A.2.1. Order 0 – resolution at the microscopic scale: the scale of the bubble

The microscopic scale is the scale corresponding to a zoom on a single bubble at a given time $t$. At this scale, the spherical coordinate ${{\boldsymbol {x}}}_{\mu }=({1}/{{\boldsymbol{\mathsf{r}}}_{eq}R({{\boldsymbol {r}}},t)}){{{\boldsymbol {x}}}}/{\varepsilon ^2}$ (see (A 3)) where, following Caflisch et al. (Reference Caflisch, Miksis, Papanicolaou and Ting1985b), the discrete version of the bubble radius $R_i$ (see e.g. (2.3)) has been replaced by a continuous counterpart $R({{\boldsymbol {r}}},t)$. Doing so, the bubble has a unitary radius and the nearby bubbles have been sent to infinity. We consider the following expansions of the solution in the gas:

(A 7)\begin{equation} \left.\begin{gathered} {{\boldsymbol{u}}}=\dfrac{1}{\varepsilon^2}\hat{\boldsymbol{u}}_{\mu}^{-2}({{\boldsymbol{x}}}_{\mu},{{\boldsymbol{r}}},t)+\dfrac{1}{\varepsilon} \hat{\boldsymbol{u}}_{\mu}^{-1}({{\boldsymbol{x}}}_{\mu},{{\boldsymbol{r}}},t)\cdots,\quad p=\hat{p}_{\mu}^0({{\boldsymbol{x}}}_{\mu},{{\boldsymbol{r}}},t)+\varepsilon \hat{p}_{\mu}^1({{\boldsymbol{x}}}_{\mu},{{\boldsymbol{r}}},t)+\cdots,\\ \rho=\hat{\rho}_{\mu}^0({{\boldsymbol{x}}}_{\mu},{{\boldsymbol{r}}},t)+\varepsilon \hat{\rho}_{\mu}^1({{\boldsymbol{x}}}_{\mu},{{\boldsymbol{r}}},t)+\cdots,\quad R=R^0({{\boldsymbol{r}}},t)+\varepsilon R^1({{\boldsymbol{r}}},t)+\cdots, \end{gathered}\right\} \end{equation}

and in the liquid (where $p=\rho$)

(A 8a,b)\begin{equation} {{\boldsymbol{u}}}=\displaystyle\frac{1}{\varepsilon^2}{\boldsymbol{u}}_{\mu}^{-2}({{\boldsymbol{x}}}_{\mu},{{\boldsymbol{r}}},t)+\frac{1}{\varepsilon} {\boldsymbol{u}}_{\mu}^{-1}({{\boldsymbol{x}}}_{\mu},{{\boldsymbol{r}}},t)+\cdots, \quad p=p_{\mu}^0({{\boldsymbol{x}}}_{\mu},{{\boldsymbol{r}}},t)+\varepsilon p_{\mu}^1({{\boldsymbol{x}}}_{\mu},{{\boldsymbol{r}}},t)+\cdots. \end{equation}

Accordingly, denoting ${{\boldsymbol {\nabla }}}_{\mu }\,=\partial /\partial \boldsymbol {x}_{\mu }$ and ${\boldsymbol {\nabla }}_{{{\boldsymbol {r}}}}=\partial /\partial \boldsymbol {r}$, the differential operators read

(A 9)\begin{equation} {\boldsymbol{\nabla}} \to \displaystyle \frac{1}{{\boldsymbol{\mathsf{r}}}_{eq}R} \frac{{{\boldsymbol{\nabla}}}_{\mu}\, }{\varepsilon^2} +{\boldsymbol{\nabla}}_{{{\boldsymbol{r}}}}- \frac{{\boldsymbol{\nabla}}_{{{\boldsymbol{r}}}} R}{R}\, {{\boldsymbol{x}}}_{\mu} \boldsymbol{\cdot} {{\boldsymbol{\nabla}}}_{\mu}\,\quad \text{and} \quad \displaystyle \frac{\partial}{\partial t} \to \displaystyle \frac{\partial}{\partial t}-\frac{\dot{R}}{R} \,{{\boldsymbol{x}}}_{\mu} \boldsymbol{\cdot} {{\boldsymbol{\nabla}}}_{\mu}\,. \end{equation}

Resolution inside the bubble – using the expansions (A 7) in (A 4) along with (A 9), the problem at the dominant order gives uniform pressure and mass density with ${{\boldsymbol {\nabla }}}_{\mu }\, \hat {p}_{\mu }^0 = 0$ and

(A 10a–c)\begin{align} \displaystyle \frac{\partial \hat{\rho}_{\mu}^0}{\partial t} +\frac{\left(\alpha + m\hat{\rho}_{\mu}^0\right)}{{\boldsymbol{\mathsf{r}}}_{eq}R^0}\,\text{div}_{\mu}\, \hat{\boldsymbol{u}}_{\mu}^{-2}=0,\quad \boldsymbol{\nabla} \land \hat{\boldsymbol{u}}_{\mu}^{-2} = 0, \quad 1+\gamma\frac{m}{\beta} \hat{p}_{\mu}^0=\left(1+\frac{m}{\alpha} \hat{\rho}_{\mu}^0\right)^\gamma, \end{align}

and initial and boundary conditions, from (A 6), of the form

(A 11)\begin{equation} \left.\begin{gathered}{}{\rm i.c.\quad at } t=0,\quad \hat{p}_{\mu}^0=\hat{\rho}_{\mu}^0=0, \quad \hat{\boldsymbol{u}}_{\mu}^{-2}=\textbf{{0}}, \quad R^0=1,\\ {\rm b.c.\quad for } |{{\boldsymbol{x}}}_{\mu}|=1,\quad \displaystyle \hat{p}_{\mu}^0=p_{\mu}^0, \quad \hat{\boldsymbol{u}}_{\mu}^{-2}\boldsymbol{\cdot} {{\boldsymbol{e}}}_r= \frac{{\boldsymbol{\mathsf{r}}}_{eq}}{m}\, \dot{R}^0, \end{gathered}\right\} \end{equation}

where ${{\boldsymbol {e}}}_r={{\boldsymbol {x}}}_{\mu }/|{{\boldsymbol {x}}}_{\mu }|$. Now, we shall express $\hat {p}_{\mu }^0({{\boldsymbol {r}}},t)$ and $\hat {\rho }_{\mu }^0({{\boldsymbol {r}}},t)$ as a function of $R^0({{\boldsymbol {r}}},t)$. By integrating the balance of mass over the gas bubble (first equation in (A 10a–c)) we get $({4{\rm \pi} }/{3})(\partial\hat {\rho }_{\mu }^0/\partial t)+(\alpha + m\hat {\rho }_{\mu }^0)({1}/{{\boldsymbol{\mathsf{r}}}_{eq}R^0})\int _{|{{\boldsymbol {x}}}_{\mu }|=1} \hat {\boldsymbol {u}}_{\mu }^{-2}\boldsymbol {\cdot } {{\boldsymbol {e}}}_r\, \text {d}s=0$. This differential equation on $\hat {\rho }_{\mu }^0$ is solved using in (A 11) (i) the boundary condition on $\hat {\boldsymbol {u}}_{\mu }^{-2}$ and (ii) the initial conditions $\hat {\rho }_{\mu }^0=0$, $R^0=1$ at $t=0$. Also, making use of the equation of state in (A 10a–c), we obtain the pressure $\hat {p}_{\mu }^0$. Eventually, the velocity $\hat {\boldsymbol {u}}_{\mu }^{-2}$ is a byproduct of $\hat {\rho }_{\mu }^0$ in (A 10a–c). The result is

(A 12a–c)\begin{equation} \left.\begin{gathered} \hat{\rho}_{\mu}^0({{\boldsymbol{r}}},t)=\frac{\alpha}{m} \left(\frac{1}{(R^0)^3}-1\right),\quad \hat{p}_{\mu}^0({{\boldsymbol{r}}},t)=\frac{\beta}{\gamma m}\left(\frac{1}{(R^0)^{3\gamma}} -1\right),\\ \hat{\boldsymbol{u}}_{\mu}^{-2}({{\boldsymbol{x}}}_{\mu},{{\boldsymbol{r}}},t)=\frac{{\boldsymbol{\mathsf{r}}}_{eq}}{m} \dot{R}^0 |{{\boldsymbol{x}}}_{\mu}|\,{{\boldsymbol{e}}}_r, \end{gathered}\right\} \end{equation}

given in dimensional form in (2.7).

Resolution in the liquid – in the liquid, we repeat the exercise, using the expansions (A 8) in (A 5) and (A 6) along with (A 9). We get at the dominant order

(A 13)\begin{equation} \left.\begin{array}{c} \displaystyle \text{div}_{\mu}\, {\boldsymbol{u}}_{\mu}^{-2}=0, \quad {{\boldsymbol{\nabla}}}_{\mu}\, \land {\boldsymbol{u}}_{\mu}^{-2} = 0,\\ \displaystyle \dfrac{\partial{\boldsymbol{u}}_{\mu}^{-2}}{\partial t} +\dfrac{m}{{\boldsymbol{\mathsf{r}}}_{eq}R^0} ({\boldsymbol{u}}_{\mu}^{-2} \boldsymbol{\cdot} {{\boldsymbol{\nabla}}}_{\mu}\,) {\boldsymbol{u}}_{\mu}^{-2}-\dfrac{\dot{R}^0}{R^0}\,{{\boldsymbol{x}}}_{\mu} \boldsymbol{\cdot} {{\boldsymbol{\nabla}}}_{\mu}\,{\boldsymbol{u}}_{\mu}^{-2}=-\dfrac{1}{{\boldsymbol{\mathsf{r}}}_{eq}R^0} {{\boldsymbol{\nabla}}}_{\mu}\, p_{\mu}^0, \end{array} \right\} \end{equation}

with initial and boundary conditions

(A 14)\begin{align} \left.\begin{gathered}{} \displaystyle \text{i.c.}\quad \text{at }t=0, \quad p_{\mu}^0=0, \quad {\boldsymbol{u}}_{\mu}^{-2}=\textbf{{0}}, \\ \displaystyle \text{b.c.}\quad \text{for } |{{\boldsymbol{x}}}_{\mu}|=1,\quad {\boldsymbol{u}}_{\mu}^{-2}\boldsymbol{\cdot} {{\boldsymbol{e}}}_r= \dfrac{{\boldsymbol{\mathsf{r}}}_{eq}}{m}\, \dot{R}^0\quad \text{and} \quad \displaystyle \lim_{|{{\boldsymbol{x}}}_{\mu}|\to +\infty} p_{\mu}^0({{\boldsymbol{x}}}_{\mu},{{\boldsymbol{r}}},t)=P^0({{\boldsymbol{r}}},t). \end{gathered}\right\} \end{align}

For the time being, the continuity of the pressure at $|{{\boldsymbol {x}}}_{\mu }|=1$ is not needed, but we have introduced the (unknown) pressure field $P^0({{\boldsymbol {r}}},t)$; its meaning will appear later on. We have also assumed that the bubble remains spherical.

From (A 13), the flow is incompressible and irrotational. Hence it is associated with a velocity potential $\phi$ and, the bubble remaining spherical, ${\phi =({A}/{|{{\boldsymbol {x}}}_{\mu }|})+B}$. Next, the boundary condition in (A 14) for ${\boldsymbol {u}}_{\mu }^{-2}={{\boldsymbol {\nabla }}}_{\mu }\, \phi$ provides $A=-{{\boldsymbol{\mathsf{r}}}_{eq}\dot {R}^0}/{m}$. Using ${\boldsymbol {u}}_{\mu }^{-2}$ in (A 13), the pressure $p_{\mu }^0$ is found to satisfy a Bernoulli equation ${{\boldsymbol {\nabla }}}_{\mu }\,(p_{\mu }^0+({{\boldsymbol{\mathsf{r}}}_{eq}^2}/{m})\,F(R^0,\dot R^0,\ddot R^0,|{{\boldsymbol {x}}}_{\mu }|))=0$, that we integrate using $p_{\mu }^0\to _{|{{\boldsymbol {x}}}_{\mu }|\to +\infty } P^0({{\boldsymbol {r}}},t)$ from (A 14). Doing so, the fields $({\boldsymbol {u}}_{\mu }^{-2},p_{\mu }^0)$ are found to be functions of $R^0$ and of their time derivatives

(A 15a,b)\begin{align} {\boldsymbol{u}}_{\mu}^{-2}({{\boldsymbol{x}}}_{\mu},{{\boldsymbol{r}}},t)=\frac{{\boldsymbol{\mathsf{r}}}_{eq}}{m}\,\frac{\dot{R}^0}{|{{\boldsymbol{x}}}_{\mu}|^2}\, {{\boldsymbol{e}}}_r, \quad p_{\mu}^0({{\boldsymbol{x}}}_{\mu},{{\boldsymbol{r}}},t)=P^0+\frac{{\boldsymbol{\mathsf{r}}}_{eq}^2}{m} \left(\frac{R^0\ddot{R}^0+2(\dot{R}^0)^2}{|{{\boldsymbol{x}}}_{\mu}|}- \frac{(\dot{R}^0)^2}{2|{{\boldsymbol{x}}}_{\mu}|^4}\right), \end{align}

given in dimensional form in (2.7). At this stage, once $R^0({{\boldsymbol {r}}},t)$ is known, the solution in the bubble is known from (A 12a–c), and the solution in the liquid is known up to the pressure field $P^0({{\boldsymbol {r}}},t)$ from (A 15a,b).

An equation of the Rayleigh–Plesset type – we shall establish that $R^0({{\boldsymbol {r}}},t)$ satisfies a continuous version of the Rayleigh–Plesset equation. To do so, we use $p_{\mu }^0=\hat {p}_{\mu }^0$ for ${|{{\boldsymbol {x}}}_{\mu }|=1}$, with $\hat {p}_{\mu }^0$ in (A 12a–c) and $p_{\mu }^0$ in (A 15a,b). We get that $R^0({{\boldsymbol {r}}},t)$ is solution of

(A 16)\begin{equation} R^0\ddot{R}^0+\frac{3}{2}(\dot{R}^0)^2+\frac{\beta}{\gamma {\boldsymbol{\mathsf{r}}}_{eq}^2}\left(1-\frac{1}{(R^0)^{3\gamma}}\right)=-\frac{m}{{\boldsymbol{\mathsf{r}}}_{eq}^2} P^0({{\boldsymbol{r}}},t) \end{equation}

and at the equilibrium $P^0=0$, the bubbles and the liquid are at rest and the acoustic pressures vanish ($R^0=1$ in (A 16) hence, ${\boldsymbol {u}}_{\mu }^{-2}=\textbf {{0}}$ and $\hat {p}_{\mu }^0=p_{\mu }^0=0$ in (A 15a,b)).

A.2.2. Order 0 – resolution at the mesoscopic scale: the scale of the screen spacing

From (A 12a–c), (A 15a,b) and (A 16), the solution at the microscopic scale is known once $P^0({{\boldsymbol {r}}},t)$ is known, and $P^0$ will be provided by the mesoscopic problem. This scale is that of the array spacing, intermediate between the bubble size and the wavelength. It is associated with ${{\boldsymbol {x}}}_{m}={{{\boldsymbol {x}}}}/{\varepsilon }$, ${{\boldsymbol {x}}}_{m}=(x_{m},{{\boldsymbol {r}}}_{m})$ defined in (A 3), and we use the expansions

(A 17)\begin{align} \left.\begin{gathered} \displaystyle{{\boldsymbol{u}}}={\boldsymbol{u}}_{m}^0({{\boldsymbol{x}}}_{m},{{\boldsymbol{r}}},t)+\varepsilon {\boldsymbol{u}}_{m}^1({{\boldsymbol{x}}}_{m},{{\boldsymbol{r}}},t)+\cdots,\\ p=p_{m}^0({{\boldsymbol{x}}}_{m},{{\boldsymbol{r}}},t)+\varepsilon p_{m}^1({{\boldsymbol{x}}}_{m},{{\boldsymbol{r}}},t)+\cdots,\quad \rho=\rho_{m}^0({{\boldsymbol{x}}}_{m},{{\boldsymbol{r}}},t)+\varepsilon \rho_{m}^1({{\boldsymbol{x}}}_{m},{{\boldsymbol{r}}},t)+\cdots, \end{gathered}\right\} \end{align}

where the $({\boldsymbol {u}}_{m}^i,p_{m}^i,\rho _{m}^i)$, $i=0,1,\ldots$ are periodic with respect to ${{\boldsymbol {r}}}_{m}\in {\boldsymbol{\mathsf{Y}}}$. In ${{\boldsymbol {x}}}_{m}$-coordinate, denoting ${\boldsymbol {\nabla }}_{{m}}\,=\partial /\partial \boldsymbol {x}_{m}$ (and ${\boldsymbol {\nabla }}_{{{\boldsymbol {r}}}}=\partial /\partial \boldsymbol {r}$ as in (A 9)), the differential operators read

(A 18a,b)\begin{equation} {\boldsymbol{\nabla}} \to \displaystyle \frac{1}{\varepsilon} {\boldsymbol{\nabla}}_{{m}}\, +{\boldsymbol{\nabla}}_{{{\boldsymbol{r}}}}, \quad \frac{\partial}{\partial t} \to \displaystyle \frac{\partial}{\partial t}. \end{equation}

Matching conditions – the matching conditions tell us that the solution of the microscopic problem has to coincide with that of the mesoscopic problem in some intermediate region, when $|{{\boldsymbol {x}}}_{\mu }|\to +\infty$ and ${{\boldsymbol {x}}}_{m} \to 0$. Since $|{{\boldsymbol {x}}}_{m}|=O({|{{\boldsymbol {x}}}|}/{\varepsilon })$ and $|{{\boldsymbol {x}}}_{\mu }|=O({|{{\boldsymbol {x}}}|}/{\varepsilon ^2})$, this intermediate region can be sought at distance $\varepsilon \sqrt {\varepsilon }$ of the bubble, in ${{\boldsymbol {x}}}$ coordinate; hence, $|{{\boldsymbol {x}}}_{\mu }|\sim {1}/{\sqrt {\varepsilon }}$ and $|{{\boldsymbol {x}}}_{m}|\sim \sqrt {\varepsilon }$. The matching conditions are obtained by identifying the expansions (A 7) and (A 17) for ${{\boldsymbol {x}}}_{\mu }={{{\boldsymbol {x}}}_{m}}/{\varepsilon {\boldsymbol{\mathsf{r}}}_{eq}R}$ and ${{\boldsymbol {x}}}_{m}\to 0$. At the dominant order, anticipating that $p_{m}^0$ does not depend on ${{\boldsymbol {x}}}_{m}$, we get

(A 19a,b)\begin{equation} \displaystyle p_{m}^0({{\boldsymbol{r}}},t)\!+\cdots \sim p_{\mu}^0\left(\frac{{{\boldsymbol{x}}}_{m}}{{\boldsymbol{\mathsf{r}}}_{eq}\varepsilon R},{{\boldsymbol{r}}},t\right)\!+\cdots, \quad \displaystyle {\boldsymbol{u}}_{m}^{0}({{\boldsymbol{x}}}_{m},{{\boldsymbol{r}}},t)+\cdots \sim \frac{1}{\varepsilon^2}{\boldsymbol{u}}_{\mu}^{-2}\left(\frac{{{\boldsymbol{x}}}_{m}}{{\boldsymbol{\mathsf{r}}}_{eq}\varepsilon R},{{\boldsymbol{r}}},t\right) \!+\cdots \end{equation}

for ${{\boldsymbol {x}}}_{m}\to \textbf {{0}}$, hence

(A 20a,b)\begin{align} \lim_{|{{\boldsymbol{x}}}_{\mu}|\to +\infty} p_{\mu}^0({{\boldsymbol{x}}}_{\mu},{{\boldsymbol{r}}},t)=p_{m}^0({{\boldsymbol{r}}},t)=P^0({{\boldsymbol{r}}},t),\quad \displaystyle {\boldsymbol{u}}_{m}^0({{\boldsymbol{x}}}_{m},{{\boldsymbol{r}}},t) \underset{ {{\boldsymbol{x}}}_{m} \to \textbf{{0}}}{\sim} \frac{{\boldsymbol{\mathsf{r}}}_{eq}^3}{m} (R^0)^2\dot{R}^0\,\frac{{{\boldsymbol{x}}}_{m}}{ |{{\boldsymbol{x}}}_{m}|^3}. \end{align}

We have used ${\boldsymbol {u}}_{\mu }^{-2}$ in (A 15a,b) and we have assumed that the terms ${\boldsymbol {u}}_{\mu }^{n}$, when increasing $n>-2$, decrease faster than ${\boldsymbol {u}}_{\mu }^{-2}\propto {1}/{|{{\boldsymbol {x}}}_{\mu }|^2}$, which is consistent with non-diverging mass flux through a spherical cap. It follows that the main results provided by the mesoscopic problem are: the pressure $\tilde {P}^0({{\boldsymbol {r}}},t)$ introduced in (A 14) is the pressure $p_{m}^0$ at $|{{\boldsymbol {x}}}_{m}|=0$ (at the mesoscopic scale, the bubbles are reduced to points), the velocity ${\boldsymbol {u}}_{m}^0$ is singular at the origin, and it encapsulates the effect of the oscillating bubble point (if $\dot R^0=0$, the bubble is not seen and ${\boldsymbol {u}}_{m}^0$ is regular). In dimensional form, these results contribute to the solution given in (2.8) at the dominant order; they will be complemented with the solution found at the following order.

The mesoscopic problem – using (A 17) in (A 5) along with (A 18a,b), we get that ${\boldsymbol {\nabla }}_{{m}}\, p_{m}^0=\textbf {{0}}$. At this scale again, the pressure $p_{m}^0$ is uniform as announced in (A 19a,b) resulting in (A 20a,b). It follows that the problem at the dominant order is set on $({\boldsymbol {u}}_{m}^0,p_{m}^1)$ with

(A 21)\begin{equation} \left.\begin{gathered}{} {\rm div}_{m}\, {\boldsymbol{u}}_{m}^{0}=\textbf{{0}},\quad \boldsymbol{\nabla} \land {\boldsymbol{u}}_{m}^{0} = \textbf{{0}},\quad\displaystyle \frac{\partial {\boldsymbol{u}}_{m}^{0}}{\partial t}=-{\boldsymbol{\nabla}}_{{{\boldsymbol{r}}}} P^0({{\boldsymbol{r}}},t)-{\boldsymbol{\nabla}}_{{m}}\, p_{m}^1,\\ \text{i.c.} \quad p_{m}^1=0, \quad {\boldsymbol{u}}_{m}^{0}=\textbf{{0}}\quad \text{at } t=0,\\ \text{b.c.} \quad \displaystyle \lim_{x_{m}\to \pm\infty} {\boldsymbol{u}}_{m}^0= \textbf{U}^{0\pm}({{\boldsymbol{r}}},t), \quad p_{m}^1, {\boldsymbol{u}}_{m}^{0} \text{ periodic on } \partial{\boldsymbol{\mathsf{Y}}}, \end{gathered}\right\} \end{equation}

and ${\boldsymbol {u}}_{m}^0$ is singular at ${{\boldsymbol {x}}}_{m}=\textbf {{0}}$ from (A 20a,b). We have three unknown macroscopic fields: $P^0$ that we reach back from the microscopic scale (see (A 14)) and $\textbf {U}^{0\pm }$ in (A 21). Note that we have assumed that $\textbf {U}^{0\pm }$ do not depend on ${{\boldsymbol {r}}}_{m}$; this will be justified in the forthcoming section.

A.2.3. Order 0 – derivation of the effective transmission conditions

We now dive into the problem at the macroscopic scale, where only the coordinate ${{\boldsymbol {x}}}$ is needed, and we consider the following expansions:

(A 22a–c)\begin{equation} \left.\begin{gathered} {{\boldsymbol{u}}}={{\boldsymbol{u}}}^0({{\boldsymbol{x}}},t)+\varepsilon {{\boldsymbol{u}}}^1({{\boldsymbol{x}}},t) +\cdots,\quad p=p^0({{\boldsymbol{x}}},t)+\varepsilon p^1({{\boldsymbol{x}}},t) +\cdots,\\ \rho=\rho^0({{\boldsymbol{x}}},t)+\varepsilon \rho^1({{\boldsymbol{x}}},t)+\cdots. \end{gathered}\right\} \end{equation}

At this scale, the bubbly screen is reduced to an interface but the boundary conditions when $x\to 0^\pm$ are still missing. These are the conditions we are looking for and they are provided by matching the solutions of the mesoscopic and macroscopic problems. The intermediate region where both solutions are valid corresponds to $x\sim \pm \sqrt {\varepsilon }$ hence $x_{m}\sim \pm {1}/{\sqrt {\varepsilon }}$. Hence, at the dominant order, the matchings read

(A 23a,b)\begin{equation} p^0(0^\pm,{{\boldsymbol{r}}},t)=\displaystyle p_{m}^0({{\boldsymbol{r}}},t)=P^0({{\boldsymbol{r}}},t),\quad {{\boldsymbol{u}}}^0(0^\pm,{{\boldsymbol{r}}},t)=\displaystyle \lim_{x_{m}\to\pm\infty} {\boldsymbol{u}}_{m}^0({{\boldsymbol{x}}}_{m},{{\boldsymbol{r}}},t)= {\boldsymbol{U}}^{0\pm}({{\boldsymbol{r}}},t), \end{equation}

where we used (A 20a,b) and (A 21). In (A 23a,b), ${\boldsymbol {U}}^{0\pm }$ appear as the limits of the acoustic velocity on both sides of the array; this justifies a posteriori that ${\boldsymbol {U}}^{0\pm }$ do not depend on ${{\boldsymbol {r}}}_{m}$. Eventually, from (A 5) written at the dominant order, $(p^0,{{\boldsymbol {u}}}^0)$ satisfy the linearized Euler equations in an irrotational flow

(A 24a–c)\begin{equation} \frac{\partial p^0}{\partial t}+{\rm div}_{\boldsymbol{x}}{{\boldsymbol{u}}}^0 =0,\quad \frac{\partial {{\boldsymbol{u}}}^0}{\partial t}=- {\boldsymbol{\nabla}}_{\boldsymbol{x}} p^0,\quad \displaystyle \boldsymbol{\nabla} \land {{\boldsymbol{u}}}^0 = 0, \end{equation}

with the boundary conditions (A 23a,b) and the initial conditions $p^0({{\boldsymbol {x}}},0)=0$, ${{{\boldsymbol {u}}}^0({{\boldsymbol {x}}},0)=\textbf {{0}}}$. From (A 23a,b), the pressure $p^0$ is continuous at $x=0$. This is not the case for the normal velocity, and to show this result, we have to come back to the problem satisfied by ${\boldsymbol {u}}_{m}^0$. The equation of incompressibility, ${\rm div}_{m}\, {\boldsymbol {u}}_{m}^0=0$ in (A 21), is integrated in the domain ${\mathcal {Y}}$ minus a ball $B_r$ of radius $r$ (in order to avoid the singularity of ${\boldsymbol {u}}_{m}^0$). Using the boundary conditions in (A 21) and the singular behaviour of ${\boldsymbol {u}}_{m}^0$ in (A 20a,b), we get $0=\int _{{\mathcal {Y}}/B_r} {\rm div}_{m}\,{\boldsymbol {u}}_{m}^0\, \textrm {d}{{\boldsymbol {x}}}_{m}$ from which $u_x^0(0^+,{{\boldsymbol {r}}},t)-u_x^0(0^-,{{\boldsymbol {r}}},t)=\int _{\partial B_r}({{\boldsymbol{\mathsf{r}}}_{eq}^3}/{m})({(R^0)^2\dot {R}^0}/{ |{{\boldsymbol {x}}}_{m}|^2})\,\text {d}{{\boldsymbol {x}}}_{m}$. The integral over $\partial B_r$ gives a finite but non-zero contribution (independent of the radius of $B_r$), and eventually we obtain a non-trivial transmission condition

(A 25a,b)\begin{equation} \displaystyle [\kern-1pt[ p^0 ]\kern-1pt]=0, \quad [\kern-1pt[ u_x^0 ]\kern-1pt]=\frac{4{\rm \pi}{\boldsymbol{\mathsf{r}}}_{eq}^3}{m}(R^0)^2\dot{R}^0. \end{equation}

Conclusion of the dominant order – at this stage the dominant order provides a model in which the bubbly screen is reduced to an effective screen across which the normal velocity experiences a jump dictated by $R^0$, and $R^0$ satisfies (A 16) with $P^0({{\boldsymbol {r}}},t)=p^0(0,{{\boldsymbol {r}}},t)$. The result is identical to that obtained in Miksis & Ting (Reference Miksis and Ting1989). We shall see that conducting the analysis at the following order makes the contribution of the bubble–bubble interaction appear.

A.3.1. Order 1 – resolution at the microscopic scale

Resolution inside the bubble – at the next order, we get from (A 4) that the pressure and mass density are still uniform, with ${{\boldsymbol {\nabla }}}_{\mu }\, \hat {p}_{\mu }^1 = 0$ and

(A 26)\begin{equation} \left.\begin{gathered} \displaystyle\frac{\partial \hat{\rho}_{\mu}^1}{\partial t} +\frac{\left(\alpha + m\hat{\rho}_{\mu}^0\right)}{{\boldsymbol{\mathsf{r}}}_{eq}R^0}\,\text{div}_{\mu}\, \hat{\boldsymbol{u}}_{\mu}^{-1} +\left(\frac{m\hat{\rho}_{\mu}^1}{{\boldsymbol{\mathsf{r}}}_{eq}R^0}-\frac{\left(\alpha + m\hat{\rho}_{\mu}^0\right)R^1}{{\boldsymbol{\mathsf{r}}}_{eq}(R^0)^2}\right)\text{div}_{\mu}\, \hat{\boldsymbol{u}}_{\mu}^{-2} =0,\\ \displaystyle\boldsymbol{\nabla} \land \hat{\boldsymbol{u}}_{\mu}^{-1} = 0,\quad\hat{p}_{\mu}^1=\frac{\beta}{\alpha}(R^0)^{3(1-\gamma)}\hat{\rho}_{\mu}^1, \end{gathered}\right\} \end{equation}

where $(\alpha + m\hat {\rho }_{\mu }^0)={\alpha }/{(R^0)^3}$ and $\text {div}_{\mu }\, \hat {\boldsymbol {u}}_{\mu }^{-2}={3{\boldsymbol{\mathsf{r}}}_{eq}\dot {R}^0}/{m}$ from (A 12a–c). The system in $(\hat {\boldsymbol {u}}_{\mu }^{-1},\hat {p}_{\mu }^1,\hat {\rho }_{\mu }^1)$ is complemented with initial and boundary conditions coming from (A 6), of the form

(A 27)\begin{equation} \left.\begin{gathered} {\rm i.c.\quad at } t=0,\quad \hat{p}_{\mu}^1=\hat{\rho}_{\mu}^1=0, \quad \hat{\boldsymbol{u}}_{\mu}^{-1}=\textbf{{0}}, \quad R^0=0,\\ {\rm b.c.\quad for } |{{\boldsymbol{x}}}_{\mu}|=1,\quad \displaystyle \hat{p}_{\mu}^1=p_{\mu}^1, \quad \hat{\boldsymbol{u}}_{\mu}^{-1}\boldsymbol{\cdot} {{\boldsymbol{e}}}_r= \frac{{\boldsymbol{\mathsf{r}}}_{eq}}{m}\, \dot{R}^1. \end{gathered}\right\} \end{equation}

By integrating the balance of mass in (A 26) along with (A 27), as we have done at order 0 ((A 10a–c) and (A 11) providing (A 12a–c)) we get an ordinary differential equation on $\hat {\rho }_{\mu }^1$ which can be solved explicitly and leads to

(A 28a–c)\begin{equation} \hat{\rho}_{\mu}^1({{\boldsymbol{r}}},t)=-\frac{3\alpha}{m}\frac{R^1}{(R^0)^4},\quad \hat{p}_{\mu}^1({{\boldsymbol{r}}},t)=-\frac{3\beta}{m}\frac{R^1}{(R^0)^{3\gamma+1}},\quad \hat{\boldsymbol{u}}_{\mu}^{-1}({{\boldsymbol{x}}}_{\mu},{{\boldsymbol{r}}},t)=\frac{{\boldsymbol{\mathsf{r}}}_{eq}}{m} \dot{R}^1 |{{\boldsymbol{x}}}_{\mu}|\,{{\boldsymbol{e}}}_r. \end{equation}

We remark that (A 12a–c) along with (A 28a–c) correspond to the Taylor expansions up to $O(\varepsilon ^2)$ of $\hat {\rho }_{\mu }={\alpha }/{m} ({1}/{R^3}-1)$, $\hat {p}_{\mu }={\beta }/{\gamma m }({1}/{R^{3\gamma }} -1)$ and $\hat {\boldsymbol {u}}_{\mu }=({{\boldsymbol{\mathsf{r}}}_{eq}}/{m})\dot {R} |{{\boldsymbol {x}}}_{\mu }|\,{{\boldsymbol {e}}}_r$. Hence, order 1 does not modify the form of the solution obtained at order 0, and given in dimensional form in (2.7).

Resolution in the liquid – next, in the liquid, we have from (A 5)

(A 29)\begin{align} \left.\begin{gathered} \displaystyle \text{div}_{\mu}\, {\boldsymbol{u}}_{\mu}^{-1}=0, \quad {{\boldsymbol{\nabla}}}_{\mu}\, \land {\boldsymbol{u}}_{\mu}^{-1} = 0,\\ \displaystyle \frac{\partial{\boldsymbol{u}}_{\mu}^{-1}}{\partial t} +\frac{m}{{\boldsymbol{\mathsf{r}}}_{eq}R^0} \left(({\boldsymbol{u}}_{\mu}^{-2} \boldsymbol{\cdot} {{\boldsymbol{\nabla}}}_{\mu}\,) {\boldsymbol{u}}_{\mu}^{-1}+ ({\boldsymbol{u}}_{\mu}^{-1} \boldsymbol{\cdot} {{\boldsymbol{\nabla}}}_{\mu}\,) {\boldsymbol{u}}_{\mu}^{-2}-\frac{R^1}{R^0} ({\boldsymbol{u}}_{\mu}^{-2} \boldsymbol{\cdot} {{\boldsymbol{\nabla}}}_{\mu}\,) {\boldsymbol{u}}_{\mu}^{-2}\right)\ldots\\ \displaystyle\ldots-\frac{\dot{R}^0}{R^0}\,{{\boldsymbol{x}}}_{\mu} \boldsymbol{\cdot} {{\boldsymbol{\nabla}}}_{\mu}\,{\boldsymbol{u}}_{\mu}^{-1}+\frac{R^1 \dot{R}^0-R^0\dot{R}^1}{(R^0)^2}\,{{\boldsymbol{x}}}_{\mu} \boldsymbol{\cdot} {{\boldsymbol{\nabla}}}_{\mu}\,{\boldsymbol{u}}_{\mu}^{-2} \displaystyle=-\frac{1}{{\boldsymbol{\mathsf{r}}}_{eq}R^0} \left({{\boldsymbol{\nabla}}}_{\mu}\, p_{\mu}^0-\frac{R^1}{R^0} {{\boldsymbol{\nabla}}}_{\mu}\, p_{\mu}^1\right), \end{gathered} \right\} \end{align}

where $({\boldsymbol {u}}_{\mu }^{-2},p_{\mu }^0)$ are known from (A 15a,b), hence the above system is set on $({\boldsymbol {u}}_{\mu }^{-1},p_{\mu }^1)$. The first two equations in (A 29) together with the boundary condition (A 27) provide the same form of the velocity as that obtained at order 0. Using further ${\boldsymbol {u}}_{\mu }^{-1}$ in the third equation of (A 29), we obtain a differential equation in $p_{\mu }^1$ which can be integrated in space. This provides

(A 30a,b)\begin{align} {\boldsymbol{u}}_{\mu}^{-1}({{\boldsymbol{x}}}_{\mu},{{\boldsymbol{r}}},t)=\frac{{\boldsymbol{\mathsf{r}}}_{eq}}{m}\,\frac{\dot{R}^1}{|{{\boldsymbol{x}}}_{\mu}|^2}\, {{\boldsymbol{e}}}_r, \quad p_{\mu}^1({{\boldsymbol{x}}}_{\mu},{{\boldsymbol{r}}},t)\!=\!\tilde{P}^1+\frac{{\boldsymbol{\mathsf{r}}}_{eq}^2}{m} \left(\frac{R^1\ddot{R}^0\!+R^0\ddot{R}^1\!+4\dot{R}^0\dot{R}^1}{|{{\boldsymbol{x}}}_{\mu}|}- \frac{\dot{R}^0\dot{R}^1}{|{{\boldsymbol{x}}}_{\mu}|^4}\right). \end{align}

Again, we recognize the second contribution of the Taylor expansions of the pressure and of the velocity in (A 15a,b). However, we shall see that $\tilde {P}^1({{\boldsymbol {r}}},t)$ in (A 30), being the equivalent of $P^0({{\boldsymbol {r}}},t)=p^0(0,{{\boldsymbol {r}}},t)$ in (A 15a,b), differs from $p^1(0,{{\boldsymbol {r}}},t)$; specifically, it contains a contribution due to the arrangement of the bubbles within the array. Once $\tilde {P}^1$ is known, the Rayleigh–Plesset equation will be obtained by equating the pressure $p_{\mu }^1$ in (A 30) and $\hat {p}_{\mu }^1$ in (A 28a–c) at the interface $|{{\boldsymbol {x}}}_{\mu }|=1$; this requires some work and it will be done in the forthcoming (A 50).

A.3.2. Order 1 – resolution at the mesoscopic scale

The mesoscopic problem associated with $(p_{m}^1,{\boldsymbol {u}}_{m}^0)$ is given by (A 21) but now, we have established that $P^0({{\boldsymbol {r}}},t)=p^0(0,{{\boldsymbol {r}}},t)$ and ${\boldsymbol {U}}^{0\pm }({{\boldsymbol {r}}},t)={{\boldsymbol {u}}}^0(0^\pm ,{{\boldsymbol {r}}},t)$, (A 23a,b). It is convenient to denote ${{\boldsymbol {u}}}^0=u^0_x{{\boldsymbol {e}}}_x+{\boldsymbol {u}}_{{{\boldsymbol {r}}}}^0$ (with ${\boldsymbol {u}}_{{{\boldsymbol {r}}}}^0\boldsymbol{\cdot} {{\boldsymbol {e}}}_x=0$). From (A 24a–c) and (A 25a,b), $u_x^0$ is discontinuous at $x=0$ while ${\boldsymbol {u}}_{{{\boldsymbol {r}}}}^0$ is continuous with $\partial _t{\boldsymbol {u}}_{{{\boldsymbol {r}}}}^0=-{\boldsymbol {\nabla }}_{{{\boldsymbol {r}}}} p^0$ (and $p^0$ is continuous). Besides, from (A 20a,b) and (A 25a,b), we know that ${\boldsymbol {u}}_{m}^0 \sim _{{{{\boldsymbol {x}}}_{m}\to \textbf {{0}}}} {1}/{4{\rm \pi} }[\kern-1pt[ u^0_x ]\kern-1pt] \,{{{\boldsymbol {x}}}_{m}}/{ |{{\boldsymbol {x}}}_{m}|^3}$. Hence denoting

(A 31a,b)\begin{equation} {{\boldsymbol{u}}}^0(0^\pm,{{\boldsymbol{r}}},t)=u^0_x(0^\pm,{{\boldsymbol{r}}},t){{\boldsymbol{e}}}_x+{\boldsymbol{u}}_{{{\boldsymbol{r}}}}^0(0,{{\boldsymbol{r}}},t) \quad \text{and}\quad u^0_x(0^\pm,{{\boldsymbol{r}}},t)=\bar{u^0_x }(0,{{\boldsymbol{r}}},t)\pm\tfrac{1}{2}[\kern-1pt[ u^0_x ]\kern-1pt], \end{equation}

the problem (A 21) can be written

(A 32)\begin{align} \left.\begin{aligned} {\rm div}_{m}\, {\boldsymbol{u}}_{m}^{0}=\textbf{0},\quad \boldsymbol{\nabla} \land {\boldsymbol{u}}_{m}^{0} = \textbf{0},\quad \displaystyle\dfrac{\partial {\boldsymbol{u}}_{m}^{0}}{\partial t}=\dfrac{\partial {\boldsymbol{u}}_{{{\boldsymbol{r}}}}^0}{\partial t} (0,{{\boldsymbol{r}}},t)-{\boldsymbol{\nabla}}_{{m}}\, p_{m}^1,\\ \begin{aligned} \text{b.c.} & \displaystyle \lim_{x_{m}\to \pm\infty} {\boldsymbol{u}}_{m}^0=\left(\bar{u^0_x }(0,{{\boldsymbol{r}}},t)\pm\dfrac{1}{2} [\kern-1pt[ u^0_x ]\kern-1pt]\right){\boldsymbol{e}}_x+{\boldsymbol{u}}_{{{\boldsymbol{r}}}}^0(0,{{\boldsymbol{r}}},t), \\ & \displaystyle(p_{m}^1, {\boldsymbol{u}}_{m}^{0}) \text{ periodic on } \partial{\boldsymbol{\mathsf{Y}}},\quad {\boldsymbol{u}}_{m}^0 \underset{ {{\boldsymbol{x}}}_{m} \to \textbf{{0}}}{\sim} \dfrac{[\kern-1pt[ u^0_x ]\kern-1pt]}{4{\rm \pi}}\, \dfrac{{{\boldsymbol{x}}}_{m}}{ |{{\boldsymbol{x}}}_{m}|^3}. \end{aligned} \end{aligned}\right\} \end{align}

The above problem set in $(\partial _t({\boldsymbol {u}}_{m}^0-{\boldsymbol {u}}_{{{\boldsymbol {r}}}}^0(0,{{\boldsymbol {r}}},t)),p_{m}^1)$ is linear with respect to the macroscopic fields $\partial _t \bar {u^0_x }(0,{{\boldsymbol {r}}},t)$ and $\partial _t[\kern-1pt[ u^0_x ]\kern-1pt]$. It follows that $(p_{m}^1,{\boldsymbol {u}}_{m}^0)$ read

(A 33)\begin{equation} \left.\begin{array}{c} \displaystyle p_{m}^1({{\boldsymbol{x}}}_{m},{{\boldsymbol{r}}},t)=-\dfrac{\partial \bar{u_x^0 }}{\partial t}(0,{{\boldsymbol{r}}},t)x_{m} -\dfrac{1}{4{\rm \pi}}\dfrac{\partial [\kern-1pt[ u_x^0 ]\kern-1pt]}{\partial t}\, G({{\boldsymbol{x}}}_{m})+P^1({{\boldsymbol{r}}},t),\\ \displaystyle {\boldsymbol{u}}_{m}^{0}({{\boldsymbol{x}}}_{m},{{\boldsymbol{r}}},t)={\boldsymbol{u}}_{{{\boldsymbol{r}}}}^0(0,{{\boldsymbol{r}}},t)+\bar{u_x^0 }(0,{{\boldsymbol{r}}},t){{\boldsymbol{e}}}_x +\dfrac{[\kern-1pt[ u_x^0 ]\kern-1pt]}{4{\rm \pi}}\, {\boldsymbol{\nabla}}_{{m}}\, G({{\boldsymbol{x}}}_{m}), \end{array}\right\}\end{equation}

where $P^1({{\boldsymbol {r}}},t)$ is an unknown (at this stage) constant at the mesoscopic scale and where the field $G$ is the Green's function of the periodic cell ${\mathcal {Y}}$ satisfying $G$, ${\boldsymbol {\nabla }}_{{m}}\, G$ periodic in $\partial {\boldsymbol{\mathsf{Y}}}$ and

(A 34)\begin{equation} \displaystyle\Delta_{m}\, G=0,\quad \text{with}\ \displaystyle \lim_{x_{m}\to \pm\infty} (G-2{\rm \pi} |x_{m}|)=0, \quad G \underset{ {{\boldsymbol{x}}}_{m} \to \textbf{{0}}}{\sim} -\frac{1}{ |{{\boldsymbol{x}}}_{m}|}. \end{equation}

It is worth noting that (A 33) leads to

(A 35)\begin{equation} \lim_{x_{m}\to\pm\infty}{\boldsymbol{u}}_{m}^{0}({{\boldsymbol{x}}}_{m},{{\boldsymbol{r}}},t)={{\boldsymbol{u}}}^0(0^\pm,{{\boldsymbol{r}}},t), \end{equation}

as imposed by the macro-to-meso matching conditions (A 23a,b). Eventually, we shall see that an important effective parameter resulting from the Green's function problem is

(A 36)\begin{equation} \delta=\lim_{{{\boldsymbol{x}}}_{m}\to \textbf{{0}}} \left(G({{\boldsymbol{x}}}_{m})+\frac{1}{ |{{\boldsymbol{x}}}_{m}|}\right). \end{equation}

A.3.3. Order 1 – derivation of the effective transmission conditions

Jump condition on the pressure – equipped with the expression (A 33) of $p_{m}^1$, the jump conditions at order 1 in the pressure can now be written. The macro-to-meso matching condition at order 1 in the pressure is given by

(A 37a,b)\begin{equation} p^0({{\boldsymbol{x}}},t)+\varepsilon p^1({{\boldsymbol{x}}},t)\underset{x\rightarrow0^\pm, x_{m}\rightarrow \pm\infty}{\sim}p_{m}^0({{\boldsymbol{x}}}_{m},{{\boldsymbol{r}}},t)+\varepsilon p_{m}^1({{\boldsymbol{x}}}_{m},{{\boldsymbol{r}}},t), \quad {{\boldsymbol{x}}}=\varepsilon{{\boldsymbol{x}}}_{m}. \end{equation}

Making the Taylor expansion of the left-hand side in (A 37) and passing to the limit when $x_{m}\to \pm \infty$ gives the order-0 matching condition (A 23a,b) and at order 1, we get

(A 38)\begin{equation} p^1(0^\pm,{{\boldsymbol{r}}},t)=\lim_{x_{m}\to \pm\infty}\left(p_{m}^1({{\boldsymbol{x}}}_{m},{{\boldsymbol{r}}},t)-x_{m}\frac{\partial p^0}{\partial x}(0^\pm,{{\boldsymbol{r}}},t) \right).\end{equation}

At the macroscopic scale, the Euler equation (A 24a–c) imposes

(A 39)\begin{equation} \frac{\partial p^0}{\partial x}(0^\pm,{{\boldsymbol{r}}},t)=-\frac{\partial \bar{u^0_x }}{\partial t}(0,{{\boldsymbol{r}}},t)\mp\frac{1}{2}\frac{\partial [\kern-1pt[ u^0_x ]\kern-1pt]}{\partial t}. \end{equation}

By using the above relation in (A 38) along with (A 33) and (A 34), we get the continuity of the pressure at the next order

(A 40)\begin{equation} p^1(0^\pm,{{\boldsymbol{r}}},t)=P^1({{\boldsymbol{r}}},t),\quad \Rightarrow [\kern-1pt[ p^1 ]\kern-1pt]=0.\end{equation}

We also obtain $P^1$, which contributes to $p_{m}^1$ in the liquid (A 33), and we shall see now that $P^1$ differs from $\tilde {P}^1$, which contributes to $p_{\mu }^1$ in the bubble (A 30), as previously said, this is the important result at this order. To do so, we have to inspect the meso-to-micro matching conditions.

Jump condition on the velocity – we use the meso-to-micro matching condition on the velocity which formally reads

(A 41)\begin{align} {\boldsymbol{u}}_{m}^{0}({{\boldsymbol{x}}}_{m},{{\boldsymbol{r}}},t)+\varepsilon {\boldsymbol{u}}_{m}^{1}({{\boldsymbol{x}}}_{m},{{\boldsymbol{r}}},t) +\cdots\underset{{{\boldsymbol{x}}}_{m}\rightarrow\textbf{{0}},|{{\boldsymbol{x}}}_{\mu}|\to \infty}{\sim} \,\frac{1}{\varepsilon^2}{\boldsymbol{u}}_{\mu}^{-2}({{\boldsymbol{x}}}_{\mu},{{\boldsymbol{r}}},t)+\frac{1}{\varepsilon}{\boldsymbol{u}}_{\mu}^{-1}({{\boldsymbol{x}}}_{\mu},{{\boldsymbol{r}}},t)+\cdots, \end{align}

where up to the second order in $\varepsilon$, ${{\boldsymbol {x}}}_{m}\sim \varepsilon {\boldsymbol{\mathsf{r}}}_{eq}(R^0+\varepsilon R ^1){{\boldsymbol {x}}}_{\mu }$, see (A 3). Using ${\boldsymbol {u}}_{\mu }^{-2}$ in (A 15a,b) and ${\boldsymbol {u}}_{\mu }^{-1}$ in (A 30), we recover the singular behaviour of ${\boldsymbol {u}}_{m}^0$ in (A 20a,b) but also that of ${\boldsymbol {u}}_{m}^1$ which reads

(A 42)\begin{equation} {\boldsymbol{u}}_{m}^{1}({{\boldsymbol{x}}}_{m},{{\boldsymbol{r}}},t) \underset{{{\boldsymbol{x}}}_{m}\rightarrow\textbf{{0}}}{\sim} \,\frac{{\boldsymbol{\mathsf{r}}}_{eq}^3}{m}\left(2R^0R^1\dot{R}^0+(R^0)^2\dot{R}^1\right) \frac{{{\boldsymbol{x}}}_{m}}{|{{\boldsymbol{x}}}_{m}|^3}.\end{equation}

We are now able to derive the jump conditions on the velocity at order 1. As we have done at order 0 to get (A 25a,b), we start with the balance of mass at order 0 at the mesoscopic scale. From (A 4) and (A 18a,b), it reads

(A 43)\begin{equation} \frac{\partial p_{m}^0}{\partial t}+{\rm div}_{m}\,{\boldsymbol{u}}_{m}^1+{\rm div}_{{{\boldsymbol{r}}}}\,{\boldsymbol{u}}_{m}^0=0,\end{equation}

that we shall integrate on ${\mathcal {Y}}_r={\mathcal {Y}}^*/ B_r$ where ${\mathcal {Y}}^*=(-x_{m}^*,x_{m}^*)\times (-\frac {1}{2},\frac {1}{2})^2$ and $B_r$ a ball of radius $r$ then pass to the limit $r\to 0$ and $x_{m}^*\rightarrow +\infty$. Firstly, we know that $p_{m}^0=p^0(0,{{\boldsymbol {r}}},t)$ from (A 23a,b) while ${\boldsymbol {u}}_{m}^0$ is given by (A 33) hence ${\rm div}_{{{\boldsymbol {r}}}}\,{\boldsymbol {u}}_{m}^0={\rm div}_{{{\boldsymbol {r}}}}\,{\boldsymbol {u}}_{{{\boldsymbol {r}}}}^0(0,{{\boldsymbol {r}}},t)+({1}/{4{\rm \pi} })\,{\boldsymbol {\nabla }}_{{{\boldsymbol {r}}}} [\kern-1pt[ u_x^0 ]\kern-1pt] \boldsymbol{\cdot} {\boldsymbol {\nabla }}_{{m}}\, G({{\boldsymbol {x}}}_{m})$. From (A 34), $G$ is an even function of ${{\boldsymbol {x}}}_{m}\boldsymbol {\cdot } {{\boldsymbol {e}}}_y$ and of ${{\boldsymbol {x}}}_{m}\boldsymbol {\cdot } {{\boldsymbol {e}}}_z$ hence $\int _{{\mathcal {Y}}_r}{\boldsymbol {\nabla }}_{{{\boldsymbol {r}}}} [\kern-1pt[ u_x^0 ]\kern-1pt] \boldsymbol {\cdot } {\boldsymbol {\nabla }}_{{m}}\, G({{\boldsymbol {x}}}_{m})\,\text {d}{{\boldsymbol {x}}}_{m}=0$. Using (A 24a–c), it results that $\int _{{\mathcal {Y}}_r}(\partial _t p_{m}^0+{\rm div}_{{{\boldsymbol {r}}}}\,{\boldsymbol {u}}_{m}^0)\,\text {d}{{\boldsymbol {x}}}_{m} =-{\mathcal {V}}({\mathcal {Y}}_r){\partial u^0_x}/{\partial x}(0,{{\boldsymbol {r}}},t)$ with ${\mathcal {V}}({\mathcal {Y}}_r)=(2x_{m}^*- ({4{\rm \pi} }/{3})r^3)$. The result of the integration of (A 43) reads

(A 44)\begin{align} &\displaystyle \int \left({{\boldsymbol{u}}_{m}^1\boldsymbol{\cdot} {{\boldsymbol{e}}}_x}_{|x_{m}^*}-x_{m}^*\frac{\partial u^0_x}{\partial x}(0,{{\boldsymbol{r}}},t)\right)\text{d} {\boldsymbol{r}}_{m}\nonumber\\ &\quad -\int \left({{\boldsymbol{u}}_{m}^1\boldsymbol{\cdot} {{\boldsymbol{e}}}_x}_{|-x_{m}^*}+x_{m}^*\frac{\partial u^0_x}{\partial x}(0,{{\boldsymbol{r}}},t)\right)\text{d} {\boldsymbol{r}}_{m}=\cdots \int_{\partial B_r} {\boldsymbol{u}}_{m}^{1}({{\boldsymbol{x}}}_{m},{{\boldsymbol{r}}},t)\boldsymbol{\cdot}{{\boldsymbol{e}}}_r\,\text{d}{{\boldsymbol{x}}}_{m}. \end{align}

It is now sufficient to use the matching condition written as

(A 45)\begin{equation} {{\boldsymbol{u}}}^1(0^\pm,{{\boldsymbol{r}}},t)=\lim_{x_{m}^*\to +\infty}\left({{\boldsymbol{u}}_{m}^1}_{|\pm x_{m}^*}\mp x_{m}^*\frac{\partial {\boldsymbol{u}}_{m}^0}{\partial x}(0^\pm,{{\boldsymbol{r}}},t) \right),\end{equation}

and the singular behaviour (A 42) of ${\boldsymbol {u}}_{m}^1$ yielding $\int _{\partial B_r} {\boldsymbol {u}}_{m}^{1}\boldsymbol {\cdot }{{\boldsymbol {e}}}_r={4{\rm \pi} {\boldsymbol{\mathsf{r}}}_{eq}^3}/{m}(2R^0R^1\dot {R}^0+(R^0)^2\dot {R}^1)$ to get

(A 46)\begin{equation} [\kern-1pt[ u_x^1 ]\kern-1pt]=\frac{4{\rm \pi}{\boldsymbol{\mathsf{r}}}_{eq}^3}{m}\left(2R^0R^1\dot{R}^0+(R^0)^2\dot{R}^1\right).\end{equation}

Again, we recognize a Taylor expansion with respect to the bubble radius of the jump on the normal velocity at the dominant order, see (A 25a,b).

A.3.4. The Rayleigh–Plesset equation at order 1

To derive the Rayleigh–Plesset equation at order 1, we use the meso-to-micro matching conditions on the pressure. It formally reads

(A 47)\begin{equation} p_{m}^{0}({{\boldsymbol{r}}},t)+\varepsilon p_{m}^{1}({{\boldsymbol{x}}}_{m},{{\boldsymbol{r}}},t)+\cdots \underset{{{\boldsymbol{x}}}_{m}\rightarrow\textbf{{0}},|{{\boldsymbol{x}}}_{\mu}|\rightarrow +\infty}{\sim} \,p_{\mu}^{0}({{\boldsymbol{x}}}_{\mu},{{\boldsymbol{r}}},t)+\varepsilon p_{\mu}^{1}({{\boldsymbol{x}}}_{\mu},{{\boldsymbol{r}}},t) +\cdots, \end{equation}

where, up to the first order in $\varepsilon$, ${{\boldsymbol {x}}}_{m}\sim \varepsilon {\boldsymbol{\mathsf{r}}}_{eq}R^0{{\boldsymbol {x}}}_{\mu }$, see (A 3). All the fields are known from (A 15a,b), (A 20a,b), (A 23a,b), (A 30), (A 33) and (A 40), but in (A 30) $\tilde {P}^1$ is still unknown. Owing to ${{\boldsymbol {x}}}_{m}\sim \varepsilon {\boldsymbol{\mathsf{r}}}_{eq}R^0{{\boldsymbol {x}}}_{\mu }$, it results that

(A 48)\begin{equation} \frac{{\boldsymbol{\mathsf{r}}}_{eq}^2}{m} \frac{R^0\ddot{R}^0+2(\dot{R}^0)^2}{|{{\boldsymbol{x}}}_{\mu}|}+\varepsilon \tilde{P}^1 +\cdots \underset{|{{\boldsymbol{x}}}_{\mu}|\rightarrow+\infty}{\sim} \frac{\partial_t [\kern-1pt[ u_x^0 ]\kern-1pt]}{4{\rm \pi}{\boldsymbol{\mathsf{r}}}_{eq}R^0|{{\boldsymbol{x}}}_{\mu}|} +\varepsilon \left(p^1_{|x=0} -\frac{\partial_t[\kern-1pt[ u_x^0 ]\kern-1pt]}{4{\rm \pi}}\, \delta\right)+\cdots. \end{equation}

With $[\kern-1pt[ u_x^0 ]\kern-1pt] ={4{\rm \pi} {\boldsymbol{\mathsf{r}}}_{eq}^3}/{m}(R^0)^2\dot {R}^0$ from (A 25a,b), it is easy to see that the above matching is consistent at the order 0 and at the order 1 it provides $\tilde {P}^1$ of the form

(A 49)\begin{equation} \tilde{P}^1=p^1_{|x=0} -\frac{\delta {\boldsymbol{\mathsf{r}}}_{eq}^3}{m}\frac{\partial }{\partial t}\left((R^0)^2\dot{R}^0\right). \end{equation}

Once $\tilde {P}^1$ is known, we use the equality of the pressures in the gas (A 28a–c) and in the liquid (A 30) at the microscale at $|{{\boldsymbol {x}}}_{\mu }|=1$. We deduce the Rayleigh–Plesset equation at the order 1

(A 50)\begin{equation} R^1\ddot{R}^0+R^0\ddot{R}^1+3\dot{R}^0\dot{R}^1 +3\frac{\beta}{{\boldsymbol{\mathsf{r}}}_{eq}^2}\frac{R^1}{(R^0)^{1+3\gamma}}-\delta {\boldsymbol{\mathsf{r}}}_{eq}\frac{\partial }{\partial t}\left((R^0)^2\dot{R}^0\right) =-\frac{m}{{\boldsymbol{\mathsf{r}}}_{eq}^2}p^1_{|x=0}.\end{equation}

The new contribution of order 1 is visible in the above relation which appears to differ from the simple Taylor expansion of (A 16) by the contribution of the term proportional to $\delta$.