A Variational Approach to Morphogenesis

Abstract

We model the process of cell fate determination of the flower Arabidopsis-thaliana employing a system of reaction–diffusion equations governed by a potential field. This potential field mimics the flower’s epigenetic landscape as defined by Waddington. It is derived from the underlying genetic regulatory network (GRN), which is based on detailed experimental data obtained during cell fate determination in the early stages of development of the flower. The system of equations has a variational structure, and we use minimax techniques (in particular the Mountain Pass Lemma) to show that the minimal energy solution of our functional is, in fact, the one that traverses the epigenetic landscape (the potential field) in the spatial order that corresponds to the correct architecture of the flower, that is, following the observed geometrical features of the meristem. This approach can generally be applied to systems with similar structures to establish a genotype to phenotype correspondence. From a broader perspective, this problem is related to phase transition models with a multiwell vector potential, and the results and methods presented here can potentially be applied in this case.

Appendix

1.1 Coordinate Reduction

We want to interpolate the discrete dynamical system to obtain a continuous one. We shall recall that each one of the components of the vectors (Sect. 2) corresponds to a specific gene of the organ that can be in one of two different states: (0 or 1). Since there are thirteen genes in the network, when doing an interpolation, we would obtain a system in a 13-dimensional space that would be very difficult and costly to solve. For this reason, we reduce the dimensionality of the system using singular value decomposition. Through a least square adjustment, we find the bidimensional plane that best fits the four points (i.e., the one that minimizes the square of the sum of the distances of each one of the fixed points to it), and we then project the 13-dimensional points to this plane. Let $e_1$ and $e_2$ be two orthonormal vectors in $\mathbb {R}^{13}$ and $\varPi =\langle e_1,e_2 \rangle$ the plane generated by these vectors. We want to find $e_1$ and $e_2$ such that the sum of the distances of each vector (fixed point) $q_1, q_2, q_3, q_4$ to the plane $\varPi$ is the least one. We have to minimize the quantity

$$\begin{aligned} S=\sum _i d^2(q_i,\varPi ) \end{aligned}$$

(10)

where $S:=S(e_1,e_2)$, and $d^2(q_i,\varPi )$ is the square of the distance of the vector $q_i$ to the plane $\varPi$. That is,

$$\begin{aligned} d^2(q_i,\varPi )=||q_i-Pq_i||^2, \end{aligned}$$

(11)

where $Pq_i$ is the projection of the vector $q_i$ to the plane $\varPi$,

$$\begin{aligned} Pq_i=(q_i\cdot e_1)e_1+(q_i\cdot e_2)e_2, \end{aligned}$$

(12)

and $(\cdot )$ denotes the scalar product between the corresponding vectors.

To minimize S, we use singular value decomposition (SVD), which is an excellent tool when working with sets of equations of matrices that are singular or numerically nearly singular.

We start by adjusting a plane to the set of vectors $\{q_1,q_2,q_3,q_4\}$. Let Q be the matrix of size $m\times n$ $(m=13, n=4)$, formed by placing the vectors $q_i$ as columns, i.e.,

$$\begin{aligned} Q=(q_1|,q_2|,\ldots |, q_4). \end{aligned}$$

(13)

Using SVD, we can decompose the matrix Q into three factors and find its singular values. Namely, we express the matrix Q as the product of three matrices: U, an orthogonal (by columns) matrix of size $m\times n$; D, a diagonal matrix of size $n\times n$, whose entries are greater or equal to zero; and finally, V, a transpose matrix of an orthogonal one, of size $n\times n$. That is,

$$\begin{aligned} Q=U\cdot D \cdot V^T. \end{aligned}$$

(14)

Since both U and V have orthogonal columns, then $U^TU=V^TV=I$ and $V\cdot V^T=I$; therefore, D will be the diagonal matrix given by:

$$\begin{aligned} D=\left[ \begin{array}{cccc} w_1 &{} 0 &{} \ldots &{} 0 \\ 0 &{} w_2 &{} \ldots &{} 0 \\ \vdots &{} \vdots &{} \ddots &{} \vdots \\ 0 &{} 0 &{} \ldots &{} w_n \end{array} \right] \\ \end{aligned}$$

where $w_i$ will be the singular values of Q and $w_i\ge 0\ \forall i$.

Having obtained the matrices U, V, and D, we will take $e_1$ and $e_2$ as the first and second columns of $V^T$ (which are mutually orthogonal), respectively. The plane formed by these orthogonal vectors is the plane that minimizes the sum of the distances of each vector $q_i$ to it.

Now we can compute the projection $Pq_i$ of each vector $q_i$ to the plane $\varPi =\varPi <e_1,e_2>$,

$$\begin{aligned} Pq_i=(q_i\cdot e_1)e_1+(q_i\cdot e_2)e_2. \end{aligned}$$

(15)

If we take $e_1$ and $e_2$ as the vectors that generate, respectively, the horizontal and vertical axis of a bidimensional coordinate system, we can compute both coordinates of each vector $q_i$ with respect to $\{e_1,e_2\}$, using the following equation

$$\begin{aligned} x_{q_i}=Pq_i\cdot e_1, \ \ \ \ y_{q_i}=Pq_i\cdot e_2 \end{aligned}$$

(16)

where x and y denote the horizontal and vertical axis, respectively.

We find that a basis that generates the plane we need is given by

$$\begin{aligned} \begin{array}{lll} e_1&{}=&{}[ -0.2202,-0.4050,-0.1848,0,-0.4050,-0.4050,0,\\ &{} &{}\,\,\, -\,0.2202,0,-0.3291,-0.4050,-0.2286,-0.2286\ ] \end{array} \end{aligned}$$

and

$$\begin{aligned} \begin{array}{lll} e_2&{}=&{}[ -0.5118,0.1032,0.6150,0,0.1032,0.1032,0,\\ &{} &{} \,\,\,-\,0.5118,0,-0.2273, 0.1032,0.0422,0.0422\ ]. \end{array} \end{aligned}$$

Table 2 Equilibrium points in the plane $u-v$. We find the plane $u-v$ that best fits the 4 points $q_i$ and project each point into it. The result is the following for points now in $\mathbb R^2$:

By projecting our original vectors into this plane, we obtain the following four points in $\mathbb R^2$, each corresponding to a flower organ (see Table 2).

Note that if we wish to stay in the positive octant, it suffices to do a translation in such a way that the four initial conditions $(u_1,v_1),(u_2,v_2),(u_3,v_3)$, and $(u_4,v_4)$ are all positive.