Appendix A : minimization of free energy functional

We consider one function \(\phi(x,y,z):=\phi(\boldsymbol{x})\) called a phase index, which is a function of position \(\boldsymbol{x}\). The following section is inspired of the action \(\mathscr{S}[q]\) and the « least action principle » to introduce a « free energy functional » \(\mathscr{F}[\phi]\). That functional is defined by the integral over the volume of a « free energy density » \(\mathcal{F}(\phi(\boldsymbol{x}),\boldsymbol{\nabla}\phi)\):

(393)\[\begin{split}\mathscr{F}[\phi] &=\int_{x_{1}}^{x_{2}}\int_{y_{1}}^{y_{2}}\int_{z_{1}}^{z_{2}}\mathcal{F}(\phi,\underbrace{\partial_{x}\phi,\partial_{y}\phi,\partial_{z}\phi}_{\equiv\boldsymbol{\nabla}\phi})\underbrace{dxdydz}_{\equiv dV}\\&=\int_{V}\mathcal{F}(\phi,\boldsymbol{\nabla}\phi)dV\end{split}\]

where \(V\) is the volume. The Euler-Lagrange equation is derived in 1D and next in 3D.

Euler-Lagrange 1D : functions \(\phi(x)\) and \(d\phi/dx\)

In 1D, the functional \(\mathcal{F}(\phi(x),\phi^{\prime}(x))\) is the « free energy density » of the system and \(\phi^{\prime}(x)\equiv d\phi/dx\). The total free energy of the system writes (space integration):

(394)\[\mathscr{F}[\phi]=\int_{x_{1}}^{x_{2}}\mathcal{F}(\phi,\phi^{\prime})dx\]

By applying the \(\delta\)-operator to Eq. (394), we obtain:

(395)\[\begin{split}\delta\mathscr{F} &=\int_{x_{1}}^{x_{2}}\delta\mathcal{F}(\phi,\phi^{\prime})dx\\&=\int_{x_{1}}^{x_{2}}\left[\frac{\partial\mathcal{F}}{\partial\phi}\delta\phi+\frac{\partial\mathcal{F}}{\partial\phi^{\prime}}\delta\phi^{\prime}\right]dx\end{split}\]

After integration by parts of the second term (with \(\delta\phi(x_{1})=\delta\phi(x_{2})=0\)) we obtain:

(396)\[\delta\mathscr{F}=\int_{x_{1}}^{x_{2}}\left[\frac{\partial\mathcal{F}}{\partial\phi}-\frac{d}{dx}\left(\frac{\partial\mathcal{F}}{\partial\phi^{\prime}}\right)\right]\delta\phi dx=0\]

Finally, as the first order condition \(\delta\mathscr{F}=0\) must be true whatever the variation \(\delta\phi\), we obtain:

(397)\[\boxed{\frac{\partial\mathcal{F}}{\partial\phi}-\frac{d}{dx}\left(\frac{\partial\mathcal{F}}{\partial\phi^{\prime}}\right)=0}\]

which is the 1D Euler-Lagrange equation for a given functional \(\mathcal{F}(\phi,\,\phi^{\prime})\) where \(\phi\) is a function of position \(x\). This is a similar equation to the one obtained with the action. The time derivative \(d/dt\) is replaced by a spatial derivative \(d/dx\).

Euler-Lagrange 3D : functions \(\phi(\boldsymbol{x})\) and \(\boldsymbol{\nabla}\phi\)

We consider now a function depending on three independent variables \(\phi(x,y,z)\equiv\phi(\boldsymbol{x})\) and one free energy density which is a function of \(\phi\), and its three derivatives \(\partial_{x}\phi\), \(\partial_{y}\phi`\) and \(\partial_{z}\phi\):

(398)\[\mathcal{F}(\phi,\partial_{x}\phi,\partial_{y}\phi,\partial_{z}\phi):=\mathcal{F}(\phi,\boldsymbol{\nabla}\phi)\]

The 3D, free energy density is a function of \(\phi\) and its gradient \(\boldsymbol{\nabla}\phi\). Eq. ([eq:EnergieLibre_F]) writes:

(399)\[\mathscr{F}[\phi]=\int_{x_{1}}^{x_{2}}\int_{y_{1}}^{y_{2}}\int_{z_{1}}^{z_{2}}\mathcal{F}(\phi,\partial_{x}\phi,\partial_{y}\phi,\partial_{z}\phi)dxdydz\]

By applying the \(\delta\) operator:

(400)\[\begin{split}\delta\mathscr{F}[\phi] &=\int_{x_{1}}^{x_{2}}\int_{y_{1}}^{y_{2}}\int_{z_{1}}^{z_{2}}\delta\mathcal{F}(\phi,\partial_{x}\phi,\partial_{y}\phi,\partial_{z}\phi)dxdydz\\&=\int_{x_{1}}^{x_{2}}\int_{y_{1}}^{y_{2}}\int_{z_{1}}^{z_{2}}\Biggl[\frac{\partial\mathcal{F}}{\partial\phi}\delta\phi+\frac{\partial\mathcal{F}}{\partial(\partial_{x}\phi)}\delta(\partial_{x}\phi)+\frac{\partial\mathcal{F}}{\partial(\partial_{y}\phi)}\delta(\partial_{y}\phi)+\frac{\partial\mathcal{F}}{\partial(\partial_{z}\phi)}\delta(\partial_{z}\phi)\Biggr]dxdydz\\&=\int_{V}\left[\frac{\partial\mathcal{F}}{\partial\phi}\delta\phi+\sum_{\alpha=x,y,z}\frac{\partial\mathcal{F}}{\partial(\partial_{\alpha}\phi)}\delta(\partial_{\alpha}\phi)\right]dV\end{split}\]

For the second term we use the Einstein summation convention for repeated indices (the summation sign is canceled) and the operators \(\delta\) and \(\partial_{\alpha}\) are commuted:

(401)\[\begin{split}\delta\mathscr{F}&=\int_{V}\frac{\partial\mathcal{F}}{\partial\phi}\delta\phi dV+\underbrace{\int_{V}\frac{\partial\mathcal{F}}{\partial(\boldsymbol{\nabla}\phi)}\boldsymbol{\nabla}(\delta\phi)dV}_{\text{integration by parts}}\\ &=\int_{V}\frac{\partial\mathcal{F}}{\partial\phi}\delta\phi dV+\underbrace{\cancel{\int_{\partial V}\boldsymbol{\mathcal{F}}\cdot\hat{\boldsymbol{n}}\delta\phi d(\partial V)}}_{\text{hyp: }\boldsymbol{\mathcal{F}}\cdot\hat{\boldsymbol{n}}=0\text{ on }\partial V}-\int_{V}\boldsymbol{\nabla}\cdot\left[\frac{\partial\mathcal{F}}{\partial(\boldsymbol{\nabla}\phi)}\right]\delta\phi dV\end{split}\]

The second term in the right-hand side, involving \(\hat{\boldsymbol{n}}\) the normal vector at the boundary, has been negected:

(402)\[\delta\mathscr{F}[\phi]=\int_{V}\left[\frac{\partial\mathcal{F}}{\partial\phi}-\partial_{\alpha}\left(\frac{\partial\mathcal{F}}{\partial(\partial_{\alpha}\phi)}\right)\right]\delta\phi dV\]

We recognize the divergence operator. Finally the Euler-Lagrange equation is:

(403)\[\boxed{\frac{\partial\mathcal{F}}{\partial\phi}-\boldsymbol{\nabla}\cdot\left(\frac{\partial\mathcal{F}}{\partial(\boldsymbol{\nabla}\phi)}\right)=0}\]

Section author: Alain Cartalade