3.1. Inviscid airflow (region 1)

In this region we have a uniform flow which is perturbed by an acoustic wave. For simplicity, we shall assume that the acoustic wave propagates along the main flow with the front of the wave being perpendicular to the plate surface. Consequently, we represent the fluid-dynamic functions in region 1 in the form of the asymptotic expansions

(3.1)\begin{equation} \left.\begin{gathered} u = 1 + \sigma _{\mu }^{{-}1/2} Re^{{-}1/8} \chi u_1 (t_{{\ast} } , x_1) + \cdots , \\ p = \sigma _{\mu }^{{-}1/2} Re^{{-}1/8} \chi p_1 (t_{{\ast} } , x_1) + \cdots , \\ h = \frac{1}{(\gamma - 1) M_{\infty }^2} + \sigma _{\mu }^{{-}1/2} Re^{{-}1/8} \chi h_1 (t_{{\ast} } , x_1) + \cdots , \\ \rho = 1 + \sigma _{\mu }^{{-}1/2} Re^{{-}1/8} \chi \rho _1 (t_{{\ast} } , x_1) + \cdots , \end{gathered}\right\} \end{equation}

with the independent variables $t_{\ast }$, $x_1$ defined by

(3.2a,b)\begin{equation} t = \sigma _{\mu }^{{-}1} Re^{{-}1/4} t_{{\ast} } , \quad x = 1 + \sigma _{\mu }^{{-}1} Re^{{-}1/4} x_1 . \end{equation}

Note that the viscosity coefficient is zero in the inviscid region. In (3.1), the amplitude of acoustic wave $\sigma _{\mu }^{-1/2} Re^{-1/8} \chi$ is chosen such that for $\chi = O (1)$ the perturbations appear to be nonlinear in the Stokes layer (region 3 in figure 1). We shall see that the frequency of the interfacial instability waves is an order $O (\sigma _{\mu } Re^{1/4})$ quantity, which explains the choice of the scaling factor in the equation for $t$ in (3.2a,b). Since the speed of propagation of acoustic waves is finite, the wavelength appears to be $O (\sigma _{\mu }^{-1} Re^{-1/4})$ which is used to scale $x$ in (3.2a,b).

Substituting (3.1) and (3.2a,b) into the Navier–Stokes equations (2.2) and assuming that $\sigma _{\mu }^{-1/2}Re^{-1/8} \chi$ is small, results in the following set of linearised Euler equations:

(3.3)\begin{equation} \left.\begin{gathered} \frac{\partial u_1}{\partial t_{{\ast} }} + \frac{\partial u_1}{\partial x_1} ={-} \frac{\partial p_1}{\partial x_1} , \\ \frac{\partial h_1}{\partial t_{{\ast} }} + \frac{\partial h_1}{\partial x_1} = \frac{\partial p_1}{\partial t_{{\ast} }} + \frac{\partial p_1}{\partial x_1} ,\\ \frac{\partial \rho _1}{\partial t_{{\ast} }} + \frac{\partial u_1}{\partial x_1} + \frac{\partial \rho _1}{\partial x_1} = 0 , \\ h_1 = \frac{\gamma }{\gamma - 1} p_1 - \frac{1}{(\gamma - 1) M_{\infty }^2} \rho _1 . \end{gathered}\right\} \end{equation}

These may be reduced to a single equation for the pressure

(3.4)\begin{equation} M_{\infty }^2 \frac{\partial ^2 p_1}{\partial t_{{\ast} }^2} + (M_{\infty }^2 - 1) M_{\infty }^2 \frac{\partial ^2 p_1}{\partial x_1^2} + 2 M_{\infty }^2 \frac{\partial ^2 p_1}{\partial t_{{\ast} } \partial x_1} = 0 . \end{equation}

By definition (3.3), the enthalpy $h_1$ depends on $M_{\infty }$, hence the factor of $M_{\infty }$ appears in the pressure governing equation (3.4).

The solution to (3.4) representing a monochromatic acoustic wave is written as

(3.5)\begin{equation} p_1 = p_a (t_{{\ast} } , x_1) = \alpha \sin (\omega t_{{\ast} } + k_a x_1) , \end{equation}

with the other fluid-dynamic functions being

(3.6a–c)\begin{equation} u_1 = M_{\infty } p_a , \quad \rho _1 = M_{\infty }^2 p_a , \quad h_1 = p_a . \end{equation}

Here, $k_a$ in the wavenumber of the acoustic wave given by

(3.7)\begin{equation} k_a ={-} \frac{M_{\infty }}{M_{\infty } + 1} \omega . \end{equation}

We shall show how this slow acoustic wave leads to the formation of a slow motion air layer, the so-called Stoke layer, in the boundary layer.

3.2. Main part of the boundary layer (region 2)

Now we shall see how the acoustic wave penetrates the boundary layer. When there is no acoustic wave, the flow inside the boundary layer is given by the compressible version of the Blasius solution

(3.8)\begin{equation} \left.\begin{gathered} u = U_0 (x , y_2) + \cdots , \quad v = Re^{{-}1/2} V_0 (x , y_2) + \cdots , \\ p = Re^{{-}1/2} P_0 (x , y_2) + \cdots , \quad \rho = \rho_0 (x , y_2) + \cdots , \\ h = h_0 (x , y_2) + \cdots , \quad \mu = \mu _0 (x , y_2) + \cdots , \end{gathered}\right\} \end{equation}

where

(3.9)\begin{equation} y_2 = Re^{1/2} y . \end{equation}

The behaviour of the leading-order terms in (3.8) were studied by various authors (see, for e.g. Ruban Reference Ruban2018). It is known that by substituting (3.8) into the Navier–Stokes equations (2.2) a boundary-value problem is derived which admits a self-similar solution for certain conditions. In the presence of the acoustic wave, asymptotic expansions (3.8) become

(3.10)\begin{equation} \left.\begin{gathered} u = U_0 (x , y_2) + \sigma _{\mu }^{{-}1/2} Re^{{-}1/8} \chi u_2 (t_{{\ast} } , x_1 , y_2) +\cdots ,\\ v = \sigma ^{1/2} Re^{{-}3/8} \chi v_2 (t_{{\ast} } , x_1 , y_2) + \cdots , \\ p = \sigma _{\mu }^{{-}1/2} Re^{{-}1/8} \chi p_2 (t_{{\ast} } , x_1 , y_2) + \cdots , \\ \rho = \rho_0 (x , y_2) + \sigma _{\mu }^{{-}1/2} Re^{{-}1/8} \chi \rho _2 (t_{{\ast} } , x_1 , y_2) + \cdots , \\ h = h_0 (x , y_2) + \sigma _{\mu }^{{-}1/2} Re^{{-}1/8} h_2 \chi (t_{{\ast} } , x_1 , y_2) + \cdots , \\ \mu = \mu _0 (x , y_2) + \sigma _{\mu }^{{-}1/2} Re^{{-}1/8} \chi \mu _2 (t_{{\ast} } , x_1 , y_2) + \cdots . \end{gathered}\right\} \end{equation}

Substituting (3.10) into the Navier–Stokes equations (2.2) and assuming that

(3.11)\begin{equation} \frac{Re^{{-}1/4}}{\sigma _{\mu }} \ll 1 , \end{equation}

we obtain the following set of equations for region 2:

(3.12)\begin{equation} \left.\begin{gathered} \frac{\partial u_2}{\partial t_{{\ast} }} + U_0 \frac{\partial u_2}{\partial x_1} + v_2 \frac{\partial U_0}{\partial y_2} ={-} \frac{1}{\rho _0} \frac{\partial p_2} {\partial x_1} , \\ \frac{\partial p_2}{\partial y_2} = 0 , \\ \frac{\partial h_2}{\partial t_{{\ast} }} + U_0 \frac{\partial h_2}{\partial x_1} + v_2 \frac{\partial h_0}{\partial y_2} = \frac{1}{\rho _0} \left( \frac{\partial p_2} {\partial t_{{\ast} }} + U_0 \frac{\partial p_2}{\partial x_1} \right) , \\ \frac{\partial \rho _2}{\partial t_{{\ast} }} + U_0 \frac{\partial \rho _2}{\partial x_1} + \rho _0 \frac{\partial u_2}{\partial x_1} + \frac{\partial \rho _0 v_2} {\partial y_2} = 0 , \\ \rho _0 h_2 + h_0 \rho _2 = \frac{\gamma }{\gamma - 1} p_2 . \end{gathered}\right\} \end{equation}

These equations show that the flow in region 2 is inviscid. Therefore, instead of the no-slip conditions we have to pose the impermeability condition

(3.13)\begin{equation} v_2 = 0 \quad \text{at} \ y_2 = 0 . \end{equation}

Notice that the pressure does not change across the boundary layer. This allows us to conclude that it coincides with the pressure in region 1

(3.14)\begin{equation} p_2 = \alpha \sin (\omega t_{{\ast} } + k_a x_1) = \frac{\alpha }{2 {\rm i}} \exp({{\rm i} (\omega t_{{\ast} } + k_a x_1)}) + ({\rm c.c.}) . \end{equation}

Since the perturbations in region 2 appear in response to the acoustic wave in region 1, we shall seek the solution to (3.12) in the form

(3.15)\begin{equation} (u_2 , v_2 , h_2 , \rho _2) = (\breve{u}_2 , \breve{v}_2 , \breve{h}_2 , \breve{\rho }_2) \exp({{\rm i} (\omega t_{{\ast} } + k_a x_1)}) + ({\rm c.c.}) , \end{equation}

where (c.c.) denotes the complex conjugate of the expression in front of it. With (3.15), (3.12) turn into

(3.16a)$$\begin{gather} ({\rm i} \omega + {\rm i} k_a U_0) \breve{u}_2 + \breve{v}_2 \frac{\partial U_0}{\partial y_2} ={-} \frac{1}{\rho _0} \frac{\alpha }{2} k_a , \end{gather}$$

(3.16b)$$\begin{gather}({\rm i} \omega + {\rm i} k_a U_0) \breve{h}_2 + \breve{v}_2 \frac{\partial h_0}{\partial y_2} = \frac{1}{\rho _0} \frac{\alpha }{2} (\omega + k_a U_0) , \end{gather}$$

(3.16c)$$\begin{gather}({\rm i} \omega + {\rm i} k_a U_0) \breve{\rho }_2 + {\rm i} k_a \rho _0 \breve{u}_2 + \frac{\partial \rho _0 \breve{v}_2}{\partial y_2} = 0 , \end{gather}$$

(3.16d)$$\begin{gather}\rho _0 \breve{h}_2 + h_0 \breve{\rho }_2 = \frac{\gamma }{\gamma - 1} \frac{\alpha }{2 {\rm i}} . \end{gather}$$

To see what happens at the bottom of region 2, we set $y_2 = 0$ in (3.16a). Using the impermeability condition (3.13) and the fact that in the Blasius boundary layer, $U_0 = 0$ at $y_2 = 0$, we find that

(3.17)\begin{equation} \breve{u}_2 = \frac{{\rm i} \alpha k_a}{2 \rho _w \omega } \quad \text{at} \ y_2 = 0 , \end{equation}

where $\rho _w$ is the gas density at the plate surface in the Blasius boundary layer. Substitution of (3.17) back into (3.15) allows us to conclude that the longitudinal velocity oscillations at the bottom of region 2

(3.18)\begin{equation} \left.u_2\right|_{y_2=0} ={-} \frac{\alpha k_a}{\rho _w \omega } \sin (\omega t_{{\ast} } + k_a x_1) . \end{equation}

Since (3.18) does not satisfy the no-slip condition, we need to introduce a viscous sublayer closer to the plate surface.

3.3. Stokes layer (region 3)

The flow in region 3 is viscous, which means that the following terms should be in balance in the longitudinal momentum equation (2.2a):

(3.19)\begin{equation} \frac{\partial u}{\partial t} \sim \frac{\partial p}{\partial x} \sim \frac{1} {Re} \frac{\partial ^2 u}{\partial y^2} . \end{equation}

We know that

(3.20a–c)\begin{equation} t \sim \sigma _{\mu }^{{-}1} Re^{{-}1/4} , \quad \Delta x \sim \sigma _{\mu }^{{-}1} Re^{{-}1/4} , \quad \Delta p \sim \sigma _{\mu }^{{-}1/2} Re^{{-}1/8} \chi . \end{equation}

Combining (3.20a–c) with (3.19), one can easily find that in region 3

(3.21a,b)\begin{equation} u \sim \sigma _{\mu }^{{-}1/2} Re^{{-}1/8} \chi , \quad y \sim \sigma _{\mu }^{{-}1/2} Re^{{-}5/8} . \end{equation}

The lateral velocity component can be now estimated based on the usual balance in the continuity equation (2.2d)

(3.22)\begin{equation} \frac{\partial u}{\partial x} \sim \frac{\partial v}{\partial y} . \end{equation}

We find that

(3.23)\begin{equation} v \sim Re^{{-}1/2} \chi . \end{equation}

Guided by (3.20a–c)–(3.23) we represent the solutions in region 3 in the form

(3.24)\begin{equation} \left.\begin{gathered} u = \sigma _{\mu }^{{-}1/2} Re^{{-}1/8} \lambda y_3 + \sigma _{\mu }^{{-}1/2} Re^{{-}1/8} \chi u_3 (t_{{\ast} } , x_1 , y_3) + \cdots ,\\ v = Re^{{-}1/2} \chi v_3 (t_{{\ast} } , x_1 , y_3) + \cdots ,\\ p = \sigma _{\mu }^{{-}1/2} Re^{{-}1/8} \chi p_3 (t_{{\ast} } , x_1 , y_3) + \cdots ,\\ \rho = \rho _w + \cdots , \quad \mu = \mu _w + \cdots , \\ y_3 = \sigma _{\mu }^{1/2} Re^{5/8} y , \end{gathered}\right\} \end{equation}

where $\lambda$ is the dimensionless skin friction coefficient. Its value is found numerically with the self-similar solution of the Blasius boundary-layer equations. If we substitute (3.24) into the longitudinal momentum equation (2.2a) and use assumption (3.11) again, then we will see that $u_3$ satisfies the following equation:

(3.25)\begin{equation} \rho _w \frac{\partial u_3}{\partial t_{{\ast} }} ={-} \frac{\partial p_3}{\partial x_1} + \mu _w \frac{\partial ^2 u_3}{\partial y_3^2} . \end{equation}

It further follows from the lateral momentum equation (2.2b) that

(3.26)\begin{equation} \frac{\partial p_3}{\partial y_3} = 0 , \end{equation}

which proves that the pressure in the Stokes layer coincides with the pressure (3.14) in region 2

(3.27)\begin{equation} p_3 = \alpha \sin (\omega t_{{\ast} } + k_a x_1) . \end{equation}

The boundary conditions for (3.25) are the condition of matching with the solution (3.18) in region 2

(3.28)\begin{equation} \left. u_3\right|_{y_3 = \infty } ={-} \frac{\alpha k_a}{\rho _w \omega } \sin (\omega t_{{\ast} } + k_a x_1) , \end{equation}

and the no-slip condition

(3.29)\begin{equation} u_3 = 0 \quad \text{at} \ y_3 = 0 . \end{equation}

The solution of the boundary-value problem (3.25)–(3.29) is written as

(3.30)\begin{align} u_3 ={-} \frac{\alpha k_a}{\rho _w \omega } \left[ \sin (\omega t_{{\ast} } + k_a x_1) - \exp\left({-\sqrt{\frac{\rho _w \omega }{2 \mu _w}} \,y_3}\right) \sin \left( \omega t_{{\ast} } + k_a x_1 - \sqrt{\frac{\rho _w \omega }{2 \mu _w}} \,y_3 \right) \right] . \end{align}

3.4. Film flow

In order to find the form of the solution in the film, one needs to use the tangential stress condition on the interface. As mentioned before, we assume the film thickness to be an order $O (L Re^{-5/8})$ quantity, in which case the interface may be defined by the equation

(3.31)\begin{equation} \hat{y} = L Re^{{-}5/8} H (t_{{\ast} } , x_1) . \end{equation}

The tangential stress produced by the airflow in region 3 is calculated, in dimensional variables, as

(3.32)\begin{equation} \hat{\tau } = \mu _{\infty } \mu _w \frac{\partial \hat{u}}{\partial \hat{y}} . \end{equation}

Using the asymptotic expansion of $u$ in (3.24) we can express (3.32) as

(3.33)\begin{equation} \hat{\tau } = \mu _{\infty } \mu _w \left( \frac{V_{\infty } \sigma _{\mu }^{{-}1/2} Re^{{-}1/8}} {L \sigma _{\mu }^{{-}1/2} Re^{{-}5/8}} \lambda + \frac{V_{\infty } \sigma _{\mu }^{{-}1/2} Re^{{-}1/8}} {L \sigma _{\mu }^{{-}1/2} Re^{{-}5/8}} \chi \frac{\partial u_3}{\partial y_3} + \cdots \right) . \end{equation}

It is easily found from (3.30) that, on the interface,

(3.34)\begin{equation} \left.\frac{\partial u_3}{\partial y_3}\right|_{y_3=0} ={-} \frac{\alpha k_a} {\sqrt{\mu _w \rho _w \omega }} \sin \left( \omega t_{{\ast} } + k_a x_1 + {\rm \pi}/ 4 \right) . \end{equation}

Let us now consider the shear stress in the film

(3.35)\begin{equation} \hat{\tau } = \left.\frac{\mu _{\infty }}{\sigma _{\mu }} \frac{\partial \hat{u}} {\partial \hat{y}}\right|_{\hat{y}=\hat{H}} . \end{equation}

If we define the lateral coordinate as

(3.36)\begin{equation} \hat{y} = L Re^{{-}5/8} y^{\prime } , \end{equation}

then (3.35) assumes the form

(3.37)\begin{equation} \hat{\tau } = \left.\frac{\mu _{\infty }}{\sigma _{\mu } L Re^{{-}5/8}} \frac{\partial \hat{u}} {\partial y^{\prime }}\right|_{y^{\prime } =H} . \end{equation}

Comparing (3.37) with (3.33) we can conclude that the asymptotic expansion of $\hat {u}$ in the film should be written as

(3.38)\begin{equation} u = \frac{\hat{u}}{V_{\infty }} = \sigma _{\mu } Re^{{-}1/8} \mu _w \lambda y^{\prime } + \sigma _{\mu } Re^{{-}1/8} \chi u_f^{\prime } (t_{{\ast} } , x_1 , y^{\prime }) + \cdots . \end{equation}

Now, using the continuity equation (2.2d) one can easily predict the form of the asymptotic expansion for the lateral velocity component

(3.39)\begin{equation} v = \sigma _{\mu }^{2} Re^{{-}1/2} \chi v_f^{\prime } (t_{{\ast} } , x_1 , y^{\prime }) + \cdots . \end{equation}

As far as the pressure is concerned, of course, it has the same form as in region 3

(3.40)\begin{equation} p = \sigma _{\mu }^{{-}1/2} Re^{{-}1/8} \chi p_f^{\prime } (t_{{\ast} } , x_1 , y^{\prime }) + \cdots . \end{equation}

Substitution of (3.38)–(3.40) and (3.36) into the longitudinal momentum equation (2.2a) reduces it to

(3.41)\begin{equation} \frac{\partial ^2 u_f^{\prime }}{\partial y^{\prime 2}} = 0 . \end{equation}

Equation (3.41) should be solved with the no-slip condition on the plate surface

(3.42)\begin{equation} u_f^{\prime } = 0 \quad \text{at} \ y^{\prime } = 0 , \end{equation}

and the shear-stress balance condition at the interface

(3.43)\begin{equation} \left.\frac{\partial u_f^{\prime}}{\partial y^{\prime }}\right|_{y^{\prime } = H} = \mu _w \left.\frac{\partial u_3}{\partial y_3}\right|_{y_3 = 0} , \end{equation}

which should be considered together with (3.34). It is easily found that the solution of (3.41)–(3.43) is written as

(3.44)\begin{equation} u_f^{\prime } ={-} \alpha k_a \sqrt{\frac{\mu _w}{\rho _w \omega }} \, y^{\prime } \sin \left( \omega t_{{\ast} } + k_a x_1 + {\rm \pi}/ 4 \right) . \end{equation}

The lateral velocity component $v_f^{\prime }$ may be now found from the continuity equation

(3.45)\begin{equation} \frac{\partial u_f^{\prime }}{\partial x_1} + \frac{\partial v_f^{\prime }} {\partial y^{\prime }} = 0 . \end{equation}

Substituting (3.44) into (3.45), and integrating the resulting equation for $v_f^{\prime }$ with the impermeability condition of the plate surface

(3.46)\begin{equation} v_f^{\prime } = 0 \quad \text{at} \ y^{\prime } = 0 , \end{equation}

we have

(3.47)\begin{equation} v_f^{\prime } = \frac{\alpha k_a^2}{2} \sqrt{\frac{\mu _w}{\rho _w \omega }} \, y^{\prime 2} \cos \left( \omega t_{{\ast} } + k_a x_1 + {\rm \pi}/ 4 \right) . \end{equation}

3.5. Interface dynamics

In order to determine the perturbations in the shape of the interface caused by the acoustic wave, we need to consider the kinematic condition on the interface surface

(3.48)\begin{equation} \frac{\partial \varPhi }{\partial t} + u \frac{\partial \varPhi }{\partial x} + v \frac{\partial \varPhi }{\partial y} = 0 . \end{equation}

Here,

(3.49)\begin{equation} \varPhi (t , x , y) = y - Re^{{-}5/8} H (t_{{\ast} } , x_1), \end{equation}

with

(3.50a,b)\begin{equation} t = \sigma _{\mu }^{{-}1} Re^{{-}1/4} t_{{\ast} } , \quad x = 1 + \sigma _{\mu }^{{-}1} Re^{{-}1/4} x_1 , \end{equation}

where the derivatives $\partial \varPhi / \partial t$ and $\partial \varPhi / \partial x$ are same order quantities. Keeping this in mind and taking into account that $u$ is small, we can disregard the second term in (3.48). Hence, combining (3.49) and (3.48), we have

(3.51)\begin{equation} - \sigma _{\mu } Re^{{-}3/8} \frac{\partial H}{\partial t_{{\ast}}} + v = 0 . \end{equation}

Here, $v$ is given by (3.39) and (3.47), which suggests that the asymptotic expansion of $H (t_{\ast } , x_1)$ should be sought in the form

(3.52)\begin{equation} H = H_0 + \sigma _{\mu } Re^{{-}1/8} \chi H_1 (t_{{\ast} } , x_1) + \cdots . \end{equation}

Substitution of (3.52) and (3.39) into (3.51) yields

(3.53)\begin{equation} \frac{\partial H_1}{\partial t_{{\ast} }} = \left. v_f^{\prime } \right|_{y^{\prime }=H_0} . \end{equation}

It remains to substitute (3.47) into (3.53) and integrate the resulting equation for $H_1$. We find that

(3.54)\begin{equation} H_1 = \frac{\alpha k_a^2}{2} \sqrt{\frac{\mu _w}{\rho _w \omega ^3}} \, H_0^2 \sin \left( \omega t_{{\ast} } + k_a x_1 + {\rm \pi}/ 4 \right) . \end{equation}

4.1. Upper tier of airflow

The form of the asymptotic expansions of the fluid-dynamic functions in region 4 may be predicted by analysing the behaviour of the solution in region 1 in the vicinity of the roughness. Comparing (4.2) with the second equation in (3.2a,b), we can see that the longitudinal coordinates in regions 1 and 6 are related as

(4.3)\begin{equation} x_1 = \sigma _{\mu } Re^{{-}1/8} x_2 . \end{equation}

Using (4.3), we can express the solution (3.1), (3.5), (3.6a–c) in region 1 in terms of the variable $x_2$ of region 6. For example, for the longitudinal velocity component, we have

(4.4)\begin{equation} u = 1 + \sigma _{\mu }^{{-}1/2} Re^{{-}1/8} \chi M_{\infty } \alpha \sin (\omega t_{{\ast} } + \sigma _{\mu } Re^{{-}1/8} k x_2) + \cdots . \end{equation}

The fact that $\sigma _{\mu } Re^{-1/8}$ is small allows us to apply the Taylor expansion to (4.4), which leads to

(4.5)\begin{equation} u = 1 + \sigma _{\mu }^{{-}1/2} Re^{{-}1/8} \chi M_{\infty } \alpha \sin (\omega t_{{\ast} }) + \sigma _{\mu }^{1/2} Re^{{-}1/4} \chi M_{\infty } \alpha k x_2 \cos (\omega t_{{\ast} }) + \cdots . \end{equation}

The last term in (4.5), which is small compared with the $O (Re^{-1/4})$ perturbations produced by the roughness (4.1), will be disregarded, and we can conclude that in region 4 the solution for $u$ should be sought in the form

(4.6a)\begin{equation} u = 1 + \sigma _{\mu }^{{-}1/2} Re^{{-}1/8} \chi M_{\infty } \alpha \sin (\omega t_{{\ast} }) + Re^{{-}1/4} u_4 (t_{{\ast} } , x_2 , y_4) + \cdots . \end{equation}

Similar arguments may be applied to the pressure $p$, enthalpy $h$ and density $\rho$, while the form of the solution for the lateral velocity component is predicted using the continuity equation (2.2d). We have

(4.6b)$$\begin{gather} v = Re^{{-}1/4} v_4 (t_{{\ast} } , x_2 , y_4) + \cdots , \end{gather}$$

(4.6c)$$\begin{gather}p = \sigma _{\mu }^{{-}1/2} Re^{{-}1/8} \chi \alpha \sin (\omega t_{{\ast} }) + Re^{{-}1/4} p_4 (t_{{\ast} } , x_2 , y_4) + \cdots , \end{gather}$$

(4.6d)$$\begin{gather}h = \frac{1}{(\gamma - 1) M_{\infty }^2} + \sigma _{\mu }^{{-}1/2} Re^{{-}1/8} \chi \alpha \sin (\omega t_{{\ast} }) + Re^{{-}1/4} h_4 (t_{{\ast} } , x_2 , y_4) + \cdots , \end{gather}$$

(4.6e)$$\begin{gather}\rho = 1 + \sigma _{\mu }^{{-}1/2} Re^{{-}1/8} \chi M_{\infty }^2 \alpha \sin (\omega t_{{\ast} }) + Re^{{-}1/4} \rho _4 (t_{{\ast} } , x_2 , y_4) + \cdots . \end{gather}$$

Here,

(4.7a–c)\begin{equation} t = \sigma _{\mu }^{{-}1} Re^{{-}1/4} t_{{\ast} } , \quad x = 1 + Re^{{-}3/8} x_2 , \quad y = Re^{{-}3/8} y_4 . \end{equation}

Substitution of (4.6), (4.7a–c) into the Navier–Stokes equations (2.2) yields the set of quasi-steady linearised Euler equations

(4.8a)$$\begin{gather} \frac{\partial u_4}{\partial x_2} ={-} \frac{\partial p_4}{\partial x_2} , \end{gather}$$

(4.8b)$$\begin{gather}\frac{\partial v_4}{\partial x_2} ={-} \frac{\partial p_4}{\partial y_4} , \end{gather}$$

(4.8c)$$\begin{gather}\frac{\partial h_4}{\partial x_2} = \frac{\partial p_4}{\partial x_2} , \end{gather}$$

(4.8d)$$\begin{gather}\frac{\partial u_4}{\partial x_2} + \frac{\partial \rho _4}{\partial x_2} + \frac{\partial v_4}{\partial y_4} = 0 , \end{gather}$$

(4.8e)$$\begin{gather}h_4 = \frac{\gamma }{\gamma - 1} p_4 - \frac{1}{(\gamma - 1) M_{\infty }^2} \rho _4 . \end{gather}$$

Standard elimination procedure allows us to reduce (4.8) to a single equation for the pressure

(4.9)\begin{equation} (1 - M_{\infty }^2) \frac{\partial ^2 p_4}{\partial x_2^2} + \frac{\partial ^2 p_4} {\partial y_4^2} = 0 . \end{equation}

We shall formulate the boundary conditions for (4.9) after completing the flow analysis in regions 5 and 6.

4.2. Lower tier of the airflow

Asymptotic expansions of the fluid-dynamic functions in the lower tier (region 6) have the form

(4.10)\begin{equation} \left.\begin{gathered} u = Re^{{-}1/8} u_6 (t_{{\ast} } , x_2 , y_6) + \cdots , \quad v = Re^{{-}3/8} v_6 (t_{{\ast} } , x_2 , y_6) + \cdots , \\ p = \sigma _{\mu }^{{-}1/2} Re^{{-}1/8} \chi \alpha \sin (\omega t_{{\ast} }) + Re^{{-}1/4} p_6 (t_{{\ast} } , x_2 , y_6) + \cdots , \\ \rho = \rho _w + \cdots , \quad \mu = \mu _w + \cdots , \end{gathered}\right\} \end{equation}

with the independent variables defined by

(4.11a–c)\begin{equation} t = \sigma _{\mu }^{{-}1} Re^{{-}1/4} t_{{\ast} } , \quad x = 1 + Re^{{-}3/8} x_2 , \quad y = Re^{{-}5/8} y_6 . \end{equation}

Substitution of (4.10), (3.47) into the Navier–Stokes equations (2.2) results in

(4.12)\begin{equation} \left.\begin{gathered} \rho _w \left( u_6 \frac{\partial u_6}{\partial x_2} + v_6 \frac{\partial u_6} {\partial y_6} \right) ={-} \frac{\partial p_6}{\partial x_2} + \mu _w \frac{\partial ^2 u_6}{\partial y_6^2} , \\ \frac{\partial u_6}{\partial x_2} + \frac{\partial v_6}{\partial y_6} = 0 . \end{gathered} \right\}\end{equation}

Now we need to formulate the boundary conditions for (4.12). On the interface, given by

(4.13)\begin{equation} y = Re^{{-}5/8} H (t_{{\ast} } , x_2) , \end{equation}

the no-slip conditions should be satisfied

(4.14)\begin{equation} u_6 = v_6 = 0 \quad \text{at} \ y_6 = H (t_{{\ast} } , x_2) . \end{equation}

We also need to formulate the condition of matching with the solution in region 3. According to (3.24), the longitudinal velocity component $u$ is represented in region 3 by the asymptotic expansion

(4.15)\begin{equation} u = \sigma _{\mu }^{{-}1/2} Re^{{-}1/8} \lambda y_3 + \sigma _{\mu }^{{-}1/2} Re^{{-}1/8} \chi u_3 (t_{{\ast} } , x_1 , y_3) + \cdots . \end{equation}

For small values of $y_3$, we have

(4.16)\begin{equation} u = \sigma _{\mu }^{{-}1/2} Re^{{-}1/8} \left( \left.\lambda + \chi \frac{\partial u_3} {\partial y_3}\right|_{y_3=0} \right) y_3 + \cdots . \end{equation}

Since $y_3 = \sigma _{\mu }^{1/2} y_6$, we can further write

(4.17)\begin{equation} u = Re^{{-}1/8} \left( \left.\lambda + \chi \frac{\partial u_3} {\partial y_3}\right|_{y_3=0} \right) y_6 + \cdots . \end{equation}

It remains to substitute (3.34) into (4.17), and we can conclude that the sought boundary condition is written as

(4.18)\begin{equation} u_6 = \varLambda (t_{{\ast} } ) (y_6 - H_0) \quad \text{at} \ x_2 ={-} \infty , \end{equation}

where

(4.19)\begin{equation} \varLambda (t_{{\ast} } ) = \lambda - \frac{\alpha k_a \chi }{\sqrt{\mu _w \rho _w \omega }} \sin \left( \omega t_{{\ast} } + k_a x_1^0 + {\rm \pi}/ 4 \right) , \end{equation}

with $x_1^0$ denoting the value of the variable $x_1$ at the location point of the roughness, and $H_0$ is the film thickness before the interaction region.

4.3. Main part of the boundary layer

It is known that the main part of the boundary layer (region 5) plays a passive role in the viscous–inviscid interaction process. It preserves the pressure unchanged across the boundary layer, and it does not contribute into the displacement effect of the boundary layer. The latter means that the streamline slope produced by the viscous sublayer remains unchanged across region 5. It was shown (see e.g. Ruban Reference Ruban2018) that the solution of (4.12) satisfying the upstream boundary condition (4.18) exhibits the following behaviour at the outer edge of the viscous sublayer:

(4.20a,b)\begin{equation} u_6 = \varLambda (t_{{\ast} } ) y_6 + A (t_{{\ast} } , x_2) + \cdots , \quad v_6 ={-} \frac{\partial A} {\partial x_2} y_6 + \cdots \quad \text{as} \ y_4 \to \infty . \end{equation}

Substituting (4.20a,b) into the asymptotic expansions of the velocity components in (4.10), we have

(4.21a,b)\begin{equation} u = Re^{{-}1/8} \varLambda (t_{{\ast} } ) y_6 + \cdots , \quad v = Re^{{-}3/8} \left( - \frac{\partial A}{\partial x_2} \right) y_6 + \cdots . \end{equation}

Hence, the streamline slope angle

(4.22)\begin{equation} \vartheta = \arctan \frac{v}{u} = Re^{{-}1/4} \left[ - \frac{1}{\varLambda (t_{{\ast} } )} \frac{\partial A}{\partial x_2} \right] + \cdots . \end{equation}

This should coincide with the streamline slope angle at the bottom of region 4. The latter is calculated using asymptotic expansion of the velocity components (4.6a), (4.6b). Restricting our attention to the leading-order terms, we have

(4.23)\begin{equation} \vartheta = \arctan \frac{v}{u} = Re^{{-}1/4} v_4 (t_{{\ast} } , x_2 , 0) + \cdots . \end{equation}

Equations (4.23) and (4.22) allow us to conclude that

(4.24)\begin{equation} v_4 (t_{{\ast} } , x_2 , 0) ={-} \frac{1}{\varLambda (t_{{\ast} } )} \frac{\partial A}{\partial x_2} . \end{equation}

If we now set $y_4 = 0$ in (4.8b) and use (4.24) on the left-hand side of (4.8b), then we will have the following boundary condition for equation (4.9):

(4.25)\begin{equation} \left.\frac{\partial p_4}{\partial y_4}\right|_{y_4=0} = \frac{1}{\varLambda (t_{{\ast} } )} \frac{\partial ^2 A}{\partial x_2^2} . \end{equation}

It should be supplemented by the condition of attenuation of the perturbations far from the roughness

(4.26)\begin{equation} p_4 \to 0 \quad \text{as} \ x_2^2 + y_4^2 \to \infty . \end{equation}

4.4. Film flow

As before, using the shear-stress continuity condition on the interface, we can confirm again that inside the film the solution should be sought in the form

(4.27)\begin{equation} \left.\begin{gathered} u = \sigma _{\mu } Re^{{-}1/8} u^{\prime } (t_{{\ast} } , x_2 , y^{\prime }) + \cdots , \quad v = \sigma _{\mu } Re^{{-}3/8} v^{\prime } (t_{{\ast} } , x_2 , y^{\prime }) + \cdots , \\ p = Re^{{-}1/4} p^{\prime } (t_{{\ast} } , x_2 , y^{\prime }) + \cdots , \quad \hat{\rho } = \frac{1}{\sigma _{\rho }} \rho _{\infty } + \cdots , \quad \hat{\mu } = \frac{1}{\sigma _{\mu }} \mu _{\infty } + \cdots , \end{gathered}\right\} \end{equation}

with the independent variables defined by

(4.28a–c)\begin{equation} t = \sigma _{\mu }^{{-}1} Re^{{-}1/4} t_{{\ast} }, \quad x = 1 + Re^{{-}3/8} x_2 , \quad y = Re^{{-}5/8} y^{\prime } . \end{equation}

By substituting (4.27), (4.28a–c) into the Navier–Stokes equations (2.2), while assuming $\sigma _{\mu } / \sigma _{\rho } \ll 1$, we arrive at a conclusion that the film flow is described by the following equations:

(4.29a)$$\begin{gather} \frac{\partial ^2 u^{\prime }}{\partial y^{\prime 2}} - \frac{\partial p^{\prime }} {\partial x_2} = 0 , \end{gather}$$

(4.29b)$$\begin{gather}\frac{\partial p^{\prime }}{\partial y^{\prime }} = 0 , \end{gather}$$

(4.29c)$$\begin{gather}\frac{\partial u^{\prime }}{\partial x_2} + \frac{\partial v^{\prime }} {\partial y^{\prime }} = 0 . \end{gather}$$

These should be solved with the no-slip conditions on the surface of the roughness (4.1)

(4.30)\begin{equation} u^{\prime } = v^{\prime } = 0 \quad \text{at} \ y^{\prime } = f (x_2) , \end{equation}

and the following conditions on the interface. Firstly, the continuity of the tangential stress requires that

(4.31a)\begin{equation} \left.\frac{\partial u^{\prime }}{\partial y^{\prime }}\right|_{y^{\prime } = H - 0} = \mu _w \left.\frac{\partial u_6}{\partial y_6}\right|_{y_6 = H + 0} . \end{equation}

Secondly, the normal stress condition is written as

(4.31b)\begin{equation} p_6 = p^{\prime } + \gamma _{{\ast} } \frac{\partial ^2 H}{\partial x_2^2} . \end{equation}

Finally, the kinematic condition has the form

(4.31c)\begin{equation} v^{\prime } = \frac{\partial H}{\partial t_{{\ast} }} + u^{\prime } \frac{\partial H} {\partial x_2} \quad \text{at} \ y^{\prime } = H (t_{{\ast} } , x_2) . \end{equation}

7.1. Interfacial instability modes

By setting the denominator of the expression in (7.13) to zero we can obtain the dispersion equation describing the instability modes of the flow. Since the airflow in the viscous sublayer has a relatively slow speed we shall assume the flow in this layer is incompressible and let $\beta \approx 1$. Considering that the Mach number is small we are able to set the following values: $\mu _w = \rho _w =1$ and $\lambda = 0.3321$. The corresponding dispersion equation for an incompressible flow is

(7.15)\begin{equation} a_3 + a_1 a_4 = 0, \end{equation}

which yields to the following expression:

(7.16)\begin{equation} \omega ={-}kH_0 \lambda+ {\rm i} \frac{\gamma_* H^3_{0} k^4}{3} - {\rm i} \dfrac{\dfrac{k^2 H^3_0}{3}- \dfrac{Ai(0) ({\rm i}k)^{5/3}H^2_0 }{2 |Ai'(0)|} }{\dfrac{\beta}{|k|} +\dfrac{({\rm i}k)^{1/3}}{3 |Ai'(0)|}\lambda^{{-}5/3}} . \end{equation}

By conducting a linear stability analysis for an incompressible flow while focusing on temporal instability we found that, without the surface tension, $\gamma =0$, the amplitude of the dispersion equation grows, however, for the case where $\gamma \neq 0$ the amplitude decays

(7.17)\begin{equation} - {\rm Im}\left\{\omega(k)\right\} ={-}\frac{1}{3} \tilde{\gamma} H_{0}^3k^4+\frac{N_{r} D_{r}+N_{i}D_{i}}{D_{r}^2+D_{i}^2}, \end{equation}

where

(7.18)\begin{equation} \left.\begin{gathered} N_{r}=k^2 H_{0}^3 |k||Ai'(0)|+\frac{3\sqrt{3}}{4}k^{5/3}H_{0}^2|k|Ai(0), \quad N_{i}={-}\frac{3}{4}k^{5/3}H_{0}^2|k|Ai(0), \\ D_{r}=3|Ai'(0)|+\frac{\sqrt{3}}{2}|k|k^{1/3} \lambda^{{-}5/3}, \quad D_{i}=\frac{1}{2}|k|k^{1/3} \lambda^{{-}5/3}. \end{gathered}\right\} \end{equation}

Figure 3 shows that the flow is unstable within the interval $[0, k_{*}]$ but becomes stable for $k>k_{*}$ and we see that the neutral wavenumber is $k_{*} \approx 1.05$.

Figure 3.

Representation of the dispersion equation with the following values: $\beta =\mu =\rho =1$ and $\lambda = 0.3321$.

7.2. Receptivity coefficient

To find the solutions of (7.11) and (7.13) we need to apply the inverse Fourier transform

(7.19)\begin{equation} \tilde{p}=\frac{1}{2{\rm \pi}} \int\limits_{-\infty }^{\infty } \left(\bar{p}^{{\star}}+ \chi \left[\bar{p}^{{\star}{\star}} {\rm e}^{{\rm i}\omega t_{*}} + ({\rm c.c.})\right] \right) {\rm e}^{{\rm i}kx_2} \,{\rm d}k. \end{equation}

Our task is to find the amplitude of the Tollmien–Schlichting wave which is described by the receptivity coefficient. This coefficient is obtained by finding the solution of the unsteady term in (7.19). First, we choose the contour of integration as a semi-circle in the upper half of the $k$-plane shown in figure 4. Then, turning our attention to the second term of the integral in (7.19)

(7.20)\begin{equation} \tilde{I}^{{\star}{\star}}=\frac{{\rm e}^{{\rm i}\omega t_{*}}}{2{\rm \pi}} \int\limits_{-\infty }^{\infty} \bar{p}^{{\star}{\star}} {\rm e}^{{\rm i}kx_{2}}\,{\rm d}k =\frac{1}{2{\rm \pi}} \int\limits_{-\infty }^{\infty} \frac{kH_0 \bar{y}_{w}(k)}{\mathcal{D} ^{{\star}{\star}}(k)}{\rm e}^{{\rm i}kx_{2}}\,{\rm d}k.\end{equation}

Figure 4.

Deformation of the contour of integration.

We can write the integral in (7.20) as

(7.21)\begin{equation} \tilde{I}^{{\star}{\star}}=\frac{1}{2{\rm \pi}} \left(\tilde{I}_{1}^{{\star}{\star}}+ \tilde{I}_{2}^{{\star}{\star}}\right), \end{equation}

where the integration interval is divided into two parts

(7.22a)$$\begin{gather} \tilde{I}_{1}^{{\star}{\star}}= \int\limits_{-\infty }^{0} \frac{kH_0 \bar{y}_{w}(k)}{\mathcal{D} ^{{\star}{\star}}(k)} {\rm e}^{{\rm i}kx_{2}}\,{\rm d}k, \end{gather}$$

(7.22b)$$\begin{gather}\tilde{I}_{2}^{{\star}{\star}}= \int\limits_{0}^{\infty } \frac{kH_0 \bar{y}_{w}(k)}{\mathcal{D} ^{{\star}{\star}}(k)} {\rm e}^{{\rm i}kx_{2}}\,{\rm d}k. \end{gather}$$

Considering the integration contour shown in figure 4, we find that the first integral, (7.22a), to be divided into three parts

(7.23a)\begin{equation} \tilde{I}_{1}^{{\star}{\star}}=\left(\int\limits_{C_{-}}+ \int\limits_{C_{-}^{\prime}}+ \int\limits_{C_{R}^{-}}\right) \left( \frac{kH_0 \bar{y}_{w}(k)}{\mathcal{D} ^{{\star}{\star}}(k)} \right) {\rm e}^{{\rm i}kx_2} \,{\rm d}k. \end{equation}

According to the Jordan lemma, the integral along the contour $C_{R}^{-}$ tends to zero and the integral along the contour $C_{-}$ retrieves the original integral $\tilde {I}_{1}^{\star }$. The integral along $C_{-}^{\prime }$ is of Laplace type and it can be evaluated at large values of $x_2$. Thus, with help of the Watson lemma, we find that the main contribution to the behaviour of the integral is represented by the following term:

(7.24a,b)\begin{equation} k\rightarrow 0, \quad \omega={O}(1), \end{equation}

the dominant term of unsteady pressure perturbation is written as

(7.25)\begin{equation} \bar{p}^{{\star}{\star}} \approx \frac{5(\sqrt{3}+{\rm i}) }{18 \beta^2}\frac{\lambda^{{\star}}}{Ai'(0)} \frac{k^{7/3} \bar{y}_{w}(k)}{(\mu_{w} \rho_{w}^2 \lambda^8)^{1/3}}. \end{equation}

With the help of Watson's lemma we find

(7.26)\begin{equation} \int\limits_{C_{-}^{\prime}} \bar{p}^{{\star}{\star}} {\rm e}^{{\rm i}\omega t_{*}} {\rm e}^{{\rm i}kx_2} \,{\rm d}k \approx \mathcal{B} \frac{{\rm e}^{{\rm i}({5 {\rm \pi}}/{3}) }}{x_2^{10/3}}\varGamma\left(\frac{10}{3}\right) \quad \text{as} \ x_2 \rightarrow \infty, \end{equation}

where

(7.27)\begin{equation} \mathcal{B} = \frac{5(\sqrt{3}+{\rm i}) }{18 \beta^2}\frac{\lambda^{{\star}}}{Ai'(0)} \left(\mu_{w} \rho_{w}^2 \lambda^8\right)^{{-}1/3} . \end{equation}

By residue theorem we find the solution of (7.23a) to be

(7.28)\begin{equation} \tilde{I}_{1}^{{\star}{\star}}=2{\rm \pi} {\rm i} \left( \frac{k_0 H_0 \bar{y}_{w}(k_0)}{{\rm d}\mathcal{D} ^{{\star}{\star}}(k_0)/{\rm d}k} \exp({{\rm i}k_0 x_2})\right)-\mathcal{B} \frac{\exp\left({{\rm i}\dfrac{5 {\rm \pi}}{3}}\right)}{x_2^{10/3}}\varGamma\left(\frac{10}{3}\right), \end{equation}

where $k_0$ is the solution to $\mathcal {D} ^{\star \star }(k)=0$ and it is displayed in figure 5(a). We find the solution of integral (7.22b) with the same procedure, except that there is no singularity inside the closed contour composed of $C^{+}_{R}$, $C^{+}$ and $C^{\prime }_{+}$. With the help of Jordan's lemma we found the value of integral along arc $C^{+}_{R}$ to be zero as the radius $R$ tends to infinity. The integral along $C_{+}$ recovers the original integral. The integral $C^{\prime }_{+}$ along the positive imaginary semi-axis $k$ is found with help of Watson's lemma for large values of $x_2$. Consequently, we find

(7.29)\begin{equation} \tilde{I}_{2}^{{\star}{\star}}={-}\mathcal{B} \frac{\exp\left({\rm i}\dfrac{5 {\rm \pi}}{3}\right)}{x_2^{10/3}}\varGamma\left(\frac{10}{3}\right).\end{equation}

By combining (7.28) and (7.29) we can write solution of (7.21) as

(7.30)\begin{equation} \tilde{I}^{{\star}{\star}}={\rm i} \left(\frac{k_0H_0 \bar{y}_{w}(k_0)}{{\rm d}\mathcal{D} ^{{\star}{\star}}(k_0)/{\rm d}k} \exp({{\rm i}k_0 x_2})\right) - \frac{\mathcal{B}}{\rm \pi} \frac{\exp\left({{\rm i}\dfrac{5 {\rm \pi}}{3}}\right)}{x_2^{10/3}}\varGamma\left(\frac{10}{3}\right).\end{equation}

The solution for the unsteady term of the pressure may be expressed in terms of the receptivity coefficient

(7.31)\begin{equation} \tilde{I}^{{\star}{\star}}= \mathcal{K}(\omega) \bar{y}_{w}(k_0) \exp({{\rm i}k_0 x_2})-\frac{\mathcal{B}}{\rm \pi} \frac{\exp\left({{\rm i}\dfrac{5 {\rm \pi}}{3}}\right)}{x_2^{10/3}}\varGamma\left(\frac{10}{3}\right), \end{equation}

where the receptivity coefficient is written as

(7.32)\begin{equation} \mathcal{K}(\omega)= \frac{{\rm i} k_0 H_0}{{\rm d}\mathcal{D} ^{{\star}{\star}}(k_0)/{\rm d}k} , \end{equation}

and $\mathcal {D} ^{\star \star }(k)$ is defined by (7.14), where $k_0$ is solution to $D(k_0)=0$. The modulus of the receptivity coefficient in terms of the frequency $\omega$ is displayed in figures 6 and 7. It is shown that in general the amplitude of the instability wave decreases and tends to a small value with increasing frequency.

Figure 5.

Value of $k_{0}$ in the complex $k$-plane which is the solution to $\mathcal {D} ^{\star \star }(k_0)=0$ for various values of the Mach number as frequency increases. (a) Incompressible flow. (b) Subsonic flow.

Figure 6.

Modulus of the receptivity coefficient for various values of initial film thickness $H_0$, surface tension $\gamma _*$ and roughness location $x_1^0$ and Mach number $M_{\infty }$; (a) $\gamma _*= x_1^0 = 0.5$, $M_{\infty }=0.1$, (b) $H_0=\gamma _*= x_1^0= 0.5$, (c) $H_0= x_1^0 =0.5$, $M_{\infty }=0.1$, (d) $H_0=\gamma _*=0.5$, $M_{\infty }=0.1$.

Figure 7.

Modulus of receptivity coefficient for subsonic flow with Prandtl number of 0.72, ratio of specific heat of 1.4 and Sutherland law coefficient of 110.4 Kelvin for all cases; (a) $H_0=\gamma _*= x_1^0 = 0.5$, (b) $H_0= x_1^0 = 0.5$, (c) $M_{\infty }=H_0=x_1^0 = 0.5$, (d) $H_0=x_1^0 = 0.5$, (e) $M_{\infty }=\gamma _*=H_0=0.5$, ( f) $H_0=0.5$.