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Why is it that at $y=0$ (at the wall), we have $v=0$ (vertical component of velocity)? Obviously $v$ cannot be negative there as there is no flow through the wall, however how do fluid particles move off the wall if they don't have positive velocity?

  • Has this got anything to do with the fact that gravity is pointing downwards? –  Apr 07 '15 at 18:46

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Individual molecules can collide with the wall and move to it and away from it, but we're talking about a continuous model of fluid motion here. If $v$ was positive at some point $(x_0,0)$ on the wall (and continuous, so positive in some rectangle $R = \{(x,y): |x - x_0| < \epsilon, 0 \le y < \delta\}$, then (taking $\delta$ sufficiently small relative to $\epsilon$) all the fluid would soon flow out of $R$, leaving a vacuum.

Robert Israel
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  • This is an inviscid flow, could there not be fluid moving along the wall which enters the rectangle as the fluid in the rectangle leaves? –  Apr 07 '15 at 20:42
  • Not enough... the left and right edges of the rectangle have length $\delta \to 0$, so the rate at which fluid enters goes to $0$ as $\delta \to 0+$, while the rate at which fluid leaves (through a side of length $\epsilon$) does not. – Robert Israel Apr 07 '15 at 21:05
  • I am a bit confused as to why you are allowed to construct the rectangle in this way. Are you using the definition of continuity (epsilon-delta)? –  Apr 07 '15 at 21:33
  • The definition of continuity implies there is some $r > 0$ such that $v > 0$ at all points within distance $r$ of $(x_0,0)$. In particular this applies in the rectangle $R$ if $\sqrt{\epsilon^2 + \delta^2} < r$. – Robert Israel Apr 07 '15 at 23:04