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\textbf{Question}

%Question 4
Two-dimensional fluid flow takes place in the first quadrant of the
$(x,y)$-plane. The stream function for the flow is given by
$$\psi(x,y) = Cxy$$
where $C$ is a positive constant.

\begin{description}
%Question 4i
\item{(i)}
Determine the flow velocity components.

%Question 4ii
\item{(ii)}
Show that the flow is irrotational and incompressible.

%Question 4iii
\item{(iii)}
Sketch the streamlines of the flow
\end{description}

The stream function $\psi$ is now regarded as the outer flow of a high
Reynolds number steady viscous flow (with no body forces) and we wish
to examine the boundary later near the wall $y=0$. Derive the
(dimensional) boundary layer equations
\begin{eqnarray*}
uu_x + vu_y &amp; = &amp; C^2x + vu_{yy}\\
u_x + v_y &amp; = &amp; 0
\end{eqnarray*}
where $u$ and $v$ are the velocity components of the flow, and give
suitable boundary conditions for these equations.

Verify that a similarity solution exists in the form
$$\psi = xf(y)$$
and determine the differential equation satisfied by $f$, giving
suitable boundary conditions.

\newpage
\textbf{Answer}

We have $\phi = Cxy$, flow in $x \geq 0$, $y \geq 0$.

\begin{description}
\item{(i)}
$\left. \begin{array}{l} u=\phi_y=Cx\\ v=-\phi_x=-Cy \end{array}
\right \}$ $\Rightarrow \underline{q}=(Cx,-Cy,0)$

\item{(ii)}
$curl \underline{q} =$ $\left | \begin{array}{ccc} \underline{i} &amp;
\underline{j} &amp; \underline {k}\\ \displaystyle
\frac{\partial}{\partial x} &amp; \displaystyle
\frac{\partial}{\partial y} &amp; \displaystyle
\frac{\partial}{\partial z}\\ Cx &amp; -Cy &amp; 0
\end{array} \right | $ $=$ $\left ( \begin{array}{c} 0\\ 0\\ 0
\end{array} \right )$ $\Rightarrow$ irrotational.

Also $\displaystyle div(\underline{q})=\frac{\partial}{\partial
x}(Cx)+\frac{\partial}{\partial y}(-Cy)+\frac{\partial}{\partial
z}(0)=C-C=0.$

\item{(iii)}
$\phi =0$ on $x=0$, $y=0$

$\phi=constant=\beta$ say $\Rightarrow xy=constant$
$\Rightarrow$hyperbolae.

\begin{center}
\epsfig{file=311-hyperfam.eps, width=70mm}
\end{center}

Now consider the Navier-Stokes equations

$\left. \begin{array}{c} \displaystyle \underline{q}_t
+(\underline{q}.\nabla)\underline{q}=\frac{-1}{\rho}\nabla p+\nu
\nabla^2\underline{q}\\ div(\underline{q})=0 \end{array} \right
\}$ $\left. \begin{array}{l} \rm{Non-dimensionalise}\ \rm{with}\\
\underline{x}=L\underline{\overline{x}},\ \
\underline{q}=U\underline{\overline{q}},\ \  p=\rho U^2
\overline{p}
\end{array} \right. $

Where L and U are a representitive length and speed.

(Dropping bars) $$\displaystyle
\frac{U^2}{L}(\underline{q}.\nabla)\underline{q}=
\frac{-U^2}{L}\nabla p + \frac{\nu U}{L^2}\nabla^2
\underline{q}=0$$ $$div(\underline{q})=0$$

Thus the momentum equation becomes
$$(\underline{q}.\nabla)\underline{q}= -\nabla
p+\frac{1}{Re}\nabla^2 \underline{q}, \ \ \ (Re=\frac{LU}{\nu})$$

Away from the boundaries in the flow, since $Re \gg 1$ the flow is
essentially inviscid so that $\phi = Cxy$. But near $y=0$ we must
rescale $y=\delta \tilde{y}$, $v=\delta \tilde{v}$. ($\delta \ll
1$)

$\Rightarrow$ \begin{eqnarray*} uu_x+\tilde{v}u_{\tilde{y}} &amp; = &amp;
-p_x + \frac{1}{Re} \left ( \begin{array}{l} \displaystyle
u_{xx}+\frac{1}{\delta^2}u_{\tilde{y}\tilde{y}}
\end{array} \right ) \\ \delta (u\tilde{v}_x
+\tilde{v}\tilde{v}_{\tilde{y}}) &amp; = &amp;  \frac{1}{Re} \left (
\begin{array}{l} \displaystyle
\delta\tilde{v}_{xx}+\frac{1}{\delta}\tilde{v}_{\tilde{y}\tilde{y}}
\end{array} \right ) \\ u_x + \tilde{v}_{\tilde{y}} &amp; = &amp; 0
\end{eqnarray*}

Now consider the size of $\delta$.

If $\delta^2Re\ll 1$ then $u_{\tilde{y}\tilde{y}}=0$ to leading
order, and this can never match with the outer flow. If
$\delta^2Re \gg 1$ then we just get back to the inviscid
equations.

$\Rightarrow \delta^2Re=1$, so $\delta=\frac{1}{\sqrt{Re}}$.

Then the leading order equations (redimensionalized) are

$\left. \begin{array}{rl} uu_x + vu+y = &amp; \frac{-1}{\rho}p_x + \nu
u_{yy}\\ 0 = &amp; p_y\\ u_x + v_y = &amp; 0 \end{array} \right \}$

Outer flow: $p+\frac{1}{2}p\underline{q}^2=constant$, \ \
$\Rightarrow p_x=-\rho uu_x$

But $u=Cx$, \ \ $\Rightarrow -p_x=\rho C^2x$

$\left. \begin{array}{rll} \Rightarrow uu_x+vu_y &amp; = &amp; C^2x+\nu
u_yy\\ u_x + v_y &amp; = &amp; 0
\end{array} \right.$

Boundary conditions, $u=v=0$ on $y=0$ (no slip), $u\rightarrow Cx$
as $y\rightarrow\infty$ (matching).

Now with $\phi = xf(y)$ $\left. \begin{array}{lll} u=xf' &amp;
u_y=xf'' &amp; u_{yy}=xf'''\\v=-f &amp; u_x=f' \end{array} \right.$

$\Rightarrow xf'^2-fxf''=C^2x+\nu xf'''$

$\Rightarrow f'^2-ff''-C^2-\nu f'''=0$

$\Rightarrow \nu f''' + ff''-f'^2 + C^2=0$

Boundary conditions: $f(0)=f'(0)=0$, \ \ $f'(\infty)=c$
\end{description}


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