Closed jlperla closed 6 years ago
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 Ref
s on the ends, in which case it will have the information for the view.
Closing this because I think the answer is confirmed.
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
and we also know that
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
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