Boundary Conditions for an inviscid fluid at a fixed boundary

[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...

 

[pasmith]

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"

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

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.

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

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...

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

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!

Following the question as written, the first thing is to define
<br /> g(x_1,x_3) = x_3 - h(x_1,x_3)<br />
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
<br /> \nabla g = \frac{\partial g}{\partial x_1} \hat{\mathbf{x}}_1 + \frac{\partial g}{\partial x_3} \hat{\mathbf{x}}_3.<br />

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...

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

 

[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?

Thanks for the reply.

 

[Turing101]

Hello, i know this is old but i would also really love to have some help with this question.
Thank you so much.