# Well-Posedness of a Hirota–Satsuma System Posed on a Half Line

Authors: Haiyan Lu
Journal: Journal of Mathematics
 (2023)

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## Abstract

This paper studies the initial boundary value problem of the Hirota–Satsuma system posed on the half line.ut−αuxxx+6uux=2βvvx,x>0,t>0,vt+vxxx+3uvx=0,x>0,t>0,ux,0=u0x,vx,0=v0x,x>0,u0,t=ft,v0,t=gt,t>0.. For −1/8<s<3/2,s≠1/2, we demonstrate that the abovementioned system is locally well-posed in Hsℝ×H1+sℝ by utilizing several analytic boundary forcing operators.


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## Body

## 1. Introduction

In 1981, Hirota and Satsuma [1] introduced(1)ut−αuxxx+6uux=2βvvx,x∈ℝ,t≥0,vt+vxxx+3uvx=0,x∈ℝ,t≥0,where α∈ℝ, β∈ℝ; u=ux,t, v=vx,t are the real functions modeling the interactions of two long waves with different dispersion relations. Nowadays, (1) is called the Hirota–Satsuma system. Additionally, Hirota and Ohta in [2] derived the Hirota–Satsuma system as a reduction of a special hierarchy of coupled bilinear equations.The initial value problem for the Hirota–Satsuma system on the whole line and periodic domain has been extensively studied. In 2005, Angulo [3] proved that the system is locally well-posed in Hperiodics0,L×Hperiodics0,L, for s≥0, when α=−1, and globally well-posed in Hperiodics0,L×Hperiodics0,L for s≥1, when α≠−1,0. Furthermore, in 2007, Panthee and Silva [4] verified that the system is locally well-posed in Hperiodics0,L×Hperiodic1+s0,L , for s≥−1/2, when α=−1, and globally well-posed in Hperiodics0,L×Hperiodic1+s0,L, for s≥−14/3, when α=−1. Moreover, in 1994, Feng [5] demonstrated that the system is locally well-posedHsℝ×Hsℝ, for s>2, and in 2022, Zhao and Lv [6] confirmed that the system is locally well-posed in Hsℝ×H1+sℝ , for s≥−1/8, when α=−1,β=1.Without losing generalization, for the remainder of this paper, we assume thatα=−1 and β=1.This study will investigate the initial boundary value problem for the Hirota–Satsuma system posed on the half line as follows:(2)ut+uxxx+6uux=2vvx,x>0,t>0,vt+vxxx+3uvx=0,x>0,t>0,ux,0=u0x,vx,0=v0x,x>0,u0,t=ft,v0,t=gt,t>0.The main result regarding the IBVP (2) can be stated as follows:Theorem 1.
Let−1/8<s<3/2,s≠1/2. Then, there exists(3)T=Tu0Hsℝ+,v0H1+sℝ+,fHs+1/3ℝ+,gHs+1/3ℝ+,and a local solution u⋅,t,v⋅,t of IBVP (2), in the distributional sense, such that(4)u⋅,t,v⋅,t∈C0,T;Hsℝ+×H1+sℝ+.Moreover, the mappingu0,v0,f,g↦u⋅,t,v⋅,t is locally Lipschitz-continuous from Hsℝ+×H1+sℝ+×Hs+1/3ℝ+×Hs+1/3ℝ+ to C0,T;Hsℝ+×H1+sℝ+.


## 2. Presenting the Solution

### 2.1. Solution to the Initial Value Problem

We define the linear unitary group(5)e−t∂x3:S′ℝ⟶S′ℝ,which associated to the linear KdV equation as(6)e−t∂x3ϕx=12π∫ℝeiξxeitξ3ϕx^dξ,so that(7)∂t+∂x3e−t∂x3ϕx,t=0,x,t∈ℝ×ℝ,e−t∂x3x,0=ϕx,x∈ℝ.Thus, if we set(8)wx,t=e−t∂x3ϕx,then w solves the linear problem(9)wtx,t+wxxxx,t=0,x,t∈ℝ∗×ℝ,ux,0=ϕx,x∈ℝ.


### 2.2. Solution to the Boundary Value Problem

First, we introduce the Duhamel boundary forcing operator suggested by Colliander and Kenig [7]. Its definition is(10)Vfx,t=3∫0te−t−t′∂x3δ0xℐ−2/3ft′dt′=3∫0tAxt−t′1/3ℐ−2/3ft′t−t′1/3dt′,which is defined for all f∈C0∞ℝ+ and A signifies the Airy function(11)Ax=12π∫ξeixξeiξ3dξ.From this definition, we observe that(12)∂t+∂x3Vfx,t=3δ0xℐ−2/3ft,x,t∈ℝ×ℝ,Vfx,0=0,x∈ℝ.Next, we introduce the generalization of operatorV used in Holmer [8].Forf∈C0∞ℝ+ and λ∈ℂ with Reλ>−3, we have(13)V+λfx,t=x−λ−1Γλ∗Vℐ−λ/3f⋅,tx,where x−λ−1/Γλ=eiπλ−x+λ−1/Γλ. Then, by using (12), we obtain(14)∂t+∂x3V+λfx,t=3x−λ−1Γλℐ−2/3−λ/3ft.For any−1/8<s<3/2, s≠1/2, we can address IBVP (2) by replacing V in (12) with V+λ for a suitable λ=λs.The proofs of the following Lemmas exhibited in this section are presented in [8].Lemma 1.
(Spatial continuity and decay properties forV+λfx,t). If f∈C0∞ℝ+ and fix t≥0. Then, we have(15)V+0f=Vf.
Forλ>−2,V+λfx,t is continuous in x for all x∈ℝ. For −2≤λ≤1 and 0≤t≤1,V+λfx,t satisfies the following decay bounds:(16)V+λfx,t≤cm,λ,fx−m,x≥0,m≥0,V+λfx,t≤cλ,fxλ−1,x≤0.Lemma 2.
(Values ofV+λfx,t at x=0). For f∈C0∞ℝ+ and Reλ>−2, we have(17)V+λf0,t=eiπλft.Now, by adopting [8], we obtain the solution to the boundary value problem by using the V+λ class. Specifically, let −1<λ<1, h∈C0∞ℝ+, if we set(18)wx,t=V+λhx,t,from Lemmas 1 and 2,(19)w0,t=ft=eiπλ1ht.Moreover, ifft is given and we set(20)ht=e−iπλft,then w solves the linear problem(21)wtx,t+wxxxx,t=0,x,t∈ℝ∗×ℝ,wx,0=0,x∈ℝ,w0,t=ft,t>0.


### 2.3. Solution to the Initial-Boundary Value Problem

From the Lemmas presented and considering(22)wx,t=e−t∂x3ϕx+V+λhx,t,where(23)ht=e−iπλf−e−t∂x3ϕ|x=0x,t,then w solves the linear problem(24)wtx,t+wxxxx,t=0,x,t∈ℝ∗×ℝ,wx,0=ϕx,x∈ℝ,w0,t=ft,t>0.


### 2.4. Nonhomogeneous Versions

We introduce the Duhamel nonhomogeneous solution operatorK associated with the KdV equation as(25)Kwx,t=∫0te−t−t′∂x3wx,t′dt′,and it follows that(26)∂t+∂x3Kwx,t=wx,t,x,t∈ℝ×ℝ,Kwx,0=0,x∈ℝ.


### 2.5. Solution to the Hirota–Satsuma System

Using (9), (21), (24), and (26), we obtain the solution of IBVP (2) as(27)ux,t=e−t∂x3u0x+V+λ1h1x,t+K−6uux+2vvxx,t,vx,t=e−t∂x3v0x+V+λ2h2x,t+K−3uvxx,t,where(28)h1t=e−iπλ1f−e−t∂x3ϕ|x=0−K−6uux+2vvx0,tx,t,h2t=e−iπλ2g−e−t∂x3ϕ|x=0−K−3uvx0,tx,t.


## 2.1. Solution to the Initial Value Problem

We define the linear unitary group(5)e−t∂x3:S′ℝ⟶S′ℝ,which associated to the linear KdV equation as(6)e−t∂x3ϕx=12π∫ℝeiξxeitξ3ϕx^dξ,so that(7)∂t+∂x3e−t∂x3ϕx,t=0,x,t∈ℝ×ℝ,e−t∂x3x,0=ϕx,x∈ℝ.Thus, if we set(8)wx,t=e−t∂x3ϕx,then w solves the linear problem(9)wtx,t+wxxxx,t=0,x,t∈ℝ∗×ℝ,ux,0=ϕx,x∈ℝ.


## 2.2. Solution to the Boundary Value Problem

First, we introduce the Duhamel boundary forcing operator suggested by Colliander and Kenig [7]. Its definition is(10)Vfx,t=3∫0te−t−t′∂x3δ0xℐ−2/3ft′dt′=3∫0tAxt−t′1/3ℐ−2/3ft′t−t′1/3dt′,which is defined for all f∈C0∞ℝ+ and A signifies the Airy function(11)Ax=12π∫ξeixξeiξ3dξ.From this definition, we observe that(12)∂t+∂x3Vfx,t=3δ0xℐ−2/3ft,x,t∈ℝ×ℝ,Vfx,0=0,x∈ℝ.Next, we introduce the generalization of operatorV used in Holmer [8].Forf∈C0∞ℝ+ and λ∈ℂ with Reλ>−3, we have(13)V+λfx,t=x−λ−1Γλ∗Vℐ−λ/3f⋅,tx,where x−λ−1/Γλ=eiπλ−x+λ−1/Γλ. Then, by using (12), we obtain(14)∂t+∂x3V+λfx,t=3x−λ−1Γλℐ−2/3−λ/3ft.For any−1/8<s<3/2, s≠1/2, we can address IBVP (2) by replacing V in (12) with V+λ for a suitable λ=λs.The proofs of the following Lemmas exhibited in this section are presented in [8].Lemma 1.
(Spatial continuity and decay properties forV+λfx,t). If f∈C0∞ℝ+ and fix t≥0. Then, we have(15)V+0f=Vf.
Forλ>−2,V+λfx,t is continuous in x for all x∈ℝ. For −2≤λ≤1 and 0≤t≤1,V+λfx,t satisfies the following decay bounds:(16)V+λfx,t≤cm,λ,fx−m,x≥0,m≥0,V+λfx,t≤cλ,fxλ−1,x≤0.Lemma 2.
(Values ofV+λfx,t at x=0). For f∈C0∞ℝ+ and Reλ>−2, we have(17)V+λf0,t=eiπλft.Now, by adopting [8], we obtain the solution to the boundary value problem by using the V+λ class. Specifically, let −1<λ<1, h∈C0∞ℝ+, if we set(18)wx,t=V+λhx,t,from Lemmas 1 and 2,(19)w0,t=ft=eiπλ1ht.Moreover, ifft is given and we set(20)ht=e−iπλft,then w solves the linear problem(21)wtx,t+wxxxx,t=0,x,t∈ℝ∗×ℝ,wx,0=0,x∈ℝ,w0,t=ft,t>0.


## 2.3. Solution to the Initial-Boundary Value Problem

From the Lemmas presented and considering(22)wx,t=e−t∂x3ϕx+V+λhx,t,where(23)ht=e−iπλf−e−t∂x3ϕ|x=0x,t,then w solves the linear problem(24)wtx,t+wxxxx,t=0,x,t∈ℝ∗×ℝ,wx,0=ϕx,x∈ℝ,w0,t=ft,t>0.


## 2.4. Nonhomogeneous Versions

We introduce the Duhamel nonhomogeneous solution operatorK associated with the KdV equation as(25)Kwx,t=∫0te−t−t′∂x3wx,t′dt′,and it follows that(26)∂t+∂x3Kwx,t=wx,t,x,t∈ℝ×ℝ,Kwx,0=0,x∈ℝ.


## 2.5. Solution to the Hirota–Satsuma System

Using (9), (21), (24), and (26), we obtain the solution of IBVP (2) as(27)ux,t=e−t∂x3u0x+V+λ1h1x,t+K−6uux+2vvxx,t,vx,t=e−t∂x3v0x+V+λ2h2x,t+K−3uvxx,t,where(28)h1t=e−iπλ1f−e−t∂x3ϕ|x=0−K−6uux+2vvx0,tx,t,h2t=e−iπλ2g−e−t∂x3ϕ|x=0−K−3uvx0,tx,t.


## 3. Main Estimates

In this section, we introduce the space trace estimates, the time trace estimates, and the Bourgain regularities.Lets∈ℝ and b<1/2; we introduce the classical Bourgain spaces Xs,b associated with ∂t+∂x3 as the completion of the Schwartz space SR2 under the norm(29)uXs,b=∫∫ξ2sτ−ξ32bu^ξ,τ2dξdτ1/2,where ξ=1+ξ.To obtain our results, we define the following auxiliary modified Bourgain spaces. LetYs,b and Zα be the completion of S′ℝ2 with respect to the norms(30)uYs,b=∫∫τ2s/3τ−ξ32bu^ξ,τ2dξdτ1/2,and(31)uZα=∫∫τ2αu^ξ,τ2dξdτ1/2.The following Lemma is an important estimate presented [9].Lemma 3.
Let−1/2<b′<b≤0 or 0≤b′<b<1/2,w∈Xs,b, and s∈ℝ. Then, we have(32)ψTwXs,b′ϕ≤cTb−b′wXs,bϕ.


### 3.1. Space Trace Estimates

Lemma 4.
Lets∈ℝ.(a)
Ifϕ∈Hsℝ, then we have(33)e−t∂x3ϕxCℝt;Hsℝx≤cϕHsℝ.(b)
Ifs−5/2<λ<s+1/2,λ<1/2 and suppf⊂0,1, then we have(34)V+λfx,tCℝt;Hsℝx≤cfH0s+1/3ℝ+.(c)
If1/2<d<1, then we have(35)ψtKwx,tCℝt;Hsℝx≤cwXs,d−1.Proof.
See reference [8].


### 3.2. Time Trace Estimates

Lemma 5.
Lets∈ℝ.(a)
For allϕ∈Hsℝ, we have(36)ψte−t∂x3ϕxCℝx;Hs+1/3ℝt≤cϕHsℝ.(b)
For all−2<λ<1, we have(37)ψtV+λfx,tCℝx;H0s+1/3ℝt+≤cfH0s+1/3ℝ+.(c)
For all1/2<d<1, we have(38)ψtKwx,tCℝx;Hs+1/3ℝt≲wXs,d−1,−1≤s≤12,wXs,d−1+wYs,d−1,s∈ℝ.Proof.
See reference [8].


### 3.3. Bourgain Regularities

Lemma 6.
Lets∈ℝ.(a)
For all0<b<1, if ϕ∈Hsℝ, we have(39)ψte−t∂x3ϕxXs,b∩Zα≤cϕHsℝ.(b)
For alls−1≤λ<s+1/2,λ<1/2,α≤s−λ+2/3, and 0≤b<1/2, we have(40)ψtV+λfx,tXs,b∩Zα≤cfH0s+1/3R+.(c)
For all1/2<b<1 and α>1/2, we have(41)ψtKwx,tXs,b∩Zα≤wXs,b−1.Proof.
See references [8, 10].


### 3.4. Bilinear Estimates

The following Lemma was proposed by Zhao and its proof can be found in [6].Lemma 7.
Ifs≥−1/8,1/2<b<9/16, then ∀b′>1/2, we have(42)u1xu2X1+s,b−1≤Cu1X1+s,b′u2Xs,b′,and(43)u1u2xXs,b−1≤Cu1Xs,b′u2Xs,b′.


## 3.1. Space Trace Estimates

Lemma 4.
Lets∈ℝ.(a)
Ifϕ∈Hsℝ, then we have(33)e−t∂x3ϕxCℝt;Hsℝx≤cϕHsℝ.(b)
Ifs−5/2<λ<s+1/2,λ<1/2 and suppf⊂0,1, then we have(34)V+λfx,tCℝt;Hsℝx≤cfH0s+1/3ℝ+.(c)
If1/2<d<1, then we have(35)ψtKwx,tCℝt;Hsℝx≤cwXs,d−1.Proof.
See reference [8].


## 3.2. Time Trace Estimates

Lemma 5.
Lets∈ℝ.(a)
For allϕ∈Hsℝ, we have(36)ψte−t∂x3ϕxCℝx;Hs+1/3ℝt≤cϕHsℝ.(b)
For all−2<λ<1, we have(37)ψtV+λfx,tCℝx;H0s+1/3ℝt+≤cfH0s+1/3ℝ+.(c)
For all1/2<d<1, we have(38)ψtKwx,tCℝx;Hs+1/3ℝt≲wXs,d−1,−1≤s≤12,wXs,d−1+wYs,d−1,s∈ℝ.Proof.
See reference [8].


## 3.3. Bourgain Regularities

Lemma 6.
Lets∈ℝ.(a)
For all0<b<1, if ϕ∈Hsℝ, we have(39)ψte−t∂x3ϕxXs,b∩Zα≤cϕHsℝ.(b)
For alls−1≤λ<s+1/2,λ<1/2,α≤s−λ+2/3, and 0≤b<1/2, we have(40)ψtV+λfx,tXs,b∩Zα≤cfH0s+1/3R+.(c)
For all1/2<b<1 and α>1/2, we have(41)ψtKwx,tXs,b∩Zα≤wXs,b−1.Proof.
See references [8, 10].


## 3.4. Bilinear Estimates

The following Lemma was proposed by Zhao and its proof can be found in [6].Lemma 7.
Ifs≥−1/8,1/2<b<9/16, then ∀b′>1/2, we have(42)u1xu2X1+s,b−1≤Cu1X1+s,b′u2Xs,b′,and(43)u1u2xXs,b−1≤Cu1Xs,b′u2Xs,b′.


## 4. Proof of the Main Theorem

Before proving the main theorem, some helpful properties of the Sobolev spaces are outlined in the results that follow. The proof of them can be found in [7].Lemma 8.
Let−1/2<s<1/2. If f∈Hsℝ, then there is a constant cs such that(44)χ0,+∞fHsℝ≤csfHsℝ.Lemma 9.
Let1/2<s<3/2. If f∈Hsℝ+ with f0=0; then, we have H0sℝ+=f∈Hsℝ+;f0=0 and there is a constant cs such that(45)χ0,+∞fH0sℝ+≤csfHsℝ+.Proof of Theorem 1.
Letu˜0,v˜0,f˜, and g˜ be the extensions of u0,v0,f, and g such that u˜0Hsℝ≤cu0Hsℝ+, v˜0H1+sℝ≤cv0H1+sℝ+,f˜Hs+1/3ℝ≤cfHs+1/3ℝ+, and g˜Hs+1/3ℝ≤cgHs+1/3ℝ+. Let b<9/16 such that Lemma 7 is valid.Step 1.
Solution mapping and spaces.
Take−1<λ1,λ2<1, h1,h2∈C0∞ℝ+. By Section 2.5 we define the operator Λ=Λ1,Λ2, which is given by(46)Λ1u,v=ψte−t∂x3u˜0x+ψtK−3ψTu2x+ψTv2xx,t+ψte−iπλ1V+λ1h1x,t,and(47)Λ2u,v=ψte−t∂x3v˜0x+ψtK−3ψTuvxx,t+ψte−iπλ2V+λ2h2x,t,where h1 and h2 are given by(48)h1t=ψtf˜t−ψte−t∂x3u˜0|x=0−ψtK−3ψTu2x+ψTv2x0,t|0,+∞,and(49)h2t=ψtg˜t−ψte−t∂x3v˜0|x=0−ψtK−3ψTuvx0,t|0,+∞.`
We considerΛ in the Banach space W=W1×W2, where(50)W1=Cℝt;Hsℝx∩Cℝx;Hs+1/3ℝt∩Xs,b∩Zα,W2=Cℝt;H1+sℝx∩Cℝx;Hs+1/3ℝt∩X1+s,b∩Zα,with norm(51)u,vW=uW1+vW2:=uCℝt;Hsℝx+uCℝx;Hs+1/3ℝt+uXs,b+uZα+vCℝt;H1+sℝx+vCℝx;Hs+1/3ℝt+vX1+s,b+vZα.Step 2.
Estimates of boundary values.
For this purpose, by using Lemmas4–6 (b), we show that h1∈H0s+1/3ℝ+ and h2∈H0s+1/3ℝ+.
By hypothesizingf∈Hs+1/3ℝ+, Lemmas 8 and 5 (a) imply(52)ψte−t∂x3u˜0|x=0|0,+∞Hs+1/3ℝ+≤ψte−t∂x3u˜0|x=0Hs+1/3ℝ≤cu˜0Hsℝ.
Now, Lemmas3, 5 (c), 7, and 8 imply(53)ψtK−3ψTu2x+ψTv2x0,t|0,+∞Hs+1/3ℝ+≤ψtKψT−3u2x+v2x0,tHs+1/3ℝ≤cψT−3u2x+v2xXs,b−1≤cTϵ−3u2x+v2xXs,b′−1≤cTϵuXs,b2+vXs,b2≤cTϵuXs,b2+vX1+s,b2,where ϵ=b′−b is adequately small.
If−1/8<s<1/2, then7/24<s+1/3<1/2, and Lemma 8 shows that Hs+1/3ℝ+=H0s+1/3ℝ+. If 1/2<s<3/2, then1/2<s+1/3<5/6 due to the compatibility condition, and Lemma 9 shows that Hs+1/3ℝ+=H0s+1/3ℝ+. Thus, equations (52) and (53) reveal that h1∈H0s+1/3ℝ+.
By combining the abovementioned, we obtain(54)h1H0s+1/3ℝ+≤cfHs+1/3ℝ++cu0Hsℝ++cTϵuXs,b2+vX1+s,b2.
Similarly, from Lemmas3, 5 (a), 5 (c), 7, and 8, we have(55)ψtg˜−ψte−t∂x3v˜0|x=0−ψtψtK−3ψTuvx0,t|0,+∞Hs+1/3ℝ+≤ψtg˜−ψte−t∂x3v˜0|x=0−ψtK−3ψTuvx0,tHs+1/3ℝ,≤cgHs+1/3ℝ++v0H1+sℝ++ψTuvxXs,b−1,≤cgHs+1/3ℝ++v0H1+sℝ++TϵuvxXs,b′−1,≤cgHs+1/3ℝ++v0H1+sℝ++TϵuXs,bvX1+s,b.
It follows that(56)ψtg˜−e−t∂x3v˜0|x=0−K−3ψTuvx0,t|0,+∞∈Hs+1/3ℝ+.
Similarly, ash1, we obtain(57)ψtg˜−e−t∂x3v˜0|x=0−K−3ψTuvx0,t|0,+∞∈H0s+1/3ℝ+.
Indeed, the abovementioned equations show thath2∈H0s+1/3ℝ+.
Thus, we obtain(58)h2H0s+1/3ℝ+≤cgHs+1/3ℝ++cv0H1+sℝ++cTϵuXs,bvX1+s,b.Step 3.
Contraction of the solution mapping.
From Lemmas3, 4, 7, and equation (54), we have(59)Λ1u,vCℝt;Hsℝx≤cu0Hsℝ++fHs+1/3ℝ++TϵuXs,b2+TϵvX1+s,b2.
From Lemmas3, 5, 7, and equation (54), we have(60)Λ1u,vCℝx;Hs+1/3ℝt≤cu0Hsℝ++fHs+1/3ℝ++TϵuXs,b2+TϵvX1+s,b2.
From Lemmas3, 6, 7, and equation (54), we have(61)Λ1u,vXs,b∩Zα≤cu0Hsℝ++fHs+1/3ℝ++TϵuXs,b2+TϵvX1+s,b2,by combining the abovementioned, we obtain(62)Λ1u,vW1≤cu0Hsℝ++fHs+1/3ℝ++TϵuXs,b2+TϵvX1+s,b2.
From Lemmas3–7, and equation (58), we have(63)Λ2u,vW2≤cv0H1+sℝ++gHs+1/3ℝ++TϵuXs,bvX1+s,b.
Inspired from [11], we have(64)Λ1u1,v1−Λ1u2,v2=ψtK−3ψTu1+u2u1−u2x+ψTv1+v2v1−v2xx,t+ψte−iπλ1V+λ1h1x,t,Λ2u1,v1−Λ2u2,v2=ψtK−3ψTu1v1x−u2v2xx,t+ψte−iπλ2V+λ2h2x,t,Λu1,v1−Λu2,v2=Λ1u1,v1−Λ1u2,v2Λ2u1,v1−Λ2u2,v2,where h1 and h2 are given by(65)h1t=−ψtK−3ψTu1+u2u1−u2x+ψTv1+v2v1−v2x0,t|0,+∞,and(66)h2t=−ψtK−3ψTu1v1x−u2v2x0,t|0,+∞.
Similarly, as abovementioned, we have(67)Λu1,v1−Λu2,v2W≤cTϵu1Xs,b+u2Xs,bu1−u2Xs,b+v1−v2X1+s,bv1X1+s,b+v2X1+s,b+v1X1+s,bu1−u2Xs,b+u2Xs,bv1−v2X1+s,b.
We set the ball ofW as(68)B=u,v∈W;uW1≤M1,vW2≤M2,where M1=2cu0Hsℝ++fHs+1/3ℝ+ and M2=2cv0H1+sℝ++gHs+1/3ℝ+.
By restrictingu,v on the ball B, by (62) and (63) and (67), we obtain(69)Λ1u,vW1≤M12+cTϵM12+M22,Λ2u,vW2≤M22+cTϵM1M2,Λu1,v1−Λu2,v2W1≤cTϵM1u1−u2W1+M2v1−v2W2,Λu1,v1−Λu2,v2W2≤cTϵM1+M2u1−u2W1+v1−v2W2.
Then, we takeT=TM1,M2 to be suitably small, such that(70)Λ1u,vW1≤M1,Λ2u,vW2≤M2,and(71)Λu1,v1−Λu2,v2W≤12u1,v1−u2,v2W.
Thus,Λ defines a contraction in W∩B, and we obtain a fixed point in u,v in B. Therefore, the IBVP (2) is solved in the sense of distributions by(72)u,v:=u|x,t∈C0,T;Hsℝ+,v|x,t∈C0,T;H1+sℝ+.


## 5. Conclusions

According to [4, 6], the asymmetry of the nonlinear component uvx makes the asymmetrical product space Hsℝ×H1+sℝ better suited for the Hirota–Satsuma system than the symmetric product space Hsℝ×Hsℝ. Thus, this paper studies the IBVP of the Hirota–Satsuma system in the asymmetrical product space. Inspired by [8, 11], we propose the following directions for further consideration:Remark 1.
The skill used in this paper is also applicable to the IBVP of the Hirota–Satsuma system posed on the left half lineℝ−=−∞,0:(73)ut−αuxxx+6uux=2βvvx,x∈−∞,0,t≥0,vt+vxxx+3uvx=0,x∈−∞,0,t≥0,ux,0=u0x,vx,0=v0x,x∈−∞,0,u0,t=ft,v0,t=g1t,vx0,t=g2t,t≥0.Remark 2.
In addition to discussing the left half line, we may consider other boundary conditions on the right half line. More specifically, letA denote the differential operator associated with the KdV equation(74)AX=−d3udx3,−d3vdx3T,where X=u,vT. Since(75)AX,X=A∗X,X=−d2udx2u¯|0∞+12dudx2|0∞−d2vdx2v¯|0∞+12dvdx2|0∞=0,with any of the following boundary values:(76)a.uxx0,t=vxx0,t=0,b.u0,t=vxx0,t=0,c.u0,t−v0,t=0,uxx0,t+vxx0,t=0,d.u0,t+vxx0,t=0,uxx0,t−v0,t=0.
For any one of the boundary values a, b, c, and d, operatorA generates an operator group by the Phillips–Lumper theorem. We will investigate those questions in the future.



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*Source: 6211563-2023-01-17.xml*