Determine solutions of the Jacobi-Hamilton problem $u_{t}+|u_x|^{2}=0$ How determine solutions of the initial value problem, $$u_{t}+|u_x|^{2}=0\qquad \mbox{in } \mathbb{R}\times(0,\infty)$$
With condition $u=0$ on $\mathbb{R}\times\{t=0\}$. Clearly one solution is $u(x,t)=0$ (as in the answers), but how determine another solution? My teacher say that there exits the following lipschitz continuous solution a.e.
$
u^{*}(x,t):=
\begin{cases}
 0&\text{if}\, |x|\geq t\\
 |x|-t&\text{if}\, |x|\leq t\\
\end{cases}
$ 
But the true is, I don't know how obtain this lipschitz continuous solution, that solves the pde a.e.
So, How I determine the solution $u^{*}$?
Thanks! 
 A: For first thing you can search solutions of the type:
$u(x,t)= \alpha(x)+\beta(t)$
In this case the equation is
$\beta_t(t)+\alpha^2_x(x)=0$
and so
$\beta_t(t)=-\alpha^2_x(x)$
In particular you can choose, for example, x=0 and you have that
$\beta_t(t)=-\alpha^2_x(0)=cost$
and so
$\beta(t)=at+b$ with a$,b\in\mathbb{R}$
Now you must resolve
$\alpha^2_x(x)=-a$
You can observe that $a\leq 0$ and in this case
$\alpha_x(x)=\sqrt(-a)$
And the solution is 
 $\alpha(x)=\sqrt(-a) x+c$
Now for every $x$ you have that $u(x,0)=0$  and  you have that 
$\sqrt(-a) x+c=b $
For x=0 you have that $c=b$ and so for $x\neq0$ you have  $ \sqrt(-a)=0$ and so there your solution is $u(x,t)=0$.
In other words the unique solution of the form $u(x,t)= \alpha(x)+\beta(t)$ is the constant function zero.
You can prove in general that the constant function zero is the unique solution of your problem.
A: Differentiate the equation w.r.t $x$ and substitute $v = u_x$ to get a first-order problem
$$ v_t + 2vv_x = 0 $$
This can be solved using the method of characteristics to get an implicit solution
$$ v = f(x - 2vt) $$
The initial condition gives
$$ v(x,0) = u_x(x,0) = f(x) = 0 $$
Therefore $u_x\equiv 0$, which implies $u\equiv 0$ is the only solution
A: The function $u^{\ast}:=\min\left(\left|x\right|-t,0\right)$ is differentiable
whenever $\left|x\right|\not=t$ and $x\not=0$. Hence it is differentiable
almost everywhere and we have
$$
\partial_{t}u^{\ast}(x,t)=\begin{cases}
0, & \left|x\right|>t\\
-1, & \left|x\right|<t
\end{cases},\,\partial_{x}u^{\ast}(x,t)=\begin{cases}
0, & \left|x\right|>t\\
-1, & t<x<0\\
1, & 0>x>t
\end{cases}.
$$
These satisfy $\partial_{t}u^{\ast}(x,t) +\left|\partial_{x}u^{\ast}\right(x,t)|^{2}=0$ where they are defined.
