Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm trying to test a PDE solver and I'm wondering if there is any 2D vector field that satisfies the following on the domain $\Omega = [0,1] \times [0,1]$:

$$\text{curl} \;\mathbf{u} = 0 \;\;\;\forall \mathbf{x} \in \Omega$$ $$\mathbf{u}\cdot\mathbf{t} = 0 \;\;\;\forall \mathbf{x} \in \partial\Omega$$

where $\mathbf{t}$ is the tangential vector to $\partial\Omega$.

i.e. I'm looking for a vector field that is conservative and also, on the boundary of the domain (the unit square) is perpendicular to the boundary.

Is this possible? I can come up with several examples for each separate condition, but none for both.

share|cite|improve this question
How continuous should your field be? What about the corners? – Christian Blatter Jul 17 '13 at 8:48
@christianblatter I'm honestly not entirely sure but I'll give anything a try at this point. – asdfghjkl Jul 17 '13 at 8:57
up vote 1 down vote accepted

Consider the function $$f(x,y):=\sin(\pi x)\>\sin(\pi y)\qquad\bigl((x,y\in\Omega\bigr)$$ and put $${\bf u}(x,y):=\nabla f(x,y)=\bigl(\pi\cos(\pi x)\>\sin(\pi y),\ \pi\sin(\pi x)\>\cos(\pi y)\bigr)\ .$$ Then ${\rm curl}({\bf u})\equiv0$ and $${\bf u}(0,y)=\bigl(\pi\sin(\pi y),0\bigr)\ne{\bf 0}\qquad(0< y<1)\ ,$$ and similarly for the other three edges of $\Omega$.

share|cite|improve this answer
Great, thanks!! – asdfghjkl Jul 17 '13 at 10:00

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.