Initial boundary value problem for nonlinear Dirac equation of Gross-Neveu type in 1+1 dimensions
aa r X i v : . [ m a t h . A P ] M a r INITIAL BOUNDARY VALUE PROBLEM FOR NONLINEARDIRAC EQUATION OF GROSS-NEVEU TYPE IN
DIMENSIONS
YONGQIAN ZHANG AND QIN ZHAO
Abstract.
This paper studies an initial boundary value problem for aclass of nonlinear Dirac equations with cubic terms and moving boundary.For the initial data with bounded L norm and the suitable boundary con-ditions, the global existence and the uniqueness of the strong solution areproved. Introduction
Consider the nonlinear Dirac equations (cid:26) i ( u t + u x ) = − mv + N ( u, v ) ,i ( v t − v x ) = − mu + N ( u, v ) , (1.1)in a domain Ω = { ( x, t ) (cid:12)(cid:12) t ≥ , x ≥ z ( t ) } for m ≥ u ( x, t = 0) , v ( x, t = 0)) = ( u ( x ) , v ( x )) , x ≥ , (1.2)and boundary condition u ( z ( t ) , t ) = λ ( t ) v ( z ( t ) , t ) , t ≥ . (1.3)The nonlinear terms take the following form N = ∂ u W ( u, v ) = αu | v | + 2 βv ( uv + uv ) , (1.4) N = ∂ v W ( u, v ) = αv | u | + 2 βu ( uv + uv ) , (1.5)with W ( u, v ) = α | u | | v | + β ( uv + uv ) , where α, β ∈ R and u, v are complex conjugate of u and v .The boundary { x = z ( t ) } , denoted by Γ B , is assumed to satisfy the follow-ing, (H1): − < z t ( t ) <
1, for t ≥ z (0) = 0. (H2): | λ ( t ) | (1 − z t ( t )) ≤ (1 + z t ( t )), for t ≥ Key words and phrases.
Nonlinear Dirac equation; Gross-Neveu model; global strongsolution; Bony type functional; Glimm type functional.( AMS subject classification. Primary: 35Q41 ; Secondary: 35L60, 35Q40) .
Here and in sequel, we denote z t = dzdt , λ t = dλdt , u t = ∂u∂t , u x = ∂u∂x etc. forsimplification.The nonlinear Dirac equation (1.1) is called Thirring equation for α = 1and β = 0, while it is called Gross-Neveu equation for α = 0 and β = 1 /
4; seefor instance [23] and [14], [20]. There are a number of works devoted to thelocal and global well-poedness of the Cauchy problem for the nonlinear Diracequation with various type of nonlinearities in different spatial dimensions(see for instance [2, 4, 6, 7, 9, 10, 11, 12, 14, 16, 20, 21, 23, 24, 25], and thereferences therein). There are also some papers on the initial boundary valueproblem (see for example [5] and [18]). In [5], motivated to study the Hawkingeffect describing the collapse of a spherically symmetric star to a Schwarzchildblack hole, Bouvier and G´erard used technique from C ∗ algebra to study theasymptotic behaviour of the global solution to (1.1),(1.2) and (1.3) with aclass of special initial data in R , where the non-characteristic boundaryis assumed to approach characteristic as t → ∞ , with | λ ( t ) | (1 − z t ( t )) =(1 + z t ( t )) for t ≥ H to initial boundary valueproblem for Thirring model in quarter plane { t > , x > } with small dataand study the scattering behaviour of solution. To our knowledge there is noresult on the well posedness of initial boundary value problem for Gross-Neveumodel with general initial data in L . Our purpose is to prove the existenceand the uniqueness in C (Ω) and in L (Ω) of global solution to (1.1-1.3).The first result is the following. Theorem 1.1.
Suppose that (H1) and (H2) hold. Let ( u , v ) ∈ C ([0 , ∞ )) with compact support in [0 , ∞ ) and satisfy the compatibility conditions as fol-low, u (0) = λ (0) v (0) (1.6) and (1 − z t (0)) u x (0) + λ (0)(1 + z t (0)) v x (0) + iλ (0) (cid:0) mu (0) − N ( u (0) , v (0)) (cid:1) − i (cid:0) mv (0) − N ( u (0) , v (0)) (cid:1) + λ t (0) v (0) = 0 . (1.7) Then (1.1-1.3) has a unique global solution ( u, v ) ∈ C (Ω) . This result could be generalized to the following case.
Theorem 1.2.
Suppose that (H1) and (H2) hold. Let ( u , v ) ∈ H ([0 , ∞ )) satisfy the compatibility conditions as follows, u (0) = λ (0) v (0) . (1.8) Then (1.1-1.3) has a unique global solution ( u, v ) ∈ H loc (Ω) ∩ C (Ω) . Moreover, ( u ( · , t ) , v ( · , t )) ∈ H ([ z ( t ) , ∞ )) for t ∈ [0 , ∞ ) . NITIAL BOUNDARY VALUE PROBLEM FOR NONLINEAR DIRAC EQUATION 3
With Theorem 1.1, we can look for the global strong solution. Here thestrong solution is defined as follows.
Definition 1.1.
A pair of measurable functions ( u, v ) is called a strong so-lution to (1.1-1.3) if there exits a sequence of classical solutions ( u ( n ) , v ( n ) ) ∈ C (Ω) to (1.1) such that u ( n ) ( z ( t ) , t ) = λ ( t ) v ( n ) ( z ( t ) , t ) , f or t > , and lim n →∞ Z N (cid:0) | u ( n ) ( x, − u | + | v ( n ) ( x, − v | (cid:1) dx = 0 , lim n →∞ Z Z K (cid:0) | u ( n ) − u | + | v ( n ) − v | (cid:1) dxdt = 0 for any compact set K ⊂ Ω and for any N > . Theorem 1.3.
Suppose that (H1) and (H2) hold. For any ( u , v ) ∈ L loc ([0 , ∞ )) ,(1.1-1.3) has a unique global strong solution ( u, v ) ∈ L loc (Ω) . Moreover, | u || v | ∈ L loc (Ω) , and ( u, v ) solves (1.1-1.3) in the following sense, Z Z Ω (cid:16) iu ( φ t + φ x ) − mvφ + N ( u, v ) φ (cid:17) dxdt = − i Z ∞ u φ ( x, dx, (1.9) Z Z Ω (cid:16) iv ( ψ t − ψ x ) − muψ + N ( u, v ) ψ (cid:17) dxdt = − i Z ∞ v ψ ( x, dx (1.10) for any ( φ, ψ ) ∈ C (Ω) with bounded support in Ω and ( φ, ψ )( z ( t ) , t ) = 0 for t ≥ . Moreover, we have the following.
Theorem 1.4.
Suppose that (H1) and (H2) hold. If ( u , v ) ∈ L ([0 , ∞ )) ,then the strong solution ( u, v ) given by Theorem 1.3 satisfies the following, ( u, v ) ∈ L (Ω ∩ ( R × [0 , T ])) , | u || v | ∈ L (Ω ∩ ( R × [0 , T ])) for any T > . Moreover, if | λ ( t ) | (1 − z t ( t )) = (1 + z t ( t )) for t ≥ , then Z ∞ z ( t ) ( | u ( x, t ) | + | v ( x, t ) | ) dx = Z ∞ ( | u ( x ) | + | v ( x ) | ) dx for almost every t ∈ [0 , ∞ ) . The remaining is organized as follows. First, in section 2, to prove Theorem1.1 and Theorem 1.2 for (1.1-1.3), we derive the equations (2.1) and (2.2)for | u | and | v | for local smooth solution ( u, v ), and apply the characteristicmethod to the equations (2.1) and (2.2) to get the pointwise bounds on | u | and | v | . Then it enables us to get the uniform L ∞ bounds on ( u, v ) in the YONGQIAN ZHANG AND QIN ZHAO domain Ω ∩ { ≤ t < T } for any T > Q ( t, ∆) and a Glimm typefunctional F ( t, ∆) = L ( t, u, ∆) + K L ( t, v, ∆) + C Q ( t, ∆) for smooth solution( u, v ) to get L estimates of nonlinear term, R R ∆ | u ( x, t ) | | v ( x, t ) | dxdt on eachcharacteristic triangle ∆. Here different from the work in [25], for the case that∆ ∩ ∂ Ω = ∅ , by the assumption (H2) we choose a suitable constant K > L norm, ddt (cid:0) L ( t, u, ∆) + K L ( t, v, ∆) (cid:1) can control the possible increasing of the functional Q ( t, ∆), and choose asuitable constant C so that F ( t, ∆) can control R R ∆ | u ( x, t ) | | v ( x, t ) | dxdt ,while for the case that ∆ ⊆ Ω same argument as in [25] can be carried outto get the control on
R R ∆ | u ( x, t ) | | v ( x, t ) | dxdt . In section 4, we considerthe difference ( U, V ) = ( u − u ′ , v − v ′ ) for two smooth solutions ( u, v ) and( u ′ , v ′ ). We first write down the equations (4.1) and (4.2) for ( U, V ), whichcontain (
U, V ), ( u, v ) and ( u ′ , v ′ ). Then we introduce a Bony type functional Q ( t, ∆) and a Glimm type functional F ( t, ∆) for | U | , | V | , | u | , | v | and | u ′ | and | v ′ | , and use it to prove the L stability estimates in Proposition 4.1.Here, as in section 3, for the case that ∆ ∩ ∂ Ω = ∅ , by the assumption (H2)we choose a suitable constant K > L norm, ddt (cid:0) L ( t, U, ∆) + K L ( t, V, ∆) (cid:1) can control the possible increasing ofthe functional Q ( t, ∆). In section 5, we first approximate the initial data(1.2) by a sequence of smooth functions. Then, by the result on the globalwellposedness for smooth solution in section 2, we can have a sequence ofglobal smooth solutions for smooth data for (1.1). With the help of the L stability estimates in section 4, we show that the sequence of global smoothsolutions converges to a strong solution in L (∆) for any triangle ∆. In section6, we complete the proof of Theorem 1.3 and Theorem 1.4.2. Global classical solution
For
T >
0, denoteΩ( T ) = { ( x, t ) | z ( t ) ≤ x < ∞ , ≤ t < T } . Classical theory on semilinear hyperbolic systems [1] gives the following localexistence result (see also [17]).
Lemma 2.1.
Suppose that the compatibility conditions (1.6) and (1.7) hold.For any ( u , v ) ∈ C ([0 , ∞ )) with compact support in [0 , ∞ ) , there exists a T ∗ > such that (1.1-1.3) has a unique solution ( u, v ) ∈ C (Ω( T ∗ )) . Our aim in this section is to extend the solution ( u, v ) globally to Ω. To thisend, let ( u , v ) ∈ C ([0 , ∞ )) with compact support and let ( u, v ) ∈ C (Ω( T ))be the solution to (1.1-1.3) for T ≥ T ∗ , taking ( u , v ) as its initial data, wehave to establish the estimates on || ( u, v ) || L ∞ (Ω( T )) in the next. Here we assumethat the compatibility conditions (1.6) and (1.7) hold for ( u , v ). NITIAL BOUNDARY VALUE PROBLEM FOR NONLINEAR DIRAC EQUATION 5
Multiplying the first equation of (1.1) by u and the second equation by v gives ( | u | ) t + ( | u | ) x = 2 m ℜ ( iuv ) + 2 ℜ ( iN u ) , (2.1)and ( | v | ) t − ( | v | ) x = 2 m ℜ ( iuv ) + 2 ℜ ( iN v ) , (2.2)which, together with the structure of nonlinear terms, leads to( | u | + | v | ) t + ( | u | − | v | ) x = 0 . (2.3)For the nonlinear terms in the righthand side of (2.1) and (2.2), we have thefollowing by direct computation. Lemma 2.2.
Let r ( x, t ) = m ( | u ( x, t ) | + | v ( x, t ) | ) + 8 | β || u ( x, t ) | | v ( x, t ) | .Then there hold the followings, (cid:12)(cid:12) m ℜ ( iuv ) + 2 ℜ ( iN u ) (cid:12)(cid:12) ≤ r ( x, t ) and (cid:12)(cid:12) m ℜ ( ivu ) + 2 ℜ ( iN v ) (cid:12)(cid:12) ≤ r ( x, t ) . And we have the estimates on the L norm of the solution as follows. Lemma 2.3.
Let E = R ∞ ( | u ( x ) | + | v ( x ) | ) dx . Then for any t ∈ [0 , T ) ,there holds the following, Z ∞ z ( t ) ( | u ( x, t ) | + | v ( x, t ) | ) dx ≤ E . (2.4) Proof.
By (1.3) and (2.3), and by assumption (H2), we have ddt Z ∞ z ( t ) ( | u ( x, t ) | + | v ( x, t ) | ) dx = | u ( z ( t ) , t ) | (1 − z t ( t )) − | v ( z ( t ) , t ) | (1 + z t ( t ))= | v ( z ( t ) , t ) | [ | λ ( t ) | (1 − z t ( t )) − (1 + z t ( t ))] ≤ , which gives the desired inequality and completes the proof. (cid:3) We consider the characteristic triangles for ( u, v ) in Ω( T ). For any a, b ∈ R with a < b and for any t ≥
0, we denote∆( a, b, t ) = { ( x, t ) (cid:12)(cid:12) a − t + t < x < b + t − t, t < t < b − a t } , see Figure 1, and, denoteΓ u ( x , t ; t ) = { ( x, t ) (cid:12)(cid:12) x = x − t + t, t ≤ t ≤ t } and Γ v ( x , t ; t ) = { ( x, t ) (cid:12)(cid:12) x = x + t − t, t ≤ t ≤ t } for t ≤ t , see Figure 2. It is obvious that Γ u ( x , t ; t ) is a characteristic line YONGQIAN ZHANG AND QIN ZHAO ★★★★★★★★★★★★ ❝❝❝❝❝❝❝❝❝❝❝❝ ( a, t ) ( b, t ) t = t ( a + b , b − a + t )∆( a, b, t ) Figure 1.
Domain ∆( a, b, t ) ★★★★★★★★★★★★ ❝❝❝❝❝❝❝❝❝❝❝❝ ( x − t + t , t ) ( x + t − t , t ) t = t ( x , t )Γ u ( x , t ; t ) Γ v ( x , t ; t ) Figure 2.
Characteristic lines Γ u and Γ v for the first equation of u in (1.1) while Γ v ( x , t ; t ) is a characteristic line forthe second equation of v in (1.1).Along these characteristic lines in Ω( T ), we have the following estimates. Lemma 2.4. If Γ v ( x , t ; t ) ⊆ Ω( T ) , then Z t t | u ( x + t − s, s ) | ds ≤ E . Here E = R ∞ ( | u | + | v | ) dx .Proof. Denote ω ( x , t ) = { ( x, t ) | z ( t ) ≤ x ≤ x + t − t, ≤ t ≤ t } . NITIAL BOUNDARY VALUE PROBLEM FOR NONLINEAR DIRAC EQUATION 7
Then taking the integration of (2.3) over ω ( x , t ) gives the following, Z x + t ( | u ( x ) | + | v ( x ) | ) dx = 2 Z t | u ( x + t − s, s ) | ds + Z x z ( t ) ( | u ( x, t ) | + | v ( x, t ) | ) dx + Z t (cid:8) ( − z t ( s )) | u ( z ( s ) , s ) | + (1 + z t ( s )) | v ( z ( s ) , s ) | (cid:9) ds ≥ Z t | u ( x + t − s, s ) | ds, where we use the boundary condition (1.3) and assumption (H2) to get thelast inequality. This implies the result and the proof is complete. (cid:3) Lemma 2.5. If Γ u ( x , t ; t ) ⊆ Ω( T ) , then Z t t | v ( x − t + s, s ) | ds ≤ E . Here E = R ∞ ( | u | + | v | ) dx .Proof. Since Γ u ( x , t ; t ) ⊆ Ω( T ), then the domain∆( x − t + t , x + t − t , t ) ⊆ Ω( T ) . Taking the integration of (2.3) over ∆( x − t + t , x + t − t , t ), we have Z t t | v ( x − t + s, s ) | ds ≤ Z x + t − t x − t + t ( | u ( x, t ) | + | v ( x, t ) | ) dx ≤ Z ∞ z ( t ) ( | u ( x, t ) | + | v ( x, t ) | ) dx ≤ E , where the last inequality is given by Lemma 2.3. The proof is complete. (cid:3) Using the above estimates on along the characteristic lines, we can get thefollowing pointwise estimates on v at first. Lemma 2.6.
For ( x, t ) ∈ Ω( T ) , | v ( x, t ) | ≤ ( | v ( x + t ) | + mE ) exp( mt + 8 | β | E ) . Here E = R ∞ ( | u | + | v | ) dx .Proof. Assumption (H1) implies thatΓ v ( x, t ; 0) ⊂ Ω( T )for any ( x, t ) ∈ Ω( T ). YONGQIAN ZHANG AND QIN ZHAO
Then, by Lemma 2.2, along Γ v ( x, t ; 0) we use the equation (2.2) to derivethat dds | v ( x + t − s, s ) | ≤ m | u ( x + t − s, s ) | + (cid:0) m +8 | β || u ( x + t − s, s ) | (cid:1) | v ( x + t − s, s ) | . Therefore dds (cid:0) | v ( x + t − s, s ) | e ( x, t, s ) (cid:1) ≤ m | u ( x + t − s, s ) | e ( x, t, s ) ≤ m | u ( x + t − s, s ) | , where e ( x, t, s ) = exp (cid:0) − ms − | β | Z s | u ( x + t − τ, τ ) | dτ (cid:1) . Taking the integration of the above from s = 0 to t , we can prove the desiredresult by Lemma 2.4. The proof is complete. (cid:3) To get the pointwise estimates on u , we look for the intersection point ofthe boundary Γ B and the characteristic line { ( x, t ) | x − t = b } for b ≤ ★★★★★★★★★★★★ x − t = b ( b,
0) (0 , t = 0( z ( p ( b )) , p ( b ))Γ B Figure 3.
Intersection of boundary and Characteristic ( b ≤ Lemma 2.7.
For any b ≤ , the equation z ( t ) − t = b has a unique solution t = p ( b ) , where p ∈ C ( −∞ , and p ′ ( s ) < for s ≤ .Proof. From assumption (H1) it follows that z t ( t ) − < t >
0, which implies that the function z ( t ) − t has a global inverse p ∈ C ( −∞ , p ′ ( s ) = 11 − z t ( p ( s )) < . Therefore the proof is complete. (cid:3)
Now we can have the following pointwise estimates on u . NITIAL BOUNDARY VALUE PROBLEM FOR NONLINEAR DIRAC EQUATION 9
Lemma 2.8. If ( x, t ) ∈ Ω( T ) with x − t ≥ , then | u ( x, t ) | ≤ ( | u ( x − t ) | + mE ) exp( mt + 8 | β | E ) . If ( x, t ) ∈ Ω( T ) with x − t < , then | u ( x, t ) | ≤ ( | λ ( t ) | + 1)( | v (2 p ( x − t ) + x − t ) | + mE ) exp(2 mt + 16 | β | E ) . Here E = R ∞ ( | u | + | v | ) dx .Proof. For ( x, t ) ∈ Ω( T ) with x − t ≥
0, the assumption (H1) implies thatΓ u ( x, t ; 0) ⊆ Ω( T ) . Then, by (2.1) and by Lemma 2.2, we have dds (cid:0) | u ( x − t + s, s ) | e ( x, t, s ) (cid:1) ≤ m | v ( x − t + s, s ) | e ( x, t, s ) ≤ m | v ( x − t + s, s ) | , (2.5)where e ( x, t, s ) = exp( − ms − | β | Z s | v ( x − t + τ, τ ) | dτ ) . Taking the integration of (2.5) from 0 to t and using Lemma 2.5, we get | u ( x, t ) | ≤ ( | u ( x − t ) | + mE ) exp( mt + 8 | β | E ) . For ( x, t ) ∈ Ω( T ) with x − t <
0, Lemma 2.7 implies that the characteristicline Γ u ( x, t ; 0) and the boundary intersect only at the point ( z (cid:0) p ( x − t ) (cid:1) , p ( x − t )).Then, by (2.1) and by Lemma 2.2, along the characteristic line Γ u ( x, t ; p ( x − t )) we have dds (cid:0) | u ( x − t + s, s ) | e ( x, t, s ) (cid:1) ≤ m | v ( x − t + s, s ) | e ( x, t, s ) ≤ m | v ( x − t + s, s ) | . where e ( x, t, s ) = exp (cid:0) − m ( s − p ( x − t )) − | β | Z sp ( x − t ) | v ( x − t + τ, τ ) | dτ (cid:1) . Taking the integration of the above from p ( x − t ) to t , we use Lemma 2.5and Lemma 2.6 to get the following, | u ( x, t ) | ≤ ( | u ( p ( x − t ) + x − t, p ( x − t )) | + mE ) exp( mt + 8 | β | E ) ≤ ( | λ ( t ) | | v ( p ( x − t ) + x − t, p ( x − t )) | + mE ) exp( mt + 8 | β | E ) ≤ ( | λ ( t ) | + 1)( | v (2 p ( x − t ) + x − t ) | + mE ) exp(2 mt + 16 | β | E ) . The proof is complete. (cid:3)
Now using the pointwise estimates on u and v , we can prove Theorem 1.1. Proof of Theorem 1.1.
For ( u , v ) ∈ H ([0 , ∞ )) ⊂ L ∞ ([0 , ∞ )), Lemma2.6 and Lemma 2.8 lead to | u ( x, t ) | + | v ( x, t ) | ≤ ( | λ ( t ) | + 2) (cid:0) || ( u , v ) || L ∞ + 2 mE (cid:1) exp(2 mt + 16 | β | E )for x ≥ z ( t ) and 0 ≤ t < T .Then by the standard theory on semilinear hyperbolic equations (see [1] forinstance), we can extend the solution ( u, v ) across the time t = T .Therefore, repeating the same argument for any time, we can extend thesolution globally to Ω. The proof is complete. (cid:3) Furthermore Theorem 1.2 follows from Theorem 1.1.
Proof of Theorem 1.2.
Let ( φ , ψ ) ∈ C ∞ c ( R ) be a pair of functions suchthat φ ( x ) = u (0) and ψ ( x ) = v (0) for x belonging to a neighbourhood ofzero. Then we choose a sequence of functions ( φ k , ψ k ) ∈ C ∞ c (0 , ∞ ) such that( u ( k )0 , v ( k )0 ) := ( φ + φ k , ψ + ψ k ) is convergent to ( u , v ) in H (0 , ∞ ) as k tendsto ∞ .It is obvious that ( u ( k )0 , v ( k )0 ) satisfies the compatibility conditions as (1.6)and (1.7). Therefore, by Theorem 1.1, the equations (1.1) has a global smoothsolution ( u ( k ) , v ( k ) ) with the initial data ( u ( k )0 , v ( k )0 ) for k ≥ k ≥ || ( u ( k ) , v ( k ) ) || L ∞ (Ω( T )) < ∞ for any T >
0, which enables us to show as in [1] and [17] that the sequence( u ( k ) , v ( k ) ) is convergent in H (Ω( T )) to a solution ( u, v ) of (1.1)-(1.3) as k tends to ∞ for any T > L ∞ (Ω( T )) ∩ H (Ω( T )) as in [1] and [17]. The proof is complete. (cid:3) Estimates on the classical solution
Consider the case that ( u , v ) ∈ C ([0 , ∞ )), and let ( u, v ) ∈ C (Ω) be theglobal solution to (1.1) with boundary condition (1.3). Here we assume thatthe compatibility condition (1.6) and (1.7) hold. Our aim in this section is toestablish the local estimates on ( u, v ).To this end, set ∆ = ∆( a, b ; t ) for simplification and assume that ∆ ∩ Ω = ∅ in this section.Let x ′ = b + a , t ′ = b − a t . Then Γ u ( x ′ , t ′ ; t ) and Γ v ( x ′ , t ′ ; t ) are the left and right edges of ∆. By (H1),Γ B and Γ u ( x ′ , t ′ ; t ) ∪ Γ v ( x ′ , t ′ ; t ) intersect at one point at the most. NITIAL BOUNDARY VALUE PROBLEM FOR NONLINEAR DIRAC EQUATION 11
We introduce a time interval as follows. Denote I ∆ = { t (cid:12)(cid:12) t ≤ t ≤ b − a t , z ( t ) ≤ b + t − t } . By Lemma 2.7, we have the following.
Lemma 3.1.
There hold the following statements. (1) If Γ B ∩ Γ v ( x ′ , t ′ ; t ) = { ( z ( τ ) , τ ) } for some τ ≥ t (Figure 4), then I ∆ = [ t , τ ] and { x (cid:12)(cid:12) ( x, t ) ∈ ∆ ∩ Ω } = [ z ( t ) , b + t − t ] . (2) If Γ B ∩ Γ u ( x ′ , t ′ ; t ) = { ( z ( τ ) , τ ) } for some τ ≥ t (Figure 5), then I ∆ = [ t , b − a + t ] and { x (cid:12)(cid:12) ( x, t ) ∈ ∆ ∩ Ω } = [ z ( t ) , b + t − t ] for t ∈ [ t , τ ] , and { x (cid:12)(cid:12) ( x, t ) ∈ ∆ ∩ Ω } = [ a − t + t, b + t − t ] for t ∈ [ τ , b − a t ] . (3) If ∆ ⊂ Ω , then I ∆ = [ t , b − a + t ] and { x (cid:12)(cid:12) ( x, t ) ∈ ∆ ∩ Ω } = [ a − t + t, b + t − t ] . ★★★★★★★★★★★★ ❝❝❝❝❝❝❝❝❝❝❝❝ ( a, t ) Γ B ( b, t ) t = t Γ u Γ v ( z ( τ ) , τ )( x ′ , t ′ ) Figure 4.
Case: Γ B ∩ Γ v = ∅ ★★★★★★★★★★★★ ❝❝❝❝❝❝❝❝❝❝❝❝ ( a, t ) ( b, t ) t = t ( z ( τ ) , τ ) ( x ′ , t ′ )Γ u Γ v Γ B Figure 5.
Case: Γ B ∩ Γ u = ∅ Now we can define the functionals for ( u, v ) on ∆ ∩ Ω as follow.
Definition 3.1.
For t ∈ I ∆ , and for any w ∈ C (Ω) , define, L ( t, w, ∆) = Z b − t + t z a ( t ) | w ( x, t ) | dx, (3.1) where z a ( t ) = max { a + t − t , z ( t ) } . Definition 3.2.
For t ∈ I ∆ , and for the solution ( u, v ) , define L ( t, ∆) = L ( t, u, ∆) + L ( t, v, ∆) and D ( t, ∆) = Z b − t + t z a ( t ) | u ( x, t ) | | v ( x, t ) | dx,Q ( t, ∆) = Z Z z a ( t ) For t ∈ I ∆ , there holds the following, L ( t, ∆) ≤ L ( t , ∆) . Proof. It suffices to prove lemma for three cases according to Lemma 3.1.Case 1: The right edge of ∆ and Γ B intersect at some point ( z ( τ ) , τ ), seeFigure 4. In this case I ∆ = [ t , τ ].Then for t ∈ [ t , τ ], z a ( t ) = z ( t ). Moreover, by (1.3) and (2.3), and byassumption (H2), we have ddt Z b − t + t z ( t ) ( | u ( x, t ) | + | v ( x, t ) | ) dx = | u ( z ( t ) , t ) | (1 − z t ( t )) − | v ( z ( t ) , t ) | (1 + z t ( t )) − ( | u ( b − t + t , t ) | + | v ( b − t + t , t ) | ) ≤ | v ( z ( t ) , t ) | [ | λ ( t ) | (1 − z t ( t )) − (1 + z t ( t ))] ≤ . This leads to the desired result.Case 2: The left edge of ∆ and Γ B intersect at some point ( z ( τ ) , τ ), seeFigure 5. Then I ∆ = [ t , t + b − a ].For t ∈ [ t , τ ], z a ( t ) = z ( t ), and in the same way as in the proof of Case 1,we can get L ( t, ∆) ≤ L ( t , ∆) . NITIAL BOUNDARY VALUE PROBLEM FOR NONLINEAR DIRAC EQUATION 13 For t ∈ [ τ , b − a + t ], z a ( t ) = a − t + t , then we can use the result for Case 2to deduce that L ( t, ∆) ≤ L ( τ , ∆) . Case 3: ∆ lies in the interior of Ω. The proof can be carried out in the sameway as in Case 1.Therefore the proof is complete. (cid:3) For any T > 0, we recall the notationΩ( T ) = { ( x, t ) | z ( t ) ≤ x < ∞ , ≤ t < T } , and have the control on the potential Q for the case that ∆ ⊂ Ω( T ) as follows. Lemma 3.3. Suppose that ∆ ⊂ Ω( T ) for T > . Then there exists constants δ > such that for the initial data satisfying L ( t , ∆) ≤ δ there holds thefollowing dQ ( t, ∆) dt + D ( t, ∆) ≤ m ( L ( t , ∆)) (3.2) for t ∈ ( t , b − a + t ) . Therefore, Q ( t, ∆) + t Z t D ( τ, ∆) dτ ≤ m ( L ( t , ∆)) ( t − t ) + Q ( t , ∆) ≤ m ( L ( t , ∆)) ( t − t ) + ( L ( t , ∆)) (3.3) for t ∈ [ t , b − a + t ] . Here δ is independent of T . The proof of Lemma 3.3 has been given in [25] and is similar to the proofof Lemma 3.4 in the next.To get the control on the potential Q near the boundary, we introduce anew functional as follows. Definition 3.3. For constants K > and C > and for t ∈ I ∆ , define F ( t, ∆) = L ( t, u, ∆) + K L ( t, v, ∆) + C Q ( t, ∆) . For any T > 0, we have the control on F near the boundary as follows. Lemma 3.4. Suppose that ∆ ⊂ R × [0 , T ] and ∆ ∩ Γ B = ∅ for T > . Thenthere exist constants δ > , K > and C > such that for L ( t , ∆) ≤ δ there hold the following, ddt F ( t, ∆) ≤ − D ( t, ∆) − | v ( z a ( t ) , t ) | + O (1) δ , (3.4) for t ∈ I ∆ with z ( t ) = a + t − t . Here the constants δ , K and C depend onlyon T ; and the bound of O (1) depends only on T . Proof. For simplification, we denote L ( t, ∆), D ( t, ∆), F ( t, ∆) and Q ( t, ∆)by L ( t ), D ( t ), F ( t ) and Q ( t ). Now it suffices to prove the lemma for twocases.Case 1: The boundary Γ B and the right edge Γ v of ∆ intersect at the point( z ( τ ) , τ ) for some τ ∈ [ t , t + b − a ], see Figure 4.Then I ∆ = [ t , τ ], z a ( t ) = z ( t ). For t ∈ [ t , τ ], by Lemma 2.2, we use (2.1),(2.2) to get ddt L ( t, u, ∆) ≤ Z b − t + t z ( t ) (cid:0) − ( | u ( x, t ) | ) x + r ( x, t ) (cid:1) dx −| u ( b − t + t , t ) | − z t ( t ) | u ( z ( t ) , t ) | ≤ (1 − z t ( t )) | u ( z ( t ) , t ) | + Z b − t + t z ( t ) r ( x, t ) dx, and ddt L ( t, v, ∆) ≤ Z b − t + t z ( t ) (cid:0) ( | v ( x, t ) | ) x + r ( x, t ) (cid:1) dx −| v ( b − t + t , t ) | − z t ( t ) | v ( z ( t ) , t ) | ≤ − (1 + z t ( t )) | v ( z ( t ) , t ) | + Z b − t + t z ( t ) r ( x, t ) dx, which lead to ddt (cid:0) L ( t, u, ∆) + K L ( t, v, ∆) (cid:1) ≤ (cid:0) (1 − z t ( t )) | λ ( t ) | − K (1 + z t ( t )) (cid:1) | v ( z ( t ) , t ) | +(1 + K ) Z b − t + t z ( t ) r ( x, t ) dx ≤ − | v ( z ( t ) , t ) | + (1 + K ) Z b − t + t z ( t ) r ( x, t ) dx ≤ − | v ( z ( t ) , t ) | + O (1)( L ( t ) + D ( t )) , where we choose K > − z t ( t )) | λ ( t ) | − K (1 + z t ( t )) < − t ∈ [0 , T ].On the other hand, by Lemma 2.2, we use (2.1), (2.2) again to get thefollowing for Q , NITIAL BOUNDARY VALUE PROBLEM FOR NONLINEAR DIRAC EQUATION 15 ddt Q ( t ) = Z Z z ( t ) Let ( u, v ) ∈ C (Ω) and ( u ′ , v ′ ) ∈ C (Ω) be two classical solutions to (1.1)with (1.3). We consider the difference between these two solutions and denote( U, V ) = ( u − u ′ , v − v ′ ) . Then, U t + U x = imV − i ( N ( u, v ) − N ( u ′ , v ′ )) ,V t − V x = imU − i ( N ( u, v ) − N ( u ′ , v ′ )) , which lead to( | U | ) t + ( | U | ) x = ℜ { imV U − i ( N ( u, v ) − N ( u ′ , v ′ )) U } , (4.1)( | V | ) t − ( | V | ) x = ℜ { imU V − i ( N ( u, v ) − N ( u ′ , v ′ )) V } , (4.2)For the nonlinear terms in the righthandsides of (4.1) and (4.2), we havefollowing by direct computations. Lemma 4.1. There exists a c ∗ > such that |ℜ { imV U − i ( N ( u, v ) − N ( u ′ , v ′ )) U }| ≤ r ( x, t ) and |ℜ { imU V − i ( N ( u, v ) − N ( u ′ , v ′ )) V }| ≤ r ( x, t ) , where r ( x, t ) = m ( | U ( x, t ) | + | V ( x, t ) | ) + c ∗ r ( x, x, t ) ,r ( x, y, t ) = | U ( x, t ) | (cid:0) | v ( y, t ) | + | v ′ ( y, t ) | (cid:1) + (cid:0) | u ( x, t ) | + | u ′ ( x, t ) | (cid:1) | V ( y, t ) | . To get the control on ( U, V ) via (4.1) and (4.2), we introduce followingfunctionals on ∆ ∩ Ω for ( U, V ) as in [25]. Here it is assume that ∆ ∩ Ω = ∅ . Definition 4.1. For ∆ = ∆( a, b, t ) and K > , C > , define L ( t, ∆) = L ( t, U, ∆) + K L ( t, V, ∆) ,D ( t, ∆) = Z b − t + t z a ( t ) r ( x, x, t ) dx,Q ( t, ∆) = Z Z z a ( t ) In addition we use the notations in Definition 3.2 for ( u, v ), and use thefollowing for ( u ′ , v ′ ), L ′ ( t, ∆) = L ( t, u ′ , ∆) + L ( t, v ′ , ∆)and D ′ ( t, ∆) = Z b − t + t z a ( t ) | u ′ ( x, t ) | | v ′ ( x, t ) | dx,Q ′ ( t, ∆) = Z Z z a ( t ) 0, we can have the estimates on F near the boundary Γ B as follows. Lemma 4.2. Suppose that ∆ ⊂ R × [0 , T ] and ∆ ∩ Γ B = ∅ . Then, thereexist constants δ > , K > and C > such that if L ( t , ∆) ≤ δ and L ′ ( t , ∆) ≤ δ then there holds the following, ddt F ( t, ∆) ≤ − D ( t, ∆) + (cid:2) O (1) + C Λ ( t, ∆) + C Λ ( t ) (cid:3) F ( t, ∆) (4.3) for t ∈ I ∆ with z ( t ) = a + t − t , where Λ ( t, ∆) = 4 mδ + 8 | β | ( D ( t, ∆) + D ′ ( t, ∆)) , and Λ ( t ) = (cid:26) (1 − z t ( t )) | λ ( t ) | ( | v ( z ( t ) , t ) | + | v ′ ( z ( t ) , t ) | ) , if a + t − t ≤ z ( t ) , , if a + t − t > z ( t ) .Here the constants K > , δ and C depend only on T .Proof. It suffices to prove lemma for two cases.Case 1: The boundary Γ B and the right edge of ∆ intersec at some point( z ( τ ) , τ ), see Figure 4. Then I ∆ = [ t , τ ], and z a ( t ) = z ( t ) for t ∈ I ∆ . For t ∈ ( t , τ ), by Lemma 2.2 and Lemma 4.1, we use (4.1) and (2.2) forboth ( u, v ) and ( u ′ , v ′ ) to derive that ddt Z Z z ( t ) Therefore (4.3) is proved for Case 1.Case 2: The boundary Γ B and the left edge of ∆ intersect at ( z ( τ ) , τ ).Then, I ∆ = [ t , t + b − a ], and z a ( t ) = z ( t ) for t ≤ t ≤ τ , z a ( t ) = a − t + t for τ ≤ t ≤ t + b − a . The proof can be carried out in the same way as in Case 1for t = τ . Thus the proof is complete. (cid:3) Remark 4.1. For the case that ∆ ⊂ Ω( T ) , we have similar estimates on F without boundary terms, see [25] for the proof, where only D ( t, ∆) makescontribution to the control on F . For the case that ∆ ∩ Γ B = ∅ , both Q ( t, ∆) and L ( t, V, ∆) are needed to give the control on F . As conclusion of the above argument, we get the stability result for smoothsolutions for any T > Proposition 4.1. Suppose that ∆ ⊂ R × [0 , T ] with b > z ( t ) , and supposethat L ( t , ∆) ≤ δ , L ′ ( t , ∆) ≤ δ . Then for t ∈ I ∆ , there holds the following Z b + t − tz a ( t ) ( | u ( x, t ) − u ′ ( x, t ) | + | v ( x, t ) − v ′ ( x, t ) | ) dx ≤ C Z b max { z ( t ) ,a } ( | u ( x, t ) − u ′ ( x, t ) | + | v ( x, t ) − v ′ ( x, t ) | ) dx, and Z Z ∆ ∩ Ω ( | uv − u ′ v ′ | ) dxdt ≤ C Z b max { z ( t ) ,a } ( | u ( x, t ) − u ′ ( x, t ) | + | v ( x, t ) − v ′ ( x, t ) | ) dx, Z Z ∆ ∩ Ω ( | uv − u ′ v ′ | ) dxdt ≤ C Z b max { z ( t ) ,a } ( | u ( x, t ) − u ′ ( x, t ) | + | v ( x, t ) − v ′ ( x, t ) | ) dx. Here the constant C depends only on T and E .Proof. It suffices to prove lemma for two cases.Case 1: a < z ( t ) < b , that is, ∆ ∩ Γ B = ∅ . Then taking the integral of (3.4)in Lemma 3.4 over I ∆ , we have R I ∆ (cid:0) D ( t, ∆) + D ′ ( t, ∆) + | v ( z a ( t ) , t ) | + | v ′ ( z a ( t ) , t ) | (cid:1) dt ≤ O (1)( L ( t , ∆) + L ′ ( t , ∆)) + O (1) δ T, which leads to Z I ∆ (cid:0) Λ ( t, ∆) + Λ ( t, ∆) (cid:1) dt ≤ C ( T ) for some constant C ( T ) > T .Therefore, we use Lemma 4.2 to deduce that F ( t ) ≤ exp( Z I ∆ [ O (1) + C Λ ( s, ∆) + C Λ ( s, ∆)] ds ) F ( t ) ≤ C ′ ( T ) Z bz ( t ) ( | u ( x, t ) − u ′ ( x, t ) | + | v ( x, t ) − v ′ ( x, t ) | ) dx for t ∈ I ∆ , and Z I ∆ D ( t, ∆) dt ≤ F ( t )+ ( Z I ∆ [ O (1) + C Λ ( s, ∆) + C Λ ( s, ∆)] ds ) T max t ≤ t ≤ t + b − a F ( t ) ≤ F ( t ) + C ′′ ( T ) max t ≤ t ≤ t + b − a F ( t ) , which lead to the result for Case 1. Here the constants C ′ ( T ) and C ′′ ( T )depend only on T .Case 2: z ( t ) < a , that is, ∆ ∩ Γ B = ∅ and ∆ ⊂ Ω( T ). Then z a ( t ) = a + t − t .The result for this case has been proved in [25], and its proof can be carriedout in the same way as above. Therefore the proof is complete. (cid:3) Convergence of global classical solutions Choose a sequence of smooth functions( u ( k )0 , v ( k )0 ) ∈ C ∞ c (0 , ∞ ) , k = 1 , , · , such that ( u ( k )0 , v ( k )0 ) → ( u , v ) in L loc (0 , ∞ )as m → ∞ . Theorem 1.1 implies that there is a sequence of classical so-lutions, ( u ( k ) , v ( k ) ) ∈ C (Ω), k = 1 , , · · · , to (1.1), which satisfy boundarycondition (1.3) and take ( u ( k )0 , v ( k )0 ) as their initial data respectively. And supp ( u ( k ) ( · , t ) , v ( k ) ( · , t )) has bounded support in R for any t ≥ k ≥ { ( u ( k ) , v ( k ) ) } ∞ k =0 on ∆( − A, A, ∩ Ω for any A > 0. To this end, we first give the estimate on L norm of solution oversmall interval [ a, b ] ∩ [ z ( t ) , A − t ] for any a and b . Lemma 5.1. There is a constant r > such that if < b − a ≤ r and b ≤ A then sup k ≥ Z min { b,A − t } max { z ( t ) ,a } ( | u ( k ) ( x, t ) | + | v ( k ) ( x, t ) | ) dx ≤ δ for t ∈ [0 , A ] with z ( t ) ≤ b . NITIAL BOUNDARY VALUE PROBLEM FOR NONLINEAR DIRAC EQUATION 23 Proof. It is obvious thatlim k →∞ Z A ( | u ( k )0 − u | + | v ( k )0 − v | ) dx = 0 . As in [25], we choose r > mA + 8 | β | j ) (cid:0) Z b max { z (0) ,a } ( | u ( k )0 ( x ) | + | v ( k )0 ( x ) | ) dx + mj ( b − a ) (cid:1) ≤ δ mA + 16 | β | j ) (cid:0) Z b max { z (0) ,a } | v ( k )0 (2 p ( x ) + x ) | dx + mj ( b − a ) (cid:1) ≤ δ | b − a | ≤ r and for k = 0 , , · · · . Here for simplification ( u (0)0 , v (0)0 ) =( u , v ).Then with the pointwise estimates along the characteristics in Lemma 2.6and Lemma 2.8, we can deduce the desired result. The proof is complete. (cid:3) Now application of Proposition 4.1 and Lemma 5.1 to any pair of smoothsolutions ( u ( k ) , v ( k ) ) and ( u ( n ) , v ( n ) ) gives the following. Lemma 5.2. Suppose that ∆( a, b, τ ) ⊂ ∆( − A, A, with < b − a ≤ r and ∆( a, b, τ ) ∩ Ω = ∅ . Then there exists a constant C ( A ) > such that Z Z ∆( a,b,τ ) ∩ Ω ( | u ( k ) − u ( n ) | + | v ( k ) − v ( n ) | ) dxdt + Z Z ∆( a,b,τ ) ∩ Ω ( | u ( k ) v ( k ) − u ( n ) v ( n ) | + | u ( k ) v ( k ) − u ( n ) v ( n ) | ) dxdt ≤ C ( A ) Z b max( z (0) ,a ) ( | u ( k ) ( x, τ ) − u ( n ) ( x, τ ) | + | v ( k ) ( x, τ ) − v ( n ) ( x, τ ) | ) dx for any k ≥ and n ≥ . Here the constant C ( A ) depends only on A and E ;the constant r > is given by Lemma 5.1. In the next, we prove the convergence of { ( u ( k ) , v ( k ) ) } ∞ k =0 on ∆( − A, A, ∩ Ωby the induction step as follows.DenoteΩ( A, τ ) = ∆( − A, A, ∩ Ω ∩ { ( x, t ) | ≤ t ≤ τ } , ≤ τ ≤ A. Lemma 5.3. Suppose that lim m,n →∞ Z Z Ω( A,τ ) ( | u ( m ) − u ( n ) | + | v ( m ) − v ( n ) | ) dxdt = 0 , ★★★★★★★★★★★★ ❝❝❝❝❝❝❝❝❝❝❝❝ ( − A, 0) ( A, t = 0( τ − A, τ ) ( A − τ, τ ) t = τ Ω( A, τ ) Figure 6. Domain Ω( A, τ )lim m,n →∞ Z Z Ω( A,τ ) ( | u ( m ) v ( m ) − u ( n ) v ( n ) | ) dxdt = 0 and lim m,n →∞ Z Z Ω( A,τ ) ( | u ( m ) v ( m ) − u ( n ) v ( n ) | ) dxdt = 0 for τ ∈ [0 , A − r ] . Then lim m,n →∞ Z Z Ω( A,τ + r ) ( | u ( m ) − u ( n ) | + | v ( m ) − v ( n ) | ) dxdt = 0 , lim m,n →∞ Z Z Ω( A,τ + r ) ( | u ( m ) v ( m ) − u ( n ) v ( n ) | ) dxdt = 0 and lim m,n →∞ Z Z Ω( A,τ + r ) ( | u ( m ) v ( m ) − u ( n ) v ( n ) | ) dxdt = 0 . Here r is given in Lemma 5.1.Proof. We choose a finite number of subintervals, [ a j , b j ], j = 1 , , · · · , J ,with [ a j , b j ] ∩ [max { τ − A, z ( τ ) } , A − τ ] = ∅ and b j − a j = 4 r , such thatΩ( A, τ + r ) \ Ω( A, τ ) ⊂ ∪ Jj =1 ∆( a j , b j , τ ′ ) , where τ ′ = τ − r , and J ≤ Ar + 1.For 1 ≤ j ≤ J , by Proposition 4.1 and Lemma 5.1, we havelim m,n →∞ Z Z ∆( a j ,b j ,τ ′ ) ∩ Ω ( | u ( m ) − u ( n ) | + | v ( m ) − v ( n ) | ) dxdt = 0 , lim m,n →∞ Z Z ∆( a j ,b j ,τ ′ ) ∩ Ω ( | u ( m ) v ( m ) − u ( n ) v ( n ) | ) dxdt = 0and lim m,n →∞ Z Z ∆( a j ,b j ,τ ′ ) ∩ Ω ( | u ( m ) v ( m ) − u ( n ) v ( n ) | ) dxdt = 0 . NITIAL BOUNDARY VALUE PROBLEM FOR NONLINEAR DIRAC EQUATION 25 Therefore we have the convergence of the sequences { ( u ( m ) , v ( m ) ) } ∞ m =1 , { u ( m ) v ( m ) } ∞ m =1 and { u ( m ) v ( m ) } ∞ m =1 in L (Ω( A, τ + r )) respectively. The proof is complete. (cid:3) Now we have the following convergence result. Proposition 5.1. There exists a ( u, v ) ∈ L loc (Ω) such that lim m →∞ || ( u ( m ) , v ( m ) ) − ( u, v ) || L (∆( − A,A, ∩ Ω) = 0 , and lim m →∞ (cid:0) || u ( m ) v ( m ) − uv || L (∆( − A,A, ∩ Ω) + || u ( m ) v ( m ) − uv || L (∆( − A,A, ∩ Ω) (cid:1) = 0 for any A > .Proof. With the induction steps given by Lemma 5.3, we havelim m,n →∞ Z Z (∆( − A,A, ∩ Ω) ( | u ( m ) − u ( n ) | + | v ( m ) − v ( n ) | ) dxdt = 0 , lim m,n →∞ Z Z (∆( − A,A, ∩ Ω) ( | u ( m ) v ( m ) − u ( n ) v ( n ) | ) dxdt = 0and lim m,n →∞ Z Z (∆( − A,A, ∩ Ω) ( | u ( m ) v ( m ) − u ( n ) v ( n ) | ) dxdt = 0 , for any A > 0. These lead to the desired result. The proof is complete. (cid:3) Proof of main results on strong solutions In the same way as in the proof of Lemma 5.3 and Proposition 5.1, we canprove the following. Proposition 6.1. Suppose that { u ( m ) j , v ( m ) j } ∞ m =1 , j = 1 , , are two sequences ofclassical solution to (1.1) satisfy boundary condition (1.3) with the following, lim m →∞ Z M ( | u ( m )1 ( x, − u ( m )2 ( x, | + | v ( m )1 ( x, − v ( m )2 ( x, | ) dx = 0 for some M > . Then, lim m →∞ Z Z ∆( − M,M, ∩ Ω ( | u ( m )1 − u ( m )2 | + | v ( m )1 − v ( m )2 | ) dxdt = 0 . Proof of Theorem1.3. The existence of solution ( u, v ) is proved by Propo-sition 5.1. Moreover, ( u, v ) satisfies (1.9) and (2.6).To prove the uniqueness, let ( u j , v j ), j = 1 , 2, be two strong solutions to(1.1-1.3), and let ( u ( m ) j , v ( m ) j ), j = 1 , u j , v j ), j = 1 , L loc (Ω). Moreover, the initial data ( u ( m ) j ( x, , v ( m ) j ( x, u , v ) for j = 1 , Then by Proposition 6.1, we havelim m →∞ Z Z ∆( − A,A, ∩ Ω ( | u ( m )1 ( x, − u ( m )2 ( x, | + | v ( m )1 ( x, − v ( m )2 ( x, | ) dx = 0 , which yields that( u , v )( x, t ) = ( u , v )( x, t ) , a.e. ( x, t ) ∈ ∆( − A, A, ∩ Ω . This leads to the uniqueness of the strong solution. The proof is complete. (cid:3) Proof of Theorem 1.4. Indeed the results hold for the classical solutions.Then by taking the limit, we can prove the result still hold for the strongsolution. The proof is complete. (cid:3) Acknowledgement This work was partially supported by NSFC Project 11421061 and by the111 Project B08018. References [1] S. Alinhac, Blowup for nonlinear hyperbolic equations, Birkhauser Boston, Inc., Boston,1995.[2] A. Bachelot, Global Cauchy problem for semilinear hyperboloc systems with nonlocalinteractions. Applications to Dirac equations, J. Math. Pures. Appl. (2006), 201-236.[3] J. M. Bony, Solution globales born´ees pour les mod`eles discrets de l’´equation de Boltz-mann, en dimension 1 d’espace, Journ´ees ”Equations aux deriv´ees partielles” (SaintJean de Monts, 1987), Exp. No XVI, 10pp., ´Ecole Polytech., Palaiseau, 1987.[4] N. Bournaveas and G. E. Zouraris, Theory and numerical approximations for a nonlinear Dirac system, ESAIM: Math. Model. Num. Analysis (4) (2012), 841-874.[5] P. Bouvier and C. G´erard, Hawking effect for toy model of interacting Fermions, Ann.Heri. Poincar´e (5) (2015), 1191-1230.[6] F. Cacciafesta, Global small solutions to the critical radial Dirac equation with potential, Nonlinear Analysis (2011), 6060-6073.[7] T. Candy, Global existence for an L critical nonlinear Dirac equation in one dimension, Adv. Differential Equations No. 7-8 (2011), 643-666.[8] C. M. Dafermos, Hyperbolic Conservation Laws in Continuum Physics , Springer-Verlag,Berlin, 2010.[9] V. Delgado, Global solution of the Cauchy problem for the (classical) coupled Maxwell-Dirac and other nonlinear Dirac equations, Proc. Amer. Math. Soc. (2)(1978), 289-296.[10] J. P. Dias and M. Figueira, Remarque sur le probl`eme de Cauchy pour une equation deDirac non lin´eaire avec masse nulle, Portucaliae Math. (4) (1988), 327-335.[11] M. Escobedo and L. Vega, A semilinear Dirac equation in H s ( R ) for s > , SIAM J.Math. Anal. (2) (1997), 338-362.[12] M. J. Esteban, M. Lewin and E. S´er´e, Variational methods in relativistic quantummechanics, Bulletin A.M.S. No. 4 (2008), 535-598.[13] J. Glimm, Solution in the large for nonlinear systems of conservation laws, Comm.Pure Appl. Math. (1965), 695-715. NITIAL BOUNDARY VALUE PROBLEM FOR NONLINEAR DIRAC EQUATION 27 [14] D. J. Gross and A. Neveu, Dynamical symmetry breaking in asymptotically free fieldtheories, Phys. Rev. D 10 (1974), 3235-3253.[15] S-Y Ha and A. E. Tzavaras, Lyapunov functionals and L -stability for discrete velocityBoltzmann equations, Commun. Math. Phys. (2003), 65-92.[16] H. Huh, Global solutions to Gross-Neveu equation, Lett. Math. Phys. (8)(2013),927-931.[17] Mizohata, The Theory of Partial Differential Equations, Cambridge University Press,New York, 1973.[18] I.P. Naumkin, Cubic nonlinear Dirac equation in a quarter plane, J. Math. Anal. Appl. (2) (2016), 1633-1664.[19] I. P. Naumkin, Initial boundary value problem for the one dimensional Thirring model, J. Differential Equations (8) (2016), 4486-4523.[20] D. Pelinovsky, Survey on global existence in the nonlinear Dirac equations in one di-mension, in: Harmonic analysis and nonlinear partial differential equations (edited byT. Ozawa and M. Sugimoto), RIMS Kˆokyˆuroku Bessatsu, B26 (2011), 37-50.[21] S. Selberg, Global existence in the critical space for the Thirring and Gross-neveu modelscoupled with the electromagnetic field, Disrete Contin. Dyn. Syst. (5) (2018), 2555-2569.[22] L. Tartar, From Hyperbolic Systems to Kinetic Theory, Springer 2008.[23] W.E. Thirring, A soluble relativistic field theory, Ann. Phys. (1958), 91-112.[24] Y. Zhang, Global strong solution to a nonlinear Dirac type equation in one dimension, Nonlinear Analysis: Theory, Method and Applications (2013), 150-155.[25] Y. Zhang and Q. Zhao, Global solution to nonlinear Dirac equation for Gross-Neveumodel in dimensions, Nonlinear Analysis: Theory, Method and Applications (2015), 82-96.[26] Y. Zhou, Uniqueness of weak solutions in 1+1 dimensional wave maps, Math. Z. (1999), 707-719. Yongqian Zhang: School of Mathematical Sciences, Fudan University,Shanghai 200433, P.R.China E-mail address : [email protected] Qin Zhao : School of Mathematical Sciences, Shanghai Jiao Tong Uni-versity, Shanghai 200240, P.R.China E-mail address ::