To show a given function is not the viscosity solution.

Question

For the equation $ F(x,u,u',u'') = -au''-1 =0$ for $ x\in (0,2)$ with $ u(0) = 0 = u(2) $ and $a(x)$ is $1$ for $x\in (0,1)$ and $2$ for $x\in [1,2)$. Need to show that the function $$ u(x) = \begin{cases} -x^2/2 + 5x/6 &\mbox{if } x \in (0,1] \\ -x^2/4 + 5x/12+1/6 & \mbox{if } x\in(1,2) \end{cases} $$ is not a viscosity solution. Here we observe $ u(1) = 1/3 $ from both sides, thus I need to show there exists $ \phi \in C^2(0,2)$ such that $ \phi(x) < u(x) $ for $ x\neq 1$ and $\phi(1) = 1/3 = u(1)$ with $ F(1,\phi(1),\phi'(1))<0$. I can not seem to be able to find such a $ \phi$. Any hints or solutions would be extremely helpful. Thank you.