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<section id="finite-element-problems-solvability-and-stability">
<span id="fe-problems"></span><h1><span class="section-number">4. </span>Finite element problems: solvability and stability<a class="headerlink" href="#finite-element-problems-solvability-and-stability" title="Link to this heading">¶</a></h1>
<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/490671279"
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=fd348dc9-840d-41ac-8906-ac8f00b77a93">watch this video on Panopto</a></p>
</div>
</details><p>In section 1, we saw the example of a finite element approximation
for Poisson’s equation in the unit square, which we now recall below.</p>
<div class="proof proof-type-definition" id="id1">
<div class="proof-title">
<span class="proof-type">Definition 4.1</span>
</div><div class="proof-content">
<p>The finite element approximation <span class="math notranslate nohighlight">\(u_h \in \mathring{V}_h\)</span> to the
solution <span class="math notranslate nohighlight">\(u_h\)</span> of Poisson’s equation is defined by</p>
<div class="math notranslate nohighlight" id="equation-l4-feprobs-0">
<span class="eqno">(4.1)<a class="headerlink" href="#equation-l4-feprobs-0" title="Link to this equation">¶</a></span>\[\int_\Omega \nabla v \cdot \nabla u_h \, d x =
\int_\Omega vf \, d x, \quad \forall v\in \mathring{V}_h.\]</div>
</div></div><p>A fundamental question is whether the solution <span class="math notranslate nohighlight">\(u_h\)</span> exists and is
unique. This question is of practical interest because if these
conditions are not satisfied, then the matrix-vector system for the
basis coefficients of <span class="math notranslate nohighlight">\(u_h\)</span> will not be solvable. To answer this
question for this approximation (and others for related equations), we
will use some general mathematical machinery about linear problems
defined on Hilbert spaces. It will turn out that this machinery will
also help us show that the approximation <span class="math notranslate nohighlight">\(u_h\)</span> converges to the exact
solution <span class="math notranslate nohighlight">\(u\)</span> (and in what sense).</p>
<section id="finite-element-spaces-and-other-hilbert-spaces">
<h2><span class="section-number">4.1. </span>Finite element spaces and other Hilbert spaces<a class="headerlink" href="#finite-element-spaces-and-other-hilbert-spaces" title="Link to this heading">¶</a></h2>
<p>In the previous sections, we introduced the concept of finite element
spaces, which contain certain functions defined on a domain <span class="math notranslate nohighlight">\(\Omega\)</span>.
Finite element spaces are examples of vector spaces (hence the use
of the word “space”).</p>
<div class="proof proof-type-definition" id="id2">
<div class="proof-title">
<span class="proof-type">Definition 4.2</span>
<span class="proof-title-name">(Vector space)</span>
</div><div class="proof-content">
<p>A vector space over the real numbers <span class="math notranslate nohighlight">\(\mathbb{R}\)</span> is a set <span class="math notranslate nohighlight">\(V\)</span>,
with an addition operator <span class="math notranslate nohighlight">\(+:V\times V\to V\)</span>, plus a scalar
multiplication operator <span class="math notranslate nohighlight">\(\times:\mathbb{R}\times V \to V\)</span>, such
that:</p>
<ol class="arabic simple">
<li><dl class="simple">
<dt>There exists a unique zero element <span class="math notranslate nohighlight">\(e\in V\)</span> such that:</dt><dd><ul class="simple">
<li><p><span class="math notranslate nohighlight">\(k\times e = e\)</span> for all <span class="math notranslate nohighlight">\(k\in \mathbb{R}\)</span>,</p></li>
<li><p><span class="math notranslate nohighlight">\(0\times v = e\)</span> for all <span class="math notranslate nohighlight">\(v \in V\)</span>,</p></li>
<li><p><span class="math notranslate nohighlight">\(e+v = v\)</span> for all <span class="math notranslate nohighlight">\(v \in V\)</span>.</p></li>
</ul>
</dd>
</dl>
</li>
<li><dl class="simple">
<dt><span class="math notranslate nohighlight">\(V\)</span> is closed under addition and multiplication, i.e.,</dt><dd><p><span class="math notranslate nohighlight">\(a\times u + v\in V\)</span> for all <span class="math notranslate nohighlight">\(u,v\in V\)</span>, <span class="math notranslate nohighlight">\(a\in \mathbb{R}\)</span>.</p>
</dd>
</dl>
</li>
</ol>
</div></div><div class="proof proof-type-lemma" id="id3">
<div class="proof-title">
<span class="proof-type">Lemma 4.3</span>
</div><div class="proof-content">
<p>Let <span class="math notranslate nohighlight">\(V\)</span> be a finite element space. Then <span class="math notranslate nohighlight">\(V\)</span> is a vector 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>First, we note that the zero function <span class="math notranslate nohighlight">\(u(x):=0\)</span> is in <span class="math notranslate nohighlight">\(V\)</span>, and
satisfies the above properties. Further, let <span class="math notranslate nohighlight">\(u,v\in V\)</span>, and
<span class="math notranslate nohighlight">\(a\in\mathbb{R}\)</span>. Then, when restricted to each triangle <span class="math notranslate nohighlight">\(K_i\)</span>,
<span class="math notranslate nohighlight">\(u+av\in P_i\)</span>. Also, for each shared mesh entity, the shared nodal
variables agree between triangles. Therefore, <span class="math notranslate nohighlight">\(u+av\in V\)</span>.</p>
</div></div><p>We now introduce bilinear forms on vector spaces. Bilinear forms are
important because they will represent the left hand side of finite
element approximations of linear PDEs.</p>
<div class="proof proof-type-definition" id="id4">
<div class="proof-title">
<span class="proof-type">Definition 4.4</span>
<span class="proof-title-name">(Bilinear form)</span>
</div><div class="proof-content">
<blockquote>
<div><p>A bilinear form <span class="math notranslate nohighlight">\(b(\cdot,\cdot)\)</span> on a vector space <span class="math notranslate nohighlight">\(V\)</span> is a mapping
<span class="math notranslate nohighlight">\(b:V\times V\to \mathbb{R}\)</span>, such that</p>
<ol class="arabic simple">
<li><p><span class="math notranslate nohighlight">\(v\to b(v,w)\)</span> is a linear map in <span class="math notranslate nohighlight">\(v\)</span> for all <span class="math notranslate nohighlight">\(w\)</span>.</p></li>
<li><p><span class="math notranslate nohighlight">\(v\to b(w,v)\)</span> is a linear map in <span class="math notranslate nohighlight">\(v\)</span> for all <span class="math notranslate nohighlight">\(w\)</span>.</p></li>
</ol>
</div></blockquote>
<p>It is a symmetric bilinear form if in addition, <span class="math notranslate nohighlight">\(b(v,w)=b(w,v)\)</span>, for all <span class="math notranslate nohighlight">\(v,w\in V\)</span>.</p>
</div></div><p>Here are two important examples of bilinear forms on finite element spaces.</p>
<div class="proof proof-type-example" id="id5">
<div class="proof-title">
<span class="proof-type">Example 4.5</span>
</div><div class="proof-content">
<p>Let <span class="math notranslate nohighlight">\(V_h\)</span> be a finite element space. The following are bilinear
forms on <span class="math notranslate nohighlight">\(V_h\)</span>,</p>
<div class="math notranslate nohighlight">
\[ \begin{align}\begin{aligned}b(u,v) &= \int_\Omega u v \, d x,\\b(u,v) &= \int_\Omega \nabla u \cdot \nabla v \, d x.\end{aligned}\end{align} \]</div>
</div></div><p>To turn a vector space into a Hilbert space, we need to select an
inner product.</p>
<div class="proof proof-type-definition" id="id6">
<div class="proof-title">
<span class="proof-type">Definition 4.6</span>
<span class="proof-title-name">(Inner product)</span>
</div><div class="proof-content">
<p>A real inner product, denoted by <span class="math notranslate nohighlight">\((\cdot,\cdot)\)</span>, is
a symmetric bilinear form on a vector space <span class="math notranslate nohighlight">\(V\)</span> with</p>
<ol class="arabic simple">
<li><p><span class="math notranslate nohighlight">\((v,v)\geq 0\)</span> <span class="math notranslate nohighlight">\(\forall v\in V\)</span>,</p></li>
<li><p><span class="math notranslate nohighlight">\((v,v)=0\iff v=0\)</span>.</p></li>
</ol>
</div></div><p>This enables the following definition.</p>
<div class="proof proof-type-definition" id="id7">
<div class="proof-title">
<span class="proof-type">Definition 4.7</span>
<span class="proof-title-name">(Inner product space)</span>
</div><div class="proof-content">
<p>We call a vector space <span class="math notranslate nohighlight">\((V, (\cdot,\cdot))\)</span> equipped with an inner product
an inner product space.</p>
</div></div><p>We now introduce two important examples of inner products for finite
element spaces.</p>
<div class="proof proof-type-definition" id="id8">
<div class="proof-title">
<span class="proof-type">Definition 4.8</span>
<span class="proof-title-name">((L^2) inner product)</span>
</div><div class="proof-content">
<p>Let <span class="math notranslate nohighlight">\(f\)</span>, <span class="math notranslate nohighlight">\(g\)</span> be two functions in <span class="math notranslate nohighlight">\(L^2(\Omega)\)</span>. The <span class="math notranslate nohighlight">\(L^2\)</span> inner
product between <span class="math notranslate nohighlight">\(f\)</span> and <span class="math notranslate nohighlight">\(g\)</span> is defined as</p>
<div class="math notranslate nohighlight">
\[( f,g)_{L^2} = \int_{\Omega} fg \, d x.\]</div>
</div></div><p>The <span class="math notranslate nohighlight">\(L^2\)</span> inner product satisfies condition 2 provided that we
understand functions in <span class="math notranslate nohighlight">\(L^2\)</span> as being equivalence classes of
functions under the relation <span class="math notranslate nohighlight">\(f\equiv g \iff \int_\Omega (f-g)^2\, d
x=0\)</span>.</p>
<div class="proof proof-type-definition" id="id9">
<div class="proof-title">
<span class="proof-type">Definition 4.9</span>
<span class="proof-title-name">((H^1) inner product)</span>
</div><div class="proof-content">
<p>Let <span class="math notranslate nohighlight">\(f\)</span>, <span class="math notranslate nohighlight">\(g\)</span> be two <span class="math notranslate nohighlight">\(C^0\)</span> finite element functions. The <span class="math notranslate nohighlight">\(H^1\)</span> inner
product between <span class="math notranslate nohighlight">\(f\)</span> and <span class="math notranslate nohighlight">\(g\)</span> is defined as</p>
<div class="math notranslate nohighlight">
\[( f,g)_{H^1} = \int_{\Omega} fg + \nabla f\cdot \nabla g \, d x.\]</div>
</div></div><p>The <span class="math notranslate nohighlight">\(H^1\)</span> inner product satisfies condition 2 since</p>
<div class="math notranslate nohighlight">
\[( f,f)_{L^2} \leq ( f,f)_{H^1}.\]</div>
<p>The Schwarz inequality is a useful tool for bounding the size of inner
products.</p>
<div class="proof proof-type-theorem" id="id10">
<div class="proof-title">
<span class="proof-type">Theorem 4.10</span>
<span class="proof-title-name">(Schwarz inequality)</span>
</div><div class="proof-content">
<p>If <span class="math notranslate nohighlight">\((V,(\cdot,\cdot))\)</span> is an inner product space, then</p>
<div class="math notranslate nohighlight">
\[|(u,v)| \leq (u,u)^{1/2}(v,v)^{1/2}.\]</div>
<p>Equality holds if and only if <span class="math notranslate nohighlight">\(u=\alpha v\)</span> for some <span class="math notranslate nohighlight">\(\alpha\in\mathbb{R}\)</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>See a course on vector spaces.</p>
</div></div><p>Our solvability conditions will make use of norms that measure the
size of elements of a vector space (the size of finite element
functions, in our case).</p>
<div class="proof proof-type-definition" id="id11">
<div class="proof-title">
<span class="proof-type">Definition 4.11</span>
<span class="proof-title-name">(Norm)</span>
</div><div class="proof-content">
<p>Given a vector space <span class="math notranslate nohighlight">\(V\)</span>, a norm <span class="math notranslate nohighlight">\(\|\cdot\|\)</span> is a function from <span class="math notranslate nohighlight">\(V\)</span> to <span class="math notranslate nohighlight">\(\mathbb{R}\)</span>, with</p>
<ol class="arabic simple">
<li><p><span class="math notranslate nohighlight">\(\|v\|\geq 0,\,\forall v \in V\)</span>,</p></li>
<li><p><span class="math notranslate nohighlight">\(\|v\| = 0 \iff v=0\)</span>,</p></li>
<li><p><span class="math notranslate nohighlight">\(\|cv\|=|c|\|v\| \,\forall c\in \mathbb{R}, v\in V\)</span>,</p></li>
<li><p><span class="math notranslate nohighlight">\(\|v+w\| \leq \|v\|+\|w\|\)</span>.</p></li>
</ol>
</div></div><p>For inner product spaces, there is a natural choice of norm.</p>
<div class="proof proof-type-lemma" id="id12">
<div class="proof-title">
<span class="proof-type">Lemma 4.12</span>
</div><div class="proof-content">
<p>Let <span class="math notranslate nohighlight">\((V,(\cdot,\cdot))\)</span> be an inner product space. Then <span class="math notranslate nohighlight">\(\|v\|=\sqrt{(v,v)}\)</span>
defines a norm on <span class="math notranslate nohighlight">\(V\)</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>From bilinearity we have</p>
<div class="math notranslate nohighlight">
\[\|\alpha v\| = \sqrt{(\alpha v,\alpha v)} = \sqrt{\alpha^2( v,v)}=|\alpha|\|v\|,\]</div>
<p>hence property 3.</p>
<p><span class="math notranslate nohighlight">\(\|v\|=( v,v)^{1/2} \geq 0\)</span>, hence property 1.</p>
<p>If <span class="math notranslate nohighlight">\(0=\|v\|=( v,v)^{1/2} \implies (v,v)=0 \implies v=0\)</span>, hence property 2.</p>
<p>We finally check the triangle inequality (property 4).</p>
<div class="math notranslate nohighlight">
\[ \begin{align}\begin{aligned}\|u+v\|^2 &= (u+v,u+v)\\&= (u,u) + 2(u,v) + (v,v)\\&= \|u\|^2 + 2(u,v) + \|v\|^2\\&\leq \|u\|^2 + 2\|u\|\|v\| + \|v\|^2 \quad \mbox{[Schwarz]},\\&= (\|u\|+\|v\|)^2,\end{aligned}\end{align} \]</div>
<p>hence <span class="math notranslate nohighlight">\(\|u+v\|\leq \|u\|+\|v\|\)</span>.</p>
</div></div><p>We introduce the following useful term.</p>
<div class="proof proof-type-definition" id="id13">
<div class="proof-title">
<span class="proof-type">Definition 4.13</span>
<span class="proof-title-name">(Normed space)</span>
</div><div class="proof-content">
<p>A vector space <span class="math notranslate nohighlight">\(V\)</span> with a norm <span class="math notranslate nohighlight">\(\|\cdot\|\)</span> is called a normed
vector space, written <span class="math notranslate nohighlight">\((V,\|\cdot\|)\)</span>.</p>
</div></div><p>To finish our discussion of Hilbert spaces, we need to review the
concept of completeness (which you might have encountered in an
analysis course). This seems not so important since finite element
spaces are finite dimensional, but later we shall consider sequences
of finite element spaces with smaller and smaller triangles, where
completeness becomes important.</p>
<p>Completeness depends on the notion of a Cauchy sequence.</p>
<div class="proof proof-type-definition" id="id14">
<div class="proof-title">
<span class="proof-type">Definition 4.14</span>
<span class="proof-title-name">(Cauchy sequence)</span>
</div><div class="proof-content">
<p>A Cauchy sequence on a normed vector space <span class="math notranslate nohighlight">\((V,\|\cdot\|)\)</span> is a
sequence <span class="math notranslate nohighlight">\(\{v_i\}_{i=1}^{\infty}\)</span> satisfying <span class="math notranslate nohighlight">\(\|v_j-v_k\|\to 0\)</span> as
<span class="math notranslate nohighlight">\(j,k\to \infty\)</span>.</p>
</div></div><p>This definition leads to the definition of completeness.</p>
<div class="proof proof-type-definition" id="id15">
<div class="proof-title">
<span class="proof-type">Definition 4.15</span>
<span class="proof-title-name">(Complete normed vector space)</span>
</div><div class="proof-content">
<p>A normed vector space <span class="math notranslate nohighlight">\((V,\|\cdot\|)\)</span> is complete if all Cauchy
sequences have a limit <span class="math notranslate nohighlight">\(v\in V\)</span> such that <span class="math notranslate nohighlight">\(\|v-v_j\|\to 0\)</span>
as <span class="math notranslate nohighlight">\(j\to\infty\)</span>.</p>
</div></div><p>Finally, we reach the definition of a Hilbert space.</p>
<div class="proof proof-type-definition" id="id16">
<div class="proof-title">
<span class="proof-type">Definition 4.16</span>
<span class="proof-title-name">(Hilbert space)</span>
</div><div class="proof-content">
<p>An inner product space <span class="math notranslate nohighlight">\((V,(\cdot,\cdot))\)</span> is a Hilbert space if
the corresponding normed space <span class="math notranslate nohighlight">\((V,\|\cdot\|)\)</span> is complete.</p>
</div></div><p>All finite dimensional normed vector spaces are complete. Hence, <span class="math notranslate nohighlight">\(C^0\)</span>
finite element spaces equipped with <span class="math notranslate nohighlight">\(L^2\)</span> or <span class="math notranslate nohighlight">\(H^1\)</span> inner products are
Hilbert spaces. Later we shall understand our finite element
spaces as subspaces of infinite dimensional Hilbert spaces.</p>
</section>
<section id="linear-forms-on-hilbert-spaces">
<span id="sec-linearforms"></span><h2><span class="section-number">4.2. </span>Linear forms on Hilbert spaces<a class="headerlink" href="#linear-forms-on-hilbert-spaces" title="Link to this heading">¶</a></h2>
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</div>
</details><p>We will now build some structures on Hilbert spaces that allow us to
discuss variational problems on them, which includes finite element
approximations such as the Poisson example discussed so far.</p>
<p>Linear functionals are important as they will represent the right-hand
side of finite element approximations of PDEs.</p>
<div class="proof proof-type-definition" id="id17">
<div class="proof-title">
<span class="proof-type">Definition 4.17</span>
<span class="proof-title-name">(Continuous linear functional)</span>
</div><div class="proof-content">
<p>Let <span class="math notranslate nohighlight">\(H\)</span> be a Hilbert space with norm <span class="math notranslate nohighlight">\(|\cdot|_H\)</span>.</p>
<ol class="arabic">
<li><p>A functional <span class="math notranslate nohighlight">\(L\)</span> is a map from <span class="math notranslate nohighlight">\(H\)</span> to <span class="math notranslate nohighlight">\(\mathbb{R}\)</span>.</p></li>
<li><p>A functional <span class="math notranslate nohighlight">\(L:H\to \mathbb{R}\)</span> is linear if <span class="math notranslate nohighlight">\(u,v\in H\)</span>, <span class="math notranslate nohighlight">\(\alpha\in \mathbb{R}\)</span> <span class="math notranslate nohighlight">\(\implies L(u+\alpha v)=L(u)+\alpha L(v)\)</span>.</p></li>
<li><p>A functional <span class="math notranslate nohighlight">\(L:H\to\mathbb{R}\)</span> is continuous if there exists <span class="math notranslate nohighlight">\(C>0\)</span> such that</p>
<div class="math notranslate nohighlight">
\[|L(u)-L(v)| \leq C\|u-v\|_H \quad \forall u,v\in H.\]</div>
</li>
</ol>
</div></div><p>It is important that linear functionals are “nice” in the following sense.</p>
<div class="proof proof-type-definition" id="id18">
<div class="proof-title">
<span class="proof-type">Definition 4.18</span>
<span class="proof-title-name">(Bounded functional)</span>
</div><div class="proof-content">
<p>A functional <span class="math notranslate nohighlight">\(L:H\to\mathbb{R}\)</span> is bounded if there exists <span class="math notranslate nohighlight">\(C>0\)</span> such that</p>
<div class="math notranslate nohighlight">
\[|L(u)| \leq C\|u\|_H, \quad \forall u\in H.\]</div>
</div></div><p>For linear functionals we have the following relationship between boundedness
and continuity.</p>
<div class="proof proof-type-lemma" id="id19">
<div class="proof-title">
<span class="proof-type">Lemma 4.19</span>
</div><div class="proof-content">
<p>Let <span class="math notranslate nohighlight">\(L:H\to\mathbb{R}\)</span> be a linear functional. Then <span class="math notranslate nohighlight">\(L\)</span> is continuous if and only if it is bounded.</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><span class="math notranslate nohighlight">\(L\)</span> bounded <span class="math notranslate nohighlight">\(\implies L(u)\leq C\|u\|_{H} \implies |L(u)-L(v)| = |L(u-v)| \leq C\|u-v\|_{H} \, \forall u,v\in H,\)</span> i.e. <span class="math notranslate nohighlight">\(L\)</span> is continuous.</p>
<p>On the other hand, <span class="math notranslate nohighlight">\(L\)</span> continuous <span class="math notranslate nohighlight">\(\implies |L(u-v)| \leq C\|u-v\|_{H}
\quad \forall u,v\in H\)</span>. Pick <span class="math notranslate nohighlight">\(v=0\)</span>, then <span class="math notranslate nohighlight">\(|L(u)| = |L(u-0)| \leq
C\|u-0\|_H = C\|u\|_H\)</span>, i.e. <span class="math notranslate nohighlight">\(L\)</span> is bounded.</p>
</div></div><p>We can also interpret bounded linear functionals as elements of a vector space.</p>
<div class="proof proof-type-definition" id="id20">
<div class="proof-title">
<span class="proof-type">Definition 4.20</span>
<span class="proof-title-name">(Dual space)</span>
</div><div class="proof-content">
<p>Let <span class="math notranslate nohighlight">\(H\)</span> be a Hilbert space. The dual space <span class="math notranslate nohighlight">\(H'\)</span> is the space of continuous (or bounded) linear functionals <span class="math notranslate nohighlight">\(L:H\to \mathbb{R}\)</span>.</p>
</div></div><p>This dual space can also be equipped with a norm.</p>
<div class="proof proof-type-definition" id="id21">
<div class="proof-title">
<span class="proof-type">Definition 4.21</span>
<span class="proof-title-name">(Dual norm)</span>
</div><div class="proof-content">
<p>Let <span class="math notranslate nohighlight">\(L\)</span> be a continuous linear functional on <span class="math notranslate nohighlight">\(H\)</span>, then</p>
<div class="math notranslate nohighlight">
\[\|L\|_{H'} = \sup_{0\neq v\in H}\frac{L(v)}{\|v\|_H}.\]</div>
</div></div><p>There is a simple mapping from <span class="math notranslate nohighlight">\(H\)</span> to <span class="math notranslate nohighlight">\(H'\)</span>.</p>
<div class="proof proof-type-lemma" id="id22">
<div class="proof-title">
<span class="proof-type">Lemma 4.22</span>
</div><div class="proof-content">
<p>Let <span class="math notranslate nohighlight">\(u\in H\)</span>. Then the functional <span class="math notranslate nohighlight">\(L_u:H\to \mathbb{R}\)</span> defined by</p>
<div class="math notranslate nohighlight">
\[L_u(v) = (u,v), \quad \forall v\in H,\]</div>
<p>is linear and continuous.</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>For <span class="math notranslate nohighlight">\(v,w\in H\)</span>, <span class="math notranslate nohighlight">\(\alpha\in\mathbb{R}\)</span> we have</p>
<div class="math notranslate nohighlight">
\[L_u(v+\alpha w) = (u,v+\alpha w) = (u,v) + \alpha(u,w) = L_u(v)
+ \alpha L_u(w).\]</div>
<p>Hence <span class="math notranslate nohighlight">\(L_u\)</span> is linear.</p>
<p>We see that <span class="math notranslate nohighlight">\(L_u\)</span> is bounded by Schwarz inequality,</p>
<div class="math notranslate nohighlight">
\[|L_u(v)| = |(u,v)| \leq C\|v\|_H \mbox{ with }C=\|u\|_H.\]</div>
</div></div><p>The following famous theorem states that the converse is also true.</p>
<div class="proof proof-type-theorem" id="id23">
<div class="proof-title">
<span class="proof-type">Theorem 4.23</span>
<span class="proof-title-name">(Riesz representation theorem)</span>
</div><div class="proof-content">
<p>For any continuous linear functional <span class="math notranslate nohighlight">\(L\)</span> on <span class="math notranslate nohighlight">\(H\)</span> there exists <span class="math notranslate nohighlight">\(u\in H\)</span> such that</p>
<div class="math notranslate nohighlight">
\[L(v) = (u,v)\quad \forall v\in H.\]</div>
<p>Further,</p>
<div class="math notranslate nohighlight">
\[\|u\|_H = \|L\|_{H'}.\]</div>
</div></div><div class="proof proof-type-proof">
<div class="proof-title">
<span class="proof-type">Proof </span>
</div><div class="proof-content">
<p>See a course or textbook on Hilbert spaces.</p>
</div></div></section>
<section id="variational-problems-on-hilbert-spaces">
<h2><span class="section-number">4.3. </span>Variational problems on Hilbert spaces<a class="headerlink" href="#variational-problems-on-hilbert-spaces" title="Link to this heading">¶</a></h2>
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</div>
</details><p>We will consider finite element methods that can be formulated in the
following way.</p>
<div class="proof proof-type-definition" id="id24">
<div class="proof-title">
<span class="proof-type">Definition 4.24</span>
<span class="proof-title-name">(Linear variational problem)</span>
</div><div class="proof-content">
<p>Let <span class="math notranslate nohighlight">\(b(u,v)\)</span> be a bilinear form on a Hilbert space <span class="math notranslate nohighlight">\(V\)</span>, and <span class="math notranslate nohighlight">\(F\)</span> be
a linear form on <span class="math notranslate nohighlight">\(V\)</span>. This defines a linear variational problem: find
<span class="math notranslate nohighlight">\(u\in V\)</span> such that</p>
<div class="math notranslate nohighlight">
\[b(u,v) = F(v), \quad \forall v\in V.\]</div>
</div></div><p>We now discuss some important examples from finite element discretisations
of linear PDEs.</p>
<div class="proof proof-type-example" id="id25">
<div class="proof-title">
<span class="proof-type">Example 4.25</span>
<span class="proof-title-name">((Pk) discretisation of (modified) Helmholtz problem with Neumann bcs)</span>
</div><div class="proof-content">
<p>For some known function <span class="math notranslate nohighlight">\(f\)</span>,</p>
<div class="math notranslate nohighlight">
\[ \begin{align}\begin{aligned}b(u,v) &= \int_\Omega uv + \nabla u \cdot \nabla v \, d x,\\F(v) &= \int_\Omega vf \, d x,\end{aligned}\end{align} \]</div>
<p>and <span class="math notranslate nohighlight">\(V\)</span> is the Pk continuous finite element space on a triangulation
of <span class="math notranslate nohighlight">\(\Omega\)</span>.</p>
</div></div><div class="proof proof-type-example" id="id26">
<div class="proof-title">
<span class="proof-type">Example 4.26</span>
<span class="proof-title-name">((Pk) discretisation of Poisson equation with partial Dirichlet bcs)</span>
</div><div class="proof-content">
<p>For some known function <span class="math notranslate nohighlight">\(f\)</span>,</p>
<div class="math notranslate nohighlight">
\[ \begin{align}\begin{aligned}b(u,v) &= \int_\Omega \nabla u \cdot \nabla v \, d x,\\F(v) &= \int_\Omega vf \, d x,\end{aligned}\end{align} \]</div>
<p>and <span class="math notranslate nohighlight">\(V\)</span> is the subspace of the Pk continuous finite element space on a
triangulation of <span class="math notranslate nohighlight">\(\Omega\)</span> such that functions vanishes on
<span class="math notranslate nohighlight">\(\Gamma_0\subseteq \partial \Omega\)</span>.</p>
</div></div><div class="proof proof-type-example" id="id27">
<div class="proof-title">
<span class="proof-type">Example 4.27</span>
<span class="proof-title-name">((Pk) discretisation of Poisson equation with pure Neumann bcs)</span>
</div><div class="proof-content">
<p>For some known function <span class="math notranslate nohighlight">\(f\)</span>,</p>
<div class="math notranslate nohighlight">
\[ \begin{align}\begin{aligned}b(u,v) &= \int_\Omega \nabla u \cdot \nabla v \, d x,\\F(v) &= \int_\Omega vf \, d x,\end{aligned}\end{align} \]</div>
<p>and <span class="math notranslate nohighlight">\(V\)</span> is the subspace of the Pk continuous finite element space on a
triangulation of <span class="math notranslate nohighlight">\(\Omega\)</span> such that functions satisfy</p>
<div class="math notranslate nohighlight">
\[\int_\Omega u \, d x = 0.\]</div>
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</div>
</details><p>We now introduce two important properties of bilinear forms that determine
whether a linear variational problem is solvable or not. The first is
continuity.</p>
<div class="proof proof-type-definition" id="id28">
<div class="proof-title">
<span class="proof-type">Definition 4.28</span>
<span class="proof-title-name">(Continuous bilinear form)</span>
</div><div class="proof-content">
<p>A bilinear form is continuous on a Hilbert space <span class="math notranslate nohighlight">\(V\)</span> if there exists a
constant <span class="math notranslate nohighlight">\(0<M<\infty\)</span> such that</p>
<div class="math notranslate nohighlight">
\[|b(u,v)| \leq M\|u\|_V\|v\|_V.\]</div>
</div></div><p>The second is coercivity.</p>
<div class="proof proof-type-definition" id="id29">
<div class="proof-title">
<span class="proof-type">Definition 4.29</span>
<span class="proof-title-name">(Coercive bilinear form)</span>
</div><div class="proof-content">
<p>A bilinear form is coercive on a Hilbert space <span class="math notranslate nohighlight">\(V\)</span> if there exists a
constant <span class="math notranslate nohighlight">\(0<\gamma<\infty\)</span> such that</p>
<div class="math notranslate nohighlight">
\[|b(u,u)| \geq \gamma\|u\|_V\|u\|_V.\]</div>
</div></div><p>These two properties combine in the following theorem providing sufficient
conditions for existence and uniqueness for solutions of linear variational
problems.</p>
<div class="proof proof-type-theorem" id="id30">
<div class="proof-title">
<span class="proof-type">Theorem 4.30</span>
<span class="proof-title-name">(Lax-Milgram theorem)</span>
</div><div class="proof-content">
<p>Let <span class="math notranslate nohighlight">\(b\)</span> be a bilinear form, <span class="math notranslate nohighlight">\(F\)</span> be a linear form, and <span class="math notranslate nohighlight">\((V,\|\cdot\|)\)</span> be a
Hilbert space. If <span class="math notranslate nohighlight">\(b\)</span> is continuous and coercive, and <span class="math notranslate nohighlight">\(F\)</span> is continuous,
then a unique solution <span class="math notranslate nohighlight">\(u\in V\)</span> to the linear variational problem exists,
with</p>
<div class="math notranslate nohighlight">
\[\|u\|_V \leq \frac{1}{\gamma}\|F\|_{V'}.\]</div>
</div></div><div class="proof proof-type-proof">
<div class="proof-title">
<span class="proof-type">Proof </span>
</div><div class="proof-content">
<p>See a course or textbook on Hilbert spaces.</p>
</div></div><p>We are going to use this result to show solvability for finite
element discretisations. In particular, we also want to know that our
finite element discretisation continues to be solvable as the maximum
triangle edge diameter <span class="math notranslate nohighlight">\(h\)</span> goes to zero. This motivates the following
definition.</p>
<div class="proof proof-type-definition" id="id31">
<div class="proof-title">
<span class="proof-type">Definition 4.31</span>
<span class="proof-title-name">(Stability)</span>
</div><div class="proof-content">
<p>Consider a sequence of triangulations <span class="math notranslate nohighlight">\(\mathcal{T}_h\)</span> with corresponding
finite element spaces <span class="math notranslate nohighlight">\(V_h\)</span> labelled by a maximum triangle diameter
<span class="math notranslate nohighlight">\(h\)</span>, applied to a variational problem with bilinear form <span class="math notranslate nohighlight">\(b(u,v)\)</span> and
linear form <span class="math notranslate nohighlight">\(L\)</span>. For each <span class="math notranslate nohighlight">\(V_h\)</span> we have a corresponding coercivity constant
<span class="math notranslate nohighlight">\(\gamma_h\)</span>.</p>
<p>If <span class="math notranslate nohighlight">\(\gamma_h \to \gamma>0\)</span>, and <span class="math notranslate nohighlight">\(\|F\|_{V'_h}\to c <\infty\)</span>, then
we say that the finite element discretisation is stable.</p>
</div></div><p>With this in mind it is useful to consider <span class="math notranslate nohighlight">\(h\)</span>-independent definitions
of <span class="math notranslate nohighlight">\(\|\cdot\|_V\)</span> (such as the <span class="math notranslate nohighlight">\(L^2\)</span> and <span class="math notranslate nohighlight">\(H^1\)</span> norms), which is why we
introduced them.</p>
</section>
<section id="solvability-and-stability-of-some-finite-element-discretisations">
<h2><span class="section-number">4.4. </span>Solvability and stability of some finite element discretisations<a class="headerlink" href="#solvability-and-stability-of-some-finite-element-discretisations" title="Link to this heading">¶</a></h2>
<p>In this section we will introduce some tools for showing coercivity
and continuity of bilinear forms, illustrated with finite element
approximations of some linear PDEs where they may be applied.</p>
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</details><p>We start with the simplest example, for which continuity and
coercivity are immediate.</p>
<div class="proof proof-type-theorem" id="id32">
<span id="thm-helm"></span>
<div class="proof-title">
<span class="proof-type">Theorem 4.32</span>
<span class="proof-title-name">(Solving the (modified) Helmholtz problem)</span>
</div><div class="proof-content">
<p>Let <span class="math notranslate nohighlight">\(b\)</span>, <span class="math notranslate nohighlight">\(L\)</span> be the forms from the Helmholtz problem, with
<span class="math notranslate nohighlight">\(\|f\|_{L^2}<\infty\)</span>. Let <span class="math notranslate nohighlight">\(V_h\)</span> be a Pk continuous finite element space
defined on a triangulation <span class="math notranslate nohighlight">\(\mathcal{T}\)</span>. Then the finite element
approximation <span class="math notranslate nohighlight">\(u_h\)</span> exists and the discretisation is stable
in the <span class="math notranslate nohighlight">\(H^1\)</span> norm.</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>First we show continuity of <span class="math notranslate nohighlight">\(F\)</span>. We have</p>
<div class="math notranslate nohighlight">
\[F(v) = \int_\Omega fv \, d x \leq \|f\|_{L^2}\|v\|_{L^2}
\leq \|f\|_{L^2}\|v\|_{H^1},\]</div>
<p>since <span class="math notranslate nohighlight">\(\|v\|_{L^2}\leq \|v\|_{H^1}\)</span>.</p>
<p>Next we show continuity of <span class="math notranslate nohighlight">\(b\)</span>.</p>
<div class="math notranslate nohighlight">
\[|b(u,v)| = |(u,v)_{H^1}| \leq 1\times\|u\|_{H^1}\|v\|_{H^1},\]</div>
<p>from the Schwarz inequality of the <span class="math notranslate nohighlight">\(H^1\)</span> inner product.
Finally we show coercivity of <span class="math notranslate nohighlight">\(b\)</span>.</p>
<div class="math notranslate nohighlight">
\[b(u,u) = \|u\|^2_{H^1} \geq 1\times\|u\|^2_{H^1},\]</div>
<p>The continuity and coercivity constants are both 1, independent
of <span class="math notranslate nohighlight">\(h\)</span>, so the discretisation is stable.</p>
</div></div><div class="proof proof-type-exercise" id="id33">
<div class="proof-title">
<span class="proof-type">Exercise 4.33</span>
</div><div class="proof-content">
<blockquote>
<div><p>Let <span class="math notranslate nohighlight">\(V\)</span> be a <span class="math notranslate nohighlight">\(C^0\)</span> finite element space on <span class="math notranslate nohighlight">\([0,1]\)</span>, defined
on a one-dimensional mesh with vertices <span class="math notranslate nohighlight">\(0=x_0<x_1<x_2<\ldots<x_{n-1}<x_n=1\)</span>. Show that
<span class="math notranslate nohighlight">\(u\in V\)</span> satisfies the fundamental theorem of calculus, <span class="math notranslate nohighlight">\(i.e.\)</span></p>
<div class="math notranslate nohighlight">
\[\int_0^1 u' \, d x = u[1] - u[0],\]</div>
</div></blockquote>
<p>where <span class="math notranslate nohighlight">\(u'\)</span> is the usual finite element derivative defined in <span class="math notranslate nohighlight">\(L^2([0,1])\)</span>
by taking the usual derivative when restricting <span class="math notranslate nohighlight">\(u\)</span> to any subinterval
<span class="math notranslate nohighlight">\([x_k,x_{k+1}]\)</span>.</p>
</div></div><div class="proof proof-type-exercise" id="id34">
<div class="proof-title">
<span class="proof-type">Exercise 4.34</span>
</div><div class="proof-content">
<p>Let</p>
<div class="math notranslate nohighlight">
\[a(u,v) = \int_0^1\left(u'v' + u'v + uv\right)\, d x.\]</div>
<p>Let <span class="math notranslate nohighlight">\(V\)</span> be a <span class="math notranslate nohighlight">\(C^0\)</span> finite element space on <span class="math notranslate nohighlight">\([0,1]\)</span> and let
<span class="math notranslate nohighlight">\(\mathring{V}\)</span> be the subspace of functions that vanish at <span class="math notranslate nohighlight">\(x=0\)</span> and
<span class="math notranslate nohighlight">\(x=1\)</span>. Using the finite element version of the fundamental theorem of
calculus above, prove that</p>
<div class="math notranslate nohighlight">
\[a(v,v) = \int_0^1\left((v')^2 + v^2\right)\, d x := \|v\|_{H^1}^2, \quad \forall v\in \mathring{V}.\]</div>
<p>Hence conclude that the bilinear form is coercive on <span class="math notranslate nohighlight">\(\mathring{V}\)</span>.</p>
</div></div><div class="proof proof-type-exercise" id="id35">
<div class="proof-title">
<span class="proof-type">Exercise 4.35</span>
</div><div class="proof-content">
<p>Consider the variational problem with bilinear form</p>
<div class="math notranslate nohighlight">
\[a(u,v) = \int_0^1 (u'v' + u'v + uv)\, d x,\]</div>
<p>corresponding to the differential equation</p>
<div class="math notranslate nohighlight">
\[-u'' + u' + u = f.\]</div>
<p>Prove that <span class="math notranslate nohighlight">\(a(\cdot,\cdot)\)</span> is continuous and coercive on a <span class="math notranslate nohighlight">\(C^0\)</span> finite element space <span class="math notranslate nohighlight">\(V\)</span> defined on <span class="math notranslate nohighlight">\([0,1]\)</span>, with respect to the <span class="math notranslate nohighlight">\(H^1\)</span> inner product.</p>
<p>Hints: for continuity, just use the triangle inequality and the
relationship between <span class="math notranslate nohighlight">\(L^2\)</span> and <span class="math notranslate nohighlight">\(H^1\)</span> norms. For coercivity, try completing the square for the integrand in <span class="math notranslate nohighlight">\(a\)</span>.</p>
</div></div><p>For the Helmholtz problem, we have</p>
<div class="math notranslate nohighlight">
\[b(u,v) = \int_\Omega uv + \nabla u\cdot \nabla v \, d x = (u,v)_{H^1},\]</div>
<p>i.e. <span class="math notranslate nohighlight">\(b(u,v)\)</span> is the <span class="math notranslate nohighlight">\(H^1\)</span> inner product of <span class="math notranslate nohighlight">\(u\)</span> and <span class="math notranslate nohighlight">\(v\)</span>, which makes the
continuity and coercivity immediate.</p>
<p>For the Poisson problem, we have</p>
<div class="math notranslate nohighlight">
\[b(u,u) = \int_\Omega |\nabla u|^2 \, d x = |u|^2_{H^1} \neq \|u\|^2_{H^1},\]</div>
<p>where we recall the <span class="math notranslate nohighlight">\(H^1\)</span> seminorm from the interpolation
section. Some additional results are required to show coercivity, as
<span class="math notranslate nohighlight">\(b(u,u)\)</span> is not the <span class="math notranslate nohighlight">\(H^1\)</span> norm squared any more. A seminorm has all
the properties of a norm except <span class="math notranslate nohighlight">\(|u|= 0 \nRightarrow u=0\)</span>, which is
precisely what is needed in the Lax-Milgram theorem.</p>
<div class="proof proof-type-exercise" id="id36">
<span id="exe-pure-neumann"></span>
<div class="proof-title">
<span class="proof-type">Exercise 4.36</span>
</div><div class="proof-content">
<blockquote>
<div><p>Let <span class="math notranslate nohighlight">\(\mathcal{T}_h\)</span> be a triangulation on the <span class="math notranslate nohighlight">\(1\times 1\)</span> unit
square domain <span class="math notranslate nohighlight">\(\Omega\)</span>, and let <span class="math notranslate nohighlight">\(V\)</span> be a <span class="math notranslate nohighlight">\(C^0\)</span> Lagrange finite
element space of degree <span class="math notranslate nohighlight">\(k\)</span> defined on <span class="math notranslate nohighlight">\(\mathcal{T}_h\)</span>. A finite element
discretisation for the Poisson equation with Neumann boundary conditions
is given by: find <span class="math notranslate nohighlight">\(u \in V\)</span> such that</p>
</div></blockquote>
<div class="math notranslate nohighlight">
\[\int_{\Omega} \nabla v \cdot \nabla u \, d x = \int_\Omega v f \, d x,
\quad \forall v \in V,\]</div>
<p>for some known function <span class="math notranslate nohighlight">\(f\)</span>. Show that the bilinear form for this
problem is not coercive in <span class="math notranslate nohighlight">\(V\)</span>.</p>
</div></div><p>For the Poisson problem, coercivity comes instead from the following
mean estimate.</p>
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</details><div class="proof proof-type-lemma" id="id37">
<div class="proof-title">
<span class="proof-type">Lemma 4.37</span>
<span class="proof-title-name">(Mean estimate for finite element spaces)</span>
</div><div class="proof-content">
<p>Let <span class="math notranslate nohighlight">\(u\)</span> be a member of a <span class="math notranslate nohighlight">\(C^0\)</span> finite element space, and define</p>
<div class="math notranslate nohighlight">
\[\bar{u} = \frac{\int_\Omega u \, d x}{\int_\Omega \, d x}.\]</div>
<p>Then there exists a positive constant <span class="math notranslate nohighlight">\(C\)</span>, independent of the
triangulation but dependent on (convex) <span class="math notranslate nohighlight">\(\Omega\)</span>, such that</p>
<div class="math notranslate nohighlight">
\[\|u-\bar{u}\|_{L^2} \leq C|u|_{H^1}.\]</div>
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</details><div class="proof proof-type-proof">
<div class="proof-title">
<span class="proof-type">Proof </span>
</div><div class="proof-content">
<p>(Very similar to the proof of the estimate for averaged Taylor polynomials.)</p>
<p>Let <span class="math notranslate nohighlight">\(x\)</span> and <span class="math notranslate nohighlight">\(y\)</span> be two points in <span class="math notranslate nohighlight">\(\Omega\)</span>. We note that
<span class="math notranslate nohighlight">\(f(s)=u(y+s(x-y))\)</span> is a <span class="math notranslate nohighlight">\(C^0\)</span>, piecewise polynomial function of
<span class="math notranslate nohighlight">\(s\)</span>. Let <span class="math notranslate nohighlight">\(s_0 = 0 < s_1 < s_2 < \ldots < s_{k-1} < s_k = 1\)</span> denote the points
where <span class="math notranslate nohighlight">\(y+s(x-y)\)</span> intersects a triangle edge or vertex. Then
<span class="math notranslate nohighlight">\(f\)</span> is a continuous function when restricted to each interval <span class="math notranslate nohighlight">\([s_i,s_{i+1}]\)</span>, <span class="math notranslate nohighlight">\(i=0,\ldots,k-1\)</span>. This means that</p>
<div class="math notranslate nohighlight">
\[ \begin{align}\begin{aligned}f(s_{i+1}) - f(s_i) &= \int_{s_i}^{s_{i+1}} f'(s) \, ds\\&= \int_{s_i}^{s_{i+1}} (x-y)\cdot\nabla u(y+s(x-y)) \, ds,\end{aligned}\end{align} \]</div>
<p>where <span class="math notranslate nohighlight">\(\nabla u\)</span> is the finite element derivative of <span class="math notranslate nohighlight">\(u\)</span>. Summing
this up from <span class="math notranslate nohighlight">\(i=0\)</span> to <span class="math notranslate nohighlight">\(i=k-1\)</span>, we obtain</p>
<div class="math notranslate nohighlight">
\[u(x)=u(y) +\int_0^1
(x-y)\cdot\nabla u(y+s(x-y))\, d s.\]</div>
<p>Then</p>
<div class="math notranslate nohighlight">
\[ \begin{align}\begin{aligned}u(x)-\bar{u} &= \frac{1}{|\Omega|}\int_{\Omega}u(x)-u(y)\, d y\\&= \frac{1}{|\Omega|}\int_{\Omega}(x-y)\cdot \int_{s=0}^1 \nabla u(y + s(x-y)) \, d s\, d y,\end{aligned}\end{align} \]</div>
<p>Therefore</p>
<div class="math notranslate nohighlight">
\[ \begin{align}\begin{aligned}\|u-\bar{u}\|^2_{L^2(\Omega)} &= \frac{1}{|\Omega|^2}\int_{\Omega}
\left(\int_\Omega (x-y)\cdot\int_{s=0}^1 \nabla u(y + s(x-y))\, d s
\, d y\right)^2 \, d x,\\&\leq \frac{1}{|\Omega|^2}\int_{\Omega}
\int_\Omega |x-y|^2 \, d y
\int_\Omega \int_{s=0}^1 |\nabla u(y + s(x-y))|^2\, d s
\, d y \, d x,\\&\leq C\int_{\Omega}
\int_\Omega \int_{s=0}^1 |\nabla u(y + s(x-y))|^2\, d s
\, d y \, d x.\end{aligned}\end{align} \]</div>
<p>We split this final quantity into two parts (to avoid singularities),</p>
<div class="math notranslate nohighlight">
\[\|u-\bar{u}\|^2_{L^2(\Omega)} \leq C(I + II),\]</div>
<p>where</p>
<div class="math notranslate nohighlight">
\[ \begin{align}\begin{aligned}I &= \int_\Omega \int_{s=0}^{1/2} \int_\Omega
| \nabla u(y+s(x-y))|^2 \, d y \, d s \, d x,\\II &= \int_\Omega \int_{s=1/2}^2 \int_\Omega
| \nabla u(y+s(x-y))|^2 \, d x \, d s \, d y,\end{aligned}\end{align} \]</div>
<p>which we will now estimate separately.</p>
<p>To evaluate <span class="math notranslate nohighlight">\(I\)</span>, change variables <span class="math notranslate nohighlight">\(y \to y' = y + s(x-y)\)</span>,
defining <span class="math notranslate nohighlight">\(\Omega'_s\subset \Omega\)</span> as the image of <span class="math notranslate nohighlight">\(\Omega\)</span>
under this transformation. Then,</p>
<div class="math notranslate nohighlight">
\[ \begin{align}\begin{aligned}I &= \int_\Omega \int_{s=0}^{1/2} \frac{1}{(1-s)^2}
\int_{\Omega'_s} |\nabla u(y')|^2 \, d y' \, d s \, d x,\\&\leq \int_\Omega \int_{s=0}^{1/2} \frac{1}{(1-s)^2}
\int_{\Omega} |\nabla u(y')|^2 \, d y' \, d s \, d x,\\&= \frac{|\Omega|}{2} |\nabla u|^2_{H^1(\Omega)}.\end{aligned}\end{align} \]</div>
<p>To evaluate <span class="math notranslate nohighlight">\(II\)</span>, change variables <span class="math notranslate nohighlight">\(x \to x' = y + s(x-y)\)</span>,
defining <span class="math notranslate nohighlight">\(\Omega'_s\subset \Omega\)</span> as the image of <span class="math notranslate nohighlight">\(\Omega\)</span>
under this transformation. Then,</p>
<div class="math notranslate nohighlight">
\[ \begin{align}\begin{aligned}II &= \int_\Omega \int_{s=1/2}^2 \frac{1}{s^2}
\int_{\Omega'_s} |\nabla u(x')|^2 \, d x' \, d s \, d y,\\&\leq \int_\Omega \int_{s=0}^{1/2} \frac{1}{s^2}
\int_{\Omega} |\nabla u(x')|^2 \, d x' \, d s \, d y,\\&=|\Omega| |\nabla u|^2_{H^1(\Omega)}.\end{aligned}\end{align} \]</div>
<p>Combining,</p>
<div class="math notranslate nohighlight">
\[\|u-\bar{u}\|^2_{L^2(\Omega)} \leq C(I + II) = \frac{3C|\Omega|}{2}
|u|_{H^1(\Omega)}^2,\]</div>
<p>which has the required form.</p>
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</details><p>The mean estimate can now be used to show solvability for the Poisson
problem with pure Neumann conditions.</p>
<div class="proof proof-type-theorem" id="id38">
<div class="proof-title">
<span class="proof-type">Theorem 4.38</span>
<span class="proof-title-name">(Solving the Poisson problem with pure Neumann conditions)</span>
</div><div class="proof-content">
<p>Let <span class="math notranslate nohighlight">\(b\)</span>, <span class="math notranslate nohighlight">\(L\)</span>, <span class="math notranslate nohighlight">\(V\)</span>, be the forms for
the pure Neumann Poisson problem, with
<span class="math notranslate nohighlight">\(\|f\|_{L^2}<\infty\)</span>. Let <span class="math notranslate nohighlight">\(V_h\)</span> be a Pk continuous finite element space
defined on a triangulation <span class="math notranslate nohighlight">\(\mathcal{T}\)</span>, and define</p>
<div class="math notranslate nohighlight">
\[\bar{V}_h = \{u\in V_h:\bar{u}=0\}.\]</div>
<p>Then for <span class="math notranslate nohighlight">\(\bar{V}_h\)</span>, the finite element approximation <span class="math notranslate nohighlight">\(u_h\)</span> exists and the
discretisation is stable in the <span class="math notranslate nohighlight">\(H^1\)</span> norm.</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>Using the mean estimate, for <span class="math notranslate nohighlight">\(u\in\bar{V}_h\)</span>, we have</p>
<div class="math notranslate nohighlight">
\[\|u\|_{L^2}^2=\|u-\underbrace{\bar{u}}_{=0}\|^2_{L^2} \leq C^2|u|_{H^1}^2.\]</div>
<p>Hence we obtain the coercivity result,</p>
<div class="math notranslate nohighlight">
\[\|u\|_{H^1}^2 = \|u\|_{L^2}^2 + |u|_{H^1}^2 \leq (1+C^2)|u|_{H^1}^2 = (1+C^2)b(u,u).\]</div>
<p>Continuity follows from Schwarz inequality,</p>
<div class="math notranslate nohighlight">
\[|b(u,v)| \leq |u|_{H^1}|v|_{H^1} \leq \|u\|_{H^1}\|v\|_{H^1}.\]</div>
</div></div><p>The coercivity constant is independent of <span class="math notranslate nohighlight">\(h\)</span>, so the approximation is stable.</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">
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</div>
</details><p>Proving the coercivity for the Poisson problem with Dirichlet or
partial Dirichlet boundary conditions requires some additional
results. We start by showing that the divergence theorem also applies
to finite element derivatives of <span class="math notranslate nohighlight">\(C^0\)</span> finite element functions.</p>
<div class="proof proof-type-lemma" id="id39">
<div class="proof-title">
<span class="proof-type">Lemma 4.39</span>
<span class="proof-title-name">(Finite element divergence theorem)</span>
</div><div class="proof-content">
<p>Let <span class="math notranslate nohighlight">\(\phi\)</span> be a <span class="math notranslate nohighlight">\(C^1\)</span> vector-valued function.
and <span class="math notranslate nohighlight">\(u\in V\)</span> be a member of a <span class="math notranslate nohighlight">\(C^0\)</span> finite element space. Then</p>
<div class="math notranslate nohighlight">
\[\int_{\Omega} \nabla\cdot(\phi u)\, d x = \int_{\partial\Omega}
\phi\cdot n u\, d S,\]</div>
<p>where <span class="math notranslate nohighlight">\(n\)</span> is the outward pointing normal to <span class="math notranslate nohighlight">\(\partial\Omega\)</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">
<div class="math notranslate nohighlight">
\[ \begin{align}\begin{aligned}\int_\Omega\nabla\cdot(\phi u)\, d x &= \sum_{K\in \mathcal{T}}
\int_K \nabla\cdot(\phi u)\, d x,\\&= \sum_{K\in \mathcal{T}} \int_{\partial K}\phi\cdot n_K u\, d S,\\&= \int_{\partial\Omega} \phi\cdot n u\, d S
+ \underbrace{\int_\Gamma \phi\cdot(n^++n^-)u\, d S}_{=0}.\end{aligned}\end{align} \]</div>
</div></div><p>This allows us to prove the finite element trace theorem, which
relates the <span class="math notranslate nohighlight">\(H^1\)</span> norm of a <span class="math notranslate nohighlight">\(C^0\)</span> finite element function to the <span class="math notranslate nohighlight">\(L^2\)</span>
norm of the function restricted to the boundary.</p>
<div class="proof proof-type-theorem" id="id40">
<div class="proof-title">
<span class="proof-type">Theorem 4.40</span>
<span class="proof-title-name">(Trace theorem for continuous finite elements)</span>
</div><div class="proof-content">
<p>Let <span class="math notranslate nohighlight">\(V_h\)</span> be a continuous finite element space, defined on a triangulation <span class="math notranslate nohighlight">\(\mathcal{T}\)</span>, on a polygonal domain <span class="math notranslate nohighlight">\(\Omega\)</span>. Then</p>
<div class="math notranslate nohighlight">
\[\|u\|_{L^2(\partial\Omega)} \leq C\|u\|_{H^1(\Omega)},\]</div>
<p>where <span class="math notranslate nohighlight">\(C\)</span> is a constant that depends only on the geometry of
<span class="math notranslate nohighlight">\(\Omega\)</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>The first step is to construct a <span class="math notranslate nohighlight">\(C^1\)</span> function <span class="math notranslate nohighlight">\(\xi\)</span> satisfying
<span class="math notranslate nohighlight">\(\xi\cdot n=1\)</span> on <span class="math notranslate nohighlight">\(\Omega\)</span>. We do this by finding a triangulation
<span class="math notranslate nohighlight">\(\mathcal{T}_0\)</span> (unrelated to <span class="math notranslate nohighlight">\(\mathcal{T}\)</span>), and defining an <span class="math notranslate nohighlight">\(C^1\)</span>
Argyris finite element space <span class="math notranslate nohighlight">\(V_0\)</span> on it. We then choose <span class="math notranslate nohighlight">\(\xi\)</span> so that
both Cartesian components are in <span class="math notranslate nohighlight">\(V_0\)</span>, satisfying the boundary
condition.</p>
<p>Then,</p>
<div class="math notranslate nohighlight">
\[ \begin{align}\begin{aligned}\|u\|_{L^2(\partial\Omega)}^2 &=
\int_{\partial\Omega} u^2 \, d S = \int_{\partial\Omega}\xi\cdot n u^2\, d S,\\&= \int_\Omega \nabla\cdot (\xi u^2)\, d x,\\&= \int_\Omega u^2 \nabla\cdot \xi + 2u\xi\cdot\nabla u \, d x,\\&\leq \|u\|_{L^2}^2\|\nabla\cdot\xi\|_{\infty} + 2|\xi|_{\infty}
\|u\|_{L^2}|u|_{H^1},\end{aligned}\end{align} \]</div>
<p>So,</p>
<div class="math notranslate nohighlight">
\[ \begin{align}\begin{aligned}\|u\|_{L^2(\partial\Omega)}^2 &\leq \|u\|_{L^2}^2\|\nabla\cdot\xi\|_{\infty} + |\xi|_{\infty}
\left(\|u\|_{L^2}^2 + |u|_{H^1}^2\right),\\&\leq C\|u\|_{H^1}^2,\end{aligned}\end{align} \]</div>
<p>where we have used the geometric-arithmetic mean inequality <span class="math notranslate nohighlight">\(2ab \leq
a^2+b^2\)</span>.</p>
</div></div><details class="sd-sphinx-override sd-dropdown sd-card sd-mb-3">
<summary class="sd-summary-title sd-card-header">
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</div>
</details><p>We can now use the trace inequality to estabilish solvability for the
Poisson problem with (full or partial) Dirichlet conditions.</p>
<div class="proof proof-type-theorem" id="id41">
<div class="proof-title">