= Solution
{c}
Here a prime denotes $d/d\sigma$. It is useful to retain terms through $O(\epsilon^2)$ 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> are
$$
r'=s'\cos\theta,\qquad z'=-s'\sin\theta.
$$
Their first-order <Taylor expansions> give
$$
\epsilon r_1'=-\epsilon\theta_1+O(\epsilon^2),\qquad
-1+\epsilon z_1'=-1-\epsilon s_1'+O(\epsilon^2),
$$
and hence
$$
\theta_1=-r_1',\qquad s_1'=-z_1'.
$$
The <principal stretches> and the first-order <constitutive equation> are
$$
\lambda_s=1+\epsilon s_1'+O(\epsilon^2),\qquad
\lambda_\varphi=1+\epsilon\frac{r_1}{a_0}+O(\epsilon^2),
\qquad
\boxed{t_1=-Az_1'+\frac B{a_0}r_1.}
$$
Because $t_s=t_\varphi=t$ exactly, the tangential <mechanical equilibrium> equation gives $\partial_s t=0$. Thus both $t_1$ and $t_2$ are spatial constants: $t_1'=t_2'=0$.
Expand the geometric factors in the normal balance:
$$
\frac{\partial\theta}{\partial s}=\epsilon\theta_1'+O(\epsilon^2),\qquad
\frac{\sin\theta}{r}=\frac1{a_0}-\epsilon\frac{r_1}{a_0^2}+O(\epsilon^2).
$$
Consequently
$$
\epsilon p=\epsilon\frac{t_1}{a_0}
+\epsilon^2\left(t_1\theta_1'+\frac{t_2}{a_0}-\frac{t_1r_1}{a_0^2}\right)
+O(\epsilon^3).
$$
Matching the first-order coefficient gives the requested relation
$$
\boxed{p=\frac{t_1}{a_0},\qquad t_1=a_0p.}
$$
Since the prescribed <pressure> has no second-order term, and $\theta_1'=-r_1''$, its second-order coefficient gives
$$
-t_1r_1''+\frac{t_2}{a_0}-\frac{t_1r_1}{a_0^2}=0.
$$
A closed <boundary value problem> for the three requested unknowns is therefore
$$
\boxed{\begin{aligned}
p a_0^2 r_1''+p r_1&=t_2,\\
A z_1'&=\frac B{a_0}r_1-a_0p,\\
t_2'&=0,
\end{aligned}\qquad
r_1(0)=r_1(L)=0,\quad z_1(0)=z_1(L)=0.}
$$
The 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 $p\ne0$, an equivalent formulation eliminates $z_1$ and displays that constraint as an <integral>:
$$
r_1''+\frac{r_1}{a_0^2}=\frac{t_2}{p a_0^2},\qquad
r_1(0)=r_1(L)=0,\qquad
\int_0^L r_1\,d\sigma=\frac{a_0^2pL}{B}.
$$
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 $r_1=z_1=t_2=0$.
Back to article page
