$u_t+u_x^2=t$
$u_{tx}+2u_xu_{xx}=0$
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. $v=F(x-2vt)$
$v(x,0)=0$ :
$F(x)=0$
$\therefore v=0$
$u_x=0$
$u(x,t)=g(t)$
$u_t=g_t(t)$
$\therefore g_t(t)=t$
$g(t)=\dfrac{t^2}{2}+C$
$\therefore u(x,t)=\dfrac{t^2}{2}+C$
$u(x,0)=0$ :
$C=0$
$\therefore u(x,t)=\dfrac{t^2}{2}$
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.
Best Answer
The equation $u_t+x^2u_x = 0$ asserts that $\langle \nabla u(t,x), (1,x^2) \rangle = 0$, and so $u$ is constant along the curves determined by $$\frac{{\rm d}x}{{\rm d}t} = \frac{x^2}{1} = x^2.$$Hence $$\frac{1}{x^2}\frac{{\rm d}x}{{\rm d}t} = 1 \implies -\frac{1}{x} = t+c \implies x = \frac{-1}{t+c},$$great. So $$u(t,x) = u\left(t, \frac{-1}{t+c} \right) = u(0,-1/c) = \cos(-1/c).$$Since $$x = \frac{-1}{t+c} \implies c = -t-\frac{1}{x} = \frac{-tx-1}{x} \implies -\frac{1}{c} = \frac{x}{1+tx},$$we get $$u(t,x) = \cos\left(\frac{x}{1+tx} \right).$$The condition $u(t,0) = 1$ is satisfied. This won't be defined for all $x \in \Bbb R$ unless $t=0$ (you'll always have trouble when $1+tx=0$). When $x \neq 0$ and $t \to \pm \infty$ it's clear that $u(t,x) \to 1$.