Two-phase with surfactant
Free energy
(516)\[\mathscr{F}[\phi,{\color{red}c}]=\underbrace{\int_{V}\left[H\phi^{2}(1-\phi)^{2}+\frac{\zeta}{2}(\boldsymbol{\nabla}\phi)^{2}\right]dV}_{\text{usual free energy for phase-field }\equiv\mathscr{F}[\phi]}+{\color{red}\int_{V}\underbrace{f_{c}(\phi,\boldsymbol{\nabla}\phi,c)}_{\text{free energy for surfactant}}}dV\]
(517)\[{\color{red}f_{c}(\phi,\boldsymbol{\nabla}\phi,c)}=\frac{1}{\beta}\underbrace{\left[c\ln(c)+(1-c)\ln(1-c)\right]}_{(I)}{\color{red}-}\underbrace{\frac{\epsilon}{2}c\left|\boldsymbol{\nabla}\phi\right|^{2}}_{(II)}+\underbrace{\frac{k}{2}c\left[\phi-\frac{1}{2}\right]^{2}}_{(III)}\]
(I): Mixing entropy, penalization of homogeneous mixtures (\(c=0\) or \(c=1\))
(II): Energetic preferency for composition at interface (negative sign because \(c\nearrow\) when \(\mathscr{F}\searrow\))
(III): Penalization of high compositions in the bulk
Derivation of 1D analytical solution
Derivation of chemical potential \(\mu_{c}\)
(518)\[\mathscr{F}[\phi,c]=\mathscr{F}[\phi]+\int\biggl\{\frac{1}{\beta}[c\ln(c)+(1-c)\ln(1-c)]-\frac{\epsilon}{2}c\left|\boldsymbol{\nabla}\phi\right|^{2}+\frac{k}{2}c\left[\phi-\frac{1}{2}\right]^{2}\biggr\} dV\]
(519)\[\begin{split}\mu_{c}(\boldsymbol{x},t)&\,\hat{=}\,\frac{\delta\mathscr{F}}{\delta c}\\
&=\frac{\partial f_{c}(\phi,\boldsymbol{\nabla}\phi,c)}{\partial c}\\
&=\frac{1}{\beta}\ln\left(\frac{c}{1-c}\right)-\frac{\epsilon}{2}\left|\boldsymbol{\nabla}\phi\right|^{2}+\frac{k}{2}\left[\phi-\frac{1}{2}\right]^{2}\end{split}\]
Derivation of bulk chemical potential at equilibrium \(\mu_{c}^{b}\)
With following hypotheses:
\[\begin{split}\begin{cases}
c_{b}\ll1 & \text{(input param)}\\
\phi_{b}=0\text{ or }\phi_{b}=1\\
\left|\boldsymbol{\nabla}\phi\right|=0
\end{cases}\end{split}\]
Bulk chemical potential
Eq. (519) becomes
(520)\[\mu_{c}^{b}=\frac{1}{\beta}\ln(c_{b})+\frac{k}{8}\]
1D analytical solution at equilibrium \(c^{eq}(x)\)
Equilibrium: equality of chemical potential \(\mu_{c}^{b}=\mu_{c}(x)\) (Eq. (519) = Eq. (520)
\[\begin{split}\frac{1}{\beta}\ln(c_{b})+\frac{k}{8}&=\frac{1}{\beta}\ln\left[\frac{c^{eq}(x)}{1-c^{eq}(x)}\right]-\frac{\epsilon}{2}\underbrace{\left|\partial_{x}\phi^{eq}\right|^{2}}_{\text{Eq. }(\ref{eq:Kernel3D})}+\frac{k}{2}\left(\phi^{eq}-\frac{1}{2}\right)^{2}\\\frac{1}{\beta}\left\{ \ln(c_{b})-\ln\left[\frac{c^{eq}(x)}{1-c^{eq}(x)}\right]\right\}&=\underbrace{-\frac{\epsilon}{2}\left[\frac{4}{W}\phi(1-\phi)\right]^{2}+\frac{k}{2}\phi\left(\phi-1\right)}_{\equiv-G(\phi)}\\
\ln\left[\frac{1-c^{eq}(x)}{c^{eq}(x)}c_{b}\right]&=-\beta G(\phi)\end{split}\]
1D analytical solution \(c^{eq}(x)\) at equilibrium
\(c^{eq}(x)\)
(521)\[c^{eq}(x)=\frac{c_{b}}{c_{b}+exp(-G(\phi))}\]
With \(G(\phi)\) defined by
(522)\[G(\phi)=\beta\phi(1-\phi)\left[\frac{8\epsilon}{W^{2}}\phi(1-\phi)+\frac{k}{2}\right]\]
Derivation of counter term for composition equation
The conservation equation for \(c\) writes:
(523)\[\partial_{t}c=-\boldsymbol{\nabla}\cdot\boldsymbol{j}_{tot}\]
where the total flux \(\boldsymbol{j}_{tot}\) is the sum of three terms:
(524)\[\boldsymbol{j}_{tot}=\boldsymbol{j}_{Adv}+\boldsymbol{j}_{Diff}+\boldsymbol{j}_{CT}\]
The advective and diffusive fluxes are defined by their classical forms:
Advective flux
\[\boldsymbol{j}_{Adv}=c\boldsymbol{u}\]
Diffusive flux
\[\boldsymbol{j}_{Diff}=-M_{c}\boldsymbol{\nabla}c\]
To derive the counter term flux \(\boldsymbol{j}_{CT}\), we consider the diffusive flux at equilibrium:
(525)\[\boldsymbol{j}_{Diff}^{eq}=-M_{c}\boldsymbol{\nabla}c^{eq}\]
By using the analytical solution (521):
\[\begin{split}\boldsymbol{j}_{Diff}^{eq}&=-M_{c}\boldsymbol{\nabla}\left[\frac{c_{b}}{c_{b}+c_{0}(s)}\right]\\
&=-M_{c}\boldsymbol{n}_{\phi}\frac{\partial}{\partial s}\left[\frac{c_{b}}{c_{b}+c_{0}(s)}\right]\\
&=-M_{c}c^{eq}(1-c^{eq})\mathcal{P}(\phi^{eq})\boldsymbol{n}_{\phi}\end{split}\]
where
(526)\[\mathcal{P}(\phi)=\beta\frac{4}{W}\phi(1-\phi)\left(1-2\phi\right)\left[\frac{k}{2}+\frac{16}{W²}\epsilon(1-\phi)\phi\right]\]
Finally, the counter flux is defined such as it cancels the diffusive flux at equilibrium: \(\boldsymbol{j}_{CT}=-\boldsymbol{j}_{Diff}^{eq}\)
Counter term flux
Counter term flux
(527)\[\boldsymbol{j}_{CT}=+M_{c}c(1-c)\mathcal{P}(\phi)\boldsymbol{n}_{\phi}\]
Definition of polynom \(\mathcal{P}(\phi)\)
(528)\[\mathcal{P}(\phi)=\beta\frac{4}{W}\phi(1-\phi)\left(1-2\phi\right)\left[\frac{k}{2}+\frac{16}{W²}\epsilon(1-\phi)\phi\right]\]
Model of incompressible two-phase flows with surfactant
Mathematical model of surfactant
Incompressible Navier-Stokes
(529)\[\boldsymbol{\nabla}\cdot\boldsymbol{u} =0,\]
(530)\[\rho\left[\frac{\partial\boldsymbol{u}}{\partial t}+\boldsymbol{\nabla}\cdot(\boldsymbol{u}\boldsymbol{u})\right] =-\boldsymbol{\nabla}p_{h}+\boldsymbol{\nabla}\cdot\left[\eta(\boldsymbol{\nabla}\boldsymbol{u}+\boldsymbol{\nabla}\boldsymbol{u}^{T})\right]+\boldsymbol{F}_{tot}\]
Conservative Allen-Cahn
(531)\[\frac{\partial\phi}{\partial t}+\boldsymbol{\nabla}\cdot(\boldsymbol{u}\phi) =\boldsymbol{\nabla}\cdot\left[M_{\phi}\left(\boldsymbol{\nabla}\phi-\frac{4}{W}\phi(1-\phi)\boldsymbol{n}_{\phi}\right)\right],\]
Transport equation
(532)\[\frac{\partial c}{\partial t}+\boldsymbol{\nabla}\cdot(c\boldsymbol{u}) =\boldsymbol{\nabla}\cdot\left[M_{c}\left(\boldsymbol{\nabla}c{\color{red}-c(1-c)\mathcal{P}(\phi)\boldsymbol{n}_{\phi}}\right)\right]\]
(533)\[{\color{red}\mathcal{P}(\phi)} =\beta\frac{4}{W}\phi(1-\phi)\left(1-2\phi\right)\left[\frac{k}{2}+\frac{16}{W{{}^2}}\epsilon(1-\phi)\phi\right]\]
Force
In Eq. (530) the total force contains the gravity force, the capillary force and the Marangoni force \(\boldsymbol{F}_{tot}=\boldsymbol{F}_{c}+\boldsymbol{F}_{g}+\boldsymbol{F}_{M}\).