# Global Strong Solutions to Some Nonlinear Dirac Equations in

Authors: Hyungjin Huh
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We study the initial value problem of some nonlinear Dirac equations which areLmℝ critical. Corresponding to the structure of nonlinear terms, global strong solutions can be obtained in different Lebesgue spaces by using solution representation formula. The
uniqueness of weak solutions is proved for the solution U∈L∞0,T;Ym+2ℝ.


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## Body

## 1. Introduction

In this study, we are interested in global strong solutions of nonlinear Dirac type equations:(1)-i(∂tU1+∂xU1)=∂U-1F(U1,U2),-i(∂tU2-∂xU2)=∂U-2F(U1,U2),
where F(U1,U2):ℂ2→ℝ is a nonlinear potential function which will be specified later.The following Thirring model with the potentialF(U1,U2)=|U1|2|U2|2 has been studied in [1–4]:
(2)-i(∂tU1+∂xU1)=|U2|2U1,-i(∂tU2-∂xU2)=|U1|2U2,U1(0,x)=u1(x),U2(0,x)=u2(x).
The initial value problem of (2) was studied in [1, 3] in terms of Sobolev space Hs(s≥1). Low regularity well-posedness was discussed in [4] showing that there exists a time T>0 and solution U∈C([0,T],Hs(ℝ)) (s>0) of the Cauchy problem (2). Their results were based on the observation of the null structure of Thirring equations and application of Xs,b spaces which is a certain subspace of C([0,T],Hs(ℝ)). They also proved global well-posedness for s>1/2 and unconditional uniqueness for s>1/4. Nonlinear Dirac equations in ℝ1+1 have been studied by several authors [5–9].In the context of Bragg grating [10, 11], the nonlinear term F takes the form
(3)F=|U1|4+|U2|4+4|U1|2|U2|2,
which gives ∂U-1F=(2|U1|2+4|U2|2)U1 and ∂U-2F=(2|U2|2+4|U1|2)U2. We may consider the sixth and higher orders. The following potential term is introduced in the context of the Bose-Einstein condensates [12]:
(4)F=(|U1|2+|U2|2)|U1|2|U2|2,
which gives ∂U-1F=(2|U1|2|U2|2+|U2|4)U1 and ∂U-2F=(2|U2|2|U1|2+|U1|4)U2.Several authors [4, 5, 7] have studied the initial value problem of the following Dirac equations with quadratic nonlinearities:
(5)-i(∂tU1+∂xU1)=Q1(U1,U2),-i(∂tU2-∂xU2)=Q2(U1,U2).
If the nonlinear term is of Q1=|U2|2, Q2=|U1|2 type (Q1=U1U-2, Q2=U-1U2, resp.), then one can obtain local well-posedness for the Sobolev space Hs(s>0) [4] (Hs(s>-1/4) [7], resp.). Note that the scaling properties of quadratic Dirac equations give the critical Sobolev exponent scr=-1/2.Now it seems natural to consider the following equations:(6)∂tU1+∂xU1=i|U1|k|U2|m-kU1,∂tU2-∂xU2=i|U2|k|U1|m-kU2,
which are generalization of the basic cases of the literature and model problem to investigate regularities of solutions according to the structure of nonlinearities. Here, m is positive integer and k=0,1,…,m.The system (6) is invariant under the scaling
(7)Uλ(t,x)=λU(λmt,λmx),
from which we deduce a scale invariant Lebesgue space Lm(ℝ). We study the initial value problem of (6) in Lebesgue space. Let us denote
(8)p(m,k)=Max(m-k,2k+1).
Note that p(m,k)≥2 except for m=1 and k=0 where p(1,0)=1. We define
(9)Yp(ℝ)=L2(ℝ)∩Lp(ℝ),
where p≥2. We call Yp as subcritical space for p>m, critical space for p=m, and supercritical space for p<m.For the initial datauj∈L2(ℝ)(j=1,2), we have proved in [2] that there exists a global strong solution U=(U1,U2) to the initial value problem (2) in the critical space
(10)U∈C([0,∞);L2(ℝ)).
Elementary and interesting approach was made. and special structure of nonlinearity was made use of in applying Fubini's theory in integration. The following is concerned with the existence of global strong solutions for the case of m≥2 or k≥1.Theorem 1.
For the initial datavj∈Yp(m,k)(ℝ)(j=1,2), there exists a global strong solution V=(V1,V2) to the initial value problem (6) which satisfies
(11)V∈C([0,∞);Yp(m,k)(ℝ)).Remark 2.
(
1
) Related with the title of this paper, we emphasize the case p(m,k)<m where global strong solution can be constructed in the super-critical space.
(
2
) Taking the embedding Yq(ℝ)↪Yr(ℝ)    (q≥r) into account, it is likely to say that the system (6) with small p(m,k) has better smoothing property than equations with large p(m,k). We can check, in the case of (4), the existence of solution V∈C([0,∞);Y5(ℝ)) because of p(4,2)=5>4=p(4,0).
(
3
) For the case of m=1 and k=0, we can construct a global strong solution V=(V1,V2) to the initial value problem (6) which satisfies
(12)V∈C([0,∞);L1(ℝ))
for the initial data vj∈L1(ℝ)(j=1,2).Our second result is concerned with the uniqueness of weak solutions. The null structure of the nonlinearity of Thirring model which isL2 critical problem was used in [2] to prove the uniqueness in Y4. To treat the general nonlinearity in (6) of which we could not find the null structure, a different approach is considered.Theorem 3.
LetU and V be solutions of (6) in the distribution sense with same initial data. Moreover, one assumes that
(13)U,V∈L∞([0,T];Ym+2(ℝ)).
Then, one has ∥(U-V)(t,·)∥L2=0 for 0≤t≤T.Theorems1 and 3 imply that if p(m,k)≥m+2 then a strong solution to the initial value problem (6) is unique and is in fact a well-posed solution.Theorem1 is proved in Section 2 and Theorem 3 in Section 2.2. We use C to denote various constants and A≲B to denote an estimate of the form A≤CB.


## 2. Proof of Theorem1

To construct global strong solution of (6), we basically follow the idea of [2] with modification. We will find a solution representation formula. Then, global strong solutions can be obtained by constructing an explicit approximation and using Fubini's theorem.


### 2.1. Representation Formula of Classical Solution

In this subsection, we considerC∞((0,T)×ℝ) solutions which satisfy (6) in the classical sense. An explicit representation of its solution is given in terms of initial data [2, 13]. It is interesting to express the solution of nonlinear partial differential equations in terms of initial data.Integrating (6) along the outgoing and ingoing characteristics, we obtain
(14)U1(t,x)=u1(x-t)×exp(i∫0t‍|U1(s,x-t+s)|k×|U2(s,x-t+s)|m-kdsi∫0t‍|U1(s,x-t+s)|k),U2(t,x)=u2(x+t)×exp(i∫0t‍|U2(s,x+t-s)|k×|U1(s,x+t-s)|m-kdsi∫0t‍|U1(s,x-t+s)|k).
Taking into account
(15)|U1(t,x)|=|u1(x-t)|,|U2(t,x)|=|u2(x+t)|,
(14) becomes
(16)U1(t,x)=u1(x-t)×exp(i2|u1(x-t)|k∫x-tx+t‍|u2(y)|m-kdy),U2(t,x)=u2(x+t)×exp(i2|u2(x+t)|k∫x-tx+t‍|u1(y)|m-kdy).
In the case of k=m, we have the following obvious representation formula of solution:
(17)U1(t,x)=u1(x-t)exp(it|u1(x-t)|m),U2(t,x)=u2(x+t)exp(it|u1(x-t)|m).


### 2.2. Global Strong Solutions inYp(ℝ)

We now present the proof of Theorem1. Let us introduce a strong solution of the initial value problem of the system (6).Definition 4.
Consider the Cauchy problem (6) with initial data v=(v1,v2)∈Yp(m,k)(ℝ)). It is said that V=(V1,V2) is a strong solution to the Cauchy problem on the time interval [0,T] provided that
(18)(V1,V2)∈C([0,T];Yp(m,k)(ℝ))
satisfy (6) in the sense of distributions. That is, for any ϕ∈C0∞((-T,T)×ℝ), we have
(19)∫0T‍∫ℝ‍V1∂tϕ+V1∂xϕ+i|V1|k|V2|m-kV1ϕdxdt+∫ℝ‍v1(x)ϕ(0,x)dx=0,(20)∫0T‍∫ℝ‍V2∂tϕ-V2∂xϕ+i|V2|k|V1|m-kV2ϕdxdt+∫ℝ‍v2(x)ϕ(0,x)dx=0.Remark 5.
We say thatV is a weak solution if V∈L∞([0,T];Yp(m,k)(ℝ)) satisfies (6) in the sense of distribution.For the initial datavj∈Yp(m,k)(ℝ)(j=1,2), we propose that the following functions Vj are the global strong solution of (6):
(21)V1(t,x)=v1(x-t)×exp(i2|v1(x-t)|k∫x-tx+t‍|v2(y)|m-kdy),V2(t,x)=v2(x+t)×exp(i2|v2(x+t)|k∫x-tx+t‍|v1(y)|m-kdy).
For the smooth function sequence {uj(n)}n=1∞  (j=1,2) which converges to vj in Yp(m,k)(ℝ) as n→∞, we consider sequence of classical solutions Uj(n) of (6). First of all, we estimate the difference of (Vj,Uj(n)). Using the representation (16) and (21), we have
(22)|(V1-U1(n))(t,x)|≤|exp(i2|u1(n)(x-t)|k∫x-tx+t‍|u2(n)(y)|m-kdy)×(v1-u1(n))(x-t)exp(i2|u1(n)(x-t)|k∫x-tx+t‍|u2(n)(y)|m-kdy)|+|v1(x-t)(e(i/2)|v1(x-t)|k∫x-tx+t‍|v2(y)|m-kdy-e(i/2)|u1(n)(x-t)|k∫x-tx+t‍|u2(n)(y)|m-kdy)|≤|(v1-u1(n))(x-t)|+12|v1(x-t)|||v1(x-t)|k∫x-tx+t‍|v2(y)|m-kdy-|u1(n)(x-t)|k∫x-tx+t‍|u2(n)(y)|m-kdy|=(A)+(B),
where we used |eix-eiy|≤|x-y| for x, y∈ℝ. The Lq norm of the first term (A) can be treated as follows:
(23)(∫ℝ‍|(v1-u1(n))(x-t)|qdx)1/q=∥v1-u1(n)∥Lq.
To estimate the Lp norm of the second term (B), we consider the following three cases. (i)
For the casek=0, we have with 2≤p≤m(24)∫ℝ‍|v1(x-t)|p|∫x-tx+t‍|v2(y)|mdy-∫x-tx+t‍|u2(n)(y)|mdy|pdx≲∥v2-u2(n)∥Lmp(∥v2∥Lmm-1+∥u2(n)∥Lmm-1)p∥v1∥Lpp.(ii)
For the casek=m, we have with 1≤p≤(2m+1)/(m+1)(25)(2t)p∫ℝ‍|v1(x-t)|p||v1(x-t)|m-|u1(n)(x-t)|m|pdx≲(2t)p∥v1∥L(m+1)pp∥v1-u1(n)∥L(m+1)pp×(∥v1∥L(m+1)p(m-1)p+∥u1(n)∥L(m+1)p(m-1)p).(iii)
For other cases1≤k≤m-1, we have with 1≤p≤p(m,k)/(k+1)(26)∫ℝ‍|v1(x-t)|p||v1(x-t)|k∫x-tx+t‍|v2(y)|m-kdy-|u1(n)(x-t)|k∫x-tx+t‍|u2(n)(y)|m-kdy|pdx≲supx(∫x-tx+t‍|v2(y)|m-kdy)p×∫ℝ‍|v1(x-t)|p||v1(x-t)|k-|u1(n)(x-t)|k|pdx+supx(∫x-tx+t‍||v2(y)|m-k-|u2(n)(y)|m-k|dy)p×∫ℝ‍|v1(x-t)|p|u1(n)(x-t)|kpdx≲supx(∫x-tx+t‍|v2(y)|m-kdy)p∥v1∥L(k+1)pp×∥v1-u1(n)∥L(k+1)pp(∥v1∥L(k+1)p(k-1)p+∥u1(n)∥L(k+1)p(k-1)p)+supx(∫x-tx+t‍||v2(y)|m-k-|u2(n)(y)|m-k|dy)p×∥v1∥L(k+1)pp∥u1(n)∥L(k+1)pkp,where
(27)supx(∫x-tx+t‍|v2(y)|m-kdy)p≲{(2t)p/2∥v2∥L2pform-k=1,∥v2∥Lm-k(m-k)pform-k≥2,supx(∫x-tx+t‍||v2(y)|m-k-|u2(n)(y)|m-k|dy)p≲{(2t)p/2∥v2-u2(n)∥L2pform-k=1,∥v2-u2(n)∥Lm-kp(∥v2∥Lm-km-k-1+∥u2(n)∥Lm-km-k-1)pform-k≥2.Now, we are ready to prove thatV=(V1,V2) is a global strong solution to the Cauchy problem (6). It is easy to check (18) by considering the representation (21). Making use of the classical solution to (6), we have
(28)∫0T‍∫ℝ‍V1∂tϕ+V1∂xϕ+i|V1|k|V2|m-kV1ϕdxdt+∫ℝ‍v1(x)ϕ(0,x)dx=∫0T‍∫ℝ‍(V1-U1(n))(∂tϕ+∂xϕ)dxdt+∫ℝ‍(v1-u1(n))ϕ(0,x)dx+∫0T‍∫ℝ‍i(|V1|k|V2|m-kV1-|U1(n)|k|U2(n)|m-kU1(n))ϕdxdt=(1)+(2)+(3).
Considering (23)–(26) with q=p=p(m,k)/(k+1), the first integral (1) can be estimated as follows:
(29)|(1)|≲Tsup0≤t≤T(∥∂tϕ(t,·)∥Lq′+∥∂xϕ(t,·)∥Lq′)×sup0≤t≤T∥(V1-U1(n))(t,·)∥Lq,
where
(30)sup0≤t≤T∥(V1-U1(n))(t,·)∥Lq≲{∥v1-u1(n)∥Lq+[∥v2-u2(n)∥L2p∥v1∥Lp(m,k)p∥u1(n)∥Lp(m,k)kp(2T)p/2∥v2∥L2p∥v1∥Lp(m,k)p×∥v1-u1(n)∥Lp(m,k)p(∥v1∥Lp(m,k)(k-1)p+∥u1(n)∥Lp(m,k)(k-1)p)+(2T)p/2∥v2-u2(n)∥L2p∥v1∥Lp(m,k)p∥u1(n)∥Lp(m,k)kp]1/pform-k=1,∥v1-u1(n)∥Lq+[∥u1(n)∥Lp(m,k)kp∥v2∥Lm-k(m-k)p∥v1∥Lp(m,k)p∥v1-u1(n)∥Lp(m,k)p×(∥v1∥Lp(m,k)(k-1)p+∥u1(n)∥Lp(m,k)(k-1)p)+∥v2-u2(n)∥Lm-kp(∥v2∥Lm-km-k-1+∥u2(n)∥Lm-km-k-1)p×∥v1∥Lp(m,k)p∥u1(n)∥Lp(m,k)kp]1/pform-k≥2.
The integral (2) can be bounded easily:
(31)∫ℝ‍(v1-u1(n))ϕ(0,x)dx≤∥v1-u1(n)∥L2∥ϕ(0,x)∥L2.
To take care of the integral (3), we decompose the integral by using representation (16) and (21) with the notation E(v1,v2,k)=exp((i/2)|v1(x-t)|k∫x-tx+t‍|v2(y)|m-kdy):
(32)∫0T‍∫ℝ‍(|V1|k|V2|m-kV1-|U1(n)|k|U2(n)|m-kU1(n))ϕ(t,x)dxdt=∫0T‍∫ℝ‍(|v1(x-t)|k-|u1(n)(x-t)|k)|v2(x+t)|m-k×v1(x-t)E(v1,v2,k)ϕ(t,x)dxdt+∫0T‍∫ℝ‍|u1(n)(x-t)|k×(|v2(x+t)|m-k-|u2(n)(x+t)|m-k)×v1(x-t)E(v1,v2,k)ϕ(t,x)dxdt+∫0T‍∫ℝ‍|u1(n)(x-t)|k|u2(n)(x+t)|m-k×(v1-u1(n))(x-t)E(v1,v2,k)ϕ(t,x)dxdt+∫0T‍∫ℝ‍|u1(n)(x-t)|k|u2(n)(x+t)|m-ku1(n)(x-t)×(E(v1,v2,k)-E(u1(n),u2(n),k))ϕ(t,x)dxdt=(I)+(II)+(III)+(IV),
where we understand that the integral (I) does not exist for k=0 and the integral (II) for k=m.Changing variablesx+t=y and x-t=s, the first integral (I) can be bounded as follows:
(33)|(I)|≤∫ℝ‍∫ss+2T‍||v1(s)|k-|u1(n)(s)|k||v2(y)|m-k×|v1(s)||ϕ(y,s)|dyds≲sups(∫ss+2T‍|v2(y)|m-k|ϕ(y,s)|dy)×∫ℝ‍|(v1-u1(n))(s)|×(|v1(s)|k-1+|u1(n)(s)|k-1)|v1(s)|ds≲sups(∫ss+2T‍|v2(y)|m-k|ϕ(y,s)|dy)∥v1-u1(n)∥Lk+1×(∥v1∥Lk+1k-1+∥u1(n)∥Lk+1k-1)∥v1∥Lk+1,
where we note that k≥1. We also have
(34)sups∫ss+2T‍|v2(y)|m-k|ϕ(y,s)|dy≤{sups∫ss+2T‍|ϕ(y,s)|dyifm-k=0,∥v2∥L2sups(∫ss+2T‍|ϕ(y,s)|2dy)1/2ifm-k=1,M∥v2∥Lm-km-kifm-k≥2,
where we denote M=supΣT|ϕ(t,x)| where ΣT={(x,t)∣x∈ℝ,0≤t≤T}.For the integral (II), we consider two cases. Note that the integral (II) disappears ifk=m. For the first case m-k=1, we have
(35)|(II)|≤∫ℝ‍∫ss+2T‍|u1(n)(s)|k||v2(y)|-|u2(n)(y)||×|v1(s)||ϕ(y,s)|dyds≲sups(∫ss+2T‍||v2(y)|-|u2(n)(y)|||ϕ(y,s)|dy)×∫ℝ‍|u1(n)(s)|k|v1(s)|ds≲sups(∫ss+2T‍|ϕ(y,s)|2dy)1/2×∥v2-u2(n)∥L2∥u1(n)∥Lk+1k∥v1∥Lk+1,
where k≥1 because m=1 and k=0 case is excluded. For the second case m-k≥2, we have
(36)|(II)|≤∫ℝ‍∫y-2Ty‍|u1(n)(s)|k||v2(y)|m-k-|u2(n)(y)|m-k|×|v1(s)||ϕ(y,s)|dsdy≲supy(∫y-2Ty‍|u1(n)(s)|k|v1(s)||ϕ(y,s)|ds)×∥v2-u2(n)∥Lm-k(∥v2∥Lm-km-k-1+∥u2(n)∥Lm-km-k-1),
where we have
(37)supy∫y-2Ty‍|u1(n)(s)|k|v1(s)||ϕ(y,s)|ds≤{∥v1∥L2supy(∫y-2Ty‍|ϕ(y,s)|2ds)1/2ifk=0,M∥u1(n)∥Lk+1k∥v1∥Lk+1ifk≥1.The integral (III) can be treated in a similar way to (II). For the casek=0, we have
(38)|(III)|≲supy(∫y-2Ty‍|(v1-u1(n))(s)||ϕ(y,s)|ds)∥u2(n)∥Lmm≲supy(∫y-2Ty‍|ϕ(y,s)|2ds)1/2×∥v1-u1(n)∥L2∥u2(n)∥Lmm.
For the case k≥1, we have
(39)|(III)|≲sups(∫ss+2T‍|u2(n)(y)|m-k|ϕ(y,s)|dy)×∥v1-u1(n)∥Lk+1∥u1(n)∥Lk+1k≲M∥u2(n)∥Lm-k∥v1-u1(n)∥Lk+1∥u1(n)∥Lk+1k.Taking into account|eix-eiy|≤|x-y| for x, y∈ℝ, the integral (IV) can be bounded for the case k=0:
(40)∫ℝ‍∫y-2Ty‍|u1(n)(s)||u2(n)(y)|m×|∫sy‍|v2(h)|mdh-∫sy‍|u2(n)(h)|mdh||ϕ(y,s)|dsdy≲supy,s(∫sy‍||v2(h)|m-|u2(n)(h)|m|dh)×∫ℝ‍|u2(n)(y)|m∫y-2Ty‍|u1(n)(s)||ϕ(y,s)|dsdy,
where supremum supy,s is taken over s∈ℝ and s≤y≤s+2T. Then, we obtain
(41)(IV)≲∥v2-u2(n)∥Lm(∥v2∥Lmm-1+∥u2(n)∥Lmm-1)×∥u2(n)∥Lmm∥u1(n)∥L2supy(∫y-2Ty‍|ϕ(y,s)|2ds)1/2.
For the case k≥1, we have
(42)∫ℝ‍∫ss+2T‍|u1(n)(s)|k+1|u2(n)(y)|m-k×||v1(s)|k∫sy‍|v2(h)|m-kdh-|u1(n)(s)|k∫sy‍|u2(n)(h)|m-kdh||ϕ|dyds≤∫ℝ‍∫ss+2T‍|u1(n)(s)|k+1|u2(n)(y)|m-k|v1(s)|k×∫sy‍||v2(h)|m-k-|u2(n)(h)|m-k|dh|ϕ(y,s)|dyds+∫ℝ‍∫ss+2T‍|u1(n)(s)|k+1|u2(n)(y)|m-k×||v1(s)|k-|u1(n)(s)|k|×∫sy‍|u2(n)(h)|m-kdh|ϕ(y,s)|dyds≲supy,s(∫sy‍||v2(h)|m-k-|u2(n)(h)|m-k|dh)×∫ℝ‍|u1(n)(s)|k+1|v1(s)|k∫ss+2T‍|u2(n)(y)|m-k|ϕ|dyds+supy,s(∫sy‍|u2(n)(h)|m-kdh)∫ℝ‍|u1(n)(s)|k+1×||v1(s)|k-|u1(n)(s)|k|∫ss+2T‍|u2(n)(y)|m-k|ϕ|dyds.
Considering s≤y≤s+2T, we can bound (IV) as follows. For the case m-k=0, we have
(43)|(IV)|≲2T∥u1(n)∥L2m+1m+1∥v1-u1(n)∥L2m+1×(∥v1∥L2m+1m-1+∥u1(n)∥L2m+1m-1)sups(∫ss+2T‍|ϕ(y,s)|dy).
For the case m-k=1, we have
(44)|(IV)|≲2T∥v2-u2(n)∥L2∥u1(n)∥L2m-1m∥v1∥L2m-1m-1×∥u2(n)∥L2sups(∫ss+2T‍|ϕ(y,s)|2dy)1/2.
For other cases (k≥1 and m-k≥2), we finally have
(45)|(IV)|≲M∥v2-u2(n)∥Lm-k(∥v2∥Lm-km-k-1+∥u2(n)∥Lm-km-k-1)×∥u2(n)∥Lm-km-k∥u1(n)∥L2k+1k+1∥v1∥L2k+1k+1+M∥u2(n)∥Lm-k2m-2k∥u1(n)∥L2k+1k+1∥v1-u1(n)∥L2k+1×(∥v1∥L2k+1k-1+∥u1(n)∥L2k+1k-1).
Then, we have |(1)|+|(2)|+|(3)|→0 as n→∞ which implies that (19) holds.


## 2.1. Representation Formula of Classical Solution

In this subsection, we considerC∞((0,T)×ℝ) solutions which satisfy (6) in the classical sense. An explicit representation of its solution is given in terms of initial data [2, 13]. It is interesting to express the solution of nonlinear partial differential equations in terms of initial data.Integrating (6) along the outgoing and ingoing characteristics, we obtain
(14)U1(t,x)=u1(x-t)×exp(i∫0t‍|U1(s,x-t+s)|k×|U2(s,x-t+s)|m-kdsi∫0t‍|U1(s,x-t+s)|k),U2(t,x)=u2(x+t)×exp(i∫0t‍|U2(s,x+t-s)|k×|U1(s,x+t-s)|m-kdsi∫0t‍|U1(s,x-t+s)|k).
Taking into account
(15)|U1(t,x)|=|u1(x-t)|,|U2(t,x)|=|u2(x+t)|,
(14) becomes
(16)U1(t,x)=u1(x-t)×exp(i2|u1(x-t)|k∫x-tx+t‍|u2(y)|m-kdy),U2(t,x)=u2(x+t)×exp(i2|u2(x+t)|k∫x-tx+t‍|u1(y)|m-kdy).
In the case of k=m, we have the following obvious representation formula of solution:
(17)U1(t,x)=u1(x-t)exp(it|u1(x-t)|m),U2(t,x)=u2(x+t)exp(it|u1(x-t)|m).


## 2.2. Global Strong Solutions inYp(ℝ)

We now present the proof of Theorem1. Let us introduce a strong solution of the initial value problem of the system (6).Definition 4.
Consider the Cauchy problem (6) with initial data v=(v1,v2)∈Yp(m,k)(ℝ)). It is said that V=(V1,V2) is a strong solution to the Cauchy problem on the time interval [0,T] provided that
(18)(V1,V2)∈C([0,T];Yp(m,k)(ℝ))
satisfy (6) in the sense of distributions. That is, for any ϕ∈C0∞((-T,T)×ℝ), we have
(19)∫0T‍∫ℝ‍V1∂tϕ+V1∂xϕ+i|V1|k|V2|m-kV1ϕdxdt+∫ℝ‍v1(x)ϕ(0,x)dx=0,(20)∫0T‍∫ℝ‍V2∂tϕ-V2∂xϕ+i|V2|k|V1|m-kV2ϕdxdt+∫ℝ‍v2(x)ϕ(0,x)dx=0.Remark 5.
We say thatV is a weak solution if V∈L∞([0,T];Yp(m,k)(ℝ)) satisfies (6) in the sense of distribution.For the initial datavj∈Yp(m,k)(ℝ)(j=1,2), we propose that the following functions Vj are the global strong solution of (6):
(21)V1(t,x)=v1(x-t)×exp(i2|v1(x-t)|k∫x-tx+t‍|v2(y)|m-kdy),V2(t,x)=v2(x+t)×exp(i2|v2(x+t)|k∫x-tx+t‍|v1(y)|m-kdy).
For the smooth function sequence {uj(n)}n=1∞  (j=1,2) which converges to vj in Yp(m,k)(ℝ) as n→∞, we consider sequence of classical solutions Uj(n) of (6). First of all, we estimate the difference of (Vj,Uj(n)). Using the representation (16) and (21), we have
(22)|(V1-U1(n))(t,x)|≤|exp(i2|u1(n)(x-t)|k∫x-tx+t‍|u2(n)(y)|m-kdy)×(v1-u1(n))(x-t)exp(i2|u1(n)(x-t)|k∫x-tx+t‍|u2(n)(y)|m-kdy)|+|v1(x-t)(e(i/2)|v1(x-t)|k∫x-tx+t‍|v2(y)|m-kdy-e(i/2)|u1(n)(x-t)|k∫x-tx+t‍|u2(n)(y)|m-kdy)|≤|(v1-u1(n))(x-t)|+12|v1(x-t)|||v1(x-t)|k∫x-tx+t‍|v2(y)|m-kdy-|u1(n)(x-t)|k∫x-tx+t‍|u2(n)(y)|m-kdy|=(A)+(B),
where we used |eix-eiy|≤|x-y| for x, y∈ℝ. The Lq norm of the first term (A) can be treated as follows:
(23)(∫ℝ‍|(v1-u1(n))(x-t)|qdx)1/q=∥v1-u1(n)∥Lq.
To estimate the Lp norm of the second term (B), we consider the following three cases. (i)
For the casek=0, we have with 2≤p≤m(24)∫ℝ‍|v1(x-t)|p|∫x-tx+t‍|v2(y)|mdy-∫x-tx+t‍|u2(n)(y)|mdy|pdx≲∥v2-u2(n)∥Lmp(∥v2∥Lmm-1+∥u2(n)∥Lmm-1)p∥v1∥Lpp.(ii)
For the casek=m, we have with 1≤p≤(2m+1)/(m+1)(25)(2t)p∫ℝ‍|v1(x-t)|p||v1(x-t)|m-|u1(n)(x-t)|m|pdx≲(2t)p∥v1∥L(m+1)pp∥v1-u1(n)∥L(m+1)pp×(∥v1∥L(m+1)p(m-1)p+∥u1(n)∥L(m+1)p(m-1)p).(iii)
For other cases1≤k≤m-1, we have with 1≤p≤p(m,k)/(k+1)(26)∫ℝ‍|v1(x-t)|p||v1(x-t)|k∫x-tx+t‍|v2(y)|m-kdy-|u1(n)(x-t)|k∫x-tx+t‍|u2(n)(y)|m-kdy|pdx≲supx(∫x-tx+t‍|v2(y)|m-kdy)p×∫ℝ‍|v1(x-t)|p||v1(x-t)|k-|u1(n)(x-t)|k|pdx+supx(∫x-tx+t‍||v2(y)|m-k-|u2(n)(y)|m-k|dy)p×∫ℝ‍|v1(x-t)|p|u1(n)(x-t)|kpdx≲supx(∫x-tx+t‍|v2(y)|m-kdy)p∥v1∥L(k+1)pp×∥v1-u1(n)∥L(k+1)pp(∥v1∥L(k+1)p(k-1)p+∥u1(n)∥L(k+1)p(k-1)p)+supx(∫x-tx+t‍||v2(y)|m-k-|u2(n)(y)|m-k|dy)p×∥v1∥L(k+1)pp∥u1(n)∥L(k+1)pkp,where
(27)supx(∫x-tx+t‍|v2(y)|m-kdy)p≲{(2t)p/2∥v2∥L2pform-k=1,∥v2∥Lm-k(m-k)pform-k≥2,supx(∫x-tx+t‍||v2(y)|m-k-|u2(n)(y)|m-k|dy)p≲{(2t)p/2∥v2-u2(n)∥L2pform-k=1,∥v2-u2(n)∥Lm-kp(∥v2∥Lm-km-k-1+∥u2(n)∥Lm-km-k-1)pform-k≥2.Now, we are ready to prove thatV=(V1,V2) is a global strong solution to the Cauchy problem (6). It is easy to check (18) by considering the representation (21). Making use of the classical solution to (6), we have
(28)∫0T‍∫ℝ‍V1∂tϕ+V1∂xϕ+i|V1|k|V2|m-kV1ϕdxdt+∫ℝ‍v1(x)ϕ(0,x)dx=∫0T‍∫ℝ‍(V1-U1(n))(∂tϕ+∂xϕ)dxdt+∫ℝ‍(v1-u1(n))ϕ(0,x)dx+∫0T‍∫ℝ‍i(|V1|k|V2|m-kV1-|U1(n)|k|U2(n)|m-kU1(n))ϕdxdt=(1)+(2)+(3).
Considering (23)–(26) with q=p=p(m,k)/(k+1), the first integral (1) can be estimated as follows:
(29)|(1)|≲Tsup0≤t≤T(∥∂tϕ(t,·)∥Lq′+∥∂xϕ(t,·)∥Lq′)×sup0≤t≤T∥(V1-U1(n))(t,·)∥Lq,
where
(30)sup0≤t≤T∥(V1-U1(n))(t,·)∥Lq≲{∥v1-u1(n)∥Lq+[∥v2-u2(n)∥L2p∥v1∥Lp(m,k)p∥u1(n)∥Lp(m,k)kp(2T)p/2∥v2∥L2p∥v1∥Lp(m,k)p×∥v1-u1(n)∥Lp(m,k)p(∥v1∥Lp(m,k)(k-1)p+∥u1(n)∥Lp(m,k)(k-1)p)+(2T)p/2∥v2-u2(n)∥L2p∥v1∥Lp(m,k)p∥u1(n)∥Lp(m,k)kp]1/pform-k=1,∥v1-u1(n)∥Lq+[∥u1(n)∥Lp(m,k)kp∥v2∥Lm-k(m-k)p∥v1∥Lp(m,k)p∥v1-u1(n)∥Lp(m,k)p×(∥v1∥Lp(m,k)(k-1)p+∥u1(n)∥Lp(m,k)(k-1)p)+∥v2-u2(n)∥Lm-kp(∥v2∥Lm-km-k-1+∥u2(n)∥Lm-km-k-1)p×∥v1∥Lp(m,k)p∥u1(n)∥Lp(m,k)kp]1/pform-k≥2.
The integral (2) can be bounded easily:
(31)∫ℝ‍(v1-u1(n))ϕ(0,x)dx≤∥v1-u1(n)∥L2∥ϕ(0,x)∥L2.
To take care of the integral (3), we decompose the integral by using representation (16) and (21) with the notation E(v1,v2,k)=exp((i/2)|v1(x-t)|k∫x-tx+t‍|v2(y)|m-kdy):
(32)∫0T‍∫ℝ‍(|V1|k|V2|m-kV1-|U1(n)|k|U2(n)|m-kU1(n))ϕ(t,x)dxdt=∫0T‍∫ℝ‍(|v1(x-t)|k-|u1(n)(x-t)|k)|v2(x+t)|m-k×v1(x-t)E(v1,v2,k)ϕ(t,x)dxdt+∫0T‍∫ℝ‍|u1(n)(x-t)|k×(|v2(x+t)|m-k-|u2(n)(x+t)|m-k)×v1(x-t)E(v1,v2,k)ϕ(t,x)dxdt+∫0T‍∫ℝ‍|u1(n)(x-t)|k|u2(n)(x+t)|m-k×(v1-u1(n))(x-t)E(v1,v2,k)ϕ(t,x)dxdt+∫0T‍∫ℝ‍|u1(n)(x-t)|k|u2(n)(x+t)|m-ku1(n)(x-t)×(E(v1,v2,k)-E(u1(n),u2(n),k))ϕ(t,x)dxdt=(I)+(II)+(III)+(IV),
where we understand that the integral (I) does not exist for k=0 and the integral (II) for k=m.Changing variablesx+t=y and x-t=s, the first integral (I) can be bounded as follows:
(33)|(I)|≤∫ℝ‍∫ss+2T‍||v1(s)|k-|u1(n)(s)|k||v2(y)|m-k×|v1(s)||ϕ(y,s)|dyds≲sups(∫ss+2T‍|v2(y)|m-k|ϕ(y,s)|dy)×∫ℝ‍|(v1-u1(n))(s)|×(|v1(s)|k-1+|u1(n)(s)|k-1)|v1(s)|ds≲sups(∫ss+2T‍|v2(y)|m-k|ϕ(y,s)|dy)∥v1-u1(n)∥Lk+1×(∥v1∥Lk+1k-1+∥u1(n)∥Lk+1k-1)∥v1∥Lk+1,
where we note that k≥1. We also have
(34)sups∫ss+2T‍|v2(y)|m-k|ϕ(y,s)|dy≤{sups∫ss+2T‍|ϕ(y,s)|dyifm-k=0,∥v2∥L2sups(∫ss+2T‍|ϕ(y,s)|2dy)1/2ifm-k=1,M∥v2∥Lm-km-kifm-k≥2,
where we denote M=supΣT|ϕ(t,x)| where ΣT={(x,t)∣x∈ℝ,0≤t≤T}.For the integral (II), we consider two cases. Note that the integral (II) disappears ifk=m. For the first case m-k=1, we have
(35)|(II)|≤∫ℝ‍∫ss+2T‍|u1(n)(s)|k||v2(y)|-|u2(n)(y)||×|v1(s)||ϕ(y,s)|dyds≲sups(∫ss+2T‍||v2(y)|-|u2(n)(y)|||ϕ(y,s)|dy)×∫ℝ‍|u1(n)(s)|k|v1(s)|ds≲sups(∫ss+2T‍|ϕ(y,s)|2dy)1/2×∥v2-u2(n)∥L2∥u1(n)∥Lk+1k∥v1∥Lk+1,
where k≥1 because m=1 and k=0 case is excluded. For the second case m-k≥2, we have
(36)|(II)|≤∫ℝ‍∫y-2Ty‍|u1(n)(s)|k||v2(y)|m-k-|u2(n)(y)|m-k|×|v1(s)||ϕ(y,s)|dsdy≲supy(∫y-2Ty‍|u1(n)(s)|k|v1(s)||ϕ(y,s)|ds)×∥v2-u2(n)∥Lm-k(∥v2∥Lm-km-k-1+∥u2(n)∥Lm-km-k-1),
where we have
(37)supy∫y-2Ty‍|u1(n)(s)|k|v1(s)||ϕ(y,s)|ds≤{∥v1∥L2supy(∫y-2Ty‍|ϕ(y,s)|2ds)1/2ifk=0,M∥u1(n)∥Lk+1k∥v1∥Lk+1ifk≥1.The integral (III) can be treated in a similar way to (II). For the casek=0, we have
(38)|(III)|≲supy(∫y-2Ty‍|(v1-u1(n))(s)||ϕ(y,s)|ds)∥u2(n)∥Lmm≲supy(∫y-2Ty‍|ϕ(y,s)|2ds)1/2×∥v1-u1(n)∥L2∥u2(n)∥Lmm.
For the case k≥1, we have
(39)|(III)|≲sups(∫ss+2T‍|u2(n)(y)|m-k|ϕ(y,s)|dy)×∥v1-u1(n)∥Lk+1∥u1(n)∥Lk+1k≲M∥u2(n)∥Lm-k∥v1-u1(n)∥Lk+1∥u1(n)∥Lk+1k.Taking into account|eix-eiy|≤|x-y| for x, y∈ℝ, the integral (IV) can be bounded for the case k=0:
(40)∫ℝ‍∫y-2Ty‍|u1(n)(s)||u2(n)(y)|m×|∫sy‍|v2(h)|mdh-∫sy‍|u2(n)(h)|mdh||ϕ(y,s)|dsdy≲supy,s(∫sy‍||v2(h)|m-|u2(n)(h)|m|dh)×∫ℝ‍|u2(n)(y)|m∫y-2Ty‍|u1(n)(s)||ϕ(y,s)|dsdy,
where supremum supy,s is taken over s∈ℝ and s≤y≤s+2T. Then, we obtain
(41)(IV)≲∥v2-u2(n)∥Lm(∥v2∥Lmm-1+∥u2(n)∥Lmm-1)×∥u2(n)∥Lmm∥u1(n)∥L2supy(∫y-2Ty‍|ϕ(y,s)|2ds)1/2.
For the case k≥1, we have
(42)∫ℝ‍∫ss+2T‍|u1(n)(s)|k+1|u2(n)(y)|m-k×||v1(s)|k∫sy‍|v2(h)|m-kdh-|u1(n)(s)|k∫sy‍|u2(n)(h)|m-kdh||ϕ|dyds≤∫ℝ‍∫ss+2T‍|u1(n)(s)|k+1|u2(n)(y)|m-k|v1(s)|k×∫sy‍||v2(h)|m-k-|u2(n)(h)|m-k|dh|ϕ(y,s)|dyds+∫ℝ‍∫ss+2T‍|u1(n)(s)|k+1|u2(n)(y)|m-k×||v1(s)|k-|u1(n)(s)|k|×∫sy‍|u2(n)(h)|m-kdh|ϕ(y,s)|dyds≲supy,s(∫sy‍||v2(h)|m-k-|u2(n)(h)|m-k|dh)×∫ℝ‍|u1(n)(s)|k+1|v1(s)|k∫ss+2T‍|u2(n)(y)|m-k|ϕ|dyds+supy,s(∫sy‍|u2(n)(h)|m-kdh)∫ℝ‍|u1(n)(s)|k+1×||v1(s)|k-|u1(n)(s)|k|∫ss+2T‍|u2(n)(y)|m-k|ϕ|dyds.
Considering s≤y≤s+2T, we can bound (IV) as follows. For the case m-k=0, we have
(43)|(IV)|≲2T∥u1(n)∥L2m+1m+1∥v1-u1(n)∥L2m+1×(∥v1∥L2m+1m-1+∥u1(n)∥L2m+1m-1)sups(∫ss+2T‍|ϕ(y,s)|dy).
For the case m-k=1, we have
(44)|(IV)|≲2T∥v2-u2(n)∥L2∥u1(n)∥L2m-1m∥v1∥L2m-1m-1×∥u2(n)∥L2sups(∫ss+2T‍|ϕ(y,s)|2dy)1/2.
For other cases (k≥1 and m-k≥2), we finally have
(45)|(IV)|≲M∥v2-u2(n)∥Lm-k(∥v2∥Lm-km-k-1+∥u2(n)∥Lm-km-k-1)×∥u2(n)∥Lm-km-k∥u1(n)∥L2k+1k+1∥v1∥L2k+1k+1+M∥u2(n)∥Lm-k2m-2k∥u1(n)∥L2k+1k+1∥v1-u1(n)∥L2k+1×(∥v1∥L2k+1k-1+∥u1(n)∥L2k+1k-1).
Then, we have |(1)|+|(2)|+|(3)|→0 as n→∞ which implies that (19) holds.


## Proof of Theorem3

Here, we prove Theorem3 which shows the uniqueness of weak solutions to (6). Before proving Theorem 3, we introduce a lemma given in [14] for easy reference. Consider the following equations on the time interval [-L,L]:
(46)(∂t+∂x)v+=f+,(∂t-∂x)v-=f-.Lemma 6.
Suppose thatf±∈L1([-L,L]×ℝ) and v±∈L∞([-L,L];L1(ℝ)) satisfy (46) in the sense of distribution. Then, one has
(47)∫-L+TL-T‍|v+(T,y)|dy≤∫-LL‍|v+(0,y)|dy+∫0T‍∫-L+tL-t‍|f+(t,y)|dydt,∫-L+TL-T‍|v-(T,y)|dy≤∫-LL‍|v-(0,y)|dy+∫0T‍∫-L+tL-t‍|f-(t,y)|dydt.Now letUj and Vj be two weak solutions of (6) with the same initial data. We define ωj=Uj-Vj. Then, we have equations for ωj:
(48)∂tω1+∂xω1=i(|U1|-|V1|)|U1|k-1|U2|m-kU1+⋯+i|V1|k(|U2|-|V2|)|U2|m-k-1U1+⋯+i|V1|k|V2|m-k-1(|U2|-|V2|)U1+i|V1|k|V2|m-kω1,∂tω2-∂xω2=i(|U2|-|V2|)|U2|k-1|U1|m-kU2+⋯+i|V2|k(|U1|-|V1|)|U1|m-k-1U2+⋯+i|V2|k|V1|m-k-1(|U1|-|V1|)U2+i|V2|k|V1|m-kω2.
Multiplying by complex conjugates, respectively, and taking the real parts, we obtain
(49)∂t|ω1|2+∂x|ω1|2=2(|U1|-|V1|)|U1|k-1|U2|m-kIm(U-1ω1)+⋯+2|V1|k|V2|m-k-1(|U2|-|V2|)Im(U-1ω1),∂t|ω2|2-∂x|ω2|2=2(|U2|-|V2|)|U2|k-1|U1|m-kIm(U-2ω2)+⋯+2|V2|k|V1|m-k-1(|U1|-|V1|)Im(U-2ω2),
where we note that Re(i|V1|k|V2|m-k|ω1|2)=0=Re(i|V2|k|V1|m-k|ω2|2). Applying Lemma 6 to (49) and considering ωj(0,x)=0, we have
(50)∫-L+TL-T‍|ω1(t,y)|2dy≲∫0T‍∥|U1|-|V1|∥Lm+2∥U1∥Lm+2k∥U2∥Lm+2∥ω1∥Lm+2+⋯+∥V1∥Lm+2k∥V2∥Lm+2m-k-1∥|U2|-|V2|∥Lm+2×∥U1∥Lm+2∥ω1∥Lm+2dt,
where we understand ∥f∥Lm+2=(∫-L+tL-t‍|f(t,x)|m+2dx)1/(m+2).To estimate∥|Uj(t,·)|-|Vj(t,·)|∥Lm+2(j=1,2) in (50) we multiply (6) by |Uj|mU¯j to obtain
(51)∂t|U1|m+2+∂x|U1|m+2=0,∂t|U2|m+2-∂x|U2|m+2=0.
With the same equations for Vj, we obtain
(52)∂t(|U1|m+2-|V1|m+2)+∂x(|U1|m+2-|V1|m+2)=0,∂t(|U2|m+2-|V2|m+2)-∂x(|U2|m+2-|V2|m+2)=0.
Applying Lemma 6 and considering |Uj(0,x)|m+2-|Vj(0,x)|m+2=0, we have
(53)∫-L+tL-t‍||Uj(t,x)|-|Vj(t,x)||m+2dx≤∫-L+tL-t‍||Uj(t,x)|m+2-|Vj(t,x)|m+2|dx≤0.
Considering the inequality (50), we conclude that ∫-L+TL-T‍|ω1(t,y)|2dy=0.



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*Source: 602753-2013-06-20.xml*