# Finding Free surface elevation in semi-infinite channel

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 ?

We know that : in the fluid mass $$\phi_{xx} + \phi_{yy} = 0$$ And at $$x=0$$ we have $$\frac{\partial \phi}{\partial x} = 0$$ And on $$y=0$$ we have $$\frac{\partial^2 \phi}{\partial t^2} + g \frac{\partial \phi}{\partial y} = 0$$ And we have initial condition $$\phi(x,y,0)=0$$ How should I write the condition on the bottom ? Is it correct if i wrote $$\frac{\partial \phi}{\partial y} = f(x,t)$$ at y = $$-h$$

• When you have crossposted, please include a link to the crosspost on both posts. – user44191 Jan 10 at 6:10
• I am very tempted to say that the main part of the question (asking about interpretation of a phrase) belongs more on english.stackexchange than on MO or Math.stackexchange. The sentence has no clear mathematical meaning. Either you have been given more context (from which there is a unique interpretation) and you should include it in your question, or (the better option) you should just go ask the person who originated that phrase to ask what exactly is meant, mathematically or physically, by that phrase. – Willie Wong Jan 10 at 15:18

What I seem to understand is that the bottom is at $$y=-h+v(x-ct)$$ where $$v$$ has compact support. That corresponds to an obstacle moving at a constant speed towards the infinite end. But it might mean something else, such as $$y=-h+v(x)\sin(\omega t)$$. In any case the condition at the (moving) bottom has to be $$\frac{\partial\phi}{\partial n}=0$$ where $$n$$ is the instantaneous normal to the bottom. Clearly the fluid motion is incompressible with velocity $$\nabla\phi$$.