Viscosity solution

Remarks

If $u$ is continuously differentiable and $u - \varphi$ has a local extremum at $x_0$, then $Du(x_0) = D\varphi(x_0)$. So a continuously differentiable $u$ is a viscosity solution exactly when it solves the equation pointwise. One gets the same notion with smooth test functions $\varphi$ instead of continuously differentiable ones. The definition depends on the sign of $F$: the equations $F = 0$ and $-F = 0$ have different viscosity solutions in general. For second-order equations $F(x, u, Du, D^2 u) = 0$, the Hessian $D^2 \varphi(x_0)$ enters as well, and $F$ is required to be degenerate elliptic, which means non-increasing in the matrix variable for this sign convention.

Example: on $U = (-1, 1)$, the function $u(x) = 1 - \abs{x}$ is a viscosity solution of $\abs{u'} - 1 = 0$. Away from $0$, $u$ is smooth with $\abs{u'} = 1$. At $0$, if $u - \varphi$ has a local maximum then $-\abs{x} \le \varphi(x) - \varphi(0)$ near $0$, so $\abs{\varphi'(0)} \le 1$ and the subsolution condition holds. No continuously differentiable $\varphi$ makes $u - \varphi$ have a local minimum at $0$, because that would need $\varphi(x) - \varphi(0) \le -\abs{x}$ near $0$, so the supersolution condition is empty there.

Non-example: $w(x) = \abs{x} - 1$ is not a viscosity solution of the same equation, although $\abs{w'} = 1$ wherever $w'$ exists. With $\varphi = 0$, the function $w - \varphi$ has a local minimum at $0$, and $\abs{\varphi'(0)} - 1 = -1 < 0$, so the supersolution condition fails.