Why does the choice of basis functions change the approximate solution in FEM even when the same space is spanned?

Question

Let's say we have the the weak form laplace problem Find $u$ $\in$ $H^1$ what satisfies \begin{equation} \int_\Omega \nabla u \cdot \nabla v \, d\Omega = \int_\Omega f v \, d\Omega \end{equation} for all $v$ $\in$ $H^1_0$

To find an approximate solution we project on a subspace $V \subset H_1$. Let's say $V$ is the space of monomials $ax+b$

There are many basis that can span the monomial space ax+b. For example the $[1,x]$ basis, the Lagrange basis $[(x-x_1)/(x_2-x_1),(x_2-x)/(x_2-x_1)]$, or the Legendre basis etc... It is the same space but with different basis. Why does the basis of choice make a difference in the approximate solution? The math statement above doesn't mention anything about the choice of basis.

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Related question: If different basis were chosen for spaces $u$ and $v$, are we doing Petrov-Galerkin? Also, consider the weak form for an advection probelem is: \begin{equation} \int_{\Omega} (\beta \cdot \nabla u) \cdot v \, d\Omega = \int_\Omega f v \, d\Omega \end{equation} And the SUPG stabiliazation is \begin{equation} \int_{\Omega} (\beta \cdot \nabla u_h) \cdot (v_h + \alpha \beta \nabla v_h) \, d\Omega = \int_\Omega f (v_h + \alpha \beta \nabla v_h) \, d\Omega \end{equation} where $\alpha$ is a stabilizing parmeter.

if $v_h$ spans the monomila space, then $v_h + \alpha \beta \nabla v_h$ also spans the same monomial space. It's the same space for both $u_h$ and the test space $v_h + \alpha \beta \nabla v_h$. Why is this refered to as a Petrov-Galerken technique when both spaces are the same? Aren't they?

Answer 1

Expanding the discussion in the comments,

1. The change of basis does not matter under the same space $V_h$, you end up with the same solutions whichever basis you use. This can be seen easily for simple problems but for real FEM problems, where the geometry gets complex and the equations involve nonlinearity, numerical integration may not be accurate as quadrature results may be different for different bases.

2. The space $V_h$ and $V_h + \alpha\beta\nabla V_h$ are not the same space. In the finite element context where $V_h$ mostly refers to piecewise linear continuous functions, $V_h + \nabla V_h$ can contain discontinuous piecewise linear functions. To see this we can consider a simple example, $f(x) = |x|$, then

\begin{equation} f(x) + \frac{df}{dx}= \begin{cases} -x-1 & \text{if } x < 0\\ x + 1 & \text{if } x > 0\\ undefined & \text{if } x = 0 \end{cases} \end{equation}

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EDIT: For the basis argument, yes, linear polynomials $V$ has two basis and we can use these to represent any linear polynomial in $[-\infty, \infty]$, however piecewise linear polynomials $V_h$ also has two basis but in a single 1d cell (or interval) and we can use those two bases to represent any linear polynomial in that interval. This allows us to approximately represent a complex nonlinear function with just linear polynomials by discretizing the interval.

Hence in the global sense, just two linear basis is not enough to approximate a nonlinear function. We could see it as if each piecewise polynomial possessing two basis functions.

Then for the above example if we consider the function $y = |x|$ in the interval $[-1, 1]$, by splitting the interval to $[-1, 0]$ and $[0, 1]$, we could see that there can be four bases possible. But by definition of the space $V_h$ we fix one value in the middle to be the same for both the intervals for continuity, but the space $V_h + \alpha\nabla V_h$ allows some more freedom in this sense than the space $V_h$.

Answer 2

The given statement makes sense when considering only the subspace V in overall sense.

Find $u \in V $, such that $a(u,v) = F(v)$ for all $ v \in V$

However in Finite Element context, we will take this formulation into a Finite dimensional subspace $V_h$ such that

Find $u_h \in V_h $, such that $a(u_h,v_h) = F(v_h)$ for all $ v_h \in V_h$

Here $v_h$ can be space of any functions, which can be taken as lagrange or legendre polynomials.

However, as per the Cea's lemma and interpolation theory, the approximated FEM solution $u_h$ is only as good as the choice of $v_h$ for a given problem

$$ ||u - u_h ||_V \leq inf_{v_h \in V_h} || u - v_h ||_V \leq C || u - \pi_hu ||_V$$

So, a different choice of $v_h$ should provide a different approximation of the FEM solution $u_h$ depending upon the regularity and nature of actual solution $u$ .

This makes more sense intuitively because, if we have a solution which is quadratic in nature, a basis function with second order will provide less error ( $ || u - v_h ||_V $ ) when compared to using a basis function with first order.