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 then we must restart from the result of 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.

$$\begin{array}{lll} \bold 0 &\approx& \bold F \bold m - \bold d \bold 0 &\approx& \bold A \bold m \end{array}$$ (15)

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

$$\begin{array}{lll} \bold 0 &\approx& \bold F \bold u \ +\ (\b... ... 0 &\approx& \bold A \bold u \ +\ \bold A \bold m_0 \end{array}$$ (16)

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 this introduces the preconditioning variable $\bold p = \bold A\bold u$ . Our regression pair becomes:

$$\begin{array}{lll} \bold 0 &\approx& \bold F \bold A^{-1} \bo... ...\\ \bold 0 &\approx& \bold p \ +\ \bold A \bold m_0 \end{array}$$ (17)

Here is the problem: Now we require both $\bold A$ and $\bold A^{-1}$ operators. In 2- and 3-dimensional spaces we don't know very many operators with an easy inverse. Indeed, that is why I found myself pushed to come up with the helix methodology of the previous chapter - because it provides invertible operators for smoothing and roughening.

Preconditioning