PDE: Determine the region above the $x$-axis for which there is a classical solution. Consider
\begin{cases}
u_t- 2 u_x = u^2 \\
u(x,0) = g(x)
\end{cases}
Determine the region above the $x$-axis for which there is a classical solution.
Answer: First, we need to solve the IVP. Using the method of characteristics, we get the following IVPS:


*

*$\begin{cases}  \dfrac{\,dt}{\,ds}=1\\ t(r,0)=0 \end{cases}$

*$\begin{cases}  \dfrac{\,dx}{\,ds}=-2\\ x(r,0)=r \end{cases}$

*$\begin{cases}  \dfrac{\,du}{\,ds}=u^2\\ u(r,0)=g(r) \end{cases}$


Solutions of the IVPS: 


*$\begin{aligned}[t]
            \dfrac{\,dt}{\,ds}=1 \\
            \,dt=\,ds \\ 
            \int \,dt=\int \,ds \\ 
            t = s + \mathcal{C}
          \end{aligned}$
where $C$ is a constant. Using $t(r,0)=0$, we find that
$\mathcal{C}=0$. 
Hence $t=s$.

*$\begin{aligned}[t]
        \dfrac{\,dx}{\,ds}=-2 \\
        \,dx=-2\,ds \\
        \int \,dx=-2\int \,ds \\
        x = -2s + \mathcal{C}
        \end{aligned}$
where $C$ is a constant. Using $x(r,0)=r$, we find that
$\mathcal{C}=r$. 
Hence $x=-2s+r$.

*$\begin{aligned}[t]
        \dfrac{\,du}{\,ds}=u^2 \\
        \dfrac{1}{u^2}\,du=\,ds \\
        \int u^{-2}\,du=\int \,ds \\
        -\dfrac{1}{u}=-u^{-1} = s + \mathcal{C} 
        \end{aligned}$
where $C$ is a constant. Using $u(r,0)=g(r)$, we find that
$\mathcal{C}=-1/g(r)$. 
Hence 
$\begin{aligned}[t]
        -\dfrac{1}{u} = s -\dfrac{1}{g(r)} \\
        \dfrac{1}{u} = \dfrac{1}{g(r)}-s \\
        \dfrac{1}{u} = \dfrac{1-s(g(r))}{g(r)} \\
        u = \dfrac{g(r)}{1-s(g(r))}
        \end{aligned}$
Note that $s=t$ (by the $1^\text{st}$ IVP) and $r=x+2t$ (by the $2^\text{nd}$ IVP). Hence 
$$u(x,t)=\dfrac{g(x+2t)}{1-t(g(x+2t))}$$
Next, we need to determine the region above the $x$-axis for which this solution is a classical solution. How would I determine this region?
 A: Next, we need to determine the region above the $x$-axis for which this solution is a classical solution 
Recall that a classical solution is a solution which is differentiable as many times as needed if you want to plug the function into the PDE (for example, if the PDE contains the term $u_{xxxx}$, then the fourth derivate $u_{xxxx}$ must exist in order for $u$ to be a classical solution).
In our case, we need to show that the first derivatives $u_t$ and $u_x$ exists.
\begin{equation*}
\begin{aligned}
u_x(x,t) & =\dfrac{[1-tg(x+2t)]g_x(x+2t)-g(x+2t)[-tg_x(x+2t)]}{(1-tg(x+2t))^2} =\dfrac{g_x(x+2t)}{(1-tg(x+2t))^2} \\
u_t(x,t) & =\dfrac{[1-tg(x+2t)]2g_t(x+2t)-g(x+2t)[-2tg_t(x+2t)]}{(1-tg(x+2t))^2} =\dfrac{2g_t(x+2t)}{(1-tg(x+2t))^2}
\end{aligned}
\end{equation*}
Recall that $g(x)\in C^1(\mathbb{R})$ (i.e. $g_x$ and $g_t$ exists). Note that $u$, $u_x$ and $u_t$ exists only if $$1-tg(x+2t)\neq 0 \iff 1\neq tg(x+2t).$$
If $t=0$, then $1=\neq 0$. Hence our solution
$$u(x,0)=g(x),$$ which we know from the original problem.
If $t\neq 0$, then $g(x+2t)\neq 1/t$. Note that $g$ takes a value of $x$ and $t$ but only returns $1/t$. Hence the region above the $x$-axis for which $$u(x,t)=\dfrac{g(x+2t)}{1-t(g(x+2t))}$$
is a classical solution is
\begin{cases}
\mathbb{R} & \text{ when } t=0\\
\mathbb{R}\backslash \{1/t: \forall t>0\} & \text{ when } t>0
\end{cases}
