1
$\begingroup$

I'm trying to do FEM for a very basic version of the linear full Stokes equations in two dimensions. Say we are working in the grid $[0,1]\times[0,1]$ in the $xy-$plane.

To solve an FEM problem for a scalar function $u(x,y)$ in the past, I have used the 2D FEM basis function definition $\psi_{ab}(x,y) = \phi_{a}(x)\phi_{b}(y)$ for $1\leq a,b \leq N-1$ with

$$ \begin{equation*} {\phi _a (x) =\left\{ \begin{array}{ll} \frac{x-x_{a-1}}{h} & x\in [(a-1)h,ah] \\ \frac{x_{a+1}-x}{h} & x \in [ah,(a+1)h] \\ 0 & \mbox{else} \end{array} \right.} \mbox{ }\mbox{ } \mbox{ }\mbox{ }\mbox{ }\mbox{ for } a = 1,...,N-1 \end{equation*} $$

  • $N = $ number of intervals we cut the full interval $[0,1]$ into (can change for $x$ and $y$ intervals)
  • $h = 1/N$
  • $x_{j} = jh$

I now want to create a basis to represent a vector function $v(x,y)\in \mathbb{R}^2$ for the 2D linear full stokes equations. My current thinking is that if I have some vector function

$$ \begin{equation*} v (x,y) =\left( \begin{array}{ll} v^1 (x,y) \\ v^2(x,y) \end{array} \right) \end{equation*} $$

I think I can represent this function as $ v = \sum _ {a,b = 1} ^ {N-1} V_{a,b} \hat{\psi}_{ab}(x,y)$ where $V_{a,b}$ are coefficients and

$$ \begin{equation*} \hat{\psi}_{ab}(x,y) =\left( \begin{array}{ll} \phi_{a}^1(x)\phi_{b}^1(y) \\ \phi_{a}^2(x)\phi_{b}^2(y) \end{array} \right) \end{equation*} $$

My thinking is if we know that $\psi(x,y)$ is a basis (since I was given the first definition I wrote), then we know that $c_{11} \psi_{11} + c_{12} \psi_{12}+...+c_{N-1,N-1} \psi_{N-1,N-1} = 0 $

if and only if

$$ c_{11} = c_{12} = c_{13}=...=c_{N-1,N-1}=0$$

So I would think that if you took a linear combination of the new basis I wrote for the vector-valued basis functions we would get $$c_{11} \hat{\psi}_{11} + c_{12} \hat{\psi}_{12}+...+c_{N-1,N-1} \hat{\psi}_{N-1,N-1} = \vec{0} $$ if and only if

$$ c_{11} = c_{12} = c_{13}=...=c_{N-1,N-1}= \left(\begin{array}{ll} 0 \\ 0 \end{array}\right) $$

Which I think shows linear independence.

  1. Is what I outlined the correct FEM basis to represent a function in $\mathbb{R}^2$?
  2. How would I even interpret the coefficients $V_{ab}$ here? Previously, in the case when I used the $\psi(x,y)$ basis functions $V_{ab}$ was a matrix. Now we get a coefficient that is a 2D vector for each combination of a and b (instead of a single number). How can I visualize this? Do I just input a vector into a matrix each time to track the coefficients?

Sorry if these are silly questions I'm struggling to fully visualize this and can't seem to find anything very helpful online.

$\endgroup$

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy