Equations in Three Space Dimensions
Yung-fu Fang & Manoussos Grillakis
Abstract. We establish a local existence result for Dirac-Klein-Gordon equations in three space dimensions, employing a null form estimate and a fixed point argument.
0. Introduction and Main Results.
In the present work, we like to study the Cauchy problem for the Dirac-Klein-Gordon equations. The unknown quantities are a spinor field
ψ : R × R3 7→ C4 and a scalar field φ : R × R3 7→ R. The evolution
equations for the fields are given below,
Dψ = φψ; (t, x) ∈ R × R3 (0.1a)
¤φ = ψψ; (0.1b)
ψ(0, x) := ψ0(x), φ(0, x) = φ0(x), φ,t(0, x) = φ1(x), (0.1c)
where D is the Dirac operator, D := −iγµ∂
µ, µ = 0, 1, 2, 3, and γµ are the
Dirac matrices, the wave operator ¤ = −∂tt+ ∆, and ψ = ψ†γ0, and † is
the complex conjugate transpose. The matrices can be written as follows. First let us define the 2 × 2 matrices σ1, σ2, and σ3,
σ1 := · 0 1 1 0 ¸ , σ2 := · 0 −i i 0 ¸ , σ3 := · 1 0 0 −1 ¸ . (0.2a) The matrices γµ are defined via
γ0 = · I2 0 0 −I2 ¸ , γj = · 0 σj −σj 0 ¸ . (0.2b) 1
The purpose of this work is to demonstrate a variant null form estimate, by employing the solution representations in Fourier transform of the DKG equations. Such estimate can improve the existence result, see [Bo]. We will take advantage of the null form structure depicted in the nonlinear term ψψ, see [KM] and [Bo].
For the DKG system, there are many conserved quantities which are not positive definite, such as the energy, the momenta, and the angular momenta. However there is a known positive conserved quantity which is the law of conservation of charge,
Z
|ψ(t)|2dx = constant, see [GS]
In ’74, Chadam and Glassey showed that the Cauchy problem for the DKG equations has a unique local solution for ψ0 ∈ H2, φ0 ∈ H2, φ1 ∈ H1, and global solution for a particular class of initial data, see [CG]. In ’81, Choquet-Bruhat proved the global existence result for the (massless) DKG equations by assuming small data, see [CB]. In ’88, Bachelot gave the global existence for DKG system with small data, see [Ba]. In ’99, Bournaveas derived a local existence for the DKG equations, based on a null form estimate, if ψ0 ∈ H
1
2, φ0 ∈ H1, φ1 ∈ L2, see [Bo].
The outline of this paper is as follows. First we derive some solutions representations in Fourier transform. Next we prove some a priori esti-mates of solutions for Dirac equation and for wave equation. Then we show a local existence result for (0.1), employing the null form estimate together with some other estimates, and a fixed point argument. Finally we show the key estimate, namely the null form estimate.
The main result in this work is as follows.
Theorem 0.1. (Local Existence) Let ² > 0. If the initial data of (0.1)
ψ0 ∈ H 1
4+²(R3), φ0 ∈ H1(R3), φ1 ∈ L2(R3), then there is a unique local solution for (0.1).
Remarks.
1. The DKG equations follow from the Lagrangian Z R3+1 n |∇φ|2− |φt|2 − ψDψ − φψψ o dxdt. (0.3)
2. The Dirac-Klein-Gordon system must be ½
Dψ = φψ;
¤φ + m2φ = ψψ, (0.4)
and the proof works for this system too.
3. Let I be the 4×4 identity matrix, bD = γ0τ +γjξj, and b¤ = τ2−|ξ|2,
thus we have bD2 = b¤I.
4. ψψ = ψ†γ0ψ = |ψ1|2 + |ψ2|2 − |ψ3|2 − |ψ4|2, where ψj are the
component functions of the vector function ψ, which take values in C. 1. Solution Representation.
In what follows, we denote by (t, x) the time-space variables and by (τ, ξ) the dual variables with respect to the Fourier transform of a given function. We will use µ = 1
4 + ², α = 1
4 + δ, and ν = ² − δ throughout the paper. We will also often skip the constant in the inequalities. For convenience, we denote the multipliers by
b E(τ, ξ) = |τ | + |ξ| + 1, (1.1a) b S(τ, ξ) = ¯ ¯ ¯|τ | − |ξ| ¯ ¯ ¯ + 1, (1.1b) c W (τ, ξ) = τ2− |ξ|2, (1.1c) b D(τ, ξ) = γ0τ + γjξj, j = 1, 2, 3, (1.1d) c M (ξ) = |ξ| + 1. (1.1e)
Notice that cW and bD are the symbols of the wave and Dirac operators
respectively. Also there is a summation over the upper and lower indeices. Consider the Dirac equation,
½
Dψ = G, (t, x) ∈ R1 × R3, ψ(0) = ψ0.
(1.2)
First by taking the Fourier transform on (1.2) over the space variable and solving the resulting ODE, we can formally write down the solution
as follows. e ψ(t, ξ) = e it|ξ| 2|ξ| D(|ξ|, ξ)γb 0ψb 0(ξ) + e−it|ξ| 2|ξ| D(|ξ|, −ξ)γb 0ψb 0(ξ) + Z t 0 ei(t−s)|ξ| 2|ξ| D(|ξ|, ξ)i eb G(s, ξ) ds + Z t 0 e−i(t−s)|ξ| 2|ξ| D(|ξ|, −ξ)i eb G(s, ξ) ds.(1.3) Rewriting the inhomogeneous terms in (1.3) gives
e ψ(t, ξ) =h e it|ξ| 2|ξ| D(|ξ|, ξ) +b e−it|ξ| 2|ξ| D(|ξ|, −ξ)b i γ0ψb0(ξ) + Z heitτ − eit|ξ| 2|ξ|(τ − |ξ|)D(|ξ|, ξ) +b eitτ − e−it|ξ| 2|ξ|(τ + |ξ|)D(|ξ|, −ξ)b i b G(τ, ξ)dτ. (1.4)
Now we split the function bG into several parts in the following manner.
Consider ba(τ ) a cut-off function equals 1 if |τ | ≤ 1
2 and equals 0 if |τ | ≥ 1, and denote by h(τ ) the Heaviside function. For simplicity, let us write
b G±(τ, ξ) := h(±τ )ba(τ ∓ |ξ|) bG(τ, ξ), (1.5a) b Gf(τ, ξ) := bG(τ, ξ) − ¡ b G+(τ, ξ) + bG−(τ, ξ) ¢ , (1.5b) b D± := bD(|ξ|, ±ξ). (1.5c)
Notice that bG± are supported in the regions {(τ, ξ) : ±τ > 0, |τ ∓|ξ|| ≤ 1}
respectively. Using the decomposition of the forcing term bG = bGf+ bG++
b
G−, the inhomogeneous term in (1.4) can be written as
Z heitτ − eit|ξ| 2|ξ|(τ − |ξ|)D(|ξ|, ξ) +b eitτ − e−it|ξ| 2|ξ|(τ + |ξ|)D(|ξ|, −ξ)b i b Gf(τ, ξ)dτ = Z eitτ D(τ, ξ)b τ2 − |ξ|2Gbf dτ − e it|ξ|Db+ 2|ξ| Z b Gf τ − |ξ|dτ − e−it|ξ|Db− 2|ξ| Z b Gf τ + |ξ|dτ, (1.6a) Z eitτ − eit|ξ| 2|ξ|(τ − |ξ|)Db+( bG++ bG−)dτ = eit|ξ|Db+ 2|ξ| Z eit(τ −|ξ|)− 1 τ − |ξ| ( bG++ ba6(τ ) bG−)dτ + Z eitτ(1 − ba6(τ )) bD+Gb− 2|ξ|(τ − |ξ|) dτ − e it|ξ|Db+ 2|ξ| Z (1 − ba6(τ )) bG− τ − |ξ| dτ, (1.6b)
where ba6(τ ) = ba(τ6) and ba is the cut-off function defined previously. Z eitτ − e−it|ξ| 2|ξ|(τ + |ξ|)Db−( bG+ + bG−)dτ = e−it|ξ|Db− 2|ξ| Z eit(τ +|ξ|)− 1 τ + |ξ| (ba6(τ ) bG+ + bG−)dτ + Z eitτ(1 − ba6(τ )) bD−Gb+ 2|ξ|(τ + |ξ|) dτ − e −it|ξ|Db− 2|ξ| Z (1 − ba6(τ )) bG+ τ + |ξ| dτ. (1.6c)
Recall the power expansion
eit(τ ±|ξ|)− 1 = ∞ X k=1 1 k!(it) k(τ ± |ξ|)k. (1.7)
Combining (1.4)-(1.7), we can give a formula for bψ, namely
b ψ(τ, ξ) = ∞ X k=0 ³ δ+(k)(τ, ξ) bA+,k(ξ) + δ−(k)(τ, ξ) bA−,k(ξ) ´ + bK(τ, ξ), (1.8)
where δ±(τ, ξ) are the delta functions supported on {τ = ±|ξ|}
respec-tively, δ(k) mean derivatives of the delta function, and
b K(τ, ξ) := D(τ, ξ)b c W (τ, ξ) b Gf+(1−ba6(τ )) bD+ b G− 2|ξ|(τ − |ξ|) + (1−ba6) bD−Gb+ 2|ξ|(τ + |ξ|) , (1.9a) b A±,0(ξ) := b D± 2|ξ| h γ0ψb 0 − Z bGf + (1 − ba6(λ)) bG∓ λ ∓ |ξ| dλ i , (1.9b) b A±,k(ξ) := b D±(−1)k 2|ξ|k! Z (λ ∓ |ξ|)k−1£Gb±+ ba6(λ) bG∓ ¤ dλ. (1.9c) Consider the wave equation,
½
¤φ = F, (t, x) ∈ R1 × R3, φ(0) = φ0, φt(0) = φ1.
(1.10) Taking Fourier transform on (1.10) and solving the resulting ODE gives
e φ(t, ξ) = cos t|ξ|bφ0(ξ) + sin t|ξ| |ξ| φb1(ξ) − Z t 0 sin (t − s)|ξ| |ξ| F (s, ξ)ds. (1.11)e
e φ(t, ξ) = e it|ξ|+ e−it|ξ| 2 φb0(ξ) + eit|ξ|− e−it|ξ| 2i|ξ| φb1(ξ) − Z eitτ − eit|ξ| 2|ξ|(|ξ| − τ )F (τ, ξ)dτ −b Z eitτ − e−it|ξ| 2|ξ|(τ + |ξ|)F (τ, ξ)dτ. (1.12)b For the homogeneous part, we rewrite it as
eit|ξ|+ e−it|ξ| 2 φb0(ξ) + eit|ξ| − e−it|ξ| 2i|ξ| φb1(ξ) = eit|ξ| 2|ξ| φb+ + e−it|ξ| 2|ξ| φb−, (1.13) where b φ± = |ξ|bφ0 ∓ ibφ1. (1.15)
Now we split bF the same manner as we did to bG. Let us write
b F±(τ, ξ) := h(±τ )ba(τ ∓ |ξ|) bF (τ, ξ), (1.16a) b Ff(τ, ξ) := bF (τ, ξ) − ¡b F+(τ, ξ) + bF−(τ, ξ) ¢ , (1.16b)
For the inhomogeneous part, we obtain Z h eitτ − eit|ξ| 2|ξ|(|ξ| − τ ) + eitτ − e−it|ξ| 2|ξ|(|ξ| + τ ) i b Ff(τ, ξ)dτ = Z eitτ Fbf |ξ|2 − τ2 dτ − eit|ξ| 2|ξ| Z b Ff |ξ| − τdτ − e−it|ξ| 2|ξ| Z b Ff |ξ| + τdτ,(1.17a) Z eitτ − eit|ξ| 2|ξ|(|ξ| − τ )( bF+ + bF−)dτ = eit|ξ| 2|ξ| Z eit(τ −|ξ|)− 1 |ξ| − τ ( bF++ ba6Fb−)dτ + Z eitτ (1 − ba6) bF− 2|ξ|(|ξ| − τ )dτ − eit|ξ| 2|ξ| Z (1 − ba6) bF− |ξ| − τ dτ, (1.17b) Z eitτ − e−it|ξ| 2|ξ|(|ξ| + τ )( bF++ bF−)dτ = e−it|ξ| 2|ξ| Z eit(τ +|ξ|)− 1 |ξ| + τ (ba6Fb++ bF−)dτ + Z eitτ (1 − ba6) bF+ 2|ξ|(|ξ| + τ )dτ − e−it|ξ| 2|ξ| Z (1 − ba6) bF+ |ξ| + τ dτ, (1.17c)
Combining (1.17a)-(1.17c), we can give a formula for bφ, namely b φ(τ, ξ) = ∞ X k=0 ³ δ+(k)(τ, ξ) bB+,k(ξ) + δ(k)− (τ, ξ) bB−,k(ξ) ´ + bL(τ, ξ), (1.18)
where δ±(τ, ξ) are the delta functions supported on {τ = ±|ξ|}
respec-tively, δ(k) mean derivatives of the delta function, and
b L(τ, ξ) := Fbf c W (τ, ξ) − (1 − ba6(τ )) bF− 2|ξ|(|ξ| − τ ) − (1 − ba6(τ )) bF+ 2|ξ|(|ξ| + τ ) , (1.19a) b B±,0(ξ) := 1 2|ξ| h b φ±+ Z bFf + (1 − ba6(λ)) bF∓ |ξ| ∓ λ dλ i , (1.19b) b B±,k(ξ) := ±(−1) k 2|ξ|k! Z (λ ∓ |ξ|)k−1£Fb±+ ba6(λ) bF∓ ¤ dλ. (1.19c) Remark. We need to localize the solutions for Dirac equation and wave equation due to the presence of the delta function.
2. Estimates.
To localize the solution in time, let b(t) be a cut-off function such that
b(t) equals 1 if |t| ≤ 1
2, and equals 0 if |t| > 1, and bT(t) = b(t/T ). For an
arbitrary function f (t, x), we have
kbbT ∗ bf kL2 = kbTf kL2 ≤ kbTkL∞kf kL2. (2.1)
For the Dirac equation (1.2), we have the following estimate.
Lemma 2.1. Let 0 ≤ ²1 < 12, α1 ≥ 0, and ψ0 ∈ Hα1. Then we have
° ° °bbT ∗ [ cMα1Sb1−²1ψ]b ° ° ° L2(R1×R3) ≤ C ³ kψ0kHα1 + ° ° °Mc α1Gb b S²1 ° ° ° L2 ´ . (2.2)
Proof. For simplicity, let ²1 = ². Without loss of generality, we prove
the special case. ° ° °bbT ∗ [ bS1−²ψ]b ° ° ° L2(R1×R3) ≤ C ³ kψ0kL2 + ° ° °Gbb S² ° ° ° L2 ´ . (2.3)
To estimate bbT ∗ [ bS1−²ψ], we apply formulae (1.8) and (1.9)s. First web compute kbbT ∗ [ bS1−²K]kb L2 ≤ Ck bS1−²Kkb L2 ≤ C ° ° ° bS1−² Db c W b Gf ° ° ° L2+ C ° ° ° bS1−²(1 − ba6) bD+Gb− 2|ξ|(τ − |ξ|) ° ° ° L2 + C ° ° ° bS1−²(1 − ba6) bD−Gb+ 2|ξ|(τ + |ξ|) ° ° ° L2 ≤ C ° ° °Gb b S² ° ° ° L2. (2.4)
For the term bbT∗ [ bS1−²δ(k)+ Ab+,k], we can mollify bS(τ, ξ) without loss of
generality such that ∂τkS(±|ξ|, ξ) = 0 if k ≥ 1. Thus we can computeb kbbT ∗ [ bS1−²δ+(k)](ξ)k2L2(dτ ) ∼ Z ³ Z bbT(τ − λ) bS(λ, ξ)1−²δ(k)(λ − |ξ|)dλ ´2 dτ ∼ Z ³ ∂k ∂λk ¡ bbT(τ − λ) bS(λ, ξ)1−²¢¯¯¯ λ=|ξ| ´2 dτ ≤ Z ³ Tk+1bb(k)¡T (τ − |ξ|)¢´2dτ ≤ T2k+1ktkbkL2 ≤ CT2k+1. (2.5) Then we calculate k bA+,0kL2 ≤kψ0kL2 + ³ Z ³ Z bGf +¡1 − ba6(τ )¢Gb− τ − |ξ| dτ ´2 dξ ´1 2 ≤kψ0kL2 + ° ° °Gb b S² ° ° ° L2 (2.6) and k bA+,kkL2 ≤1 k! ³ Z ³ Z (τ − |ξ|)k−1£Gb+ + ba6Gb− ¤ (τ, ξ)dτ ´2 dξ ´1 2 ≤2k k! ³ Z Z ¯¯ ¯ bG++ ba6Gb− ¯ ¯ ¯2(τ, ξ)dτ dξ´ 1 2 ≤ 2k k! ° ° °Gb b S² ° ° ° L2. (2.7) Therefore we have kbbT ∗ [ bS1−²δ+Ab+,0]kL2 ≤T 1 2 ³ kψ0kL2 + ° ° °Gb b S² ° ° ° L2 ´ , kbbT ∗ [ bS1−²δ+(k)Ab+,k]kL2 ≤Tk+ 1 2 2 k k! ° ° °Gbb S² ° ° ° L2. (2.8)
The calculation for the term bbT ∗ [ bS1−²δ−(k)Ab−,k] is analogous. Combine
the above results we complete the proof. ¤
Consider ½
¤Iφ = D(Dφ) = Iψψ,
Dφ(0) = −iγ0φ
1 − iγj∂jφ0.
(2.9) For the above equation, we have the following estimate and its proof is analogous to that of Lemma 2.1.
Corollary 2.1. Let δ > 0. Then we have ° ° ° bS1+2δ2 Dφc ° ° ° L2 ≤ kDφ(0)kL2 + ° ° ° c ψψ b S1−2δ2 ° ° ° L2. (2.10)
Consider two Dirac equations, ½
Dψj = Gj, j = 1, 2, ψj(0) = ψ0j.
(2.11) We have the key estimate whose proof will be presented in the last section. Lemma 2.2. (Null Form Estimate) Let δ > 0, and ψ1, ψ2 be the solu-tions for (2.11). If ψ0j ∈ Hα, we have
° ° °Eb δT (ψ\ 1ψ2) b S12 ° ° ° L2 ≤ C(T ) ³ kψ01kHα + ° ° °Mc αGb 1 b Sδ2 ° ° ° L2 ´ · ³ kψ02kHα + ° ° °Mc αGb 2 b Sδ2 ° ° ° L2 ´ , (2.12) where T (ψψ) is a localization of ψψ.
The Fourier transform of the quadratic expression T (ψψ) is written as the sum of the following terms.
¡ δ∓Ab±,0 ¢ ∗¡δ±Ab±,0 ¢ + X k+l>0 bbT ∗ ¡ δ∓(k)Ab±,k ¢ ∗¡δ±(l)Ab±,l ¢ , (2.13a) ¡ δ∓Ab±,0 ¢ ∗¡δ∓Ab∓,0 ¢ + X k+l>0 bbT ∗ ¡ δ∓(k)Ab±,k ¢ ∗¡δ∓(l)Ab∓,l ¢ , (2.13b) X k bbT∗ ³¡ δ∓(k)Ab±,k ¢ ∗¡Kb1+ bK2 ¢ +¡ bK1+ bK2 ¢ ∗¡δ(k)± Ab±,k ¢´ , (2.13c) b K1 ∗ bK1 + bK1 ∗ bK2 + bK2∗ bK1+ bK2 ∗ bK2. (2.13d)
Lemma 2.3. Let ²2 be any real number, ²3 > 0 and φ be the solution of (1.10). If φ0 ∈ H1+²2 and φ1 ∈ H²2, then ° ° °bbT ∗ h b E1+²2Sb12+²3φb i°° ° L2 ≤ C ³ kφ0kH1+²2 + kφ1kH²2 + ° ° ° Eb ²2Fb b S12−²3 ° ° ° L2 ´ . (2.14) Proof. For simplicity, we consider the case when ²1 = ²2 = ². To
esti-mate bbT ∗
£b
E1+²Sb12+²φb¤ in the L2-norm, we invoke the formulae (1.19).
First we compute kbbT ∗ £b E1+²Sb12+²Lb¤kL2 ≤ k bE1+²Sb12+²Lkb L2 ≤ ° ° °Eb 1+²Sb1 2+²Fbf c W ° ° ° L2+ ° ° °Eb 1+²Sb1 2+²(1−ba6) bF− 2|ξ|(|ξ| − τ ) ° ° ° L2+ ° ° °Eb 1+²Sb1 2+²(1−ba6) bF+ 2|ξ|(|ξ| + τ ) ° ° ° L2≤ ° ° °Eb ²Fb b S12−² ° ° ° L2. (2.15)
For the term bbT∗
£b
E1+²Sb12+²δ(k) + Bb+,k
¤
, we can mollify bE bS(τ, ξ) without
loss of generality such that ∂τkS(±|ξ|, ξ) = 0 if k ≥ 1. Thus we computeb
° °bbT ∗ £b E1+²Sb12+²δ(k) + ¤ (ξ)°°2L2(dτ ) = Z ¯ ¯ ¯ Z bbT(τ − λ) bE1+²Sb 1 2+²(λ, ξ)δ(k)(λ − |ξ|)dλ ¯ ¯ ¯2dτ = Z ¯ ¯ ¯ ∂ k ∂λk ³ bbT(τ − λ) bE1+²Sb 1 2+²(λ, ξ) ´¯¯ ¯ λ=|ξ| ¯ ¯ ¯2dτ ∼ Z ¯ ¯ ¯Tk+1bb(k)¡T (τ − |ξ|)¢¯¯¯2(|ξ| + 1)2+2²dτ ≤ T2k+1ktkbkL2(|ξ| + 1)2+2² ≤ CT2k+1(|ξ| + 1)2+2², (2.16a) which implies ° °bbT ∗ £ b E1+²Sb12+²δ(k) + Bb+,k ¤° ° L2 ≤ CTk+ 1 2 ³ Z (|ξ| + 1)2+2²¯¯ bB+,k(ξ) ¯ ¯2 dξ ´1 2 . (2.16b) To estimate the above integral, we first focus on the region where |ξ| > 1.
We have the following bounds: Z ¯ ¯ ¯Z cM ²Fb f(λ, ξ) ¯ ¯|ξ| − λ¯¯ dλ ¯ ¯ ¯2dξ ≤ Z Z 1 ¯ ¯|ξ| − λ¯¯1+2²dλ Z ¯¯¯ cM²Fb f(λ, ξ) ¯ ¯ ¯2 ¯ ¯|ξ| − λ¯¯1−2² dλ dξ ≤ C ° ° °Eb ²Fb f b S12−² ° ° °2 L2 (2.17) and in the same vein
Z ¯ ¯ ¯Z cM ²¡1 − ba 6(λ) ¢b F−(λ, ξ) ¯ ¯|ξ| − λ¯¯ dλ ¯ ¯ ¯2dξ ≤ C ° ° °Eb ²Fb − b S12−² ° ° °2 L2. (2.18) Hence we get ° °bbT ∗ £ b E1+²Sb12+²δ+Bb+,0¤°° L2(L2(|ξ|>1)) ≤ CT12 ³ kφ0kH1+² + kφ1kH² + ° ° ° Eb ²Fb b S12−² ° ° ° L2 ´ (2.19) and ° °bbT ∗ £ b E1+²Sb12+²δ(k) + Bb+,k ¤° ° L2(L2(|ξ|>1)) ≤ CTk+12 c k k! ³ Z Z b E2²¯¯ bF+ + ba6Fb− ¯ ¯2 (τ, ξ)dλ dξ´ 1 2 ≤ CTk+12 c k k! ° ° °Eb ²Fb b S12−² ° ° ° L2. (2.20)
The calculation for the term bbT ∗
£b
E1+²Sb12+²δ(k) − Bb−,k
¤
is analogous. For region |ξ| ≤ 1, we consider bbT ∗
£ b E1+²Sb12+²(δ(k) + Bb+,k+ δ(k)− Bb−,k) ¤ . This is clear from the derivation of the solution representation which indicates that the solution is actually not singular along the cones.
bbT ∗ £ b E1+²Sb12+²(δ(k) + Bb+,k+ δ−(k)Bb−,k) ¤ (τ, ξ) ∼ Tk+1¡|ξ| + 1¢1+²£ ctkb¡T (τ − |ξ|)¢Bb +,k(ξ) + ctkb ¡ T (τ + |ξ|)¢Bb−,k(ξ) ¤ = Tk+1¡|ξ| + 1¢1+²£ ctkb¡T (τ − |ξ|)¢− ctkb¡T (τ + |ξ|)¢¤Bb +,k(ξ) + Tk+1¡|ξ| + 1¢1+²tckb¡T (τ + |ξ|)¢£Bb +,k(ξ) + bB−,k(ξ) ¤ . (2.21)
Under the restriction of |ξ| ≤ 1, we have c tkb¡T (τ − |ξ|)¢− ctkb¡T (τ + |ξ|)¢∼ T [tk+1b¡T (τ − (1 − 2θ)|ξ|)¢|ξ|, (2.22) b B+,0(ξ) + bB−,0(ξ) ∼ bφ0 + Z b Ff |ξ|2 − λ2 dλ, (2.23) and b B+,k(ξ) + bB−,k(ξ) ∼ 1 (k − 1)! Z ¡ λ − (1 − 2θ)|ξ|¢k−2( bF+ + ba6Fb−) dλ. (2.24) Combine the above results we complete the proof. ¤
Since a Dirac equation implies a wave equation, thus we consider ½
¤ψ = D(Dψ) = (Dφ)ψ + φ2ψ,
ψ(0) = ψ0, ∂tψ(0) = iγ0(iγj∂jψ0+ φ0ψ0).
(2.25a) We have the following estimate whose proof is analogous to that of Lemma 2.3.
Corollary 2.3. Let δ > 0. Then we have ° ° bEµSb1+2δ2 ψb°° L2 ≤ C(T ) ³ kψ0kHµ+ kφ0kH1kψ0kHµ+ ° ° °(Dφ)ψ + d\ φ2ψ b E1−µSb1−2δ2 ° ° ° L2 ´ . (2.25b) We will also need some technical lemmas.
Lemma 2.4. (Hardy-Littlewood-Polya) Let r = 2−1
p− 1 q. Then we have Z R1×R1 f (s)g(t) |s − t|r dsdt ≤ Ckf kLpkgkLq. (2.26)
Lemma 2.5. Let f (t, x) be any function. Assume that ²1 ≥ 0, r1 =
2 1 + ²1, 0 ≤ ²2 ≤ 1 2, and r2 = 2 1 − 2²2. Then we have ° ° °bdTf b S²1 ° ° ° L2 ≤ CkbTf kLr1(L2), (2.27a) kf kLr2([0,T ];L2) ≤ Ck bS²2f kb L2. (2.27b)
Proof. They are dual estimates if ²1 = 2²2, hence we only show (2.27a).
For simplicity, let ²1 = ². Through some analysis, we have Fτ−1¡Sb−²¢(t, ξ) ∼ e ±it|ξ| |t|1−² 2, (2.28) which implies Fτ−1³ dbTf b S² ´ (t, ξ) ∼ Z e±i(t−s)|ξ| |t − s|1−² 2( gbTf )(s, ξ)ds. (2.29)
Invoke (2.29), duality argument, and Hardy-Littlewood-Polya inequality, then we can compute
¯ ¯ ¯ D bTf, g E¯¯ ¯ = ¯ ¯ ¯D dbTf b S² , bS ²bgE¯¯¯ = ¯ ¯ ¯ Z Z Z e±i(t−s)|ξ| |t − s|1−² 2( gbTf )(s, ξ)dsF −1 τ ( bS²bg)(t, ξ) dtdξ ¯ ¯ ¯ ≤ Z k( gbTf )(s)kL2kFτ−1( bS²bg)(t)kL2 |t − s|1−² 2 dsdt ≤Ck gbTf kLr(L2)kFτ−1( bS²bg)kL2(L2) = CkbTf kLr(L2)k bS²bgkL2. (2.30)
This completes the proof of (2.27a). ¤
Lemma 2.6. Assume that 0 < γ < 1, 0 < β < 1
2, 0 < 2β − γ < 1, p = 2 1 − γ, and r = 2 1 + γ − 2β. Then we have kf kLr([0,T ];Lp) ≤ Ck bEγSbβf kb L2, (2.31a) and k bE−γSb−βbgkL2 ≤ CkgkLr0([0,T ];Lp0), (2.31b)
where r0 and p0 are dual exponents of r and p respectively.
Proof. Since the two estimates are dual, hance we only show (2.31a) Let
g be an arbitrary function. First, we compute |hf, gi| =¯¯h bEγSbβf , bb E−γSb−βbgi¯¯ ≤ ° ° ° bEγSbβfb ° ° ° L2 ¯ ¯ ¯hbg, bE−2γSb−2βbgi ¯ ¯ ¯ 1 2 . (2.32)
To prove the estimate (2.31), we employ the dyadic (Paley-Littlewood) decomposition for the function g(t, x). Let us introduce an auxiliary func-tion bβ(ξ) which is supported in the region {1
2 < |ξ| < 2}. Denote b βk(ξ) := bβ( ξ 2k), bη = 1 − ∞ X k=1 b βk, ϕ := bb β−1+ bβ0+ bβ1, (2.33a) b g0 := bηbg, ϕb0(ξ) := bη(ξ 2) , (2.33b) b gk := bβk(ξ)bg, ϕbk(ξ) := bϕ( ξ 2k), k = 1, 2, 3, · · ·. (2.33c)
Notice that bgk(τ, ξ) = bϕk(ξ)bgk(τ, ξ). Through some analysis, we have
Fτ−1¡Eb−2γSb−2β¢(t, ξ) ∼ C
|t|1−2β
e±it|ξ|
(|ξ| + 1)2γ`(t), (2.34a)
where |`(t)| ≤ C, and a classic estimate gives ¯ ¯ ¯Fξ−1(e±it|ξ|ϕ)(t, x)b ¯ ¯ ¯ ≤ C(ϕ) |t| (2.34b) Let us write K(t)gk(x) = F−1 ³ b E−2γSb−2βbgk ´ (t, x). (2.35)
Thus the estimates (2.34a) and (2.34b) imply, for k = 0, 1, 2, · · ·,
kK(t)gkkL∞ ≤ Z 22(1−γ)k |t − s|2(1−β)kgk(s)kL1ds, (2.36) kK(t)gkkL2 ≤ Z 2−2γk |t − s|1−2βkgk(s)kL2ds. (2.37)
The proof of (2.37) is straight forward and we will show (2.36) at the end of the paper. Assuming the validity of the above two estimates, the interpolation between them gives
kK(t)gkkLp ≤
Z
C
where p = 2 1 − γ.
Now we invoke (2.38) and (2.26) to compute ¯ ¯ ¯hbgk, bE−2γSb−2βgbki ¯ ¯ ¯ = ¯ ¯ ¯ Z gk(t, x)F−1 ³ b E−2γSb−2βbg k ´ (t, x)dtdx ¯ ¯ ¯ ≤ ¯ ¯ ¯ Z kgk(t)kLp0kK(t)gkkLpdt ¯ ¯ ¯ ≤ Z kgk(t)kLp0kgk(s)kLp0 |t − s|1+γ−2β dsdt ≤ kgkk 2 Lr0(Lp0). (2.39) Finally we obtain |hf, gi| ≤ ° ° ° bEγSbβfb ° ° ° L2 ³X∞ k=0 ¯ ¯ ¯hbgk, bE−2γSb−2βbgki ¯ ¯ ¯ ´1 2 ≤ ° ° ° bEγSbβfb ° ° ° L2 ³X∞ k=0 kgkk2Lr0(Lp0) ´1 2 ≤ ° ° ° bEγSbβfb ° ° ° L2kgkLr0(Bp0,20 ). (2.40)
This completes the proof. ¤
Remark. Through an analogous argument, we can actually obtain a better estimate
kf kLr([0,T ];Lp) ≤ Ck cMγSbβf kb L2,
where r, p, γ, and β satisfy the same conditions as those in Lemma 2.6. 3. Local Existence.
Now we are ready to prove the local existence for the (DKG) equations. Proof of Theorem 0.1. Consider a Picard iteration scheme for the Dirac equation
Dψk+1 = bTφkψk; (3.1a) ψk+1(0, x) := ψ0(x), (3.1b)
and for the Klein-Gordon equation
¤φk+1 = T (ψk+1ψk+1); (3.1c)
φk+1(0, x) = φ0(x), φk+1,t (0, x) = φ1(x), (3.1d)
where T (ψψ) is defined in (2.13a-d), which is a localization of ψψ. Iteration scheme introduces a map T1 defined by
T1(φk, ψk) = (φk+1, ψk+1). (3.2a)
We want to show that T1 is a contraction under the norm N (φ, ψ) =°° bE1−νSb1+2δ2 φb°° L2 + ° ° bEµSb1+2δ 2 ψb°° L2 (3.2b)
and the condition 0 < δ < ². For convenience, we call
J(0) = kφ0kH1 + kφ1kL2 + kψ0k2Hµ + 1. (3.3)
First we employ (2.31a) to obtain the following estimates.
k∂αφkLr1(Lp1) ≤ k bEγSbβMcαφkb L2 ≤ ° ° bE1−νSb1+2δ 2 φb ° ° L2, (3.4a) where r1 = 8 3 − 4² and p1 = 8 1 + 4². kψkLr2(Lp2)≤ k bEγSbβψkb L2 ≤ ° ° bEµSb1+2δ2 ψb°° L2, (3.4b) where r2 = 8 1 + 4² and p2 = 8 3 − 4². kφkLr3(Lp3) ≤ k bEγSbβφkb L2 ≤ ° ° bE1−νSb1+2δ 2 φb ° ° L2, (3.4c) where r3 = 2 1 − ν and p3 = 2 ν. k∂αψkLr4(Lp4) ≤ k bEγSbβMcαφkb L2 ≤ ° ° bEµSb1+2δ2 φb°° L2, (3.4d) where r4 = 2 ν and p4 = 2 1 − ν.
we apply (2.14), (2.12), and (2.27) to compute N (T1φ) = ° ° bE1−νSb1+2δ2 Td 1φ ° ° L2 ≤ C ³ J(0) + ° ° ° [ T ψψ b EνSb1−2δ2 ° ° ° L2 ´ ≤ C³J(0) + ° ° °Mc α\b Tφψ b Sδ2 ° ° °2 L2 ´ ≤ C³J(0) + ° ° ° cMαφψf ° ° °2 L2+δ4 ([0,T ];L2) ´ . (3.5)
To bound the term above, we observe that c
M (ξ + η) ≤ cM (ξ) + cM (η), (3.6) which implies that
c
Mαφψ ≤c ¡Mcαφb¢∗ bψ + bφ ∗¡Mcαψb¢. (3.7) For convenience, we use the notation
∂α
xf := Fξ−1
¡ c
Mαfe¢. (3.8)
Then we invoke (3.4a-d) to obtain ° ° ° cMαφψf ° ° ° L2+δ4 ([0,T ];L2) ≤ ° ° °¡∂xαφ¢ψ ° ° ° L2+δ4 ([0,T ];L2)+ ° ° °φ¡∂xαψ¢°°° L2+δ4 ([0,T ];L2) ≤T δ8 ° °∂α xφ ° ° Lr1(Lp1) ° °ψ°° Lr2(Lp2)+ T δ 4 ° °φ°° Lr3(Lp3) ° °∂α xψ ° ° Lr4(Lp4) ≤T δ8 ° ° bE1−νSb1+2δ 2 φb ° ° L2 ° ° bEµSb1+2δ 2 ψb ° ° L2. (3.9)
Hence, using (3.5) and (3.9), we have
N (T1(φ)) ≤ C
¡
For the spinor field ψ, we first invoke (2.27b), (2.10), and (2.12) to derive kDφkL∞([0,T ];L2) ≤ k bS 1+2δ 2 Dφkc L2 ≤ kDφ(0)kL2 + ° ° ° c ψψ b S1−2δ2 ° ° ° L2 ≤kφ0kH1 + kφ1kL2 + kψ0k2Hα + ° ° °Mc αφψc b Sδ2 ° ° °2 L2 ≤J(0) + T δ4N4(φ, ψ) (3.11)
Next we apply (3.4b) and (2.27) to compute
k(Dφ)ψkLr0([0,T ];Lp0)≤kψkLr2(Lp2)kDφk L1−2δ2 (L2) ≤T 1−2δ2 N (ψ)kDφk L∞([0,T ];L2) (3.12) and kφ2ψkLr0([0;T ];Lp0) ≤ kψkLr2(Lp2)kφk2 L1−2δ4 (L4) ≤ T δN3(φ, ψ), (3.13) where r0 = 8 5 + 4² − 8δ and p 0 = 8 7 − 4².
Finally we employ (2.25b) and (2.31b) to get
N (T1ψ) = ° ° bEµSb1+2δ2 Td 1ψ ° ° L2 ≤C(T )³kψ0kHµ + kφ0kH1kψ0kHµ + ° ° °(Dφ)ψ + d\ φ2ψ b E1−µSb1−2δ2 ° ° ° L2 ´ ≤J(0) + k(Dφ)ψ + φ2ψkLr0([0;T ];Lp0) ≤C(T )¡J(0) + TδN5(φ, ψ)¢ (3.14)
Thus we combine the above results to get
N (T1(ψ, φ)) ≤ C(T )
¡
J(0) + Tδ4N5(φ, ψ)¢.
Choosing sufficiently large L, for suitable T , we have
N (ψ, φ) ≤ L =⇒ N¡T1(ψ, φ)
¢
provided that
C(J(0) + Tδ4L5) ≤ L. (3.16)
Now we consider the difference T1(ψ, φ) − T1(ψ0, φ0). Base on the
ob-servations
2(ψψ − ψ0ψ0) = (ψ − ψ0)(ψ + ψ0) + (ψ + ψ0)(ψ − ψ0), (3.17a)
2(φψ − φ0ψ0) = (φ − φ0)(ψ + ψ0) + (φ + φ0)(ψ − ψ0), (3.17b) Employing (2.14), (2.12), and (3.17), we first calculate
N (T1φ − T1φ0) = ° ° bE1−νSb1+2δ 2 F(T1φ − T1φ0) ° ° L2 ≤ C ³°° °F(T (ψ − ψ 0)(ψ + ψ0)) b EνSb1−2δ2 ° ° ° L2 + ° ° °F(T (ψ + ψ 0)(ψ − ψ0)) b EνSb1−2δ2 ° ° ° L2 ´ ≤ C ³° ° °Mc αF(b T(φ − φ0)(ψ + ψ0)) b Sδ2 ° ° ° L2+ ° ° °Mc αF(b T(φ + φ0)(ψ − ψ0)) b Sδ2 ° ° ° L2 ´ · · ³ J(0) + ° ° °Mc αF(b T(φψ + φ0ψ0)) b Sδ2 ° ° ° L2 ´ ≤ CTδ8¡k bE1−νSb 1+2δ 2 φ − φ\0kL2 + k bEµSb 1+2δ 2 ψ − ψ\0kL2 ¢ L(J(0) + Tδ8L2) ≤ CTδ8L3¡N (φ − φ0) + N (ψ − ψ0)¢ (3.18) Analogously, we get N (T1ψ − T1ψ0) = ° ° bEµSb1+2δ 2 F(T1ψ − T1ψ0) ° ° L2 ≤ CTδL4 ³ k bE1−νSb1+2δ2 φ − φ\0k L2 + k bEµSb 1+2δ 2 ψ − ψ\0k L2 ´ . (3.19)
Combining (3.18)and (3.19), we have
N¡T1(ψ − ψ0, φ − φ0)
¢
≤ CTδ8L4N (ψ − ψ0, φ − φ0). (3.20)
Therefore for suitable T , we obtain
N¡T1(ψ − ψ0, φ − φ0) ¢ ≤ 1 2N (ψ − ψ 0, φ − φ0), (3.21) provided that CTδ8L4 ≤ 1 2. (3.22)
We can conclude that the map T1 is indeed a contraction with respect to
4. Null Form Estimate.
In this section, we demonstrate the proof of the key estimate.
Lemma 2.2. ( Null Form Estimate) Let δ > 0, and ψ1, ψ2 be the solu-tions for (2.11). If the initial data ψ0j ∈ Hα, j = 1, 2, then we have
° ° °Eb δT (ψ\ 1ψ2) b S12 ° ° ° L2 ≤ C(T ) ³ kψ01kHα+ ° ° °Mc αGb 1 b Sδ2 ° ° ° L2 ´³ kψ02kHα+ ° ° °Mc αGb 2 b Sδ2 ° ° ° L2 ´ . (4.1)
The proof for the estimate is based on the duality argument and it will be given in a number of steps. Without loss of generality, we assume that
ψ1 = ψ2, and prove the following case:
° ° °Eb δT (ψψ)\ b S12 ° ° ° L2 ≤ C(T ) ³ kψ0kHα + ° ° °Mc αGb b Sδ2 ° ° ° L2 ´2 . (4.2)
Recall that the notations: b E(τ, ξ) := |τ | + |ξ| + 1, S(τ, ξ) :=b ¯ ¯ ¯|τ | − |ξ| ¯ ¯ ¯ + 1, (4.3a) c W (τ, ξ) := τ2 − |ξ|2, D(τ, ξ) := γb 0τ + γ1ξ, (4.3b) b D+ := bD(|ξ|, +ξ), Db− := bD(|ξ|, −ξ). (4.3c)
The formula for bψ, as in (1.8), for the Dirac equation (1.2) is given by
b ψ(τ, ξ) = ∞ X k=0 ³ δ+(k)(τ, ξ) bA+,k(ξ) + δ−(k)(τ, ξ) bA−,k(ξ) ´ + bK(τ, ξ), (4.4) where δ±(τ, ξ) are the delta functions supported on {τ = ±|ξ|}
respec-tively, δ(k) mean derivatives of the delta function, and b K(τ, ξ) := D(τ, ξ)b c W (τ, ξ) b Gf+ (1−ba6(τ )) bD+Gb− 2|ξ|(τ − |ξ|) + (1−ba6) bD−Gb+ 2|ξ|(τ + |ξ|) , (4.5a) b A±,0(ξ) := b D± 2|ξ| h γ0ψb0 − Z bGf + (1 − ba6(λ)) bG∓ λ ∓ |ξ| dλ i , (4.5b) b A±,k(ξ) := b D±(−1)k 2|ξ|k! Z (λ ∓ |ξ|)k−1£Gb±+ ba6(λ) bG∓ ¤ dλ. (4.5c)
Moreover we write b A±,k(ξ) := b D± 2|ξ|fb±,k(ξ), (4.6) and split bK = bK1+ bK2, where
b K1 := b D(τ, ξ) c W (τ, ξ) b Gf; Kb2 := b1 b D+Gb−+ b2Db−Gb+ b E bS , (4.7)
and b1, b2 are bounded functions. T (ψψ) is the sum of the following\
terms. ¡ δ∓Ab±,0 ¢ ∗¡δ±Ab±,0 ¢ + X k+l>0 bbT ∗ ¡ δ(k)∓ Ab±,k ¢ ∗¡δ(l)± Ab±,l ¢ , (4.8a) ¡ δ∓Ab±,0 ¢ ∗¡δ∓Ab∓,0 ¢ + X k+l>0 bbT ∗ ¡ δ(k)∓ Ab±,k ¢ ∗¡δ(l)∓ Ab∓,l ¢ , (4.8b) X k bbT ∗ ³¡ δ(k)∓ Ab±,k ¢ ∗¡Kb1 + bK2 ¢ +¡ bK1 + bK2 ¢ ∗¡δ±(k)Ab±,k ¢´ , (4.8c) b K1 ∗ bK1 + bK1∗ bK2+ bK2∗ bK1+ bK2 ∗ bK2. (4.8d) Notice that [ A†±,k(ξ) = bA†±,k(−ξ); fd±,k+ (ξ) = bf±,k† (−ξ), (4.9a) [ A±,k(ξ) = bf±,k† (−ξ) b D± |ξ| γ 0; K(τ, ξ) = bb K†(−τ, −ξ)γ0, (4.9b) and b ψ(τ, ξ) = ∞ X k=0 ³ δ(k)− (τ, ξ)bA+,k(ξ) + δ+(k)(τ, ξ)bA−,k(ξ) ´ + bK(τ, ξ), (4.10)
Lemma 4.1. Let k + l > 0 and δ ≥ 0. The following estimates hold ° ° °Eb 2δ¡δ ∓Ab±,0 ¢ ∗¡δ∓Ab∓,0 ¢ b S12 ° ° ° L2 ≤ Ckf±,0kHαkf∓,0kHα, (4.11a) ° ° °Eb 2δbb T ∗ ¡ δ∓(k)Ab±,k ¢ ∗¡δ∓(l)Ab∓,l ¢ b S12 ° ° ° L2 ≤ C(k + l)Tk+l−12kf±,kkHαkf∓,lkHα. (4.11b)
Proof. We only demonstrate the case (4.11b) since the proof for (4.11a) is analogous. Let us call
b
Z±,k ≡ δ±(k)Ab±,k = δ±(k)
b
D±
2|ξ|fb±,k. (4.12) Using duality, we demonstrate the case (−, +), while the case (+, −) is being similar. We first compute the following term
b D−(ξ)γ0Db+(η) = γ0(|ξ||η| − ξ · η) + γ0γjγkξjηk ¯ ¯ ¯ j6=k− γ j(|η|ξ j − |ξ|ηj). (4.13) Thus ¯ ¯ ¯bTZ−,kZ+,l, g® ¯¯¯ = ¯ ¯ ¯ Z b f−,k† (−ξ)D(|ξ|, −ξ)γb 0D(|η|, η)b |ξ||η| fb+,l(η) \tk+lbTg(|ξ| + |η|, ξ + η) dξdη ¯ ¯ ¯ ≤kf−,kkHαkf+,lkHα ³Z |ξ||η| − ξ · η |ξ|1+2α|η|1+2α ¯ ¯ \tk+lb Tg(|ξ| + |η|, ξ + η) ¯ ¯2 dξdη ´1 2 . (4.14) To estimate the above integral, we change the variables by
|ξ| + |η| = τ, ξ + η = z, ξ = ρω, |ω| = 1, (4.15) thus we can rewrite it as
Z E(τ, z)¯¯ \tk+lb Tg(τ, z) ¯ ¯2 dτ dz, (4.16) where E(τ, z) = Z |ξ||η| − ξ · η |ξ|1+2α|η|1+2αρ 2dρ dτdω. (4.17)
Throughout some computations, we have
E(τ, z) ≤ Cτ − |z| τ4δ , (4.18) ° °(|τ| + 1)1 2t\k+lbT°° L1 ≤ C(k + l)Tk+l− 1 2kbk H1. (4.19)
With the aid of the above inequalities and the observation ¯ ¯|τ| − |ξ|¯¯ + 1 (|τ | + |ξ| + 1)4δ ≤ C ¯ ¯|τ − σ| − |ξ|¯¯ + 1 (|τ − σ| + |ξ| + 1)4δ ¡ |σ| + 1¢1−4δ, (4.20) we can estimate k Sb 1 2 b E2δ \ tk+lb TgkL2 ≤ ° °(|τ| + 1)1 2t\k+lbT ° ° L1 ° ° ° Sb 1 2 b E2δbg ° ° ° L2 ≤C(k + l)Tk+l−12 ° ° ° Sb 1 2 b E2δbg ° ° ° L2. (4.21)
This completes the proof. ¤
Lemma 4.2. Let k + l > 0 and δ > 0. The following estimates hold ° ° °Eb 2δ¡δ ∓Ab±,0 ¢ ∗¡δ±Ab±,0 ¢ b S12 ° ° ° L2 ≤ Ckf±,0kH αkf±,0kHα, (4.22a) ° ° °Eb 2δbb T ∗ ¡ δ∓(k)Ab±,k ¢ ∗¡δ±(l)Ab±,l ¢ b S12 ° ° ° L2 ≤ C(k + l)Tk+l−12kf±,kkHαkf±,lkHα. (4.22b)
Proof. We only demonstrate the case (4.22b) since the proof for (4.22a) is analogous. Using duality, we demonstrate the case (+, +), while the case (−, −) is being similar. We first compute the fractional term
b D+(ξ)γ0Db+(η) = γ0(|ξ||η| + ξ · η) − γ0γjγkξjηk ¯ ¯ ¯ j6=k− γ j(|η|ξ j + |ξ|ηj). (4.23) Thus, in the same manner we have
¯ ¯ ¯bTZ+,kZ+,l, g® ¯¯¯ = ¯ ¯ ¯ Z b f+,k† (−ξ)D(|ξ|, ξ)γb 0D(|η|, η)b |ξ||η| fb+,l(η) \tk+lbTg(−|ξ| + |η|, ξ + η) dξdη ¯ ¯ ¯ ≤kf+,kkHαkf+,lkHα· ³ Z |ξ||η| + ξ · η |ξ|1+2α|η|1+2α ¯ ¯ \tk+lb Tg(−|ξ| + |η|, ξ + η) ¯ ¯2 dξdη ´1 2 . (4.24)
To estimate the above integral, we change the variables by
|ξ| − |η| = τ, ξ + η = z, ξ = ρω, |ω| = 1, (2.25) thus we can rewrite it as
Z H(τ, z)¯¯ \tk+lb Tg(−τ, z) ¯ ¯2 dτ dz, (2.26) where H(τ, z) = Z |ξ||η| + ξ · η |ξ|1+2α|η|1+2αρ 2dρ dτdω. (2.27)
Throughout some computations, we have
H(τ, z) ≤ C|z| − τ |z|4δ , (4.28) ° °(|τ| + 1)1 2t\k+lbT ° ° L1 ≤ C(k + l)T k+l−1 2kbkH1. (4.29)
With the aid of the above inequalities and the observation ¯ ¯|τ| − |ξ|¯¯ + 1 (|τ | + |ξ| + 1)4δ ≤ C ¯ ¯|τ − σ| − |ξ|¯¯ + 1 (|τ − σ| + |ξ| + 1)4δ ¡ |σ| + 1¢1−4δ, (4.30) we can estimate k Sb 1 2 b E2δ \ tk+lb TgkL2 ≤ ° °(|τ| + 1)1 2t\k+lbT°° L1 ° ° ° Sb 1 2 b E2δbg ° ° ° L2 ≤C(k + l)Tk+l−12 ° ° ° Sb 1 2 b E2δbg ° ° ° L2. (4.31)
This completes the proof. ¤
Remark. Invoke the dyadic decomposition, in fact, we can show that the
estimates (4.22) hold for δ = 0.
Lemma 4.3. Let δ1 > 0. The following estimates hold kf±,0kHα ≤ C ³ kψ0kHα + ° ° °Mc αGb b S12−δ1 ° ° ° L2 ´ , (4.32a) kf±,kkHα ≤ C2 k k! ³ k cMαGb+kL2 + k cMαGb−kL2 ´ . (4.32b) The proof for the Lemma 4.3 is straight forward so that we skip it. Notice that, in the (4.32b), bS ∼ 1 on the support of bG±.
Lemma 4.4. With the notation above, the following estimate holds ° ° °Eb δKb 1∗ bK1 b S12 ° ° ° L2 ≤ C ° ° °Mc αGb f b Sδ2 ° ° °2 L2. (4.33)
Proof. For simplicity, we write bG := bGf and bK := bK1. We use dyadic
decomposition to handle this case. Assume that
b G = ∞ X k=0 b G±,±,k, (4.34)
where bG±,±,k(τ, ξ) is supported in one of the following types of regions:
Σ+,+ := {(τ, ξ) : τ > 0, +2k−1 < τ − |ξ| < +2k+1}, (4.35a)
Σ+,− := {(τ, ξ) : τ > 0, −2k+1 < τ − |ξ| < −2k−1}, (4.35b)
Σ−,+ := {(τ, ξ) : τ < 0, +2k−1 < τ + |ξ| < +2k+1}, (4.35c)
Σ−,− := {(τ, ξ) : τ < 0, −2k+1 < τ + |ξ| < −2k−1}. (4.35d)
They are forward and backward cones with thickness about 2k, also half of them are truncated. The decomposition of bG induces a decomposition
for bK, namely b K±,±,k = b D c W b G±,±,k. (4.36)
Let g be an arbitrary function. we compute D b Kk∗ bKl, bg E = Z b K±,±,k ∗ bK±,±,l(−τ, −ξ)bg(−τ, −ξ)dτ dξ = Z b K±,±,k(−τ − σ, −ξ − η) bK±,±,l(σ, η)bg(−τ, −ξ) dσdηdτ dξ = Z b K±,±,k† (τ + σ, ξ + η)γ0Kb±,±,l(σ, η)bg(−τ, −ξ) dσdηdτ dξ, (4.37)
we have 16 cases resulted from (4.35a-d) and (4.36b). Due to the support of the integrand, the integral is integrated over one of the sets of (τ, σ, ξ, η):
{τ + σ > 0, σ > 0, τ + σ − |ξ + η| ∼ ±2k, σ − |η| ∼ ±2l}, (4.38a) {τ + σ < 0, σ < 0, τ + σ + |ξ + η| ∼ ±2k, σ + |η| ∼ ±2l}, (4.38b) {τ + σ < 0, σ > 0, τ + σ + |ξ + η| ∼ ±2k, σ − |η| ∼ ±2l}, (4.38c) {τ + σ > 0, σ < 0, τ + σ − |ξ + η| ∼ ±2k, σ + |η| ∼ ±2l}. (4.38d) We label them as Σk,l[(±, ±); (±, ±)], (4.39)
and denote by Σk,l without specifying which one precisely. We also use
b
Kk for abbreviation of bK±,±,k and bGk for bG±,±,k .
We first compute b D(τ + σ, −ξ − η)γ0D(σ, η)b = £γ0(τ + σ) − γj(ξj + ηj) ¤ γ0£γ0σ + γjηj ¤ = γ0£(τ + σ)σ − (ξ + η) · η¤+ γ0γjγk(ξj + ηj)ηk ¯ ¯ ¯ j6=k + γj£(τ + σ)ηj − σ(ξj + ηj) ¤ . (4.40) Thus, we have ¯ ¯ ¯ D b Kk∗ bKl, bg E ¯ ¯ ¯ = ¯ ¯ ¯ Z b G†k(τ + σ, ξ + η)γ 0(τ + σ) − γj(ξ j + ηj) (τ + σ)2− (ξ + η)2 γ 0γ0σ + γjηj σ2 − η2 Gbl(σ, η)· b g(−τ, −ξ)dσdηdτ dξ ¯ ¯ ¯ ≤Ck cMαGbkkL2k cMαGblkL2 ³ Z Ik,l(τ, ξ) ¯ ¯bg(−τ, −ξ)¯¯2 dτ dξ ´1 2 , (4.41) where Ik,l(τ, ξ) is given by Ik,l(τ, ξ) := Z Dk,l c M−2α(ξ + η) cM−2α(η)Q(τ, σ, ξ, η) c W2(τ + σ, ξ + η)cW2(σ, η) dσdη, (4.42)
and Q is given by the expression
Q(τ, σ, ξ, η) :=
£
(τ + σ)σ − (ξ + η) · η¤2+¯¯(ξ + η) × η¯¯2 +¯¯(τ + σ)η − σ(ξ + η)¯¯2,
(4.43) and Dk,l(τ, ξ) is a slice of Σk,l for fixed (τ, ξ), i.e.
Dk,l(τ, ξ) := {(σ, η) : (τ, σ, ξ, η) ∈ Σk,l}. (4.44)
They are the intersections of cones with thickness 2k and cones with thickness 2l.
We distinguish the cases into two sets,
Σk,l[(±, ·); (±, ·)] and Σk,l[(±, ·); (∓, ·)], (4.45)
due to the fact that for the first set it is the intersection of two forward cones or two backward cones; for the second set it is the intersection of a forward cone and a backward cone. Therefore presumably the computa-tions for the 8 cases in each set are similar.
Cases H. We have the following estimate ° ° °Eb δKb +,·,k ∗ bK+,·,l b S12 ° ° ° L2 ≤ C 2δ2(k+l) ° ° °Mc αGb +,·,k b Sδ2 ° ° ° L2 ° ° °Mc αGb +,·,l b Sδ2 ° ° ° L2, ° ° °Eb δKb −,·,k ∗ bK−,·,l b S12 ° ° ° L2≤ C 2δ2(k+l) ° ° °Mc αGb −,·,k b Sδ2 ° ° ° L2 ° ° °Mc αGb −,·,l b Sδ2 ° ° ° L2. (4.46) In these cases, we have (τ + σ)σ > 0. Take the case
b
K+,+,k∗ bK+,+,l, (4.47)
as an example and in which Dk,l = {(σ, η) : τ + σ > 0, σ > 0, τ + σ − |ξ + η| ∼ 2k, σ − |η| ∼ 2l, (τ, σ, ξ, η) ∈ Σk,l[(+, +); (+, +)]}. In the
ησ-space, this is the region of the intersection of two forward cones, which is unbounded.
To manage the integral, we change the variables:
h = |ξ + η| − |η|, 2X = h cosh ζ, (4.48a)
2Y =p|ξ|2− h2sinh ζ cos ϕ, 2Z =p|ξ|2 − h2sinh ζ sin ϕ. (4.48b)
This is due to the fact that for fixed ξ and h, η stays on an hyperboloid. The coordinates (X, Y, Z) identity the vector η as follows η = (X, Y, Z) − (|ξ|/2, 0, 0) and ξ + η = (X, Y, Z) + (|ξ|/2, 0, 0). Hence dη = dXdY dZ =
Jdhdαdϕ, where J is the Jacobian given by
8J = 8∂(X, Y, Z)
∂(h, ζ, ϕ) =
¡
|ξ|2cosh2ζ − h2¢| sinh ζ| = 4|ξ + η||η|| sinh ζ|.
(4.49) Throughout some algebraic manipulations, the Q can be rewritten as
2Q =(τ + σ − |ξ + η|)2(σ + |η|)2+ (τ + σ + |ξ + η|)2(σ − |η|)2+
4(τ + σ)σ(|ξ|2− h2). (4.50)
Now the integral Ik,l can be split into three parts according to (4.50). To
estimate each term separately, we consider a further dyadic decomposition b
Gk,m(τ, ξ) is supported in {τ > 0, τ − |ξ| ∼ 2k, |ξ| ∼ 2m},
b
Gl,n(τ, ξ) is supported in {τ > 0, τ − |ξ| ∼ 2l, |ξ| ∼ 2n}.
Therefore the integral in (4.29b) is over the region Dk,lm,n(τ, ξ) which is the set of (σ, η):
{τ + σ > 0, τ > 0, τ + σ − |ξ + η| ∼ 2k, σ − |η| ∼ 2l, |ξ + η| ∼ 2m, |η| ∼ 2n}.
(4.51) For simplicity, we will assume k ≥ l, m ≥ n, and n > k while the other case is similar.
Claim. The size of the set Dm,nk,l is bounded as follows
¯ ¯Dm,n k,l ¯ ¯ = Z Dm,nk,l dσdη ≤ C22n2k+l. (4.52)
Also E(τ, ξ) ≥ 2b k for k À l and
Z
Dk,lm,n
|ξ|2− h2dσdη ≤ C22k+l2m+2nS.b (4.53)
The proof of the claim will be given in the appendix. Assuming the claim, we can estimate the first part:
Ik,l,m,n1 (τ, ξ) := Z Dk,lm,n c M−2α(ξ + η) cM−2α(η)(τ + σ − |ξ + η|)2(σ + |η|)2 c W2(τ + σ, ξ + η)cW2(σ, η) dσdη = Z Dk,lm,n c M−2α(ξ + η) cM−2α(η) (τ + σ + |ξ + η|)2(σ − |η|)2dσdη ≤ C 22l Z Dk,lm,n (|ξ + η| + 1)−2α(|η| + 1)−2α (2k+ |ξ + η|)2 dσdη ≤ C 22l 2−2αn 2(2+2α)m Z Dm,nk,l dσdη ≤ C 22l 2−2αn 2(2+2α)m2 2n2k+l ≤ C 2−k+l 2(2−2α)n 2(2+2α)m b S22δm b E2δ ≤ C 2−k+l 1 2(2−2α)(m−n)24αm−2δm b S b E2δ ≤ C 22δm+l 1 2(2−2α)(m−n)2m−k b S b E2δ. (4.54a)
For the second part, we obtain
Ik,l,m,n2 (τ, ξ) := Z Dk,lm,n c M−2α(ξ + η) cM−2α(η)(τ + σ + |ξ + η|)2(σ − |η|)2 c W2(τ + σ, ξ + η)cW2(σ, η) dσdη = Z Dk,lm,n c M−2α(ξ + η) cM−2α(η) (τ + σ − |ξ + η|)2(σ + |η|)2dσdη ≤ C 22k 2−2αm 2(2+2α)n Z Dk,lm,n dσdη ≤ C 22k 2−2αm 2(2+2α)n2 2n2k+l ≤ C 2k−l 1 22α(m+n) b S22δm b E2δ ≤ C 2k−l 1 212m+2αn b S b E2δ. (4.54b)
For the third part, we get Ik,l,m,n3 (τ, ξ) := Z Dk,lm,n c M−2α(ξ + η) cM−2α(η)(τ + σ)σ¡|ξ|2− h2¢ c W2(τ + σ, ξ + η)cW2(σ, η) dσdη ≤ C 22k+2l Z Dm,nk,l c M−2α(ξ + η) cM−2α(η)(τ + σ)σ¡|ξ|2 − h2¢ (τ + σ + |ξ + η|)2(σ + |η|)2 dσdη ≤ C 22k+2l Z Dm,nk,l |ξ|2− h2 (|ξ + η| + 1)1+2α(|η| + 1)1+2α dσdη ≤ C 22k+2l 1 2(1+2α)(m+n) Z Dm,nk,l |ξ|2 − h2 dσdη ≤ C 22k+2l 1 2(1+2α)(m+n) 2 2k+l2m+2nSb22δm b E2δ ≤C 2l 1 212(m−n)+2δn b S b E2δ. (4.54c)
Combine the above results , we have ° ° °Eb δKb +,+,k∗ bK+,+,l b S12 ° ° ° L2 ≤ X m,n ° ° °Eb δKb +,+,k,m∗ bK+,+,l,n b S12 ° ° ° L2 ≤X m,n 1 214|m−n| C 2δ2(k+l) ° ° °Mc αGb +,+,k,m b Sδ2 ° ° ° L2 ° ° °Mc αGb +,+,l,n b Sδ2 ° ° ° L2 ≤ C 2δ2(k+l) ° ° °Mc αGb +,+,k b Sδ2 ° ° ° L2 ° ° °Mc αGb +,+,l b Sδ2 ° ° ° L2. (4.55)
Cases E. We have the following estimate ° ° °Eb δKb −,·,k ∗ bK+,·,l b S12 ° ° ° L2 ≤ C 2δ2(k+l) ° ° °Mc αGb −,·,k b Sδ2 ° ° ° L2 ° ° °Mc αGb +,·,l b Sδ2 ° ° ° L2, ° ° °Eb δKb +,·,k∗ bK−,·,l b Sα ° ° ° L2≤ C 2δ2(k+l) ° ° °Mc αGb +,·,k b Sδ2 ° ° ° L2 ° ° °Mc αGb −,·,l b Sδ2 ° ° ° L2. (4.56) In these cases, we have (τ + σ)σ < 0. We will only demonstrate the case of
b
and in which Dk,l = {(σ, η) : τ + σ + |ξ + η| ∼ 2k, σ − |η| ∼ 2l, (τ, σ, ξ, η) ∈
Σk,l[(−, +); (+, +)]}. In this case τ + σ < 0 and σ > 0. In ησ-space,
this is the region of the intersection of a forward cone with a truncated backward cone, which is bounded.
To manage the integral, we change the variables:
e = |ξ + η| + |η|, 2X = e cos θ, (4.58a)
2Y = pe2 − |ξ|2sin θ cos ϕ, 2Z =pe2− |ξ|2sin θ sin ϕ. (4.58b)
This is due to the fact that for fixed ξ and h, η stays on an ellipsoid. The coordinates (X, Y, Z) identity the vector η as follows η = (X, Y, Z) − (|ξ|/2, 0, 0) and ξ + η = (X, Y, Z) + (|ξ|/2, 0, 0). Hence dη = dXdY dZ =
Jdedθdϕ, where J is the Jacobian given by
8J = 8∂(X, Y, Z)
∂(e, θ, ϕ) =
¡
e2 − |ξ|2cos2θ¢| sin θ| = 4|ξ + η||η|| sin θ|. (4.59)
Throughout some algebraic manipulations, the Q can be rewritten as 2Q =(τ + σ + |ξ + η|)2(σ + |η|)2+ (τ + σ − |ξ + η|)2(σ − |η|)2+
4(τ + σ)σ(e2 − |ξ|2). (4.60)
Now the integral Ik,l can be split into three parts according to (4.60). For
simplicity, we will assume k ≥ l, while the other case is similar. Claim. Z dσ ∼ 2l, σ ≥ 2l, |ξ + η| ≥ 2k, E(τ, ξ) ≥ 2b k, (4.61) Z π 0 (e − |ξ| cos θ)1−2αsin θ (e + |ξ| cos θ)1+2α dθ ≤ C (e2− |ξ|2)2α, (4.62) Z −τ +2k+1−2l−1 −τ +2k−1−2l+1 1 (e2− |ξ|2)2αde ≤ C 22δkEb2δ, (4.63) Z π 0 (e + |ξ| cos θ)1−2α (e − |ξ| cos θ)2α sin θ dθ ≤ C e4α−1. (4.64)
The proof of the claim will be given in the appendix. Now for the first part, we can estimate
Ik,l1 (τ, ξ) := Z Dk,l c M−2α(ξ + η) cM−2α(η)(τ + σ + |ξ + η|)2(σ + |η|)2 c W2(τ + σ, ξ + η)cW2(σ, η) dσdη ≤ 1 22l Z Dk,l (|ξ + η| + 1)−2α(|η| + 1)−2α (τ + σ − |ξ + η|)2 dσdη ≤ 1 22l Z Dk,l 1 |ξ + η|2+2α(|η| + 1)2αdσdη ≤C 2l Z |ξ + η||η|| sin θ| |ξ + η|2+2α(|η| + 1)2αdϕdθde ≤C 2l Z (e − |ξ| cos θ)1−2α| sin θ| (e + |ξ| cos θ)1+2α dθde ≤C 2l Z −τ +2k+1−2l−1 −τ +2k−1−2l+1 1 (e2− |ξ|2)2αde ≤ C 2l22δkEb2δ. (4.65a) The argument for the possible cases of −τ + 2k−1 − 2l+1 < |ξ|, is not
harder than that in (4.65a). For the second part, we derive
Ik,l2 (τ, ξ) := Z Dk,l c M−2α(ξ + η) cM−2α(η)(τ + σ − |ξ + η|)2(σ − |η|)2 c W2(τ + σ, ξ + η)cW2(σ, η) dσdη ≤ C 22k Z Dk,l (|ξ + η| + 1)−2α(|η| + 1)−2α (σ + |η|)2 dσdη ≤ C 22k−l Z e Dk,l 1 (2l+ |η|)2(|ξ + η| + 1)2α(|η| + 1)2αdη ≤ C 22k Z |ξ + η||η|| sin θ| (2l+ |η|)(|ξ + η| + 1)2α(|η| + 1)2αdϕdθde ≤ C 22k Z |ξ + η|1−2α (|η|)2α | sin θ| dθde ≤ C 22k Z (e + |ξ| cos θ)1−2α
(e − |ξ| cos θ)2α | sin θ| dθde ≤ C 22k Z −τ +2k+1−2l−1 −τ +2k−1−2l+1 1 e4α−1 de ≤ C 2k 1 b E4δ. (4.65b)
Again the argument for the possible cases of −τ + 2k−1 − 2l+1 < |ξ|, is
not harder than that in (4.65b). For the third part, we have
Ik,l3 (τ, ξ) := Z Dk,l c M−2α(ξ + η) cM−2α(η)|τ + σ|σ(e2 − |ξ|2) c W2(τ + σ, ξ + η)cW2(σ, η) dσdη ≤ C 22k+2l Z Dk,l c M−2α(ξ + η) cM−2α(η)(e2 − |ξ|2) |τ + σ − |ξ + η||(σ + |η|) dσdη ≤ C 22k+l Z e Dk,l c M−2α(ξ + η) cM−2α(η)(e2 − |ξ|2) |ξ + η|(2l+ |η|) dη ≤ C 22k+l Z (e2 − |ξ|2)| sin θ| (|ξ + η| + 1)2α(|η| + 1)2α dθde ≤ C 22k+l Z (e2− |ξ|2)| sin θ| (e2− |ξ|2cos2θ)2α dθde ≤ C 22k+l Z −τ +2k+1−2l−1 −τ +2k−1−2l+1 e2 − |ξ|2 e4α de ≤ C 22k+l Z −τ +2k+1−2l−1 −τ +2k−1−2l+1 e − |ξ| e4α−1 de ≤ C 2l 1 22δk b S b E2δ. (4.65c) Now we return to the proof of (4.33). Combine (4.46), (4.56), we get ¯ ¯ ¯KkKl, g® ¯¯¯ ≤C ° ° ° cMαGbk ° ° ° L2 ° ° ° cMαGbl ° ° ° L2 ³ Z Ik,l(τ, ξ) ¯ ¯bg(−τ, −ξ)¯¯2 dτ dξ ´1 2 ≤ C 212l 1 2δk ° ° ° cMαGbk ° ° ° L2 ° ° ° cMαGbl ° ° ° L2k b S12 b EδbgkL2 ≤ C 2(1 2−δ2)l+δ2k ° ° °Mc αGb k b Sδ2 ° ° ° L2 ° ° °Mc αGb l b Sδ2 ° ° ° L2k b S12 b EδbgkL2. (4.66) Finally, we have ° ° °Eb δ b S12 b K ∗ bK ° ° ° L2 ≤ X k,l ° ° °Eb δ b S12 b Kk∗ bKl ° ° ° L2 ≤X k,l C 2(1 2−δ2)l+δ2k ° ° °Mc αGb k b Sδ2 ° ° ° L2 ° ° °Mc αGb l b Sδ2 ° ° ° L2 ≤ C ° ° °Mc αGb b Sδ2 ° ° °2 L2. (4.67)
This completes the proof. ¤
Lemma 4.5. For j = 1, 2 and k = 0, 1, 2, · · ·. The following estimates hold ° ° °Eb δ b S12 bbT ∗ ¡ δ∓(k)Ab±,k ¢ ∗¡Kbj ¢°° ° L2 ≤ C(k + 1)T k−1 2kf±,kkHα ° ° °Mc αGb b Sδ2 ° ° ° L2, (4.68) ° ° °Eb δ b S12 bbTKbj ∗ ¡ δ(k)± Ab±,k ¢°° ° L2 ≤ C(k + 1)T k−1 2kf±,kkHα ° ° °Mc αGb b Sδ2 ° ° ° L2, (4.69) ° ° °Eb δ b S12 bbT ∗ bK1 ∗ bK2 ° ° ° L2 ≤ C ° ° °Mc αGb b Sδ2 ° ° °2 L2, (4.70) ° ° °Eb δ b S12 bbT ∗ bK2∗ bKj ° ° ° L2 ≤ C ° ° °Mc αGb b Sδ2 ° ° °2 L2, (4.71)
The proof of Lemma 4.5 is a repetition of the arguments presented in Lemmas 4.1, 4.2, and 4.4, so that we omit it.
Proof of Claim.
Notice that since h = |ξ + η| − |η| = −(τ + σ − |ξ + η|) + (σ − |η|) + τ, thus we have Proof of (4.61): 2l−1+ |η| < σ < 2l+1+ |η| implies Z dσ ∼ 2l. (4.72) Proof of (4.52): |Dk,lm,n| ≤ Z Dk,lm,n dσdη ≤ 2l Z e Dk,lm,n dη ≤ 2l Z |ξ + η||η|| sinh ζ| dϕdζdh ≤2l2m+n Z Z ζ2 ζ1 | sinh ζ| dζdh ≤2l2m+n Z cosh ζ2− cosh ζ1dh ≤2l2m+n Z τ −2k−1+2l+1 τ −2k+1+2l−1 h + 2|η2| |ξ| − h + 2|η1| |ξ| dh ≤2l2m+n2 n 2m Z τ −2k−1+2l+1 τ −2k+1+2l−1 dh ≤ C2k+l22n. (4.73)
