2.1. Governing equations and boundary conditions

Consider surface-tension-driven flow of an incompressible liquid in a Plateau border with constant density $\rho$ , neglecting the effects of gravity. As mentioned in the Introduction, measurements obtained by Bouret et al. (Reference Bouret, Cohen, Fraysse, Argentina and Raufaste2016), found that the Reynolds number is large, approximately 100, hence we will assume here that the fluid is inviscid. The time scale of interest is much smaller than time scale of drainage of fluid out of the Plateau border and into the thin films (Koehler et al. Reference Koehler, Hilgenfeldt and Stone2004a , Reference Koehler, Hilgenfeldt and Stoneb ); therefore, we will neglect the flow out of the Plateau border and into the thin films.

We assume the flow is irrotational so that the fluid velocity is given by $\boldsymbol{u}=\boldsymbol{\nabla } \phi$ , where $\phi (x,y,z^*,t)$ is the fluid potential, $(x,y,z^*)$ denote Cartesian coordinates and $t$ is time. The $z^*$ -axis is oriented along the direction of the flow, while the $x$ - and $y$ -axes lie in the transverse cross-section of the Plateau border. The $y$ -axis is chosen to be normal to one of the Plateau-border surfaces at the mid-point and directed outward.

The representation of the velocity field as a gradient of the fluid potential substituted into the incompressibility condition ( $\boldsymbol{\nabla }\boldsymbol{\cdot }\boldsymbol{u}=0$ ), yields Laplace’s equation

(2.1) \begin{align} \boldsymbol{\nabla} ^2\phi \equiv \frac {\partial ^2\phi }{\partial x^2}+\frac {\partial ^2\phi }{\partial y^2}+\frac {\partial ^2\phi }{\partial z^{*2}}=0 \ \ \ \mbox{for } x,y,z^*\in \varOmega , \end{align}

where $\varOmega$ represents the region of the Plateau border that will be defined further on. Integrating the Euler equations of motion and employing a standard vector calculus identity leads to the Bernoulli equation

(2.2) \begin{align} \frac {\partial \phi }{\partial t}+\frac {1}{2}|\boldsymbol{\nabla } \phi |^2+\frac {p}{\rho }=B(t) \ \ \ \mbox{for } x,y,z^*\in \varOmega , \end{align}

where $p=p(x,y,z^*,t)$ is the fluid pressure and $B(t)$ is the Bernoulli constant, which in general depends on time. Since the flow is surface-tension-driven, we also have the Young–Laplace equation (Batchelor Reference Batchelor1967)

(2.3) \begin{align} p-p_0=\gamma (\boldsymbol{\nabla }\boldsymbol{\cdot }\hat {\boldsymbol{n}}), \ \ \ \mbox{for } x,y,z^*\in \partial \varOmega , \end{align}

where $p_0$ is the assumed constant air pressure outside the Plateau border which can be taken to be zero without loss of generality, $\partial \varOmega$ is the unknown boundary of the Plateau border to be determined, $\gamma$ is the constant surface tension coefficient and $\hat {\boldsymbol{n}}$ is the outward-pointing unit normal to the free surface. Since both the Bernoulli equation and the Young–Laplace equation are evaluated at the boundary of $\varOmega$ (i.e. the free surface), we can eliminate pressure from the Bernoulli equation, to yield

(2.4) \begin{align} \frac {\partial \phi }{\partial t}+\frac {1}{2}|\boldsymbol{\nabla } \phi |^2+\frac {\gamma }{\rho }(\boldsymbol{\nabla }\boldsymbol{\cdot }\hat {\boldsymbol{n}})=B(t) \ \ \ \mbox{for } x,y,z^*\in \partial \varOmega . \end{align}

A further boundary condition is required, the kinematic condition, which enforces that the normal velocity of the free surface is equal to the normal component of the fluid velocity at the free surface

(2.5) \begin{align} \boldsymbol{\nabla }\phi \boldsymbol{\cdot }\hat {\boldsymbol{n}} = \frac {\partial \boldsymbol{r}}{\partial t}\boldsymbol{\cdot }\hat {\boldsymbol{n}} \ \ \ \mbox{for } x,y,z^* \in \partial \varOmega , \end{align}

where $\boldsymbol{r}$ is position vector of a fluid particle on the free surface.

Figure 2.

Cross-section of a Plateau border centred at the origin $O$ .

2.2. Formulation of the equations of motion in the alternative coordinate system

A cross-section of a single Plateau border can be split into three equally sized regions, namely I, II and III, due to symmetry about the origin $O$ , as illustrated in figure 2. The region I is bounded by a thick circular arc with radius $R(z^*,t)$ , centred at $\nu R(z^*,t)$ . Region I is separated from regions II and III by dashed straight lines on the left- and right-hand sides of the $y$ -axis, extending from the origin to points where the cusps occur, as depicted in figure 2. The flow in regions II and III can be obtained by rotating the solution in region I by $2\pi /3$ and $4\pi /3$ counter-clockwise, respectively, about the origin $O$ .

Figure 3.

Coordinate system $(r,\theta )$ in the cross-section of the Plateau border at a fixed $z$ . Note that the figure is not to scale, being dilated on the $y$ -axis. Two examples of $(r,\theta )$ are shown, corresponding to the cases $\theta \lt \pi /2$ and $\theta \gt \pi /2$ . The positive $z$ -direction points into the page.

To describe the geometry of a Plateau border, we extend the work of Argentina et al. (Reference Argentina, Cohen, Bouret, Fraysse and Raufaste2015), which provides a parametrisation of the free surface of a Plateau border. We introduce a new coordinate system $(r,\theta ,z)$ that describes the position of a fluid particle in a Plateau border, where independent variables $(r,\theta )$ are illustrated in figure 3. The $z$ -axis points into the page in figure 3. The $(r,\theta )$ variables describe the location of the fluid particle on a cross-sectional slice of a Plateau border, i.e. at a fixed $z$ . From figure 3, we observe that

(2.6) \begin{align} x=r\cos \theta , \ \ \ y=\nu R(z,t)-r\sin \theta , \ \ \ z^*=z, \end{align}

where $\nu =2/\sqrt {3}$ (see Argentina et al. Reference Argentina, Cohen, Bouret, Fraysse and Raufaste2015 for details). We note that the origin of this coordinate system depends on $z$ and $t$ . The variable $r$ measures the distance from the point $\nu R(z,t)$ to a point in the Plateau border, and $\theta$ measures the angle between the horizontal and the line subtended by the joining of the point $\nu R(z,t)$ to a point in the Plateau border.

We observe that the free surface is independent of $\theta$ so that

(2.7) \begin{align} r=R(z,t), \end{align}

defines the location of the free surface. We note that when $r=R(z,t)$ , (2.6) coincides with the parametrisation of the free surface of a Plateau border as given in Argentina et al. (Reference Argentina, Cohen, Bouret, Fraysse and Raufaste2015). From figure 3, it follows that the Cartesian variables take on the following ranges inside the Plateau border region we are considering (i.e. region I):

(2.8) \begin{align} -\frac {R(z^*\!,t)}{2}\leqslant x \leqslant \frac {R(z^*\!,t)}{2}, \ \ \ \frac {\sqrt {3}}{3}|x| \leqslant y \leqslant \nu R(z^*\!,t)-\sqrt {R^2(z^*\!,t)-x^2}, \ \ -\infty \lt z^*\!\lt \infty . \end{align}

The inequalities (2.8) define region I of a Plateau border, and therefore, $\varOmega$ . From (2.6) and (2.8), it follows that $(r,\theta ,z)$ take on the following values inside region I of the Plateau border:

(2.9) \begin{align} R(z,t) \leqslant r\leqslant \nu R(z,t), \ \ \ \ \frac {\pi }{3}\leqslant \theta \leqslant \frac {2\pi }{3}, \ \ \ \ -\infty \lt z\lt \infty . \end{align}

We define the $(r,\theta ,z)$ in the coordinate system for the Plateau border as

(2.10) \begin{align} r&=\sqrt {x^2+(y-\nu R(z^*,t))^2}, \end{align}

(2.11) \begin{align} \theta &= \left \{\begin{array}{ccc} \arctan\! \left(\dfrac {\nu R(z^*,t)-y}{x} \right) & : & x\gt 0 ,\\[6pt] \dfrac {\pi }{2} & : & x=0, \\[6pt] \pi +\arctan\! \left( \dfrac {\nu R(z^*,t)-y}{x} \right) & : & x\lt 0, \end{array} \right . \end{align}

(2.12) \begin{align} z &=z^*, \end{align}

where $r\gt 0$ since it is a measure of distance. We note that $\nu R(z^*,t)-y\gt 0$ in region I of the Plateau border, and thus from (2.11) it can be seen that $\theta$ is continuous at $x=0$ . Moreover, it can be shown that $\theta$ is differentiable at $x=0$ by considering limits.

The position, $\boldsymbol{r}$ , of a fluid element in the Plateau border is defined by the coordinates $(r,\theta ,z)\equiv (x^1,x^2,x^3)$ as

(2.13) \begin{align} \boldsymbol{r}(r,\theta ,z)=r\cos \theta \hat {\boldsymbol{i}}+(R(z,t)\nu -r\sin \theta )\hat {\!\boldsymbol{j}}+z\hat {\boldsymbol{k}}, \end{align}

where $\hat {\boldsymbol{i}},\hat {\!\boldsymbol{j}},\hat {\boldsymbol{k}}$ are the Cartesian basis vectors pointing in the positive $x,y,z^*$ directions, respectively. The basis vectors are (see Simmonds Reference Simmonds2012) $\boldsymbol{e}_i= {\partial \boldsymbol{r}}/{\partial x^i}$ , i.e.

(2.14) \begin{align} \boldsymbol{e}_1 &\equiv \boldsymbol{e}_r =\frac {\partial \boldsymbol{r}}{\partial r}=\cos \theta \hat {\boldsymbol{i}}-\sin \theta \hat {\!\boldsymbol{j}},\\[-6pt] \nonumber \end{align}

(2.15) \begin{align} \boldsymbol{e}_2 & \equiv \boldsymbol{e}_{\theta }=\frac {\partial \boldsymbol{r}}{\partial \theta }=-r\sin \theta \hat {\boldsymbol{i}}-r\cos \theta \hat {\!\boldsymbol{j}},\\[-6pt] \nonumber \end{align}

(2.16) \begin{align} \boldsymbol{e}_3 & \equiv \boldsymbol{e}_{z}=\frac {\partial \boldsymbol{r}}{\partial z}=\nu \frac {\partial \!R}{\partial z}\hat {\!\boldsymbol{j}}+\hat {\boldsymbol{k}}, \\[9pt] \nonumber \end{align}

which are neither all orthogonal nor all unit vectors. We note that there is a similarity between the alternative coordinate system $(r,\theta ,z)$ and a cylindrical coordinate system, as seen from (2.14) and (2.15), except $\boldsymbol{e}_z\neq \hat {\boldsymbol{k}}$ . Is is also worth noting that they can be normalised as

(2.17) \begin{align} \boldsymbol{\hat {e}}_r = \boldsymbol{e}_r , \quad \boldsymbol{\hat {e}}_\theta = \frac {\boldsymbol{e}_{\theta }}{r}, \quad \boldsymbol{\hat {e}}_z =\frac {\boldsymbol{e}_{z}}{\sqrt {\left (\nu \dfrac {\partial\! R}{\partial z}\right )^{\!2}+1}}. \end{align}

The corresponding inverse relations to (2.14)–(2.16) are

(2.18) \begin{align} \hat {\boldsymbol{i}} & = \cos \theta \boldsymbol{e}_r-\frac {\sin \theta }{r}\boldsymbol{e}_{\theta },\\[-12pt]\nonumber \end{align}

(2.19) \begin{align} \hat {\!\boldsymbol{j}} & = -\sin \theta \boldsymbol{e}_r-\frac {\cos \theta }{r}\boldsymbol{e}_{\theta },\\[-12pt]\nonumber \end{align}

(2.20) \begin{align} \hat {\boldsymbol{k}} & =\nu \sin \theta \frac {\partial\! R}{\partial z}\boldsymbol{e}_r+\frac {\nu \cos \theta }{r}\frac {\partial\! R}{\partial z}\boldsymbol{e}_{\theta }+\boldsymbol{e}_z. \\[9pt] \nonumber \end{align}

Using the multi-variable chain rule and (A1) from Appendix A, Laplace’s equation (2.1) becomes

(2.21) \begin{align} \boldsymbol{\nabla} ^2\phi = & \frac {\partial ^2\phi }{\partial r^2}+\frac {1}{r}\frac {\partial \phi }{\partial r}+\frac {1}{r^2}\frac {\partial ^2 \phi }{\partial \theta ^2}+\frac {\partial ^2\phi }{\partial z^2} \notag \\ & +\frac {\partial ^2\phi }{\partial r^2}\left ( \frac {\partial\! R}{\partial z}\right )^{\!2}\nu ^2\sin ^2\theta +\frac {\partial ^2\phi }{\partial r\partial \theta }\!\left (\frac {\partial\! R}{\partial z}\right )^{\!2}\frac {2\nu ^2\cos \theta \sin \theta }{r}+\frac {\partial \phi }{\partial r}\!\left ( \frac {\partial\! R}{\partial z}\right )^{\!2}\!\frac {\nu ^2\cos ^2\theta }{r} \notag \\ & -2\frac {\partial \phi }{\partial \theta }\left ( \frac {\partial\! R}{\partial z}\right )^{\!2}\frac {\nu ^2\sin \theta \cos \theta }{r^2}+\frac {\partial ^2\phi }{\partial \theta ^2}\left ( \frac {\partial\! R}{\partial z} \right )^{\!2}\frac {\nu ^2\cos ^2\theta }{r^2}+2\frac {\partial ^2\phi }{\partial z\partial r}\frac {\partial\! R}{\partial z}\nu \sin \theta \notag \\ & +\frac {\partial \phi }{\partial r}\frac {\partial ^2R}{\partial z^2}\nu \sin \theta +2\frac {\partial ^2\phi }{\partial z\partial \theta }\frac {\partial\! R}{\partial z}\frac {\nu \cos \theta }{r}+\frac {\partial \phi }{\partial \theta }\frac {\partial ^2R}{\partial z^2}\frac {\nu \cos \theta }{r}=0, \ \ \ \notag \\ & \mbox{for } \ \ \ R(z,t)\leqslant r \leqslant \nu R(z,t), \ \ \frac {\pi }{3}\leqslant \theta \leqslant \frac {2\pi }{3}, \ \ \mbox{and} \ -\infty \lt z\lt \infty . \end{align}

The $|\boldsymbol{\nabla }\phi |^2$ term appearing in the Bernoulli condition (2.4) is more easily calculated in the Cartesian basis vectors. Thus, from (A1), we have

(2.22) \begin{align} \boldsymbol{\nabla }\phi &\equiv \frac {\partial \phi }{\partial x}\hat {\boldsymbol{i}}+\frac {\partial \phi }{\partial y}\skew 2\hat {\!\boldsymbol{j}}+\frac {\partial \phi }{\partial z^*}\hat {\boldsymbol{k}} \notag \\ &= \left ( \cos \theta \frac {\partial \phi }{\partial r}-\frac {\sin \theta }{r}\frac {\partial \phi }{\partial \theta } \right )\!\hat {\boldsymbol{i}}+\left (-\sin \theta \frac {\partial \phi }{\partial r}-\frac {\cos \theta }{r}\frac {\partial \phi }{\partial \theta } \right )\!\skew 2\hat {\!\boldsymbol{j}} \notag \\ & \quad +\left (\nu \sin \theta \frac {\partial\! R}{\partial z}\frac {\partial \phi }{\partial r}+\frac {\nu \cos \theta }{r}\frac {\partial r}{\partial z}\frac {\partial \phi }{\partial \theta }+\frac {\partial \phi }{\partial z} \right )\!\hat {\boldsymbol{k}}. \end{align}

Using (2.22), $|\boldsymbol{\nabla }\phi |^2$ is given by

(2.23) \begin{align} |\boldsymbol{\nabla }\phi |^2 &= \left ( \frac {\partial \phi }{\partial r}\right )^{\!2}+\frac {1}{r^2}\left ( \frac {\partial \phi }{\partial \theta }\right )^{\!2}+\left ( \frac {\partial\! R}{\partial z}\right )^{\!2}\left (\frac {\partial \phi }{\partial r}\right )^{\!2}\nu ^2\sin ^2\theta +\left (\frac {\partial\! R}{\partial z}\right )^{\!2} \left (\frac {\partial \phi }{\partial \theta }\right )^{\!2}\frac {\nu ^2\cos ^2\theta }{r^2} \notag \\ &\quad +\left (\frac {\partial \phi }{\partial z}\right )^{\!2} +\frac {2\nu ^2\cos \theta \sin \theta }{r}\left ( \frac {\partial\! R}{\partial z}\right )^{\!2}\frac {\partial \phi }{\partial r}\frac {\partial \phi }{\partial \theta }+2\nu \sin \theta \frac {\partial\! R}{\partial z}\frac {\partial \phi }{\partial r}\frac {\partial \phi }{\partial z} \notag \\ &\quad +\frac {2\nu \cos \theta }{r}\frac {\partial\! R}{\partial z}\frac {\partial \phi }{\partial \theta }\frac {\partial \phi }{\partial z}. \end{align}

2.3. Curvature of the free surface

The material surface that defines the free surface of a Plateau border is

(2.24) \begin{align} F(x,y,z^*,t)\equiv R(z^*,t)^2-x^2-(y-\nu R(z^*,t))^2=0. \end{align}

Hence the outward-pointing unit normal vector is

(2.25) \begin{align} \hat {\boldsymbol{n}}= \frac {\boldsymbol{\nabla } F}{|\boldsymbol{\nabla } F|}= \dfrac {{-}\cos \theta \hat {\boldsymbol{i}}+\sin \theta \skew 2\hat {\!\boldsymbol{j}}-\dfrac {\partial\! R}{\partial z}(\nu \sin \theta -1)\hat {\boldsymbol{k}} }{\sqrt {1+\left (\dfrac {\partial\! R}{\partial z} \right )^{\!2}(\nu \sin \theta -1)^2}},\end{align}

and its divergence is, after some algebra,

(2.26) \begin{align} \boldsymbol{\nabla }\boldsymbol{\cdot }\hat {\boldsymbol{n}} =\dfrac {(1-\nu \sin \theta )\dfrac {\partial ^2R}{\partial z^2}R-1+(2\nu \sin \theta -\nu ^2-1)\left (\dfrac {\partial\! R}{\partial z}\right )^{\!2}}{R\Bigg [1+(\nu \sin \theta -1)^2\left ( \dfrac {\partial\! R}{\partial z}\right )^{\!2}\Bigg ]^{3/2}}. \end{align}

The Bernoulli condition, after substituting (2.26) and (2.23) into (2.4), is

(2.27) \begin{align} &\frac {\partial \phi }{\partial t}+\frac {1}{2}\Bigg (\! \left ( \frac {\partial \phi }{\partial r}\right )^{\!2}+\frac {1}{r^2}\left ( \frac {\partial \phi }{\partial \theta }\right )^{\!2}+\left ( \frac {\partial\! R}{\partial z}\frac {\partial \phi }{\partial r}\right )^{\!2}\nu ^2\sin ^2\theta +\left (\frac {\partial\! R}{\partial z}\right )^{\!2} \left (\frac {\partial \phi }{\partial \theta }\right )^{\!2}\frac {\nu ^2\cos ^2\theta }{r^2} \notag \\ & +\left (\frac {\partial \phi }{\partial z}\right )^{\!2} +\frac {2\nu ^2\cos \theta \sin \theta }{r}\left ( \frac {\partial\! R}{\partial z}\right )^{\!2}\frac {\partial \phi }{\partial r}\frac {\partial \phi }{\partial \theta }+2\nu \sin \theta \frac {\partial\! R}{\partial z}\frac {\partial \phi }{\partial r}\frac {\partial \phi }{\partial z}+\frac {2\nu \cos \theta }{r}\frac {\partial\! R}{\partial z}\frac {\partial \phi }{\partial \theta }\frac {\partial \phi }{\partial z}\! \Bigg ) \notag \\ & +\frac {\gamma }{\rho }\left (\dfrac {(1-\nu \sin \theta )\dfrac {\partial ^2R}{\partial z^2}R-1+(2\nu \sin \theta -\nu ^2-1)\left (\dfrac {\partial\! R}{\partial z}\right )^{\!2}}{R\Bigg [1+(\nu \sin \theta -1)^2\left ( \dfrac {\partial\! R}{\partial z}\right )^{\!2}\Bigg ]^{3/2}} \right )=B(t), \end{align}

on $r=R(z,t)$ .

2.4. Kinematic boundary condition

We note from (2.13) that $ {\partial \boldsymbol{r}}/{\partial t}$ , where $\boldsymbol{r}$ is the position of the fluid particle on the free surface, is given by

(2.28) \begin{align} \frac {\partial \boldsymbol{r}}{\partial t}=\frac {\partial\! R}{\partial t}\cos \theta \hat {\boldsymbol{i}}+\frac {\partial\! R}{\partial t}(\nu -\sin \theta )\skew 2\hat {\!\boldsymbol{j}}. \end{align}

From (2.22), (2.25) and (2.28) the kinematic boundary condition (2.5) becomes, after some algebra,

(2.29) \begin{align} & \frac {\partial\! R}{\partial t}(1-\nu \sin \theta )-\frac {\partial \phi }{\partial r} \notag \\ & \ \ \ \ +\frac {\partial\! R}{\partial z}(1-\nu \sin \theta )\left (\nu \sin \theta \frac {\partial\! R}{\partial z}\frac {\partial \phi }{\partial r}+\frac {\nu \cos \theta }{r}\frac {\partial\! R}{\partial z}\frac {\partial \phi }{\partial \theta }+\frac {\partial \phi }{\partial z}\right )=0 \ \ \ \mbox{on } r=R(z,t). \end{align}

2.5. Symmetry boundary condition

To establish another boundary condition, consider a cross-sectional slice of a Plateau border. Imagine the fluid velocity vector at the origin $O$ , pointing radially outwards (in the $\boldsymbol{e}_r$ direction). If the velocity vector has non-zero components in the $\hat {\boldsymbol{i}}$ and $\skew 2\hat {\!\boldsymbol{j}}$ directions, rotating the cross-section of the Plateau border by integer multiples of $2\pi /3$ would lead to an inconsistency. By symmetry, the fluid velocity of the Plateau border must be identical to the fluid velocity of the rotated Plateau border. The only way to satisfy this for any integer multiple of $2\pi /3$ is if the components of fluid velocity vector in $\hat {\boldsymbol{i}}$ and $\skew 2\hat {\!\boldsymbol{j}}$ directions at the origin are zero. Mathematically, this means

(2.30) \begin{align} \boldsymbol{u}\boldsymbol{\cdot } \hat {\boldsymbol{i}}=0, \ \ \ \mbox{and } \boldsymbol{u}\boldsymbol{\cdot }\skew 2\hat {\!\boldsymbol{j}}=0, \ \ \ \ \mbox{on } r=\nu R(z,t), \end{align}

where the origin $O$ is defined by $r=\nu R(z,t)$ in region I of the Plateau border. The fluid velocity vector in the alternative coordinate system, expressed in terms of the $\{\boldsymbol{e}_r,\boldsymbol{e}_{\theta },\boldsymbol{e}_z\}$ vectors, is

(2.31) \begin{align} \boldsymbol{u}=u^r\boldsymbol{e}_r+u^{\theta }\boldsymbol{e}_{\theta }+u^z\boldsymbol{e}_z, \end{align}

where $u^r,u^{\theta },u^z$ are the velocity components in the $\boldsymbol{e}_r,\boldsymbol{e}_{\theta }$ and $\boldsymbol{e}_z$ directions, respectively.

After some work (see Appendix A) we find that the symmetry condition at $r=\nu R(z,t)$ is given by

(2.32) \begin{align} u^r\cos \theta -u^{\theta }r\sin \theta =0, \ \ \ \ \mbox{and } \ \ \ {-}u^r\sin \theta -u^{\theta }r\cos \theta +\nu \frac {\partial\! R}{\partial z}u^z=0. \end{align}

To determine the components of the fluid velocity in terms of the fluid potential, we substitute (2.18)–(2.20) into (2.22), to obtain

(2.33) \begin{align} u^r & =\left (1+\left (\frac {\partial\! R}{\partial z}\right )^{\!2}\nu ^2\sin ^2\theta \right )\frac {\partial \phi }{\partial r}+\frac {\nu ^2\cos \theta \sin \theta }{r}\left (\frac {\partial\! R}{\partial z} \right )^{\!2}\frac {\partial \phi }{\partial \theta }+\nu \sin \theta \frac {\partial\! R}{\partial z}\frac {\partial \phi }{\partial z}, \end{align}

(2.34) \begin{align} u^{\theta } & =\frac {\nu ^2\cos \theta \sin \theta }{r}\left (\frac {\partial\! R}{\partial z} \right )^{\!2}\frac {\partial \phi }{\partial r}+\frac {1}{r^2}\left (1+\left ( \frac {\partial\! R}{\partial z} \right )^{\!2}\nu ^2\cos ^2\theta \right )^{\!2}\frac {\partial \phi }{\partial \theta } +\frac {\nu \cos \theta }{r}\frac {\partial\! R}{\partial z}\frac {\partial \phi }{\partial z}, \end{align}

(2.35) \begin{align} u^z & =\nu \sin \theta \frac {\partial\! R}{\partial z}\frac {\partial \phi }{\partial r}+\frac {\nu \cos \theta }{r}\frac {\partial\! R}{\partial z}\frac {\partial \phi }{\partial \theta }+\frac {\partial \phi }{\partial z}. \end{align}

The symmetry condition expressed in terms of the fluid potential, after substituting (2.33)–(2.35) into the symmetry condition (2.32), is

(2.36) \begin{align} \frac {\partial \phi }{\partial r}\cos \theta -\frac {1}{r}\frac {\partial \phi }{\partial \theta }\sin \theta =0, \quad \mathrm{and} \quad \frac {\partial \phi }{\partial r}\sin \theta +\frac {1}{r}\frac {\partial \phi }{\partial \theta }\cos \theta =0, \ \ \ \ \mbox{on } r=\nu R(z,t). \end{align}

This completes the derivation of the boundary conditions.

3.1. Non-dimensionalisation and long-wave speed

We now derive the non-dimensional dispersion relation by non-dimensionalising all lengths in units of the unperturbed $r$ -coordinate, $a$ . We perform the non-dimensionalisation as follows:

(3.18) \begin{align} r=ar', \ \ \ \ z=az', \ \ \ \ t=\sqrt {\frac {a^3\rho }{\gamma }}t', \end{align}

where primed quantities denote non-dimensional ones. Substituting (3.18) into (3.13), we obtain the non-dimensional dispersion relation

(3.19) \begin{align} s^2=-k\left ( \frac {1}{1-\nu }-k^2 \right )\frac { {I}_1(k){K}_1(k\nu )-{I}_1(k\nu ){K}_1(k)}{{I}_0(k){K}_1(k\nu )+{I}_1(k\nu ){K}_0(k)}. \end{align}

By writing $s=-\mathrm{i}kc$ , where $c$ is the phase speed of the wave, and substituting in (3.19) we obtain

(3.20) \begin{align} c^2=\frac {1}{k}\left ( \frac {1}{1-\nu }-k^2 \right )\frac { {I}_1(k){K}_1(k\nu )-{I}_1(k\nu ){K}_1(k)}{{I}_0(k){K}_1(k\nu )+{I}_1(k\nu ){K}_0(k)}. \end{align}

The leading-order terms of the asymptotic expansions for small arguments of the modified Bessel functions are (Olver Reference Olver2010)

(3.21) \begin{align} {I}_{0}(\chi )\sim 1, \ \ \ {I}_1(\chi )\sim \frac {1}{2}\chi , \ \ \ \ {K}_0(\chi )\sim {-}\!\log \chi , \ \ \ {K}_{1}(\chi )\sim \frac {1}{\chi }, \ \ \ \ \mbox{as } \chi \to 0. \end{align}

Substituting these asymptotic expansions (3.21) into the dispersion relation (3.20), we find the long-wave speed and its dimensional counterpart (denoted with $\star$ ) to be

(3.22) \begin{align} c_0=\sqrt {\frac {\nu +1}{2}}\approx 1.0308, \ \ \ \ \ \ \ c_0^{\star }=\sqrt {\frac {\gamma }{\rho a}\frac {\nu +1}{2}}. \end{align}

A plot of the dispersion relation (3.20), showing $c$ against the wavenumber $k$ , is depicted in figure 4. From figure 4, we observe the minimum is located at $k=0$ , which is confirmed by numerical investigation.

Figure 4.

Plot of dispersion relation (3.19): dimensionless phase speed $c$ against the wavenumber $k$ .

Different types of solitary waves can bifurcate from either the long-wave limit or the minimum phase speed (see Dauxois & Peyrard Reference Dauxois and Peyrard2006). However, we have established a minimum phase speed for finite $k$ does not exist in our model. This suggests that, when we proceed to weakly nonlinear theory, we should use a perturbation expansion for $k\ll 1$ , i.e. in the long-wave limit.

The absence of a minimum phase speed for finite $k$ implies that pure solitary waves, such as whose observed by Bouret et al. (Reference Bouret, Cohen, Fraysse, Argentina and Raufaste2016), must travel with phase speed slightly less than the long-wave speed. The reasoning behind this is that if the phase speed were slightly larger than the long-wave limit, linear capillary waves would resonate with the fully nonlinear wave, resulting in a generalised solitary wave (Hunter & Vanden-Broeck Reference Hunter and Vanden-Broeck1983; Vanden-Broeck Reference Vanden-Broeck2010) – a solitary wave with non-decaying oscillatory tails. This is indeed confirmed by our weakly nonlinear theory (as discussed in the next section).

Finally, it is important to note that our theoretical predictions differ from those reported in previous studies (Argentina et al. Reference Argentina, Cohen, Bouret, Fraysse and Raufaste2015; Bouret et al. Reference Bouret, Cohen, Fraysse, Argentina and Raufaste2016; Le Doudic et al. Reference Le Doudic, Perrard and Pham2021). Our analysis predicts the long-wave speed to be $\sqrt {\gamma (\nu +1)/(2\rho a)}$ , whereas previous authors reported the long-wave speed as $\sqrt {\gamma /(2\rho a)}$ . This difference may arise from the fact that in those studies they model a slightly different problem of a Plateau border formed by holding films attached to a triangular prism frame, while in our case we assumed that the Plateau border is formed at the intersection of three bubbles or liquid foams. Hence, we neglect the physical effect of any holding film attached to a prism frame.

4.1. Dimensionless equations of motion

We assume that the characteristic horizontal length scale in the $z$ -direction, denoted by $\ell$ , is long compared with the amplitude of the undisturbed free surface of the Plateau border $r=a$ (where $a$ is a constant). This allows us to define a slenderness ratio

(4.1) \begin{align} \epsilon =\frac {a}{\ell }\ll 1. \end{align}

Following Wang, Papageorgiou & Vanden-Broeck (Reference Wang, Papageorgiou and Vanden-Broeck2015), we non-dimensionalise the problem using the following scalings:

(4.2) \begin{align} \phi =\sqrt {\frac {\gamma \ell ^2}{\rho a}}\phi ', \ \ \ \ r=ar', \ \ \ \ z=\ell z', \ \ \ \ R =\textit{aR}', \ \ \ \ t=\sqrt {\frac {\rho a \ell ^2}{\gamma }}t', \end{align}

where primes denote non-dimensional quantities. As in Argentina et al. (Reference Argentina, Cohen, Bouret, Fraysse and Raufaste2015) and § 3, we average the azimuthal coordinate ( $\theta$ ) over a given cross-section. Henceforth, in this section, we assume the conditions given by (3.1).

Substituting (4.2) into Laplace’s equation (2.21), the kinematic condition (2.29), the Bernoulli condition (2.27) and the symmetry condition (2.36), we obtain the following equations (after dropping the primes for brevity): Laplace’s equation

(4.3) \begin{align} &\frac {\partial ^2\phi }{\partial r^2}+\frac {1}{r}\frac {\partial \phi }{\partial r}+\epsilon ^2\frac {\partial ^2\phi }{\partial z^2}+2\epsilon ^2\nu \frac {\partial\! R}{\partial z}\frac {\partial ^2\phi }{\partial r\partial z}+\epsilon ^2\nu ^2\frac {\partial ^2\phi }{\partial r^2}\left (\frac {\partial\! R}{\partial z}\right )^{\!2}\nu ^2+\epsilon ^2\nu \frac {\partial ^2R}{\partial z^2}\frac {\partial \phi }{\partial r}=0, \end{align}

for $R(z,t)\leqslant r\leqslant \nu R(z,t)$ , the Bernoulli equation

(4.4) \begin{align} & \epsilon ^2\frac {\partial \phi }{\partial t}+\frac {1}{2}\Bigg ( \left (\frac {\partial \phi }{\partial r}\right )^{\!2}+\epsilon ^2\nu ^2\left (\frac {\partial\! R}{\partial z}\frac {\partial \phi }{\partial r}\right )^{\!2} +\epsilon ^2\left (\frac {\partial \phi }{\partial z}\right )^{\!2}+2\epsilon ^2\nu \frac {\partial\! R}{\partial z}\frac {\partial \phi }{\partial r}\frac {\partial \phi }{\partial z}\Bigg ) \notag \\ &\quad +\dfrac {(1-\nu )\epsilon ^4\dfrac {\partial ^2R}{\partial z^2}R-\epsilon ^2-(\nu -1)^2\epsilon ^4\left (\dfrac {\partial\! R}{\partial z}\right )^{\!2}}{R\Bigg [1+(\nu -1)^2\epsilon ^2\left ( \dfrac {\partial\! R}{\partial z}\right )^{\!2}\Bigg ]^{3/2}}+\epsilon ^2=0 \ \ \ \ \mbox{on } r=R(z,t). \end{align}

The kinematic boundary condition is

(4.5) \begin{align} \epsilon ^2\frac {\partial\! R}{\partial t}(1-\nu )-\frac {\partial \phi }{\partial r}+\epsilon ^2\frac {\partial\! R}{\partial z}(1-\nu )\left (\nu \frac {\partial\! R}{\partial z}\frac {\partial \phi }{\partial r}+\frac {\partial \phi }{\partial z} \right )=0 \ \ \ \ \mbox{on } r=R(z,t), \end{align}

and the symmetry condition is

(4.6) \begin{align} \frac {\partial \phi }{\partial r}=0 \ \ \ \ \mbox{on } r=\nu R(z,t). \end{align}

We now seek a solution to (4.3)–(4.6) for $0\lt \epsilon \ll 1$ . Therefore, we look for a a perturbation solution for small $\epsilon$ with weakly nonlinear free surface deformation, so we write

(4.7) \begin{align} R(z,t)=1+\epsilon ^2\eta (z,t), \end{align}

and assume the expansion for the velocity potential is of the form

(4.8) \begin{align} \phi (r,z,t)=\epsilon ^2\phi _0(r,z,t)+\epsilon ^4\phi _1(r,z,t)+\epsilon ^6\phi _2(r,z,t)+O(\epsilon ^8). \end{align}

We apply Taylor expansions to the symmetry condition (4.6) about $r=\nu$ and then substitute the expansion for the velocity potential into the resulting equation. This yields the following equations at each order:

(4.9) \begin{align} & O (\epsilon ^2) : \ \ \ \ \ \frac {\partial \phi _0}{\partial r}\Big |_{r=\nu } =0, \end{align}

(4.10) \begin{align} & O (\epsilon ^4) : \ \ \ \ \ \frac {\partial \phi _1}{\partial r}\Big |_{r=\nu }+\nu \eta \frac {\partial ^2\phi _0}{\partial r^2} =0, \end{align}

(4.11) \begin{align} & O(\epsilon ^6) : \ \ \ \ \ \ \frac {\partial \phi _2}{\partial r}\Big |_{r=\nu }+\nu \eta \frac {\partial ^2\phi _1}{\partial r^2}\Big |_{r=\nu }+\frac {\nu ^2\eta ^2}{2}\frac {\partial ^3\phi _0}{\partial r^3}\Big |_{r=\nu } =0. \end{align}

4.2. Asymptotic solutions of Laplace’s equation

Substituting the perturbation expansion for the velocity potential (4.8) into Laplace’s equation (4.3), and collecting terms in a series of $\epsilon ^2$ , we find at each order

(4.12) \begin{align} & O(\epsilon ^2) : \ \ \ \ \ \frac {\partial ^2\phi _0}{\partial r^2}+\frac {1}{r}\frac {\partial \phi _0}{\partial r}=0, \end{align}

(4.13) \begin{align} & O (\epsilon^4) : \ \ \ \ \ \frac {\partial ^2\phi _1}{\partial r^2}+\frac {1}{r}\frac {\partial \phi _1}{\partial r}=-\frac {\partial ^2\phi _0}{\partial z^2}, \end{align}

(4.14) \begin{align} & O (\epsilon ^6) : \ \ \ \ \ \frac {\partial ^2\phi _2}{\partial r^2}+\frac {1}{r}\frac {\partial \phi _2}{\partial r}=-\frac {\partial ^2\phi _1}{\partial z^2}-2\nu \frac {\partial \eta }{\partial z}\frac {\partial ^2\phi _0}{\partial r\partial z}-\nu \frac {\partial ^2\eta }{\partial z^2}\frac {\partial \phi _0}{\partial r}. \\[9pt] \nonumber \end{align}

Noting that

(4.15) \begin{align} \frac {\partial ^2}{\partial r^2}+\frac {1}{r}\frac {\partial }{\partial r}\equiv \frac {1}{r}\frac {\partial }{\partial r}\left (r\frac {\partial }{\partial r}\right )\!, \end{align}

we can integrate to find the solution at each order

(4.16) \begin{align} \phi _n= \int \frac {1}{r}\left (\int r\phi _{n-1} {\rm d} r\right ) {\rm d} r+A_n(z,t)\ln r+B_n(z,t), \ \ \ \ \mbox{for } n=0,1,2, \end{align}

where $\phi _{-1}=0$ and $A_n(z,t),B_n(z,t)$ are unknown functions of integration which generally depend on $z$ and $t$ . The function $B_n(z,t)$ is an unknown that will govern the weakly nonlinear evolution equation for a Plateau border. Let us denote $B_0(z,t)\equiv B(z,t)$ . With $\phi _{-1}=0$ and the recurrence relations (4.16), after some integration and application of the boundary conditions (4.9)–(4.11), we find

(4.17) \begin{align} \phi _0(z,t) &=B(z,t), \end{align}

(4.18) \begin{align} \phi _1(r,z,t) &=\left (-\frac {r^2}{4}+\frac {\nu ^2}{2}\ln r \right )\frac {\partial ^2 B}{\partial z^2}+D(z,t), \end{align}

(4.19) \begin{align} \phi _2(r,z,t) &=\left (\frac {r^4}{64}-\frac {\nu ^2}{8}[r^2(\ln r-1)]+\ln r\left (\frac {\nu ^4\ln \nu }{4}-\frac {3\nu ^4}{16} \right ) \right )\frac {\partial ^4 B}{\partial z^4}\notag \\ &\quad +\nu ^2\eta (z,t)\frac {\partial ^2 B}{\partial z^2}\ln r+\frac {\partial ^2D}{\partial z^2}\left (-\frac {r^2}{4}+\frac {\nu ^2}{2}\ln r \right )+F(z,t), \end{align}

where $B(z,t),D(z,t),F(z,t)$ are arbitrary functions of integration. Following Wang et al. (Reference Wang, Papageorgiou and Vanden-Broeck2015), we introduce the canonical KdV scaling

(4.20) \begin{align} \xi =z-\mathcal{C}t, \ \ \ \ \ \tau =\epsilon ^2t, \end{align}

where $\mathcal{C}$ is the wave speed that is to be determined. The derivatives with respect to $z$ and $t$ in the equations of motion are then rewritten in terms of the derivative with respect to $\xi$ and $\tau$ using the transformations

(4.21) \begin{align} \frac {\partial }{\partial z}=\frac {\partial }{\partial \xi }, \ \ \ \ \ \ \ \frac {\partial }{\partial t}=-\mathcal{C}\frac {\partial }{\partial \xi }+\epsilon ^2\frac {\partial }{\partial \tau }. \end{align}

Substituting the change of variables (4.20) with the transformations (4.21) and the expansion for $R(z,t)$ , (4.7), into the Bernoulli (4.4), we find the Bernoulli equation becomes

(4.22) \begin{align} &-\!\mathcal{C}\epsilon ^2\frac {\partial \phi }{\partial \xi }-\epsilon ^4\frac {\partial \phi }{\partial \tau }+\frac {1}{2}\left [\left (\frac {\partial \phi }{\partial r}\right )^{\!2}+\epsilon ^6\nu ^2\left (\frac {\partial \eta }{\partial \xi }\frac {\partial \phi }{\partial r} \right )^{\!2}+\epsilon ^2\left ( \frac {\partial \phi }{\partial \xi }\right )^{\!2}+2\epsilon ^4\nu \frac {\partial \eta }{\partial \xi }\frac {\partial \phi }{\partial r}\frac {\partial \phi }{\partial \xi } \right ] \notag \\ &\quad +\dfrac {(1-\nu )\epsilon ^6\dfrac {\partial ^2\eta }{\partial \xi ^2}(1+\epsilon ^2\eta (\xi ,\tau ))-\epsilon ^2-(\nu -1)^2\epsilon ^8\left (\dfrac {\partial \eta }{\partial \xi }\right )^{\!2}}{(1+\epsilon ^2\eta (\xi ,\tau ))\sqrt {1+(\nu -1)^2\epsilon ^6\left (\dfrac {\partial \eta }{\partial \xi }\right )^{\!2}}}+\epsilon ^2=0 \notag \\ & \hspace {23em} \mbox{on } r=1+\epsilon ^2\eta (\xi ,\tau ). \end{align}

Similarly, the kinematic condition (4.5) becomes

(4.23) \begin{align} &{-}\epsilon ^4\mathcal{C}\frac {\partial \eta }{\partial \xi }(1-\nu )+\epsilon ^6\frac {\partial \eta }{\partial \tau }(1-\nu )-\frac {\partial \phi }{\partial r} \notag \\ & \quad +\epsilon ^2\frac {\partial \eta }{\partial \xi }(1-\nu )\left ( \nu \epsilon ^4\frac {\partial \eta }{\partial \xi }\frac {\partial \phi }{\partial r}+\epsilon ^2\frac {\partial \phi }{\partial \xi } \right )=0 \ \ \ \ \mbox{on } r=1+\epsilon ^2\eta (\xi ,\tau ). \end{align}

We also note the asymptotic solution to Laplace’s equation in the new variables $(\xi ,\tau )$ is given by

(4.24) \begin{align} \phi _0(\xi ,\tau ) &=B(\xi ,\tau ), \end{align}

(4.25) \begin{align} \phi _1(r,\xi ,\tau ) &=\left (-\frac {r^2}{4}+\frac {\nu ^2}{2}\ln r \right )\frac {\partial ^2 B}{\partial \xi ^2}+\theta _2(\xi ,\tau ), \end{align}

(4.26) \begin{align} \phi _2(r,\xi ,\tau ) &=\left (\frac {r^4}{64}-\frac {\nu ^2}{8}[r^2(\ln r-1)]+\ln r\left (\frac {\nu ^4\ln \nu }{4}-\frac {3\nu ^4}{16} \right ) \right )\frac {\partial ^4 B}{\partial \xi ^4} \notag \\ &\quad +\nu ^2\eta (z,t)\frac {\partial ^2 B}{\partial \xi ^2}\ln r+\frac {\partial ^2 D}{\partial \xi ^2}\left (-\frac {r^2}{4}+\frac {\nu ^2}{2}\ln r \right )+\theta _4(\xi ,\tau ), \end{align}

where we denote

(4.27) \begin{align} \theta _2(\xi ,\tau )\equiv D(z,t), \ \ \ \ \ \ \ \theta _4(\xi ,\tau )\equiv F(z,t). \end{align}

We expand in Taylor series about $r=1$ the Bernoulli condition (4.22), to move the terms in the equation evaluated at the unknown free surface $r=1+\epsilon ^2\eta (\xi ,\tau )$ to the known location of $r=1$ . Then substituting the perturbation expansion for the velocity potential (4.8) yields, at orders $O(\epsilon ^4)$ and $O(\epsilon ^6)$ , respectively,

(4.28) \begin{align} -\mathcal{C}\frac {\partial \phi _0}{\partial \xi }\Big |_{r=1}+\eta =0, \end{align}

and

(4.29) \begin{align} -\mathcal{C}\frac {\partial \phi _1}{\partial \xi }\Big |_{r=1}+\frac {\partial \phi _0}{\partial \tau }\Big |_{r=1}+\frac {1}{2}\left (\frac {\partial \phi _0}{\partial \xi }\Big |_{r=1} \right )^{\!2}-\eta ^2+(1-\nu )\frac {\partial ^2\eta }{\partial \xi ^2}=0. \end{align}

Similarly, for the kinematic condition we obtain at orders $O(\epsilon ^4)$ and $O(\epsilon ^6)$ , respectively,

(4.30) \begin{align} -\mathcal{C}\frac {\partial \eta }{\partial \xi }(1-\nu )-\frac {\partial \phi _1}{\partial r}\Big |_{r=1}=0, \end{align}

and

(4.31) \begin{align} \frac {\partial \eta }{\partial \tau }(1-\nu )-\eta \frac {\partial ^2\phi _1}{\partial r^2}\Big |_{r=1}-\frac {\partial \phi _2}{\partial r}\Big |_{r=1}+(1-\nu )\frac {\partial \eta }{\partial \xi }\frac {\partial \phi _0}{\partial \xi }\Big |_{r=1}=0. \end{align}

4.3. Leading-order solution $O(\epsilon ^4)$

Upon substituting the asymptotic solutions to the Laplace equations (4.24)–(4.26) into the Bernoulli condition (4.28) at $O(\epsilon ^4)$ , we obtain

(4.32) \begin{align} \mathcal{C}\frac {\partial \eta }{\partial \xi }=\frac {\nu +1}{2}\frac {\partial ^2 B}{\partial \xi ^2}. \end{align}

Similarly, for the kinematic condition at $O(\epsilon ^4)$ (4.30), we obtain

(4.33) \begin{align} -\mathcal{C}\frac {\partial\! B}{\partial \xi }+\eta =0. \end{align}

Eliminating $B(\xi ,\tau )$ between (4.32) and (4.33), we find

(4.34) \begin{align} \mathcal{C}^2=\frac {\nu +1}{2}. \end{align}

This expression for $\mathcal{C}$ represents the long-wave speed. As expected, the value of $\mathcal{C}$ agrees with long-wave speed (3.22) from linear theory in § 3.

4.4. First-order solution at $O(\epsilon ^6)$

Proceeding similarly to the derivation of the leading-order solution, we substitute the solutions to the Laplace equations (4.24)–(4.26) into the Bernoulli (4.29) and kinematic conditions (4.31) at $O(\epsilon ^6)$ , to obtain

(4.35) \begin{align} &\frac {\partial\! B}{\partial \tau }-\eta ^2+(1-\nu )\frac {\partial ^2\eta }{\partial \xi ^2}+\frac {1}{2}\left (\frac {\partial\! B}{\partial \xi } \right )^{\!2}+\frac {\mathcal{C}}{4}\frac {\partial ^3 B}{\partial \xi ^3}-\mathcal{C}\frac {\partial \theta _2}{\partial \xi }=0, \end{align}

(4.36) \begin{align} &\frac {\partial \eta }{\partial \tau }(1-\nu )+\eta \left (\frac {1-\nu ^2}{2}\frac {\partial ^2 B}{\partial \xi ^2}\right )-\frac {\partial ^4 B}{\partial \xi ^4}\left (\frac {1}{16}+\frac {\nu ^2}{8}+\frac {\nu ^4\ln \nu }{4}-\frac {3\nu ^4}{16} \right ) \notag \\ & \quad +\frac {1-\nu ^2}{2}\frac {\partial ^2\theta _2}{\partial \xi ^2}+(1-\nu )\frac {\partial \eta }{\partial \xi }\frac {\partial\! B}{\partial \xi }=0, \end{align}

respectively. Differentiating (4.35) with respect to $\xi$ , we find

(4.37) \begin{align} \frac {\partial ^2 B}{\partial \xi \partial \tau }-2\eta \frac {\partial \eta }{\partial \xi }+(1-\nu )\frac {\partial ^3\eta }{\partial \xi ^3}+\frac {\partial ^2 B}{\partial \xi ^2}\frac {\partial\! B}{\partial \xi }+\frac {\mathcal{C}}{4}\frac {\partial ^4 B}{\partial \xi ^4}-\mathcal{C}\frac {\partial ^2\theta _2}{\partial \xi ^2}=0. \end{align}

We can now eliminate the unknown arbitrary function of integration $\theta _2(\xi ,\tau )$ from (4.36) and (4.37). Then, substituting the leading-order solution, namely: $ {\partial\! B}/{\partial \xi }= {\eta }/{\mathcal{C}}$ into the resulting equation, we obtain

(4.38) \begin{align} \frac {\partial \eta }{\partial \tau }+\alpha \eta \frac {\partial \eta }{\partial \xi }+\beta \frac {\partial ^3\eta }{\partial \xi ^3}=0, \end{align}

where

(4.39) \begin{align} \alpha &=\frac {3-\nu }{4\mathcal{C}}\approx 0.44446, \end{align}

(4.40) \begin{align} \beta &=\frac {1}{8\mathcal{C}(\nu -1)}\left ( \nu ^4\ln \nu -\frac {3}{4}\nu ^4-2\nu ^3+3\nu ^2+2\nu -\frac {9}{4} \right )\approx -0.075835, \end{align}

(4.41) \begin{align} \mathcal{C}^2 & =\frac {\nu +1}{2}\approx 1.0774. \end{align}

The coefficients of the nonlinear and dispersive terms in the evolution equation for a Plateau border (4.38) are given to five significant figures. Equation (4.38) is the single evolution equation for the waves travelling on the Plateau border and is a form of the KdV equation.

The solutions of (4.38) are well-known and can be calculated analytically using the substitution $\eta =( {3s}/{\alpha })\textrm {sech}^2\,\vartheta$ , to yield

(4.42) \begin{equation} \eta (\xi ,\tau )=\frac {3s}{\alpha }\textrm {sech}^2\left (\frac {1}{2}\sqrt {\frac {s}{\beta }}(\xi -s\tau )\right )\!, \end{equation}

where $s=\alpha \eta _0/3$ and $\eta _0$ is the amplitude of the wave. Based on the coefficients of the nonlinear and dispersive terms, $\alpha$ and $\beta$ , the evolution (4.38) only admits depression solitary waves with $s\lt 0$ . This means a solitary wave travels with speed slightly less than the long-wave speed, i.e. $(\mathcal{C}+s\epsilon ^2)$ . A plot of the solution (4.42) for $s=-0.05$ is depicted in figure 5.

Figure 5.

Plot of solution (4.42) to the KdV (4.38) for $s=-0.05$ .