How to look for the given Cauchy Problem? Let $u=u(x,t)$ be the solution of the Cauchy problem 
$$\frac{\partial u}{\partial t}+ \bigg(\frac{\partial u}{\partial x}\bigg)^2=1 \quad x\in \mathbb{R},t>0$$
$$u(x,0)=-x^2 \quad \quad x \in \mathbb{R}$$
Then 


*

*$u(x,t)$ exists for all $x \in \mathbb{R}$ and $t>0$

*$|u(x,t)| \to \infty$ as $t \to t^*$ for some $t^*>0$ and $x\neq 0$.

*$u(x,t)\leq 0$ for all $x \in \mathbb{R}$ and for all $t<\frac{1}{4}$

*$u(x,t)>0$ for all $x \in \mathbb{R}$ and for all $0<t<\frac{1}{4}$
As- 2, 4
From Charpit's Auxiliary equation, we have $$u(x, t) = cx+(1-c^2)t$$ But from this solution on putting $u(x, 0) = -x^2$ constant $c$ comes out to be variable but $c$ is a constant. Where i am wrong?
 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. $v=F(x-2vt)$
$v(x,0)=-2x$ :
$F(x)=-2x$
$\therefore v=-2(x-2vt)$
$v=-2x+4vt$
$v=\dfrac{2x}{4t-1}$
$u_x=\dfrac{2x}{4t-1}$
$u(x,t)=\dfrac{x^2}{4t-1}+g(t)$
$u_t=-\dfrac{4x^2}{(4t-1)^2}+g_t(t)$
$\therefore-\dfrac{4x^2}{(4t-1)^2}+g_t(t)+\dfrac{4x^2}{(4t-1)^2}=1$
$g_t(t)=1$
$g(t)=t+C$
$\therefore u(x,t)=\dfrac{x^2}{4t-1}+t+C$
$u(x,0)=-x^2$ :
$C=0$
$\therefore u(x,t)=\dfrac{x^2}{4t-1}+t$
A: $$\begin{cases}
\frac{\partial u}{\partial t}+ \bigg(\frac{\partial u}{\partial x}\bigg)^2=1 \\
u(x,0)=-x^2 
\end{cases}$$
Change of function :$\quad u(x,t)=v(x,t)+t \quad\to\quad \frac{\partial v}{\partial t}+ \bigg(\frac{\partial v}{\partial x}\bigg)^2=0$
Method of separation of variables : $\quad v(x,t)=f(x)g(t) \quad\to\quad
fg'+f'^2g^2=0$
$$\frac{f'^2}{f}=-\frac{g'}{g^2}=\lambda \quad\to\quad \begin{cases}
g(t)=\frac{1}{\lambda t+C} \\
f(x)=\frac{\lambda}{4}x^2
\end{cases}$$
$$v(x,t)=\frac{\frac{\lambda}{4}x^2}{\lambda t+C}=\frac{x^2}{4 t+(4C/\lambda)}$$
$$u(x,t)=t+\frac{x^2}{4 t+(4C/\lambda)}$$
With condition $u(x,0)=x^2=0+\frac{x^2}{0+(4C/\lambda)} \quad\to\quad (4C/\lambda)=1$
$$u(x,t)=t+\frac{x^2}{4 t+1}$$
