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\textbf{Question}

Define the Reynolds number $Re$ for a viscous flow. YOU MAY ASSUME
that a suitable non-dimensional form of the steady Navier-Stokes
equations for use in studying boundary layer theory is
\begin{eqnarray*}
(\un{q}.\nabla)\un{q} &amp; = &amp; -\nabla p + \frac{1}{Re} \nabla^2
\un{q}\\ \nabla . \un{q} &amp; = &amp; 0
\end{eqnarray*}
where the fluid pressure and velocity are denoted respectively by
$p$ and $\un{q}=(u,v)^T$, and lengths, pressures and velocities
have been non-dimensionalised using $L$, $\rho U_{\infty}^2$ and
$U_{\infty}$ respectively. (Here $\rho$ denotes the constant fluid
density, and $L$ and $U_{\infty}$ are a typical length and speed
in the flow.)

Starting from these equations, derive the non-dimensional boundary
layer equations
\begin{eqnarray*}
uu_x + vu_y &amp; = &amp; -p_x + u_{yy}\\ u_x + v_y &amp; = &amp; 0
\end{eqnarray*}
for two-dimensional steady incompressible flow at high Reynolds
number past a flat plate situated at $y=0$.

Now assume that the horizontal speed of the external flow field is
given (in dimensional variables) by $$U(x) = U_{\infty} \left (
\frac{x}{L} \right )^m,$$ where $U_{\infty}$ and $m$ are
constants. By defining a stream function $\psi(x,y)$ which
satisfies $u=\psi_y$, $v=-\psi_x$, show that $\psi$ satisifies
$$\psi_y\psi_{xy} -\psi_x\psi_{yy} = m x^{2m-1} + \psi_{yyy}.$$

Give boundary conditions for $\psi$ for this flow.

By assuming a solution of the form
\begin{eqnarray*}
\psi(x,y) &amp; =&amp; x^{\frac{m+1}{2}} f(\eta) \\ \eta &amp; = &amp;
yx^{\frac{m-1}{2}}
\end{eqnarray*}
where $'=d/d\eta$, show that $f$ satisfies the Falkner-Skan
equation $$f'''+ \left ( \frac{m+1}{2} \right ) f f'' + m(1-f'^2)
=0,$$ and give suitable boundary conditions for $f$.



\textbf{Answer}

ASSUME $\left.
\begin{array}{rl} (\underline{q}.\nabla)\underline{q} &amp;  \displaystyle = -\nabla p+
\frac{1}{Re}\nabla^2\underline{q}\\ div(\underline{q}) &amp; = 0
\end{array} \right \}$

$\ \ \ \ \ \displaystyle Re=\frac{LU}{\nu}
\begin{array}{c}\rm{L:Typical \ length \ \ U:Typical \ speed}\\
\nu\rm{:Kinematic \ viscosicty}
\end{array}$

Now further rescale $y=\epsilon Y$, $v=\epsilon V$ (thin layer)

$\Rightarrow \begin{array}{rcl} uu_x+Vu_y &amp; = &amp; \displaystyle
-p_x+\frac{1}{Re}(u_{xx}+\frac{1}{\epsilon^2}u_{YY})\\
\displaystyle \epsilon(uV_x+VV_Y) &amp; = &amp; \displaystyle
\frac{-1}{\epsilon}p_Y+\frac{1}{Re}(\epsilon
V_{xx}+\frac{1}{\epsilon}V_{YY})\\ u_x+V_Y &amp; = &amp; 0 \end{array} $

Now consider the relative size of $\epsilon$ and $Re$. If
$Re\epsilon^2 \gg 1$ then for $Re \gg 1$ the first equation just
gives Euler (no good). Also, if $Re\epsilon^2 \ll 1$ then it just
gives $u_{YY}=0 \Rightarrow u=A(x)Y$ which $\to\infty$ as
$Y\to\infty \Rightarrow$ need $Re\epsilon^2=O(1)$. Thus for $Re
\gg 1$ get $p_Y=0$ in the second equation $\Rightarrow p=p(x)$
alone, and

$uu_x+Vu_Y=-p_x+u_{YY}$, $u_x+V_Y=0$ or, back in N/D (unscaled)
variables

$\left. \begin{array}{rcl} uu_x+vu_y &amp; = &amp; -p_x + u_{yy}\\ u_x+v_y
&amp; = &amp; 0 \end{array} \right \}$

Now in the outer flow $ \displaystyle U(x)=U_{\infty}\left (
\frac{x}{L} \right )^m$. BUt here the flow is inviscid and so
$p+\frac{1}{2}\rho U^2 = constant$ (Bernoulli)

$\Rightarrow p_x+\rho UU_x = 0$

i.e. $ \displaystyle p_x = -\rho U_{\infty}^2 \left ( \frac{x}{L}
\right )^m m\frac{1}{L}\left ( \frac{x}{L} \right )^{m-1}$

$\Rightarrow$ scaling p with $\rho U_{\infty}^2$, $x$ with $L$
gives (N/D)

$p_x=-mx^{2m-1}$

Now e have $uu_x+vu_y=mx^{2m-1}+u_{yy}$ and setting $u=\psi_y$,
$v=-\psi_x$ we see that $u_x+v_y=0$.

The momentum equation now gives
$$\psi_y\psi_{xy}-\psi_x\psi_{yy}=ms^{2m-1}+\psi_{yyy}$$

B/C's:- (no slip) $\psi=\psi_y$ at $y=0$ $$\psi_y=x^m \ \rm{as} \
y\to\infty \ (\rm{MATCHING})$$

So now try $ \displaystyle \psi=x^{\frac{m+1}{2}}f(\eta)$,
$\eta=yx^{\frac{m-1}{2}}$

$\psi_y=x^mf'$, $\psi_{yy}=x^{\frac{3m-1}{2}}f''$,
$\psi_{yyy}=x^{\frac{4m-2}{2}}f'''$

$ \displaystyle \psi_x=\left ( \frac{m+1}{2} \right ) x^{\left (
\frac{m-1}{2} \right ) }f+x^{\left ( \frac{m+1}{2} \right )}y
\left ( \frac{m-1}{2} \right ) x^{\frac{m-3}{2}}f'$

$ \displaystyle \psi_{yx}=mx^{m-1}f'+x^my\left ( \frac{m-1}{2}
\right ) x^{\frac{m-3}{2}}f''$

\begin{eqnarray*}
\Rightarrow &amp; &amp; x^mf'\left(mx^{m-1}f'+x^{\frac{3m-3}{2}}y\left (
\frac{m-1}{2} \right ) f'' \right )\\ &amp; &amp; -
x^{\frac{3m-1}{2}}f''\left ( \left ( \frac{m+1}{2} \right )
x^{\frac{m-1}{2}}f+x^{\frac{2m-2}{2}}y \left ( \frac{m-1}{2}
\right ) f' \right )\\ &amp; = &amp; mx^{2m-1}+x^{2m-1}f'''
\end{eqnarray*}

$ \displaystyle \Rightarrow mx^{2m-1}f'^2-\left ( \frac{m+1}{2}
\right )x^{2m-1}ff''=mx^{2m-1}+x^{2m-1}f'''$

$ \displaystyle \Rightarrow mf'^2-\left ( \frac{m+1}{2}\right )
ff''=m+f'''$

i.e. $ \displaystyle f'''+\left ( \frac{m+1}{2} \right )
ff''+m(1-f'^2) = 0$

B/C's:- $\begin{array}{lr} f(0)=f'(0)=0 &amp; (\rm{no \ slip})\\
f'(\infty)=1 &amp;(\rm{MATCHING}) \end{array} $


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