5
$\begingroup$

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.


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?

$\endgroup$
12
  • 2
    $\begingroup$ The short answer is: Your example is too simple. In practice your PDE generally contains non-linear expressions, which can't be a priori represented by the basis, e.g. the RHS is non-rational, the simulation domain is curved or the boundary condition may contain non-polynomial expressions. Moreover, you have to distinguish between the weak formulation using the exact integral expressions and the later version with discrete and finite quadrature rules. I hope someone will expand this a little bit. $\endgroup$
    – ConvexHull
    Jun 18, 2023 at 16:14
  • $\begingroup$ Exactly this. In your example, you aren't approximating anything. Approximation comes in when integrals can't be computed exactly (quadrature results can be different for different bases) and nonlinearities must be applied pointwise to combinations of basis functions. This is where the difference comes from $\endgroup$
    – whpowell96
    Jun 19, 2023 at 2:38
  • 2
    $\begingroup$ The choice of basis does not actually make a difference. You say that it does, but can you actually show that it does? $\endgroup$ Jun 20, 2023 at 2:49
  • $\begingroup$ @WolfgangBangerth My question was mainly motivated by the SUPG example. If $v_h$ belongs to the space of polynomials of degree n then $V_h+\alpha \beta \nabla v_h$ also belongs to the same polynomial space of degree n. Where's the Petrov in Petrov-Galerkin? $\endgroup$ Jun 20, 2023 at 7:57
  • 2
    $\begingroup$ @CuteCompute No, that's not true. If you have piecewise linear continuous elements, for example, then $\nabla v_h$ is discontinuous and the whole things is also only piecewise polynomial if $\alpha\beta$ is polynomial as well. $\endgroup$ Jun 20, 2023 at 17:06

2 Answers 2

4
$\begingroup$

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}


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 $V_h$.

$\endgroup$
8
  • $\begingroup$ Thanks. OKay let's say I want to interpolate the function $f(x)=2x$ between $[-1,1]$. I can use the Lagrange basis $V_h$ $\{ \dfrac{1-\xi}{2} , \dfrac{1+\xi}{2}\}$ and write $f(x) = 2 \dfrac{1-\xi}{2} - 2 \dfrac{1+\xi}{2}$. What if I wanted to interpolate function $g(x)=|x|$?. Can I use the space $V_h + \alpha \nabla V_h$ to interpolate $g(x)$? How so? $\endgroup$ Jun 21, 2023 at 15:25
  • $\begingroup$ As $V_h$ is the space of piecewise polynomial functions, you can do that by splitting the line $[-1, 1]$ into two $[-1, 0]$ and $[0, 1]$ and fitting two polynomials (one for $f(x) = -x$ and another for $f(x) = x$), however the whole function will be continuous at $x = 0$ $\endgroup$ Jun 21, 2023 at 16:29
  • 1
    $\begingroup$ I understand what you're all saying but, I want to see the $V_h+\nabla V_h$ space do the thing that the $V_h$ space can't. $\endgroup$ Jun 22, 2023 at 17:05
  • $\begingroup$ Well, it can contain the above discontinuous function, how would be able to do that with $V_h$? $\endgroup$ Jun 22, 2023 at 18:17
  • $\begingroup$ How can I do that with $V_h+\alpha \nabla V_h$ if the basis of $V_h$ are $\{ \dfrac{1-\xi}{2}, \dfrac{1+\xi}{2}\}$? $\endgroup$ Jun 23, 2023 at 9:59
0
$\begingroup$

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.

$\endgroup$
3
  • 4
    $\begingroup$ I am not quite sure your answer matches the question. $\endgroup$
    – ConvexHull
    Jun 19, 2023 at 9:03
  • $\begingroup$ There seems to be a lot of confusion about the question itself. I don't think I explained well and I have been trying to do a better job at it but I don't think I can. My question is partly this: in FEM we project on a space to get an approximate solution right? okay let's chose the space of monomials ax+b. I can span the space in different ways. I can span it with [1,x] basis or with the Lagrange basis? Why does it matter what basis I am choosing? Seems like I can span the space which ever way I want, why does it matter how I span the space? $\endgroup$ Jun 19, 2023 at 9:54
  • $\begingroup$ No i don't think there is a confusion. You explained it very well. The point is, you get the same result, because your solution is already a polynomial. Use a simple trigonometric function and you will see differences. $\endgroup$
    – ConvexHull
    Jun 19, 2023 at 11:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.