5.3 Model for a Contractile Fiber
217
With the stress fiber taken as incompressible (J = 1), the Lagrange multiplier
p is identically zero because the 1D form for W a yields no transverse stresses. The
Cauchy stress, provided by Eq. (5.11) 2 , is
σ = σ a = λ
∗ ∂W a
∂λ ∗ = 2c a λ
∗
λ
∗
− 1
= 2c a
λ
K
λ
K
− 1
,
(5.15)
and (3.99) gives
P =
J
λ
σ =
2c a
K
λ
K
− 1
.
(5.16)
Substituting this relation and (5.13) into (5.14) yields
u
− α
2 u = 0,
(5.17)
where
α
2
=
kK 2
2c a
.
The solution to the above differential equation can be written in the form
u(X) = C 1 sinh αX + C 2 cosh αX.
Two boundary conditions are needed to determine the constants C 1 and C 2 .
Obviously, the displacement is zero at X = 0. The end X = L encounters
a compressive stress from the nucleus when u is positive. Thus, the boundary
conditions are
u(0) = 0
P (L) = −k n u(L)/A 0 .
(5.18)
To write the second condition in terms of u alone, we substitute Eqs. (5.13)
and (5.16) to obtain
u
(L) = K − 1 − βu(L),
where
β =
k n K 2
2c a A 0
.