Past exam of the mathematical mechanical biology course of the University of Oxford / 2023 / 2 / b

Take $s=\sigma$ up to an irrelevant additive constant, $r=a$ constant, $z=z_0-\sigma$, and $\theta =\pi /2$. These satisfy the geometric relations and give the principal stretchesThe constitutive equations therefore produce constant membrane tensionsThe tangential mechanical equilibrium equations hold automatically, while the normal equation reduces toEvery positive root of this algebraic equation provides a cylindrical solution with no axial stretch ratio change. There is no further constraint on $t_s$ in the interior equations when end boundary conditions are discarded. In particular, a closed-end pressure balance would impose additional axial loading data that are absent here. Existence of a positive root for an arbitrary specified $P$ depends on the given constitutive equation.

Here a prime denotes $d/d\sigma$. It is useful to retain terms through $O({ϵ}^2)$ in the normal mechanical equilibrium equation, even though only first-order shape corrections are sought. The undeformed elastic membrane is tensionless, so a first-order change of curvature contributes to normal force only at second order.

The geometric equations written in reference arc length areTheir first-order Taylor expansions giveand henceThe principal stretches and the first-order constitutive equation areBecause $t_s=t_{\phi }=t$ exactly, the tangential mechanical equilibrium equation gives ${\partial }_{s}t=0$. Thus both $t_1$ and $t_2$ are spatial constants: $t_1^{\prime }=t_2^{\prime }=0$.

Expand the geometric factors in the normal balance:ConsequentlyMatching the first-order coefficient gives the requested relationSince the prescribed pressure has no second-order term, and ${\theta }_1^{\prime }=-r_1^{\prime \prime }$, its second-order coefficient givesA closed boundary value problem for the three requested unknowns is thereforeThe first equation has order two, the second order one, and the last order one, so the four boundary conditions match the total order. The unknown constant $t_2$ is selected by the axial endpoint constraint, rather than by an additional specified traction.

For $p\ne 0$, an equivalent formulation eliminates $z_1$ and displays that constraint as an integral:After solving this system, recover $z_1$ by integration from $z_1(0)=0$. No second derivatives of the unspecified function $f$ are required: they enter the next-order shape corrections through the second-order constitutive equation, whereas the displayed normal balance already determines the leading shape. The case $p=0$ has the undeformed solution $r_1=z_1=t_2=0$.