I know the viscosity solution for $ |u'| = 1$ with $u(0) = u(1) = 0$ is $\frac{1}{2} - |x - \frac{1}{2}|$, but just wondering what will I get if I use the vanishing viscosity method?
For a similar equation $u' = 1$ I can solve the ode and get $u_{\epsilon} = \frac{e^{\frac{x}{\epsilon}} - 1}{1 - e^{\frac{1}{\epsilon}}} + x$, then as $\epsilon$ goes to $0$, $u_{\epsilon}$ will go to $x, x\in [0,1)$ and $0$ at $x = 1$ which is not a viscosity solution.
But how shall I deal with the absolute value in the vanishin viscosity ode?
Thank you!
You can get an explicit solution in this case as well. Here is a hint: By symmetry, you can consider the equation $-\varepsilon u''(x) + u'(x)^2 = 1$ for $x\in [0,1/2]$ with $u(0)=0$ and $u'(1/2)=0$. Then solve it explicitly like in the other case you quoted. You may also find it interesting to change the sign of the viscosity term, by solving instead $\varepsilon u''(x) + u'(x)^2 = 1$. You will get different solutions in each case.