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 (0.1a)
¤φ = ψψ; (0.1b)
ψ(0, x) := ψ0(x), φ(0, x) = φ0(x), φt(0, x) = φ1(x), (0.1c)
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 full advantage of the null form structure depicted in the nonlinear term ψψ, which has been observed for possessing such structure, see [KM] and [Bo].
2000 Mathematics Subject Classif ication. 35L70 1
For the DKG system, there are many conserved quantities which are not positive definite, such as the energy. Therefore they are not applica-ble to derive a priori 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, based on 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 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 result for (0.1), employing the null form estimate together with other estimates derived previously, and a fixed point argument. Fi-nally 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
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)
and the proof works for this system too.
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 of a given function. We will use α = 1
4 − ² 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 + γ1ξ (1.1d) c M (ξ) = |ξ| + 1 (1.1e)
Notice that cW 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) = 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 eit|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) 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)
where ba6(τ ) = ba(τ6) and ba is the cut-off function defined previously. 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+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)
Consider the wave equation, ½
¤φ = F, (t, x) ∈ R1 × R1, φ(0) = φ0, φt(0) = φ1.
(1.10) 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. (1.11)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| ≤ 12, 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)
Lemma 2.1. If ψ0 ∈ H−α, then we have
° ° °bbT ∗ [ cM−αSb 3 4ψ]b ° ° ° L2(R1×R1) ≤ C ³ kψ0kH−α + ° ° ° Gb c MαSb14 ° ° ° L2 ´ . (2.2)
Proof. Without loss of generality, we prove the special case. ° ° °bbT ∗ [ bS 3 4ψ]b ° ° ° L2(R1×R1) ≤ C ³ kψ0kL2 + ° ° °bGb S14 ° ° ° L2 ´ . (2.3)
To estimate bbT ∗ [ bS
3
4ψ], we apply formulae (1.8) and (1.9)s. First web
compute kbbT ∗ [ bS 3 4K]kb L2 ≤ k bS 3 4Kkb L2 ≤ ° ° ° bS34 b D c W b Gf ° ° ° L2+ ° ° ° bS34 (1 − ba6) bD+ b G− 2|ξ|(τ − |ξ|) ° ° ° L2 + ° ° ° bS34 (1 − ba6) bD− b G+ 2|ξ|(τ + |ξ|) ° ° ° L2 ≤ C ° ° °bGb S14 ° ° ° L2. (2.4)
For the term bbT ∗ [ bS
3 4δ(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 ∗ [ bS 3 4δ(k) + ](ξ)k2L2(dτ ) ∼ Z ³ Z bbT(τ − λ) bS(λ, ξ) 3 4δ(k)(λ − |ξ|)dλ ´2 dτ ∼ Z ³ ∂k ∂λk ¡ bbT(τ − λ) bS(λ, ξ) 3 4¢¯¯¯ 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(dξ) ≤kψ0kL2 + ³ Z ³ Z bGf +¡1 − ba6(τ )¢Gb− τ − |ξ| dτ ´2 dξ ´1 2 ≤kψ0kL2 + ° ° ° Gb b S14 ° ° ° L2 (2.6) and k bA+,kkL2(dξ)≤ 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 S14 ° ° ° L2. (2.7) Therefore we have kbbT ∗ [ bS 3 4δ+Ab+,0]k L2 ≤T 1 2 ³ kψ0kL2 + ° ° ° Gb b S14 ° ° ° L2 ´ , kbbT ∗ [ bS 3 4δ(k) + Ab+,k]kL2 ≤Tk+ 1 2 2 k k! ° ° ° bGb S14 ° ° ° L2. (2.8)
The calculation for the term bbT∗ [ bS
3 4δ(k)
− Ab−,k] is analogous. Combine the
above results we complete the proof. ¤
Consider two Dirac equations, ½
Dψj = Gj, j = 1, 2, ψj(0) = ψ0j.
(2.9) For the solutions of (2.9), we have the following key estimate whose proof will be presented in the last section.
Lemma 2.2. (Null Form Estimate) Let α = 1
4 − ², ² > 0, and ψ1, ψ2 be
the solutions for (2.9). If ψ0j ∈ H−α, we have
° ° °( \bTbψ1ψ2) EαSbα ° ° ° L2 ≤ C(T ) ³ kψ01kH−α + ° ° ° Gb1 c MαSb14 ° ° ° L2 ´ · ³ kψ02kH−α + ° ° ° Gb2 c MαSb14 ° ° ° L2 ´ . (2.10)
For the wave equation (1.10), we have the following estimate.
Lemma 2.3. Let φ be the solution of (1.10). If φ0 ∈ H1−2α and φ1 ∈ H−2α, then ° °bbT ∗ h c M−α( bE bS)1−αφbi°°L2 ≤ C ³ kφ0kH1−2α + kφ1kH−2α + ° ° ° Fb c MαEbαSbα ° ° ° L2 ´ . (2.11)
Proof. Without loss of generality, we show the following special case: ° °bbT ∗ £ ( bE bS)1−αφb¤°°L2 ≤ C ³ kφ0kH1−α+ kφ1kH−α+ ° ° ° bFb EαSbα ° ° ° L2 ´ . (2.12) To estimate bbT ∗ £
( bE bS)1−αφb¤ in the L2-norm, we invoke the formulae
(1.18) and (1.19). First we compute
kbbT ∗ £ ( bE bS)1−αLb¤kL2 ≤ k( bE bS)1−αLkb L2 ≤ ° ° °( bE bS) 1−αFb f c W ° ° ° L2+ ° ° °( bE bS) 1−α(1− ba 6) bF− 2|ξ|(|ξ| − τ ) ° ° ° L2+ ° ° °( bE bS) 1−α(1− ba 6) bF+ 2|ξ|(|ξ| + τ ) ° ° ° L2≤ ° ° ° Fb b EαSbα ° ° ° L2. (2.13)
For the term bbT ∗
£
( bE bS)1−αδ+(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 ∗ £ ( bE bS)1−αδ(k)+ ¤(ξ)°°2L2(dτ ) = Z ¯ ¯ ¯ Z bbT(τ − λ)( bE bS)1−α(λ, ξ)δ(k)(λ − |ξ|)dλ ¯ ¯ ¯2dτ = Z ¯ ¯ ¯ ∂ k ∂λk ³ bbT(τ − λ)( bE bS)1−α(λ, ξ) ´¯¯ ¯ λ=|ξ| ¯ ¯ ¯2dτ ∼ Z ¯ ¯ ¯Tk+1bb(k)¡T (τ − |ξ|)¢¯¯¯2(|ξ| + 1)2−2αdτ ≤ T2k+1ktkbkL2(|ξ| + 1)2−2α ≤ CT2k+1(|ξ| + 1)2−2α, (2.14) which implies ° °bbT ∗ £ ( bE bS)1−αδ+(k)Bb+,k ¤° ° L2 ≤ CT k+1 2 ³ Z (|ξ| + 1)2−2α¯¯ bB+,k(ξ) ¯ ¯2 dξ´ 1 2 . (2.15) To estimate the above integral, we first focus on the region where |ξ| > 1. Due to the observation that on the supports of (1 − ba6) bF− and bFf, the
following inequality holds b
E2αSb2α =¡|λ|+|ξ|+1¢2α¡¯¯|λ|−|ξ|¯¯+1¢2α ≤ (|ξ|+1)2α¯¯λ−|ξ|¯¯4α, (2.16)
we have the following bounds: Z ¯ ¯ ¯ Z b Ff(λ, ξ) (|ξ| + 1)α¯¯|ξ| − λ¯¯ dλ ¯ ¯ ¯2dξ ≤ Z Z ||ξ|−λ|≥1 2 1 ¯ ¯|ξ| − λ¯¯1+4²dλ Z ¯¯¯ bFf(λ, ξ)¯¯¯2 (|ξ| + 1)2α¯¯|ξ| − λ¯¯1−4²dλ dξ ≤C ° ° ° bFbf EαSbα ° ° °2 L2 (2.17)
and in the same vein Z ¯ ¯ ¯ Z ¡ 1 − ba6(λ) ¢b F−(λ, ξ) (|ξ| + 1)α¯¯|ξ| − λ¯¯ dλ ¯ ¯ ¯2dξ ≤ C ° ° ° bFb− EαSbα ° ° °2 L2. (2.18)
Hence we get ° °bbT ∗ £ ( bE bS)1−αδ+Bb+,0 ¤° ° L2(L2(|ξ|>1)) ≤ CT 12 ³ kφ0kH1−α + kφ1kH−α+ ° ° ° Fb b EαSbα ° ° ° L2 ´ (2.19) and ° °bbT ∗ £ ( bE bS)1−αδ+(k)Bb+,k ¤°° L2(L2(|ξ|>1)) ≤ CTk+12 c k k! ³ Z Z ¯¯ bF++ ba6Fb−¯¯2 (τ, ξ) (|ξ| + 1)2α dλ dξ ´1 2 ≤ CTk+12 c k k! ° ° ° Fb b EαSbα ° ° ° L2. (2.20)
The calculation for the term bbT ∗
£
( bE bS)1−αδ−(k)Bb−,k
¤
is analogous. For the region |ξ| ≤ 1, we consider bbT∗
£
( bE bS)1−α(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 ∗ £ ( bE bS)1−α(δ+(k)Bb+,k+ δ−(k)Bb−,k) ¤ (τ, ξ) ∼ Tk+1¡|ξ| + 1¢1−α£ ctkb¡T (τ − |ξ|)¢Bb +,k(ξ) + ctkb ¡ T (τ + |ξ|)¢Bb−,k(ξ) ¤ = Tk+1¡|ξ| + 1¢1−α£ ctkb¡T (τ − |ξ|)¢− ctkb¡T (τ + |ξ|)¢¤Bb +,k(ξ) + Tk+1¡|ξ| + 1¢1−αtckb¡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|, (2.23) b B+,0(ξ) + bB−,0(ξ) ∼ bφ0 + Z b Ff |ξ|2 − λ2 dλ, (2.24) and b B+,k(ξ) + bB−,k(ξ) ∼ 1 (k − 1)! Z ¡ λ − (1 − 2θ)|ξ|¢k−2( bF+ + ba6Fb−) dλ. (2.25) Combine the above results we complete 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. (2.26)
Lemma 2.5. Let f (t, x) and g(t, x) be any functions such that f ∈
Lq(L2(R)) and bSβbg ∈ L2(L2(R)). Assume that δ ≥ 0, q = 8 5 − 4δ, 1
r =
1
2 − β, and 2 ≤ r < ∞. Then we have ° ° °bbbT ∗ bf S14−δ ° ° ° L2 ≤ CkbTf kLq(L2), (2.27) kgkLr(L2) ≤ Ck bSβbgkL2(L2). (2.28)
Proof. The proofs for (2.27) and (2.28) are analogous. Therefore we will only prove the case of (2.28).
Taking the inverse Fourier transform in the time variable over the iden-tity b g = 1 b Sβ b Sβbg (2.29) gives e g(t, ξ) = Z e±i(t−s)|ξ| |t − s|1−βF −1 τ ( bSβbg)(s, ξ) ds. (2.30)
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). (2.31)
3. Local Existence.
Now we are ready to prove the local existence for the (DKG) equations. Proof of Theorem 0.1. Consider the DKG problem
Dψ = bTφψ; (t, x) ∈ R × R1 (3.1a)
¤φ = bTψψ; (3.1b)
ψ(0, x) := ψ0(x), φ(0, x) = φ0(x), φt(0, x) = φ1(x), (3.1c)
Iteration scheme induces a map T defined by
T (ψk, φk) = (ψk+1, φk+1). (3.2a) We want to show that T is a contraction under the norm
N (ψ, φ) =°° cM−αSb34ψb°°
L2 +
°
° cM−α( bE bS)1−αφb°°
L2. (3.2b)
For convenience, we call
J(0) = kφ0kH1−2α + kφ1kH−2α + kψ0k2H−α + 1. (3.3)
First we apply (2.11), (2.10), and (2.27) to compute ° ° cM−α( bE bS)1−αT φc°° L2 ≤ C ³ J(0) + ° ° ° \ bTψψ c MαEbαSbα ° ° ° L2 ´ ≤ C³J(0) + ° ° ° \bTφψ c MαSb14 ° ° °2 L2 ´ ≤ C ³ J(0) + ° ° °bTφψf c Mα ° ° °2 L85([0,T ],L2) ´ ≤ C ³ J(0) + T18 ° ° °bTφψf c Mα ° ° °2 L2([0,T ],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)kH1−2αkψ(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 (2.27) and (2.28) to obtain ° ° °bTMc−αφψf ° ° ° L2 ≤ CkφkL4([0,T ],H1−2α)kψkL4([0,T ],H−α) ≤ C°° bS14Mc1−2αφb°° L2 ° ° bS1 4Mc−αψb°° L2 ≤ C°° cMαSb14Mc1−αφb ° ° L2 ° ° bS3 4Mc−αψb ° ° L2 ≤ C°° cM−α( bE bS)1−αφb°°L2 ° ° cM−αSb3 4ψb ° ° L2. (3.7) Next we want to bound the term involved with bψ. The estimate (2.2)
implies that ° ° ° cM−αSb34T ψd ° ° ° L2 ≤ C ³ kψ0kH−α + ° ° ° \bTφψ c MαSb14 ° ° ° L2 ´ . (3.9)
Hence, using (3.4), (3.7), and (3.9), we have
N (T (ψ, φ)) ≤ C¡J(0) + T18N4(ψ, φ)¢. (3.10)
Choosing sufficiently large L, for suitable T , we have
N (ψ, φ) ≤ L =⇒ N¡T (ψ, φ)¢≤ L, (3.11) provided that
C(J(0) + T18L4) ≤ 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 (2.11), (2.10), and (3.13), we first calculate ° ° cM−α( bE bS)1−αF(T φ − T φ0)°° L2 ≤C³kF(bT(ψ − ψ0)(ψ + ψ0)) c MαEbαSbα kL2 + k F(bT(ψ + ψ0)(ψ − ψ0)) c MαEbαSbα kL2 ´ ≤C³°° F(bT(φ − φ0)(ψ + ψ0)) c MαSb14 ° ° L2 + ° ° F(bT(φ + φ0)(ψ − ψ0)) c MαSb14 ° ° L2 ´ · ³ I.D. +°° F(bT(φψ + φ 0ψ0)) c MαSb14 ° ° L2 ´ ≤CT81¡k cM−α( bE bS)1−αφ − φ\0kL2 + k cM−αSb 3 4ψ − ψ\0kL2 ¢ L(I.D. + L2) ≤CT18L3¡k cM−α( bE bS)1−αφ − φ\0k L2 + k cM−αSb 3 4ψ − ψ\0k L2 ¢ (3.15) Analogously, we get ° ° cM−αSb3 4F(T ψ − T ψ0) ° ° L2 ≤ CT18L ³ k cM−α( bE bS)1−αφ − φ\0k L2 + k cM−αSb 3 4ψ − ψ\0kL2 ´ . (3.16)
Combining (3.15)and (3.16), we have
N¡T (ψ − ψ0, φ − φ0)¢≤ CT18L3N (ψ − ψ0, φ − φ0). (3.17)
Therefore for suitable T , we obtain
N¡T (ψ − ψ0, φ − φ0)¢≤ 1 2N (ψ − ψ 0, φ − φ0), (3.18) provided that CT18L3 ≤ 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 ]
To bound φ we apply the following formula, 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| f ψψ(s, ξ)ds. (3.25)
Then we invoke (3.21), (2.10) (for α = 0), (2.27), and (3.24) to compute
kφ(t)k H12+δ ≤kφ0kH21+δ + kφ1kH− 12+δ + Z t 0 kbTψψ(s)kH− 12+δds ≤J(0) + T12 ° ° ° \ bTψψ c M12−δ ° ° ° L2 ≤J(0) + T12k\bTψψk L2 ≤J(0) + T12 ° ° °\bTbφψ S14 ° ° °2 L2 ≤J(0) + TρkbTφψk2L2 ≤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 proof of the key estimate.
Lemma 2.2. ( Null Form Estimate) Let α = 14− ², ² > 0, and ψ1, ψ2 be the solutions for the Dirac equations (2.9). If the initial data ψ0j ∈ H−α, j = 1, 2, then we have ° ° ° \ bTψ1ψ2 b EαSbα ° ° ° 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
° ° ° \ bTψψ b EαSbα ° ° ° 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 ° ° °bbT ∗ ¡ δ(k)∓ Ab±,k ¢ ∗¡δ(l)∓ Ab∓,l ¢ b EαSbα ° ° ° L2 ≤ C(k + l + 1)Tk+l−2αkf±,kkH−αkf∓,lkH−α. (4.11)
Proof. Let us call b
Z±,k ≡ δ(k)± Ab±,k = δ±(k)
b
D±
|ξ| fb±,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)
and observe that, for ξη < 0, ¯ ¯|ξ| + |η|¯¯ + |ξ + η| + 1 ∼ max{|ξ|, |η|} + 1, (4.14a) ¯ ¯ ¯¯¯|ξ| + |η|¯¯ − |ξ + η| ¯ ¯ ¯ + 1 ∼ min{|ξ|, |η|} + 1. (4.14b) 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 ¯ ¯ ¯ ≤Ckf−,kkH−αkf+,lkH−α· ³ Z (|ξ| + 1)2α(|η| + 1)2α¯¯ \tk+lb Tg(|ξ| + |η|, ξ + η) ¯ ¯2 dξdη ´1 2 ≤Ckf−,kkH−αkf+,lkH−αk bEαSbαtk+l\bTgkL2, (4.15)
Through some computations, we have ° ° \tk+lb T ° ° L1 ≤ C(k + l)T k+lkbk H1, (4.16a) ° °|τ|2αt\k+lb T ° ° L1 ≤ C(k + l)Tk+l−2αkbkH1, (4.16b)
provided that α < 1
4. With the aid of the above and the observation b E(τ, ξ) ≤ |τ − λ| + bE(λ, ξ), S(τ, ξ) ≤ |τ − λ| + bb S(λ, ξ), (4.16c) we can estimate k bEαSbαtk+l\b TgkL2 ≤³°° \tk+lb T ° ° L1 + ° °|τ|2αt\k+lb T ° ° L1 ´ k bEαSbαbgkL2 ≤C(k + l + 1)Tk+l−2αk bEαSbαbgkL2. (4.17)
This completes the proof. ¤
Lemma 4.2. Let α < 1
4. The following estimate holds ° ° °bbT ∗ ¡ δ(k)∓ Ab±,k ¢ ∗¡δ(l)± Ab±,l ¢ b EαSbα ° ° ° L2 ≤ C(k + l + 1)Tk+l−2αkf±,kkH−αkf±,lkH−α. (4.18)
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, (4.19a)
and observe that, for ξη > 0, ¯ ¯ − |ξ| + |η|¯¯ + |ξ + η| + 1 ∼ max{|ξ|, |η|} + 1, (4.19b) ¯ ¯ ¯¯¯ − |ξ| + |η|¯¯ − |ξ + η| ¯ ¯ ¯ + 1 ∼ min{|ξ|, |η|} + 1. (4.19c) 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 ¯ ¯ ¯ ≤Ckf+,kkH−αkf+,lkH−α· ³ Z (|ξ| + 1)2α(|η| + 1)2α¯¯ \tk+lb Tg(−|ξ| + |η|, ξ + η) ¯ ¯2 dξdη ´1 2 ≤Ckf+,kkH−αkf+,lkH−αk bEαSbαtk+l\bTgkL2 ≤C(k + l + 1)Tk+l−2αkf+,kkH−αkf+,lkH−αk bEαSbαgkb L2, (4.20)
¤ Lemma 4.3. Let δ > 0. The following estimates hold
kf±,0kH−α ≤ C ³ kψ0kH−α + ° ° ° Gb c MαSb12−δ ° ° ° L2 ´ , (4.21a) kf±,kkH−α ≤ C 1 k!k b G± c MαSb12−δ kL2. (4.21b)
The proof for the Lemma 4.3 is straight forward so that we skip it. Notice that, in the (4.21b), bS ∼ 1 on the support of bG±.
Lemma 4.4. With the notation above, the following estimate holds ° ° °bbT ∗ bbK1 ∗ bK1 EαSbα ° ° ° 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 types of regions:
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£γ0σ + 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)γ0(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 distinguish 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 b EαSbα ° ° ° 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 b EαSbα ° ° ° 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
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
other has thickness of 2l. It is mostly bounded, except for the extreme
case which is when one cone moves along the other cone such that the intersection region is unbounded. Denote the set eDkl to be the projection
of the set Dk,l onto the η-axis. When the set Dk,l is bounded, two facts, |ξ| = |ξ + η| + |η| and | eDkl| ≤ C2k, are available and will be used in the
following estimates.
For the first part, 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 ≤C 2l Z e Dk,l (|ξ + η| + 1)2α(|η| + 1)2α (2k+ |ξ + η|)2 dη. (4.33a)
Consider the case when |ξ + η| ≥ |η|, we get
Ik,l1 (τ, ξ) ≤C 2l Z e Dk,l (|ξ + η| + 1)2α(|η| + 1)2α (2k + |ξ + η|)2 dη ≤C 2l Z e Dk,l (1 + |η|)2α (2k+ |ξ + η|)2dη bE 2α(τ, ξ) ≤C 2l 1 2(1−2α)kEb 2αSb2α. (4.33b)
The extreme case is that when one of the cones moves along the other, say down right, this will not cause any trouble. Here, the region Dk,l is
unbounded. For the case |ξ + η| ≤ |η|, we get
Ik,l1 (τ, ξ) ≤C 2l Z e Dk,l (|ξ + η| + 1)2α(|η| + 1)2α (2k + |ξ + η|)2 dη ≤C 2l Z e Dk,l (1 + |ξ + η|)2α (2k+ |ξ + η|)2dη bE 2α(τ, ξ) ≤C 2l 1 2(1−2α)kEb 2αSb2α. (4.33c)
Again the extreme case when one of the cones moves along the other cone, say up right, will not cause trouble. Here, the region Dk,l is unbounded.
For the second part, we obtain
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−l Z e Dk,l (|ξ + η| + 1)2α(|η| + 1)2α (2l+ |η|)2 dη. (4.34a)
Consider the case when |ξ + η| ≥ |η|, we get
I2 k,l(τ, ξ) ≤ C 22k−l Z e Dk,l (|ξ + η| + 1)2α(|η| + 1)2α (2l+ |η|)2 dη ≤ C 22k−l Z e Dk,l (|η| + 1)2α (2l+ |η|)2 dη bE 2α ≤C 2k 1 2(1−2α)lEb 2αSb2α. (4.34b)
For the case |ξ + η| ≤ |η|, we get
Ik,l2 (τ, ξ) ≤ C 22k−l Z e Dk,l (|ξ + η| + 1)2α(|η| + 1)2α (2k+ |ξ + η|)2 dη ≤C 2l Z e Dk,l (1 + |ξ + η|)2α (2l+ |η|)2 dη bE 2α(τ, ξ) ≤C 2k 1 2(1−2α)lEb 2αSb2α. (4.34c)
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 (|ξ + η| + 1)2α(|η| + 1)2α(τ + σ)σ|ξ + η||η| (τ + σ + |ξ + η|)2(σ + |η|)2 dσdη ≤ C 22k+l Z e Dk,l (|ξ + η| + 1)2α(|η| + 1)2α|ξ + η||η| (2k+ |ξ + η|)(2l+ |η|) dη. (4.35a)
Consider the case when |ξ + η| ≥ |η|, we have Ik,l3 (τ, ξ) ≤ C 22k+l Z e Dk,l (|ξ + η| + 1)2α(|η| + 1)2α|ξ + η||η| (2k+ |ξ + η|)(2l+ |η|) dη ≤ C 22k+l Z e Dk,l (|η| + 1)2α|ξ + η||η| (2k+ |ξ + η|)(2l+ |η|) dη bE 2α ≤ C 22k+l Z e Dk,l (|η| + 1)2αdη bE2α ≤C 2l 1 2(1−2α)kEb 2αSb2α. (4.35b)
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. For the case |ξ + η| ≤ |η|, we get
Ik,l3 (τ, ξ) ≤ C 22k+l Z e Dk,l (|ξ + η| + 1)2α(|η| + 1)2α|ξ + η||η| (2k+ |ξ + η|)(2l+ |η|) dη ≤ C 22k+l Z e Dk,l (|ξ + η| + 1)2α|ξ + η||η| (2k+ |ξ + η|)(2l+ |η|) dη bE 2α ≤ C 22k+l Z e Dk,l (|ξ + η| + 1)2αdη bE2α ≤C 2l 1 2(1−2α)kEb 2αSb2α. (4.35c)
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.
Cases E. We have the following estimate ° ° ° b K−,·,k ∗ bK+,·,l b EαSbα ° ° ° L2 ≤ C 22l 1 2(1 2−α)k ° ° °Gb−,·,k c Mα ° ° ° L2 ° ° °Gb+,·,l c Mα ° ° ° L2, (4.36a) ° ° ° b K+,·,k∗ bK−,·,l b EαSbα ° ° ° L2 ≤ C 22l 1 2(1 2−α)k ° ° °Gb+,·,k c Mα ° ° ° L2 ° ° °Gb−,·,l c Mα ° ° ° L2. (4.36b)
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−
Take the case of
b
K−,+,k∗ bK+,+,l,
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 truncated backward cone with a forward 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 ≤ C 22l Z Dk,l (|ξ + η| + 1)2α(|η| + 1)2α (τ + σ − |ξ + η|)2 dσdη. (4.38a)
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) ≤ C 2l
1 2(1−2α)kEb
2αSb2α. (4.38b)
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 ≤C2l 22k Z e Dk,l (|ξ + η| + 1)2α (2l+ |η|)2+2α dη. (4.39a)
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) ≤ C 2l
1 2(1−2α)kEb
2αSb2α. (4.39b)
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. (4.40a) 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) ≤ C 2l
1 2(1−2α)kEb
2αSb2α. (4.40b)
Now we return to the proof of (4.22). Combine (4.31), (4.36), 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 bE αSbαbgk L2 ≤ C 2(1 4−α)k+l4 ° ° ° Gbk c MαSb14 ° ° ° L2 ° ° ° Gbl c MαSb14 ° ° ° L2k bE αSbαbgk L2. (4.41) Finally, we have ° ° ° b K ∗ bK b EαSbα ° ° ° L2 ≤ X k,l ° ° ° b Kk∗ bKl b EαSbα ° ° ° L2 ≤X k,l C 2(1 4−α)k+4l ° ° ° Gbk c MαSb14 ° ° ° L2 ° ° ° Gbl c MαSb14 ° ° ° L2 ≤ C ° ° ° Gb c MαSb14 ° ° °2 L2. (4.42)
This completes the proof. ¤
Lemma 4.5. For j = 1, 2 and k = 0, 1, 2, · · ·. The following estimates hold ° ° °bbT ∗ ¡ δ(k)∓ Ab±,k ¢ ∗¡Kbj ¢ b EαSbα ° ° ° L2 ≤ C(k + 1)T k−1 2kf±,kk H−α ° ° ° Gb c MαSb14 ° ° ° L2, (4.43a) ° ° °bbT b Kj ∗ ¡ δ±(k)Ab±,k ¢ b EαSbα ° ° ° L2 ≤ C(k+1)T k−1 2kf±,kkH−α ° ° ° Gb c MαSb14 ° ° ° L2, (4.43b) ° ° °bbT ∗ bbK1 ∗ bK2 EαSbα ° ° ° L2 ≤ C ° ° ° Gb c MαSb14 ° ° °2 L2, (4.43c) ° ° °bbT ∗ bbK2∗ bKj EαSbα ° ° ° L2 ≤ C ° ° ° Gb c MαSb14 ° ° °2 L2, (4.43d)
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.
Acknowledgement. The author wants to express his gratitude to M.
Gril-lakis for his encouragement and inspiring conversation, and also to Chi-kun Lin for his help.
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Department of Math, Cheng Kung University, Tainan 701 Taiwan
