## Mathematical Biology

Mar 28, 2014 - Y. Wang et al. Fig. 2 Part of cell division images θ 2 h a. Fig. 3 The division of .... r + cos(θ/2) = sin(θ/2) + cos(θ/2), for the 3rd case. Here, the half ...

Mathematical Biology

J. Math. Biol. DOI 10.1007/s00285-014-0784-9

The fencing problem and Coleochaete cell division Yuandi Wang · Mingya Dou · Zhigang Zhou

Received: 28 July 2013 / Revised: 28 March 2014 © Springer-Verlag Berlin Heidelberg 2014

Abstract The findings in this study suggest that the solution of a boundary value problem for differential equation system can be used to discuss the fencing problem in mathematics and Coleochaete, a green algae, cell division. This differential equation model in parametric expression is used to simulate the two kinds of cell division process, one is for the usual case and the case with a “dead” daughter cell. Keywords Fencing problem · Mathematical model · Differential equation · Cell division Mathematics Subject Classification (2000)

34B60 · 49S05 · 92B05

1 Introduction The mathematical problem of fencing has some connection with the division of cells in the two-dimensional case.

Electronic supplementary material The online version of this article (doi:10.1007/s00285-014-0784-9) contains supplementary material, which is available to authorized users. The research was supported partly by a grant of The First-Class Discipline of Universities in Shanghai. Y. Wang (B) · M. Dou Department of Mathematics, Shanghai University, Shanghai 200444, China e-mail: [email protected] M. Dou e-mail: [email protected] Z. Zhou College of Aqua-life Sciences and Technology, Shanghai Ocean University, Shanghai 201306, China e-mail: [email protected]

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The fencing problem resumes in how to divide a domain into parts with an optimal boundary. Some story say, that a farmer has a flat field and his two sons will each inherit half of the field. The farmer wishes to build a fence to divide the field into two equal parts so that each son inherits the same area. Fences are expensive to build, so naturally the farmer wishes to build the shortest fence he can. Wiener (1914) resolved mathematically the problem of “the shortest line dividing an area in a given ratio”. By using geometric methods, Wiener had proved that “the shortest line passing through two given points on the boundary of a given circle, dividing the area of the circle in a given ratio, is an arc of a circle”. Wiener wrote at the end of his short paper, “it is almost self-evident that the shortest line to divide a convex area in a given ratio is a single arc of a circle, but this I have not yet been able to prove”. Goldberg (1969) proposed the following problem: Given a convex quadrilateral, find the shortest curve which divides it into two equal areas. The fencing problem is still unsolved. Wiener’s shortest-line conjecture went unproved for almost sixty years, but in 1973, Richard Joss announced that he had proved it (Notices A.M.S., June 1973, Abstract 705-D1, p.A-461) (Klamkin 1974; Croft et al. 1991). However, Klamkin (1992) wrote: Since Joss’ proof has not been published, it should still be considered as an open problem. Coleochaete orbicularis can grow as a flat disc, comprising a single layer of cells on a flat substrate. Coleocheate scutata is a microscopic fresh-water green alga with simple anatomical features that allow for accurate quantification of morphogenetic processes. It has two kinds of different structures (Li and Bi 1998), one is a disc, the other one is linear and semicircular radial. In this work we investigate the former case. Its branches grow from a little to the surrounding flat, dense form a disk-shaped planar body structure. This configuration is typically composed of a layer of cells, composed by 2–3-layer structures. In addition to the Coleochaete own characteristics, the interest of this work is on Coleochaete having a strong relation with the origin of land plants (Graham 1984; Jiang and Zhang 1988; Karol et al. 2001). Coleochaete is close and very similar to the ancestors of the now-extinct Embryophyte. There are some results on this topic such as Besson and Dumais (2011), Dupuy et al. (2010) and Shi and Wang (2012) in recent years. Image analysis techniques also were used to extract precise models for cell growth geometry. Shi and Wang (2012) discussed Coleochaete cell division process from the perspective of image simulation, mathematical morphology, and C-V model for image processing and image fusion method. In plants, empirical evidence suggests that tensional forces within the cytoskeleton cause cells to divide along the plane that minimizes the surface area of the cell plate (Erreras rule) while creating daughter cells of equal size. Besson and Dumais (2011) used the differential model to discuss the cell evolution. Motivated by the idea of Besson and Dumais (2011), a general differential model in parametric equation has been deduced by the variational method, and the boundary condition is also obtained. In our work we establish a link between geometry and plant cell. The model could describe dividing flat domains with more complex boundaries, especially for some non-convex domains. It can simulate a general planar cell’s division. Geometrically, this model gives also us an idea to find the solution to the fencing problem.

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The fencing problem and Coleochaete cell division Fig. 1 dividing a planar domain into two equal parts P x, y

L C

P x, y

The remainder of this article is organized as follows. The parametric form model is deduced in the next section. The analysis for experimental data is arranged in Sect. 3 and the simulation discussion is at Sect. 4 while the conclusion section is the last one.

2 Parametric equation form 

x = ϕ(t), (t ∈ [0, T ]), and with y = ψ(t) its area ||, where C is a simple closed smooth (may be by part) curve (see Fig. 1). Now, we want to find a shortest curve dividing the domain into two equal parts with the same area. Rectangular and circular are simple cases. For the case when C is a circle, the diameter of the disc is intuitively the shortest curve. And if the boundary C is a rectangular, the shortest curve halving the domain should be the line segment paralleling to the short side. For the general case, let a point P(x, y) ∈ C, and two points P∗ (x∗ , y∗ ) and P ∗ (x ∗ , y ∗ ) on C corresponding, respectively, to t = t∗ and t ∗ with t∗ < t ∗ . Notice 1 xdy − ydx in Fig. 1. that P(x(0), y(0)) = P(x(T ), y(T )) and || = 2 C Now assume L : (x, y) = (α(t), β(t)), (t ∈ [t∗ , t ∗ ]) to be the shortest curve dividing  into two equal parts. It surrounds a half domain together with C ∗ , a part of C to form a boundary. Now consider the Lagrange function for the conditional maximum problem:

Let  be a planar domain with the boundary C :

t ∗  J (L) = t∗

α  2 (t) + β  2 (t) dt

   1 || . +λ xdy − ydx − 2 C ∗ ∪(−L) 2

(2.1)

Assume the curve L here is the shortest one to cut the domain equally. Then, for any perturbation curve L 1 : (x, y) = (α1 (t), β1 (t)), (t ∈ [t∗ , t ∗ ]),

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it changes L into L ε := L + εL 1 with a parameter ε, so, t∗ (ε)

J (L ε ) =

⎛ (α  + εα1 )2 + (β  + εβ1 )2 dt +

λ⎜ ⎝ 2



⎞ ⎟ xdy − ydx − ||⎠ .

C ∗ ∪L ε

t∗ (ε)

(2.2) For convenience, we introduce the wronskian determinant: W ( f, g)(t) = f (t)g  (t) − f  (t)g(t). Computing the derivative with respect to ε, we obtain that t∗ (ε)

(α  + εα1 )α1 + (β  + εβ1 )β1  dt 2     2 (α + εα1 ) + (β + εβ1 ) t∗ (ε)   dt ∗ (ε)  + (α  + εα1 )2 + (β  + εβ1 )2  ∗ · t (ε) dε   dt∗ (ε)  · − (α  + εα1 )2 + (β  + εβ1 )2  t∗ (ε) dε λ λ ∗ ∗ + W (ϕ, ψ)(t (ε)) · t (ε) − W (ϕ, ψ)(t∗ (ε)) · t∗  (ε) 2 2 t∗ (ε) λ t=t ∗ (ε) − λ {β  + εβ1 , −(α  + εβ1 )}·{α1 , β1 }dt + {β, −α}·{α1 , β1 }|t=t∗ (ε) 2

d J (L ε ) = dε

t∗ (ε)

 λ − W (α+εα1 , β +εβ1 )(t ∗ (ε))t ∗  (ε)−W (α+εα1 , β +εβ1 )(t∗ (ε))t∗  (ε) . 2 We denote t ∗ (0) = t ∗ and t∗ (0) = t∗ . At the saddle point, the derivative goes to zero. In fact, we have t ∗

α  α1 + β  β1  dt α2 + β 2 t∗   + α  2 (t ∗ ) + β  2 (t ∗ ) · t ∗  (0) − α  2 (t∗ ) + β  2 (t∗ ) · t∗  (0)  λ W (ϕ, ψ)(t ∗ ) · t ∗  (0) − W (ϕ, ψ)(t∗ ) · t∗  (0) + 2 t ∗ λ ∗ − λ {β  , −α  } · {α1 , β1 }dt + {β, −α} · {α1 , β1 }|t=t t=t∗ 2

d J (L + εL 1 )|ε=0 = 0= dε

t∗

λ λ − W (α, β)(t ∗ ) · t ∗  (0) + W (α, β)(t∗ ) · t∗  (0). 2 2

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Integrating by part, one can get that ⎛ ⎞ t ∗ , β } d {α 0 = − {α1 , β1 } · ⎝  + λ{β  , −α  }⎠ dt dt 2 2   α +β t∗ ⎞ ⎛ t ∗ , β } {α λ  + {β, −α}⎠ · {α1 , β1 } + ⎝ t∗ 2 2 2 α + β     λ W (ϕ, ψ)(t ∗ ) − W (α, β)(t ∗ ) · t ∗  (0) + α  2 (t ∗ ) + β  2 (t ∗ ) + 2   λ α  2 (t∗ ) + β  2 (t∗ ) + (W (ϕ, ψ)(t∗ ) − W (α, β)(t∗ )) · t∗  (0). − 2

(2.3)

Take note that on the cross connect points, we have 

P ∗ (x ∗ , y ∗ ) : {ϕ(t ∗ (ε)), ψ(t ∗ (ε))} = {α(t ∗ (ε)) + εα1 (t ∗ (ε)), β(t ∗ (ε)) + εβ1 (t ∗ (ε))}; P∗ (x∗ , y∗ ) : {ϕ(t∗ (ε)), ψ(t∗ (ε))} = {α(t∗ (ε)) + εα1 (t∗ (ε)), β(t∗ (ε)) + εβ1 (t∗ (ε))}.

By calculating the derivatives, ϕ  (t ∗ (ε))t ∗  (ε) = (α  (t ∗ (ε)) + α1 (t ∗ (ε)))t ∗  (ε) + α1 (t ∗ (ε)). one has t ∗  (ε) =

α1 (t ∗ (ε))  ∗ ϕ (t (ε)) − α  (t ∗ (ε)) − α1 (t ∗ (ε))

=

β1 (t ∗ (ε)) .  ∗ ψ (t (ε)) − β  (t ∗ (ε)) − β1 (t ∗ (ε)) (2.4)

Similarly, we obtain that α1 (t∗ (ε)) β1 (t∗ (ε)) =  . ψ  (t∗ (ε)) − β  (t∗ (ε)) − β1 (t∗ (ε)) ϕ (t∗ (ε)) − α  (t∗ (ε)) − α1 (t∗ (ε)) (2.5) Here, the two Eqs. (2.4) and (2.5) on t ∗  (ε) and t∗  (ε) should be two compatibility conditions. Since the vector {α1 , β1 } is arbitrary, Eq. (2.3) gives that t∗  (ε) =

d {α  , β  }  + λ{β  , −α  } = 0. dt 2 2   α +β

(2.6)

It is easy to find that {α(t), β(t)} =

1 {cos Ct, sin Ct} + {C1 , C2 } λ

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can solve the equation system (2.6). On the other hand, putting the Eqs. (2.4)–(2.5) with ε = 0 into the Eq. (2.3), then ⎛

⎞ t ∗ , β } {α λ  0 = ⎝ + {β, −α}⎠ · {α1 , β1 } t∗ 2 α2 + β 2    λ ∗ ∗ 2 ∗ 2 ∗   W (ϕ, ψ)(t ) − W (α, β)(t ) · t ∗  (0) + α (t ) + β (t ) + 2   λ α  2 (t∗ ) + β  2 (t∗ ) + (W (ϕ, ψ)(t∗ ) − W (α, β)(t∗ )) · t∗  (0) − 2 ⎞ ⎛ t ∗ , β } {α λ  = ⎝ + {β, −α}⎠ · {α1 , β1 } + (t ∗ , λ) − (t∗ , λ). (2.7) t∗ 2 2 2   α +β Here, 

(η, λ) =

α  2 (η) + β  2 (η) +

λ 2

(W (ϕ, ψ)(η) − W (α, β)(η))

ψ  (η) − β  (η)

· β1 (η).

One can use ϕ  (η) − α  (η) (η = t ∗ or t∗ ) to express t ∗  (0) and t∗  (0) if ψ  (η) − β  (η) = 0, and can analyze it similarly. Hence, the boundary condition of system (2.6) is as follows λ α  (t ∗ ) + β(t ∗ ), at t ∗ : 0 =  2 α  2 (t ∗ ) + β  2 (t ∗ ) ⎞ ⎛  (t ∗ ) β λ ∗ − α(t )⎠ (ψ  (t ∗ ) − β  (t ∗ )) 0 = ⎝ 2 α  2 (t ∗ ) + β  2 (t ∗ )   λ W (ϕ, ψ)(t ∗ ) − W (α, β)(t ∗ ) ; + α  2 (t ∗ ) + β  2 (t ∗ ) + 2 λ α  (t∗ ) at t∗ : 0 =  + β(t∗ ), 2 α  2 (t∗ ) + β  2 (t∗ ) ⎞ ⎛  (t ) β λ ∗ − α(t∗ )⎠ (ψ  (t∗ ) − β  (t∗ )) 0 = ⎝ 2 2 2 α  (t∗ ) + β  (t∗ )  λ + α  2 (t∗ ) + β  2 (t∗ ) + (W (ϕ, ψ)(t∗ ) − W (α, β)(t∗ )) . 2

123

(2.8)

(2.9) (2.10)

(2.11)

The fencing problem and Coleochaete cell division

By combining α(η) = ϕ(η) and β(η) = ψ(η) (η = t∗ , t ∗ ) with (2.8) and (2.10), (2.9) and (2.11) become  β  (η)ψ  (η)+α  2 (η) λ  0=  + W (ϕ, ψ)(η) − W (α, β)(η) − ϕ(η)ψ  (η)+ϕ(η)β  (η) 2 α  2 (η)+β  2 (η) 1 = (α  (η)ϕ  (η) + β  (η)ψ  (η)). 2 2 α  (η) + β  (η) That is, at the points η = t∗ and t ∗ , 

α  (η) 2

2

α (η) + β (η)

+

λ ψ(η) = 0, {α  (η), β  (η)} · {ϕ  (η), ψ  (η)} = 0. 2

(2.12)

Consequently, we get the following theorem. Theorem 1 The shortest curve dividing a planar domain with a smooth (may be piecewisely) boundary C is a circle satisfying the following boundary value problem for differential equations: ⎧ d {α  , β  } ⎪ ⎪  + λ{β  , −α  } = 0, in (t∗ , t ∗ ); ⎪ ⎪ ⎨ dt α  2 + β  2 α λ ⎪ ⎪  + ψ = 0, {α  , β  } · {ϕ  , ψ  } = 0, at t = t∗ and t ∗ . ⎪ ⎪ ⎩ 2 2 2 α + β  Here, the constant λ is the curvature of the circle. This result can be used to find the solution to the fencing problem, especially for some non-convex domain with complex boundaries. And the equation system (2.6) and their boundary value conditions (2.12) are also consistent with the Cartesian coordinate form by Besson and Dumais (2011). The procedure to find the shortest curve can begin at any point P ∗ on the boundary C, and then seek out the circle (or line) L intersecting vertically the boundary at that point. This circle L should also cross vertically the boundary C at another point P∗ . Then pick out all segments which divide the domain equally into two parts. The last step is to compare the length for all the segments L. 3 Experimental results Let’s see some pictures of Coleochaete cell dividing process (Fig. 2). In order to find a clearer analysis on the fission, firstly, we discuss two simple cases of isosceles triangle domain and sector domain by using Theorem 1.

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Fig. 2 Part of cell division images

θ 2

h

a

Fig. 3 The division of triangle

3.1 Equally dividing for isosceles triangle The shortest curve dividing the isosceles triangle into two same area parts is an arc perpendicular to the waist line(s). There are two cases, the angle of π/6 being a critical one. When the top angle is not bigger than π/6, the center of arc is at the top vertex. When the top angle is bigger than π/6, the center of arc is at the left or the right vertex. Actually, assume the top angle of an isosceles triangle is θ . The base line has a length of a and the height is h. It is not difficult to get that a line l being parallel to the base line can dichotomize the triangle. This short line l is longer than that arc like the left two pictures in Fig. 3, i.e.,  the length of l = the length of arc

a/(2h) ∈ (1, 1.01168), arctan a/(2h)

 π as θ ∈ 0, . 6

Similarly, the height h is also longer than the arc with the center at the left or the right base vertex (the right two pictures in Fig. 3), i.e.,

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The fencing problem and Coleochaete cell division

1.002 1.000

LONGTHOF ARC LONGTHOF LINE

1.002 1.000

0.998

0.998

0.996

0.996

0.994

0.994

0.992

0.992

5

10

15

20

5

LONGTHOF ARC LONGTHOF LINE

10

15

20

Fig. 4 The division of polygon

 the height h = the length of arc

tan(π − θ/2) ∈ (1, 1.05), π − θ/2

as θ ∈

π 6

,

π . 2

3.2 Equally dividing for regular polygon For a regular k-sided polygon, one can find a shortest curve to equally divide it. • k = 2n. The shortest dividing curve is a segment of a line which goes through the center of the circle and is perpendicular to the two bottom sides parallel to each other. • k = 2n + 1. The shortest dividing curve is an arc which has a center outside. The case k = 2n + 1 can be discussed in two types, one is for k = 4n + 1 and the other is for k = 4n − 1. A segment of the axis of symmetry or a piece of the line, can separate the polygon equally, but it is not the shortest curve. The shortest one is an arc whose center is on the axis of symmetry. The upper three images in Fig. 4 show the dividing, and the growth-curve in the lower two images is the ratio of the length of shortest arc over the length of dividing line. The detail is expressed as follows. ⎧         π 2nπ π 2nπ ⎪ 2 0.75sin 2π sin (2n−1)π ⎪ 2− k k +cos( k )sec k k ⎪ , k = 4n + 1; la ⎨ 1+cos( πk )  =        (2n−1)π π (2n−1)π π ⎪ ll 2 0.75sin 2π sin 2(n−1)π ⎪ 2− k k +cos( k )sec k k ⎪ ⎩ , k = 4n − 1; 1+cos( πk ) where “la” is the abbreviation for the length of arc and “ll” is for the length of line.

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LENGTH OF CURVE

l

2.0

O θ 2

1.5

α

r 1.0

α

0.5 θ

0.5

1.0

1.5

2.0

2.5

3.0

Fig. 5 The division of union of an isosceles triangle and a half circle

3.3 Equally dividing for the union of an isosceles triangle and a half circle Now, we should like to divide equally the union of an isosceles triangle and a circle whose diameter is the basis of the isosceles triangle. Suppose the isosceles triangle has a top angle of θ and hypotenuses of unit length 1. Its basis has a length of r = sin(θ/2). This r is the radius of the lower circle (the left part of Fig. 5). There are three types of dividing curves. The first dividing curve is a part of a circle with center at the top point and radius of l. The second one is a part of a circle centering below the union with radius of R. The last one is the center line with the length of r + cos(θ/2). The three dividing curves are the bold curves in Fig. 5. The right part shows the length of these three dividing curves. The length of the dividing curve (line) is equal to ⎧    ⎪ 2 ⎪ ⎨lθ = θ sin θ + π sin (θ/2) ,

for the 1st case;

2Rα = 2α cot α sin(θ/2), ⎪ ⎪ ⎩r + cos(θ/2) = sin(θ/2) + cos(θ/2),

for the 2nd case; for the 3rd case.

Here, the half angle α is an inverse function of θ coming from the following function θ = −2 arctan

2 , 4α cot 2 α − 4 cot α − 4α + π

since θ is strictly decreasing when θ > 0. This result is consistent with the example 3.1 by Miori et al. (2004). 3.4 Equally dividing for a sector Let the shadow domain in Fig. 6 be the area between a circle of radius R and a second one of radius r , here their lengths are expressed by the notations L and l respectively. The area S of the shadow sector is

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The fencing problem and Coleochaete cell division Fig. 6 The division of sector

L

l

R-r

R r

θ

S=

1 2 (R − r 2 )θ. 2

Since Rθ = L , r θ = l, then θ = Therefore, the area becomes S=

L −l L +l L +l ,R +r = = · (R − r ). R −r θ L −l

L +l L −l L +l 1 (R − r ) · (R − r ) · = (R − r ). 2 L −l R −r 2

We define a “radial division” when dividing by a radius, and a “tangent division” when dividing by a concentric arc. It is that the centric angle θ , two arcs determining that the shadow domain has a radial division or a tangent division. The value  2 θ∗ = · (R − r ). (3.1) R2 + r 2 is critical for the two kinds of division. The tangent division happens as θ < θ ∗ , and the radial division happens as θ > θ ∗ . The following figure shows the growth process for the northeast part of a cell in Fig. 2 within 160 h. The solid and the broken curves express the growth and the division process respectively. The left picture in Fig. 8 demonstrates the distribution of two kinds of division. The square is radial and the circle is tangent. The right picture in Fig. 8 is about the area of the “mother cell” when the division happens. We take every point in the left part of Fig. 7 as a centric point of a bivariate normal distribution. The fitting results show that the cell area is on two hyperbolas, the lower one is for small area. As the time goes up, the area goes from the lower one up to the upper hyperbola. Moreover, the area distribution can be found in Fig. 9, which has two accumulation points. This result is confirmed by the right part in Fig. 8. And, Fig. 10 shows the centric angle θ in experimental data which is consistent with the theoretic result θ ∗ in (3.1). It describes the average values of θ/θ ∗ , the lower one being 0.66622 for tangent division and the upper one being 1.37679 for radial division. The detailed data for Fig. 10 can be found in Table 1.

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5th gen 4th gen

200

1st gen

3rd gen

150

2nd gen

100

50

0

0

16

32

48

64

80

96

112

128

144

160

Fig. 7 Experimental data for cell division in Fig. 2 40

2nd period

1500

30 1000 20

1st period 500

10

0

0

10

20

30

0

40

0

10

20

30

40

Fig. 8 Radial and tangent division distribution

50

2nd period

40

30

20

10

1st period

0 0

10

20

Fig. 9 Area curve fitting for experimental data

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30

40

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The fencing problem and Coleochaete cell division 3

3

2

2

TANGENT

1

0

5

10

15

20

25

30

35

1

1

0

0.2

0.4

0.6

0.8

1.0

1

Fig. 10 The centric angle distribution for experimental data

4 Model growth Now we turn out to simulate the processing of Coleochaete cell dividing. In plants, empirical evidence shows that green alga cells divide along the plane that minimizes the boundary length of the cell plate while creating daughter cells of equal size (Besson and Dumais 2011). It is supposed that initially, the cell shape is a disk, and it grows to the double size and then divides into two parts. By the theory in Sect. 2, the shape of these two parts is a half disk. For convenience, we call the single cell as the first generation, the two half disks as the second generation, and so on. Obviously, all the third generation cells are of the shape of one-fourth disc. So, the first three generations come simply.

4.1 Uniform growth and division The cells continuously grow, and when the whole area becomes eight (the radius of √ disk is r4 = 2 2), the cell number becomes to eight. For simplicity, let’s see only the one-fourth disk in the first quadrant for x and y both bigger than zero. Considering the symmetry and the randomness, one may assume the dividing curve L 4 (see the left in Fig. 11) is an arc with the center (0, y4 ) being at y-axis. L 4 : x 2 + (y − y4 )2 = R4 2 intersects the circle C4 : x 2 + y 2 = 8 perpendicularly at point D4 (dx4 , d y4 ). At D4 , it has,  ⇒ x + yy  = 0, C4 : x 2 + y 2 = 8, 2 2 2 L 4 : x + (y − y4 ) = R4 , ⇒ x + (y − y4 )y  = 0. Notice that the variables y in C4 and L 4 are different.

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3.0 5

y 2.8

R4

4 2

2

2.6 3 2.4 D4

yo4

2 2.2

C4 L4

2

1

2

x O

x

1

2

3

4

5

2.0 0.0

0.5

1.0

1.5

2.0

2.5

3.0

Fig. 11 Eight equally portions schematically (left). The fourth dividing (center)

Combining these equations, one gets ⎧ ⎨−2yy4 + y4 2 = R4 2 − 8, −x −x ⎩derivative ralation · = −1; y (y − y4 )

 ⇒ intersect

yy4 = 8; y4 2 − R4 2 = 8.

And goes down, √    (2 2R4 , 8) 2  . the center of L 4 : (0, y4 ) = 0, 8 + R4 , D4 : (dx4 , d y4 ) = 8 + R4 2 The shadow part between C4 and L 4 in the left of Fig. 11 has an area of dx4

8 − x2 −



8 + R2 −



R2 − x 2



dx = π.

0

That is, √ 2 2 √ R2 π = 4 arcsin √ arctan − 2R. + 2 R 8 + R2 R

For the radius R4 , it follows R4 ∈ (4.09384, 4.09385), and ˙ 4.97589, D4 (dx4 , d y4 ) = ˙ (2.32704, 1.60775), y40 = ˙ 0.882054, y4 = and the length of the arc L 4 in the left of Fig. 11 is about 2.4751. As the disk becomes bigger and bigger, the upper part (the center in Fig. 11) goes to split-up first. There are three possibilities of division, and the fencing curves have varied lengths. The detailed data are as follows (the right of Fig. 11).

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The fencing problem and Coleochaete cell division No.

Center

The length of arc

1 

(0,0)

2.828

2.73284

2 

(0,4.82322)

2.95983

2.69376

3 

(5.0212,2.2614)

3.9779

2.5643

Fig. 12 The uniform division process

Fig. 13 The non-uniform division process simulation

Calculating down, the division data in the first quadrant are in Table 2 in Appendix. The first several divisions like Fig. 12, and the motion picture is given in the attached file “cmovie1.avi”.

4.2 Non-uniform growth and division Now we see that there is a daughter cell dead after the third generation appears. The first four quarters come like the case of the uniform division, and the dead cell never grows up and splits up. This non-uniform division becomes to be complicated. The experimental picture Fig. 2 shows that it has a dead cell. We assume the quarter in the forth quadrant will not split up. The other three quarters grow continuously (see Fig. 13). The dead one slows down the expanding of its two neighbors in the first and third quadrants. The quarter in the second quadrant goes normally and it splits first. Then, the other two go to be split. One can compute the rest of dividing process by using the differential equation

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Y. Wang et al.

model, see Fig. 13 and the motion picture “cmovie2.avi”. The simulation data can be found in Tables 3, 4 of Appendix.

5 Conclusion Coleochaete planar cell division is complex due to various physical factors, but the mechanism complies with some basic principles. These principles include the shortest membrane dividing two daughter cells. In this paper, we establish the boundary value problem of differential equations by omitting some physical and random factors, and use this model to discuss the cell division process. This discussion method is more reasonable than the simple simulation by only using image evolution. The parametric equation form of differential model can be used to investigate the cell division in which the cell boundary is more complicated. This model can avoid that kind of difficulty caused by the multivalent for complex boundary. This method is to be used to imitate the dividing process for two cases, i.e., uniform and a kind of non-uniform segmentation. This imitation is similar to the real experimental process to a certain degree. And this method can also be used to discuss the cell division in high dimensional space. Another feature of this work is to obtain some connection between a classical mathematical problem; the fencing problem, and the cellular fission. This study tells us that many mathematical methods can be applied to discussing the problem of plant. Acknowledgments The authors would like to thank the Editor/referee for the valuable comments on our submitted manuscript. And we also would like to thank Prof. Heping Ma and Dr. Hamdi Zorgati for their suggestion. The author Zhigang Zhou thanks Dr. Haseloff in the University of Cambridge for helping to get these Coleochaete images.

6 Appendix 6.1 Experimental data The picture Fig. 2 is only a selection of all experiment records of 280 h. The radial division and the tangent division in the experiment can be showed in the following Table 1. The exception is with “†”. The notation of θ ∗ is the theoretic value coming from (3.1), and θ is the measurement in the experiment. All of natations has the meaning as in Fig. 6.

6.2 Uniform cell division The notation of θ1 in Table 2 is the angle between x-axis and the beginning side θ2 is between x-axis and the ending side. And θ3 = (θ2 − θ1 )/2 is the angle for radial dividing line. We denote by r the radius of the disc, and by “Center” and R the center and the radius of dividing arc respectively. The letter of “R” denotes the radial division and “T” is for tangent.

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The fencing problem and Coleochaete cell division Table 1 The check list of the radial (R) and the tangent (T) division in experiment No.

θ(◦ )

1 2

θ∗

θ/θ ∗

R

r

R −r

130

2.26893

10

1

9

16.12

1.26628

1.7918

R

65

1.13446

32

2

30

25.71

1.32304

0.8575†

R

L

Direction

3

60

1.04720

18

8.5

9.5

14.73

0.67482

1.5518

R

4

80

1.39626

20

8.5

11.5

21.44

0.74828

1.8660

R

17.1

0.66287

1.3165

R

0.63236

1.3800

R

0.60440

1.4439

R

5

50

0.87266

25

12

13

6

50

0.87266

26

13

13

7

50

0.87266

27

14

13

18.76

8

44

0.76794

26.5

16.5

10

18.34

0.45296

1.6954

R

9

36

0.62832

26.5

17

19.5

13.98

0.42666

1.4726

R

10

38

0.66323

31.5

18.5

13

17.12

0.50319

1.3180

R

11

32

0.55851

37

21.5

15.5

16.89

0.51216

1.0905

R

12

30

0.52360

35

23

12

15.49

0.40515

1.2924

R

13

35

0.61087

35

25.5

9.5

18.696

0.31020

1.9693

R

14

20

0.34907

47

26

21

13.25

0.55284

0.6314†

R

15

20

0.34907

70

53

17

21.66

0.27378

1.2750

R

16

12

0.20944

73

60

13

13.99

0.19453

1.0766

R

17

88

1.53589

22

1

21

23.9

1.34834

1.1391

T

18

60

1.04720

15

1.5

13.5

11.16

1.26628

0.8270

T

19

70

1.22173

13

2

11

11.35

1.18255

1.0331†

T

20

35

0.61087

21

2

19

9.11

1.27357

0.4796

T

21

33

0.57596

22

2

20

8.99

1.28018

0.4499

T

22

30

0.52360

23

3

20

8.58

1.21924

0.4294

T

23

45

0.78540

23

6.5

16.5

13.27

0.97616

0.8046

T

24

25

0.43633

23

8

15

7.51

0.87099

0.5010

T

25

28

0.48869

30

10

20

10.92

0.89429

0.5465

T

26

28

0.48869

28.5

11

17.5

10.55

0.81001

0.6033

T

27

19

0.33161

33

12.5

20.5

13.93

0.82144

0.4037

T

28

25

0.43633

33

13

20

10.94

0.79733

0.5472

T

29

20

0.34907

33

16

17

9.05

0.65545

0.5326

T

30

17

0.29671

33

16

17

7.69

0.65545

0.4527

T

31

24

0.41888

58

16.5

41.5

17.85

0.97313

0.4304

T

32

36

0.62832

30

17

13

15.312

0.53309

1.1786†

T

33

27

0.47124

35

20

15

13.426

0.52616

0.8956

T

34

18

0.31416

43

20

23

10.53

0.68578

0.4581

T

35

18

0.31416

37

21

16

9.45

0.53178

0.5908

T

36

20

0.34907

45

21

24

12.25

0.68338

0.5108

T

37

14

0.24435

37

21.5

15.5

7.39

0.51216

0.4771

T

38

17

0.29671

41

22

19

9.76

0.57740

0.5139

T

39

18

0.31416

42

23

19

10.63

0.56105

0.5599

T

40

32

0.55851

42

23

19

18.9

0.56105

0.9955

T

41

19

0.33161

42

25

17

11.46

0.49180

0.6743

T

42

16

0.27925

44

26

18

10.09

0.49801

0.5607

T

43

16

0.27925

54

27

27

11.92

0.63236

0.4416

T

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Y. Wang et al. Table 1 continued No.

θ(◦ )

R

r

R −r

L

θ∗

θ/θ ∗

Direction

44

20

0.34907

57

30

23

15.89

0.59271

0.5889

T T

45

18

0.31416

42

31

11

11.59

0.29796

1.0544†

46

20

0.34907

65

35

32

18.21

0.57461

0.6075

T

47

22

0.38397

64

39

25

20.34

0.47167

0.8141

T

48

17

0.29671

70

45

25

17.45

0.42480

0.6985

T

49

17

0.29671

70

46

24

17.56

0.40515

0.7323

T

50

16

0.27925

73

48

25

17.24

0.40461

0.6902

T

51

17

0.29671

72

48

24

18.146

0.39217

0.7566

T

52

20

0.34907

72

49

23

21.48

0.37342

0.9348

T

53

18

0.31416

78

50.5

27.5

20.63

0.41848

0.7507

T

Table 2 A simulation for the uniform cell dividing data No.

r

Direction

Center

R

(θ1 ,θ2 )

1

2.828

T

(0,4.976)

2

3.808

T

(5.021,2.261)

3.978

3

4.289

T

(0,0)

2.828

4

4.954

T

(0,0)

3.808

(0.605,1.230)

5

5.365

T

(0,0)

4.289

(0,0.605)

(0,0)

3.809

θ3

4.094

6

5.741

T

7

5.881

R

(0.605,1.230)

0.626/2

8

6.260

R

(0,0.605)

0.605/2

9

7.170

T

(0,0)

5.741

(1.230,1.571)

(1.230,1.571)

10

7.394

T

(0,0)

5.881

(0.605,1.230)

11

7.744

T

(0,0)

6.260

(0,0.605)

12

8.354

T

(0,0)

7.170

(1.230,1.571)

13

8.646

T

(0,0)

7.394

(0.605,1.230)

(0,0)

7.744

14

8.986

T

15

9.398

R

(1.230,1.571)

0.341/2

16

9.738

R

(0.605,1.231)

0.626/4

17

10.076

R

(0,0.605)

0.605/4

18

11.190

T

(0,0)

9.398

19

11.619

T

(0,0)

9.738

20

11.963

T

(0,0)

10.076

(0,0.605)

(1.230,1.571) (0.605,1.230) (0,0.605)

21

12.733

T

(0,0)

11.190

(1.230,1.571)

22

13.235

T

(0,0)

11.619

(0.605,1.230)

23

13.590

T

(0,0)

11.963

(0,0.605)

24

14.108

T

(0,0)

12.733

(1.230,1.571)

26

14.674

T

(0,0)

13.235

(0.605,1.230)

25

15.042

T

(0,0)

13.590

(0,0.605)

27

15.360

T

(0,0)

14.108

(1.230,1.571)

28

15.984

T

(0,0)

14.674

(0.605,1.230)

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The fencing problem and Coleochaete cell division Table 3 A simulation for the non-uniform dividing in the first quadrant No.

r

Center-1

r1

Center-2

r2

(θ1 ,θ2 )

Direction

1

2.258

(0,10.001)

8.663

2

3.142

(0,1.048)

2.258

T

3

3.645

(0,1.048)

2.258

T

4

3.827

(0.254,1.570)

R

5

4.580

(−0.142,0.254)

R

6

4.920

(0.254,1.570)

R

7

5.882

8

6.070

(3.894,0.343)

(0,1.048)

4.580

(−0.142,0.254)

T

9

6.580

(0,1.048)

4.919

(0.254,1.570)

T

Arched

Arched

10

6.908

(10.46,−0.44)

5.594

7.261

(0,1.048)

6.070

(−0.142,0.254)

T

12

7.900

(0,1.048)

6.580

(0.254,1.570)

T

13

8.020

14

8.282 8.913

16

9.028

17

9.82

18

10.017

2.442

line

11

15

(4.414,0.377)

1.925

(4.976,0.39)

3.001

(4.414,0.377)

2.442

Arched

T

(−0.142,0.254)

R

(5.426,0.41)

3.451

(2.20,−5.74)

6.02

Arched

(5.882,0.417)

3.904

(5.426,0.41)

3.45

Arched

T

(0,1.048)

8.282

(−0.142,0.254)

T

(0.254,1.570)

19

10.592

20

10.941

(6.27,0.425)

4.291

(2.222,−9.883) (0,1.048)

10.21 9.028

21

11.493

(0,1.048)

10.017

R

Arched (0.254,1.570)

T

(−0.142,0.254)

T

22

11.760

(6.86,0.435)

4.876

(4.414,0.377)

2.442

Arched

T

23

11.821

(6.88,0.435)

4.896

(6.270,0.425)

4.291

Arched

T

(6.977,0.436)

4.996

24

12.062

Arched

T

25

12.566

(0,1.048)

10.941

(0.254,1.57)

T

26

12.800

(0,1.048)

11.493

(−0.142,0.254)

T

Table 4 The angle of dividing line for the radial non-uniform division in the first quadrant

(6.270,0.425)

4.291

No.

r

θ3

4

3.82738

1.31581/2

5

4.57951

0.39583

6

4.91926

1.31581/4

14

8.28203

0.39583/2

16

9.02778

1.31581/8

6.3 Non-uniform cell division The notation r in Table 3 and Table 4 is the center of the arc in the first quadrant, “Center-1” and r1 are the center and the radius of the southeast to the first quadrant respectively. “Center-2” and r2 are the center and the radius of dividing arc respectively.

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The radius R of piece of circle in the second quadrant is equal to r + 1.04812. θ1 and θ2 are the angles of the beginning side and the ending side for the sector in the first quadrant respectively. The sector’s center is at the point (0, 1.04812). The “Arched” is the half circle in the southeast to the first quadrant. θ3 in Table 4 is the angle of the dividing line in the sector. References Besson S, Dumais J (2011) Universal rule for the symmetric division of plant cells. Proc Natl Acad Sci USA 108(15):6294–6299 Croft HT, Falconer KJ, Guy RK (1991) A26-Dividing up a piece of land by a short fence. Unsolved problems in intuitive mathematics, vol II, Unsolved problems in geometry Springer-Verlag, New York, pp 37–38 Dupuy L, Mackenzie J, Haseloff J (2010) Coordination of plant cell division and expansion in a simple morphogenetic system. Proc Natl Acad Sci USA 107(6):2711–2716 Goldberg M (1969) Elementary problems: E2185. Am Math Mon 76(7):825–825 Graham LE (1984) Coleochaete and the origin of land plants. Am J Bot 71(4):603–608 Jiang WS, Zhang FD (1988) The origin of terrestrial plant life history. Plants 3:45–46 (in Chinese) Karol KG, McCourt RM, Cimino MT, Delwiche CF (2001) The closest living relatives of land plants. Science 294(14):2351–23 Klamkin MS (1974) Minimal curve for fixed area. Am Math Mon 81:903–904 Klamkin MS (1992) Book reviews. SIAM Rev 34(2):335–338 Li SH, Bi LJ (eds) (1998) Chinese freshwater algae chi, vol 5. Science Press, Beijing, pp 82–86 (in Chinese) Miori C, Peri C, Gomis SS (2004) On fencing problems. J Math Anal Appl 300(2):464–476 Shi BL, Wang YD (2012) Evolution modeling on growth image of cells. Comput Simul 29(12):307–311 (in Chinese) Wiener N (1914) The shortest line dividing an area in a given ratio. Proc Cam Philos Soc 18:56–58

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