---
abstract: |
    The phenomenon of steady streaming, or acoustic streaming, is an important physical phenomenon studied extensively in the literature. Its mathematical formulation involves the Navier-Stokes equations, thus due to the complexity of these equations is usually studied heuristically using formal perturbation expansions. It turns out that the Burgers equation formulated on the half line provides a simple model of the above phenomenon. The physical situation corresponds to the solution of the Dirichlet problem on the half-line, which decays as $x
    \rightarrow \infty$ and which is [*time periodic*]{}. We show that the Dirichlet problem, where the usual prescription of the initial condition is now replaced by the requirement of the time periodicity, yields a well posed problem. Furthermore, we show that the solution of this problem tends to the “inner” and “outer” solutions obtained by the perturbation expansions.
author:
- |
    A.S. Fokas$^1$ and J.T. Stuart$^2$\
    $^1$Department of Applied Mathematics and Theoretical Physics\
    University of Cambridge,\
    Cambridge, CB30WA, UK\
    t.fokas@damtp.cam.ac.uk\
    $^2$Department of Mathematics\
    Imperial College\
    London, SW72BZ, UK\
    t.stuart@imperial.ac.uk
date: October 2004
title: |
    The Time Periodic Solution of the Burgers Equation\
    on the Half-Line and an Application to Steady Streaming
---

Introduction
============

The phenomenon of “steady streaming”, or “acoustic streaming” as it is sometimes called, dates back to the 19th century with the work of Kundt on the circulations of air in tubes (see Rayleigh 1883). There are related phenomena of circulations in the flows under water waves with a free surface (Longuet-Higgins 1953, 1960; Hunt and Johns 1963). Following Rayleigh, Schlichting (1932) did both experimental and theoretical work on a particular case of the flow induced by a circular cylinder oscillating along a diameter.

The nature of the phenomenon, which is common to all the papers mentioned above, may be explained as follows. If the flow of fluid (liquid or gas) is periodic in time but with zero mean, and is parallel to a rigid surface, then the governing equations ensure that a thin periodic boundary layer is formed in the neighborhood of that surface. If the time frequency is $\omega$-radians per second and $\nu$ is the kinematic viscosity then the thickness of this layer is of order $(\nu/w)^{1/2}$. This boundary layer is known as a Stokes layer. A case of particular simplicity is that in which the flow is parallel to a flat rigid surface, whose velocity is $U_w\cos w t$ parallel to itself, $t$ being the time and $U_w$ a constant. Then the velocity in the fluid neighboring the surface is $$u = U_we^{-\eta} \cos(\omega t -\eta), \eqno (1.1)$$ where $$\eta = y(w/2\nu)^{1/2}, \eqno (1.2)$$ $y$ being the coordinate normal to the surface. This is the prototype Stokes layer (Stokes 1851), see also (Stuart 1963, Benney 1964). If a Galilean tranfsormation is imposed such that the surface is at rest but the fluid has velocity $U \cos wt$ parallel to that surface as $y$ and $\eta$ tend to infinity, then the velocity in the fluid is $$U[\cos wt - e^{-\eta} \cos (wt-\eta)]. \eqno (1.3)$$

In the above description, the fluid motion is intrinsically linear so that the simple form of the Stokes layer emerges exactly: the nonlinear terms are identically zero for (1.1) and (1.3).

The situation is quite different, however, if the velocity (as $y$ and $\eta$ tend to infinity) has the form $$U(\xi)\cos wt, \eqno (1.4)$$ where $\xi$ is the coordinate parallel to the surface: the latter may be curved and its curvature is neglected if the Stokes layer thickness $(\nu/w)^{1/2}$ is small compared with a typical length d: thus $\nu/wd^2$ is supposed to be small. The nonlinear term in the Navier-Stokes equation, namely $u \cdot \nabla u$, where $u$ is the vector velocity, is not zero however. It has terms of two types: (i) there are terms proportional to $\exp(2iwt)$ and to $\exp(-2iwt)$ which generate corresponding flow components; the equations that govern these components do have solutions which tend to zero at the edge of the Stokes layer as is required, (ii) a mean term is also generated, which drives the steady streaming; we can refer to this mean term, which is an effect of rectification, as a Reynolds stress (or, rather a derivative of the Reynolds stress). An equation can be obtained for the mean flow generated by this Reynolds stress but, somewhat paradoxically, it is not possible to obtain a solution whose component of velocity that is parallel to the surface, tends to zero as the edge of the Stokes layer is approached. Rather the best that can be achieved is to ensure that the solution is finite for that component of velocity.

Indeed it is found that, as the edge of the Stokes layer is approached, the velocity component approaches $$\frac{-3}{4w} U(\xi) \frac{dU(\xi)}{d\xi}. \eqno (1.5)$$

Since $\xi$ has a typical length $d$, this velocity component has scale $U^2_0/wd$, where $U_0$ is the scale of $U(\xi)$. Now in problems of the Navier-Stokes equations, the famous Reynolds number plays a significant role; it is the product of a velocity and a distance, divided by the kinematic viscosity. In the present case we have $$R_s = (U^2_0/wd)(d/\nu) = U^2_0/w\nu. \eqno (1.6)$$ This parameter is known as the steady-streaming Reynolds number. Stuart (1963, 1966) showed the importance of this concept for the calculation of the flow forced by (1.5) outside the Stokes layer.

In the papers of Rayleigh (1883) and Schlichting (1932), it is implicit that the parameter $R_s$ is small, and the calculation of the flow outside the Stokes layer is performed on that basis. Even so, Schlichting’s paper makes it clear that he was aware that his theory for small values of $R_s$ is inappropriate as an explanation of his experimental work, for which $R_s =250$.

Stuart (1963, 1966) showed how an “outer-boundary-layer theory”, which is valid for large values of $R_s$, can be used to obtain a solution for which (1.5) applies at the edge of the Stokes layer but which tends to zero as the distance form the surface tends to infinity. Implicit in that work is the idea that this “outer” (or "steady-streaming’) boundary layer is much thicker than the Stokes layer, indeed by a factor $R_s^{1/2}$. Following Stuart’s work, Riley (1998) has pursued problems of this type in much detail.

A problem can be posed that shows many of the characteristics that are outlined above, but without the complications of the full Navier-Stokes equations. The relevant equation is a Burgers type equation for $u(x,t)$, namely $$u_t + \beta (u-k \beta)u_x = \frac{1}{2} u_{xx}, \eqno (1.7)$$ where $\beta$ is a small parameter which plays a role analogous to $R_s^{-1/4}$ above, $R_s$ being large and $k$ is $O(1)$. The boundary conditions are $$x =0: \quad u(0,t)=\cos t,$$ $$x \rightarrow \infty: \quad u(\infty,t) \rightarrow 0, \eqno (1.8)$$ and $$u(x,t) = u(x,t+2\pi).$$ The Cole-Hopf transformation $$u-k\beta = -\theta_x/\beta\theta \eqno (1.9)$$ gives $$\theta_t - \frac{1}{2}\theta_{xx} = 0, \eqno (1.10)$$ with the boundary conditions $$x=0: \quad \theta_x(0,t) + \beta(\cos t- k\beta)\theta(0,t)=0,$$ $$x\rightarrow \infty: \quad \theta_x(\infty,t) / \theta(\infty, t) = k\beta^2, \eqno (1.11)$$ $$\theta_x(x,t)/\theta(x,t) = \theta_x(x,t+2\pi)/\theta(x,t+2\pi).$$ The equation (1.10) is simply stated, but the boundary and periodicity conditions are complicated. We return to a discussion of this problem in later sections.

In the meantime we note that a formal solution to (1.7), (1.8) can be obtained by expanding $u$ in a power series in $\beta$ $$u=u_0(x,t) +\beta u_1(x,t) + \beta^2u_2(x,t)+\cdots,$$ and it is quickly found that $$u_0 = e^{-x}\cos(t-x). \eqno (1.12)$$ At order $\beta$, the $u_1$ terms involving $e^{2it}$ and $e^{-2it}$ give no difficulty, but the steady (or rectified) term has to satisfy $$u_{1xx} =-e^{-2x};$$ the solution which is bounded as $x\rightarrow \infty$ and which is zero at $x=0$ is $$u_1 = \frac{1}{4} (1-e^{-2x}). \eqno (1.13)$$ We note that as $x\rightarrow \infty$, $u\rightarrow
\frac{1}{4}\beta$ plus higher order terms. A re-scaling of (1.7) for the steady part of the solution with $$u(x) = \beta v(x), \quad x = z/2\beta^2$$ yields $$(v-k)v_z = v_{zz}, \eqno (1.14)$$ with the boundary conditions $$z\rightarrow 0: \quad v\rightarrow 1/4,$$ $$z\rightarrow \infty: \quad v\rightarrow 0. \eqno (1.15)$$ The solution of (1.14) subject to (1.15) is $$v = 2k/[(8k-1)e^z +1]. \eqno (1.16)$$

We note that the length scale of this “outer” region is $(2\beta^2)^{-1}$ times the scale of the “inner” region of (1.12) and (1.13). Also $z\rightarrow 0$ in (1.15) corresponds to $x
\rightarrow \infty$ in (1.13), in the sense of $\beta \rightarrow
0$ with $x$ fixed and $\beta \rightarrow 0$ with $z$ fixed giving an equivalence.

This heuristic argument will be justified in later sections by solving explicitly Burgers equation (1.7) with the conditions (1.8). We note that the mathematical novelty of this problem is that it is posed on the half-line and that it requires periodicity in $t$. We recall that the usual Dirichlet problem for the Burgers equation on the half-line was analysed by Calogero and De Lillo. In their formulation, in addition to the Dirichlet boundary condition, one also specifies the initial condition $u(x,0)$. In the present situation, instead of specifying $u(x,0)$, one requires periodicity.

In section 3 we will analyse the time periodic solution of the Burgers equation on the half-line with two different boundary conditions at the origin: we specify either (a) the integral of $u(x,t)$, or (b) $u(0,t)$. The physical problem corresponds to the case (b), however we have also included case (a), because for this case the coefficients $A_n$ of the associated Fourier series (given by the representation (3.3)) can be computed explicitly. For case (b) the coefficients $A_n$ satisfy a [*second*]{} order difference equation, see equation (3.4b). We will show that this equation has a [*unique*]{} solution by imposing the condition that $A_n \rightarrow 0$ as $n \rightarrow \infty$ (which is needed for the convergence of the series). Although we cannot give the explicit form of $A_n$ in general, we will compute $A_n$ in the case that $\beta$ is small. In this case we will show that the associated series converges and that the solution $u(x,t)$ tends to the “inner” and “outer” solutions obtained by the perturbation expansion, see equations (5.15) and (5.16).

From Burgers to the Heat Equation
=================================

[**Proposition 2.1.**]{} Let $\beta$ and $c$ be positive constants. Let the real-valued function $u(x,t)$ solve

$\displaystyle{ \ \ \qquad \ \ \qquad  \ \ \ \ \qquad \ \ \qquad \
\ u_t = \frac{1}{2}u_{xx} - \beta uu_x + cu_x, \quad 0<x<\infty,
\quad t>0}$, (2.1a)

.1in $\displaystyle{ \ \ \qquad \ \ \qquad  \ \ \ \ \qquad
\ \ \qquad \ \ u\rightarrow 0}$ as $x \rightarrow \infty$, (2.1b)

.1in $\displaystyle{\ \ \qquad \ \ \qquad  \ \ \ \ \qquad \
\ \qquad \ \ u(x,t)}$ is $2\pi$ periodic in $t$, (2.1c)

.1in and one of the following boundary conditions: $$\int^\infty_0 u(x,t)dx = \cos t, \quad t>0, \eqno (2.2a)$$ or $$u(0,t) = \cos t, \quad t>0. \eqno (2.2b)$$ Define $\varphi(x,t)$ by $$\varphi(x,t) = e^{-\beta \int^x_\infty u(\xi,t)d\xi} - 1. \eqno (2.3)$$ Then $\varphi(x,t)$ is a real-valued function and solves

.1in $\displaystyle{ \ \ \qquad \ \ \qquad  \ \ \ \ \qquad
\ \ \qquad \ \ \varphi_t = \frac{1}{2} \varphi_{xx} + c\varphi_x,
\quad 0<x<\infty, \quad t>0}$, (2.4a)

.1in $\displaystyle{ \ \ \qquad \ \ \qquad  \ \ \ \ \qquad
\ \ \qquad \ \ \varphi \rightarrow 0}$ as $x\rightarrow \infty$, (2.4b)

.1in $\displaystyle{ \ \ \qquad \ \ \qquad  \ \ \ \ \qquad
\ \ \qquad \ \ \varphi(x,t)}$ is $2\pi$ periodic in $t$, (2.4c)

.1in and satisfies one of the following boundary conditions, respectively: $$\varphi(0,t) = e^{\beta \cos t}-1, \quad t>0, \eqno (2.5a)$$ or $$\varphi_x(0,t) + \beta \cos t\varphi(0,t) + \beta \cos t =0, \quad t>0. \eqno (2.5b)$$

#### Proof

The function $\varphi$ is well defined and $\varphi \rightarrow 0$ as $x \rightarrow \infty$.

If $E$ is defined by $$E \Doteq \exp \left[ - \beta \int^x_{\infty}
u(\xi,t)d\xi \right],$$ then

.1in $\displaystyle{ \ \ \qquad \ \ \qquad  \ \ \ \ \qquad
\ \ \qquad \ \ \varphi_x =-\beta uE},$

.1in $\displaystyle{ \ \ \qquad \ \ \qquad  \ \ \ \ \qquad
\ \ \qquad \ \ \varphi_{xx} =- \beta u_xE + \beta^2u^2E},$

.1in $\displaystyle{ \ \ \qquad \ \ \qquad  \ \ \ \ \qquad
\ \ \qquad \ \ \varphi_t = -\beta \int^x_\infty u_td\xi},$\
thus

.1in $\displaystyle{ \ \ \qquad \ \ \qquad  \ \ \ \ \qquad
\ \ \qquad \ \ \varphi_t - \frac{1}{2} \varphi_{xx} - c\varphi_x =
\left( \int^x_\infty u_td\xi - \frac{1}{2} u_x + \frac{1}{2} \beta
u^2-cu\right) E \beta =0}.$\
Also $$\varphi
(x,t+2\pi) = e^{-\beta \int^x_\infty u(\xi,t+2\pi)d\xi} -1 =
\varphi(x,t).$$ Equation (2.2a) implies $$\varphi(0,t) = e^{\beta \cos t}-1.$$ Furthermore, $$\varphi_x(0,t) =- \beta u(0,t) \exp\left[ -\beta
\int^0_\infty u(\xi,t)d\xi\right],$$ $$\varphi (0,t) = \exp \left[ -\beta\int^0_\infty
u(\xi,t)d\xi\right],$$ hence $$\varphi_x(0,t) = -\beta u(0,t)\varphi(0,t),$$ which implies (2.5b).

The Solution for $u(x,t)$
=========================

[**Proposition 3.1.**]{} Assume that the positive constants $\beta$ and $c$, are such that $\lambda_n$ defined by $$\lambda_n \Doteq -c + (c^2+2in)^{\frac{1}{2}}, \quad n = 1,2,\cdots \eqno (3.1)$$ has 1 value with Re$\lambda_n <0$. Then the solution of (2.1)-(2.2) is given by $$u(x,t) = -\frac{1}{\beta} \frac{\varphi_x(x,t)}{1+\varphi(x,t)}, \eqno (3.2)$$ where $$\varphi(x,t) = \frac{1}{2\pi} \left\{ A_0e^{\lambda_0x} + \sum^\infty_1 \left(A_ne^{\lambda_nx +int} + {\bar A_n} e^{-{\bar
\lambda_n}
x -int} \right)\right\}, \eqno (3.3)$$ and $\lambda_0, \{
A_j\}^\infty_0$ are defined as follows: $$\lambda_0 =-2c;$$ in the case of the boundary condition (2.2a) $$A_n = \int^{2\pi}_0 e^{-int} \left[ e^{\beta \cos t}-1\right] dt, \eqno (3.4a)$$ whereas in the case of the boundary condition (2.2b) $$2\lambda_n A_n + \beta(A_{n+1} + A_{n-1}) + 2\pi\beta\delta_{n,1} =0, \quad n =0,1,2,\cdots. \eqno (3.4b)$$

#### Proof.

Equation (2.3) implies equation (3.2). Thus the problem reduces to solving an initial-boundary value problem for the heat equation (2.4)-(2.5).

Let $$\hat\varphi(x,n) =\int^{2\pi}_0\varphi(x,t)e^{-int}dt, \eqno (3.5)$$ $$\varphi(x,t) = \frac{1}{2\pi} \sum^\infty_{n=-\infty} \hat\varphi(x,n) e^{int} = \frac{1}{2\pi} \left\{ \hat \varphi(x,0) + \sum^\infty_1 \left(\hat\varphi (x,n)e^{int} + \hat\varphi(x,-n)e^{-int}\right)\right\}.$$ Equation (3.5) and reality imply $$\hat\varphi(x,-n) = \overline{\hat \varphi(x,n)}$$.

Thus $$\varphi(x,t) = \frac{1}{2\pi} \left\{ \hat\varphi(x,0) + \sum^\infty_1 (\hat\varphi(x,n)e^{int} + \overline{\hat \varphi(x,n)}e^{-int})\right\}. \eqno (3.6)$$ Equations (3.5) and (2.4a) imply $$\hat\varphi_{xx} + 2c\hat\varphi_x = 2\int^{2\pi}_0 \varphi_t(x,t)e^{-int}dt = 2\left[\left. \varphi(x,t)e^{-int}\right|^{2\pi}_0 + in\hat\varphi\right].$$ Thus using periodicity, we find $$\hat \varphi_{xx} + 2c\hat\varphi_x - 2in\hat\varphi = 0. \eqno (3.7)$$

#### $n=0$:

$$\hat\varphi = c_1 + A_0e^{\lambda_0x}, \quad \lambda_0 =-2c;$$ the boundness requirement as $x\rightarrow \infty$ implies $c_1=0$. Thus $$\hat\varphi(x,0) = A_0e^{\lambda_0x}. \eqno (3.8a)$$

#### $n=1,2,\cdots$:

$$\hat\varphi(x,n) = A_ne^{\lambda_nx}, \quad \lambda^2_n + 2c\lambda_n - 2in =0. \eqno (3.8b)$$ Substituting (3.8) into (3.6) we find (3.3). If the boundary condition (2.2a) is valid, then $$A_n = \hat\varphi(0,n) = \int^{2\pi}_0\varphi(0,t)e^{-int}dt, \quad n=0,1,2,\cdots,$$ which is equation (3.4a).

If the boundary condition (2.2b) is valid, then multiplyng this equation by $e^{-int}$ and integrating from 0 to $2\pi$ we find $$\int^{2\pi}_0 e^{-int} \left[ 2\varphi_x(0,t) + \beta(e^{it} + e^{-it})\varphi(0,t) + \beta(e^{it} + e^{-it})\right] =0, \quad n=0,1,\cdots.$$ Thus $$2\hat\varphi_x(0,n) + \beta[\hat \varphi(0,n+1)+\hat\varphi(0,n-1)] + 2\pi\beta\delta_{n,1} =0, n=0,1,\cdots.$$ Using $$\hat\varphi_x(0,n) = \lambda_nA_n,$$ the above equation becomes (3.4b).

The Boundary Value Problem (2.1)-(2.2a) with $c=\beta^2k$ as $\beta\rightarrow 0$
=================================================================================

Let $$c=\beta^2k. \eqno (4.1)$$ The solution of this problem is given by (3.2) and (3.3) where $A_n$ is defined by (3.4a). The integral (3.4a) can be computed explicitly. For simplicity we only compute it as $\beta
\rightarrow 0$: $$A_n = \int^{2\pi}_0 e^{-int} \left[ \beta\cos t + \frac{\beta^2}{2!} (\cos t)^2 + O(\beta^3)\right]dt, \quad n=0,1,\cdots$$ $$= \oint_{|z|=1} \frac{1}{iz^{n+1}} \left[ \frac{\beta}{2} (z + \frac{1}{z}) + \frac{\beta^2}{8} \left( z^2 + \frac{1}{z^2} + 2\right) + O(\beta^3)\right] dz$$ $$= \frac{\beta}{2i} \oint_{|z|=1} \frac{1}{z^{n+1}} \left( z + \frac{1}{z}\right) dz + \frac{\beta^2}{8i} \oint_{|z|=1} \frac{1}{z^{n+1}} \left( z^2 + 2 + \frac{1}{z^2}\right) dz + O(\beta^3).$$ Thus $$A_n = 2i\pi \left[ \frac{\beta}{2i} \delta_{n,1} + \frac{\beta^2}{8i} \left( 2\delta_{n,0} + \delta_{n,2} \right)\right] + O(\beta^3).$$ Hence $$A_1 = \pi\beta + O(\beta^3), \quad A_0 = \frac{\beta^2\pi}{2} + O(\beta^3), \quad A_2 = \frac{\pi\beta^2}{4} + O(\beta^3).$$ Also $$\lambda_n = -\sqrt{2} e^{\frac{i\pi}{4}} \sqrt{n} - k\beta^2 + O(\frac{\beta^4}{\sqrt{n}}).$$ Therefore equation (3.3) implies $$\varphi(x,t) = \frac{1}{2\pi} \left\{ \frac{\beta^2\pi}{2} e^{-2\beta^2kx} + \pi\beta e^{-k\beta^2x} \left[ e^{-(1+i)x+it} + e^{-(1-i)x-it}\right] \right.$$ $$\left. + \frac{\pi\beta^2}{4}e^{-k\beta^2x} \left[ e^{-\sqrt{2}(1+i)x+2it} + e^{-\sqrt{2}(1-i)x-2it}\right] \right\} + O(\beta^3)$$

#### 1. $x = O(1)$

$$u = -\frac{[-(1+i)e^{-(1+i)x+it} - (1-i)e^{-(1-i)x-it}] + \frac{\beta}{4} [ -\sqrt{2} (1+i)e^{-\sqrt{2}(1+i)x+2it)} - \sqrt{2}(1-i) e^{-\sqrt{2}(1-i)x-2it}] + O(\beta^2)}{ 2 + \beta [e^{-(1+i)x+it} + e^{-(1-i)x-it}] + O(\beta^2)},$$ or $$u = \frac{1}{2} \left[ (1+i) e^{-(1+i)x+it} - (1-i) e^{-(1-i)x-it} \right] + \frac{\beta}{4}
\left\{ \frac{(1+i)e^{-\sqrt{2}(1+i)x+2it}}{\sqrt{2}} + \frac{
(1-i)e^{-\sqrt{2}(1-i)x-2it}}{\sqrt{2}} \right. \eqno (4.2)$$ $$-2e^{-2x} - (1+i)\left. e^{-2(1+i)x+2it} - (1-i)e^{-2(1-i)x-2it} \right\} + O(\beta^2)$$

#### 2. $x=O(\beta^{-2})$

$$u = \frac{2\beta^3k}{\beta^2 + 4e^{2\beta^2kx}}. \eqno (4.3)$$

The Analysis of (2.1)-(2.2b) with $c=\beta^2k$, $k\neq 1/8$, as $\beta \rightarrow 0$
=====================================================================================

The solution of the linear homogeneous equation (3.4b), with the requirement that $$A_n \rightarrow 0 \ \ {\mathrm{as}} \ \ n \rightarrow \infty \eqno (5.1)$$ is unique. This solution is given by $$A_n = -2\pi \beta F_1+(-1)^nA_0F_1\cdots F_n, \eqno (5.2)$$ where $F_n$ is defined by $$F_n = \frac{1}{ \frac{2\lambda_n}{\beta} - \frac{1}{ \frac{2\lambda_{n+1}}{\beta} - \frac{1}{ \frac{2\lambda_{n+2}}{\beta} \cdots}}} \eqno (5.3)$$ and the real constant $A_0$ is determined from $$-4cA_0 + \beta(A_1 + \bar A_1) =0 . \eqno (5.4)$$ Indeed, the definition $F_n$ implies $$F_n = \frac{1}{\frac{2\lambda_n}{\beta} - F_{n+1}}.$$ Solving this equation for $2\lambda_n/\beta$ we find $$\frac{2\lambda_n}{\beta} = F_{n+1} + \frac{1}{F_n}.$$ Substituting this expression into equation (3.4b) we obtain $$A_{n+1} + F_{n+1}A_n + \frac{1}{F_n} (A_n + F_nA_{n-1}) = -2\pi\beta\delta_{n,1}, \quad n=0,1,2,\cdots \eqno (5.5)$$ Letting $$G_n \Doteq A_n + F_nA_{n-1},\eqno (5.6)$$ equation (5.5) becomes $$G_{n+1} + \frac{1}{F_n} G_n =-2\pi\beta\delta_n,1. \eqno (5.7)$$ Since $F_n\sim \frac{\beta}{2\lambda_n}$, $n\rightarrow \infty$, it follows that there does [**not**]{} exist a homogeneous solution of equation (5.7) that decays as $n\rightarrow \infty$. Thus $G_n
=-2\pi\beta F_1\delta_n,1$, and equation (5.6) yields $$A_n + F_nA_{n-1} =-2\pi\beta F_1\delta_{n,1}.$$ The unique solution of this equation is given by (5.2). Evaluating equation (3.4b) at $n=0$ and using $A_{-1} = \bar A_1$ we find (5.4).

Equation (5.2) implies that $$A_n \sim (-1)^nA_0 \left( \frac{\beta}{2}\right)^n \prod^n_1 \frac{1}{\lambda_j}, \quad n \rightarrow \infty, \eqno (5.8)$$ thus the series (3.3) converges.

#### The asymptotic behavior as $\beta \rightarrow 0$

  The definition of $\lambda_n$ (equation (3.1)) together with the requirement that Re $\lambda_n<0$, imply that if $c=\beta^2k$, then $$\lambda_n =-\sqrt{n}(1+i) + O(\beta^2). \eqno (5.9)$$ Equations (5.2), (5.3) yield $$A_0 = a_0 + O(\beta), A_1 = \beta a_1 + O(\beta^2), A_2 = \beta^2 a_2 + O(\beta^3), \cdots . \eqno (5.10)$$ Equation (3.4b) with $n=0,1,$ becomes $$n=0: \quad A_1 + \bar A_1 - 4k\beta A_0 = 0 \eqno (5.11)$$ $$n = 1: \quad 2\lambda_1 A_1 + \beta(A_2 + A_0) + 2\pi\beta =0. \eqno (5.12)$$ Substituting equations (5.10) into (5.11), (5.12), we find $$a_1 + \bar a_1 = 4ka_0 \eqno (5.13a)$$ $$-2(1+i)a_1 + a_0 + 2\pi =0. \eqno (5.13b)$$ Equation (5.13b) yields $$a_1 = \frac{(a_0+2\pi)(1-i)}{4}. \eqno (5.14a)$$ Substituting this expression in equation (5.13a) and using that $a_0$ is real, we find $$a_0 = \frac{2\pi}{8k-1}. \eqno (5.14b)$$ Substituting (5.10) into the expression for $\varphi(x,t)$ (equation (3.3)) we obtain $$\varphi(x,t) = \frac{1}{2\pi} \left\{ \frac{2\pi}{8k-1} e^{-2\beta^2kx} + \frac{4k\pi}{8k-1} e^{-k\beta^2x} \left[ (1-i)e^{-(1+i)x+it} + (1+i)e^{-(1-i)x-it}\right] \beta \right\} + O(\beta^2).$$

#### 1. $x =O(\beta^{-2})$

$$u = \frac{2\beta k}{1+(8k-1)e^{2\beta^2kx}}. \eqno (5.15)$$

#### 2. $x=O(1)$

$$u(x,t) = \frac{1}{2} \left[ e^{-(1+i)x+it} + e^{-(1-i)x-it} \right] + O(\beta). \eqno (5.16)$$

Appendix {#appendix .unnumbered}
========

#### The Analysis of equation (3.4b)

  Let $A_n$ satisfy the linear homogeneous difference equation $$A_{n+1} + \frac{2\Lambda_n}{\beta} A_n + A_{n-1} =0, \quad n=2,3,\cdots \eqno (A.1)$$ where $\beta$ is a constant and $\Lambda_n\rightarrow\infty$ monotonically as $n\rightarrow \infty$, for example $\Lambda_n =
\lambda_n$, where $\lambda_n$ is defined by equation (3.1). We shall show that $$A_n \sim \left( -\frac{\beta}{2}\right)^{n-1} \left[ c_1 \prod^{n-1}_1 \frac{\Lambda_j}{\Lambda^2_{j-1}} + c_2 \prod^{n-1}_1 \Lambda_j\right], \quad n \rightarrow \infty. \eqno (A.2)$$ Indeed, we first make the change of variables $$A_n= B_n \prod^{n-1}_{j=1} \Lambda^2_j. \eqno (A.3)$$ Thus equation (A.1) becomes $$B_{n+1} + \frac{2}{\beta \Lambda_n} B_n + \frac{B_{n-1}}{\Lambda^2_{n-1}\Lambda^2_n} = 0, \quad n = 2,3,\cdots \eqno (A.4)$$ We write this equation in matrix form, $$\Psi_n = V_n \Psi_{n-1}, V_n = \left( \begin{array}{ll}
0 & 1 \\ \\
- \frac{1}{\Lambda^2_{n-1}\Lambda_n^2} & - \frac{2}{\beta
\Lambda_n}
\end{array} \right), \quad \Psi_n = \left( \begin{array}{ll} B_n
\\ \\ B_{n+1} \end{array} \right). \eqno (A.5)$$ The eigenvalues and the eigenvectors of the matrix $V_n$ are $$\mu^\pm_n = - \frac{1}{\beta \Lambda_n} \pm \sqrt{ \frac{1}{\beta^2\Lambda_n^2} - \frac{1}{\Lambda^2_n \Lambda^2_{n-1} }}, \quad \left(
\begin{array}{ll}
1 \\ \\ \mu^\pm_n \end{array} \right). \eqno (A.6)$$ Using the gauge transformation $$\Psi_n = T_n\Phi_n, \quad T_n =
\left( \begin{array}{ll}
1 & 1 \\ \\
\mu^+_n & \mu^-_n  \end{array} \right), \eqno (A.7)$$ we find $$\Phi_n = M_n T_n^{-1} T_{n-1} \Phi_{n-1}, \quad M_n =
\left( \begin{array}{ll}
\mu^+_n & 0\\ \\
0 & \mu^-_n \end{array} \right). \eqno (A.8)$$ Equation (A.6a) implies $$\mu^+_n = - \frac{\beta}{2\Lambda_n\Lambda_{n-1}^2} \left( 1 + O(\frac{1}{\Lambda_n^2})\right),$$ $$\mu^-_n = - \frac{2}{\beta \Lambda_n} \left( 1 + O(\frac{1}{\Lambda_n^2})\right). \eqno (A.9)$$ These estimates and the explicit form of $T_n$ imply that $$T^{-1}_nT_{n-1} = I + O(\frac{1}{\Lambda_n^3}),$$ thus the WKB approximation of equation (A.8a) yields the two equations $$f^+_n = \mu^+_n f_{n-1}, \quad f^-_n = \mu^-_nf_{n-1}.$$ We recall that the general solution of the first order difference equation $F_{n+1} = g_nF_n$, is given by $$F_n = g_1 \ldots g_{n-1}.$$ Thus to the leading order as $n\rightarrow\infty$, $$f^+_n = \prod^{n-1}_1 \left( - \frac{\beta}{2\Lambda_j\Lambda_{j-1}^2} \right) = (- \frac{\beta}{2})^{n-1} \prod^{n-1}_1 \frac{1}{\Lambda_j\Lambda_{j-1}^2},$$ $$f^-_n = \prod^{n-1}_1 \left( - \frac{\beta}{2\Lambda_j}\right) = (-\frac{\beta}{2})^{n-1} \prod^{n-1}_1 \frac{1}{\Lambda_j}.$$ Hence using (A.7) and (A.3) we find (A.2).

Acknowledgments {#acknowledgments .unnumbered}
===============

ASF is grateful to A.A. Kapaev for useful suggestions.

[99]{}

D.J. Benney, The flow iduced by a disc oscillating in its own plane, J. Fluid Mech. [**18**]{}, 385-391 (1964).

F. Calogero and S. De Lillo, The Burgers equation of the semiline with general boundary conditions at the origin, J. Math Phys. [**23**]{}, 99-105 (1991).

J.N. Hunt and B. Johns, Currents induced by tides and gravity waves, Tellus [**15**]{}, 343-351 (1963).

M.S. Longuet-Higgins, Mass transport in water waves, Phil. Trans. A [**245**]{}, 535-581 (1953).

M.S. Longuet-Higgins, Mass transport in the boundary layer at a free oscillating surface, J. Fluid Mech. [**8**]{}, 293-306 (1960).

Lord Rayleigh, On the circulations of air observed in Kundt’s tubes and on some allied acoustical problems, Phil. Trans. A [**175**]{}, 1-71 (1883). Also, Scientific Papers [**2**]{}, 239-257.

N. Riley, Acoustic streaming, Theor. Comp. Fluid Dyn. [**10**]{}, 349-356 (1998).

H. Schlichting, Berechnung ebner periodischer Grenzschichtströmungen, Phys. Z. [**33**]{} 327-335 (1932).

G.G. Stokes, On the effect of the interval friction of fluids on the motion of pendulums, Trans. Camb. Phil. Soc. [**9**]{}, 8-106 (1851).

J.T. Stuart, Unsteady boundary layers. In [*Laminar Boundary Layers*]{}, (ed. L. Rosenhead), pp. 347-408. Oxford: Clarendon Press (1963).

J.T. Stuart, Double boundary layers in oscillatory viscous flow, J. Fluid Mech. [**24**]{}, 673-687 (1966).
