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\textbf{Question}

A two-dimensional viscous unsteady flow takes place in the
$(x,y)$-plabe. Ther are no body forces. The fluid is incompressible
and hsas constant density $\rho$ and constant dynamic viscosity
$\mu$. Show that, when the Reynolds number Re is much less than one,
the stream function $\phi (x,y)$ satisfies the biharmonic equation
$$\nabla^4 \phi = 0.$$

Explain briefly why, even for unsteady flwo, there are no time
derivatives in this equation and comment on how temporal changes would
enter the problem for a truly unsteady flow.

Steady low Reynolds number two-dimensional flow takes place in a wedge
of semi-angle $\alpha$, the flow beinh driven by a shearing mechanism
far away from the corner of the wege. Give the boundary condtions that
must be satisfied by the stream function $\phi$ and its derivatives on
the wedge walls $\theta = \pm \alpha$, where $(r,\theta00$ are plane
polar coordinates. Verify that a flow with stream function
$$\phi (r, \theta) = r^{\lambda} (A \cos \lambda\theta + B\cos
(\lambda-2)\theta )$$
is possible for non-zero $A$ and $B$ only if the constant $\lambda$
satisfies
$$(\lambda -2)\tan((\lambda -2)\alpha) = \lambda \tan \lambda\alpha.$$

[If you wish you may use, without proof, the fact that in cylindrical polar
coordinates $(r,\theta, z)$
$$\nabla^2 = \frac{1}{r}\frac{\partial}{\partial r} \left ( r
\frac{\partial}{\partial r} \right ) + \frac{1}{r^2}
\frac{\partial^2}{\partial \theta^2} + \frac{\partial^2}{\partial z^2}.]$$

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\textbf{Answer}

We have 2/D unsteady flow in the $(x,y)$-plane, so as usual the
continuity equation is automatically satisfied by $\psi = \psi
(x,y,t)$ where $u=\psi_y$, $v=-\psi_x$.

The Navier-Stokes momentum equations are $$\displaystyle
\underline{q}_t +(\underline{q}.\nabla)\underline{q}=
-\frac{1}{\rho}\nabla p+\nu \nabla^2\underline{q}$$

Non-dimensionalising using
\begin{eqnarray*} \underline{q} &amp; = &amp; U_{\infty}\underline{q}'\\ \underline{x}
 &amp; = &amp; L\underline{x}'\\ p &amp; = &amp; \mu \frac{U_{\infty}}{l}p' \end{eqnarray*}

where $U_{\infty}$, $L$ are a typical velocity and length in the flow,
we find that (with $t=(L/U_{\infty}t'))$

(dropping primes) $$\displaystyle
\frac{U_{\infty}^2}{L}(\underline{q}_t+
(\underline{q}.\nabla)\underline{q}) =  -\frac{\mu U_{\infty}}{\rho
L^2}\nabla^2 \underline{q}$$
$\Rightarrow Re(\underline{q}_t+(\underline{q}.\nabla)\underline{q})=
-\nabla p + \nabla^2\underline{q}\ \ \ \ (Re=LU_{\infty}/\nu).$

So for $Re \ll 1$ we get, to lowest order, $$\nabla p = \nabla^2 \underline{q}.$$

$\Rightarrow curl\nabla p = curl\nabla^2\underline{q}= \nabla^2
curl\underline{q} = 0$

Now $curl\underline{q}= \left | \begin{array}{ccc} \underline{i} &amp;
\underline{j} &amp; \underline{k}\\ \partial_x &amp; \partial_y &amp; \partial_z\\
\psi_x &amp; -\psi_y &amp; 0 \end{array}\right | = \left ( \begin{array}{c}
0\\ 0\\ -\psi_{xx}- \psi_{yy} \end{array} \right )$

%\Rightarrow \nabla^2(-\psi_{xx}-\psi_{yy}) = 0$ i.e. $$-\nabl^2(\nabla^2\psi)=0$$
$\Rightarrow$ for slow flow $\nabla^4\psi=0$

This equation contains no time derivatives as all inertia has vanished
to lowest order. In a truly unsteady flow the time derivative would
manifest itself via the boundary conditions.

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

Now we must solve $\nabla^4\psi=0$ ($r\ge 0,\ -\alpha\le\theta\le\alpha$).

Suitable B/C's are no slip:- $$\psi=\psi_{\theta}=0 \ \ \ \rm{on} \
\theta=\pm \alpha$$
[also symmetry conditions could be used]

So try $\psi=r^{\lambda}(A\cos\lambda\theta +B\cos(\lambda-2)\theta)$

$\begin{array}{rcl} \nabla^2\psi &amp; = &amp; \displaystyle \lambda(\lambda
-1)r^{\lambda -2}(A\cos\lambda\theta +B\cos(\lambda -2)\theta\\
&amp; &amp; + \frac{1}{r}\lambda r^{\lambda -1}(A\cos\lambda\theta +B\cos(\lambda
-2)\theta)\\
&amp; &amp; \displaystyle +\frac{r^{\lambda}}{r^2}(-\lambda^2
A\cos\lambda\theta -(\lambda -2)^2B\cos(\lambda -2)\theta )\\ &amp; = &amp;
(\lambda(\lambda -1)+ \lambda) r^{\lambda-2}(A\cos\lambda\theta
-B\cos(\lambda -2)\theta)\\ &amp; &amp; + r^{\lambda-2}(-\lambda^2
A\cos\lambda\theta- (\lambda -2)^2 B\cos(\lambda -2)\theta)\\
&amp; = &amp; B(\lambda^2 -(\lambda -2)^2)\cos(\lambda -2)\theta(r^{\lambda
-2})
\end{array}$

\begin{eqnarray*} \rm{so}\ \ \nabla^4\psi &amp; \propto &amp;  (\lambda -2)(\lambda
-3)r^{\lambda -4}\cos(\lambda -2)\theta +(\lambda -2)r^{\lambda
-4}\cos(\lambda - 2)\\ &amp; &amp; -r^{\lambda -4}(\lambda -2)^2\cos(\lambda
-2)\theta\\ &amp; = &amp; (\lambda -2)(\lambda -3 + 1- \lambda + 2)\\ &amp; = &amp; 0
\end{eqnarray*}

Thus $$\nabla^4 \psi =0$$

B/C's:- ($\psi$ is even, so need only impose at $\theta=+\alpha$)

$\begin{array}{ccc} \psi = 0 &amp; \rm{at} &amp; \theta=\alpha \ :-\\
\psi_{\theta}=0 &amp; \rm{at} &amp; \theta=\alpha \ :- \end{array} \ \
\begin{array}{lcl} 0 &amp; = &amp; A\cos\lambda\alpha + B\cos(\lambda
-2)\alpha\\ 0 &amp; = &amp; A\lambda\sin\lambda\alpha +(\lambda
-2)B\sin(\lambda -2)\alpha \end{array}$

These homgeneous equations have no non-zero solution unless the
determinant of the coefficients is zero.

i.e. need $\left | \begin{array}{cc} \cos\lambda\alpha &amp; \cos(\lambda
-2)\alpha\\ \lambda\sin\lambda\alpha &amp; (\lambda -2)\sin(\lambda
-2)\alpha \end{array} \right | =0$

$\Rightarrow(\cos\lambda\alpha)(\lambda -2)\sin(\lambda -2) -
\cos(\lambda -2)\alpha (\lambda\sin\alpha) =0$
$$\Rightarrow (\lambda-2)\tan(\lambda-2)\alpha =\lambda\tan\lambda\alpha$$


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