Meaning of "$\triangledown$u*ñ=0 on the boundary"

322 Views Asked by At

I'm doing homework for my PDE class, I'm coming across this notation and I don't what the ñ means: $\triangledown$u* ñ=0. I have tried to google it, but unfortunately questions like this don't really lend themselves for googling.

The full description of the problem in the book is:

On a rectangle ( $0 < x <L, 0 < y <H$ ) consider $\triangledown^2$u= Q(x,y) with $\triangledown$u* ñ=0 on the boundary.

My first thought was that it just meant that $\triangledown$u should equal 0 on the boundary, but then the book could have just stated that.

Just to be clear, I know how to deal with these equations, I just need to be sure I understand the boundary coundition correctly.

1

There are 1 best solutions below

1
On BEST ANSWER

It's a Neumann boundary condition. Here $n$ stands for the outward-pointing unit normal vector on the boundary of your domain and $\nabla u \cdot n = \sum_{i=1}^2 \partial_{x_i} u n_i$ is the dot product of the gradient, $\nabla u$, and the vector $n$.