Argyris.Basis.tex

%In this subsection we derive the basis for the Argyris element using the
%standard Cartesian coordinates on a reference triangle $\hat{K}$ as opposed to
%the area coordinates on a general triangle $K$ as used by \cite{Okabe}.
%The transformation developed by \cite{Dominguez08}, which will be discussed in \autoref{sse:Trans}, allows for all computations to be done on the reference triangle
%and so the bulk of the discussion will be concerned with developing the basis
%functions on $\hat{K}$ and the transformation.
\input{Ref.Triangle.tex}

In most references, which present the Argyris element, only the constraints
found in \autoref{tab:Constraints} are presented. Thus, for completeness, we
derive and explicitly state the the basis functions for the Argyris FE in this
subsection. The Argyris triangle has 21 degrees of freedom and therefore has 21
basis functions per triangle.  Additionally, the basis for Argyris in reference
coordinates (depicted in \autoref{fig:RefTriangle}) belong to the space
$\mathbb{P}_5$, which has the standard monomial basis
\begin{equation*}
	\left\{
    1, x, y, x^2, xy, y^2, x^3, x^2y, xy^2, y^3, x^4, x^3y,
    x^2y^2, xy^3, y^4, x^5, x^4y, x^3y^2, x^2y^3, xy^4, y^5
  \right\}.
\end{equation*}
Thus, the $i^{th}$ basis function for the Argyris triangle can be written as
\begin{equation}
	\begin{split}
	\hat{\varphi}_i(\hat{x},\hat{y}) = m^i_1 + m^i_2 \hat{x} + m^i_3 \hat{x}^2 + m^i_4 \hat{x}^3 + m^i_5 \hat{x}^4 + m^i_6 \hat{x}^5 + m^i_7 \hat{y} + m^i_8
 	\hat{y}^2 + m^i_9 \hat{y}^3 \\
 + m^i_{10} \hat{y}^4 + m^i_{11} \hat{y}^5 + m^i_{12}  \hat{x} \hat{y} + m^i_{13} \hat{x} \hat{y}^2 + m^i_{14} \hat{x} \hat{y}^3 + m^i_{15}
 	\hat{x} \hat{y}^4 + m^i_{16} \hat{x}^2 \hat{y} \\
 + m^i_{17} \hat{x}^2 \hat{y}^2 + m^i_{18}\hat{x}^2 \hat{y}^3 + m^i_{19} \hat{x}^3 \hat{y} + m^i_{20}\hat{x}^3 \hat{y}^2 + m^i_{21} \hat{x}^4 \hat{y}.
 \end{split}
	\label{eqn:Basis}
\end{equation}

Now, consider the reference triangle $\hat{K}$ in \autoref{fig:RefTriangle} with
vertices numbered counterclockwise $1,\, 2,\text{ and } 3$, i.e.
$(\hat{x}_1,\hat{y}_1)=(0,0),\, (\hat{x}_2,\hat{y}_2)=(1,0),\text{ and } (\hat{x}_3,\hat{y}_3)=(0,1)$.
Additionally, let the vector $v_i$ represent the $i^{th}$ edge with
\begin{equation*}
  v_1 = [\hat{x}_2-\hat{x}_1,\hat{y}_2-\hat{y}_1]^T, \quad v_2=[\hat{x}_3-\hat{x}_1,\hat{y}_3-\hat{y}_1]^T \text{ and } v_3=[\hat{x}_3
  -\hat{x}_2,\hat{y}_3-\hat{y}_2]^T.
\end{equation*}
For book keeping purposes, we will use the convention that the $i^{th}$ normal
vector is the rotation of the $v_i$ counter-clockwise $90^\circ$. This is the
same convention used by Dominguez et. al. \cite{Dominguez08}. Therefore, the
$i^{th}$ basis function can be found using the restriction in \autoref{tab:Constraints}.

\input{Basis.Table.tex}

Now, let $z$ be the vector containing the monomial basis for $\mathbb{P}_5$, i.e.
\small{
\begin{equation*}
  z=\left[
    1, \hat{x}, \hat{y}, \hat{x}^2, \hat{x}\hat{y}, \hat{y}^2, \hat{x}^3, \hat{x}^2\hat{y}, \hat{x}\hat{y}^2, \hat{y}^3, \hat{x}^4, \hat{x}^3\hat{y},
    \hat{x}^2\hat{y}^2, \hat{x}\hat{y}^3, \hat{y}^4, \hat{x}^5, \hat{x}^4\hat{y}, \hat{x}^3\hat{y}^2, \hat{x}^2\hat{y}^3, \hat{x}\hat{y}^4, \hat{y}^4
  \right]^{T}.
\end{equation*}}
Then the $i^{th}$ Argyris basis function on $\hat{K}$ is given by
\begin{equation*}
  \hat{\varphi}_i = M_i z,
\end{equation*}
where $M_i$ is the $i^{th}$ row of the matrix $M$. Therefore, the evaluation of
$\hat{\varphi}_i(\hat{x},\hat{y})$ comes down to the matrix-vector multiplication
\begin{equation*}
  \hat{\varphi}_i(\hat{x},\hat{y}) = M_i z(\hat{x},\hat{y}).
\end{equation*}

To determine the matrix $M$ we must solve the linear system
\begin{equation*}
  ZM^T=I_{21}
\end{equation*}
that results from the constraints in \autoref{tab:Constraints}. Here
\begin{equation*}
  \begin{split}Z=[z(0,0), z(0,1), z(1,0),
  z_{\hat{x}}(0,0), z_{\hat{y}}(0,0),
  z_{\hat{x}}(1,0), z_{\hat{y}}(1,0),
  z_{\hat{x}}(0,1), z_{\hat{y}}(0,1), \\
  z_{\hat{x}\hat{x}}(0,0), z_{\hat{x}\hat{y}}(0,0), z_{\hat{y}\hat{y}}(0,0),
  z_{\hat{x}\hat{x}}(1,0), z_{\hat{x}\hat{y}}(1,0), z_{\hat{y}\hat{y}}(1,0), \\
  z_{\hat{x}\hat{x}}(0,1), z_{\hat{x}\hat{y}}(0,1), z_{\hat{y}\hat{y}}(0,1),
  z_{\hat{y}}(\nicefrac{1}{2},0), -z_{\hat{x}}(0,\nicefrac{1}{2}), \\
  -\frac{1}{\sqrt{2}}(z_{\hat{x}}(\nicefrac{1}{2},\nicefrac{1}{2}) +
  z_{\hat{y}}(\nicefrac{1}{2},\nicefrac{1}{2}))]^T.\end{split}
\end{equation*}
$I_{21}$ is a $21\times 21$ identity matrix and $M$ is the matrix containing the
coefficients for all 21 basis functions. Therefore solving this system will
result in a matrix, $M$, that contains the coefficients for the Argyris basis
functions for the reference triangle as in \eqref{eqn:Basis}. Thus, the matrix
$M$ can be seen in \autoref{tab:Coefficients}.
\input{Coefficients.tex}
Therefore, the Argyris basis functions, on the reference element $\hat{K}$, are
\input{Basis.Functions.tex}
\begin{remark}
  Throughout our search for an explicit statement of the Argyris basis functions
  we only came across two reference, \cite{Wiki} and \cite{FEM++}, both of which are
  online references. However, this is not to mean that a reference containining
  the explicit statement of the Argyris basis functions doesn't exist, but it
  certainly eluded the author of this thesis.
\end{remark}