Past exam of the perturbation methods and applied complex variables course of the University of Oxford / 2009 / B / 4

Write the complex potential of a two-dimensional potential flow as $w=\varphi +i\psi$, and the complex velocity as $q=w^{\prime }=u-iv$. The lower solid wall has $\psi =0$. The upper solid wall and its continuing free streamline have constant $\psi$ equal to the far-field volume flux per unit depth, namely $a$. Consequently the potential domain is the infinite stripThe points $A,E$ correspond to $\varphi \to -\infty$, $B,C$ to $\varphi \to +\infty$, and $D$ is $w=ia$ by the chosen potential normalization.

In the hodograph plane, the horizontal wall has $q$ real positive, the vertical wall has $q$ imaginary positive because the physical velocity is downward, and the fluid free boundary has $|q|=1$. The upstream reservoir velocity tends to zero, the outgoing velocity is one, and the velocity at $D$ is $i$. Thus the hodograph is the first-quadrant unit quarter disk, with $A,E$ at $q=0$, $B,C$ at $q=1$, and $D$ at $q=i$.

For the potential strip, exponentiation followed by a Möbius transformation givesSince $e^{\pi w/a}$ lies in the upper half-plane, so does $\zeta$. The map sends $A,E$ to $-1$, $D=ia$ to infinity, and $B,C$ to zero.

For the first-quadrant hodograph plane, successively square $q$, apply $(q^2-1)/(q^2+1)$, and then use a negative square:The first map takes the quarter disk into the upper half of the unit disk. The fractional transformation takes this to the second quadrant, whose negative square is the upper half-plane. The boundary points $0,i,1$ map to $-1,\infty ,0$, as required. Uniqueness of the normalized conformal map therefore identifies the two expressions:

On the free streamline choose $q=e^{-i\theta }$ with $-\pi /2\le \theta <0$. The lower endpoint of this parameter interval is $D$, and $\theta \to 0^{-}$ approaches $C$. The fractional expression in the map isHence $e^{\pi w/a}=-{\csc }^2\theta$, and the branch on the upper edge of the potential strip isDifferentiate to obtain $dw/d\theta =-2a\cot \theta /\pi$. Since $dw/dz=q$, the chain rule givesThe sign and interval are consistent with the free surface moving to the right and downward from $D$.

Separate the real and imaginary parts of the derivative and integrate from $D=(0,1)$:Because $\cos s\cot s=\csc s-\sin s$ and $\sin s\cot s=\cos s$, these integrals areAs $\theta \to 0^{-}$, $x\to +\infty$ and $y\to 1-2a/\pi$. This must equal the outgoing stream depth $a$, so

The solution has a vertical tangent at $D$ and approaches the horizontal line $y=a$ downstream, agreeing with both solid-wall and free-streamline conditions.