Why is the solution to the Blasius boundary layer problem self-similar? In every course or textbook that I encountered so far, the authors transform the Navier-Stokes equations of the Blasius boundary layer problem into the Blasius ODE. The problem with many of those texts is that they claim that the solution is self-similar and give very weak justifications, or none at all and that really bothers me as I'm trying to understand rigorously why such a transformation is possible.
Does anyone know of references to look at for a rigorous proof of the above claim, or at least an experience showing that the plot of $\frac{u}{U_{\infty}}$ vs $\frac{y}{\sqrt{x}}$ doesn't change when changing $x$ ?
 A: The wiki page on Blasius boundary layers is a useful and thorough resource in this case. 
Blasius boundary layers arise in steady, laminar 2D flow over a semi-infinite plate oriented parallel to the flow. In this scenario, the Navier-Stokes equations are particularly simple and amount to a leading-order balance between inertia and viscous forces. The similarity solutions are derived from scaling arguments of these equations. It's not worth me repeating the derivation of the ode as the wiki page is very thorough. 
A: The  self  similarity  condition  is  applied  when  deriving  the  Blasius  equation,  implying  that the  streamwise  velocity  $u(x,  y)$  across  a  boundary  layer  is  a  function  of  the  derivative of  $f(\eta)$  scaled  by  the  flow  $U$  as  follows $$u(x,  y)  =  Uf'(\eta)$$  The  self-similar  streamwise  profile  derives  from  the  following stream  function.  $$\psi(x,y)  =(U\nu  x)^{\frac{1}{2}}f(\eta)$$  To  see  this  clearly,  we differentiate  the  streamfunction  with  respect  to  $x$,  finding  the  $x$  component  of  the  velocity. $$u(x,y)  =  \frac{\partial  \psi}{\partial  y}  =  U(U\nu  x)^{\frac{1}{2}}\,\,f'(\eta)\frac{d\eta}{dy}  =  U(U \nu  x)^{\frac{1}{2}}\,\,f'(\eta)\bigg(\frac{U}{\nu  x}\bigg)^\frac{1}{2}$$  Which  simplifies  to  the desired equation.
A: If there are no solutions of that form you wont find any.
