Write up the Gaussian elimination explanation for the creation of Q

[jlperla]

Part of this is checking the rank of [L; B]

[jlperla]

How many rows are needed for the jump diffusion, for example

[jlperla]

Separate into the linear and affine case. If the Q_L is the linear part, then how to get Q_b?

Is Q doing both the restriction and the gaussian elimination?

[jlperla]

Basically, is it the product of R and a bunch of elementary matrices (which would also explain why it isn't square).

[jlperla]

Some rough notes from discussions on slack to review. Only minor condensing:

[L; B] u_hat = [0; b] solving that would mean your solution has to satisfy the BCs What Sheehan pointed out before is that you can do Gaussian elimination on just a few rows

Jesse Perla [2:46 PM] Basically, get rid of the B, right?

Chris Rackauckas [2:46 PM] call that resulting operator Ltilde yeah, basically get rid of the B you don't getL, you getLtilde and that'sLtildeu = 0 where it operates on things in the interior that satisfy the BCs and spits out things in the interior that satisfy the BCs Jesse Perla [2:50 PM] So lets look at the linear case, where the boundary condition isB u = 0 In that case, the connection between what we have written down asQand theLtildeis that: Ltilde = LQ`

Chris Rackauckas [2:52 PM] Yes, the derivative operators are naturally defined as L which spit out something only in the interior but they are defined from the closure to the interior.

Jesse Perla [2:52 PM] L right now is a M X Mbar and Q is Mbar * M`

Jesse Perla [2:56 PM] What I am trying to do is mentally map from the [L; B] u_hat = [p; 0] with u = R u_hat to the L Q u = p case, That mapping is where we are having some trouble. I think we are finally understanding why this is connected to gaussian elimination, and how that would generalize to the jump-diffusion case.

Jesse Perla [3:03 PM] Lets say I do gaussian elimination on [L; B] u_hat = [p; 0] to get rid of the B. Which can be represented as the product of elementary matrices... so part of Q is exactly that product of intermediate matrices?

Chris Rackauckas [3:03 PM] yes

Jesse Perla [3:03 PM] But then somehow the restirction R is begin applied (in the Q) to go from u_hat to u.

Chris Rackauckas [3:03 PM] the non-identity part of Q is those elementary matrices.

But then somehow the restirction R is begin applied (in the Q) to go from u_hat to u.

What do you mean by that?

Jesse Perla [3:04 PM] Lets say I am doing gaussian elimination on [L; B] u_hat = [p; 0] So I do all sorts of elimination steps to get it to [L; 0] u_hat = [p; 0]... correct? And I "record" those elimination steps as a sequence of elementary matrices. When you said there are guassian elimination steps, this is what you meant, right?

Chris Rackauckas [3:05 PM] yes

Jesse Perla [3:06 PM] OK, so lets call the product of those elementary matrices E. Then what is the next algebra step here? E * [L; B] * u_hat = E * [p; 0] ?

Chris Rackauckas [3:07 PM] E * [L; B] * u_hat = E * [p; 0]

Jesse Perla [3:07 PM] Perfect.

Chris Rackauckas [3:07 PM] [L; 0] * u_hat = E * [p; 0]

Jesse Perla [3:07 PM] Now, R * u_hat = u

Chris Rackauckas [3:07 PM] notice [L; 0] * u_hat = L*R*u_hat = L*u (edited) wait do I have that in the wrong order? yes I do R*L*u_hat = E* p you're supposed to get that because of the appended zeros there must be a better way to show/explain that wait a second.

Jesse Perla [3:10 PM] I think that Q is the product of the E and something... just not sure what the something is.

Chris Rackauckas [3:10 PM] yes I think it's like E^(-1) but somehow it's in the wrong spot order of operations wise.

Jesse Perla [3:11 PM] You don't think there is an R in there anywhere?

Chris Rackauckas [3:11 PM] Well if you multiply by R in the right spots you get rid of the ;0 stuff since then it's all in the interior.