# 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 $v|_T=0$.

## 1 Answer

Let's combine your analysis with some content of the following reference:

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.
Can we make sense of it? Yes we can, but only if our triangle is extended in the way depicted below:

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.

Now we jump to another reference to show that the scheme may be correct indeed, but not so good:

With Labrujère's Problem, the continuity equation is completed with an equation for irrotational flow; the whole is discretized at conforming (linear) triangles and the equations are solved in a least squares sense. The result is that the numerical solution does converge to the analytical solution but that convergence is terribly slow. This is what I meant by "essentially correct, but not so good".
One can also suggest that the least squares residual is too far from being zero for a non-zero velocity field, which is perhaps what the authors of the book are trying to say.