350
7 Remodeling
from t = 0 (with k n − = 0.5 day −1 omitted for clarity). The red curves show results
for incremental turnover, as computed using the discrete form of Eq. (7.14) given by
J
n (t) = J
n
0 e
−k n − t
+
i
k
n + J
n
0 e
−k n − (t−τ i ) H (t − τ i ) )τ,
(7.16)
where the τ i are deposition times, and H (x) is the Heaviside step function defined
in Eq. (6.122). For illustration, we take τ = τ i+1 − τ i = 0.1 day. At each
t = τ i , an increment is added to J n that then decays exponentially in time. For
each value of k n − , the discrete solution (red curve) generally follows the continuous
solution (blue curve). Of course, the discrete curves approach the continuous curves
as τ decreases. For k n − = k n + = 1 day −1 (or k n − = k n + in general), the decay in
J n cancels the incremental increase, keeping J n unchanged. For k n − = 2 day −1
(or k n − > k n + in general), the decay is greater than the added increment, and J n
decreases.
The dashed curves at the bottom of Fig. 7.4b show the individual contributions
to the total J n . For k n − = 2 day −1 , the decay of the initial constituent volume
(green curve) and the incremental deposition for every other τ i are shown. The total
response (red staggered curve) is the sum of these individual curves. These results
indicate that the overall decrease in J n is caused mainly by degradation of the initial
volume, while turnover continually adds volume to the bar, eventually establishing a
new homeostatic condition at a smaller volume.
7.3.2 Kinematics of Turnover
The preceding analysis of turnover-induced changes in volume does not consider
the mechanical effects of G&R. Even in 1D, the full Humphrey-Rajagopal theory
can be quite complicated, as it generally involves multiple fiber families undergoing
turnover at different rates. Understanding the kinematics of deformation and growth
for fiber families undergoing turnover is essential to understanding the theory.
Elastic Deformation Consider isovolumic deformation of a bar composed entirely
of elastic fibers (Fig. 7.5). Four instants of time are shown. At t = 0, the bar is in a
state of homeostatic equilibrium with all fibers having stretch ratio λ 0 = λ 0 relative
to their ZSS. Fiber 0 has just been created, stretched by λ 0 , and deposited in the
bar, which also is subjected to stretch λ 0 . The homeostatic state is chosen as the
reference configuration for all stretch ratios except λ 0 .
At t = τ 1 , the bar is stretched further by λ 1 relative to the reference state. Since
fiber 0 is attached to the surrounding fibers and deforms with the bar, its total stretch
ratio is now λ 0 = λ 0 λ 1 , while fiber 1 is freshly laid down with stretch ratio λ 1 = λ 0 .
Similarly, fibers 2 and 3 are added with stretch λ 0 at times τ 2 and τ 3 , respectively,
as the bar continues stretch to λ 2 and then λ 3 . Following the deformation of the bar,
7 Remodeling
from t = 0 (with k n − = 0.5 day −1 omitted for clarity). The red curves show results
for incremental turnover, as computed using the discrete form of Eq. (7.14) given by
J
n (t) = J
n
0 e
−k n − t
+
i
k
n + J
n
0 e
−k n − (t−τ i ) H (t − τ i ) )τ,
(7.16)
where the τ i are deposition times, and H (x) is the Heaviside step function defined
in Eq. (6.122). For illustration, we take τ = τ i+1 − τ i = 0.1 day. At each
t = τ i , an increment is added to J n that then decays exponentially in time. For
each value of k n − , the discrete solution (red curve) generally follows the continuous
solution (blue curve). Of course, the discrete curves approach the continuous curves
as τ decreases. For k n − = k n + = 1 day −1 (or k n − = k n + in general), the decay in
J n cancels the incremental increase, keeping J n unchanged. For k n − = 2 day −1
(or k n − > k n + in general), the decay is greater than the added increment, and J n
decreases.
The dashed curves at the bottom of Fig. 7.4b show the individual contributions
to the total J n . For k n − = 2 day −1 , the decay of the initial constituent volume
(green curve) and the incremental deposition for every other τ i are shown. The total
response (red staggered curve) is the sum of these individual curves. These results
indicate that the overall decrease in J n is caused mainly by degradation of the initial
volume, while turnover continually adds volume to the bar, eventually establishing a
new homeostatic condition at a smaller volume.
7.3.2 Kinematics of Turnover
The preceding analysis of turnover-induced changes in volume does not consider
the mechanical effects of G&R. Even in 1D, the full Humphrey-Rajagopal theory
can be quite complicated, as it generally involves multiple fiber families undergoing
turnover at different rates. Understanding the kinematics of deformation and growth
for fiber families undergoing turnover is essential to understanding the theory.
Elastic Deformation Consider isovolumic deformation of a bar composed entirely
of elastic fibers (Fig. 7.5). Four instants of time are shown. At t = 0, the bar is in a
state of homeostatic equilibrium with all fibers having stretch ratio λ 0 = λ 0 relative
to their ZSS. Fiber 0 has just been created, stretched by λ 0 , and deposited in the
bar, which also is subjected to stretch λ 0 . The homeostatic state is chosen as the
reference configuration for all stretch ratios except λ 0 .
At t = τ 1 , the bar is stretched further by λ 1 relative to the reference state. Since
fiber 0 is attached to the surrounding fibers and deforms with the bar, its total stretch
ratio is now λ 0 = λ 0 λ 1 , while fiber 1 is freshly laid down with stretch ratio λ 1 = λ 0 .
Similarly, fibers 2 and 3 are added with stretch λ 0 at times τ 2 and τ 3 , respectively,
as the bar continues stretch to λ 2 and then λ 3 . Following the deformation of the bar,