Modifying the problem. Rather than consider the PDE $u_{xt}+u\ u_{xx}+\frac{1}{2}u_{x}^2=0$ with initial condition $u(x,0)=u_0(x)$ as asked above, I will consider the following variant.
$$
\text{Solve }u_{xy}+u\ u_{xx} + u_x^2=0\text{ subject to }u(x,0)=f(x).\qquad(\star)
$$
There are three differences between this question and that which was asked originally.
- The coefficient of $u_x^2$ has changed from $\frac{1}{2}$ to $1$.
- The variable $t$ has been renamed to $y$.
- The initial function $u_0(x)$ has been renamed to $f(x)$.
Only (1) represents a significant modification of the problem. It makes the solution more tractable and enables it to be found using an elementary application of the method of characteristics. For these reasons, it is conceivable that this was the intended question.
Note: I will not delve into regularity of the solutions in this answer.
Reduction to a first order quasilinear PDE. Write the equation as
$$
\frac{\partial}{\partial x}\left(u_y+u\ u_x\right)=0.
$$
Thus $(\star)$ is equivalent to
$$
u_y+u\ u_x=g(y),\qquad u(x,0)=f(x),\qquad (\star\star)
$$
where $g(y)$ is an arbitrary function of $y$ (with sufficient regularity).
Method of characteristics.
Perhaps the simplest formulation of the method of characteristics is for quasilinear first order PDEs. These are PDEs of the form
$$a(x,y,u)u_x+b(x,y,u)u_y=c(x,y,u).$$
To solve this equation, one regards the solution as a surface $z=u(x,y)$ in $xyz$-space. Let $s$ parametrize the initial curve $\bigl(s,0,f(s)\bigr)$ and let $t$ be a second parameter, which can be thought of as the distance flowed along a characteristic curve emanating from $\bigl(s,0,f(s)\bigr)$.
The characteristic equations are then
$$
\frac{dx}{dt}=a(x,y,z),\quad \frac{dy}{dt}=b(x,y,z),\quad \frac{dz}{dt}=c(x,y,z).
$$
Returning to our equation $(\star\star)$, this reduces to $a(x,y,u)=u$ and $b(x,y,u)=1$ and $c(x,y,u)=g(y)$.
Thus
$$
\frac{dx}{dt}=z,\quad \frac{dy}{dt}=1,\quad \frac{dz}{dt}=g(y)
$$
with initial conditions $x(0)=s$ and $y(0)=0$ and $z(0)=f(s)$.
The solution to this system is
$$
x=s+zt,\quad y=t,\quad z=f(s)+h(t),
$$
where $h(t)$ is the antiderivative of $g(t)$ satisfying $h(0)=0$. Since $g$ was arbitrary, so is $h$ given $h(0)=0$.
The solution. Now we eliminate all occurrences of $t$ by replacing them with $y$, then eliminate $s$ by writing $s=x-zy$. Finally, replace $z$ with $u$ to obtain the implicit equation
$$
\boxed{u=f(x-uy)+h(y)},
$$
where $h(y)$ is any sufficiently regular function satisfying $h(0)=0$. This is an implicit equation for the general solution of $(\star)$.
TL;DR. Change the $\frac{1}{2}$ in the original question to $1$ to obtain a PDE solvable by the method of characteristics.
Characteristic lines are paths that carry information. If two characteristics collide, there will be two contradictory accounts of the solution there, and the continuum description will break down. At that point, usually $\partial_xu$ becomes infinite.
From the solution given, we know that along a characteristic $$\frac{dx}{dt}=u,\qquad\frac{du}{dt}=-\frac{u}{2}$$ so that the location of the characteristic originating at $x_0$ is
$$x=x_0+2u_0(1-\exp{(-t/2)})$$
and$$\frac{dx}{dx_0}=1+2\frac{du_0}{dx_0}(1-\exp{(-t/2)})$$
We want to know the time and place where this first vanishes. This will be where $\frac{du_0}{dx_0}$ takes its most negative value, of $-1$ at $x_0=\pi$, and when
$$1-2(1-\exp{(-t/2)})=0$$ which gives you $t_c$.
It might be instructive if you used Mathematica or something to plot the characteristic paths. You would see how they intersect and how they then produce a double covering of the plane.
Best Answer
$u_t+u_x^2=t$
$u_{tx}+2u_xu_{xx}=0$
Let $v=u_x$ ,
Then $v_t+2vv_x=0$ with $v(x,0)=0$
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. $v=F(x-2vt)$
$v(x,0)=0$ :
$F(x)=0$
$\therefore v=0$
$u_x=0$
$u(x,t)=g(t)$
$u_t=g_t(t)$
$\therefore g_t(t)=t$
$g(t)=\dfrac{t^2}{2}+C$
$\therefore u(x,t)=\dfrac{t^2}{2}+C$
$u(x,0)=0$ :
$C=0$
$\therefore u(x,t)=\dfrac{t^2}{2}$