[A complimentary Cc of this posting was NOT [per weedlist] sent to
Ilya Zakharevich
], who wrote in article :
Conjectu consider a family of rk=2 vector bundles on P1 parameterized
by S which is isomorphic to O(n) (+) O(n') outside of subvariety S'
of S, and degenerates into O(N) (+) O(N') on S'; Nn,n'. Then the
direct image to S is not flat if codim S' = 2 and n,n' = -1, N -1.

[*] [I do not know what happens if one drops n,n' = -1, N -1.]

[Heuristically, I would expect that if direct image is flat, so is the
higher direct image (to "preserve Euler characteristic")
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

Apparently, it was a kind of circular reasoning...

take Ker(xA + yB) with matrices A,B depending on a,b:

1 0 0 0 0 0 1 0 0 0
0 1 0 0 a 0 0 1 0 0
0 0 1 0 b 0 0 0 1 0

- and I would twist by O(1) - while to closely follow your example,
the twist would be by O(2). (Since I prefer Ker to Coker, the picture
is kinda dual to yours.) So O(-2) (+) O(1) would deform into O(-1) (+) O(0).

Should not one be able to do it without computer algebra? :-(

It turns out that the calculation is very simple indeed - and bye-bye
the conjecture! (It looks like the direct image is flat for both
twistings...) One hope which remains (to use the machinery described
in my preceeding message to understand the original counterexample) is
that a deformation of O(-2) (+) O(1) into O(-1) (+) O(0) obtained via
Ker is not isomorphic to one obtained via Coker...

============================================ Calculation:

Twist by O(1): to calculate ( Ker(xA+yB) )(1), one needs to find
linear functions V(x,y) = x v1 + y v2 of (x,y) such that (xA + yB)V = 0.
When (a,b) != (0,0), one gets that V is proportional to V0 with

v1 = -b^2 e4 + a (e5 - a e2 - b e3)
v2 = a^2 e1 + b e5.

So the direct image is {P(a,b) (x v1 + y v2)} with a regular scalar
function P; so it is iso to the structure sheaf.

Twist by O(2): to calculate ( Ker(xA+yB) )(2), one needs to find
quadratic functions Q(x,y) of (x,y) such that (xA + yB)V = 0. When
(a,b) != (0,0), one gets x V0(x,y), y V0(x,y), and

Q0 = x^2 (e5 - a e2 - b e3) + xy (a e1 + b e2) + y^2 b e2;

all 3 quadratic functions are linearly independent (at least when a,b
!= 0), so the direct image is {x P1(a,b) V0 + y P2(a,b) V0 + P3(a,b) Q0}.

So AFAICS, the direct image is 3 copies of the structure sheaf...

Puzzled,
Ilya