Modelling Alkaline Band Formation in Chara corallina · regions and a fuchsia color indicates alkaline regions. [20] under water than on land. The di usion of CO 2 di uses 10,000

C.S. Verbeek

Modelling Alkaline Band Formation in

Chara corallina

Master thesis, February 18, 2016

Supervisor: Dr. S.C. Hille

Mathematical Institute of Leiden University

Preface

This MSc project is part of modelling and experimental activities from the PlantBioDynamics Lab (PBDL) at the Institute of Biology Leiden (IBL), concentrat-ing on band formation in aqueous plant, e.g., Chara corallina. The biologicalorigins of band formation are still unclear. By means of mathematical mod-elling of the knows and hypothesised processes we aimed at understanding theunderlying dynamics of this phenomenon.

Unfortunately there are no conclusive results in this thesis, but it contains aconsolidation of different aspects that were looked at. Despite the inconclusiveresults a hint has been made towards the possibility of Turing patterns. Themathematical expressions are still too complicated to be properly analysed.They need to be properly simplified.

A diversity of details needed to be looked at:

• Rate expressions for symporters

• Unstirred layer

• Proton Flux Corrections

Acknowledgement to Kees Boot who offered support by pointing out thatproton symporters have not been reported to exist (so far) in any plant species,neither OH–-symporters, which were hypothesized in the literature [15]. How-ever, Na+-symporters have been reported to exist and this was motivation forimplementation of such symporters in the models. Therefore this is the firsttime that sodium symporters are considered in the subject of band formationin Chara corallina.

We hope that the results in this thesis will contribute to further research onthis matter.

Contents

1 Introduction 31.1 The Studied Biological Phenomenon . . . . . . . . . . . . . . . . 31.2 Uptake of Inorganic Carbon by Aqueous Plants . . . . . . . . . . 3

1.2.1 Inorganic Carbon in Aqueous Solutions . . . . . . . . . . 41.2.2 Diffusional Transport through the Unstirred Layer . . . . 61.2.3 Cross-Membrane Transport . . . . . . . . . . . . . . . . . 71.2.4 Photosynthesis . . . . . . . . . . . . . . . . . . . . . . . . 7

1.3 pH Homeostasis . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2 Models for a HCO–3-Symporter 9

2.1 A Detailed Six-State Model . . . . . . . . . . . . . . . . . . . . . 92.2 A Four-State Model for the H+/HCO–

3-Symporter . . . . . . . . 122.3 The Na+/HCO–

3-Symporter . . . . . . . . . . . . . . . . . . . . . 14

3 Initial ODE-Models for Inorganic Carbon Uptake 163.1 Existence of Steady States for the Generic Model . . . . . . . . . 173.2 Stability of Steady States for the Generic Model . . . . . . . . . 213.3 Reduced Model Derived from Time Scale Separation . . . . . . . 223.4 Existence of Steady States for the Reduced Model . . . . . . . . 223.5 Stability of Steady Sates for the Reduced Model . . . . . . . . . 313.6 Discussion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

4 Diffusion-Limited Supply of Inorganic Carbon - the UnstirredLayer 334.1 Diffusion in Total Carbon Concentration . . . . . . . . . . . . . . 334.2 Steady State Solution . . . . . . . . . . . . . . . . . . . . . . . . 35

4.2.1 Inorganic Carbon at the Membrane - Sodium Case . . . . 374.2.2 Inorganic Carbon at the Membrane - Proton Case . . . . 41

4.3 Convergence to Steady State . . . . . . . . . . . . . . . . . . . . 444.3.1 Inactive Sodium Symporters . . . . . . . . . . . . . . . . 444.3.2 Active Sodium Symporters . . . . . . . . . . . . . . . . . 504.3.3 Linearised Stability with Symporter Active . . . . . . . . 50

4.4 The Modelling Effect of Cylindrical Coordinates on the CarbonSupply . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

5 Derivation of Corrected Dynamics 545.1 Intracellular Dynamics . . . . . . . . . . . . . . . . . . . . . . . . 545.2 Extracellular Dynamics . . . . . . . . . . . . . . . . . . . . . . . 60

6 Towards a Turing Pattern? 63

A Separation Constant λ is Real and Strictly Positive. 65A.1 Case λ = 0: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65A.2 Case λ ∈ C \ {0}: . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

1

B Needed Calculations to Compare C?T (h) and fCO2(h). 68B.1 Behaviour of C?T (h) at h = 0. (Proton Symporter Case) . . . . . 68B.2 Behaviour of C?T (h) at h = 0. (Sodium Symporter Case) . . . . . 72B.3 Value of C?T (∞). (Proton Symporter Case) . . . . . . . . . . . . 73B.4 Value of C?T (∞). (Sodium Symporter Case) . . . . . . . . . . . . 74

2

1 Introduction

This thesis is motivated by a biological problem and therefore it is necessaryto include content explaining the origin of assumptions and models discussed inlater paragraphs. This will be done in this section.

1.1 The Studied Biological Phenomenon

Chara corallina is a fresh water algae. In light the cell surface of Chara corallinais largely acid and the extent of the alkaline areas varies considerably. The ad-dition of NaHCO–

3 to the bathing solution causes intense alkaline banding at thesurface of these cells. A quantitative investigation of the pH banding in Characorallina shows the formation of pH banding after 2 hours of illumination (seeFigure 1). Measurements in alkaline sites show that the maximum pH was at-tained after approximately 1 hour after illumination after an dark-equilibrationperiod. When switched from light to dark there is an immediate decline in pHand the pH steadied off at the dark level within 1 hour. Measurements in anacid site show that there is an initial rise in pH and subsequent acidificationthat is less rapid then the alkalinization in the alkaline site. [14]

Figure 1: Typical pH traces showing values obtained after 2 h illumination (•)and dark cell wall values following 30-min dark pre-treatments (N). The brokenlines indicate the pH of the bathing solution. [14]

Similar research has been performed where phenol red was used to indicatethe location of the acid and alkaline bands. Figure 2 shows photo’s from thisresearch.

1.2 Uptake of Inorganic Carbon by Aqueous Plants

Algae, including Chara corallina, are restricted to aqueous environments. Phys-ical restrictions on light availability and gas exchange are much more profound

3

Figure 2: A cell from Chara corallina where acid and alkaline bands are visu-alised by using a pH indicator dye, phenol red. A yellow color indicates acidicregions and a fuchsia color indicates alkaline regions. [20]

under water than on land. The diffusion of CO2 diffuses 10,000 times slower inwater than in air [3, 4]. When dissolved in water, inorganic carbon equilibratesto a species distribution on a pH-dependent basis. The result is a major limita-tion in the availability of CO2 in aqueous environments and hence a limitationon photosynthesis of Chara corallina.

1.2.1 Inorganic Carbon in Aqueous Solutions

In this thesis we will frequently use the the term dissolved inorganic carbon.Dissolved inorganic carbon is the total of all the inorganic carbon species. Thedifferent dissolved inorganic carbon species are CO2, HCO–

3, CO2–3 and H2CO3.

The distribution of these inorganic carbon species in an aqueous solution isdependent on the pH-value, since they are transformed into each other accordingto the chemical reactions

CO2 + H2O H+ + HCO−3 ,

HCO−3 H+ + CO2−3 ,

H2CO3 H+ + HCO−3 ,

with equilibrium constants

Ka1=

[H+][HCO−3 ]

[CO2]= 1.14 · 10−6M [13, 16], (1a)

Ka2 =[H+][CO2−

3 ]

[HCO−3 ]= 4.68 · 10−11M [13, 16], (1b)

Ka3=

[H+][HCO−3 ]

[H2CO3]= 3.55 · 10−4M [12, 16]. (1c)

In Figure 3 you can find these fractions of the different species of inorganiccarbon plotted against pH. This gives a clear picture of the effect of pH on thedistribution of dissolved inorganic carbon in a solution.

The exact expressions for the different fractions as function of H+ concen-tration, h, are

fCO2(h) =Ka3

h2

Ka1Ka3h+Ka3h2 +Ka1h

2 +Ka1Ka2Ka3

,

fHCO−3 (h) =Ka1

Ka3h

Ka1Ka3h+Ka3h2 +Ka1h

2 +Ka1Ka2Ka3

,

fCO2−3 (h) =

Ka1Ka2

Ka3

Ka1Ka3h+Ka3h2 +Ka1h

2 +Ka1Ka2Ka3

,

fH2CO3(h) =Ka1

h2

Ka1Ka3h+Ka3h2 +Ka1h

2 +Ka1Ka2Ka3

.

4

0 2 4 6 8 10 12 140

10

20

30

40

50

60

70

80

90

100

pH

Con

centr

ati

onp

erce

nta

ge

HCO–3

CO2

CO2–3

H2CO3

Figure 3: Inorganic carbon distribution against pH. For each species, the fractionof dissolved inorganic carbon of total carbon concentration is shown.

A derivation can be found in [16]. Note that

fH2CO3(h) =Ka1

Ka3

fCO2 ≈ 1.14

3.55· 10−2fCO2 .

Hence it is obvious that fH2CO3 is always very small and therefore negligible inmost cases.

An interesting finding is the pH value where we have the maximum amountof usable inorganic carbon, i.e., the species of inorganic carbon that can crossthe cell membrane, CO2 and HCO–

3. Calculate the derivative of fCO2(h′) +

fHCO−3 (h′):

d

dh′fCO2(h′) =

Ka1K2a3

(h′)2 + 2Ka1Ka2K2a3h′

(Ka1Ka3h′ +Ka3(h′)2 +Ka1(h′)2 +Ka1Ka2Ka3)2

d

dh′fHCO−3 (h′) =

K2a1Ka2K

2a3−Ka1K

2a3

(h′)2 −K2a1Ka3(h′)2)


d

dh′(fCO2(h′) + fHCO−3 (h′)) =

K2a1Ka2K

2a3

+ 2Ka1Ka2K2a3h′ −K2

a1Ka3(h′)2


Searching for local minima or maxima we get

Ka1(h′)2 − 2Ka2Ka3h′ −Ka1Ka2Ka3 = 0.

Solving this using the abc-formula gives

h′± =2Ka2Ka3 ±

√(2Ka2Ka3)2 + 4K2

a1Ka2Ka3

2Ka1

.

5

0 h′+

0

Ka3

Ka3+Ka1

h′

fCO2(h′) + fHCO−3 (h′)

Figure 4: Sketch of fCO2(h′) + fHCO−3 (h′).

Note thath′+ · h′− = −Ka2Ka3 < 0.

Hence there is one unique positive solution and so there is only one minimum

or maximum for h′ > 0 (at h′ = h′+). Since fCO2(h′) + fHCO−3′(h′) ↓ 0 as h′ ↓ 0

this has to be a maximum (see Figure 4). Substituting the values of Ka1, Ka2

and Ka3gives

h′+ ≈ 1.14 · 10−7

which is equivalent to a pH value of approximately 6.8.

1.2.2 Diffusional Transport through the Unstirred Layer

Chara corallina is a fresh water algae. Due to its aqueous environment, theability to access CO2 is limited as opposed to plants that are exposed to air. Thediffusion coefficients of gases in water are lower than the diffusion coefficients ofgases in air. On top of this diffusive boundary layers develop on all surfaces. Thethickness of this layer is adjacent to a plants leaves and is of the same order ofmagnitude in water and air. In this case we are not looking at leaves, but at cellsof Chara corallina. However, it is most likely that there is a diffusive boundarylayer on the surface of these cells as well, but this layer may be of a differentmagnitude than the diffusive boundary layer on leaves. Further on in this thesiswe will refer to this diffusive boundary layer as the unstirred layer. Theselimiting effects result in aqueous plants, like Chara corallina, having a depletedsupply of CO2. This restriction has lead to certain plants gaining adaptivefeatures to reduce this restriction. Among these features is the supplementaryuse of HCO–

3, since in many fresh waters the pool of HCO–3 is several fold higher

than of CO2. Near the cell surface HCO–3 is converted to CO2 in acid bands.

As shown in Figure 3, the fraction of CO2 is higher if the environment is moreacid. [17]

In Table 1 you can find several values for the diffusion coefficients of thedifferent inorganic carbon species in water. The diffusion coefficient of CO2 in

6

Solute D (·10−5cm2/s)

CO2 1.9 [3]2.02 [24]

HCO–3 1.17 [24]

1.000 [18] (sea water)1.18 [11] (sea water)

CO2–3 0.81 [24]

0.955 [11] (sea water)H2CO3 not found

H+ 7 [1]

Table 1: Diffusion coefficients(D) in water at 25◦C.

air at 25◦C is 0.162 cm2/s [4]. As mentioned earlier, this diffusion coefficient ismuch higher in air than the diffusion coefficient in water. Roughly 104 timeshigher.

Another important diffusion coefficient is that of protons in water. Forcompletion this is also put in Table 1.

1.2.3 Cross-Membrane Transport

In our model we assume that the cross-membrane transport of dissolved inor-ganic carbon is by cross-membrane diffusion of CO2 and by HCO–

3-symporters.However, in the literature there seems to be cases where there is CO2 trans-portation in the membrane too [19]. Note that we have in our model that thereis no cross-membrane transport of any form of dissolved inorganic carbon besideCO2 and HCO–

3.The existence of HCO–

3-symporters in cells of Chara corallina is unclear. Inliterature H+/HCO–

3-symporters were postulated [7] and so we will model thislater on in this thesis. However, another possibility is that there are Na+/HCO–

3-symporters present in cells of Chara corallina. Such symporters have beenproven to exist [19], but not for cells of Chara corallina. However, it is morelikely for Na+/HCO–

3-symporters to be present in cells of Chara corallina, sincethere is no evidence on the existence of H+/HCO–

3-symporters in the literatureso far. Therefore this possibility will be modelled further on in this thesis aswell.

In our models we assume that the transporters have no preference for acertain direction (inward or outward) and hence the net flux will be determinedonly by the concentrations of the relevant ions on either side of the membrane.

1.2.4 Photosynthesis

The net process of photosynthesis in water is often described simply as thefixation of HCO–

3 catalysed by several enzymes driven by light and resulting inproduction of organic matter, O2 and OH–:

HCO−3 + H2O→ CH2O + O2 + OH− [17]

This rapid interconversion from bicarbonate and protons to carbon dioxide

7

and water can be catalysed by carbon anhydrases. Carbonic anhydrase in air-grown Chara corallina was detected mainly in the intracellular fraction [21].

In our model we consider that bicarbonate is utilized for photosynthesis bycarbon anhydrases. The catalysed reaction will be considered to be linear inthe concentration of HCO–

3 and the uptake of carbon by the chloroplasts willconsist only of this to CO2 converted HCO–

3 by carbon anhydrases. We mightexpect a leakage of CO2 back in to the cytoplasm, but to avoid a too complexmodel, we assume all converted HCO–

3 will be used for photosynthesis.In Figure 3 you can see that when the intracellular pH is near 8, that intra-

cellular dissolved inorganic carbon consist almost entirely of HCO–3. In the next

paragraph in Figure 5 it is shown that the for an external pH between 5 and 8the vacuolar pH is around 8. Therefore we consider only HCO–

3 to be used forphotosynthesis and not CO2.

These descriptions are very similar to the generalized model for thecyanobacterial CO2 concentrating mechanism in [2].

1.3 pH Homeostasis

The maintenance of a stable acid-base status within biological tissue is a fun-damental homeostatic process in all organisms. In cells of Chara corallina thishomeostasis is threatened by the cross-membrane transport of protons throughthe H+/HCO–

3-symporter or through the H+/Na+-antiporters when we assumeNa+/HCO–

3-symporters to be present in cells of Chara corallina instead. Theseantiporters regulate the Na+ homeostasis.

In our model we assume that Chara corallina regulates its intracellular acid-base homeostasis with proton pumps, H+-ATPase. If the inside of the cell is tooacid, the proton pump will pump out the surplus of protons, making the insideof the cell less acid. If the inside of the cell is too alkaline, the proton pumpwill work in the other direction and pumps more protons into the cell to makeit more acid. This more a less implies a constant intracellular H+ concentrationif we assume this regulation of the pH homeostasis is fast enough.

The average pH in Chara corallina has been measured as a function of pHin aqueous environment [23]. In Figure 5 measurements are shown.

8

Figure 5: Intracellular pH in Chara from DMO distribution, as a function ofexternal pH. Squares, cytoplasmic pH; circles, vacuolar pH: open symbols, light;filled symbols, dark. S.E.M., smaller than symbol size except where indicated.Arrowed symbols indicate pH values too small to measure, i.e. less than about5.5.[23]

2 Models for a HCO–3-Symporter

The most important models used in this thesis are based on the model found inthe bachelor thesis of Mijke Carlier [16]. In this section there will be a detaileddescription of the model she used and after that the alternate derived modelswill be introduced and worked out in detail.

2.1 A Detailed Six-State Model

The model found in [16] considers H+/HCO–3-symporters to be present in the cell

membrane. These H+/HCO–3-symporters can be in six different states. These

states are combinations of the direction it’s facing (facing either the interior orexterior of the cell) and the bounding of H+ and HCO–

3 (both not bound, onlyH+ bound or both bound). See Figure 6 for a cartoon of the six different states.This gives us the six state diagram shown in Figure 7.

9

Figure 6: Cartoon of the six different states of the hypothetical symporter asin the bachelor thesis of Mijke Carlier [16]. Ext is the exterior side of themembranem and In the interior side (cytoplasm).

Figure 7: Diagram of the six states.

10

The corresponding differential equations are

dx1

dt= −k12[H+]extx1 + k21x2 + k61x6 − k16x1,

dx2

dt= k12[H+]extx1 − k21x2 − k23[HCO−3 ]extx2 + k32x3,

dx3

dt= k23[HCO−3 ]extx2 − k32x3 − k34x3 + k43x4,

dx4

dt= k34x3 − k43x4 − k45x4 + k54[HCO−3 ]inx5,

dx5

dt= k45x4 − k54[HCO−3 ]inx5 − k56x5 + k65[H+]inx6,

dx6

dt= k56x5 − k65[H+]inx6 − k61x6 + k16x1,

where kij are the rate constants from state i to j and xi(t) the fraction of

symporters in state i at time t. So it also holds that∑6i=1 xi = 1.

To get the expression for the total net inward flux of HCO–3 (and H+) through

all symporters at steady state based on this model, we look at

N(k45xss4 − k54x

ss5 ),

where N denotes the number of symporters on the membrane and xssi denotesthe fraction of symporters in state i at steady state. King-Altman’s diagram-matic method [6, 10] is used to find expressions for xss4 and xss5 . Substitutingthese expressions gives

JHCO−3in = N · k0[H+]ext[HCO−3 ]ext − k1[H+]in[HCO−3 ]in

Σ

= N · k0hfHCO−3 (h)Cext

T − k1h′fHCO−3 (h′)C in

T

Σ(2)

as the inward net number flux of HCO–3 at steady state, where k0 and k1 are con-

stants derived in the steps of the applied method, h and h′ the extracellular and

intracellular proton concentration respectively, fCO2(h), fCO2(h′), fHCO−3 (h)

and fHCO−3 (h′) denote the fractions of CO2 inside and outside the cell andHCO–

3 inside and outside the cell respectively,

CextT = [CO2]ext + [HCO−3 ]ext + [CO2−

3 ]ext + [H2CO3]ext

and

C inT = [CO2]in + [HCO−3 ]in + [CO2−

3 ]in + [H2CO3]in

denote the extracellular and intracellular concentration of total dissolved inor-ganic carbon respectively and

Σ = k2 + k3[HCO−3 ]ext + k4[HCO−3 ]in + k5[H+]ext + k6[H+]in

+ k7[HCO−3 ]ext[H+]ext + k8[HCO−3 ]in[H+]in + k9[HCO−3 ]ext[H

+]in

+ k10[HCO−3 ]in[H+]ext + k11[HCO−3 ]ext[H+]ext[HCO−3 ]in

+ k12[HCO−3 ]ext[H+]ext[H

+]in + k13[HCO−3 ]in[H+]in[HCO−3 ]ext

+ k14[HCO−3 ]in[H+]in[H+]ext + k15[HCO−3 ]in[H+]in[HCO−3 ]ext[H+]ext

(3)

11

denotes the sum of all directed diagrams, where all the ki’s are again constantsderived in the steps of this method. The expression for Σ is awkward, to saythe least, so we shall spend some time to obtain a reduction of complexity.

2.2 A Four-State Model for the H+/HCO–3-Symporter

Recall we assumed a symporter with six different states model. See Figure 6.In Figure 7 the corresponding six-state diagram is shown. With the assumptionthat the binding of [H+] with the symporter is fast in comparison with thebinding of [HCO−3 ] and the symporter switching sides. So instead of state 1 and2 separated we now get a state that combines these two.However both states still exist and so we have to look at the equilibrium of

Sym + H+ k+−−⇀↽−−k−

Sym ·H+. (4)

Let x+1 denote the fraction of symporters facing outwards with only [H+] bound

and x−1 the fraction of empty symporters facing outwards. It follows that x+1 +

x−1 = x1 is the fraction of symporters facing outwards without [HCO−3 ] bound.We get

dx+1

dt= k+[H+]extx

−1 − k−x

+1 .

For the equilibrium it must hold that

k+[H+]extx−1 − k−x

+1 = 0.

According to reaction (4) we get the following:

x+1

[H+]extx−1

=k+

k−=: Keq

x+1 = Keq · [H+]extx

−1

x+1 = Keq · [H+]ext(x1 − x+

1 )

x+1 =

Keq · [H+]ext

1 +Keq · [H+]extx1

In a similar way we get

x+4 =

Keq · [H+]in

1 + Keq · [H+]inx4,

where x+4 denotes the fraction of symporters facing inwards with only [H+]

bound and x4 is the fraction of symporters facing inwards without [HCO−3 ]bound. The corresponding diagram can be found in Figure 8, where state 1 and4 correspond to x1 and x4 respectively. This corresponds to lumping togetherx1 and x2 (state 1 and 2) and lumping together x5 and x6 (state 5 and 6) inthe six-state model (see Figure 7).

12

Figure 8: Diagram of the four states.

We get the following differential equations:

dx1

dt= −k12[HCO−3 ]extx

+1 + k21x2 + k41x

−4 − k14x

−1

= −k12[HCO−3 ]extKeq[H

+]ext

1 +Keq[H+]extx1 + k21x2 + k41(x4 −

Keq[H+]in

1 + Keq[H+]inx4)

− k14(x1 −Keq[H

+]ext

1 +Keq[H+]extx1)

= −k12Keq[H

+]ext

1 +Keq[H+]ext[HCO−3 ]extx1 + k21x2 + k41

1

1 + Keq[H+]inx4

− k141

1 +Keq[H+]extx1

dx2

dt= k12[HCO−3 ]extx

+1 − k21x2 + k32x3 − k23x2

= k12Keq[H

+]ext

1 +Keq[H+]ext[HCO−3 ]extx1 − k21x2 + k32x3 − k23x2

dx3

dt= −k32x3 + k23x2 − k34x3 + k43[HCO−3 ]inx

+4

= −k32x3 + k23x2 − k34x3 + k43Keq[H

+]in

1 + Keq[H+]in[HCO−3 ]inx4

dx4

dt= −k41x

−4 + k14x

−1 + k34x3 − k43[HCO−3 ]inx

+4

= −k411

1 + Keq[H+]inx4 + k14

1

1 +Keq[H+]extx1 + k34x3

− k43Keq[H

+]in

1 + Keq[H+]in[HCO−3 ]inx4

Now we calculate the inward flux of [HCO−3 ] at steady state, Jssin .

Jssin = N · (Jss34 − Jss43)

= N · (k34xss3 − k43x

+ss4 ),

where N is the number of symporters.

13

After applying King-Altman’s diagrammatic method [6, 10] we get

Jssin =N

Σ

(k12

Keq · [H+]ext

1 +Keq · [H+]ext[HCO−3 ]extk23k34k41

1

1 + Keq[H+]in

−k141

1 +Keq[H+]extk43

Keq · [H+]in

1 + Keq · [H+]in[HCO−3 ]ink32k21

),

=N

Σ

1

(1 +Keq[H+]ext)(1 + Keq[H+]in)(k0Keq · [H+]ext[HCO−3 ]ext − k1Keq · [H+]in[HCO−3 ]in

),

=N

Σ

(k0Keq · [H+]ext[HCO−3 ]ext − k1Keq · [H+]in[HCO−3 ]in

).

It follows that

JHCO−3in = N · k0Keq · h · fHCO−3 (h) · Cext

T − k1Keq · h′ · fHCO−3 (h′) · C inT

Σ. (5)

Considering all possible directed diagrams, we find that Σ can be written as

Σ =k21

1 +Keq · [H+]ext+k3

1

1 + Keq · [H+]in+k4

1

1 +Keq · [H+]ext[HCO−3 ]in

Keq [H+]in

1 + Keq [H+]in

+ k51

1 + Keq · [H+]in[HCO−3 ]ext

Keq [H+]ext

1 +Keq [H+]ext+ k6

Keq · [H+]ext

1 +Keq · [H+]ext[HCO−3 ]ext

+ k7Keq · [H+]in

1 + Keq · [H+]in[HCO−3 ]in

+ k8Keq · [H+]ext

1 +Keq · [H+]ext[HCO−3 ]ext

Keq · [H+]in

1 + Keq · [H+]in[HCO−3 ]in

and so

Σ = k2(1 + Keqh′) + k3(1 +Keqh) + Keqh

′fHCO−3 (h′)C inT (k4 + k7(1 +Keqh))

+KeqhfHCO−3 (h)Cext

T (k5 + k6(1 + Keqh′))

+ k8Keqh′fHCO−3 (h′)C in

T ·KeqhfHCO−3 (h)Cext

T . (6)

Assuming that h′ and CextT are constant, we get

Σ(h,C inT ) = κ0 + κ1Keq · h+ κ2Keq · h · fHCO−3 (h)

+ C inT

(κ3 + κ4Keq · h+ κ5Keq · h · fHCO−3 (h)

). (7)

2.3 The Na+/HCO–3-Symporter

Before we continue with these findings, it’s important to note that there is noevidence of the existence of a H+/HCO–

3-symporter in the biological literature.Therefore we will also consider a symporter where sodium replaces the protons.There is also no evidence for the presence of this HCO–

3/Na+-symporter in cellsof Chara corallina, but it is certain that it does exist in cyanobacteria. From nowon we may refer to these symporters as proton symporter and sodium symporterfor better readability.

14

Repeating the steps taken previously, where we replace the assumption ofproton symporters with sodium symporters, we get

JHCO−3in =

N

Σ

(k0Keq · [Na+]ext[HCO−3 ]ext − k1Keq · [Na+]in[HCO−3 ]in

),

= N · k0Keq · η · fHCO−3 (h) · CextT − k1Keq · η′ · fHCO−3 (h′) · C in

T

Σ, (8)

where η and η′ represent the extracellular and intracellular sodium concentrationrespectively and

Σ = k2(1 + Keqη′) + k3(1 +Keqη) + Keqη

′fHCO−3 (h′)C inT (k4 + k7(1 +Keqη))

+KeqηfHCO−3 (h)Cext

T (k5 + k6(1 + Keqη′))

+ k8Keqη′fHCO−3 (h′)C in

T ·KeqηfHCO−3 (h)Cext

T (9)

or with the assumption that the intracellular proton concentration, h′, and theextracellular dissolved inorganic carbon concentration, Cext

T , are constant

Σ(h,C inT ) = κ0 + κ1Keq · η + κ2Keq · η · fHCO−3 (h)

+ C inT

(κ3 + κ4Keq · η + κ5Keq · η · fHCO−3 (h)

). (10)

This expression constitutes a fair simplification compared to (3).

15

3 Initial ODE-Models for Inorganic Carbon Up-take

The structure of our system modelling the inorganic carbon uptake is the fol-lowing:

VindC in

T

dt= J

CO2

in + JHCO−3in − JHCO−3

φ (11a)

Voutdh

dt= fHCO−3 (h′)J

CO2

in − fCO2(h′)JHCO−3in (11b)

The equation (11a) is very straightforward. The first two terms are the numberinfluxes of the different inorganic carbon species. Note that there are no termsfor the influx of CO2–

3 and H2CO3, since these species cannot cross the cellmembrane of Chara corallina. The last term is the uptake of HCO–

3 inside thecell by the chloroplasts to use for the process of photosynthesis. Vin is theintracellular (cytoplasmic) volume.

We assume free cross-membrane diffusion of CO2 and so we get an expressionlinear in the difference of intracellular and extracellular CO2 concentration:

JCO2

in = PCO2A(fCO2(h)CextT − fCO2(h′)C in

T ) (12)

Here PCO2 is the permeability coefficient of CO2 through the membrane and Ais the total surface area of the cell membrane.

For symplicity we assume the uptake of HCO–3 by the chloroplast to be be

linear in the concentration of HCO–3 (see Paragraph 1.2.4):

JHCO−3φ = fHCO−3 (h′)C in

T φ. (13)

The parameter φ will depend on the light intensity, the more light, the higherφ, in a saturating fashion.

Recall that the expression for JHCO−3in is dependent on which type of sym-

porter is assumed (see Paragraph 1.2.3) and has already been determined forboth cases in the previous section.

Equation (11b) is less straightforward. Recall that we assumed protonpumps to be present in the cell membrane of cells of Chara corallina to regu-late the pH homeostasis inside the cells. Therefore we consider the intracellularproton concentration, h′, to be constant. This means that when CO2 entersthe cell there will be a surplus of this specie according to the intracellular pro-ton concentration. (See Figure 3) This surplus of CO2 (equal to the influx of

CO2, JCO2

in ) will react with water to form H+ and HCO–3. In order to get the

right fractions of inorganic carbon species fHCO−3 (h′)JCO2

in will react with wa-ter. Therefore the intracellular proton concentration would rise, but the protonpumps will maintain the pH homeostasis by pumping out the surplus of pro-

tons. This results in an increase of fHCO−3 (h′)JCO2

in in the extracellular protonconcentration, h. This explains the first term in (11b).

The second term can be explained in a similar fashion. The influx of HCO–3

will result in a surplus of HCO–3 inside the cell, J

HCO−3in . Note that in both

the cases of HCO–3-symporters effectively there will be transported a proton

16

with every HCO–3 ion across the membrane. We consider this threat to the

pH homeostasis to be solved immediately by the proton pumps. Effectivelythis doesn’t change the proton concentration anywhere. So we only have todeal with the surplus of HCO–

3 inside the cell. This is solved by a reaction of

fCO2(h′)JHCO−3in HCO–

3 ions reaction with protons to form CO2 and water. Thiswould cause a depletion of the intracellular proton concentration, but the protonpumps will maintain the pH homeostasis by pumping in protons to compensate

the proton deficit inside the cell. This results in an decrease of fCO2(h′)JHCO−3in

in the extracellular proton concentration, h. This explains the second term in(11b).

We shall elaborate on this derivation in Section 5. In [16], essentially thisreasoning was followed.

3.1 Existence of Steady States for the Generic Model

Before we make any assumption on what type of symporter is present in cells ofChara corallina, we will look at the generic Model (11) to see which conditionshave to hold in order to have at least one steady state. From Model (11) weimmediately get that at steady state

fHCO−3 (h′)JCO2

in (h?, C?T ) = fCO2(h′)JHCO−3in (h?, C?T ),

(14a)

JCO2

in (h?, C?T ) + JHCO−3in (h?, C?T )− JHCO−3

φ (h?, C?T ) = 0. (14b)

From this we obtain the following conditions:

Lemma 3.1. For the generic model (11) it holds at steady state, (h,C inT ) =

(h?, C?T ), thatC?T = c0(φ)fCO2(h?), (15)

where

c0(φ) =CextT

fCO2(h′)

(1 + fHCO

−3 (h′)

fCO2 (h′)+fHCO−3 (h′)

φ

PCO2A

) .This holds when assuming proton symporters and assuming sodium symporterto be present in cells of Chara corallina.

Proof. First we eliminate JHCO−3in in (14a) by using (14b) and noting that

fHCO−3 (h′)

fCO2(h′)=Ka1

h′.

We get (1 +

Ka1

h′

)J

CO2

in = JHCO−3φ .

The expressions for JCO2

in and JHCO−3φ , (12) and (13) respectively, are the same

for both type of symporters, so without specifying which symporter we assume tobe present in cells of Chara corallina we get after substituting these expressions

PCO2A(fCO2(h?)CextT − fCO2(h′)C?T )

(1 +

Ka1

h′

)= fHCO−3 (h′)C?Tφ.

17

So

C?T =PCO2AfCO2(h?)Cext

T

(Ka1

h′ + 1)

PCO2A(fCO2(h′) + fHCO−3 (h′)) + fHCO−3 (h′)φ

=CextT fCO2(h?)

fCO2(h′)

(1 + fHCO

−3 (h′)


φ

PCO2A

)= c0(φ)fCO2(h?),

where

c0(φ) =CextT

fCO2(h′)

(1 + fHCO

−3 (h′)


φ

PCO2A

) .

Equation (15) provides a first condition that need to hold for a state of (11).A second is derived from (14a). We first observe:

Lemma 3.2. At steady state (h,C inT ) = (h?, C?T ), Σ(h?, C?T ) satisfies

Σ = A,

where A > 0 and independent of h? and C?T when we assume proton symportersto be present in cells of Chara corallina, and

Σ(h?, C?T ) =N(A1 − h?A2)

h?A0,

where A0, A1, A2 > 0 and independent of h? and C?T , when we assume sodiumsymporters to be present instead.

Proof. We have to make a distinction between the assumption that proton sym-porters are present in cells of Chara corallina and the assumption that sodiumsymporters are present instead. When we assume proton symporters to be

present, then we get after substituting the appropriate expressions for JCO2

in

and JHCO−3in together with the result from Lemma 3.1 into (14a)

fHCO−3 (h′)

fCO2(h′)PCO2A(fCO2(h?)Cext

T − fCO2(h′)c0(φ)fCO2(h?))

=N

Σ(k0Keqh

?fHCO−3 (h?)CextT − k1Keqh

′fHCO−3 (h′)c0(φ)fCO2(h?)),

Ka1

h′PCO2A(Cext

T − c0(φ)fCO2(h′))

=N

Σ

(k0Keqh

?Ka1

h?CextT − k1Keqh

′fHCO−3 (h′)c0(φ)

),

1

h′PCO2A(Cext

T − c0(φ)fCO2(h′))

=N

Σ

(k0KeqC

extT − k1Keqf

CO2(h′)c0(φ)).

18

This can be written as

A0 =A1

Σ,

where A0, A1 > 0 (if φ > 0) and independent of h? and C?T . So Σ has to be

independent of h? and C?T as well. Hence, we can write Σ = A, where A > 0 andindependent of h? and C?T , when we assume proton symporters to be present incells of Chara corallina.

When we assume sodium symporters to be present in cells of Chara corallina,

then we get after substituting the appropriate expressions for JCO2

in and JHCO−3in

together with the result from Lemma 3.1 into (14a)

fHCO−3 (h′)

fCO2(h′)PCO2A(fCO2(h?)Cext

T − fCO2(h′)c0(φ)fCO2(h?))

=N

Σ(k0Keqηf

HCO−3 (h?)CextT − k1Keqη

′fHCO−3 (h′)c0(φ)fCO2(h?)),

Ka1

h′PCO2A(Cext

T − c0(φ)fCO2(h′))

=N

Σ

(k0Keqη

Ka1

h?CextT − k1Keqη

′fHCO−3 (h′)c0(φ)

).

This can be written as

A0 =N

Σ

(A1

h?−A2

),

where A0, A1, A2 > 0 and independent of h? and C?T . So in this case Σ is notconstant. Instead we get

Σ =N(A1 − h?A2)

h?A0

when we assume sodium symporters to be present in cells of Chara corallina.

We are now in a position to prove the main result of this paragraph:

Proposition 3.3. There is a unique non-zero steady state (h?, C?T ) to (11).

Proof. The expression for Σ can be found in (7) and (10) when assuming protonsymporters and sodium symporters to be present in cells of Chara corallinarespectively. Together with our result from Lemma 3.2 we get that at steadystate, (h,C in

T ) = (h?, C?T ), it holds that

C?T =Σ(h?, C?T )− (κ0 + κ1Keqh

? + κ2Keqh?fHCO−3 (h?))

κ3 + κ4Keqh? + κ5Keqh?fHCO−3 (h?)

=A− (κ0 + κ1Keqh

? + κ2KeqKa1fCO2(h?))

κ3 + κ4Keqh? + κ5KeqKa1fCO2(h?)

, (16)

19

Σ−κ0

κ3

−κ1

κ4

−NA2+A0(κ0+κ1Keqη)A0(κ3+κ4Keqη)

h?

c0(φ)fCO2(h?)

(16)

(17)

Figure 9: Visualization of the unique steady state when having condition (15)when we don’t make any assumptions on the type of HCO–

3-symporter presentin cells of Chara corallina (blue) and additional conditions when we do assumeproton symporters to be present in cells of Chara corallina (green) or assumesodium symporters to be present in cells of Chara corallina instead (red).

when we assume proton symporters to be present in cell of Chara corallina and

C?T =

N(A1−h?A2)h?A0

− (κ0 + κ1Keqη + κ2KeqηfHCO−3 (h?))

κ3 + κ4Keqη + κ5KeqηfHCO−3 (h?)

=N(A1 − h?A2)− h?A0(κ0 + κ1Keqη + κ2Keqηf

HCO−3 (h?))

h?A0(κ3 + κ4Keqη + κ5KeqηfHCO−3 (h?))

=NA1 − h?(NA2 +A0(κ0 + κ1Keqη))−A0κ2KeqKa1

ηfCO2(h?)

A0(h?(κ3 + κ4Keqη) + κ5KeqηKa1fCO2(h?))

, (17)

when we assume sodium symporters to be present instead. Let us denote theright-hand sides in (16) and (17) by Ch2 (h?) and Cη2 (h?) respectively. Note thatin both cases the numerator is a decreasing expression in h? and the denominator

is an increasing expression in h?, sodCh2 (h?)dh? < 0 and

dCη2 (h?)dh? < 0. Also note

that in the case of proton symporters we have

Ch2 (0) =Σ− κ0

κ3and lim

h?→∞Ch2 (h?) = −κ1

κ4< 0

and in the case of sodium symporters we have

limh?↓0

Cη2 (h?) =∞ and limh?→∞

Cη2 (h?) = −NA2 +A0(κ0 + κ1Keqη)

A0(κ3 + κ4Keqη)< 0.

Together with the result from Lemma 3.1 it follows that there is exactly onesteady state for both the assumption that proton symporters are present incells of Chara corallina and the assumption that sodium symporters are presentinstead (see Figure 9).

20

3.2 Stability of Steady States for the Generic Model

The generic model (11) can be written as(dCin

T

dtdhdt

)=

(V −1

in F1(C inT , h)

V −1outF2(C in

T , h)

),

where

F1(C inT , h) = J

CO2


φ ,

F2(C inT , h) = fHCO−3 (h′)J

CO2

in − fCO2(h′)JHCO−3in .

Note that we have a vector field that is at least C1. To say something about thestability of the steady state, (C?T , h), we want to compute the Jacobian matrixof the vector field, DF (C?T , h

?):(V −1

in∂F1

∂CinT

V −1in

∂F1

∂h

V −1out

∂F2

∂CinT

V −1out

∂F2

∂h

)(18)

We have

∂JCO2

in

∂h= PA

dfCO2(h)

dhCextT ,

∂JCO2

in

∂C inT

= −PAfCO2(h′),

∂JHCO−3φ

∂h= 0,

∂JHCO−3φ

∂C inT

= fHCO−3 (h′)φ,

and when we assume sodium symporters to be present in cells of Chara corallina

∂JHCO

−3

in

∂h= N

k0Keqηdf

HCO−3 (h)

dhCextT

Σ −(k0Keqηf

HCO−3 (h)Cext

T − k1Keqη′fHCO

−3 (h′)Cin

T

)∂Σ∂h

Σ2,

∂JHCO

−3

in

∂CinT

= N

−k1Keqη′fHCO

−3 (h′)Σ −

(k0Keqηf

HCO−3 (h)Cext

T − k1Keqη′fHCO

−3 (h′)Cin

T

)∂Σ∂CinT

Σ2.

Let λ1 and λ2 be the eigenvalues of (18) in (C inT , h) = (C?T , h

?). (C?T , h?)

is linearly stable if and only if Re λi < 0 for i = 1, 2. We know that for a2× 2-matrix, M , it holds that

Tr(M) = λ1 + λ2 and Det(M) = λ1λ2.

Hence (C?T , h?) is linearly stable if and only if Det(DF (C?T , h

?)) > 0 andTr(DF (C?T , h

?)) < 0. Due to many unknown parameters it is difficult to say ifthese conditions are met and therefore also difficult to say anything about thestability of (C?T , h

?). Therefore we will adjust this model to a more manageableone.

21

3.3 Reduced Model Derived from Time Scale Separation

To say more about the stability of the steady states, we are going to get a reducedmodel derived from time scale separation. Assume that the intracellular carbondynamics are fast compared to the extracellular proton dynamics, i.e., C in

T is atquasi-steady state. From (11a) we get

JHCO−3in = J

HCO−3φ − JCO2

in , (19)

which implies C inT at quasi-steady state as function of h. Let us write C?T =

C?T (h) for this value. Substituting (19) into (11b) gives

Voutdh

dt= fHCO−3 J

CO2

in − fCO2(h′)(JHCO−3φ − JCO2

in )

= (fCO2(h′) + fHCO−3 (h′))JCO2

in − fCO2(h′)JHCO−3φ

= (fCO2(h′) + fHCO−3 (h′))PCO2A(fCO2(h)CextT − fCO2(h′)C?T (h))

− fCO2(h′)fHCO−3 (h′)C?T (h)φ

= (fCO2(h′) + fHCO−3 (h′))PCO2ACextT fCO2(h)

− [(fCO2(h′) + fHCO−3 (h′))PCO2A+ fHCO−3 (h′)φ]fCO2(h′)C?T (h).

So the reduced model is given by

Voutdh

dt= αfCO2(h)− β(φ)C in

T , (20)

withα = (fHCO−3 (h′) + fCO2(h′))PCO2ACext

T

and

β(φ) =(

(fHCO−3 (h′) + fCO2(h′))PCO2A+ fHCO−3 (h′)φ)fCO2(h′).

3.4 Existence of Steady States for the Reduced Model

For a steady state (h,C inT ) = (h?, C?T ) for the reduced model (20) it must hold

that

fCO2(h?) =β(φ)

αC?T . (21)

Since the exact expressions in this equation have a lot of unknown constantswe will investigate when there exist intersections of the left hand side and righthand side. First we need C?T as a function of h.

Note that in both cases Σ(h,C?T ) is of the form g0(h) + g1(h)C?T , where inthe case of proton symporters we have

gh0 (h) = κ0 + κ1Keqh+ κ2KeqhfHCO−3 (h),

gh1 (h) = κ3 + κ4Keqh+ κ5KeqhfHCO−3 (h)


gη0 (h) = κ0 + κ1Keqη + κ2KeqηfHCO−3 (h),

gη1 (h) = κ3 + κ4Keqη + κ5KeqηfHCO−3 (h).

22

When we write g0(h) and g1(h), read gh0 (h) and gh1 (h) when we assume protonsymporters to be present in cells of Chara corallina and gη0 (h) and gη1 (h) when we

assume sodium symporters instead. After replacing JHCO−3in , J

CO2

in and JHCO−3φ

with the appropriate expressions ((5), (12) and (13) respectively) we get fromequation (19) that

gh1 (h)(fCO2(h′)PCO2A+ fHCO−3 (h′)φ)(C?T )2

+(gh0 (h)(fCO2(h′)PCO2A+ fHCO−3 (h′)φ)− gh1 (h)PCO2AfCO2(h)Cext

T

+Nk1Keqh′fHCO−3 (h′)

)C?T

− gh0 (h)PCO2AfCO2(h)CextT −Nk0Keqhf

HCO−3 (h)CextT = 0

and

gη1 (h)(fCO2(h′)PCO2A+ fHCO−3 (h′)φ)(C?T )2

+(gη0 (h)fCO2(h′)PCO2A+ fHCO−3 (h′)φ)− gη1 (h)PCO2AfCO2(h)Cext

T

+Nk1 · Keqη′fHCO−3 (h′)

)C?T

− gη0 (h)PCO2AfCO2(h)CextT −Nk0Keqηf

HCO−3 (h)CextT = 0

in the case of proton symporters and sodium symporters respectively. Boththese equations can be written as

a(h)(C?T )2 + b(h)C?T + c(h) = 0, (22)

where in the case of proton symporters we have

ah(h) = gh1 (h)(fCO2(h′)PCO2A+ fHCO−3 (h′)φ)

bh(h) = gh0 (h)(fCO2(h′)PCO2A+ fHCO−3 (h′)φ)− gh1 (h)PCO2AfCO2(h)CextT

+Nk1Keqh′fHCO−3 (h′),

ch(h) = −gh0 (h)CextT PCO2AfCO2(h)−Nk0Keqhf

HCO−3 (h)CextT (23a)


aη(h) = gη1 (h)(fCO2(h′)PCO2A+ fHCO−3 (h′)φ)

bη(h) = gη0 (h)(fCO2(h′)PCO2A+ fHCO−3 (h′)φ)− gη1 (h)PCO2AfCO2(h)CextT

+Nk1Keqη′fHCO−3 (h′),

cη(h) = −gη0 (h)CextT PCO2AfCO2(h)−Nk0Keqηf

HCO−3 (h)CextT . (24a)

When we write a(h), b(h) and c(h), read ah(h), bh(h) and ch(h) when we assumeproton symporters to be present in cells of Chara corallina and aη(h), bη(h)and cη(h) when we assume sodium symporters instead. Let D(h) = b(h)2 −4a(h)c(h), then

C?T (h) =−b(h) +

√D(h)

2a(h). (25)

23

Since there are a lot of unknown constants it is hard to give the behaviourof C?T (h) as function h in all detail. Therefore we are going to look at thebehaviour of C?T (h) at h = 0 and at C?T (∞) := lim

h→∞C?T (h) and deduce from

this information the existence of a steady state. In fact, (21) shows that we

have to prove that the graphs of h 7→ fCO2(h) and h 7→ β(φ)α C?T (h) intersect.

However, at h = 0, both C?T (h) = 0 and fCO2(h) = 0. So the behaviour dependson the derivatives there:

Lemma 3.4. When we assume proton symporters to be present in cells ofChara corallina we have

C?T (0) = fCO2(0) = 0,

C?T′(0) = fCO2 ′(0) = 0,

C?T′′(0) =

κ0PCO2A 2

Ka1Ka2CextT + 2Nk0

KeqKa2

CextT

κ0γ +Nk1KeqKa1fCO2(h′)

,

fCO2 ′′(0) =2

Ka1Ka2

. (26)

Proof. Appendix B.1.

Lemma 3.5. When we assume sodium symporters to be present in cells ofChara corallina we have

C?T (0) = fCO2(0) = 0, (27a)

C?T′(0) > fCO2 ′(0) = 0. (27b)

Proof. Appendix B.2.

Lemma 3.6. When we assume proton symporters to be present in cells ofChara corallina we have

limh→∞

β(φ)

αC?T (h) =

1 + fHCO−3 (h′)φ

(fHCO−3 (h′)+fCO2 (h′))PCO2A

1 + fHCO−3 (h′)φ

fCO2 (h′)PCO2A

Ka3

Ka3+Ka1

.

Proof. In Appendix B.3 it is shown that

limh→∞

C?T (h) =Ka3

Ka3+Ka1

PCO2ACextT

fCO2(h′)PCO2A+ fHCO−3 (h′)φ.

We have

β(φ)

α=fCO2(h′)

CextT

(1 +

fHCO−3 (h′)φ

(fHCO−3 (h′) + fCO2(h′))PCO2A

).

24

So it follows that

limh→∞

β(φ)

αC?T (h) =

fCO2(h′)

CextT

1 +fHCO

−3 (h′)φ

(fHCO

−3 (h′) + f

CO2(h′))PCO2A

·Ka3

Ka3 +Ka1

PCO2ACext

T

fCO2(h′)PCO2A + f

HCO−3 (h′)φ

=

1 +fHCO

−3 (h′)φ

(fHCO

−3 (h′) + f

CO2(h′))PCO2A

Ka3

Ka3+Ka1

fCO2(h′)PCO2A

fCO2(h′)PCO2A + f

HCO−3 (h′)φ

=

1 +fHCO

−3 (h′)φ

(fHCO

−3 (h′) + f

CO2(h′))PCO2A

1 +

fHCO

−3 (h′)φ

fCO2(h′)PCO2A

−1

Ka3

Ka3+Ka1

Lemma 3.7. When we assume sodium symporters to be present in cells ofChara corallina we have

limh→∞

β(φ)

αC?T (h) <

1 + fHCO−3 (h′)φ


1 + fHCO−3 (h′)φ

fCO2 (h′)PCO2A

Ka3

Ka3+Ka1

.

(The exact expression has been determined in Appendix B.4 (Equations (82)and (83)), but it is too complicated to use for further analysis. Therefore weonly state the inequality in this Lemma.)

Proof. In Appendix B.4 it is shown that

limh→∞

C?T (h) <Ka3

Ka3+Ka1

PCO2ACextT


We have

β(φ)

α=fCO2(h′)

CextT

(1 +

fHCO−3 (h′)φ

(fHCO−3 (h′) + fCO2(h′))PCO2A

).

So it follows that

limh→∞

β(φ)

αC?T (h) <

1 +fHCO

−3 (h′)φ

(fHCO

−3 (h′) + f

CO2(h′))PCO2A

1 +

fHCO

−3 (h′)φ

fCO2(h′)PCO2A

−1

Ka3

Ka3 +Ka1

From Lemma 3.4 it is not immediately clear whether

β(φ)

αC?T (h) > fCO2(h)

orβ(φ)

αC?T (h) < fCO2(h)

near h = 0. Therefore we have to compare β(φ)α C?T

′′(h) and fCO2 ′′(h):

25

Lemma 3.8. When we assume proton symporters to be present in cells ofChara corallina, there exists h such that

β(φ)

αC?T (h) > fCO2(h)

when h′ > h andβ(φ)

αC?T (h) < fCO2(h)

when h′ < h.

Proof. When we assume proton symporters to be present in cells of Chara coral-lina we have

β(φ)C?T′′(0)

α=

((fCO2(h′) + fHCO−3 (h′))PCO2A+ fHCO−3 (h′)φ)fCO2(h′)

(fCO2(h′) + fHCO−3 (h′))PCO2ACextT

·κ0P

CO2A 2Ka1Ka2

CextT + 2Nk0

KeqKa2

CextT


=

(1 +

fHCO−3 (h′)φ

(fCO2(h′) + fHCO−3 (h′))PCO2A

)

· 2

Ka1Ka2

κ0PCO2A+Nk0KeqKa1

κ0

(PCO2A+

Ka1

h′ φ)

+Nk1KeqKa1

.

Using k0 = k1 (thermodynamic constraints [22]) and Keq = Keq we get

β(φ)C?T′′(0)

α=

(1 +

fHCO−3 (h′)φ

(fCO2(h′) + fHCO−3 (h′))PCO2A

)

· 2

Ka1Ka2

κ0PCO2A+Nk0KeqKa1

κ0

(PCO2A+

Ka1

h′ φ)

+Nk0KeqKa1

=2

Ka1Ka2

(1 +

fHCO−3 (h′)φ

r

)q

q + κ0Ka1

h′ φ, (28)

where

q = κ0PCO2A+Nk0KeqKa1

,

r = (fHCO−3 (h′) + fCO2(h′))PCO2A.

Now we know from (26) and (28) that we have to determine whether thefraction

r + fHCO−3 (h′)φ

r· q

q + κ0Ka1

h′ φ

26

is bigger than 1, equal to 1 or smaller than 1. Rewriting this fraction gives

r + fHCO−3 (h′)φ

r· q

q + κ0Ka1

h′ φ=qr + qfHCO−3 (h′)φ

qr + rκ0Ka1

h′ φ

=1 + fHCO

−3 (h′)φr

1 +κ0Ka1

φ

qh′

.

It’s clear that for φ = 0, this is equal to 1. So if we can determine the sign of

d

dφ

1 + fHCO−3 (h′)φr

1 +κ0Ka1

φ

qh′

(29)

we are done. This expression is of the form

d

dφ

[1 + aφ

1 + bφ

],

where a, b > 0 and independent of φ. Using the quotient rule we get

d

dφ

[1 + aφ

1 + bφ

]=a(1 + bφ)− b(1 + aφ)

(1 + bφ)2

=a− b

(1 + bφ)2.

We can see that the sign of this expression is the sign of a− b. Now from (29)we can see that we are interested in the sign of

fHCO−3 (h′)

r− κ0Ka1

qh′.

Plugging back in the expressions for r and q, this becomes

fHCO−3 (h′)

(fCO2(h′) + fHCO−3 (h′))PCO2A− κ0Ka1

h′(κ0PCO2A+Nk0KeqKa1).

Rewriting this expression gives

fHCO−3 (h′)


h′(κ0PCO2A+Nk0KeqKa1)

=fHCO−3 (h′)


h′(κ0PCO2A+Nk0KeqKa1)

=Ka1

(h′ +Ka1)PCO2A− Ka1

h′(PCO2A+ N

κ0k0KeqKa1

)= Ka1

h′(PCO2A+ N

κ0k0KeqKa1

)− PCO2A(h′ +Ka1

)

h′(PCO2A+ N

κ0k0KeqKa1

)(h′ +Ka1

)PCO2A

= K2a1

h′ Nκ0k0Keq − PCO2A

h′(PCO2A+ N

κ0k0KeqKa1

)(h′ +Ka1)PCO2A

27

1

h

fCO2(h)β(φ)α C?T (h) (h′ big enough)

β(φ)α C?T (h) (h′ small enough)

Figure 10: The blue graph is the sketch of fCO2(h). The red and green graph are

sketches of β(φ)α C?T (h) when the interior of the cell is acidic enough and alkaline

enough respectively.

Now it’s enough to know the sign of h′ Nκ0k0Keq − PCO2A.

Another approach is to say that the sign of the derivative is independentof φ and so we can determine it by showing where the expression between thebrackets in (29) goes to as φ→∞. We are not showing the steps here, but thisapproach also ends up in questioning the sign of h′ Nκ0

k0Keq − PCO2A.There are too many unknown constants in this expression to say whether

this is positive or negative, but due to the dependency of h′ we can say thatthis is positive if h′ is big enough (pH-value inside the cell low enough) andnegative if h′ is small enough (pH-value inside the cell high enough). Also there

is a unique value h′ = h when it equals 0. Figures 10 and 11 show the differentpossibilities.

Proposition 3.9. There exists an odd number of steady states when we assumesodium symporters to be present in cells of Chara corallina. There exists an oddnumber of steady states when we assume proton symporters to be present in cellsof Chara corallina when certain conditions are met. Otherwise there exists aneven number of steady states (including 0).

Proof. Since

1 + fHCO−3 (h′)φ


1 + fHCO−3 (h′)φ

fCO2 (h′)PCO2A

< 1,

it is clear from Lemma 3.6 and 3.7 that

limh→∞

β(φ)

αC?T (h) <

Ka3

Ka3+Ka1

= limh→∞

fCO2(h),

when we assume proton symporters or sodium symporters to be present in cellsof Chara corallina. Hence, the existence of a steady state can only be assuredwhen near h = 0

β(φ)

αC?T (h) > fCO2(h).

28

0.1× 10−11 0.5× 10−11 1× 10−11

1

2

3

4

·10−6

h (M)

fCO2(h)β(φ)α C?T (h) (h′ big enough)

β(φ)α C?T (h) (h′ small enough)

Figure 11: The blue graph is the actual graph of fCO2(h) near h = 0. The reden green graph are again sketches like in Figure 10.

From Lemma 3.5 it’s immediately clear that this holds when we assume sodiumsymporters to be present in cells of Chara corallina. From Lemma 3.8 we getthat this holds when we assume proton symporters to be present instead if theinterior of the cell is acidic enough (h′ big enough). In Figure 12 you can seethat there exists at least one steady state in these cases.

In the cases where near h = 0 it holds that

β(φ)

αC?T (h) > fCO2(h),

there exist steady states. If we order the steady states

0 < h?0 < h?1 < . . . < h?n,

then we can conclude that for h?0 < h < h?1 it holds that

β(φ)

αC?T (h) < fCO2(h)

and then again for h?1 < h < h?2 it holds that

β(φ)

αC?T (h) > fCO2(h).

Continuing like this gives that for h?i < h < h?i+1 it holds that

β(φ)


when i is odd andβ(φ)

αC?T (h) < fCO2(h),

29

1

h

fCO2(h)β(φ)α C?T (h) (sodium symporters)

β(φ)α C?T (h) (h′ big enough & proton symporters)

Figure 12: The blue graph is a sketch of fCO2(h). The green graph is a sketch ofβ(φ)α C?T (h) when we assume sodium symporters to be present in cells of Chara

corallina and the red graph is a sketch of β(φ)α C?T (h) when we assume proton

symporters instead with the interior of the cell acidic enough.

when i is even. Therefore n has to be even in order to have

β(φ)

αC?T (h) < fCO2(h)

for h > h?n. So there exists an odd number of steady states for the cases wherenear h = 0 it holds that

β(φ)


i.e., when we assume sodium symporters to be present in cells of Chara corallinaand when we assume proton symporters to be present instead with the interiorof the cell acidic enough.

For the case where near h = 0 it holds that near h = 0

β(φ)


we can follow a similar elaboration and conclude that n has to be odd in orderto have

β(φ)

αC?T (h) < fCO2(h)

for h > h?n. So there exists an even number of steady states (including 0) forthe cases where near h = 0 it holds that

β(φ)


i.e., when we assume proton symporters to be present in cells of Chara corallinawith the interior of the cell alkaline enough.

Now we have proven the existence of steady states in some cases and analysedhow much steady states there can be in all cases, we are interested in the stabilityof these steady states.

30

3.5 Stability of Steady Sates for the Reduced Model

It was shown that when we assume sodium symporters to be present in cells ofChara corallina that there is always at least one steady state. Recall from (20)that we had

dh

dt= αfCO2(h)− β(φ)C?T (h).

At the intersection in Figure 12 this yields dhdt = 0, so we say the intersection is

at the steady state h?. Note that for h / h? we have dhdt < 0 and for h ' h? we

have dhdt > 0. So the steady state h? is unstable.

It was also shown that there can be more than one steady state, so if weorder the steady states like before,

h?0 < h?1 < h?2 < . . . < h?n,

where n is even, then these findings are about h?i when i is even.Note that the inequalities are opposite for any subsequent steady state and

so h?i is stable when i is odd.All this is under the assumption that there are no tangent points.It was also shown that when we assume proton symporters to be present in

cells of Chara corallina that there is at least one steady state when the interior ofthe cell is acidic enough, i.e., h′ is big enough. The observations in the previouscase also hold for this case.

The only remaining case is when we assume proton symporters to be presentin cells of Chara corallina and that the interior of the cell is not acidic enough,i.e., h′ not big enough. It’s not certain that there exists a steady state in thiscase, but if there exists steady states, then there are an even number of steadystates. If we order the steady states

h?0 < h?1 < h?2 < . . . < h?n,

then h?i is stable when i is even and unstable when i is odd. This is under theassumption that there are no tangent points.

3.6 Discussion

In this section different approaches and assumptions has been taken. All withtheir upsides and downsides.

The generic model has a unique steady state no matter if we assume pro-ton symporters to be present in cells of Chara corallina or sodium symportersinstead. The downside is that is difficult (or even impossible) to determine thestability of this steady state. This is caused by all the unknown constants inthe symporter expressions.

The reduced model had at least one steady state when we assumed sodiumsymporters to be present in cells of Chara corallina. When we assumed protonsymporters to be present in cells of Chara corallina, we had the additionalcondition that the interior of the cell had to be acidic enough for a steady stateto exist. In the cases where we had the existence of at least one steady state, weshowed that there was surely one unstable steady state. In all cases the numberof steady states is unclear.

31

In this section we assumed CextT to be constant. The slow diffusivity of inor-

ganic carbon species in an aqueous solution may however result in a significantchange of the inorganic carbon concentration at the surface of the cell. Sincethe cell is acting as a sink for inorganic carbon, due to carbon uptake for pho-tosynthesis, the inorganic carbon concentration at the surface of the cell willbe smaller than the inorganic carbon concentration far out in the environment.This depletion of inorganic carbon is dependent on the uptake of inorganic car-bon by the cell (and therefore also on the pH value at the cell and intracellularinorganic carbon concentration) and we get Cext

T (C inT , h) instead of it being con-

stant.

32

4 Diffusion-Limited Supply of Inorganic Carbon- the Unstirred Layer

Chara corallina is a fresh water alga. The carbon supply to the cells of Characorallina is limited due to its aqueous environment, since the diffusion of dis-solved inorganic carbon through unstirred water is relatively slow. If the supplyof carbon to the cells is not fast enough to keep up with the carbon uptake ofthe cell, the unstirred water is considered to be a limiting factor.

It has been shown that around leaves of aqueous plants and algae there isa thin layer of unstirred water, even if the leaf or alga is in streaming water[9]. This layer is called the unstirred layer. In the experimental set-up for bandformation for Chara corallina, the whole Petri dish forms the unstirred layer.

We assume that the unstirred water around the cell has a finite length, L,and it will be referred to as the unstirred layer. Another assumption is thatthe concentration of carbon outside the unstirred layer, u, is constant. Thepurpose of this section is to determine the decrease of total inorganic carbonconcentration near the cell due to limited diffusion at steady state (with constantuptake by the cell assumed).Moreover, we want to find the time scale on whichthis steady state is approached.

4.1 Diffusion in Total Carbon Concentration

In the extracellular medium inorganic carbon is present in different forms. Themost important ones being carbon dioxide and bicarbonate. For now we ignorethis fact and assume there is one form of carbon in this medium (and hencethere is one diffusivity constant for carbon). We will refer to this form of carbonas total carbon, since it can be thought of as the total of the different forms ofcarbon when these forms would have the same diffusivity constant. However, westill consider the carbon to be in the form of carbon dioxide or bicarbonate whenwe look at the uptake of carbon through the cell membrane. We do this becausein our model these two forms of carbon pass the cell membrane in a totallydifferent manner. Carbon dioxide can diffuse freely through the membrane andbicarbonate must cross the membrane through symporters. Inside the cell weassume a constant concentration of total inorganic carbon, C in

T .With these assumptions we get the following initial/boundary-value prob-

lem: ut = Duxx, on (0, L)

Dux(0, t) = 1A

(J

CO2

in + JHCO−3in

), t ≥ 0

u(L, t) = u, t ≥ 0u(x, 0) = u, for x ∈ (0, L)

(30)

Here u represents the carbon concentration, JCO2

in and JHCO−3in are the number

fluxes of CO2 and HCO–3 through the membrane respectively given by (12) and

(5)/(8). As before D is the diffusivity constant for the total carbon and A isthe area of the cell membrane.

Recall that we have two possible symporters transporting HCO–3 through the

cell membrane. This effects the expression for JHCO−3in but not the expression

33

for JCO2

in . The expression for the latter in both cases is

JCO2

in = PCO2 ·A(fCO2(h)CextT − fCO2(h′)C in

T )

= PCO2 ·A · fCO2(h′)

(fCO2(h)

fCO2(h′)CextT − C in

T

)= PA

(RCext

T − C inT

)where PCO2 is the permeability coefficient of CO2 through the membrane, P =

PCO2 · fCO2(h′) and R = fCO2 (h)

fCO2 (h′)(accumulation ratio). Cext

T and C inT are the

total carbon concentration at the exterior and interior of the cell. Note thatCextT = u(0, t) and can be used interchangeably so we can pick the one that suits

the situation the best. For consistency in notation we also introduce ui = C inT

so we get interchangeable pairs (CextT , C in

T ) and (u(0, t), ui).Recall from (5) and (8) that when we assume proton symporters to be present

in cells of Chara corallina we get

JHCO−3in = ρs ·A ·

k0KeqhfHCO−3 (h)Cext

T − k1Keqh′fHCO−3 (h′)C in

T

Σ

and when we assume sodium symporters to be present instead we get

JHCO−3in = ρs ·A ·

k0KeqηfHCO−3 (h)Cext

T − k1Keqη′fHCO−3 (h′)C in

T

Σ.

Here ρs is the surface density of HCO–3-symporters. So ρs · A = N denotes the

number of symporters in the cell membrane.Substituting these expressions into (30) (and using (u(0, t), ui) as notation)

we get

ut = DuxxDux(0, t) = P (Ru(0, t)− ui)

+ρsk0Keqhf

HCO−3 (h)u(0,t)−k1Keqh

′fHCO−3 (h′)ui

Σ

(proton

symporters

)Dux(0, t) = P (Ru(0, t)− ui)

+ρsk0Keqηf

HCO−3 (h)u(0,t)−k1Keqη


Σ

(sodium

symporters

)u(L, t) = uu(x, 0) = u

(31)

The specific expressions of JHCO−3in are not changing through the various steps

we are going to take, so we introduce

S(h, u(0, t)) = ρsk0Keqhf

HCO−3 (h)u(0, t)− k1Keqh′fHCO−3 (h′)ui

Σ=J

HCO−3in

A(32)

in the case of proton symporters and

S(h, u(0, t)) = ρsk0Keqηf

HCO−3 (h)u(0, t)− k1Keqη′fHCO−3 (h′)ui

Σ=J

HCO−3in

A(33)

34

Figure 13: Visualisation of all assumptions with the steady state solution (blue),the initial condition (red) and a sketch of the solution for some t > 0. ’UL’ isthe unstirred layer.

in the case of proton symporters. Note that Σ in (32) and (33) are different ((6)and (9) respectively). Now we get in both cases

ut = DuxxDux(0, t) = P (Ru(0, t)− ui) + S(h, u(0, t))u(L, t) = uu(x, 0) = u

. (34)

To reduce the number of parameters, we take ξ = xL and τ = tD

L2 , then we getuτ = uξξDL uξ(0, τ) = P (Ru(0, τ)− ui) + Su(1, τ) = uu(ξ, 0) = u

. (35)

4.2 Steady State Solution

Let u?(ξ) be the steady state solution of (35), then u?ξξ(ξ) = 0, so

u?(ξ) = c1ξ + c2,

where c1, c2 ∈ R (see Figure 13). The boundary conditions require

u?ξ(0) =PL

D(Ru?(0)− ui) +

L

DS and u?(1) = u.

Therefore

c1 =PL

D(Rc2 − ui) +

L

DS,

c1 + c2 = u.

35

Substituting the first of these equations into the second and c2 = u − c1 intothe first equation gives

PL

D(Rc2 − ui) +

L

DS + c2 = u,

c2(PLR

D+ 1) = u+

PL

Dui −

L

DS,

c2 =u+ PL

D ui − LDS

PLRD + 1

and

c1 =PL

D(R(u− c1)− ui) +

L

DS,

c1(PLR

D+ 1) =

PL

D(Ru− ui) +

L

DS,

c1 =PLD (Ru− ui) + L

DSPLRD + 1

.

Introduce ν = PLRD . We get

c1 =(νu− ν

Rui) + LDS

ν + 1, (36a)

c2 =u+ ν

Rui −LDS

ν + 1. (36b)

Note that S is dependent on c2, so these are not explicit solutions yet.

Lemma 4.1. There exists a unique positive solution c2 to (36b).

Proof. We have

c2 =u+ ν

Rui −LDS

ν + 1,

with the expression for S is as in (32) or (33). In these expressions we have Σfor which we have found the expressions in (6) or (9). For solving (36b) we arenot interested in the dependency of S or Σ on anything other than c2. So inboth cases (proton and sodium symporters) we can write

Σ = α+ βc2,

S(c2) =−γ + δc2α+ βc2

,

where α, β, γ, δ > 0 are independent of c2. After substituting this into (36b) weget

c2 =u+ ν

Rui −LD−γ+δc2α+βc2

ν + 1,

(ν + 1)c2(α+ βc2) = (α+ βc2)(u+

ν

Rui

)− L

D(−γ + δc2),

0 = (ν + 1)βc22 +

[(ν + 1)α− β

(u+

ν

Rui

)+L

Dδ

]c2

− α(u+

ν

Rui

)− L

Dγ.

36

We can write the last equation as

Ac22 +Bc2 + C = 0

and solve it with the abc-formula. We get

c±2 = − B

2A±

√(B

2A

)2

− C

A.

Note that c±2 ∈ R and so there is a unique positive solution if and only ifc+2 · c

−2 < 0. We have

c+2 · c−2 =

C

A

=−α

(u+ ν

Rui)− L

Dγ

(ν + 1)β.

Note that all parameters are positive and since α, β, γ > 0 it follows that thenumerator is negative and the denominator is positive and so c+2 ·c

−2 < 0. Hence

there is a unique positive solution to (36b).

Remark. It follows that there is a unique solution c1 ∈ R too.

We now want to derive an expression for c2 = u?(0), as this is the steadystate total inorganic carbon concentration at the cell membrane, when u isthe inorganic carbon concentration ’far out’ in the environment and ui is theintracellular constant concentration of inorganic carbon. We have to distinguishthe sodium symporter and proton symporter case.

4.2.1 Inorganic Carbon at the Membrane - Sodium Case

Recall

JCO2

in := PA(RfCO2(h)u(0, τ)− ui),

JHCO−3in := NA

k0KeqηfHCO−3 (h)u(0, τ)− k1Keqη


Σ

=Cs(rsf

HCO−3 (h)u(0, τ)− ui)σ0 + σ1ui + σ2fHCO−3 (h)u(0, τ) + σ3fHCO−3 (h)uiu(0, τ)

= CsRs(h)u(0, τ)− ui

a(ui) + b(h;ui)u(0, τ).

where

Rs(h) = rsfHCO−3 (h) =

k0Keqη

k1Keqη′· f

HCO−3 (h)

fHCO−3 (h′),

Cs = NAk1Keqη′fHCO−3 (h′),

a(ui) = σ0 + σ1ui,

b(h;ui) = (σ2 + σ3ui)fHCO−3 (h).

37

We can write

JHCO−3in = C(h;ui)

u(0, τ)− α(h;ui)

u(0, τ) + β(h;ui), (37)

where

α(h;ui) =ui

Rs(h), β(h;ui) =

a(ui)

b(h;ui)and C(h;ui) =

CsRs(h)

b(h;ui).

The only root of (37) is u(0, τ) = α and the value ofJ

HCO−3

in

C for u(0, τ) = 0 is−αβ . There is a vertical asymptote at u(0, τ) = −β. Note that

α

β=

uiRs(h)

· b(h;ui)

a(ui)=ui(σ2 + σ3ui)

rs(σ0 + σ1ui)(38)

is independent of extracellular acidity, h. Similarly,

C =Csrs

σ2 + σ3ui

is also independent on h. Therefore the graph of (37) will always go throughthe point (

u?(0),J

HCO−3in

C

)= (0, c),

where −αβ = c ∈ R, for all h.

In an acidic environment the fraction of HCO−3 is very small and thereforeα will be large. In an alkaline environment the fraction of HCO−3 is larger andtherefore α will be closer to ui

rs. Since α

β is not dependent on the acidity of the

environment, the same must hold for β (which will be closer to σ0+σ1uiσ2+σ3ui

in analkaline environment).In Figure 14 you can find plots of

JHCO−3in

C=u(0, τ)− αu(0, τ) + β

for a more acidic environment (red) and an alkaline environment (blue).For the steady state we get

u?(ξ) = c1ξ + c2, (39a)

u?ξ(0) =L

D

(P (Ru?(0)− ui) + C(h;ui)

u?(0)− αu?(0) + β

), (39b)

u?(1) = u. (39c)

It follows that

u?ξ(0) = c1

=L

D

(P (Ru?(0)− ui) + C(h;ui)

u?(0)− αu?(0) + β

),

u?(0) = c2

= (c1 + c2)− c1= u− c1.

38

−β −β α α

−αβ

1

u(0,τ)

JHCO

−3

in

C

Figure 14: Plots ofJ

HCO−3

in

C = u(0,τ)−αu(0,τ)+β for an alkaline environment (blue) and a

more acidic region (red).

Solving gives:

u?(0) = u− L

D

(P (Ru?(0)− ui) + C(h;ui)

u?(0)− αu?(0) + β

)u?(0)(ν + 1) = u+

LP

Dui −

LC

D

u?(0)− αu?(0) + β

= µ− λu?(0)− αu?(0) + β

u?(0)(ν + 1)(u?(0) + β) = µ(u?(0) + β)− λ(u?(0)− α)

0 = u?(0)2(ν + 1) + u?(0) [β(ν + 1)− µ+ λ]− βµ− αλ

Here we have taken

ν =LPR

D, µ = u+

LP

Dui and λ =

LC

D.

This can be solved with the abc-formula where the discriminant is

D = [β(ν + 1)− µ+ λ]2

+ 4(ν + 1)(βµ+ αλ)

= β2(ν + 1)2 + µ2 + λ2 − 2βµ(ν + 1) + 2βλ(ν + 1)− 2λµ+ 4βµ(ν + 1)

+ 4αλ(ν + 1)

= β2(ν + 1)2 + µ2 + λ2 + 2βµ(ν + 1) + 2βλ(ν + 1)− 2λµ+ 4αλ(ν + 1)

= (β(ν + 1) + µ)2 + λ [λ+ 2β(ν + 1)− 2µ+ 4α(ν + 1)]

39

and our solution is

u?(0) =1

2(ν + 1)

[(β(ν + 1)− µ+ λ)

(−1 +

√1 +

4(ν + 1)(βµ+ αλ)

(β(ν + 1)− µ+ λ)2

)]

≈ β(ν + 1)− µ+ λ

2(ν + 1)

(−1 + 1 +

2(ν + 1)(βµ+ αλ)

(β(ν + 1)− µ+ λ)2

)=β(ν + 1)− µ+ λ

2(ν + 1)

(2(ν + 1)(βµ+ αλ)

(β(ν + 1)− µ+ λ)2

)=

βµ+ αλ

β(ν + 1)− µ+ λ,

where we used the Tayor expansion of√

1 + x near x = 0, i.e.,

√1 + x ∼ 1 +

x

2.

This approximation is valid only if

4(ν + 1)(βµ+ αλ)

(β(ν + 1)− µ+ λ)2� 1.

We would like to know the effect of the transporter on the dissolved inorganiccarbon concentration at the outside of the cell. So we introduce u?(0) which isthe concentration of dissolved inorganic carbon at the outside of the cell when

there is no (transport of carbon through the) symporter, i.e., JHCO−3in = 0.

The ratio of the concentrations is

u?(0)

u?(0)=

βµ+ αλ

β(ν + 1)− µ+ λ· ν + 1

µ

=β + αλµ

β + λ−µν+1

=1 + α

βλµ

1 + λ−µβ(ν+1)

=1 + α

βλµ

1 + µα(ν+1)

αβ

(λµ − 1

) .Recall from (38) that α

β is independent of h, so the h-dependence of this ratio

is limited to 1α . There are no further conclusions to be made, because of the

unknown parameters in this expression.

Instead of looking for an exact solution we can approximateJ

HCO−3

in

C = u?(0)−αu?(0)+β

by 1βu

?(0)− αβ , then our problem (39) becomes

u?(ξ) = c1ξ + c2, on (0, 1),

u?ξ(0) = LD

(P (Ru?(0)− ui) + C

(1βu

?(0)− αβ

)),

u?(1) = u.

(40)

40

−β −β α α

−αβ

1

u(0,τ)

JHCO

−3

in

C

Figure 15: Estimates ofJ

HCO−3

in

C = u(0,τ)−αu(0,τ)+β for an alkaline environment (blue)

and a more acidic region (red).

and we get

u?(0) = u− L

D

(P (Ru?(0)− ui) + C

(1

βu?(0)− α

β

)),

0 = u+LP

Dui +

LCα

Dβ− u?(0)

(1 +

LPR

D+LC

Dβ

),

u?(0) =u+ LP

D ui + LCαDβ

1 + LPRD + LC

Dβ

=u+ k1ui + k2

1 + k1R+ k2α−1

where k1 = LPD and k2 = LC

Dαβ . For acidic regions we expect this to be a

small overestimate and in alkaline regions an underestimate for which the sizeis unknown. In Figure 15 you can find the plots of Figure 14 with the approxi-mations.

4.2.2 Inorganic Carbon at the Membrane - Proton Case

Recall

JCO2

in := PA(RfCO2(h)u(0, τ)− ui),

JHCO−3in := NA

k0KeqhfHCO−3 (h)u(0, τ)− k1Keqh


Σ

=Cs(rshf

HCO−3 (h)u(0, τ)− ui)

σ0 + σ1h+ σ2ui + σ3hui + σ4hfHCO−3 (h)u(0, τ) + σ5hfHCO−3 (h)uiu(0, τ)

= CsRs(h)u(0, τ)− ui

a(h;ui) + b(h;ui)u(0, τ).

41

where

Rs(h) = rshfHCO−3 (h) =

k0Keq

k1Keq

· hfHCO−3 (h)

h′fHCO−3 (h′),

Cs = NAk1Keqh′fHCO−3 (h′),

a(h;ui) = σ0 + σ2ui + h(σ1 + σ3ui),

b(h;ui) = (σ4 + σ5ui)hfHCO−3 (h).

We can write

JHCO−3in = C(h;ui)

u(0, τ)− α(h;ui)

u(0, τ) + β(h;ui), (41)

where

α(h;ui) =ui

Rs(h), β(h;ui) =

a(h;ui)

b(h;ui)and C(h;ui) =

CsRs(h)

b(h;ui).

The only root of (41) is u(0, τ) = α and the value ofJ

HCO−3

in

C for u(0, τ) = 0is −αβ . There is a vertical asymptote at u(0, τ) = −β. Note that

α

β=

uiRs(h)

· b(h;ui)

a(ui)=

ui(σ4 + σ5ui)

rs(σ0 + σ2ui + h(σ1 + σ3ui))

is not dependent on h, but

C =Csrs

σ4 + σ5uiis still not dependent on h. Therefore the graph of (41) will not have any ’anchorpoints’ like in the case for sodium symporters.

Note that α(h;ui) and α(h;ui)β(h;ui)

decrease as h increases. The behaviour of

β(h;ui) is not immediately apparent. We can rewrite β(h;ui) as

σ0 + σ2ui(σ4 + σ5ui)Ka1f

CO2(h)+

σ1 + σ3ui

(σ4 + σ5ui)fHCO−3 (h).

The first term decreases from∞ to a constant larger than 0 as h increases from0 to ∞. The second term has a minimum and goes to ∞ as h → 0 or h → ∞.This is enough to conclude there is a minimum for β(h;ui).

In Figure 16 you can find plots of

JHCO−3in

C=u(0, τ)− αu(0, τ) + β

for a more acidic environment (red), an alkaline environment (blue) and a en-vironment in between (where the value of β is smaller than in both the othercases) (green). In Figure 17 you can find the plots of Figure 16 with the ap-proximations.

For the steady state we get again problem (39). Solving this will go ina similar way as before. If we use the same approximation as in the sodiumcase, we get again (40). However, in this case we expect the solution to be anunderestimate for acidic regions and an overestimate for alkaline regions. It’smore difficult to say anything about the size of the error, since we do not havethe anchor point, (0,−αβ ), like we did before (see Figure 17).

42

−β −β−β ααα

−αβ

−αβ

−αβ

1

u(0,τ)

JHCO

−3

in

C

Figure 16: Plots ofJ

HCO−3

in

C = u(0,τ)−αu(0,τ)+β for an alkaline environment (blue) and a

more acidic region (red) and one in between (green).

−β −β−β ααα

−αβ

−αβ

−αβ

1

u(0,τ)

JHCO

−3

in

C

Figure 17: Estimates ofJ

HCO−3

in

C = u(0,τ)−αu(0,τ)+β for an alkaline environment (blue)

and a more acidic region (red) and one in between (green).

43

4.3 Convergence to Steady State

First we look at the case when the sodium symporters are inactive. In theorythis can be achieved by fixing experimentally the sodium concentration at zeroor make it very low. In this case we get problem (34) with S = 0. Then weconsider the case where S 6= 0. We show that the steady state u? then is linearlystable by reduction to the case S = 0 essentially.

4.3.1 Inactive Sodium Symporters

Theorem 4.2. When S = 0, u? is stable and the solution converges to u? atexponential rate λ = λ0

DL2 . It satisfies

π2

4

D

L2< λ < π2 D

L2.

Proof. The proof of this theorem takes this entire section.

If we set S = 0 and let v = u− u?, then from (35) we get

vτ = vξξ (42a)

vξ(0, τ) =PLR

Dv(0, τ) (42b)

v(1, τ) = 0 (42c)

v(ξ, 0) = c1(1− ξ). (42d)

Now we apply separation of variables. Assume v(ξ, τ) = X(ξ)T (τ), then

X(ξ)T ′(τ) = X ′′(ξ)T (τ),

T ′(τ)

T (τ)=X ′′(ξ)

X(ξ)= −λ, (43)

where λ ∈ C is the separation constant. We get

X ′′(ξ) + λX(ξ) = 0, (44)

T ′(τ) + λT (τ) = 0. (45)

Proposition 4.3. There exists a non-zero solution Xλ to

X ′′ + λX = 0

and the boundary conditions

X(1) = 0 and X ′(0) = νX(0), (ν > 0),

if and only if λ > 0.

Proof. Appendix A.

For λ > 0 we have

X(ξ) = c3ei√λξ + c4e

−i√λξ, (46)

X(1) = 0,

X ′(0) =PLR

DX(0)

= νX(0),

44

andT (τ) = c5e

−λτ

where c3, c4 ∈ C, c5 ∈ R and ν = PLRD .

Substitution of the boundary conditions gives

c3ei√λ + c4e

−i√λ = 0, (47)

i√λ(c3 − c4) = ν(c3 + c4). (48)

With X(ξ) ∈ R we can say

c3ei√λξ + c4e

−i√λξ = c3e

−i√λξ + c4e

i√λξ, (49)

c3 = c4. (50)

Substitution of (50) into (47) gives

c3 = −c3e−2i√λ,

c23 = −|c3|2e−2i√λ

= |c3|2eiπe−2i√λ,

c3 = ±|c3|eiπ2 e−i

√λ

= ±re−i√λ+iπ2 , (51)

where r = |c3|. It follows from (50) that

c4 = ∓rei√λ+iπ2

= ±rei√λ−iπ2 . (52)

Substitution of (51) and (52) into (48) gives

i√λ(±r(

(e−i√λ+iπ2 + ei

√λ+iπ2

))= ν

(±r(e−i√λ+iπ2 − ei

√λ+iπ2

)),

i√λ(

(e−i√λ + ei

√λ)

= ν(e−i√λ − ei

√λ),

2i√λ cos(

√λ) = −2iν sin(

√λ),

−1

ν

√λ = tan(

√λ). (53)

See Figure 18 for a sketch of the solutions of (53). We see there is an infiniteincreasing sequence (λn) ⊂ (0,∞) of solutions to (53) with λn →∞ as n→∞.

Substitution of (51) and (52) into (46) gives

X(ξ) = ±r(e−i√λ+iπ2 ei

√λξ − ei

√λ+iπ2 e−i

√λξ)

= ±ri(ei√λ(ξ−1) − e−i

√λ(ξ−1)

)= ±2ri2 sin(

√λ(ξ − 1))

= ∓2r sin(√λ(ξ − 1))

= ±2r sin(√λ(1− ξ)).

45

π 2π 3π

√λ0

√λ1

√λ2

√λ

Figure 18: Visualization of solutions of (53). The blue graph is tan√λ and the

red graph is − 1ν

√λ.

From (53) we know all λn satisfy

ν sin(√λn) +

√λn cos(

√λn) = 0. (54)

The corresponding eigenfunctions are

φn(ξ) = kn sin(√λn(1− ξ)).

We use theory from [8]. We can fix kn by normalizing:∫ 1

0

φ2n(ξ) dξ = 1

k2n

∫ 1

0

sin2(√λn(1− ξ)) dξ = 1

k2n

[ξ

2+

sin(2√λn(1− ξ))

4√λn

]1

ξ=0

= 1

k2n

(1

2− sin(2

√λn)

4√λn

)= 1

k2n

(1

2− 2 sin(

√λn) cos(

√λn)

4√λn

)= 1

k2n

(1

2+

√λn cos2(

√λn)

2ν√λn

)= 1

k2n

(1

2+

cos2(√λn)

2ν

)= 1

We take

kn =

√2ν

ν + cos2(√λn)

.

So

φn(ξ) =

(2ν

ν + cos2(√λn)

) 12

sin(√λn(1− ξ)).

46

We now get

v(ξ, τ) =

∞∑n=0

µn

(2ν

ν + cos2(√λn)

) 12

sin(√λn(1− ξ))e−λnτ ,

where µn are Fourier-like coefficients of expansion in orthonormal basis. We candetermine µn by substituting the initial condition of (42):

v(ξ, 0) =

∞∑n=0

µn

(2ν

ν + cos2(√λn)

) 12

sin(√λn(1− ξ))

c1(1− ξ) =

∞∑n=0

µnφn(ξ)

∫ 1

0

c1(1− ξ)φm(ξ) dξ =

∞∑n=0

µn

∫ 1

0

φn(ξ)φm(ξ) dξ

= µm

∫ 1

0

(φm(ξ))2 dξ

= µm

So

µn =

∫ 1

0

c1(1− ξ)(

2ν

ν + cos2(√λn)

) 12

sin(√λn(1− ξ)) dξ.

The expression for v is

v(ξ, τ) =

∞∑n=0

µne−λτ sin(

√λn(1− ξ)), (55)

where

µn := µn

(2ν

ν + cos2(√λn)

) 12

=

(2ν

ν + cos2(√λn)

)∫ 1

0

c1(1− ξ) sin(√λn(1− ξ)) dξ

=c1λn

2ν

ν + cos2(√λn)

(sin(√λn)−

√λn cos(

√λn))

(53)=

c1λn· 2ν cos(

√λn)

ν + cos2(√λn)

(−1

ν

√λn −

√λn)

= − c1√λn

(ν + 1)2 cos(

√λn)

ν + cos2(√λn)

.

So the solution for (35) is

u(ξ, τ) = c1ξ + c2 +

∞∑n=0

µne−λτ sin(

√λn(1− ξ)),

where c1 and c2 are known from (36). It follows that the solution of (31) is

u(x, t) =c1Lx+ c2 +

∞∑n=0

µne−λnDt

L2 sin(√

λn

(1− x

L

)). (56)

47

We can estimate |µn| from above:

|µn| ≤|c1|√λn· 2(ν + 1)

ν≤ |c1|

π2 + nπ

· 2(ν + 1)

ν

Note that for n large that µn becomes small. In fact we will show that (|µn|) ⊂R+ a decreasing sequence (see Proposition 4.6). In order to do that we firstneed two Lemma’s.

Lemma 4.4. The equation in λ,

−1

ν

√λ = tan

√λ (57)

has for each n ∈ N0 exactly one positive solution√λn ∈ ( 2n+1

2 π, (n+ 1)π) andall solutions λ to (57) are of the form λ = λn, n ∈ N0.

Proof. Since the left hand side of (57) is negative, it must hold that tan√λn < 0.

Since tan√λn < 0 if and only if

√λn ∈ ( 2n+1

2 π, (n+ 1)π) for some n ∈ N0, wecan conclude that solutions of (57) does not exist anywhere else.

Note that for each n ∈ N0, tanx → −∞ as x ↓ 2n+12 π and tanx → 0 as

x ↑ (n+1)π. It follows from the intermediate value theorem that there is at leastone solution in ( 2n+1

2 π, (n + 1)π). For each n ∈ N0, tanx is strictly increasingon ( 2n+1

2 π, (n + 1)π) and − 1νx is strictly decreasing. It follows that there is

exactly one solution√λn ∈ ( 2n+1

2 π, (n+ 1)π).

Lemma 4.5. | cos√λn| > | cos

√λn+1| for all n ∈ N0.

Proof. Note that tan(√λn + π) = tan

√λn and − 1

ν (√λn + π) < − 1

ν

√λn for all

n ∈ N0. So

tan(√λn + π) = tan

√λn

= −1

ν

√λn

> −1

ν(√λn + π).

Note that√λn + π ∈ ( 2n′+1

2 π, (n′ + 1)π), where n′ = n + 1. Since tanx is

strictly increasing on this interval, − 1νx strictly decreasing and tan(

√λn+π) >

− 1ν (√λn + π) it follows from Lemma 4.4 that

√λn′ ∈ ( 2n′+1

2 π,√λn + π). From

this it follows that δn :=√λn − 2n+1

2 π is decreasing as n increases.Note that for each n ∈ N0, | cosx| is a strictly increasing function on

( 2n+12 π, (n+ 1)π). It follows that

| cos√λn| = | cos(

√λn + π)|

> | cos√λn′ |

for all n ∈ N0.

Proposition 4.6. |µn| is decreasing.

48

Proof. |µn| is decreasing if |µn| − |µn+1| > 0 for all n ∈ N0. We have

|µn| − |µn+1| = 2|c1|(ν + 1)

(| cos(

√λn)|

√λn(ν + cos2(

√λn))

−| cos(

√λn+1)|√

λn+1(ν + cos2(√λn+1))

)= 2|c1|(ν + 1)·

√λn+1 · | cos(

√λn)| · (ν + cos2(

√λn+1)) −

√λn · | cos(

√λn+1)| · (ν + cos2(

√λn))√

λn√λn+1(ν + cos2(

√λn))(ν + cos2(

√λn+1))

.So (|µn|) ⊂ R+ is a decreasing sequence if√λn+1·| cos(

√λn)|·(ν+cos2(

√λn+1))−

√λn·| cos(

√λn+1)|·(ν+cos2(

√λn)) > 0. (58)

From Lemma 4.5 it follows that (58) is true if√λn · | cos(

√λn)| · (ν + cos2(

√λn+1))−

√λn · | cos(

√λn+1)| · (ν + cos2(

√λn)) ≥ 0,

| cos(√λn)| · (ν + cos2(

√λn+1))− | cos(

√λn+1)| · (ν + cos2(

√λn)) ≥ 0,

ν(| cos√λn| − | cos

√λn+1|) + | cos

√λn cos

√λn+1|(| cos

√λn+1| − | cos

√λn|) ≥ 0,

| cos√λn cos

√λn+1| ≤ ν.

And again from Lemma 4.5 we know this holds if

| cos2√λn| ≤ ν.

So (|µn|) is a decreasing sequence if

cos2√λ0 ≤ ν. (59)

From (53) we get that

ν =−√λ0

tan√λ0

and (59) becomes

cos2√λ0 ≤

−√λ0

tan√λ0

,

cos2√λ0 ≤ −

√λ0

cos√λ0

sin√λ0

=√λ0

∣∣∣∣cos√λ0

sin√λ0

∣∣∣∣ ,| cos

√λ0 sin

√λ0| ≤

√λ0,

1

2| sin 2

√λ0| ≤

√λ0.

This is obviously true, since√λ0 >

1

2π >

1

2≥ 1

2| sin 2

√λ0|.

So (|µn|) is a decreasing sequence.

Now we know that |µ0| > |µn| and 0 < λ0 < λn for all n ∈ N0. So

|µ0|e−λ0Dt

L2 > |µn|e−λnDt

L2

for all n ∈ N0. Now it’s clear from (56) and Lemma 4.4 that Theorem 4.2 holds.

49

4.3.2 Active Sodium Symporters

If we take into account a nonzero contribution of the symporter, then our prob-lem becomes

ut = DuxxDux(0, t) = P (Ru(0, t)− ui) + S(h, u(0, t))u(L, t) = uu(x, 0) = u

(60)

where, depending on the type of symporter, the expression for S is as in (32)or (33) and the expression for Σ is as in (6) or (9), P = PCO2fCO2(h′) and

R = fCO2 (h)

fCO2 (h′). So the problem has a non-linear boundary condition. There seem

to be techniques to arrive at a semi-explicit solution to our problem (60), similarto the series expansion (55). However, we expect to obtain even less manageableexpressions than for the case studied above, i.e., case with inactive symporter.

4.3.3 Linearised Stability with Symporter Active

Let v(ξ, τ) = u(ξ, τ) − u?(ξ), then we get (42) where (42b) changes due to theactive symporter:

vξ(0, τ) =PLR

Dv(0, τ) +

L

DA(S(u(0, τ))− S(u?(0)))

≈ PLR

Dv(0, τ) +

L

DAS′(u?(0))v(0, τ). (61)

This approximation is ’good’ when u(0, τ) is ’close’ to u?(0).

Proposition 4.7. u? is locally linearly stable.

Proof. When we replace (42b) in Problem (42) with (61), we get a similar prob-lem as Problem (42). In fact if we would replace R in (42b) by R+ 1

P S′(u?(0))

we get Problem (61). Just like R, R+ 1P S′(u?(0)) is constant, so provided that

ν′ = PLRD + L

DS′(u?(0)) = ν+ L

DS′(u?(0)) > 0 we get the same results as before

with Problem (42).

4.4 The Modelling Effect of Cylindrical Coordinates onthe Carbon Supply

The model (30) on page 33 is based on Cartesian coordinates, however thecylindrical alternative would be a better match to reality (since a cell is shapedmore or less cylindrical). Therefore we investigate the effect on the decrease ofcarbon through the unstirred layer based on a cylindrical model and compareit to the previous result.

We have a cylindrical cell with length H. So the ’bottom’ of the cell is atz = 0 and the ’top’ of the cell is at z = H. The radius of the cell is r1 andthe thickness of the unstirred layer is L. So the unstirred layer starts at thecell (r = r1) and ends at r = r1 + L. We assume azimuthal symmetric initialconditions and boundary conditions, hence resulting in a azimuthal symmet-ric concentration, u, throughout the unstirred layer for all t > 0. Therefore∂u∂φ |(r,φ,z,t) = 0 for all (r, φ, z) ∈ [unstirred layer] and t > 0.

50

In cylindrical coordinates diffusion is given by

∂u

∂t= D

[1

r

∂

∂r

(r∂u

∂r

)+

1

r2

∂2u

∂φ2+∂2u

∂z2

]which reduces to

∂u

∂t= D

[1

r

∂

∂r

(r∂u

∂r

)+∂2u

∂z2

]= D

[1

r

∂

∂r+∂2u

∂r2+∂2u

∂z2

].

Our model becomes

ut = D[1

rur + urr + uzz],

Dur(r1, z, t) =1

A(J

CO2

in + JHCO−3in ),

u(r1 + L, z, t) = u,

u(r, z, 0) = u.

Note that in this model the boundary conditions and initial conditions are alsoheight invariant. So if we don’t specify boundary conditions at z = 0 and z = Hwe can assume uz = 0 for all t > 0 and we get

ut = D[1

rur + urr],

Dur(r1, t) =1

A(J

CO2

in + JHCO−3in ),

u(r1 + L, t) = u,

u(r, 0) = u.

The expression for the fluxes JCO2

in and JHCO−3in are known and introducing

S =J

HCO−3

in

A gives

ut = D[1

rur + urr],

Dur(r1, t) = P (Ru(r1, t)− ui) + S,

u(r1 + L, t) = u,

u(r, 0) = u.

Take ρ = rr1+L and τ = tD

(r1+L)2 . We get

uτ = uρρ +1

ρuρ,

D

r1 + Luρ(

r1

r1 + L, τ) = P (Ru(

r1

r1 + L, τ)− ui) + S,

u(1, τ) = u,

u(ρ, 0) = u.

51

Let u?(ρ) be the steady state solution. Introduce r2 = r1 + L. We get

u?ρρ +1

ρu?ρ = 0,

u?ρ(r1

r2) =

Pr2

D

(Ru?

(r1

r2

)− ui

)+r2

DS,

u?(1) = u.

Introduce v = u?ρ, thenρvρ + v = 0.

v(ρ) has to be some polynomial (one term), so take v = ρλ. Substitution gives

ρλρλ−1 + ρλ = 0,

λρλ + ρλ = 0,

λ = −1.

So v = c1ρ−1 and hence u? = c1 ln(ρ) + c2. The boundary condition gives

c1 ln(1) + c2 = u, so c2 = u. Determining c1 is not that easy and we won’t get

a readable expression for it. However, we need to have c1 · ln(r1r2

)< 0, since

we want to have a reduction of total carbon concentration at the cell and notan increase. So we need c1 > 0. With some calculations we find that there is aunique positive solution to c1 if

−[δP

D(Ru− ui) +

β

D

]< 0,

where we used S(c1) = αc1+βγc1+δ (α, γ, δ > 0, β S 0). This is likely to hold, since

PA(u − ui) describes the inward flux of CO2 when the concentration at theexterior of the cell would be u, which is higher than the actual concentrationand therefore more likely to have a positive value. A similar reasoning can bemade for

β = N(k0Keqηf

HCO−3 (h)u− k1Keqη′fHCO−3 (h′)ui

)although we have to be careful with the dependency on η.

For the cylindrical model we have found

u?(r) = c1 ln

(r

r1 + L

)+ u, (62)

where c1 > 0. While for the Cartesian model we had

u?(x) = c1x

L+ u,

where c1 < 0. Note that the logarithmic term in (62) is negative for the relevantvalues r1 ≤ r ≤ r1 + L.

In order to compare these results we have to express the first one as a solutionin x = r − r1:

u?(x) = c1 ln

(x+ r1

r1 + L

)+ u

= c1 ln

(1− L− x

L+ r1

)+ u.

52

Taking y = L−xL+r1

we get the following Taylor expansion around y = 0 (x =L):

ln(1− y) = −

( ∞∑k=1

yk

k

)

ln

(1− L− x

L+ r1

)= −

( ∞∑k=1

1

k

(L− xL+ r1

)k)

≈ −∞∑k=1

1

k

(L

L+ r1

)k+x

L·∞∑k=1

(L

L+ r1

)kSo we get an approximation for our cylindrical result

u?(x) ≈ c1x

L

∞∑k=1

(L

L+ r1

)k− c1

∞∑k=1

1

k

(L

L+ r1

)k+ u

The difference at the cell (x = 0) compared to the Cartesian model is

c1

∞∑k=1

1

k

(L

L+ r1

)k= −c1 ln

(r1

r1 + L

).

r1 is a fixed value, but the parameter L is not known. We only know L > 0,so

0 < −c1 ln

(r1

r1 + L

).

So the difference is small when L ≈ 0 and goes to ∞ as L goes to ∞.This means that if the unstirred layer is thin, the Cartesian model is a goodapproximation, but as the unstirred layer becomes thicker the approximation ofthe Cartesian model gets worse.

53

5 Derivation of Corrected Dynamics

We assume that all reactions are very fast compared to all the forms of trans-port through the membrane. This means that it holds for the lowest orderapproximation that

[HCO−3 ][H+]

[CO2]=k−1k+

1

=: K1 and[CO2−

3 ][H+]

[HCO−3 ]=k+

2

k−2=: K2

everywhere. Note that K1 = Ka1and K2 = Ka2

as in (1) on page 4. Theseequations are deduced from the equilibrium equations

H+ + HCO−3k+

1−−⇀↽−−k−1

CO2 + H2O and HCO−3k+

2−−⇀↽−−k−2

CO2−3 + H+.

Recall that dissolved inorganic carbon is the sum of the carbon species CO2,HCO–

3, CO2–3 and H2CO3. However, since the fraction of H2CO3 is very low for

the relevant pH-values (see Figure 3), we consider

CT = [CO2] + [HCO−3 ] + [CO2−3 ], (63)

to denote the concentration of dissolved inorganic carbon.In the current model the only processes affecting the total concentration of

inorganic carbon inside the cell are the transport through the cell membraneand photosynthesis. Therefore

dCTdt

=1

Vin[J

CO2


φ ]. (64)

The expressions for JCO2

in , JHCO−3in and J

HCO−3φ are already known ((12),(5)/(8)

and (13) respectively). Here Vin denotes the intracellular volume relevant toinorganic carbon uptake into the cell and the uptake by the chloroplasts, i.e.,the volume of the cortical plasm.

To simplify notation, put

h = [H+],

A = [CO2],

B = [HCO−3 ],

C = [CO2−3 ],

for the concentrations of the indicated compounds in the cortical plasm. Fornow we assume that these are spatially homogeneous.

5.1 Intracellular Dynamics

First we ignore that there is transportation of protons through the membrane.We determine the change of protons inside the cell when there is no transporta-tion of protons through the membrane in absence of any proton buffering inthe cortical plasm. In the model we assume that inorganic carbon is taken upby chloroplasts in its HCO–

3 form mainly. In there it is converted to CO2 bycarbonic anhydrase for use in the photosynthesis process. After that we use the

54

assumption of a constant intracellular proton concentration to conclude that theefflux of protons of the cell is equal to the change inside the cell when we ignorethe transportation of protons through the membrane. With these assumptionswe get

dh

dt= k−1 A− k

+1 hB + k+

2 B − k−2 hC,

dA

dt=J

CO2

in

Vin+ k+

1 hB − k−1 A,

dB

dt=J

HCO−3in

Vin−J

HCO−3φ

Vin+ k−1 A− k

+1 hB + k−2 hC − k

+2 B,

dC

dt= k+

2 B − k−2 hC.

Using (63) and (64) we can write this system as

dh

dt= k−1 A− k

+1 h(CT −A− C) + k+

2 (CT −A− C)− k−2 hC,

dA

dt=J

CO2

in

Vin+ k+

1 h(CT −A− C)− k−1 A,

dC

dt= k+

2 (CT −A− C)− k−2 hC,

dCTdt

=J

CO2


φ

Vin=JCVin

.

The inorganic carbon fluxes have the following dependence of A, C, CT andh:

JCO2

in = A(RCCO2ext −A),

JHCO−3in = Sym(CT −A− C,C

HCO−3ext ),

JHCO−3φ = φ(CT −A− C).

Hence a suitable choice for the slow time scale would be τs = AVin

. However,

we assume JHCO−3in and J

HCO−3φ to be of the same order of speed and so there are

more suitable choices. For now we leave the precise choice of τs in the middleand keep it more general, but keep in mind it has the same order of speed asthis specific choice.

Let τs be the slow time scale and τ = tτs

. Now we can take

ε1 =1

k−1 τsand ε2 =

1

k+2 τs

=k−1k+

2

ε1.

55

Our system becomes

ε1dh

dτ= A−K−1

1 h(CT −A− C) +ε1

ε2(CT −A− C)− k−2

k−1hC,

ε1dA

dτ=A(RC

CO2ext −A)

k−1 Vin

+K−11 h(CT −A− C)−A,

ε2dC

dτ= (CT −A− C)−K−1

2 hC,

dCTdτ

=JCVin

τs.

Putting ε2 = αε1 where α > 0, we get

ε1dh

dτ= A−K−1

1 h(CT −A− C) +1

α(CT −A− C)− 1

αK−1

2 hC,

ε1dA

dτ= ε1τs

A(RCCO2ext −A)

Vin+K−1

1 h(CT −A− C)−A,

ε1dC

dτ=

1

α(CT −A− C)− 1

αK−1

2 hC,

dCTdτ

=τsVin

[A(RCCO2ext −A) + Sym(CT −A− C,C

HCO−3ext )− φ(CT −A− C)].

Substituting expansions

h = h0 + ε1h1 + . . . ,

A = A0 + ε1A1 + . . . ,

C = C0 + ε1C1 + . . . ,

CT = C0T + ε1C

1T + . . . ,

gives the following O(ε01) and O(ε1

1) system:O(1) :

0 = αA0 − αK−11 h0(C0

T −A0 − C0) + (C0T −A0 − C0)−K−1

2 h0C0 (65a)

0 = K−11 h0(C0

T −A0 − C0)−A0 (65b)

0 = (C0T −A0 − C0)−K−1

2 h0C0 (65c)

dC0T

dτ=

τsVin

[A(RCCO2ext −A0) + Sym(C0

T −A0 − C0, CHCO−3ext )− φ(C0

T −A0 − C0)]

(65d)

56

O(ε1) :

dh0

dτ= A1 −K−1

1 (h0(C1T −A1 − C1) + h1(C0

T −A0 − C0))

+1

α(C1

T −A1 − C1)− 1

αK−1

2 (h0C1 + h1C0) (66a)

dA0

dτ=A(RC

CO2ext −A0)τsVin

+K−11 (h0(C1

T −A1 − C1) + h1(C0T −A0 − C0))−A1 (66b)

dC0

dτ=

1

α(C1

T −A1 − C1)− 1

αK−1

2 (h0C1 + h1C0) (66c)

dC1T

dτ= . . . (66d)

Notice that (65a) is automatically satisfied if (65b) and (65c) hold. (66d) isadded for completeness of the system, but the expression is omitted since it willnot be used. From (66a)-(66c) we get

dh0

dτ=dC0

dτ− dA0

dτ+A(RC

CO2ext −A0)τsVin

. (67)

From (65b) and (65c) it follows that

A0

K−11 h0

= K−12 h0C0 = C0

T −A0 − C0,

A0 = K−11 K−1

2 h20C0. (68)

SodA0

dτ= K−1

1 K−12

(2h0

dh0

dτC0 + h2

0

dC0

dτ

).

Substituting this into (67) gives

dh0

dτ=dC0

dτ−K−1

1 K−12

(2h0

dh0

dτC0 + h2

0

dC0

dτ

)+A(RC

CO2ext −K−1

1 K−12 h2

0C0)τsVin

,

(1 + 2K−11 K−1

2 h0C0)dh0

dτ= (1−K−1

1 K−12 h2

0)dC0

dτ

+A(RC

CO2ext −K−1

1 K−12 h2

0C0)τsVin

. (69)

Now we have to get rid of dC0

dτ in this expression. From (65c) we get

C0 = (C0T −A0)(1 +K−1

2 h0)−1.

Substituting (68) into this equation gives

C0 = (C0T −K−1

1 K−12 h2

0C0)(1 +K−12 h0)−1,

(1 +K−12 h0)C0 = C0

T −K−11 K−1

2 h20C0,

C0 =C0T

1 +K−12 h0 +K−1

1 K−12 h2

0

. (70)

57

Put

fA(h0) = K−11 K−1

2 h20F−1, (71a)

fB(h0) = K−12 h0F

−1, (71b)

fC(h0) = F−1, (71c)

where F = 1 + K−12 h0 + K−1

1 K−12 h2

0 and note that these expression are very

similar to fCO2(h), fHCO−3 (h) and fCO2−3 (h) respectively. In fact

fA(h) = fCO2(h)

(1 +

K−1a2K−1a3h2

1 +K−1a2 K

−1a1 h

2 +K−12 h

),

fB(h) = fHCO−3 (h)

(1 +

K−1a2K−1a3h2

1 +K−1a2 K

−1a1 h

2 +K−12 h

),

fC(h) = fCO2−3 (h)

(1 +

K−1a2K−1a3h2

1 +K−1a2 K

−1a1 h

2 +K−12 h

).

The expression between the parentheses, increases from 1 to 1 +Ka1

Ka3≈ 1 +

0.32 · 10−2 as h increases from 0 to ∞. So this expression is always very closeto one and the difference may therefore be neglected. Hence fA, fB and fC can

effectively be interpreted as fCO2 , fHCO−3 and fCO2−3 . It follows that

dC0

dτ=

dC0T

dτ (1 +K−12 h0 +K−1

1 K−12 h2

0)− C0T (K−1

2 + 2K−11 K−1

2 h0)dh0

dτ

(1 +K−12 h0 +K−1

1 K−12 h2

0)2.

Substituting this together with (71) back into (69) gives(1 + 2fA(h0)

C0T

h0

)dh0

dτ=

(fC(h0)− fA(h0))

[dC0

T

dτ− C0

T

h0(fB(h0) + 2fA(h0))

dh0

dτ

]+A(RC

CO2ext − fA(h0)C0

T )τsVin

,

(1 + 2fA(h0)

C0T

h0+C0

T

h0(fC(h0)− fA(h0))(fB(h0) + 2fA(h0)

)dh0

dτ=

(fC(h0)− fA(h0))dC0

T

dτ+A(RC

CO2ext − fA(h0)C0

T )τsVin

.

Introduce

γ(C0T , h0) := 1 +

C0T

h0

(2fA(h0) + (fC(h0)− fA(h0))(fB(h0) + 2fA(h0))

).

We get

γ(C0T , h0)

dh0

dτ= (fC(h0)− fA(h0))

dC0T

dτ+A(RC

CO2ext − fA(h0)C0

T )τs

Vin,

dh0

dτ=

1

γ(C0T , h0)

[(fC(h0)− fA(h0))

dC0T

dτ+A(RC

CO2ext − fA(h0)C0

T )τs

Vin

].

58

Recall τ = tτs

, so

dh0

dt=

1

γ(C0T , h0)

[(fC(h0)− fA(h0))

dC0T

dt+A(RC

CO2ext − fA(h0)C0

T )

Vin

],

wheredC0

T

dt is known from (65d). Substituting (68) and (70) in this equationyields an expression only dependent on h0 and C0

T :

dC0T

dt=

1

Vin[A(RC

CO2ext − fA(h0)C0

T )

+ Sym(C0T fB(h0)), C

HCO−3ext )

− φC0T fB(h0)]

We have

Sym(C0T fB(h0), C

HCO−3ext ) =

BRC

HCO−3ext − C0

T fB(h0)

D0 +D1C0T fB(h0) +D2C

HCO−3ext +D3C

HCO−3ext C0

T fB(h0), (72)

where from (8) and (9) we have

B = Nk1Keqη′,

R =k0Keqη

k1Keqη′,

D0 = k2(1 + Keqη′) + k3(1 +Keqη),

D1 = Keqη′(k4 + k7(1 +Keqη

′)),

D2 = Keqη(k5 + k6(1 + Keqη′)),

D3 = k8Keqη′Keqη.

So we get

dh0

dt=

1

γ(C0T , h0)

[(fC(h0)− fA(h0) + 1)

A(RCCO2ext − fA(h0)C0

T )

Vin

+ (fC(h0)− fA(h0))

(B

RCHCO

−3

ext − C0T fB(h0)

D0 +D1C0T fB(h0) +D2C

HCO−3

ext +D3CHCO

−3

ext C0T fB(h0)

− φC0T fB(h0)

)]. (73)

Note that we ignored the transport of protons through the membrane. How-ever, the proton concentration inside the cell is assumed to be constant (pHhomeostasis). Therefore there must be some kind of proton transportationthrough the membrane. We already assumed H+/Na+ antiporters and pro-ton pumps to be present in the membrane of the cell. Effectively there is nochange in the Na+ concentration (since we assumed that every Na+ crossing

59

the membrane will be returned immediately with the antiporter). Therefore wecan simply talk about a net flux of protons through the membrane, Jh, whichis equal to the change of protons as shown in (73). Note that a positive signmeans that more protons leave the cell than enter it.

5.2 Extracellular Dynamics

A similar approach can be taken to investigate the dynamics in the unstirredlayer. We will address the different ions with the same names as in the previousparagraph. Note, however, that there is a difference in the location of the ions.In this paragraph we consider the ions to be in the unstirred layer. We get forx ∈ (0, L)

∂h

∂t= Dh

∂2h

∂x2+ k−1 A− k

+1 hB + k+

2 B − k−2 hC,

∂A

∂t= DA

∂2A

∂x2+ k+

1 hB − k−1 A,

∂B

∂t= DB

∂2B

∂x2+ k−1 A− k

+1 hB + k−2 hC − k

+2 B,

∂C

∂t= DC

∂2C

∂x2+ k+

2 B − k−2 hC.

With boundary conditions

∂h

∂x(0, t) = Jh + k−1 A− k

+1 hB + k+

2 B − k−2 hC,

∂A

∂x(0, t) = −JCO2

in + k+1 hB − k

−1 A,

∂B

∂x(0, t) = −JHCO−3

in + k−1 A− k+1 hB + k−2 hC − k

+2 B,

∂C

∂x(0, t) = k+

2 B − k−2 hC,

h(L, t) = APWh,

A(L, t) = APWA,

B(L, t) = APWB ,

C(L, t) = APWC ,

where Jh represents the outward proton flux through the membrane which isequal to the expression in (73).

Using (63) and

∂CT∂t

= DA∂2A

∂x2+DB

∂2B

∂x2+DC

∂2C

∂x2

60

we get

∂h

∂t= Dh

∂2h

∂x2+ k−1 A− k

+1 h(CT −A− C) + k+

2 (CT −A− C)− k−2 hC,

for x ∈ (0, L),

∂A

∂t= DA

∂2A

∂x2+ k+

1 h(CT −A− C)− k−1 A, for x ∈ (0, L),

∂C

∂t= DC

∂2C

∂x2+ k+

2 (CT −A− C)− k−2 hC, for x ∈ (0, L),

∂CT∂t

= DA∂2A

∂x2+DB

∂2

∂x2(CT −A− C) +DC

∂2C

∂x2, for x ∈ (0, L).

Like before we substitute expansions

h = h0 + ε1h1 + . . . ,

A = A0 + ε1A1 + . . . ,

C = C0 + ε1C1 + . . . ,

CT = C0T + ε1C

1T + . . .

and take

ε1 =1

k−1 τsand ε2 =

1

k+2 τs

and assume ε2 = αε1. Let τ = tτs

, then we get

ε1∂h

∂τ= ε1τsDh

∂2h

∂x2+A−K−1

1 h(CT −A− C) +1

α(CT −A− C)− 1

αK−1

2 hC,

ε1∂A

∂τ= ε1τsDA

∂2A

∂x2+K−1

1 h(CT −A− C)−A,

ε1∂C

∂τ= ε1τsDC

∂2C

∂x2+

1

α(CT −A− C)− 1

αK−1

2 hC,

∂CT∂τ

= τs

[DA

∂2A

∂x2+DB

∂2

∂x2(CT −A− C) +DC

∂2C

∂x2

].

We get the following O(1) and O(ε1) system:O(1) :

0 = αA0 − αK−11 h0(C0

T −A0 − C0) + (C0T −A0 − C0)−K−1

2 h0C0 (74a)

0 = K−11 h0(C0

T −A0 − C0)−A0 (74b)

0 = (C0T −A0 − C0)−K−1

2 h0C0 (74c)

dC0T

dτ= τs

[(DA −DB)

∂2A0

∂x2+ (DC −DB)

∂2C0

∂x2+DB

∂2C0T

∂x2

](74d)

61

O(ε1) :

∂h0

∂τ= τsDh

∂2h0

∂x2+A1 −K−1

1 (h0(C1T −A1 − C1) + h1(C0

T −A0 − C0))

+1

α(C1

T −A1 − C1)− 1

αK−1

2 (h0C1 + h1C0) (75a)

∂A0

∂τ= τsDA

∂2A0

∂x2+K−1

1 (h0(C1T −A1 − C1) + h1(C0

T −A0 − C0))−A1

(75b)

∂C0

∂τ= τsDC

∂2C0

∂x2+

1

α(C1

T −A1 − C1)− 1

αK−1

2 (h0C1 + h1C0) (75c)

∂C1T

∂τ= . . . (75d)

(75d) is added for completeness of the system, but the expression is omittedsince it will not be used. Equations (74a)-(74c) are the same as equations(65a)-(65c). So (68) and (70) both hold again. From (75a)-(75c) we get

∂h0

∂τ= τs

[Dh

∂2h0

∂x2+DA

∂2A0

∂x2−DC

∂2C0

∂x2

]− ∂A0

∂τ+∂C0

∂τ.

Note that this is very similar to (67). The only terms where there are derivativesof h0 to τ appearing are −∂A0

∂τ and ∂C0

∂τ and these are also the only terms in(67) where this happens. The remaining term in (67) is in this case replaced by

τs[Dh∂2h0

∂x2 +DA∂2A0

∂x2 −DC∂2C0

∂x2 ]. From this it follows that

∂h0

∂t=

1

γ(C0T , h0)

[(fC(h0)− fA(h0))

∂C0T

∂t+Dh

∂2h0

∂x2+DA

∂2A0

∂x2−DC

∂2C0

∂x2

],

where∂C0

T

∂t is known from (74d). To get an expression in h0 and C0T we have to

substitute the following:

∂2A0

∂x2= K−1

1 K−12

([2∂2h0

∂x2h0 + 2

(∂h0

∂x

)2]C0 + 4

∂h0

∂xh0∂C0

∂x+ h2

0

∂2C0

∂x2

),

C0 = fC(h0)C0T ,

∂C0

∂x=∂fC(h0)

∂xC0T + fC(h0)

∂C0T

∂x,

∂2C0

∂x2=∂2fC(h0)

∂x2C0T + 2

∂fC(h0)

∂x

∂C0T

∂x+ fC(h0)

∂2C0T

∂x2,

where

∂fC(h0)

∂x= −fC(h0)

1

h0(fB(h0) + 2fA(h0))

∂h0

∂x,

∂2fC(h0)

∂x2= fC(h0)

(2

1

h20

(fB(h0) + 2fA(h0))2

(∂h0

∂x

)2

− 21

h20

fA(h0)

(∂h0

∂x

)2

+1

h0(fB(h0) + 2fA(h0))

∂2h0

∂x2

).

62

6 Towards a Turing Pattern?

In this thesis we only looked at models with at most one spatial dimensionvariable, corresponding to the direction perpendicular to the surface of the cell.This was reasonable under the assumption that solutions where homogeneous inthe longitudinal direction of the cell and rotation invariant. However, in order toget results about the pattern formation along the cell surface, the longitudinaldirection should be taken into account as well and therefore an extra spatialdimension variable is needed.

Taking into account this direction with the variable y, our previous findingswill essentially hold for fixed y. However, diffusion of the different substances,i.e., dissolved inorganic carbon and protons, in the y-direction will play a rolein the dynamics. In Table 1 you can see that the diffusion coefficient of protonsare much bigger than the diffusion coefficients of all the dissolved inorganiccarbon species. This anomalously high diffusion rate is often attributed to theGrotthuss mechanism, by which H3O+ is able to shuttle a proton to a watermolecule in its first solvation shell, which can then in turn shuttle to yet anothermolecule. This mechanism makes excess protons able to diffuse much faster asopposed to conventional diffusion. [5]

This big difference in the diffusivity gives rise to the question whether thealkaline bands forming at the surface of the cells of Chara corallina is a Turingpattern. Ideally we would integrate this into the most detailed model (Section5). Then obviously all intracellular substances will be dependent on y and tand all extracellular substances will be dependent on x, y and t. Hence, forthe intracellular dynamics we get PDE’s instead of ODE’s and we get an extraterm in all equations representing the diffusion in the cell (in the y-directiononly) and for the intracellular dynamics we get the Laplace operator instead ofthe second derivative. Taking similar steps we would end up with a model ofthree equations, i.e., two equations describing the intracellular and extracellulartotal inorganic carbon dynamics and one equation describing the extracellularproton dynamics. Initial inspection revealed that expressions will become lessmanageable than they are now sooner than they would become more manageableand analysing this adjusted model is expected to be quite a challenge.

In order to make this model more manageable, a simplification would beneeded. A possibility is to omit the x-direction (in the extracellular dynamics)and just use concentrations of the substances at the cell surface. This could bethe concentrations outside the unstirred layer or a depleted concentration dueto limiting effect of the unstirred layer (Section 4). However we would still havethree equations in this model.

Another possibility is taking our initial ODE-models (Section 3) as a startingpoint. The advantage of using this model is that we had existence of a uniquesteady state and only two equations in this model. We get something like

∂C inT

∂t= DCT

∂2C inT

∂y2+LyVin

(J

CO2


φ

), (76a)

∂h

∂t= Dh

∂2h

∂y2+

LyVout

Jh, (76b)

where Ly is the length of the cell and Jh is the expression in (73). Equations(76a)-(76b) are complemented by zero-flux boundary conditions at y = 0 and

63

Figure 19: The difference between slow diffusion of dissolved inorganic carboninside the cell (blue) and fast diffusion of protons outside the cell (red) canpossibly explain the alkaline band formation at the surface of the cell of Characorallina (green).

y = Ly. Note that in this model we took Jh from the previous section (Equation(73)) instead of the expression in Section 3. In this case Turing patterns mayform due to the slow diffusion of inorganic carbon inside the cell and fast diffu-sion of protons at the surface of the cell (Figure 19). Now we expect that theremust be a stable homogeneous steady state which possibly becomes unstable byadjusting parameters φ and η/η′, since Dh

DCT� 1.

A more detailed analysis of the stability of the steady state of the genericmodel (11) is thus required. Also a corresponding dispersion relation should be

obtained. The complicated expression for JHCO−3in (the symporter flux) makes

this yet another challenge.A possible approach to obtain a positive result on the existence of a Tur-

ing pattern could be to replace the reaction terms in (76a)-(76b) by a simplermath expression having similar characteristics. One could investigate what char-acteristics of the non-linear reaction term are important for obtaining Turingpatterns and then see to what extend the explicit expressions in (76a)-(76b)meet these conditions. The time constraint of this master project prevented usfrom pursuing this direction of research.

64

A Separation Constant λ is Real and StrictlyPositive.

We call a solution of the form v(ξ, τ) = X(ξ)T (τ) separated. The separationconstant, λ, introduced in (43), must be real and positive in order to have non-trivial solutions for vλ. In this appendix it is shown that in any other case,there are no non-trivial solutions for vλ. System (34) has a real-valued solution.So either Xλ(ξ)Tλ(τ) is real itself for all ξ and τ , or Xλ(ξ)Tλ(τ) is not realfor all ξ and τ , but then also Xλ(ξ)Tλ(τ) is a solution, so Re(Xλ(ξ)Tλ(τ)) andIm(Xλ(ξ)Tλ(τ)) are real-valued solutions.

A.1 Case λ = 0:

If λ = 0 we getX ′′λ(ξ) = 0.

The solution for this ordinary differential equation is

Xλ(ξ) = aξ + b.

Recall the boundary conditions

Xλ(1) = 0,

X ′λ(0) = νXλ(0),

where ν > 0. Substituting the boundary conditions yields

a+ b = 0,

a = νb.

The only solution to this system is a = 0 and b = 0. Therefore Xλ ≡ 0 andhence vλ ≡ 0. So when λ = 0 there are no non-trivial separated solutions forvλ.

A.2 Case λ ∈ C \ {0}:Assume vλ = XλTλ is a non-zero solution, then we get

X ′′λ(ξ) + λXλ(ξ) = 0. (77)

Let z1 ∈ C be a solution to the characteristic equation

z2 + λ = 0.

The other solution is then z2 = −z1 ∈ C. The general solution for the ordinarydifferential equation (77) is

Xλ(ξ) = aez1ξ + be−z1ξ,

with a, b ∈ C. Substituting the boundary conditions yields

aez1 + be−z1 = 0,

(a− b)z1 = ν(a+ b).

65

ζ

From the first equation in this system we get

b = −ae2z1 .

Substituting this into the second equation of the system gives

z1(a+ ae2z1) = ν(a− ae2z1).

From this it follows that

a(z1(1 + e2z1) + ν(e2z1 − 1)

)= 0.

If a = 0 we would get the trivial solution vλ ≡ 0 (which is real), so

z1(e2z1 + 1) + ν(e2z1 − 1) = 0.

We know e2z1 +1 6= 0, since then e2z1−1 = −2 and it would follow that −2ν = 0,which is clearly not true. So we can write

z1 = ν1− e2z1

1 + e2z1

= νe−z1 − ez1e−z1 + ez1

.

Introduce ζ = iz1, then

−iζ = νeiζ − e−iζ

eiζ + e−iζ

= iν tan ζ,

so ζ must solve the transcendental equation

−1

νζ = tan ζ. (78)

Lemma A.1. Any solution ζ ∈ C to (78), must be real.

66

Proof.

Im(tan ζ) = −Re(i tan ζ)

= −Re

(eiζ − e−iζ

eiζ + e−iζ

)= −Re

(eiζ − e−iζ

eiζ + e−iζ· e

iζ + e−iζ

eiζ + e−iζ

).

We have

(eiζ − e−iζ)(eiζ + e−iζ) = (eiζ − e−iζ)(e−iζ + eiζ)

= ei(ζ−ζ) + ei(ζ+ζ) − e−i(ζ+ζ) − e−i(ζ−ζ)

= e−2Im ζ − e2Im ζ + e2iRe ζ − e−2iRe ζ

= e−2Im ζ − e2Im ζ + cos(2Re ζ) + i sin(2Re ζ)

− cos(−2Re ζ)− i sin(−2Re ζ)

= e−2Im ζ − e2Im ζ + 2i sin(2Re ζ).

So

Im(tan(ζ)) =e2Im ζ − e−2Im ζ

|eiζ + e−iζ |2=

2 sinh(2Im ζ)

|eiζ + e−iζ |2.

Note that sinh(x) > 0 for x > 0 and sinh(x) < 0 for x < 0. So if Im ζ > 0, thenIm(tan(ζ)) > 0, which contradicts (78). Similarly Im ζ < 0 is not possible, soIm ζ = 0.

Corollary A.2. z1 (and z2) must be purely imaginary, i.e., z1, z2 ∈ iR.

Corollary A.3. A non-trivial separated solution exists if and only if λ > 0.

Proof. z1 and z2 = −z1 satisfy z2 + λ = 0. So for all z

z2 + λ = (z − z1)(z − z2)

= z2 − (z1 + z2)z + z1z2

= z2 + z1z2.

Hence λ = z1z2.Form Corollary A.2 we know that z1, z2 ∈ iR, so we can write

λ = (ix)(−ix)

= x2 ≥ 0.

The case λ = 0 has already been excluded, so λ > 0.

67

B Needed Calculations to Compare C?T (h) and

fCO2(h).

In this appendix you find the calculations needed for the results in Paragraph3.3. The calculations with the assumption of proton symporters present in cellsof Chara corallina and the assumption of sodium symporters present in cells ofChara corallina are treated separately and not alongside each other. Howeverthe calculations are very similar and therefore it might be convenient to readthese parts simultaneously when interested in both parts.

B.1 Behaviour of C?T (h) at h = 0. (Proton Symporter

Case)

In this paragraph we will compare C?T (h)) and fCO2(h) near h = 0, when weassume proton symporters to be present in cells of Chara corallina. In order tothis we need

a(0) = κ3(fCO2(h′)PCO2A+ fHCO−3 (h′)φ)

b(0) = κ0(fCO2(h′)PCO2A+ fHCO−3 (h′)φ) +Nk1Keqh′fHCO−3 (h′)

c(0) = 0

D(0) =(κ0(fCO2(h′)PCO2A+ fHCO−3 (h′)φ) +Nk1Keqh

′fHCO−3 (h′))2

=(κ0(fCO2(h′)PCO2A+ fHCO−3 (h′)φ) +Nk1KeqKa1

fCO2(h′))2

It follows that

C?T (0) =−b(0) +

√b(0)2

2a(h)= 0.

So C?T (0) = fCO2(0) = 0. It is still unclear whether C?T (h) is greater or smallerthan fCO2(h) near h = 0, so we need to look at the first derivative of C?T (h)and fCO2(h) at h = 0. We have

fCO2 ′(h) =

(2Ka3h(Ka1Ka3h+Ka3h

2 +Ka1h2 +Ka1Ka2Ka3)

−Ka3h2(Ka1

Ka3+ 2Ka3

h+ 2Ka1h)

)· 1

(Ka1Ka3

h+Ka3h2 +Ka1

h2 +Ka1Ka2

Ka3)2

=Ka1

K2a3h2 + 2Ka1

Ka2K2a3h

(Ka1Ka3h+Ka3h2 +Ka1h

2 +Ka1Ka2Ka3)2,

fCO2 ′(0) = 0

and

C?T′(h) =

(−b′(h) + D′(h)

2√D(h)

)a(h)− a′(h)(−b(h) +

√D(h))

2a(h)2. (79)

68

We need

a′(h) = g′1(h)γ,

b′(h) = g′0(h)γ − g′1(h)PCO2AfCO2(h)CextT

− g1(h)PCO2AfCO2 ′(h)CextT ,

c′(h) = −g′0(h)PCO2AfCO2(h)CextT − g0(h)PCO2AfCO2 ′(h)Cext

T

−Nk0KeqfHCO−3 (h)Cext

T −Nk0KeqhfHCO−3 ′(h)Cext

T ,

D′(h) = 2b(h)b′(h)− 4(a′(h)c(h) + a(h)c′(h)),

with

g′0(h) = κ1Keq + κ2(fHCO−3 (h) · h)′,

g′1(h) = κ4Keq + κ5(fHCO−3 (h) · h)′,

and where γ = (fCO2(h′)PCO2A+ fHCO−3 (h′)φ).We get

a′(0) = g′1(0)γ

= κ4Keqγ,

b′(0) = g′0(0)γ − g′1(0)PCO2AfCO2(0)Cext

T

− g1(0)PCO2AfCO2 ′(0)CextT

= κ1Keqγ,

c′(0) = −g′0(0)PCO2AfCO2(0)CextT − g0(0)PCO2AfCO2(0)Cext

T

−Nk0KeqfHCO−3 (0)Cext

T

= 0.

Since c(0) = 0 and c′(0) = 0 we get D(0) = b(0)2 and D′(0) = 2b(0)b′(0).Substituting this into (79) gives

C?T′(0) =

(−b′(0) + 2b(0)b′(0)

2b(0)

)a(0)− a′(0)(−b(0) + b(0))

2a(0)2

= 0.

Again it is still unclear whether C?T (h) is greater or smaller than fCO2(h) nearh = 0, so we need to look at the second derivative of C?T (h) and fCO2(h) at

69

h = 0. We have

fCO2 ′′(h) =2K2

a1Ka2K

3a3h− 2K2

a1K2a3h3 − 2Ka1K

3a3h3

(Ka1Ka3

h+Ka3h2 +Ka1

h2 +Ka1Ka2

Ka3)3

+2K2

a1K2a2K3a3 − 2K2

a1Ka2

K2a3h2 − 2Ka1

Ka2K3a3h2

(Ka1Ka3

h+Ka3h2 +Ka1

h2 +Ka1Ka2

Ka3)3

=2(h+Ka2

)(K2a1Ka2

K3a3−K2

a1K2a3h2 −Ka1

K3a3h2)

(Ka1Ka3h+Ka3h2 +Ka1h

2 +Ka1Ka2Ka3)3

=2(h+Ka2

)Ka1K2a3

(Ka1Ka2

Ka3−Ka1

h2 −Ka3h2)

(Ka1Ka3

h+Ka3h2 +Ka1

h2 +Ka1Ka2

Ka3)3

,

fCO2 ′′(0) =2K2

a1K2a2K3a3

(Ka1Ka2

Ka3)3

=2

Ka1Ka2

and

C?T′′(h) =

2a(h)2((−b′(h) + D′(h)

2√

D(h)

)a(h)− a′(h)

(−b(h) +

√D(h)

))′4a(h)4

−

((−b′(h) + D′(h)

2√

D(h)

)a(h)− a′(h)

(−b(h) +

√D(h)

))4a(h)a′(h)

4a(h)4

=

a(h)2((−b′(h) + D′(h)

2√

D(h)

)a(h)− a′(h)

(−b(h) +

√D(h)

))′2a(h)4

−

((−b′(h) + D′(h)

2√

D(h)

)a(h)− a′(h)

(−b(h) +

√D(h)

))a(h)a′(h)

a(h)4.

So

C?T′′(0) =

((−b′(h) + D′(h)

2√

D(h)

)a(h)− a′(h)

(−b(h) +

√D(h)

))′∣∣∣∣h=0

2a(0)2

−

((−b′(0) + D′(0)

2√

D(0)

)a(0)− a′(0)

(−b(0) +

√D(0)

))a(0)a′(0)

a(0)4

=1

2a(0)2

((−b′(h) +

D′(h)

2√D(h)

)a(h)− a′(h)

(−b(h) +

√D(h)

))′∣∣∣∣∣h=0

=1

2a(0)2·

(−b′(0)a′(0)− b′′(0)a(0)− D′(0)2a(0)

4D(0)32

+D′′(0)a(0)

2√D(0)

+D′(0)a′(0)

2√D(0)

− a′′(0)(−b(0) +√D(0))− a′(0)

(−b′(0) +

D′(0)

2√D(0)

))

=

−b′′(0)− D′(0)2

4D(0)32

+ D′′(0)

2√

D(0)

2a(0).

70

We have

D′′(h) = 2b(h)b′′(h) + 2b′(h)2 − 4a′′(h)c(h)− 8a′(h)c′(h)− 4a(h)c′′(h).

SoD′′(0) = 2b(0)b′′(0) + 2b′(0)2 − 4a(0)c′′(0).

We need

b′′(h) = g′′0 (h)γ − g′′1 (h)PCO2AfCO2(h)CextT − 2g′1(h)PCO2AfCO2 ′(h)Cext

T

− g1(h)PCO2AfCO2 ′′(h)CextT ,

c′′(h) = −g′′0 (h)PCO2AfCO2(h)CextT − 2g′0(h)PCO2AfCO2 ′(h)Cext

T

− 2Nk0KeqfHCO−3 ′(h)Cext

T −Nk0KeqhfHCO−3 ′′(h)Cext

T

− 2Nk0KeqfHCO−3 ′(0)Cext

T .

So we get

b′′(0) = g′′0 (0)γ − g′′1 (0)PCO2AfCO2(0)CextT − 2g′1(0)PCO2AfCO2 ′(0)Cext

T

− g1(0)PCO2AfCO2 ′′(0)CextT ,

c′′(0) = −g′′0 (0)PCO2AfCO2(0)CextT − 2g′0(0)PCO2AfCO2 ′(0)Cext

T

− g0(0)PCO2AfCO2 ′′(0)CextT ,

with

g′′0 (h) = κ2(fHCO−3 (h) · h)′′,

g′′0 (0) = 2κ2

Ka2

,

g′′1 (h) = κ5(fHCO−3 (h) · h)′′,

g′′1 (0) = 2κ5

Ka2

.

We have

fCO2(0) = 0

fCO2 ′(0) = 0

fCO2 ′′(0) =2

Ka1Ka2

After substitution we get

b′′(0) = 2κ2

Ka2

γ − κ3PCO2A

2

Ka1Ka2

CextT ,

c′′(0) = −κ0PCO2A

2

Ka1Ka2

CextT − 2Nk0

Keq

Ka2

CextT .

71

So

C?T′′(0) =

−b′′(0)− 4b(0)2b′(0)2

4b(0)3 + 2b(0)b′′(0)+2b′(0)2−4a(0)c′′(0)2b(0)

2a(0)

=−b′′(0)− b′(0)2

b(0) + b′′(0) + b′(0)2

b(0) − 2a(0)c′′(0)b(0)

2a(0)

= −c′′(0)

b(0)

=κ0P

CO2A 2Ka1

Ka2CextT + 2Nk0

KeqKa2

CextT

κ0γ +Nk1Keqh′fHCO−3 (h′)

=κ0P

CO2A 2Ka1Ka2

CextT + 2Nk0

KeqKa2

CextT


.

Now we know that fCO2 ′′(0) 6= C?T′′(0), since C?T

′′(0) is still dependent on h′,while fCO2 ′′(0) is not. Hence C?T (h) is either greater or smaller than fCO2(h).The elaborations on this matter are continued in 3.3.

B.2 Behaviour of C?T (h) at h = 0. (Sodium Symporter

Case)

In this paragraph we will compare C?T (h)) and fCO2(h) near h = 0, when weassume sodium symporters to be present in cells of Chara corallina. Introduce

D(h) = b(h)2 − 4a(h)c(h),

then our solution to (22) is

C?T (h) =−b(h) +

√D(h)

2a(h)

and hence

C?T (0) =−b(0) +

√D(0)

2a(0).

Since fCO2(0) = 0 and fHCO−3 (0) = 0 we have c(0) = 0 and therefore D(0) =b(0)2. So it follows that C?T (0) = 0.

Now we have to look at the derivative at h = 0.

C?T′(h) =

(−b′(h) + D′(h)

2√D(h)

)a(h)− a′(h)(−b(h) +

√D(h))

2a(h)2,

whereD′(h) = 2b(h)b′(h)− 4(a′(h)c(h) + a(h)c′(h)).

72

It follows that

C?T′(0) =

(−b′(0) + 2b(0)b′(0)−4a(0)c′(0)

2b(0)

)a(0)

2a(0)2

=−b′(0) + b(0)− 4a(0)c′(0)

b(0)

2a(0)

= −2c′(0)

b(0).

We have

c′(h) = −(g′0(h)fCO2(h) + g0(h)fCO2 ′(h))PCO2ACextT −Nk0Keqηf

HCO−3 ′(h)CextT

and so

c′(0) = Nk0Keqη1

Ka2

CextT (0).

It follows that

C?T′(0) = 2

Nk0Keqη1

Ka2CextT

(κ0 + κ1Keqη)(fCO2 (h′)PCO2A+ fHCO−3 (h′)φ) +Nk1Keqη′fHCO−3 (h′)

=2Nk0KeqηC

extT

Ka2 (κ0+κ1Keqη)(fCO2 (h′)PCO2A+ fHCO−3 (h′)φ)+Ka2Nk1Keqη′fHCO−3 (h′)

> 0.

B.3 Value of C?T (∞). (Proton Symporter Case)

In this paragraph we will compare C?T (h)) and fCO2(h) as h → ∞, when weassume proton symporters to be present in cells of Chara corallina. Recall thatwe wrote

C?T (h) =−b(h) +

√D(h)

2a(h),

where a(h), b(h) and c(h) are as in (23). We can rewrite this as

C?T (h) = − b(h)

2a(h)+

√b(h)2

4a(h)2− 4a(h)c(h)

4a(h)2(80)

= − b(h)

2a(h)+

√1

4

(b(h)

a(h)

)2

− c(h)

a(h). (81)

From this last expression it is clear that in order to know more about C?T (∞)we only want to know more about

limh→∞

− b(h)

a(h)and lim

h→∞− c(h)

a(h).

73

−b(h)

a(h)=g1(h)PCO2AfCO2 (h)Cext

T − g0(h)(fCO2 (h′)PCO2A+ fHCO−3 (h′)φ)−Nk0Keqh

′fHCO−3 (h′)

g1(h)(fCO2 (h′)PCO2A+ fHCO−3 (h′)φ)

= fCO2 (h)

PCO2ACextT

fCO2 (h′)PCO2A+ fHCO−3 (h′)φ

−g0(h)

g1(h)−

1

g1(h)

Nk0Keqh′fHCO

−3 (h′)


,

−c(h)

a(h)=g0(h)Cext

T PCO2AfCO2 (h) +Nk0KeqhfHCO

−3 (h)Cext

T


=g0(h)fCO2 (h)

g1(h)

PCO2ACextT


+Ka1

fCO2 (h)

g1(h)

Nk0KeqCextT


We know that

limh→∞

fCO2(h) =Ka3

Ka3+Ka1

, limh→∞

g0(h)

g1(h)=κ1

κ4and lim

h→∞

1

g1(h)= 0.

It follows that

limh→∞

− b(h)

a(h)=

Ka3

Ka3 +Ka1

PCO2ACextT

fCO2(h′)PCO2A+ fHCO−3 (h′)φ− κ1

κ4,

limh→∞

− c(h)

a(h)=

Ka3

Ka3 +Ka1

κ1

κ4

PCO2ACextT


Substituting this back into the expression for C?T (h) gives

C?T (∞) =Ka3

Ka3+Ka1

PCO2ACextT

2(fCO2(h′)PCO2A+ fHCO−3 (h′)φ)− κ1

2κ4

+

( Ka3

Ka3+Ka1

PCO2ACextT


2κ4

)2

+Ka3

Ka3 +Ka1

κ1

κ4

PCO2ACextT

fCO2(h′)PCO2A+ fHCO−3 (h′)φ

] 12

=Ka3

Ka3+Ka1

PCO2ACextT


2κ4

+Ka3

Ka3+Ka1

PCO2ACextT

2(fCO2(h′)PCO2A+ fHCO−3 (h′)φ)+

κ1

2κ4

=Ka3

Ka3 +Ka1

PCO2ACextT


B.4 Value of C?T (∞). (Sodium Symporter Case)

In this paragraph we will compare C?T (h)) and fCO2(h) as h → ∞, when weassume sodium symporters to be present in cells of Chara corallina. Recall thatwe wrote

C?T (h) =−b(h) +

√D(h)

2a(h),

74

where a(h), b(h) and c(h) are as in (24). We can rewrite this as (81) on page73. Hence we are interested in the same limits.

−b(h)

a(h)=g1(h)PCO2AfCO2 (h)Cext

T − g0(h)(fCO2 (h′)PCO2A+ fHCO−3 (h′)φ)−Nk0Keqη

′fHCO−3 (h′)


= fCO2 (h)

PCO2ACextT


−g0(h)

g1(h)−

1

g1(h)

Nk0Keqη′fHCO

−3 (h′)


,

−c(h)

a(h)=g0(h)Cext

T PCO2AfCO2 (h) +Nk0KeqηfHCO

−3 (h)Cext

T


=g0(h)fCO2 (h)

g1(h)

PCO2ACextT


+fHCO

−3 (h)

g1(h)

Nk0KeqηCextT


We know that

limh→∞

fCO2(h) =Ka3

Ka3+Ka1

, limh→∞

fHCO−3 (h) = 0,

limh→∞

g0(h)

g1(h)=κ0 + κ1Keqη

κ3 + κ4Keqηand lim

h→∞

1

g1(h)=

1

κ3 + κ4Keqη.

It follows that

limh→∞

− b(h)

a(h)=

Ka3

Ka3+Ka1

PCO2ACextT

fCO2(h′)PCO2A+ fHCO−3 (h′)φ− κ0 + κ1Keqη

κ3 + κ4Keqη

− 1

κ3 + κ4Keqη

Nk0Keqη′fHCO−3 (h′)

fCO2(h′)PCO2A+ fHCO−3 (h′)φ,

limh→∞

− c(h)

a(h)=

Ka3

Ka3 +Ka1

κ0 + κ1Keqη

κ3 + κ4Keqη

PCO2ACextT


Substituting this back into the expression for C?T (h) gives

C?T (∞) =

1

2

Ka3

Ka3+Ka1

PCO2ACext

T

fCO2(h′)PCO2A + f

HCO−3 (h′)φ

−κ0 + κ1Keqη

κ3 + κ4Keqη−

1

κ3 + κ4Keqη

Nk0Keqη′fHCO

−3 (h′)

fCO2(h′)PCO2A + f

HCO−3 (h′)φ

+

1

4

Ka3

Ka3+Ka1

PCO2ACext

T

fCO2(h′)PCO2A + f

HCO−3 (h′)φ

−κ0 + κ1Keqη

κ3 + κ4Keqη−

1

κ3 + κ4Keqη

Nk0Keqη′fHCO

−3 (h′)

fCO2(h′)PCO2A + f

HCO−3 (h′)φ

2

+Ka3

Ka3 +Ka1

κ0 + κ1Keqη

κ3 + κ4Keqη

PCO2ACext

T

fCO2(h′)PCO2A + f

HCO−3 (h′)φ

12

. (82)

Another way to write this is

C?T (∞) =

1

2

Ka3

Ka3+Ka1

PCO2ACext

T

fCO2(h′)PCO2A + f

HCO−3 (h′)φ

−κ0 + κ1Keqη

κ3 + κ4Keqη−

1

κ3 + κ4Keqη

Nk0Keqη′fHCO

−3 (h′)

fCO2(h′)PCO2A + f

HCO−3 (h′)φ

+

1

4

Ka3

Ka3 +Ka1

PCO2ACext

T

fCO2(h′)PCO2A + f

HCO−3 (h′)φ

+κ0 + κ1Keqη

κ3 + κ4Keqη+

1

κ3 + κ4Keqη

Nk0Keqη′fHCO

−3 (h′)

fCO2(h′)PCO2A + f

HCO−3 (h′)φ

2

−Ka3

Ka3+Ka1

1

κ3 + κ4Keqη

Nk0Keqη′fHCO

−3 (h′)

fCO2(h′)PCO2A + f

HCO−3 (h′)φ

PCO2ACext

T

fCO2(h′)PCO2A + f

HCO−3 (h′)φ

12

.

(83)

75

From these two expressions we can see that

Ka3

Ka3+Ka1

PCO2ACext

T

fCO2(h′)PCO2A + f

HCO−3 (h′)φ

−κ0 + κ1Keqη

κ3 + κ4Keqη−

1

κ3 + κ4Keqη

Nk0Keqη′fHCO

−3 (h′)

fCO2(h′)PCO2A + f

HCO−3 (h′)φ

< C?T (∞) <Ka3

Ka3+Ka1

PCO2ACextT


76

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78

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