# FEM Basis for functions of two variables in $\mathbb{R}^2$ (Applied to Linear Full Stokes Equations)

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.