Question on free surface elevation of water wave

Question

A semi-infinite channel of finite depth is occupied by an ideal fluid layer initially at rest . the vertical finite end of the channel is fixed and only a part of the horizontal bottom , with finite support , is set in a bounded motion .

Find the resulting free surface elevation at any subsequent instant of time .

My question is what does it mean that horizontal bottom with finite support ?

Is that means that we have a wave maker on the horizontal bottom ?

score 2 · Answer 1 · answered Jan 09 '19 at 22:11

2

The support of a function is the part of the domain where the function value is nonzero. What this is saying is that a finite section of the bottom of the channel is set in motion. If $x$ is along the channel, maybe the section from $x=0$ to $x=1$ is alternately lifted and dropped, so the bottom there has elevation say $y=A\sin \omega t$. This would be produced by having a plate of that size on a piston, moving up and down with amplitude $A$ and frequency $\frac \omega{2\pi}$. The rest of the bottom of the channel would be fixed at $y=0$

answered Jan 09 '19 at 22:11

Ross Millikan

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suppose that the depth of the channel is $h$ Is it coorect if i say that the condition on the bottom $\frac{\partial \phi}{\partial y} = f(x,t)$ at y = -h – Mahmoud Hassan Jan 09 '19 at 22:20
or $\frac{\partial \phi }{\partial y} = f(x) \sin \omega t$ – Mahmoud Hassan Jan 09 '19 at 22:21
What is $\phi?$ – Ross Millikan Jan 09 '19 at 22:24
The velocity potential for ideal fluid we can write $\vec{V}=\vec{\nabla \phi}$ So if $\vec{V}= u \hat{i} + v\hat{j}$ which implies that $u=\frac{\partial \phi}{\partial x}$ and $v=\frac{\partial \phi}{\partial y}$ – Mahmoud Hassan Jan 09 '19 at 22:47
I can't help you with any of that. – Ross Millikan Jan 09 '19 at 22:48
Ok , Thank you ! – Mahmoud Hassan Jan 09 '19 at 22:51

Question on free surface elevation of water wave

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