For a scalar variable $u$, the weak form of the following Laplacian:
$$\nabla^2 u =0 $$ with the assumption that $v$ vanishes at the boundary is $$\int \nabla v . \nabla u \, d\Omega = 0 $$ which is something you are most probably familiar with, but if $u$ is a vector, I can't seem to get a weak form which is consistent I'll explain.

Consider the following vector Laplacian equation: $$ \nabla^2 \boldsymbol{u}=\boldsymbol{0}$$ now this can be broken down into two equations in 2D: \begin{aligned} \nabla^2 {u_1}=0 \\ \nabla^2 {u_2}=0 \end{aligned} we can get the weak from of each equation as above as: \begin{aligned} \int \nabla v_1 . \nabla u_1 \, d\Omega = 0 \\ \int \nabla v_2 . \nabla u_2 \, d\Omega = 0 \end{aligned} which is straight forward as well.

Now my issue is this. If I want to keep the Laplacian in vector form and find it's weak form I would write: $$ \int \nabla {v} . \nabla \boldsymbol{u} \, d\Omega = \boldsymbol{0} $$ ($v$ being a scalar here) and expand the system to give: \begin{aligned} \int \nabla v . \nabla u_1 \, d\Omega = 0 \\ \int \nabla v . \nabla u_2 \, d\Omega = 0 \end{aligned} which is inconsistent with what I wrote earlier. In the above weak expanded system I could multiply each equation of my system with a different test function, $v_1$ and $v_2$ , but I can't do this now which feels weird. I also tried to multiply with a vector test function which gives something like this: $$ \int \nabla \boldsymbol{v} . \nabla \boldsymbol{u} \, d\Omega = 0 $$ but it also seems off since grad v and grad u are matrices and would lead to a matrix when multiplied but the right hand side is a zero scalar. It is weird that this fails.

So can someone tell me the weak from of a vector valued Laplacian?


1 Answer 1


Your reasoning for vector type equation is correct when you wrote it down for each individual component. What occurs is that you need to write your interpolation basis function and your test function as vectors.

Hence, starting from: $$\nabla^2 \mathbf{u} =0 $$ in strong integral form is: $$\int \mathbf{v} \cdot \nabla^2 \mathbf{u} \, d\Omega = 0 $$

where $\mathbf{v}$ and $\mathbf{u}$ are vectors. This could alternatively be written as: $$\int \mathbf{v} \cdot (\nabla \cdot (\nabla \mathbf{u})) \, d\Omega = 0 $$

Here it is important to understand that $\nabla \mathbf{u}$ is a second order tensor. You could write it in Einstein's notation as: $$ \nabla \mathbf{u} = \partial_i u_j $$ Note that in tensor form, this is a bit different than what we generally write as the gradient or the jacobian of a vector-valued function. It is actually its transpose. See for example the book by Bird, Stuart & Lightfoot for a clear definition of this.

With the assumption that $\mathbf{v}$ vanishes at the boundary and by integrating by part you will get: $$\int \nabla \mathbf{v} : \nabla \mathbf{u} \, d\Omega = 0 $$

where $:$ is the double dot product (or double contraction), for which you can find an extended discussion here or again in the book by Bird, Stuart & Lightfoot.

To be brief, the double dot product for two second-order tensors $\mathbf{a}$ and $\mathbf{b}$ is given by: $$ c = a_{ij} b_{ji} $$ where it is implicitly assumed that repeated indices are summed over. The result $c$ is a scalar. If you are not familiar with Einstein's notation, the BSL book reference above has a nice section dedicated to it. There is even a nice wiki page The integral above can be written in Einstein's notation as: $$\int (\partial_i v_j) (\partial_j u_i) \, d\Omega = 0 $$

You could also write it in term of matrix transpose, but I find it significantly easier to explain this concept in Einstein's notation. As soon as you start to manipulate second-order or higher order tensors, it becomes much easier to use this approach.

  • 2
    $\begingroup$ A short version of this answer is to stay in vector form and apply the Divergence Theorem which is the vector form of integration by parts to move a derivative off the trial functions and onto the test functions (which is your real goal). The notation is all here, and I've given you some different terms. $\endgroup$
    – Bill Barth
    Commented May 25, 2021 at 14:13

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