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A Fluid Boundary Layer Mathy Question

  1. Oct 13, 2016 #1

    joshmccraney

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    Hi PF!

    So after scaling Navier-Stokes for a flow over a flat plate we ultimately arrive at ##f f'' + f''' = 0## subject to ##f(0)=0##, ##f'(0)=0##, and ##f'(\infty) = 1## where independent variable is ##\eta##. The source I was reading is trying to reduce this BVP to an IVP. Thus they suggest for some solution ##F(\eta)##, ##CF(C \eta)## also is a solution. Then we have $$1 = \lim_{\eta \to \infty} f'(\eta) = C^2 \lim_{\eta \to \infty} F'(C \eta) \implies C = \left( \lim_{\eta \to \infty} F'(\eta) \right)^{-1/2}$$. But this is where it get's strange. They then say "if we specify ##F''(0) = 1##... but how can they do this? We know ##f''(0) = C^3 F''(0) \implies F''(0) = f''(0) C^{-3}## but ##C## has already been specified.

    The link to this is here: http://web.mit.edu/fluids-modules/www/highspeed_flows/ver2/bl_Chap2.pdf around eq. (3.48)

    Any help at understanding this would be awesome!
     
  2. jcsd
  3. Oct 13, 2016 #2

    pasmith

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    Let [itex]f[/itex] be the solution of the BVP [itex]ff'' + f''' = 0[/itex] with [itex]f(0) = f'(0) = 0[/itex] and [itex]f'(\infty) = 1[/itex].

    Now for each [itex]C > 0[/itex] define a function [itex]g_C(\eta) = Cf(C\eta)[/itex]. Then [itex]g_C[/itex] is the solution of the BVP [itex]g_Cg_C'' + g_C''' = 0[/itex] with [itex]g_C(0) = g_C'(0) = 0[/itex] and [itex]g_C'(\infty) = C^2[/itex]. (This is not the original BVP unless [itex]C = 1[/itex], and by definition [itex]g_1 = f[/itex] anyway.)

    Consider the IVP [itex]hh'' + h''' = 0[/itex] subject to [itex]h(0) = h'(0) = 0[/itex] and [itex]h''(0) = 1[/itex]. Knowing [itex]h[/itex] we can find [itex]f[/itex], because [itex]h = g_C[/itex] where [itex]C^2 = L = \lim_{x \to \infty} h'(x)[/itex] (assuming, of course, that this limit exists). Hence by definition of [itex]g_C[/itex] we have [tex]f(\eta) = C^{-1}g_C(C^{-1}\eta) = L^{-1/2}h(L^{-1/2}\eta).[/tex]
     
  4. Oct 13, 2016 #3

    joshmccraney

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    Gotcha, I think this is making sense! I have a corollary question then, though I can post as a separate thread if that's more appropriate. I am trying to transform this equation to transform this BVP ##y'' + 6 y^{2/3} = 0## subject to ##y'(0)=0## and ##y(1)=0## into an IVP so I can numerically solve it. Right not when I use NDsolve in mathematica I get no output. Any insight on when this transformation is possible, and when it is, how to go about doing it?

    I noticed the BVP is invariant when ##y(x) = \lambda^n Y(\lambda^{-n/6} x)##. I'm just unsure how to proceed.
     
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