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On a Cahn–Hilliard equation for the growth and division of chemically active droplets modelling protocells

Published online by Cambridge University Press:  06 November 2025

Harald Garcke
Affiliation:
Fakultät für Mathematik, Universität Regensburg, Regensburg, Germany
Kei Fong Lam*
Affiliation:
Department of Mathematics, Hong Kong Baptist University, Kowloon Tong, Hong Kong
Robert Nürnberg
Affiliation:
Department of Mathematics, University of Trento, Trento, Italy
Andrea Signori
Affiliation:
Department of Mathematics, Politecnico di Milano, Milano, Italy
*
Corresponding author: Kei Fong Lam; Email: akflam@hkbu.edu.hk
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Abstract

The Cahn–Hilliard model with reaction terms can lead to situations in which no coarsening is taking place and, in contrast, growth and division of droplets occur which all do not grow larger than a certain size. This phenomenon has been suggested as a model for protocells, and a model based on the modified Cahn–Hilliard equation has been formulated. We introduce this equation and show the existence and uniqueness of solutions. Then, formally matched asymptotic expansions are used to identify a sharp interface limit using a scaling of the reaction term, which becomes singular when the interfacial thickness tends to zero. We compute planar solutions and study their stability under non-planar perturbations. Numerical computations for the suggested model are used to validate the sharp interface asymptotics. In addition, the numerical simulations show that the reaction terms lead to diverse phenomena such as growth and division of droplets in the obtained solutions, as well as the formation of shell-like structures.

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Papers
Creative Commons
Creative Common License - CCCreative Common License - BY
This is an Open Access article, distributed under the terms of the Creative Commons Attribution licence (https://creativecommons.org/licenses/by/4.0/), which permits unrestricted re-use, distribution and reproduction, provided the original article is properly cited.
Copyright
© The Author(s), 2025. Published by Cambridge University Press
Figure 0

Figure 1. Convergence experiment for a moving front in $(0,1)^2$. We compare the true solution $q$ of the sharp interface problem with the discrete approximations $q_h$ of the Cahn–Hilliard equation for $\varepsilon =(2^k\pi )^{-1}$, $k=2,4,6$.

Figure 1

Table 1. Convergence experiment for a moving front in $(0,1)^2$, over the time interval $[0,1]$. We also display the experimental order of convergence (EOC)

Figure 2

Table 2. Convergence experiment for a moving front in $(0,1)^3$, over the time interval $[0,1]$. We also display the experimental order of convergence (EOC)

Figure 3

Figure 2. Convergence experiment for a moving front in $(0,1)^3$. We compare the true solution $q$ of the sharp interface problem with the discrete approximations $q_h$ of the Cahn–Hilliard equation for $\varepsilon =(2^k\pi )^{-1}$, $k=2,3,5$.

Figure 4

Figure 3. ($\varepsilon =\frac 1{32\pi }$, $\Omega =(0,1)^2$) evolution for $\beta =0.1$, $S_-=8$, $S_+ = -8$. We show the solution at times $t=0,0.1,1,2,10$.

Figure 5

Figure 4. ($\varepsilon =\frac 1{16\pi }$, $\Omega = (0,4) \times (0,2)$) evolution for $\beta =0.1$, $S_-=1.5$, $S_+ = -1.5$. We show the solution at times $t=0,1,2,5,10$.

Figure 6

Figure 5. ($\varepsilon =\frac 1{32\pi }$, $\Omega = (0,4) \times (0,1)$) evolution for $\beta =0.1$, $S_-=3$, $S_+ = -3$. We show the solution at times $t=0,1,2,5,10$.

Figure 7

Figure 6. ($\varepsilon =\frac {1}{64\pi }$, $\Omega = (0,2)^2$) evolution for $\beta =0.01$, $S_-=1.3$, $S_+ = 1.3$. We show the solution at times $t=0, 1, 2, 3, 20$.

Figure 8

Figure 7. ($\varepsilon =\frac 1{16\pi }$, $\Omega =(0,1)^3$) evolution for $\beta =0.1$, $m_\pm = 0.2$, $S_-=4.5$, $S_+ = -4.5$. We show the solution at times $t=0,0.1,0.5,1,1.5$.

Figure 9

Figure 8. ($\varepsilon =\frac 1{16\pi }$, $\Omega =(0,4)^2$) evolution for $\beta =0.002$, $S_-=0.25$, $S_+ = -4$. We show the solution at times $t=0.1,0.5,1,2,10,100$. In our numerical computations, the solution at the final time appears to be a steady state.

Figure 10

Figure 9. ($\varepsilon =\frac 1{16\pi }$, $\Omega =(0,4)^2$) evolution for $\beta =0.02$, $S_-=0.25$, $S_+ = -4$. We show the solution at times $t=0.1,0.5,1,2,10,100$. In our numerical computations, the solution at the final time appears to be a steady state.

Figure 11

Figure 10. ($\varepsilon =\frac 1{16\pi }$, $\Omega =(0,4)^2$) evolution for $\beta =0.02$, $S_-=0.25$, $S_+ = -4$. We show the solution at times $t=0,1,10,100$. In our numerical computations, the solution at the final time appears to be a steady state.

Figure 12

Figure 11. ($\varepsilon =\frac 1{8\pi }$, $\Omega =(0,1)^3$) evolution for $\beta =0.02$, $S_-=0.25$, $S_+ = -4$. We show the solution at times $t=0.1,0.2,0.5,1,5$.

Figure 13

Figure 12. ($\varepsilon =\frac 1{8\pi }$, $\Omega =(0,1)^3$) evolution for $\beta =0.02$, $S_-=1$, $S_+ = -4$. We show the solution at times $t=0.01,0.05,0.1,0.5,5$.

Figure 14

Figure 13. ($\varepsilon =\frac 1{8\pi }$, $\Omega =(0,1)^3$) evolution for $\beta =0.1$, $S_-=1$, $S_+ = -4$. We show the solution at times $t=0.02,0.1,0.5,1,5$.

Figure 15

Figure 14. ($\varepsilon =\frac 1{8\pi }$, $\Omega =(0,4)^3$) evolution for $\beta =0.1$, $S_- = 0.8$, $S_+= -10$, $m_-=0.2$, $m_+=0.5$, $\rho _\pm = 0.6$. We show the solution at times $t=0,1,5$. In our numerical computations, the solution at the final time appears to be a steady state.

Figure 16

Figure 15. ($\varepsilon =\frac 1{32\pi }$, $\Omega =(0,2)^2$) evolution for $\beta =0.002$, $S_-=0.25$, $S_+ = -4$. We show the solution at times $t=0,0.2,1,2,5$.

Figure 17

Figure 16. ($\varepsilon =\frac 1{32\pi }$, $\Omega =(0,1)^2$) evolution for $\beta =0.002$, $S_-=0.25$, $S_+ = -4$. We show the solution at times $t=0,0.2,1,2,5$.

Figure 18

Figure 17. ($\varepsilon =\frac 1{16\pi }$, $\Omega =(0,2)^2$) evolution for $\beta =0.02$, $S_-=1$, $S_+ = -4$. We show the solution at times $t=0,2,4,6,8,10$.

Figure 19

Figure 18. ($\varepsilon =\frac 1{16\pi }$, $\Omega =(0,8)^2$) evolution for $\beta =0.02$, $S_-=1$, $S_+ = -4$. We show the solution at times $t=0,2,4,6,8,10$.

Figure 20

Figure 19. ($\varepsilon =\frac 1{16\pi }$, $\Omega =(0,2)^3$) evolution for $\beta =0.002$, $S_-=0.25$, $S_+ = -4$. We show the solution at times $t=0,0.3,1,2,5$.

Figure 21

Figure 20. ($\varepsilon =\frac 1{16\pi }$, $\Omega =(0,2)^3$) evolution for $\beta =0.002$, $S_-=0.25$, $S_+ = -4$. We show the solution at times t = 0, 0.3, 0.5, 1, 1.5, 1.6, 2, 5, where the visualisations in the two rows differ.

Figure 22

Figure 21. ($\varepsilon =\frac 1{16\pi }$, $\Omega = (0,8)^3$) evolution for $\beta =0.1$, $S_-=0.8$, $S_+ = -10$, $m_-=0.2$, $m_+=0.5$, $\rho _-=0.2$, $\rho _+=0.1$, $L=-1$. We show the solution at times $t=0,1,1.3,1.4,2,3.5$.

Figure 23

Figure 22. ($\varepsilon =\frac 1{16\pi }$, $\Omega = (0,8)^3$) evolution for $\beta =0.1$, $S_-=0.8$, $S_+ = -10$, $m_-=0.2$, $m_+=0.5$, $\rho _-=0.2$, $\rho _+=0.1$, $L=-1$. We show the solution at times $t=0,1.4,1.8,2,3,4$.

Figure 24

Figure 23. The evolution from Figure 22 at time $t=4.4$ viewed from the front, from the side and from above.