Fluid Boundary Layer Mathy Question

[member 428835]

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!

 

[pasmith]

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

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

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

 

[member 428835]

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.