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The biharmonic equation is ${\nabla }^{4}f=0$, where the biharmonic operator is the square of the Laplacian. In planar Stokes flow, taking the curl of the momentum equation and using a Cartesian streamfunction produces this equation. Its decaying spatial Fourier series modes have form $(A+By)e^{-|k|y}{e}^{ikx}$.

If $H,G$ are holomorphic functions, both real and imaginary parts of $\bar{z}H(z)+G(z)$ solve the biharmonic equation. In Wirtinger derivatives, $\Delta =4{\partial }_z{\partial }_{\bar{z}}$, so $\Delta (\bar{z}H+G)=4H^{\prime }$ is harmonic and its next Laplacian is zero. This construction gives local representations of planar solutions of the biharmonic equation; it is useful for the biharmonic stream function for planar Stokes flow.

For a stationary cylinder of radius $a$ in the far-field shear $(\Gamma y,0)$, the stream function is $\psi =(\Gamma /4)[r^2-2a^2\log (r/a)-a^2-(r^2-2a^2+a^4/r^2)\cos 2\phi ]$. It solves the biharmonic equation and the no-slip boundary condition. The fluid exerts torque $-2\pi \mu \Gamma a^2$ per unit length, because the angular part of the surface traction averages to zero and the mean tangential traction is $-\mu \Gamma$.

The undisturbed ambient solution is smooth and free of body forces throughout the ball $r\le a$. Its Stokes equations implyTaking the divergence gives ${\nabla }^2{p}_{\infty }=0$, and a further Laplacian therefore givesEach velocity component obeys the biharmonic equation in the whole ball.

Expand a component of ${\mathbf{u}}_{\infty }$ at the centre and average over $\mathbf{x}=a\mathbf{n}$. The odd spherical moments vanish, whileThus the spherical mean of a biharmonic function givesThe truncation is exact, not just a small-sphere approximation: all terms containing two or more Laplacians vanish. One can avoid any Taylor-convergence issue by letting $M(r)=⟨{\mathbf{u}}_{\infty }(r\mathbf{n})⟩$. Since ${\nabla }^2{\mathbf{u}}_{\infty }$ is harmonic, its spherical mean is ${\nabla }^2{\mathbf{u}}_{\infty }(0)$, soIntegrating gives $M(r)={\mathbf{u}}_{\infty }(0)+r^2{\nabla }^2{\mathbf{u}}_{\infty }(0)/6$ for every admissible $r$.

Combining with the force-free result of part c proves Faxén's first law in the requested form:The force-free qualification carries over. Local absence of ambient body forces and regularity inside the ball are also needed; if forcing or a singularity lies inside it, the harmonic-pressure and biharmonic-mean argument does not apply unchanged.

Source paper: www.whatdotheyknow.com/request/oxford_physics_past_papers_2/response/2340696/attach/9/Physics%20Papers%202020.pdf?cookie_passthrough=1

The cover dates this paper to 11 June 2021 despite the attachment filename. Solutions identify a false printed biharmonic equation identity, retain the stipulated drop curvature approximation explicitly, and state the conditions needed for the final Joukowski map.

Write ${\omega }_h=\dot{\alpha }$, positive on opening. The no-slip boundary condition equates the fluid velocity with each plate's velocity:The upper condition is a moving-wall no-penetration boundary condition, not zero normal velocity in the laboratory frame. At the hinge the velocity tends to zero. With the arbitrary constant of the streamfunction chosen so that $\psi (r,0)=0$, these conditions becomeA changing angle cannot give a literally steady flow in fixed coordinates. We interpret the stated steadiness as the quasi-steady Stokes approximation at each instantaneous $\alpha$.

The printed expression is not biharmonic. To test it without reproducing the source formula, write its bracket as $G_p(\theta )$ and its streamfunction as ${\psi }_p=-\Omega r^2{G}_p/2$. For any angular function $G$, the Laplacian givesAt the concrete angle $\alpha =\pi /4$, differentiation of the printed bracket givesFor example, at the interior point $\theta =\pi /8$,The printed bracket does satisfy $G_p(0)=G_p^{\prime }(0)=G_p^{\prime }(\alpha )=0$ and $G_p(\alpha )=1$, so its boundary velocities would require $\Omega ={\omega }_h$. Satisfying these boundary conditions does not cure its nonzero biharmonic operator. The stated series is likewise not a general biharmonic basis: its $r^2\theta \sin 2\theta$ term already fails the displayed test. The separated polar solutions of the biharmonic equation contain additional radial powers; for the required $r^2$ dependence, the angular basis is $1,\theta ,\sin 2\theta ,\cos 2\theta$.

The corrected slowly opening Stokes wedge similarity solution follows by setting $\psi =-{\omega }_h{r}^{2}G(\theta )/2$. The biharmonic equation gives $G^{(4)}+4G^{\prime \prime }=0$. The two lower-wall conditions reduce its general solution toThe upper-wall condition $G^{\prime }(\alpha )=0$ fixes the ratio of $A$ and $B$, and $G(\alpha )=1$ fixes their scale. An expression that also remains finite at $\alpha =\pi /2$ isFor $0<\alpha <\pi$, its denominator is positive: the derivative of $\sin \alpha -\alpha \cos \alpha$ is $\alpha \sin \alpha >0$. Direct differentiation verifies all four boundary conditions and the biharmonic equation. The resulting velocity components are $v_r=-\dot{\alpha }r{G}^{\prime }/2$ and $v_{\theta }=\dot{\alpha }rG$.

For a fixed angle of order one, $U(r)\sim |\dot{\alpha }|r$, so the local Reynolds number is ${\mathrm{Re}}_r\sim |\dot{\alpha }|r^2/\nu$. One breakdown length is thereforeA second is the geometric length $l_2\sim \min (L,W)$, where $L$ is the distance to the radial ends and $W$ the spanwise width. At $r\sim l_2$, matching to the surrounding bath or three-dimensional end flow matters. Thus $r\ll \min (l_1,l_2)$ is required for this local two-dimensional Stokes flow, with angle-dependent factors in the inertial estimate. Ideal infinite plates remove the geometric cutoff but not the inertial one.

There is also a temporal restriction: the relevant viscous diffusion time must be short compared with the time $T_s$ over which the shape and its velocity field change appreciably. Using $r$ as a conservative diffusion length gives $r\ll \sqrt{\nu T_s}$. For order-one angles and constant $\dot{\alpha }$, $T_s\sim |\dot{\alpha }{|}^{-1}$ gives the same scale as $l_1$; narrow angles require using their actual gap and angular factors rather than extrapolating this order-one-angle estimate.

Given an admissible velocity field, recover the pressure gradient from the Stokes equation, $\nabla p=\mu {\nabla }^2\mathbf{v}$, and integrate it, fixing the additive constant by matching to the bath pressure near the open ends. The biharmonic equation ensures the compatibility of this integration. For a freely moving hinge, the laboratory velocity field also contains the unknown rigid translation and rotation. Solve the interior and exterior Stokes flows with these wall velocities, then impose zero total external force and torque on the complete apparatus if its motor supplies only internal actuation. The pinned solution alone cannot determine that free motion.

At a fixed angle, the actuation velocity scale is $U\sim L|\dot{\alpha }|$. The Stokes equation gives $\Delta p\sim \mu U/L\sim \mu |\dot{\alpha }|$. Integrating normal traction over a plate of length $L$ gives a force per unit width of scalewith a dimensionless signed coefficient depending on geometry and exterior matching. Normal viscous stress has the same dimensional scale and is included in $C$. The force reverses when the actuation reverses. A narrow gap can make $|C|$ large.

More specifically, the corrected local streamfunction gives, up to a reference pressure,Indeed, inserting $\mathbf{v}=(-\dot{\alpha }r{G}^{\prime }/2,\dot{\alpha }rG)$ into the vector Laplacian gives $p_r=-\mu \dot{\alpha }(G^{\prime \prime \prime }+4G^{\prime })/(2r)=\mu \dot{\alpha }K/r$ and $p_{\theta }=0$. For $\alpha <\pi /2$, opening produces reduced pressure toward the hinge when matched at $r_{\ast }\sim L$. The logarithmic singularity is integrable: ${\int }_0^L\log (L/r)dr=L$, consistent with the stated force scale. A real hinge regularizes the singularity; the local formula is not a full end-matched solution.

The scallop theorem applies to a single periodically actuated hinge in a stationary Newtonian fluid with negligible fluid and body inertia. At each shape, Linearity of Stokes flow makes its rigid velocity proportional to $\dot{\alpha }$. Closing retraces the sequence of shapes and reverses the corresponding motion, irrespective of different opening and closing rates. The hinge may move during each half-cycle, butA motor can generate instantaneous forces and consume energy through viscous dissipation without producing sustained swimming. Additional independently controlled shape changes, appreciable inertia, or a fluid with memory would require a different analysis.

Write $P(x)$ for the outer pressure extrapolated to the wall. The normal balance in the thin boundary layer is $p_y={\overline{F}}_y$, givingThe spatially varying magnetic pressure deficit drives a tangential pressure gradient. For fixed forcing as $\omega \to \infty$, layer inertia is small compared with viscous dissipation; explicitly the ratio is $U{\delta }^2/(\nu b)$. The leading tangential balance isThe outer gradient $P^{\prime }$ contributes only at higher inner order. Matching the leading outer shear gives $u_y\to 0$, and the physical no-slip boundary condition gives $u(x,0)=0$. Two integrations yield the magnetic skin-layer streaming slip:For the present field amplitude,Using incompressibility and $v(x,0)=0$ also givesHence $v/u=O(\delta /b)$ in the layer, except at symmetry zeros where a componentwise ratio is inappropriate. In the matching region $\delta \ll y\ll b$, $u\sim U$ and $v\sim -y{U}^{\prime }$, up to the smaller displacement term $\delta U^{\prime }/2$.

The effective outer boundary conditions are tangential slip $u_{\mathrm{bulk}}(x,0)=U(x)$ and zero leading normal velocity $v_{\mathrm{bulk}}(x,0)=0$. These describe the bulk flow extrapolated through the unresolved magnetic skin layer; the actual wall remains at rest. For $x>0$ the slip is negative, and for $x<0$ it is positive. Fluid converges along the wall toward the wire, turns upward near $x=0$, and returns outward farther above the wall.

The following original streamline sketch uses the small-Reynolds number bulk Stokes flow for this slip. It illustrates that circulation without claiming that the unspecified bulk Reynolds number fixes a unique complete flow. With $X=x/b$, $Y=y/b$, $Z=X+iY$, a dimensionless stream function isIts factor in brackets is holomorphic in $Y>0$, so $\Psi$ solves the biharmonic equation; its wall derivative is ${\Psi }_Y(X,0)=-X/(1+X^2{)}^3$. Thus it satisfies both effective wall conditions and gives an exact creeping-flow illustration of the derived slip.

Incompressibility gives $r^{-1}(ru{)}_r+r^{-1}{v}_{\varphi }=0$, permitting a stream function with $u={\psi }_{\varphi }/r$, $v=-{\psi }_r$. Its axial vorticity is $-\Delta \psi$. Taking the curl of the Stokes flow momentum equation eliminates pressure and gives $\Delta$ of that vorticity zero, hence ${\Delta }^2\psi =0$.

At an impermeable constant-angle plane, $v=0$ means $\psi$ is constant along that boundary. Fix that constant to zero. A rigid boundary also requires ${\psi }_{\varphi }=0$. The tangential stress is $\mu (-{\psi }_{rr}+{\psi }_r/r+{\psi }_{\varphi \varphi }/r^2)$; along an impermeable boundary its first two terms vanish, so a stress-free boundary requires ${\psi }_{\varphi \varphi }=0$.

Write $\psi =(S/\mu )r^{2}F(\varphi )$. The biharmonic equation becomes $F^{\prime \prime \prime \prime }+4F^{\prime \prime }=0$. The boundary conditions are $F(-\alpha )=F^{\prime }(-\alpha )=F(0)=0$ and $F^{\prime \prime }(0)=1$. ConsequentlyThe requested $f$ is $F/\mu$. Evaluating $u(r,0)=(Sr/\mu )F^{\prime }(0)$ givesIts small-angle ratio is $\alpha /4+O({\alpha }^3)$, so $U$ and $S$ have the same sign for sufficiently small positive angle. At the first nonzero critical root, $2{\alpha }_c\simeq 4.493409$, the denominator vanishes while the numerator does not. The pure $r^2$ similarity solution is then resonant and cannot satisfy the forcing; logarithmic radial terms or additional global boundary data are needed. Just beyond this angle, and up to $\alpha <\pi$, the denominator is negative and numerator positive, so the formal surface flow reverses relative to $S$. This is not a contradiction of viscous dissipation: the wedge is radially unbounded and its radial-end energy flux has not been prescribed. A physical finite-domain flow needs those outer conditions, and the Stokes approximation also fails sufficiently far out when the velocity grows with $r$.

Use the Cartesian streamfunction convention $u_x={\partial }_y\Psi$, $u_y=-{\partial }_x\Psi$. In the creeping-flow regime implicit in the question's biharmonic equation hint, the Stokes equations are $\mu {\nabla }^2\mathbf{u}=\nabla p$, $\nabla \cdot \mathbf{u}=0$. Taking their curl eliminates pressure and givesThe fluid occupies the region above the actual sheet; the perturbation calculation expands its boundary about $y=0$. In the swimming frame the material velocity is $(0,-\omega y_0\cos (kx-\omega t))$. The vertical kinematic condition and the supplied Navier slip boundary condition therefore give, at $y=y_0\sin (kx-\omega t)$,At infinity, ${\Psi }_y\to U$ and ${\Psi }_x\to 0$, with bounded velocity and no imposed shear; all fields are periodic in $x$ with period $2\pi /k$. The sign is important: a sheet translating at $-U{\mathbf{e}}_x$ sees the remote fluid translating at $+U{\mathbf{e}}_x$.

Set $X=kx$, $Y=ky$, $\tau =\omega t$, $\psi =k^2\Psi /\omega$, and $U^{\ast }=kU/\omega$. The two dimensionless parameters are the small-slope approximation parameter $ϵ=k{y}_0$ and dimensionless slip length $\delta =k\gamma$. With $\theta =X-\tau$, the dimensionless biharmonic stream function for planar Stokes flow satisfiesExpand $\psi =ϵ{\psi }_1+{ϵ}^2{\psi }_2+\cdots$ and $U^{\ast }=ϵU_1+{ϵ}^2{U}_2+\cdots$. At first order,The decaying first harmonic of the biharmonic equation has the form $(A+BY)e^{-Y}\sin \theta$. The vertical condition fixes $A=1$ and excludes a cosine component. The horizontal condition givesThus $(1+2\delta )(B-1)=0$ and $B=1$ for every $\delta \ge 0$. The spatially averaged first-order streamfunction is affine in $Y$; its Navier slip boundary condition forces $U_1=0$. ThereforeThis first-order slip independence of a transverse sheet is independent of slip length. Its surface tangential velocity and surface shear are both zero: $({\psi }_1{)}_Y(0)=0$ and $({\psi }_1{)}_{YY}(0)-({\psi }_1{)}_{XX}(0)=0$. This explains why the first-order streamfunction agrees with the no-slip boundary condition.

To find the slip-enhanced swimming speed of a transverse sheet, expand only the Navier slip boundary condition through second order. Put $Q_j=({\psi }_j{)}_{YY}-({\psi }_j{)}_{XX}$. Taylor expansion at the displaced boundary yieldsDirect differentiation givesHenceAverage over $X$. The zero Fourier series mode of a biharmonic streamfunction is a cubic polynomial in $Y$; bounded velocity at infinity removes the quadratic and cubic terms. Thus its velocity is the constant $U_2$, and its averaged shear $⟨Q_2{⟩}_X$ is zero. The mean boundary velocity determines Taylor-sheet swimming speed principle then givesThe even remainder follows because changing the sign of the amplitude is equivalent to a half-period phase shift. In dimensional form,For the Navier-slip Taylor swimming sheet, the ratio of speed to the no-slip value is $1+2\delta >1$ whenever $\delta >0$: slip increases the swimming speed. The expansion is for fixed $\delta$ as $ϵ\to 0$; it is not a uniform prediction of arbitrarily large speed if the slip length is allowed to diverge with the inverse amplitude.

True. Two applications of Green's identities, or two integrations by parts, giveThe biharmonic equation makes the left side zero, while the two clamped boundary conditions $u={\partial }_{n}u=0$ annihilate the boundary integral. Consequentlyso $\Delta u=0$. Apply the first of Green's identities once more with $\varphi =\psi =u$: because $u=0$ on the boundary, it gives ${\int }_V|\nabla u{|}^2=0$. By path connectedness, $u$ is constant, and the zero boundary value implies $\boxed{u=0}$. This argument assumes the classical differentiability needed for the displayed integrations by parts.

Apply the Laplacian in polar coordinates twice to the proposed separated function. For any real exponent $m$,so the biharmonic equation becomesIts four distinct roots are $m=-2,0,2,4$. The separated polar solutions of the biharmonic equation therefore give the complete radial family in this angular sector,Equivalently, its radial equation is a fourth-order Cauchy-Euler equation; four distinct powers supply four linearly independent solutions, so no logarithmic terms are required.

Interior. Continuity with value zero at the origin removes both $r^{-2}\cos 2\theta$ and $\cos 2\theta$. The latter depends on the approach direction even though it is bounded. The two boundary conditions at unit radius then give $C+D=0$ and $2C+4D=1$. ConsequentlyThis extends smoothly through the origin: $r^2\cos 2\theta =x^2-y^2$ and $r^4\cos 2\theta =(x^2+y^2)(x^2-y^2)$ are polynomials. Its radial derivative at $r=1$ is $(2r^3-r)\cos 2\theta =\cos 2\theta$.

Exterior. The condition on the radial derivative at infinity excludes the $r^2$ and $r^4$ terms. A nonzero $B\cos 2\theta$ is permitted, since the required decay concerns the radial derivative. Now $A+B=0$ and $-2A=1$, givingIndeed ${\partial }_r{\psi }_{\mathrm{out}}=r^{-3}\cos 2\theta \to 0$, and its value at unit radius is zero. These are particular interior and exterior solutions for the biharmonic streamfunction; no condition requiring $\psi$ itself to vanish at infinity was imposed.

For angular Fourier modes of integer order $n\ge 2$, the biharmonic equation in polar coordinates has radial basis $r^n,r^{-n},r^{n+2},r^{2-n}$. Apply the Laplacian twice to obtain the displayed indicial equation. For $n=0$, repeated roots give the radial basis $1,\log r,r^2,r^2\log r$; for $n=1$, the basis is $r^{-1},r,r\log r,r^3$. These are Cauchy-Euler equations. Regularity at the origin and boundary conditions select the admissible powers. For $n=2$, a constant radial factor still multiplies $\cos 2\theta$ or $\sin 2\theta$, so it is bounded but fails continuity at the origin.

For a fixed wall at $\theta =0$ and a wall rotating at $\dot{\alpha }$ at $\theta =\alpha$, the quasi-steady Stokes approximation with the no-slip boundary condition givesIndeed the biharmonic equation for $r^{2}G$ is $G^{(4)}+4G^{\prime \prime }=0$; the angular basis is $1,\theta ,\sin 2\theta ,\cos 2\theta$. The four wall conditions are $G(0)=G^{\prime }(0)=G^{\prime }(\alpha )=0$, $G(\alpha )=1$. The outward radial volume flux per unit width is $-r^2\dot{\alpha }/2$, precisely the negative rate of increase of wedge area. The associated local pressure is $p=\mu \dot{\alpha }[2\cos \alpha /(\sin \alpha -\alpha \cos \alpha )]\log (r/r_{\ast })+C$. Its logarithmic singularity is integrable, although the ideal hinge requires a microscopic regularization.