# An Integral Equation Formalism for Integrating a Nonlinear Initial-Boundary Value Problem for a Boussinesq Equation

Authors: T. S. Jang
Journal: Mathematical Problems in Engineering
 (2020)

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## Abstract

In this paper, a new nonlinear initial-boundary value problem for a Boussinesq equation is formulated. And a coupled system of nonlinear integral equations, equivalent to the new initial-boundary value problem, is constructed for integrating the initial-boundary value problem, but which is inherently different from other conventional formulations for integral equations. For the numerical solutions, successive approximations are applied, which leads to a functional iterative formula. A propagating solitary wave is simulated via iterating the formula, which is in good agreement with the known exact solution.


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## Body

## 1. Introduction

We begin with formulating a nonlinear initial-boundary value problem for the nonlinear partial differential equation (PDE) [1]:(1)utt−c02uxx+12u2xt=13h02uxxtt,x>0,in which constant h0 denotes water depth and constant c0 characteristic velocity, i.e.,(2)c0=gh0,with gravitation acceleration denoted by g. The nonlinear PDE (1) describes evolution of nonlinear dispersive waves in shallow water propagating in a uniform channel of water depth h0 under gravitation g, where u in (1) represents depth-average velocity in horizontal direction x. In fact, the model equation (1) is chosen for mathematical simplicity, as a starting point in this paper, which is the time evolution equation of the lowest order mathematical model for nonlinear and dispersive long wave; however, the physics represented by (1) is clearly included in higher order Boussinesq’s models [2]. Generally, the differential equations of the Boussinesq type such as (1) indicate essential model equations in physics and applied mathematics, the time evolution of which is an issue of scientific as well as engineering importance.Next, we would like to solve the nonlinear initial-boundary value problem formulated above through the construction of an (equivalent) integral equation formalism, a primary purpose of this paper. To this end, we first build an auxiliary (initial-boundary value) problem, associated with the (original) initial-boundary value problem. Thereby, it is possible to build (regular) coupled nonlinear integral equations of second kind, being equivalent to the present initial-boundary value problem. Applying the method of successive approximation to the integral equations, we arrive at a functional iterative formula. It is a semianalytic and derivative-free recurrence relation, which generates a functional sequence of improving approximate solutions of the initial-boundary value problem. It is emphasized that the only thing needed for iterating the recurrence is simply the usual numerical regular (not singular) integration in each step of iteration. By iterating the recurrence, we simulate, for example, a moving solitary wave, which agrees with the analytical solution well.In referring to the Boussinesq type equations such as (1), a considerable number of studies has been conducted over the past few decades, but most of them are limited to just their “initial” value problems (e.g., [3]), while the present study concerns the initial-“boundary” value problem as discussed at the beginning of this section. To name a few, El-Zoheiry [4] performed a numerical study of the classical Boussinesq equation using a three-level iterative scheme based on the compact implicit method, where a solitary wave solution of the equation was examined for the accuracy and efficiency. Wang et al. [5] analyzed the improved Boussinesq equation using an energy-preserving finite volume element method. Finally, in a recent year, Jang [6] investigated the classical Boussinesq equation by proposing a new dispersion-relation preserving (DRP) method and further improved the convergence characteristic of the DRP method for integrating the Boussinesq equation [7].


## 2. Initial-Boundary Value Problem and Auxiliary Problem

We consider the nonlinear PDE offourth order (1), as discussed in the preceding section, but on the half interval ℝ+=0,∞, with the dispersion relation [6, 7] between wave frequency ωΒ and wave number k:(3)ωB2=c02k21+h02k2/3.For PDE (1), we shall formulate a (initial-boundary value) problem subject to initial conditions:(4)ux,0=0,utx,0=0,0<x<∞,together with boundary condition at x=0+:(5)u0+,t=Ut,t>0,and at (positive) infinity,(6)u,ux⟶0,asx⟶∞.Associated with the initial-boundary value problem formulated above, we will next build anauxiliary problem.Definition 1.
As a counterpart of the original initial-boundary problem of (1) with (4)–(6), let an auxiliary initial-boundary value problem for uAux be defined as(7)uAuxtt−c02uAuxxx+12uAux2xt=13h02uAuxxxtt,−∞<x<0,t>0,subject to initial conditions at t=0:(8)uAuxx,0=0,uAuxtx,0=0,−∞<x<0,as well as, boundary conditions at x=0−:(9)uAux0−,t=−Ut,t>0,and at negative infinity:(10)uAux,uAuxx⟶0,asx⟶−∞.Remark 1.
The auxiliary initial-boundary value problem in Definition1, defined on the half interval ℝ−=−∞,0, is in fact constructed in such a way that(11)−uAuxx,t=u−x,t,−∞<x<0,t>0.
The original and auxiliary problems may be combined to form a new problem through the next definition.Definition 2.
u
Ext is a function from ℝ×0,T to ℝ for T∈ℝ+ satisfying a new (extended) initial-boundary value problem, being governed by the same Boussinesq PDE (1) but in a weak (derivative) sense:(12)uExttt−c02uExtxx+12uExt2xt=13h02uExtxxtt,−∞<x<∞,t>0,together with (null) initial conditions(13)uExtx,0=0,uExttx,0=0,−∞<x<∞,jump discontinuity at x=0(14)uExt0+,t=Ut,uExt0−,t=−Ut,t>0,and localized conditions(15)uExt,uExtx⟶0,asx⟶±∞.Remark 2.
Forx>0, the new initial-boundary value problem for uExt in Definition 2 reduces to the original initial-boundary value problem for u; while, for x<0, the problem for uExt remains the same as the auxiliary problem for uAux.
The above remarks immediately give the following lemma.Lemma 1.
u
Ext in Definition 2 is the odd extension of u with respect to x in the original problem of (1) with (4)–(6); thus, uExt is antisymmetric with respect to x=0 (See Figure 1).Figure 1
Two boundary velocities and their resulting (velocity) wave profiles.Proof.
u
Ext
x
,
t becomes ux,t when x>0 from Remark 2, i.e., uExtx,t=ux,t, for x>0. However, for x<0,(16)uExtx,t=uAuxx,t=−u−x,t,due to Remarks 1 and 2. This completes the proof.It would be instructive to note that, as indicated in Figure1, the original initial-boundary value problem for ux>0 corresponds to a physical problem, where a wavemaker, located at x=0, generates a propagating (velocity) wave with specified wavemaker-velocity Ut.


## 3. Integral Formalism foruExt

This section is devoted to the construction of an integral formulation which is equivalent to the extended initial-boundary value problem (appearing in Definition2) in the previous section.


### 3.1. Pseudodissipative Differential Formulation

We start by introducing two pseudoparameters ofα,β>0 in (12), adding the sum α⋅uExtt+β⋅uExt to both sides of (12) and arranging, which leaves us with(17)uExttt−c02uExtxx−13h02uExtxxtt+α⋅uExtt+β⋅uExt=φ,−∞<x<∞,and a forcing term φ with the functional form(18)φ≡−12uExt2xt+α⋅uExtt+β⋅uExt.Notice that the pair of (17) and (18) is a pseudodissipative differential formulation, but which is mathematically still equivalent to the (extended) Boussinesq equation (12); both the parameters, α,β>0, are directly related to a pseudodissipation (which will be discussed later). Thereby, we would like to build an equivalent (nonlinear) pseudodissipative integral formalism in Section 3.5. We begin with integral transforms for (17), which are explained in the next section.


### 3.2. Image ofuExt under Laplace–Fourier Transform

We first discover the image of Laplace transformL of uExt in (17), with respect to time t,(19)LuExt≡∫0∞uExtx,t⋅e−stdt≡uExt∗x,s,for a (Laplace transform) parameter s. Here, we assume that uExt is Laplace transformable.Lemma 2.
Given two positive pseudoparameters,α,β>0, PDE (17) for uExt can be converted into the following ordinary differential equation (ODE) for uExt∗:(20)s2⋅uExt∗−c02d2uExt∗dx2−h023d2dx2s2⋅uExt∗+αs⋅uExt∗+β⋅uExt∗=φ∗,for a parameter s.Proof.
We remind the differentiation properties of Laplace transform(21)uExt∗t=s⋅uExt∗−uExtx,0=s⋅uExt∗,(22)uExt∗tt=s2⋅uExt∗−s⋅uExtx,0−uExttx,0=s2⋅uExt∗,due to the null initial conditions (13). Applying Laplace transform to (17), combined with the properties of (21) and (22), leads to (20), which proves the lemma.Next, we denote Fourier transformF of uExt∗x,s, with respect to space x, by(23)FuExt∗=∫−∞∞uExt∗x,s⋅eikxdx≡u¯Ext∗k,s,for a parameter k and its inverse by F−1, i.e.,(24)uExt∗x,s=F−1u¯Ext∗=12π∫−∞∞u¯Ext∗k,s⋅e−ikxdk,in which uExt∗x,s and u¯Ext∗k,s are assumed to be Fourier transformable and inverse Fourier transformable, respectively. Then, applying Fourier transform F to ODE (20), we find that(25)s2⋅u¯Ext∗−c02⋅Fd2uExt∗dx2−h023s2⋅Fd2uExt∗dx2+αs⋅u¯Ext∗+β⋅u¯Ext∗=φ¯∗,which becomes an algebraic equation for u¯Ext∗ because of the following lemma.Lemma 3.
The boundary velocityU in (5) appears in Fourier transform of the second derivative, d2uExt∗/dx2, as follows:(26)Fd2uExt∗dx2=−ik2u¯Ext∗+2ikU∗,where U∗ stands for Laplace transform of (5), i.e.,(27)U∗s≡uExt∗x=0+,s.Proof.
At first, we consider,(28)FduExt∗dx=∫−∞∞ddxuExt∗x,s⋅eikxdx=∫0∞ddxuExt∗x,s⋅eikxdx+∫−∞0ddxuExt∗x,s⋅eikxdx=uExt∗x,s⋅eikx0+∞−ik∫0∞uExt∗x,s⋅eikxdx+uExt∗x,s⋅eikx−∞0−−ik∫−∞0uExt∗x,s⋅eikxdx=−ik∫−∞∞uExt∗x,s⋅eikxdx−2U∗s=−ik⋅FuExt∗−2U∗s,where integration by parts is used with (14).
Similarly, we further have(29)Fd2uExt∗dx2=∫−∞∞d2dx2uExt∗x,s⋅eikxdx=∫0∞d2dx2uExt∗x,s⋅eikxdx+∫−∞0d2dx2uExt∗x,s⋅eikxdx=ddxuExt∗x,s⋅eikx0+∞−ik∫0∞ddxuExt∗x,s⋅eikxdx+ddxuExt∗x,s⋅eikx−∞0−−ik∫−∞0ddxuExt∗x,s⋅eikxdx=−ik∫−∞∞ddxuExt∗x,s⋅eikxdx=−ik⋅FduExt∗dx.
Here, note that(30)ddxuExt∗x,s⋅eikx0+∞+ddxuExt∗x,s⋅eikx−∞0−=−ddxuExt∗0+,s+ddxuExt∗0−,s∵localized conditions15,=limε1,ε2⟶+0L−uExtxε1,t+uExtx−ε2,t=limε1,ε2⟶+0L−uxε1,t+uxε2,t,=L−ux0+,t+ux0+,t=0,because of the property of odd extension by Lemma 1, i.e.,(31)uExtxx,t=uxx,t,forx>0,and the chain rule(32)uExtxx,t=∂∂x−u−x,t=∂∂−xu−x,t=∂∂XuX,t=uxx,tx=X=ux−x,t,forx<0,from (16).
Plugging (28) into (29) gives(33)Fd2uExt∗dx2=−ik−ik⋅FuExt∗−2U∗s=−ik2u¯Ext∗+2ikU∗,which completes the proof.With the use of Lemma3, (25) is rewritten as(34)s2⋅u¯Ext∗−c02−ik2u¯Ext∗+2ikU∗−h023s2⋅−ik2u¯Ext∗+2ikU∗+αs⋅u¯Ext∗+β⋅u¯Ext∗=φ¯∗,which can be solved for u¯Ext∗, i.e.,(35)u¯Ext∗=11+k2h02/3⋅φ¯∗+U∗s⋅2ikc02+h02/32iks2s2+c02k2+sα+β/1+k2h02/3.Even though the above expression presents the (desired) exact result for the image ofuExt under the Laplace-Fourier transform, however, the content seems to be hard to see. To overcome the difficulty, the next section discusses an appropriate arrangement of (35) suitable for finding the image of uExt under Laplace–Fourier transform.


### 3.3. Pseudodissipative Parameter and Frequency

This section gives an appropriate decomposition of the algebraic expression foru¯Ext∗ in (35) derived in the previous section, being realized by introducing a new concept of pseudodissipation. The new concept will make it more efficient (i.e., concise and clear) to analyze the problem both physically as well as mathematically. We need some definitions.Definition 3 (see [6]).
Givenβ>0, we define a frequency function ω˜B:ℝ⟶ℝ+ such that(36)ω˜B2k≡c02k2+β1+k2h02/3>0.
If we useω˜B in the above Definition, the denominator of the second fraction in the right hand of (35) arranges to(37)s2+c02k2+sα+β1+k2h02/3=s2+sα1+k2h02/3+ω˜B2=s2+sαω˜B2c02k2+β+ω˜B2=s2+2sαω˜B22c02k2+β+ω˜B2=s2+2sαω˜B22c02k2+β+αω˜B22c02k2+β2+ω˜B2−αω˜B22c02k2+β2=s+αω˜B22c02k2+β2+ω˜B21−α2ω˜B24c02k2+β2,and (35) may be rewritten as(38)u¯Ext∗=11+k2h02/3⋅φ¯∗+U∗s⋅2ikc02+h02/3s2s+αω˜B2/2c02k2+β2+ω˜B21−α2ω˜B2/4c02k2+β2,by substituting (37) into (35).Definition 4.
ζ is defined as a (dimensionless) pseudodissipative parameter (dependent of frequency) whose absolute value is less than unity, relying on α,β>0:(39)−1<ζ≡αω˜B2c02k2+β<1.Definition 5.
Givenζ<1, ω˜BpD is a function ω˜BpD:ℝ⟶ℝ+, termed as a pseudodissipative frequency, such that(40)ω˜BpD≡ω˜B1−ζ2,where ω˜B denotes the frequency function in Definition 3.
The next lemma is immediately followed by the above definitions.Lemma 4.
LetΑ and Β be denoted by(41)Α=φ¯∗+2ikc02⋅U∗s1+k2h02/3,Β=2ikh02⋅U∗s31+k2h02/3,respectively. Then, the expression of u¯Ext∗ in (38) may be shortened to a simpler form via ζ and ω˜BpD of Definitions 4 and 5, respectively, i.e.,(42)u¯Ext∗=A+s2⋅Bs+ζω˜B2+ω˜BpD2=As+ζω˜B2+ω˜BpD2+s2⋅Bs+ζω˜B2+ω˜BpD2.Proof.
Writing (38) in terms of A and B in the above as follows:(43)u¯Ext∗=A+s2⋅Bs+αω˜B2/2c02k2+β2+ω˜B21−α2ω˜B2/4c02k2+β2,we are led to result (42) due to Definitions 4 and 5. This completes the proof.


### 3.4. Inverse Laplace–Fourier Transform

To begin with, we investigate the inverse Laplace transformL−1 of u¯Ext∗ in (42) (i.e., L−1u¯Ext∗=u¯Ext) based on the results in the preceding section and the (well known) identities (from mathematical table):(44)Leatsinbt=bs−a2+b2,Leatcosbt=s−as−a2+b2,fora,b≥0.Lemma 5.
The first fraction involvingA in the right hand of (42) can be expressed as(45)As+ζω˜B2+ω˜BpD2=Aω˜BpD⋅Lexp−ζω˜B⋅t⋅sinω˜ΒpD⋅t.Proof.
It is noted that(46)As+ζω˜B2+ω˜BpD2=Aω˜BpD⋅ω˜BpDs+ζω˜B2+ω˜BpD2=Aω˜BpD⋅Lexp−ζω˜B⋅t⋅sinω˜BpD⋅t,where (44) is used. This immediately proves the lemma.Lemma 6.
The second fraction involvingB in the right hand of (42) decomposes into(47)s2⋅Bs+ζω˜B2+ω˜BpD2=B−2Bζω˜B⋅Lexp−ζω˜B⋅t⋅cosω˜BpD⋅t+Bω˜B2ζ2−11−ζ2⋅Lexp−ζω˜B⋅t⋅sinω˜BpD⋅t.Proof.
LettingF denote the second fraction involving B, we have(48)F=s2⋅Bs+ζω˜B2+ω˜BpD2=B⋅s+ζω˜B2+ω˜BpD2−2sζω˜B−ζ2ω˜B2−ω˜BpD2s+ζω˜B2+ω˜BpD2=B⋅1−2sζω˜B+ζ2ω˜B2+ω˜BpD2s+ζω˜B2+ω˜BpD2=B−2Bζω˜B⋅s+ζω˜B−ζω˜Bs+ζω˜B2+ω˜BpD2−B⋅ζ2ω˜B2s+ζω˜B2+ω˜BpD2−B⋅ω˜BpD2s+ζω˜B2+ω˜BpD2=B−2Bζω˜B⋅s+ζω˜Bs+ζω˜B2+ω˜BpD2+B⋅ζ2ω˜B2s+ζω˜B2+ω˜BpD2−B⋅ω˜BpD2s+ζω˜B2+ω˜BpD2.
Here, we note that(49)−2Bζω˜B⋅s+ζω˜Bs+ζω˜B2+ω˜BpD2=−2Bζω˜B⋅Lexp−ζω˜B⋅t⋅cosω˜BpD⋅t,(50)B⋅ζ2ω˜B2s+ζω˜B2+ω˜BpD2=Bζ2ω˜B2ω˜BpD⋅ω˜BpDs+ζω˜B2+ω˜BpD2=Bζ2ω˜B1−ζ2⋅Lexp−ζω˜B⋅t⋅sinω˜BpD⋅t,(51)−B⋅ω˜BpD2s+ζω˜B2+ω˜BpD2=−Bω˜BpD⋅ω˜BpDs+ζω˜B2+ω˜BpD2=−Bω˜BpD⋅Lexp−ζω˜B⋅t⋅sinω˜BpD⋅t,from (40) and (44). And, (50) can be combined to (51), yielding(52)Bζ2ω˜B1−ζ2⋅Lexp−ζω˜B⋅t⋅sinω˜BpD⋅t−Bω˜BpD⋅Lexp−ζω˜B⋅t⋅sinω˜BpD⋅t=Bζ2ω˜B1−ζ2−ω˜BpD⋅Lexp−ζω˜B⋅t⋅sinω˜BpD⋅t=Bζ2ω˜B1−ζ2−ω˜B1−ζ21−ζ2⋅Lexp−ζω˜B⋅t⋅sinω˜BpD⋅t∵ω˜BpD=ω˜B1−ζ2=ω˜B1−ζ21−ζ2=Bω˜B2ζ2−11−ζ2⋅Lexp−ζω˜B⋅t⋅sinω˜BpD⋅t.
Therefore, substituting (49) and (52) into F in (48) leads to (47), completing the proof.It will be convenient, if we denote byfi, i=1,2,3,4,(53)f1k,s=Aω˜BpD⋅Lexp−ζω˜B⋅t⋅sinω˜BpD⋅t,(54)f2k,s=B,(55)f3k,s=−2Bζω˜B⋅Lexp−ζω˜B⋅t⋅cosω˜BpD⋅t,(56)f4k,s=Bω˜B2ζ2−11−ζ2⋅Lexp−ζω˜B⋅t⋅sinω˜BpD⋅t,so that u¯Ext∗ in (42) can be simply represented as the sum of fi’s from Lemmas 5 and 6, i.e.,(57)u¯Ext∗=∑i=14fik,s.Lemma 7.
Givenfi, i=1,2,3,4, in (53)–(56), their inverse Laplace transforms are calculated as(58)L−1f1=1ω˜BpD1+k2h02/3∫0texp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τ⋅φ¯k,τdτ+2ikc02ω˜BpD1+k2h02/3∫0texp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τ⋅Uτdτ,(59)L−1f2=2ikh0231+k2h02/3⋅Ut,(60)L−1f3=−4ikh02ζω˜B31+k2h02/3∫0texp−ζω˜B⋅t−τ⋅cosω˜BpD⋅t−τ⋅Uτdτ,(61)L−1f4=2ikh02ω˜B2ζ2−131+k2h02/31−ζ2∫0texp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τ⋅Uτdτ.Proof.
To prove (58), note that L−1f1 has the explicit expression:(62)L−1f1=L−1Aω˜BpD⋅Lexp−ζω˜B⋅t⋅sinω˜BpD⋅t∵53=L−1φ¯∗+2ikc02⋅U∗sω˜BpD1+k2h02/3⋅Lexp−ζω˜B⋅t⋅sinω˜BpD⋅t∵41=1ω˜BpD1+k2h02/3⋅φ¯k,t∘exp−ζω˜B⋅t⋅sinω˜BpD⋅t+2ikc02ω˜BpD1+k2h02/3⋅Ut∘exp−ζω˜B⋅t⋅sinω˜BpD⋅t=1ω˜BpD1+k2h02/3∫0texp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τ⋅φ¯k,τdτ+2ikc02ω˜BpD1+k2h02/3∫0texp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τ⋅Uτdτ,where the notation ∘ denotes Laplace convolution and L−1L is the identity. This completes the proof of (58).□
The inverse Laplace transform off2 is written as (simply through B in (41))(63)L−1f2=L−1B∵54=L−12ikh02⋅U∗s31+k2h02/3∵41=2ikh02⋅Ut31+k2h02/3,by linearity of L−1, which proves (59).Similarly as above,(64)L−1f3=L−1−2Bζω˜B⋅Lexp−ζω˜B⋅t⋅cosω˜BpD⋅t∵54=−4ikh02ζω˜B31+k2h02/3⋅L−1U∗s∘L−1Lexp−ζω˜B⋅t⋅cosω˜BpD⋅t∵41=−4ikh02ζω˜B31+k2h02/3⋅Ut∘exp−ζω˜B⋅t⋅cosω˜BpD⋅t=−4ikh02ζω˜B31+k2h02/3∫0texp−ζω˜B⋅t−τ⋅cosω˜BpD⋅t−τ⋅Uτdτ,which completes the proof of (60).□Finally,L−1f4 has the form(65)L−1f4=L−1Bω˜B2ζ2−11−ζ2⋅Lexp−ζω˜B⋅t⋅sinω˜BpD⋅t∵56=2ikh02ω˜B2ζ2−131+k2h02/31−ζ2⋅L−1U∗s∘L−1Lexp−ζω˜B⋅t⋅sinω˜BpD⋅t∵41=2ikh02ω˜B2ζ2−131+k2h02/31−ζ2⋅Ut∘exp−ζω˜B⋅t⋅sinω˜BpD⋅t=2ikh02ω˜B2ζ2−131+k2h02/31−ζ2∫0texp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τ⋅Uτdτ.This finishes proving (61).□We are now in a position to take the inverse Laplace–Fourier transform ofu¯Ext∗ in (42).Lemma 8.
The inverse Laplace–Fourier transforms offi, i=1,2,3,4, in (53)–(56), are evaluated in the forms(66)F−1L−1f1=1π∫0t∫0∞∫−∞∞exp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τω˜BpD1+k2h02/3⋅coskξ−x⋅φξ,τdξdkdτ+1π∫0t∫0∞2kc02ω˜BpD1+k2h02/3⋅exp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τ⋅sinkx⋅Uτdkdτ,(67)F−1L−1f2=sgnx⋅Ut⋅exp−x3h0,(68)F−1L−1f3=−1π∫0t∫0∞4kh02ζω˜B31+k2h02/3⋅exp−ζω˜B⋅t−τ⋅cosω˜BpD⋅t−τ⋅sinkx⋅Uτdkdτ,(69)F−1L−1f4=1π∫0t∫0∞2kh02ω˜B2ζ2−131+k2h02/31−ζ2⋅exp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τ⋅sinkx⋅Uτdkdτ.Proof.
To prove (66), we take the inverse Fourier transform for (58), leaving(70)F−1L−1f1=12π∫−∞∞1ω˜BpD1+k2h02/3∫0texp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τ⋅φ¯k,τdτ⋅e−ikxdk∵24+12π∫−∞∞2ikc02ω˜BpD1+k2h02/3⋅∫0texp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τ⋅Uτdτ⋅e−ikxdk=12π∫0t∫−∞∞exp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τω˜BpD1+k2h02/3⋅∫−∞∞φξ,τ⋅eikξdξ⋅e−ikxdkdτ+12π∫0t∫−∞∞2ikc02ω˜BpD1+k2h02/3⋅exp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τ⋅Uτ⋅e−ikxdkdτ=12π∫0t∫−∞∞∫−∞∞exp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τω˜BpD1+k2h02/3⋅eikξ−x⋅φξ,τdξdkdτ+12π∫0t∫−∞∞2ikc02ω˜BpD1+k2h02/3⋅exp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τ⋅e−ikx⋅Uτdkdτ.
Here, we note that theF−1L−1f1 remains real because the real parts of the integrands in the last two integrals in the above are even in k, whereas the imaginary parts are odd. We are thus led to (66), completing the proof.For (67), we have(71)F−1L−1f2=12π∫−∞∞2ikh0231+k2h02/3⋅Ut⋅e−ikxdk∵59and24=12π∫−∞∞2ikh0231+k2h02/3⋅Ut⋅coskx−isinkxdk=1π∫0∞2kh0231+k2h02/3⋅sinkx⋅Utdk,where we have used the fact that the imaginary part of the integrand in the above is odd in k while the real part is even.An analytic expression is obtained for the above integral, that is,(72)F−1L−1f2=Ut⋅2π∫0∞h02/3⋅ksinkx1+k2h02/3dk=Ut⋅2π⋅π2xx⋅exp−x3h0=xx⋅Ut⋅exp−x3h0,by noting from (mathematical integral) table that(73)∫0∞bksinkx1+bk2dk=π2xxexp−xb,forb≥0.This proves (67).□For the proof of (68), we calculate F−1L−1f3, being written as(74)F−1L−1f3=12π∫−∞∞−4ikh02ζω˜B31+k2h02/3∫0texp−ζω˜B⋅t−τ⋅cosω˜BpD⋅t−τ⋅Uτdτ⋅e−ikxdk∵60and24=−1π∫0t∫0∞4kh02ζω˜B31+k2h02/3⋅exp−ζω˜B⋅t−τ⋅cosω˜BpD⋅t−τ⋅sinkx⋅Uτdkdτ.The above integral becomes real, due to the fact that the real part of the integrand is even ink, on the other hand, the imaginary part is odd. This leads to (68), completing the proof.□To show that (69) is true, recall the Fourier inverse (24) together with (61), i.e.,(75)F−1L−1f4=12π∫−∞∞2ikh02ω˜B2ζ2−131+k2h02/31−ζ2⋅∫0texp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τ⋅Uτdτ⋅e−ikxdk=12π∫0t∫−∞∞2ikh02ω˜B2ζ2−131+k2h02/31−ζ2⋅exp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τ⋅Uτ⋅e−ikxdkdτ=1π∫0t∫0∞2kh02ω˜B2ζ2−131+k2h02/31−ζ2⋅exp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τ⋅sinkx⋅Uτdkdτ,where the imaginary part in the above vanishes because the imaginary part of the integrand (in the second line in the above) is odd in k, but the real part of the integrand is even.□


### 3.5. Integral Formulation InvolvinguExt

Recalling that theuExt in Definition 2 is recovered formally through the identities(76)uExt=F−1L−1u¯Ext∗=∑i=14F−1L−1fi,because of (57), we readily arrive at the following theorem from the concerning lemmas without proof, but with a short note that the sum of the last line of (66) and (69) becomes(77)1π∫0t∫0∞2kc02ω˜BpD1+k2h02/3⋅exp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τ⋅sinkx⋅Uτdkdτ+1π∫0t∫0∞2kh02ω˜B2ζ2−131+k2h02/31−ζ2⋅exp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τ⋅sinkx⋅Uτdkdτ=1π∫0t∫0∞2kc02ω˜BpD1+k2h02/3⋅exp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τ⋅sinkx⋅Uτdkdτ+1π∫0t∫0∞2kh02ω˜B22ζ2−131+k2h02/3ω˜B1−ζ2⋅exp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τ⋅sinkx⋅Uτdkdτ=2π∫0t∫0∞3kc02+kh02ω˜B22ζ2−13ω˜BpD1+k2h02/3⋅exp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τ⋅sinkx⋅Uτdkdτ,where (40) is utilized.Theorem 1.
LetuExt be satisfied by the Boussinesq equation (12) subject to (13)–(15); i.e., uExt is a solution for the (extended) initial-boundary value problem in Definition 2. Then, the following integral relation holds(78)uExt=2π∫0t∫0∞3kc02+kh02ω˜B22ζ2−13ω˜BpD1+k2h02/3⋅exp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τ⋅sinkx⋅Uτdkdτ+sgnx⋅Ut⋅exp−x3h0−1π∫0t∫0∞4kh02ζω˜B31+k2h02/3⋅exp−ζω˜B⋅t−τ⋅cosω˜BpD⋅t−τ⋅sinkx⋅Uτdkdτ+1π∫0t∫0∞∫−∞∞exp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τω˜BpD1+k2h02/3⋅coskξ−x⋅φξ,τdξdkdτ,which is considered as an (basic) integral formalism for uExt because φ involves uExt via (18).


## 3.1. Pseudodissipative Differential Formulation

We start by introducing two pseudoparameters ofα,β>0 in (12), adding the sum α⋅uExtt+β⋅uExt to both sides of (12) and arranging, which leaves us with(17)uExttt−c02uExtxx−13h02uExtxxtt+α⋅uExtt+β⋅uExt=φ,−∞<x<∞,and a forcing term φ with the functional form(18)φ≡−12uExt2xt+α⋅uExtt+β⋅uExt.Notice that the pair of (17) and (18) is a pseudodissipative differential formulation, but which is mathematically still equivalent to the (extended) Boussinesq equation (12); both the parameters, α,β>0, are directly related to a pseudodissipation (which will be discussed later). Thereby, we would like to build an equivalent (nonlinear) pseudodissipative integral formalism in Section 3.5. We begin with integral transforms for (17), which are explained in the next section.


## 3.2. Image ofuExt under Laplace–Fourier Transform

We first discover the image of Laplace transformL of uExt in (17), with respect to time t,(19)LuExt≡∫0∞uExtx,t⋅e−stdt≡uExt∗x,s,for a (Laplace transform) parameter s. Here, we assume that uExt is Laplace transformable.Lemma 2.
Given two positive pseudoparameters,α,β>0, PDE (17) for uExt can be converted into the following ordinary differential equation (ODE) for uExt∗:(20)s2⋅uExt∗−c02d2uExt∗dx2−h023d2dx2s2⋅uExt∗+αs⋅uExt∗+β⋅uExt∗=φ∗,for a parameter s.Proof.
We remind the differentiation properties of Laplace transform(21)uExt∗t=s⋅uExt∗−uExtx,0=s⋅uExt∗,(22)uExt∗tt=s2⋅uExt∗−s⋅uExtx,0−uExttx,0=s2⋅uExt∗,due to the null initial conditions (13). Applying Laplace transform to (17), combined with the properties of (21) and (22), leads to (20), which proves the lemma.Next, we denote Fourier transformF of uExt∗x,s, with respect to space x, by(23)FuExt∗=∫−∞∞uExt∗x,s⋅eikxdx≡u¯Ext∗k,s,for a parameter k and its inverse by F−1, i.e.,(24)uExt∗x,s=F−1u¯Ext∗=12π∫−∞∞u¯Ext∗k,s⋅e−ikxdk,in which uExt∗x,s and u¯Ext∗k,s are assumed to be Fourier transformable and inverse Fourier transformable, respectively. Then, applying Fourier transform F to ODE (20), we find that(25)s2⋅u¯Ext∗−c02⋅Fd2uExt∗dx2−h023s2⋅Fd2uExt∗dx2+αs⋅u¯Ext∗+β⋅u¯Ext∗=φ¯∗,which becomes an algebraic equation for u¯Ext∗ because of the following lemma.Lemma 3.
The boundary velocityU in (5) appears in Fourier transform of the second derivative, d2uExt∗/dx2, as follows:(26)Fd2uExt∗dx2=−ik2u¯Ext∗+2ikU∗,where U∗ stands for Laplace transform of (5), i.e.,(27)U∗s≡uExt∗x=0+,s.Proof.
At first, we consider,(28)FduExt∗dx=∫−∞∞ddxuExt∗x,s⋅eikxdx=∫0∞ddxuExt∗x,s⋅eikxdx+∫−∞0ddxuExt∗x,s⋅eikxdx=uExt∗x,s⋅eikx0+∞−ik∫0∞uExt∗x,s⋅eikxdx+uExt∗x,s⋅eikx−∞0−−ik∫−∞0uExt∗x,s⋅eikxdx=−ik∫−∞∞uExt∗x,s⋅eikxdx−2U∗s=−ik⋅FuExt∗−2U∗s,where integration by parts is used with (14).
Similarly, we further have(29)Fd2uExt∗dx2=∫−∞∞d2dx2uExt∗x,s⋅eikxdx=∫0∞d2dx2uExt∗x,s⋅eikxdx+∫−∞0d2dx2uExt∗x,s⋅eikxdx=ddxuExt∗x,s⋅eikx0+∞−ik∫0∞ddxuExt∗x,s⋅eikxdx+ddxuExt∗x,s⋅eikx−∞0−−ik∫−∞0ddxuExt∗x,s⋅eikxdx=−ik∫−∞∞ddxuExt∗x,s⋅eikxdx=−ik⋅FduExt∗dx.
Here, note that(30)ddxuExt∗x,s⋅eikx0+∞+ddxuExt∗x,s⋅eikx−∞0−=−ddxuExt∗0+,s+ddxuExt∗0−,s∵localized conditions15,=limε1,ε2⟶+0L−uExtxε1,t+uExtx−ε2,t=limε1,ε2⟶+0L−uxε1,t+uxε2,t,=L−ux0+,t+ux0+,t=0,because of the property of odd extension by Lemma 1, i.e.,(31)uExtxx,t=uxx,t,forx>0,and the chain rule(32)uExtxx,t=∂∂x−u−x,t=∂∂−xu−x,t=∂∂XuX,t=uxx,tx=X=ux−x,t,forx<0,from (16).
Plugging (28) into (29) gives(33)Fd2uExt∗dx2=−ik−ik⋅FuExt∗−2U∗s=−ik2u¯Ext∗+2ikU∗,which completes the proof.With the use of Lemma3, (25) is rewritten as(34)s2⋅u¯Ext∗−c02−ik2u¯Ext∗+2ikU∗−h023s2⋅−ik2u¯Ext∗+2ikU∗+αs⋅u¯Ext∗+β⋅u¯Ext∗=φ¯∗,which can be solved for u¯Ext∗, i.e.,(35)u¯Ext∗=11+k2h02/3⋅φ¯∗+U∗s⋅2ikc02+h02/32iks2s2+c02k2+sα+β/1+k2h02/3.Even though the above expression presents the (desired) exact result for the image ofuExt under the Laplace-Fourier transform, however, the content seems to be hard to see. To overcome the difficulty, the next section discusses an appropriate arrangement of (35) suitable for finding the image of uExt under Laplace–Fourier transform.


## 3.3. Pseudodissipative Parameter and Frequency

This section gives an appropriate decomposition of the algebraic expression foru¯Ext∗ in (35) derived in the previous section, being realized by introducing a new concept of pseudodissipation. The new concept will make it more efficient (i.e., concise and clear) to analyze the problem both physically as well as mathematically. We need some definitions.Definition 3 (see [6]).
Givenβ>0, we define a frequency function ω˜B:ℝ⟶ℝ+ such that(36)ω˜B2k≡c02k2+β1+k2h02/3>0.
If we useω˜B in the above Definition, the denominator of the second fraction in the right hand of (35) arranges to(37)s2+c02k2+sα+β1+k2h02/3=s2+sα1+k2h02/3+ω˜B2=s2+sαω˜B2c02k2+β+ω˜B2=s2+2sαω˜B22c02k2+β+ω˜B2=s2+2sαω˜B22c02k2+β+αω˜B22c02k2+β2+ω˜B2−αω˜B22c02k2+β2=s+αω˜B22c02k2+β2+ω˜B21−α2ω˜B24c02k2+β2,and (35) may be rewritten as(38)u¯Ext∗=11+k2h02/3⋅φ¯∗+U∗s⋅2ikc02+h02/3s2s+αω˜B2/2c02k2+β2+ω˜B21−α2ω˜B2/4c02k2+β2,by substituting (37) into (35).Definition 4.
ζ is defined as a (dimensionless) pseudodissipative parameter (dependent of frequency) whose absolute value is less than unity, relying on α,β>0:(39)−1<ζ≡αω˜B2c02k2+β<1.Definition 5.
Givenζ<1, ω˜BpD is a function ω˜BpD:ℝ⟶ℝ+, termed as a pseudodissipative frequency, such that(40)ω˜BpD≡ω˜B1−ζ2,where ω˜B denotes the frequency function in Definition 3.
The next lemma is immediately followed by the above definitions.Lemma 4.
LetΑ and Β be denoted by(41)Α=φ¯∗+2ikc02⋅U∗s1+k2h02/3,Β=2ikh02⋅U∗s31+k2h02/3,respectively. Then, the expression of u¯Ext∗ in (38) may be shortened to a simpler form via ζ and ω˜BpD of Definitions 4 and 5, respectively, i.e.,(42)u¯Ext∗=A+s2⋅Bs+ζω˜B2+ω˜BpD2=As+ζω˜B2+ω˜BpD2+s2⋅Bs+ζω˜B2+ω˜BpD2.Proof.
Writing (38) in terms of A and B in the above as follows:(43)u¯Ext∗=A+s2⋅Bs+αω˜B2/2c02k2+β2+ω˜B21−α2ω˜B2/4c02k2+β2,we are led to result (42) due to Definitions 4 and 5. This completes the proof.


## 3.4. Inverse Laplace–Fourier Transform

To begin with, we investigate the inverse Laplace transformL−1 of u¯Ext∗ in (42) (i.e., L−1u¯Ext∗=u¯Ext) based on the results in the preceding section and the (well known) identities (from mathematical table):(44)Leatsinbt=bs−a2+b2,Leatcosbt=s−as−a2+b2,fora,b≥0.Lemma 5.
The first fraction involvingA in the right hand of (42) can be expressed as(45)As+ζω˜B2+ω˜BpD2=Aω˜BpD⋅Lexp−ζω˜B⋅t⋅sinω˜ΒpD⋅t.Proof.
It is noted that(46)As+ζω˜B2+ω˜BpD2=Aω˜BpD⋅ω˜BpDs+ζω˜B2+ω˜BpD2=Aω˜BpD⋅Lexp−ζω˜B⋅t⋅sinω˜BpD⋅t,where (44) is used. This immediately proves the lemma.Lemma 6.
The second fraction involvingB in the right hand of (42) decomposes into(47)s2⋅Bs+ζω˜B2+ω˜BpD2=B−2Bζω˜B⋅Lexp−ζω˜B⋅t⋅cosω˜BpD⋅t+Bω˜B2ζ2−11−ζ2⋅Lexp−ζω˜B⋅t⋅sinω˜BpD⋅t.Proof.
LettingF denote the second fraction involving B, we have(48)F=s2⋅Bs+ζω˜B2+ω˜BpD2=B⋅s+ζω˜B2+ω˜BpD2−2sζω˜B−ζ2ω˜B2−ω˜BpD2s+ζω˜B2+ω˜BpD2=B⋅1−2sζω˜B+ζ2ω˜B2+ω˜BpD2s+ζω˜B2+ω˜BpD2=B−2Bζω˜B⋅s+ζω˜B−ζω˜Bs+ζω˜B2+ω˜BpD2−B⋅ζ2ω˜B2s+ζω˜B2+ω˜BpD2−B⋅ω˜BpD2s+ζω˜B2+ω˜BpD2=B−2Bζω˜B⋅s+ζω˜Bs+ζω˜B2+ω˜BpD2+B⋅ζ2ω˜B2s+ζω˜B2+ω˜BpD2−B⋅ω˜BpD2s+ζω˜B2+ω˜BpD2.
Here, we note that(49)−2Bζω˜B⋅s+ζω˜Bs+ζω˜B2+ω˜BpD2=−2Bζω˜B⋅Lexp−ζω˜B⋅t⋅cosω˜BpD⋅t,(50)B⋅ζ2ω˜B2s+ζω˜B2+ω˜BpD2=Bζ2ω˜B2ω˜BpD⋅ω˜BpDs+ζω˜B2+ω˜BpD2=Bζ2ω˜B1−ζ2⋅Lexp−ζω˜B⋅t⋅sinω˜BpD⋅t,(51)−B⋅ω˜BpD2s+ζω˜B2+ω˜BpD2=−Bω˜BpD⋅ω˜BpDs+ζω˜B2+ω˜BpD2=−Bω˜BpD⋅Lexp−ζω˜B⋅t⋅sinω˜BpD⋅t,from (40) and (44). And, (50) can be combined to (51), yielding(52)Bζ2ω˜B1−ζ2⋅Lexp−ζω˜B⋅t⋅sinω˜BpD⋅t−Bω˜BpD⋅Lexp−ζω˜B⋅t⋅sinω˜BpD⋅t=Bζ2ω˜B1−ζ2−ω˜BpD⋅Lexp−ζω˜B⋅t⋅sinω˜BpD⋅t=Bζ2ω˜B1−ζ2−ω˜B1−ζ21−ζ2⋅Lexp−ζω˜B⋅t⋅sinω˜BpD⋅t∵ω˜BpD=ω˜B1−ζ2=ω˜B1−ζ21−ζ2=Bω˜B2ζ2−11−ζ2⋅Lexp−ζω˜B⋅t⋅sinω˜BpD⋅t.
Therefore, substituting (49) and (52) into F in (48) leads to (47), completing the proof.It will be convenient, if we denote byfi, i=1,2,3,4,(53)f1k,s=Aω˜BpD⋅Lexp−ζω˜B⋅t⋅sinω˜BpD⋅t,(54)f2k,s=B,(55)f3k,s=−2Bζω˜B⋅Lexp−ζω˜B⋅t⋅cosω˜BpD⋅t,(56)f4k,s=Bω˜B2ζ2−11−ζ2⋅Lexp−ζω˜B⋅t⋅sinω˜BpD⋅t,so that u¯Ext∗ in (42) can be simply represented as the sum of fi’s from Lemmas 5 and 6, i.e.,(57)u¯Ext∗=∑i=14fik,s.Lemma 7.
Givenfi, i=1,2,3,4, in (53)–(56), their inverse Laplace transforms are calculated as(58)L−1f1=1ω˜BpD1+k2h02/3∫0texp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τ⋅φ¯k,τdτ+2ikc02ω˜BpD1+k2h02/3∫0texp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τ⋅Uτdτ,(59)L−1f2=2ikh0231+k2h02/3⋅Ut,(60)L−1f3=−4ikh02ζω˜B31+k2h02/3∫0texp−ζω˜B⋅t−τ⋅cosω˜BpD⋅t−τ⋅Uτdτ,(61)L−1f4=2ikh02ω˜B2ζ2−131+k2h02/31−ζ2∫0texp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τ⋅Uτdτ.Proof.
To prove (58), note that L−1f1 has the explicit expression:(62)L−1f1=L−1Aω˜BpD⋅Lexp−ζω˜B⋅t⋅sinω˜BpD⋅t∵53=L−1φ¯∗+2ikc02⋅U∗sω˜BpD1+k2h02/3⋅Lexp−ζω˜B⋅t⋅sinω˜BpD⋅t∵41=1ω˜BpD1+k2h02/3⋅φ¯k,t∘exp−ζω˜B⋅t⋅sinω˜BpD⋅t+2ikc02ω˜BpD1+k2h02/3⋅Ut∘exp−ζω˜B⋅t⋅sinω˜BpD⋅t=1ω˜BpD1+k2h02/3∫0texp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τ⋅φ¯k,τdτ+2ikc02ω˜BpD1+k2h02/3∫0texp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τ⋅Uτdτ,where the notation ∘ denotes Laplace convolution and L−1L is the identity. This completes the proof of (58).□
The inverse Laplace transform off2 is written as (simply through B in (41))(63)L−1f2=L−1B∵54=L−12ikh02⋅U∗s31+k2h02/3∵41=2ikh02⋅Ut31+k2h02/3,by linearity of L−1, which proves (59).Similarly as above,(64)L−1f3=L−1−2Bζω˜B⋅Lexp−ζω˜B⋅t⋅cosω˜BpD⋅t∵54=−4ikh02ζω˜B31+k2h02/3⋅L−1U∗s∘L−1Lexp−ζω˜B⋅t⋅cosω˜BpD⋅t∵41=−4ikh02ζω˜B31+k2h02/3⋅Ut∘exp−ζω˜B⋅t⋅cosω˜BpD⋅t=−4ikh02ζω˜B31+k2h02/3∫0texp−ζω˜B⋅t−τ⋅cosω˜BpD⋅t−τ⋅Uτdτ,which completes the proof of (60).□Finally,L−1f4 has the form(65)L−1f4=L−1Bω˜B2ζ2−11−ζ2⋅Lexp−ζω˜B⋅t⋅sinω˜BpD⋅t∵56=2ikh02ω˜B2ζ2−131+k2h02/31−ζ2⋅L−1U∗s∘L−1Lexp−ζω˜B⋅t⋅sinω˜BpD⋅t∵41=2ikh02ω˜B2ζ2−131+k2h02/31−ζ2⋅Ut∘exp−ζω˜B⋅t⋅sinω˜BpD⋅t=2ikh02ω˜B2ζ2−131+k2h02/31−ζ2∫0texp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τ⋅Uτdτ.This finishes proving (61).□We are now in a position to take the inverse Laplace–Fourier transform ofu¯Ext∗ in (42).Lemma 8.
The inverse Laplace–Fourier transforms offi, i=1,2,3,4, in (53)–(56), are evaluated in the forms(66)F−1L−1f1=1π∫0t∫0∞∫−∞∞exp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τω˜BpD1+k2h02/3⋅coskξ−x⋅φξ,τdξdkdτ+1π∫0t∫0∞2kc02ω˜BpD1+k2h02/3⋅exp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τ⋅sinkx⋅Uτdkdτ,(67)F−1L−1f2=sgnx⋅Ut⋅exp−x3h0,(68)F−1L−1f3=−1π∫0t∫0∞4kh02ζω˜B31+k2h02/3⋅exp−ζω˜B⋅t−τ⋅cosω˜BpD⋅t−τ⋅sinkx⋅Uτdkdτ,(69)F−1L−1f4=1π∫0t∫0∞2kh02ω˜B2ζ2−131+k2h02/31−ζ2⋅exp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τ⋅sinkx⋅Uτdkdτ.Proof.
To prove (66), we take the inverse Fourier transform for (58), leaving(70)F−1L−1f1=12π∫−∞∞1ω˜BpD1+k2h02/3∫0texp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τ⋅φ¯k,τdτ⋅e−ikxdk∵24+12π∫−∞∞2ikc02ω˜BpD1+k2h02/3⋅∫0texp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τ⋅Uτdτ⋅e−ikxdk=12π∫0t∫−∞∞exp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τω˜BpD1+k2h02/3⋅∫−∞∞φξ,τ⋅eikξdξ⋅e−ikxdkdτ+12π∫0t∫−∞∞2ikc02ω˜BpD1+k2h02/3⋅exp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τ⋅Uτ⋅e−ikxdkdτ=12π∫0t∫−∞∞∫−∞∞exp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τω˜BpD1+k2h02/3⋅eikξ−x⋅φξ,τdξdkdτ+12π∫0t∫−∞∞2ikc02ω˜BpD1+k2h02/3⋅exp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τ⋅e−ikx⋅Uτdkdτ.
Here, we note that theF−1L−1f1 remains real because the real parts of the integrands in the last two integrals in the above are even in k, whereas the imaginary parts are odd. We are thus led to (66), completing the proof.For (67), we have(71)F−1L−1f2=12π∫−∞∞2ikh0231+k2h02/3⋅Ut⋅e−ikxdk∵59and24=12π∫−∞∞2ikh0231+k2h02/3⋅Ut⋅coskx−isinkxdk=1π∫0∞2kh0231+k2h02/3⋅sinkx⋅Utdk,where we have used the fact that the imaginary part of the integrand in the above is odd in k while the real part is even.An analytic expression is obtained for the above integral, that is,(72)F−1L−1f2=Ut⋅2π∫0∞h02/3⋅ksinkx1+k2h02/3dk=Ut⋅2π⋅π2xx⋅exp−x3h0=xx⋅Ut⋅exp−x3h0,by noting from (mathematical integral) table that(73)∫0∞bksinkx1+bk2dk=π2xxexp−xb,forb≥0.This proves (67).□For the proof of (68), we calculate F−1L−1f3, being written as(74)F−1L−1f3=12π∫−∞∞−4ikh02ζω˜B31+k2h02/3∫0texp−ζω˜B⋅t−τ⋅cosω˜BpD⋅t−τ⋅Uτdτ⋅e−ikxdk∵60and24=−1π∫0t∫0∞4kh02ζω˜B31+k2h02/3⋅exp−ζω˜B⋅t−τ⋅cosω˜BpD⋅t−τ⋅sinkx⋅Uτdkdτ.The above integral becomes real, due to the fact that the real part of the integrand is even ink, on the other hand, the imaginary part is odd. This leads to (68), completing the proof.□To show that (69) is true, recall the Fourier inverse (24) together with (61), i.e.,(75)F−1L−1f4=12π∫−∞∞2ikh02ω˜B2ζ2−131+k2h02/31−ζ2⋅∫0texp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τ⋅Uτdτ⋅e−ikxdk=12π∫0t∫−∞∞2ikh02ω˜B2ζ2−131+k2h02/31−ζ2⋅exp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τ⋅Uτ⋅e−ikxdkdτ=1π∫0t∫0∞2kh02ω˜B2ζ2−131+k2h02/31−ζ2⋅exp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τ⋅sinkx⋅Uτdkdτ,where the imaginary part in the above vanishes because the imaginary part of the integrand (in the second line in the above) is odd in k, but the real part of the integrand is even.□


## 3.5. Integral Formulation InvolvinguExt

Recalling that theuExt in Definition 2 is recovered formally through the identities(76)uExt=F−1L−1u¯Ext∗=∑i=14F−1L−1fi,because of (57), we readily arrive at the following theorem from the concerning lemmas without proof, but with a short note that the sum of the last line of (66) and (69) becomes(77)1π∫0t∫0∞2kc02ω˜BpD1+k2h02/3⋅exp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τ⋅sinkx⋅Uτdkdτ+1π∫0t∫0∞2kh02ω˜B2ζ2−131+k2h02/31−ζ2⋅exp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τ⋅sinkx⋅Uτdkdτ=1π∫0t∫0∞2kc02ω˜BpD1+k2h02/3⋅exp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τ⋅sinkx⋅Uτdkdτ+1π∫0t∫0∞2kh02ω˜B22ζ2−131+k2h02/3ω˜B1−ζ2⋅exp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τ⋅sinkx⋅Uτdkdτ=2π∫0t∫0∞3kc02+kh02ω˜B22ζ2−13ω˜BpD1+k2h02/3⋅exp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τ⋅sinkx⋅Uτdkdτ,where (40) is utilized.Theorem 1.
LetuExt be satisfied by the Boussinesq equation (12) subject to (13)–(15); i.e., uExt is a solution for the (extended) initial-boundary value problem in Definition 2. Then, the following integral relation holds(78)uExt=2π∫0t∫0∞3kc02+kh02ω˜B22ζ2−13ω˜BpD1+k2h02/3⋅exp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τ⋅sinkx⋅Uτdkdτ+sgnx⋅Ut⋅exp−x3h0−1π∫0t∫0∞4kh02ζω˜B31+k2h02/3⋅exp−ζω˜B⋅t−τ⋅cosω˜BpD⋅t−τ⋅sinkx⋅Uτdkdτ+1π∫0t∫0∞∫−∞∞exp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τω˜BpD1+k2h02/3⋅coskξ−x⋅φξ,τdξdkdτ,which is considered as an (basic) integral formalism for uExt because φ involves uExt via (18).


## 4. Integral Equation foru

In this section, we will derive a nonlinear integral equation foru, being equivalent to the original initial-boundary value problem of (1) with (4)–(6). The derivation of the integral equation is accomplished directly by the use of the integral formalism for uExt established in the preceding section.


### 4.1. Preliminaries

Here, we need to focus on examining the triple integral, denoted byI, in (78) after substituting (18), i.e.,(79)I≡1π∫0t∫0∞∫−∞∞Kx−ξ,t−τ,k⋅−12uExt2ξτ+α⋅uExtτ+β⋅uExtdξdkdτ,with K denoted by(80)Kx−ξ,t−τ,k≡exp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τω˜BpD1+k2h02/3⋅coskξ−x.The integration domain of the above integralI can be reduced by recalling that uExt is the odd extension of u with respect to x as was pointed out in Lemma 1, making the I depend upon u(instead of uExt).Lemma 9.
The integralI in (79) can be regarded as a functional of u(instead of uExt) through the identity(81)I=1π∫0t∫0∞∫0∞K˜x,t−τ,k;ξ⋅12u2ξτ−α⋅uτ−β⋅udξdkdτ,where(82)K˜x,t−τ,k;ξ=−2sinkξ⋅sinkxω˜BpD1+k2h02/3⋅exp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τ.Proof.
We writeI in (79) as the sum of two integrals:(83)I=1π∫0t∫0∞∫−∞0Kx−ξ,t−τ,k⋅−12−u−ξ,τ2ξτ+α⋅−u−ξ,ττ+β⋅−u−ξ,τdξdkdτ+1π∫0t∫0∞∫0∞Kx−ξ,t−τ,k⋅−12u2ξτ+α⋅uτ+β⋅udξdkdτ,based on the fact that uExt is the odd extension of u with respect to x by Lemma 1.
Next, let us take a look at the first integral, denoted byJ, in the above and use the change of variables σ=−ξ. We then have(84)J=1π∫0t∫0∞∫−∞0Kx−ξ,t−τ,k⋅−12u2−ξ,τξτ+α⋅−u−ξ,ττ+β⋅−u−ξ,τdξdkdτ=1π∫0t∫0∞∫∞0Kx+σ,t−τ,k⋅12u2σ,τστ−α⋅uτσ,τ−β⋅uσ,τ−dσdkdτ=1π∫0t∫0∞∫0∞Kx+σ,t−τ,k⋅12u2σ,τστ−α⋅uτσ,τ−β⋅uσ,τdσdkdτ=1π∫0t∫0∞∫0∞Kx+ξ,t−τ,k⋅12u2ξ,τξτ−α⋅uτξ,τ−β⋅uξ,τdξdkdτ,where we utilize the chain rule(85)∂∂ξu2−ξ,ττ=∂∂τ∂∂ξu2σ,τ=∂∂τ∂∂σu2σ,τ∂σ∂ξ=−u2σ,τστ,and the fact that σ is a dummy variable.
Substituting (84) into (83) yields(86)I=1π∫0t∫0∞∫0∞Kx+ξ,t−τ,k⋅12u2ξ,τξτ−α⋅uτξ,τ−β⋅uξ,τdξdkdτ+1π∫0t∫0∞∫0∞Kx−ξ,t−τ,k⋅−12u2ξτ+α⋅uτ+β⋅udξdkdτ=1π∫0t∫0∞∫0∞Kx+ξ,t−τ,k−Kx−ξ,t−τ,k⋅12u2ξτ−α⋅uτ−β⋅udξdkdτ,with a note on(87)K˜x,t−τ,k;ξ=Kx+ξ,t−τ,k−Kx−ξ,t−τ,k,=exp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τω˜BpD1+k2h02/3⋅cosk−ξ−x−cosξ−x,=−2sinkξ⋅sinkxω˜BpD1+k2h02/3⋅exp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τ,because cosa+b−cosa−b=−2sinasinb for a,b∈ℝ. This completes the proof.


### 4.2. Constructing Integral Equation

Using the result presented in the preceding section, we finally arrive at a nonlinear integral equation identical with theoriginal initial-boundary value problem, i.e., the Boussinesq equation (1) subject to (4)–(6). We start with integration by parts.Lemma 10.
Given a kernelK˜ in (82), the following integral, denoted by I1, is calculated as(88)I1=1π∫0t∫0∞∫0∞K˜x,t−τ,k;ξ⋅12u2ξτdξdkdτ=−1π∫0t∫0∞∫0∞kcoskξ⋅sinkxω˜BpD1+k2h02/3⋅exp−ζω˜B⋅t−τ⋅ζω˜B⋅sinω˜BpD⋅t−τ−ω˜BpD⋅cosω˜BpD⋅t−τ⋅u2dξdkdτ.Proof.
By integration by parts with respect to spatial variablex,(89)I1=1π∫0t∫0∞∫0∞K˜x,t−τ,k;ξ⋅12u2ξτdξdkdτ=1π∫0t∫0∞K˜x,t−τ,k;ξ⋅12u2τξ=0ξ=∞−∫0∞K˜ξx,t−τ,k;ξ⋅12u2τdξdkdτ.
The above is shortened to(90)I1=−1π∫0t∫0∞∫0∞K˜ξx,t−τ,k;ξ⋅12u2τdξdkdτ,because the far field condition (6) implies that u vanishes as ξ⟶∞ and K˜x,t−τ,k;ξ=0=0 by (82).
Next, we interchange the order of integration in (90) due to Fubini’s theorem as(91)Ι1=−1π∫0∞∫0∞∫0tK˜ξx,t−τ,k;ξ⋅12u2τdτdξdk.
Then, we apply integration by parts with respect to time, leading to(92)I1=−1π∫0∞∫0∞K˜ξx,t−τ,k;ξ⋅12u2τ=0τ=t−∫0tK˜ξτx,t−τ,k;ξ⋅12u2dτdξdk=1π∫0∞∫0∞∫0tK˜ξτx,t−τ,k;ξ⋅12u2dτdξdk=−1π∫0t∫0∞∫0∞kcoskξ⋅sinkxω˜BpD1+k2h02/3⋅exp−ζω˜B⋅t−τ⋅ζω˜B⋅sinω˜BpD⋅t−τ−ω˜BpD⋅cosω˜BpD⋅t−τ⋅u2dξdkdτ,because u vanishes at t=0 from initial condition (4), K˜ξx,t−t,k;ξ=0 from the derivative (93)K˜ξx,t−τ,k;ξ=−2kcoskξ⋅sinkxω˜BpD1+k2h02/3⋅exp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τ,K˜ξτx,t−τ,k;ξ=−2kcoskξ⋅sinkxω˜BpD1+k2h02/3⋅exp−ζω˜B⋅t−τ⋅ζω˜B⋅sinω˜BpD⋅t−τ−ω˜BpD⋅cosω˜BpD⋅t−τ.
This completes the proof.Lemma 11.
The following triple integral, termedI2, involving ut is evaluated as(94)I2=1π∫0t∫0∞∫0∞K˜x,t−τ,k;ξ⋅−α⋅uτdξdkdτ=1π∫0t∫0∞∫0∞−2sinkξ⋅sinkxω˜BpD1+k2h02/3⋅exp−ζω˜B⋅t−τ⋅ζω˜B⋅sinω˜BpD⋅t−τ−ω˜BpD⋅cosω˜BpD⋅t−τ⋅α⋅udξdkdτ,in which the kernel K˜ is given in (82).Proof.
We begin by rewriting the triple integralI2 as(95)I2=1π∫0∞∫0∞K˜x,t−τ,k;ξ⋅−α⋅uτ=0τ=t−∫0tK˜τx,t−τ,k;ξ⋅−α⋅udτdξdk,due to Fubini’s theorem, where integration by parts with respect to time is used with interchange of the order of integration. Here, notice that u equals zero when t=0 from initial condition (4) and K˜x,t−t,k;ξ=0 from (82).
On the other hand, the differentiation of (82) with respect to τ gives(96)K˜τx,t−τ,k;ξ=−2sinkξ⋅sinkxω˜BpD1+k2h02/3⋅exp−ζω˜B⋅t−τ⋅ζω˜B⋅sinω˜BpD⋅t−τ−ω˜BpD⋅cosω˜BpD⋅t−τ.
This results in(97)I2=1π∫0∞∫0∞∫0tK˜τx,t−τ,k;ξ⋅α⋅udτdξdk=1π∫0∞∫0∞∫0t−2sinkξ⋅sinkxω˜BpD1+k2h02/3⋅exp−ζω˜B⋅t−τ⋅ζω˜B⋅sinω˜BpD⋅t−τ−ω˜BpD⋅cosω˜BpD⋅t−τ⋅α⋅udτdξdk,as required, which completes the proof.Remark 3.
Noting that the integralI in (81) decomposes into the sum(98)I=I1+I2+1π∫0t∫0∞∫0∞K˜x,t−τ,k;ξ⋅−β⋅udξdkdτ,from (88) and (94), (81) may arrange to(99)I=−1π∫0t∫0∞∫0∞kcoskξ⋅sinkxω˜BpD1+k2h02/3⋅exp−ζω˜B⋅t−τ⋅ζω˜B⋅sinω˜BpD⋅t−τ−ω˜BpD⋅cosω˜BpD⋅t−τ⋅u2dξdkdτ−1π∫0t∫0∞∫0∞2sinkξ⋅sinkxω˜BpD1+k2h02/3⋅exp−ζω˜B⋅t−τ⋅ζω˜B⋅sinω˜BpD⋅t−τ−ω˜BpD⋅cosω˜BpD⋅t−τ⋅α⋅udξdkdτ+1π∫0t∫0∞∫0∞2sinkξ⋅sinkxω˜BpD1+k2h02/3⋅exp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τ⋅β⋅udξdkdτ,by Lemmas 10 and 11.
The following theorem is immediately directed from Theorem1 plus Remarks 2 and 3.Theorem 2.
Letζ be a dissipative parameter in (39) depending on two pseudoparameters α,β>0, and let ω˜B and ω˜BpD be the two pseudofrequencies in (36) and (40), respectively. Further, let u be a solution for the original Boussinesq equation (1) subject to (4)–(6). Then, u satisfies the following nonlinear integral equation:(100)ux,t=2π∫0t∫0∞3c02+h02ω˜B22ζ2−13ω˜BpD1+k2h02/3⋅k⋅exp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τ⋅sinkx⋅Uτdkdτ+Ut⋅exp−x3h0−1π∫0t∫0∞4kh02ζω˜B⋅sinkx31+k2h02/3⋅exp−ζω˜B⋅t−τ⋅cosω˜BpD⋅t−τ⋅Uτdkdτ−1π∫0t∫0∞∫0∞kcoskξ⋅sinkxω˜BpD1+k2h02/3⋅exp−ζω˜B⋅t−τ⋅ζω˜B⋅sinω˜BpD⋅t−τ−ω˜BpD⋅cosω˜BpD⋅t−τ⋅u2dξdkdτ−1π∫0t∫0∞∫0∞2sinkξ⋅sinkxω˜BpD1+k2h02/3⋅exp−ζω˜B⋅t−τ⋅ζω˜B⋅sinω˜BpD⋅t−τ−ω˜BpD⋅cosω˜BpD⋅t−τ⋅α⋅udξdkdτ+1π∫0t∫0∞∫0∞2sinkξ⋅sinkxω˜BpD1+k2h02/3⋅exp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τ⋅β⋅udξdkdτ.Corollary 1.
The nonlinear integral equation (100) in Theorem 2 reduces to(101)ux,t=2π∫0t∫0∞3c02−h02ωB23ωB1+k2h02/3⋅k⋅sinωB⋅t−τ⋅sinkx⋅Uτdkdτ+Ut⋅exp−x3h0+1π∫0t∫0∞∫0∞kcoskξ⋅sinkx1+k2h02/3⋅cosωB⋅t−τ⋅u2dξdkdτ,as α,β⟶0.Proof.
We first note that the last two integral terms in (100) vanish, as α,β⟶0. It thus suffices to show that(102)3c02+h02ω˜B22ζ2−13ω˜BpD1+k2h02/3⟶3c02−h02ω˜B23ω˜B1+k2h02/3⟶3c02−h02ωB23ωB1+k2h02/3,asα,β⟶0,in the first integral in (100), which is immediate because(103)ζ=α2c02k2+β⋅c02k2+β1+k2h02/3⟶0,fork>0,asα⟶0,by (36) and (39). And then, the integrand in the fourth term in (100) has an asymptotic form:(104)kcoskξ⋅sinkxω˜BpD1+k2h02/3⋅exp−ζω˜B⋅t−τ⋅ζω˜B⋅sinω˜BpD⋅t−τ−ω˜BpD⋅cosω˜BpD⋅t−τ⟶kcoskξ⋅sinkxωB1+k2h02/3⋅−ωB⋅cosωB⋅t−τ=−kcoskξ⋅sinkx1+k2h02/3⋅cosωB⋅t−τ,as α,β⟶0. This completes the proof.Remark 4.
The kernels in integral equation (100) have singularities of (simple) poles at(105)k=±βc0i,±3h02−1±α2,±3h0i,±−3c02+βh02±9c04+6βc02h02+β2h04−12βc02h02+3α2c02h022c02h02.
All the singularities become off the real lineℝ with a proper choice of α,β>0, which tend further to(106)k=0,±3h0i,as α,β⟶0.


## 4.1. Preliminaries

Here, we need to focus on examining the triple integral, denoted byI, in (78) after substituting (18), i.e.,(79)I≡1π∫0t∫0∞∫−∞∞Kx−ξ,t−τ,k⋅−12uExt2ξτ+α⋅uExtτ+β⋅uExtdξdkdτ,with K denoted by(80)Kx−ξ,t−τ,k≡exp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τω˜BpD1+k2h02/3⋅coskξ−x.The integration domain of the above integralI can be reduced by recalling that uExt is the odd extension of u with respect to x as was pointed out in Lemma 1, making the I depend upon u(instead of uExt).Lemma 9.
The integralI in (79) can be regarded as a functional of u(instead of uExt) through the identity(81)I=1π∫0t∫0∞∫0∞K˜x,t−τ,k;ξ⋅12u2ξτ−α⋅uτ−β⋅udξdkdτ,where(82)K˜x,t−τ,k;ξ=−2sinkξ⋅sinkxω˜BpD1+k2h02/3⋅exp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τ.Proof.
We writeI in (79) as the sum of two integrals:(83)I=1π∫0t∫0∞∫−∞0Kx−ξ,t−τ,k⋅−12−u−ξ,τ2ξτ+α⋅−u−ξ,ττ+β⋅−u−ξ,τdξdkdτ+1π∫0t∫0∞∫0∞Kx−ξ,t−τ,k⋅−12u2ξτ+α⋅uτ+β⋅udξdkdτ,based on the fact that uExt is the odd extension of u with respect to x by Lemma 1.
Next, let us take a look at the first integral, denoted byJ, in the above and use the change of variables σ=−ξ. We then have(84)J=1π∫0t∫0∞∫−∞0Kx−ξ,t−τ,k⋅−12u2−ξ,τξτ+α⋅−u−ξ,ττ+β⋅−u−ξ,τdξdkdτ=1π∫0t∫0∞∫∞0Kx+σ,t−τ,k⋅12u2σ,τστ−α⋅uτσ,τ−β⋅uσ,τ−dσdkdτ=1π∫0t∫0∞∫0∞Kx+σ,t−τ,k⋅12u2σ,τστ−α⋅uτσ,τ−β⋅uσ,τdσdkdτ=1π∫0t∫0∞∫0∞Kx+ξ,t−τ,k⋅12u2ξ,τξτ−α⋅uτξ,τ−β⋅uξ,τdξdkdτ,where we utilize the chain rule(85)∂∂ξu2−ξ,ττ=∂∂τ∂∂ξu2σ,τ=∂∂τ∂∂σu2σ,τ∂σ∂ξ=−u2σ,τστ,and the fact that σ is a dummy variable.
Substituting (84) into (83) yields(86)I=1π∫0t∫0∞∫0∞Kx+ξ,t−τ,k⋅12u2ξ,τξτ−α⋅uτξ,τ−β⋅uξ,τdξdkdτ+1π∫0t∫0∞∫0∞Kx−ξ,t−τ,k⋅−12u2ξτ+α⋅uτ+β⋅udξdkdτ=1π∫0t∫0∞∫0∞Kx+ξ,t−τ,k−Kx−ξ,t−τ,k⋅12u2ξτ−α⋅uτ−β⋅udξdkdτ,with a note on(87)K˜x,t−τ,k;ξ=Kx+ξ,t−τ,k−Kx−ξ,t−τ,k,=exp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τω˜BpD1+k2h02/3⋅cosk−ξ−x−cosξ−x,=−2sinkξ⋅sinkxω˜BpD1+k2h02/3⋅exp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τ,because cosa+b−cosa−b=−2sinasinb for a,b∈ℝ. This completes the proof.


## 4.2. Constructing Integral Equation

Using the result presented in the preceding section, we finally arrive at a nonlinear integral equation identical with theoriginal initial-boundary value problem, i.e., the Boussinesq equation (1) subject to (4)–(6). We start with integration by parts.Lemma 10.
Given a kernelK˜ in (82), the following integral, denoted by I1, is calculated as(88)I1=1π∫0t∫0∞∫0∞K˜x,t−τ,k;ξ⋅12u2ξτdξdkdτ=−1π∫0t∫0∞∫0∞kcoskξ⋅sinkxω˜BpD1+k2h02/3⋅exp−ζω˜B⋅t−τ⋅ζω˜B⋅sinω˜BpD⋅t−τ−ω˜BpD⋅cosω˜BpD⋅t−τ⋅u2dξdkdτ.Proof.
By integration by parts with respect to spatial variablex,(89)I1=1π∫0t∫0∞∫0∞K˜x,t−τ,k;ξ⋅12u2ξτdξdkdτ=1π∫0t∫0∞K˜x,t−τ,k;ξ⋅12u2τξ=0ξ=∞−∫0∞K˜ξx,t−τ,k;ξ⋅12u2τdξdkdτ.
The above is shortened to(90)I1=−1π∫0t∫0∞∫0∞K˜ξx,t−τ,k;ξ⋅12u2τdξdkdτ,because the far field condition (6) implies that u vanishes as ξ⟶∞ and K˜x,t−τ,k;ξ=0=0 by (82).
Next, we interchange the order of integration in (90) due to Fubini’s theorem as(91)Ι1=−1π∫0∞∫0∞∫0tK˜ξx,t−τ,k;ξ⋅12u2τdτdξdk.
Then, we apply integration by parts with respect to time, leading to(92)I1=−1π∫0∞∫0∞K˜ξx,t−τ,k;ξ⋅12u2τ=0τ=t−∫0tK˜ξτx,t−τ,k;ξ⋅12u2dτdξdk=1π∫0∞∫0∞∫0tK˜ξτx,t−τ,k;ξ⋅12u2dτdξdk=−1π∫0t∫0∞∫0∞kcoskξ⋅sinkxω˜BpD1+k2h02/3⋅exp−ζω˜B⋅t−τ⋅ζω˜B⋅sinω˜BpD⋅t−τ−ω˜BpD⋅cosω˜BpD⋅t−τ⋅u2dξdkdτ,because u vanishes at t=0 from initial condition (4), K˜ξx,t−t,k;ξ=0 from the derivative (93)K˜ξx,t−τ,k;ξ=−2kcoskξ⋅sinkxω˜BpD1+k2h02/3⋅exp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τ,K˜ξτx,t−τ,k;ξ=−2kcoskξ⋅sinkxω˜BpD1+k2h02/3⋅exp−ζω˜B⋅t−τ⋅ζω˜B⋅sinω˜BpD⋅t−τ−ω˜BpD⋅cosω˜BpD⋅t−τ.
This completes the proof.Lemma 11.
The following triple integral, termedI2, involving ut is evaluated as(94)I2=1π∫0t∫0∞∫0∞K˜x,t−τ,k;ξ⋅−α⋅uτdξdkdτ=1π∫0t∫0∞∫0∞−2sinkξ⋅sinkxω˜BpD1+k2h02/3⋅exp−ζω˜B⋅t−τ⋅ζω˜B⋅sinω˜BpD⋅t−τ−ω˜BpD⋅cosω˜BpD⋅t−τ⋅α⋅udξdkdτ,in which the kernel K˜ is given in (82).Proof.
We begin by rewriting the triple integralI2 as(95)I2=1π∫0∞∫0∞K˜x,t−τ,k;ξ⋅−α⋅uτ=0τ=t−∫0tK˜τx,t−τ,k;ξ⋅−α⋅udτdξdk,due to Fubini’s theorem, where integration by parts with respect to time is used with interchange of the order of integration. Here, notice that u equals zero when t=0 from initial condition (4) and K˜x,t−t,k;ξ=0 from (82).
On the other hand, the differentiation of (82) with respect to τ gives(96)K˜τx,t−τ,k;ξ=−2sinkξ⋅sinkxω˜BpD1+k2h02/3⋅exp−ζω˜B⋅t−τ⋅ζω˜B⋅sinω˜BpD⋅t−τ−ω˜BpD⋅cosω˜BpD⋅t−τ.
This results in(97)I2=1π∫0∞∫0∞∫0tK˜τx,t−τ,k;ξ⋅α⋅udτdξdk=1π∫0∞∫0∞∫0t−2sinkξ⋅sinkxω˜BpD1+k2h02/3⋅exp−ζω˜B⋅t−τ⋅ζω˜B⋅sinω˜BpD⋅t−τ−ω˜BpD⋅cosω˜BpD⋅t−τ⋅α⋅udτdξdk,as required, which completes the proof.Remark 3.
Noting that the integralI in (81) decomposes into the sum(98)I=I1+I2+1π∫0t∫0∞∫0∞K˜x,t−τ,k;ξ⋅−β⋅udξdkdτ,from (88) and (94), (81) may arrange to(99)I=−1π∫0t∫0∞∫0∞kcoskξ⋅sinkxω˜BpD1+k2h02/3⋅exp−ζω˜B⋅t−τ⋅ζω˜B⋅sinω˜BpD⋅t−τ−ω˜BpD⋅cosω˜BpD⋅t−τ⋅u2dξdkdτ−1π∫0t∫0∞∫0∞2sinkξ⋅sinkxω˜BpD1+k2h02/3⋅exp−ζω˜B⋅t−τ⋅ζω˜B⋅sinω˜BpD⋅t−τ−ω˜BpD⋅cosω˜BpD⋅t−τ⋅α⋅udξdkdτ+1π∫0t∫0∞∫0∞2sinkξ⋅sinkxω˜BpD1+k2h02/3⋅exp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τ⋅β⋅udξdkdτ,by Lemmas 10 and 11.
The following theorem is immediately directed from Theorem1 plus Remarks 2 and 3.Theorem 2.
Letζ be a dissipative parameter in (39) depending on two pseudoparameters α,β>0, and let ω˜B and ω˜BpD be the two pseudofrequencies in (36) and (40), respectively. Further, let u be a solution for the original Boussinesq equation (1) subject to (4)–(6). Then, u satisfies the following nonlinear integral equation:(100)ux,t=2π∫0t∫0∞3c02+h02ω˜B22ζ2−13ω˜BpD1+k2h02/3⋅k⋅exp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τ⋅sinkx⋅Uτdkdτ+Ut⋅exp−x3h0−1π∫0t∫0∞4kh02ζω˜B⋅sinkx31+k2h02/3⋅exp−ζω˜B⋅t−τ⋅cosω˜BpD⋅t−τ⋅Uτdkdτ−1π∫0t∫0∞∫0∞kcoskξ⋅sinkxω˜BpD1+k2h02/3⋅exp−ζω˜B⋅t−τ⋅ζω˜B⋅sinω˜BpD⋅t−τ−ω˜BpD⋅cosω˜BpD⋅t−τ⋅u2dξdkdτ−1π∫0t∫0∞∫0∞2sinkξ⋅sinkxω˜BpD1+k2h02/3⋅exp−ζω˜B⋅t−τ⋅ζω˜B⋅sinω˜BpD⋅t−τ−ω˜BpD⋅cosω˜BpD⋅t−τ⋅α⋅udξdkdτ+1π∫0t∫0∞∫0∞2sinkξ⋅sinkxω˜BpD1+k2h02/3⋅exp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τ⋅β⋅udξdkdτ.Corollary 1.
The nonlinear integral equation (100) in Theorem 2 reduces to(101)ux,t=2π∫0t∫0∞3c02−h02ωB23ωB1+k2h02/3⋅k⋅sinωB⋅t−τ⋅sinkx⋅Uτdkdτ+Ut⋅exp−x3h0+1π∫0t∫0∞∫0∞kcoskξ⋅sinkx1+k2h02/3⋅cosωB⋅t−τ⋅u2dξdkdτ,as α,β⟶0.Proof.
We first note that the last two integral terms in (100) vanish, as α,β⟶0. It thus suffices to show that(102)3c02+h02ω˜B22ζ2−13ω˜BpD1+k2h02/3⟶3c02−h02ω˜B23ω˜B1+k2h02/3⟶3c02−h02ωB23ωB1+k2h02/3,asα,β⟶0,in the first integral in (100), which is immediate because(103)ζ=α2c02k2+β⋅c02k2+β1+k2h02/3⟶0,fork>0,asα⟶0,by (36) and (39). And then, the integrand in the fourth term in (100) has an asymptotic form:(104)kcoskξ⋅sinkxω˜BpD1+k2h02/3⋅exp−ζω˜B⋅t−τ⋅ζω˜B⋅sinω˜BpD⋅t−τ−ω˜BpD⋅cosω˜BpD⋅t−τ⟶kcoskξ⋅sinkxωB1+k2h02/3⋅−ωB⋅cosωB⋅t−τ=−kcoskξ⋅sinkx1+k2h02/3⋅cosωB⋅t−τ,as α,β⟶0. This completes the proof.Remark 4.
The kernels in integral equation (100) have singularities of (simple) poles at(105)k=±βc0i,±3h02−1±α2,±3h0i,±−3c02+βh02±9c04+6βc02h02+β2h04−12βc02h02+3α2c02h022c02h02.
All the singularities become off the real lineℝ with a proper choice of α,β>0, which tend further to(106)k=0,±3h0i,as α,β⟶0.


## 5. Application of Integral Formalism

In this section, we shall apply successive approximations to integral equation (100) constructed in the previous section. This produces an iterative strategy, whereby we investigate iterative numerical solutions of a solitary wave. Furthermore, we will examine how the two pseudoparameters α,β introduced artificially in this paper effect the convergence of the iterative solutions. For this reason, the two introduced parameters can be viewed as control ones for the iterative strategy.


### 5.1. Successive Approximations

Numerical solutions for the original initial-boundary value problem of the Boussinesq equation (1):(107)utt−c02uxx+12u2xt=13h02uxxtt,x>0,subject to (4)–(6) may be found by applying successive approximations of the Banach fixed point theorem to (100). The theorem guarantees the existence and uniqueness of fixed points of certain self-operators of metric spaces, providing a constructive method, namely, the successive approximations, to find those fixed points [6–11]; in the present study, we choose the same space as the usual metric space [6–11] for the use of the theorem.We bear in mind that (100) can be thought of a fixed form and we will seek its fixed point (solution) u through a recurrence relation with the pseudoparameters α,β for n=0,1,2,…, as expressed as(108)un+1x,t=2π∫0t∫0∞3c02+h02ω˜B22ζ2−13ω˜BpD1+k2h02/3⋅k⋅exp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τ⋅sinkx⋅Uτdkdτ+Ut⋅exp−x3h0−1π∫0t∫0∞4kh02ζω˜B⋅sinkx31+k2h02/3⋅exp−ζω˜B⋅t−τ⋅cosω˜BpD⋅t−τ⋅Uτdkdτ−1π∫0t∫0∞∫0∞kcoskξ⋅sinkxω˜BpD1+k2h02/3⋅exp−ζω˜B⋅t−τ⋅ζω˜B⋅sinω˜BpD⋅t−τ−ω˜BpD⋅cosω˜BpD⋅t−τ⋅un2dξdkdτ−1π∫0t∫0∞∫0∞2sinkξ⋅sinkxω˜BpD1+k2h02/3⋅exp−ζω˜B⋅t−τ⋅ζω˜B⋅sinω˜BpD⋅t−τ−ω˜BpD⋅cosω˜BpD⋅t−τ⋅α⋅undξdkdτ+1π∫0t∫0∞∫0∞2sinkξ⋅sinkxω˜BpD1+k2h02/3⋅exp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τ⋅β⋅undξdkdτ,with (zero) initial guess(109)u0=0.It is noted that (108) is the recurrence directly obtained from (100).The recurrence relation (108) and (109) is a new iterative formula for solving the original initial-boundary value problem, which defines recursively a functional sequence un of (108) for n=0,1,2,….


### 5.2. Numerical Solutions of a Moving Solitary Wave

It is worth discussing that the original governing equation (1) may be nondimensionalized as (see Appendix)(110)ut′t′′=ux′x′′−u′2x′t′+ux′x′t′t′′,via appropriate dimensionless variables(111)u′=u2gh0,t′=t3gh0,x′=x3h0.The dimensionless PDE (110) is known to allow a moving solitary wave solution in the form [12, 13](112)uexact′x′,t′=a′⋅sech2a′6c′x′−c′t′−x0′,constrained by amplitude dispersion(113)c′=a′3+1+a′29,in which the constants a′, c′, and x0′ denote dimensionless amplitude, dimensionless wave speed, and initial dimensionless position of the solitary wave, respectively. Here, (112) and (113) can be written in original physical variables(114)uexactx,t=2ac03h0⋅sech2ac03ch03x−c2⋅t−x0,c2=c0ah03+1+a23h02,where a and c are such that(115)a′=a3h0,c′=c2gh0,consistent with (110); in fact, it is straightforward to check that (114) satisfies (1) by direct substitution of (114) into (1).Our aim here is to simulate the moving solitary wave (114) numerically, which enables us to demonstrate that the iterative strategy (108) works well for (114). For the simulation, we attempt to iterate (108) with initial guess (109) using the null initial conditions (4) as well as the boundary condition at x=0 specified by(116)Ut=2ac03h0⋅sech2ac03ch03−c2⋅t−x0,from (114). Here, the Simpson and trapezoidal rules are employed (in space and time, respectively), by setting h0=0.2m and g=9.80665m/s2. The space and time domains are selected as 0<x<25m (or 0<x′<216.51) and 0<t<18sec (or 0<t′<218.31), respectively, with the increments, Δx=0.10m and Δt=0.05sec; we choose x0=−12m(or x0′=−103.92).Figure2 shows typical convergence behaviors for the iterative wave solutions with two amplitudes, i.e., 0.00115 m (or a′=0.01) and 0.00231 m (or a′=0.02), where two pseudoparameters are taken as α=0.0700rad/sec and β=0.0490rad/sec2, which are (alternatively) nondimensionalized as α′=10−2 and β′=10−3, respectively, according to the (dimensionless) scaling(117)α′=αh0g,β′=βh0g.Figure 2
Convergence behaviors of iterative solitary wave solutions with two dimensionless amplitudes. (a)a′=0.01. (b) a′=0.02.
(a)
(b)The first iteration gives a numerical wave solution valid for the initial stage of time evolution and an almost converged solution seems to be reached atn=10, which clearly shows a translating (moving) solitary wave.We next inspect quantitative errors for the iterative wave solutions. Figure3 plots the nature of convergence of (108) for various β′’s, when α′ is specified as α′=10−2,10−1 (which corresponds to α=0.0700,0.7002rad/sec, respectively). Here, errors are estimated by L∞:(118)L∞=maxxuexact−un,where L∞ represents L-infinity norm taken over by the space variable x only with reference time tref being kept fixed; tref=18sec (i.e., end time). We see that the L∞ decreases as the number of iterations increases. For α′=10−2 and 10−1, convergence characteristics depend on the pseudoparameter β′; to be specific, the smaller β′ would contribute to a higher convergence rate, with a specified α′. This tendency is similar to that of Jang [6]. Finally, Table 1 tabulates the numerical values of errors plotted in Figure 3.Figure 3
Errors and convergence nature of (108) for the moving solitary wave (112). (a) a′=0.01. (b) a′=0.02.
(a)
(b)Table 1
Values of numerical error ofL∞ plotted in Figure 3.
α
′
β
′
L
∞
a
′
=
0.01
a
′
=
0.02
10
−
2
1.0
×
10
−
3
5.77e − 6
3.82e − 5
5.0
×
10
−
3
4.65e − 6
3.67e − 5
1.0
×
10
−
2
6.87e − 6
3.06e − 5
10
−
1
1.0
×
10
−
3
9.33e − 6
4.26e − 5
5.0
×
10
−
3
1.04e − 5
4.45e − 5
1.0
×
10
−
2
9.10e − 6
4.27e − 5


## 5.1. Successive Approximations

Numerical solutions for the original initial-boundary value problem of the Boussinesq equation (1):(107)utt−c02uxx+12u2xt=13h02uxxtt,x>0,subject to (4)–(6) may be found by applying successive approximations of the Banach fixed point theorem to (100). The theorem guarantees the existence and uniqueness of fixed points of certain self-operators of metric spaces, providing a constructive method, namely, the successive approximations, to find those fixed points [6–11]; in the present study, we choose the same space as the usual metric space [6–11] for the use of the theorem.We bear in mind that (100) can be thought of a fixed form and we will seek its fixed point (solution) u through a recurrence relation with the pseudoparameters α,β for n=0,1,2,…, as expressed as(108)un+1x,t=2π∫0t∫0∞3c02+h02ω˜B22ζ2−13ω˜BpD1+k2h02/3⋅k⋅exp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τ⋅sinkx⋅Uτdkdτ+Ut⋅exp−x3h0−1π∫0t∫0∞4kh02ζω˜B⋅sinkx31+k2h02/3⋅exp−ζω˜B⋅t−τ⋅cosω˜BpD⋅t−τ⋅Uτdkdτ−1π∫0t∫0∞∫0∞kcoskξ⋅sinkxω˜BpD1+k2h02/3⋅exp−ζω˜B⋅t−τ⋅ζω˜B⋅sinω˜BpD⋅t−τ−ω˜BpD⋅cosω˜BpD⋅t−τ⋅un2dξdkdτ−1π∫0t∫0∞∫0∞2sinkξ⋅sinkxω˜BpD1+k2h02/3⋅exp−ζω˜B⋅t−τ⋅ζω˜B⋅sinω˜BpD⋅t−τ−ω˜BpD⋅cosω˜BpD⋅t−τ⋅α⋅undξdkdτ+1π∫0t∫0∞∫0∞2sinkξ⋅sinkxω˜BpD1+k2h02/3⋅exp−ζω˜B⋅t−τ⋅sinω˜BpD⋅t−τ⋅β⋅undξdkdτ,with (zero) initial guess(109)u0=0.It is noted that (108) is the recurrence directly obtained from (100).The recurrence relation (108) and (109) is a new iterative formula for solving the original initial-boundary value problem, which defines recursively a functional sequence un of (108) for n=0,1,2,….


## 5.2. Numerical Solutions of a Moving Solitary Wave

It is worth discussing that the original governing equation (1) may be nondimensionalized as (see Appendix)(110)ut′t′′=ux′x′′−u′2x′t′+ux′x′t′t′′,via appropriate dimensionless variables(111)u′=u2gh0,t′=t3gh0,x′=x3h0.The dimensionless PDE (110) is known to allow a moving solitary wave solution in the form [12, 13](112)uexact′x′,t′=a′⋅sech2a′6c′x′−c′t′−x0′,constrained by amplitude dispersion(113)c′=a′3+1+a′29,in which the constants a′, c′, and x0′ denote dimensionless amplitude, dimensionless wave speed, and initial dimensionless position of the solitary wave, respectively. Here, (112) and (113) can be written in original physical variables(114)uexactx,t=2ac03h0⋅sech2ac03ch03x−c2⋅t−x0,c2=c0ah03+1+a23h02,where a and c are such that(115)a′=a3h0,c′=c2gh0,consistent with (110); in fact, it is straightforward to check that (114) satisfies (1) by direct substitution of (114) into (1).Our aim here is to simulate the moving solitary wave (114) numerically, which enables us to demonstrate that the iterative strategy (108) works well for (114). For the simulation, we attempt to iterate (108) with initial guess (109) using the null initial conditions (4) as well as the boundary condition at x=0 specified by(116)Ut=2ac03h0⋅sech2ac03ch03−c2⋅t−x0,from (114). Here, the Simpson and trapezoidal rules are employed (in space and time, respectively), by setting h0=0.2m and g=9.80665m/s2. The space and time domains are selected as 0<x<25m (or 0<x′<216.51) and 0<t<18sec (or 0<t′<218.31), respectively, with the increments, Δx=0.10m and Δt=0.05sec; we choose x0=−12m(or x0′=−103.92).Figure2 shows typical convergence behaviors for the iterative wave solutions with two amplitudes, i.e., 0.00115 m (or a′=0.01) and 0.00231 m (or a′=0.02), where two pseudoparameters are taken as α=0.0700rad/sec and β=0.0490rad/sec2, which are (alternatively) nondimensionalized as α′=10−2 and β′=10−3, respectively, according to the (dimensionless) scaling(117)α′=αh0g,β′=βh0g.Figure 2
Convergence behaviors of iterative solitary wave solutions with two dimensionless amplitudes. (a)a′=0.01. (b) a′=0.02.
(a)
(b)The first iteration gives a numerical wave solution valid for the initial stage of time evolution and an almost converged solution seems to be reached atn=10, which clearly shows a translating (moving) solitary wave.We next inspect quantitative errors for the iterative wave solutions. Figure3 plots the nature of convergence of (108) for various β′’s, when α′ is specified as α′=10−2,10−1 (which corresponds to α=0.0700,0.7002rad/sec, respectively). Here, errors are estimated by L∞:(118)L∞=maxxuexact−un,where L∞ represents L-infinity norm taken over by the space variable x only with reference time tref being kept fixed; tref=18sec (i.e., end time). We see that the L∞ decreases as the number of iterations increases. For α′=10−2 and 10−1, convergence characteristics depend on the pseudoparameter β′; to be specific, the smaller β′ would contribute to a higher convergence rate, with a specified α′. This tendency is similar to that of Jang [6]. Finally, Table 1 tabulates the numerical values of errors plotted in Figure 3.Figure 3
Errors and convergence nature of (108) for the moving solitary wave (112). (a) a′=0.01. (b) a′=0.02.
(a)
(b)Table 1
Values of numerical error ofL∞ plotted in Figure 3.
α
′
β
′
L
∞
a
′
=
0.01
a
′
=
0.02
10
−
2
1.0
×
10
−
3
5.77e − 6
3.82e − 5
5.0
×
10
−
3
4.65e − 6
3.67e − 5
1.0
×
10
−
2
6.87e − 6
3.06e − 5
10
−
1
1.0
×
10
−
3
9.33e − 6
4.26e − 5
5.0
×
10
−
3
1.04e − 5
4.45e − 5
1.0
×
10
−
2
9.10e − 6
4.27e − 5


## 6. Discussions and Concluding Remarks

This work involves aninitial value problem for the Boussinesq equation (1) for depth-average velocity together with an additional constraint of boundary condition, namely, an initial-boundary value problem for (1). In this paper, we have considered a nontrivial boundary condition, while being in contrast with other studies, where the null (or trivial) boundary condition has been usually imposed [6, 14, 15]. Physically, the nontrivial boundary condition considered here may be related with the wavemaker problem in fluid mechanics [16].One of the main purposes of this paper is to construct an integral equation formalism, equivalent to the initial value problem for (1). Here, the derived formalism is required to be regular and different from the usual existing integral equation ones (e.g., see [17]). For the construction, as a counterpart of the original initial-boundary value problem (defined on the positive real axis), we have first built an auxiliary initial-boundary value problem (defined on the negative real axis). We also have introduced two pseudoparameters on artificial dissipation to formulate a pseudodissipative formulation for the initial-boundary value problem for (1) [6]. This results in formulating a two-parameter family of the regular nonlinear integral equations, which is different from conventional integral-equation formalisms.By the method of successive approximations, the regular integral equations derived above have immediately led to an iterative formula for solving the initial-boundary value problem. Further, the formula has enabled us to simulate the propagating solitary wave as illustrated in the numerical experiment. The (numerically) simulated results are compared to the exact solution and the agreement is shown to be excellent.



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