Hello Claire,
remember that it is not the stress that is conserved but its divergence:
\nabla \cdot \sigma' - \nabla P + \rho g = 0
For example, in 1D, using the SSA approximations, we get
d/dx (2 sigma'_xx) = d/dx (4 mu dvx/dx) = rho g ds/dx
and the right hand side is independent of n (it is the driving stress). Using the same equation, the effective strain rate for 1D SSA is simply |eps_xx|. Assuming that eps_xx>0 to simplify things, we get
d/dx( B eps_xx1/n -1 * eps_xx ) =d/dx( B eps_xx1/n )= rho g ds/dx
and so, if ds/dx and B are constants and eps_xx(0) = 0 as a boundary condition, we get:
eps_xx = ( 1/(B) rho g ds/dx * x)n
again, B is large (1e+8 SI), and surface slopes are small (e.g. 0.01). So we are taking the power of a number <<1. As n increases from 3 to 4, we do expect eps_xx to get smaller and so v_x to be smaller too.
If you have a numerical experiment ready from another model, I am more than happy to look into it, but I don't see anything unexpected at this point. Feel free to email me (Mathieu.Morlighem@Dartmouth.edu) if you would like to follow up!
Cheers
Mathieu