Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 1 / 39B (Fluid Dynamics II)

The inner cylinder is $x^2+(y-a{)}^2=a^2$. With $x=r\cos \theta$ and $y=r\sin \theta$, this becomes $r^2-2ar\sin \theta =0$, so away from the touching pointAt this boundary the outward radial vector from the cylinder center is $(\sin 2\theta ,-\cos 2\theta )$, so the counterclockwise unit tangent is $\mathbf{t}=(\cos 2\theta ,\sin 2\theta )$. Since ${\mathbf{e}}_r\cdot \mathbf{t}=\cos \theta$ and ${\mathbf{e}}_{\theta }\cdot \mathbf{t}=\sin \theta$, the no-slip boundary condition givesFor a streamfunction in polar coordinates, no penetration through a stationary material curve makes $\psi$ constant along that curve. Both cylinders touch at the origin, so continuity forces their constants to agree. Since adding a constant does not alter velocity, choose their common value as $\boxed{\psi =0}$.

The outer circle obeys $x^2+(y-2a{)}^2=(2a{)}^2$, and henceIts counterclockwise tangent is again $(\cos 2\theta ,\sin 2\theta )$. Its angular velocity is $-\Omega /4$ and radius is $2a$, so its signed tangential speed is $-a\Omega /2$. The boundary conditions areThe first follows both from impermeability and from the common streamfunction value forced by the contact point.

With $B=0$, the stated biharmonic stream function for planar Stokes flow isRequiring it to vanish at $r=2a\sin \theta$ and $r=4a\sin \theta$ givesThus $A=-6aC$ and $D=8a^{2}C$. Writing $C=\alpha$ yieldsUsing $u_r={\psi }_{\theta }/r$ and $u_{\theta }=-{\psi }_r$, differentiation shows that $u_r\cos \theta +u_{\theta }\sin \theta$ equals $\alpha$ on the inner boundary and $-\alpha /2$ on the outer boundary. Both wall conditions hold precisely whenThis is the Stokes flow between touching counter-rotating cylinders.

At an interior stagnation point, ${\psi }_r={\psi }_{\theta }=0$. The first equation isSince $\sin \theta >0$ inside the fluid, $r=2\sqrt{2}a\sin \theta$. AlsoOn the preceding locus the bracket is nonzero, so $\cos \theta =0$. ThereforeFor $\Omega >0$, $\psi <0$ in the interior, and its nested nonzero level curves are closed recirculating streamlines around this unique stagnation point.