( 2x-1 \right)+4\left( t+1 \right)}{{{\left( t+1 \right)}^{2}}}-\frac{3}{2} \right)}^{2}}}{1-\left( 1+{{\left( \frac{2\left( 2x-1 \right)+4\left( t+1 \right)}{{{\left( t+1 \right)}^{2}}}-\frac{3}{2} \right)}^{2}} \right)\ln \left( \frac{1}{2}\left( t+1 \right) \right)}$$ 
 Which is a great big mess (pretty much anyway you write it) but can be verified to solve the initial condition and the PDE (I suggest you use a CAS to verify it).
