# Function in piecewise linear finite element space which satisfies the divergence-free condition is the zero function

In book: The Mathematical Theory of Finite Element Methods, Susanne C.Brenner. Page 285. It says that

"It is reasonable to ask why one would want to use nonconforming finite elements. There are situations where nonconforming methods are clearly desirable. One example is the incompressible fluid flow problem. If one uses the vector conforming, piecewise linear Lagrange finite element, the only function in $V_h$ which satisfies the divergence-free condition is the zero function on generic meshes."

I don't know why "the only function in $V_h$ which satisfies the divergence-free condition is the zero function on generic meshes." Could anyone help me on this? Thank you very much.

By definition, I know that in 2-D case, let $v\in V_h$, then for any triangle $T$, there exists $a_1,b_1,c_1,a_2,b_2,c_2\in\mathbb{R}$ such that $$v|_T=(a_1x+b_1y+c_1,a_2x+b_2y+c_2)$$ then $$0=\mathrm{div}(v|_T)=a_1+b_2$$ I don't how to obtain that $v|_T=0$.

For an arbitrary function $\,T\,$ at the linear triangle we have: $$T - T_1 = A.(x - x_1) + B.(y - y_1)$$ with: $$A = [ (y_3 - y_1).(T_2 - T_1) - (y_2 - y_1).(T_3 - T_1) ] / \Delta \\ B = [ (x_2 - x_1).(T_3 - T_1) - (x_3 - x_1).(T_2 - T_1) ] / \Delta$$ and: $$\Delta = (x_2 - x_1).(y_3 - y_1) - (x_3 - x_1).(y_2 - y_1)$$ Translated to your case for velocity components $\,(u,v)$ : $$\begin{cases} u - u_1 = a_1.(x - x_1) + b_1.(y - y_1) \\ v - v_1 = a_2.(x - x_1) + b_2.(y - y_1) \end{cases}$$ with: $$\begin{cases} a_1 = [ (y_3 - y_1).(u_2 - u_1) - (y_2 - y_1).(u_3 - u_1) ] / \Delta \\ b_1 = [ (x_2 - x_1).(u_3 - u_1) - (x_3 - x_1).(u_2 - u_1) ] / \Delta \end{cases} \\ \begin{cases} a_2 = [ (y_3 - y_1).(v_2 - v_1) - (y_2 - y_1).(v_3 - v_1) ] / \Delta \\ b_2 = [ (x_2 - x_1).(v_3 - v_1) - (x_3 - x_1).(v_2 - v_1) ] / \Delta \end{cases}$$ Thus, according to your correct analysis: $$\mathrm{div}(v|_T) \times \Delta=(a_1+b_2)\times \Delta = 0 \quad \Longrightarrow$$ $$[ (y_3 - y_1).(u_2 - u_1) - (y_2 - y_1).(u_3 - u_1) ] +\\ [ (x_2 - x_1).(v_3 - v_1) - (x_3 - x_1).(v_2 - v_1) ] = 0$$ $$[ (x_3 - x_2).v_1 - (y_3 - y_2).u_1] +\\ [(x_1 - x_3).v_2 - (y_1 - y_3).u_2] +\\ [(x_2 - x_1).v_3 - (y_2 - y_1).u_3] = 0$$ $$(\vec{r}_3-\vec{r}_2) \times \vec{w}_1 + (\vec{r}_1-\vec{r}_3) \times \vec{w}_2 + (\vec{r}_2-\vec{r}_1) \times \vec{w}_3 = 0$$ Where $\vec{r}=(x,y)\,$ and $\vec{w}=(u,v)$ . Cross products of edges with velocities are recognized.
In the original (fat) triangle the velocities (${\bf\color{red}{\mbox{red}}}$) are opposite to the edges of their cross product. But there exists a (thin) triangle which is similar to the original. In the latter triangle the velocities are in the middle of the edges, which makes the cross products equal to fluxes. And because of the similarity, the discretized equations (with scaling factor $2$) remain exactly the same. For this reason, there is no doubt that the scheme as proposed in the question by the OP is essentially correct.