Preconditioning

Need for an invertible preconditioner

It is important to use regularization to solve many examples. It is important to precondition, because in practice, computer power is often a limiting factor. It is important to be able to begin from a nonzero starting solution, because in nonlinear problems we must restart from an earlier solution. Putting all three requirements together leads to a little problem. It turns out the three together lead us to needing a preconditioning transformation that is invertible. Let us see why this is so.

$$\begin{array}{lll} \bold 0 &\approx& \bold F \bold m - \bold d \ \bold 0 &\approx& \bold A \bold m \end{array}$$ (28)

First, we change variables from $\bold m$ to $\bold u = \bold m - \bold m_0$ . Clearly, $\bold u$ starts from $\bold u_0=0$ , and $\bold m = \bold u + \bold m_0$ . Then, our regression pair becomes:

$$\begin{array}{lll} \bold 0 &\approx & \bold F \bold u + (\... ...0 &\approx & \bold A \bold u + \bold A \bold m_0 \end{array}$$ (29)

This result differs from the original regression in only two minor ways, (1) revised data, and (2) a little more general form of the regularization, the extra term $\bold A \bold m_0$ .

Now, let us introduce preconditioning. From the regularization, we see preconditioning introduces the preconditioning variable $\bold p = \bold A\bold u$ . Our regression pair becomes:

$$\begin{array}{lll} \bold 0 &\approx & \bold F \bold A^{-1} \b... ... \bold 0 &\approx & \bold p + \bold A \bold m_0 \end{array}$$ (30)

Here is the problem: We now require both $\bold A$ and $\bold A^{-1}$ operators. In 2- and 3-dimensional spaces, we do not know very many operators with an easy inverse. That reason is why I found myself pushed to come up with the helix methodology of Chapter --because it provides invertible operators for smoothing and roughening.

Preconditioning