The difference star representing
is unchanged when the mesh is interleaved simultaneously on *t* and *x*.
Finding the missing values on the *t*-axis is straightforward
by any interpolation method.
Finding the missing values on the *x*-axis seems straightforward
by the CG method.
Those with wave-equation solving experience
might suggest a systematic procedure
to have *v*(*t*,*x*)=0 vanish exactly on the new traces.
I'll outline the procedure to give confidence
that the least-squares task need not be a horrific one,
despite the huge number of unknowns.
The differencing star is .Suppose the right side is placed on a known trace,
the left side multiplies an unknown trace.
The unknown trace can be found by solving a bi-diagonal set of
simultaneous equations.
This will involve recursion up or down the time axis
depending on the sign of *p*.
Since *p* will typically vary with *t* and take both signs in the range,
well-known methods are needed.
Hopefully the CG method takes care of this in a fairly automatic way.
This deserves investigation.
Potentially the CG method requires as many iterations as there
are data points,
so this could be prohibitive
when the task is set up in a huge window of data.
If the CG method is slow here,
we'll gain some valuable insights on how to speed it up
because we think we understand the mechanisms so well in this example.
On the other hand,
the CG method does not try for zero error when extrapolating
from right to left, but for minimum averaged error when
extrapolating both ways.
Unfortunately, I have burned myself out on programming,
so I have no examples to show you.

1/13/1998