Past exam of the fluids and waves course of the University of Oxford / 2021 (A10, A15292W1)

Solutions to all three questions of the University of Oxford Part A fluids and waves paper, sat on 24 June 2021. Source: original released PDF, bundle pages 236–238; paper pages 1–3.

The velocity of a fluid element is the time derivative of its position with its Lagrangian coordinates $X,Y$ held fixed. Direct differentiation givesReplacing the Lagrangian coordinates by the current Eulerian coordinates therefore gives the velocity fieldThe distinction between following a fluid element and differentiating at fixed Eulerian coordinates is what connects the given deformation map to this velocity field.

For a planar incompressible flow, the divergence of the velocity field must vanish. HereEquivalently, the Jacobian determinant of the deformation map isso every material area is preserved. However, the out-of-plane vorticity isSince $\Omega >0$, this vorticity never vanishes. The motion is an incompressible flow but is not an irrotational flow; its local spin is clockwise.

Adopt the streamfunction convention $u={\psi }_y$, $v=-{\psi }_x$. Integrating the first component of the velocity field in $y$ gives $\psi =-xy+\frac{1}{2}\Omega e^{-2t}{y}^2+f(x,t)$. The second component then requires $f_x=\Omega e^{2t}x$. Thus, omitting an arbitrary function of time,At a fixed instant, the streamlines are the connected curves $\psi =K$. To classify this quadratic form, introduce the invertible scaled and rotated coordinatesThenFor $\Omega >1$, both coefficients are positive. Every $K>0$ therefore gives an ellipse, and $K=0$ is the stagnation point at the origin. For $0<\Omega <1$, the quadratic form is an indefinite quadratic form: each $K\ne 0$ gives a hyperbola. The level $K=0$ consists of the straight separatriceswhose rays approach or leave the origin; the origin itself is a stagnation point. At the transition $\Omega =1$, $\psi =d^2$, so the nonstationary streamlines are parallel straight lines and $d=0$ is a line of stagnation points. These are instantaneous streamlines; the time dependence means that they need not coincide with pathlines.

Use $w=\varphi +i\psi$ for the complex potential of a two-dimensional potential flow, so its complex velocity is $q=w^{\prime }=u-iv$. The incident uniform flow and line source haveThe coefficient of the complex logarithm is one because a line source of strength $Q$ has complex potential of a two-dimensional potential flow $(Q/(2\pi ))\log (z-b)$, and here $Q=2\pi$. Applying the Milne-Thomson circle theorem to the unit circle givesSince $1/z-b=-b(z-1/b)/z$, locally consistent branches of the complex logarithm giveChanges of branch only alter the constant in a local expression; $w^{\prime }$ is single valued away from its singularities. The added terms represent a two-dimensional potential doublet at the origin, an image line source at $1/b$, and a compensating line sink at the origin. The image line source and line sink have equal and opposite strengths, so no net volume flux enters the cylinder.

On $|z|=1$, the Milne-Thomson circle theorem makes $w=f(z)+\overline{f(z)}$ up to a constant. Consequently the streamfunction is constant there, establishing the no-penetration boundary condition. All line source strengths are real, and the two-dimensional potential doublet has zero circulation, so the circulation around the cylinder is zero. There is no additional line vortex term.

Let $F_x,F_y$ denote the force per unit length exerted by the fluid on the cylinder, and let $\rho$ be the constant mass density. For a counterclockwise contour $C$ around the unit circle, the Blasius force theorem givesAn efficient use of the residue theorem finds the sum of the enclosed residues by subtracting the external pole at $b$ from the sum of all finite residues. At large $z$,The sum of all finite residues of $q^2$ is therefore $2U$. At $b$, write $q=(z-b{)}^{-1}+H(z)$, with $H$ a holomorphic function near $b$. ThenThusUsing $a=1/b$ and $1/(b-a)-1/b=1/[b(b^2-1)]$ givesThe vanishing vertical force also follows from reflection symmetry about the real axis. This force on a unit cylinder due to a uniform flow and an external source uses the counterclockwise orientation in the Blasius force theorem to fix the sign of the horizontal force.

Since $b>1$, the denominator inis positive. The positive root of $U{b}^2-b-U=0$ isThe other root is negative, so the horizontal force points in the positive $x$ direction for $1<b<b_{\ast }$, vanishes at $b=b_{\ast }$, and points in the negative $x$ direction for $b>b_{\ast }$. In particular, the nearby line source pulls the cylinder towards itself as $b\downarrow 1$, while at large $b$ the horizontal force is $F_x\sim -2\pi \rho U/b^2$.

The Joukowski map is a holomorphic function for $z\ne 0$, withA holomorphic function is a local conformal map wherever its derivative is nonzero. Hence the two finite nonzero critical points areThey lie on the boundary of the fluid region. At $z=0$ the displayed formula has a pole and is undefined as a map to the finite complex plane; zero is outside the fluid region. Thus $g$ is locally conformal everywhere in the fluid.

For $z=x+iy$,Both factors are positive in the fluid region, so its image is contained in the upper half-plane. The boundary correspondence is also explicit: on the upper unit semicircle, $z=e^{i\theta }$ gives $g(z)=2\cos \theta \in [-2,2]$, while the real rays $x>1$ and $x<-1$ map onto the rays $\zeta >2$ and $\zeta <-2$.

To prove the image is the whole upper half-plane, fix any $\mathrm{Im}\zeta >0$. The equation $g(z)=\zeta$ isIts roots have product one. Neither root can have unit modulus, since that would make $\zeta =z+1/z$ real. Exactly one root therefore has modulus greater than one. For that root the formula for $\mathrm{Im}g$ forces $y>0$, so it belongs to the fluid region. Moreover, if $g(z_1)=g(z_2)$, thenTwo distinct roots in the fluid region would require $z_1{z}_2=1$, impossible when both moduli exceed one. Thus the Joukowski map of the upper exterior of a unit disk is a bijection from the fluid region onto the upper half-plane.

The quadratic formula gives two algebraic possibilities. To select one consistent holomorphic square root, setwhere each factor is the principal square root. Both arguments lie in the upper half-plane, so neither crosses its branch cut. This choice satisfies $S(\zeta {)}^2={\zeta }^2-4$ and $S(\zeta )\sim \zeta$ at large $\zeta$ in the upper half-plane. The required inverse function isFor $\zeta =iR$, $R>0$, this gives $z=i(R+\sqrt{R^2+4})/2$, whose modulus exceeds one. As $\zeta$ varies continuously in the upper half-plane, this root cannot cross $|z|=1$, since that would make $\zeta$ real. Hence the selected root always lies in the fluid region. The other root is its reciprocal and lies inside the unit disk. Simply writing the principal square root of ${\zeta }^2-4$ would not specify this branch correctly throughout the upper half-plane.

The line vortex is mapped toIts imaginary part is positive because $c$ lies in the fluid region. A conformal map preserves the circulation of the local line vortex, so its strength in the mapped complex plane is still $\Gamma$.

Take positive $\Gamma$ to mean counterclockwise circulation. A line vortex at ${\zeta }_0$ has complex potential of a two-dimensional potential flow $\Gamma \log (\zeta -{\zeta }_0)/(2\pi i)$. The method of images imposes the no-penetration boundary condition on the real axis by placing an image line vortex of strength $-\Gamma$ at ${\bar{\zeta }}_0$. ThusFor real $\zeta$, the numerator and denominator have equal modulus, so the streamfunction isThe real axis is therefore a streamline. The image line vortex lies outside the mapped fluid region, while the complex velocitydecays as $|\zeta |\to \infty$. This selects the flow due to the line vortex and the boundary, without superposing an imposed background velocity field.

The physical complex potential of a two-dimensional potential flow is $w(z)=W(g(z))$. Applying the chain rule gives the complex velocityThe physical velocity is $(\mathrm{Re}{w}^{\prime },-\mathrm{Im}{w}^{\prime })$. A useful equivalent expression follows fromFactoring the image expression in the same way makes the $-1/z$ terms cancel, leavingThis represents the physical line vortex at $c$, an image line vortex of the same strength at $c^{-1}$, and image point vortices of opposite strength at $\bar{c}$ and ${\bar{c}}^{-1}$. All three image locations are outside the fluid region. Both boundary pieces inherit the constant streamfunction of the real axis in the mapped complex plane.

A free point vortex is advected by the regular part of the velocity field at its position. Its own singular complex velocity $\Gamma /[2\pi i(z-c)]$ must first be subtracted. The image expression therefore givesEquivalently, $\dot{c}$ is the complex conjugate of the boxed expression. This is the instantaneous velocity, evaluated with the current value of $c$.

To see why mapping the image velocity alone would miss a term, expand the mapped physical singularity:Thus the same point-vortex drift under a conformal map can be writtenThe finite term involving $g^{\prime \prime }$ is part of the physical regular velocity field.

If $c=id$ with $d>1$, direct substitution into the image expression givesThe bracket is strictly positive, and the result is real. Hence a positive line vortex initially moves horizontally to the right, a negative line vortex moves to the left, and the initial vertical velocity is zero. As $d\to \infty$, the leading term $\Gamma /(4\pi d)$ is the familiar drift of a point vortex above a straight wall.

For an irrotational flow, write $\mathbf{u}=\nabla \varphi$. The incompressible flow condition then gives the Laplace equationAt leading order in a small disturbance this equation holds on $-h<y<0$. Including gravity, the Unsteady Bernoulli equation isChoose the time-dependent additive freedom in the velocity potential so that $B(t)=p_0/\rho$. Neglecting surface tension, pressure continuity at the free surface gives the exact dynamic conditionThe free surface is a material boundary: applying the material derivative to $y-\eta (x,t)=0$ yields the kinematic boundary condition for a free-surface graphFor a small disturbance, the Taylor expansion from $y=\eta$ to $y=0$ changes the first-order velocity potential and its partial derivatives only at second order. Both $|\nabla \varphi {|}^2$ and ${\varphi }_x{\eta }_x$ are also second-order terms. The linearized free-surface boundary conditions are thereforeThe fixed bed has the no-penetration boundary conditionThe required approximation is small surface elevation, small wave steepness, and small velocity perturbations. The stated $p_1\ll p_0$ alone does not ensure this approximation near resonance.

Seek the travelling forced component of the surface elevation in the form $\eta =A\cos \chi$, where $\chi =x-ct$ and $A$ is a signed coefficient. The forcing has wavenumber one. Separation of variables in the Laplace equation, followed by the bed no-penetration boundary condition, givesSince ${\eta }_t=Ac\sin \chi$, the surface kinematic boundary condition requires $B\sinh h=Ac$, soAt $y=0$, ${\varphi }_t=-A{c}^2\coth h\cos \chi$. The dynamic member of the linearized free-surface boundary conditions therefore becomesConsequently the signed response and the nonnegative amplitude areThe travelling-pressure response of finite-depth surface gravity waves is opposite in phase to the imposed pressure below the resonant speed and in phase above it. These expressions describe the travelling forced component; freely propagating surface gravity waves can be superposed to meet additional initial conditions.

The denominator of the travelling response vanishes atFor the positive speeds in the paper, a finite travelling response exists whenever $c\ne c^{\ast }$. At $c=c^{\ast }$, the dynamic condition would reduce to $p_1/\rho =0$, which is impossible. The finite-depth dispersion relation of a surface gravity wave with wavenumber $k$ is ${\omega }^2=gk\tanh (kh)$. At $k=1$, its phase speed $\omega /k$ equals $c^{\ast }$: the moving pressure pattern drives a natural surface gravity wave at its own speed, producing resonance.

The restriction concerns a bounded travelling response. To distinguish this from existence of a solution of an initial value problem, set $T=\tanh h$, $P=p_1/\rho$, and ${\omega }_{\ast }^2=gT$. Ifthe linearized free-surface boundary conditions reduce to the forced harmonic oscillatorFor example, if the forcing is switched on at $t=0$ from a flat surface with zero initial velocity, then, for $c\ne {\omega }_{\ast }$,This satisfies $H(0)=\dot{H}(0)=0$ and contains both free oscillations and the forced component. At $c={\omega }_{\ast }$ a solution of the initial value problem still exists, but its particular solution containswhich grows linearly with time. Thus the ideal inviscid flow model predicts resonant growth rather than a finite constant amplitude. Additional free oscillations can set the initial conditions but cannot remove that secular growth.

For $0<c<c^{\ast }$, the nonnegative amplitude isIts first two derivatives are positive in this interval:The graph is increasing and convex, starts with a horizontal tangent at the limiting value $a(0)=p_1/(\rho g)$, and has a vertical asymptote at $c=c^{\ast }$. Therefore the hydrostatic lower bound for a travelling-pressure response isBecause the paper assumes $c>0$, this is an infimum, approached as $c\downarrow 0$, rather than an attained minimum. If the stationary limit $c=0$ is admitted, equality holds: the hydrostatic equilibrium condition is $\rho g\eta =-p_1\cos x$. Near resonance,The divergence describes the linear model; sufficiently close to $c^{\ast }$ the small-disturbance assumptions fail. The lower bound and sketch refer to the travelling forced component, since adding free oscillations can change the instantaneous surface elevation.