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.
For the first method. You get two (uniparametric) families of surfaces ($\\u+\dfrac{1}{x}=c_1$ and $\dfrac{u^2}{2}-y=c_2$), the intersection of wich is a (two parameter) family of curves. Consider some, fixed by now, values of $c_1$ and $c_2$. They determine a curve into the surface of a solution brung about by some boundary conditions. So is, $c_1$ is not free, it has to correspond to some $c_2$ to make the curve pass for some point the boundary conditions impose. But it has to be so for every value of $c_1$ or, in other words, there must exist some functional relation between them: $c_1=f(c_2)$.
The second method works almost the same, but in this case we work directly with parametric equations for the surface (two parameters and three coordinates). Your equations are ok, but not the solution. It must be:
$$x(t)=\frac{-1}{t+x_0}\\u(t)=t+u_0\\y(t)=\frac{(t+u_0)^2}{2}+y_0$$
We can set $x_0=0$ and operate, yelding:
$$x(t)=\frac{-1}{t};t=\frac{-1}{x}\\u=t+u_0=-\dfrac{1}{x}+u_0;u_0=u+\dfrac{1}{x}\\y=\frac{(t+u_0)^2}{2}+y_0=\dfrac{u^2}{2}+y_0;y_0=-\left(\dfrac{u^2}{2}-y\right)$$
We have now exactly a family of curves depending on two parameters. The same argument used in the first method works here: we have some functional relation (too brung about by the boundary conditions) between these parameters: $u_0=g(y_0)$. And we recover the same general solution as with the first method: $u+\dfrac{1}{x}=g\left(\dfrac{u^2}{2}-y\right)$
In this case, the boundary conditions are $u(x,1-x)=0$, enough to determine $f$ in $u+\dfrac{1}{x}=f\left(\dfrac{u^2}{2}-y\right)$
$$\dfrac{1}{x}=f(-(1-x))$$
Making $w=-(1-x)$, we get $x=w+1$ and $f(w)=\dfrac{1}{w+1}$
Finally $u+\dfrac{1}{x}=\dfrac{1}{\dfrac{u^2}{2}-y+1}$
Best Answer
Well, you need to use the initial data, i.e. the values of $u(s)|_{s=0}$: $$ u(x(s),t(s))|_{s=0}=u(x_0,0)=\cosh(2x_0), $$ which gives $t(s)=s$, $x(s)=cs+x_0$ and $u(s)=ks+\cosh(2x_0)$. The latter rewrites as $$ u(x,t)=kt+\cosh(2(x-ct)) $$ by using $s=t$ and $x_0=x-ct$.