Solution of Cauchy problem $$\begin{cases}
\frac{\partial u}{\partial t} + u\frac{\partial u}{\partial x} = 0, &x\in\mathbb{R},\ t>0\\
u(x,0) = u_0(x), &x\in\mathbb{R}.
\end{cases}$$
Which choices of the following functions for $u_0$ yield a $C^1$ Solution $u(x,t)$ for all  $x\in\mathbb{R}$  and $t>0$
$1)$  $u_0 (x) = \frac{1}{1+ x^2}$
$2)$  $u_0(x) =  x$
$3)$  $u_0(x) = 1+x^2$
$4)$ $u_0(x) = 1+2x$
My answer key says the answer is $2$ and $4$. My  question is why we should reject option $1$ and $3$?
$$\frac{dt}{1}  =\frac{dx}{u}= \frac{du}{0}$$  
gives me  $u=c_1$  and  $ut-x=c_2$ where $c_1$ and $c_2$ are constants.  Now  by condition given in Option $1$ I get a relation between constants  I.e  
$$c_1= \frac{1}{1+(c_2)^2}.$$ 
Then I have the solution  
$$u = \frac{1}{1+ (ut-x)^2}.$$ 
Then why  we reject option $1$? 
 A: Let us start by solving the problem from scratch and hope the proof will answer your question.
We will solve the problem using the method of characteristics. Observe we have
\begin{align}
\frac{\text{d}}{\text{d}t}u(t, x(t)) = \frac{\partial u}{\partial t}+\frac{\partial u}{\partial x}\frac{\partial x}{\partial t} = \frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x} =0
\end{align}
if we have
\begin{align}
\frac{\partial x}{\partial t} = u(t, x(t)).
\end{align}
Then we have $u(t, x(t)) =\text{const}$ and in particular we have $u(t, x(t)) = u(0, x_0)=u_0(x_0)$.  
Moreover, we see that $x(t) = tu_0(x_0)+x_0$.
Case 1: In this case, we have $u_0(x) = \frac{1}{1+x^2}$ which means
\begin{align}
x = \frac{t}{1+x_0^2}+x_0. 
\end{align}
Sketching the above family of characteristic we see that they all intersect the line $x=t$ which means the formation of shocks. Hence $u_0(x)= (1+x^2)^{-1}$ can't lead to global solutions.
Case 3: In this case, observe we have
\begin{align}
x= t(1+x_0^2)+x_0.
\end{align}
Sketching the above family of characteristic we see that half the characteristics intersects $x=t$ at some positive $t>0$ for all $x_0>0$. Hence again, we have a shock formation. 
I will leave the other two cases for the reader.  
