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<section id="finite-element-spaces-local-to-global">
<h1><span class="section-number">2. </span>Finite element spaces: local to global<a class="headerlink" href="#finite-element-spaces-local-to-global" title="Link to this heading">¶</a></h1>
<p>In this section, we discuss the construction of general finite element
spaces. Given a triangulation <span class="math notranslate nohighlight">\(\mathcal{T}\)</span> of a domain <span class="math notranslate nohighlight">\(\Omega\)</span>, finite
element spaces are defined according to</p>
<ol class="arabic simple">
<li><p>the form the functions take (usually polynomial) when restricted to each cell (a triangle, in the case considered so far),</p></li>
<li><p>the continuity of the functions between cells.</p></li>
</ol>
<p>We also need a mechanism to explicitly build a basis for the finite
element space. We first do this by looking at a single cell, which we
call the local perspective. Later we will take the global perspective,
seeing how function continuity is enforced between cells.</p>
<section id="ciarlet-s-finite-element">
<h2><span class="section-number">2.1. </span>Ciarlet’s finite element<a class="headerlink" href="#ciarlet-s-finite-element" title="Link to this heading">¶</a></h2>
<details class="sd-sphinx-override sd-dropdown sd-card sd-mb-3">
<summary class="sd-summary-title sd-card-header">
<span class="sd-summary-text">A video recording of the following material is available here.</span><span class="sd-summary-state-marker sd-summary-chevron-right"><svg version="1.1" width="1.5em" height="1.5em" class="sd-octicon sd-octicon-chevron-right" viewBox="0 0 24 24" aria-hidden="true"><path d="M8.72 18.78a.75.75 0 0 1 0-1.06L14.44 12 8.72 6.28a.751.751 0 0 1 .018-1.042.751.751 0 0 1 1.042-.018l6.25 6.25a.75.75 0 0 1 0 1.06l-6.25 6.25a.75.75 0 0 1-1.06 0Z"></path></svg></span></summary><div class="sd-summary-content sd-card-body docutils">
<div class="vimeo docutils container">
<iframe src="https://player.vimeo.com/video/490693460"
frameborder="0" allow="autoplay; fullscreen"
allowfullscreen></iframe></div>
<p class="sd-card-text">Imperial students can also <a class="reference external" href="https://imperial.cloud.panopto.eu/Panopto/Pages/Viewer.aspx?id=e79807c3-c73b-42ec-b5f9-ac8d00c7b4c6">watch this video on Panopto</a></p>
</div>
</details><p>The first part of the definition is formalised by Ciarlet’s definition
of a finite element.</p>
<div class="proof proof-type-definition" id="id1">
<div class="proof-title">
<span class="proof-type">Definition 2.1</span>
<span class="proof-title-name">(Ciarlet’s finite element)</span>
</div><div class="proof-content">
<p>Let</p>
<ol class="arabic simple">
<li><p>the element domain <span class="math notranslate nohighlight">\(K\subset \mathbb{R}^n\)</span> be some bounded closed set with piecewise smooth boundary,</p></li>
<li><p>the space of shape functions <span class="math notranslate nohighlight">\(\mathcal{P}\)</span> be a finite dimensional space of functions on <span class="math notranslate nohighlight">\(K\)</span>, and</p></li>
<li><p>the set of nodal variables <span class="math notranslate nohighlight">\(\mathcal{N}=(N_0,\ldots,N_k)\)</span> be a basis for the dual space <span class="math notranslate nohighlight">\(P'\)</span>.</p></li>
</ol>
<p>Then <span class="math notranslate nohighlight">\((K,\mathcal{P},\mathcal{N})\)</span> is called a finite element.</p>
</div></div><p>For the cases considered in this course, <span class="math notranslate nohighlight">\(K\)</span> will be a polygon such as a triangle, square, tetrahedron or cube, and <span class="math notranslate nohighlight">\(P\)</span> will be a space of polynomials. Here, <span class="math notranslate nohighlight">\(P'\)</span> is the dual space to <span class="math notranslate nohighlight">\(P\)</span>, defined as the space of linear functions from <span class="math notranslate nohighlight">\(P\)</span> to <span class="math notranslate nohighlight">\(\mathbb{R}\)</span>. Examples of dual functions to <span class="math notranslate nohighlight">\(P\)</span> include:</p>
<ol class="arabic simple">
<li><p>The evaluation of <span class="math notranslate nohighlight">\(p\in P\)</span> at a point <span class="math notranslate nohighlight">\(x\in K\)</span>.</p></li>
<li><p>The integral of <span class="math notranslate nohighlight">\(p\in P\)</span> over a line <span class="math notranslate nohighlight">\(l\in K\)</span>.</p></li>
<li><p>The integral of <span class="math notranslate nohighlight">\(p\in P\)</span> over <span class="math notranslate nohighlight">\(K\)</span>.</p></li>
<li><p>The evaluation of a component of the derivative of <span class="math notranslate nohighlight">\(p\in P\)</span> at a point <span class="math notranslate nohighlight">\(x\in K\)</span>.</p></li>
</ol>
<div class="proof proof-type-exercise" id="id2">
<div class="proof-title">
<span class="proof-type">Exercise 2.2</span>
</div><div class="proof-content">
<p>Show that the four examples above are all linear functions from <span class="math notranslate nohighlight">\(P\)</span> to <span class="math notranslate nohighlight">\(\mathbb{R}\)</span>.</p>
</div></div><div class="proof proof-type-exercise" id="id3">
<div class="proof-title">
<span class="proof-type">Exercise 2.3</span>
</div><div class="proof-content">
<p>For a domain <span class="math notranslate nohighlight">\(K\)</span> and shape space <span class="math notranslate nohighlight">\(P\)</span>, is the following
functional a nodal variable? Explain your answer.</p>
<div class="math notranslate nohighlight">
\[N_0(p) = \int_K p^2 \,d x.\]</div>
</div></div><p>Ciarlet’s finite element provides us with a standard way to define a basis for the <span class="math notranslate nohighlight">\(P\)</span>, called the nodal basis.</p>
<details class="sd-sphinx-override sd-dropdown sd-card sd-mb-3">
<summary class="sd-summary-title sd-card-header">
<span class="sd-summary-text">A video recording of the following material is available here.</span><span class="sd-summary-state-marker sd-summary-chevron-right"><svg version="1.1" width="1.5em" height="1.5em" class="sd-octicon sd-octicon-chevron-right" viewBox="0 0 24 24" aria-hidden="true"><path d="M8.72 18.78a.75.75 0 0 1 0-1.06L14.44 12 8.72 6.28a.751.751 0 0 1 .018-1.042.751.751 0 0 1 1.042-.018l6.25 6.25a.75.75 0 0 1 0 1.06l-6.25 6.25a.75.75 0 0 1-1.06 0Z"></path></svg></span></summary><div class="sd-summary-content sd-card-body docutils">
<div class="vimeo docutils container">
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frameborder="0" allow="autoplay; fullscreen"
allowfullscreen></iframe></div>
<p class="sd-card-text">Imperial students can also <a class="reference external" href="https://imperial.cloud.panopto.eu/Panopto/Pages/Viewer.aspx?id=9d432870-a298-4e6e-b495-ac8d00cc6411">watch this video on Panopto</a></p>
</div>
</details><div class="proof proof-type-definition" id="id4">
<div class="proof-title">
<span class="proof-type">Definition 2.4</span>
<span class="proof-title-name">((local) nodal basis)</span>
</div><div class="proof-content">
<p>Let <span class="math notranslate nohighlight">\((K,\mathcal{P},\mathcal{N})\)</span> be a finite element. The nodal basis is the basis <span class="math notranslate nohighlight">\(\{\phi_0,\phi_1,\ldots,\phi_k\}\)</span> of <span class="math notranslate nohighlight">\(\mathcal{P}\)</span> that is dual to <span class="math notranslate nohighlight">\(\mathcal{N}\)</span>, i.e.</p>
<div class="math notranslate nohighlight">
\[N_i(\phi_j) = \delta_{ij}, \quad 0\leq i,j \leq k.\]</div>
</div></div><p>We now introduce our first example of a Ciarlet element.</p>
<div class="proof proof-type-definition" id="id5">
<span id="d-lagrange"></span>
<div class="proof-title">
<span class="proof-type">Definition 2.5</span>
<span class="proof-title-name">(The 1-dimensional Lagrange element)</span>
</div><div class="proof-content">
<p>The 1-dimensional Lagrange element <span class="math notranslate nohighlight">\((K,\mathcal{P},\mathcal{N})\)</span> of
degree <span class="math notranslate nohighlight">\(k\)</span> is defined by</p>
<ol class="arabic simple">
<li><p><span class="math notranslate nohighlight">\(K\)</span> is the interval <span class="math notranslate nohighlight">\([a,b]\)</span> for <span class="math notranslate nohighlight">\(-\infty<a<b<\infty\)</span>.</p></li>
<li><p><span class="math notranslate nohighlight">\(\mathcal{P}\)</span> is the (<span class="math notranslate nohighlight">\(k+1\)</span>)-dimensional space of degree <span class="math notranslate nohighlight">\(k\)</span> polynomials on <span class="math notranslate nohighlight">\(K\)</span>,</p></li>
<li><p><span class="math notranslate nohighlight">\(\mathcal{N}=\{N_0,\ldots,N_k\}\)</span> with</p></li>
</ol>
<div class="math notranslate nohighlight">
\[N_i(v) = v(x_i), \, x_i = a + (b-a)i/k, \quad \forall v\in \mathcal{P},\,
i=0,\ldots,k.\]</div>
</div></div><div class="proof proof-type-exercise" id="id6">
<span id="exe-1d-lagrange-basis"></span>
<div class="proof-title">
<span class="proof-type">Exercise 2.6</span>
</div><div class="proof-content">
<p>Show that the nodal basis for <span class="math notranslate nohighlight">\(\mathcal{P}\)</span> is given by</p>
<div class="math notranslate nohighlight">
\[\phi_i(x) = \frac{\prod_{j=0,j\ne i}^k (x-x_j)}{\prod_{j=0,j\ne i}^k (x_i-x_j)}, \quad i=0,\ldots,k.\]</div>
</div></div></section>
<section id="vandermonde-matrix-and-unisolvence">
<h2><span class="section-number">2.2. </span>Vandermonde matrix and unisolvence<a class="headerlink" href="#vandermonde-matrix-and-unisolvence" title="Link to this heading">¶</a></h2>
<details class="sd-sphinx-override sd-dropdown sd-card sd-mb-3">
<summary class="sd-summary-title sd-card-header">
<span class="sd-summary-text">A video recording of the following material is available here.</span><span class="sd-summary-state-marker sd-summary-chevron-right"><svg version="1.1" width="1.5em" height="1.5em" class="sd-octicon sd-octicon-chevron-right" viewBox="0 0 24 24" aria-hidden="true"><path d="M8.72 18.78a.75.75 0 0 1 0-1.06L14.44 12 8.72 6.28a.751.751 0 0 1 .018-1.042.751.751 0 0 1 1.042-.018l6.25 6.25a.75.75 0 0 1 0 1.06l-6.25 6.25a.75.75 0 0 1-1.06 0Z"></path></svg></span></summary><div class="sd-summary-content sd-card-body docutils">
<div class="vimeo docutils container">
<iframe src="https://player.vimeo.com/video/490693083"
frameborder="0" allow="autoplay; fullscreen"
allowfullscreen></iframe></div>
<p class="sd-card-text">Imperial students can also <a class="reference external" href="https://imperial.cloud.panopto.eu/Panopto/Pages/Viewer.aspx?id=f3fb54fb-f83e-41b8-a537-ac8d00d03589">watch this video on Panopto</a></p>
</div>
</details><p>More generally, It is useful computationally to write the nodal basis
in terms of another arbitrary basis <span class="math notranslate nohighlight">\(\{\psi_i\}_{i=0}^k\)</span>. This
transformation is represented by the Vandermonde matrix.</p>
<div class="proof proof-type-definition" id="id7">
<span id="def-vandermonde"></span>
<div class="proof-title">
<span class="proof-type">Definition 2.7</span>
<span class="proof-title-name">(Vandermonde matrix)</span>
</div><div class="proof-content">
<p>Given a dual basis <span class="math notranslate nohighlight">\(\mathcal{N}\)</span> and a basis <span class="math notranslate nohighlight">\(\{\psi_i\}_{i=0}^k\)</span>,
the Vandermonde matrix is the matrix <span class="math notranslate nohighlight">\(V\)</span> with coefficients</p>
<div class="math notranslate nohighlight">
\[V_{ij} = N_j(\psi_i).\]</div>
</div></div><p>This relationship is made clear by the following lemma.</p>
<div class="proof proof-type-lemma" id="id8">
<span id="lemma-vandermonde"></span>
<div class="proof-title">
<span class="proof-type">Lemma 2.8</span>
</div><div class="proof-content">
<p>The expansion of the nodal basis <span class="math notranslate nohighlight">\(\{\phi_i\}_{i=0}^k\)</span> in terms
of another basis <span class="math notranslate nohighlight">\(\{\psi_i\}_{i=0}^k\)</span> for <span class="math notranslate nohighlight">\(\mathcal{P}\)</span>,</p>
<div class="math notranslate nohighlight">
\[\phi_i(x) = \sum_{j=0}^k \mu_{ij}\psi_j(x),\]</div>
<p>has coefficients <span class="math notranslate nohighlight">\(\mu_{ij}\)</span>, <span class="math notranslate nohighlight">\(0\leq i,j\leq k\)</span> given by</p>
<div class="math notranslate nohighlight">
\[\mu = V^{-1},\]</div>
<p>where <span class="math notranslate nohighlight">\(\mu\)</span> is the corresponding matrix.</p>
</div></div><div class="proof proof-type-proof">
<div class="proof-title">
<span class="proof-type">Proof </span>
</div><div class="proof-content">
<p>The nodal basis definition becomes</p>
<div class="math notranslate nohighlight">
\[\delta_{ij} = N_j(\phi_i) = \sum_{l=0}^k\mu_{il}N_j(\psi_l) = \sum_{l=0}^k \mu_{il}V_{lj} = (\mu V)_{ij},\]</div>
<p>where <span class="math notranslate nohighlight">\(\mu\)</span> is the matrix with coefficients <span class="math notranslate nohighlight">\(\mu_{ij}\)</span>, and <span class="math notranslate nohighlight">\(V\)</span> is the matrix with coefficients <span class="math notranslate nohighlight">\(N_j(\psi_i)\)</span>.</p>
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</div>
</details><div class="proof proof-type-exercise" id="id9">
<div class="proof-title">
<span class="proof-type">Exercise 2.9</span>
</div><div class="proof-content">
<p>Consider the following finite element.</p>
<ul>
<li><p><span class="math notranslate nohighlight">\(K\)</span> is the interval <span class="math notranslate nohighlight">\([0,1]\)</span>.</p></li>
<li><p><span class="math notranslate nohighlight">\(P\)</span> is the quadratic polynomials on <span class="math notranslate nohighlight">\(K\)</span>.</p></li>
<li><p>The nodal variables are:</p>
<div class="math notranslate nohighlight">
\[N_0[p] = p(0), \, N_1[p]=p(1), \, N_2=\int_0^1 p(x) \,d x.\]</div>
</li>
</ul>
<p>Find the corresponding nodal basis.</p>
</div></div><p>Given a triple <span class="math notranslate nohighlight">\((K,\mathcal{P},\mathcal{N})\)</span>, it is necessary to
verify that <span class="math notranslate nohighlight">\(\mathcal{N}\)</span> is indeed a basis for <span class="math notranslate nohighlight">\(\mathcal{P}'\)</span>,
i.e. that the Ciarlet element is well-defined. Then the nodal basis is
indeed a basis for <span class="math notranslate nohighlight">\(\mathcal{P}\)</span> by construction. The following lemma
provides a useful tool for checking this.</p>
<div class="proof proof-type-lemma" id="id10">
<span id="dual-condition"></span>
<div class="proof-title">
<span class="proof-type">Lemma 2.10</span>
<span class="proof-title-name">(dual condition)</span>
</div><div class="proof-content">
<p>Let <span class="math notranslate nohighlight">\(K,\mathcal{P}\)</span> be as defined above, and let <span class="math notranslate nohighlight">\(\{N_0,N_1,\ldots,N_k\}\in \mathcal{P}'\)</span>. Let <span class="math notranslate nohighlight">\(\{\psi_0,\psi_1,\ldots,\psi_k\}\)</span> be a basis for <span class="math notranslate nohighlight">\(\mathcal{P}\)</span>.</p>
<p>Then the following three statements are equivalent.</p>
<ol class="arabic">
<li><p><span class="math notranslate nohighlight">\(\{N_0,N_1,\ldots,N_k\}\)</span> is a basis for <span class="math notranslate nohighlight">\(\mathcal{P}'\)</span>.</p></li>
<li><p>The Vandermonde matrix with coefficients</p>
<div class="math notranslate nohighlight">
\[V_{ij} = N_j(\psi_i), \, 0\leq i,j\leq k,\]</div>
<p>is invertible.</p>
</li>
<li><p>If <span class="math notranslate nohighlight">\(v\in\mathcal{P}\)</span> satisfies <span class="math notranslate nohighlight">\(N_i(v)=0\)</span> for <span class="math notranslate nohighlight">\(i=0,\ldots,k\)</span>, then <span class="math notranslate nohighlight">\(v\equiv 0\)</span>.</p></li>
</ol>
</div></div><div class="proof proof-type-proof">
<div class="proof-title">
<span class="proof-type">Proof </span>
</div><div class="proof-content">
<p>Let <span class="math notranslate nohighlight">\(\{N_0,N_1,\ldots,N_k\}\)</span> be a basis for <span class="math notranslate nohighlight">\(\mathcal{P}'\)</span>. This is
equivalent to saying that given element <span class="math notranslate nohighlight">\(E\)</span> of <span class="math notranslate nohighlight">\(\mathcal{P}'\)</span>, we
can find basis coefficients <span class="math notranslate nohighlight">\(\{e_i\}_{i=0}^k\in \mathbb{R}\)</span> such
that</p>
<div class="math notranslate nohighlight">
\[E = \sum_{i=0}^k e_iN_i.\]</div>
<p>This in turn is equivalent to being able to find a vector
<span class="math notranslate nohighlight">\(e=(e_0,e_1,\ldots,e_k)^T\)</span> such that</p>
<div class="math notranslate nohighlight">
\[b_i = E(\psi_i) = \sum_{j=0}^k e_j N_j(\psi_i) = \sum_{j=0}^k e_jV_{ij},\]</div>
<p>i.e. the equation <span class="math notranslate nohighlight">\(V{e}={b}\)</span> is solvable. This means that (1) is
equivalent to (2).</p>
<p>On the other hand, we may expand any <span class="math notranslate nohighlight">\(v\in \mathcal{P}\)</span> according to</p>
<div class="math notranslate nohighlight">
\[v(x) = \sum_{i=0}^k f_i \psi_i(x).\]</div>
<p>Then</p>
<div class="math notranslate nohighlight">
\[N_i(v)=0 \iff \sum_{j=0}^k f_jN_i(\psi_j) = 0, \quad i=0,1,\ldots,k,\]</div>
<p>by linearity of <span class="math notranslate nohighlight">\(N_i\)</span>. So (3) is equivalent to</p>
<div class="math notranslate nohighlight">
\[ \sum_{j=0}^k f_jN_i(\psi_j) = 0, \quad i=0,1,\ldots,k \implies
f_j=0, \, j=0,1,\ldots,k,\]</div>
<p>which is equivalent to <span class="math notranslate nohighlight">\(V^T\)</span> being invertible, which is equivalent to
<span class="math notranslate nohighlight">\(V\)</span> being invertible, and so (3) is equivalent to (2).</p>
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</div>
</details><p>This result leads us to introducing the following terminology.</p>
<div class="proof proof-type-definition" id="id11">
<div class="proof-title">
<span class="proof-type">Definition 2.11</span>
<span class="proof-title-name">(Unisolvence.)</span>
</div><div class="proof-content">
<p>We say that <span class="math notranslate nohighlight">\(\mathcal{N}\)</span> determines <span class="math notranslate nohighlight">\(\mathcal{P}\)</span> if it satisfies
condition 3 of <a class="reference internal" href="#dual-condition"><span class="std std-numref">Lemma 2.10</span></a>. If
this is the case, we say that <span class="math notranslate nohighlight">\((K,\mathcal{P},\mathcal{N})\)</span> is
unisolvent.</p>
</div></div><p>We can now go and directly apply this lemma to the 1D Lagrange elements.</p>
<div class="proof proof-type-corollary" id="id12">
<div class="proof-title">
<span class="proof-type">Corollary 2.12</span>
</div><div class="proof-content">
<p>The 1D degree <span class="math notranslate nohighlight">\(k\)</span> Lagrange element is a finite element.</p>
</div></div><div class="proof proof-type-proof">
<div class="proof-title">
<span class="proof-type">Proof </span>
</div><div class="proof-content">
<p>Let <span class="math notranslate nohighlight">\((K,\mathcal{P},\mathcal{N})\)</span> be the degree <span class="math notranslate nohighlight">\(k\)</span> Lagrange
element. We need to check that <span class="math notranslate nohighlight">\(\mathcal{N}\)</span> determines
<span class="math notranslate nohighlight">\(\mathcal{P}\)</span>. Let <span class="math notranslate nohighlight">\(v\in \mathcal{P}\)</span> with <span class="math notranslate nohighlight">\(N_i(v)=0\)</span> for all
<span class="math notranslate nohighlight">\(N_i\in \mathcal{N}\)</span>. This means that</p>
<div class="math notranslate nohighlight">
\[v(a+(b-a)i/k)=0, \, i=,0,1,\ldots,k,\]</div>
<p>which means that <span class="math notranslate nohighlight">\(v\)</span> vanishes at <span class="math notranslate nohighlight">\(k+1\)</span> points in <span class="math notranslate nohighlight">\(K\)</span>. Since <span class="math notranslate nohighlight">\(v\)</span> is
a degree <span class="math notranslate nohighlight">\(k\)</span> polynomial, it must be zero by the fundamental theorem
of algebra.</p>
</div></div><div class="proof proof-type-exercise" id="id13">
<div class="proof-title">
<span class="proof-type">Exercise 2.13</span>
</div><div class="proof-content">
<p>Consider the following proposed finite element.</p>
<ul>
<li><p><span class="math notranslate nohighlight">\(K\)</span> is the interval <span class="math notranslate nohighlight">\([0,1]\)</span>.</p></li>
<li><p><span class="math notranslate nohighlight">\(P\)</span> is the linear polynomials on <span class="math notranslate nohighlight">\(K\)</span>.</p></li>
<li><p>The nodal variables are:</p>
<div class="math notranslate nohighlight">
\[N_0[p] = p(0.5), N_1=\int_0^1 p(x) \,d x.\]</div>
</li>
</ul>
<p>Is this finite element unisolvent? Explain your answer.</p>
</div></div></section>
<section id="d-and-3d-finite-elements">
<h2><span class="section-number">2.3. </span>2D and 3D finite elements<a class="headerlink" href="#d-and-3d-finite-elements" title="Link to this heading">¶</a></h2>
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</div>
</details><p>We would like to construct some finite elements with 2D and 3D domains
<span class="math notranslate nohighlight">\(K\)</span>. The fundamental theorem of algebra does not directly help us
there, but the following lemma is useful when checking that
<span class="math notranslate nohighlight">\(\mathcal{N}\)</span> determines <span class="math notranslate nohighlight">\(\mathcal{P}\)</span> in those cases.</p>
<div class="proof proof-type-lemma" id="id14">
<div class="proof-title">
<span class="proof-type">Lemma 2.14</span>
</div><div class="proof-content">
<p>Let <span class="math notranslate nohighlight">\(p(x):\mathbb{R}^d\to\mathbb{R}\)</span> be a polynomial of degree <span class="math notranslate nohighlight">\(k\geq 1\)</span>
that vanishes on a hyperplane <span class="math notranslate nohighlight">\(\Pi_L\)</span> defined by</p>
<div class="math notranslate nohighlight">
\[\Pi_L = \left\{ x: L(x)=0\right\},\]</div>
<p>for a non-degenerate affine function <span class="math notranslate nohighlight">\(L(x):\mathbb{R}^d\to
\mathbb{R}\)</span>. Then <span class="math notranslate nohighlight">\(p(x)=L(x)q(x)\)</span> where <span class="math notranslate nohighlight">\(q(x)\)</span> is a polynomial of
degree <span class="math notranslate nohighlight">\(k-1\)</span>.</p>
</div></div><div class="proof proof-type-proof">
<div class="proof-title">
<span class="proof-type">Proof </span>
</div><div class="proof-content">
<p>Choose coordinates (by shifting the origin and applying a linear
transformation) such that <span class="math notranslate nohighlight">\(x=(x_1,\ldots,x_d)\)</span> with <span class="math notranslate nohighlight">\(L(x)=x_d\)</span>, so
<span class="math notranslate nohighlight">\(\Pi_L\)</span> is defined by <span class="math notranslate nohighlight">\(x_d=0\)</span>. Then the general form for a
polynomial is</p>
<div class="math notranslate nohighlight">
\[P(x_1,\ldots,x_d) = \sum_{i_d=0}^k
\left(\sum_{|i_1+\ldots+i_{d-1}|\leq
k-i_d}c_{i_1,\ldots,i_{d-1},i_d} x_d^{i_d}\prod_{l=1}^{d-1}
x_{l}^{i_l}\right),\]</div>
<p>Then, <span class="math notranslate nohighlight">\(p(x_1,\ldots,x_{d-1},0)=0\)</span> for all <span class="math notranslate nohighlight">\((x_1,\ldots,x_{d-1})\)</span>,
so</p>
<div class="math notranslate nohighlight">
\[0 = \left(\sum_{|i_1+\ldots+i_{d-1}|\leq k}c_{i_1,\ldots,i_{d-1},0} \prod_{l=1}^{d-1}x_{l}^{i_l}\right)\]</div>
<p>which means that</p>
<div class="math notranslate nohighlight">
\[c_{i_1,\ldots,i_{d-1},0} = 0, \quad \forall |i_1+\ldots+i_{d-1}|\leq k.\]</div>
<p>This means we may rewrite</p>
<div class="math notranslate nohighlight">
\[ \begin{align}\begin{aligned}P(x) = {L(x)}\underbrace{\left(\sum_{i_d=1}^k\sum_{|i_1+\ldots+i_{d-1}|\leq k - i_d}c_{i_1,\ldots,i_{d-1},i_d} x_d^{i_d-1}\prod_{l=1}^{d-1} x_{l}^{i_l}\right)},\\P(x) = \underbrace{x_d}_{L(x)}\underbrace{\left(\sum_{i_d=0}^{k-1}\sum_{|i_1+\ldots+i_{d-1}|\leq k - i_d}c_{i_1,\ldots,i_{d-1},i_d} x_d^{i_d-1}\prod_{l=1}^{d-1} x_{l}^{i_l}\right)}_{Q(x)},\end{aligned}\end{align} \]</div>
<p>with <span class="math notranslate nohighlight">\(\deg(Q)=k-1\)</span>.</p>
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</div>
</details><div class="proof proof-type-exercise" id="id15">
<div class="proof-title">
<span class="proof-type">Exercise 2.15</span>
</div><div class="proof-content">
<p>The following polynomial vanishes on the line <span class="math notranslate nohighlight">\(y=1-x\)</span>. Show that
it satisfies the result of the previous theorem.</p>
<div class="math notranslate nohighlight">
\[x^{5} + 5 x^{4} y - x^{4} + 6 x^{3} y^{2} - 4 x^{3} y - 2 x^{2}
y^{3} - 2 x^{2} y^{2} - 3 x y^{4} + 4 x y^{3} + y^{5} - y^{4}\]</div>
</div></div><p>Equipped with this tool we can consider some finite elements in two
dimensions.</p>
<div class="proof proof-type-definition" id="id16">
<div class="proof-title">
<span class="proof-type">Definition 2.16</span>
<span class="proof-title-name">(Lagrange elements on triangles)</span>
</div><div class="proof-content">
<p>The triangular Lagrange element of degree <span class="math notranslate nohighlight">\(k\)</span>
<span class="math notranslate nohighlight">\((K,\mathcal{P},\mathcal{N})\)</span>, denoted <span class="math notranslate nohighlight">\(Pk\)</span>, is defined as follows.</p>
<ol class="arabic simple">
<li><p><span class="math notranslate nohighlight">\(K\)</span> is a (non-degenerate) triangle with vertices <span class="math notranslate nohighlight">\(z_1\)</span>, <span class="math notranslate nohighlight">\(z_2\)</span>, <span class="math notranslate nohighlight">\(z_3\)</span>.</p></li>
<li><p><span class="math notranslate nohighlight">\(\mathcal{P}\)</span> is the space of degree <span class="math notranslate nohighlight">\(k\)</span> polynomials on <span class="math notranslate nohighlight">\(K\)</span>.</p></li>
<li><p><span class="math notranslate nohighlight">\(\mathcal{N}=\left\{N_{i,j}:0\leq i \leq k, \, 0\leq j \leq i\right\}\)</span> defined by <span class="math notranslate nohighlight">\(N_{i,j}(v)=v(x_{i,j})\)</span> where</p></li>
</ol>
<div class="math notranslate nohighlight">
\[x_{i,j} = z_1 + (z_2-z_1)\frac{i}{k} + (z_3-z_1)\frac{j}{k}.\]</div>
</div></div><p>We illustrate this for the cases <span class="math notranslate nohighlight">\(k=1,2\)</span>.</p>
<div class="proof proof-type-example" id="id17">
<div class="proof-title">
<span class="proof-type">Example 2.17</span>
<span class="proof-title-name">(P1 elements on triangles)</span>
</div><div class="proof-content">
<p>The nodal basis for P1 elements is point evaluation at the three vertices.</p>
</div></div><div class="proof proof-type-example" id="id18">
<div class="proof-title">
<span class="proof-type">Example 2.18</span>
<span class="proof-title-name">(P2 elements on triangles)</span>
</div><div class="proof-content">
<p>The nodal basis for P2 elements is point evaluation at the three
vertices, plus point evaluation at the three edge centres.</p>
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</div>
</details><p>We now need to check that that the degree <span class="math notranslate nohighlight">\(k\)</span> Lagrange element is a
finite element, i.e. that <span class="math notranslate nohighlight">\(\mathcal{N}\)</span> determines <span class="math notranslate nohighlight">\(\mathcal{P}\)</span>. We will
first do this for <span class="math notranslate nohighlight">\(P1\)</span>.</p>
<div class="proof proof-type-lemma" id="id19">
<span id="p1unisolve"></span>
<div class="proof-title">
<span class="proof-type">Lemma 2.19</span>
</div><div class="proof-content">
<p>The degree <span class="math notranslate nohighlight">\(1\)</span> Lagrange element on a triangle <span class="math notranslate nohighlight">\(K\)</span> is a finite element.</p>
</div></div><div class="proof proof-type-proof">
<div class="proof-title">
<span class="proof-type">Proof </span>
</div><div class="proof-content">
<p>Let <span class="math notranslate nohighlight">\(\Pi_1\)</span>, <span class="math notranslate nohighlight">\(\Pi_2\)</span>, <span class="math notranslate nohighlight">\(\Pi_3\)</span> be the three lines containing the
vertices <span class="math notranslate nohighlight">\(z_2\)</span> and <span class="math notranslate nohighlight">\(z_3\)</span>, <span class="math notranslate nohighlight">\(z_1\)</span> and <span class="math notranslate nohighlight">\(z_3\)</span>, and <span class="math notranslate nohighlight">\(z_1\)</span> and <span class="math notranslate nohighlight">\(z_3\)</span>
respectively, and defined by <span class="math notranslate nohighlight">\(L_1=0\)</span>, <span class="math notranslate nohighlight">\(L_2=0\)</span>, and <span class="math notranslate nohighlight">\(L_3=0\)</span>
respectively. Consider a linear polynomial <span class="math notranslate nohighlight">\(p\)</span> vanishing at <span class="math notranslate nohighlight">\(z_1\)</span>,
<span class="math notranslate nohighlight">\(z_2\)</span>, and <span class="math notranslate nohighlight">\(z_3\)</span>. The restriction <span class="math notranslate nohighlight">\(p|_{\Pi_1}\)</span> of <span class="math notranslate nohighlight">\(p\)</span> to <span class="math notranslate nohighlight">\(\Pi_1\)</span> is
a linear function vanishing at two points, and therefore <span class="math notranslate nohighlight">\(p=0\)</span> on
<span class="math notranslate nohighlight">\(\Pi_1\)</span>, and so <span class="math notranslate nohighlight">\(p=L_1(x)Q(x)\)</span>, where <span class="math notranslate nohighlight">\(Q(x)\)</span> is a degree 0
polynomial, i.e. a constant <span class="math notranslate nohighlight">\(c\)</span>. We also have</p>
<div class="math notranslate nohighlight">
\[0 = p(z_1) = cL_1(z_1) \implies c=0,\]</div>
<p>since <span class="math notranslate nohighlight">\(L_1(z_1)\neq 0\)</span>, and hence <span class="math notranslate nohighlight">\(p(x)\equiv 0\)</span>. This means
that <span class="math notranslate nohighlight">\(\mathcal{N}\)</span> determines <span class="math notranslate nohighlight">\(\mathcal{P}\)</span>.</p>
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</details><div class="proof proof-type-exercise" id="id20">
<div class="proof-title">
<span class="proof-type">Exercise 2.20</span>
</div><div class="proof-content">
<p>Let <span class="math notranslate nohighlight">\(K\)</span> be a rectangle, <span class="math notranslate nohighlight">\(P\)</span> be the polynomial space spanned by
<span class="math notranslate nohighlight">\(\{1, x, y, xy\}\)</span>, let <span class="math notranslate nohighlight">\(\mathcal{N}\)</span> be the set of dual elements
corresponding to point evaluation at each vertex of the
rectangle. Show that <span class="math notranslate nohighlight">\(\mathcal{N}\)</span> determines the finite element.</p>
</div></div><p>This technique can then be extended to degree 2.</p>
<div class="proof proof-type-lemma" id="id21">
<div class="proof-title">
<span class="proof-type">Lemma 2.21</span>
</div><div class="proof-content">
<p>The degree <span class="math notranslate nohighlight">\(2\)</span> Lagrange element is a finite element.</p>
</div></div><div class="proof proof-type-proof">
<div class="proof-title">
<span class="proof-type">Proof </span>
</div><div class="proof-content">
<p>Let <span class="math notranslate nohighlight">\(p\)</span> be a degree <span class="math notranslate nohighlight">\(2\)</span> polynomial with <span class="math notranslate nohighlight">\(N_i(p)\)</span> for all of the
degree <span class="math notranslate nohighlight">\(2\)</span> dual basis elements. Let <span class="math notranslate nohighlight">\(\Pi_1\)</span>, <span class="math notranslate nohighlight">\(\Pi_2\)</span>, <span class="math notranslate nohighlight">\(\Pi_3\)</span>,
<span class="math notranslate nohighlight">\(L_1\)</span>, <span class="math notranslate nohighlight">\(L_2\)</span> and <span class="math notranslate nohighlight">\(L_3\)</span> be defined as for the proof of Lemma
. <span class="math notranslate nohighlight">\(p|_{\Pi_1}\)</span> is a degree 2 scalar polynomial vanishing
at 3 points, and therefore <span class="math notranslate nohighlight">\(p=0\)</span> on <span class="math notranslate nohighlight">\(\Pi_1\)</span>, and so
<span class="math notranslate nohighlight">\(p(x)=L_1(x)Q_1(x)\)</span> with <span class="math notranslate nohighlight">\(\deg(Q_1)=1\)</span>. We also have <span class="math notranslate nohighlight">\(0=p|_{\Pi_2}
=L_1Q_1|_{\Pi_2}\)</span>, so <span class="math notranslate nohighlight">\(Q_1|_{\Pi_2}=0\)</span> and we conclude that
<span class="math notranslate nohighlight">\(p(x)=cL_1(x)L_2(x)\)</span>. Finally, <span class="math notranslate nohighlight">\(p\)</span> also vanishes at the midpoint of
<span class="math notranslate nohighlight">\(L_3\)</span>, so we conclude that <span class="math notranslate nohighlight">\(c=0\)</span> as required.</p>
</div></div><p>The technique extends further to degree 3.</p>
<div class="proof proof-type-exercise" id="id22">
<div class="proof-title">
<span class="proof-type">Exercise 2.22</span>
</div><div class="proof-content">
<p>Show that the degree <span class="math notranslate nohighlight">\(3\)</span> Lagrange element is a finite element.</p>
</div></div><p>Going beyond degree 3, we have more than 1 nodal variable taking point
evaluation inside the triangle. To deal with this, we use the nested
triangular structure of the Lagrange triangle.</p>
<div class="proof proof-type-lemma" id="id23">
<span id="lem-degk-unisolve"></span>
<div class="proof-title">
<span class="proof-type">Lemma 2.23</span>
</div><div class="proof-content">
<p>The degree <span class="math notranslate nohighlight">\(k\)</span> Lagrange element is a finite element for <span class="math notranslate nohighlight">\(k>3\)</span>.</p>
</div></div><div class="proof proof-type-proof">
<div class="proof-title">
<span class="proof-type">Proof </span>
</div><div class="proof-content">
<p>We prove by induction. Assume that the degree <span class="math notranslate nohighlight">\(k-3\)</span> Lagrange
element is a finite element. Let <span class="math notranslate nohighlight">\(p\)</span> be a degree <span class="math notranslate nohighlight">\(k\)</span> polynomial
with <span class="math notranslate nohighlight">\(N_i(p)\)</span> for all of the degree <span class="math notranslate nohighlight">\(k\)</span> dual basis elements. Let
<span class="math notranslate nohighlight">\(\Pi_1\)</span>, <span class="math notranslate nohighlight">\(\Pi_2\)</span>, <span class="math notranslate nohighlight">\(\Pi_3\)</span>, <span class="math notranslate nohighlight">\(L_1\)</span>, <span class="math notranslate nohighlight">\(L_2\)</span> and <span class="math notranslate nohighlight">\(L_3\)</span> be defined as for
the proof of <a class="reference internal" href="#p1unisolve"><span class="std std-numref">lemma 2.19</span></a>. The restriction
<span class="math notranslate nohighlight">\(p|_{\Pi_1}\)</span> is a degree <span class="math notranslate nohighlight">\(k\)</span> polynomial in one variable that
vanishes at <span class="math notranslate nohighlight">\(k+1\)</span> points, and therefore <span class="math notranslate nohighlight">\(p(x)=L_1(x)Q_1(x)\)</span>, with
<span class="math notranslate nohighlight">\(\deg(Q_1)=k-1\)</span>. <span class="math notranslate nohighlight">\(p\)</span> and therefore <span class="math notranslate nohighlight">\(Q\)</span> also vanishes on <span class="math notranslate nohighlight">\(\Pi_2\)</span>, so
<span class="math notranslate nohighlight">\(Q_1(x)=L_2(x)Q_2(x)\)</span>.</p>
<p>Repeating the argument
again means that <span class="math notranslate nohighlight">\(p(x)=L_1(x)L_2(x)L_3(x)Q_3(x)\)</span>, with <span class="math notranslate nohighlight">\(\deg(Q_3)=k-3\)</span>.
<span class="math notranslate nohighlight">\(Q_3\)</span> must vanish on the remaining points in the interior of <span class="math notranslate nohighlight">\(K\)</span>, which
are arranged in a smaller triangle <span class="math notranslate nohighlight">\(K'\)</span> and correspond to the evaluation
points for a degree <span class="math notranslate nohighlight">\(k-3\)</span> Lagrange finite element on <span class="math notranslate nohighlight">\(K'\)</span>. From
the inductive hypothesis, and using the results for <span class="math notranslate nohighlight">\(k=1,2,3\)</span>, we conclude
that <span class="math notranslate nohighlight">\(Q_3\equiv=0\)</span>, and therefore <span class="math notranslate nohighlight">\(p\equiv0\)</span> as required.</p>
</div></div></section>
<section id="some-more-exotic-elements">
<h2><span class="section-number">2.4. </span>Some more exotic elements<a class="headerlink" href="#some-more-exotic-elements" title="Link to this heading">¶</a></h2>
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</details><p>We now consider some finite elements that involve derivative
evaluation. The Hermite elements involve evaluation of first
derivatives as well as point evaluations.</p>
<div class="proof proof-type-definition" id="id24">
<div class="proof-title">
<span class="proof-type">Definition 2.24</span>
<span class="proof-title-name">(Cubic Hermite elements on triangles)</span>
</div><div class="proof-content">
<p>The cubic Hermite element is defined as follows:</p>
<ol class="arabic simple">
<li><p><span class="math notranslate nohighlight">\(K\)</span> is a (nondegenerate) triangle,</p></li>
<li><p><span class="math notranslate nohighlight">\(\mathcal{P}\)</span> is the space of cubic polynomials on <span class="math notranslate nohighlight">\(K\)</span>,</p></li>
<li><p><span class="math notranslate nohighlight">\(\mathcal{N}=\{N_1,N_2,\ldots,N_{10}\}\)</span> defined as follows:</p>
<ul class="simple">
<li><p><span class="math notranslate nohighlight">\((N_1,\ldots,N_3)\)</span>: evaluation of <span class="math notranslate nohighlight">\(p\)</span> at vertices,</p></li>
<li><p><span class="math notranslate nohighlight">\((N_4,\ldots,N_9)\)</span>: evaluation of the gradient of <span class="math notranslate nohighlight">\(p\)</span> at the 3 triangle vertices.</p></li>
<li><p><span class="math notranslate nohighlight">\(N_{10}\)</span>: evaluation of <span class="math notranslate nohighlight">\(p\)</span> at the centre of the triangle.</p></li>
</ul>
</li>
</ol>
</div></div><p>It turns out that the Hermite element is insufficient to guarantee
functions with continuous derivatives between triangles. This problem
is solved by the Argyris element.</p>
<div class="proof proof-type-definition" id="id25">
<div class="proof-title">
<span class="proof-type">Definition 2.25</span>
<span class="proof-title-name">(Quintic Argyris elements on triangles)</span>
</div><div class="proof-content">
<p>The quintic Argyris element is defined as follows:</p>
<ol class="arabic simple">
<li><p><span class="math notranslate nohighlight">\(K\)</span> is a (nondegenerate) triangle,</p></li>
<li><p><span class="math notranslate nohighlight">\(\mathcal{P}\)</span> is the space of quintic polynomials on <span class="math notranslate nohighlight">\(K\)</span>,</p></li>
<li><p><span class="math notranslate nohighlight">\(\mathcal{N}\)</span> defined as follows:</p>
<ul class="simple">
<li><p>evaluation of <span class="math notranslate nohighlight">\(p\)</span> at 3 vertices,</p></li>
<li><p>evaluation of gradient of <span class="math notranslate nohighlight">\(p\)</span> at 3 vertices,</p></li>
<li><p>evaluation of Hessian of <span class="math notranslate nohighlight">\(p\)</span> at 3 vertices,</p></li>
<li><p>evaluation of the gradient normal to 3 triangle edges.</p></li>
</ul>
</li>
</ol>
</div></div></section>
<section id="global-continuity">
<h2><span class="section-number">2.5. </span>Global continuity<a class="headerlink" href="#global-continuity" title="Link to this heading">¶</a></h2>
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</details><p>Next we need to know how to glue finite elements together to form
spaces defined over a triangulation (mesh). To do this we need to
develop a language for specifying connections between finite element
functions between element domains.</p>
<div class="proof proof-type-definition" id="id26">
<div class="proof-title">
<span class="proof-type">Definition 2.26</span>
<span class="proof-title-name">(Finite element space)</span>
</div><div class="proof-content">
<p>Let <span class="math notranslate nohighlight">\(\mathcal{T}\)</span> be a triangulation made of triangles <span class="math notranslate nohighlight">\(K_i\)</span>, with
finite elements <span class="math notranslate nohighlight">\((K_i,\mathcal{P}_i,\mathcal{N}_i)\)</span>. A space <span class="math notranslate nohighlight">\(V\)</span> of
functions on <span class="math notranslate nohighlight">\(\mathcal{T}\)</span> is called a finite element space if for
each <span class="math notranslate nohighlight">\(u\in V\)</span>, and for each <span class="math notranslate nohighlight">\(K_i\in\mathcal{T}\)</span>, <span class="math notranslate nohighlight">\(u|_{K_i}\in
\mathcal{P}_i\)</span>.</p>
</div></div><p>Note that the set of finite elements do not uniquely determine a
finite element space, since we also need to specify continuity
requirements between triangles, which we will do in this chapter.</p>
<div class="proof proof-type-definition" id="id27">
<div class="proof-title">
<span class="proof-type">Definition 2.27</span>
<span class="proof-title-name">(Finite element space)</span>
</div><div class="proof-content">
<p>A finite element space <span class="math notranslate nohighlight">\(V\)</span> is a <span class="math notranslate nohighlight">\(C^m\)</span> finite element space if <span class="math notranslate nohighlight">\(u\in
C^m\)</span> for all <span class="math notranslate nohighlight">\(u\in V\)</span>.</p>
</div></div><p>The following lemma guides use in how to inspect the continuity of
finite element functions.</p>
<div class="proof proof-type-lemma" id="id28">
<span id="cty"></span>
<div class="proof-title">
<span class="proof-type">Lemma 2.28</span>
<span class="proof-title-name">(Continuity lemma)</span>
</div><div class="proof-content">
<p>Let <span class="math notranslate nohighlight">\(\mathcal{T}\)</span> be a triangulation on <span class="math notranslate nohighlight">\(\Omega\)</span>, and let
<span class="math notranslate nohighlight">\(V\)</span> be a finite element space defined on <span class="math notranslate nohighlight">\(\mathcal{T}\)</span>.
The following two statements are equivalent.</p>
<ol class="arabic simple">
<li><p><span class="math notranslate nohighlight">\(V\)</span> is a <span class="math notranslate nohighlight">\(C^m\)</span> finite element space.</p></li>
<li><p>The following two conditions hold.</p></li>
</ol>
<blockquote>
<div><ul class="simple">
<li><p>For each vertex <span class="math notranslate nohighlight">\(z\)</span> in <span class="math notranslate nohighlight">\(\mathcal{T}\)</span>, let <span class="math notranslate nohighlight">\(\{K_i\}_{i=1}^m\)</span> be the set of triangles that contain <span class="math notranslate nohighlight">\(z\)</span>. Then <span class="math notranslate nohighlight">\(u|_{K_1}(z)=u|_{K_2}(z)=\ldots = u|_{K_m}(z)\)</span>, for all functions <span class="math notranslate nohighlight">\(u\in V\)</span>, and similarly for all of the partial derivatives of degrees up to <span class="math notranslate nohighlight">\(m\)</span>.</p></li>
<li><p>For each edge <span class="math notranslate nohighlight">\(e\)</span> in <span class="math notranslate nohighlight">\(\mathcal{T}\)</span>, let <span class="math notranslate nohighlight">\(K_1\)</span>, <span class="math notranslate nohighlight">\(K_2\)</span> be the two triangles containing <span class="math notranslate nohighlight">\(e\)</span>. Then <span class="math notranslate nohighlight">\(u|_{K_1}(z) = u|_{K_2}(z)\)</span>, for all points <span class="math notranslate nohighlight">\(z\)</span> on the interior of <span class="math notranslate nohighlight">\(e\)</span>, and similarly for all of the partial derivatives of degrees up to <span class="math notranslate nohighlight">\(m\)</span>.</p></li>
</ul>
</div></blockquote>
</div></div><div class="proof proof-type-proof">
<div class="proof-title">
<span class="proof-type">Proof </span>
</div><div class="proof-content">
<p><span class="math notranslate nohighlight">\(V\)</span> is polynomial on each triangle <span class="math notranslate nohighlight">\(K\)</span>, so continuity at points on
the interior of each triangle <span class="math notranslate nohighlight">\(K\)</span> is immediate. We just need to
check continuity at points on vertices, and points on the interior
of edges, which is equivalent to the two parts of the second
condition.</p>
</div></div><p>This means that we just need to guarantee that the polynomial
functions and their derivatives agree at vertices and edges (similar
ideas extend to higher dimensions). We achieve this by assigning nodal
variables (and their associated nodal basis functions) appropriately
to vertices, edges etc. of each triangle <span class="math notranslate nohighlight">\(K\)</span>. First we need to
introduce this terminology.</p>
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</details><div class="proof proof-type-definition" id="id29">
<div class="proof-title">
<span class="proof-type">Definition 2.29</span>
<span class="proof-title-name">(local and global mesh entities)</span>
</div><div class="proof-content">
<p>Let <span class="math notranslate nohighlight">\(K\)</span> be a triangle. The local mesh entities of <span class="math notranslate nohighlight">\(K\)</span> are the
vertices, the edges, and <span class="math notranslate nohighlight">\(K\)</span> itself. The global mesh entities of a
triangulation <span class="math notranslate nohighlight">\(\mathcal{T}\)</span> are the vertices, edges and triangles
comprising <span class="math notranslate nohighlight">\(\mathcal{T}\)</span>.</p>
</div></div><p>Having made this definition, we can now talk about how nodal variables
can be assigned to local mesh entities in a geometric decomposition.</p>
<div class="proof proof-type-definition" id="id30">
<div class="proof-title">
<span class="proof-type">Definition 2.30</span>
<span class="proof-title-name">(local geometric decomposition)</span>
</div><div class="proof-content">
<p>Let <span class="math notranslate nohighlight">\((K,\mathcal{P},\mathcal{N})\)</span> be a finite element. We say that
the finite element has a (local) geometric decomposition if each
dual basis function <span class="math notranslate nohighlight">\(N_i\)</span> can be associated with a single mesh
entity <span class="math notranslate nohighlight">\(w\in W\)</span> such that for any <span class="math notranslate nohighlight">\(f\in\mathcal{P}\)</span>, <span class="math notranslate nohighlight">\(N_i(f)\)</span> can be
calculated from <span class="math notranslate nohighlight">\(f\)</span> and derivatives of <span class="math notranslate nohighlight">\(f\)</span> evaluated on <span class="math notranslate nohighlight">\(w\)</span>.</p>
</div></div><div class="proof proof-type-exercise" id="id31">
<div class="proof-title">
<span class="proof-type">Exercise 2.31</span>
</div><div class="proof-content">
<blockquote>
<div><p>Consider the finite element defined by:</p>
<ol class="arabic">
<li><p><span class="math notranslate nohighlight">\(K\)</span> is the unit interval <span class="math notranslate nohighlight">\([0,1]\)</span></p></li>
<li><p><span class="math notranslate nohighlight">\(P\)</span> is the space of quadratic polynomials on <span class="math notranslate nohighlight">\(K\)</span>,</p></li>
<li><p>The nodal variables are:</p>
<div class="math notranslate nohighlight">
\[N_0[v] = v(0), N_1[v] = v(1), N_2[v] = \int_0^1 v(x)\,d x.\]</div>
</li>
</ol>
</div></blockquote>
<p>Find the corresponding nodal basis for <span class="math notranslate nohighlight">\(P\)</span> in terms of the monomial
basis <span class="math notranslate nohighlight">\(\{1, x, x^2\}\)</span>. Provide the <span class="math notranslate nohighlight">\(C^0\)</span> geometric decomposition for
the finite element (demonstrating that it is indeed <span class="math notranslate nohighlight">\(C^0\)</span>).</p>
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</details><p>To discuss <span class="math notranslate nohighlight">\(C^m\)</span> continuity, we need to introduce some further
vocabulary about the topology of <span class="math notranslate nohighlight">\(K\)</span>.</p>
<div class="proof proof-type-definition" id="id32">
<div class="proof-title">
<span class="proof-type">Definition 2.32</span>
<span class="proof-title-name">(closure of a local mesh entity)</span>
</div><div class="proof-content">
<p>Let <span class="math notranslate nohighlight">\(w\)</span> be a local mesh entity for a triangle. The closure of <span class="math notranslate nohighlight">\(w\)</span> is
the set of local mesh entities contained in <span class="math notranslate nohighlight">\(w\)</span> (including <span class="math notranslate nohighlight">\(w\)</span>
itself).</p>
</div></div><p>This allows us to define the degree of continuity of the local
geometric decomposition.</p>
<div class="proof proof-type-definition" id="id33">
<div class="proof-title">
<span class="proof-type">Definition 2.33</span>
<span class="proof-title-name">((C^m) geometric decomposition)</span>
</div><div class="proof-content">
<blockquote>
<div><p>Let <span class="math notranslate nohighlight">\((K,\mathcal{P},\mathcal{N})\)</span> be a finite element with
geometric decomposition <span class="math notranslate nohighlight">\(W\)</span>. We say that <span class="math notranslate nohighlight">\(W\)</span> is a <span class="math notranslate nohighlight">\(C^0\)</span> geometric
decomposition if, for each local mesh entity <span class="math notranslate nohighlight">\(w\)</span>, there exists
<span class="math notranslate nohighlight">\(\mathcal{N}_w \subset \mathcal{N}\)</span> such that</p>
<ol class="arabic simple">
<li><p>All <span class="math notranslate nohighlight">\(N\in \mathcal{N}_w\)</span> are associated to elements in the closure of <span class="math notranslate nohighlight">\(w\)</span> in <span class="math notranslate nohighlight">\(W\)</span>,</p></li>
<li><p><span class="math notranslate nohighlight">\((w,\mathcal{P}|_w,\mathcal{N}_w)\)</span> is a finite element, where <span class="math notranslate nohighlight">\(\mathcal{P}|_w\)</span> is the restriction of <span class="math notranslate nohighlight">\(\mathcal{P}\)</span> to <span class="math notranslate nohighlight">\(w\)</span>.</p></li>
</ol>
<p>We additionally say that <span class="math notranslate nohighlight">\(W\)</span> is a <span class="math notranslate nohighlight">\(C^1\)</span> geometric decomposition if
for each local mesh entity <span class="math notranslate nohighlight">\(w\)</span>, there exists
<span class="math notranslate nohighlight">\(\mathcal{N}_w \subset \mathcal{N}\)</span> such that</p>
<ol class="arabic simple">
<li><p>All <span class="math notranslate nohighlight">\(N\in \mathcal{N}_w\)</span> are associated to elements in the closure of <span class="math notranslate nohighlight">\(w\)</span> in <span class="math notranslate nohighlight">\(W\)</span>,</p></li>
<li><p><span class="math notranslate nohighlight">\((w,\nabla\mathcal{P}|_w,\mathcal{N}_w)\)</span> is a finite element,</p></li>
</ol>
<p>where <span class="math notranslate nohighlight">\(\nabla\mathcal{P}|_w\)</span> is the restriction of <span class="math notranslate nohighlight">\(\nabla\mathcal{P}\)</span> to <span class="math notranslate nohighlight">\(w\)</span>, and</p>
</div></blockquote>
<div class="math notranslate nohighlight">
\[\nabla\mathcal{P} = \{ u : u = \nabla v, \, v \in \mathcal{P}\}.\]</div>
<p>This idea extends to <span class="math notranslate nohighlight">\(C^m\)</span> finite elements in an analogous way.</p>
</div></div><p>The idea behind this definition is that if two triangles <span class="math notranslate nohighlight">\(K_1\)</span> and
<span class="math notranslate nohighlight">\(K_2\)</span> are joined at a vertex $v$, with finite elements
<span class="math notranslate nohighlight">\((K_1,\mathcal{P}_1, \mathcal{N}_1)\)</span> and <span class="math notranslate nohighlight">\((K_2, \mathcal{P}_2,
\mathcal{N}_2)\)</span>, then the nodal variables <span class="math notranslate nohighlight">\(\mathcal{N}_{1,v}\)</span> and
<span class="math notranslate nohighlight">\(\mathcal{N}_{1,v}\)</span> can be chosen so that <span class="math notranslate nohighlight">\(f\)</span> (and for <span class="math notranslate nohighlight">\(m=1\)</span>, the derivatives of <span class="math notranslate nohighlight">\(f\)</span>) has the same values at <span class="math notranslate nohighlight">\(v\)</span> in both <span class="math notranslate nohighlight">\(K_1\)</span> and <span class="math notranslate nohighlight">\(K_2\)</span>.</p>
<p>Similarly, if <span class="math notranslate nohighlight">\(K_1\)</span> and <span class="math notranslate nohighlight">\(K_2\)</span> are joined at an edge <span class="math notranslate nohighlight">\(e\)</span>, then
if the corresponding <span class="math notranslate nohighlight">\(\mathcal{N}_{1,e}\)</span> and <span class="math notranslate nohighlight">\(\mathcal{N}_{2,e}\)</span> nodal
variables associated with that edge agree when applied to <span class="math notranslate nohighlight">\(u\)</span>, then
<span class="math notranslate nohighlight">\(u\)</span> will be <span class="math notranslate nohighlight">\(C^m\)</span> continuous through that edge. We just need to define
these correspondences.</p>
<p>We explore this definition through a couple of exercises.</p>
<div class="proof proof-type-exercise" id="id34">
<div class="proof-title">
<span class="proof-type">Exercise 2.34</span>
</div><div class="proof-content">
<p>Show that the Lagrange elements of degree <span class="math notranslate nohighlight">\(k\)</span> have <span class="math notranslate nohighlight">\(C^0\)</span> geometric decompositions.</p>
</div></div><div class="proof proof-type-exercise" id="id35">
<span id="exer-argyris"></span>
<div class="proof-title">
<span class="proof-type">Exercise 2.35</span>
</div><div class="proof-content">
<p>Show that the Argyris element has a <span class="math notranslate nohighlight">\(C^1\)</span> geometric decomposition.</p>
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</details><p>We now use the geometric decomposition to construct global finite
element spaces over the whole triangulation (mesh). We just need to
define what it means for elements of the nodal variables from the
finite elements of two neighbouring triangles to “correspond”.</p>
<p>We start by considering spaces of functions that are discontinuous
between triangles, before defining <span class="math notranslate nohighlight">\(C^m\)</span> continuous subspaces.</p>
<div class="proof proof-type-definition" id="id36">
<div class="proof-title">
<span class="proof-type">Definition 2.36</span>
<span class="proof-title-name">(Discontinuous finite element space)</span>
</div><div class="proof-content">
<p>Let <span class="math notranslate nohighlight">\(\mathcal{T}\)</span> be a triangulation, with finite elements
<span class="math notranslate nohighlight">\((K_i,P_i,\mathcal{N}_i)\)</span> for each triangle <span class="math notranslate nohighlight">\(K_i\)</span>. The associated
discontinuous finite element space <span class="math notranslate nohighlight">\(V\)</span>, is defined as</p>
<div class="math notranslate nohighlight">
\[V = \left\{u: u|_{K_i} \in P_i, \, \forall K_i \in \mathcal{T}\right\}.\]</div>
<p>This defines families of discontinuous finite element spaces.</p>
</div></div><div class="proof proof-type-example" id="id37">
<div class="proof-title">
<span class="proof-type">Example 2.37</span>
<span class="proof-title-name">(Discontinuous Lagrange finite element space)</span>
</div><div class="proof-content">
<p>Let <span class="math notranslate nohighlight">\(\mathcal{T}\)</span> be a triangulation, with Lagrange elements of
degree <span class="math notranslate nohighlight">\(k\)</span>, <span class="math notranslate nohighlight">\((K_i,P_i,\mathcal{N}_i)\)</span>, for each triangle <span class="math notranslate nohighlight">\(K_i\in
\mathcal{T}\)</span>. The corresponding discontinuous finite element space,
denoted <span class="math notranslate nohighlight">\(Pk\)</span> DG, is called the discontinuous Lagrange finite element
space of degree <span class="math notranslate nohighlight">\(k\)</span>.</p>
</div></div><p>Next we need to associate each nodal variable in each element to a
vertex, edge or triangle of the triangulation <span class="math notranslate nohighlight">\(\mathcal{T}_h\)</span>,
i.e. the global mesh entitles. The following definition explains how
to choose this association.</p>
<div class="proof proof-type-definition" id="id38">
<div class="proof-title">
<span class="proof-type">Definition 2.38</span>
<span class="proof-title-name">(Global (C^m) geometric decomposition)</span>
</div><div class="proof-content">
<p>Let <span class="math notranslate nohighlight">\(\mathcal{T}\)</span> be a triangulation with finite elements
<span class="math notranslate nohighlight">\((K_i,\mathcal{P}_i,\mathcal{N}_i)\)</span>, each with a <span class="math notranslate nohighlight">\(C^m\)</span> geometric
decomposition. Assume that for each global mesh entity <span class="math notranslate nohighlight">\(w\)</span>, the
<span class="math notranslate nohighlight">\(n_w\)</span> triangles containing <span class="math notranslate nohighlight">\(w\)</span> have finite elements
<span class="math notranslate nohighlight">\((K_i,\mathcal{P}_i,\mathcal{N}_i)\)</span> each with <span class="math notranslate nohighlight">\(M_w\)</span> dual basis
functions associated with <span class="math notranslate nohighlight">\(w\)</span>. Further, each of these basis
functions can be enumerated <span class="math notranslate nohighlight">\(N^w_{i,j}\in\mathcal{N}_i\)</span>,
<span class="math notranslate nohighlight">\(j=1,\ldots,M_w\)</span>, such that
<span class="math notranslate nohighlight">\(N^w_{1,j}(u|_{K_1})=N^w_{2,j}(u|_{K_2})=\ldots =
N^w_{n_w,j}(u|_{K_n}), \quad, j=1,\ldots,M_w\)</span>, for all functions
<span class="math notranslate nohighlight">\(u\in C^m(\Omega)\)</span>.</p>
<p>This combination of finite elements on <span class="math notranslate nohighlight">\(\mathcal{T}\)</span> together with
the above enumeration of dual basis functions on global mesh
entities is called a global <span class="math notranslate nohighlight">\(C^m\)</span> geometric decomposition.</p>
</div></div><p>Now we use this global <span class="math notranslate nohighlight">\(C^m\)</span> geometric decomposition to build a
finite element space on the triangulation.</p>
<div class="proof proof-type-definition" id="id39">
<div class="proof-title">
<span class="proof-type">Definition 2.39</span>
<span class="proof-title-name">(Finite element space from a global (C^m) geometric decomposition)</span>
</div><div class="proof-content">
<p>Let <span class="math notranslate nohighlight">\(\mathcal{T}\)</span> be a triangulation with finite elements
<span class="math notranslate nohighlight">\((K_i,\mathcal{P}_i,\mathcal{N}_i)\)</span>, each with a <span class="math notranslate nohighlight">\(C^m\)</span> geometric
decomposition, and let <span class="math notranslate nohighlight">\(\hat{V}\)</span> be the corresponding
discontinuous finite element space. Then the global <span class="math notranslate nohighlight">\(C^m\)</span>
geometric decomposition defines a subspace <span class="math notranslate nohighlight">\(V\)</span> of <span class="math notranslate nohighlight">\(\hat{V}\)</span>
consisting of all functions that <span class="math notranslate nohighlight">\(u\)</span> satisfy
<span class="math notranslate nohighlight">\(N^w_{1,j}(u|_{K_1})=N^w_{2,j}(u|_{K_2})=\ldots = N^w_{n_w,j}(u|_{K_{n_w}}), \quad j=1,\ldots,M_w\)</span> for all mesh entities <span class="math notranslate nohighlight">\(w\in\mathcal{T}\)</span>.</p>
</div></div><p>The following result shows that the global <span class="math notranslate nohighlight">\(C^m\)</span> geometric
decomposition is a useful definition.</p>
<div class="proof proof-type-lemma" id="id40">
<div class="proof-title">
<span class="proof-type">Lemma 2.40</span>
</div><div class="proof-content">
<p>Let <span class="math notranslate nohighlight">\(V\)</span> be a finite element space defined from a global <span class="math notranslate nohighlight">\(C^m\)</span> geometric decomposition. Then <span class="math notranslate nohighlight">\(V\)</span> is a <span class="math notranslate nohighlight">\(C^m\)</span> finite element space.</p>
</div></div><div class="proof proof-type-proof">
<div class="proof-title">
<span class="proof-type">Proof </span>
</div><div class="proof-content">
<p>From the local <span class="math notranslate nohighlight">\(C^m\)</span> decomposition, functions and derivatives up
to degree <span class="math notranslate nohighlight">\(m\)</span> on vertices and edges are uniquely determined from
dual basis elements associated with those vertices and edges, and
from the global <span class="math notranslate nohighlight">\(C^m\)</span> decomposition, the agreement of dual basis
elements means that functions and derivatives up to degree <span class="math notranslate nohighlight">\(m\)</span>
agree on vertices and edges, and hence the functions
are in <span class="math notranslate nohighlight">\(C^m\)</span> from <a class="reference internal" href="#cty"><span class="std std-numref">Lemma 2.28</span></a>.</p>
</div></div><p>We now apply this to a few examples, which can be proved as exercises.</p>
<div class="proof proof-type-example" id="id41">
<div class="proof-title">
<span class="proof-type">Example 2.41</span>
</div><div class="proof-content">
<p>The finite element space built from the <span class="math notranslate nohighlight">\(C^0\)</span> global decomposition
built from degree <span class="math notranslate nohighlight">\(k\)</span> Lagrange element is called the degree <span class="math notranslate nohighlight">\(k\)</span> continuous Lagrange finite element space, denoted <span class="math notranslate nohighlight">\(Pk\)</span>.</p>
</div></div><div class="proof proof-type-example" id="id42">
<div class="proof-title">
<span class="proof-type">Example 2.42</span>
</div><div class="proof-content">
<p>The finite element space built from the <span class="math notranslate nohighlight">\(C^1\)</span> global decomposition
built from the quintic Argyris element is called the Argyris finite
element space.</p>
</div></div><p>In this section, we have built a theoretical toolbox for the
construction of finite element spaces. In the next section, we move on
to studying how well we can approximate continuous functions as finite
element functions.</p>
</section>
</section>
<div class="clearer"></div>
</div>
</div>
</div>
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