Motility Switching and Front–Back Synchronisation in Polarised Cells

Abstract

The combination of protrusions and retractions in the movement of polarised cells leads to understand the effect of possible synchronisation between the two ends of the cells. This synchronisation, in turn, could lead to different dynamics such as normal and fractional diffusion. Departing from a stochastic single cell trajectory, where a “memory effect” induces persistent movement, we derive a kinetic-renewal system at the mesoscopic scale. We investigate various scenarios with different levels of complexity, where the two ends of the cell move either independently or with partial or full synchronisation. We study the relevant macroscopic limits where we obtain diffusion, drift-diffusion or fractional diffusion, depending on the initial system. This article clarifies the form of relevant macroscopic equations that describe the possible effects of synchronised movement in cells, and sheds light on the switching between normal and fractional diffusion.

Appendix

1.1 A Miscellaneous

(I) We compute the escape and arrival rates introduced in (63) by using the characteristic solutions (61) and (62). We start from

$$\begin{aligned} j_{\alpha ^-}(t,x)=\int _0^{t}p(k){\alpha ^-}(t,x,k)\mathrm{d}x \end{aligned}$$

which, by using (3) and (62), can be rewritten as:

$$\begin{aligned} j_{\alpha ^-}(t,x)&=\int _0^{t}\phi (k){\alpha ^-}(t-k,x+vk,0)\mathrm{d}k+{\tilde{\alpha }^-_0(x+vk)\phi (k)}\nonumber \\&= \int _0^{t} \phi (t-s)e^{v(t-s)\partial _x}{\alpha ^-}(s,x,0)\mathrm{d}s \, +{\tilde{\alpha }^-_0(x+vk)\phi (k)}. \end{aligned}$$

(101)

The last equality is obtained using the change of variables $k=t-s$ along with the following Taylor expansion:

$$\begin{aligned} e^{v(t-s)\partial _x}f(x)&=\sum _{m=0}^\infty \frac{(v(t-s) \partial _x)^m}{m!}f(x)\\&=\sum _{m=0}^\infty \frac{1}{m!}(v(t-s))^m\partial _x^mf(x)=f(x+v(t-s)). \end{aligned}$$

Analogously we can obtain (64) for $j_{\alpha ^+}(t,x)$.

(II) Now we aim to derive the expression (88). From (84) we write, after multiplying both sides by $\bar{\psi }^+\bar{\psi }^-$

$$\begin{aligned} \bar{\psi }^+\,\bar{\psi }^-\,\bar{\rho }\,=\,\bar{\psi }^+\,\bar{\phi }^- \,\bar{\tilde{\alpha }}^-\,+\,\bar{{\alpha }}^+_0\,\bar{\psi }^+ \,\bar{\psi }^-\,+\,\bar{\psi }^-\,\bar{\phi }^+\,\bar{\tilde{\alpha }}^+ \,\bar{\psi }^-\,+\,\bar{{\alpha }}^-_0\,\bar{\psi }^-\,\bar{\psi }^+. \end{aligned}$$

(102)

Using the initial conditions $\alpha ^+_{0_\varepsilon }=\varepsilon ^z\delta (x)$ and ${{\alpha }}^-_0(x)=0$, we obtain (88). Now, we introduce the scaling to (74) and we write

$$\begin{aligned} \bar{\psi }^\pm&=-\frac{a}{1-\mu }-\frac{a^2\lambda _\pm }{(1-\mu )(2-\mu )}+a^{\alpha ^+}\lambda _\pm ^{\mu -1}\Gamma (-\mu +1)+{\mathcal {O}}(a^3\lambda _\pm ^2)\ ,\\ \bar{\phi }^\pm&=1+\frac{a\lambda _\pm }{1-\mu }+a^\mu \lambda _\pm ^\mu +O(\lambda ^{\mu +1}). \end{aligned}$$

Hence from here we compute

$$\begin{aligned} \bar{\psi }^+\bar{\phi }^-&=-\frac{a}{1-\mu }-\frac{a^2\lambda _+}{(1-\mu )(2-\mu )}+a^\mu \lambda _+^{\mu -1}\Gamma (-\mu +1)-\frac{a^2\lambda _-}{(1-\mu )^2}+{\mathcal {O}}(a^{\mu +1})\ ,\\ \bar{\psi }^-\bar{\phi }^+&=-\frac{a}{1-\mu }-\frac{a^2\lambda _-}{(1-\mu )(2-\mu )}+a^\mu \lambda _-^{\mu -1}\Gamma (-\mu +1)-\frac{a^2\lambda _+}{(1-\mu )^2}+{\mathcal {O}}(a^{\mu +1})\ ,\\ \bar{\psi }^+\bar{\psi }^-&=\frac{a^2}{(1-\mu )^2}-a^{\mu +1}\frac{\Gamma (\mu +1)}{1-\mu }(\lambda _-^{\mu -1}+\lambda ^{\mu -1}_+)+{\mathcal {O}}(a^3). \end{aligned}$$

Substituting these three quantities in (102), we arrive at (89).

(III) Finally, we are going to work only with the fractional operators. Following (Ferrari 2017, 2018) we have

$$\begin{aligned} \mathbb {D}_-^{\mu -1}\tilde{\alpha }^-&=\frac{-1}{\Gamma (2-\mu )}\frac{\partial }{\partial x}\int _x^\infty \frac{\tilde{\alpha }^-(s)}{(s-x)^{\mu -1}}\mathrm{d}s\\&=\frac{\mu -1}{\Gamma (2-\mu )}\int _0^\infty \frac{\tilde{\alpha }^-(x)-\tilde{\alpha }^-(x+s)}{s^{\mu }}\mathrm{d}s\\ \mathbb {D}_+^{\mu -1}\tilde{\alpha }^+&=\frac{1}{\Gamma (2-\mu )}\frac{\partial }{\partial x}\int ^x_{-\infty }\frac{\tilde{\alpha }^+(s)}{(s-x)^{\mu -1}}\mathrm{d}s\\&=\frac{\mu -1}{\Gamma (2-\mu )}\int _0^\infty \frac{\tilde{\alpha }^+(x)-\tilde{\alpha }^+(x-s)}{s^{\mu }}\mathrm{d}s. \end{aligned}$$

The above relation is true if $\tilde{\alpha }^+,\tilde{\alpha }^-\in C^1(\mathbb {R})$ and $\tilde{\alpha }^+,\tilde{\alpha }^- =o(|x|^{\mu -2-\epsilon })$, $x\rightarrow +\infty$ for $\epsilon >0$ (equivalence between Marchaud derivative and Riemann–Liouville derivative).

Now we are going to use the fact that the sum $\mathbb {D}_-^{\mu -1}f+\mathbb {D}^{\mu -1}_+f$ gives the fractional Laplace operator in one dimension, also known as the Riesz derivative,

$$\begin{aligned}&\mathbb {D}_-^{\mu -1}\tilde{\alpha }^-+\mathbb {D}_+^{\mu -1}\tilde{\alpha }^+\\&\quad =\frac{\mu -1}{\Gamma (2-\mu )}\Bigl (\int _0^\infty \frac{\tilde{\alpha }^-(x)-\tilde{\alpha }^-(x+s)}{s^\mu }\mathrm{d}s+\int _0^\infty \frac{\tilde{\alpha }^+(x)-\tilde{\alpha }^+(x-s)}{s^\mu }\mathrm{d}s \Bigr )\\&\quad =\frac{\mu -1}{\Gamma (2-\mu )}\Bigl (\int _{-\infty }^0 \frac{\tilde{\alpha }^-(x)-\tilde{\alpha }^-(x-s)}{|s|^\mu }\mathrm{d}s+\int _0^\infty \frac{\tilde{\alpha }^+(x)-\tilde{\alpha }^+(x-s)}{s^\mu }\mathrm{d}s \Bigr )\\&\quad =\frac{\mu -1}{2\Gamma (2-\mu )}\int _{-\infty }^\infty \frac{\tilde{\alpha }^-(x)+\tilde{\alpha }^+(x)-\tilde{\alpha }^-(x-s)-\tilde{\alpha }^+(x-s)}{|s|^\mu }\mathrm{d}s\\&\quad =\frac{\mu -1}{2\Gamma (2-\mu )}\int _{-\infty }^\infty \frac{\rho (x)-\rho (x-s)}{|s|^\mu }\mathrm{d}s\\&\quad =\frac{\mu -1}{2\Gamma (2-\mu )}\frac{1}{c(1,\frac{\mu -1}{2})}\Bigl (-\frac{\mathrm{d}^2}{\mathrm{d}x^2} \Bigr )^{\frac{\mu -1}{2}}\rho (t,x)\ , \end{aligned}$$

where $c(1,\frac{\mu -1}{2})$ is a normalisation constant.