Solving or knowing something about a non-linear PDE which is "almost" linear? Let $a>0$ be fixed. I have the following PDE: $u=u(t,x)$, $t\in [0,1]$, $x\in \mathbb{R}$,
$$-\partial_t u = |\partial_x u| + \frac{1}{2}\partial_x^2 u, \quad u(1,x)=\delta_a(x)-\delta_{-a}(x),\quad \lim_{|x|\to \infty}u(t,x)=0.$$
Observe the absolute value. If there was no absolute value the PDE would be linear parabolic and a solution is possible.
The equation is not linear but "almost". Is there any trick or way to find or have an intuition of the solution? Actually, I am only interested in the behaviour of $u$ in the region $\mathbb{R}\setminus [-a,a]$.
I would like to know whether $\partial_x u(t,x)<0$ on $\mathbb{R}\setminus [-a,a]$ for every $t\in [0,1]$. In other words, whether $x\mapsto u(t,x)$ is decreasing on $\mathbb{R}\setminus [-a,a]$ for every $t\in [0,1]$.
Any ideas? Thanks :D
 A: As a rule, all the problems with the modules to be addressed by intervals:
$$u=\begin{cases}
\dot u,\text{ if }\dot u_x>0\\
\ddot u, \text{ if }\ddot u_x<0,
\end{cases}$$
$$-\dot u_t= \dot u_x +\dfrac12\dot u_{xx},$$ 
$$-\ddot u_t= -\ddot u_x +\dfrac12\ddot u_{xx},$$ 
The peculiarity of PDE case is that at the points of transition through zero of the module should be "stapled" solutions obtained by means of boundary conditions of the form
$$\begin{cases}\dot u_y(x_i, t_i\pm0) = \ddot u_y(x_i, t_i\mp0)\\
\dot u(x_i\pm0, t_i) = \ddot u(x_i\mp0, t_i)\\
\dot u_x(x_i\pm0, t_i) = \ddot u_x(x_i\mp0, t_i)\end{cases}.$$
Thus, the domain is typed solutions from several fragments, united by the conditions of "stapling".
The problem relates to a parabolic type, in which the solution for a certain time $t$ depends only on past values and it does not depend on the subsequent ones. This makes it possible to break the task into a series of time steps, each of which is easy to demarcate areas $\dot u$ and $\ddot u$. This means that the problem allows a simple implementation on computing grids.
