# Vector space of approximation (Finite element method)

I am following the book Topics in Functional analysis and applications by S. Kesavan. The text introduces the finite element method as:

Let $V$ be a real Hilbert space and let $a: V × V → \mathbb{R}$ be a $V$-elliptic and continuous bilinear form. Let $f \in V$. Then by the Lax-Milgram theorem there exists a unique $u \in V$ such that$$a(u, v) = (f, v). \quad \forall v \in V \tag{3.7.1}$$ We have seen in Section 3.2 several examples of this set-up. We now turn to the approximation of the solution $u$. The Galerkin method described in Section 3.4 gives us an idea as to how to proceed to do this.

Let $h > 0$ be a parameter. To it we associate $V_h$ which is a finite dimensional subspace of $V$. Now consider the following problem: Find $u_h \in V_h$ such that$$a(u_h, v_h) = (f, v_h). \quad \forall v_h \in V_h \tag{3.7.2}$$ Since $a$ is continuous and $V$-elliptic, it is so on $V_h$ as well and by the Lax-Milgram theorem, $u_h$ exists uniquely.

In Section 3.2, we have seen several examples of the program set out in Step 1 above. Thus we will have a bounded domain $\mit Ω$ and a problem set out in a space $V$ which will usually be a subspace of $H^1({\mit Ω})$ (if the problem is of the second order) or $H^2({\mit Ω})$ (for the fourth order problems).

The basic idea to construct the spaces $V_h$ is to partition the domain into smaller entities, say, triangles. Assume, for simplicity, that $\mit Ω$ is a polygonal domain. A triangulation $\mathscr{T}_h$ of $\mit Ω$ is a partition of $\mit Ω$ into triangles (i.e. “$n$”-simplices in $\mathbb{R}^n$) such that each “side” is either part of the boundary $\mit Γ$ or is a “side” of an adjacent triangle. Thus we do not allow triangles as shown in Fig. 13.

Thus we have a triangulation as follows

We denote by $K$ any generic triangle in $\mathscr{T}_h$, The “$h$” now stands for$$h = \max_{K \in \mathscr{T}_h} \operatorname{diam}(K)$$ so that as $h → 0$ the triangles are smaller and smaller and we get a fine mesh on $\mit Ω$.

Now we say that $V_h$ must be a space of piecewise polynomials with respect to $\mathscr{T}_h$. That is we fix an integer $k$ and if $P_k$ stands for polynomials of degree $\leqslant k$ in $n$ variables, then we wish that if $v \in V_h$ then $v|_K \in P_k$ for every $K \in \mathscr{T}_h$. Thus $V_h$ is clearly finite dimensional and once $k$ is fixed, the dimension of $V_h$ can be increased by just refining the given triangulation. Thus as $h → 0$, $N(h) → ∞$.

1. Can someone explain me the last paragraph of the last page where the text mentions about the subspace $V_h$ as the space of polynomials?
2. What is $\left.v\right|_K$?

$\Omega$ is the domain of the PDE we're trying to solve. When they say to partition $\Omega$ into a mesh of triangles, they really mean to discretize $\Omega$ as the vertices of said triangles (call them "nodes"). They want to find an approximate solution $u_h \in V_h \subset V$ that is at least defined at every node in our discretization of $\Omega$. For points in $\Omega$ that are between nodes, $u_h$ is interpolated as a polynomial of order $k$. Thus $V_h$ (which contains $u_h$) is a space of piecewise polynomials "with respect" to the triangles $\mathscr{T}_h$. That is, each $v \in V_h$ is an entire piecewise-polynomial function defined everywhere on $\Omega$ and they use the symbol $v|_K \in P_k$ to denote the "piece" of it that is "over" (interpolates through) a specific triangle $K \in \mathscr{T}_h$. To clarify, if our mesh contains $N$ triangles then for all $\omega \in \Omega$ we have, $$v(\omega) = \begin{cases} v|_{K_1}(\omega) \in P_k & \text{for } \omega \in K_1 \subset \Omega\\ \vdots \\ v|_{K_N}(\omega) \in P_k & \text{for } \omega \in K_N \subset \Omega \end{cases}$$
• 1. $v \in V_h$ is defined on $\Omega$? 2.If $1$ is true, then as $\left.v\right|_K \in P_k$ and $\Omega=\cup_{i\in\mathbb{N}} K_i$, deg$(v)\le k$. Are we sure about that? 3. What is the relation between $V, \Omega$ and $V_h,\Omega$? Commented Jun 12, 2018 at 4:09
• $V$ is an infinite-dimensional Hilbert space of functions defined on $\Omega$, and $V_h$ is a finite-dimensional subspace of $V$. I agree that the mesh should be chosen such that the union of all the triangles yields the entire domain. I don't see how these ideas are conflicting. Commented Jun 12, 2018 at 4:19
• I never said we have a contradiction. The text didn't mention specifically that $V$ is a real Hilbert space of "functions". Now, since $V_h$ is finite dimensional space of functions, are we sure all of these are polynomials or can be approximated by polynomials? Commented Jun 12, 2018 at 4:26
• For $(3.7.1)$ and $(3.7.2)$ to hold, $V$ only needs to be a Hilbert space. However, after those statements are made, the text proceeds to explain the finite element method for solving problems of the form $(3.7.2)$ where specifically $V$ is a subspace of $H^p(\Omega)$ which is an infinite-dimensional space of functions on $\Omega$. When we say that $V_h$ is a "finite-dimensional" subspace of $V$, we mean that it can be defined by a finite set of parameters (like $P^k$ can). Commented Jun 12, 2018 at 5:07
• The choice to let $V_h$ be piecewise polynomials (with respect to a some polygonal domain mesh) is not required for the finite element method, but is traditional and effective, since they are linear in their parameters and easy to integrate. Commented Jun 12, 2018 at 5:07