Conclusions About Solution of Cauchy Problem 
Let $u=u(x,t)$ be the solution of the Cauchy problem
  $$\frac{\partial u}{\partial t}+\left(\frac{\partial u}{\partial x}\right)^2=1, x\in\Bbb R, t>0$$
  $$u(x,0)=-x^2$$
Then which of the following is/are true?
  
  
*
  
*$u(x,t)$ exists for all $x\in\Bbb R$ and $t>0$.
  
*$|u(x,t)|\to\infty$ as $t\to t^*$ for some $t^*>0$ and $x\ne0$.
  
*$u(x,t)\le0$ for all $x\in\Bbb R$ and for all $t<\frac14$.
  
*$u(x,t)>0$ for all $x\in\Bbb R$ and for all $0<t<\frac14$.
  

Attempt: I tried to solve the problem, and hence derive conclusions.
Using Charpit's
$$\frac{dt}1=\frac{dx}q=\frac{du}1=\frac{dp}0=\frac{dq}0$$
I got $p=c_1,q=c_2$, and leads to $u=cx+(1-c^2)t+d$ and I'm clueless about how to plug in initial condition.
Thanks in advance.
 A: Hint:
$\dfrac{\partial u}{\partial t}+\left(\dfrac{\partial u}{\partial x}\right)^2=1$
$\dfrac{\partial^2u}{\partial t\partial x}+2\dfrac{\partial u}{\partial x}\dfrac{\partial^2u}{\partial x^2}=0$
Let $v=\dfrac{\partial u}{\partial x}$ ,
Then $\dfrac{\partial v}{\partial t}+2v\dfrac{\partial v}{\partial x}=0$ with $v(x,0)=-2x$
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dt}{ds}=1$ , letting $t(0)=0$ , we have $t=s$
$\dfrac{dv}{ds}=0$ , letting $v(0)=v_0$ , we have $v=v_0$
$\dfrac{dx}{ds}=2v=2v_0$ , letting $x(0)=f(v_0)$ , we have $x=2v_0s+f(v_0)=2vt+f(v)$ , i.e. $v=F(x-2vt)$
$v(x,0)=-2x$ :
$F(x)=-2x$
$\therefore v=-2(x-2vt)$
$v=-2x+4vt$
$v=\dfrac{2x}{4t-1}$
$u_x=\dfrac{2x}{4t-1}$
$u(x,t)=\dfrac{x^2}{4t-1}+g(t)$
$u_t=-\dfrac{4x^2}{(4t-1)^2}+g_t(t)$
$\therefore-\dfrac{4x^2}{(4t-1)^2}+g_t(t)+\dfrac{4x^2}{(4t-1)^2}=1$
$g_t(t)=1$
$g(t)=t+C$
$\therefore u(x,t)=\dfrac{x^2}{4t-1}+t+C$
$u(x,0)=-x^2$ :
$C=0$
$\therefore u(x,t)=\dfrac{x^2}{4t-1}+t$
