I have the eikonal equation:
$$\left|\frac{du}{dx}(x)\right| = 0, \, x\in \Omega, \quad u(x) = f(x), \, x\in \Gamma.$$
Does the above have the same solution as:
$$\frac{du}{dx}(x) = 0, \, x\in \Omega, \quad u(x) = f(x), \, x\in \Gamma$$
My intuition is that if I look at the time evolution the two seem to be different, in the sense that $|u_x|=u_t$ would have fronts propagating in either direction from the boundary, while $u_x=u_t$ (or $-u_x=u_t$) would propagate those only in one direction. Is this a correct interpretation. But if the above is the case wouldn't the steady states (i.e. for $T\rightarrow\infty$) differ?
I have done some numerical simulations and wanted to make sure that the solutions I get are not artifacts of my discretisation but arise from the model. I prescribe Dirichlet data pointwise $u(x_i) = f(x_i)$, and I am able to recover piecewise constant interpolation as the solution in all cases. Taking different time evolutions results in different solutions for the steady state ($T\rightarrow \infty$):
$$u_t=u_x \implies u(x)=f(x_i), \, x\in [x_i, x_{i+1})$$ $$u_t=-u_x \implies u(x)=f(x_i), \, x\in (x_{i-1}, x_i]$$ $$u_t=|u_x| \implies u(x)=f(x_i), \, x\in \left(\frac{x_{i-1}+x_i}{2}, \frac{x_{i}+x_{i+1}}{2}\right)$$
To me this implies that the three equations $u_t = \pm u_x$ and $u_t = |u_x|$ are in fact different and the difference arises in what happens at the boundaries, which then propagates to form the steady state.
The steady states have discontinuities so to me it looks like the first is taking a right derivative at the boundaries, the second a left derivative, and the third some combination of the two.