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\textbf{Question}

Verify thet, if $U$ is a constant, then the stream function
$\phi(x,y)=Uxy$ represents two-dimensional inviscid flwo with no bsy
forces in the region $\{ (x,y): (x\ge 0),(y \ge 0) \}$ bounded by
solid walls at $x=0$ and $y=0$. Derive the (dimensional) boundary
layer equations for flow near to the solid wall $y=0$ in the form
\begin{eqnarray*}
uu_x+vu_y & = & xU^2+\nu u_{yy}\\
u_x + v_y & = & 0
\end{eqnarray*}
where the fluid velocity $\un{q}$ is given by $(u,v)$ and $\nu$
denotes the constant kinematic viscosity of the fluid, giving suitable
boundary conditions for these equations.

Show that these equations possess a similarity solution of the form
\begin{eqnarray*}
\phi & = & \sqrt{U\nu} x^k f(\mu)\\
\mu & = & \sqrt{\frac{Uy^2}{\nu}}
\end{eqnarray*}
provided $k$ is suitably chosen. Obtain the ordinary differential
equation satisfied by $f$ and state boundary conditions that $f$ must
satisfy. Show further that, when $\mu\to 0$, $f \sim -\mu^3/6$.

\newpage
\textbf{Answer}

If $\psi=Uxy$ then $\nabla^2\psi=0+0=0$

Also \begin{eqnarray*} u=\psi_y & = & Ux\\ v=-\psi_x & = & -Uy
\end{eqnarray*}
so $u=0$ on $x=0$ and $v=0$ on $y=0$ (nv if they wanted to they
could also show that this satisfies the Euler equations with
$p=-\frac{\rho U^2}{2}(x^2+y^2)+constant$)

Now use the N/S equations in component form:-

Scale and non-dimensionalise using
\begin{eqnarray*} u & = & Uu'\\ v & = & \delta Uv'\\ x & = & Lx'\\
y & = & \delta Ly'\\ p & = & \rho U_{\infty}^2p'
\end{eqnarray*}
to look in the boundary layer near $y=0$ where $\delta \ll 1$ is
to be found.

$\left. \begin{array}{lrcl} \Rightarrow & \displaystyle
\frac{U^2}{L}(uu_x+vu_y) & = & \displaystyle -\frac{U^2}{L}p_x +
\nu \left ( \frac{U}{L^2}u_{xx}+\frac{U}{L^2\delta^2}u_{yy} \right
)\\ & \displaystyle \frac{\delta U^2}{L}(uv_x+vv_y) & = &
\displaystyle -\frac{U^2}{L\delta}p_y +\nu \left ( \frac{\delta
U}{L^2}v_xx+ \frac{U}{L^2 \delta} \right )\\ & u_x+v_y & = & 0
\end{array} \right \}$

(all bars dropped)

So re-arranging
\begin{eqnarray*} uu_x+vu_y & = &
-p_x+\frac{1}{Re} \left ( u_{xx}+\frac{1}{\delta^2}u_{yy} \right
)\\ \delta (uv_x+vv_y) & = & -\frac{1}{\delta}p_y+\frac{1}{Re}
\left ( \delta v_{xx}+\frac{1}{\delta}v_{yy} \right )\\ u_x+v_y &
= & 0
\end{eqnarray*}

The only chance for a non-trivial balance that retains a 2nd-order
system is to choose $\delta^2re=O(1)$.

So take $\delta=\frac{1}{\sqrt{Re}} \ \ \Rightarrow$ to lowest
order

$\left. \begin{array}{rl} uu_x+vu_y & = -p_x +u_{yy}\\ 0 & =
-p_y\\ u_x+v_y & = 0 \end{array} \right ) \longrightarrow$ (1)

In the outer flow (dimensionally) $p+\frac{1}{2}\rho
\underline{q}^2=constant$

Now $\underline{q}=(U_x,-U_y)$ so outside the boundary layer (near
$y=0$)

$\begin{array}{crcl} &\underline{q} & \sim &
U_x\underline{\hat{e}}_x\\ \Rightarrow & p_x +\rho(U_x)U & = & 0\\
\Rightarrow & -p_x/\rho & = & U^2x \end{array}$

So redimensionalising (1) gives

$\left. \begin{array}{lr} uu_x+vu_y & = U^2x+u_{yy}\nu \\ u_x+v_y
& = 0 \end{array} \right \}$

\begin{tabular}{lcr} B/C's & (No slip) & $u=v=0$ at $y=0$\\ &
(Matching) & $u\to U_x$ as $y\to\infty$ \end{tabular}

Similarity solutions:- use
\begin{eqnarray*} \psi & = & \sqrt{U\nu}x^kf{\eta}\\ \eta & = &
(Uy^2/\nu)^{\frac{1}{2}} = \sqrt{\frac{U}{\nu}}y \end{eqnarray*}

$\begin{array}{lrcl} \Rightarrow & u & = & Ux^kf'\\ & u_y & = &
 U\sqrt{\frac{U}{\nu}}x^kf''\\ & u_{yy} & = &
\displaystyle \frac{U^2}{\nu}f'''x^k\\ & v & = &
-\sqrt{U\nu}kx^{k-1}f\\ &u_x & = & kUx^{k-1}f' \end{array}$

\begin{eqnarray*}
\Rightarrow
Ux^kf'(kUx^{k-1}f')-\sqrt{U\nu}kx^{k-1}f\frac{U^{\frac{3}{2}}}{\sqrt{\nu}}x^kf''
& = & U^2x+\frac{U^2}{\nu}x^kf'''\nu
\\kU^2x^{2k-1}f'^2-kU^2x^{2k-1}ff'' & = & U^2x +
\frac{U^2}{\nu}\nu x^kf'''
\end{eqnarray*}

Obviously the similarity solution can only work if $k=1$, in which
case the ODE is $$U^2f'^2-U^2ff''=U^2+U^2f'''$$ i.e.
$$f'''+ff''+1-f'^2=0$$

\begin{tabular}{lcr} Suitable B/C's & (No slip) & $f'(0)=f(0)=0$\\ &
(Matching) & $f'(\infty)=1$\end{tabular}

Now consider the ODE for small $\eta$. Since for small values of
$\eta$, $f$ and $f'$ are small, the leading order balance in the
equation will be $$f'''+1=0$$
\begin{eqnarray*} \Rightarrow f'' & \sim & -\eta + A\\ f' & \sim &
-\eta^2/2 +A\eta +B\\ f & \sim & -\eta^3/6 +A\eta^2/2 +B\eta + C
\end{eqnarray*}

But since $f(0)=f'(0)=)$, $A=B=0$. Also $f$ is a stream function
so we can $\underline{\rm{take}}\ C=0$ $$\displaystyle \Rightarrow
f \sim -\frac{\eta^3}{6} \ \ \ (\eta\sim 0)$$


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