Function approximation by global functions¶

Many successful numerical solution methods for differential equations, including the finite element method, aim at approximating the unknown function by a sum

$$\begin{equation} u(x) \approx \sum_{i=0}^N c_i{\psi}_i(x), \label{fem:u} \tag{1} \end{equation}$$

where ${\psi}_i(x)$ are prescribed functions and $c_0,\ldots,c_N$ are unknown coefficients to be determined. Solution methods for differential equations utilizing (1) must have a principle for constructing $N+1$ equations to determine $c_0,\ldots,c_N$. Then there is a machinery regarding the actual construction of the equations for $c_0,\ldots,c_N$, in a particular problem. Finally, there is a solve phase for computing the solution $c_0,\ldots,c_N$ of the $N+1$ equations.

Especially in the finite element method, the machinery for constructing the discrete equations to be implemented on a computer is quite comprehensive, with many mathematical and implementational details entering the scene at the same time. From an ease-of-learning perspective it can therefore be wise to introduce the computational machinery for a trivial equation: $u=f$. Solving this equation with $f$ given and $u$ of the form (1), means that we seek an approximation $u$ to $f$. This approximation problem has the advantage of introducing most of the finite element toolbox, but without involving demanding topics related to differential equations (e.g., integration by parts, boundary conditions, and coordinate mappings). This is the reason why we shall first become familiar with finite element approximation before addressing finite element methods for differential equations.

First, we refresh some linear algebra concepts about approximating vectors in vector spaces. Second, we extend these concepts to approximating functions in function spaces, using the same principles and the same notation. We present examples on approximating functions by global basis functions with support throughout the entire domain. That is, the functions are in general nonzero on the entire domain. Third, we introduce the finite element type of basis functions globally. These basis functions will later, in Function approximation by finite elements, be used with local support (meaning that each function is nonzero except in a small part of the domain) to enhance stability and efficiency. We explain all the details of the computational algorithms involving such functions. Four types of approximation principles are covered: 1) the least squares method, 2) the $L_2$ projection or Galerkin method, 3) interpolation or collocation, and 4) the regression method.

Approximation of vectors¶

We shall start by introducing two fundamental methods for determining the coefficients $c_i$ in (1). These methods will be introduced for approximation of vectors. Using vectors in vector spaces to bring across the ideas is believed to be more intuitive to the reader than starting directly with functions in function spaces. The extension from vectors to functions will be trivial as soon as the fundamental ideas are understood.

The first method of approximation is called the least squares method and consists in finding $c_i$ such that the difference $f-u$, measured in a certain norm, is minimized. That is, we aim at finding the best approximation $u$ to $f$, with the given norm as measure of "distance". The second method is not as intuitive: we find $u$ such that the error $f-u$ is orthogonal to the space where $u$ lies. This is known as projection, or in the context of differential equations, the idea is also well known as Galerkin's method. When approximating vectors and functions, the two methods are equivalent, but this is no longer the case when applying the principles to differential equations.

Approximation of planar vectors¶

Let $\boldsymbol{f} = (3,5)$ be a vector in the $xy$ plane and suppose we want to approximate this vector by a vector aligned in the direction of another vector that is restricted to be aligned with some vector $(a,b)$. Figure depicts the situation. This is the simplest approximation problem for vectors. Nevertheless, for many readers it will be wise to refresh some basic linear algebra by consulting a textbook. Familiarity with fundamental operations on inner product vector spaces are assumed in the forthcoming text.

Approximation of a two-dimensional vector in a one-dimensional vector space.

We introduce the vector space $V$ spanned by the vector $\boldsymbol{\psi}_0=(a,b)$:

$$\begin{equation} V = \mbox{span}\,\{ \boldsymbol{\psi}_0\}{\thinspace .} \label{_auto1} \tag{2} \end{equation}$$

We say that $\boldsymbol{\psi}_0$ is a basis vector in the space $V$. Our aim is to find the vector

$$\begin{equation} \label{uc0} \tag{3} \boldsymbol{u} = c_0\boldsymbol{\psi}_0\in V \end{equation}$$

which best approximates the given vector $\boldsymbol{f} = (3,5)$. A reasonable criterion for a best approximation could be to minimize the length of the difference between the approximate $\boldsymbol{u}$ and the given $\boldsymbol{f}$. The difference, or error $\boldsymbol{e} = \boldsymbol{f} -\boldsymbol{u}$, has its length given by the norm

$$||\boldsymbol{e}|| = (\boldsymbol{e},\boldsymbol{e})^{\frac{1}{2}},$$

where $(\boldsymbol{e},\boldsymbol{e})$ is the inner product of $\boldsymbol{e}$ and itself. The inner product, also called scalar product or dot product, of two vectors $\boldsymbol{u}=(u_0,u_1)$ and $\boldsymbol{v} =(v_0,v_1)$ is defined as

$$\begin{equation} (\boldsymbol{u}, \boldsymbol{v}) = u_0v_0 + u_1v_1{\thinspace .} \label{_auto2} \tag{4} \end{equation}$$

Remark. We should point out that we use the notation $(\cdot,\cdot)$ for two different things: $(a,b)$ for scalar quantities $a$ and $b$ means the vector starting at the origin and ending in the point $(a,b)$, while $(\boldsymbol{u},\boldsymbol{v})$ with vectors $\boldsymbol{u}$ and $\boldsymbol{v}$ means the inner product of these vectors. Since vectors are here written in boldface font there should be no confusion. We may add that the norm associated with this inner product is the usual Euclidean length of a vector, i.e.,

$$\|\boldsymbol{u}\| = \sqrt{(\boldsymbol{u},\boldsymbol{u})} = \sqrt{u_0^2 + u_1^2}$$

The least squares method¶

We now want to determine the $\boldsymbol{u}$ that minimizes $||\boldsymbol{e}||$, that is we want to compute the optimal $c_0$ in (3). The algebra is simplified if we minimize the square of the norm, $||\boldsymbol{e}||^2 = (\boldsymbol{e}, \boldsymbol{e})$, instead of the norm itself. Define the function

$$\begin{equation} E(c_0) = (\boldsymbol{e},\boldsymbol{e}) = (\boldsymbol{f} - c_0\boldsymbol{\psi}_0, \boldsymbol{f} - c_0\boldsymbol{\psi}_0) {\thinspace .} \label{_auto3} \tag{5} \end{equation}$$

We can rewrite the expressions of the right-hand side in a more convenient form for further use:

$$\begin{equation} E(c_0) = (\boldsymbol{f},\boldsymbol{f}) - 2c_0(\boldsymbol{f},\boldsymbol{\psi}_0) + c_0^2(\boldsymbol{\psi}_0,\boldsymbol{\psi}_0){\thinspace .} \label{fem:vec:E} \tag{6} \end{equation}$$

This rewrite results from using the following fundamental rules for inner product spaces:

$$\begin{equation} (\alpha\boldsymbol{u},\boldsymbol{v})=\alpha(\boldsymbol{u},\boldsymbol{v}),\quad \alpha\in\mathbb{R}, \label{fem:vec:rule:scalarmult} \tag{7} \end{equation}$$

$$\begin{equation} (\boldsymbol{u} +\boldsymbol{v},\boldsymbol{w}) = (\boldsymbol{u},\boldsymbol{w}) + (\boldsymbol{v}, \boldsymbol{w}), \label{fem:vec:rule:sum} \tag{8} \end{equation}$$

$$\begin{equation} \label{fem:vec:rule:symmetry} \tag{9} (\boldsymbol{u}, \boldsymbol{v}) = (\boldsymbol{v}, \boldsymbol{u}){\thinspace .} \end{equation}$$

Minimizing $E(c_0)$ implies finding $c_0$ such that

$$\frac{\partial E}{\partial c_0} = 0{\thinspace .}$$

It turns out that $E$ has one unique minimum and no maximum point. Now, when differentiating (6) with respect to $c_0$, note that none of the inner product expressions depend on $c_0$, so we simply get

$$\begin{equation} \frac{\partial E}{\partial c_0} = -2(\boldsymbol{f},\boldsymbol{\psi}_0) + 2c_0 (\boldsymbol{\psi}_0,\boldsymbol{\psi}_0) {\thinspace .} \label{fem:vec:dEdc0:v1} \tag{10} \end{equation}$$

Setting the above expression equal to zero and solving for $c_0$ gives

$$\begin{equation} c_0 = \frac{(\boldsymbol{f},\boldsymbol{\psi}_0)}{(\boldsymbol{\psi}_0,\boldsymbol{\psi}_0)}, \label{fem:vec:c0} \tag{11} \end{equation}$$

which in the present case, with $\boldsymbol{\psi}_0=(a,b)$, results in

$$\begin{equation} c_0 = \frac{3a + 5b}{a^2 + b^2}{\thinspace .} \label{_auto4} \tag{12} \end{equation}$$

For later, it is worth mentioning that setting the key equation (10) to zero and ordering the terms lead to

$$(\boldsymbol{f}-c_0\boldsymbol{\psi}_0,\boldsymbol{\psi}_0) = 0,$$

or

$$\begin{equation} (\boldsymbol{e}, \boldsymbol{\psi}_0) = 0 {\thinspace .} \label{fem:vec:dEdc0:Galerkin} \tag{13} \end{equation}$$

This implication of minimizing $E$ is an important result that we shall make much use of.

The projection method¶

We shall now show that minimizing $||\boldsymbol{e}||^2$ implies that $\boldsymbol{e}$ is orthogonal to any vector $\boldsymbol{v}$ in the space $V$. This result is visually quite clear from Figure (think of other vectors along the line $(a,b)$: all of them will lead to a larger distance between the approximation and $\boldsymbol{f}$). Then we see mathematically that $\boldsymbol{e}$ is orthogonal to any vector $\boldsymbol{v}$ in the space $V$ and we may express any $\boldsymbol{v}\in V$ as $\boldsymbol{v}=s\boldsymbol{\psi}_0$ for any scalar parameter $s$ (recall that two vectors are orthogonal when their inner product vanishes). Then we calculate the inner product

$$\begin{align*} (\boldsymbol{e}, s\boldsymbol{\psi}_0) &= (\boldsymbol{f} - c_0\boldsymbol{\psi}_0, s\boldsymbol{\psi}_0)\\ &= (\boldsymbol{f},s\boldsymbol{\psi}_0) - (c_0\boldsymbol{\psi}_0, s\boldsymbol{\psi}_0)\\ &= s(\boldsymbol{f},\boldsymbol{\psi}_0) - sc_0(\boldsymbol{\psi}_0, \boldsymbol{\psi}_0)\\ &= s(\boldsymbol{f},\boldsymbol{\psi}_0) - s\frac{(\boldsymbol{f},\boldsymbol{\psi}_0)}{(\boldsymbol{\psi}_0,\boldsymbol{\psi}_0)}(\boldsymbol{\psi}_0,\boldsymbol{\psi}_0)\\ &= s\left( (\boldsymbol{f},\boldsymbol{\psi}_0) - (\boldsymbol{f},\boldsymbol{\psi}_0)\right)\\ &=0{\thinspace .} \end{align*}$$

Therefore, instead of minimizing the square of the norm, we could demand that $\boldsymbol{e}$ is orthogonal to any vector in $V$, which in our present simple case amounts to a single vector only. This method is known as projection. (The approach can also be referred to as a Galerkin method as explained at the end of the section Approximation of general vectors.)

Mathematically, the projection method is stated by the equation

$$\begin{equation} (\boldsymbol{e}, \boldsymbol{v}) = 0,\quad\forall\boldsymbol{v}\in V{\thinspace .} \label{fem:vec:Galerkin1} \tag{14} \end{equation}$$

An arbitrary $\boldsymbol{v}\in V$ can be expressed as $s\boldsymbol{\psi}_0$, $s\in\mathbb{R}$, and therefore (14) implies

$$(\boldsymbol{e},s\boldsymbol{\psi}_0) = s(\boldsymbol{e}, \boldsymbol{\psi}_0) = 0,$$

which means that the error must be orthogonal to the basis vector in the space $V$:

$$(\boldsymbol{e}, \boldsymbol{\psi}_0)=0\quad\hbox{or}\quad (\boldsymbol{f} - c_0\boldsymbol{\psi}_0, \boldsymbol{\psi}_0)=0,$$

which is what we found in (13) from the least squares computations.

Approximation of general vectors¶

Let us generalize the vector approximation from the previous section to vectors in spaces with arbitrary dimension. Given some vector $\boldsymbol{f}$, we want to find the best approximation to this vector in the space

$$V = \hbox{span}\,\{\boldsymbol{\psi}_0,\ldots,\boldsymbol{\psi}_N\} {\thinspace .}$$

We assume that the space has dimension $N+1$ and that basis vectors $\boldsymbol{\psi}_0,\ldots,\boldsymbol{\psi}_N$ are linearly independent so that none of them are redundant. Any vector $\boldsymbol{u}\in V$ can then be written as a linear combination of the basis vectors, i.e.,

$$\boldsymbol{u} = \sum_{j=0}^N c_j\boldsymbol{\psi}_j,$$

where $c_j\in\mathbb{R}$ are scalar coefficients to be determined.

The least squares method¶

Now we want to find $c_0,\ldots,c_N$, such that $\boldsymbol{u}$ is the best approximation to $\boldsymbol{f}$ in the sense that the distance (error) $\boldsymbol{e} = \boldsymbol{f} - \boldsymbol{u}$ is minimized. Again, we define the squared distance as a function of the free parameters $c_0,\ldots,c_N$,

$$E(c_0,\ldots,c_N) = (\boldsymbol{e},\boldsymbol{e}) = (\boldsymbol{f} -\sum_jc_j\boldsymbol{\psi}_j,\boldsymbol{f} -\sum_jc_j\boldsymbol{\psi}_j) \nonumber$$

$$\begin{equation} = (\boldsymbol{f},\boldsymbol{f}) - 2\sum_{j=0}^N c_j(\boldsymbol{f},\boldsymbol{\psi}_j) + \sum_{p=0}^N\sum_{q=0}^N c_pc_q(\boldsymbol{\psi}_p,\boldsymbol{\psi}_q){\thinspace .} \label{fem:vec:genE} \tag{15} \end{equation}$$

Minimizing this $E$ with respect to the independent variables $c_0,\ldots,c_N$ is obtained by requiring

$$\frac{\partial E}{\partial c_i} = 0,\quad i=0,\ldots,N {\thinspace .}$$

The first term in (15) is independent of $c_i$, so its derivative vanishes. The second term in (15) is differentiated as follows:

$$\begin{equation} \frac{\partial}{\partial c_i} 2\sum_{j=0}^N c_j(\boldsymbol{f},\boldsymbol{\psi}_j) = 2(\boldsymbol{f},\boldsymbol{\psi}_i), \label{_auto5} \tag{16} \end{equation}$$

since the expression to be differentiated is a sum and only one term, $c_i(\boldsymbol{f},\boldsymbol{\psi}_i)$, contains $c_i$ (this term is linear in $c_i$). To understand this differentiation in detail, write out the sum specifically for, e.g, $N=3$ and $i=1$.

The last term in (15) is more tedious to differentiate. It can be wise to write out the double sum for $N=1$ and perform differentiation with respect to $c_0$ and $c_1$ to see the structure of the expression. Thereafter, one can generalize to an arbitrary $N$ and observe that

$$\begin{equation} \frac{\partial}{\partial c_i} c_pc_q = \left\lbrace\begin{array}{ll} 0, & \hbox{ if } p\neq i\hbox{ and } q\neq i,\\ c_q, & \hbox{ if } p=i\hbox{ and } q\neq i,\\ c_p, & \hbox{ if } p\neq i\hbox{ and } q=i,\\ 2c_i, & \hbox{ if } p=q= i{\thinspace .}\\ \end{array}\right. \label{_auto6} \tag{17} \end{equation}$$

Then

$$\frac{\partial}{\partial c_i} \sum_{p=0}^N\sum_{q=0}^N c_pc_q(\boldsymbol{\psi}_p,\boldsymbol{\psi}_q) = \sum_{p=0, p\neq i}^N c_p(\boldsymbol{\psi}_p,\boldsymbol{\psi}_i) + \sum_{q=0, q\neq i}^N c_q(\boldsymbol{\psi}_i,\boldsymbol{\psi}_q) +2c_i(\boldsymbol{\psi}_i,\boldsymbol{\psi}_i){\thinspace .}$$

Since each of the two sums is missing the term $c_i(\boldsymbol{\psi}_i,\boldsymbol{\psi}_i)$, we may split the very last term in two, to get exactly that "missing" term for each sum. This idea allows us to write

$$\begin{equation} \frac{\partial}{\partial c_i} \sum_{p=0}^N\sum_{q=0}^N c_pc_q(\boldsymbol{\psi}_p,\boldsymbol{\psi}_q) = 2\sum_{j=0}^N c_i(\boldsymbol{\psi}_j,\boldsymbol{\psi}_i){\thinspace .} \label{_auto7} \tag{18} \end{equation}$$

It then follows that setting

$$\frac{\partial E}{\partial c_i} = 0,\quad i=0,\ldots,N,$$

implies

$$- 2(\boldsymbol{f},\boldsymbol{\psi}_i) + 2\sum_{j=0}^N c_i(\boldsymbol{\psi}_j,\boldsymbol{\psi}_i) = 0,\quad i=0,\ldots,N{\thinspace .}$$

Moving the first term to the right-hand side shows that the equation is actually a linear system for the unknown parameters $c_0,\ldots,c_N$:

$$\begin{equation} \sum_{j=0}^N A_{i,j} c_j = b_i, \quad i=0,\ldots,N, \label{fem:approx:vec:Np1dim:eqsys} \tag{19} \end{equation}$$

where

$$\begin{equation} A_{i,j} = (\boldsymbol{\psi}_i,\boldsymbol{\psi}_j), \label{_auto8} \tag{20} \end{equation}$$

$$\begin{equation} b_i = (\boldsymbol{\psi}_i, \boldsymbol{f}){\thinspace .} \label{_auto9} \tag{21} \end{equation}$$

We have changed the order of the two vectors in the inner product according to (9):

$$A_{i,j} = (\boldsymbol{\psi}_j,\boldsymbol{\psi}_i) = (\boldsymbol{\psi}_i,\boldsymbol{\psi}_j),$$

simply because the sequence $i$-$j$ looks more aesthetic.

The Galerkin or projection method¶

In analogy with the "one-dimensional" example in the section Approximation of planar vectors, it holds also here in the general case that minimizing the distance (error) $\boldsymbol{e}$ is equivalent to demanding that $\boldsymbol{e}$ is orthogonal to all $\boldsymbol{v}\in V$:

$$\begin{equation} (\boldsymbol{e},\boldsymbol{v})=0,\quad \forall\boldsymbol{v}\in V{\thinspace .} \label{fem:approx:vec:Np1dim:Galerkin} \tag{22} \end{equation}$$

Since any $\boldsymbol{v}\in V$ can be written as $\boldsymbol{v} =\sum_{i=0}^N c_i\boldsymbol{\psi}_i$, the statement (22) is equivalent to saying that

$$(\boldsymbol{e}, \sum_{i=0}^N c_i\boldsymbol{\psi}_i) = 0,$$

for any choice of coefficients $c_0,\ldots,c_N$. The latter equation can be rewritten as

$$\sum_{i=0}^N c_i (\boldsymbol{e},\boldsymbol{\psi}_i) =0{\thinspace .}$$

If this is to hold for arbitrary values of $c_0,\ldots,c_N$, we must require that each term in the sum vanishes, which means that

$$\begin{equation} (\boldsymbol{e},\boldsymbol{\psi}_i)=0,\quad i=0,\ldots,N{\thinspace .} \label{fem:approx:vec:Np1dim:Galerkin0} \tag{23} \end{equation}$$

These $N+1$ equations result in the same linear system as (19):

$$(\boldsymbol{f} - \sum_{j=0}^N c_j\boldsymbol{\psi}_j, \boldsymbol{\psi}_i) = (\boldsymbol{f}, \boldsymbol{\psi}_i) - \sum_{j=0}^N (\boldsymbol{\psi}_i,\boldsymbol{\psi}_j)c_j = 0,$$

and hence

$$\sum_{j=0}^N (\boldsymbol{\psi}_i,\boldsymbol{\psi}_j)c_j = (\boldsymbol{f}, \boldsymbol{\psi}_i),\quad i=0,\ldots, N {\thinspace .}$$

So, instead of differentiating the $E(c_0,\ldots,c_N)$ function, we could simply use (22) as the principle for determining $c_0,\ldots,c_N$, resulting in the $N+1$ equations (23).

The names least squares method or least squares approximation are natural since the calculations consists of minimizing $||\boldsymbol{e}||^2$, and $||\boldsymbol{e}||^2$ is a sum of squares of differences between the components in $\boldsymbol{f}$ and $\boldsymbol{u}$. We find $\boldsymbol{u}$ such that this sum of squares is minimized.

The principle (22), or the equivalent form (23), is known as projection. Almost the same mathematical idea was used by the Russian mathematician Boris Galerkin to solve differential equations, resulting in what is widely known as Galerkin's method.

Approximation principles¶

Let $V$ be a function space spanned by a set of basis functions ${\psi}_0,\ldots,{\psi}_N$,

$$V = \hbox{span}\,\{{\psi}_0,\ldots,{\psi}_N\},$$

such that any function $u\in V$ can be written as a linear combination of the basis functions:

$$\begin{equation} \label{fem:approx:ufem} \tag{24} u = \sum_{j\in{\mathcal{I}_s}} c_j{\psi}_j{\thinspace .} \end{equation}$$

That is, we consider functions as vectors in a vector space - a so-called function space - and we have a finite set of basis functions that span the space just as basis vectors or unit vectors span a vector space.

The index set ${\mathcal{I}_s}$ is defined as ${\mathcal{I}_s} =\{0,\ldots,N\}$ and is from now on used both for compact notation and for flexibility in the numbering of elements in sequences.

For now, in this introduction, we shall look at functions of a single variable $x$: $u=u(x)$, ${\psi}_j={\psi}_j(x)$, $j\in{\mathcal{I}_s}$. Later, we will almost trivially extend the mathematical details to functions of two- or three-dimensional physical spaces. The approximation (24) is typically used to discretize a problem in space. Other methods, most notably finite differences, are common for time discretization, although the form (24) can be used in time as well.

The least squares method¶

Given a function $f(x)$, how can we determine its best approximation $u(x)\in V$? A natural starting point is to apply the same reasoning as we did for vectors in the section Approximation of general vectors. That is, we minimize the distance between $u$ and $f$. However, this requires a norm for measuring distances, and a norm is most conveniently defined through an inner product. Viewing a function as a vector of infinitely many point values, one for each value of $x$, the inner product of two arbitrary functions $f(x)$ and $g(x)$ could intuitively be defined as the usual summation of pairwise "components" (values), with summation replaced by integration:

$$(f,g) = \int f(x)g(x)\, {\, \mathrm{d}x} {\thinspace .}$$

To fix the integration domain, we let $f(x)$ and ${\psi}_i(x)$ be defined for a domain $\Omega\subset\mathbb{R}$. The inner product of two functions $f(x)$ and $g(x)$ is then

$$\begin{equation} (f,g) = \int_\Omega f(x)g(x)\, {\, \mathrm{d}x} \label{fem:approx:LS:innerprod} \tag{25} {\thinspace .} \end{equation}$$

The distance between $f$ and any function $u\in V$ is simply $f-u$, and the squared norm of this distance is

$$\begin{equation} E = (f(x)-\sum_{j\in{\mathcal{I}_s}} c_j{\psi}_j(x), f(x)-\sum_{j\in{\mathcal{I}_s}} c_j{\psi}_j(x)){\thinspace .} \label{fem:approx:LS:E} \tag{26} \end{equation}$$

Note the analogy with (15): the given function $f$ plays the role of the given vector $\boldsymbol{f}$, and the basis function ${\psi}_i$ plays the role of the basis vector $\boldsymbol{\psi}_i$. We can rewrite (26), through similar steps as used for the result (15), leading to

$$\begin{equation} E(c_i, \ldots, c_N) = (f,f) -2\sum_{j\in{\mathcal{I}_s}} c_j(f,{\psi}_i) + \sum_{p\in{\mathcal{I}_s}}\sum_{q\in{\mathcal{I}_s}} c_pc_q({\psi}_p,{\psi}_q){\thinspace .} \label{_auto10} \tag{27} \end{equation}$$

Minimizing this function of $N+1$ scalar variables $\left\{ {c}_i \right\}_{i\in{\mathcal{I}_s}}$, requires differentiation with respect to $c_i$, for all $i\in{\mathcal{I}_s}$. The resulting equations are very similar to those we had in the vector case, and we hence end up with a linear system of the form (19), with basically the same expressions:

$$\begin{equation} A_{i,j} = ({\psi}_i,{\psi}_j), \label{fem:approx:Aij} \tag{28} \end{equation}$$

$$\begin{equation} b_i = (f,{\psi}_i){\thinspace .} \label{fem:approx:bi} \tag{29} \end{equation}$$

The only difference from (19) is that the inner product is defined in terms of integration rather than summation.

The projection (or Galerkin) method¶

As in the section Approximation of general vectors, the minimization of $(e,e)$ is equivalent to

$$\begin{equation} (e,v)=0,\quad\forall v\in V{\thinspace .} \label{fem:approx:Galerkin} \tag{30} \end{equation}$$

This is known as a projection of a function $f$ onto the subspace $V$. We may also call it a Galerkin method for approximating functions. Using the same reasoning as in (22)-(23), it follows that (30) is equivalent to

$$\begin{equation} (e,{\psi}_i)=0,\quad i\in{\mathcal{I}_s}{\thinspace .} \label{fem:approx:Galerkin0} \tag{31} \end{equation}$$

Inserting $e=f-u$ in this equation and ordering terms, as in the multi-dimensional vector case, we end up with a linear system with a coefficient matrix (28) and right-hand side vector (29).

Whether we work with vectors in the plane, general vectors, or functions in function spaces, the least squares principle and the projection or Galerkin method are equivalent.

Example of linear approximation¶

Let us apply the theory in the previous section to a simple problem: given a parabola $f(x)=10(x-1)^2-1$ for $x\in\Omega=[1,2]$, find the best approximation $u(x)$ in the space of all linear functions:

$$V = \hbox{span}\,\{1, x\}{\thinspace .}$$

With our notation, ${\psi}_0(x)=1$, ${\psi}_1(x)=x$, and $N=1$. We seek

$$u=c_0{\psi}_0(x) + c_1{\psi}_1(x) = c_0 + c_1x,$$

where $c_0$ and $c_1$ are found by solving a $2\times 2$ linear system. The coefficient matrix has elements

$$\begin{equation} A_{0,0} = ({\psi}_0,{\psi}_0) = \int_1^21\cdot 1\, {\, \mathrm{d}x} = 1, \label{_auto11} \tag{32} \end{equation}$$

$$\begin{equation} A_{0,1} = ({\psi}_0,{\psi}_1) = \int_1^2 1\cdot x\, {\, \mathrm{d}x} = 3/2, \label{_auto12} \tag{33} \end{equation}$$

$$\begin{equation} A_{1,0} = A_{0,1} = 3/2, \label{_auto13} \tag{34} \end{equation}$$

$$\begin{equation} A_{1,1} = ({\psi}_1,{\psi}_1) = \int_1^2 x\cdot x\,{\, \mathrm{d}x} = 7/3{\thinspace .} \label{_auto14} \tag{35} \end{equation}$$

The corresponding right-hand side is

$$\begin{equation} b_1 = (f,{\psi}_0) = \int_1^2 (10(x-1)^2 - 1)\cdot 1 \, {\, \mathrm{d}x} = 7/3, \label{_auto15} \tag{36} \end{equation}$$

$$\begin{equation} b_2 = (f,{\psi}_1) = \int_1^2 (10(x-1)^2 - 1)\cdot x\, {\, \mathrm{d}x} = 13/3{\thinspace .} \label{_auto16} \tag{37} \end{equation}$$

Solving the linear system results in

$$\begin{equation} c_0 = -38/3,\quad c_1 = 10, \label{_auto17} \tag{38} \end{equation}$$

and consequently

$$\begin{equation} u(x) = 10x - \frac{38}{3}{\thinspace .} \label{_auto18} \tag{39} \end{equation}$$

Figure displays the parabola and its best approximation in the space of all linear functions.

Best approximation of a parabola by a straight line.

Implementation of the least squares method¶

Symbolic integration¶

The linear system can be computed either symbolically or numerically (a numerical integration rule is needed in the latter case). Let us first compute the system and its solution symbolically, i.e., using classical "pen and paper" mathematics with symbols. The Python package sympy can greatly help with this type of mathematics, and will therefore be frequently used in this text. Some basic familiarity with sympy is assumed, typically symbols, integrate, diff, expand, and simplify. Much can be learned by studying the many applications of sympy that will be presented.

Below is a function for symbolic computation of the linear system, where $f(x)$ is given as a sympy expression f involving the symbol x, psi is a list of expressions for $\left\{ {{\psi}}_i \right\}_{i\in{\mathcal{I}_s}}$, and Omega is a 2-tuple/list holding the limits of the domain $\Omega$:

Observe that we exploit the symmetry of the coefficient matrix: only the upper triangular part is computed. Symbolic integration, also in sympy, is often time consuming, and (roughly) halving the work has noticeable effect on the waiting time for the computations to finish.

Notice.

We remark that the symbols in sympy are created and stored in a symbol factory that is indexed by the expression used in the construction and that repeated constructions from the same expression will not create new objects. The following code illustrates the behavior of the symbol factory:

Fall back on numerical integration¶

Obviously, sympy may fail to successfully integrate $\int_\Omega{\psi}_i{\psi}_j{\, \mathrm{d}x}$, and especially $\int_\Omega f{\psi}_i{\, \mathrm{d}x}$, symbolically. Therefore, we should extend the least_squares function such that it falls back on numerical integration if the symbolic integration is unsuccessful. In the latter case, the returned value from sympy's integrate function is an object of type Integral. We can test on this type and utilize the mpmath module to perform numerical integration of high precision. Even when sympy manages to integrate symbolically, it can take an undesirably long time. We therefore include an argument symbolic that governs whether or not to try symbolic integration. Here is a complete and improved version of the previous function least_squares:

The function is found in the file approx1D.py.

Plotting the approximation¶

Comparing the given $f(x)$ and the approximate $u(x)$ visually is done with the following function, which utilizes sympy's lambdify tool to convert a sympy expression to a Python function for numerical computations:

The modules='numpy' argument to lambdify is important if there are mathematical functions, such as sin or exp in the symbolic expressions in f or u, and these mathematical functions are to be used with vector arguments, like xcoor above.

Both the least_squares and comparison_plot functions are found in the file approx1D.py. The comparison_plot function in this file is more advanced and flexible than the simplistic version shown above. The file ex_approx1D.py applies the approx1D module to accomplish the forthcoming examples.

Perfect approximation¶

Let us use the code above to recompute the problem from the section Example of linear approximation where we want to approximate a parabola. What happens if we add an element $x^2$ to the set of basis functions and test what the best approximation is if $V$ is the space of all parabolic functions? The answer is quickly found by running

Now, what if we use ${\psi}_i(x)=x^i$ for $i=0,1,\ldots,N=40$? The output from least_squares gives $c_i=0$ for $i>2$, which means that the method finds the perfect approximation.

The residual: an indirect but computationally cheap measure of the error.¶

When attempting to solve a system $A c = b$, we may question how far off a start vector or a current approximation $c_0$ is. The error is clearly the difference between $c$ and $c_0$, $e=c-c_0$, but since we do not know the true solution $c$ we are unable to assess the error. However, the vector $c_0$ is the solution of the an alternative problem $A c_0 = b_0$. If the input, i.e., the right-hand sides $b_0$ and $b$ are close to each other then we expect the output of a solution process $c$ and $c_0$ to be close to each other. Furthermore, while $b_0$ in principle is unknown, it is easily computable as $b_0 = A c_0$ and does not require inversion of $A$. The vector $b - b_0$ is the so-called residual $r$ defined by

$$r = b - b_0 = b - A c_0 = A c - A c_0 .$$

Clearly, the error and the residual are related by

$$A e = r .$$

While the computation of the error requires inversion of $A$, which may be computationally expensive, the residual is easily computable and do only require a matrix-vector product and vector additions.

Orthogonal basis functions¶

Approximation of a function via orthogonal functions, especially sinusoidal functions or orthogonal polynomials, is a very popular and successful approach. The finite element method does not make use of orthogonal functions, but functions that are "almost orthogonal".

Ill-conditioning¶

For basis functions that are not orthogonal the condition number of the matrix may create problems during the solution process due to, for example, round-off errors as will be illustrated in the following. The computational example in the section Perfect approximation applies the least_squares function which invokes symbolic methods to calculate and solve the linear system. The correct solution $c_0=9, c_1=-20, c_2=10, c_i=0$ for $i\geq 3$ is perfectly recovered.

Suppose we convert the matrix and right-hand side to floating-point arrays and then solve the system using finite-precision arithmetics, which is what one will (almost) always do in real life. This time we get astonishing results! Up to about $N=7$ we get a solution that is reasonably close to the exact one. Increasing $N$ shows that seriously wrong coefficients are computed. Below is a table showing the solution of the linear system arising from approximating a parabola by functions of the form $u(x)=c_0 + c_1x + c_2x^2 + \cdots + c_{10}x^{10}$. Analytically, we know that $c_j=0$ for $j>2$, but numerically we may get $c_j\neq 0$ for $j>2$.

exact	sympy	numpy32	numpy64
9	9.62	5.57	8.98
-20	-23.39	-7.65	-19.93
10	17.74	-4.50	9.96
0	-9.19	4.13	-0.26
0	5.25	2.99	0.72
0	0.18	-1.21	-0.93
0	-2.48	-0.41	0.73
0	1.81	-0.013	-0.36
0	-0.66	0.08	0.11
0	0.12	0.04	-0.02
0	-0.001	-0.02	0.002

The exact value of $c_j$, $j=0,1,\ldots,10$, appears in the first column while the other columns correspond to results obtained by three different methods:

• Column 2: The matrix and vector are converted to the data structure mpmath.fp.matrix and the mpmath.fp.lu_solve function is used to solve the system.

• Column 3: The matrix and vector are converted to numpy arrays with data type numpy.float32 (single precision floating-point number) and solved by the numpy.linalg.solve function.

• Column 4: As column 3, but the data type is numpy.float64 (double precision floating-point number).

We see from the numbers in the table that double precision performs much better than single precision. Nevertheless, when plotting all these solutions the curves cannot be visually distinguished (!). This means that the approximations look perfect, despite the very wrong values of the coefficients.

Increasing $N$ to 12 makes the numerical solver in numpy abort with the message: "matrix is numerically singular". A matrix has to be non-singular to be invertible, which is a requirement when solving a linear system. Already when the matrix is close to singular, it is ill-conditioned, which here implies that the numerical solution algorithms are sensitive to round-off errors and may produce (very) inaccurate results.

The reason why the coefficient matrix is nearly singular and ill-conditioned is that our basis functions ${\psi}_i(x)=x^i$ are nearly linearly dependent for large $i$. That is, $x^i$ and $x^{i+1}$ are very close for $i$ not very small. This phenomenon is illustrated in Figure. There are 15 lines in this figure, but only half of them are visually distinguishable. Almost linearly dependent basis functions give rise to an ill-conditioned and almost singular matrix. This fact can be illustrated by computing the determinant, which is indeed very close to zero (recall that a zero determinant implies a singular and non-invertible matrix): $10^{-65}$ for $N=10$ and $10^{-92}$ for $N=12$. Already for $N=28$ the numerical determinant computation returns a plain zero.

The 15 first basis functions $x^i$, $i=0,\ldots,14$.

On the other hand, the double precision numpy solver does run for $N=100$, resulting in answers that are not significantly worse than those in the table above, and large powers are associated with small coefficients (e.g., $c_j < 10^{-2}$ for $10\leq j\leq 20$ and $c_j<10^{-5}$ for $j > 20$). Even for $N=100$ the approximation still lies on top of the exact curve in a plot (!).

The conclusion is that visual inspection of the quality of the approximation may not uncover fundamental numerical problems with the computations. However, numerical analysts have studied approximations and ill-conditioning for decades, and it is well known that the basis $\{1,x,x^2,x^3,\ldots,\}$ is a bad basis. The best basis from a matrix conditioning point of view is to have orthogonal functions such that $(\psi_i,\psi_j)=0$ for $i\neq j$. There are many known sets of orthogonal polynomials and other functions. The functions used in the finite element method are almost orthogonal, and this property helps to avoid problems when solving matrix systems. Almost orthogonal is helpful, but not enough when it comes to partial differential equations, and ill-conditioning of the coefficient matrix is a theme when solving large-scale matrix systems arising from finite element discretizations.

Fourier series¶

A set of sine functions is widely used for approximating functions (note that the sines are orthogonal with respect to the $L_2$ inner product as can be easily verified using sympy). Let us take

$$V = \hbox{span}\,\{ \sin \pi x, \sin 2\pi x,\ldots,\sin (N+1)\pi x\} {\thinspace .}$$

That is,

$${\psi}_i(x) = \sin ((i+1)\pi x),\quad i\in{\mathcal{I}_s}{\thinspace .}$$

An approximation to the parabola $f(x)=10(x-1)^2-1$ for $x\in\Omega=[1,2]$ from the section Example of linear approximation can then be computed by the least_squares function from the section Implementation of the least squares method:

Figure (left) shows the oscillatory approximation of $\sum_{j=0}^Nc_j\sin ((j+1)\pi x)$ when $N=3$. Changing $N$ to 11 improves the approximation considerably, see Figure (right).

Best approximation of a parabola by a sum of 3 (left) and 11 (right) sine functions.

There is an error $f(0)-u(0)=9$ at $x=0$ in Figure regardless of how large $N$ is, because all ${\psi}_i(0)=0$ and hence $u(0)=0$. We may help the approximation to be correct at $x=0$ by seeking

$$\begin{equation} u(x) = f(0) + \sum_{j\in{\mathcal{I}_s}} c_j{\psi}_j(x) {\thinspace .} \label{_auto21} \tag{49} \end{equation}$$

However, this adjustment introduces a new problem at $x=1$ since we now get an error $f(1)-u(1)=f(1)-0=-1$ at this point. A more clever adjustment is to replace the $f(0)$ term by a term that is $f(0)$ at $x=0$ and $f(1)$ at $x=1$. A simple linear combination $f(0)(1-x) + xf(1)$ does the job:

$$\begin{equation} u(x) = f(0)(1-x) + xf(1) + \sum_{j\in{\mathcal{I}_s}} c_j{\psi}_j(x) {\thinspace .} \label{_auto22} \tag{50} \end{equation}$$

This adjustment of $u$ alters the linear system slightly. In the general case, we set

$$u(x) = B(x) + \sum_{j\in{\mathcal{I}_s}} c_j{\psi}_j(x),$$

and the linear system becomes

$$\sum_{j\in{\mathcal{I}_s}}({\psi}_i,{\psi}_j)c_j = (f-B,{\psi}_i),\quad i\in{\mathcal{I}_s}{\thinspace .}$$

The calculations can still utilize the least_squares or least_squares_orth functions, but solve for $u-b$:

Figure shows the result of the technique for ensuring the right boundary values. Even 3 sines can now adjust the $f(0)(1-x) + xf(1)$ term such that $u$ approximates the parabola really well, at least visually.

Best approximation of a parabola by a sum of 3 (left) and 11 (right) sine functions with a boundary term.

Orthogonal basis functions¶

The choice of sine functions ${\psi}_i(x)=\sin ((i+1)\pi x)$ has a great computational advantage: on $\Omega=[0,1]$ these basis functions are orthogonal, implying that $A_{i,j}=0$ if $i\neq j$. In fact, when $V$ contains the basic functions used in a Fourier series expansion, the approximation method derived in the section Approximation principles results in the classical Fourier series for $f(x)$.

With orthogonal basis functions we can make the least_squares function (much) more efficient since we know that the matrix is diagonal and only the diagonal elements need to be computed:

As mentioned in the section Implementation of the least squares method, symbolic integration may fail or take a very long time. It is therefore natural to extend the implementation above with a version where we can choose between symbolic and numerical integration and fall back on the latter if the former fails:

This function is found in the file approx1D.py. Observe that we here assume that $\int_\Omega{\varphi}_i^2{\, \mathrm{d}x}$ can always be symbolically computed, which is not an unreasonable assumption when the basis functions are orthogonal, but there is no guarantee, so an improved version of the function above would implement numerical integration also for the A[i,i] term.

Numerical computations¶

Sometimes the basis functions ${\psi}_i$ and/or the function $f$ have a nature that makes symbolic integration CPU-time consuming or impossible. Even though we implemented a fallback on numerical integration of $\int f{\varphi}_i {\, \mathrm{d}x}$, considerable time might still be required by sympy just by attempting to integrate symbolically. Therefore, it will be handy to have a function for fast numerical integration and numerical solution of the linear system. Below is such a method. It requires Python functions f(x) and psi(x,i) for $f(x)$ and ${\psi}_i(x)$ as input. The output is a mesh function with values u on the mesh with points in the array x. Three numerical integration methods are offered: scipy.integrate.quad (precision set to $10^{-8}$), mpmath.quad (about machine precision), and a Trapezoidal rule based on the points in x (unknown accuracy, but increasing with the number of mesh points in x).

Here is an example on calling the function:

Remark. The scipy.integrate.quad integrator is usually much faster than mpmath.quad.

Interpolation¶

The interpolation (or collocation) principle¶

The principle of minimizing the distance between $u$ and $f$ is an intuitive way of computing a best approximation $u\in V$ to $f$. However, there are other approaches as well. One is to demand that $u(x_{i}) = f(x_{i})$ at some selected points $x_{i}$, $i\in{\mathcal{I}_s}$:

$$\begin{equation} u(x_{i}) = \sum_{j\in{\mathcal{I}_s}} c_j {\psi}_j(x_{i}) = f(x_{i}), \quad i\in{\mathcal{I}_s}{\thinspace .} \label{_auto24} \tag{52} \end{equation}$$

We recognize that the equation $\sum_j c_j {\psi}_j(x_{i}) = f(x_{i})$ is actually a linear system with $N+1$ unknown coefficients $\left\{ {c}_j \right\}_{j\in{\mathcal{I}_s}}$:

$$\begin{equation} \sum_{j\in{\mathcal{I}_s}} A_{i,j}c_j = b_i,\quad i\in{\mathcal{I}_s}, \label{_auto25} \tag{53} \end{equation}$$

with coefficient matrix and right-hand side vector given by

$$\begin{equation} A_{i,j} = {\psi}_j(x_{i}), \label{_auto26} \tag{54} \end{equation}$$

$$\begin{equation} b_i = f(x_{i}){\thinspace .} \label{_auto27} \tag{55} \end{equation}$$

This time the coefficient matrix is not symmetric because ${\psi}_j(x_{i})\neq {\psi}_i(x_{j})$ in general. The method is often referred to as an interpolation method since some point values of $f$ are given ($f(x_{i})$) and we fit a continuous function $u$ that goes through the $f(x_{i})$ points. In this case the $x_{i}$ points are called interpolation points. When the same approach is used to approximate differential equations, one usually applies the name collocation method and $x_{i}$ are known as collocation points.

Given $f$ as a sympy symbolic expression f, $\left\{ {{\psi}}_i \right\}_{i\in{\mathcal{I}_s}}$ as a list psi, and a set of points $\left\{ {x}_i \right\}_{i\in{\mathcal{I}_s}}$ as a list or array points, the following Python function sets up and solves the matrix system for the coefficients $\left\{ {c}_i \right\}_{i\in{\mathcal{I}_s}}$:

The interpolation function is a part of the approx1D module.

We found it convenient in the above function to turn the expressions f and psi into ordinary Python functions of x, which can be called with float values in the list points when building the matrix and the right-hand side. The alternative is to use the subs method to substitute the x variable in an expression by an element from the points list. The following session illustrates both approaches in a simple setting:

A nice feature of the interpolation or collocation method is that it avoids computing integrals. However, one has to decide on the location of the $x_{i}$ points. A simple, yet common choice, is to distribute them uniformly throughout the unit interval.

Example¶

Let us illustrate the interpolation method by approximating our parabola $f(x)=10(x-1)^2-1$ by a linear function on $\Omega=[1,2]$, using two collocation points $x_0=1+1/3$ and $x_1=1+2/3$:

The resulting linear system becomes

$$\left(\begin{array}{ll} 1 & 4/3\\ 1 & 5/3\\ \end{array}\right) \left(\begin{array}{l} c_0\\ c_1\\ \end{array}\right) = \left(\begin{array}{l} 1/9\\ 31/9\\ \end{array}\right)$$

with solution $c_0=-119/9$ and $c_1=10$. Figure (left) shows the resulting approximation $u=-119/9 + 10x$. We can easily test other interpolation points, say $x_0=1$ and $x_1=2$. This changes the line quite significantly, see Figure (right).

Approximation of a parabola by linear functions computed by two interpolation points: 4/3 and 5/3 (left) versus 1 and 2 (right).

Lagrange polynomials¶

In the section Fourier series we explained the advantage of having a diagonal matrix: formulas for the coefficients $\left\{ {c}_i \right\}_{i\in{\mathcal{I}_s}}$ can then be derived by hand. For an interpolation (or collocation) method a diagonal matrix implies that ${\psi}_j(x_{i}) = 0$ if $i\neq j$. One set of basis functions ${\psi}_i(x)$ with this property is the Lagrange interpolating polynomials, or just Lagrange polynomials. (Although the functions are named after Lagrange, they were first discovered by Waring in 1779, rediscovered by Euler in 1783, and published by Lagrange in 1795.) Lagrange polynomials are key building blocks in the finite element method, so familiarity with these polynomials will be required anyway.

A Lagrange polynomial can be written as

$$\begin{equation} {\psi}_i(x) = \prod_{j=0,j\neq i}^N \frac{x-x_{j}}{x_{i}-x_{j}} = \frac{x-x_0}{x_{i}-x_0}\cdots\frac{x-x_{i-1}}{x_{i}-x_{i-1}}\frac{x-x_{i+1}}{x_{i}-x_{i+1}} \cdots\frac{x-x_N}{x_{i}-x_N}, \label{fem:approx:global:Lagrange:poly} \tag{56} \end{equation}$$

for $i\in{\mathcal{I}_s}$. We see from (56) that all the ${\psi}_i$ functions are polynomials of degree $N$ which have the property

$$\begin{equation} {\psi}_i(x_s) = \delta_{is},\quad \delta_{is} = \left\lbrace\begin{array}{ll} 1, & i=s,\\ 0, & i\neq s, \end{array}\right. \label{fem:inter:prop} \tag{57} \end{equation}$$

when $x_s$ is an interpolation (collocation) point. Here we have used the Kronecker delta symbol $\delta_{is}$. This property implies that $A$ is a diagonal matrix, i.e., $A_{i,j}=0$ for $i\neq j$ and $A_{i,j}=1$ when $i=j$. The solution of the linear system is then simply

$$\begin{equation} c_i = f(x_{i}),\quad i\in{\mathcal{I}_s}, \label{_auto28} \tag{58} \end{equation}$$

and

$$\begin{equation} u(x) = \sum_{j\in{\mathcal{I}_s}} c_i {\psi}_i(x) = \sum_{j\in{\mathcal{I}_s}} f(x_{i}){\psi}_i(x){\thinspace .} \label{_auto29} \tag{59} \end{equation}$$

We remark however that (57) does not necessarily imply that the matrix obtained by the least squares or projection methods is diagonal.

The following function computes the Lagrange interpolating polynomial ${\psi}_i(x)$ on the unit interval (0,1), given the interpolation points $x_{0},\ldots,x_{N}$ in the list or array points:

The next function computes a complete basis, ${\psi}_0,\ldots,{\psi}_N$, using equidistant points throughout $\Omega$:

When x is a sym.Symbol object, we let the spacing between the interpolation points, h, be a sympy rational number, so that we get nice end results in the formulas for ${\psi}_i$. The other case, when x is a plain Python float, signifies numerical computing, and then we let h be a floating-point number. Observe that the Lagrange_polynomial function works equally well in the symbolic and numerical case - just think of x being a sym.Symbol object or a Python float. A little interactive session illustrates the difference between symbolic and numerical computing of the basis functions and points:

That is, when used symbolically, the Lagrange_polynomials_01 function returns the symbolic expression for the Lagrange functions while when x is a numerical value the function returns the value of the basis function evaluate in x. In the present example only the second basis function should be 1 in the mid-point while the others are zero according to (57).

Approximation of a polynomial¶

The Galerkin or least squares methods lead to an exact approximation if $f$ lies in the space spanned by the basis functions. It could be of interest to see how the interpolation method with Lagrange polynomials as the basis is able to approximate a polynomial, e.g., a parabola. Running

shows the result that up to N=4 we achieve an exact approximation, and then round-off errors start to grow, such that N=15 leads to a 15-degree polynomial for $u$ where the coefficients in front of $x^r$ for $r>2$ are of size $10^{-5}$ and smaller. As the matrix is ill-conditioned and we use floating-point arithmetic, we do not obtain the exact solution. Still, we get a solution that is visually identical to the exact solution. The reason is that the ill-conditioning causes the system to have many solutions very close to the true solution. While we are lucky for N=15 and obtain a solution that is visually identical to the true solution, ill-conditioning may also result in completely wrong results. As we continue with higher values, N=20 reveals that the procedure is starting to fall apart as the approximate solution is around 0.9 at $x=1.0$, where it should have been $1.0$. At N=30 the approximate solution is around $5\cdot10^8$ at $x=1$.

Successful example¶

Trying out the Lagrange polynomial basis for approximating $f(x)=\sin 2\pi x$ on $\Omega =[0,1]$ with the least squares and the interpolation techniques can be done by

Figure shows the results. There is a difference between the least squares and the interpolation technique but the difference decreases rapidly with increasing $N$.

Approximation via least squares (left) and interpolation (right) of a sine function by Lagrange interpolating polynomials of degree 3.

Less successful example¶

The next example concerns interpolating $f(x)=|1-2x|$ on $\Omega =[0,1]$ using Lagrange polynomials. Figure shows a peculiar effect: the approximation starts to oscillate more and more as $N$ grows. This numerical artifact is not surprising when looking at the individual Lagrange polynomials. Figure shows two such polynomials, $\psi_2(x)$ and $\psi_7(x)$, both of degree 11 and computed from uniformly spaced points $x_{i}=i/11$, $i=0,\ldots,11$, marked with circles. We clearly see the property of Lagrange polynomials: $\psi_2(x_{i})=0$ and $\psi_7(x_{i})=0$ for all $i$, except $\psi_2(x_{2})=1$ and $\psi_7(x_{7})=1$. The most striking feature, however, is the dominating oscillation near the boundary where $\psi_2>5$ and $\psi_7=-10$ in some points. The reason is easy to understand: since we force the functions to be zero at so many points, a polynomial of high degree is forced to oscillate between the points. This is called Runge's phenomenon and you can read a more detailed explanation on Wikipedia.

Remedy for strong oscillations¶

The oscillations can be reduced by a more clever choice of interpolation points, called the Chebyshev nodes:

$$\begin{equation} x_{i} = \frac{1}{2} (a+b) + \frac{1}{2}(b-a)\cos\left( \frac{2i+1}{2(N+1)}pi\right),\quad i=0\ldots,N, \label{_auto30} \tag{60} \end{equation}$$

on the interval $\Omega = [a,b]$. Here is a flexible version of the Lagrange_polynomials_01 function above, valid for any interval $\Omega =[a,b]$ and with the possibility to generate both uniformly distributed points and Chebyshev nodes:

All the functions computing Lagrange polynomials listed above are found in the module file Lagrange.py.

Figure shows the improvement of using Chebyshev nodes, compared with the equidistant points in Figure. The reason for this improvement is that the corresponding Lagrange polynomials have much smaller oscillations, which can be seen by comparing Figure (Chebyshev points) with Figure (equidistant points). Note the different scale on the vertical axes in the two figures and also that the Chebyshev points tend to cluster more around the element boundaries.

Another cure for undesired oscillations of higher-degree interpolating polynomials is to use lower-degree Lagrange polynomials on many small patches of the domain. This is actually the idea pursued in the finite element method. For instance, linear Lagrange polynomials on $[0,1/2]$ and $[1/2,1]$ would yield a perfect approximation to $f(x)=|1-2x|$ on $\Omega = [0,1]$ since $f$ is piecewise linear.

Interpolation of an absolute value function by Lagrange polynomials and uniformly distributed interpolation points: degree 7 (left) and 14 (right).

Illustration of the oscillatory behavior of two Lagrange polynomials based on 12 uniformly spaced points (marked by circles).

Interpolation of an absolute value function by Lagrange polynomials and Chebyshev nodes as interpolation points: degree 7 (left) and 14 (right).

Illustration of the less oscillatory behavior of two Lagrange polynomials based on 12 Chebyshev points (marked by circles). Note that the y-axis is different from [Figure](#fem:approx:global:Lagrange:fig:abs:Lag:unif:osc).

How does the least squares or projection methods work with Lagrange polynomials? We can just call the least_squares function, but sympy has problems integrating the $f(x)=|1-2x|$ function times a polynomial, so we need to fall back on numerical integration.

Illustration of an approximation of the absolute value function using the least square method .

Bernstein polynomials¶

An alternative to the Taylor and Lagrange families of polynomials are the Bernstein polynomials. These polynomials are popular in visualization and we include a presentation of them for completeness. Furthermore, as we will demonstrate, the choice of basis functions may matter in terms of accuracy and efficiency. In fact, in finite element methods, a main challenge, from a numerical analysis point of view, is to determine appropriate basis functions for a particular purpose or equation.

On the unit interval, the Bernstein polynomials are defined in terms of powers of $x$ and $1-x$ (the barycentric coordinates of the unit interval) as

$$\begin{equation} B_{i,n} = \binom{n}{i} x^i (1-x)^{n-i}, \quad i=0, \ldots, n . \label{bernstein:basis} \tag{61} \end{equation}$$

The nine functions of a Bernstein basis of order 8.

The nine functions of a Lagrange basis of order 8.

The Figure shows the basis functions of a Bernstein basis of order 8. This figure should be compared against Figure, which shows the corresponding Lagrange basis of order 8. The Lagrange basis is convenient because it is a nodal basis, that is; the basis functions are 1 in their nodal points and zero at all other nodal points as described by (57). However, looking at Figure we also notice that the basis function are oscillatory and have absolute values that are significantly larger than 1 between the nodal points. Consider for instance the basis function represented by the purple color. It is 1 at $x=0.5$ and 0 at all other nodal points and hence this basis function represents the value at the mid-point. However, this function also has strong negative contributions close to the element boundaries where it takes negative values lower than $-2$. For the Bernstein basis, all functions are positive and all functions output values in $[0,1]$. Therefore there is no oscillatory behavior. The main disadvantage of the Bernstein basis is that the basis is not a nodal basis. In fact, all functions contribute everywhere except $x=0$ and $x=1$.

Notice.

We have considered approximation with a sinusoidal basis and with Lagrange or Bernstein polynomials, all of which are frequently used for scientific computing. The Lagrange polynomials (of various order) are extensively used in finite element methods, while the Bernstein polynomials are more used in computational geometry. The Lagrange and the Bernstein families are, however, but a few in the jungle of polynomial spaces used for finite element computations and their efficiency and accuracy can vary quite substantially. Furthermore, while a method may be efficient and accurate for one type of PDE it might not even converge for another type of PDE. The development and analysis of finite element methods for different purposes is currently an intense research field and has been so for several decades. FEniCS has implemented a wide range of polynomial spaces and has a general framework for implementing new elements. While finite element methods explore different families of polynomials, the so-called spectral methods explore the use of sinusoidal functions or polynomials with high order. From an abstract point of view, the different methods can all be obtained simply by changing the basis for the trial and test functions. However, their efficiency and accuracy may vary substantially.

Approximation of functions in higher dimensions¶

All the concepts and algorithms developed for approximation of 1D functions $f(x)$ can readily be extended to 2D functions $f(x,y)$ and 3D functions $f(x,y,z)$. Basically, the extensions consist of defining basis functions ${\psi}_i(x,y)$ or ${\psi}_i(x,y,z)$ over some domain $\Omega$, and for the least squares and Galerkin methods, the integration is done over $\Omega$.

As in 1D, the least squares and projection/Galerkin methods lead to linear systems

$$\begin{align*} \sum_{j\in{\mathcal{I}_s}} A_{i,j}c_j &= b_i,\quad i\in{\mathcal{I}_s},\\ A_{i,j} &= ({\psi}_i,{\psi}_j),\\ b_i &= (f,{\psi}_i), \end{align*}$$

where the inner product of two functions $f(x,y)$ and $g(x,y)$ is defined completely analogously to the 1D case (25):

$$\begin{equation} (f,g) = \int_\Omega f(x,y)g(x,y) dx dy . \label{_auto31} \tag{64} \end{equation}$$

2D basis functions as tensor products of 1D functions¶

One straightforward way to construct a basis in 2D is to combine 1D basis functions. Say we have the 1D vector space

$$\begin{equation} V_x = \mbox{span}\{ \hat{\psi}_0(x),\ldots,\hat{\psi}_{N_x}(x)\} \label{fem:approx:2D:Vx} \tag{65} {\thinspace .} \end{equation}$$

A similar space for a function's variation in $y$ can be defined,

$$\begin{equation} V_y = \mbox{span}\{ \hat{\psi}_0(y),\ldots,\hat{\psi}_{N_y}(y)\} \label{fem:approx:2D:Vy} \tag{66} {\thinspace .} \end{equation}$$

We can then form 2D basis functions as tensor products of 1D basis functions.

Tensor products.

Given two vectors $a=(a_0,\ldots,a_M)$ and $b=(b_0,\ldots,b_N)$, their outer tensor product, also called the dyadic product, is $p=a\otimes b$, defined through

$$p_{i,j}=a_ib_j,\quad i=0,\ldots,M,\ j=0,\ldots,N{\thinspace .}$$

In the tensor terminology, $a$ and $b$ are first-order tensors (vectors with one index, also termed rank-1 tensors), and then their outer tensor product is a second-order tensor (matrix with two indices, also termed rank-2 tensor). The corresponding inner tensor product is the well-known scalar or dot product of two vectors: $p=a\cdot b = \sum_{j=0}^N a_jb_j$. Now, $p$ is a rank-0 tensor.

Tensors are typically represented by arrays in computer code. In the above example, $a$ and $b$ are represented by one-dimensional arrays of length $M$ and $N$, respectively, while $p=a\otimes b$ must be represented by a two-dimensional array of size $M\times N$.

Tensor products can be used in a variety of contexts.

Given the vector spaces $V_x$ and $V_y$ as defined in (65) and (66), the tensor product space $V=V_x\otimes V_y$ has a basis formed as the tensor product of the basis for $V_x$ and $V_y$. That is, if $\left\{ {\psi}_i(x) \right\}_{i\in{\mathcal{I}_x}}$ and $\left\{ {\psi}_i(y) \right\}_{i\in {\mathcal{I}_y}}$ are basis for $V_x$ and $V_y$, respectively, the elements in the basis for $V$ arise from the tensor product: $\left\{ {\psi}_i(x){\psi}_j(y) \right\}_{i\in {\mathcal{I}_x},j\in {\mathcal{I}_y}}$. The index sets are $I_x=\{0,\ldots,N_x\}$ and $I_y=\{0,\ldots,N_y\}$.

The notation for a basis function in 2D can employ a double index as in

$${\psi}_{p,q}(x,y) = \hat{\psi}_p(x)\hat{\psi}_q(y), \quad p\in{\mathcal{I}_x},q\in{\mathcal{I}_y}{\thinspace .}$$

The expansion for $u$ is then written as a double sum

$$u = \sum_{p\in{\mathcal{I}_x}}\sum_{q\in{\mathcal{I}_y}} c_{p,q}{\psi}_{p,q}(x,y){\thinspace .}$$

Alternatively, we may employ a single index,

$${\psi}_i(x,y) = \hat{\psi}_p(x)\hat{\psi}_q(y),$$

and use the standard form for $u$,

$$u = \sum_{j\in{\mathcal{I}_s}} c_j{\psi}_j(x,y){\thinspace .}$$

The single index can be expressed in terms of the double index through $i=p (N_y+1) + q$ or $i=q (N_x+1) + p$.

Example on polynomial basis in 2D¶

Let us again consider an approximation with the least squares method, but now with basis functions in 2D. Suppose we choose $\hat{\psi}_p(x)=x^p$, and try an approximation with $N_x=N_y=1$:

$${\psi}_{0,0}=1,\quad {\psi}_{1,0}=x, \quad {\psi}_{0,1}=y, \quad {\psi}_{1,1}=xy {\thinspace .}$$

Using a mapping to one index like $i=q (N_x+1) + p$, we get

$${\psi}_0=1,\quad {\psi}_1=x, \quad {\psi}_2=y,\quad{\psi}_3 =xy {\thinspace .}$$

With the specific choice $f(x,y) = (1+x^2)(1+2y^2)$ on $\Omega = [0,L_x]\times [0,L_y]$, we can perform actual calculations:

$$\begin{align*} A_{0,0} &= ({\psi}_0,{\psi}_0) = \int_0^{L_y}\int_{0}^{L_x} 1 dxdy = L_{x} L_{y}, \\ A_{0,1} &= ({\psi}_0,{\psi}_1) = \int_0^{L_y}\int_{0}^{L_x} x dxdy = \frac{L_{x}^{2} L_{y}}{2}, \\ A_{0,2} &= ({\psi}_0,{\psi}_2) = \int_0^{L_y}\int_{0}^{L_x} y dxdy = \frac{L_{x} L_{y}^{2}}{2}, \\ A_{0,3} &= ({\psi}_0,{\psi}_3) = \int_0^{L_y}\int_{0}^{L_x} x y dxdy = \frac{L_{x}^{2} L_{y}^{2}}{4}, \\ A_{1,0} &= ({\psi}_1,{\psi}_0) = \int_0^{L_y}\int_{0}^{L_x} x dxdy = \frac{L_{x}^{2} L_{y}}{2}, \\ A_{1,1} &= ({\psi}_1,{\psi}_1) = \int_0^{L_y}\int_{0}^{L_x} x^{2} dxdy = \frac{L_{x}^{3} L_{y}}{3}, \\ A_{1,2} &= ({\psi}_1,{\psi}_2) = \int_0^{L_y}\int_{0}^{L_x} x y dxdy = \frac{L_{x}^{2} L_{y}^{2}}{4}, \\ A_{1,3} &= ({\psi}_1,{\psi}_3) = \int_0^{L_y}\int_{0}^{L_x} x^{2} y dxdy = \frac{L_{x}^{3} L_{y}^{2}}{6}, \\ A_{2,0} &= ({\psi}_2,{\psi}_0) = \int_0^{L_y}\int_{0}^{L_x} y dxdy = \frac{L_{x} L_{y}^{2}}{2}, \\ A_{2,1} &= ({\psi}_2,{\psi}_1) = \int_0^{L_y}\int_{0}^{L_x} x y dxdy = \frac{L_{x}^{2} L_{y}^{2}}{4}, \\ A_{2,2} &= ({\psi}_2,{\psi}_2) = \int_0^{L_y}\int_{0}^{L_x} y^{2} dxdy = \frac{L_{x} L_{y}^{3}}{3}, \\ A_{2,3} &= ({\psi}_2,{\psi}_3) = \int_0^{L_y}\int_{0}^{L_x} x y^{2} dxdy = \frac{L_{x}^{2} L_{y}^{3}}{6}, \\ A_{3,0} &= ({\psi}_3,{\psi}_0) = \int_0^{L_y}\int_{0}^{L_x} x y dxdy = \frac{L_{x}^{2} L_{y}^{2}}{4}, \\ A_{3,1} &= ({\psi}_3,{\psi}_1) = \int_0^{L_y}\int_{0}^{L_x} x^{2} y dxdy = \frac{L_{x}^{3} L_{y}^{2}}{6}, \\ A_{3,2} &= ({\psi}_3,{\psi}_2) = \int_0^{L_y}\int_{0}^{L_x} x y^{2} dxdy = \frac{L_{x}^{2} L_{y}^{3}}{6}, \\ A_{3,3} &= ({\psi}_3,{\psi}_3) = \int_0^{L_y}\int_{0}^{L_x} x^{2} y^{2} dxdy = \frac{L_{x}^{3} L_{y}^{3}}{9} {\thinspace .} \end{align*}$$

The right-hand side vector has the entries

$$\begin{align*} b_{0} &= ({\psi}_0,f) = \int_0^{L_y}\int_{0}^{L_x}1\cdot (1+x^2)(1+2y^2) dxdy\\ &= \int_0^{L_y}(1+2y^2)dy \int_{0}^{L_x} (1+x^2)dx = (L_y + \frac{2}{3}L_y^3)(L_x + \frac{1}{3}L_x^3)\\ b_{1} &= ({\psi}_1,f) = \int_0^{L_y}\int_{0}^{L_x} x(1+x^2)(1+2y^2) dxdy\\ &=\int_0^{L_y}(1+2y^2)dy \int_{0}^{L_x} x(1+x^2)dx = (L_y + \frac{2}{3}L_y^3)({\frac{1}{2}}L_x^2 + \frac{1}{4}L_x^4)\\ b_{2} &= ({\psi}_2,f) = \int_0^{L_y}\int_{0}^{L_x} y(1+x^2)(1+2y^2) dxdy\\ &= \int_0^{L_y}y(1+2y^2)dy \int_{0}^{L_x} (1+x^2)dx = ({\frac{1}{2}}L_y^2 + {\frac{1}{2}}L_y^4)(L_x + \frac{1}{3}L_x^3)\\ b_{3} &= ({\psi}_3,f) = \int_0^{L_y}\int_{0}^{L_x} xy(1+x^2)(1+2y^2) dxdy\\ &= \int_0^{L_y}y(1+2y^2)dy \int_{0}^{L_x} x(1+x^2)dx = ({\frac{1}{2}}L_y^2 + {\frac{1}{2}}L_y^4)({\frac{1}{2}}L_x^2 + \frac{1}{4}L_x^4) {\thinspace .} \end{align*}$$

There is a general pattern in these calculations that we can explore. An arbitrary matrix entry has the formula

$$\begin{align*} A_{i,j} &= ({\psi}_i,{\psi}_j) = \int_0^{L_y}\int_{0}^{L_x} {\psi}_i{\psi}_j dx dy \\ &= \int_0^{L_y}\int_{0}^{L_x} {\psi}_{p,q}{\psi}_{r,s} dx dy = \int_0^{L_y}\int_{0}^{L_x} \hat{\psi}_p(x)\hat{\psi}_q(y)\hat{\psi}_r(x)\hat{\psi}_s(y) dx dy\\ &= \int_0^{L_y} \hat{\psi}_q(y)\hat{\psi}_s(y)dy \int_{0}^{L_x} \hat{\psi}_p(x) \hat{\psi}_r(x) dx\\ &= \hat A^{(x)}_{p,r}\hat A^{(y)}_{q,s}, \end{align*}$$

where

$$\hat A^{(x)}_{p,r} = \int_{0}^{L_x} \hat{\psi}_p(x) \hat{\psi}_r(x) dx, \quad \hat A^{(y)}_{q,s} = \int_0^{L_y} \hat{\psi}_q(y)\hat{\psi}_s(y)dy,$$

are matrix entries for one-dimensional approximations. Moreover, $i=p N_x+q$ and $j=s N_y+r$.

With $\hat{\psi}_p(x)=x^p$ we have

$$\hat A^{(x)}_{p,r} = \frac{1}{p+r+1}L_x^{p+r+1},\quad \hat A^{(y)}_{q,s} = \frac{1}{q+s+1}L_y^{q+s+1},$$

and

$$A_{i,j} = \hat A^{(x)}_{p,r} \hat A^{(y)}_{q,s} = \frac{1}{p+r+1}L_x^{p+r+1} \frac{1}{q+s+1}L_y^{q+s+1},$$

for $p,r\in{\mathcal{I}_x}$ and $q,s\in{\mathcal{I}_y}$.

Corresponding reasoning for the right-hand side leads to

$$\begin{align*} b_i &= ({\psi}_i,f) = \int_0^{L_y}\int_{0}^{L_x}{\psi}_i f\,dxdx\\ &= \int_0^{L_y}\int_{0}^{L_x}\hat{\psi}_p(x)\hat{\psi}_q(y) f\,dxdx\\ &= \int_0^{L_y}\hat{\psi}_q(y) (1+2y^2)dy \int_0^{L_y}\hat{\psi}_p(x) (1+x^2)dx\\ &= \int_0^{L_y} y^q (1+2y^2)dy \int_0^{L_y}x^p (1+x^2)dx\\ &= (\frac{1}{q+1} L_y^{q+1} + \frac{2}{q+3}L_y^{q+3}) (\frac{1}{p+1} L_x^{p+1} + \frac{1}{p+3}L_x^{p+3}) \end{align*}$$

Choosing $L_x=L_y=2$, we have

$$A = \left[\begin{array}{cccc} 4 & 4 & 4 & 4\\ 4 & \frac{16}{3} & 4 & \frac{16}{3}\\ 4 & 4 & \frac{16}{3} & \frac{16}{3}\\ 4 & \frac{16}{3} & \frac{16}{3} & \frac{64}{9} \end{array}\right],\quad b = \left[\begin{array}{c} \frac{308}{9}\\\frac{140}{3}\\44\\60\end{array}\right], \quad c = \left[ \begin{array}{r} -\frac{1}{9}\\ - \frac{2}{3} \frac{4}{3} \\ 8 \end{array}\right] {\thinspace .}$$

Figure illustrates the result.

Approximation of a 2D quadratic function (left) by a 2D bilinear function (right) using the Galerkin or least squares method.

The calculations can also be done using sympy. The following code computes the matrix of our example

We remark that sympy may even write the output in LaTeX or C++ format using the functions latex and ccode.

Implementation¶

The least_squares function from the section Orthogonal basis functions and/or the file approx1D.py can with very small modifications solve 2D approximation problems. First, let Omega now be a list of the intervals in $x$ and $y$ direction. For example, $\Omega = [0,L_x]\times [0,L_y]$ can be represented by Omega = [[0, L_x], [0, L_y]].

Second, the symbolic integration must be extended to 2D:

provided integrand is an expression involving the sympy symbols x and y. The 2D version of numerical integration becomes

The right-hand side integrals are modified in a similar way. (We should add that mpmath.quad is sufficiently fast even in 2D, but scipy.integrate.nquad is much faster.)

Third, we must construct a list of 2D basis functions. Here are two examples based on tensor products of 1D "Taylor-style" polynomials $x^i$ and 1D sine functions $\sin((i+1)\pi x)$:

The complete code appears in approx2D.py.

The previous hand calculation where a quadratic $f$ was approximated by a bilinear function can be computed symbolically by

We may continue with adding higher powers to the basis:

For $N_x\geq 2$ and $N_y\geq 2$ we recover the exact function $f$, as expected, since in that case $f\in V$, see the section Perfect approximation.

Extension to 3D¶

Extension to 3D is in principle straightforward once the 2D extension is understood. The only major difference is that we need the repeated outer tensor product,

$$V = V_x\otimes V_y\otimes V_z{\thinspace .}$$

In general, given vectors (first-order tensors) $a^{(q)} = (a^{(q)}_0,\ldots,a^{(q)}_{N_q})$, $q=0,\ldots,m$, the tensor product $p=a^{(0)}\otimes\cdots\otimes a^{m}$ has elements

$$p_{i_0,i_1,\ldots,i_m} = a^{(0)}_{i_1}a^{(1)}_{i_1}\cdots a^{(m)}_{i_m}{\thinspace .}$$

The basis functions in 3D are then

$${\psi}_{p,q,r}(x,y,z) = \hat{\psi}_p(x)\hat{\psi}_q(y)\hat{\psi}_r(z),$$

with $p\in{\mathcal{I}_x}$, $q\in{\mathcal{I}_y}$, $r\in{\mathcal{I}_z}$. The expansion of $u$ becomes

$$u(x,y,z) = \sum_{p\in{\mathcal{I}_x}}\sum_{q\in{\mathcal{I}_y}}\sum_{r\in{\mathcal{I}_z}} c_{p,q,r} {\psi}_{p,q,r}(x,y,z){\thinspace .}$$

A single index can be introduced also here, e.g., $i=rN_xN_y + qN_x + p$, $u=\sum_i c_i{\psi}_i(x,y,z)$.

Use of tensor product spaces.

Constructing a multi-dimensional space and basis from tensor products of 1D spaces is a standard technique when working with global basis functions. In the world of finite elements, constructing basis functions by tensor products is much used on quadrilateral and hexahedra cell shapes, but not on triangles and tetrahedra. Also, the global finite element basis functions are almost exclusively denoted by a single index and not by the natural tuple of indices that arises from tensor products.