# Dirichlet energy 2

Let $M = (V,E,F)$  be a triangulated domain in the plane where $V$ denotes the set of vertices, $E$ the set of edges and $F$ the set of triangles.  We consider the set of functions on the vertices  :
\begin{align*} \mathbb{R}^V:= \left\{ \begin{array}{c}  f : & V  \rightarrow \mathbb{R} \\  & i \mapsto f_i  \end{array}\right\}.\end{align*}
By interpolation to a piecewise linear function we get:  $\hat{f}: M \rightarrow \mathbb{R}$.
$\mathbb{R}^V \overset{\mbox{interpol.}}{\rightarrow} W_{PL}:=\left\{ \hat{f}:M\rightarrow \mathbb{R} \, \big \vert \,\left. \hat{f}\right|_{\sigma} \mbox{ is affine for all } \sigma \in F\right\}.$
In the last lecture the Dirichlet energy of  $\hat{f} \in W_{PL}$ was defined as:
$E_D(\hat{f}) = \frac{1}{2} \int_M \left| \mbox{grad} \, \tilde{f} \right|^2 = \frac{1}{4} \sum_{(i,j) \in \hat{E}} \left( \cot(\alpha_{ij}) + \cot{\beta_{ij}}\right) \left(f_i-f_j \right)^2.$
$E_D : W_{PL} \rightarrow \mathbb{R}$ is a quadratic form, which can be represented by matrix $A \in \mathbb{R}^{V \times V}:$
$E_D(f) = \left( f_1, \ldots , f_N \right) \left( \begin{array}{ccc} a_{11} & \ldots & a_{1N} \\ \vdots & \ddots & \vdots \\ a_{N1} & \ldots & a_{NN} \end{array} \right) \left( \begin{array}{ccc} f_1 \\ \vdots \\ f_N \end{array} \right),$
where $N:=|V|$ and
$a_{ij}:=\left\{ \begin{array}{rcl} -\frac{1}{4}\left( \cot(\alpha_{ij}) + \cot{\beta_{ij}} \right) & \mbox{for} & i\neq j, \,\{i,j\} \in E, \\ \frac{1}{4} \sum_{(i,k) \in \hat{E}} \left(\cot(\alpha_{ik}) + \cot{\beta_{ik}} \right) & \mbox{for} & i=j, \\ 0 & \mbox{else}. & \end{array}\right.$
One can easily check that: $A\left( \begin{array}{c} 1 \\ \vdots \\1\end{array}\right)=0$ and even further there holds: $\mbox{ker } A = \mathbb{R}\left( \begin{array}{c} 1 \\ \vdots \\1\end{array}\right).$

In order to get a better understanding of the Dirichlet energy we recap some facts on quadratic forms.
There is a natural relation between symmetric bilinear forms and quadratic forms:

Proposition: Let $W$ be a vectorspace and $b:W \times W \rightarrow \mathbb{R}$ a bilinear symmetric form then
\begin{align*} q: W \rightarrow \mathbb{R}, \\ q(w):= b(w,w),\end{align*}
is the corresponding quadratic form. $b$ is completely determined by $q$.

Proof: \begin{align*} & q(w_1+w_2) = b(w_1+w_2,w_1+w_2)=q(w_1) + 2b(w_1,w_2) + q(w_2), \\ \Rightarrow \quad \quad & b(w_1,w_2) = \frac{1}{2} \left(q(w_1+w_2) – q(w_1) – q(w_2).\right)\end{align*}

$\square$

For a vectorspace $W$ the dualspace is defined as $W^{\ast}:= \left\{ w^{\ast}: W \rightarrow \mathbb{R} \, \big \vert \,w^{\ast} \mbox{ is linear }\right\}.$ For a basis of $W$ $\left\{w_1, \ldots ,w_n \right\}$ the corresponding dual basis of $W^{\ast}$ $\left\{w_1^{\ast}, \ldots ,w_n^{\ast} \right\}$ is defined by:
$w_i^{\ast}(w_j) = \delta_{ij}.$
Note that there is a natural identification of $W^{\ast \ast}$ and $W$, i.e. the dual space of the dual space is the original space.
For a linear map $A:W \rightarrow V$ between vectorspaces $W,V$ the adjoint map is defined as:
\begin{align*}A^{\ast} : V^{\ast} &\rightarrow W^{\ast} \\ \left( A^{\ast}v^{\ast}\right)(w) & := v^{\ast}\left( Aw\right) \quad \quad \forall v^{\ast} \in V^{\ast}\mbox{ and } \forall w \in W.\end{align*}

If $b: W \times W \rightarrow \mathbb{R}$  is a bilinear form, we can define a map $A: W \rightarrow W^{\ast}$ by:
$\left(Aw_1\right)(w_2):= b(w_1,w_2).$
Conversely a  map $A: W \rightarrow W^{\ast}$ defines a bilinear form on $W$. If the bilinear form $b$ is non degenerated $A$ is an isomorphism. In this case $b$ gives us an identification of elements of $W$ and $W^{\ast}.$ A typical example is given by an euclidean vectorspace $\left( W,\langle \cdot , \cdot \rangle \right)$. There we can identify an element $w \in W$ with $w^{\ast} \in W^{\ast}$ by:
$w^{\ast} = \langle w , \cdot \rangle.$

Proposition: With the identification above we have: $b: W \times W \rightarrow \mathbb{R}$ is symmetric if and only if $A: W \rightarrow W^{\ast}$ is self-adjoint.

Proof: First note that for $A: W \rightarrow W^{\ast}$ we have $A^{\ast}: \underbrace{W^{\ast \ast}}_{= W} \rightarrow W^{\ast}.$
$b(w_1,w_2) = \left(Aw_1\right)(w_2) = \underbrace{w_2}_{\in W^{\ast \ast}}\left( Aw_1\right) = \left( A^{\ast}w_2\right)(w_1)$
Therefore we get: $b(w_1,w_2)=b(w_2,w_1) \Leftrightarrow A^{\ast} = A.$

$\square$

Now lets get back to the Dirichlet energy. The quadratic form $E_D$ on $W_{PL}$ can be identified with a symmetric bilinear map $\tilde{E_D}: W_{PL} \times W_{PL} \rightarrow \mathbb{R}$, which again corresponds to linear map $\hat{E_D} : W_{PL} \rightarrow W_{PL} ^{\ast}$. Together with the interpolation map we obtain the following diagram:
$\begin{array}{lcl} W_{PL} & \overset{\hat{E_D}}{\longrightarrow} & W_{PL}^{\ast} \\ \uparrow \mbox{interpol.} & & \downarrow \mbox{interpol.}^{\ast} \\ \mathbb{R}^V & \overset{A}{\longrightarrow} & \mathbb{R}^{V\ast} \end{array}$

If $\overset{\circ}{V}$ denotes the set of interior vertices and $V^{\partial}$ the one of boundary vertices we can give a discrete analog of the Dirichlet boundary problem in the following way:

Theorem: Given $g:V^{\partial} \rightarrow \mathbb{R}$ there is a unique $f:V \rightarrow \mathbb{R}$ such that:
$\left\{ \begin{array}{ll} \left(A f \right)_i = 0 \quad \forall i \in \overset{\circ}{V}, \\ \left. f\right|_{V^{\partial}} = g.\end{array} \right.$

A function $f:V \rightarrow \mathbb{R}$ that solves the above Dirichlet problem is called harmonic.

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