Consider the vector Poisson equation

\begin{align} -\Delta u &= f \text{ in } \Omega \subset \mathbb{R}^d \tag{1}\\ u\cdot v &= 0 \text{ on } \partial \Omega \tag{2} \end{align}

where $v = n$ or $v = \tau$ where $n$ is normal to $\partial \Omega$ and $\tau$ is tangent to the boundary and $d = 2,3$.

Before imposing any boundary conditions, the weak form of this equation is: We seek $u\in (H^1(\Omega))^d$ such that for all $v \in (H^1_0(\Omega))^d$

$$(\nabla u, \nabla v) - \int_{\partial \Omega} n^T \cdot \nabla u \cdot v \,ds= (f,v)$$

Now, normally, if we have (2) for both $n$ and $\tau$, then we can impose the homogeneous condition by noticing the boundary term is exactly zero. However, if we only have one of them, then I'm at a loss. Here is what I've tried: We rewrite $u = u_n + u_\tau = (u\cdot n)n + (u\cdot \tau) \tau$ and then whichever component is zero (say we want $u_n = 0$) we substitute that into our boundary integral to get

$$ (\nabla u, \nabla v) - \int_{\partial \Omega} n^T \cdot \nabla u_\tau \cdot v \,ds= (f,v) $$

However, it's not clear to me that $\nabla u_\tau$ makes any sense. For example, if the boundary in question happens to be the interval $x\in[0,1]$ $\nabla u_\tau$ would have to make sense of $$\frac{\partial u_\tau}{\partial y}$$

which does not seem to have any intuitive interpretation when the tangent vector has no dependence on $y$. Does this in fact, have a good interpretation? If not how can this boundary condition be applied?

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    $\begingroup$ I'm not sure $\tau$ will give you the right boundary condition here. Note that in 3D you have two tangential directions. You should probably instead look at $(I-n^T n)u$ which is "more dimension invariant". $\endgroup$
    – Jap88
    Jul 6, 2022 at 0:53
  • $\begingroup$ Yes, that's right. Thank you. Although, an answer which is able to deal even just with the 2D case would be enough help for now probably. $\endgroup$ Jul 6, 2022 at 12:28

1 Answer 1


The issue is that the equation $$ \mathbf{u} \cdot \mathbf{n}=0$$ or $$ \mathbf{u} \cdot \mathbf{t}=0$$ is a constraint equation rather than a normal flux, which is what enters naturally in the weak form of the equation.

The weak form derivation goes something like this. Multiply by $\mathbf{v}$: $$-\int_\Omega \nabla \cdot \nabla \mathbf{u} \cdot \mathbf{v} dV=\int_\Omega \mathbf{f} \cdot \mathbf{v} dV$$ Now, perform partial integration: $$-\int_{\partial\Omega} (\mathbf{n} \cdot \nabla \mathbf{u}) \cdot \mathbf{v}dS+\int_\Omega \nabla \mathbf{u} \cdot \nabla \mathbf{v} dV=\int_\Omega \mathbf{f} \cdot \mathbf{v} dV$$ The interesting term here is the boundary integral: $$-\int_{\partial\Omega} (\mathbf{n} \cdot \nabla \mathbf{u}) \cdot \mathbf{v}dS$$ which shows that we control the flux/force $\mathbf{g}$ on the boundary on the following form: $$-\int_{\partial\Omega} \mathbf{g} \cdot \mathbf{v}dS$$ On strong form this becomes: $$\mathbf{n} \cdot \nabla \mathbf{u}=\mathbf{g}$$ So the weak form shows that, on the boundary, we do not get direct control over $\mathbf{u}$ and its normal and tangential projections. Instead we only get "access" to the normal projection of the gradient: $$\mathbf{a}_n=\mathbf{n} \cdot \nabla \mathbf{u}$$ (Note that this vector can be "leaning" and doesn't have to be in the normal direction.) One formal way to get control over constraints is to add a Lagrange multiplier to the boundary flux. Something like: $$\mathbf{n} \cdot \nabla \mathbf{u}=\mathbf{P} \vec{\lambda}$$ where $\mathbf{P}$ is a matrix, and find a vector $\vec{\lambda}$ so that the constraint is fulfilled. This vector represents the flux that you need in order to fulfill the constraint (to every constraint there is a corresponding flux). Now we are getting into theory of Lagrange multipliers.

Let's now partition the boundary up in two (possibly more as needed) parts. One $\partial \Omega_1$ with a Neumann condition and one $\partial \Omega_2$ with a constraint (general Dirichlet) condition.

We can now use the following energy functional (Dirichlet's principle):

$$E=\frac{1}{2}\int_\Omega \nabla \mathbf{u} \cdot \nabla \mathbf{u} dV-\int_{\Omega} \mathbf{f} \cdot \mathbf{u} dV-\int_{\partial \Omega_1} \mathbf{g} \cdot \mathbf{u} dS-\int_{\partial \Omega_2} (P\mathbf{u}-b) \cdot \mathbf{\lambda} dS$$

where $\lambda=(\lambda_1,\lambda_2)$ is a Lagrange multiplier vector.

To simplify matters, let's only consider part of the functional: $$E=\frac{1}{2}\int_\Omega \nabla \mathbf{u} \cdot \nabla \mathbf{u} dV-\int_{\partial \Omega_2} (P\mathbf{u}-b) \cdot \mathbf{\lambda} dS$$

Now, consider the stationarity of the first variation of this functional: $$\delta E=\int_\Omega \nabla \mathbf{u} \cdot \nabla \delta\mathbf{u} dV-\int_{\partial \Omega_2} P\delta \mathbf{u} \cdot \mathbf{\lambda} dS-\int_{\partial \Omega_2} (P\mathbf{u}-b) \cdot \delta \mathbf{\lambda} dS=0$$

If we set $\mathbf{v}=\delta \mathbf{u}=(\delta u_1,\delta u_2)=(v_1,v_2)$ and $\mu=\delta \lambda=(\mu_1,\mu_2)$ this becomes: $$\int_\Omega \nabla \mathbf{u} \cdot \nabla \mathbf{v} dV-\int_{\partial \Omega_2} P\mathbf{v} \cdot \mathbf{\lambda} dS-\int_{\partial \Omega_2} (P\mathbf{u}-b) \cdot \mathbf{\mu} dS=0$$

If we now have constraints in both tangential and normal direction, assuming 2D and $\mathbf{b}=0$, then we have the constraint relationship:

$$P\mathbf{u}=\begin{bmatrix} n_x & n_{y}\\ t_{x} & t_{y}\\ \end{bmatrix} \begin{bmatrix} u_1\\ u_2\\ \end{bmatrix}= \begin{bmatrix} 0\\ 0\\ \end{bmatrix}$$

Let's say we now only have a normal constraint, then we get: $$P\mathbf{u}=\begin{bmatrix} n_x & n_{y}\\ 0 & 0\\ \end{bmatrix} \begin{bmatrix} u_1\\ u_2\\ \end{bmatrix}= \begin{bmatrix} n_x u_1+n_y u_2\\ 0\\ \end{bmatrix}=\begin{bmatrix} 0\\ 0\\ \end{bmatrix} $$ Here, the constraint matrix is:

$$A=\begin{bmatrix} n_x & n_{y}\\ 0 & 0\\ \end{bmatrix}$$

The two Lagrange multiplier terms are now:

$$\int_{\partial \Omega_2} P\mathbf{v} \cdot \mathbf{\lambda} dS=\int_{\partial \Omega_2} (n_x v_1+n_y v_2) \lambda_1 dS$$


$$\int_{\partial \Omega_2} P\mathbf{u} \cdot \mathbf{\mu} dS=\int_{\partial \Omega_2} (n_x u_1+n_y u_2) \mu_1 dS$$

Note that whether you put the constraint in the first or second row of the constraint matrix is somewhat arbitrary.

For the equation that is not acted on by the Lagrange multiplier you can apply a Neumann condition.

I didn't find any quite satisfactory references. These are some that might be helpful: https://users.oden.utexas.edu/~oden/Dr._Oden_Reprints/1982-008.finite_element.pdf

Also the appendix of this book: https://www.amazon.com/Incompressible-Finite-Element-Isothermal-Laminar/dp/0471492507

These references are far from perfect. I will put in more here if I find any.

  • $\begingroup$ Thank you for the time and effort to answer my question. It's good to know that I'm not missing something obvious and that there is some simple solution. Still, it's not great that to impose this boundary condition one has to increase the size of the system. Although, I suppose $\lambda$ should only be defined on the boundary, so perhaps this is not too bad. I'm somewhat aware of this approach, or something very similar in W. Layton, Weak imposition of “no-slip” conditions in finite element methods, Computers & Mathematics with Applications, Volume 38, Issues 5–6, 1999, Pages 129-142. $\endgroup$ Jul 7, 2022 at 3:48
  • $\begingroup$ @Chessnerd321 You don't have to use Lagrange multipliers. Instead you can also use the relationship $n \cdot u$ to eliminate equations from the original system, since it is a (linear) relationship for $u$. I think this is the more common method and should be plenty of written material on it. (Please do accept my answer above if you find it useful:-) Txs $\endgroup$
    – Jap88
    Jul 7, 2022 at 11:37
  • $\begingroup$ Thanks for the additional into. To be honest, I feel that your answer is only a partial answer to my question. It essentially outlines the same derivation I gave in the question and then eludes to a method. For me to accept it, I would need a little more (if still cursory) explanation or one or two precise references about how one comes up with this $\mathbf{P}$ or about how one "use[s] the relationship $n\cdot u$ to eliminate equations from the original system". I feel that that, in particular, is actually the heart of my question. $\endgroup$ Jul 7, 2022 at 16:04
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    $\begingroup$ @Chessnerd321That is understandable. You are right in that it is quite difficult to find good references. I can definitely write more on this interesting topic. I will get back to writing in a day or two when I can allocate some time, and I will include some references. $\endgroup$
    – Jap88
    Jul 8, 2022 at 12:51

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