Boundary Layer Thickness and Big O Notation Watch

Calculator878
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I have a solution for the velocity function of a Stokes layer flow problem as follows:


u(y,t) = U exp{-y(s^-0.5)} cos(wt - ys^-0.5)

Ucoswt - is speed of wall causing the flow
s - constant
w - frequency
t - time


the thickness of the flow is y=O[s^0.5], why? how is this found?
and what does it mean for y to be big O "this"?
I looked up the definition for big O but can't understand it, or interpret it into limits

(Original post by Farhan.Hanif93)
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Farhan.Hanif93
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(Original post by Calculator878)
I have a solution for the velocity function of a Stokes layer flow problem as follows:


u(y,t) = U exp{-y(s^-0.5)} cos(wt - ys^-0.5)

Ucoswt - is speed of wall causing the flow
s - constant
w - frequency
t - time


the thickness of the flow is y=O[s^0.5], why? how is this found?
and what does it mean for y to be big O "this"?
I looked up the definition for big O but can't understand it, or interpret it into limits
For a Stokes layer flow with velocity function solution of the form u(y,t) = Ue^{-\kappa y}\cos (\omega t - \kappa y), the Stokes boundary-layer thickness is given by \delta = \dfrac{2\pi}{\kappa} and by definition, we must have y\leq \delta (that is, rather obviously, the distance from the plate must remain within the thickness of the flow).

Informally, big O notation is used to provide a bound on how quickly one quantity can grow relative to another, and we say f(x)= O(g(x)) as x gets big if f(x) can grow no more quickly than g(x) for all values of x beyond a certain point. Similarly, we know that y cannot grow to be greater than the thickness of the flow and hence we say that y=O(\delta) (as t gets big). The more formal definition should be on wikipedia and isn't too hard to get your head around, so I do advise that you look at it again and try to understand what this means mathematically.

One other thing to note is that constant multipliers do not affect the order of the function (again, look to the formal definition to see why this is the case), i.e.:
\delta = \dfrac{2\pi}{\kappa} \Rightarrow y = O(2\pi/\kappa) = O(1/\kappa)
In your case, \kappa = s^{-1/2}, and it hence follows that y=O(s^{1/2}).

[A couple of slight niggles - I'm not around on TSR as much these days so tagging me in posts isn't always likely to yield a quick response; and questions of study help are often better placed here].
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