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.
With the full circle you get contradictory conditions on the same characteristic (e.g.: on $x=y$ must be $u(x,x)=\sqrt 2/2$ and $u(x,x)=-\sqrt 2/2$, obviously it's impossible), so it has not solutions (not "the solution has to be unique")
But you don't fix the problem giving values for a half circle. In this case occurs that the solution is not unique! With these boundary conditions $u$ is not determined along, e.g. $y=x+3$ as it doesn't cross that half circle. So said, the curve is too short to give value for $u$ along every characteristic curve ($y=x+c_1$)
Best Answer
A graphical representation of the problem in the $x$-$t$ plane may be relevant (see this post). The method of characteristic provides the set of lines $x=t-t_0+x_0$ along which $u$ is constant. Here, the value $u=x_0$ is specified along the unit circle ${x_0}^2+{t_0}^2=1$. Characteristics coming from the region $x_0+t_0< 0$ of the unit circle will cross the circle again in the region $x_0+t_0 >0$ where another boundary-value is specified. Hence, the problem is ill-posed.
To avoid this, we restrict the boundary data to the region $x_0+t_0\leq 0$ of the unit circle. One should note that characteristic curves become parallel to the unit circle at the extremities of this half-circle. To see what happens, we express the characteristic curves as $x-t-x_0=-t_0$ with $t_0=\pm\sqrt{1-{x_0}^2}$. Squaring this identity, we have $$ (x-t)^2 -2x_0(x-t) + {x_0}^2=1-{x_0}^2 $$ so that $$ x_0 = \frac{x-t \pm \sqrt{2-(x-t)^2} }{2} = u(x,t), $$ for $t-\sqrt{2} \leq x\leq t+ \sqrt{2}$. At the extremities of the domain, we have $x-t = \pm\sqrt{2}$. Differentiating the expression of $u$ along these curves would require to differentiate a square root in the vicinity of zero, which is not possible.