Past exam of the options II course of the University of Oxford / 2009 / K (Fluid Dynamics and Waves)

Orient the body boundary counterclockwise. Its outward body normal satisfies $nds=(dy,-dx)$, so the pressure force has complex formBy steady Bernoulli's equation, $p=p_{\infty }+\rho (U_{\infty }^2-|q{|}^2)/2$, where $q=w^{\prime }=u-iv$. Constant terms integrate to zero around a closed curve. On an impermeable boundary, velocity is tangent, so $qdz=\overline{q}d\overline{z}$. Therefore $|q{|}^{2}d\overline{z}=q^{2}dz$, yielding the Blasius force theoremFor a slit body, the contour encloses both faces and the tips, interpreted by a limiting exterior contour.

Choose the analytic branch $s(z)=\sqrt{z^2-a^2}$ outside the slit with $s(z)\sim z$ at infinity. It may be written $\sqrt{z-a}\sqrt{z+a}$ using the principal individual square roots. For $-a<x<a$, approaching from above gives $s_{+}(x)=i\sqrt{a^2-x^2}$, and from below gives $s_{-}(x)=-i\sqrt{a^2-x^2}$.

NowThere is no $z^{-1}$ term in $q^2$. Since $q^2$ is analytic in the exterior, enlarging the contour to a large circle shows its integral is zero. ThusThis total-force argument retains the slit-tip contribution automatically.

On the upper face, $q_{+}(x)=\cos \alpha -x\sin \alpha /\sqrt{a^2-x^2}$; on the lower face, $q_{-}(x)=\cos \alpha +x\sin \alpha /\sqrt{a^2-x^2}$. For $0<\alpha \le \pi /2$, their zeros areThey are distinct face points even at $\alpha =\pi /2$, when both have coordinate zero. At $\alpha =0$, $w=z$ and $q=1$ everywhere in the exterior: there are no velocity-zero stagnation points. The limiting positions at the two tips must not be substituted into a singular expression and treated as zeros.

For the sketch at $\alpha =\pi /4$, the far-field velocity is $(1,1)/\sqrt{2}$, and streamlines are levels of $\mathrm{Im}w$. The plate is the zero-stream-function boundary, with the two stagnation points at $\pm a/\sqrt{2}$ on the indicated faces.

For a short channel slice, the fluid volume per unit length is $b(H+\eta )$ and the axial volume flux is $b(H+\eta )u$. The quasi-one-dimensional conservation of mass equation is thereforeNeglecting products of perturbations gives ${\eta }_t+(H/b)(bu{)}_x=0$. This uses the section-averaged approximation stipulated in the problem; small transverse velocities ensure impermeability of varying-width walls but are not leading axial terms.

The Euler equations are ${\partial }_t\mathbf{u}+(\mathbf{u}\cdot \nabla )\mathbf{u}=-{\rho }^{-1}\nabla p-g{\mathbf{e}}_z$. For irrotational flow, $\mathbf{u}=\nabla \varphi$ and $(\mathbf{u}\cdot \nabla )\mathbf{u}=\nabla (|\nabla \varphi {|}^2/2)$, soAt the zero-pressure free surface, linearization about rest gives ${\varphi }_t+g\eta =C(t)$; differentiate in $x$ to get $u_t+g{\eta }_x=0$. Differentiating the mass equation in time and substituting givesPut $c_0=\sqrt{gH}$. For the exponential width, a time-harmonic amplitude satisfiesWriting $N=e^{-\mu x/2}V$ removes the first derivative and gives $V^{\prime \prime }+k^{2}V=0$, where $k^2={\omega }^2/c_0^2-{\mu }^2/4$. A nonzero real wavenumber, and hence a spatially travelling phase, exists precisely whenFor this case an outward-going solution matching the oscillating boundary isSubstitution in both linear equations verifies it, and $\mathrm{Im}(U(0)e^{i\omega t})=u_0\sin \omega t$. Below the cutoff the spatial exponents are real, giving evanescent responses; at equality the repeated-root solution has no travelling spatial phase.

This is the cutoff frequency in an exponentially widening shallow-water channel.