Dirac-Klein-Gordon Equations in one Space Dimension
Yung-fu Fang
Abstract. We establish local and global existence results for Dirac-Klein-Gordon equations in one space dimension, 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 × R1 7→ C4 and a scalar field φ : R × R1 7→ R. The evolution
equations for the fields are given below,
Dψ = φψ; (t, x) ∈ R × R1 (0.1, a)
¤φ = ψψ; (0.1, b)
ψ(0, x) := ψ0(x), φ(0, x) = φ0(x), φ,t(0, x) = φ1(x), (0.1, c)
where D is the Dirac operator, D := −iγµ∂
µ, µ = 0, 1, and γµ are the
Dirac matrices, the wave operator ¤ = −∂tt+ ∂xx, and ψ = ψ†γ0, and †
is the complex conjugate transpose.
The purpose of this work is to demonstrate the usefulness of a null form estimate, by employing the solution representations in Fourier transform of the DKG equations. We will take advantage of the null form structure depicted in the nonlinear term ψψ, which has been observed by [KM] and [Bo]. We interpret the null form in a way that is different from that given in Bournaveas’ paper [Bo].
For the DKG system, there are many conserved quantities which are not positive definite, such as the energy. Therefore they are not applicable to derive estimates. However the known positive conserved quantity is the law of conservation of charge,
Z
|ψ(t)|2dx = constant (0.2)
which leads to the global existence result, once the local existence result is established, see [Bo] and [F2].
In ’73, Chadam showed that the Cauchy problem for the DKG equa-tions has a global unique solution for ψ0 ∈ H1, φ0 ∈ H1, φ1 ∈ L2, see
[C]. In ’93, Zheng proved that there exists a global weak solution to the Cauchy problem of a modified DKG equations, based on the technique of compensated compactness, with ψ0 ∈ L2, φ0 ∈ H1, φ1 ∈ L2, see [Z]. In
’00, Bournaveas derived a new proof of a global existence for the DKG equations, via a null form estimate, if ψ0 ∈ L2, φ0 ∈ H1, φ1 ∈ L2, see
[B]. In ’02, Fang gave a direct proof for (0.1), based on a variant null form estimate, which is more straight forward, and the result is parallel to Bournaveas’, see [F2]
The outline of this paper is as follows. First we derive some solu-tions representasolu-tions in Fourier transform, depending on various purposes. Next we prove some a priori estimates of solutions for Dirac equation and for wave equation. Then we show a local and global results for (0.1), employing the null form estimate together with other estimates derived previously, 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 0 < ² ≤ 1
4 and 0 < δ ≤ 2². If the
initial data of (0.1) ψ0 ∈ H− 1
4+², φ0 ∈ H12+δ, φ1 ∈ H−12+δ, then there is a unique local solution for (0.1).
Theorem 0.2. (Global Existence) Let δ > 0. If the initial data of (0.1)
ψ0 ∈ L2, φ0 ∈ H 1
2+δ, φ1 ∈ H−12+δ, then there is a unique global solution for (0.1).
Remarks.
1. The DKG equations follow from the Lagrangian Z R1+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)
3. bD2 = b¤I, where I is the 4 × 4 identity matrix.
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. The case δ = 0 is critical in the following sense. Assuming that the ini-tial data (φ0, φ1) are in H
1
2 × H−12 does not imply that φ(t, ·) is bounded.
In fact, it is a BMO function. One of the motivations for proving the existence of global solution with low regularity, is based on an observa-tion made by Grillakis, which is that the initial data of (0.1): ψ0 ∈ L2, φ0 ∈ H
1
2, φ1 ∈ H−12, is a right space for the existence of an invariant
measure, see [B] and [Ku], resulted from the DKG equations. 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. We will use α = 1
4 − ² throughout the paper. We will also often skip the constant in the inequalities. For convenience, we also 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 + γ1ξ (1.1d) c M (ξ) = |ξ| + 1 (1.1e)
c
W and bD are the symbols of the wave and Dirac operators respectively.
Consider the Dirac equation, ½
Dψ = G, (t, x) ∈ R1 × R1, 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) = eit|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, b
a6(τ ) = ba( τ
6), 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
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++ ba(τ ) bG−)dτ + Z eitτ(1 − ba(τ )) bD+Gb− 2|ξ|(τ − |ξ|) dτ − e it|ξ|Db+ 2|ξ| Z (1 − ba(τ )) bG− τ − |ξ| dτ, (1.6b) 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| (ba(τ ) bG+ + bG−)dτ + Z eitτ(1 − ba(τ )) bD−Gb+ 2|ξ|(τ + |ξ|) dτ − e −it|ξ|Db− 2|ξ| Z (1 − ba(τ )) bG+ τ + |ξ| dτ, (1.6c)
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+Gb− 2|ξ|(τ − |ξ|) + (1−ba6) bD−Gb+ 2|ξ|(τ + |ξ|) , (1.9a) b A±,0(ξ) := b D± 2|ξ| h γ0ψb0 − 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)
Now we split ψ in a different manner. Consider the cut-off function bb(τ) equals 1 if |τ| < R, and equals 0 if |τ| > 2R. Let bb(τ) + bc(τ) = 1. Applying (1.4), we can give the following formula for bψ.
b ψ(τ, ξ) = ∞ X k=0 ³ δ+(k)(τ, ξ) bU+,k(ξ) + δ−(k)(τ, ξ) bU−,k(ξ) ´ + bU (τ, ξ), (1.12) where b U±,0(ξ) := b D± 2|ξ| h γ0ψb0− Z b c(λ ∓ |ξ|) λ ∓ |ξ| Gdλb i , (1.12a) b U±,k(ξ) := b D±(−1)k 2|ξ|k! Z (λ ∓ |ξ|)k−1bb(λ ∓ |ξ|) bGdλ, (1.12b) b U (τ, ξ) :=h bD+bc(τ − |ξ|) 2|ξ|(τ − |ξ|) + b D−bc(τ + |ξ|) 2|ξ|(τ + |ξ|) i b G(τ, ξ). (1.12c) Consider the wave equation,
½
¤φ = F, (t, x) ∈ R1 × R1, φ(0) = φ0, φt(0) = φ1.
(1.13) Taking Fourier transform on (1.13) 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. (1.14a)e e φ(t, ξ) = e it|ξ|+ e−it|ξ| 2 φb0(ξ) + eit|ξ|− e−it|ξ| i2|ξ| φb1(ξ)+ −1 2|ξ| Z eitτ − eit|ξ| τ − |ξ| F (τ, ξ)dτ +b 1 2|ξ| Z eitτ − e−it|ξ| τ + |ξ| F (τ, ξ)dτ. (1.14b)b
For convenience, we denote the following notations by b v±,0(ξ) :=1 2φb0(ξ)± 1 i2|ξ|φb1(ξ)± 1 2|ξ| Z b c(λ∓|ξ|) λ ∓ |ξ| F (λ, ξ)dλ, (1.17a)b b v±,k(ξ) := ∓ (−1)k 2|ξ|k! Z (λ ∓ |ξ|)k−1bb(λ ∓ |ξ|) bF (λ, ξ)dλ, (1.17b) b v(τ, ξ) := −1 2|ξ| h bc(τ − |ξ|) τ − |ξ| − b c(τ + |ξ|) τ + |ξ| i b F (τ, ξ). (1.17c)
Combine (1.14b) and (1.17), and invoke the cut-off function, we have b φ(τ, ξ) = ∞ X k=0 ³ b V+,k(τ, ξ) + bV−,k(τ, ξ) ´ + bV (τ, ξ) + ∞ X k=0 b Nk(τ, ξ) + bN (τ, ξ), (1.18) where b V±,k(τ, ξ) = ¡ 1 − ba(ξ)¢δ±(k)(τ, ξ)bv±,k(ξ), (1.19a) b V (τ, ξ) =¡1 − ba(ξ)¢v(τ, ξ),b (1.19b) b Nk(τ, ξ) = ba(ξ) h δ+(k)(τ, ξ)bv+,k(ξ) + δ−(k)(τ, ξ)bv−,k(ξ) i , (1.19c) b N (τ, ξ) = ba(ξ)bv(τ, ξ). (1.19d)
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 ϕ(t) be a cut-off function such that
ϕ(t) equals 1 if |t| ≤ 1
2, and equals 0 if |t| > 1, and ϕT(t) = ϕ(t/T ). Notice
that, for an arbitrary function f (t, x), we have
k bϕT ∗ bf kL2 = kϕTf kL2 ≤ kϕTkL∞kf kL2. (2.1)
For the Dirac equation (1.2), using (1.12), we denote bΨT by
b ΨT(τ, ξ) = bϕT ∗ ∞ X k=0 ³ δ+(k)Ub+,k+ δ−(k)Ub−,k ´ (τ, ξ) + bU (τ, ξ), (2.2)
Lemma 2.1. Let ² > 0 and T R ∼ 1. If ψ0 ∈ H−α, then we have
° ° ° bS12Mc−αΨbT ° ° ° L2(R1×R1) ≤ C ³ kψ0kH−α + T² ° ° ° Gb c MαSb12−² ° ° ° L2 ´ . (2.3)
Proof. Applying formulae (1.12)s, we can derive the following bounds: ° ° ° bS12Mc−αϕbT∗¡δ±Ub±,0¢°°° L2≤C ³ kψ0kH−α+T² ° ° ° Gb c MαSb12−² ° ° ° L2 ´ , (2.4a) ° ° ° bS12Mc−αϕbT ∗¡δ(k) ± Ub±,k ¢°° ° L2 ≤ C (4RT )k k! T ²°°° Gb c MαSb12−² ° ° ° L2, (2.4b) ° ° ° bS12Mc−αUb ° ° ° L2(R1×R1) ≤ CT ²°°° Gb c MαSb12−² ° ° ° L2, (2.4c)
Combine (2.4a, b, c), we have (2.3). ¤
Consider two Dirac equations, ½
Dψj = Gj, j = 1, 2, ψj(0) = ψ0j.
(2.6) For the solutions of (2.6), we have the following key estimate whose proof will be presented in the last section.
Lemma 2.2. (Null Form Estimate) Let ² > 0, and ψ1, ψ2 be the solutions for (2.6). If ψ0j ∈ H−α, we have ° ° °( \ϕTψ1ψ2) c MαSb2α ° ° ° L2 ≤ C(T ) ³ kψ01kH−α + ° ° ° Gb1 c MαSb14 ° ° ° L2 ´ · ³ kψ02kH−α + ° ° ° Gb2 c MαSb14 ° ° ° L2 ´ . (2.7)
For the wave equation (1.13), we denote bΦT by
b ΦT(τ, ξ) = bϕT∗ ∞ X k=0 ( bV+,k+ bV−,k)(τ, ξ)+ bV (τ, ξ)+ bϕT∗ ∞ X k=0 b Nk(τ, ξ)+ bN (τ, ξ). (2.9) Thus we have the following estimate.
Lemma 2.3. Let ² > 0, δ ≥ 0, T R ∼ 1, and φ be the solution of (1.13).
If φ0 ∈ H 1 2+δ and φ1 ∈ H−12+δ, then ° ° bS1 2Mc12+δΦbT ° ° L2 ≤ C ³ kφ0kH1 2+δ + kφ1kH− 12+δ + T ²°°° Fb c M12−δSb12−² ° ° ° L2 ´ . (2.10)
Proof. Applying the formula (1.18), we can derive the following bounds: ° ° bS1 2Mc12+δϕbT ∗ bV±,0 ° ° L2 ≤C ³ kφ0kH1 2+δ + kφ1kH− 12+δ + T ²°°° Fb c M12−δSb12−² ° ° ° L2 ´ , (2.11a) ° ° bS1 2Mc12+δϕbT ∗ bN0°° L2 ≤C³kφ0kH1 2+δ + kφ1kH− 12+δ + T ²°°° Fb c M12−δSb12−² ° ° ° L2 ´ , (2.11b) ° ° bS1 2Mc12+δϕbT ∗ bV±,k ° ° L2 ≤ C (4RT )k k! T ²°°° Fb c M12−δSb12−² ° ° ° L2, (2.11c) ° ° bS1 2Mc12+δϕbT ∗ bNk ° ° L2 ≤ C (4RT )k k! T ²°°° Fb c M12−δSb12−² ° ° ° L2, (2.11d) ° ° bS1 2Mc12+δVb ° ° L2 ≤ CT ²°°° Fb c M12−δSb12−² ° ° ° L2, (2.11e) ° ° bS1 2Mc12+δNb ° ° L2 ≤ CT ²°°° Fb c M12−δSb12−² ° ° ° L2. (2.11f)
Combine (2.11a-f), we conclude the proof. ¤
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. (2.13)
Lemma 2.5. Let f (t, x) and g(t, x) be any functions such that f ∈
Lq(L2(Rn)) and bSβbg ∈ L2(L2(Rn)). Assume that ² > 0, 1
q = 1 − ²,
1
r =
1
2 − β, and 2 ≤ r < ∞. Then we have ° ° ° bfb S12−² ° ° ° L2(L2(Rn)) ≤ Ckf kLq(L2(Rn)), (2.14a) kgkLr(L2(Rn)) ≤ Ck bSβgkb L2(L2(Rn)). (2.14b)
Proof. The proofs for (2.14a) and (2.14b) are analogous. Therefore we will only prove the case of (2.14b).
Taking the inverse Fourier transform in the time variable over the iden-tity b g = 1 b SβSb βbg (2.18) gives e g(t, ξ) = Z e±i(t−s)|ξ| |t − s|1−βF −1 τ ( bSβbg)(s, ξ) ds. (2.19)
Then we use duality and Hardy-Littlewood-Polya inequality to compute ¯ ¯ ¯ D g, ϕ E¯ ¯ ¯ = ¯ ¯ ¯ D e g, eϕ E¯ ¯ ¯ = ¯ ¯ ¯ Z Z Z e±i(t−s)|ξ| |t − s|1−βF −1 τ ( bSβg)(s, ξ) ds eb ϕ(t, ξ) dtdξ ¯ ¯ ¯ ≤ Z kF−1 τ ( bSβbg)(s)kL2k eϕ(t)kL2 |t − s|1−β dsdt ≤CkFτ−1( bSβbg)kL2k eϕkLr0(L2) = Ck bSβbgkL2kϕkLr0(L2). (2.20)
This completes the proof of (2.14b). ¤
3. Local and Global Existence.
Now we are ready to prove the local existence for the (DKG) equations. Proof of Theorem 0.1. Consider the DKG equations
½
Dψ = ϕTφψ,
¤φ = ϕTψψ,
(3.1) and the map T defined by
T (ψ, φ) = (ΨT, ΦT). (3.2a)
We want to show that T is a contraction under the norm
N (ψ, φ) =°° bS12Mc−αψb°° L2 + ° ° bS1 2Mc12+δφb°° L2. (3.2b)
For convenience, we call
J(0) = kφ0kH1
2+δ + kφ1kH− 12+δ + kψ0k 2
First we apply (2.10) and (2.7) to compute ° ° bS1 2Mc12+δΦbT ° ° L2 ≤ C ³ J(0) + T² ° ° ° \ ϕTψψ c M12−δSb12−² ° ° ° L2 ´ ≤ C³J(0) + T² ° ° ° ϕ\Tφψ c MαSb14 ° ° °2 L2 ´ . (3.4)
To bound the term above, we first compute ° ° ° cM−αφψ(t)f ° ° ° L2 ∼ kGα∗ (φψ)(t)kL2 ≤ kφ(t)kL∞kGα∗ ψ(t)kL2 ≤ kφ(t)k H12+δkψ(t)kH−α, (3.5)
where Gα(x) is an L1-function with the following property:
c
Gα(ξ) ∼ (1 + |ξ|)−α, (3.6)
see [S], then we invoke (2.14a,b) and obtain ° ° ° ϕ\Tφψ c MαSb14 ° ° ° L2 ≤ C ° ° °ϕ^Tφψ c Mα ° ° ° Lq(L2) ≤ kφk L2q(H12+δ)kψkL2q(H−α) ≤ ° ° bS1 2Mc12+δφb°° L2 ° ° bS1 2Mc−αψb°° L2. (3.7) Thus we get ° ° ° ϕ\Tφψ c MαSb12−² ° ° ° L2 ≤ CN 2(ψ, φ). (3.8)
Next we want to bound the term involved with bΨT. The estimate (2.3)
implies that ° ° ° bS12Mc−αΨbT ° ° ° L2(R1×R1) ≤ C ³ kψ0kH−α + T² ° ° ° ϕ\Tφψ c MαSb14 ° ° ° L2 ´ . (3.9) Hence, using (3.3), (3.8), and (3.5), we have
N (T (ψ, φ)) ≤ C¡J(0) + T²N4(ψ, φ)¢. (3.10) Choosing sufficiently large L, for suitable T , we have
provided that
C(J(0) + T²L4) ≤ L. (3.12) Now we consider the difference T (ψ, φ) − T (ψ0, φ0). Base on the obser-vations ψψ − ψ0ψ0 = 1 2(ψ − ψ0)(ψ + ψ 0) + 1 2(ψ + ψ0)(ψ − ψ 0), (3.13a) φψ − φ0ψ0 = 1 2(φ − φ 0)(ψ + ψ0) + 1 2(φ + φ 0)(ψ − ψ0), (3.13b)
Employing (2.10), (2.7), (3.13) and (3.8), we first calculate ° ° bS1 2Mc12+δF(ΦT − Φ0 T) ° ° L2 ≤CT² ³ kF((ψ − ψ0)(ψ + ψ0)) c M12−δSb12−² kL2 + kF((ψ + ψ 0)(ψ − ψ0)) c M12−δSb12−² kL2 ´ ≤CT²³°° F((φ − φ 0)(ψ + ψ0)) c MαSb14 ° ° L2 + ° ° F((φ + φ0)(ψ − ψ0)) c MαSb14 ° ° L2 ´ · ³ J(0) +°° F(φψ + φ 0ψ0) c MαSb14 ° ° L2 ´ ≤CT²¡k bS12Mc12+δφ − φ\0k L2 + k bS 1 2Mc−αψ − ψ\0k L2 ¢ L(J(0) + L2) ≤CT²L3¡k bS12Mc12+δφ − φ\0k L2 + k bS 1 2Mc−αψ − ψ\0k L2 ¢ (3.15) Analogously, we get ° ° bS1 2Mc−α(ΨT\− Ψ0 T) ° ° L2 ≤ CT²L ³ k bS12Mc−αψ − ψ\0k L2 + k bS 1 2Mc12+δφ − φ\0k L2 ´ . (3.16)
Combining (3.15)and (3.16), we have
N¡T (ψ − ψ0, φ − φ0)¢≤ CT²L3N (ψ − ψ0, φ − φ0). (3.17)
Therefore for suitable T , we obtain
N¡T (ψ − ψ0, φ − φ0)¢≤ 1
2N (ψ − ψ
provided that
CT²L3 ≤ 1
2. (3.19)
We can conclude that the map T is indeed a contraction with respect to the norm N , thus it has a unique fixed point. ¤
We now prove the global existence.
Proof of Theorem 0.2. From the law of conservation of charge, we have
sup
[0,T ]
kψ(t)kL2 = kψ0kL2. (3.20)
To bound φ we apply the formula (1.14a), 2φ(t, x) = (φ0(x + t) + φ0(x − t)) + Z x+t x−t φ1(y) dy + Z t 0 Z x+t−s x−t+s ψψ(s, y)dyds. (3.21) First we write φ = φL+ φN, the homogeneous and inhomogeneous parts
of the solution, then we obtain
kφL(t)kL∞ ≤kφL(t)k H12+δ ≤kφ0kH1 2+δ + kφ1kH− 12+δ ≤ J(0), (3.22) and kφN(t)kL∞ ≤ Z t 0 Z x+t−s x−t+s ¯ ¯ ¯ψψ(s, y) ¯ ¯ ¯dyds ≤ CT kψ0k2L2. (3.23)
Combine (3.22) and (3.23), we get
kφ(t)kL∞ ≤ C(T, J(0)). (3.24)
Take Fourier transform of the solution φ(t), we have e φ(t, ξ) = cos t|ξ|bφ0(ξ) + sin t|ξ| |ξ| φb1(ξ) + Z t 0 sin (t − s)|ξ| |ξ| ^ ϕTψψ(s, ξ)ds. (3.25)
Then we invoke (2.7), (2.14a), (for ² = 1 4), and (3.23) to compute kφ(t)k H12+δ ≤kφ0kH21+δ + kφ1kH− 12+δ + Z t 0 kϕTψψ(s)kH− 12+δds ≤J(0) + T12 ° ° ° \ ϕTψψ c M12−δ ° ° ° L2 ≤J(0) + T12k \ϕTψψkL2 ≤J(0) + T12 ° ° °ϕ\Tφψ b S14 ° ° °2 L2 ≤J(0) + TρkϕTφψkL2 ≤J(0) + Tρ Z T 0 kφ(t)k2L∞kψ(t)k2L2dt ≤C¡T, J(0)¢, (3.26)
where ρ is some positive number. The calculation for kφt(t)kH− 12+δ is
analogous. Thus the above bounds ensure us to proceed the construction
of solution beyond T . ¤
4. Null Form Estimate.
In this section, we demonstrate the key estimate.
Lemma 2.2. ( Null Form Estimate) Let ² > 0 and ψ1, ψ2 be the solutions for the Dirac equations (2.6). If the initial data ψ0j ∈ H−
1 4+², j = 1, 2, then we have ° ° ° \ ϕTψ1ψ2 c MαSb2α ° ° ° L2 ≤ C(T ) ³ kψ01kH−α + ° ° ° Gb1 c MαSb14 ° ° ° L2 ´ · ³ kψ02kH−α + ° ° ° Gb2 c MαSb14 ° ° ° 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: if ψ is a solution of the Dirac equation (1.2), then
° ° ° \ ϕTψψ c MαSb2α ° ° ° L2 ≤ C(T ) ³ kψ0kH−α+ ° ° ° Gb c MαSb14 ° ° ° 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+ b G− 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. The Fourier transform of the quadratic
expression, cψψ = bψ ∗ bψ, can be written as the sum of the following terms.
X k,l ¡ δ∓(k)Ab±,k ¢ ∗¡δ±(l)Ab±,l ¢ , (4.8a) X k,l ¡ δ∓(k)Ab±,k ¢ ∗¡δ∓(l)Ab∓,l ¢ , (4.8b) X k ¡ δ∓(k)Ab±,k ¢ ∗¡Kb1 + bK2 ¢ +¡ bK1+ bK2 ¢ ∗X k ¡ 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 α < 1
4. The following estimate holds ° ° °ϕbT ∗ ¡ δ(k)∓ fd±,k† Db± |ξ| γ0 ¢ ∗¡δ(l)∓ Db∓ |ξ| fb∓,l ¢ c MαSb2α ° ° ° L2 ≤ C(k + l + 1)Tk+l−12kf±,kk H−αkf∓,lkH−α. (4.11)
Proof. Let us call b Z±,k ≡ δ±(k) b D± 2|ξ|fb±,k = δ (k) ± Ab±,k. (4.12)
Using duality, we demonstrate the case (−, +), while the case (+, −) is being similar. We first compute the fractional term
b D(|ξ|, −ξ)γ0D(|η|, η)b |ξ||η| = ½ 0, if ξη > 0, 2γ0 ± 2γ1, if ξη < 0. (4.13)
Throughout elementary analysis we have the bound: (1 + |ξ|)α(1 + |η|)α (1 + |ξ + η|)α³¯¯¯¯¯|ξ| + |η|¯¯ − |ξ + η|¯¯¯ + 1´2α ≤ C, (4.14) for ξη < 0. Thus ¯ ¯ ¯ϕTZ−,kZ+,l, g® ¯¯¯ = ¯ ¯ ¯ Z b f−,k† (−ξ)D(|ξ|, −ξ)γb 0D(|η|, η)b |ξ||η| fb+,l(η) \tk+lϕTg(|ξ| + |η|, ξ + η) dξdη ¯ ¯ ¯ ≤Ckf−,kkH−αkf+,lkH−αk cMαSb2αtk+l\ϕTgkL2, (4.15a)
and through some computations, we have
k cMαSb2αtk+l\ϕ
TgkL2 ≤ C(k + l + 1)Tk+l− 1
2k cMαSb2αbgkL2. (4.15b)
Lemma 4.2. Let α < 1
4. The following estimate holds ° ° °ϕbT ∗ ¡ δ(k)∓ fd±,k† Db± |ξ| γ0 ¢ ∗¡δ(l)± Db± |ξ| fb±,l ¢ c MαSb2α ° ° ° L2 ≤ C(k + l + 1)Tk+l−12kf±,kkH−αkf±,lkH−α. (4.16)
Proof. Using duality, we demonstrate the case (+, +), while the case (−, −) is being similar. We first compute the fractional term
b D(|ξ|, ξ)γ0D(|η|, η)b |ξ||η| = ½ 0, if ξη < 0, 2γ0 ∓ 2γ1, if ξη > 0. (4.17)
Throughout elementary analysis we have the bound: (1 + |ξ|)α(1 + |η|)α (1 + |ξ + η|)α³¯¯¯¯¯ − |ξ| + |η|¯¯ − |ξ + η| ¯ ¯ ¯ + 1 ´2α ≤ C (4.18) for ξη > 0. Thus ¯ ¯ ¯ϕTZ+,kZ+,l, g® ¯¯¯ = ¯ ¯ ¯ Z b f+,k† (−ξ)D(|ξ|, ξ)γb 0D(|η|, η)b |ξ||η| fb+,l(η) \tk+lϕTg(−|ξ| + |η|, ξ + η) dξdη ¯ ¯ ¯ ≤Ckf+,kkH−αkf+,lkH−αk cMαSb2αtk+l\ϕTgkL2. (4.19)
This together with (4.15b) complete the proof. ¤ Lemma 4.3. With the notation above, the following estimates hold
kf±,0kH−α ≤ C ³ kψ0kH−α+ ° ° ° Gb c MαSb14 ° ° ° L2 ´ , (4.20) kf±,kkH−α ≤ C 1 k!k b G± c MαSb14 kL2. (4.21)
The proof for the Lemma 4.3 is straight forward so that we skip it. Notice that, in the (4.21), bS ∼ 1 on the support of bG±.
Lemma 4.4. With the notation above, the following estimate holds ° ° °ϕbT ∗ bK1 ∗ bK1 c MαSb2α ° ° ° L2 ≤ C ° ° ° Gbf c MαSb14 ° ° °2 L2. (4.22)
Proof. For simplicity, we write bG := bGf and bK := bK1. We use dyadic
decomposition to handle this case. Assume that
b G = ∞ X k=1 b G±,±,k, (4.23)
where bG±,±,k(τ, ξ) is supported in one of the following four types of
re-gions:
Σ+,+ := {(τ, ξ) : τ > 0, +2k−1 < τ − |ξ| < +2k+1}, (4.24a)
Σ+,− := {(τ, ξ) : τ > 0, −2k+1 < τ − |ξ| < −2k−1}, (4.24b)
Σ−,+ := {(τ, ξ) : τ < 0, +2k−1 < τ + |ξ| < +2k+1}, (4.24c)
Σ−,− := {(τ, ξ) : τ < 0, −2k+1 < τ + |ξ| < −2k−1}. (4.24d)
The decomposition of bG induces a decomposition for bK, namely
b K±,±,k = b D c W b G±,±,k. (4.25a)
To compute the convolution in (4.22), b K±,±,k∗ bK±,±,l(−τ, −ξ) = Z b K±,±,k(−τ − σ, −ξ − η) bK±,±,l(σ, η) dσdη = Z b K±,±,k† (τ + σ, ξ + η)γ0Kb±,±,l(σ, η) dσdη, (4.25b)
we have 16 cases resulted from (4.24a-d) and (4.25b) as follows. {(τ, σ, ξ, η) : τ + σ > 0, σ > 0, τ + σ − |ξ + η| ∼ ±2k, σ − |η| ∼ ±2l} (4.26a) {(τ, σ, ξ, η) : τ + σ < 0, σ < 0, τ + σ + |ξ + η| ∼ ±2k, σ + |η| ∼ ±2l} (4.26b) {(τ, σ, ξ, η) : τ + σ < 0, σ > 0, τ + σ + |ξ + η| ∼ ±2k, σ − |η| ∼ ±2l} (4.26c) {(τ, σ, ξ, η) : τ + σ > 0, σ < 0, τ + σ − |ξ + η| ∼ ±2k, σ + |η| ∼ ±2l} (4.26d) We label them as Σk,l[(±, ±); (±, ±)], (4.27)
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 .
Let g be an arbitrary function. We first compute £ γ0(τ + σ) − γ1(ξ + η)¤γ0£γ0τ + γ1η¤ =γ0£(τ + σ)σ − (ξ + η)η¤+ γ1£(τ + σ)η − σ(ξ + η)¤. (4.28) Thus, we have ¯ ¯ ¯ D b Kk∗ bKl, bg E ¯ ¯ ¯ = ¯ ¯ ¯ Z b G†k(τ + σ, ξ + η)γ 0(τ + σ) − γ1(ξ + η) (τ + σ)2 − (ξ + η)2 γ 0γ0σ + γ1η σ2 − η2 Gbl(σ, η)· b g(−τ, −ξ)dσdηdτ dξ ¯ ¯ ¯ ≤CkGbk c MαkL2k b Gl c MαkL2 ³ Z Ik,l(τ, ξ) ¯ ¯bg(−τ, −ξ)¯¯2 dτ dξ´ 1 2 , (4.29a) where Ik,l(τ, ξ) is given by Ik,l(τ, ξ) := Z Dk,l c M2α(ξ + η) cM2α(η)Q(τ, σ, ξ, η) c W2(τ + σ, ξ + η)cW2(σ, η) dσdη, (4.29b)
and Q is given by the expression
Q(τ, σ, ξ, η) :=£(τ + σ)σ − (ξ + η)η¤2 +£(τ + σ)η − σ(ξ + η)¤2, (4.29c)
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}. (4.29d)
We need to sort the cases into two sets,
Σk,l[(±, ·); (±, ·)] and Σk,l[(±, ·); (∓, ·)], (4.30)
due to the fact that the computation for the 8 cases in each set is similar. For simplicity, we will assume k ≥ l, while the other case is similar. Cases H. We have the following estimate
° ° ° b K+,·,k∗ bK+,·,l c MαSb2α ° ° ° L2 ≤ C 22l 1 2(1 2−α)k ° ° °Gb+,·,k c Mα ° ° ° L2 ° ° °Gb+,·,l c Mα ° ° ° L2, (4.31a) ° ° ° b K−,·,k∗ bK−,·,l c MαSb2α ° ° ° L2 ≤ C 22l 1 2(1 2−α)k ° ° °Gb−,·,k c Mα ° ° ° L2 ° ° °Gb−,·,l c Mα ° ° ° L2, (4.31b)
In these cases, we have (τ + σ)σ > 0. Throughout some algebraic manipulation, the expression Q can be written as
2Q =(τ + σ − |ξ + η|)2(σ + |η|)2+ (τ + σ + |ξ + η|)2(σ − |η|)2+ 8(τ + σ)σ£|ξ + η||η| − (ξ + η)η¤. (4.32) Take the case of
b
K+,+,k∗ bK+,+,l, (4.33)
as an example and in which Dk,l = {(η, σ) : τ + σ − |ξ + η| ∼ 2k, σ − |η| ∼
2l, (τ, σ, ξ, η) ∈ Σk,l[(+, +); (+, +)]}. In this case τ + σ > 0 and σ > 0. In
the ησ-plane, this is the region of the intersection of two forward cones. One has the thickness of 2k and the translation of (−ξ, −τ ), while the
case which is when one cone moves along the other cone such that the intersection region is unbounded.
For the first part, we distinguish three cases: |ξ + η| ≤ |η|, |ξ + η| ≥ |η|, and the extreme case. For the first two cases, we have
Ik,l1 (τ, ξ) := Z Dk,l c M2α(ξ + η) cM2α(η)(τ + σ − |ξ + η|)2(σ + |η|)2 c W2(τ + σ, ξ + η)cW2(σ, η) dσdη = Z Dk,l c M2α(ξ + η) cM2α(η) (τ + σ + |ξ + η|)2(σ − |η|)2dσdη ∼1 2l Z e Dk,l c M2α(ξ + η) cM2α(η) (2k + |ξ + η|)2 dη ≤1 2l Z e Dk,l 1 (2k+ |ξ + η|)2−2αdη cM 2α(ξ) ≤ C 2(1−2α)k+lMc 2α(ξ) bS4α(τ, ξ). (4.34a)
For the extreme case, we obtain
Ik,l1 (τ, ξ) ∼1 2l Z e Dk,l c M2α(ξ + η) cM2α(η) (2k + |ξ + η|)2 dη ≤1 2l Z e Dk,l 1 (2k+ |ξ + η|)2−4αdη ≤ C 2(1−2α)k+lMc 2α(ξ) bS4α(τ, ξ). (4.34b)
For the second part, again we distinguish three cases: |ξ + η| ≤ |η|,
|ξ + η| ≥ |η|, and the extreme case. For the first two cases, we get Ik,l2 (τ, ξ) := Z Dk,l c M2α(ξ + η) cM2α(η)(τ + σ + |ξ + η|)2(σ − |η|)2 c W2(τ + σ, ξ + η)cW2(σ, η) dσdη = Z Dk,l c M2α(ξ + η) cM2α(η) (τ + σ − |ξ + η|)2(σ + |η|)2dσdη ∼ 1 22k−l Z e Dk,l c M2α(ξ + η) cM2α(η) (2l+ |η|)2 dη ≤ 1 22k−l Z e Dk,l 1 (2l+ |η|)2−2αdη cM 2α(ξ) ≤ C 2k+(1−2α)lMc 2α(ξ) bS4α(τ, ξ). (4.35a)
For the extreme case, we have Ik,l2 (τ, ξ) ∼ 1 22k−l Z e Dk,l c M2α(ξ + η) cM2α(η) (2l+ |η|)2 dη ≤ 1 22k−l Z e Dk,l 1 (2l+ |η|)2−4αdη cM 2α(ξ) ≤ C 2k+(1−2α)lMc 2α(ξ) bS4α(τ, ξ). (4.35b)
For the third part, we get
Ik,l3 (τ, ξ) := Z Dk,l c M2α(ξ + η) cM2α(η)(τ + σ)σ£|ξ + η||η| − (ξ + η)η¤ c W2(τ + σ, ξ + η)cW2(σ, η) dσdη ≤ C 22k+2l Z Dk,l c M2α(ξ + η) cM2α(η)(τ + σ)σ|ξ + η||η| (τ + σ + |ξ + η|)2(σ + |η|)2 dσdη ≤ C 22k+2l Z Dk,l c M2α(ξ + η) cM2α(η)dσdη ≤ C 2(1−2α)k+lMc 2α(ξ) bS4α(τ, ξ). (4.36a)
The extreme case will not cause trouble since ξ + η and η are of the same sign except on a bounded region, i.e. £|ξ + η||η| − (ξ + η)η¤= 0 except on a bounded region. Let us denote the small region by R.
Ik,l3 (τ, ξ) ≤ C 22k+2l Z R c M2α(ξ + η) cM2α(η)(τ + σ)σ|ξ + η||η| (τ + σ + |ξ + η|)2(σ + |η|)2 dσdη ≤ C 22k+2l Z R c M2α(ξ + η) cM2α(η)dσdη ≤ C 22k+2l Z R dσdη22αkMc2α(ξ) ≤ C 2(1−2α)k+lMc 2α(ξ) bS4α(τ, ξ). (4.36b)
Cases E. We have the following estimate ° ° ° b K+,·,k∗ bK−,·,l c MαSb2α ° ° ° L2 ≤ C 22l 1 2(1 2−α)k ° ° °Gb+,·,k c Mα ° ° ° L2 ° ° °Gb−,·,l c Mα ° ° ° L2, (4.37a) ° ° ° b K−,·,k ∗ bK+,·,l c MαSb2α ° ° ° L2 ≤ C 22l 1 2(1 2−α)k ° ° °Gb−,·,k c Mα ° ° ° L2 ° ° °Gb+,·,l c Mα ° ° ° L2, (4.37b)
In these cases, we have (τ + σ)σ < 0. Throughout some algebraic manipulation, the expression Q can be written as
2Q =(τ + σ + |ξ + η|)2(σ + |η|)2+ (τ + σ − |ξ + η|)2(σ − |η|)2−
8(τ + σ)σ£|ξ + η||η| + (ξ + η)η¤. (4.38)
Take the case of
b
K−,+,k∗ bK+,+,l, (4.39)
as an example and in which Dk,l = {(η, σ) : 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-plane, this is the region of the intersection of a forward cone with a
truncated backward cone. One has the thickness of 2k and the translation
of (−ξ, −τ ), while the other has thickness of 2l. It is bounded for all cases.
We still have the extreme case which is when one cone moves along the other cone, though the region of intersection can be as large as possible, nevertheless it is bounded.
Again for the first part, we can estimate
Ik,l1 (τ, ξ) := Z Dk,l c M2α(ξ + η) cM2α(η)(τ + σ + |ξ + η|)2(σ + |η|)2 c W2(τ + σ, ξ + η)cW2(σ, η) dσdη = Z Dk,l c M2α(ξ + η) cM2α(η) (τ + σ − |ξ + η|)2(σ − |η|)2dσdη ≤ 1 22l Z Dk,l (|ξ + η| + 1)2α(|η| + 1)2α (τ + σ − |ξ + η|)2 dσdη. (4.40a)
To estimate the above integral, we separate the cases for |ξ + η| ≥ |η|,
|ξ + η| ≤ |η|, and the extreme case. Throughout some calculations, in
each case, we have
Ik,l1 (τ, ξ) ≤ 1 2l
1
2(1−2α)kMc
For the second part, we derive Ik,l2 (τ, ξ) := Z Dk,l c M2α(ξ + η) cM2α(η)(τ + σ − |ξ + η|)2(σ − |η|)2 c W2(τ + σ, ξ + η)cW2(σ, η) dσdη = Z Dk,l c M2α(ξ + η) cM2α(η) (τ + σ + |ξ + η|)2(σ + |η|)2dσdη ≤ C 22k Z Dk,l (|ξ + η| + 1)2α(|η| + 1)2α (σ + |η|)2 dσdη ≤C2 l 22k Z e Dk,l (|ξ + η| + 1)2α (2l+ |η|)2+2α dη. (4.41a)
To estimate the above integral, we separate the cases for |ξ + η| ≥ |η|,
|ξ + η| ≤ |η|, and the extreme case. Throughout some calculations, in
each case, we have
Ik,l2 (τ, ξ) ≤ 1 2l
1
2(1−2α)kMc
2αSb4α. (4.41b)
For the third part, we have
Ik,l3 (τ, ξ) := Z Dk,l c M2α(ξ + η) cM2α(η)|τ + σ|σ£|ξ + η||η| + (ξ + η)η¤ c W2(τ + σ, ξ + η)cW2(σ, η) dσdη ≤ C 22k+2l Z Dk,l c M2α(ξ + η) cM2α(η)|τ + σ|σ|ξ + η||η| (τ + σ − |ξ + η|)2(σ + |η|)2 dσdη ≤ C 22k+2l Z Dk,l (|ξ + η| + 1)2α(|η| + 1)2α dσdη. (4.42a)
To estimate the above integral, we separate the cases for |ξ + η| ≥ |η|,
|ξ + η| ≤ |η|, and the extreme case. Notice that for the extreme case, we
have |ξ + η||η| + (ξ + η)η = 0 except on a small part of the region of the intersection. Throughout some calculations, in each case, we have
Ik,l3 (τ, ξ) ≤ 1 2l
1
2(1−2α)kMc
Now we return to the proof of (4.22). Combine the above, we get ¯ ¯ ¯KkKl, g® ¯¯¯ ≤C ° ° ° Gbk c Mα ° ° ° L2 ° ° ° Gbl c Mα ° ° ° L2 ³ Z Ik,l(τ, ξ) ¯ ¯bg(−τ, −ξ)¯¯2 dτ dξ ´1 2 ≤C 22l 1 2(1 2−α)k ° ° ° Gbk c Mα ° ° ° L2 ° ° ° Gbl c Mα ° ° ° L2k cM αSb2αbgk L2 ≤ C 214l 1 2²k ° ° ° Gbk c MαSb14 ° ° ° L2 ° ° ° Gbl c MαSb14 ° ° ° L2k cM αSb2αbgk L2. (4.43) Finally, we have ° ° ° b K ∗ bK c MαSb2α ° ° ° L2 ≤ X k,l ° ° ° b Kk∗ bKl c MαSb2α ° ° ° L2 ≤X k,l C 2²k+1 4l ° ° ° Gbk c MαSb14 ° ° ° L2 ° ° ° Gbl c MαSb14 ° ° ° L2 ≤ C ° ° ° Gb c MαSb14 ° ° °2 L2. (4.44)
This completes the proof. ¤
The estimates for the remaining cases are given in the following Lemma. Lemma 4.5. For j = 1, 2 and k = 0, 1, 2, · · ·. The following estimates
hold ° ° °ϕbT ∗ ¡ δ(k)∓ fd±,k† Db± |ξ| γ0 ¢ ∗¡Kbj ¢ c MαSb2α ° ° ° L2 ≤ C(k + 1)Tk−12kf±,kkH−α ° ° ° Gb c MαSb14 ° ° ° L2, (4.45a) ° ° °ϕbT ∗ bKj ∗ ¡ δ(k)± Db± |ξ| fb±,k ¢ c MαSb2α ° ° ° L2 ≤ C(k + 1)Tk−12kf±,kk H−α ° ° ° Gb c MαSb14 ° ° ° L2, (4.45b) ° ° °ϕbT ∗ bK1 ∗ bK2 c MαSb2α ° ° ° L2 ≤ C ° ° ° Gb c MαSb14 ° ° °2 L2, (4.45c) ° ° °ϕbT ∗ bK2∗ bKj c MαSb2α ° ° ° L2 ≤ C ° ° ° Gb c MαSb14 ° ° °2 L2, (4.45d)
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.
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Department of Math, Cheng Kung University, Tainan 701 Taiwan
