Questions for the non-linear PDE $2u_t=\log(-u_{xx})$ Consider the PDE as follows :
$$2u_t=\log(-u_{xx}), \quad \forall (t,x)\in [0,1)\times (-1,1)$$
with the terminal and boundary conditions
$$u(1,x)=0,\quad \forall -1<x<1 \quad\quad \mbox{and} \quad\quad u(t,\pm 1)=0,\quad \forall 0\le t<1.$$
Is this PDE well posed? How can we solve it numerically?
Any answers, comments and references are highly appreciated!
 A: This is a quick qualitative comment on the possible numerical difficulty to approximate this PDE, say, using an explicit finite difference approximation $u_h$ to approximate $u$ in time:
$$
2\frac{u_h(t_{n+1}, x) - u_h(t_h, x)}{\Delta t} = \log \big(-\partial_{xx} u_h(t_h, x)\big).
$$
Judging by the terminal condition of $u$, it is not difficult to see that this equation resembles the Heat eq and we want $u$ to quickly decay to 0. If the initial condition is assumed to be $u_0>0$ and $0<-\partial_{xx} u_0(\cdot) <1$ but close to 1, i.e., $u_{xx}+1$ is positive and is close 0, $\log(-u_{xx})$ will be negative and driving $u$ toward $0$.
As time goes by,
if we assume that $u$ is decaying and converging to a stationary point, $u_{xx}$ will converge to $0^+$. However, the log will make $u_t$ diverge which contradicts the ansatz. Therefore, most likely this equation will develop a shock and is only well-posed in a small neighborhood of certain time range. If an explicit finite difference is used, I recommend setting the time step to be decaying $\propto e^{-\gamma t}$ with $\gamma$ a tweakable parameter.
A: It appears that the singularity comes from the terminal condition. My feeling is that we can replace $u(1,x)=0$ by $u(1,x)=\epsilon f(x)$ for $\epsilon>0$ and some concave function $f$ s.t. $f(\pm 1)=0$, e.g. $f(x)=1-x^2$ or $f(x)=\cos(\pi x/2)$. Therefore, the numerical solution can be visualized. More details will be added afterwards.
