# Force generation by a cylindrical cell under stationary osmolytes synthesis

Wei-Yuan Kong  Laboratoire Matière et Systèmes Complexes, Université Paris
Diderot CNRS UMR 7057, 10 Rue Alice Domont et Léonie Ducquet, 75205 Paris
Cedex 13, France  Antonio Mosciatti Jofré  Laboratoire Matière et Systèmes
Complexes, Université Paris Diderot CNRS UMR 7057, 10 Rue Alice Domont et
Léonie Ducquet, 75205 Paris Cedex 13, France  Manon Quiros  PMMH, CNRS, ESPCI
Paris, Université PSL, Sorbonne Université, Université de Paris, F-75005,
Paris, France  Marie-Béatrice Bogeat-Triboulot  Université de Lorraine,
AgroParisTech, INRAE, UMR Silva, 54000 Nancy, France  Evelyne Kolb  PMMH,
CNRS, ESPCI Paris, Université PSL, Sorbonne Université, Université de Paris,
F-75005, Paris, France  Etienne Couturier  Laboratoire Matière et Systèmes
Complexes, Université Paris Diderot CNRS UMR 7057, 10 Rue Alice Domont et
Léonie Ducquet, 75205 Paris Cedex 13, France

###### Abstract

Turgor is the driving force of plant growth, making possible for roots to
overcome soil resistance or for stems to counteract gravity. Maintaining a
constant growth rate while avoiding the cell content dilution, which would
progressively stop the inward water flux, imposes the production or import of
osmolytes in proportion to the increase of volume. We coin this phenomenon
stationary osmoregulation. The article explores the quantitative consequences
of this hypothesis on the interaction of a cylindrical cell growing axially
against an obstacle.

An instantaneous axial compression of a pressurized cylindrical cell generates
a force and a pressure jump which both decrease toward a lower value once
water has flowed out of the cell to reach the water potential equilibrium. In
a first part, the article derives analytical formula for these force and over-
pressure both before and after relaxation. In a second part, we describe how
the coupling of the Lockhart’s growth law with the stationary osmoregulation
hypothesis predicts a transient slowdown in growth due to contact before a re-
acceleration in growth. We finally compare these predictions with the output
of an elastic growth model which ignores the osmotic origin of growth: models
only match in the early phase of contact for high stiffness obstacle.

##  1 Introduction

Plant growth requires water fluxes that are generated by gradients of water
potential between the growing cells and the water source. The water potential
gradient is tightly regulated by the growing cell through both osmoregulation
and cell wall relaxation, which acts on bulk modulus. These regulations modify
the cell internal pressure, called turgor pressure [1]. The maintenance of
turgor pressure and thus of osmotic pressure require osmolyte synthesis or
import to compensate for the increase of the cell volume. We coin this process
”stationary osmoregulation”. The generated driving force is sufficient to
counteract the resistance of the surrounding fluids (air, water, tissue or
soil). When a growing plant organ encounters a rigid obstacle, before touch-
induced regulations [2] have time to be effective and osmotic pressure to be
altered, the pressure pattern in the plant is modified by the resistance
opposed by the obstacle. On a longer time scale, hard obstacles are
counteracted by the active modification of the cell turgor, through the
increase of osmotic pressure, as shown in roots [3], [4]. Turgor can in
certain conditions relax cell walls through the Lockhart’s law, which tells
that the growth rate allowed by cell wall relaxation is proportional to the
pressure above a threshold [5]. Lockhart’s model can even be a quantitative
tool to predict how the force caused by obstacle modifies the growth dynamics
[6], [7].
Cylindrical organs, omnipresent in the vegetal world, range from unicellular
cylindrical internodes of characeae, to multicellular plant stems and roots.
The present note is focused on a single cylindrical cell presenting a surface
extending homogeneously in the axial direction, which is coined monoaxial
diffuse growth [8]. We quantify how an obstacle will modify the cell wall
stress pattern and the cell internal pressure as well as the growth dynamics.
More generally our model can be applied to other ”walled” cells of organisms
other than plants, such as the fungus Phycomyces blackeslaneus or the Gramm
positive bacteria Bacillus subtilis: these organisms present a cell wall and a
positive regulation of growth by turgor similar to the one observed in plants
[9], [10]. Patterns of cell wall stress in multicellular cylindrical organs
are more complex. Nonetheless, for plants, the material property distribution
is often simple. The stiffest part is the outermost epidermis, which bears
most of the tensile stresses induced by turgor, making the mechanical analysis
of giant cellular internode broadly applicable to the multicellular
cylindrical organs (stems, hypocotyles …) [11]. The main difference between a
single cylindrical cell and a cylindrical organ rather lays on water potential
gradients: for single cell, the water potential gradient necessary to move
water into the cell during growth is small compared to the internal pressure
[12] while for multicellular organs, in general the water potential gradient
cannot be neglected in front of the internal pressure [13]. Nevertheless the
present results modeled for a single cell can serve as a valuable guide to
interpret the effect of an obstacle on the growth of cylindrical multicellular
organs.
Two thought experiments are explored theoretically and solved analytically:
the first experiment consists in compressing the cell to impose
instantaneously a smaller length than the natural length (Figure 1a) and the
second experiment consists in the progressive loading of the cell by its
proper axial growth (Figure 1b). Without an osmolyte content increase through
import or internal production, the water inflow in the growing cell would
dilute the osmolytes and decrease turgor. Our model quantifies how the force
generated by a stationary osmolyte production/import and the elastic force
(cell wall, obstacle) retroact on growth through the Lockhart’s law.
Parameters can be directly estimated from pressure probe literature [5]. The
growth experiment is an opportunity to compare our physiological model to a
phenomenological approach, coined morpho-elasticity, based on an analogy with
metal thermo-elasticity [14].

Figure 1: Schema of the two thought experiments. a. From left to right:
Plasmolyzed cell, turgid cell, compressed turgid cell before relaxation and
compressed turgid cell after relaxation. $R^{*}$ (resp. $L^{*}$) are the
plasmolyzed radius (resp. the plasmolyzed length). $R_{0}$ (resp. $L_{0}$) are
the turgid radius (resp. the turgid length) before compression. $L_{ind}$ is
the length of the compressed cell. $R_{br}$ is the turgid radius after
compression and before relaxation. $R_{ar}$ is the turgid radius after
compression and after relaxation. b. From left to right: Schema of a
plasmolyzed cell and the obstacle (in violet), schema of the turgid cell once
the growth has pushed the obstacle. $R^{*}$ (resp. $L^{*}$) are the
plasmolyzed radius (resp. the plasmolysed length). $L_{c}$ is the length of
the turgid cell at contact. $R$ (resp. $L$) are the turgid radius (resp. the
turgid length) during the contact and $F$ is the force exerted by the
obstacle.

##  2 Determination of the apparent stiffness by a compression experiment

Figure 2: Relationship between parameters. $+$ stands for the pressure before
relaxation and plain $\circ$ for the data after relaxation. The color code of
the lines tells the Poisson ratio $\nu$ (red $\nu=0$, green $\nu=0.25$, blue
$\nu=0.5$). Black color is used when the variable is independent of $\nu$. a.
Normalized pressure $P/P_{0}$ b. Normalized radius $R/R_{0}$ c. Normalized
force $F/(2\pi ER^{*}h)$ vs relative longitudinal deformation due to the
compression $(L_{0}-L_{ind})/L^{*}$.

We aim to measure the apparent stiffness of a cylindrical cell of turgor
$P_{0}$ by compressing the cell to a given length and measuring the resulting
force. It corresponds experimentally to a compression of the cell with a force
sensor of infinite stiffness. We start with a cylindrical cell without turgor,
whose plasmolyzed length and radius are $L^{*}$ and $R^{*}$ respectively and
with a cell wall thickness $h\ll R^{*}$. The cell wall is mechanically
characterized by a Young’s modulus $E$ and a Poisson ratio $\nu$. The cell is
inflated at turgor $P_{0}$ and the resulting turgid length and turgid radius
are $L_{0}$ and $R_{0}$ respectively. According to the the elastic theory of
thin shells (valid for wall thickness much smaller than cylinder radius and
for small strains), $h$ stays constant (at first order) and the cell wall
stresses before compression are given by $\sigma_{LL,0}=P_{0}R_{0}/(2h)$ ,
$\sigma_{\Theta\Theta,0}=P_{0}R_{0}/h$ and the radius and the length are
provided by the two classical formula for cylindrical cells [15]:

$R_{0}=\frac{R^{*}}{1-P_{0}R^{*}(2-\nu)/(2Eh)},$
$L_{0}=\frac{L^{*}(1-(1+\nu)P_{0}R^{*}/(2Eh))}{1-P_{0}R^{*}(2-\nu)/(2Eh)},$

We suppose the cell to be at water potential equilibrium which gives the
following osmotic (molar) content [16]: $N_{i}=\pi
L_{0}R_{0}^{2}(\Pi_{ext}+P_{0})/(\mathcal{R}T),$ $\Pi_{ext}$ being the
external osmotic pressure and $\mathcal{R}$ the perfect gas constant. We
latter suppose $\Pi_{ext}=0$. Considering the high membrane hydraulic
conductivity and thus the short relaxation duration, we suppose $N_{i}$, the
number of moles of osmolytes to be constant during the relaxation.
An instantaneous compression experiment consists in imposing a new length,
$L_{ind}$ to the cell; as water has no time to move out of the cell, the
volume is at first the same as before the compression, which imposes the
radius before relaxation to be: $R_{br}=R_{0}\sqrt{L_{0}/L_{ind}}.$ Stresses
before relaxation read: $\sigma_{LL,br}=P_{br}R_{br}/(2h)-F_{br}/(2\pi
R_{br}h)$ , $\sigma_{\Theta\Theta,br}=P_{br}R_{br}/h$. For small elastic
deformation, the mechanical equilibrium reads:

$\left(\begin{array}[]{c}L_{ind}/L^{*}-1\\\
R_{br}/R^{*}-1\end{array}\right)=\left(\begin{array}[]{cc}1&-\nu\\\
-\nu&1\end{array}\right)\left(\begin{array}[]{c}P_{br}R_{br}/(2Eh)-F_{br}/(2\pi
ER_{br}h)\\\ P_{br}R_{br}/(Eh)\end{array}\right)$ (1)

$P_{br}$, being the pressure before relaxation and $F_{br}$, the force before
the relaxation.
Consistently with the linear elasticity hypothesis, the second order term can
be neglected, which gives the following simple formula (See Annex 6.1):

$\displaystyle R_{br}$ $\displaystyle=$ $\displaystyle
R_{0}+R^{*}\frac{(L_{0}-L_{ind})}{2L^{*}},$ $\displaystyle P_{br}$
$\displaystyle=$ $\displaystyle
P_{0}+\frac{Eh}{R^{*}}\frac{1-2\nu}{2(1-\nu^{2})}\frac{L_{0}-L_{ind}}{L^{*}}$
$\displaystyle F_{br}$ $\displaystyle=$ $\displaystyle(2\pi
ER^{*}h)\left(\frac{5}{4}-\nu\right)\frac{L_{0}-L_{ind}}{L^{*}(1-\nu^{2})}.$

The apparent stiffness $k_{br}$ of the cell before relaxation, defined from
$F_{br}=k_{br}\,(L_{0}-L_{ind})$ reads:

$k_{br}=\left(\frac{5/4-\nu}{1-\nu^{2}}\right)\frac{2\pi ER^{*}h}{L_{c}^{*}}.$

Once water has moved out of the cell and the water potential equilibrium is
reached, the pressure and radius after relaxation ($P_{ar}$ and $R_{ar}$) are
linked by:

$P_{ar}=\mathcal{R}TN_{i}/(\pi R_{ar}^{2}L_{ind}).$

Substituting the expression for $P_{ar}$ in the mechanical equilibrium
provides:

$\left(\begin{array}[]{c}L_{ind}/L^{*}-1\\\
R_{ar}/R^{*}-1\end{array}\right)=\left(\begin{array}[]{cc}1&-\nu\\\
-\nu&1\end{array}\right)\left(\begin{array}[]{c}\mathcal{R}TN_{i}/(2Eh\pi
R_{ar}L_{ind})-F_{ar}/(2\pi ER_{ar}h)\\\ \mathcal{R}TN_{i}/(Eh\pi
R_{ar}L_{ind})\end{array}\right)$ (2)

Neglecting the second order term yields the radius, force and pressure after
relaxation (See Annex 6.2):

$\displaystyle R_{ar}$ $\displaystyle=$ $\displaystyle R_{0}+\nu
R^{*}\frac{(L_{0}-L_{ind})}{L^{*}}$ $\displaystyle P_{ar}$ $\displaystyle=$
$\displaystyle P_{0}\left(1+(1-2\nu)\frac{(L_{0}-L_{ind})}{L^{*}}\right)$
$\displaystyle F_{ar}$ $\displaystyle=$ $\displaystyle(2\pi
ER^{*}h)\frac{(L_{0}-L_{ind})}{L^{*}}.$

The apparent stiffness after relaxation reads:

$k_{ar}=\frac{2\pi ER^{*}h}{L_{c}^{*}}.$ (3)

For a one centimeter long Chara corallina internodal cell, the calculation of
the stiffness gives a value of $k_{ar}=3800\,N\,m^{-1}$ (See 6.5 for the
parameters). A similar stiffness was derived in [17] by physical argument.

The forces before and after relaxation depend solely on the Young’s modulus,
the thickness of the cell wall, the plasmolyzed radius and the Poisson ratio,
not on the initial turgor. $\Delta P$ (resp. $\Delta F$) the difference of
pressure (resp. force) before and after relaxation are zero for $\nu=0.5$ and
maximal for $\nu=0$:

$\displaystyle\Delta P$ $\displaystyle=$
$\displaystyle(1-2\nu)\left(\frac{Eh}{2R^{*}(1-\nu^{2})}-P_{0}\right)\frac{L_{0}-L_{ind}}{L^{*}},$
$\displaystyle\Delta F$ $\displaystyle=$ $\displaystyle(2\pi
ER^{*}h)\frac{\left(\frac{1}{2}-\nu\right)^{2}}{1-\nu^{2}}\frac{L_{0}-L_{ind}}{L^{*}}.$

The linear elasticity hypothesis implies: $\frac{P_{0}R^{*}}{Eh}\ll 1$. The
pressure jump after relaxation (ie. $P_{ar}>P_{0}$) is a second order term,
not visible on Figure 2a .

##  3 Growth interaction with an obstacle under stationary osmolyte synthesis

Growth can be introduced in the model by adding two equations, one for the
production or import of osmolytes and one for the cell wall relaxation. Other
processes such as the change of mechanical properties of the cell wall due to
maturation will not be considered herein. The growth of giant internodal cell
is one-dimensional and exponential. We coin ”stationary osmoregulation” an
exponential osmolyte production, which compensate the dilution due to the
exponential growth: $N_{i}=N_{i}(0)\exp(\gamma t)$, $\gamma$ being the
osmolyte production rate. For giant internodal cells, the water potential
gradient necessary to drive water into the cell can be neglected in front of
the absolute value of cell turgor and osmotic pressure [12]; $P$, the turgor
pressure satisfies the following relation:

$P=(\mathcal{R}TN_{i})/(\pi LR^{2})-\Pi_{ext}.$ (4)

For matter of simplicity we suppose the external osmotic pressure $\Pi_{ext}$
to be $0$. As the cell is always supposed to be at the water potential
equilibrium, we call cell stiffness the stiffness of the cell after relaxation
(Formula 3):

$k_{cell}=k_{ar}.$

The obstacle can be described as a spring of stiffness $k$. At the time of
contact set at $t=0$, $L_{c}$ is the turgid length of the cell and $L_{c}^{*}$
is the plasmolyzed length. At $t>0$, $L$ (respectively $L^{*}$) stands for the
turgid (resp. plasmolyzed) length and the force is $F=k(L-L_{c})$. The
stresses read: $\sigma_{LL}=PR/(2h)-F/(2\pi Rh)$ and
$\sigma_{\Theta\Theta}=PR/h.$ Substituting the pressure expression provides:

$\sigma_{LL}=(\mathcal{R}TN_{i})/(2\pi hRL)-F/(2\pi Rh),$ (5)
$\sigma_{\Theta\Theta}=(\mathcal{R}TN_{i})/(\pi hRL).$ (6)

The mechanical equilibrium is supposed to be always satisfied: solving (4, 5,
6) while neglecting second order terms following linear elasticity hypothesis
(see Annex 6.3) provides the longitudinal and orthoradial strains in the cell
wall:

$\epsilon_{LL}=\frac{(1-2\nu)\mathcal{R}TN_{i}/(2Eh\pi
R^{*}L^{*})-k(L^{*}-L_{c})/(2\pi ER^{*}h)}{1+k(2L^{*}-L_{c})/(2\pi ER^{*}h)},$
(7) $\epsilon_{\Theta\Theta}=(1-\nu^{2})\mathcal{R}TN_{i}/(Eh\pi
R^{*}L^{*})-\nu\epsilon_{LL}.$ (8)

In some cylindrical organs at a late stage of development (stems, internodes
…), the cylindrical organ grows solely in length. The growth allowed by the
relaxation dynamics of the cell wall will be introduced by an equation on the
dynamics of the plasmolyzed length:

$\frac{dL^{*}}{dt}=f(L^{*},t)$ (9)

The function $f$ could correspond to different scenari. In the simplest one,
there is no retroaction of the force on relaxation ($f$ is proportional to the
rest length) but this scenario does not describe the phenomenology of the
growth slowdown after contact. The next step of complexity is the Lockhart’s
law which tells that growth rate is proportional to the pressure above a
threshold $f_{P}=L^{*}m_{P}(P-Y_{P})_{+}$ [18]. Lockhart’s law can also be
formulated in cell wall strains
$f_{\epsilon}=L^{*}m_{\epsilon}(\epsilon_{LL}-Y_{\epsilon})_{+}$ [19] or
stresses $f_{\sigma}=L^{*}m_{\sigma}(\sigma_{LL}-Y_{\sigma})_{+}$ [20]. As a
compression tends to slightly increase the turgor after relaxation (See
Section 2), the traditional pressure formulation of the Lockhart’s law would
predict a slight acceleration of the growth during the contact, which is
contrary to experimental observations. Herein $f_{\sigma}$, the stress
formulation will be used as it has been recently proven that it can
quantitatively predict how the maize root growth slows down when encountering
an axial rigid obstacle [7]. The equation for the stress formulation reads:

$\frac{dL^{*}}{dt}=m_{\sigma}E\left(\frac{(1-2\nu)\mathcal{R}TN_{i}/(2Eh\pi
R^{*})-kL^{*}(L^{*}-L_{c})/(2\pi ER^{*}h)}{1+k(2L^{*}-L_{c})/(2\pi
ER^{*}h)}+\frac{\nu\mathcal{R}TN_{i}}{\pi
ER^{*}h}-\frac{Y_{\sigma}L^{*}}{E}\right)_{+}.$ (10)

$\gamma$, the osmolyte production rate is chosen such as the pressure remains
constant before the contact. As the cell grows solely in length, the volume
increase rate equals the length increase rate before the contact:

$\frac{dN_{i}/dt}{N_{i}}=\frac{dL^{*}/dt}{L^{*}},$

which provides $\gamma$:

$\gamma\approx\frac{f(L_{c}^{*},0)}{L_{c}^{*}}.$ (11)

The following non-dimensional variables:

$\hat{t}=m_{\sigma}Et,\ \hat{L}^{*}=\frac{L^{*}}{L_{c}^{*}},\
\hat{P}=\frac{\mathcal{R}TN_{i}}{2\pi R^{*}hL_{c}^{*}E},\
\hat{P}_{c}=\frac{\mathcal{R}TN_{i}(0)}{2\pi R^{*}hL_{c}^{*}E},\
\hat{Y}_{\sigma}=\frac{Y_{\sigma}}{E},$ (12) $\hat{k}=\frac{k}{k_{cell}},\
\hat{\gamma}=\left(\hat{P_{c}}-\hat{Y}_{\sigma}\right)_{+}$ (13)

can be used to rewrite (10) (See Annex 6.5):

$\frac{d\hat{L}^{*}}{d\hat{t}}=\hat{f}(\hat{L}^{*},\hat{t})$ (14)

with:

$\hat{f}(\hat{L}^{*},\hat{t})=\left(\frac{(1-2\nu)\hat{P}_{c}\exp(\hat{\gamma}\hat{t})-\hat{k}\hat{L}^{*}(\hat{L}^{*}-1-(1-2\nu)\hat{P}_{c})}{1+\hat{k}(2\hat{L}^{*}-1-(1-2\nu)\hat{P}_{c})}+2\nu\hat{P}_{c}\exp(\hat{\gamma}\hat{t})-\hat{Y}_{\sigma}\hat{L}^{*}\right)_{+}.$
(15)

$\hat{P}$, the normalized pressure corresponds at first order to the
longitudinal elastic strain before contact for a Poisson ratio $\nu=0$.
Linear elasticity supposes small strains which correspond to a normalized
pressure $\hat{P}_{c}$ of less than a few percent. The normalized Lockhart’s
threshold $\hat{Y}_{\sigma}$ is also less than a few percent because the cell
is supposed to grow before the contact occurs ($Y<P_{c}$).
The equation was numerically solved with Matlab function ode23s (See 6.5 for
the parameters of the ODE which corresponds to an internodal cell Chara
corallina whose length at contact with the obstacle is $1\,cm$.

Figure 3: (a) Normalized plasmolyzed length increment, (b) normalized pressure
and (c) normalized force vs normalized time for different normalized stiffness
$\hat{k}=1/100$ (violet), $\hat{k}=1/10$ (blue), $\hat{k}=1/5$ (green),
$\hat{k}=1$ (yellow), $\hat{k}=5$ (red). Other parameters are detailed in the
annex (6.5). Plain lines, $+$ symbol, dotted line stand for the numerical
solution, the analytical solution respectively and the morphoelastic solution.
The upper plot is a zoom on the initial behavior whereas the lower plot is
integrated until one of the strains equals $3\%$.

The initial growth dynamics can be studied by solving analytically the
linearized problem associated with (14) (See Annex 6.5) which provides a
combination of two exponential functions:

$\hat{L}^{*}(\hat{t})\approx
1+\hat{\beta}\left(\exp\left(\hat{\gamma}\hat{t}\right)-1\right)+\Delta\hat{L}^{*}\left(1-\exp\left(-\hat{\gamma}_{i}\hat{t}\right)\right),$
(16)

and:

$\hat{\gamma}_{i}=\frac{\hat{k}}{1+\hat{k}}+\hat{Y}_{\sigma},\
\hat{\beta}=\frac{\hat{P_{c}}\left(1+2\nu\hat{k}\right)}{(1+\hat{k})\hat{P_{c}}+\hat{k}}.\
\Delta\hat{L}^{*}=(1-\hat{\beta})\frac{\hat{\gamma}}{\hat{\gamma}_{i}}$

It corresponds to a normalized turgid length:

$\hat{L}(\hat{t})\approx\hat{L}^{*}(\hat{t})(1+\epsilon_{LL}(\hat{t})),$ (17)

which after substitution reads:

$\hat{L}(\hat{t})\approx
1+(1-2\nu)\hat{P}_{c}+\frac{\hat{P}_{c}(\exp(\hat{\gamma}\hat{t})-1)(1-2\nu+\hat{k})}{(1+\hat{k})((1+\hat{k})\hat{P_{c}}+\hat{k})}+\Delta\hat{L}\left(1-\exp\left(-\hat{\gamma}_{i}\hat{t}\right)\right)$

with :

$\Delta\hat{L}=\frac{1-\hat{\beta}}{1+\hat{k}}\frac{\hat{\gamma}}{\hat{\gamma}_{i}}.$

It corresponds to a normalized force:

$\hat{F}(\hat{t})\approx\hat{k}(\hat{L}-\hat{L}_{c})$ (18)

which reads after substitution of $\hat{L}$ and $\hat{L}_{c}$:

$\hat{F}(\hat{t})\approx\frac{\hat{k}}{1+\hat{k}}\left(\frac{\hat{P}_{c}(\exp(\hat{\gamma}\hat{t})-1)(1-2\nu+\hat{k})}{((1+\hat{k})\hat{P_{c}}+\hat{k})}+(1-\hat{\beta})\frac{\hat{\gamma}}{\hat{\gamma}_{i}}\left(1-\exp\left(-\hat{\gamma}_{i}\hat{t}\right)\right)\right).$
(19)

In the limit $\hat{k}\gg 1$, and after the initial relaxation phase
($\hat{\gamma}\hat{t}>3$), the normalized force is independent of the
stiffness and follows the normalized pressure (See Figure 3 lower panel):

$\hat{F}(\hat{t})\approx\hat{P}_{c}(\exp(\hat{\gamma}\hat{t})-1).$ (20)

For various organs (maize roots [21], characeae internodes [5] $\cdots$),
$(P-Y)/Y$ lays between $1/5$ and $1/10$, it implies that $\gamma/\gamma_{i}$
is in general less than $1/5$. The linearized solution for the plasmolyzed
length (equation (16)) is the sum of a positive exponential (characteristic
time $1/\hat{\gamma}$) and a negative exponential with a smaller
characteristic time, $1/\hat{\gamma_{i}}$ : on short time scale (Figure 3(a)
upper plot) the solution relaxes toward a plateau following the negative
exponential while on longer time it follows the positive exponential (Figure
3(a) lower plot) dictated by the osmolyte production rate. For high
stiffnesses
($\beta\approx\frac{\hat{P_{c}}\left(2\nu\right)}{(1+\hat{P_{c}})}$ which is
second order term according to the linear elasticity hypothesis) the transient
plateau is more pronounced than for low stiffnesses ($\beta\approx 1$) as the
respective weight of the two exponential functions in the sum shifts from
almost $0$ to $1$ (Figure 3(a)). Analytical and numerical solutions match very
well. Regardless of stiffness, the plasmolyzed length keeps diverging
exponentially for high $t$ (See Annex 6.6):

$L^{*}\sim_{t\to\infty}\frac{2\nu N_{i}(t)}{\pi(R^{*})^{2}(P_{0}+Eh/R^{*})}.$
(21)

At $t=0$, the turgid cell is under positive longitudinal elastic strain, while
for high $t$ (equation (21)) longitudinal strains are strongly negative
$\epsilon_{LL}=-50\%$, outside the hypothesis of linear elasticity. In order
to ensure that the numerical model output remains in the linear elasticity
hypothesis, the maximal integration time (Figure 3) was chosen such that both
strains remain small ($<3\%$).
Due to the constant osmolyte production rate and the slow-down of growth rate
after contact, the pressure keeps increasing with time (Figure 3 (b)). For an
obstacle of low stiffness, ($k\leqslant k_{cell}/10$), the stationary
osmoregulation is more effective and the pressure remains fairly constant in
the early phase of the interaction with the obstacle (Figure 3 (b)(violet
curve)) while for an obstacle of high stiffness ($k\geqslant k_{cell}$), the
pressure increases from the beginning of the interaction (Figure 3 (b)(red
curve)). For a $1\ cm$ Chara corallina internodal cell, an increase of $1$ bar
is reached in approximately $5$ hours for an obstacle stiffness of $38\
N.m^{-1}$ ($k_{cell}/100$) and in $90\ min$ for a stiffness of $3800\ Nm^{-1}$
($k_{cell}$). The force presents two successive regimes: a first phase of
steep rise (Figure 3 (c) upper panel) followed by a second phase of weaker
increase resulting from the turgor pressure increase dynamics (Figure 3 (c)
lower panel). For a $1\,cm$ Chara corallina internodal cell and for high
stiffness obstacle, the transition occurs for a force of a few tenth of $mN$
after a time lag of a tenth of minutes.

##  4 Comparison with an elastic growth model

The growing cell is now modeled as a growing spring in series with one spring
for the obstacle. This morphoelastic model is similar to a model developed for
the growth of bacteria in a gel [17]; we present it here for comparison with
the physiological model developed above. The analytical solution of the
compression experiment provides the cell spring stiffness. Growth is
phenomenologically introduced by increasing the rest cell length (coined
target length $L_{tar}$). The target length increase rate and the negative
force retro-acting on growth can be incorporated in the model thanks to two
phenomenological constants calibrated with the analytical solution. Elastic
model totally ignores water fluxes; it is thus not easy to incorporate the
stationary osmoregulation hypothesis.
The elastic growth model (See Annex 6.7) provides an expression for the
temporal evolution of the non-dimensionalized observed length, $L_{obs}$ after
the contact:

$\displaystyle\hat{L}_{obs}(\hat{t})=\frac{\hat{k}\hat{L}_{c}}{1+\hat{k}}+\frac{\hat{L}_{c}\hat{\gamma}_{el}\exp(\hat{\gamma}_{el}\hat{t})}{(1+\hat{k})\hat{\gamma}+\hat{k}\hat{L}_{c}\exp(\hat{\gamma}_{el}\hat{t})},$

with:
$\hat{\gamma}_{el}=\hat{P}_{c}\left(1+(1-2\nu)\frac{\hat{k}}{1+\hat{k}}\right)-\hat{Y}_{\sigma}+\frac{\hat{k}}{1+\hat{k}}$.
The solution saturates for an increment:

$\displaystyle\Delta\hat{L}_{obs}=\frac{\hat{P}_{c}-\hat{Y}_{\sigma}}{\hat{k}}.$
(22)

The elastic growth model predicts a saturation of the observed (turgid) length
which corresponds to a transient behavior of the physiological growth model
more pronounced at high obstacle stiffness. In general both the relaxation
rate ($\hat{\gamma}_{i}=\hat{k}/(1+\hat{k})+\hat{Y}_{\sigma}$) and the length
increment differ with the elastic growth model outputs:

$\Delta\hat{L}=\frac{\hat{k}+\hat{k}\hat{P_{c}}\left(1-2\nu\right)}{(1+\hat{k})\hat{P}_{c}+\hat{k}}\frac{\hat{P}_{c}-\hat{Y}_{\sigma}}{\hat{k}+(1+\hat{k})\hat{Y}_{\sigma}}.$

As $\hat{P}_{c}\ll 1$, and $\hat{Y}_{\sigma}\ll 1$, $\gamma_{i}$ equals
$\gamma_{el}$ and $\Delta L$ equals $\Delta L_{obs}$ at the first order of
approximation supposing $k\geqslant k_{cell}$; both models coincide in the
early phase of the interaction ( a few minutes for a $1\ cm$ long Chara
corralina internodal cell for an obstacle stiffness superior to $3800\
Nm^{-1}$). The morphoelastic model predict the same asymptotic force
independently of the obstacle stiffness (Figure 3 (c)).

##  5 Conclusion

An apical compression is the simplest way to probe the mechanics of a
cylindrical cell. The article derives the forces (before and after relaxation)
exerted by the sensor on the compressed cell and it shows that the forces are
independent on the pressure at the first order. The initial pressure,
equilibrated by the tension in the cell wall, does not contribute to the
forces. The forces only depend on the surface modulus $Eh$ multiplied by the
radius and the force drop observed during the relaxation is Poisson ratio
dependent; the force drop is null when the cell wall is an incompressible
material, that is for a Poisson ratio of 0.5, and is maximal for a zero
Poisson ratio. A compression also induces a turgor pressure jump which drops
strongly during relaxation due to water outflow: before the relaxation, the
pressure jump is proportional to the surface modulus divided by the radius
while after the relaxation, the pressure jump is proportional to the initial
pressure. According to the linear elasticity hypothesis $PR/Eh$ is inferior to
$1\%$, meaning the pressure jump after relaxation can be neglected in front of
the pressure jump before relaxation. As in the case of isotropic poroelastic
gels, the ratio between the force before and the force after relaxation only
depends on the Poisson ratio. However the ratio value for cylindrical cells
$\left(\frac{5}{4}-\nu\right)/(1-\nu^{2})$ is different from that for gels
$2(1-\nu)$ [22]; interestingly in both cases the ratio equals one for an
incompressible material. As a perspective, the model should be refined to
describe the anisotropic properties of plant cell wall which could induce a
stronger drop of the force or of the pressure during the relaxation.

The model explores how a stationary osmoregulation process coupled with the
stress-formulated Lockhart’s law predicts the dynamics of turgor pressure and
cell length throughout the contact of a cell growing against an obstacle. The
phenomenology of the growth against a stiff obstacle is reproduced: a
transient growth arrest (Figure 3) followed by a second phase with a slower
variation of growth velocity which remains strictly positive (See [7] Figure 6
top left panel for forces superior to $0.04N$ ). Interestingly this regime
corresponds to the growth behavior of Bacillus subtilis, a Gramm positive
bacteria, in gels with an agarose concentration ranging between $1\%$ and
$8\%$ [17]. The article also provides the opportunity to compare a
physiological model to a more phenomenological approach, coined
morphoelasticity, which neglects the osmotic origin of the growth. Both models
are calibrated to be equivalent before the contact with obstacle. The
morphoelastic model and the physiological model are not equivalent in general;
they only coincids for high obstacle stiffness ($k\geqslant k_{cell}$) on
short time scale ($\gamma t\ll 1$). The physiological model could be extended
to describe the invasive growth of fungi (hyphe) or of some plant organ
(pollen tube, root hairs …) [23] by adapting the framework to apical growth;
the analytical approach is solvable as analytical solutions linking the force
to the apical deformation are available [24].

A contact and more generally a force exerted on a cell is known to induce many
biological regulations. Some regulations tend to stop growth (touch responses)
[2], [25], [26] while other regulations on the contrary tend to favor growth
despite the cell wall tension drop, such as cell wall loosening [27] or
through the increase of osmotic pressure [3], [4]). The deviations from the
quantitative predictions of our model are a good base to quantify how these
active responses act and differ from the stationary osmoregulation.

Acknowledgement. We thank Bruno Moulia for a stimulating discussion on
stationary osmoregulation.

##  6 Annex

###  6.1 Force and pressure before relaxation

We start with the equations of mechanical equilibrium before compression of
the cell. The longitudinal $\epsilon_{LL,0}$ and orthoradial
$\epsilon_{\Theta\Theta,0}$ strains in the cell wall can be written as:

$\displaystyle\left(\begin{array}[]{c}\epsilon_{LL,0}\\\
\epsilon_{\Theta\Theta,0}\end{array}\right)=\left(\begin{array}[]{c}(1-2\nu)P_{0}R^{*}(1+\epsilon_{\Theta\Theta,0})/(2Eh)\\\
(2-\nu)P_{0}R^{*}(1+\epsilon_{\Theta\Theta,0})/(2Eh)\end{array}\right)$ (27)

with $\epsilon_{LL,0}=L_{0}/L^{*}-1$ and
$\epsilon_{\Theta\Theta,0}=R_{0}/R^{*}-1$. At the first order in strain, (27)
rewrites:

$\displaystyle\left(\begin{array}[]{c}\epsilon_{LL,0}\\\
\epsilon_{\Theta\Theta,0}\end{array}\right)=\left(\begin{array}[]{c}(1-2\nu)P_{0}R^{*}/(2Eh)\\\
(2-\nu)P_{0}R^{*}(2Eh)\end{array}\right)$ (32)

leading to:

$\displaystyle R_{0}$ $\displaystyle=$ $\displaystyle
R^{*}(1+(2-\nu)P_{0}R^{*}/(2Eh)),$ (33) $\displaystyle L_{0}$ $\displaystyle=$
$\displaystyle L^{*}(1+(1-2\nu)P_{0}R^{*}/(2Eh)).$ (34)

Then we write the equations following the instantaneous compression before the
relaxation occurs. The volume is conserved which implies:

$\displaystyle R_{br}$ $\displaystyle=$ $\displaystyle
R_{0}\sqrt{L_{0}/L_{ind}}.$

At first order,

$\displaystyle R_{br}=R_{0}+R^{*}\frac{(L_{0}-L_{ind})}{2L^{*}}$

substituting with the values of (33,34), it gives:

$\displaystyle
R_{br}=R^{*}\left(1+\left(2-\nu+\frac{1-2\nu}{2}\right)P_{0}R^{*}/(2Eh)-\frac{1}{2}(L_{ind}/L^{*}-1)\right)$
(35)

Developing the formula (1) of the mechanical equilibrium, it gives:

$\displaystyle\left(\begin{array}[]{c}L_{ind}/L^{*}-1\\\
R_{br}/R^{*}-1\end{array}\right)=\left(\begin{array}[]{c}(1-2\nu)P_{br}R^{*}/(2Eh)-F_{br}/(2\pi
ER^{*}h)\\\ (2-\nu)P_{br}R^{*}/(2Eh)+\nu F_{br}/(2\pi
ER^{*}h)\end{array}\right)$

The force $F_{br}$ can be expressed:

$\displaystyle F_{br}$ $\displaystyle=$ $\displaystyle(\pi
ER_{br}h)\frac{(-2+\nu)(L_{ind}/L^{*}-1)+(1-2\nu)(R_{br}/R^{*}-1)}{1-\nu^{2}}$

Substituting the expression for $R_{br}$ (35) gives:

$\displaystyle F_{br}$ $\displaystyle=$ $\displaystyle(\pi
ER_{br}h)\frac{(-2+\nu)(L_{ind}/L^{*}-1)+(1-2\nu)\left(\left(2-\nu+\frac{1-2\nu}{2}\right)P_{0}R^{*}/(2Eh)-\frac{1}{2}(L_{ind}/L^{*}-1)\right)}{1-\nu^{2}}$

which can be simplified by reintroducing $L_{0}$:

$\displaystyle F_{br}=(\pi
ER^{*}h)\left(\frac{5}{2}-2\nu\right)\frac{(L_{0}-L_{ind})}{L^{*}(1-\nu^{2})}$
(37)

The mechanical equilibrium also provides the pressure:

$\displaystyle
P_{br}=(Eh)\frac{(R_{br}/R^{*}-1)+\nu(L_{ind}/L^{*}-1)}{(1-\nu^{2})R^{*}}$

Substituting the expression for $R_{br}$ and rearranging the term to make
appear $L_{0}$ gives:

$\displaystyle
P_{br}=P_{0}+Eh\frac{(1-2\nu)}{2(1-\nu^{2})}\frac{L_{0}-L_{ind}}{L^{*}R^{*}}.$
(38)

###  6.2 Force and pressure after relaxation

The equation (2) can be rewritten as:

$\displaystyle\left(\begin{array}[]{c}\epsilon_{LL,ind}\\\
\epsilon_{\Theta\Theta,ar}\end{array}\right)$ $\displaystyle=$
$\displaystyle\left(\begin{array}[]{c}(1-2\nu)\mathcal{R}TN_{i}/(2Eh\pi
R^{*}(1+\epsilon_{\Theta\Theta,ar})L_{ind})-F_{ar}/(2\pi
ER^{*}(1+\epsilon_{\Theta\Theta,ar})h)\\\ (2-\nu)\mathcal{R}TN_{i}/(2Eh\pi
R^{*}(1+\epsilon_{\Theta\Theta,ar})L_{ind})+\nu F_{ar}/(2\pi
ER^{*}(1+\epsilon_{\Theta\Theta,ar})h)\end{array}\right)$ (43)

with $\epsilon_{LL,ind}=L_{ind}/L^{*}-1$ and
$\epsilon_{\Theta\Theta,ar}=R_{ar}/R^{*}-1$. Multiplying the both sides of
(43) by $(1+\epsilon_{\Theta\Theta,ar})$ and truncating at the first order in
strain gives:

$\displaystyle\left(\begin{array}[]{c}\epsilon_{LL,ind}\\\
\epsilon_{\Theta\Theta,ar}\end{array}\right)$ $\displaystyle=$
$\displaystyle\left(\begin{array}[]{c}(1-2\nu)\mathcal{R}TN_{i}/(2Eh\pi
R^{*}L_{ind})-F_{ar}/(2\pi ER^{*}h)\\\ (2-\nu)\mathcal{R}TN_{i}/(2Eh\pi
R^{*}L_{ind})+\nu F_{ar}/(2\pi ER^{*}h)\end{array}\right)$

which can be expanded in:

$\displaystyle\left(\begin{array}[]{c}L_{ind}/L^{*}-1\\\
R_{ar}/R^{*}-1\end{array}\right)$ $\displaystyle=$
$\displaystyle\left(\begin{array}[]{c}(1-2\nu)\mathcal{R}TN_{i}/(2Eh\pi
R^{*}L_{ind})-F_{ar}/(2\pi ER^{*}h)\\\ (2-\nu)\mathcal{R}TN_{i}/(2Eh\pi
R^{*}L_{ind})+\nu F_{ar}/(2\pi ER^{*}h)\end{array}\right).$ (49)

Substracting (32) from (49) gives:

$\displaystyle\left(\begin{array}[]{c}(L_{ind}-L_{0})/L^{*}\\\
(R_{ar}-R_{0})/R^{*}-1\end{array}\right)$ $\displaystyle=$
$\displaystyle\left(\begin{array}[]{c}-F_{ar}/(2\pi ER^{*}h)\\\ \nu
F_{ar}/(2\pi ER^{*}h)\end{array}\right).$

which provides easily:

$\displaystyle R_{ar}$ $\displaystyle=$ $\displaystyle R_{0}+\nu
R^{*}\frac{(L_{0}-L_{ind})}{L^{*}},$ (51) $\displaystyle F_{ar}$
$\displaystyle=$ $\displaystyle(2\pi ER^{*}h)\frac{(L_{0}-L_{ind})}{L^{*}}.$
(52)

The apparent stiffness after relaxation $k_{ar}$ equals $(2\pi
ER^{*}h)/L^{*}$. The pressure after relaxation reads:

$\displaystyle P_{ar}=\mathcal{R}TN_{i}/(\pi R_{ar}^{2}L_{ind})$

which at the first order in strain:

$\displaystyle
P_{ar}=P_{0}\left(1+(1-2\nu)\frac{(L_{0}-L_{ind})}{L^{*}}\right).$

###  6.3 First order solution for the osmotic equilibrium and the mechanical
equilibrium with an obstacle

Substituting (4) in (5,6) gives:

$\left(\begin{array}[]{c}L/L^{*}-1\\\
R/R^{*}-1\end{array}\right)=\left(\begin{array}[]{cc}1&-\nu\\\
-\nu&1\end{array}\right)\left(\begin{array}[]{c}\mathcal{R}TN_{i}/(2Eh\pi
RL)-k/(2\pi ERh)(L-L_{c})_{+}\\\ \mathcal{R}TN_{i}/(Eh\pi
RL)\end{array}\right)$ (53)

Introducing the small deformations $\epsilon_{LL}$ and
$\epsilon_{\Theta\Theta}$ gives:

$\left(\begin{array}[]{c}\epsilon_{LL}\\\
\epsilon_{\Theta\Theta}\end{array}\right)=\left(\begin{array}[]{cc}1&-\nu\\\
-\nu&1\end{array}\right)\left(\begin{array}[]{c}\mathcal{R}TN_{i}/(2Eh\pi
R^{*}L^{*}(1+\epsilon_{LL})(1+\epsilon_{\Theta\Theta}))-k(L^{*}(1+\epsilon_{LL})-L_{c})_{+}/(2\pi
ER^{*}(1+\epsilon_{\Theta\Theta})h)\\\ \mathcal{R}TN_{i}/(Eh\pi
R^{*}L^{*}(1+\epsilon_{LL})(1+\epsilon_{\Theta\Theta}))\end{array}\right)$
(54)

Multiplying both sides by $(1+\epsilon_{LL})(1+\epsilon_{\Theta\Theta})$
gives:

$\left(\begin{array}[]{c}\epsilon_{LL}(1+\epsilon_{LL})(1+\epsilon_{\Theta\Theta})\\\
\epsilon_{\Theta\Theta}(1+\epsilon_{LL})(1+\epsilon_{\Theta\Theta})\end{array}\right)=\left(\begin{array}[]{c}(1-2\nu)\mathcal{R}TN_{i}/(2Eh\pi
R^{*}L^{*})-k(1+\epsilon_{LL})(L^{*}(1+\epsilon_{LL})-L_{c})_{+}/(2\pi
ER^{*}h)\\\ (2-\nu)\mathcal{R}TN_{i}/(2Eh\pi R^{*}L^{*})+\nu
k(1+\epsilon_{LL})(L^{*}(1+\epsilon_{LL})-L_{c})_{+}/(2\pi
ER^{*}h)\end{array}\right)$ (55)

At the first order in $\epsilon_{LL}$ and $\epsilon_{\Theta\Theta}$:

$\left(\begin{array}[]{c}\epsilon_{LL}\\\
\epsilon_{\Theta\Theta}\end{array}\right)=\left(\begin{array}[]{c}(1-2\nu)\mathcal{R}TN_{i}/(2Eh\pi
R^{*}L^{*})-k(L^{*}(1+2\epsilon_{LL})-L_{c}(1+\epsilon_{LL}))_{+}/(2\pi
ER^{*}h)\\\ (2-\nu)\mathcal{R}TN_{i}/(2Eh\pi R^{*}L^{*})+\nu
k(L^{*}(1+2\epsilon_{LL})-L_{c}(1+\epsilon_{LL}))_{+}/(2\pi
ER^{*}h)\end{array}\right)$ (56)

It rewrites:

$\left(1+k(2L^{*}-L_{c}))/(2\pi
ER^{*}h)\right)\epsilon_{LL}=(1-2\nu)\mathcal{R}TN_{i}/(2Eh\pi
R^{*}L^{*})-k(L^{*}-L_{c})/(2\pi ER^{*}h)$

which yields:

$\epsilon_{LL}=\frac{(1-2\nu)\mathcal{R}TN_{i}/(2Eh\pi
R^{*}L^{*})-k(L^{*}-L_{c})/(2\pi ER^{*}h)}{1+k(2L^{*}-L_{c})/(2\pi ER^{*}h)}.$
(57)

Exploiting:

$\epsilon_{\Theta\Theta}+\nu\epsilon_{LL}=(2-2\nu^{2})\mathcal{R}TN_{i}/(2Eh\pi
R^{*}L^{*}),$ (58)

it provides $\epsilon_{\Theta\Theta}$:

$\epsilon_{\Theta\Theta}=(1-\nu^{2})\mathcal{R}TN_{i}/(Eh\pi
R^{*}L^{*})-\nu\epsilon_{LL}.$ (59)

The longitudinal stress reads:

$\sigma_{LL}/E=(\epsilon_{LL}+\nu\epsilon_{\Theta\Theta})/(1-\nu^{2})=\epsilon_{LL}+\nu\mathcal{R}TN_{i}/(Eh\pi
R^{*}L^{*}).$ (60)

###  6.4 First order expression of the Lockhart’s model

Substituting (60) in the time derivative of the plasmolyzed length
corresponding to the stress-based formulation Lockhart’s law gives:

$f_{\sigma}=m_{\sigma}E\left(\frac{(1-2\nu)\mathcal{R}TN_{i}/(2Eh\pi
R^{*})-kL^{*}(L^{*}-L_{c})/(2\pi ER^{*}h)}{1+k(2L^{*}-L_{c})/(2\pi
ER^{*}h)}+\frac{\nu\mathcal{R}TN_{i}}{\pi
ER^{*}h}-\frac{Y_{\sigma}L^{*}}{E}\right)_{+}.$ (61)

The growth ODE (10) thus reads:

$\frac{dL^{*}}{dt}=m_{\sigma}E\left(\frac{(1-2\nu)\mathcal{R}TN_{i}/(2Eh\pi
R^{*})-kL^{*}(L^{*}-L_{c})/(2\pi ER^{*}h)}{1+k(2L^{*}-L_{c})/(2\pi
ER^{*}h)}+\frac{\nu\mathcal{R}TN_{i}}{\pi
ER^{*}h}-\frac{Y_{\sigma}L^{*}}{E}\right)_{+}.$

At $t=0$, the contact time, the force exerted by the obstacle is zero and the
classical formula (34) for a turgid cylinder is applicable:

$L_{c}=L^{*}_{c}(1+(1-2\nu)\mathcal{R}TN_{i}(0))/(2\pi R^{*}hL_{c}^{*}E)).$

Introducing the variables $\hat{P}_{c}=\mathcal{R}TN_{i}(0)/(2\pi
R^{*}hL_{c}^{*}E)$, $\hat{Y}_{\sigma}=Y_{\sigma}/E$, and substituting with the
expression for $L_{c}$ provides:

$\frac{dL^{*}}{dt}=m_{\sigma}E\left(\frac{(1-2\nu)\hat{P}_{c}\exp(\gamma
t)L^{*}_{c}-kL^{*}(L^{*}-L^{*}_{c}(1+(1-2\nu)\hat{P}_{c}))/(2\pi
ER^{*}h)}{1+k(2L^{*}-L^{*}_{c}(1+(1-2\nu)\hat{P}_{c}))/(2\pi
ER^{*}h)}+2\nu\hat{P}_{c}\exp(\gamma
t)L^{*}_{c}-\hat{Y}_{\sigma}L^{*}\right)_{+}.$ (62)

The obstacle stiffness can be non-dimensionalized with $k_{ar}=2\pi
R^{*}hE/L^{*}_{c}$, the cell stiffness after relaxation at contact (3), by
introducing $\hat{k}=k/k_{ar}$:

$\frac{dL^{*}}{dt}=m_{\sigma}E\left(\frac{(1-2\nu)\hat{P}_{c}\exp(\gamma
t)L^{*}_{c}-\hat{k}L^{*}(L^{*}/L^{*}_{c}-(1+(1-2\nu)\hat{P}_{c}))}{1+\hat{k}(2L^{*}/L^{*}_{c}-(1+(1-2\nu)\hat{P}_{c}))}+2\nu\hat{P}_{c}\exp(\gamma
t)L^{*}_{c}-\hat{Y}_{\sigma}L^{*}\right)_{+}.$ (63)

The growth ODE (63) can be rewritten with the non-dimensional time and length
variables $\hat{t}=m_{\sigma}Et$ and $\hat{L}^{*}=L^{*}/L_{c}^{*}$:

$\frac{d\hat{L}^{*}}{d\hat{t}}=\left(\frac{(1-2\nu)\hat{P}_{c}\exp(\hat{\gamma}\hat{t})-\hat{k}\hat{L}^{*}(\hat{L}^{*}-(1+(1-2\nu)\hat{P}_{c}))}{1+\hat{k}(2L^{*}-(1+(1-2\nu)\hat{P}_{c}))}+2\nu\hat{P}_{c}\exp(\hat{\gamma}\hat{t})-\hat{Y}_{\sigma}\hat{L}^{*}\right)_{+}.$
(64)

with: $\hat{\gamma}=\left(\hat{P_{c}}-\hat{Y}_{\sigma}\right)_{+}$.
It corresponds to a non-dimensional growth ODE function (14):

$\hat{f}_{\sigma}(\hat{L}^{*},\hat{t})=\left(\frac{(1-2\nu)\hat{P}_{c}\exp(\hat{\gamma}\hat{t})-\hat{k}\hat{L}^{*}(\hat{L}^{*}-1-(1-2\nu)\hat{P}_{c})}{1+\hat{k}(2\hat{L}^{*}-1-(1-2\nu)\hat{P}_{c})}+2\nu\hat{P}_{c}\exp(\hat{\gamma}\hat{t})-\hat{Y}_{\sigma}\hat{L}^{*}\right)_{+}.$
(65)

The longitudinal elastic strain (57) can be rewritten in term of the non-
dimensional variables:

$\epsilon_{LL}=\frac{(1-2\nu)\hat{P}_{c}/\hat{L}^{*}\exp(\hat{\gamma}\hat{t})-\hat{k}(\hat{L}^{*}-1-(1-2\nu)\hat{P}_{c})}{1+\hat{k}(2\hat{L}^{*}-1-(1-2\nu)\hat{P}_{c})}.$
(66)

The $\hat{L}^{*}$ derivative of $\hat{f}_{\sigma}$ reads:

$\partial_{\hat{L}^{*}}\hat{f}_{\sigma}=-\left(\frac{\hat{k}\left(2\hat{L}^{*}-1-(1-2\nu)\hat{P}_{c}+2\hat{L}^{*}\epsilon_{LL}\right)}{1+\hat{k}\left(2\hat{L}^{*}-1-(1-2\nu)\hat{P}_{c}\right)}+\hat{Y}_{\sigma}\right).$
(67)

making use of (66) to introduce $\epsilon_{LL}$.
For $\hat{t}>0$ and a length which equals the contact length, the normalized
length derivative reads:

$\hat{f}_{\sigma}(\hat{L}^{*}=1,\hat{t})=\left(\frac{(1-2\nu)\hat{P}_{c}\exp(\hat{\gamma}\hat{t})+\hat{k}(1-2\nu)\hat{P}_{c}}{1+\hat{k}(1-(1-2\nu)\hat{P}_{c})}+2\nu\hat{P}_{c}\exp(\hat{\gamma}\hat{t})-\hat{Y}_{\sigma}\right)_{+},$
(68)
$\partial_{\hat{L}^{*}}\hat{f}_{\sigma}(\hat{L}^{*}=1,\hat{t})=-\left(\frac{\hat{k}\left(1-(1-2\nu)\hat{P}_{c}+2\epsilon_{LL}\right)}{1+\hat{k}\left(1-(1-2\nu)\hat{P}_{c}\right)}+\hat{Y}_{\sigma}\right).$
(69)

Following the hypothesis of linear elasticity, longitudinal elastic strain
$\epsilon_{LL}=(1-2\nu)\hat{P}_{c}$ is small. Keeping the dominant order
provides:

$\hat{f}_{\sigma}(\hat{L}^{*}=1,\hat{t})=\left(\exp(\hat{\gamma}\hat{t})\hat{P}_{c}\left(\frac{1+2\nu\hat{k}}{1+\hat{k}}\right)+\hat{P}_{c}\left(\frac{(1-2\nu)\hat{k}}{1+\hat{k}}\right)-\hat{Y}_{\sigma}\right)_{+},$
(70)
$\partial_{\hat{L}^{*}}\hat{f}_{\sigma}(\hat{L}^{*}=1,\hat{t})=-\left(\frac{\hat{k}}{1+\hat{k}}+\hat{Y}_{\sigma}\right).$
(71)

###  6.5 Analytical solution of the linearized problem

The equation (64) linearized at $(\hat{L}^{*}=1,\hat{t})$ reads:

$\frac{d\hat{L}^{*}}{d\hat{t}}\approx\hat{f}_{\sigma}\left(1,\hat{t}\right)+(\partial_{\hat{L}^{*}}\hat{f}_{\sigma})\left(1,\hat{t}\right)\left(\hat{L}^{*}-1\right).$
(72)

(70) tells $\partial_{\hat{L}^{*}}\hat{f}_{\sigma}$ is independent of the time
at first order. The solution of (72) reads:

$\hat{L}^{*}=1+\exp\left(\partial_{\hat{L}^{*}}\hat{f}_{\sigma}(1,0)\hat{t}\right)\int_{u=0}^{\hat{t}}\hat{f}_{\sigma}\left(1,u\right)\exp\left(-\partial_{\hat{L}^{*}}\hat{f}_{\sigma}(1,0)u\right)du.$
(73)

Once substituted with (72), the integral reads:

$\int_{u=0}^{\hat{t}}\left(\exp\left((\hat{\gamma}+\hat{\gamma}_{i})u\right)\hat{P}_{c}\left(\frac{1+2\nu\hat{k}}{1+\hat{k}}\right)+\exp\left(\hat{\gamma}_{i}u\right)\left(\hat{P}_{c}\left(\frac{(1-2\nu)\hat{k}}{1+\hat{k}}\right)-\hat{Y}_{\sigma}\right)\right)du$
(74)

with $\hat{\gamma}=\hat{P}_{c}-\hat{Y}_{\sigma}$ and
$\hat{\gamma}_{i}=\frac{\hat{k}}{1+\hat{k}}+\hat{Y}_{\sigma}$. The solution
can be explicited:

$\hat{L}^{*}(\hat{t})\approx
1+\hat{\beta}\left(\exp\left(\hat{\gamma}\hat{t}\right)-1\right)+(1-\hat{\beta})\frac{\hat{\gamma}}{\hat{\gamma}_{i}}\left(1-\exp\left(-\hat{\gamma}_{i}\hat{t}\right)\right),$
(75)

with:

$\hat{\beta}=\frac{\hat{P_{c}}\left(1+2\nu\hat{k}\right)}{(1+\hat{k})\hat{P_{c}}+\hat{k}}.$

For a Chara corallina internode of length at contact $L_{c}=1cm$, the
mechanical parameters are of the following order of magnitude [20]:

$E=1GPa,\ h=10\mu m,\ \nu=0.3,\ R^{*}=500\mu m.$

The Lockhart’s parameters expressed in pressure read:

$P=0.55MPa,\ Y_{P}=0.5MPa,\ m_{P}=2.6000\times 10^{-5}(MPa)s^{-1},$

which correpond to Lockhart’s parameters expressed in stresses obtained with
these formula:

$\sigma_{LL}=\frac{PR}{2h},\ Y_{\sigma}=\frac{YR}{2h},\
m_{\sigma}=\frac{2hm}{R}.$

It corresponds to a cell stiffness after relaxation: $k_{ar}=\pi\times
10^{3}N/m$.

###  6.6 Asymptotic behavior of the Lockhart’s model with an obstacle

The time derivative of (10) gives:

$\frac{d^{2}L^{*}}{dt^{2}}=\partial_{t}f_{\sigma}(L^{*},t)+\partial_{L}^{*}f_{\sigma}(L^{*},t)\frac{dL^{*}}{dt}.$

Let suppose $\frac{dL^{*}}{dt}$ equals zero for a given $t_{zero}$:

$\frac{d^{2}L^{*}}{dt^{2}}(t_{zero})=\partial_{t}f_{\sigma}(L^{*}(t_{zero}),t_{zero})$

The time derivative reads:

$\partial_{t}f_{\sigma}(L^{*}(t_{zero}),t_{zero})=m_{\sigma}\gamma\frac{\mathcal{R}TN_{i}(t_{zero})}{2h\pi
R^{*}L^{*}}\left(\frac{1+2\nu k(2L^{*}-L_{c})/(2\pi
ER^{*}h)}{1+k(2L^{*}-L_{c})/(2\pi ER^{*}h)}\right),$

which implies:

$\frac{d^{2}L^{*}}{dt^{2}}(t_{zero})>0.$

$L^{*}$ is thus strictly increasing with no fixed point; $L^{*}$ asymptotic
limit reads:

$\lim_{t\to\infty}L^{*}=+\infty.$

The longitudinal stress reads:

$\sigma_{LL}=\left(\frac{(1-2\nu)\mathcal{R}TN_{i}/(2Eh\pi
R^{*}L^{*})-k(L^{*}-L_{c})/(2\pi R^{*}h)}{1+k(2L^{*}-L_{c})/(2\pi
ER^{*}h)}+\frac{\nu\mathcal{R}TN_{i}}{\pi R^{*}hL^{*}}\right).$

When $t\to\infty$ it can be rewritten:

$\sigma_{LL}=-\frac{E}{2}+\frac{\nu\mathcal{R}TN_{i}}{\pi
ER^{*}hL^{*}}+o_{t\to\infty}\left(\frac{\mathcal{R}TN_{i}}{\pi
R^{*}hL^{*}}\right).$

(61) can be rewritten:

$\frac{dL^{*}}{dt}=m_{\sigma}\left(-\frac{EL^{*}}{2}+\frac{\nu\mathcal{R}TN_{i}}{\pi
R^{*}h}-Y_{\sigma}L^{*}+o_{t\to\infty}\left(\frac{\mathcal{R}TN_{i}}{\pi
R^{*}h}\right)\right)_{+}.$ (76)

Let write the time derivative of $\rho=L^{*}/N_{i}$:

$\frac{d\rho}{dt}=\frac{1}{N_{i}}\frac{dL^{*}}{dt}-\frac{L^{*}}{N_{i}^{2}}\frac{dN_{i}}{dt}$

Substituting the value of $N_{i}$ and of $\dot{L^{*}}$:

$\frac{d\rho}{dt}=m_{\sigma}\left(-\frac{E}{2}\rho+\frac{\nu\mathcal{R}T}{\pi
R^{*}h}-Y_{\sigma}\rho+o_{t\to\infty}\left(\frac{\mathcal{R}T}{\pi
R^{*}h}\right)\right)_{+}-\gamma\rho$

Substituting the value of $\gamma=m\sigma(P_{0}R^{*}/(2h)-Y_{\sigma})$ gives:

$\frac{d\rho}{dt}=m_{\sigma}\left(-\frac{E}{2}-P_{0}R^{*}/(2h)\right)\rho+m_{\sigma}\frac{\nu\mathcal{R}T}{\pi
R^{*}h}+o_{t\to\infty}\left(m_{\sigma}\frac{\mathcal{R}T}{\pi R^{*}h}\right)$
(77)

Let take a very high $t_{o}$, and integrate the ODE from the time $t_{o}$
(77):

$\rho=\exp\left(m_{\sigma}\left(-\frac{E}{2}-P_{0}R^{*}/(2h)\right)(t-t_{o})\right)\left(\rho_{0}-\frac{2\nu\mathcal{R}T}{\pi
R^{*}h(Eh+P_{0}R^{*})}\right)+\frac{2\nu\mathcal{R}T}{\pi
R^{*}h(Eh+P_{0}R^{*})}$ (78)

Finally, $L^{*}$ has a simple equivalent for high $t$:

$L^{*}\sim_{t\to\infty}\frac{2\nu\mathcal{R}TN_{i}(t)}{\pi
R^{*}(P_{0}R^{*}+Eh)},$ (79)

it corresponds to a value of elastic strain outside of the range of linear
elasticity:

$\lim_{t\to\infty}\epsilon_{LL}=-1/2,$ (80)

and an asymptotic pressure:

$\lim_{t\to\infty}P=\frac{P_{0}+Eh/R^{*}}{2}.$ (81)

###  6.7 Elastic growth model

The contact experiment is modeled by one spring of a constant stiffness
$k_{cell}$ and an increasing target length $L_{tar}$ and one spring of
constant rest length $L_{c}$ and a constant stiffness $k$. The observed length
$L_{obs}$ is provided by the mechanical equilibrium:

$k_{cell}(L_{obs}-L_{tar})+k(L_{obs}-L_{c})=0,$ (82)

thus:

$L_{obs}=\frac{kL_{c}+k_{cell}L_{tar}}{k_{cell}+k}.$ (83)

To retrieve the phenomenology, we suppose the target length evolution rate
decreases linearly with the force:

$\dot{L}_{tar}=L_{tar}f(L_{tar})$ (84)

with $f(L_{tar})=(c_{1}-c_{2}k(L_{obs}-L_{c}))$ the growth rate. $c_{1}$
corresponds to the growth rate before the contact:

$c_{1}=m_{\sigma}\left(\frac{P_{c}R}{2h}-Y_{\sigma}\right).$ (85)

$c_{2}$ can be estimated by considering the growth rate tangent at the contact
time:

$c_{2}=\frac{(\frac{\dot{L}}{L}-c_{1})}{F}.$ (86)

Approximating $dL/dt/L$ by $dL^{*}/dt/L^{*}$ at a contact time which is
reasonable according to the analytical solution for the Lockhart model:

$c_{2}=\frac{m_{\sigma}}{2\pi Rh}.$ (87)

Substituting (83) in (84) gives:

$\frac{dL_{tar}}{dt}=L_{tar}\left(m_{\sigma}\left(\frac{P_{c}R}{2h}-Y_{\sigma}\right)-\frac{m_{\sigma}}{2\pi
Rh}\frac{k_{cell}k}{k_{cell}+k}(L_{tar}-L_{c})\right)_{+}.$ (88)

The equation can be non-dimensionalized using the following non-dimensional
variables:

$\hat{t}=m_{\sigma}Et,\ \hat{L}_{tar}=\frac{L_{tar}}{L_{c}^{*}},\
\hat{P}_{c}=\frac{\mathcal{R}TN_{i}(0)}{2\pi R^{*}hL_{c}^{*}E},\
\hat{Y}_{\sigma}=\frac{Y_{\sigma}}{E},$ (89) $\hat{k}=\frac{k}{k_{cell}},\
\hat{\gamma}=\left(\hat{P_{c}}-\hat{Y}_{\sigma}\right)_{+},\
\hat{\gamma}_{el}=\hat{\gamma}+\frac{\hat{k}}{1+\hat{k}}\hat{L}_{c}.$ (90)

The non-dimensionalized equation reads:

$\frac{d\hat{L}_{tar}}{d\hat{t}}=\hat{L}_{tar}\left(\hat{\gamma}_{el}-\frac{\hat{k}}{1+\hat{k}}\hat{L}_{tar}\right)_{+}.$
(91)

It rewrites:

$\frac{d\hat{L}_{tar}}{d\hat{t}}\left(\frac{1}{\hat{L}_{tar}}+\frac{\frac{\hat{k}}{1+\hat{k}}}{\hat{\gamma}_{el}-\frac{\hat{k}}{1+\hat{k}}\hat{L}_{tar}}\right)=\hat{\gamma}_{el}.$

We suppose the contact takes place at time $t=0$; it implies
$L_{tar}(0)=L_{c}$. The expression (91) can be integrated:

$\displaystyle\log(\hat{L}_{tar}/\hat{L}_{c})-\log\left(\frac{\hat{\gamma}_{el}-\frac{\hat{k}}{1+\hat{k}}\hat{L}_{tar}}{\hat{\gamma}}\right)=\hat{\gamma}_{el}\hat{t}.$

Finally:

$\displaystyle\hat{L}_{tar}(\hat{t})=\frac{\hat{L}_{c}\hat{\gamma}_{el}\exp(\hat{\gamma}_{el}\hat{t})}{\hat{\gamma}+\frac{\hat{k}\hat{L}_{c}}{1+\hat{k}}\exp(\hat{\gamma}_{el}\hat{t})}.$
(92)

The solution saturates for an increment:

$\displaystyle\Delta\hat{L}_{tar}=\frac{\hat{\gamma}_{el}(1+\hat{k})}{\hat{k}}-\hat{L}_{c}.$

Substituting with the value of $\gamma$ and $\Delta\hat{L}_{tar}$ rewrites:

$\displaystyle\Delta\hat{L}_{tar}=\frac{\hat{\gamma}(1+\hat{k})}{\hat{k}}.$

Combining (92, 83) provides the observed length dynamics:

$\displaystyle\hat{L}_{obs}(\hat{t})=\frac{\hat{k}\hat{L}_{c}}{1+\hat{k}}+\frac{\hat{L}_{c}\hat{\gamma}_{el}\exp(\hat{\gamma}_{el}\hat{t})}{(1+\hat{k})\hat{\gamma}+\hat{k}\hat{L}_{c}\exp(\hat{\gamma}_{el}\hat{t})},$
(93)

which saturates for a non-dimensionalized observed length increment:

$\displaystyle\Delta\hat{L}_{obs}=\frac{\hat{\gamma}}{\hat{k}}.$ (94)

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