On Tue, Dec 13, 2016 at 2:50 AM, Matthew Knepley <[email protected]> wrote:
> On Mon, Dec 12, 2016 at 2:29 AM, Praveen C <[email protected]> wrote: > >> Hello Matt >> >> I have attached the detailed output. >> >> Fenics automatically computes Jacobian, so I think Jacobian should be >> correct. I am not able to run the Fenics code without giving the Jacobian. >> I am currently writing a C code where I can test this. >> >> This equation is bit weird. Its like this >> >> u_t = ( K u_x)_x >> >> K = u / sqrt(u_x^2 + eps^2) >> > > I do not understand how to show parabolicity in this case. However, I have > a more fundamental misunderstanding. In your code, I see > > R0 = idt*(u-uold)*v*dx + K0*ux*vx*dx > > for the residual, which looks like > > R0 = u_t + K u_x > > to me. Where is the extra derivative you show above? > > Hello Matt vx is derivative of test function. So the last term is a second order differential operator. The PDE in full looks like this u_t = u_x^2/sqrt(u_x^2 + eps^2) + eps^2 * u * u_xx / (u_x^2 + eps^2)^1.5 With u > 0, the last term is parabolic. But numerically, I solve it in divergence form. Best praveen
