= Solution
The <higher-order Euler-Lagrange equation> for a functional depending on $h_x$ and $h_{xx}$ gives
$$
k_ch_{xxxx}-\sigma h_{xx}=0.
$$
With the <membrane elastic length>
$$
\xi=\sqrt{\frac{k_c}{\sigma}},
$$
the general free-membrane profile is
$$
h(x)=A+Bx+Ce^{x/\xi}+De^{-x/\xi}.
$$
Two integrations by parts, followed by use of the field equation, turn the energy of any free interval $[a,b]$ into the <boundary term>
$$
E_m[a,b]=\frac12
\left[k_c(h_{xx}h_x-h_{xxx}h)+\sigma hh_x\right]_a^b.
$$
Take the undeformed membrane to have $h,h_x\to0$ at infinity. In the small-contact approximation, the cylindrical profile is
$$
h_c(x)=h_c(0)-\frac{x^2}{2R},
\qquad h_c'(\delta_o)=-\frac{\delta_o}{R}.
$$
Continuity of height and slope at the right contact point and exponential decay then give
$$
h(x)=\frac{\xi\delta_o}{R}e^{-(x-\delta_o)/\xi},
\qquad x\geq\delta_o,
$$
with its reflected copy on the left. The two free tails carry
$$
E_{\rm out}=2\cdot\frac12\int_{\delta_o}^{\infty}
\left(k_ch_{xx}^2+\sigma h_x^2\right)dx
=\frac{k_c\delta_o^2}{\xi R^2}.
$$
Within the contact, $h_{xx}=-1/R$. Since $\delta_o\ll\xi$, the tension contribution there is smaller than the bending contribution by $O(\delta_o^2/\xi^2)$, so
$$
E_{\rm contact}=\frac{k_c\delta_o}{R^2},
\qquad
E_a=-2\mathcal U\delta_o.
$$
Writing the <dimensionless adhesion strength> as $U=R^2\mathcal U/k_c$, the total energy is
$$
E(\delta_o)=\frac{k_c}{R^2}
\left[\frac{\delta_o^2}{\xi}+(1-2U)\delta_o\right].
$$
Its stationary point is
$$
\boxed{\delta_o=\xi\left(U-\frac12\right)}.
$$
It is an admissible bound state only for $U>1/2$; otherwise the constrained minimum is the unbound state $\delta_o=0$. Substitution gives
$$
\boxed{E_{\min}=-\frac{k_c\xi}{R^2}
\left(U-\frac12\right)^2}.
$$
Back to article page
