For an axisymmetric surface $r=a(z,t)$, twice the <mean curvature> with the outward-normal convention is
$$
\kappa=\frac1{a\sqrt{1+a_z^2}}
-\frac{a_{zz}}{(1+a_z^2)^{3/2}}.
$$
Writing $a=a_0+\eta$ and retaining linear terms gives
$$
\boxed{\kappa=\frac1{a_0}
-\frac{\eta}{a_0^2}-\eta_{zz}},
\qquad
\kappa'= -\frac{\eta}{a_0^2}-\eta_{zz}.
$$

At the unperturbed boundary $r=a_0$, the linearized <kinematic boundary condition>, tangential-stress condition, and <Young–Laplace equation> are
$$
u=\eta_t,
\qquad
\sigma_{rz}=\mu(u_z+w_r)=0,
\qquad
p'-2\mu u_r=\gamma\kappa'.
$$
For a normal mode $e^{ikz+st}$ these become $u(a_0)=s\eta$, $\sigma_{rz}(a_0)=0$, and
$$
p'(a_0)-2\mu u_r(a_0)
=\gamma(k^2-a_0^{-2})\eta.
$$

The <Papkovich–Neuber representation> of body-force-free <Stokes flow> is
$$
2\mu\mathbf u=\nabla(\mathbf x\mathbin\cdot\boldsymbol\Phi+\chi)-2\boldsymbol\Phi,
\qquad
p=\nabla\mathbin\cdot\boldsymbol\Phi,
$$
where $\boldsymbol\Phi$ and $\chi$ are harmonic.

Let $x=kr$ and $E=e^{ikz+st}$. Substitution of the given radial vector potential and scalar potential gives
$$
\boxed{
u=\frac{E}{2\mu}
\left[P(xI_1'(x)-I_1(x))+QI_0'(x)\right]},
$$
$$
\boxed{
w=\frac{iE}{2\mu}
\left[PxI_1(x)+QI_0(x)\right]},
\qquad
p=PkI_0(x)E.
$$
Using the <modified Bessel function> identity $(xI_1)'=xI_0$, the tangential stress simplifies to
$$
\boxed{
\sigma_{rz}=ikE
\left[PxI_1'(x)+QI_0'(x)\right]}.
$$
The stress-free condition at $r=a$ therefore yields
$$
\boxed{PxI_1'(x)+QI_0'(x)=0
\quad\text{at }x=ka}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 329 / 3 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-329/3/a/solution