Mixing
Question on free surface elevation of water wave
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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 ?
fluid-dynamics
asked Jan 9 at 21:57
topspintopspin
727413
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add a comment |
$begingroup$
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 ?
fluid-dynamics
asked Jan 9 at 21:57
topspintopspin
727413
$endgroup$
add a comment |
$begingroup$
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 ?
fluid-dynamics
asked Jan 9 at 21:57
topspintopspin
727413
$endgroup$
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 ?
fluid-dynamics
fluid-dynamics
asked Jan 9 at 21:57
topspintopspin
727413
asked Jan 9 at 21:57
topspintopspin
727413
edited Jan 10 at 5:50
topspin
asked Jan 9 at 21:57
topspintopspin
727413
asked Jan 9 at 21:57
topspintopspin
727413
asked Jan 9 at 21:57
topspintopspin
727413
727413
add a comment |
add a comment |
1 Answer
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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 $y=Asin omega t$. This would be produced by having a plate of that size on a piston, moving up and down with amplitude $A$ and frequency $frac omega{2pi}$. The rest of the bottom of the channel would be fixed at $y=0$
answered Jan 9 at 22:11
Ross MillikanRoss Millikan
295k23198371
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$begingroup$
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
$endgroup$
– topspin
Jan 9 at 22:20
$begingroup$
or $frac{partial phi }{partial y} = f(x) sin omega t$
$endgroup$
– topspin
Jan 9 at 22:21
$begingroup$
What is $phi?$
$endgroup$
– Ross Millikan
Jan 9 at 22:24
$begingroup$
The velocity potential for ideal fluid we can write $vec{V}=vec{nabla phi}$ So if $vec{V}= u hat{i} + vhat{j}$ which implies that $u=frac{partial phi}{partial x}$ and $v=frac{partial phi}{partial y}$
$endgroup$
– topspin
Jan 9 at 22:47
$begingroup$
I can't help you with any of that.
$endgroup$
– Ross Millikan
Jan 9 at 22:48
|
show 1 more comment
Your Answer
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1 Answer
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1 Answer
1
active
oldest
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active
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active
oldest
votes
$begingroup$
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 $y=Asin omega t$. This would be produced by having a plate of that size on a piston, moving up and down with amplitude $A$ and frequency $frac omega{2pi}$. The rest of the bottom of the channel would be fixed at $y=0$
answered Jan 9 at 22:11
Ross MillikanRoss Millikan
295k23198371
$endgroup$
$begingroup$
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
$endgroup$
– topspin
Jan 9 at 22:20
$begingroup$
or $frac{partial phi }{partial y} = f(x) sin omega t$
$endgroup$
– topspin
Jan 9 at 22:21
$begingroup$
What is $phi?$
$endgroup$
– Ross Millikan
Jan 9 at 22:24
$begingroup$
The velocity potential for ideal fluid we can write $vec{V}=vec{nabla phi}$ So if $vec{V}= u hat{i} + vhat{j}$ which implies that $u=frac{partial phi}{partial x}$ and $v=frac{partial phi}{partial y}$
$endgroup$
– topspin
Jan 9 at 22:47
$begingroup$
I can't help you with any of that.
$endgroup$
– Ross Millikan
Jan 9 at 22:48
|
show 1 more comment
$begingroup$
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 $y=Asin omega t$. This would be produced by having a plate of that size on a piston, moving up and down with amplitude $A$ and frequency $frac omega{2pi}$. The rest of the bottom of the channel would be fixed at $y=0$
answered Jan 9 at 22:11
Ross MillikanRoss Millikan
295k23198371
$endgroup$
$begingroup$
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
$endgroup$
– topspin
Jan 9 at 22:20
$begingroup$
or $frac{partial phi }{partial y} = f(x) sin omega t$
$endgroup$
– topspin
Jan 9 at 22:21
$begingroup$
What is $phi?$
$endgroup$
– Ross Millikan
Jan 9 at 22:24
$begingroup$
The velocity potential for ideal fluid we can write $vec{V}=vec{nabla phi}$ So if $vec{V}= u hat{i} + vhat{j}$ which implies that $u=frac{partial phi}{partial x}$ and $v=frac{partial phi}{partial y}$
$endgroup$
– topspin
Jan 9 at 22:47
$begingroup$
I can't help you with any of that.
$endgroup$
– Ross Millikan
Jan 9 at 22:48
|
show 1 more comment
$begingroup$
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 $y=Asin omega t$. This would be produced by having a plate of that size on a piston, moving up and down with amplitude $A$ and frequency $frac omega{2pi}$. The rest of the bottom of the channel would be fixed at $y=0$
answered Jan 9 at 22:11
Ross MillikanRoss Millikan
295k23198371
$endgroup$
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 $y=Asin omega t$. This would be produced by having a plate of that size on a piston, moving up and down with amplitude $A$ and frequency $frac omega{2pi}$. The rest of the bottom of the channel would be fixed at $y=0$
answered Jan 9 at 22:11
Ross MillikanRoss Millikan
295k23198371
answered Jan 9 at 22:11
Ross MillikanRoss Millikan
295k23198371
answered Jan 9 at 22:11
Ross MillikanRoss Millikan
295k23198371
answered Jan 9 at 22:11
Ross MillikanRoss Millikan
295k23198371
295k23198371
$begingroup$
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
$endgroup$
– topspin
Jan 9 at 22:20
$begingroup$
or $frac{partial phi }{partial y} = f(x) sin omega t$
$endgroup$
– topspin
Jan 9 at 22:21
$begingroup$
What is $phi?$
$endgroup$
– Ross Millikan
Jan 9 at 22:24
$begingroup$
The velocity potential for ideal fluid we can write $vec{V}=vec{nabla phi}$ So if $vec{V}= u hat{i} + vhat{j}$ which implies that $u=frac{partial phi}{partial x}$ and $v=frac{partial phi}{partial y}$
$endgroup$
– topspin
Jan 9 at 22:47
$begingroup$
I can't help you with any of that.
$endgroup$
– Ross Millikan
Jan 9 at 22:48
|
show 1 more comment
$begingroup$
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
$endgroup$
– topspin
Jan 9 at 22:20
$begingroup$
or $frac{partial phi }{partial y} = f(x) sin omega t$
$endgroup$
– topspin
Jan 9 at 22:21
$begingroup$
What is $phi?$
$endgroup$
– Ross Millikan
Jan 9 at 22:24
$begingroup$
The velocity potential for ideal fluid we can write $vec{V}=vec{nabla phi}$ So if $vec{V}= u hat{i} + vhat{j}$ which implies that $u=frac{partial phi}{partial x}$ and $v=frac{partial phi}{partial y}$
$endgroup$
– topspin
Jan 9 at 22:47
$begingroup$
I can't help you with any of that.
$endgroup$
– Ross Millikan
Jan 9 at 22:48
$begingroup$
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
$endgroup$
– topspin
Jan 9 at 22:20
$begingroup$
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
$endgroup$
– topspin
Jan 9 at 22:20
$begingroup$
or $frac{partial phi }{partial y} = f(x) sin omega t$
$endgroup$
– topspin
Jan 9 at 22:21
$begingroup$
or $frac{partial phi }{partial y} = f(x) sin omega t$
$endgroup$
– topspin
Jan 9 at 22:21
$begingroup$
What is $phi?$
$endgroup$
– Ross Millikan
Jan 9 at 22:24
$begingroup$
What is $phi?$
$endgroup$
– Ross Millikan
Jan 9 at 22:24
$begingroup$
The velocity potential for ideal fluid we can write $vec{V}=vec{nabla phi}$ So if $vec{V}= u hat{i} + vhat{j}$ which implies that $u=frac{partial phi}{partial x}$ and $v=frac{partial phi}{partial y}$
$endgroup$
– topspin
Jan 9 at 22:47
$begingroup$
The velocity potential for ideal fluid we can write $vec{V}=vec{nabla phi}$ So if $vec{V}= u hat{i} + vhat{j}$ which implies that $u=frac{partial phi}{partial x}$ and $v=frac{partial phi}{partial y}$
$endgroup$
– topspin
Jan 9 at 22:47
$begingroup$
I can't help you with any of that.
$endgroup$
– Ross Millikan
Jan 9 at 22:48
$begingroup$
I can't help you with any of that.
$endgroup$
– Ross Millikan
Jan 9 at 22:48
|
show 1 more comment
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