# Boundary Conditions for an inviscid fluid at a fixed boundary

1. Mar 24, 2014

### Type1civ

This is my first post so I hope this in the right place. I am fairly sure this is quite a straight forward question but I having trouble working out the details of it.

"State the boundary conditions for an inviscid fluid at an impermeable fixed boundary

$$x_3-h(x_1,x_3)=0$$

where we do not permit normal velocity but permit non-zero tangential velocity"

So my attempt is quite limited as I feel if I could get started I'd be able to make some progress. But I believe that since the wall is stationary then the velocity of the fluid at the wall must also be stationary. And since it says that normal velocity is zero i guess this means that in the $x_3$ direction the velocity is always zero.
so:
$$\mathbf{u}\cdot \mathbf{\hat x}_3 =0$$
Then the velocity of the fluid only has components in the other two directions...
I was then also thinking that the $\hat x_1$ component of velocity wouldn't change with $x_1$ but I am not really sure why I think that. I am just getting a bit confused as you might be able to tell, any help would be really appreciated!

Edit: I am also not sure why my tex commands aren't working so if somebody could point out what I am doing wrong that would also be great...

Last edited: Mar 24, 2014
2. Mar 24, 2014

### pasmith

That boundary seems an unnecessary complicated way to write $g(x_1,x_3) = 0$. Does the question actually say
$$x_3 - h(x_1, x_2) = 0$$
which would make sense as the upper boundary of a fluid layer of variable depth?

For an inviscid fluid only the normal component must be stationary relative to the wall.

If the normal to the boundary were in the $x_3$ direction this would be correct, but that's not necessarily the case.

Following the question as written, the first thing is to define
$$g(x_1,x_3) = x_3 - h(x_1,x_3)$$
so that the boundary is at $g(x_1,x_3) = 0$. The boundary is then a contour of $g$, so the direction normal to the boundary is given by
$$\nabla g = \frac{\partial g}{\partial x_1} \hat{\mathbf{x}}_1 + \frac{\partial g}{\partial x_3} \hat{\mathbf{x}}_3.$$

The closing tag is [/tex] not [\tex].

3. Mar 24, 2014

### Type1civ

Yeah, sorry! It is:
$$x_3-h(x_1,x_2)=0$$

So then in this case my direction normal would be:
$$\nabla_h=\hat x_3 -\frac{\partial h}{\partial x_1} \hat x_1 -\frac{\partial h}{\partial x_2} \hat x_2$$
and I can dot this with the velocity of the fluid to get zero. So that:
$$\mathbf{u}\cdot\nabla_h= \mathbf{u}\cdot\hat x_3 -\mathbf{u}\cdot \frac{\partial h}{\partial x_1} \hat x_1 - \mathbf{u}\cdot\frac{\partial h}{\partial x_2} \hat x_2=0$$
This surely isn't the whole whole answer though is it?