Solving this non-linear PDE (which reminds of a linear parabolic PDE)

Problem: consider the following PDE: $$-u_t=\mbox{sign}(u) u_x+ \frac{1}{2}u_{xx},$$ with some boundary condition $u(T,x)=\delta_a(x)-\delta_{-a}(x)$, $a>0$ fixed, being $\mbox{sign}(u)\in \{-1,1\}$ the sign of $u$ and $\delta$ the Dirac-function. The final condition could also be taken general $u(T,x)=\Phi(x)$ with $\Phi$ regular if it helps.

Here, $u$ is a function $u:[0,T]\times \mathbb{R}\rightarrow \mathbb{R}$.

Observe that if we change $\mbox{sign}(u)$ by simply a function $f(x)$ of $x$, this would make things easier or even if $f$ was constant, a semi-explicit solution using series could be found. However, $\mbox{sign}(u)$ is in some sense "simple" since it just changes sign from -1 to 1 and so on.

Question: Could there be a chance to find a (semi-explicit) solution or at least, properties of the sign of $u$? i.e. the regions where $u$ is positive, negative, etc. Is there any trick one could use for this specific kind of PDE?

Thanks for any tips or ideas!

• For $u(T,x)=\delta_y(x)$ the solution $u$ will be the fundamental solution of the corresponding parabolic equation $u_t=u_x+ \frac{1}{2}u_{xx}$ with unversed time since it is positive for $t>0$. – Andrew May 13 '16 at 20:05
• Yes you are right, I thought about that as well, if you start positive then you remain positive and then the equation can be reduced to the heat equation by tranforming the function. Nevertheless, I have a more complex starting condition, namely $u(T,x)=\delta_{a}-\delta_{-a}$, $a>0$ fixed. Just imagine we approach this initial condition by a function $\Phi$ which is positive on the right, negative on the left and $\Phi(0)=0$. Would a similar argument work then as well? – Martingalo May 13 '16 at 20:58
• Derivatives $u_{xx}$ and $u_t$ cannot be both continuous at points where $u=0$ and $u_x\ne0$. How the equation is understood then? – Andrew May 13 '16 at 21:43
• Hopefully $u=0$ on a null measure set and the equation might be understood in a weak sense indeed, due to the discontinuities popping up from $\mbox{sign}(u)$. – Martingalo May 13 '16 at 21:45
• Let $G(x,y,t)$ be the Green's function of the first BVP $u_t=u_x+ \frac{1}{2}u_{xx}$, $x>0$, $u|_{x=0}=0$. Put $u(x,t)=G(x,a,t)$ for $x\ge0$ and Put $u(x,t)=-G(-x,a,t)$ for $x<0$. Then $u$ satisfy the initial condition. Also $u$ is positive for $x>0$, $t>0$; $u_{xx}$ and $u_t$ change sign after the mapping $x\to-x$, but $u_x(-x,t)=u(x,t)$. So $u$ satisfy the equation when $x\ne0$. But $u_{xx}$ is not continuous on the line $x=0$. – Andrew May 13 '16 at 22:09

Green's function generally is defined in the corresponding domain only. So what I meant can be written as $u(t,x)=G^+(x,a,t)1_{\{x>0\}}-G^-(-x,a,t)1_{\{x<0\}}$, where $G^+(x,a,t)$ is the solution of $−u_t=u_x+1/2u_{xx}$, $x>0$ and $u(T,x)=δ_a$, $u|_{x=0}=0$; $G^−(x,a,t)$ is the solution of $−u_t=-u_x+1/2u_{xx}$ and $u(T,x)=δ_a$, $u|_{x=0}=0$.
The solution $u$ and its derivatives $u_x$ and $u_t$ are continuous for $t<T$ but, as I've mentioned in the comments, $u_{xx}$ is not continuous on the line $x=0$.