Appendix B: solutions of one-dimensional Euler-Lagrange

In this appendix we detail the main stages to obtain the solution of:

(465)\[\zeta\frac{d^{2}\phi}{dx^{2}}-H\frac{dg(\phi)}{d\phi}=0\]

We demonstrate that the solution has a hyperbolic tangent profile for several choices of double-wells g(phi).

Equivalent form of 1D Euler-Lagrange

First, we multiply Eq. (465) by \(d\phi/dx\):

(466)\[\zeta\frac{d\phi}{dx}\frac{d^{2}\phi}{dx^{2}}-H\frac{d\phi}{dx}\frac{dg(\phi)}{d\phi}=0\]

The first term is the derivative of \(d\phi/dx\) squared, whereas the second one becomes a derivative wrt \(x\):

(467)\[\frac{\zeta}{2}\frac{d}{dx}\left(\frac{d\phi}{dx}\right)^{2}-H\frac{dg(\phi)}{dx}=0\]

We obtain a first result that will be used for the calculation of the surface tension in Appendix C: calculation of surface tension:

(468)\[\boxed{\frac{\zeta}{2}\left(\frac{d\phi}{dx}\right)^{2}=Hg(\phi)}\]

The second result derives from that relation by keeping the positive root \(d\phi/dx=\sqrt{2Hg(\phi)/\zeta}\). We obtain one relation between \(dx\) and \(d\phi\):

(469)\[dx=\frac{1}{\sqrt{2Hg(\phi)/\zeta}}d\phi\]

That relation will be useful for variable change by transforming the integral on \(x\) into integral on \(\phi\). That relation is equivalent to:

(470)\[\boxed{\frac{d\phi}{dx}=\sqrt{2\frac{H}{\zeta}g(\phi)}}\]

Case \(g_1(\phi)=\phi^2(1-\phi)^2\)

If we choose \(g(\phi)=g_1(\phi)=\phi^2(1-\phi)^2\), then we obtain:

(471)\[\frac{d\phi}{dx}=\sqrt{2\frac{H}{\zeta}}\phi(1-\phi)\]

L’équation s’écrit simplement \(d\phi/\phi(1-\phi)=Cdx\) with \(C=\sqrt{2H/\zeta}\), that, once integrated is equal to: \(\ln(\phi)-\ln(1-\phi)=Cx\). By applying the exponential on both sides and making appear the factor 2 inside the right-hand side, we obtain: \(\phi/(1-\phi)=\exp\left[2(Cx/2)\right]\). Next we compute separately \(\exp\left[2(Cx/2)\right]-1\) and \(\exp\left[2(Cx/2)\right]+1\). We express those two expressions with \(\phi\) and we take the ratio (let us remind that \(\tanh(X)=(e^{2X}-1)/(e^{2X}+1)\)). We find: \(2\phi-1=\tanh(Cx/2)\). Finally by replacing \(C\) by \(C=\sqrt{2H/\zeta}\), we obtain:

(472)\[\phi(x)=\frac{1}{2}\left[1+\tanh\left(\frac{1}{2}\sqrt{\frac{2H}{\zeta}}x\right)\right]\]

We make appear inside the hyperbolic tangent function the simplified form \(2x/W\), we obtain:

(473)\[\boxed{\phi(x)=\frac{1}{2}\left[1+\tanh\left(\frac{2x}{W}\right)\right]}\]

where the interface width is defined by:

(474)\[\boxed{W=\sqrt{\frac{8\zeta}{H}}}\]

Case \(g_{2}(\phi)=(\phi_{l}-\phi)^{2}(\phi-\phi_{g})^{2}\)

We start from Eq. (470) by using \(g_{2}(\phi)=(\phi_{l}-\phi)^{2}(\phi-\phi_{g})^{2}\) where \(\phi_{g}<\phi<\phi_{l}\). After integration of \(d\phi/(\phi_{l}-\phi)(\phi-\phi_{g})=Cdx\), we obtain \(\ln(\phi-\phi_{g})-\ln(\phi_{l}-\phi)=(\phi_{l}-\phi_{g})Cx\). We make appear the ratio \((\phi-\phi_{g})/(\phi_{l}-\phi)\) inside a unique logarithmic function, next we apply the exponential function to both members of equality with a factor 2 in the right-hand side: \((\phi-\phi_{g})/(\phi_{l}-\phi)=\exp\left[2(\phi_{l}-\phi_{g})Cx/2\right]\).

We set \(X=(\phi_{l}-\phi_{g})Cx/2\), to make appear the hyperbolic tangent function (like in previous section), we calculate separately \(e^{2X}-1\) and \(e^{2X}+1\) and we make the ratio. We obtain:

(475)\[\frac{e^{2X}-1}{e^{2X}+1}=\tanh\left[\frac{(\phi_{l}-\phi_{g})}{2}Cx\right]=\frac{2\phi-(\phi_{l}+\phi_{g})}{\phi_{l}-\phi_{g}}\]

Finally we obtain:

(476)\[\phi(x)=\frac{\phi_{l}+\phi_{g}}{2}+\frac{\phi_{l}-\phi_{g}}{2}\tanh\left[\frac{(\phi_{l}-\phi_{g})}{2}\sqrt{\frac{2H}{\zeta}}x\right]\]

We make appear inside the hyperbolic tangent function the simplified expression 2x/W:

(477)\[\boxed{\phi(x)=\frac{\phi_{l}+\phi_{g}}{2}+\frac{\phi_{l}-\phi_{g}}{2}\tanh\left[\frac{2x}{W}\right]}\]

with the interface width defined by:

(478)\[\boxed{W=\frac{4}{(\phi_{l}-\phi_{g})}\sqrt{\frac{\zeta}{2H}}}\]

Case \(g_{3}(\phi)=(\phi^{\star}-\phi)^{2}(\phi^{\star}+\phi)^{2}\)

This is a particular case of \(g_{2}(\phi)\) with \(\phi_{l}=\phi^{\star}\) and \(\phi_{g}=-\phi^{\star}\). With those variable changes Eq. (477) becomes

(479)\[\boxed{\phi(x)=\phi^{\star}\tanh\left[\frac{2x}{W}\right]}\]

with an interface width defined by

(480)\[\boxed{W=\frac{1}{\phi^{\star}}\sqrt{\frac{2\zeta}{H}}}\]

Proof

We check by starting from Eq. (470) en utilisant \(g_{3}(\phi)=(\phi^{\star}-\phi)^{2}(\phi^{\star}+\phi)^{2}\) where \(-\phi^{\star}<\phi<\phi^{\star}\). By integrating \(d\phi/(\phi^{\star}-\phi)(\phi^{\star}+\phi)=Cdx\), we obtain \(\ln(\phi^{\star}+\phi)-\ln(\phi^{\star}-\phi)=2\phi^{\star}Cx\). After application of exponential function: \(\exp\left[2\phi^{\star}Cx\right]=(\phi^{\star}+\phi)/(\phi^{\star}-\phi)\). Relation à partir de laquelle on en déduit simplement \(\phi\):

\[\phi(x)=\phi^{\star}\tanh\left(\phi^{\star}\sqrt{\frac{2H}{\zeta}}x\right)\]

With simple form \(2x/W\) inside the hyperbolic tangent:

\[\phi(x)=\phi^{\star}\tanh\left(\frac{2x}{W}\right)\]

with

\[W=\frac{1}{\phi^{\star}}\sqrt{\frac{2\zeta}{H}}\]

Section author: Alain Cartalade