Diffuse interface models in fluid mechanics

and (ii) there is an equivalent of the Laplace relationwhere the “pressure” P 0 is defined byP 0 (c i ) − P 0 (c i ) = 2 σRP 0 (c) ˆ= c dF 0dc − F 0 (c)In this case, the “pressure” P 0 does not have the interpretation of a physical pressure.A Taylor expansion of these equilibrium conditions of a spherical inclusion around the equilibriumstate of a planar interface allow to show thatµ s e ≃ µ eq + 1 2 σc i − c e R(61)It is worth noting that the form of this equilibrium condition is actually similar to the Gibbs-Thompson condition (12).This equilibrium condition shows in particular that the value of the chemical potential is proportionalto the curvature of the interface. This explains how an interface tends to get spherical.Indeed, let us consider an closed interface whose shape is initially irregular. If, at each point ofthe interface, the interfacial zone is at local thermodynamic equilibrium, the above equilibriumcondition is satisfied locally. These means in particular that, along the interface, the chemicalpotential is not uniform. According to the Cahn-Hilliard equation, this yields a diffusion massflux and therefore a mass diffusion. This mass diffusion makes the overall system evolve and,since the Cahn-Hilliard equation is thermodynamically coherent, this evolution tends to makethe system get closer to an equilibrium state. If we consider a spherical inclusion at equilibrium,the chemical potential along the interface is constant, therefore no mass flux exist and the systemkeeps at rest.3.2 The Cahn-Hilliard equationIn the previous section, we derived the equilibrium condition for a Cahn-Hilliard fluid. The issueis to determine how the system behaves out of equilibrium. To start this analysis, we will firstsimplify the system by assuming that the velocity of the mixture is null. In this case, the evolutionof the concentration evolves through a mass diffusion equation that reads∂c∂t = −∇ · jwhere j is the diffusion mass flux, whose expression must be determined.The evolution of the free energy of the mixture is given by∂F∂t = −∇ · q − ∆ fwhere q is the heat flux and ∆ f is the energy dissipation. The second law of thermodynamicsimposes that ∆ f ≥ 0.Using the same developments as those presented in section 2.2 for the van der Waals model,one finds that[∆ f = −j · ∇ (µ − ∇ · φ) − ∇ · q + φ ∂c]∂t − j (µ − ∇ · φ)whereµ ˆ= ∂F∂c23