feat(poly): major work on preconditioner for block form of lane emden equation

working on a "smart" schur compliment preconditioner for the block form of the lane emden equation. Currently this is stub and should not be considered usable

docs/derivations/laneEmdenVariationalBlockForm.aux

@@ -1,9 +1,11 @@
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-\newlabel{tab:mfem_mixed_integrators}{{1}{4}{}{table.1}{}}
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+\@writefile{toc}{\contentsline {subsection}{\numberline {2.2}Preconditioner}{4}{}\protected@file@percent }
+\@writefile{toc}{\contentsline {subsection}{\numberline {2.3}A Few Quick Notes}{5}{}\protected@file@percent }
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+\@writefile{lot}{\contentsline {table}{\numberline {1}{\ignorespaces Selection of MFEM Mixed Bilinear Form Integrators}}{6}{}\protected@file@percent }
+\newlabel{tab:mfem_mixed_integrators}{{1}{6}{}{table.1}{}}
+\gdef \@abspage@last{6}

docs/derivations/laneEmdenVariationalBlockForm.tex

@@ -122,6 +122,114 @@ We can then set this up as a matrix operation
     \end{bmatrix}
 \end{align}
 
+From this form we can easily see that the residual matrix is
+
+\begin{align}
+    R &= \begin{bmatrix}
+        f(\bar{\theta}) - M\bar{\phi} \\
+        D\bar{\phi} - Q\bar{\theta}
+    \end{bmatrix}
+\end{align}
+
+\subsection{The Jacobian}
+We need to define the Jacobian of this system of equations so that we can use it
+in our Newton-Raphson method. Generally the Jacobian is the matrix of partial derivitives wrt. the state vector. We will let our state vector, $X$, be
+\begin{align}
+    X = \begin{bmatrix}
+        \bar{\theta} \\
+        \bar{\phi}
+    \end{bmatrix}
+\end{align}
+
+So then the Jacobian is
+\begin{align}
+    J &= \begin{bmatrix}
+        \frac{\partial}{\partial \theta}\left(f(\theta) - M\phi\right) & \frac{\partial}{\partial \phi}\left(f(\theta) - M\phi\right) \\
+        \frac{\partial}{\partial \theta}\left(D\phi - Q\theta\right) & \frac{\partial}{\partial \phi}\left(D\phi - Q\theta\right)
+    \end{bmatrix} \\
+    J &= \begin{bmatrix}
+        \frac{df}{d\theta} - \phi\frac{\partial M}{\partial \theta} & -M-\phi\frac{\partial M}{\partial \phi} \\
+        -Q - \theta\frac{\partial Q}{\partial \theta} & D + \phi\frac{\partial D}{\partial \phi} - \theta\frac{\partial Q}{\partial \phi}
+    \end{bmatrix}
+\end{align}
+
+Finally, we know that the matrices $M$, $D$, and $Q$ are constant with respect to $\theta$ and $\phi$. Therefore, we can drop the partial derivatives with respect to $\theta$ and $\phi$ from the Jacobian. This gives us
+\begin{align}
+  \mathbf{J} &= \begin{bmatrix}
+        \frac{df}{d\theta} & -M \\
+        -Q & D
+    \end{bmatrix}
+\end{align}
+
+\noindent In a fully assembled, distritized form this will look like
+
+\begin{align}
+  \mathbf{J} = \begin{bmatrix} \frac{df}{d\theta}_{00}          & \dots  & \frac{df}{d\theta}_{0n_{\theta}}          & -M_{00}                  & \dots            & -M_{0n_{\phi}} \\
+  \vdots          & \ddots &                          & \vdots                   & \ddots           & \\
+    \frac{df}{d\theta}_{n_{\theta}0} &        & \frac{df}{d\theta}_{n_{\theta}n_{\theta}} & -M_{n_{\theta}0}         &                  & -M_{n_{\theta}n_{\phi}} \\
+  -Q_{00}         & \dots  & -Q_{0n_{\theta}}         & D_{00}                   & \dots            & D_{0n_{\phi}} \\
+  \vdots          & \ddots &                          & \vdots                   & \ddots           & \\
+  -Q_{n_{\phi}0}  &        & -Q_{n_{\phi}n_{\theta}}  & D_{n_{\phi}0}            &                  & D_{n_{\phi}n_{\phi}}
+  \end{bmatrix}
+\end{align}
+
+\noindent Where $N_{dof}^{\theta} = n_{\theta}$ is the number of degrees of freedom on $\theta$, and $N_{dof}^{\phi} = n_{\phi}$ is the number of degrees of freedom on $\phi$. Note how the Jacobian is a matrix of size $\left(N_{dof}^{\theta} + N_{dof}^{\phi} \times N_{dof}^{\theta} + N_{dof}^{\phi}\right)$
+
+\subsection{Preconditioner}
+Due to the eventual size of these matrices we would like to be able to solve
+each step in this using a memory efficiet approach. Krylov solvers, such as
+GMRES, allow for matrix free iterative solutions (as long the concept of
+multiplication is defined). However, for these systems to be well formed for
+such solvers it is useful for us to use a preconditioner. However, this is a
+somewhat strongly coupled system where we cannot simply use the inverse
+diagonals of the matrix as a preconditioner. Instead, to encode the coupling we
+will use Schur's Compliment. Each Newton iteration we solve the equation
+\begin{align}
+  \mathbf{J}\Delta \vec{x} = \vec{b}
+\end{align}
+If we expand this out
+\begin{align}
+  \begin{bmatrix} \mathbf{\dot{f}} & -\mathbf{M} \\
+    -\mathbf{Q} & \mathbf{D}
+  \end{bmatrix}\begin{bmatrix} \theta \\
+    \phi
+    \end{bmatrix} = \begin{bmatrix} b_{0} \\
+  b_{1}\end{bmatrix}
+\end{align}
+We can pull out the first equation from this system
+\begin{align}
+  \mathbf{\dot{f}}\theta - \mathbf{M}\phi &= b_{0} \\
+  \theta &= \mathbf{\dot{f}}^{-1}b_{0} + \mathbf{\dot{f}}^{-1}\mathbf{M}\phi
+\end{align}
+Then if we pull out the second equation from the system
+\begin{align}
+  -\mathbf{Q}\theta + \mathbf{D}\phi &= b_{1} \\
+  -\mathbf{Q}\left(\mathbf{\dot{f}}^{-1}b_{0} + \mathbf{\dot{f}}^{-1}\mathbf{M}\phi\right) + \mathbf{D}\phi &= b_{1}
+\end{align}
+rearanging terms a bit
+\begin{align}
+  -\mathbf{Q}\mathbf{\dot{f}}^{-1}b_{0}-\mathbf{Q}\mathbf{\dot{f}}^{-1}\mathbf{M}\phi + \mathbf{D}\phi &= b_{1} \\
+  \left(\mathbf{D} - \mathbf{Q}\mathbf{\dot{f}}^{-1}\mathbf{M}\right)\phi &= b_{1} + \mathbf{Q}\mathbf{\dot{f}}^{-1}b_{0}
+\end{align}
+The term $\mathbf{D}-\mathbf{Q}\mathbf{\dot{f}}^{-1}\mathbf{M}$ is Schur's Compliment for this system, and we will represent this by the symbol $\mathbf{\tilde{S}}$. We can use Schur's Compilment to precondition our equation if we let the preconditioner be of the form
+\begin{align}
+  \mathbf{P} = \begin{bmatrix} \mathbf{\dot{f}}^{-1} & 0 \\
+    0 & \mathbf{\tilde{S}}^{-1}
+  \end{bmatrix}
+\end{align}
+So then the preconditioned equation which can be more easily solved by some Krylov solver (such as GMRES) is
+\begin{align}
+  \mathbf{P}\mathbf{J}\Delta \vec{x} = \mathbf{P}\vec{b}
+\end{align}
+
+It is easy to see here that for this system to be solvable / well defined both
+$\mathbf{\tilde{S}}$ and $\mathbf{\dot{f}}$ need to be invertable
+matrices. They are both easily shown to be square (with $\mathbf{\tilde{S}}$
+having a size $\left(N_{dof}^{\phi}\times N_{dof}^{\phi}\right)$ and
+$\mathbf{\dot{f}}$ having a size $\left(N_{dof}^{\theta}\times
+N_{dof}^{\theta}\right)$).
+
+
 \subsection{A Few Quick Notes}
 A few notes on the dimensions of $\mathbf{M}$, $\mathbf{Q}$, $\mathbf{D}$, and $f(\bar{\theta})$.
 \begin{itemize}