Past exam of the mathematics course of the University of Cambridge / 2010 / ii / Paper 4 / 37A (Fluid Dynamics II)

For axisymmetric incompressible flow without swirl, the Stokes streamfunction satisfiesSubstitution into the divergence in spherical coordinates makes incompressibility an identity. Its only vorticity component isTaking the curl of the Stokes equation eliminates pressure and gives ${\nabla }^2\mathbf{\omega }=0$. For an axisymmetric azimuthal vector, its scalar component obeys $(\Delta -1/(R^2{\sin }^2\theta )){\omega }_{\varphi }=0$. Direct differentiation verifiesApply this to $h={\mathcal{D}}^2\Psi$ to obtain

For $\Psi =f(R){\sin }^2\theta$, the angular differentiation givesTrying $f=R^m$ in the fourth-order equation gives the indicial productwhose four distinct roots are $-1,1,2,4$. This proves that the stated four-term expression is the general separated solution of this angular type.

Put $z=b/a$ and $F(s)=A/s+Bs+C{s}^2+D{s}^4$. The velocity formulas become $u_R=2UF(s)\cos \theta /s^2$ and $u_{\theta }=-U{F}^{\prime }(s)\sin \theta /s$. The four boundary conditions therefore give

The $C$ term produces a nondecaying velocity at infinity and the $D$ term a growing one, so decay forces $C=D=0$. The first two boundary equations become $A+B=1/2$ and $-A+B=0$, givingThe remaining velocity terms decay respectively as $R^{-3}$ and $R^{-1}$.

Divide the two outer boundary equations respectively by $z$ and leave the derivative equation as written:Their difference gives $Cz+3D{z}^3=2A/z^2$. As $z\to \infty$, the inner equations require $A,B\to 1/4$. The two outer equations at leading order thus give $Cz+D{z}^3=-1/4$ and $Cz+3D{z}^3=0$, soThe inner equations also give $A+B=1/2-C-D$ and $B-A=-2C-4D$. Solving and substituting yieldsReplacing $z$ by $b/a$ gives all four requested approximations.