Operator composition question for constant multiplication

[jlperla]

I just want to double-check the domains for multiplicative constants.

Consider the stochastic process $d x_t = \tilde{\sigma}(x_t) d W_t$. For simplicity, assume it is reflected at 0 and 2.

We know that the infinitesimal generator is $\tilde{L} = \tilde{\sigma}(x)^2/2 \partial_{xx}$.

The question I have is to make sure we are precisely is how compose the multiplicative $\tilde{\sigma}(x)^2/2$ term... especially if $\tilde{\sigma}(x)$ is not defined below 0. Lets say that we let $\Delta x = 1$ so that the interior is $x \equiv$ [0, 1, 2] and the extension is $\bar{x} \equiv$ [-1, 0, 1, 2, 3].

In all cases, the discretized boundary operator is

B =
1  -1  0   0  0
0   0  0  -1  1

and we also know that

Q =
1 0 0
1 0 0
0 1 0
0 0 1
0 0 1

---

First the easy case. is if $\tilde{\sigma}(x) = \sqrt{2}$, then we know that the discretization in central differences of the $\partial_{xx}$ operator is

L_2 =
	1 -2   1 0   0
	0  1  -2  1  0
	0  0   1  -2 1	

and then $L = L_2$ and there is an appropriate $L Q$ for the composed operator which we can apply to $u \in R^3$.

Next, let $\tilde{\sigma}(x) = \sqrt{2}(1+x)$ and $\sigma(x)$ is undefined for $x$ outside of the $[0,2]$ domain. Then just to make sure we are thinking this through correctly... define the vector of this applied to the domain (and not the extended domain, since it is meaningless) as
$$
\sigma =\sqrt{2} \begin{bmatrix} 1\ 2\ 3 \end{bmatrix}
$$
Then put this into a diagonal matrix,
$$
C_1 = diag(\sigma.^2/2) \in R^{3\times 3}
$$

Finally, to compose this, we have
$$
L = C L_2
$$
Do I have that correct? Does this generalize, that if the domain is of size $M$ and the extended domain is of size $\bar{M}$ that all multiplicative constants are diagonal and of size $M \times M$?

Finally, is there any case in the composition of operators where you do would not want this as a diagonal, and what is the interpretation if you don't? This may be helpful to figure out before we make it lazy

[ChrisRackauckas]

Finally, is there any case in the composition of operators where you do would not want this as a diagonal, and what is the interpretation if you don't?

This can happen in spectral discretizations since the basis isn't local but global.

But this is one of the reasons that a pre-restriction seems to be the easier way to go. This can actually be easily achieved with a view. We will want the result of Q to be a special type anyways so that way we are just allocating two Refs on the ends, in which case it will have the information for the view.

[jlperla]

Closing this because I think the answer is confirmed.