text:
Detached-eddy simulation (DES) is used to study the massively separated flow over a rounded-corner square. The configuration is an idealization of the flow around a forebody cross section rotating at high angle of attack. Simulations are performed at sub- and supercritical Reynolds numbers, between which experimental measurements show a reversal of the side force. DES predictions are evaluated using experimental measurements and contrasted with unsteady Reynolds-averaged Navier–Stokes (URANS) results. The computations are also subjected to a moderate grid refinement, a doubling of the spanwise period, an enlargement of the domain in the other directions, and the removal of any explicit turbulence model. The sub- and supercritical flows are computed at Reynolds numbers of $105$ and $8×105$, respectively, and with the freestream at $10deg$ angle of attack. Boundary-layer separation characteristics (laminar or turbulent) are established via the initial and boundary conditions of the eddy viscosity. Following boundary layer detachment, a chaotic and three-dimensional wake rapidly develops. For the supercritical flow, the pressure distribution is close to the measured values and both the streamwise and side forces are in adequate agreement with measurements. For the subcritical flow, DES side-force predictions do not follow the experimental measurements far enough to achieve reversal.

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text:
# Rank nullity inequality fom T^2

1. Mar 5, 2008

### rsa58

1. The problem statement, all variables and given/known data
Let T be a linear operator on a vector space V (finite dimensional). show that
nullity(T^2)<= 2nullity(T)

2. Relevant equations
the rank nullity theorem.

3. The attempt at a solution
i take T to be a lin. transf. from ker(T^2) to V, that is i am restricting the domain of T.

well, i show kerT is a subset of ker(T^2) then i write
nullity(T^2)= nullity(T(Ker(T^2)) + dimT(ker(T^2))= dimkerT +dimT(ker(T^2))

and this last equality i am not sure of. then i show T(kerT^2) is a subset of kerT and the theorem follows from the equality which i am not sure of. the proff. said to use the fact that kerT is a subset of ker(T^2). but i don't see how this helps. i started by examining T(kerT^2) and said that oh... okay i get it now. since kerT is a subset of kerT^2 there could be some vectors ouside of the kerT in this latter set. but that still makes the kernel of T(kerT^2) equal to the kerT since it is the kernel of {0, T(v1), ... T(vn)}. is this correct, please respond i have wasted so much time on this damn thing.

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