Desperate on Viscosity (Sub)solutions

Question

The HJ equation is: $H(u'(x))+1=0, x\in(-1,1)$. And $H:\mathbb{R}\rightarrow\mathbb{R},H(p)=\min_{a\in[-1,1]}ap,p\in\mathbb{R}$.
The question is show that $u(x)=1-|x|,x\in(-1,1)$ is a viscosity solution.
I know that it's a supersolution since there is no element in $D^-u(x)$, but why it is also a subsolution?
I mean for any $\phi(x)$ above $u(x)$, the derivative of it at point x=0 must be in [-1,1], then for any $p\in[-1,1]$, H should have $H\geq-1$, which makes $H+1\geq0$. Then it doesn't satisfy the subsolution definition.
I don't know where I am wrong. Please, any hint?

Answer 1

The solution is wrong. Your equation is

$$-|u'(x)| + 1 = 0.$$

The viscosity solution satisfying $u(-1)=0=u(1)$ is $u(x) = |x|-1$. To see this, we only have to check the viscosity solution conditions at $x=0$, since the equation is satisfied classically elsewhere. At $x=0$, there are no test functions touching from above, so $D^+$ is empty. Anything touching from below (i.e. $D^-$) has derivative in the range $[-1,1]$, which yields the supersolution condition $-|u'(x)| + 1\geq 0$.

If you switch the sign of the equation and consider

$$|u'(x)| - 1 = 0$$

then the solution is $u(x)=1-|x|$. This is the peculiar thing about viscosity solutions; the solutions of $H=0$ and $-H=0$ can be different. It is natural if you think about vanishing viscosity, since in one case you regularize by adding $\varepsilon \Delta u$, while in the other case you add $-\varepsilon \Delta u$.