= Solution
Reflection symmetry makes the two outer free tails identical. Their combined energy is
$$
E_o=\frac{k_c\delta_o^2}{\xi R^2}.
$$
Between the cylinders the free profile is even. Up to an irrelevant additive height it has the form
$$
h_i(x)=C\cosh(x/\xi),
\qquad |x|\leq L-\delta_i.
$$
Slope matching at $x=L-\delta_i$ gives
$$
C=\frac{\xi\delta_i}{R\sinh[(L-\delta_i)/\xi]}.
$$
Direct integration, or the boundary expression from part (a), yields
$$
E_i=\frac{k_c\delta_i^2}{\xi R^2}
\coth\frac{L-\delta_i}{\xi}.
$$
At the retained order $\delta_i/\xi\ll1$, replace the argument by $L/\xi$. The four contact halves contribute bending plus adhesion energy
$$
E_c=\frac{k_c}{R^2}(1-2U)(\delta_o+\delta_i).
$$
Thus
$$
E_2=\frac{k_c}{\xi R^2}
\left[\delta_o^2+\delta_i^2\coth(L/\xi)
-2\xi\left(U-\frac12\right)(\delta_o+\delta_i)
\right].
$$
Independent minimization gives
$$
\delta_o=\xi\left(U-\frac12\right),
\qquad
\delta_i=\xi\left(U-\frac12\right)\tanh(L/\xi),
$$
and hence
$$
E_2(L)=-\frac{k_c\xi}{R^2}
\left(U-\frac12\right)^2
\left[1+\tanh(L/\xi)\right].
$$
At infinite separation the energy is twice the one-cylinder minimum. The <membrane-mediated interaction potential> is therefore
$$
\boxed{
V(L)=E_2(L)-E_2(\infty)
=\frac{k_c\xi}{R^2}
\left(U-\frac12\right)^2
\left[1-\tanh(L/\xi)\right]
}.
$$
It is positive and decreases monotonically to zero, so the two cylinders repel. The physical cause is the overlap of their exponentially relaxing membrane deformations.
Back to article page
