Impact of interfacial slip on the stability of liquid two-layer polymer films

Abstract

In this work we derive systems of coupled thin-film equations for immiscible liquid polymer layers on a solid substrate. We take into account slip between liquids and solids and also slip between both liquids. On the scale of tens of nanometres, such two-layer systems are susceptible to instability and may rupture and dewet due to intermolecular forces. The stability of the two-layer system and its significant dependence on the order of magnitude of slip is investigated via these thin-film models. With weak slip at both the liquid–liquid and liquid–solid interfaces and polymer layers of comparable thickness, the dispersion relation typically shows two local maxima, one in the long-wave regime and the other at moderate wavenumbers. The former is associated with perturbations that mainly affect the gas–liquid interface and the latter with larger relative perturbation amplitudes at the liquid–liquid interface. Increasing the slip at the liquid–liquid interface generally favours the long-wave regime and can in fact revert the mode of the instability and thus significantly change the spinodal patterns. Moreover, the maxima shift to small wavenumbers for increasing slip.

Appendices

Appendix 1: Derivation of thin-film model for strong slip at liquid–liquid and solid–liquid interfaces

We expand the variables in (2.12) as

$$\begin{aligned}&\!\!\!\left( u_1, w_1,u_2, w_2\right) = \left( {u_1^{(0)}}, {w_1^{(0)}},{u_2^{(0)}}, {w_2^{(0)}}\right) + \varepsilon ^2_{\ell }\, \left( {u_1^{(1)}}, {w_1^{(1)}}, {u_2^{(1)}}, {w_2^{(1)}}\right) + O\left( \varepsilon ^{4}_{\ell }\right) ,\end{aligned}$$

(6.1a)

$$\begin{aligned}&\!\!\!\left( p_1, p_2\right) = \left( {p_1^{(0)}}, {p_2^{(0)}}\right) + \varepsilon ^2_{\ell }\, \left( {p_1^{(1)}}, {p_2^{(1)}}\right) + O\left( \varepsilon ^{4}_{\ell }\right) ,\end{aligned}$$

(6.1b)

$$\begin{aligned}&\!\!\!\left( h_1, h_2\right) = \left( {h_1^{(0)}}, {h_2^{(0)}}\right) + \varepsilon ^2_{\ell }\, \left( {h_1^{(1)}}, {h_2^{(1)}}\right) + O\left( \varepsilon ^{4}_{\ell }\right) \!. \end{aligned}$$

(6.1c)

and consider in turn the leading- and next-order problem in $\varepsilon _{\ell }$.

1.1 Leading-order problem

$$\begin{aligned} 0&= {\partial _{zz}}{u_2^{(0)}}, \end{aligned}$$

(6.2a)

$$\begin{aligned} 0&= -{\partial _z}{p_2^{(0)}}+ {\partial _{zz}}{w_2^{(0)}}, \end{aligned}$$

(6.2b)

$$\begin{aligned} 0&= {\partial _x}{u_2^{(0)}}+{\partial _z}{w_2^{(0)}}, \end{aligned}$$

(6.2c)

$$\begin{aligned} 0&= {\partial _{zz}}{u_1^{(0)}}, \end{aligned}$$

(6.2d)

$$\begin{aligned} 0&= -{\partial _z}{p_1^{(0)}}+\mu {\partial _{zz}}{w_1^{(0)}}, \end{aligned}$$

(6.2e)

$$\begin{aligned} 0&= {\partial _x}{u_1^{(0)}}+{\partial _z}{w_1^{(0)}}. \end{aligned}$$

(6.2f)

For the boundary condition at the free surface $z={h_2^{(0)}}$ we get for the normal, tangential and kinematic conditions, respectively,

$$\begin{aligned}&{p_2^{(0)}}-\phi '({h^{(0)}}) + {\partial _{xx}}{h_2^{(0)}}- 2\left( {\partial _z}{w_2^{(0)}}-{\partial _z}{u_2^{(0)}}\,{\partial _x}{h_2^{(0)}}\right) =0, \end{aligned}$$

(6.2g)

$$\begin{aligned}&\!\!\!{\partial _z}{u_2^{(0)}}= 0,\end{aligned}$$

(6.2h)

$$\begin{aligned}&\!\!\!{\partial _t}{h_2^{(0)}}={w_2^{(0)}}-{u_2^{(0)}}{\partial _x}{h_2^{(0)}}. \end{aligned}$$

(6.2i)

For the boundary condition at the free liquid–liquid interface $z={h_1^{(0)}}$ we get for the normal, tangential and kinematic conditions, respectively,

$$\begin{aligned}&\!\!\!{p_1^{(0)}}-{p_2^{(0)}}+\phi '({h^{(0)}}) + \sigma \,{\partial _{xx}}{h_1^{(0)}}- 2\,\left[ \left( \mu {\partial _z}{w_1^{(0)}}-{\partial _z}{w_2^{(0)}}\right) -\left( \mu {\partial _z}{u_1^{(0)}}-{\partial _z}{u_2^{(0)}}\right) {\partial _x}{h_1^{(0)}}\right] =0, \end{aligned}$$

(6.2j)

$$\begin{aligned}&\!\!\!{\partial _z}\left( \mu {u_1^{(0)}}-{u_2^{(0)}}\right) = 0,\end{aligned}$$

(6.2k)

$$\begin{aligned}&\!\!\!{\partial _t}{h_1^{(0)}}= {w_1^{(0)}}-{u_1^{(0)}}{\partial _x}{h_1^{(0)}}. \end{aligned}$$

(6.2l)

The impermeability condition at $z={h_1^{(0)}}$ between the two liquid layers is given by

$$\begin{aligned}&\left( {w_2^{(0)}}-{w_1^{(0)}}\right) -\left( {u_2^{(0)}}-{u_1^{(0)}}\right) {\partial _x}{h_1^{(0)}}=0. \end{aligned}$$

(6.2m)

The slip condition at the liquid–liquid interface $z={h_1^{(0)}}$ is

$$\begin{aligned} {u_2^{(0)}}={\beta _1}\,\frac{\mu +1}{\mu }\,{\partial _z}{u_2^{(0)}}. \end{aligned}$$

(6.2n)

For the boundary conditions at the substrate we assume impermeability and no slip:

$$\begin{aligned}&\!\!\!{w_1^{(0)}}=0, \end{aligned}$$

(6.2o)

$$\begin{aligned}&\!\!\!{\partial _z}{u_1^{(0)}}=0. \end{aligned}$$

(6.2p)

From (6.2d), (6.2p) and (6.2a), (6.2h) we conclude

$$\begin{aligned}&{u_1^{(0)}}= {u_1^{(0)}}(x,t), \end{aligned}$$

(6.3a)

$$\begin{aligned}&{u_2^{(0)}}= {u_2^{(0)}}(x,t), \end{aligned}$$

(6.3b)

and thus the horizontal velocity components are independent of $z$. Using this in (6.2f), (6.2o) and (6.2c), (6.2m) we find

$$\begin{aligned}&\!\!\!{w_1^{(0)}}=-z{\partial _x}{u_1^{(0)}}, \end{aligned}$$

(6.4a)

$$\begin{aligned}&\!\!\!{w_2^{(0)}}=-\left( z-{h_1^{(0)}}\right) {\partial _x}{u_2^{(0)}}-{\partial _x}{u_1^{(0)}}{h_1^{(0)}}+({u_2^{(0)}}-{u_1^{(0)}}){\partial _x}{h_1^{(0)}}. \end{aligned}$$

(6.4b)

Using (6.2e), (6.2j) and (6.2b), (6.2g) we find

$$\begin{aligned}&\!\!\!{p_1^{(0)}}+2\mu {\partial _x}{u_1^{(0)}}+{\partial _{xx}}{h_2^{(0)}}+\sigma {\partial _{xx}}{h_1^{(0)}}=0, \end{aligned}$$

(6.5a)

$$\begin{aligned}&\!\!\!{p_2^{(0)}}-\phi '\left( {h^{(0)}}\right) +2{\partial _x}{u_2^{(0)}}+{\partial _{xx}}{h_2^{(0)}}=0, \end{aligned}$$

(6.5b)

hence also independent of $z$.

1.2 Next-order problem

To close the problem to leading order and determine an equation for ${u_1^{(0)}}$ and ${u_2^{(0)}}$, we need to consider the problem to next order. We will formulate here only the equations that are required to accomplish the task of fixing these leading-order velocity components.

The next-order upper and lower layer equations in the bulk are

$$\begin{aligned}&0 = -{\partial _x}{p_2^{(0)}}+ {\partial _{xx}}{u_2^{(0)}}+ {\partial _{zz}}{u_2^{(1)}},\end{aligned}$$

(6.6a)

$$\begin{aligned}&0 = -{\partial _z}{p_2^{(1)}}+ {\partial _{xx}}{w_2^{(0)}}+ {\partial _{zz}}{w_2^{(1)}},\end{aligned}$$

(6.6b)

$$\begin{aligned}&0 = {\partial _x}{u_2^{(1)}}+{\partial _z}{w_2^{(1)}},\end{aligned}$$

(6.6c)

$$\begin{aligned}&0 = -{\partial _x}{p_1^{(0)}}+ \mu {\partial _{xx}}{u_1^{(0)}}+ \mu {\partial _{zz}}{u_1^{(1)}},\end{aligned}$$

(6.6d)

$$\begin{aligned}&0 = -{\partial _z}{p_1^{(1)}}+ \mu {\partial _{xx}}{w_1^{(0)}}+ \mu {\partial _{zz}}{w_1^{(1)}},\end{aligned}$$

(6.6e)

$$\begin{aligned}&0 = {\partial _x}{u_1^{(1)}}+{\partial _z}{w_1^{(1)}}. \end{aligned}$$

(6.6f)

The next-order tangential stress boundary conditions at liquid–gas interface $z={h_2^{(0)}}$ are

$$\begin{aligned} \!\!\!{\partial _z}{u_2^{(1)}}+{\partial _x}{w_2^{(0)}}-4{\partial _x}{u_2^{(0)}}{\partial _x}{h_2^{(0)}}=0, \end{aligned}$$

(6.6g)

$$\begin{aligned} \!\!\!{\partial _z}\left( \mu {u_1^{(1)}}-{u_2^{(1)}}\right) +{\partial _x}\left( \mu {w_1^{(0)}}-{w_2^{(0)}}\right) -4{\partial _x}\left( \mu {u_1^{(0)}}-{u_2^{(0)}}\right) {\partial _x}{h_1^{(0)}}=0. \end{aligned}$$

(6.6h)

At the liquid–liquid interface $z={h_1^{(0)}}$ we have

$$\begin{aligned} \!\!\!{u_2^{(0)}}-{u_1^{(0)}}= {\beta _1}\frac{\mu }{\mu +1}\left[ {\partial _z}{u_2^{(1)}}+{\partial _x}{w_2^{(0)}}-4{\partial _x}{u_2^{(0)}}{\partial _x}{h_1^{(0)}}\right] , \end{aligned}$$

(6.6i)

and at the solid substrate $z=0$,

$$\begin{aligned}&\!\!\! {u_1^{(0)}}=\beta _1{\partial _z}{u_1^{(1)}}. \end{aligned}$$

(6.6j)

In the preceding equations, we have already made use of the fact that ${u_1^{(0)}}$ and ${u_2^{(0)}}$ do not depend on $z$ and dropped all derivatives of these variables with respect to $z$.

Integrating now (6.6a) and (6.6d), we obtain

$$\begin{aligned} {\partial _z}\left. {u_2^{(1)}}\right| _{{h_2^{(0)}}}-{\partial _z}\left. {u_2^{(1)}}\right| _{{h_1^{(0)}}} =- h^{(0)} \left( -{\partial _x}{p_2^{(0)}}+{\partial _{xx}}{u_2^{(0)}}\right) , \end{aligned}$$

(6.7)

$$\begin{aligned} {\partial _z}\left. {u_1^{(1)}}\right| _{{h_2^{(0)}}}-{\partial _z}\left. {u_1^{(1)}}\right| _{{h_1^{(0)}}} =- {h_1^{(0)}}\left( -{\partial _x}{p_1^{(0)}}+\mu {\partial _{xx}}{u_1^{(0)}}\right) . \end{aligned}$$

(6.8)

The pressure terms on the right-hand side can be eliminated by using (6.5a) and (6.5b). The terms on the left-hand side can be expressed in terms of the leading-order solutions ${u_2^{(0)}}$, ${u_2^{(0)}}$, ${h_1^{(0)}}$ and ${h_2^{(0)}}$ by first using (6.6g), (6.6h), (6.6i) and (6.6j), then eliminating the occurring ${w_2^{(0)}}$ and ${w_1^{(0)}}$ via the solutions (6.4a) and (6.4b). This yields Eqs. (3.8a) and (3.8c). The other two equations, (3.8b) and (3.8d), are obtained by integrating (6.2c) and (6.2f) and using the conditions (6.2i), (6.2l), (6.2m), (6.2p).

Appendix 2: Dispersion relations for Stokes equations

We start the stability analysis with the Stokes equations

$$\begin{aligned} 0&= -{\partial _x}p_n +\mu _n\left( {\partial _{xx}}u_n +{\partial _{zz}}u_n\right) ,\end{aligned}$$

(7.1a)

$$\begin{aligned} 0&= -{\partial _z}p_n +\mu _n\left( {\partial _{xx}}w_n +{\partial _{zz}}w_n\right) ,\end{aligned}$$

(7.1b)

$$\begin{aligned} 0&= {\partial _x}u_n + {\partial _z}w_n \end{aligned}$$

(7.1c)

and the boundary conditions from the previous sections. To simplify our problem, we introduce the stream functions $\Psi _1$ and $\Psi _2$ with

$$\begin{aligned} u_n=\frac{{\partial }\Psi _n}{{\partial }z},\quad w_n=-\frac{{\partial }\Psi _n}{{\partial }x}\quad \quad (n=1,2). \end{aligned}$$

(7.2)

Plugging this into the Stokes equations we obtain two biharmonic equations:

$$\begin{aligned} {\partial }_x^4 \Psi _n +2{\partial }_x^2 {\partial }_z^2 \Psi _n+ {\partial }_z^4 \Psi _n=0\quad \quad (n=1,2). \end{aligned}$$

(7.3)

Linear stability is carried out by introducing small perturbations

$$\begin{aligned} \left( \Psi _n,\ h_1-h_1^0,\ h_2-h_2^0,\ p_n-p_n^0,\ \phi -\phi _0\right) =\delta \left( \psi _n(z), \chi , 1, \Pi _n(z), (1-\chi )\left. \phi '\right| _{h^0}\right) \! \exp ({\mathrm {i}}kx+\omega t) \end{aligned}$$

(7.4)

around the base state

$$\begin{aligned} \Psi _n=0,\quad h_n= h^0_n,\quad h^0=h_2^0-h_1^0,\quad p_n=p_n^0,\quad \phi |_{h^0}=\phi _0, \end{aligned}$$

(7.5)

where $\omega$ and $k$ are the growth coefficient and the wavenumber, and obtain

$$\begin{aligned} {\partial }^4_z \psi _n-2k^2 {\partial }_z^2 \psi _n + k^4 \psi _n=0, \end{aligned}$$

(7.6)

with the general solutions

$$\begin{aligned} \psi _n(z)=u_{n1}\exp (kz)+u_{n2}z\exp (kz)+u_{n3} \exp (-kz)+u_{n4}z\exp (-kz), \end{aligned}$$

(7.7)

where the coefficients are determined using the boundary conditions. First, slip at $z=0$ leads to

$$\begin{aligned} (k-{b\,}k^2)u_{11}+(1-2{\,b\,}k)u_{12}-(k+{b\,}k^2)u_{13}+(1+2{\,b\,}k)u_{14}=0, \end{aligned}$$

(7.8)

while impermeability simply reads

$$\begin{aligned} u_{11}+u_{13}=0. \end{aligned}$$

(7.9)

At the free surface $z=h_2$ the kinematic condition becomes

$$\begin{aligned} k\hbox {e}^{kh_2^0}u_{21}+kh_2^0\hbox {e}^{kh_2^0}u_{22}+k\hbox {e}^{-kh_2^0}u_{23} +kh_2^0\hbox {e}^{-kh_2^0}u_{24}=\hbox {i}\omega \end{aligned}$$

(7.10)

the tangential stress condition

$$\begin{aligned} k\hbox {e}^{kh_2^0}u_{21}+(kh_2^0+1)\hbox {e}^{kh_2^0}u_{22}+k\hbox {e}^{-kh_2^0}u_{23} +(kh_2^0-1)\hbox {e}^{-kh_2^0}u_{24}=0. \end{aligned}$$

(7.11)

At the liquid–liquid interface $z=h_1$, the kinematic condition now reads

$$\begin{aligned} k\hbox {e}^{kh_1^0}u_{11}+kh_1^0\hbox {e}^{kh_1^0}u_{12}+k\hbox {e}^{-kh_1^0}u_{13} +kh_1^0\hbox {e}^{-kh_1^0}u_{14} = \hbox {i}\chi \omega \end{aligned}$$

(7.12)

the tangential stress condition

$$\begin{aligned}&\!\!\!\mu _1[k\hbox {e}^{kh_1^0}u_{11}+(kh_1^0+1)\hbox {e}^{kh_1^0}u_{12} +k\hbox {e}^{-kh_1^0}u_{13}+(kh_1^0-1)\hbox {e}^{-kh_1^0}u_{14}] \nonumber \\&-\,\mu _2[k\hbox {e}^{kh_1^0}u_{21}+(kh_1^0+1)\hbox {e}^{kh_1^0}u_{22} +k\hbox {e}^{-kh_1^0}u_{23}+(kh_1^0-1)\hbox {e}^{-kh_1^0}u_{24}]=0. \end{aligned}$$

(7.13)

The impermeability condition is equivalent to

$$\begin{aligned} k\hbox {e}^{kh_1^0}u_{21}+kh_1^0\hbox {e}^{kh_1^0}u_{22} +k\hbox {e}^{-kh_1^0}u_{23}+kh_1^0\hbox {e}^{-kh_1^0}u_{24}=\hbox {i}\chi \omega \end{aligned}$$

(7.14)

and, finally, the slip condition

$$\begin{aligned}&\!\!\!k\hbox {e}^{kh_1^0}u_{11}+(kh_1^0+1)\hbox {e}^{kh_1^0}u_{12} -k\hbox {e}^{-kh_1^0}u_{13}-(kh_1^0-1)\hbox {e}^{-kh_1^0}u_{14} \nonumber \\&+\,k(2b_*k-1)\hbox {e}^{kh_1^0}u_{21}+(kh_1^0+1) (2b_*k-1)\hbox {e}^{kh_1^0}u_{22}\nonumber \\&+\,k(2b_*k+1)\hbox {e}^{-kh_1^0}u_{23}+(kh_1^0-1) (2b_*k+1)\hbox {e}^{-kh_1^0}u_{24}=0, \end{aligned}$$

(7.15)

where $b_*=(1+\mu _2/\mu _1){b_1}$. For the solution of this algebraic system we use the remaining boundary conditions, namely the normal stress conditions at $z=h_2$

$$\begin{aligned} {\mathrm {i}}\mu _2[2k^2\hbox {e}^{kh_2^0}u_{21}+2k^2h_2^0\hbox {e}^{kh_2^0}u_{22} -2k^2\hbox {e}^{-kh_2^0}u_{23}-2k^2h_2^0\hbox {e}^{-kh_2^0}u_{24}] = \sigma _2 k^2+(1-\chi )\left. \phi ^{\prime \prime }\right| _{h^0}, \end{aligned}$$

(7.16)

and at $z=h_1$

$$\begin{aligned}&\!\!\!\hbox {i}\mu _1[2k^2\hbox {e}^{kh_1^0}u_{11}+2k^2h_1^0\hbox {e}^{kh_1^0} u_{12}-2k^2\hbox {e}^{-kh_1^0}u_{13}-2k^2h_1^0\hbox {e}^{-kh_1^0}u_{14}] \nonumber \\&-\,\hbox {i}\mu _2[2k^2\hbox {e}^{kh_1^0}u_{21}+2k^2h_1^0\hbox {e}^{kh_1^0}u_{22} -2k^2\hbox {e}^{-kh_1^0}u_{23}-2k^2h_1^0\hbox {e}^{-kh_1^0}u_{24}] =\sigma _1 \chi k^2-(1-\chi )\left. \phi ^{\prime \prime }\right| _{h^0}. \end{aligned}$$

(7.17)