Consider a terminal-boundary value problem for $v: (t,x,y)\in [0,T]\times \mathbb R^2_+\to \mathbb R\ni v(t,x,y)$:

$$ \begin{cases} v_t + \max(v_x,v_y)+ \frac 1 2 (v_{xx}+v_{yy})=0, & \forall (t,x,y)\in (0,T)\times (0,\infty)^2\\ \\ v(T,x,y)={\bf 1}_{\{x>0\}}+{\bf 1}_{\{y>0\}}\equiv g(x,y), & \forall (x,y)\in (0,\infty)^2 \\ \\ \displaystyle v(t,x,0)=\int_0^{T-t} \frac{x\exp(-(x-s)^2/2s)}{\sqrt{2\pi s^3}}ds\equiv f(t,x), & \forall (t,x)\in (0,T)\times (0,\infty)\\ \displaystyle v(t,0,y)=\int_0^{T-t} \frac{y\exp(-(y-s)^2/2s)}{\sqrt{2\pi s^3}}ds\equiv f(t,y), & \forall (t,y)\in (0,T)\times (0,\infty). \end{cases} $$ Is the above problem well posed (in any sense)? More precisely, I look for a solution $v$ that is spatially symmetric, i.e. $v(t,x,y)=v(t,y,x)$ and $0\le v\le 2$?

PS. An analysis for more general functions $g, f$ are welcome. In particular, two alternative problems below are also of interest for me. The PDE and terminal condition are the same, with different boundary conditions:

$$ \begin{cases} \displaystyle v(t,x,0)={\bf 1}_{\{x>c\}}f(t,x-c), & \forall (t,x)\in (0,T)\times (0,\infty)\\ \displaystyle v(t,0,y)={\bf 1}_{\{x>c\}}f(t,y-c), & \forall (t,y)\in (0,T)\times (0,\infty). \end{cases} $$

and

$$ \begin{cases} \displaystyle v(t,x,0)= f(t,\alpha x), & \forall (t,x)\in (0,T)\times (0,\infty)\\ \displaystyle v(t,0,y)= f(t,\beta y), & \forall (t,y)\in (0,T)\times (0,\infty), \end{cases} $$

where $c>0$, $0<\alpha, \beta<1$ are constants.